$$100-97+94-91+88-85+\\cdots +4-1$$
\n$$=\\left( 100+94+88+\\cdots +4 \\right)-\\left( 97+91+85+\\cdots +1 \\right)$$
\n$$=\\frac{\\left( 100+4 \\right)\\times 17}{2}-\\frac{\\left( 97+1 \\right)\\times 17}{2}$$
\n$$=\\left( 104-98 \\right)\\times \\frac{17}{2}$$
\n$$=6\\times \\frac{17}{2}$$
\n$$=51$$．
\n故选$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3291", "queId": "8c0064d3ec4546fa9d8d8cf858123b7e", "competition_source_list": ["2005年第3届创新杯六年级竞赛初赛第10题", "2005年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "有五个砝码，其中重$$1$$克的砝码$$1$$个，重$$3$$克的砝码$$2$$个，重$$5$$克的砝码$$2$$个。如果规定砝码只能放在天平的同一边，那么在$$1\\sim17$$克所有整数克的重量中，不能称出的两个重量是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$克和$$9$$克 "}], [{"aoVal": "B", "content": "$$2$$克和$$11$$克 "}], [{"aoVal": "C", "content": "$$2$$克和$$13$$克 "}], [{"aoVal": "D", "content": "$$2$$克和$$15$$克 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["我们只用分析所给四个选项中的重量能否称出，而$$9=5+3+1\\text{，}11=5+5+1\\text{，}13=5+5+3$$都可称出，选$$D$$。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2518", "queId": "22fa6501a31e41319a1941678ef43fdb", "competition_source_list": ["2020年希望杯二年级竞赛模拟第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "请你根据数串的规律再横线上填上正确的答案：3,6,9,12~\\uline{~~~~~~~~~~}~，18. ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->思想->数形结合思想"], "answer_analysis": ["后一个数等于前一个数$$+$$ 3 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1920", "queId": "a12889b5e75941d2873393dfc5a0177d", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "今年哥哥$$16$$岁，弟弟$$8$$岁，问年前，哥哥的年龄是弟弟的$$3$$倍． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["从题干中可以知道，今年哥哥$$16$$岁，弟弟$$8$$岁，二者相差，$$16-8=8$$（岁）， 那么题目问几年前，哥哥的年龄是弟弟的$$3$$倍， 也就是二者年龄差应该是弟弟的$$2$$倍，$$8\\div 2=4$$（岁），$$8-4=4$$（年）， 即$$4$$年前，哥哥的年龄是弟弟的$$3$$倍． 我们可以验证一下，四年前，哥哥的年龄是：$$16-4=12$$（岁），弟弟的年龄是：$$8-4=4$$（岁）． $$12$$岁正好是$$4$$岁的$$3$$倍． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1016", "queId": "03257410e3be488f80e86d90b23ed285", "competition_source_list": ["2019年浙江杭州滨江区杭州江南实验学校五年级竞赛模拟（江南杯）第12题1分"], "difficulty": "0", "qtype": "single_choice", "problem": "学校举行``六一欢乐大卖场''活动，$$505$$班同学批发了一些小玩具参加活动，共卖出玩具$$22$$个，最便宜的售价$$10.5$$元$$/$$个，最贵的售价$$15.5$$元$$/$$个，这些玩具大约卖出元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$340-400$$ "}], [{"aoVal": "B", "content": "$$200-230$$ "}], [{"aoVal": "C", "content": "$$230-350$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["假设都按照最低价$$10.5$$一个，$$22$$个总价是$$231$$元，假设都按照最高价$$15.5$$元一个，总价是$$341$$元，所以这些玩具的总价不少于$$230$$元，也不高于$$340$$元．据此解答． $$10.5\\times22=231$$（元）， $$15.5\\times22=341$$（元）， 所以这些玩具的总价不少于$$230$$元，也不高于$$340$$元． 故答案为∶$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2057", "queId": "d8d03079467946c7b57a0116cdd60e61", "competition_source_list": ["2015年全国世奥赛竞赛A卷第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "在编号为$$1$$，$$2$$，$$3$$的三个相同的杯子里，分别盛着半杯液体．$$1$$好杯中溶有$$100$$克糖，$$2$$号杯中是水，$$3$$号杯中溶有$$100$$克盐．先将$$1$$号杯中液体的一半及$$3$$号杯中液体的$$\\frac{1}{4}$$倒入$$2$$号杯，然后搅匀．再从$$2$$号杯倒出所盛液体的$$\\frac{2}{7}$$到$$1$$号杯．这时$$1$$号杯含盐量与含糖量之比是~\\uline{~~~~~~~~~~}~．（写成最简整数比） ", "answer_option_list": [[{"aoVal": "A", "content": "$$2:9$$ "}], [{"aoVal": "B", "content": "$$1:9$$ "}], [{"aoVal": "C", "content": "$$1:8$$ "}], [{"aoVal": "D", "content": "$$1:7$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->比例应用题->复杂的比例问题"], "answer_analysis": ["第一次倒完后，$$1$$号杯中剩下$$50$$克糖，$$2$$号杯中有$$50$$克糖和$$25$$克盐．此时把$$2$$号杯的$$\\frac{2}{7}$$倒入$$1$$号杯，那么$$2$$号杯的糖和盐都倒了$$\\frac{2}{7}$$，即糖倒了$$50\\times \\frac{2}{7}=\\frac{100}{7}$$（克），盐倒了$$25\\times \\frac{2}{7}=\\frac{50}{7}$$（克）．那么$$1$$号杯的糖就变成了$$50+\\frac{100}{7}=\\frac{450}{7}$$（克），盐$$\\frac{50}{7}$$克． 所以含盐量和含糖量之比为$$\\frac{50}{7}:\\frac{450}{7}=1:9$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2738", "queId": "ff8080814518d5240145191ee20103f9", "competition_source_list": ["2014年全国迎春杯四年级竞赛复赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "对于任何自然数，定义$$n!=1\\times 2\\times 3\\times \\cdots \\times n$$，如$$8!=1\\times2\\times 3\\times \\cdots \\times 8$$；那么，算式：$$2014!+2013!-2012!+2011!+\\cdots -4!+3!-2!+1!$$，计算结果的个位数字是（~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->尾数特征->末一位数"], "answer_analysis": ["因为从$$5!=1\\times 2\\times 3\\times 4\\times 5$$ 开始，都含有$$2$$和$$5$$，那么个位必然是$$0$$，那么我们只要计算$$-4!+3!-2!+1!$$的个位数字即可，得出个位数字为$$1$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1283", "queId": "1b6f00c74c024851b5f48f5e6a24377c", "competition_source_list": ["2019年全国小学生数学学习能力测评四年级竞赛复赛第8题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "游乐场游玩，每人第一个小时付款$$15$$元，以后每小时付款$$8$$元，军军和弟弟游玩后共付了$$62$$元，他们在游乐场玩了个小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["除去第$$1$$小时以外的费用：$$62-15\\times2=32$$（元）； 每人除了第一小时以外所花的时间：$$32\\div 2\\div 8=2$$（小时）； 共时长：$$2+1=3$$（小时）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "28", "queId": "031197a17061487db7badbb2648908d1", "competition_source_list": ["2021年新希望杯二年级竞赛初赛第30题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "谎言岛有一半的人只在星期三、星期五、星期六说谎，另一半的人只在星期二、星期四、星期日说谎．某一天，岛上的所有人都说：``我明天说真话．''那么，这一天是． ", "answer_option_list": [[{"aoVal": "A", "content": "星期二 "}], [{"aoVal": "B", "content": "星期三 "}], [{"aoVal": "C", "content": "星期五 "}], [{"aoVal": "D", "content": "星期六 "}], [{"aoVal": "E", "content": "星期日 "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["前一半$$7$$天的情况：真、真、假、真、假、假、真， 后一半$$7$$天的情况：真、假、真、假、真、真、假， $$\\text{A}$$选项：若这一天是星期二，则前一半人在今天说的是真话，那么明天应该说真话才合理，但星期三他们说的是假话，相互矛盾．后一半人在今天说的是假话，那么明天应该说假话才合理，但是星期三他们说的是真话，也相互矛盾，故$$\\text{A}$$错误； $$\\text{B}$$选项：若这一天是星期三，则前一半人在今天说的是假话，那么明天应该说假话才合理，但星期四他们说的是真话，相互矛盾．后一半人在今天说的是真话，那么明天应该说真话才合理，但是星期四他们说的是假话，也相互矛盾，故$$\\text{B}$$错误； $$\\text{C}$$选项：若这一天是星期五，则前一半人在今天说的是假话，那么明天应该说假话才合理，星期六他们说的正好是假话，符合．后一半人在今天说的是真话，那么明天应该说真话才合理，星期六他们说的正好是真话，符合，故$$\\text{C}$$正确； $$\\text{D}$$选项：若这一天是星期六，则前一半人在今天说的是假话，那么明天应该说假话才合理，但星期日他们说的是真话，相互矛盾．后一半人在今天说的是真话，那么明天应该说真话才合理，但是星期日他们说的是假话，也相互矛盾，故$$\\text{D}$$错误； E选项：若这一天是星期日，则前一半人在今天说的是真话，那么明天应该说真话才合理，星期一他们说的是真话，符合．后一半人在今天说的是假话，那么明天应该说假话才合理，但是星期一他们说的是真话，相互矛盾，故$$\\text{E}$$错误． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "752", "queId": "88856aef562c4200ac34e26b6c173c0d", "competition_source_list": ["2014年第15届上海中环杯小学低年级二年级竞赛初赛第16题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一种靶子，上面有$$1$$环、$$3$$环、$$5$$环、$$7$$环和$$9$$环表示射中的环数． 甲说：``我打了$$5$$枪，每枪都中靶，共中了$$35$$环．'' 乙说：``我打了$$6$$枪，每枪都中靶，共中了$$36$$环．'' 丙说：``我打了$$3$$枪，每枪都中靶，共中了$$24$$环．'' 丁说：``我打了$$4$$枪，只有$$1$$枪没中靶，共中了$$21$$环．'' 已知他们四人中，只有一人说了假话．那么说假话的是． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "丁 "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾"], "answer_analysis": ["单双数加减的性质是：单数个单数和为单，双数个单数和为双． 甲：$$9+9+9+1+7=35$$，正确． 乙：$$9+9+9+1+1+7=36$$，正确． 丙：三个单数的和为单数，结果$$24$$为双数，错误． 丁：$$7+7+7+0=21$$，正确． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3277", "queId": "5a6baa9d60b74ffebe8bbdef38d05c48", "competition_source_list": ["2007年第5届创新杯六年级竞赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$2007$$盏亮着的灯，各有一个拉线开关控制着，拉一下拉线开关灯会由亮变灭，再拉一下又由灭变亮，现按其顺序将灯编号为$$1$$，$$2$$，$$\\cdots$$，$$2007$$，然后将编号为$$2$$的倍数的灯线都拉一下，再将编号为$$3$$的倍数的灯线都拉一下，最后将编号为$$5$$的倍数的灯线都拉一下，三次拉完后亮着的灯有盏． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1004$$ "}], [{"aoVal": "B", "content": "$$1002$$ "}], [{"aoVal": "C", "content": "$$1000$$ "}], [{"aoVal": "D", "content": "$$998$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["∵有$$2007$$盏亮着的电灯，现按其顺序编号为$$1$$，$$2$$，$$\\cdots$$，$$2007$$， ∴编号为$$2$$的倍数的灯有$$(2007-1)\\div2=1003$$（只）， 编号为$$3$$的倍数的灯有$$2007\\div3=669$$（只）， 编号为$$5$$的倍数的灯的有$$(2007-2)\\div5=401$$（只）， 其中既是$$3$$的倍数也是$$5$$的倍数有$$(2007-12)\\div15=133$$（只）， 既是$$2$$的倍数也是$$3$$的倍数有$$(2007-3)\\div6=334$$（只）， 既是$$2$$的倍数也是$$5$$的倍数有$$(2007-7)\\div10=200$$（只）， 既是$$2$$的倍数也是$$5$$的倍数，还是$$3$$的倍数有$$(2007-27)\\div30=66$$（只）， 只拉$$1$$次的：$$1003-334-200+66=535$$， $$669-334-133+66=268$$， $$401-200-133+66=134$$， 拉$$3$$次的$$66$$， 所以亮的就是$$2007-535-268-134-66=1004$$（只）． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1934", "queId": "aa560020cf464d41bfadb4edc70c3a78", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "甲瓶盐水浓度为$$5 \\%$$，乙瓶盐水浓度为$$10 \\%$$，两瓶混合后盐水的浓度（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "是$$5 \\%$$ "}], [{"aoVal": "B", "content": "是$$10 \\%$$ "}], [{"aoVal": "C", "content": "是$$7.5 \\%$$ "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["如果甲瓶盐水为$$a$$升，则含盐$$5a \\%$$，如果乙瓶盐水为$$b$$升，则含盐$$10 \\%b$$，依照浓度的计算公式，混合后盐水的浓度是：$$\\frac{5 \\%a+10 \\%b}{a+b}$$，因为不知$$a$$和$$b$$的数值，所以两瓶混合后盐水的浓度不能确定． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1557", "queId": "ff8080814502fa2401450775bb5d0aa1", "competition_source_list": ["2014年全国迎春杯四年级竞赛复赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "小元和小芳合作进行一项$$10000$$字的打字作业，但他们都非常马虎，小元每打$$10$$个字，就会打错$$1$$个；小芳每打字$$10$$个，就会打错$$2$$个，最后，当两人完成工作时，小元打正确的字数恰好是小芳打正确的字数的$$2$$倍，那么，两人打正确的字共有（~~~~~~~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5000$$ "}], [{"aoVal": "B", "content": "$$7320$$ "}], [{"aoVal": "C", "content": "$$8000$$ "}], [{"aoVal": "D", "content": "$$8640$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->已知工时反推->两人合作"], "answer_analysis": ["我们可以理解为小元每打$$10$$份的字数就会打错$$1$$份，小芳每打$$10$$份的字数就会打错$$2$$份，即小芳打$$5$$份的字数只能正确$$4$$份． 假设小元打的总字数为$$80$$份，那么他正确的为$$8\\times 9=72$$份，根据题意可知小芳正确的为$$72\\div 2=36$$份， 那么小芳打字的总分数为$$36\\div 4\\times 5=45$$份．小元和小芳的总字数份为$$80+45=125$$份，共$$10000$$字，即每份字数为$$10000\\div125=80$$． 小元和小芳共正确的字数为$$(72+36)\\times 80=8640$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1282", "queId": "629f89421d444e8f8b7a124bf4538e77", "competition_source_list": ["2016年陈省身杯六年级竞赛国际青少年数学解题能力展示活动第11题"], "difficulty": "3", "qtype": "single_choice", "problem": "文具店以每支$$8$$元的价格购进一批钢笔，并标价每支$$20$$元出售．当售出这批钢笔的$$\\frac{3}{4}$$时发现，不仅已经收回了全部成本，且获利$$224$$元．那么这批钢笔共有~\\uline{~~~~~~~~~~}~支． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$32$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$64$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "课内体系->能力->运算求解"], "answer_analysis": ["成本$$:$$利润$$:$$售价$$=8:12:20=2:3:5$$，卖出$$\\frac{3}{4}$$后，售价为$$\\frac{3}{4}\\times 5=\\frac{15}{4}$$(份)，此时获利份数为$$\\frac{15}{4}-2=\\frac{7}{4}$$(份)，全成本为$$224\\div \\frac{7}{4}\\times 2=256$$（元），钢笔共有$$256\\div 8=32$$(支)． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "734", "queId": "3748cf45c9de4b53a0a03c675453b1b8", "competition_source_list": ["2016年第14届全国创新杯小学高年级五年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个整数，除以$$3$$余数是$$2$$，除以$$5$$余数是$$3$$，除以$$7$$余数是$$4$$，这个数可能是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$67$$ "}], [{"aoVal": "B", "content": "$$73$$ "}], [{"aoVal": "C", "content": "$$158$$ "}], [{"aoVal": "D", "content": "$$22$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["设这个数为$$a$$，则$$\\left { \\begin{matrix}a\\div 3\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 2 a\\div 5\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 3 a\\div 7\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 4 \\end{matrix} \\right.$$，整理得$$\\left { \\begin{matrix}a\\div 3\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 11 a\\div 5\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 3 a\\div 7\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 11 \\end{matrix} \\right.$$，所以$$a=21k+11$$ 所以当$$k=7$$时，$$a=158$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "993", "queId": "0548a2551c534d1daef0cc71b97a7308", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$26$$个小朋友排队做操，排在涛涛前面的有$$10$$人，排在涛涛后面的有人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["除了涛涛前面的人和涛涛本身，剩下的就是排在涛涛后面的人， 所以共有：$$26-10-1=15$$（人）． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1538", "queId": "4087c73c729549e4aa282363fe0f76f0", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明今年$$14$$岁，小强今年$$8$$岁，$$2$$年后，小明比小强大岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题基本关系->年龄差"], "answer_analysis": ["从题干中可以知道，小明今年$$14$$岁，小强今年$$8$$岁， 那么今年二者的年龄差是$$14-8=6$$（岁）； 无论经过几年，小明和小强的年龄差还是不变的； 所以$$2$$年后，小明比小强大$$6$$岁； 可以换另一种方法检验一下：$$2$$年后小明$$16$$岁，小强$$10$$岁；$$16-10=6$$（岁）． 所以选择$$\\text{B}$$选项． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "491", "queId": "e1263b323754425ebb457db711514639", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "（3分）在垒球比赛中，若赢$$1$$场得$$3$$分，平$$1$$场得$$1$$分，输$$1$$场不得分．每个队都与其他队交锋$$4$$场，这时四个参赛队的总积分为：$$A$$队$$22$$分，$$B$$队$$19$$分，$$C$$队$$14$$分，$$D$$队$$12$$分．那么有场比赛为平局． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->逻辑分析"], "answer_analysis": ["对于赛况分析试题，尤其对于与分数相关的试题，最重要的是思维方式，本题如果从整体上来考虑比赛所产生的总分值，问题将迎刃而解，依题意可知比赛总场次为$$24$$场比赛之中，若平局则将会让所有队伍的总分增加$$2$$分（比赛双方均得$$1$$分），若出现了胜败，则所有队伍的总分增加$$3$$分，而现在所有队伍获得的总分值为：$$22+19+14+12=67$$（分），$$24$$场比赛，有$$3$$分，有$$2$$分，总分为$$67$$分，可当做鸡兔同笼问题解答，易得平局有$$5$$场． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "445", "queId": "e05104d6577e498b8bdd505b82ac5abc", "competition_source_list": ["2008年二年级竞赛学而思杯"], "difficulty": "1", "qtype": "single_choice", "problem": "古时候某国有两座城，一座``真城''，一座``假城''，真城的人都说真话，假城的人都说假话.一天，一个国外游客来到其中的一座城，他向遇到的一位该国国民提了一个问题，就明白了自己到的是真城还是假城.以下（ ）最可能是游客提的问题. ", "answer_option_list": [[{"aoVal": "A", "content": "你是真城的人吗? "}], [{"aoVal": "B", "content": "你是假城的人吗? "}], [{"aoVal": "C", "content": "你是说真话的人吗? "}], [{"aoVal": "D", "content": "你是说假话的人吗? "}], [{"aoVal": "E", "content": "你是这座城的人吗? "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理"], "answer_analysis": ["对于A，B，C，D，不论问哪个城里的人，很明显得到的是相同的答案，无法判断真假，只有问E时，真城里的人会回答``是''，而假城里的人的答案是``不是''. "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1740", "queId": "6145fde6be8d42abbb324a70fe516de8", "competition_source_list": ["2012年第10届创新杯四年级竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "两位篮球运动员的体重分别为$$75$$千克和$$87.5$$千克，第$$3$$位运动员的体重介于这两者之间，下列哪一个不可能是这$$3$$位运动员的平均体重？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$80.5$$ "}], [{"aoVal": "B", "content": "$$81.5$$ "}], [{"aoVal": "C", "content": "$$82.5$$ "}], [{"aoVal": "D", "content": "$$83.5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["排除法： $$80.5\\times 3-(75+87.5)=241.5-162.5=79$$； $$81.5\\times 3-(75+87.5)=244.5-162.5=82$$； $$82.5\\times 3-(75+87.5)=247.5-162.5=85$$； $$83.5\\times 3-(75+87.5)=250.5-162.5=88\\textgreater87.5$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "465", "queId": "c9dda81a33f44f8bb8c34025b7ae45dd", "competition_source_list": ["2010年五年级竞赛创新杯", "2010年六年级竞赛创新杯"], "difficulty": "3", "qtype": "single_choice", "problem": "要想用天平称出$$1\\sim40$$克所有整数克的质量，如果允许两边放砝码，至少用（ ）个砝码。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["若只有一个砝码，那么这个砝码有放左、放右和不放$$3$$种不同选择，其中左、右都不放的情况不计，剩下的情况会出现左右重复，只能称出$$\\frac{3-1}{2}=1$$（种）不同的质量；取$$1$$克，可以称出$$1$$克； 若有两个砝码，共有$$3\\times 3=9$$（种）放法，最多称出$$\\frac{{{3}^{2}}-1}{2}=4$$（种）不同的质量；取$$1$$克和$$3$$克，即可称出$$1$ $4$$克； 若有三个砝码，共有$${{3}^{3}}=27$$（种）放法，最多称出$$\\frac{{{3}^{3}}-1}{2}=13$$（种）不同的质量；取$$1$$克、$$3$$克和$$9$$克，即可称出$$1$ $13$$克； 若有四个砝码，共有$${{3}^{4}}=81$$（种）放法，最多称出$$\\frac{{{3}^{4}}-1}{2}=40$$（种）不同的质量．取$$1$$克、$$3$$克、$$9$$克和$$27$$克，即可称出$$1$ $40$$克． 由此我们只需$$1$$克、$$3$$克、$$9$$克、$$27$$克这样的$$4$$个砝码，即可称出$$1$ $40$$克所有整数克的质量． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1641", "queId": "914492a0de9d4509aa9dd207c900bc41", "competition_source_list": ["2015年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "桌上有编号为$$1\\sim 20$$的$$20$$张卡片，小明每次取出$$2$$张卡片，要求一张卡片的编号是另一张卡片的$$2$$倍多$$2$$，则小明最多取出（ ）张卡片。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->乘除法应用->除法应用"], "answer_analysis": ["解：设另一张卡号是$$x$$。则： $$2x+2\\leqslant 20$$ $$2x+2-2\\leqslant 20-2$$ $$2x\\leqslant 18$$ $$2x\\div 2\\leqslant 18\\div 2$$ $$x\\leqslant 9$$ 又因为$$x\\geqslant 1$$ $$9\\times 2=18$$（张） $$\\because 4$$、$$6$$、$$8$$即可以做为一倍量的数，也可以作为$$2$$倍量多$$2$$的数， $$\\therefore $$即总共可以取出：$$18-3\\times 2=12$$（张）。 答：小明最多可以取出$$12$$张卡片。 故选：$$A$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1573", "queId": "ff8080814518d5240145190941300312", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "奶奶折一个纸鹤用$$3$$分钟，每折好一个需要休息$$1$$分钟，奶奶从$$2$$时$$30$$分开始折，她折好第$$5$$个纸鹤时已经到了（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$时$$45$$分 "}], [{"aoVal": "B", "content": "$$2$$时$$49$$分 "}], [{"aoVal": "C", "content": "$$2$$时$$50$$分 "}], [{"aoVal": "D", "content": "$$2$$时$$53$$分 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["$$3\\times 5+1\\times 4=19$$，折好第$$5$$个时已经到了$$2$$时$$49$$分． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2810", "queId": "6078a63ffcd241b68ad97f9f5a7907bc", "competition_source_list": ["2019年华夏杯一年级���赛复赛（华南赛区）第16题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "陈老师按小奥、小林、小匹、小克这个顺序派发$$20$$粒糖果给这$$4$$位小朋友，每次一粒，那么最后一次糖会派给哪位小朋友？ ", "answer_option_list": [[{"aoVal": "A", "content": "小奥 "}], [{"aoVal": "B", "content": "小林 "}], [{"aoVal": "C", "content": "小匹 "}], [{"aoVal": "D", "content": "小克 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数与运算->数的运算的实际应用（应用题）->整数的简单实际问题->加法的实际应用"], "answer_analysis": ["小克$$20$$颗糖发放给$$4$$个小朋友，每发完一次就少$$4$$个糖，$$20=4+4+4+4+4$$． $$20$$颗糖刚好可以发$$5$$个轮回，最后一颗糖刚好发到小克，所以答案是小克． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1220", "queId": "3da244afc15143e182441a9c76d6fe2a", "competition_source_list": ["2014年全国迎春杯五年级竞赛初赛第3题"], "difficulty": "0", "qtype": "single_choice", "problem": "（5分）一辆大卡车一次最多可以装煤$$2.5$$吨，现在要一次运走$$48$$吨煤，那么至少需要辆这样的大卡车． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["$$48\\div 2.5=19\\cdots 0.5$$，$$19+1=20$$（辆）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "209", "queId": "4c7692da34b74807ac24febc1229a0b8", "competition_source_list": ["2011年全国希望杯四年级竞赛初赛第16题"], "difficulty": "2", "qtype": "single_choice", "problem": "王强步行去公园，回来时坐车，往、返用了一个半小时，如果他来回都步行，则需要$$2$$个半小时，那么，他来回都坐车，则需~\\uline{~~~~~~~~~~}~分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["一个半小时是$$90$$分钟，$$2$$个半小时是$$2\\times 60+30=150$$分钟．易知王强步行单程需要$$150\\div 2=75$$分钟，则坐车单程要$$90-75=15$$分钟，所以来回都坐车要$$15\\times 2=30$$分钟． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3421", "queId": "d7733d6409ad4b6ba55131ce03af17e4", "competition_source_list": ["2018年第8届北京学而思综合能力诊断二年级竞赛年度教学质量第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "五一到了，花花想要出去旅游，她有$$2$$顶帽子、$$3$$条项链、$$5$$种头花、$$2$$双鞋、$$3$$条裙子，每类物品中各选一类进行搭配．那么，一共有种不同的搭配方法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$200$$ "}], [{"aoVal": "B", "content": "$$160$$ "}], [{"aoVal": "C", "content": "$$180$$ "}], [{"aoVal": "D", "content": "$$170$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算求解", "拓展思维->拓展思维->计数模块->加乘原理->乘法原理->物品搭配"], "answer_analysis": ["$$2\\times 2\\times 3=12$$（种）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2105", "queId": "ebc55cce51d540668553066fffa571d0", "competition_source_list": ["2017年河南郑州联合杯竞赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "青蛙从井底向井口跳，井深$$15$$米，青蛙每次向上跳$$6$$米，又会滑下来$$3$$米，这样青蛙需要跳（~ ）次才可以出井． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$次 "}], [{"aoVal": "B", "content": "$$4$$次 "}], [{"aoVal": "C", "content": "$$5$$次 "}], [{"aoVal": "D", "content": "$$6$$次 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->蜗牛爬井问题->蜗牛爬井求天数（爬出去类）"], "answer_analysis": ["第一次跳时，向上跳$$6$$米，又下滑$$3$$米，此时距离井底$$3$$米，第二次跳完距离井底$$6$$米，第三次跳完距离井底$$9$$米，第四次先向上跳$$6$$米，此时青蛙跳出了井． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2478", "queId": "1599856f795f4ab2963e1d566c6a2ccd", "competition_source_list": ["2020年希望杯六年级竞赛模拟第25题"], "difficulty": "1", "qtype": "single_choice", "problem": "比较大小： $$1+\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\frac{1}{{{4}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}}$$~\\uline{~~~~~~~~~~}~$$2$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\textgreater$$ "}], [{"aoVal": "B", "content": "$$\\textless{}$$ "}], [{"aoVal": "C", "content": "$$=$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为$$\\frac{1}{{{2}^{2}}}\\textless{}\\frac{1}{1\\times 2}$$，$$\\frac{1}{{{3}^{2}}}\\textless{}\\frac{1}{2\\times 3}$$，$$\\frac{1}{{{4}^{2}}}\\textless{}\\frac{1}{3\\times 4}$$， 所以$$\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\frac{1}{{{4}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}}\\textless{}\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{2019\\times 2020}$$， 因为$$\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\frac{1}{3\\times 4}\\cdots +\\frac{1}{2019\\times 2020}$$ $$=1-\\frac{1}{2}+\\frac{1}{2}-\\frac{1}{3}+\\frac{1}{3}-\\frac{1}{4}+\\cdots +\\frac{1}{2019}-\\frac{1}{2020}$$ $$=1-\\frac{1}{2020}$$ $$=\\frac{2019}{2020}$$， 所以$$\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}}\\textless{}\\frac{2019}{2020}$$， 所以$$1+\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\frac{1}{{{4}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}}\\textless{}2$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "597", "queId": "6b24da2c7391400081fb0ff2a5b3aa57", "competition_source_list": ["2020年广东广州羊排赛五年级竞赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个四位数的千位是最小的质数，百位是最小的合数，十位是最小的自然数．这个四位数是$$3$$的倍数，则它的个位数字可能是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$2$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->因数与倍数基础", "Overseas Competition->知识点->数论模块->质数与合数->质数与合数判定"], "answer_analysis": ["最小质数是$$2$$，所以千位是$$2$$； 最小合数是$$4$$，所以百位是$$4$$；最小的自然数是$$0$$，所以十位是$$0$$，这个四位数是$$3$$的倍数，那么这个四位数各个位数之和是$$3$$的倍数，$$2+4+0=6$$． $$\\text{A}$$选项：$$6+5=11$$，不是$$3$$的整数倍，故$$\\text{A}$$错误； $$\\text{B}$$选项：$$6+6=12$$，是$$3$$的整数倍，故$$\\text{B}$$正确； $$\\text{C}$$选项：$$6+7=13$$，不是$$3$$的整数倍，故$$\\text{C}$$错误； $$\\text{D}$$选项：$$6+8=14$$，不是$$3$$的整数倍，故$$\\text{D}$$错误； $$\\text{E}$$选项：$$6+2=8$$，不是$$3$$的整数倍，故$$\\text{E}$$错误． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1733", "queId": "bf6caaf45c044ca4accdeecc12f9c345", "competition_source_list": ["2020年新希望杯二年级竞赛决赛（8月）第4题", "2020年新希望杯二年级竞赛初赛（个人战）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$1$$斤菠萝可以做$$2$$个菠萝蛋糕，那么要做$$10$$个菠萝蛋糕，需要斤菠萝． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->乘除法应用->除法应用"], "answer_analysis": ["根据$$1$$斤菠萝可以做$$2$$个菠萝蛋糕，那么要做$$10$$个菠萝蛋糕，需要菠萝$$10\\div2=5$$（斤）．故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "381", "queId": "6f26597099f24f42acb20ca89c9a2818", "competition_source_list": ["2017年IMAS小学高年级竞赛（第一轮）第15题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在算式$$\\overline{ab}+ \\overline{cd}= \\overline{ef}$$ 中， $$\\overline{ab}、 \\overline{cd}、 \\overline{ef}$$ 各代表一个二位数．且$$a$$、$$b$$、$$c$$、$$d$$、$$e$$、$$f$$六个数码两两不同．请问$$\\overline{ef}$$的最小可能值是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$34$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$39$$ "}], [{"aoVal": "E", "content": "$$41$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["因$$a+c\\geqslant 1+2=3$$．故$$e\\geqslant 3$$，因此取$$e=3$$，此时可判断出$$a$$、$$c$$分别为$$1$$与$$2$$，且知$$b$$、$$d$$均不能为$$0$$，否则数码$$f$$会与$$b$$或$$d$$相等，故$$b+d\\geqslant 4+5=9$$，即$$f=9$$，例如$$14+25=39$$、$$15+24=39$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2555", "queId": "62a6616b10194309912fc66792abcd8d", "competition_source_list": ["2019年浙江杭州滨江区杭州江南实验学校五年级竞赛模拟（江南杯）第6题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$9\\div 11$$的商的小数部分第$$50$$位上的数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "课内体系->能力->数据处理"], "answer_analysis": ["列竖式 故$$9\\div 11=0.\\dot{8}\\dot{1}$$． 周期为$$2$$故$$50\\div 2=25$$， 故第$$50$$位为$$1$$． 故答案为：$$\\text{B}$$选项． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1347", "queId": "478d3e6c4aea43fa835715a1f7e96e68", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "幼儿园老师给小朋友分糖果，每个小朋友分$$5$$颗糖果，就多出$$24$$颗；每个小朋友分$$7$$颗糖果，就少$$16$$颗糖果．一共有糖果颗． ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$124$$ "}], [{"aoVal": "C", "content": "$$126$$ "}], [{"aoVal": "D", "content": "$$130$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["两次分糖剩余之差：$$24+16=40$$（颗）， 每次分糖数之差：$$7-5=2$$（颗）． 小朋友人数：$$40\\div2=20$$（人）， 糖果总数：$$20\\times5+24=124$$（颗）． 所以选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2932", "queId": "a057a12b8efe42f690a0f0c1a50969a0", "competition_source_list": ["2017年全国美国数学大联盟杯小学高年级五年级竞赛初赛第36题"], "difficulty": "2", "qtype": "single_choice", "problem": "小罗星期一工作了$$2$$个小时．在那之后的每一天，他的工作时间是前一天的两倍．请问他从星期一到星期四总共工作了多少小时？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$28$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["这是一个等比数列问题，$${{a}_{n}}={{a}_{1}}{{q}^{n-1}}={{2}^{n}}$$，$${{S}_{n}}=\\frac{{{a}_{1}}\\left( 1-{{q}^{n}} \\right)}{1-q}=\\frac{2\\left( 1-{{2}^{n}} \\right)}{1-2}=2\\left( {{2}^{n}}-1 \\right)=30$$，所以选$$D$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2882", "queId": "e8f3d290583b471c830124ea9c437e15", "competition_source_list": ["2017年河南郑州豫才杯四年级竞赛初赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "聪聪在数学课上总能专心听讲，每次完成练习之后，总要进行自检自查，发现错误能够及时更正过来．这次他在检查一道除法笔算时，发现自己把除数$$65$$写成了$$56$$，得到的商是$$15$$，余数是$$11$$．请你根据以上信息选择正确的结果． ", "answer_option_list": [[{"aoVal": "A", "content": "商是$$15$$，余数是$$20$$ "}], [{"aoVal": "B", "content": "商是$$13$$，余数是$$6$$ "}], [{"aoVal": "C", "content": "商是$$24$$，余数是$$11$$ "}], [{"aoVal": "D", "content": "商是$$15$$，余数是$$2$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$56\\times 15+11=851$$，$$851\\div 65=13\\cdots \\cdots 6$$，所以$$\\text{B}$$选项正确． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1655", "queId": "ff8080814767499401477b3b55cc15bb", "competition_source_list": ["2013年全国华杯赛小学高年级竞赛初赛B卷第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "某日，甲学校买了$$56$$千克水果糖，每千克$$8.06$$元．过了几日，乙学校也需要买同样的$$56$$千克水果糖，不过正好赶上促销活动，每千克水果糖降价$$0.56$$元，而且只要买水果糖都会额外赠送$$5 \\%$$同样的水果糖．那么乙学校将比甲学校少花（~ ~ ~ ~）元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$51.36$$ "}], [{"aoVal": "C", "content": "$$31.36$$ "}], [{"aoVal": "D", "content": "$$10.36$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["甲学���花钱$$56\\times 8.06=451.36$$元；乙学校要买糖$$56\\div (1+5 \\%)=\\frac{160}{3}$$kg，单价$$8.06-0.56=7.5$$元，甲学校花钱$$\\frac{160}{3}\\times7.5=400$$元；乙学校将比甲学校少花$$51.36$$元． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2682", "queId": "5ab31d10e4764e8c8560806baf7ec117", "competition_source_list": ["2022年第九届鹏程杯四年级竞赛初赛第3题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：（1+2+3+···+29+30）$\\times6-6\\times128=\\left( \\right)$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2022$$ "}], [{"aoVal": "B", "content": "$$2021$$ "}], [{"aoVal": "C", "content": "$$2020$$ "}], [{"aoVal": "D", "content": "$$2019$$ "}], [{"aoVal": "E", "content": "$$2018$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数乘除->整数乘除法巧算"], "answer_analysis": ["无 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "893", "queId": "d6d49a392b294375a8ca78fcf994262f", "competition_source_list": ["2013年第25届广东广州五羊杯六年级竞赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知一个素数的三倍与另一个素数的五倍的和是$$133$$，则这两个素数和为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$38$$ "}], [{"aoVal": "B", "content": "$$43$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$53$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["显然两个素数中至少有一个是偶数，而素数中只有$$2$$是偶数． 将$$2$$代入，可得另一素数为$$41$$，故和为$$43$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3304", "queId": "3c3c0d6ae897454ab801254e069dc3e1", "competition_source_list": ["2016年河南郑州联合杯小学高年级六年级竞赛复赛第17题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "五一大假期间，甲商场以打五折的方式优惠，乙商场以每满$$200$$送$$100$$元优惠券的措施优惠，妈妈打算花$$1000$$元购物，去（~ ）商场比较合算． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "甲、乙均可 "}], [{"aoVal": "D", "content": "不知道 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["经济问题；甲：$$1000$$元可以买到$$1000\\div 50 \\% =2000$$元，乙：$$1000\\div 200=5$$，$$100\\times 5=500$$元，$$1000$$元可以买到：$$1000+500=1500$$元，$$2000\\textgreater1500$$，所以去甲商场合算． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "337", "queId": "6e3bd81aa1404e508056ac65434053ce", "competition_source_list": ["2019年第24届YMO六年级竞赛决赛第7题3分"], "difficulty": "3", "qtype": "single_choice", "problem": "有一个$$12$$级的楼梯．某人每次能登上$$1$$级或$$2$$级或$$3$$级，现在他要从地面登上第$$12$$级，有种不同的方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$149$$ "}], [{"aoVal": "B", "content": "$$244$$ "}], [{"aoVal": "C", "content": "$$264$$ "}], [{"aoVal": "D", "content": "$$274$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->归纳递推->斐波那契数列递推"], "answer_analysis": ["设第$$n$$级有$${{a}_{n}}$$种登的方式， 则第$$n-3$$，$$n-2$$，$$n-1$$级再分别登$$3$$，$$2$$，$$1$$级可到第$$n$$级， 则$${{a}_{n}}={{a}_{n-3}}+{{a}_{n-2}}+{{a}_{n-1}}$$， 而$${{a}_{1}}=1$$，$${{a}_{2}}=1+1=2$$，$${{a}_{3}}=1+1+2=4$$， 故$${{a}_{4}}=1+2+4=7$$， $${{a}_{5}}=2+4+7=13$$， $${{a}_{6}}=4+7+13=24$$， $${{a}_{7}}=7+13+24=44$$， $${{a}_{8}}=13+24+44=81$$， $${{a}_{9}}=24+44+81=149$$， $${{a}_{10}}=44+81+149=274$$． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "221", "queId": "634d324416b945a19d65585ef1ec0736", "competition_source_list": ["2014年第25届亚太杯四年级竞赛决赛第18题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "小董并非既懂英文又懂法语．如果上述判断为真，那么下述判定为真的是． ", "answer_option_list": [[{"aoVal": "A", "content": "小董懂英文但不懂法语． "}], [{"aoVal": "B", "content": "小董懂法语但不懂英文． "}], [{"aoVal": "C", "content": "小董既不懂英文也不懂法语． "}], [{"aoVal": "D", "content": "如果小董懂英文，小董一定不懂法语． "}], [{"aoVal": "E", "content": "如果小董不懂法语，那么他一定懂英文�� "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["逻辑推理． 小董并非既懂英文又懂法语． 则有以下$$3$$种情况： $$(1)$$小董懂英文但不懂法语． $$(2)$$小董懂法语但不懂英文． $$(3)$$小董既不懂英文也不懂法语． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3100", "queId": "f45eb434a57f416fbf21dc522fee4c23", "competition_source_list": ["2014年第10届全国新希望杯小学高年级六年级竞赛复赛第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "对自然数$$n$$进行如下操作：如果$$n$$是偶数，就把它除以$$2$$，如果$$n$$是奇数，就把它加上$$7$$．现在对$$154$$进行有限次操作，得到的结果不可能是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["154按照操作依次为$$77$$，$$84$$，$$42$$，$$21$$，$$28$$，$$14$$，$$7$$，$$14$$，7\\ldots\\ldots，不可能得到$$11$$． ", "154和$$7$$都是$$7$$点倍数，每次操作后最终结果都是$$7$$的倍数，不可能得到11.
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1050", "queId": "d4fd9d727c534fdd9a5a1d7fb6023fd2", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛B卷第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "学生问老师的年龄，老师说：``当我是你这么大的时候，你刚$$3$$岁，当你是我这么大的时候，我已经$$39$$岁了，''这位老师今年岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$26$$ "}], [{"aoVal": "C", "content": "$$27$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["根据年龄差不会变这一特征，从年龄差入手，年龄差$$+$$老师现在的年龄$$=39$$， 所以老师$$+$$学生$$=42$$，设老师今年岁数为$$x$$， 则学生的岁数是$$42-x$$岁，再根据年龄差$$+$$老师现在的年龄$$=39$$， 列出方程解决问题， $$x-(42---x)+x=39$$， $$3x---42=39$$， $$3x=39+42$$， $$3x=81$$， $$x=27$$， 也就是老师今年$$27$$岁， 所以选择$$\\text{C}$$选项． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1331", "queId": "42ef1153cc1e49d2a405ff5ab85d019a", "competition_source_list": ["2017年河南郑州豫才杯五年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "欢欢在今年的国庆节$$8$$天假期里，计划先读完老师推荐的青少年版的《红楼梦》，字数总计约为$$20$$万字，她的阅读速度可以达到$$500$$字/分钟，她需要平均每天坚持阅读（~ ）分钟左右，才能完成阅读计划． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$500$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$200000\\div 500\\div 8=50$$（分钟），所以她需要平均每天坚持阅读$$50$$分钟才能完成阅读计划． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "385", "queId": "96bda4b0141b43bc8e12fe59c6d60073", "competition_source_list": ["2019年第24届YMO五年级竞赛决赛第6题3分", "2020年第24届YMO五年级竞赛决赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小$$Y$$、小$$M$$、小$$O$$、小$$T$$四人中只有$$1$$人会开车．小$$Y$$说：``我会开''．小$$M$$说：``我不会开''．小$$O$$说：``小$$Y$$不会开''．小$$T$$什么也没说．已知小$$Y$$、小$$M$$、小$$O$$三人只有一人说了真话．会开车的是． ", "answer_option_list": [[{"aoVal": "A", "content": "小$$Y$$ "}], [{"aoVal": "B", "content": "小$$M$$ "}], [{"aoVal": "C", "content": "小$$O$$ "}], [{"aoVal": "D", "content": "小$$T$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["对比发现，小$$O$$与小$$Y$$说的矛盾，相互对立， 则小$$O$$与小$$Y$$必一对一错， 则小$$M$$说假话，则小$$M$$会开车，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1074", "queId": "0f55a722f9a544ddafcf00b266282d95", "competition_source_list": ["2008年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "两根电线杆之间相隔115米，在它们之间等距离增加22根电线杆后，第2根和第16根电线杆之间相隔 ( )米. ", "answer_option_list": [[{"aoVal": "A", "content": "68 "}], [{"aoVal": "B", "content": "70 "}], [{"aoVal": "C", "content": "71 "}], [{"aoVal": "D", "content": "72 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题"], "answer_analysis": ["24根电线杆之间有23个间隔，每个间隔的长是$$115\\div 23=5$$(米)，又第2根和第16根电线杆之间有14个间隔，所以，它们相隔$$5\\times 14=70$$(米) "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "484", "queId": "b850000f15514a08992ffaf46c7d12b4", "competition_source_list": ["2020年第24届YMO三年级竞赛决赛第8题3分", "2017年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$至$$10$$这$$10$$个整数中，至少取个数，才能保证其中有两个数的和等于$$10$$。（把和为$$10$$的两个数看成是一双袜子，利用抽屉原理来做） ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->简单抽屉原理"], "answer_analysis": ["先找到和为$$10$$的组合：$$(1$$，$$9)$$、$$(2$$，$$8)$$、$$(3$$，$$7)$$、$$(4$$，$$6)$$，剩下$$5$$、$$10$$，若前面四组数中各取一个，后面$$5$$、$$10$$全取，这样还不能满足题目要求的两数和为$$10$$，如果再取剩下四个中的任意一个，必定会凑出和为$$10$$的组合，故最少要取$$6+1=7$$（个）。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3430", "queId": "fc407edef8184f58ae240303ce42114c", "competition_source_list": ["2012年第8届全国新希望杯五年级竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "某十字路口的交通信号灯，黄灯亮$$3$$秒，绿灯亮$$9$$秒，红灯亮$$24$$秒，那么某一时刻亮绿灯的可能性为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{12}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->计数求概率"], "answer_analysis": ["亮绿灯的可能性为：$$9\\div (3+9+24)=\\frac{1}{4}$$，故选$$\\text{B}$$． ~ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3439", "queId": "ceef9867d23e4dd9ace97ece28971669", "competition_source_list": ["2012年第10届创新杯四年级竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "整数$$2012$$具有如下的性质：它是$$4$$的倍数，它的各位数字和为$$5$$．在具有这两个性质的整数中，按由小到大顺序排列，$$2012$$是第个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["计数------分类枚举． 一位数：$$0$$个， 两位数：$$32$$，共$$1$$个， 三位数：$$104$$、$$140$$、$$212$$、$$320$$、$$500$$，共$$5$$个， 四位数：$$1004$$、$$1040$$、$$1112$$、$$1220$$、$$1400$$、$$2012\\cdots $$ 故$$2012$$为第$$12$$个． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2077", "queId": "dd94a0a3269843598a48014e32466c04", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（五）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "现有甲、乙两个仓库，甲仓库的货物是乙仓库的$$2.5$$倍．如果甲仓库减少$$100$$吨，乙仓库增加$$80$$吨，则两个仓库的货物一样重，两个仓库的货物一共有吨． ", "answer_option_list": [[{"aoVal": "A", "content": "$$400$$ "}], [{"aoVal": "B", "content": "$$420$$ "}], [{"aoVal": "C", "content": "$$440$$ "}], [{"aoVal": "D", "content": "$$460$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["甲仓库比乙仓库多$$100+80=180$$(吨)，乙仓库：$$180\\div (2.5-1)=120$$（吨），甲仓库：$$120+180=300$$（吨），$$120+300=420$$（吨）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2973", "queId": "a9b4639778b54679972496c3c1d0f969", "competition_source_list": ["2014年第10届全国新希望杯五年级竞赛复赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "欣欣、希希、望望和贝贝四人共给希望小学捐赠了$$49$$本课外书，其中欣欣捐赠的课外书本数是希希和望望捐赠的课外书本数之和，希希捐赠的课外书本数是望望和贝贝捐赠的课件书本数之和．如果贝贝捐赠了$$3$$本课外书，那么欣欣捐赠了（~~~ ）本课外书． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$23$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["欣+希+望+贝=$$49$$ 欣=希+望=$$2$$望+$$3$$ 希=望+贝=望+$$3$$ 则$$4$$望+$$9=49$$，所以望=$$10$$， 那么欣欣捐了$$2\\times 10+3=23$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1670", "queId": "9a84c66d5f004d55a4ba4ca886da2658", "competition_source_list": ["2020年新希望杯六年级竞赛（2月）第17题"], "difficulty": "1", "qtype": "single_choice", "problem": "在一个奇怪的动物村庄里只住着猫和狗，有$$20 \\% $$的狗认为它们自己是猫，有$$20 \\% $$的猫认为它们自己是狗，其余的猫和狗都是正常的，所有的猫和狗中，有$$36 \\% $$认为自己是猫，如果这个奇怪的动物村庄里狗比猫多$$140$$只，那么狗有只． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$180$$ "}], [{"aoVal": "C", "content": "$$80$$ "}], [{"aoVal": "D", "content": "$$220$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$80 \\%$$的猫认为自己是猫，$$20 \\%$$的狗认为自己是猫，猫狗混合后有$$36 \\%$$的认为自己是猫，所以狗：猫$$=\\left(80 \\%-36 \\%\\right):\\left(36 \\%-20 \\%\\right)=11:4$$， 狗：$$140\\div \\left( 11-4 \\right)\\times 11=220$$（只）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "765", "queId": "68c7685cb8aa4d418419879aae54f2c9", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第22题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\underbrace{99\\cdots99}_{2021个}\\times\\underbrace{99\\cdots99}_{2021个}+1\\underbrace{99\\cdots99}_{2021个}$$，能被$$10^{}n$$整除，不能被$$10^{n+1}$$整除，则$$n=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2020$$ "}], [{"aoVal": "B", "content": "$$2021$$ "}], [{"aoVal": "C", "content": "$$4040$$ "}], [{"aoVal": "D", "content": "$$4042$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->整除->整除特征->整除特征综合"], "answer_analysis": ["$$\\underbrace{99\\cdots99}_{2021个}\\times\\underbrace{99\\cdots99}_{2021个}+1\\underbrace{99\\cdots99}_{2021个}$$ $$=\\underbrace{99\\cdots99}_{2021个}\\times\\underbrace{99\\cdots99}_{2021个}+1\\underbrace{00\\cdots00}_{2021个}+\\underbrace{99\\cdots99}_{2021个}$$ $$=\\underbrace{99\\cdots99}_{2021个}\\times(\\underbrace{99\\cdots99}_{2021个}+1)+1\\underbrace{00\\cdots00}_{2021个}$$ $$=\\underbrace{99\\cdots99}_{2021个}\\times 1\\underbrace{00\\cdots00}_{2021个}+1\\underbrace{00\\cdots00}_{2021个}$$ $$=(\\underbrace{99\\cdots99}_{2021个}+1)\\times 1\\underbrace{00\\cdots00}_{2021个}$$ $$=1\\underbrace{00\\cdots00}_{2021个}+\\times 1\\underbrace{00\\cdots00}_{2021个}$$ $$=1\\underbrace{00\\cdots00}_{4042个}$$ $$n$$最大取$$4042$$． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1698", "queId": "9611ffc45cfa43d1b9383b16546a1b10", "competition_source_list": ["2009年全国迎春杯小学中年级四年级竞赛初赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "老师买了同样数目的田格本、横线本和练习本．他发给每个同学$$1$$个田格本、$$3$$个横线本和$$5$$个练习本．这时横线本还剩$$24$$个，那么田格本和练习本共剩了~\\uline{~~~~~~~~~~}~个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$54$$ "}], [{"aoVal": "D", "content": "$$56$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["同倍数变化问题．把$$1$$个田格本和$$5$$个练习本捆绑成一组，那么每发$$3$$个横格本，就发一组田格本和练习本，田格本和练习本的总量是横线本的$$2$$倍，每次发的数量也是$$2$$倍，所以剩下的也是$$2$$倍，即$$48$$本． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1624", "queId": "52ca58de9a024ff3b74c267e240c7916", "competition_source_list": ["2013年第9届全国新希望杯小学高年级五年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "在期末考试中，方方五科的总分是$$445$$分，除数学外的四科平均分是$$87.25$$分，方方的数学是（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$95$$分 "}], [{"aoVal": "B", "content": "$$96$$分 "}], [{"aoVal": "C", "content": "$$97$$分 "}], [{"aoVal": "D", "content": "$$98$$分 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型"], "answer_analysis": ["$$445-87.25\\times 4=96$$．~ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "967", "queId": "f5048b4ef467401b925e9f27b852ece2", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在整数$$1$$、$$2$$、$$3$$、$$\\cdots \\cdots 999$$、$$1000$$中，所有偶数之和比所有奇数之和多． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$500$$ "}], [{"aoVal": "D", "content": "$$1000$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$1$$到$$1000$$中，一共有$$500$$个奇数，$$500$$个偶数， 奇数之和$$=1+3+5+\\cdots \\cdots +997+999$$， 偶数之和$$=2+4+6+\\cdots \\cdots +998+1000$$， 可以用比每个奇数大$$1$$的偶数去减该奇数，正好所有的奇数都可以找到唯一比它大$$1$$的偶数， 所有偶数之和比所有奇数之和多$$500$$， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3143", "queId": "09ef30e2f87641fa950e0e8d329ac6e7", "competition_source_list": ["2013年小机灵杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "罗马数字是由罗马人发明的，它一共由（ ）个数字组成。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->加法原理"], "answer_analysis": ["解：罗马数字是由罗马人发明的，它一共由$$7$$个数字组成； 故选:C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1232", "queId": "4ff40ea4baa74c22af81246b414424fc", "competition_source_list": ["六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "有两根同样长的钢管，第一根用去$$\\frac{3}{10}$$米，第二根用去$$\\frac{3}{10}$$，比较这两根钢管剩下部分的长度，结果是( ) ", "answer_option_list": [[{"aoVal": "A", "content": "第一根钢管剩下的部分长些 "}], [{"aoVal": "B", "content": "第二根钢管剩下的部分长些 "}], [{"aoVal": "C", "content": "两根钢管剩下的部分同样长 "}], [{"aoVal": "D", "content": "以上三种结果都有可能 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["当两根钢管都是$$1$$米时，则剩下的钢管相等；当两根钢管都大于$$1$$米时，则用去$$\\frac{3}{10}$$米的那一根剩下的多；当两根钢管都小于$$1$$米时，用去$$\\frac{3}{10}$$的那一根剩下的多。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3293", "queId": "defb04cc83bc444ab4855ff5c37e78f5", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛初赛模拟"], "difficulty": "1", "qtype": "single_choice", "problem": "六个小朋友排成一排照相，其中有四个男生和两个女生，两个女生必须站在一起而且不能站在边上，则一共有~\\uline{~~~~~~~~~~}~种不同的排列方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$144$$ "}], [{"aoVal": "D", "content": "$$240$$ "}]], "knowledge_point_routes": ["拓展思维->能力->抽象概括"], "answer_analysis": ["$$\\text{P}_{4}^{4}\\times \\text{P}_{3}^{2}=4\\times 3\\times 2\\times 3\\times 2=144$$． 另一种方法：先捆绑，再插空，$$2!\\times 4!\\times C_{4}^{3}=144$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3208", "queId": "3037baf8c9944deb9eb8974698a96b9b", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级四年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "一共有四种硬币，penny（一美分），nickel（五美分），dime（十美分），quarter（二十五美分），每种硬币你可以取任意多个（也可以是$$0$$个）放在你的袋子里（在放这些硬币之前，你的袋子是空的），规则是你袋子例的硬币无法凑成一美元（一美元$$=100$$美分）．你的袋子里最多可以有多少钱？ ", "answer_option_list": [[{"aoVal": "A", "content": "＄$$0.90$$ "}], [{"aoVal": "B", "content": "＄$$0.99$$ "}], [{"aoVal": "C", "content": "＄$$1.19$$ "}], [{"aoVal": "D", "content": "＄$$1.29$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$3$$个quarter（二十五美分）$$+4$$个dime（十美分）$$+4$$个penny（一美分）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3033", "queId": "c12412b0521b420e8b2803884a2ea2db", "competition_source_list": ["2018~2019学年浙江杭州西湖区杭州市行知小学五年级上学期期中期中竞赛第19题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果甲数$$\\times 1.2=$$乙数$$\\div 1.2$$（甲数、乙数$$\\ne 0$$），那么． ", "answer_option_list": [[{"aoVal": "A", "content": "甲数$$=$$乙数 "}], [{"aoVal": "B", "content": "甲数$$\\textgreater$$乙数 "}], [{"aoVal": "C", "content": "甲数$$\\textless{}$$乙数 "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->小数->小数乘除->小数除法运算"], "answer_analysis": ["$$1.2=\\frac{6}{5}$$，即甲数$$\\times \\frac{6}{5}=$$乙数$$\\times \\frac{5}{6}$$（且甲、乙$$\\ne 0$$），根据积的特性，当积一定时，一个因数越大，则另一个因数越小； 那么$$\\frac{6}{5}\\textgreater\\frac{5}{6}$$，则甲$$\\textless{}$$乙． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "345", "queId": "84d92520b8804f53ab2be91393d55dd4", "competition_source_list": ["2014年迎春杯四年级竞赛初赛", "2014年迎春杯三年级竞赛初赛", "2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$20$$间房间，有的开着灯，有的关着灯，在这些房间里的人都希望与大多数房间保持一致。现在，从第一间房间的人开始，如果其余$$19$$间房间的灯开着的多，就把灯打开，否则就把灯关上，如果最开始开灯与关灯的房间各$$10$$间，并且第一间的灯开着。那么， 这$$20$$间房间里的人轮完一遍后，关着灯的房间有（ ）间。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["解：因为最开始开灯和关灯的房间各是$$10$$间，由于第一间的灯是开着的， 所以，第一间人看到的，开灯的$$9$$间，关灯的$$10$$间， 之后，他就关灯， 以后无论开灯的出来看，还是关灯的出来看，始终关灯的多， 即：一轮结束，灯全部会关闭。 故选：A。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2835", "queId": "a86f33a95fc24cfebfd1f16c2b4b7378", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（三）第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "下列说法正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "一个分数的分母越小，它的分数单位就越小 "}], [{"aoVal": "B", "content": "分数都比整数小 "}], [{"aoVal": "C", "content": "假分数都大于$$1$$ "}], [{"aoVal": "D", "content": "$$2$$米的$$\\frac{1}{3}$$和$$1$$米的$$\\frac{2}{3}$$一样长 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$2$$米的$$\\frac{1}{3}$$和$$1$$米的$$\\frac{2}{3}$$一样长． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2967", "queId": "c4d0b3b5732949fca460651b6e45a280", "competition_source_list": ["2016年全国学而思杯小学高年级六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "找规律：有一组式子：$$a^{2}$$，$$-\\dfrac{{{a}^{3}}}{2}$$，$$\\dfrac{{{a}^{4}}}{3}$$，$$-\\dfrac{{{a}^{5}}}{4}$$，$$\\cdots$$，从左到右数的第$$10$$个式子是下面算式的第( ~ ~ ~ ~)个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\dfrac{{{a}^{11}}}{10}$$ "}], [{"aoVal": "B", "content": "$$-\\dfrac{{{a}^{11}}}{10}$$ "}], [{"aoVal": "C", "content": "$$-\\dfrac{{{a}^{10}}}{11}$$ "}], [{"aoVal": "D", "content": "$$-\\dfrac{{{a}^{11}}}{11}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["分别观察分子、分母的规律，以及符号的规律．从左到右数的第$$10$$个式子是$$-\\dfrac{{{a}^{11}}}{10}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "802", "queId": "5bc30212ca964e3bbd97831632eea516", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛A卷第7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "一块砖的长、宽、高分别为$$18$$厘米、$$12$$厘米、$$6$$厘米，要堆成一个正方体，至少需要这样的砖块． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->几何模块->立体图形->长方体与正方体->长方体与正方体基本概念运用->正方体的基本概念"], "answer_analysis": ["正方体的棱长是$$18$$、$$12$$、$$6$$的倍数， 其中$$36$$是它们的最小公倍数， $$36\\div 18=2$$（层）， $$36\\div 12=3$$（层）， $$36\\div 6=6$$（层）， 需要砖块为：$$2\\times 3\\times 6=36$$（块）． 故答案选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2465", "queId": "11699cb51a6c4589b2bf092fe2ae694d", "competition_source_list": ["2019年浙江杭州滨江区杭州江南实验学校五年级竞赛模拟（江南杯）第6题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$9\\div 11$$的商的小数部分第$$50$$位上的数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$9$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数的认识->认、读、写数->小数->循环小数", "拓展思维->思想->对应思想"], "answer_analysis": ["列竖式 故$$9\\div 11=0.\\dot{8}\\dot{1}$$． 周期为$$2$$故$$50\\div 2=25$$， 故第$$50$$位为$$1$$． 故答案为：$$\\text{B}$$选项． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1129", "queId": "98beb05ae828404c9bffb7dbea15e240", "competition_source_list": ["2017年河南郑州豫才杯四年级竞赛初赛第14题"], "difficulty": "1", "qtype": "single_choice", "problem": "在班级图书角里，聪聪特别喜欢《哈利波特与魔法石》，字数总计达$$200$$千字，他打算在国庆的$$7$$天假期里看完，他的阅读速度可以达到每分钟$$300$$字．按照这个计划，他平均每天要读（~ ）才能保证读完这本书． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$分钟~~~ "}], [{"aoVal": "B", "content": "$$50$$分钟~~~ "}], [{"aoVal": "C", "content": "$$1$$小时$$30$$分钟 "}], [{"aoVal": "D", "content": "$$100$$分钟 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->双归一问题"], "answer_analysis": ["$$200000\\div 300\\div 7\\approx 100$$分钟． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "768", "queId": "ff8080814518d5240145192480f704c5", "competition_source_list": ["2014年全国迎春杯五年级竞赛初赛第8题", "2014年全国迎春杯六年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$4$$个质数的积是它们和的$$11$$倍，则它们的和为（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$46$$ "}], [{"aoVal": "B", "content": "$$47$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "没有符合条件的数 "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["由已知条件，$$4$$个质数中一定有$$11$$，那么则满足$$a\\times b\\times c=a+b+c+11$$，其中$$a$$、$$b$$、$$c$$都是质数．若$$a$$、$$b$$、$$c$$都是奇数，那么等式左边是奇数，右边为偶数，矛盾．若$$a$$、$$b$$、$$c$$中有$$1 $$个偶数，那么一定是$$2$$．即$$a\\times b\\times 2=a+b+2+11$$，此时，根据奇偶性，$$a$$、$$b$$中也必有一个偶数为$$2$$，解得$$a$$、$$b$$、$$c$$、$$d$$为$$2$$、$$2$$、$$5$$、$$11$$和为$$20$$．选项中ABC均不符合条件，故选D． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2120", "queId": "ec11d2d42e05470a851a48ef537807ab", "competition_source_list": ["2015年第13届全国创新杯六年级竞赛第6题", "小学高年级六年级其它2015年数学思维能力等级测试初试第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "六年级（一）班三名同学，结伴骑自行车从学校出发到东湖磨山春游，已经走了全程的$$\\frac{2}{7}$$，如果再行$$9$$千米，已行路程和剩下路程之比为$$5:2$$，那学校到东湖磨山的路程是（~ ）千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["第一次走了$$\\frac{2}{7}$$，第二次一共走了$$\\frac{5}{7}$$，则全程是$$9\\div \\left( \\frac{5}{7}-\\frac{2}{7} \\right)=21$$千米． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1738", "queId": "7312aff182924b72a2092744292a6f02", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第12题4分", "2018年IMAS小学高年级竞赛（第一轮）第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "\\uline{小杨}在期末考试中﹐语文与数学两科的平均分数为$$97$$分，而英语考了$$94$$分，请问他三科的平均分数为多少分？（~ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$94$$ "}], [{"aoVal": "B", "content": "$$94.5$$ "}], [{"aoVal": "C", "content": "$$95$$ "}], [{"aoVal": "D", "content": "$$95.5$$ "}], [{"aoVal": "E", "content": "$$96$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["可知\\uline{小杨}在期末考试中，语文与数学两科的总分为$$97\\times 2=194$$分， 所以三科的总分为$$194+94=288$$分， 即三科的平均分为$$288\\div 3=96$$分． 故选$$\\text{E}$$． ", "可知小杨在期末考试中，语文与数学两科的平均分数为$$97$$分，而英语考了$$94$$分，
\n所以语文与数学两科总共比英语多了$$\\left( 97-94 \\right)\\times 2=6$$分，
\n所以三科的总平均分数比英语多了$$6\\div 3=2$$分，
\n即三科的平均分为$$94+2=96$$分．
\n故选$$\\text{E}$$．
"], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1246", "queId": "f5b9913961ec42b888bac23828b8b3f7", "competition_source_list": ["2013年第25届广东广州五羊杯六年级竞赛第10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "莉莉在期末考试中语数英的平均分是$$85$$分，已知她三科的成绩均不相同，且分数都是整数．最高分的科目语文比最低分的科目英语高$$4$$分，则莉莉的英语考了分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$82$$ "}], [{"aoVal": "B", "content": "$$83$$ "}], [{"aoVal": "C", "content": "$$84$$ "}], [{"aoVal": "D", "content": "$$85$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["三科总分为$$85\\times 3=255$$（分）． 设最低分的科目英语为$$x$$分， 则有$$x+4\\textgreater255-2x-4\\textgreater x$$， 解得$$\\begin{cases}x\\textless{}84 x\\textgreater82 \\end{cases}$$， 所以$$x=83$$（分）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2993", "queId": "8f2a4f3724184d30a2ce768c0824fa02", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "3", "qtype": "single_choice", "problem": "计算$$\\frac{1}{1\\times 2\\times 3}+\\frac{1}{2\\times 3\\times 4}+\\frac{1}{3\\times 4\\times 5}+...+\\frac{1}{8\\times 9\\times 10}$$=~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{9}{10}$$ "}], [{"aoVal": "B", "content": "$$\\frac{9}{20}$$ "}], [{"aoVal": "C", "content": "$$\\frac{11}{45}$$ "}], [{"aoVal": "D", "content": "$$\\frac{22}{45}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->归纳总结->归纳推理"], "answer_analysis": ["因为$$\\frac{1}{n\\left( n+1 \\right)\\left( n+2 \\right)}=\\frac{1}{2}\\times \\frac{\\left( n+2 \\right)-n}{n\\left( n+1 \\right)\\left( n+2 \\right)}=\\frac{1}{2}\\left[ \\frac{1}{n\\left( n+1 \\right)}-\\frac{1}{\\left( n+1 \\right)\\left( n+2 \\right)} \\right]$$ 所以原式$$=\\frac{1}{2}\\times \\left( \\frac{1}{1\\times 2}-\\frac{1}{2\\times 3}+\\frac{1}{2\\times 3}-\\frac{1}{3\\times 4}+\\frac{1}{3\\times 4}-\\frac{1}{4\\times 5}...+\\frac{1}{8\\times 9}-\\frac{1}{9\\times 10} \\right)$$ $$=\\frac{1}{2}\\times \\left[ \\frac{1}{2}-\\frac{1}{90} \\right]=\\frac{11}{45}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3229", "queId": "3556a2d16ab448259b1b8ebf87fd2740", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$600$$的因数有很多，这些因数中有个是$$6$$的倍数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->因数与倍数基础"], "answer_analysis": ["$$600$$的因数有$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$8$$，$$10$$，$$12$$，$$15$$，$$20$$，$$24$$，$$25$$，$$30$$，$$40$$，$$50$$，$$60$$，$$75$$，$$100$$，$$120$$，$$150$$，$$200$$，$$300$$，$$600$$，其中是$$6$$的倍数的有$$6$$，$$12$$，$$24$$，$$30$$，$$60$$，$$120$$，$$150$$，$$300$$，$$600$$这$$9$$个． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2582", "queId": "2cb5f63a7a8641b09383b4c0857b61a8", "competition_source_list": ["2019年广东深圳全国小学生数学学习能力测评六年级竞赛初赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "糖水质量是$$900$$克，糖与水的质量比是$$1:9$$，若再加入$$100$$克糖，则糖与水的质量比是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9:11$$ "}], [{"aoVal": "B", "content": "$$1:10$$ "}], [{"aoVal": "C", "content": "$$19:81$$ "}], [{"aoVal": "D", "content": "$$19:100$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["依题意得，糖水质量是$$900$$克，糖与水的质量比是$$1:9$$，那么糖的质量是$$90$$克，水的质量是$$900-90=810$$（克），又加入了$$100$$克糖，现在糖与水的质量比是：$$(90+100):810=190:810=19:81$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1013", "queId": "02d40baa7e964de5adefae753b2dd973", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（二）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "某货运公司运送一批货物，原计划安排$$18$$辆小卡车和$$12$$辆大卡车刚过运$$4$$次，已知$$2$$辆大卡车与$$5$$辆小卡车装的重量相同，现在只能派出$$8$$辆小卡车，需运次才能把货物运完． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->列方程解应用题->列方程解应用题等量代换->等量代换传递型"], "answer_analysis": ["$$12$$辆大卡车装的重量等于$$12\\div 2\\times 5=30$$辆小卡车，则原计划运$$1$$次的货物现在要运$$\\left( 18+30 \\right)\\div 8=6$$（次），所以一共需运$$6\\times 4=24$$次． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3215", "queId": "6bbb925aea634ba68d6b497215983c58", "competition_source_list": ["2004年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$50$$束鲜花中，$$16$$束有马蹄莲，$$21$$束有白兰花，$$15$$束有月季花，有$$7$$束中既有月季花又有马蹄莲，有$$8$$束中既有马蹄莲又有白兰花，有$$10$$束中既有月季花又有白兰花，还有$$5$$束鲜花中，月季花、马蹄莲、白兰花都有。则这$$50$$束鲜花中，上述三种花都没有的花束有（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$束 "}], [{"aoVal": "B", "content": "$$18$$束 "}], [{"aoVal": "C", "content": "$$19$$束 "}], [{"aoVal": "D", "content": "$$20$$束 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->容斥原理->三量容斥"], "answer_analysis": ["由容斥原理：$$50-\\left( 16+15+21-7-8-10+5 \\right)=18$$（束），选$$B $$。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1485", "queId": "5636e4ca92fa46778dda4efcdd1f91d9", "competition_source_list": ["2012年第10届全国创新杯小学高年级六年级竞赛第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "~有$$n$$个自然数（数可以重复）其中包括$$2012$$，不包括$$0$$，这$$n$$个自然数的平均数是$$572$$，如果去掉$$2012$$后，剩下（$$n-1$$）个数的平均数是$$412$$，那么这$$n$$个数最大的数可以是（~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2012$$ "}], [{"aoVal": "B", "content": "$$4024$$ "}], [{"aoVal": "C", "content": "$$3700$$ "}], [{"aoVal": "D", "content": "$$3800$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["根据题意可以得到：$$2012+412\\left( n-1 \\right)=572n$$，$$n=10$$，十个数总和为$$5720$$，要使最大的数尽量大，则其他的尽量小，所以最大数最大值为 $$5720-2012-1\\times 8=3700$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "270", "queId": "c35c69c9d1d54610a3cbdba04a7b550b", "competition_source_list": ["2014年全国学而思杯一年级竞赛第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$5$$名分别来自美国、俄国、中国、日本国和韩国的运动员参加冬奥会滑雪决赛，比赛结束后： 美国运动员说：俄国运动员紧跟在我后面； 俄国运动员说：我才不是最后一名； 中国运动员说：我比日本国人和美国人都快； 韩国运动员说：有三个人比我先到达终点． 那么， 哪个国家运动员是第一名？ ", "answer_option_list": [[{"aoVal": "A", "content": "美国 "}], [{"aoVal": "B", "content": "俄国 "}], [{"aoVal": "C", "content": "中国 "}], [{"aoVal": "D", "content": "日本 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数学广角->推理->解决简单逻辑推理问题"], "answer_analysis": ["中国比日本、美国快，美国比俄国快，所以日本、美国、俄国都不是第一，而韩国也不是第一个到的，所以中国是第一． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1056", "queId": "2a290e7bb9ba4275a2b40c7488b86997", "competition_source_list": ["2003年六年级竞赛创新杯", "2003年第1届创新杯六年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "有两根钢管，第一根用去$$\\frac{3}{10}$$米，第二根用去$$\\frac{3}{10}$$，比较这两根钢管剩下部分的长度，结果是( ) ", "answer_option_list": [[{"aoVal": "A", "content": "第一根钢管剩下的部分长些 "}], [{"aoVal": "B", "content": "第二根钢管剩下的部分长些 "}], [{"aoVal": "C", "content": "两根钢管剩下的部分同样长 "}], [{"aoVal": "D", "content": "以上三种结果都有可能 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题"], "answer_analysis": ["现在只知道第一根钢管用去$$\\frac{3}{10}$$米，第二根钢管用去$$\\frac{3}{10}$$，并不知道这两只钢管的长度，它们可能很长，也可能很短，这样就可以任意由我们来假设这两根钢管的长度。我们先来看看选项$$A$$是否成立，$$A$$要求第一根钢管剩下的部分长些，那我们可以假设第一根钢管的长为$$10$$米，第二根钢管的长为$$1$$米，显然这样就可以满足选项$$A$$。再来看看选项$$B$$，$$B$$要求第二根钢管剩下的部分长些，那我们可以假设第一根钢管的长为$$1$$米，第二根钢管剩下的部分同样长，我们假设第一根钢管的长为$$1$$米，那么用去$$\\frac{3}{10}$$米后还剩下$$\\frac{7}{10}$$米，若要第二根钢管剩下的长度也为$$\\frac{7}{10}$$米，只要第二根钢管原来的长为$$\\frac{7}{10}\\div \\left( 1-\\frac{3}{10} \\right)=1$$（米）即可，选项$$C$$也可以满足。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "156", "queId": "a260b3b829ef40db83cdb03306a4afc0", "competition_source_list": ["2017年第13届湖北武汉新希望杯小学高年级五年级竞赛决赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙、丁四人进行象棋比赛，并决出了一、二、三、四名．已知：①乙比丙的名次靠前；②甲和丁经常一起打篮球；③第一名和第三名在这次比赛时才认识；④第二名不会骑自行车，也不喜欢打篮球；⑤甲和丙每天一起骑自行车上班． 获得第四名的是谁？ ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "丁 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理", "Overseas Competition->知识点->组合模块->逻辑推理->条件型逻辑推理->表格法"], "answer_analysis": ["由②④⑤推断出乙为第二名，由于甲丁一起打球，甲丙一起上班，再由③推断出丙丁一个是第一名，另一个为第三名，则第四名为甲． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1447", "queId": "7138f7afdb6a41f2ae1cf364599d9d09", "competition_source_list": ["2009年第7届创新杯四年级竞赛初赛第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "黑板上写有一个数，男同学从黑板前走过时，把它乘$$3$$再减去$$14$$，擦去原数，换上答案；女同学从黑板前走过时，把它乘$$2$$再减去$$7$$，擦去原数，换上答案． 全班$$25$$名男生和$$15$$名女生都走过以后，老师把最后的数乘$$5$$，减去$$5$$，结果是$$30$$． 那么黑板上最初的数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["全体同学走后，黑板上的数是$$\\left(30+5\\right)\\div5=7$$；最后一名学生走过之前，黑板上的数是$$\\left(7+7\\right)\\div2=7$$，或$$\\left(7+14\\right)\\div3=7$$，总之，最后一名学生（即第$$40$$名学生）走过之前，黑板上的数还是$$7$$． 同理，第$$39$$名学生来到之时，黑板上的数还是$$7\\cdots \\cdots $$由此可知，第$$1$$名学生到来之时，黑板上的数还是$$7$$，即黑板上最初的数是$$7$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1840", "queId": "81b1619d26fb4f8cbcc718a64acdad55", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛三年级竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "（2018 YMO, Grade 3, Question\\#4） According to this set of number, $$3$$，$$1$$，$$2$$，$$3$$，$$1$$，$$2$$，$$\\cdots \\cdots $$, what will be the $$53$$\\textsuperscript{rd} one? ． 有一列数：$$3$$，$$1$$，$$2$$，$$3$$，$$1$$，$$2$$，$$\\cdots \\cdots $$那么第$$53$$个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["这列数按照``$$3$$，$$1$$，$$2$$''的周期排列，则，$$53\\div 3=17$$（组）$$\\cdots \\cdots 2$$（个）， 所以第$$53$$个数位这个周期的第$$2$$个数，即为$$1$$． 故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1236", "queId": "5491c0b36fb84f4a8a9b2adbda948a8e", "competition_source_list": ["2015年第13届全国创新杯小学高年级五年级竞赛复赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "三部同样的抽水机同时抽水，抽干一池水需用$$15$$小时，五部这样的抽水机抽干这一池子水需用（~ ~ ~ ）小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["一台功效$$\\frac{1}{45}$$，五台需要$$9$$小时． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "319", "queId": "b5f6c07698d74220b313b32a8551d650", "competition_source_list": ["2003年六年级竞赛创新杯", "2003年第1届创新杯六年级竞赛初赛第7题"], "difficulty": "0", "qtype": "single_choice", "problem": "一个长方体木块的长、宽、高分别是5厘米、4厘米、3厘米，如果用它锯成一个尽量大的正方体，那么体积比原来减少的百分数是（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "40\\% "}], [{"aoVal": "B", "content": "45\\% "}], [{"aoVal": "C", "content": "55\\% "}], [{"aoVal": "D", "content": "60\\% "}]], "knowledge_point_routes": ["拓展思维->拓展思维->几何模块->立体图形->长方体与正方体->长方体与正方体基本概念运用->正方体的体积"], "answer_analysis": ["正方形的棱长最大为3厘米，正方形的体积最大为$$3\\times 3\\times 3=27$$(立方厘米)，原来长方形体积为$$5\\times 4\\times 3=60$$(立方厘米)，那么正方形的体积比原来的体积减少$$\\left( 60-27 \\right)\\div 60\\times 100 \\%=55 \\%$$. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "974", "queId": "fa19e75a53a64413a347919c9a23fe77", "competition_source_list": ["2017年第17届世奥赛六年级竞赛决赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "通过下面算式考察个位上数字是$$5$$的数的平方： $${{15}^{2}}=225$$可写成$$100\\times 1\\times \\left( 1+1 \\right)+25$$ $${{25}^{2}}=625$$可写成$$100\\times 2\\times \\left( 2+1 \\right)+25$$ $${{35}^{2}}=1225$$可写成$$100\\times 3\\times \\left( 3+1 \\right)+25$$ $${{45}^{2}}=2025$$可写成$$100\\times 4\\times \\left( 4+1 \\right)+25$$ ······ 请你推测$${{\\left( 10n+5 \\right)}^{2}}$$可写成． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100n\\left( n+1 \\right)+25$$ "}], [{"aoVal": "B", "content": "$$100\\left( n+1 \\right)+25$$ "}], [{"aoVal": "C", "content": "$$100{{n}^{2}}+25$$ "}], [{"aoVal": "D", "content": "$$100{{n}^{2}}+100n+5$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["$${{\\left( 10n+5 \\right)}^{2}}=100n\\left( n+1 \\right)+25=100{{n}^{2}}+100n+25$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2632", "queId": "a2c11c1bed4e493c934cfaa1f69e6835", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$\\frac{{{5}^{2}}+1}{{{5}^{2}}-1}+\\frac{{{7}^{2}}+1}{{{7}^{2}}-1}+\\frac{{{9}^{2}}+1}{{{9}^{2}}-1}+...+\\frac{9{{9}^{2}}+1}{9{{9}^{2}}-1}$$=~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$48\\frac{6}{25}$$ "}], [{"aoVal": "C", "content": "$$46\\frac{6}{25}$$ "}], [{"aoVal": "D", "content": "$$\\frac{6}{25}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->裂项与通项归纳->分数裂项->分数裂差->两分数间接裂差"], "answer_analysis": ["原式$$=\\frac{{{5}^{2}}-1+2}{{{5}^{2}}-1}+\\frac{{{7}^{2}}-1+2}{{{7}^{2}}-1}+\\frac{{{9}^{2}}-1+2}{{{9}^{2}}-1}+...+\\frac{{{99}^{2}}-1+2}{{{99}^{2}}-1}$$ $$=1+\\frac{2}{{{5}^{2}}-1}+1+\\frac{2}{{{7}^{2}}-1}+1+\\frac{2}{{{9}^{2}}-1}+...+1+\\frac{2}{{{99}^{2}}-1}$$ $$=48+\\frac{2}{4\\times 6}+\\frac{2}{6\\times 8}+\\frac{2}{8\\times 10}+...+\\frac{2}{98\\times 100}$$ $$=48+\\left( \\frac{1}{4}-\\frac{1}{6} \\right)+\\left( \\frac{1}{6}-\\frac{1}{8} \\right)+\\left( \\frac{1}{8}-\\frac{1}{10} \\right)+...+\\left( \\frac{1}{98}-\\frac{1}{100} \\right)$$ $$=48+\\frac{1}{4}-\\frac{1}{100}$$ $$=48\\frac{6}{25}$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1443", "queId": "4ce7f76537774ebcbf5952b2004fff68", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛六年级竞赛决赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有含糖量为$$7 \\%$$的糖水$$600$$克，要使其含糖量加大到$$10 \\%$$，需要再加入克糖． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["不变量为水，则水：$$600\\times (1-7 \\%)=558$$（克）， 后总：$$558\\div (1-10 \\%)=620$$（克）， 加糖：$$620-600=20$$（克）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "980", "queId": "0df34268b70a4544bbf791ccb41c00d2", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛B卷第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一件工作，甲单独做用的时间比乙单独做少$$\\frac{1}{3}$$，甲和乙工作效率的比是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4:3$$ "}], [{"aoVal": "B", "content": "$$3:4$$ "}], [{"aoVal": "C", "content": "$$3:2$$ "}], [{"aoVal": "D", "content": "$$2:3$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["我们把乙的时间看做单位``$$1$$''，则甲的工作时间就是$$1-\\frac{2}{3}$$，然后分别求出他们的工作效率，进一步求出答案． $$1\\div \\left( 1-\\frac{1}{3} \\right)\\div (1\\div 1)$$$$=\\frac{3}{2}\\div 1$$$$=3:2$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3066", "queId": "b8c8520c17c74665b71065e9bd0ce0af", "competition_source_list": ["2014年陕西西安小升初西工大附中入学真卷（六）第1题3分", "2017年河南郑州东风杯竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\frac{a}{b}$$（$$a\\textgreater2$$）是一个真分数，下面各分数中最大的一个是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{a\\times 2}{b\\times 2}$$~~ "}], [{"aoVal": "B", "content": "$$\\frac{a-2}{b-2}$$~~~~~~~ "}], [{"aoVal": "C", "content": "$$\\frac{a\\div 2}{b\\div 2}$$~~~~~~~~~~~ "}], [{"aoVal": "D", "content": "$$\\frac{a+2}{b+2}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数基础->分数的性质"], "answer_analysis": ["根据分数的基本性质可知，$$\\frac{a\\times 2}{b\\times 2}$$和$$\\frac{a\\div 2}{b\\div 2}$$与$$\\frac{a}{b}$$的大小相等．$$\\frac{a}{b}$$的分子和分母同时加$$2$$得到的分数比$$\\frac{a}{b}$$大．$$\\frac{a}{b}$$的分子和分母同时减$$2$$得到的分数比$$\\frac{a}{b}$$小．所以$$\\frac{a+2}{b+2}$$最大． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2289", "queId": "c4909e6fe9624da8b3ad855554ce690c", "competition_source_list": ["2017年IMAS小学高年级竞赛（第一轮）第16题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "小华从早上$$9:00$$到公司上班，下午$$5:00$$下班．请问在此期间分针转过的度数比时针转过的度数多了多少度？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$1200$$ "}], [{"aoVal": "C", "content": "$$1320$$ "}], [{"aoVal": "D", "content": "$$2640$$ "}], [{"aoVal": "E", "content": "$$2880$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->时钟问题->时钟上的追及问题->已知时间求角度"], "answer_analysis": ["小华上班时间为$$8$$小时， 此期间分针转了$$8$$圈， 即$$8\\times 360=2880$$度， 而时针每小时转$$30$$度， $$8$$小时共转了$$8\\times 30=240$$度， 分针转过的度数比时针转过的度数多了$$2880-240=2640$$度， 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1815", "queId": "edaa60f19a744655870085ecf372dc3e", "competition_source_list": ["2018年全国小学生数学学习能力测评六年级竞赛初赛第8题3分", "2018年四川成都锦江区四川师范大学附属第一实验中学小升初模拟8第3题3分", "2018年湖南长沙小升初数学入学试卷第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "王师傅加工一批零件，$$\\frac{1}{2}$$小时加工了这批零件的$$\\frac{3}{8}$$，全部加工完还需要小时．（2018.SDYZ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$1\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{10}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{6}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->简单工程问题->基本工程问题"], "answer_analysis": ["首先根据王师傅加工一批零件，$$\\frac{1}{2}$$小时加工了这批零件的$$\\frac{3}{8}$$，工作效率$$=$$工作量$$\\div $$工作时间，求出每小时加工这批零件的几分之几；求出剩下的工作量，然后根据工作时间$$=$$工作量$$\\div $$工作效率，求出全部加工完还需要多少小时即可． 解；$$\\frac{3}{8}\\div \\frac{1}{2}=\\frac{3}{4}$$ $$(1-\\frac{3}{8})\\div \\frac{3}{4}$$ $$=\\frac{5}{8}\\div \\frac{3}{4}$$ $$=\\frac{5}{6}$$（小时） 答:全部加工完还需要$$\\frac{5}{6}$$小时． 故选$$\\rm D$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3009", "queId": "fc16173dbd8c47a392df6c54329c7dfe", "competition_source_list": ["2019年亚洲国际数学奥林匹克公开赛（AIMO）四年级竞赛决赛第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在以下那一道除法中﹐除数不能整除被除数?（以英文字母作答） In which of the following divisions， the dividend is not divisible by the divisor? ", "answer_option_list": [[{"aoVal": "A", "content": "$$45825\\div 33$$ "}], [{"aoVal": "B", "content": "$$29484\\div 36$$ "}], [{"aoVal": "C", "content": "$$41552\\div 56$$ "}], [{"aoVal": "D", "content": "$$39715\\div 65$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数乘除->整数除法运算->表外除法计算"], "answer_analysis": ["$$\\text{B}$$答案为$$819$$，$$\\text{C}$$答案为$$742$$，$$\\text{D}$$答案为$$611$$，$$BCD$$都能被整除． 故选$$\\text{A}$$. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2102", "queId": "e724a20e8f4145ae8906f41c1d3eb1ca", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（三）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "祖父年龄$$67$$岁，$$3$$个孙子的年龄分别是$$15$$岁，$$13$$岁，$$9$$岁．多少年后$$3$$个孙子的年龄之和等于祖父的年龄？~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题基本关系->年龄和"], "answer_analysis": ["设$$x$$年后$$3$$个孙子的年龄之和等于祖父的年龄． $$67+x=15+13+9+3x$$，$$x=15$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "679", "queId": "551c1e1c3958413caad7bd88af3ba0c7", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（三）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "房间里有一盏灯亮着，突然停电了，第$$1$$个人进来后拉了一下开关，第$$2$$个人进来后拉了$$2$$下开关，第$$3$$个人进来后拉了$$3$$下开关$$\\cdots \\cdots $$，第$$2014$$个人进来后拉了$$2014$$下开关，来电后，这盏灯是着的． ", "answer_option_list": [[{"aoVal": "A", "content": "开 "}], [{"aoVal": "B", "content": "关 "}], [{"aoVal": "C", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->拓展��维->数论模块->奇数与偶数->奇数与偶数的加减规律"], "answer_analysis": ["第$$2014$$个人进来后，一共拉了开关$$\\left( 1+2+3+\\ldots \\ldots +2014 \\right)$$下，这$$2014$$个加数中有$$2014\\div 2=1007$$个偶数，任意数量的偶数之和一定是偶数；有$$1007$$个奇数，奇数个奇数的和一定是奇数．因为偶数$$+$$奇数$$=$$奇数，而这盏灯开始是亮着的，则拉奇数下开关，灯一定是关着的． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1221", "queId": "1e73db58a2d14c44adb4e6bbf2ebe9fb", "competition_source_list": ["2017年河南郑州联合杯六年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$10$$克的盐放入$$100$$克的水中，盐占盐水的：（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{10}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{11}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{9}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{8}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->求分率"], "answer_analysis": ["盐水共$$10+100=110$$克，则盐占盐水的$$\\frac{10}{110}=\\frac{1}{11}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3476", "queId": "fe245e05216e47bbb208bb4ee97a6ac5", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛B卷第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列描述正确的有句． （$$1$$）$$9$$个连续偶数的平均数是$$90$$，这些数中最小的一个是$$2$$，最大的是$$18$$． （$$2$$）暗室里有红、绿、黄三种颜色的袜子若干只，为确保取出一双相同颜色的袜子，最少要取$$4$$只． （$$3$$）某班共有学生$$48$$人，其中$$27$$人会游泳，$$25$$人会骑自行车，有$$12$$人既不会游泳也不会骑自行车，那么这个班既会游泳又会骑自行车的有$$16$$人． （$$4$$）一个楼梯共有$$10$$级，如果每次能向上迈一级或两级，登上这$$10$$级楼梯，一共有$$89$$种不同的走法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["完成本题要根据每个题目的内容分别进行分析，才能确定正确的选项有几个． （$$1$$）相邻两个偶数相差$$2$$，由此可设这$$9$$个连续的偶数中的中间的那个为$$x$$，则这$$9$$个连续偶数的和为： $$x-2\\times 4+x-2\\times 3+\\cdots +x+x+2+\\cdots +x+2\\times 4$$，则其平均数为$$9x\\div 9=x$$，即$$90$$则最小的为$$90-2\\times 4=82$$，最大的为$$90+2\\times 4=98$$．所以，$$9$$个连续偶数的平均数是$$90$$，这些数中最小的一个是$$2$$，最大的是$$18$$是错误的； （$$2$$）暗室中共有$$3$$种不同颜色的袜子，最差情况是取出三只后，每种颜色各一只，此时只要再取出一只即能确保取出一双相同颜色的袜子，即最少取出$$3+1=4$$只；正确． （$$3$$）由题意可知，游泳或骑自行车会其中至少一项的有$$48-12=36$$人，这$$36$$人中，不会游泳的有$$36-27=9$$人，不会骑自行车的有$$36-25=11$$人，则班既会游泳又会骑自行车的有$$36-(9+11)=16$$人．正确； （$$4$$）如果有一级，则有一种走法：$$1$$；如有两级，则有$$2$$级走法即：$$1$$，$$1$$，如果有$$3$$级有$$3$$种：$$111$$，$$12$$，$$21$$；四级有$$5$$种：$$1111$$，$$22$$，$$121$$，$$211$$，$$112$$；$$\\cdots $$，由此可以发现，从第$$3$$个数开始，每一个数都等于它前面的$$2$$个数之和．共有$$10$$级，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$34$$，$$55$$，$$89$$．则到第十级共有$$89$$种不同走法．正确． 所以，（$$1$$）（$$3$$）（$$4$$）的描述都是正确的，共$$3$$句． 故选$$\\text{C}$$ ． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1137", "queId": "2abfbcbb42134f3888d252378c13bde9", "competition_source_list": ["2006年第4届创新杯四年级竞赛初赛A卷第9题", "2006年四年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁四人拿出同样多的钱，合伙订购同样规格的若干件货物，货物买来后，甲、乙、丙分别比丁多拿了3件，7件，14件，乙付给丁14元，那么，丙应付给丁( ). ", "answer_option_list": [[{"aoVal": "A", "content": "28元 "}], [{"aoVal": "B", "content": "56元 "}], [{"aoVal": "C", "content": "70元 "}], [{"aoVal": "D", "content": "112元 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模��->平均数问题->移多补少"], "answer_analysis": ["出同样多的钱应拿同样多的货，多拿货就要多出钱，以丁为标准，甲、乙、丙比丁共多$$3+7+14=24$$件，这24件本应平均分配，即每人要拿$$24\\div 4=6$$件(每人所得件数本应是丁现有件数加6)，而乙比丁实际多拿7件，也就是说乙比原计划多拿$$7-6=1$$件，乙要多拿出14元，则每件14元.丙比计划多拿$$14-6=8$$件，这样丙应再拿出8件的钱，即$$14\\times 8=112$$元.又因为丁比计划少拿6件，应取回$$6\\times 14=84$$元，但丁已经收了乙的14元，所以丙还应付给丁$$84-14=70$$元，丙另外的钱，3件共$$112-70=42$$(元)应给甲，故选C "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2604", "queId": "b07881beeba842b9813e25f8a1f71a3f", "competition_source_list": ["2017年河南郑州豫才杯四年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "聪聪是谁?他今年$$10$$周岁，是一名酷酷的小男生，就读于光明小学四年级（$$1$$）班．他爱运动，也爱阅读，是个小机灵鬼，凡是总爱问个究竟．他今年几岁？对，$$10$$岁！如果换个说法呢？估一估，以下哪个选项最接近他的年龄．（~ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$个月 "}], [{"aoVal": "B", "content": "$$700$$个星期 "}], [{"aoVal": "C", "content": "$$36500$$天~~~~~~~~~ "}], [{"aoVal": "D", "content": "$$87600$$小时 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$10$$岁相当于$$120$$个月，$$480$$个星期，$$3650$$天，$$87600$$小时． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2491", "queId": "1e324e666ed7445691a5da4f4db4bc33", "competition_source_list": ["2017年四川成都六年级竞赛“全能明星”选拔赛第1题2分", "2017年四川成都锦江区四川师范大学附属第一实验中学小升初模拟（五）第1题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$\\frac{8}{9}\\times \\left[ \\frac{3}{4}-\\left( \\frac{7}{16}-\\frac{1}{4} \\right) \\right]$$的值为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$0.5$$ "}], [{"aoVal": "D", "content": "$$1.5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$\\frac{8}{9}\\times \\left[ \\frac{3}{4}-\\left( \\frac{7}{16}-\\frac{1}{4} \\right) \\right]=0.5$$． ", "$$\\frac{8}{9}\\times \\left[ \\frac{3}{4}-\\left( \\frac{7}{16}-\\frac{1}{4} \\right) \\right]$$
\n$$=\\frac{8}{9}\\times \\left( \\frac{3}{4}-\\frac{7}{16}+\\frac{1}{4} \\right)$$
\n$$=\\frac{8}{9}\\times \\frac{9}{16}$$
\n$$=\\frac{1}{2}$$
\n$$=0.5$$
\n\n"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1213", "queId": "3002ae4e87374f8ba69f0a2d0393e587", "competition_source_list": ["2006年第4届创新杯四年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "张师傅加工一批零件，原计划每天加工$$80$$个，$$5$$天加工完，实际张师傅只用$$4$$天就加工完了，实际每天比原计划多加工零件( )个. ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->单归一问题"], "answer_analysis": ["实际每天加工的个数为$$80\\times 5\\div 4=100$$(个)；每天比计划增加的个数是$$100-80=20$$ (个)，故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2717", "queId": "889fe50bc8384e9f984992e76ba041ca", "competition_source_list": ["2008年六年级竞赛创新杯", "2008年第6届创新杯六年级竞赛初赛B卷第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "要使$$\\frac{x}{15}$$是最简真分数，$$\\frac{x}{7}$$是假分数，则符合条件的所有自然数$$x$$的和是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "53 "}], [{"aoVal": "B", "content": "40 "}], [{"aoVal": "C", "content": "39 "}], [{"aoVal": "D", "content": "15 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数基础->分数的分类"], "answer_analysis": ["$$x$$可取7，8，11，13，14.其和为53. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1960", "queId": "a1acb6bd069545d9ae353b3d1c72259f", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛��年级竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一串数：$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$34$$，$$55$$，$$89$$$$\\cdots \\cdots $$，其中第一个数$$2$$，第二个数$$3$$，从第三个数起，每个数是前两个数的和．那么在这串数中，第$$2019$$个数被$$3$$除后所得余数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["将这一串数写成除以$$3$$的余数，则为：$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0$$，$$1$$，$$1$$，$$2$$，$$0$$，$$2\\cdots \\cdots $$ 所以重复的为：``$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0$$，$$1$$，$$1$$''， 故周期为$$8$$．$$2019\\div 8=252$$（组）$$\\cdots \\cdots 3$$（个），则答案为$$2$$，故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2712", "queId": "7ae6bcd5aa934170a931f6e3a9a70015", "competition_source_list": ["2008年第6届创新杯四年级竞赛初赛A卷第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$20082008\\times 2007-20072007\\times 2008=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2007$$ "}], [{"aoVal": "B", "content": "$$2008$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$20072008$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde2008\\times 20072007-2007\\times 20082008$$ $$=2008\\times 2007\\times 10001-2007\\times 2008\\times 10001$$ $$=0$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1831", "queId": "a9432232674e4cd7808e9ed2f34046fe", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛B卷第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "英语试卷总分是$$120$$分，淘气四次考试的平均成绩是$$105$$分．为了使平均成绩尽快达到$$110$$分，他至少需要再考次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["根据题意分析可知，总分数$$\\div $$考的次数$$=$$平均分数；可设他至少要考$$x$$次才能尽快达到$$110$$分以上，那么 淘气的总分数为：$$120x+105\\times 4$$，淘气考试的次数为：$$x+4$$，淘气平均分应为$$\\geqslant $$$$100$$，根据公式列方程解答即可． 设淘气至少要考$$x$$次才能尽快达到$$110$$分以上， $$\\left( 120x+105\\times 4 \\right)$$$$\\div \\left( x+4 \\right)\\geqslant 110$$， $$120x+420\\geqslant 110x+440$$， $$120x-110x\\geqslant 440-420$$， $$10x\\geqslant 20$$， $$x\\geqslant 2$$． 即再考$$2$$次满分平均分可达到$$110$$分以上． 故选答案$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1656", "queId": "7b6ea61b8fbb464d8390a40df28cef42", "competition_source_list": ["2020年新希望杯四年级竞赛初赛第26题", "2020年希望杯四年级竞赛模拟第26题"], "difficulty": "1", "qtype": "single_choice", "problem": "【$$2020$$四年级卷第$$26$$题】小和尚、高和尚和胖和尚三人每天轮流到山下打水上山．小和尚第一次打水是星期一，那么，他第$$50$$次打水是． ", "answer_option_list": [[{"aoVal": "A", "content": "星期一 "}], [{"aoVal": "B", "content": "星期二 "}], [{"aoVal": "C", "content": "星期三 "}], [{"aoVal": "D", "content": "星期四 "}], [{"aoVal": "E", "content": "星期五 "}], [{"aoVal": "F", "content": "星期六 "}], [{"aoVal": "G", "content": "星期日 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["根据题目可知，打水分别按照小和尚，高和尚和胖和尚的顺序进行的，也就是说以每$$3$$天为一个周期进行循环的，所以小和尚第$$50$$次打水应该经过了$$49$$个周期多$$1$$天． 也就是$$3\\times 49+1=148$$（天），小和尚第一次打水是星期一， 那么经过$$148$$天应该是$$148\\div 7=21$$（周）$$\\cdots \\cdots 1$$（天），所以小和尚第$$50$$次打水仍然是星期一． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1012", "queId": "09eb4bfdc810468c98624920e0a3051b", "competition_source_list": ["其它改编自2015年全国希望杯六年级竞赛初赛第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "六年级甲班的女生人数是男生人数的$$\\frac{10}{9}$$倍，新年联欢会中，$$\\frac{2}{5}$$的女生和$$\\frac{1}{3}$$的男生参加了演出，则参加演出的人数占全班人数的~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{7}{19}$$ "}], [{"aoVal": "D", "content": "$$\\frac{10}{19}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->求分率"], "answer_analysis": ["设女生人数为$$10$$份，男生人数为$$9$$份，则参加演出的人数为$$\\frac{2}{5}\\times 10+\\frac{1}{3}\\times 9=7$$，占全班人数的$$\\frac{7}{10+9}=\\frac{7}{19}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "660", "queId": "34c984ebc5704bf186ca7db25cd5d736", "competition_source_list": ["2004年第2届创新杯六年级竞赛复赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个两位数，它的十位数字加上个位的$$7$$倍，还是等于这个两位数，这样的两位数有． ", "answer_option_list": [[{"aoVal": "A", "content": "一个 "}], [{"aoVal": "B", "content": "两个 "}], [{"aoVal": "C", "content": "三个 "}], [{"aoVal": "D", "content": "四个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->位值原理运用->计算中的位值原理"], "answer_analysis": ["设这个两位数为$$\\overline{ab}$$，依题意有$$\\overline{ab}=a+7b$$，即$$10a+b=a+7b$$，整理得$$a=\\frac{2}{3}b$$，又$$a$$，$$b$$均为一位数且$$a$$不为$$0$$，那么$$b$$只能为$$3$$，$$6$$，$$9$$对应的$$a$$分别为$$2$$，$$4$$，$$6$$，所以这样的两位数有$$23$$，$$46$$，$$69$$三个，故选$$\\text{C}$$． ", "设十位数是$$x$$，个位数字是$$y$$，
\n根据题意，得$$x+7y=10x+y$$，即$$3x=2y$$，
\n而$$1\\leqslant x\\leqslant 9$$，$$0\\leqslant y\\leqslant 9$$，且$$x$$，$$y$$都是整数，
\n根据条件同时满足的$$x$$，$$y$$的值有：$$2$$，$$3$$；$$4$$，$$6$$；$$6$$，$$9$$共$$3$$组．
\n即这样的两位数是$$23$$，$$46$$，$$69$$共$$3$$个．
\n故选：$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2339", "queId": "019d3a64e84c40a5b85007a9d4f9920f", "competition_source_list": ["2021年第8届鹏程杯四年级竞赛初赛第5题4分", "2021年第8届鹏程杯五年级竞赛初赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$1-2+3-4+5-6+\\cdots -\\cdots +2019-2020+2021=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1010$$ "}], [{"aoVal": "B", "content": "$$1011$$ "}], [{"aoVal": "C", "content": "$$2020$$ "}], [{"aoVal": "D", "content": "$$2021$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之分组法"], "answer_analysis": ["分组运算，$$2$$和$$3$$一组，$$4$$和$$5$$一组，$$\\cdots \\cdots 2020$$和$$2021$$一组，每组结果都是$$1$$，一共有$$2020\\div2=1010$$组，所以和是$$1010$$，前面还剩下一个$$1$$，$$1+1010=1011$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2265", "queId": "84e0b31d7df4444eaea6859f787a46c7", "competition_source_list": ["2012年IMAS小学高年级竞赛第一轮检测试题第7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁各有一只手表． （$$1$$）甲的手表快了$$10$$分钟，但他以为慢了$$5$$分钟； （$$2$$）乙的手表慢了$$5$$分钟，但他以为快了$$10$$分钟； （$$3$$）丙的手表快了$$5$$分钟，但他以为快了$$3$$分钟； （$$4$$）丁的手表慢了$$5$$分钟，但他以为慢了$$10$$分钟． 用他们的手表，每个人都认为自己恰好能准时到达学校，请问谁会迟到？ ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "丁 "}], [{"aoVal": "E", "content": "都不会 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["甲早到$$5-(-10)=15$$分钟，丙早到$$5-3=2$$分钟，丁早到$$(-5)-(-10)=5$$分钟，乙迟到$$10-(-5)=15$$分钟． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "438", "queId": "fbd573f06db94e74a2169bc3453a086d", "competition_source_list": ["2017年北京学而思杯四年级竞赛年度教学质量测评第20题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "有一条长度为$$8$$的绳子，将它对折$$3$$次，梓琪同学用剪刀从正中间剪开，得到一些短绳子．那么长度为$$1$$的绳子有段． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->归纳递推"], "answer_analysis": ["略 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "227", "queId": "ff46b2a8155d4d77b82eec56d0b63f13", "competition_source_list": ["2018年全国小学生数学学习能力测评四年级竞赛初赛第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个整数，与$$1$$、$$2$$、$$3$$这三个数通过加减乘除运算（可以加括号）组成算式，其结果等于$$24$$，那么这个数就称为``可用数''．在$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$、$$10$$、$$11$$、$$12$$这$$9$$个数中，``可用数''有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["$$4\\times\\left(1+2+3\\right)=24$$，$$5\\times\\left(2+3\\right)-1=24$$， $$6\\times 2\\times\\left(3-1\\right)=24$$，$$7\\times 3+2+1=24$$， $$8\\times 3\\div\\left(2-1\\right)=24$$，$$9\\times 3-2-1=24$$， $$10\\times 2+3+1=24$$，$$11\\times 2+3-1=24$$， $$12\\times\\left(3+1\\right)\\div 2=24$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "532", "queId": "fe1592fec28044afa06ca5e8888eec82", "competition_source_list": ["2018年第17届春蕾杯一年级竞赛初赛第14题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "四耳穴学校也举办了一场足球比赛，皮皮、蛋君和大强所在的三个队伍获得不同的奖牌．根据下面三句话，猜一猜他们分别获得什么奖牌？ 皮皮：``我分到的不是金牌．'' 蛋君：``我们队不是第四名．'' 大强：``他们俩的队伍一支获得金牌，一支获得银牌．'' 请问，蛋君所在的队伍获得什么奖牌？ ", "answer_option_list": [[{"aoVal": "A", "content": "金 "}], [{"aoVal": "B", "content": "银 "}], [{"aoVal": "C", "content": "铜 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["推理题目 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "722", "queId": "514d88517036428f86e757ebfb156b53", "competition_source_list": ["2021年新希望杯六年级竞赛初赛第28题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "迷糊老师今天上课讲解高斯记号，告诉我们$$\\left[ a \\right]$$表示不大于$$a$$的最大整数，例如$$\\left[ 1.1 \\right]=1$$，$$\\left[ 3 \\right]=3$$，然后计算：$$\\left[ \\frac{1}{7} \\right]+\\left[ \\frac{3}{7} \\right]+\\left[ \\frac{5}{7} \\right]+\\cdots +\\left[ \\frac{2019}{7} \\right]+\\left[ \\frac{2021}{7} \\right]=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$145504$$ "}], [{"aoVal": "B", "content": "$$135184$$ "}], [{"aoVal": "C", "content": "$$145164$$ "}], [{"aoVal": "D", "content": "$$135524$$ "}], [{"aoVal": "E", "content": "$$145584$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->高斯记号->高斯记号的复杂应用"], "answer_analysis": ["列出一些分数找规律： （$$1$$）$$\\left[ \\frac{1}{7} \\right]$$、$$\\left[ \\frac{3}{7} \\right]$$、$$\\left[ \\frac{5}{7} \\right]$$都为$$0$$； （$$2$$）$$\\left[ \\frac{7}{7} \\right]$$、$$\\left[ \\frac{9}{7} \\right]$$、$$\\left[ \\frac{11}{7} \\right]$$、$$\\left[ \\frac{13}{7} \\right]$$都为$$1$$； （$$3$$）$$\\left[ \\frac{15}{7} \\right]$$、$$\\left[ \\frac{17}{7} \\right]$$、$$\\left[ \\frac{19}{7} \\right]$$都为$$2$$； （$$4$$）$$\\left[ \\frac{21}{7} \\right]$$、$$\\left[ \\frac{23}{7} \\right]$$、$$\\left[ \\frac{25}{7} \\right]$$、$$\\left[ \\frac{27}{7} \\right]$$都为$$3$$； （$$5$$）$$\\left[ \\frac{29}{7} \\right]$$、$$\\left[ \\frac{31}{7} \\right]$$、$$\\left[ \\frac{33}{7} \\right]$$都为$$4$$； $$\\cdots $$； （$$287$$）$$\\left[ \\frac{2009}{7} \\right]$$、$$\\left[ \\frac{2011}{7} \\right]$$、$$\\left[ \\frac{2013}{7} \\right]$$、$$\\left[ \\frac{2015}{7} \\right]$$都为$$287$$； （$$288$$）$$\\left[ \\frac{2017}{7} \\right]$$、$$\\left[ \\frac{2019}{7} \\right]$$、$$\\left[ \\frac{2021}{7} \\right]$$都为$$288$$， 所以题目式$$=4\\times \\left( 1+3+5+\\cdots +287 \\right)+3\\times \\left( 2+4+\\cdots +288 \\right)$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=4\\times \\left( 2+4+6+\\cdots +288-144 \\right)+3\\times \\left( 2+4+\\cdots +288 \\right)$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=7\\times \\left( 2+4+6+\\cdots +288 \\right)-4\\times 144$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=7\\times 2\\times \\left( 1+2+\\cdots +144 \\right)-576$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=7\\times 2\\times \\frac{\\left( 1+144 \\right)\\times 144}{2}-576$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=145584$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2925", "queId": "8e3fef665580483d8a733b96956002bc", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛A卷第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$f\\left( m,n \\right)=a{{m}^{3}}+b{{n}^{3}}$$，若$$f\\left( 1,2 \\right)=a\\times {{1}^{3}}+b\\times {{2}^{3}}=4$$，那么$$\\textasciitilde f\\left( 2,4 \\right)$$的值应为~ ~~ ~ ~ ~ ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$34$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["已知$$f\\left( m,n \\right)=a{{m}^{3}}+b{{n}^{3}}$$，$$f\\left( 1,2 \\right)=a\\times {{1}^{3}}+b\\times {{2}^{3}}=a+8b=4$$， 那么$$f\\left( 2,4 \\right)=a\\times {{2}^{3}}+b\\times {{4}^{3}}=8a+64b=8\\times \\left( a+8b \\right)=8\\times 4=32$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "754", "queId": "3c45ef6542654659ac04603c61e0b261", "competition_source_list": ["2022年小学高年级六年级竞赛（小学奥数5-2-5整除与分类计数综合专项训练）第6题", "2022年小学高年级六年级竞赛（小学奥数5-2-5整除与分类计数综合专项训练）"], "difficulty": "3", "qtype": "single_choice", "problem": "在$$1$$、$$2$$、$$3$$、$$4$$\\ldots\\ldots$$2007$$这$$2007$$个数中有~\\uline{~~~~~~~~~~}~个自然数$$a$$能使$$2008+a$$能被$$2007-a$$整除。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数"], "answer_analysis": ["【详解】 要使得$$2008+a$$能被$$2007-a$$整除，我们可以将条件等价的转化为只要让$$\\frac{2008+a}{2007-a}$$是一个整数即可。下面是一个比较难的技巧，我们知道若$$a$$可以使得$$\\frac{2008+a}{2007-a}$$是一个整数，那么$$a$$也同样可以使得$$\\frac{2008+a}{2007-a}+1=\\frac{2008+a+2007-a}{2007-a}=\\frac{4015}{2007-a}$$是一个整数，这样只要$$2007-a$$是$$4015$$的约数即可，将$$4015$$分解可知其共有$$8$$个因数，其中$$4015$$是最大的一个，但是显然没有可以让$$2007-a$$等于$$4015$$的$$a$$的值，其余的$$7$$个均可以有对应的$$a$$的值，所以满足条件的$$a$$的取值共有$$7$$个。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3194", "queId": "66dd97fb66d04c5f964aa337539ca1fc", "competition_source_list": ["2010年第8届创新杯六年级竞赛初赛第3题4分", "2018年四川成都武侯区成都西川中学小升初面试真卷（四）第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个骰子六个面上写着$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$，将它投掷两次，则面朝上的两个数字之和为$$3$$的倍数的可能性是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{6}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["数字和是$$3$$的倍数有：$$（1，2），（1，5），（2，4），（3，6），（3，3），（4，5）$$，则$$\\frac{6\\times 2}{6\\times 6}=\\frac{1}{3}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1931", "queId": "9ccec85772724f31bca78aac0e84ae7d", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "$$2019$$年元旦是星期二，$$2019$$年的国庆节是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["从$$2019$$年的$$1$$月$$1$$日至$$2019$$年的$$10$$月$$1$$日共：$$31+28+31+30+31+30+31+31+30+1=274$$（天）， 根据``二、三、四、五、六、日、一''的周期规律：$$274\\div 7=39$$（周）$$\\cdots \\cdots 1$$（天）． 所以$$10$$月$$1$$日是这个周期的第$$1$$天，即星期二． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2770", "queId": "52ccb806e2444b7d902d7881174c3f8d", "competition_source_list": ["2020年新希望杯三年级竞赛初赛（团战）第35题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面哪个算式的计算结果是偶数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 784-455 \\right)\\times 39+44\\times 11$$ "}], [{"aoVal": "B", "content": "$$11\\times 1+22\\times 2+33\\times 3+44\\times 4+55\\times 5+66\\times 6+77\\times 7+88\\times 8+99\\times 9$$ "}], [{"aoVal": "C", "content": "$$11\\times 2+22\\times 2+33\\times 3+44\\times 4+55\\times 5+66\\times 6+77\\times 7+88\\times 8+99\\times 9$$ "}], [{"aoVal": "D", "content": "$$123\\times 456+789$$ "}], [{"aoVal": "E", "content": "$$2\\times 4\\times 6\\times \\cdots \\times 2018\\times 2020-1\\times 3\\times 5\\times \\cdots \\times 2017\\times 2019$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数乘法巧算之提取公因数（普通型）"], "answer_analysis": ["暂无 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "306", "queId": "a81219b29ece4bd48574356eacaa5029", "competition_source_list": ["2011年全国华杯赛竞赛初赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "老师问$$5$$名学生：``昨天你们有几个人复习数学了？'' 张：``没有人．''李：``一个人．''王：``二个人．''赵：``三个人．''刘：``四个人．'' 老师知道，他们昨天下午有人复习，也有人没复习，复习了的人说的都是真话，没复习的人说的都是假话．那么，昨天这$$5$$个人中复习数学的有（~ ~ ~ ~）个人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾"], "answer_analysis": ["任何两人说的话都不能同时为真，所以最多有一个人说的是真话，如果有一个人复习了，那么李说的是真话，符合题意；如果没有人复习了，那么张说的是真话，矛盾． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1763", "queId": "dfa90a1671214df38a72a5ab88f633e8", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（一）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "三位采购员定期去某市场采购，小王每$$8$$天去一次，大刘每$$5$$天去一次，老李每$$6$$天去一次，三人星期二第一次在这里碰面，下次碰面将在星期． ", "answer_option_list": [[{"aoVal": "A", "content": "二 "}], [{"aoVal": "B", "content": "三 "}], [{"aoVal": "C", "content": "四 "}], [{"aoVal": "D", "content": "五~ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$8$$、$$5$$、$$6$$的最小公倍数是$$120$$，这三人相会的周期是$$120$$天，$$120\\div 7=17$$（周）$$\\cdots\\cdots1$$（天），下次碰面将在星期三． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1747", "queId": "b1c8dadc7c524f5dbf346cb915000cc9", "competition_source_list": ["2016年创新杯六年级竞赛训练题（二）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "哈利波特制作加强型魔法药剂``生死水''（这是一种效力很强的安眠药，由水仙根粉末和艾草浸液配成，``生死水''的浓度是指水仙根粉末占整个药剂的百分比）．他首先在普通型``生死水''中加入一定量的艾草浸液，使``生死水''的浓度变为$$9 \\%$$；再加入同等量的水仙根粉末，这时``生死水''的浓度变为$$23 \\%$$．那么普通型``生死水''的浓度为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10 \\%$$ "}], [{"aoVal": "B", "content": "$$11 \\%$$ "}], [{"aoVal": "C", "content": "$$12 \\%$$ "}], [{"aoVal": "D", "content": "$$15 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设普通型``生死水''的浓度为$$x \\%$$，初始重量为$$100$$，连续两次加入的艾草浸液和水仙根粉末重量都是$$a$$，那么$$\\left { \\begin{matrix} \\dfrac{x}{100+a}=9 \\% \\dfrac{x+a}{100+2a}=23 \\% \\end{matrix}\\Rightarrow \\left { \\begin{matrix} 100x-9a=900 100x+54a=2300 \\end{matrix}\\Rightarrow x=11 \\right. \\right.$$，故普通型``生死水''的浓度为$$11 \\%$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2768", "queId": "ff8080814694a4fc014694fcff0801a6", "competition_source_list": ["2012年全国华杯赛小学高年级竞赛初赛第1题", "2017年全国小升初八中入学备考课程"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$$\\left[ \\left( 0.8+\\frac{1}{5} \\right)\\times 24+6.6 \\right]\\div \\frac{9}{14}-7.6=$$（~ ~ ~ ~~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数运算->分小四则混合运算"], "answer_analysis": ["$$\\left[ \\left( 0.8+\\frac{1}{5} \\right)\\times 24+6.6 \\right]\\div \\frac{9}{14}-7.6$$ $$=\\left[ (0.8+0.2)\\times 24+6.6 \\right]\\div \\frac{9}{14}-7.6$$ $$=\\left[ 1\\times 24+6.6 \\right]\\div \\frac{9}{14}-7.6$$ $$=30.6\\div \\frac{9}{14}-7.6$$ $$=30.6\\times \\frac{14}{9}-7.6$$ $$=47.6-7.6$$ $$=40$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2146", "queId": "7461c3e158b247d7aedafc04f8ae3c0d", "competition_source_list": ["2014年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "某学校组织一次远足活动，计划$$10$$点$$10$$分从甲地出发，$$13$$点$$10$$分到达乙地，但出发晚了$$5$$分钟，却早到达了$$4$$分钟。甲乙两地之间的丙地恰好是按照计划时间到达的，那么到达丙地的时间是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$点$$40$$分 "}], [{"aoVal": "B", "content": "$$11$$点$$50$$分 "}], [{"aoVal": "C", "content": "$$12$$点 "}], [{"aoVal": "D", "content": "$$12$$点$$10$$分 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题"], "answer_analysis": ["根据题意，实际走比计划走全程省了$$5+4=9$$（分钟）。将全程平均分成$$9$$等份，则在每一等份中实际走比计划走省$$1$$分钟。所以当走完$$5$$等份时，到达的地点时间实际和计划的时间相同，此时是丙地即在全程的$$\\frac{5}{9}$$处。计划走全程用$$3$$小时即$$180$$分钟，则走到丙地用$$180\\times \\frac{5}{9}=100$$（分钟）。$$10$$点$$10$$分出发走$$100$$分钟是$$11$$点$$50$$分。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2441", "queId": "219bb6f5aead4ba3933e8ec598386b28", "competition_source_list": ["2008年五年级竞赛创新杯", "2008年第6届创新杯五年级竞赛初赛A卷第9题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "快慢两列火车相向而行，快车长50米，慢车长80米，快车速度是慢车的2倍，如果坐在慢车上的人见快车驶过窗口的时间是5秒，那么坐在快车上的人见慢车驶过窗口的时间是( )秒. ", "answer_option_list": [[{"aoVal": "A", "content": "6 "}], [{"aoVal": "B", "content": "7 "}], [{"aoVal": "C", "content": "8 "}], [{"aoVal": "D", "content": "10 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比和比例->比例->正比例与反比例"], "answer_analysis": ["记快车、慢车的速度为$${{v}_{\\text{快}}}$$米/秒、$${{v}_{\\text{慢}}}$$米/秒，且$${{v}_{\\text{快}}}={{v}_{\\text{慢}}}$$，快、慢两列火车相向而行，则相对速度为$${{v}_{\\text{快}}}+{{v}_{\\text{慢}}}=3{{v}_{\\text{慢}}}$$（米/秒）.设坐在快火车上的人见慢车驶过的时间是$$x$$秒.则$$\\begin{cases}\\frac{50}{3{{v}_{\\text{慢}}}}=5 \\frac{80}{3{{v}_{\\text{慢}}}}=x \\end{cases}$$ 即$$\\begin{cases}80=x\\times 3{{v}_{\\text{慢}}}\\left( 1 \\right) 50=5\\times {{v}_{\\text{慢}}}\\left( 2 \\right) \\end{cases}$$$$\\left( 2 \\right)\\div \\left( 1 \\right)$$ 得$$\\frac{50}{80}=\\frac{5}{x}$$.解得$$x=8$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1158", "queId": "2aea0cb99cf2476d97b40a86bc4c1036", "competition_source_list": ["2020年希望杯二年级竞赛模拟第15题"], "difficulty": "0", "qtype": "single_choice", "problem": "鸭妈妈带着小鸭们在池塘里游玩，黄小鸭发现：有$$2$$只小鸭在它的前面，$$3$$只小鸭在它的后面．池塘里共有几只小鸭？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}], [{"aoVal": "E", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意分析可知，用黄小鸭前面的数量加上后面的数量再加上自己所以一共有： $$2+3+1=6$$（只）小鸭， 故答案为：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2561", "queId": "42c1ba38c4ba476f88f3cc5c618d31ae", "competition_source_list": ["2017年河南郑州豫才杯竞赛第10题", "2017年河南郑州小升初豫才杯第二场第10题"], "difficulty": "3", "qtype": "single_choice", "problem": "古希腊数学家把$$1$$，$$3$$，$$6$$，$$10$$，$$15$$，$$21$$，$$\\cdots \\cdots $$称为三角形数，它有一定的规律性，则第$$99$$个三角形数和第$$97$$个三角形数的差为 ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$99$$ "}], [{"aoVal": "B", "content": "$$197$$ "}], [{"aoVal": "C", "content": "$$97$$ "}], [{"aoVal": "D", "content": "$$196$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["分析可知，三角形数的规律为第$$n$$个数比第$$n-1$$个数大$$n$$（$$n$$大于$$1$$），则第$$99$$个三角形数比第$$98$$个三角形数大$$99$$，第$$98$$个三角形数比第$$97$$个三角形数大$$98$$，所以所求的差为$$99+98=197$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "55", "queId": "0b6eb76b77e9451ea2f3bca0c95c4383", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁和戊参加$$100$$米比赛，比赛结束后丁说：``我比乙跑得快。''丙说：``戊在我前面冲过终点线。''甲说：``我的名次排在丁的前面，丙的后面。''请根据他们的话排出名次（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "戊$$\\textgreater$$丙$$\\textgreater$$丁$$\\textgreater$$甲$$\\textgreater$$乙 "}], [{"aoVal": "B", "content": "甲$$\\textgreater$$乙$$\\textgreater$$丙$$\\textgreater$$丁$$\\textgreater$$戊 "}], [{"aoVal": "C", "content": "乙$$\\textgreater$$丁$$\\textgreater$$甲$$\\textgreater$$丙$$\\textgreater$$戊 "}], [{"aoVal": "D", "content": "戊$$\\textgreater$$丙$$\\textgreater$$甲$$\\textgreater$$丁$$\\textgreater$$乙 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["解：根据分析，由丁说的可得``丁$$\\textgreater$$乙''，根据丙说的可得``戊$$\\textgreater$$丙''， 根据甲说的可得``丙$$\\textgreater$$甲$$\\textgreater$$丁''，综合可得``戊$$\\textgreater$$丙$$\\textgreater$$甲$$\\textgreater$$丁$$\\textgreater$$乙''。 故选：D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3193", "queId": "15e7c56860744089b21f7bd13d4fd783", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$15$$个玻璃球分成数量不同的$$2$$堆，共有种不同的分法．（每堆至少分$$1$$个） ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->整数分拆->整数拆分应用->加法拆数（应用）"], "answer_analysis": ["根据题意可得： $$15=1+2+3+9$$；$$15=1+2+4+8$$； $$15=1+2+5+7$$；$$15=1+3+4+7$$； $$15=1+3+5+6$$；$$15=2+3+4+6$$； 一共有$$6$$种． 答：共有$$6$$种不同的分法． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "553", "queId": "09982d2d1a6f437bbfc45ba748715bcf", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛复赛第7题"], "difficulty": "3", "qtype": "single_choice", "problem": "给出一列数：$$23+m$$，$$23+2m$$，$$23+3m$$，$$\\cdot \\cdot \\cdot $$，$$23+2023m$$，这$$2023$$个数的和除以$$14$$的余数是（其中$$m$$为正整数）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["$$23\\times 2023+(1+2+\\cdot \\cdot \\cdot +2023)m$$除以$$14$$的余数， $$23\\div 14$$余$$9$$， $$23\\div 14$$余$$7$$， $$1+2+\\cdot \\cdot \\cdot +2023=2023\\times 1012$$刚好是$$14$$的倍数， 那么只用看$$23\\times 2023$$除以$$14$$的余数是$$7$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3201", "queId": "2b8323d7b40d4678bf8af1bad0b7d233", "competition_source_list": ["2014年IMAS小学高年级竞赛第二轮检测试题第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明将$$27$$个苹果分给若干位小朋友，这些小朋友得到的苹果数是一些连续的正整数，请问 这些小朋友最多有多少位？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}], [{"aoVal": "E", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为$$2+3+4+5+6+7$$ $$=5+9+13$$ $$=14+13$$ $$=27$$（个） 所以最多有$$6$$位小朋友． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2482", "queId": "15d058581e4e4de0b6e3386af14ea11a", "competition_source_list": ["2017年华杯赛小学高年级竞赛（深圳营一）第4题10分"], "difficulty": "3", "qtype": "single_choice", "problem": "对于整数$$n\\geqslant 3$$，用$$\\varphi \\left( n \\right)$$表示所有小于$$n$$的素数的乘积，求满足条件$$\\varphi \\left( n \\right)=22n-32$$的所有正整数$$n$$，$$n$$为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$n\\textgreater11$$ "}], [{"aoVal": "B", "content": "$$n\\textless{}11$$ "}], [{"aoVal": "C", "content": "$$n=11$$ "}], [{"aoVal": "D", "content": "$$n$$有多个值 "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["若$$n\\textgreater11$$，则$$11$$整除$$\\varphi \\left( n \\right)$$，但$$11$$不能整除$$22n-32$$． 因此，$$n\\textgreater11$$不符合要求．故，$$n\\leqslant 11$$． 若$$7\\textless{}n\\leqslant 11$$，则$$\\varphi \\left( n \\right)=2\\times 3\\times 5\\times 7=210$$， 由$$210=22n-32$$， 得$$n=11$$． 若$$5\\textless{}6\\leqslant 7$$，则$$\\varphi \\left( n \\right)=2\\times 3\\times 5=30$$， 由$$30=22n-32$$，得正整数$$n$$不存在． 若$$3\\textless{}n\\leqslant 5$$，则$$\\varphi \\left( n \\right)=2\\times 3=6$$， 由$$6=22n-32$$，得正整数$$n$$不存在． 若$$n=3$$，则$$\\varphi \\left( n \\right)=2$$，由$$2=22n-32$$， 得正整数$$n$$不存在． ∴满足条件的正整数$$n$$只有$$1$$个，$$n=11$$． 解法二：由$$\\varphi \\left( n \\right)=22n-32$$，得$$\\varphi \\left( n \\right)-1024=22\\left( n-48 \\right)$$． 由于$$\\varphi \\left( n \\right)$$是偶数，但不是$$4$$的倍数，因此，$$n-48$$是奇数． 若$$n-48\\geqslant 3$$，则$$n-48$$含有奇数的素数因子$$p$$， 即$$p$$为奇素数，且$$p$$整除$$n-48$$． 由$$n-48\\textless{}n$$知，$$p$$整除$$\\varphi \\left( n \\right)$$．由此$$p$$整除$$1024$$矛盾． 故，$$n-48\\textless{}3$$，即$$n\\leqslant 49$$，且$$n$$为奇数． ∵$$n\\leqslant 49$$时，$$22n-32\\leqslant 22\\times 49-32=1046$$， ∴$$\\varphi \\left( n \\right)\\leqslant 1046$$． 又$$2\\times 3\\times 5\\times 7=210$$，$$2\\times 3\\times 5\\times 7\\times 11=210\\times 11\\textgreater1046$$． ∴$$n\\leqslant 11$$，即$$n=3，$$5$$，$$7$$，$$9$$，11$$． 将$$n=3，$$5$$，$$7$$，$$9$$，11$$分别代入$$\\varphi \\left( n \\right)=22n-32$$验证， $$n=3$$时，$$\\varphi \\left( 3 \\right)=2$$，$$22n-32=34$$，不符合要求． $$n=$$5时，$$\\varphi \\left( 5 \\right)=2\\times 3=6$$，$$22n-32=78$$，不符合要求． $$n=7$$时，$$\\varphi \\left( 7 \\right)=2\\times 3\\times 5=30$$，$$22n-32=122$$，不符合要求． $$n=9$$时，$$\\varphi \\left( 9 \\right)=2\\times 3\\times 5\\times 7=210$$，$$22n-32=166$$，不符合要求． $$n=11$$时，$$\\varphi \\left( 11 \\right)=2\\times 3\\times 5\\times 7=210$$，$$22n-32=210$$，符合要求． ∴满足条件的正整数$$n$$只有$$1$$个，$$n=11$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1231", "queId": "8b136ca998cc41a196d8eda98cad7fb9", "competition_source_list": ["2016年第21届全国华杯赛小学中年级四年级竞赛初赛B卷", "2019年四川成都锦江区四川师范大学附属第一实验中学小升初（八）第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "库里是美国NBA勇士队当家球星, 在过去的$$10$$场比赛中已经得了$$333$$分的高分, 他在第$$11$$场得（ ）分就能使前$$11$$场的平均得分达到$$34$$分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$41$$ "}], [{"aoVal": "D", "content": "$$47$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["由前$$11$$场的总分减去前$$10$$场的总分即为第$$11$$场的得分：$$34\\times 11-333=374-333=41$$分． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2999", "queId": "d7527b0ef227431983da56358e036b82", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级三年级竞赛初赛第39题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$9+99+999+9999+\\cdots +9999999999$$（最后一项有$$10$$位数字）的和的十位数是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["十位数是$$0$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3441", "queId": "bce64defa2f04a00932de1baa7960bf0", "competition_source_list": ["2017年河南郑州小升初豫才杯第二场第13题", "2017年河南郑州豫才杯竞赛第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "一次数学小测试只有两道题，结果全班有$$10$$人全对，第一题有$$25$$人做对，第二题有$$18$$人做错，那么两题都做错的有（~ ）人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["第一题做对的$$25$$人中，有$$10$$人全对，则有$$25-10=15$$人是只做对第一题，也是做错第二题的；已知第二题总共有$$18$$人做错，那么多的$$3$$人就是两题都错的．故选$$C$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3481", "queId": "fe7acf7d1cd547e4a229caf51baf04be", "competition_source_list": ["2016年创新杯六年级竞赛训练题（二）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "三个大于$$1000$$的正整数满足:其中任意两个数之和的个位数字都等于第三个数的个位数字，那么这$$3$$个数之积的末尾$$3$$位数字有（~ ）种可能数值． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["满足这样的尾数有（$$0$$，$$0$$，$$0$$），（$$0$$，$$5$$，$$5$$），故末三位可能为$$000$$，$$250$$，$$500$$，$$750$$共$$4$$种． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1349", "queId": "3e8b9289189742c7ae2331d494a0df99", "competition_source_list": ["2017年河南郑州联合杯竞赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "青蛙从井底向井口跳，井深$$15$$米，青蛙每次向上跳$$6$$米，又会滑下来$$3$$米，这样青蛙需要跳次才可以出井． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["第一次跳时，向上跳$$6$$米，又下滑$$3$$米，此时距离井底$$3$$米，第二次跳完距离井底$$6$$米，第三次跳完距离井底$$9$$米，第四次先向上跳$$6$$米，此时青蛙跳出了井． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1988", "queId": "ca6f5d00158140059ce44eedbed58799", "competition_source_list": ["2017年全国迎春杯六年级竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "侠客岛的人，原来有$$\\frac{2}{3}$$是卧底，后来卧底中有$$20 \\% $$的人被驱离出岛，而不是卧底的人有$$\\frac{1}{5}$$转变成了卧底．如果侠客岛上现在还有$$728$$人，那么现在侠客岛上有~\\uline{~~~~~~~~~~}~人是卧底（没有其他人入岛）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$320$$ "}], [{"aoVal": "B", "content": "$$410$$ "}], [{"aoVal": "C", "content": "$$504$$ "}], [{"aoVal": "D", "content": "$$610$$ "}], [{"aoVal": "E", "content": "$$728$$ "}]], "knowledge_point_routes": ["海外竞赛体系->知识点->应用题模块->分百应用题", "拓展思维->拓展思维->应用题模块->分百应用题->转化单位1->统一单位1"], "answer_analysis": ["分数应用题，既考查了量率对应的思想，也考查了单位$$1$$变化问题的处理，非常全面． 由于卧底的$$20 \\% $$被驱离出岛，相当于原来的$$\\frac{2}{3}\\times 20 \\% =\\frac{2}{15}$$被驱离，那么$$728$$人相当于原来人数的$$1-\\frac{2}{15}=\\frac{13}{15}$$，则原来人数为$$728\\div \\frac{13}{15}=840$$人，那么现在还剩下的卧底有$$840\\times \\frac{2}{3}\\times \\left( 1-20 \\% ~\\right)+840\\times \\left( 1-\\frac{2}{3} \\right)\\times \\frac{1}{5}=504$$人． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "745", "queId": "4d7568924daa4f409f0de1114044a6a0", "competition_source_list": ["2017年第13届湖北武汉新希望杯五年级竞赛决赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "（人教版）下列说法正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "互质的两个数没有公因数 "}], [{"aoVal": "B", "content": "$$12$$和$$18$$的最小公倍数是$$72$$ "}], [{"aoVal": "C", "content": "$$24$$和$$30$$的最大公因数是$$6$$ "}], [{"aoVal": "D", "content": "两个质数的和一定是偶数 "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["$$\\text{A}$$选项：互质的两个数公因数为$$1$$，$$\\text{B}$$选项：$$12$$和$$18$$的最小公倍数是$$36$$，$$\\text{D}$$选项：$$2+3=5$$为奇数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2078", "queId": "d48887ed07834122bc7f02518c1e19a5", "competition_source_list": ["2017年河南郑州豫才杯小学高年级五年级竞赛初赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "游览景区期间，爸爸将车放在停车场，收费标准为：不超过$$1$$小时收费$$3$$元，每多停半小时加收$$1.5$$元．爸爸最终一共交了$$13.5$$元的停车费，他们的车在停车场最多停了（~ ）小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$小时 ~~~ "}], [{"aoVal": "B", "content": "$$4.5$$小时~~ "}], [{"aoVal": "C", "content": "$$5$$小时 "}], [{"aoVal": "D", "content": "$$6$$小时 "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->经济问题->分段计价问题"], "answer_analysis": ["设最多在停车场停了$$x$$小时，$$3+\\left( x-1 \\right)\\times 2\\times 1.5=13.5$$，解得$$x=4.5$$，所以他们的车在停车场最多停$$4.5$$小时． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "13", "queId": "2541c84fa7484b3e8c8d27c91d3d183b", "competition_source_list": ["2020年广东广州羊排赛四年级竞赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "六支队伍进行单循环赛，每两队都要赛一场．如果赛平，每队各得$$1$$分，否则胜队得$$2$$分，负队得$$0$$分．那么，打完所有比赛后，六支队伍的总得分是分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["在一场比赛中，如果一胜一负， 则胜得$$2$$分，负得$$0$$分， 总分为$$2+0=2$$（分）， 如果赛平，则总分为$$1+1=2$$（分）， 即在一场比赛中，无论结果如何，比赛总分是不变的，都是$$2$$分， $$6$$支队伍进行单循环赛，共有比赛：$$5+4+3+2+1=15$$（场）， 所以六支队伍的总得分是：$$15\\times 2=30$$（分）， 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1215", "queId": "39112d81c8bc4bc9a686de85e3e0d728", "competition_source_list": ["2016年河南郑州联合杯小学高年级六年级竞赛复赛第14题2分", "2018年四川成都锦江区四川师范大学附属第一实验中学小升初模拟12第8题3分", "2014年四川成都小升初七中嘉祥外国语学校第30题", "小学高年级六年级上学期其它北师大版53天天练"], "difficulty": "1", "qtype": "single_choice", "problem": "一个玻璃瓶内原有盐水，盐的重量是水的$$\\frac{1}{11}$$，加��$$15$$克盐后，盐的重量占盐水总量的$$\\frac{1}{9}$$，瓶内原有盐水（~ ）克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$480$$ "}], [{"aoVal": "B", "content": "$$300$$~ "}], [{"aoVal": "C", "content": "$$360$$~ "}], [{"aoVal": "D", "content": "$$440$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->浓度问题->浓度基本题型->已知溶液浓度求溶质"], "answer_analysis": ["浓度问题；盐的重量占盐水重量的$$\\frac{1}{9}$$，则盐占水的$$\\frac{1}{8}$$， $$15\\div \\left( \\frac{1}{8}-\\frac{1}{11} \\right)$$ $$=15\\div\\frac{3}{88}$$ $$=440$$． 故选$$\\text{D}$$. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2296", "queId": "b73790d29ea74beaa474c19e672686d9", "competition_source_list": ["2016年创新杯五年级竞赛训练题（一）第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "一条船往返于甲乙两港之间，由甲至乙是顺水行驶，由乙至甲是逆水行驶．已知船在静水中的速度为$$8$$千米/小时，平时逆水航行所用时间是顺水航行所用的时间的$$2$$倍．某天恰逢暴雨，水流速度是原来的两倍，这条船往返共用了$$9$$小时．问：甲、乙两地相距千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$25$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["平时逆水航行所用时间是顺水航行所用时间的$$2$$倍，所以顺水航行速度是逆水航行的$$2$$倍，即$${{V}_{水}}+8=2\\times \\left( 8-{{V}_{水}} \\right)$$，解得：$${{V}_{}}=\\frac{8}{3}$$，所以水速为$$\\frac{8}{3}$$千米/小时，变为原来的$$2$$倍后变为$$\\frac{16}{3}$$千米/小时．设甲、乙两地相距$$S$$千米，则：$$\\frac{S}{\\frac{16}{3}+8}+\\frac{S}{8-\\frac{16}{3}}=9$$ 解得：$$S=20$$，所以甲、乙两地相距$$20$$千米． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "834", "queId": "5cadd11f9f7a4859914151ee792491a5", "competition_source_list": ["2006年第4届创新杯五年级竞赛初赛A卷第2题", "2006年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "一个电子钟，每9分钟亮一次灯，每到整点响一次铃，中午12点时，电子钟恰好又亮灯又响铃，问下次既亮灯又响铃是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "2点 "}], [{"aoVal": "B", "content": "3点 "}], [{"aoVal": "C", "content": "4点 "}], [{"aoVal": "D", "content": "5点 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->两数的最小公倍数"], "answer_analysis": ["整点的分钟数$$60n$$应为9的倍数，即$$n$$最小为3，所以12点后最早为3点时既亮灯又响铃. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2329", "queId": "004ebef5f5ed4a3abe0867742d37c88c", "competition_source_list": ["2014年全港小学数学挑战赛四年级竞赛初赛A卷第13题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "（2014 AIMO, Grade 4, Question\\#13) The sum of $$5$$ consecutive odd numbers is $$145$$, which is the largest one among these numbers? $$5$$个连续奇数的和是$$145$$，求当中最大的数． ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$31$$ "}], [{"aoVal": "B", "content": "$$33$$ "}], [{"aoVal": "C", "content": "$$35$$ "}], [{"aoVal": "D", "content": "$$37$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["因为这$$5$$个奇数是连续的，所以每相邻两个奇数之间相差$$2$$，用$$5$$个连续奇数的和除以$$5$$，可得到最中间的奇数，最大的数与最中间的奇数相差$$4$$，所以用最中间的奇数加上$$4$$即可得到当中最大的奇数，故列式为：$$145\\div 5+4=29+4=33$$，即最大的奇数是$$33$$．故答案为：$$33$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "512", "queId": "ef5e066f20f9419fa9fec355c9ed0f82", "competition_source_list": ["2016年全国学而思杯一年级竞赛样卷第7题"], "difficulty": "0", "qtype": "single_choice", "problem": "张、黄、李分别是三位小朋友的姓．根据下面三句话，请你猜一猜，三位小朋友各姓什么? 已知： $$\\left( 1 \\right)$$甲不姓张； $$\\left( 2 \\right)$$姓黄的不是丙； $$\\left( 3 \\right)$$甲和乙正在听姓李的小朋友唱歌． ", "answer_option_list": [[{"aoVal": "A", "content": "甲姓张，乙姓黄，丙姓李． "}], [{"aoVal": "B", "content": "甲姓黄，乙姓李，丙姓���． "}], [{"aoVal": "C", "content": "甲姓黄，乙姓张，丙姓李． "}], [{"aoVal": "D", "content": "甲姓李，乙姓张，丙姓黄． "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->表格法"], "answer_analysis": ["甲姓黄，乙姓张，丙姓李． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2915", "queId": "7d0e160cf4b948c38a9c23f3aa4e358f", "competition_source_list": ["其它改编自2012年全国希望杯六年级竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "在小数$$3.1415926$$的两个数字上方加$$2$$个循环点，得到循环小数，这样的循环小数中，最小的是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3.\\dot{1}41592\\dot{6}$$ "}], [{"aoVal": "B", "content": "$$3.1\\dot{4}1592\\dot{6}$$ "}], [{"aoVal": "C", "content": "$$3.14\\dot{1}592\\dot{6}$$ "}], [{"aoVal": "D", "content": "$$3.14159\\dot{2}\\dot{6}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数感认知->小数数字加工"], "answer_analysis": ["要使小数最小，则循环节开始的几位尽量小，因此从$$1$$开始循环，下一位为$$4$$，依次往下，最小的为$$3.\\dot{1}41592\\dot{6}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1374", "queId": "554df635cf5b4b3185c49a64e9597705", "competition_source_list": ["2016年第16届世奥赛六年级竞赛决赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "高速智能化办公是当代企业的发展趋势，市面上各种办公软件也是五花八门．甲方要传输一分文件给乙方，若单用$$A$$软件传输，需$$10$$分钟；若单用$$B$$软件传输，需$$8$$分钟；若同时用$$A$$、$$B$$软件传输，$$A$$、$$B$$软件每分钟共少传输$$0.2$$页．已知$$A$$、$$B$$同时传输，$$5$$分钟传完了整份文件，那么这份文件的页数为页． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题->方程法解其他问题"], "answer_analysis": ["找到功能做效率的等量关系列出方程，设这份文件总共有$$x$$页，$$\\frac{x}{10}+\\frac{x}{8}-0.2=\\frac{x}{5}$$，解得$$x=8$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2989", "queId": "bbff3dfc8e2c4253b3da13e789922574", "competition_source_list": ["2020年第24届YMO四年级竞赛决赛第9题3分", "2019年第24届YMO四年级竞赛决赛第9题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$5$$，$$a$$，$$b$$，$$c$$，$$4035$$这五个数成等差数列，则$$b=$$． We know that $$5$$, $$a$$, $$b$$, $$c$$, and $$4035$$ are five numbers in an equal series, then $$b = $$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2017$$ "}], [{"aoVal": "B", "content": "$$2018$$ "}], [{"aoVal": "C", "content": "$$2019$$ "}], [{"aoVal": "D", "content": "$$2020$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->言语化数学原理"], "answer_analysis": ["$$5$$，$$a$$，$$b$$，$$c$$，$$4035$$是等差数列， 则$$5$$，$$b$$，$$4035$$也是等差数列， 则$$b=\\left( 4035+5 \\right)\\div 2$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=4040\\div 2$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=2020$$． 选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3289", "queId": "63bef6464e6848b8a61255b434fc6194", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$42$ children line up for the autumn outing. Counting from front to back, $Y$ is the $22$\\textsuperscript{nd}. Counting from back to front, $M$ is the $22$\\textsuperscript{nd}. How many children are there between $Y$ and $M$? $$42$$个小朋友排成一队去秋游，从排头往后数，小$$Y$$是第$$22$$个，从排尾往前数，小$$M$$是第$$22$$个，小$$Y$$和小$$M$$中间有个人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->计数模块->加乘原理->排队问题"], "answer_analysis": ["根据题意分析可知，小$$Y$$后面有$$42-22=20$$（���），$$22-20-1=1$$（人）．那么小$$M$$应该在小$$Y$$的正前方，所以小$$Y$$和小$$M$$之间有$$0$$个人． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2833", "queId": "7bc8314a40cc41a5b4d3be203578ff57", "competition_source_list": ["2016年全国小学生数学学习能力测评四年级竞赛初赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "两个数相乘，如果一个因数增加$$3$$，积就增加$$54$$；如果另一个因数减少$$4$$，积就减少$$96$$，原来两个因数的积是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$432$$ "}], [{"aoVal": "B", "content": "$$434$$ "}], [{"aoVal": "C", "content": "$$436$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数乘除->整数乘法运算->表外乘法计算"], "answer_analysis": ["一个因数为$$54\\div 3=18$$，另一个因数为$$96\\div 4=24$$， 所以原来两数的积为$$18\\times 24=432$$． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3475", "queId": "e73942628a8849ec9b0475a475419cfa", "competition_source_list": ["2016年创新杯六年级竞赛训练题（三）第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "用$$8$$、$$9$$、$$10$$、$$15$$中的任意两个数组成互质数，可组成（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$对 "}], [{"aoVal": "B", "content": "$$2$$对 "}], [{"aoVal": "C", "content": "$$3$$对 "}], [{"aoVal": "D", "content": "$$4$$对 "}]], "knowledge_point_routes": ["拓展思维->能力->抽象概括"], "answer_analysis": ["按照互质的定义枚举判断即可． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3320", "queId": "5fda18a397a34a4ca7fc75c2ace45c44", "competition_source_list": ["2017年第22届全国华杯赛小学中年级竞赛初赛第3题10分", "2021年陕西西安六年级下学期小升初模拟分类专题五十五（计数原理 排列组合 容斥原理 抽屉原理）第2题", "2019年陕西西安碑林区西北工业大学附属中学小升初入学真卷六第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明行李箱锁的密码是由两个数字$$8$$与$$5$$构成的三位数．某次旅行，小明忘记了密码，他最少要试次，才能确保打开箱子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由题知，密码由三位数构成，且只包括$$8$$与$$5$$两个数字． 以$$5$$为百位数字，有$$558$$、$$585$$、$$588$$三种情况； 以$$8$$为百位数字，有$$855$$、$$885$$、$$858$$三种情况． 所以最少要试$$3+3=6$$（次），才能确保打开箱子． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1714", "queId": "69cabd0e97b342fcabfa96c990b0f56e", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（五）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "一根绳子剪去全长的$$20 \\% $$后，又接上$$20$$米，接上后的长度是接上前的$$125 \\% $$，那么原来的绳子长（~ ）米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$44$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$125$$ "}], [{"aoVal": "D", "content": "$$150$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->量率对应求单位1"], "answer_analysis": ["$$20\\div \\left( 125 \\% -1 \\right)=80$$（米），$$80\\div \\left( 1-20 \\% \\right)=100$$（米）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "760", "queId": "df1bdc287d8b46c293d5a2c204217da9", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛复赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "给出一列数：$$23+m$$，$$23+2m$$，$$23+3m$$，$$\\cdot \\cdot \\cdot $$，$$23+2015m$$，这$$2015$$个数的和除以$$14$$的余数是（其中$$m$$为正整数）．（注意写过程） ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数的性质->余数性质综合"], "answer_analysis": ["$$23\\times 2015+(1+2+\\cdot \\cdot \\cdot +2015)m$$除以$$14$$的余数，由于$$1+2+\\cdot \\cdot \\cdot +2015=2015\\times 1008$$是$$14$$的倍数，那么只用看$$23\\times 2015$$除以$$14$$的余数是$$5$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2494", "queId": "1e404817621145e297927366f0cecb7c", "competition_source_list": ["2013年全国美国数学大联盟杯小学高年级竞赛初赛第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "在前$$100$$个正整数中，$$8$$的倍数个数与$$4$$的倍数个数之比是多少？ Of the first $$100$$ positive whole numbers, the ratio of the number of multiples of $$8$$ to the number of multiples of $$4$$ is ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2:1$$ "}], [{"aoVal": "B", "content": "$$12:25$$ "}], [{"aoVal": "C", "content": "$$13:25$$ "}], [{"aoVal": "D", "content": "$$1:2$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->计算模块->比和比例"], "answer_analysis": ["$4$的倍数有$$25$$个 $8$的倍数有$$12$$个 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2447", "queId": "2abe4d2652c7431a9b6e3161568fe826", "competition_source_list": ["走美杯三年级竞赛", "走美杯六年级竞赛"], "difficulty": "0", "qtype": "single_choice", "problem": "计算：$$1+2+3+\\cdots +50+\\cdots +3+2+1=$$（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$2500$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->金字塔数列"], "answer_analysis": ["原式$$=50\\times50=2500$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2", "queId": "0097cb8eaf6448ee990b0613c9ee36bb", "competition_source_list": ["2015年第11届全国新希望杯五年级竞赛复赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "星星、希希、望望、贝贝四个人参加了``新希望杯''全国数学大赛，他们的得分情况是：望望的分数没有希希的分数高，星星的分数比贝贝的分数高，希希的分数不是最高的，贝贝的分数没有希希的分数高，他们中分数最高的是（~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "星星 "}], [{"aoVal": "B", "content": "希希 "}], [{"aoVal": "C", "content": "望望 "}], [{"aoVal": "D", "content": "贝贝 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->表格法"], "answer_analysis": ["星星和希希比贝贝高，希希比望望高，希希不是最高，那么只能是星星． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2784", "queId": "d1775673f25c4c92b2197841d00e4cb8", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（一）"], "difficulty": "2", "qtype": "single_choice", "problem": "~已知一个数列按照一定规律排列，如：$$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$34$$，$$\\ldots \\ldots $$，那么这个数列第$$2017$$个数除以$$5$$的余数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["从数列的第一个数开始余数分别是$$1$$,$$1$$,$$2$$,$$3$$,$$0$$,$$3$$,$$3$$,$$1$$,$$4$$,$$0$$,$$4$$,$$4$$,$$3$$,$$2$$,$$0$$,$$2$$,$$2$$,$$4$$,$$1$$,$$0$$.周期为$$20$$ $$2017\\div 5=100$$（组）$$\\ldots \\ldots 17$$（个），所以余数是此数列第$$17$$个余数，是$$2$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1484", "queId": "f17aab6f17334076aa6f857aa635e64c", "competition_source_list": ["2016年创新杯六年级竞赛训练题（二）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "五个数$$a$$， $$b$$，$$c$$，$$d$$，$$e$$每次去掉一个数，将其余得四个平均数，这样算了五次，结果分别为：$$7$$、$$7.5$$、$$8$$、$$8.5$$，那么原来的五个数的平均数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$7.5$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$8.5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型"], "answer_analysis": ["$$\\left( 7+7.5+8+8.5+9 \\right)\\div 5=8$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2242", "queId": "e844fa1e48c24583a9975d5e96f6ec8e", "competition_source_list": ["2015年第11届全国新希望杯五年级竞赛复赛第6题", "2016年创新杯五年级竞赛训练题（四）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "★★★★甲、乙两船在静水中的速度相同，它们分别从相距$$60$$千米的两港同时出发相向而行，$$2$$小时后相遇，如果两船的速度各增加$$5$$千米/小时，再次从两港同时出发相向而行，那么，它们再次相遇的地点就与前一次的相遇地点相距$$0.45$$千米，则水流的速度是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.7$$千米/小时 "}], [{"aoVal": "B", "content": "$$1.4$$干米/小时 "}], [{"aoVal": "C", "content": "$$0.9$$千米/小时 "}], [{"aoVal": "D", "content": "$$1.8$$千米/小时 "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["原来的速度和是船速和$$60\\div 2=30$$，那么船速是$$30\\div 2=15$$千米/时，第二次时间为$$60\\div 40=1.5$$小时．两船两次相向运动的速度差都是$$2$$个水速，路程差为$$0.45\\times 2=0.9千米，那么$$0.9$$除以（2-1.5）等于$$1.8$$，$$1.8$$除以$$2$$等于$$0.9$$，故水速为$$0.9$$千米/时． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "97", "queId": "58cfd5b230a741c083c6577cba5103aa", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第15题"], "difficulty": "3", "qtype": "single_choice", "problem": "老师把某两位数的六个不同因数分别告诉了$$A\\sim F$$六个聪明诚实的同学． $$A$$和$$B$$同时说：我知道这个数是多少了． $$C$$和$$D$$同时说：听了他们的话，我也知道这个数是多少了． $$E$$：听了他们的话，我知道我的数一定比$$F$$的大． $$F$$：我拿的数的大小在$$C$$和$$D$$之间． 那么六个人拿的数之和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$141$$ "}], [{"aoVal": "B", "content": "$$152$$ "}], [{"aoVal": "C", "content": "$$171$$ "}], [{"aoVal": "D", "content": "$$175$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["（1）这个数的因数个数肯定不低于$$6$$个（假定这个数为$$N$$，且拿到的$$6$$个数从大到小分别是$$ABCDEF$$~） （2）有两个人同时第一时间知道结果，这说明以下几个问题： 第一种情况：有一个人知道了最后的结果，这个结果是怎么知道的呢？很简单，他拿到的因数在$$50\\sim 99$$之间（也就是说$$A$$ 的$$2$$倍是$$3$$位数，所以$$A$$其实就是$$N$$） 第二种情况：有一个人拿到的不是最后结果，但是具备以下条件： 这个数的约数少于$$6$$个，比如：有人拿到$$36$$，但他不能判断$$N$$究竟是$$36$$还是$$72$$． 这个数小于$$50$$，不然这个数就只能也是$$N$$了． 这个数大于$$33$$，比如：有人拿到$$29$$，那么他不能断定$$N$$ 是$$58$$还是$$87$$；(这里有个特例是$$27$$，因为$$27\\times 2=54$$，因数个数不少于$$6$$个；$$27\\times 3=81$$，因数个数少于$$6$$个，所以如果拿到$$27$$可以判断$$N$$只能为$$54$$） 这个数还不能是是质数，不然不存在含有这个因数的两位数． 最关键的是，这两人的数是$$2$$倍关系 但是上述内容并不完全正确，需要注意还有一些``奇葩''数：$$17$$、$$19$$、$$23$$也能顺利通过第一轮． 因此，这两个人拿到的数有如下可能： （$$54$$，$$27$$）（$$68$$，$$34$$）（$$70$$，$$35$$）（$$76$$，$$38$$）（$$78$$，$$39$$）（$$92$$，$$46$$）（$$98$$，$$49$$） （3）为了对比清晰，再来把上面所有的情况的因数都列举出来： （$$54$$，$$27$$，$$18$$，$$9$$，$$6$$，$$3$$，$$2$$，$$1$$） （$$68$$，$$34$$，$$17$$，$$4$$，$$2$$，$$1$$）（$$\\times$$） （$$70$$，$$35$$，$$14$$，$$10$$，$$7$$，$$5$$，$$2$$，$$1$$） （$$76$$，$$38$$，$$19$$，$$4$$，$$2$$，$$1$$）（$$\\times$$） （$$78$$，$$39$$，$$26$$，$$13$$，$$6$$，$$3$$，$$2$$，$$1$$） （$$92$$，$$46$$，$$23$$，$$4$$，$$2$$，$$1$$）（$$\\times$$） （$$98$$，$$49$$，$$14$$，$$7$$，$$2$$，$$1$$） 对于第一轮通过的数，用红色标注，所以$$N$$不能是$$68$$、$$76$$、$$92$$中的任意一个． 之后在考虑第二轮需要通过的两个数． 用紫色标注的$$6$$、$$3$$、$$2$$、$$1$$，因为重复使用，如果出现了也不能判断$$N$$是多少，所以不能作为第二轮通过的数． 用绿色标注的$$14$$和$$7$$也不能作为第二轮通过的数，这样$$N$$也不是$$98$$． 那么通过第二轮的数只有黑色的数． 所以$$N$$ 只能是$$54$$、$$70$$、$$78$$中的一个． 再来观察可能满足$$E$$和$$F$$所说的内容： （$$54$$，$$27$$，$$18$$，$$9$$，$$6$$，$$3$$，$$2$$，$$1$$） （$$70$$，$$35$$，$$14$$，$$10$$，$$7$$，$$5$$，$$2$$，$$1$$） （$$78$$，$$39$$，$$26$$，$$13$$，$$6$$，$$3$$，$$2$$，$$1$$） 因为$$F$$说他的数在$$C$$和$$D$$之间，发现上面的数据只有当$$N=70$$的时候，$$F=7$$，在$$CD$$$$(10$$和$$5)$$之间，是唯一满足条件的一种情况． 又因为$$E$$ 确定自己比$$F$$的大，那么他拿到的数一定是该组中剩余数里最大的．所以$$E$$拿到的是$$14$$（$$N=70$$~）． 所以$$N=70$$，六个人拿的数之和为：$$70+35+14+10+7+5=141$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "541", "queId": "f543a9b7d68348fa906a0f795317ac0b", "competition_source_list": ["2016年IMAS小学高年级竞赛第一轮检测试题第13题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知袋子中装有$$n$$颗小球，依次编号为$$1$$、$$2$$、$$3$$、$$\\cdots$$、$$n$$，每次都从袋子中取出两颗球，把它们的编号相加并记下结果，然后把它们放回袋子内．重复抽取直到袋子中每一对小球都被取到为止，记录中恰好有$$215$$种不同的数值，请问$$n$$的值是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$105$$ "}], [{"aoVal": "C", "content": "$$108$$ "}], [{"aoVal": "D", "content": "$$109$$ "}], [{"aoVal": "E", "content": "$$215$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->归纳递推->汉诺塔递推"], "answer_analysis": ["每两个数相加，和最小为$$1+2=3$$，最大为$$n+(n-1)=2n-1$$，现在共有$$215$$种不同的数值，即最大的数值为$$215+3-1=217$$，$$217=2n-1$$，$$n=109$$． 最大值和最小值之间有多少数字就是多少种可能性．$$2n -1-3=215$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1661", "queId": "848d23e3fc414895ae1381c20f883406", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "大李和小李比赛吃葡萄，两人一共吃$$68$$个葡萄，大李比小李的$$4$$倍少$$2$$个，大李吃了个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$52$$ "}], [{"aoVal": "C", "content": "$$54$$ "}], [{"aoVal": "D", "content": "$$56$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["解：设小李吃$$x$$个葡萄，则大李吃（$$68-x$$）个葡萄， $$68-x=4x-2$$， $$68+2=4x+x$$， $$70=5x$$， $$\\textasciitilde\\textasciitilde x=14$$， ∴$$68-x=68-14=54$$（个）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1412", "queId": "3ac90452c59d43279b45ea86208d245a", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "小白兔有$$15$$个萝卜，小灰兔给了它$$5$$个萝卜后，它俩的萝卜就一样多了，小灰兔原来有个萝卜． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->移多补少->不等变相等（无图）"], "answer_analysis": ["小灰兔给了小白兔$$5$$个萝卜后， 此时小白兔和小灰兔都有$$15+5=20$$（个）萝卜， 则小灰兔原来有萝卜：$$20+5=25$$（个）． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1107", "queId": "1ce39922d78d4651a44ad09b30465ec9", "competition_source_list": ["2016年全国小学生数学学习能力测评四年级竞赛复赛第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "爸爸以均匀的速度在马路边散步，他从第一根电线杆到第$$12$$根电线杆，用了$$8$$分钟，照这样的速度，再走$$8$$分钟，他会走到第根电线杆． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$23$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$(12-1)\\div 8\\times (8+8)+1$$ $$=11\\times 2+1$$ $$=23$$（根）． 答：他会走到第$$23$$根电线杆． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "281", "queId": "ff8080814502fa24014507544dd409f6", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第11题"], "difficulty": "2", "qtype": "single_choice", "problem": "你能根据以下的线索找出百宝箱的密码吗？ （1）密码是一个���位数； （2）密码既是$$3$$的倍数又是$$25$$的倍数； （3）这个密码在$$20000000$$到$$30000000$$之间； （4）百万位与十万位上的数字相同； （5）百位数字比万位数字小$$2$$； （6）十万位、万位、千位上数字组成的三位数除以千万位、百万位上数字组成的两位数，商是$$25$$． 依据上面的条件，推理出这个密码应该是（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25526250$$ "}], [{"aoVal": "B", "content": "$$26650350$$ "}], [{"aoVal": "C", "content": "$$27775250$$ "}], [{"aoVal": "D", "content": "$$28870350$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->答案（数字）正误问题"], "answer_analysis": ["将A、B、C、D 逐一代入检验．只有B 满足$$（1）$$、$$（2）$$、$$（3）$$、$$（4）$$、$$（5）$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1113", "queId": "1cf79a066fdb45f39e4ce13e873d94fc", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "毛毛用围棋子摆成一个三层空心方阵，最内一层有围棋子$$28$$个．毛毛摆这个方阵共用围棋子~\\uline{~~~~~~~~~~}~个。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$ "}], [{"aoVal": "B", "content": "$$108$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$132$$ "}]], "knowledge_point_routes": ["拓展思维->能力->图形认知"], "answer_analysis": ["由于相邻两层数量相差$$8$$，已知最内层有$$28$$个棋子 则 第二层棋子有：$$28+8=36$$（个）， 第三层棋子有：$$36+8=44$$（个）， 所以三层一共有$$28+36+44=108$$（个）． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3350", "queId": "5c663aa25d114aa4b2b4a7933ba359be", "competition_source_list": ["2006年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "一根长木棍上刻有三种刻度，第一种刻度将木棍十等分，第二种刻度将木棍十二等分，第三种刻度将木棍十五等分。如果沿刻度线将木棍锯开，木棍总共被锯成（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$段 "}], [{"aoVal": "B", "content": "$$24$$段 "}], [{"aoVal": "C", "content": "$$28$$段 "}], [{"aoVal": "D", "content": "$$30$$段 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->两数的最小公倍数"], "answer_analysis": ["在没有重合点时，内点数为$$\\left( 10-1 \\right)+\\left( 12-1 \\right)+\\left( 15-1 \\right)=34$$，$$\\left[ 10,12,15 \\right]=60$$，不妨设木棍长$$60$$厘米； 则第一种刻度线分割的长度为$$6$$厘米； 则第二种刻度线分割的长度为$$5$$厘米； 则第三种刻度线分割的长度为$$4$$厘米； 第一种与第二种刻度线重合处为$$\\left[ 6,5 \\right]=30$$厘米，全棍重合$$1$$次； 第二种与第三种刻度线重合处为$$\\left[ 5,4 \\right]=20$$厘米，全棍重合$$2$$次； 第三种与第一种刻度线重合处为$$\\left[ 4,6 \\right]=12$$厘米，全棍重合$$4$$次； 则总内点数为$$34-7=27$$，于是总段数为$$27+1=28$$（段）。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3416", "queId": "d74812de0b844828a63e8ed964881583", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "芳草地小学四年级有$$58$$人学钢琴，$$43$$人学画画，$$37$$人既学钢琴又学画画，问只学钢琴和只学画画的分别有多少人？ ", "answer_option_list": [[{"aoVal": "A", "content": "6，15 "}], [{"aoVal": "B", "content": "6，21 "}], [{"aoVal": "C", "content": "15，6 "}], [{"aoVal": "D", "content": "24，8 "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["解包含与排除题，画图是一种很直观、简捷的方法，可以帮助解决问题，画图时注意把不同的对象与不同的区域对应清楚．建议教师帮助学生画图分析，清楚的分析每一部分的含义． 如图，$$A$$圆表示学画画的人，$$B$$圆表示学钢琴的人，$$C$$表示既学钢琴又学画画的人，图中$$A$$圆不含阴影的部分表示只学画画的人，有：$$43-37=6$$人，图中$$B$$圆不含阴影的部分表示只学钢琴的人，有：$$58-37=21$$人． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "412", "queId": "8a8db137f6414480a47fb82bdf35093f", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$是三个不同的自然数，且$$a\\times b\\times c=210$$ ．求$$a+b+c$$的最大值和最小值．． ", "answer_option_list": [[{"aoVal": "A", "content": "$$108$$；$$27$$ "}], [{"aoVal": "B", "content": "$$108$$；$$18$$ "}], [{"aoVal": "C", "content": "$$120$$；$$12$$ "}], [{"aoVal": "D", "content": "$$120$$；$$18$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["三个不同的自然数$$a$$，$$b$$，$$c$$的乘积是$$210$$，当三个数越接近，它们的和越小；反之，它们的和越大．因为$$210=1\\times 2\\times 3\\times 5\\times 7$$，所以当$$a$$，$$b$$，$$c$$分别取$$1$$，$$2$$，$$105$$时，它们的和最大，为$$1+2+105=108$$；当$$a$$，$$b$$，$$c$$分别取$$5$$，$$6$$，$$7$$时，它们的和最小，为$$5+6+7=18$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2230", "queId": "be9a530f3ac74833b63764c655e17299", "competition_source_list": ["2011年北京五年级竞赛", "2019年广东广州海珠区中山大学附属中学小升初第42题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙丙三人同时从东村去西村，甲骑自行车每小时比乙快$$12$$公里，比丙快$$15$$公里，甲行$$3.5$$小时到达西村后立刻返回．在距西村$$30$$公里处和乙相聚，问：丙行了多长时间和甲相遇？ ", "answer_option_list": [[{"aoVal": "A", "content": "2.8小时 "}], [{"aoVal": "B", "content": "5.6小时 "}], [{"aoVal": "C", "content": "3.6小时 "}], [{"aoVal": "D", "content": "1.5小时 "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->实践应用"], "answer_analysis": ["在距西村$$30$$公里处和乙相聚，则甲比乙多走$$60$$公里，而甲骑自行车每小时比乙快$$12$$公里，所以，甲乙相聚时所用时间是$$60 \\div 12 = 5$$（小时），所以甲从西村到和乙相聚用了$$5 - 3.5 = 1.5$$（小时），所以，甲速是$$30 \\div 1.5 = 20$$（公里/小时），所以，丙速是$$20 - 15 = 5$$（公里/小时），东村到西村的距离是：$$20 \\times 3.5 = 70$$（公里），所以，甲丙相遇时间是：$$\\left( {2 \\times 70} \\right) \\div \\left( {20 + 5} \\right) = 5.6$$$$($$小时$$)$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "336", "queId": "d1a6712272ad400cae4b766ebe14da2b", "competition_source_list": ["2017年第17届世奥赛六年级竞赛决赛第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "地球因为自转的关系，所以每个国家都有时差．例：纽约比韩国晚$$13$$个小时，韩国$$4$$月$$28$$日晚$$6$$点时纽约是$$4$$月$$28$$日凌晨$$5$$点．思思暑假的时候要去纽约找杰克，在纽约的时候每周星期一、三、五早上$$10$$点给父母打电话，思思是$$7$$月$$15$$日星期一早上$$7$$点（韩国时间）出发去纽约．$$8$$月$$2$$日星期五早上$$10$$点（韩国时间）回到韩国．思思在纽约跟父母一共打了（~ ）个电话．（注：到纽约的飞机要坐$$14$$个小时） ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->时间问题->时间计算"], "answer_analysis": ["到达纽约时间，韩国时间是$$7$$月$$15$$日$$21:00$$，纽约时间是$$7$$月$$15$$日$$8：00$$；纽约出发时间，韩国时间是$$8$$月$$1$$日$$20:00$$，纽约时间是$$8$$月$$1$$日$$7:00$$．依次可知思思在纽约一共给父母打了$$8$$个电话． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2458", "queId": "fa456e30fd3540ff82619de4326d3093", "competition_source_list": ["五年级竞赛创新杯", "六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "在下面四个算式中，得数最大的是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "$$(\\frac{1}{17}-\\frac{1}{19})\\div 20$$ "}], [{"aoVal": "B", "content": "$$(\\frac{1}{15}-\\frac{1}{21})\\div 60$$ "}], [{"aoVal": "C", "content": "$$(\\frac{1}{13}-\\frac{1}{23})\\div 100$$ "}], [{"aoVal": "D", "content": "$$(\\frac{1}{11}-\\frac{1}{25})\\div 140$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->整数比较大小"], "answer_analysis": ["A，B，C，D依次为$$\\frac{1}{17\\times 19\\times 10}，\\frac{1}{15\\times 21\\times 10}，\\frac{1}{13\\times 23\\times 10}，\\frac{1}{11\\times 25\\times 10}$$，和同近积大，$$11\\times 25\\times 10$$最小，所以$$\\frac{1}{11\\times 25\\times 10}$$最大. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "602", "queId": "0cada157fbed469fa6fdba1eff33e882", "competition_source_list": ["2013年美国数学大联盟杯小学高年级竞赛初赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\sqrt{4\\times 9\\times 16}=$$．（请换算成$$2$$进制数） ", "answer_option_list": [[{"aoVal": "A", "content": "$$111$$ "}], [{"aoVal": "B", "content": "$$1101$$ "}], [{"aoVal": "C", "content": "$$1110$$ "}], [{"aoVal": "D", "content": "$$11000$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$\\sqrt{4\\times 9\\times 16}$$，可以看成$$\\sqrt{{{2}^{2}}\\times {{3}^{2}}\\times {{4}^{2}}}$$，也就是$$2\\times 3\\times 4=24$$， 所以答案应该是$$24$$． 但是答案都是$$2$$进制表示，$$24$$的$$2$$进制为$$11000$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "614", "queId": "0d77e31235c24a87a9d618eecfc72523", "competition_source_list": ["2005年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$1$$，$$2$$，$$3$$，$$5$$，$$7$$，$$8$$这$$6$$个数字中，有（ ）个数字不可能是一个整数平方的个位数。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$个 "}], [{"aoVal": "B", "content": "$$2$$个 "}], [{"aoVal": "C", "content": "$$3$$个 "}], [{"aoVal": "D", "content": "$$4$$个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->完全平方数->平方数的尾数特征"], "answer_analysis": ["一个完全平方数的末尾只能是$$0$$、$$1$$、$$4$$、$$5$$、$$6$$、$$9$$，所以在$$1$$、$$2$$、$$3$$、$$5$$、$$7$$、$$8$$中，$$2$$、$$3$$、$$7$$、$$8$$这$$4$$个数不是一个整数的平方的个位数。 故选：D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3090", "queId": "e1d55053f0464d648e3bde4f7b5d1003", "competition_source_list": ["2013年全国学而思杯二年级竞赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$3+5+7+9+11+13+15+17=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$70$$ "}], [{"aoVal": "C", "content": "$$80$$ "}], [{"aoVal": "D", "content": "$$90$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之分组法"], "answer_analysis": ["此题分为两种思路：一种可以直接使用凑整的方法，共有$$4$$对，$$20\\times ~4=80$$；另一种可以使用等差数列求和的方法：$$\\left( 3+17 \\right)\\times ~8\\div2=80$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2180", "queId": "274c12107f734fcd981669e97221bb12", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛六年级竞赛决赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两个学生放学回家，甲要比乙多走$$\\frac{1}{5}$$的路，而乙走的时间比甲少$$\\frac{1}{11}$$，甲、乙两人速度的比是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11:12$$ "}], [{"aoVal": "B", "content": "$$12:11$$ "}], [{"aoVal": "C", "content": "$$11:13$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["甲、乙路程比$$=6:5$$， 甲、乙时间比$$=11:10$$， 故甲、乙的速度比为$$\\frac{6}{11}:\\frac{5}{10}=12:11$$． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2541", "queId": "b03c966d65284527817ccc1cfc044f43", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（一）"], "difficulty": "1", "qtype": "single_choice", "problem": "将分数$$\\frac{1}{7}$$写成循环小数，那么小数点后的第$$2017$$位数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->循环小数->循环小数的周期问题"], "answer_analysis": ["$$\\frac{1}{7}=0.\\dot{1}4285\\dot{7}$$，循环节有$$6$$个数字，$$2017\\div 6=336\\cdots \\cdots 1$$，所以小数点后第$$2017$$位数字是$$1$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "775", "queId": "5b5b45fc57694040b1429699e32daaa0", "competition_source_list": ["2016年全国AMC六年级竞赛8第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$N$$是一个两位数数字．当$N$除以$9$，余数为$1$.当$N$除以$10$，余数为$3$.当$N$除以$11$时，余数是多少? The number $$N$$ is a two-digit number. When $$N$$ is divided by $$9$$, the remainder is$$1$$. When $$N$$ is divided by $$10$$, the remainder is $$3$$. $$What$$ is the remainder when $$N$$ is divided by $$11$$? ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->数论模块->余数问题->中国剩余定理"], "answer_analysis": ["翻译：$$N$$是一个两位数，$$N$$除以$$9$$余$$1$$，除以$$10$$ 余$$3$$，那么$$N$$除以$$11$$余． $$N\\div 9=\\cdots \\cdots 1$$，且个位是$$3$$，所以$$N=9\\times 8+1=73$$（$$N$$是一个两位数），$$73\\div 11=6\\cdots \\cdots 7$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1222", "queId": "22bc48df53a3459cac96a1df00814a96", "competition_source_list": ["2020年新希望杯六年级竞赛（2月）第13题", "2020年希望杯六年级竞赛模拟第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "海尔兄弟被困在一个无人岛上，他们要做一个独木舟逃出这个无人岛，哥哥单独做要$$6$$小时完成，弟弟单独做要$$9$$小时完成．如果按照哥哥、弟弟、哥哥、弟弟$$\\cdots \\cdots $$的顺序交替工作，每人工作$$1$$小时后交换，那么需要~\\uline{~~~~~~~~~~}~小时能做好独木舟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6.5$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$7.5$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$8.5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["哥哥工效：$$\\frac{1}{6}$$，弟弟工效：$$\\frac{1}{9}$$， 则$$1\\div \\left( \\frac{1}{6}+\\frac{1}{9} \\right)=\\frac{18}{5}=3$$（周期）$$\\cdots \\cdots \\frac{3}{5}$$， $$1-3\\times \\left( \\frac{1}{6}+\\frac{1}{9} \\right)=\\frac{1}{6}$$， $$\\frac{1}{6}\\div\\frac{1}{6}=1$$（小时）， 共$$3\\times 2+1=7$$（小时）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2434", "queId": "4635d244e94549e9a1b3c6733e9d9f75", "competition_source_list": ["2016年第12届全国新希望杯五年级竞赛决赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个分数的分子和分母相差$$60$$，约成最简分数是$$\\frac{4}{9}$$，这个分数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{64}{69}$$ "}], [{"aoVal": "B", "content": "$$\\frac{42}{102}$$ "}], [{"aoVal": "C", "content": "$$\\frac{60}{135}$$ "}], [{"aoVal": "D", "content": "$$\\frac{48}{108}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数感认知->分数数字加工"], "answer_analysis": ["分子、分母缩小相同倍数，分子分母之间差也缩小相同的倍数，所以$$60\\div 5=12$$．则$$\\frac{4}{9}=\\frac{4\\times 12}{9\\times 12}=\\frac{48}{108}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2823", "queId": "6e0e9d20e68b41b58c2ab007183d6ea7", "competition_source_list": ["2013年全国美国数学大联盟杯小学高年级竞赛初赛第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "【闯关7】 $${{2}^{3}}\\times {{2}^{5}}\\times {{2}^{7}}\\div {{2}^{11}}=$$ ． ", "answer_option_list": [[{"aoVal": "A", "content": "$${{4}^{13}}$$ "}], [{"aoVal": "B", "content": "$${{2}^{26}}$$ "}], [{"aoVal": "C", "content": "$${2}^{4}$$ "}], [{"aoVal": "D", "content": "$${{16}^{1155}}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->乘方->乘方的运算->乘方的幂运算"], "answer_analysis": ["$$2^{3}\\times2^{5}\\times2^{7}\\div2^{11}=2^{3+5+7-11}=2^{4}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1936", "queId": "b7d27b21eed747ccbf14d0fcbd854e42", "competition_source_list": ["2014年全国学而思杯二年级竞赛第3题"], "difficulty": "0", "qtype": "single_choice", "problem": "妈妈带了一些钱去买书，刚好够买一套《窗边的小豆豆》；这些钱如果买$$10$$元一本的《安徒生童话》可以买$$4$$本，那么，一套《窗边的小豆豆》的价格是~\\uline{~~~~~~~~~~}~元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$40$$ "}], [{"aoVal": "E", "content": "$$50$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->倍的基本计算->两��倍", "海外竞赛体系->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["$$10\\times 4=40$$（元）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "977", "queId": "0df17bd74d204f19b45a9a917a3a768b", "competition_source_list": ["2009年全国迎春杯六年级竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一批图书总数在$$1000$$本以内，若按$$24$$本书包成一捆，则最后一捆差$$2$$本；若按$$28$$本书包成一捆，最后一捆还是差$$2$$本书；若按$$32$$本包一捆，则最后一捆是$$30$$本．那么这批图书共有~\\uline{~~~~~~~~~~}~本． ", "answer_option_list": [[{"aoVal": "A", "content": "答案请写在答题卡上 "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->盈亏问题->盈亏基本类型->亏亏问题"], "answer_analysis": ["答案请写在答题卡上 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2493", "queId": "ab7f6c565a2a4ac3bad374a7429e48a6", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第17题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "观察： $$74\\times 6=444$$， $$74\\times 12=888$$， $$74\\times $$$$=444888$$ 请问括号中应填填入下列哪一个数才可以使得算式正确？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$444$$ "}], [{"aoVal": "D", "content": "$$888$$ "}], [{"aoVal": "E", "content": "$$6012$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$444888=444000+888=74\\times 6000+74\\times 12=74\\times 6012$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3237", "queId": "42fe45d7fccc4a1892665330e6f0177d", "competition_source_list": ["2009年全国迎春杯小学中年级竞赛复赛第7题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "（10分）过年了，妈妈买了$$7$$件不同的礼物，要送给亲朋好友的$$5$$个孩子每人一件．其中姐姐的儿子小强想从智力拼图和遥控汽车中选一个，朋友的女儿小玉想从学习机和遥控汽车中选一件．那么妈妈送出这$$5$$件礼物共有~\\uline{~~~~~~~~~~}~种方法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$180$$ "}], [{"aoVal": "B", "content": "$$160$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$80$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计数模块->加乘原理->乘法原理->物品搭配"], "answer_analysis": ["假如给小强的是智力拼图，则有$$2\\times 5\\times 4\\times 3=120$$种方法． 假如给小强的是遥控汽车，则有$$1\\times 5\\times 4\\times 3=60$$种方法． 总共有$$120+60=180$$种方法． 若将遥控汽车给小强，则学习机要给小玉，此时另外$$3$$个孩子在剩余$$5$$件礼物中任选$$3$$件，有$$5\\times 4\\times 3=60$$种方法;若将遥控汽车给小玉，则智力拼图要给小强，此时也有$$60$$种方法;若遥控汽车既不给小强、也不给小玉，则智力拼图要给小强，学习机要给小玉，此时仍然有$$60$$种方法. 所以共有$$60+60+60=180$$种方法. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "931", "queId": "af700d87872d42b99c96b96d9c50ff2b", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第14题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果一个正整数（大于$$0$$的自然数）能够表示为两个自然数（自然数包括$$0$$）的平方差，就称这个正整数为``鹏程数''．例如$$3$$是一个``鹏程数''，因为$$3=2^{2}-1^{2}$$，$$16$$也是一个``鹏程数''，因为$$16=4^{2}-0^{2}$$．现在将所有``鹏程数''由小到大排序：$$1$$，$$3$$，$$4$$，$$5$$，$$7$$，$$8$$，$$9$$，$$11$$，$$12$$，$$\\cdots $$，则第$$2021$$个``鹏程数''是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2694$$ "}], [{"aoVal": "B", "content": "$$2695$$ "}], [{"aoVal": "C", "content": "$$2696$$ "}], [{"aoVal": "D", "content": "$$2697$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["完全平方数除以$$4$$只能余$$0$$或$$1$$，两个完全平方数相减，其差除以$$4$$余数只能是$$0$$或$$1$$或$$3$$， 即除以$$4$$余$$2$$的数不能写成两个完全平方数的差． 由于$$(n+1)^{2}-(n-1)^{2}=4n$$ $$(2n+1)^{2}-(2n)^{2}=4n+1$$， $$(2n+2)^{2}-(2n+1)^{2}=4n+3$$， 所以，除$$4n+2$$外，均可表示成两个数平方差， 即每$$4$$个连续的自然数中，有一个不是``鹏程数''， 所以第$$2021$$个``鹏程数''为：$$2695$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1430", "queId": "43dded80564d4547ae77fb5c842e6c1b", "competition_source_list": ["2017年第15届全国希望杯小学中年级四年级竞赛第1试试题第19题"], "difficulty": "3", "qtype": "single_choice", "problem": "袋子中有黑、白两种颜色的棋子，黑子的个数是白子个数的$$2$$倍，每次从袋中同时取出$$3$$个黑子和$$2$$个白子．某次取完后，白子剩下$$1$$个，黑子剩下$$31$$个，则袋中原有黑子~\\uline{~~~~~~~~~~}~个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$119$$ "}], [{"aoVal": "C", "content": "$$118$$ "}], [{"aoVal": "D", "content": "$$117$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题"], "answer_analysis": ["黑子个数$$=$$白子个数$$\\times 2$$；黑子个数$$=$$次数$$\\times 3+31$$，白子个数$$=$$次数$$\\times 2+1$$； 设取的次数为$$x$$次，则白子有($$2x+1$$)个，黑子有($$3x+31$$)个， 由题意得$$2\\times (2x+1)=3x+31$$， 解得$$x=29$$， 则袋子中原有黑子$$3\\times 29+31=118$$（个）． 假设黑子每次取的个数也是白子的$$2$$倍，即黑子每次取$$2\\times2=4$$（个），白子每次取$$2$$个，则： $$(31-1\\times2)\\div(2\\times2-3)$$ $$=29\\div1$$ $$=29$$（次） $$3\\times29+31$$ $$=87+31$$ $$=118$$（个）． 答：袋中原有黑子 $$118$$个． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1830", "queId": "9ba11ad6d9ff4b93a4134a542a652d35", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级三年级竞赛初赛第40题"], "difficulty": "1", "qtype": "single_choice", "problem": "约翰的钱比吉尔多$$20$$美元，他们两个总共有$$40$$美元．约翰有美元 . ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$(40+20)\\div 2=30$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3150", "queId": "61bb0f03218b40349d536f44d354c832", "competition_source_list": ["2011年第9届全国创新杯小学高年级六年级竞赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "~连接武汉与广州的高铁，中途只停六个车站，那么应该印刷（~ ）种车票． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$42$$ "}], [{"aoVal": "C", "content": "$$56$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["加上首位两站共$$8$$个站，共有$C_{8}^{2}=28$种组合，考虑车票的次序共有：$$28×2=56$$种． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3453", "queId": "e1abda06f465463d95de2ed9855a7610", "competition_source_list": ["2017年全国华杯赛竞赛初赛模拟题1第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "用红色和黄色给正方体的$$6$$个面染色，每个面必须染色，染色后经过旋转和翻转后相同的算同一种，共有（ ）种不同染色方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["分类计数 用$$1$$种颜色染色，则有$$2$$种不同的染色方式； 用$$2$$种颜色染色，再分类如下计数： （$$1$$）仅$$1$$个面是红色，有$$1$$种不同的染色方式； （$$2$$）仅$$2$$个面是红色，相对和相邻，有$$2$$种不同的染色方式； （$$3$$）仅$$3$$个面是红色，其中有$$2$$个相对，经适当转动，固定为红色上底面和红色下底面，仅有$$1$$种不同的染色方式：$$3$$个红色面两两相邻，仅有$$1$$种不同的染色方式； （$$4$$）有$$4$$个红色面，即有$$2$$个黄色面，不同染色方式的个数同（$$2$$）； （$$5$$）有$$5$$个红色面，即有$$1$$个黄色面，不同染色方式的个数同（$$1$$）． 共有$$10$$种不同的染色方式． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1712", "queId": "60e1cb38a4c843e2899f86e389cbc9f1", "competition_source_list": ["2013年第9届全国新希望杯小学高年级五年���竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "王伯去水果店买水果，如果买$$4$$千克梨和$$7$$千克苹果，要付款$$84$$元；如果买$$4.5$$千克梨和$$7$$千克苹果，要付款$$87.5$$元，那么买$$1$$千克梨要付款（~ ~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3.5$$元 "}], [{"aoVal": "B", "content": "$$5$$元 "}], [{"aoVal": "C", "content": "$$6$$元 "}], [{"aoVal": "D", "content": "$$7$$元 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->多元一次方程解应用题->小数系数方程（组）解题"], "answer_analysis": ["设买$$x$$千克梨，$$y$$千克苹果 $$\\begin{cases}4x+7y=84 4.5x+7y=87.5 \\end{cases}\\Rightarrow 0.5x=3.5\\Rightarrow x=7$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2002", "queId": "bd074f3f332f4644b16cb35edfb5f840", "competition_source_list": ["2015年全国美国数学大联盟杯五年级竞赛初赛第35题"], "difficulty": "1", "qtype": "single_choice", "problem": "汤姆有一件花了$$64$$美金买来的衬衫，他打算以比原价高出$$25 \\%$$的价格出售，他会卖出多少钱? ", "answer_option_list": [[{"aoVal": "A", "content": "＄$$16$$ "}], [{"aoVal": "B", "content": "＄$$32$$ "}], [{"aoVal": "C", "content": "＄$$48$$ "}], [{"aoVal": "D", "content": "＄$$80$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$64\\times(1+25 \\%)=80$$，最后答案是$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2377", "queId": "0b0a59568d074f319abe908bb6d9bb11", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "如果关于$$x$$的方程组$$\\left { \\begin{matrix}ax+3y=7 3x+6y=2 \\end{matrix} \\right.$$有解，则$$a$$不等于（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->一元一次方程->整数系数方程"], "answer_analysis": ["如果$$a=\\frac{3}{2}$$，则有$$2=3x+6y=2\\times \\left( \\frac{3}{2}x+3y \\right)=14$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2067", "queId": "eb2281832a0e436db4755306eda87fc4", "competition_source_list": ["2005年六年级竞赛创新杯", "2005年第3届创新杯六年级竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙三个小运动员参加100米赛跑，当甲到达终点时，乙离终点还有5米;当乙到达终点时，丙离终点还有5米;那么当甲到达终点时，丙离终点还有( ). ", "answer_option_list": [[{"aoVal": "A", "content": "10米 "}], [{"aoVal": "B", "content": "9.75米 "}], [{"aoVal": "C", "content": "9.25米 "}], [{"aoVal": "D", "content": "10.25米 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题"], "answer_analysis": ["当甲到达终点时，乙离终点还有5米，那么甲、乙的速度比是甲:乙$$=100:\\left( 100-5 \\right)=100:95=20:19$$，当乙到达终点时，丙还有5米，那么乙、丙的速度比是乙:丙$$=100:\\left( 100-5 \\right)=20:19$$，现把这两个比例中乙的份数变为一样的，那么甲:乙$$=20:19=400:380$$，乙:丙$$=20:19=380:361$$，从而甲:乙:丙$$=400:380:361$$，即甲:丙$$=400:361$$，所以当甲到达终点跑了100米时，丙跑了$$100\\times \\frac{361}{400}=90.25$$(米)，离终点还有$$100-90.25=9.75$$(米)，选$$B$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "579", "queId": "1355890f2a3c44c9b4279733c4ab3154", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "妈妈在家开着灯做饭，突然停电了．爸爸回家按了$$4$$下开关，薇儿回家又按了$$5$$下开关．当来电的时候，灯泡是~\\uline{~~~~~~~~~~}~的． ", "answer_option_list": [[{"aoVal": "A", "content": "亮 "}], [{"aoVal": "B", "content": "不亮 "}], [{"aoVal": "C", "content": "不能确定 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["开单数次数，改变原来灯的状态，开双数次，不改变原来的状态．总共开的次数为：$$4+5=9$$（次）， 所以改变了原来的状态，原来是开着的， 所以现在不亮． 故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1142", "queId": "264f2e4f31b74547acb3309a623a6d8d", "competition_source_list": ["2011年全国走美杯三年级竞赛初赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "灰太狼给儿子买了一本叫《捕羊宝典$$300$$篇》的书，这本书共$$301$$页，这本书的页码共用了~\\uline{~~~~~~~~~~}~个数字． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2916$$ "}], [{"aoVal": "B", "content": "$$856$$ "}], [{"aoVal": "C", "content": "$$795$$ "}], [{"aoVal": "D", "content": "$$660$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->分类讨论思想"], "answer_analysis": ["把页码进行分类： ①页码是一位数的是$$1\\tilde{ }9$$，共$$9$$个数字； ②页码是两位数的是$$10\\tilde{ }99$$，共$$90\\times 2=180$$个数字； ③页码是三位数的是$$100\\tilde{ }301$$，共$$202\\times 3=606$$个数字；所以，总共有``数字''个数为$$9+180+606=795$$个． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1034", "queId": "9897d38bbde84255870cb1b76eb88e0e", "competition_source_list": ["2016年第12届全国新希望杯小学高年级六年级竞赛复赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "``新希望杯''吉祥物若打九折，可以盈利$$60$$元；若打八折出售，可以盈利$$46$$元，则该吉祥物的成本是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$65$$元 "}], [{"aoVal": "B", "content": "$$66$$元 "}], [{"aoVal": "C", "content": "$$67$$元 "}], [{"aoVal": "D", "content": "$$68$$元 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->多元一次方程解应用题->小数系数方程（组）解题"], "answer_analysis": ["假设原价为$$x$$元，进价为$$y$$元，则$$\\left { \\begin{matrix}0.9x-y=60 0.8x-y=46 \\end{matrix} \\right.$$，解得$$\\left { \\begin{matrix}x=140 y=66 \\end{matrix} \\right.$$，即成本为$$66$$元． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "310", "queId": "d172938451f24c2ab7965a6cd377a050", "competition_source_list": ["2015年第13届全国创新杯小学高年级五年级竞赛初赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "将``$$OPQRST$$''连续接下去可得到；``$$OPQRSTOPQRST\\cdot \\cdot \\cdot $$''，从左至右第$$2015$$个字母应该是（5分） ", "answer_option_list": [[{"aoVal": "A", "content": "$$S$$ "}], [{"aoVal": "B", "content": "$$Q$$ "}], [{"aoVal": "C", "content": "$$O$$ "}], [{"aoVal": "D", "content": "$$T$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->字母规律->数字与字母结合"], "answer_analysis": ["$$2015\\div 6\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 5$$，则$$2015$$个数应为$$S$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1546", "queId": "4096675af058435db197e30c0af6466d", "competition_source_list": ["2020年第1届广东深圳超常思维竞赛五年级竞赛初赛第15题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "某地有甲、乙两个水池，今要在那里做基本建设，虽值雨季，也要先将池水抽出以便开工．已知甲池的面积是$$2$$亩，用$$3$$辆水车可于$$2$$天将水抽尽；若改用$$2$$辆水车便要$$4$$天才可将水抽尽．乙池的面积是甲池的$$3$$倍，要用$$6$$天将水抽尽，则需要用辆水车． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$5$$ "}], [{"aoVal": "E", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->多元一次方程解应用题->整数系数多元一次方程组解应用题"], "answer_analysis": ["设甲池原有水量为$$v$$，每天降于甲池的雨量为$$x$$，那么$$\\frac{v+2x}{3\\times 2}$$及$$\\frac{v+4x}{2\\times 4}$$都是表示一辆水车一天抽出甲池的水量，即$$\\frac{v+2x}{6}=\\frac{v+4x}{8}$$，得$$v=4x$$， 所以$$x=\\frac{v}{4}$$． 由此可知一辆水车一天抽甲池的水量为$$\\frac{v+2x}{6}=\\frac{v+\\frac{v}{2}}{6}=\\frac{v}{4}$$， 从上面两式，得知降于甲池一天的雨水，恰被一辆水车抽尽． 因乙池的面积是甲池的$$3$$倍，故要当天抽尽乙池当天的雨水，恰需要$$3$$辆水车．又因甲池原有水量为$$v$$，那么乙池原有水量为$$3v$$．已知一辆水车抽甲池的水量为$$\\frac{v}{4}$$，那么一辆水车抽尽乙池原有水量，必须$$3v\\div \\frac{v}{4}=12$$天．今要$$6$$天抽尽乙池原有水量，就需要$$2$$辆水车．故要$$6$$天抽尽乙池的雨量和原有水量，共需要用$$3+2=5$$辆水车． 故选：$$\\text{D}$$． （算数解法）（$$1$$）因甲池的水（包括下的雨水），用$$3$$辆水车，$$2$$天抽尽；若一天抽完，要用$$6$$辆水车．同样，用$$2$$辆水车，$$4$$天抽尽；若一天抽尽，要用$$8$$辆水车． 因为多经过两天，多下了两天雨，所以多用$$8-6=2$$辆水车， 所以，甲池每天的雨量要用一辆水车才能抽尽． 若只抽甲池原存水量，用$$3-1=2$$辆水车，$$2$$天抽尽；用一辆水车，$$4$$天抽尽， 所以每辆水车每天抽甲池原存水量的$$\\frac{1}{4}$$. （$$2$$）因乙池是甲池的$$3$$倍，所以抽当天下的雨水，需要$$3$$辆水车． 又每辆水车是抽乙池原存水量的$$\\frac{1}{4}\\div 3=\\frac{1}{12}$$．今要$$6$$天抽尽乙池的水，每天应抽出$$\\frac{1}{6}$$． 所以只抽乙池原存水量，每天需$$\\frac{1}{6}\\div \\frac{1}{12}=2$$辆水车， 所以抽乙池雨水和原存水量共用的水车是$$3+2=5$$辆这就是说，$$5$$辆水车在$$6$$天抽尽乙池的水． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "105", "queId": "3449325fd55246e08d7476067cca6b8b", "competition_source_list": ["2015年全国学而思杯小学低年级竞赛学前组第2题", "2015年第5届全国学而思综合能力诊断学前班竞赛第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "现在是下午$$2$$时，灰太狼准备在$$1$$小时后去喝下午茶，请问灰太狼几点钟喝下午茶？ ", "answer_option_list": [[{"aoVal": "A", "content": "下午$$2$$时 "}], [{"aoVal": "B", "content": "下午$$3$$时 "}], [{"aoVal": "C", "content": "下午$$4$$时 "}], [{"aoVal": "D", "content": "我不知道 "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->时间问题->认识钟表"], "answer_analysis": ["现在是下午$$2$$时，$$1$$小时后是下午$$3$$时． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2734", "queId": "ff8080814502fa2401450bea93a51ca7", "competition_source_list": ["2014年全国迎春杯四年级竞赛复赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个$$12$$项的等差数列，公差是$$2$$，且前$$8$$项的和等于后$$4$$项的和，那么，这个数列的第二项是（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["根据题意得$$({{a}_{1}}+{{a}_{8}})\\times 8\\div 2=({{a}_{9}}+{{a}_{12}})\\times4\\div 2$$，因为$${{a}_{8}}={{a}_{1}}+14$$，$${{a}_{9}}={{a}_{1}}+16$$，$${{a}_{12}}={{a}_{1}}+22$$， 所以$$({{a}_{1}}+{{a}_{1}}+14)\\times 8\\div2=({{a}_{1}}+16+{{a}_{1}}+22)\\times 4\\div 2$$，解得$${{a}_{1}}=5$$，因此$${{a}_{2}}=5+2=7$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2563", "queId": "1fc02823eda14413b6376433d0201f13", "competition_source_list": ["2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$123456789$$除以$$10000$$的商转化为小数以后，十位和千分位的积是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$48$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->计算模块->小数", "拓展思维->拓展思维->计算模块->小数->小数基础->小数点的移动规律"], "answer_analysis": ["$$123456789$$除以$$10000$$，小数点朝前移动四位，答案是$$12345.6789$$，此时，十位是$$4$$，千分位是$$8$$，积为$$32$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1061", "queId": "2ed6c11e1a9f4defbd0d65f9981d4038", "competition_source_list": ["2014年第26届广东广州五羊杯六年级竞赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有糖水若干克，第一次加水若干，浓度变为$$4 \\%$$；然后又加入同样多的水，浓度变为$$3 \\%$$；第三次再加入同样多的水，这时浓度变为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1 \\%$$ "}], [{"aoVal": "B", "content": "$$2 \\%$$ "}], [{"aoVal": "C", "content": "$$2.4 \\%$$ "}], [{"aoVal": "D", "content": "$$2.8 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["原浓度为 $$4 \\%=\\frac{4}{100}=\\frac{12}{300}\\overrightarrow{加水}$$ 后来浓度为$$3 \\%=\\frac{3}{100}=\\frac{12}{400}$$，可理解为原有$$12$$克糖（只加水，前后糖不变），原来有糖水$$300$$克，加$$100$$克水，变成$$400$$克糖��．最后再加$$100$$克水，浓度变成$$\\frac{12}{500}=\\frac{24}{1000}=2.4 \\%$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2735", "queId": "ff8080814502fa2401450bea998d1ca9", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "老师在黑板上将从$$1$$ 开始的计数连续地写下去：$$1$$，$$3$$，$$5$$，$$7$$，$$9$$，$$11$$\\ldots\\ldots 写好后，擦去了其中的两个数，将这些奇数隔成了$$3$$ 段，如果前两段的和分别是$$961$$ 和$$1001$$，那么，老师擦去的两个奇数之和是（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$154$$ "}], [{"aoVal": "B", "content": "$$156$$ "}], [{"aoVal": "C", "content": "$$158$$ "}], [{"aoVal": "D", "content": "$$160$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的加减规律"], "answer_analysis": ["从$$1$$开始连续奇数相加应该等于项数的平方．因为$$961={{31}^{2}}$$，所以擦去的第一个奇数为$$31\\times2-1+2=63$$． 设第二段有$$n$$个数，$$[65+65+2(n-1)]\\times n\\div 2=1001=7\\times 11\\times 13$$，经尝试，$$n=13$$．所以擦去的第二个数为$$65+2\\times(13-1)+2=91$$． 两个数的和为$$63+91=154$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3449", "queId": "c64cf3a18a704dc7a237e8802e90b2c5", "competition_source_list": ["2012年全国美国数学大联盟杯小学高年级竞赛初赛第34题"], "difficulty": "1", "qtype": "single_choice", "problem": "小龙每月读书的数量恰好都为互不相同的质数．问：下面哪个数不能作为小龙三个月该书收量的总和？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->整数分拆->简单拆分->加法拆数（指定个数）"], "answer_analysis": ["$$2+3+5=10$$ $$2+3+7=12$$ $$3+5+7=15$$ 只有$$13$$不行． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "31", "queId": "5d063f2767bb4bfda193a9d8aaee1f84", "competition_source_list": ["2019年希望杯五年级竞赛模拟第33题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\text{NBA}$$篮球比赛，勇士、掘金、爵士、开拓者、火箭五支球队获前五名，甲、乙、丙、丁、戊对具体名次进行猜测： 甲：第$$1$$名是勇士，第$$3$$名是掘金； 乙：第$$3$$名是勇士，第$$5$$名是爵士； 丙：第$$4$$名是火箭，第$$2$$名是勇士； 丁：第$$2$$名是掘金，第$$5$$名是开拓者； 戊：第$$3$$名是开拓者，第$$5$$名是火箭， 甲、乙、丙、丁、戊每个人只猜对了一半，并且每个名次都有人说对，那么开拓者获得了第~\\uline{~~~~~~~~~~}~名． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["由题意可知每人只猜对一半，并且每个名次都有人说对，因此可以通过假设法来解决．假设甲说的第一名是勇士是对的，第三名是掘金是错的．由此推出乙说的第三名是勇士是错的，第五名是爵士是对的．丙说的第二名是勇士是错的，第四名是火箭是对的．戊说的第$$5$$名是火箭是错的，第三名是开拓者是对的．丁说的第五名开拓者是错的，第二名是掘金是对的．因此可以得出队伍的名次依次为：勇士、掘金、开拓者、火箭、爵士，开拓者获得第三名． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "343", "queId": "b6280bd2fe1e4a668d99b1cf1e691465", "competition_source_list": ["2004年第3届全国希望杯四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "箱子里有$$13$$个红球、$$10$$个黄球、$$15$$个蓝球，从中取出（ ~~）个球，才能保证三个颜色的球都至少有$$4$$个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$38$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["箱子里有红球$$13$$个、黄球$$10$$个、蓝球$$15$$个，最差的情况是：取出的球为$$15$$个蓝球、$$13$$个红球、$$3$$个黄球这$$31$$个球，此时只���再取出一个必为黄色；所以至少要从中取出$$31 + 1 = 32$$（个）球，才能保证三种颜色的球都至少有$$4$$个． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2947", "queId": "81edfa88f28a470399bcecea91f4fb6d", "competition_source_list": ["2015年全国美国数学大联盟杯小学高年级五年级竞赛初赛第29题"], "difficulty": "2", "qtype": "single_choice", "problem": "数列$$2016$$, $$225$$, $$141$$, $$66$$, $$432$$, $$99$$, $$1458$$, $$\\cdots$$中，后一项等于前一项各数位的立方和$$.$$ 则该数列的第$$100$$项为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$153$$ "}], [{"aoVal": "B", "content": "$$351$$ "}], [{"aoVal": "C", "content": "$$370$$ "}], [{"aoVal": "D", "content": "$$371$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->计算模块->数列与数表->数列规律->多重数列"], "answer_analysis": ["Calculate following the rule. $$1458-702-351-153-153-153-\\cdots $$ All of the terms since the $$10$$th term are $$153$$. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "286", "queId": "90f4bb70692b42bfb3eb046682a9568e", "competition_source_list": ["2017年第20届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙、丁比赛羽毛球，每两人要赛一场．结果甲胜了丁，并且甲、乙、丙三人胜得场数相同．丁胜了场． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$场 "}], [{"aoVal": "B", "content": "$$1$$场 "}], [{"aoVal": "C", "content": "$$2$$场 "}], [{"aoVal": "D", "content": "$$3$$场 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["一共赛了$$4\\times (4-1)\\div2=6$$场，每人各有$$3$$场比赛， 因为甲，乙，丙三人胜的场数相同，若甲，乙，丙各胜$$1$$场， 则丁胜$$6-1\\times3=3$$场，即丁全胜，不合题意（甲胜了丁）． 若甲，乙，丙各胜$$2$$场，则丁胜$$6-2\\times3=0$$场，即丁全输，符合题意． 所以，丁胜了$$0$$场． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1055", "queId": "0eee928faf1f4a9eb0b3f8f4b0dc32c6", "competition_source_list": ["2014年走美杯二年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "少先队员排队去参观科技馆，从排头数起小明是第$$10$$个；从排尾数起，小英是第$$13$$个。小明的前面就是小英，这队少先队员共有（ ）人。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$23$$ "}], [{"aoVal": "C", "content": "$$22$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->应用题模块排队问题"], "answer_analysis": ["解：$$10-1+13-1=21$$（人） 答：这队少先队员共有$$21$$人。 故答案选：D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "324", "queId": "bf2ac8ceae5649a894c2a0f5c5b6b6e3", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第15题"], "difficulty": "3", "qtype": "single_choice", "problem": "老师把某两位数的六个不同因数分别告诉了$$A\\sim F$$六个聪明诚实的同学． $$A$$和$$B$$同时说：我知道这个数是多少了． $$C$$和$$D$$同时说：听了他们的话，我也知道这个数是多少了． $$E$$：听了他们的话，我知道我的数一定比$$F$$的大． $$F$$：我拿的数的大小在$$C$$和$$D$$之间． 那么六个人拿的数之和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$141$$ "}], [{"aoVal": "B", "content": "$$152$$ "}], [{"aoVal": "C", "content": "$$171$$ "}], [{"aoVal": "D", "content": "$$175$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->答案（数字）正误问题"], "answer_analysis": ["AB看到自己手里的数，就说我知道了。这里必须理解是什么情况。绝不单单是，其中一个数很大，不能乘以$$2$$，乘以$$2$$就超出$$100$$，而是利用了一个条件$$-\\/-\\/-\\/-\\/-\\/-\\/-$$六个不同的因数 AB拿的数必须具有这样的特点，他手中的数要么乘$$2$$就会超出两位数，且这个数至少有$$6$$个因数，要么乘以一个数后，至少有$$6$$个因数，他手中的数不能有$$6$$个或者$$6$$个以上的因数，否则他不能确定这个数是自己手中的数还是要乘以一个数得到的数。（一个是最大的，一个是次大的） 这样我们根据枚举就能枚举出所有可以满足的情况 枚举如下， $2\\times 3\\times 13=78, 3\\times 13=39$ 可能 $2\\times 3\\times 17=102 $排除 $2\\times 2\\times 13=52, 2\\times 13=26$ 因为$$26\\times 3=78$$ 所以排除 $2\\times 2\\times 17=68, 2\\times 17=34$ 可能 $2\\times 2\\times 19=76, 2\\times 19=38$ 可能 $2\\times 2\\times 23=92, 2\\times 23=46$ 可能 $2\\times 2\\times 29$ 排除 $2\\times 3\\times 3\\times 3=54, 3\\times 3\\times 3=27$ 可能（这一点容易丢） $3\\times 3\\times 5=45$ 因为$$45\\times 2$$ 排除 $3\\times 3\\times 11=99, 3\\times 11=33$，因为$$33\\times 2=66$$ 排除 $2\\times 5\\times 7=70, 5\\times 7=35$ 可能 $2\\times 3\\times 3\\times 5=90, 5\\times 9=45$ 排除因为$$45$$就已经$$6$$个因数了，不能确定是$$45$$还是90 $2\\times 7\\times 7=98, 7\\times 7=49$ 可能 这样就剩下⑦个数了。同时我将他们的因数都写出来。 78 39 26 13 6 3 2 1 一共$$8$$个因数 76 38 19 4 2 1~~一共$$6$$个因数 68 34 17 4 2 1~~一共$$6$$个因数 92 46 23 4 2 1~~一共$$6$$个因数 54 27 18 9 6 3 2 1 一共$$8$$个因数 70 35 14 10 7 5 2 1 一共$$8$$个因数 98 49 14 7 2 1~~~一共$$6$$个因数 $$C$$和$$D$$能从$$AB$$知道，可以很``容易''判断出这$$7$$个数，但此时$$C$$和$$D$$也说自己知道了。很显然，他们手里不可能是上面重复的数，如7,6，$$4$$ ， 3， 2， 1。否则无法确定老师手中的数。 既然能确定，那我们就能排除掉98,因为除了98, 49，没有$$2$$个数能给$$C$$和$$D$$了。 同理，也可以排除76~~68 和92。 如果是$$78$$，$$E$$ 就是$$13$$后的第一个数$$6$$，如果是$$70$$，由于还有一个$$7$$，所以$$E$$拿的一定是$$14$$，这样才能确保$$E$$一定比$$F$$的大。如果是$$54$$，$$E$$只能是6. 所以可以知道$$C$$和$$D$$之间必须有一个数存在才行，这样的话，这个数就能判断只能是$$70$$了，$$E$$是$$14$$，那$$C$$和$$D$$就是$$10$$和$$5$$，$$F$$就是7 答案就是70+35+14+10+7+5=141 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "153", "queId": "308781fbba334fb6a22da9b943571d9f", "competition_source_list": ["2020年希望杯二年级竞赛模拟第17题"], "difficulty": "1", "qtype": "single_choice", "problem": "小皮、小舒和小贝是同班同学，他们中一个是班长，一个是学习委员，一个是体育委员． （2020年希望杯二年级竞赛模拟数学试卷） 现在知道： ①小皮的年龄比体育委员的年龄大； ②小舒比学习委员的年龄大； ③小贝和学习委员年龄不同． 那么小皮、小舒和小贝分别担任． ", "answer_option_list": [[{"aoVal": "A", "content": "班长，学习委员，体育委员 "}], [{"aoVal": "B", "content": "学习委员，体育委员，班长 "}], [{"aoVal": "C", "content": "学习委员，班长，体育委员 "}], [{"aoVal": "D", "content": "班长，体育委员，学习委员 "}], [{"aoVal": "E", "content": "体育委员，班长，学习委员 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["由②小舒比学习委员的年龄大；和③小贝和学习委员年龄不同．可知小舒和小贝都不是学习委员，所以学习委员是小皮； 已知小舒比小皮大，并且小皮的年龄比体育委员的年龄大，所以小舒不是体育委员，所以小舒是班长；那么小贝就是体育委员． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1512", "queId": "b58205faa6654a7d8fc5d2ee1140facb", "competition_source_list": ["2018年湖北武汉新希望杯小学高年级五年级竞赛训练题（三）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$42$$ has~\\uline{~~~~~~~~~~}~factors. $$42$$的因数共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["$$42=2\\times 3\\times 7$$，$$(1+1)\\times (1+1)\\times (1+1)=8$$（个）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3217", "queId": "2c11f4c1415d4322aa04db1834702f9d", "competition_source_list": ["2020年希望杯一年级竞赛模拟第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$4$$人拥抱，每两人之间都拥抱一次，一共要拥抱次．（不能重复计数） ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->加法原理->握手碰杯问题"], "answer_analysis": ["因为由题干��知，一共有$$4$$个人，每两人之间需要握一次手，我们可以将这$$4$$个人分别用$$A$$、$$B$$、$$C$$和$$D$$来指代，则所有的握手情况为：$$AB$$，$$AC$$，$$AD$$，$$BC$$，$$BD$$和$$CD$$，所以一共要握$$6$$次，故本题答案为$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2362", "queId": "1c26f7dfd8834cc0a1b96f332562dc75", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛A卷第8题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "分数单位是$$\\frac{1}{10}$$的最简分数共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "无数 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数基础->分数的性质"], "answer_analysis": ["依据分数单位的意义，即把单位``$$1$$''平均分成若干份取一份的数，叫做分数单位，以及最简分数的概念，即可进行解答． 因为$$\\frac{9}{10}$$、 $$\\frac{7}{10}$$、$$\\frac{3}{10}$$、$$\\frac{11}{10}\\cdots $$的分数单位都是$$\\frac{1}{10}$$，也就是说只要分母是$$10$$的最简分数，它的分数单位都是$$\\frac{1}{10}$$，这样的分数有无数个． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1098", "queId": "ec42755dc8594603a34577b570bf2f57", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级三年级竞赛初赛第38题"], "difficulty": "1", "qtype": "single_choice", "problem": "一次考试中，总共有$$40$$道题，学生答对一道题获得$$4$$分，答错一道题或者未作答扣$$1$$分．如果小明在考试中获得$$95$$分，那么他答对了多少道题？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$26$$ "}], [{"aoVal": "C", "content": "$$27$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$40-(40\\times 4-95)\\div (4+1)=27$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1284", "queId": "2c3b3fd9837c4ecb831dab376df0df9f", "competition_source_list": ["小学中年级三年级上学期其它", "2017年全国华杯赛小学中年级竞赛初赛模拟第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两人在春节一共得$$200$$元压岁钱，甲给乙$$40$$元，还比乙多$$10$$元，那么，原来甲得了~\\uline{~~~~~~~~~~}~元压岁钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$145$$ "}], [{"aoVal": "B", "content": "$$140$$ "}], [{"aoVal": "C", "content": "$$125$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->知识点->应用题模块->和差倍问题->和差问题->二量和差问题->明和暗差"], "answer_analysis": ["因为甲给乙$$40$$元，还比乙多$$10$$元，所以甲比乙多$$40\\times 2+10=90$$（元）， 所以甲：$$(200+90)\\div 2=145$$（元）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3138", "queId": "0186b270da514430a0e1b7ef2577bb0f", "competition_source_list": ["2016年全国小学生数学学习能力测评五年级竞赛初赛第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$1$$角硬币分正面与反面．拿$$2$$个$$1$$角硬币一起投掷一次，得到一个正面和一个反面的可能性为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{8}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->典型问题->抛硬币"], "answer_analysis": ["全部情况有：$$2\\times 2=4$$（种）， 最终得到一个正面一个反面可有以下$$2$$种情况： 正反、反正， 故所求概率为$$\\frac{2}{4}$$． 故答案为：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2228", "queId": "7a6179be6b67404fb527e196db4ce14f", "competition_source_list": ["2011年第7届全国新希望杯六年级竞赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列时刻中，时针和分针所成的角最接近$$30{}^{}\\circ $$是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3:27$$ "}], [{"aoVal": "B", "content": "$$4:17$$ "}], [{"aoVal": "C", "content": "$$5:14$$ "}], [{"aoVal": "D", "content": "$$6:22$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->��程模块->时钟问题->时钟上的追及问题->已知时间求角度"], "answer_analysis": ["略． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "988", "queId": "098745fcef13492aa54b98a2f6b2d845", "competition_source_list": ["2017年全国小学生数学学习能力测评六年级竞赛复赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "两件衣服都按$$80$$元出售，其中一件赚了$$\\frac{1}{4}$$，另一件亏了$$\\frac{1}{4}$$，那么两件衣服合算在一起，结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "赚了 "}], [{"aoVal": "B", "content": "亏了 "}], [{"aoVal": "C", "content": "不赚不亏 "}], [{"aoVal": "D", "content": "无法比较 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["可把衣服的成本价看作是单位``$$1$$''，根据分数除法的意义，列式求出衣服的成本价，再分别求出赚的钱数和亏的钱数，再进行比较．据此解答． 本题的关键是求出每件衣服的成本价，再求出每件衣服赚的钱数和亏的钱数，再进行比较． $$80\\div \\left( 1+\\frac{1}{4} \\right)$$ $$=80\\div \\frac{5}{4}$$ $$=64$$（元）， $$80\\div \\left( 1-\\frac{1}{4} \\right)$$ $$=80\\div \\frac{3}{4}$$ $$=106\\frac{2}{3}$$（元）， $$106\\frac{2}{3}-80-\\left( 80-64 \\right)$$ $$=26\\frac{2}{3}-16$$ $$=10\\frac{2}{3}$$（元）， 所以两件衣服合算在一起亏了$$10\\frac{2}{3}$$元． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3241", "queId": "289a8f82b344418ba4c68d465419b956", "competition_source_list": ["2011年四年级竞赛创新杯", "2011年第9届创新杯四年级竞赛初赛第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "2011的各位数字的和为4，具有这种性质的四位数共有( )个 ", "answer_option_list": [[{"aoVal": "A", "content": "10 "}], [{"aoVal": "B", "content": "15 "}], [{"aoVal": "C", "content": "20 "}], [{"aoVal": "D", "content": "21 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->加乘原理综合"], "answer_analysis": ["枚举法:$$4=4+0+0+0=3+1+0+0=2+1+1+0=2+2+0+0=1+1+1+1$$，对应4，0，0，0的四位数有1个，对应3，1，0，0的四位数有6个，对应2，1，1，0的四位数有9个，对应2，2，0，0的四位数有3个，对应1，1，1，1的四位数有1个，故$$1+6+9+3+1=20$$(个)或者也可用隔板法 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "632", "queId": "198bcf3792e448048aae2e781a78215a", "competition_source_list": ["2017年希望杯六年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "若质数$$a,b$$满足$$5a+b=2027$$，则$$a+b=$$（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$115$$ "}], [{"aoVal": "B", "content": "$$2015$$ "}], [{"aoVal": "C", "content": "$$2017$$ "}], [{"aoVal": "D", "content": "$$2019$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数->质数与合数判定"], "answer_analysis": ["解：依题意可知： 两数和为奇数，那么一定有一个偶数。偶质数是$$2$$。 当$$b=2$$时，$$5a+2=2027,a=405$$不符合题意。 当$$a=2$$时，$$10+b=2027,b=2017$$符合题意， $$a+b=2+2017=2019$$。 故答案为：D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2650", "queId": "43eaf3d0cdf14429b1d3addd89daadcf", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（一）第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "下列说法正确的是（~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "女生人数与全班人数的比是$$4:9$$，男生人数与女生人数的比是$$4:5$$ "}], [{"aoVal": "B", "content": "比的前项和后项同时乘一个相同的数，比值不变 "}], [{"aoVal": "C", "content": "最简单的整数比，就是比的前项和后项都是质数的比 "}], [{"aoVal": "D", "content": "如果$$\\text{A:B=C}$$，那么$$\\text{A}$$是比的前项，$$\\text{B}$$是比的后项，$$\\text{C}$$是比的比值 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比和比例->比->比的认识"], "answer_analysis": ["$$A$$项，男生人数与女生人数的比是$$5:4$$； $$B$$项，同时乘的数不能是$$0$$； $$C$$项，最简比的前后项是互质数，不一定都是质数． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2677", "queId": "f176a6c9f22a4903b8e75253cab2b4d8", "competition_source_list": ["2003年第1届创新杯五年级竞赛复赛第3题", "2003年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "小马虎在计算一道除法算式时，把除数$$4.13$$错写成$$41.3$$，结果所得到的商比正确的商减少$$2.52$$，这道除法算式的被除数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.11564$$ "}], [{"aoVal": "B", "content": "$$1.1564$$ "}], [{"aoVal": "C", "content": "$$11.564$$ "}], [{"aoVal": "D", "content": "$$115.64$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["除数$$4.13$$变成$$41.3$$扩大到原来的$$10$$倍，被除数没有变，那么商缩小到原来的$$10$$倍，又现在的商比正确的少$$2.52$$，那么现在的商为:$$2.52\\div \\left( 10-1 \\right)=0.28$$，所以被除数为$$0.28\\times 41.3=11.564$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2170", "queId": "2fa5ac6819e84479b1e9ae8e01f4ae2e", "competition_source_list": ["2018年湖北武汉锦奥杯小学高年级五年级竞赛初赛湖北赛区第16题10分"], "difficulty": "3", "qtype": "single_choice", "problem": "一个渔民划着小船顺流而下，在途中甲地一个用来装鱼的带盖空塑料桶（不会沉入水中）掉落水中后随水漂流而下，渔民毫不知情继续划着小船往前行进了若干分钟，才发现丢了桶，于是立即返回寻找，最终在距离甲地$$4.05$$千米的地方找回桶．已知渔民从丢掉桶到发现桶丢了这段时间划着小船往前行进了$$6525$$米，水流的速度为$$45$$米/分，求小船在静水中的速度． ", "answer_option_list": [[{"aoVal": "A", "content": "$$80$$米/分 "}], [{"aoVal": "B", "content": "$$100$$米/分 "}], [{"aoVal": "C", "content": "$$150$$米/分 "}], [{"aoVal": "D", "content": "$$180$$米/分 "}]], "knowledge_point_routes": ["知识标签->拓展思维->行程模块->流水行船问题->基本流水行船问题->水中坠物"], "answer_analysis": ["从丢桶到找到桶，桶一共顺流而下的时间：$$4050\\div 45=90$$（分钟）， 渔民从丢桶到发现桶丢了的时间等于渔民从返回找桶到找到桶的时间$$90\\div 2=45$$（分钟） 小船顺流而下的速度：$$6525\\div 45=145$$（米/分） 小船在静水中的速度：$$145-45=100$$（米/分） "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1122", "queId": "1d10b06c3ad247b8bc57de360fcbe284", "competition_source_list": ["2014年全国学而思杯二年级竞赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "为了唤起人们对读书的热爱和对儿童图书的关注，$$4$$月有两个与读书有关的特别节日：$$2014$$年$$4$$月$$2$$日（星期三）是国际儿童图书日，那么，$$2014$$年$$4$$月$$23$$日世界读书日是在星期~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}], [{"aoVal": "E", "content": "五 "}]], "knowledge_point_routes": ["海外竞赛体系->知识点->应用题模块->分百应用题->认识单位1", "拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["$$2$$、$$9$$、$$16$$、$$23 $$星期三；或者$$(23-2+1)\\div7=3$$（周）$$\\cdots 1$$（天），周三那天． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2171", "queId": "794f43f877ad4ccba237264b4b8b1c0b", "competition_source_list": ["2005年五年级竞赛创新杯", "2005年第3届创新杯五年级竞赛初赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "小明步行上学，每分钟行$$70$$米，离家$$12$$分钟后，爸爸发现小明的文具盒忘在家中，爸爸带着文具盒，立即骑自行车以每分钟$$280$$米的速度去追小明，那么爸爸出发( )分钟后追上小明． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->两人相遇与追及问题->相遇问题->同时出发相向而行"], "answer_analysis": ["爸爸从家出发去追小明时，小明已经离家$$70\\times 12=840$$(米)，那么爸爸追上小明需要的时间为$$840\\div \\left( 280-70 \\right)=4$$(分钟)，选B． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1866", "queId": "a081350edf94445283506844a64acb08", "competition_source_list": ["2017年IMAS小学高年级竞赛（第二轮）第1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$80$$个三角排成一列﹐然后依照下面的规律涂上黑色或白色﹐请问涂上黑色的三角形总共比涂上白色的三角形多几个？ ▲▲▲$$\\triangle \\triangle $$▲▲▲$$\\triangle \\triangle $$▲▲▲$$\\triangle \\triangle \\cdots \\cdots $$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}], [{"aoVal": "E", "content": "$$32$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["由图可知，从第一个三角形开始，以每五个三角形为一个周期，每个周期内有$$3$$个黑色三角形与$$2$$个白色三角形，即黑色三角形比白色三角形多$$1$$个．可知$$80$$个三角形共有$$16$$个周期， 所以黑色三角形比白色三角形总共多$$16$$个． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "885", "queId": "9b8e079dce3b43a4b22e865909ae0f5a", "competition_source_list": ["小学中年级三年级其它第九讲最小公倍数与最大公约数", "2016年创新杯六年级竞赛训练题（二）第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙三个非零自然数满足：甲和乙的最大公因数恰有$$1$$个因数，乙和丙的最大公因数恰有$$2$$个因数，丙和甲的最大公因数恰有$$3$$个因数．那么，甲、乙、丙三数之和的最小值是~\\uline{~~~~~~~~~~}~（中上） ", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$22$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$\\left( 甲，乙\\right)=1$$，则甲、乙互质；$$\\left(乙，丙 \\right)=$$质数；$$\\left(甲，丙 \\right)=$$质数的平方，最小的质数的平方为$$4$$，$$\\therefore $$丙最小：$$3\\times ~4=12$$，甲：$$4$$，乙$$3$$．$$12+4+3=19$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "453", "queId": "8f2b58cadbfe47b88d90bf6b0ad3139b", "competition_source_list": ["2004年五年级竞赛创新杯", "2011年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$A$$，$$B$$，$$C$$三个盒子，分别装有红、黄、蓝三种颜色的小球各一种，将它们分给甲、乙、丙三个人，已知甲没有得到$$A$$盒；乙没有得到$$B$$盒，也没有得到黄球；$$A$$盒中没有装红球，$$B$$盒中装着蓝球，则丙得到的盒子编号与小球的颜色分别是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$A$$，黄 "}], [{"aoVal": "B", "content": "$$B$$，蓝 "}], [{"aoVal": "C", "content": "$$C$$，红 "}], [{"aoVal": "D", "content": "$$B$$，黄 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->表格法"], "answer_analysis": ["根据题意，因为乙没有得到$$B$$盒，而$$B$$盒中装有蓝球，所以乙没有得到蓝球；又乙没有得到黄球，所以乙得到的是红球，又$$A$$盒中装的不是红球，乙没有得到$$B$$盒，所以乙得到的$$C$$盒红球，又甲没有得到$$A$$盒，所以甲得到的是$$B$$盒蓝球，所以丙得到的为$$A$$盒黄球，选$$A$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "467", "queId": "d2f6a81bc34c4a009e91e9cf18c944b3", "competition_source_list": ["2019年第24届YMO三年级竞赛决赛第8题3分", "2017年第22届全国华杯赛小学中年级竞赛初赛第2题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$至$$10$$这$$10$$个整数中．至少取（~ ）个数，才能保证其中有两个数的和等于$$10$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["将和为$$10$$的两个数作为一组，有$$\\left( 1,9 \\right)$$，$$\\left( 2,8 \\right)$$，$$\\left( 3,7 \\right)$$，$$\\left( 4,6 \\right)$$、$$\\left( 5,10 \\right)$$，考虑极端情况，若在前$$4$$组中各取一个数字，再取$$5$$和$$10$$，不会有两数和为$$10$$，若再在剩下的数中取任何一个，都会使两数凑成$$10$$，所以至少需要取$$7$$个数． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3087", "queId": "d8b2b7d50467402a88926f649d25900b", "competition_source_list": ["2015年IMAS小学高年级竞赛第一轮检测试题第1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "（2015 IMAS，Question\\#1） 请问算式$$32\\times 37\\times 75$$的值为多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$88075$$ "}], [{"aoVal": "B", "content": "$$88800$$ "}], [{"aoVal": "C", "content": "$$88200$$ "}], [{"aoVal": "D", "content": "$$74000$$ "}], [{"aoVal": "E", "content": "$$80800$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想", "Overseas Competition->知识点->计算模块->整数->整数乘除->整数乘除法巧算"], "answer_analysis": ["$$32\\times 37\\times 75$$ $$=4\\times 8\\times 37\\times 3\\times 25$$ $$=(4\\times 25)\\times (37\\times 3)\\times 8$$ $$=100\\times 111\\times 8$$ $$=88800$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1973", "queId": "b833cba3b77e44a7a7e467580fae3354", "competition_source_list": ["2009年第7届创新杯六年级竞赛初赛第2题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "美是一种感觉，本应没有什么客观的标准，但在自然界里，物体形状的比例却提供了在匀称与协调上的一种美感的参考，在数学上，这个比例称为黄金分割．在人体下躯干（由脚底至肚脐的长度）与身高的比例上，肚脐是理想的黄金分割点，也就是说，若此比值越接近$$0.618$$，就越给别人一种美的感觉．如果某女士身高为$$1.60\\text{m}$$，下躯干与身高的比为$$0.60$$，为了追求美，她想利用高跟鞋达到这一效果，那么她选的高跟鞋的高度约为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2.5\\text{cm}$$ "}], [{"aoVal": "B", "content": "$$5.1\\text{cm}$$ "}], [{"aoVal": "C", "content": "$$7.5\\text{cm}$$ "}], [{"aoVal": "D", "content": "$$8.2\\text{cm}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题->方程法解比例问题"], "answer_analysis": ["根据已知条件得下半身长是$$160\\times 0.6=96\\text{cm}$$， 设选的高跟鞋的高度是$$x\\text{cm}$$，则根据黄金分割的定义得： $$\\frac{96+x}{160+x}=0.618$$， 解得：$$x\\approx 7.5\\text{cm}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1620", "queId": "c834fff57a944940809feb1204651052", "competition_source_list": ["2018年IMAS小学高年级竞赛（第二轮）第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "四名同学相约去爬山，往返共花费$$50$$元车资．在山上，他们每人各购买了一瓶$$5$$元的饮料．请问平均每人各总共花费多少元？（~ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$12.5$$ "}], [{"aoVal": "B", "content": "$$13.75$$ "}], [{"aoVal": "C", "content": "$$17.5$$ "}], [{"aoVal": "D", "content": "$$30$$ "}], [{"aoVal": "E", "content": "$$55$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["四个人总共花费了$$50+5\\times 4=70$$元， 故平均每人花费$$70\\div 4=17.5$$元． 故选$$\\text{C}$$． ", "每人坐车的平均花费为$$50\\div 4=12.5$$元，
\n平均每人各总共花费$$12.5+5=17.5$$元．
\n故选$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1937", "queId": "ee576482e05b433dbe891875bad39aa7", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第5题", "2018年四川成都锦江区四川师范大学附属第一实验中学小升初（五）第5~0题3分", "2019年四川成都锦江区四川师范大学附属第一实验中学小升初（八）第6题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "动物园的饲养员把一堆桃子分给若干只猴子，如果每只猴子分$$6$$个，剩$$57$$个桃子；如果每只猴子分$$9$$个，就有$$5$$只猴子一个也分不到，还有一只猴子只分到$$3$$个．那么，有个桃子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$216$$ "}], [{"aoVal": "B", "content": "$$324$$ "}], [{"aoVal": "C", "content": "$$273$$ "}], [{"aoVal": "D", "content": "$$301$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["每只猴子多分了$$3$$个，分了$$5\\times 9+(9-3)+57=108$$ （个），那么共$$108\\div 3=36$$（只）猴子．共$$36\\times6+57=273$$（个）桃子． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1883", "queId": "a9d41e5beab94520beebb1f0c6312b26", "competition_source_list": ["2006年六年级竞赛创新杯", "2006年五年级竞赛创新杯", "2006年第4届创新杯四年级竞赛复赛第6题", "2006年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "在一场篮球比赛中，其中一个队总是保持场上有$$5$$名队员，在场外还有$$3$$名候补队员（场上的队员可以随时替换），在整场比赛中这$$8$$个队员每人上场时间相同，如果这场比��持续的时间是$$48$$分钟，那么每个队员上场的时间是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$分钟 "}], [{"aoVal": "B", "content": "$$24$$分钟 "}], [{"aoVal": "C", "content": "$$30$$分钟 "}], [{"aoVal": "D", "content": "$$36$$分钟 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->公式类"], "answer_analysis": ["$$48\\times 5\\div 8=30$$（分钟） "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2140", "queId": "0b1da5e525a64ed7a825c8e7fd38f71a", "competition_source_list": ["2017年河南郑州联合杯竞赛第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "小明是六年级（$$1$$）班的，六年级（$$1$$）班的教室在$$6$$楼，每两层楼之间的楼梯都有$$16$$级，则小明每次去学校从地面到班里要走（~ ）级楼梯． ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$级 "}], [{"aoVal": "B", "content": "$$88$$级 "}], [{"aoVal": "C", "content": "$$80$$级 "}], [{"aoVal": "D", "content": "$$90$$级 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块"], "answer_analysis": ["$$16\\times \\left( 6-1 \\right)=80$$级． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "259", "queId": "7f2cf61995b74d2191197593dab47b2d", "competition_source_list": ["2013年全国学而思杯三年级竞赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "小何、小琳和小俊参加了一次数独比赛，赛后，他们对比赛结果进行了预测． 小何说：``我是第$$1$$，小琳是第$$2$$，小俊是第$$3$$．'' 小琳说：``我是第$$1$$，小何是第$$2$$，小俊是第$$3$$．'' 小俊说：``我是第$$1$$，小琳是第$$2$$，小何是第$$3$$．'' 如果他们$$3$$人中，有$$1$$人$$3$$句话都预测正确，其余两人都只预测正确了$$1$$句话，那么，$$3$$人的名次按小何、小琳、小俊的顺序组成的$$3$$位数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$123$$ "}], [{"aoVal": "B", "content": "$$132$$ "}], [{"aoVal": "C", "content": "$$213$$ "}], [{"aoVal": "D", "content": "$$231$$ "}], [{"aoVal": "E", "content": "$$312$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->组合模块->逻辑推理", "拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾"], "answer_analysis": ["方法一：由于$$3$$人中有且仅有$$1$$人全预测正确，故可枚举全正确的人的情况： 若小何全正确，则小琳只有第$$3$$句正确，小俊只有第$$2$$句正确，符合要求； 若小琳全正确，则小俊全错，不符合要求； 若小俊全正确，则小琳全错，不符合要求； 方法二：注意到小琳和小俊的预测全都不同，故知全正确的人不可能在这两人之中（否则另一个人就全错，不符合要求） 综上，小何全正确，答案为$$123$$. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3159", "queId": "17f92f772cd04d28a3fe537c8c5cdc06", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第7题3分", "五年级其它"], "difficulty": "1", "qtype": "single_choice", "problem": "一个书架上有故事书、科技书、画册、字典四种书籍共$$35$$本，每种书籍的本数互不相同．其中故事书和科技书共有$$17$$本，科技书和画册共有$$16$$本．有一种书籍有$$9$$本，那么这种书籍是． ", "answer_option_list": [[{"aoVal": "A", "content": "故事书 "}], [{"aoVal": "B", "content": "科技书 "}], [{"aoVal": "C", "content": "画册 "}], [{"aoVal": "D", "content": "字典 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["故事书和科技书共有$$17$$本， 则画册和字典共有$$35-17=18$$本， 则画册与字典都不可能是$$9$$本，否则画册与字典都是$$9$$本， 则只能是故事书和科技书中的一类为$$9$$本，故故事书为$$9$$本， 则科技书为$$17-9=8$$本， 则画册为$$16-8=8$$本，矛盾， 故科技书为$$9$$本， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3057", "queId": "afa675b6c5d64a04b2e17c25bcb8b525", "competition_source_list": ["2021年新希望杯六年级竞赛初赛第11题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "小糊涂遇到一个问题：比较$$\\frac{99}{100}$$，$$ \\frac{100}{101}$$，$$\\frac{199}{201}$$的大小．他感到很迷糊，请你帮他找到正确的答案． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{199}{201}$$ "}], [{"aoVal": "B", "content": "$$\\frac{199}{201}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{99}{100}$$ "}], [{"aoVal": "C", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{100}{101}$$ "}], [{"aoVal": "D", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater\\frac{99}{100}$$ "}], [{"aoVal": "E", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{99}{100}\\textgreater{} \\frac{199}{201}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数感认知->分数数字加工"], "answer_analysis": ["$$\\frac{99}{100}=1- \\frac{1}{100}=1- \\frac{2}{200}$$， $$\\frac{100}{101}=1- \\frac{1}{101}=1- \\frac{2}{202}$$， $$\\frac{199}{201}=1- \\frac{2}{201}$$， 因为$$\\frac{2}{202}\\textless{} \\frac{2}{201}\\textless{} \\frac{2}{200}$$， 所以$$1- \\frac{2}{202}\\textgreater1- \\frac{2}{201}\\textgreater1- \\frac{2}{200}$$， 即$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{99}{100}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3024", "queId": "a1972f22318d49b49b9ec2ecb81ddf06", "competition_source_list": ["2016年全国华杯赛小学高年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "算式$$\\underbrace{999\\cdots 9}_{2016个}\\times \\underbrace{999\\cdots 9}_{2016个}$$的结果中含有个数字$$0$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2017$$ "}], [{"aoVal": "B", "content": "$$2016$$ "}], [{"aoVal": "C", "content": "$$2015$$ "}], [{"aoVal": "D", "content": "$$2014$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->公式类运算->完全平方公式"], "answer_analysis": ["$${{\\left( {{10}^{2016}}-1 \\right)}^{2}}=\\left( {{10}^{2016}}-2 \\right)\\times {{10}^{2016}}+1=\\underbrace{999\\cdots 99}_{2015}8\\underbrace{000\\cdots 00}_{2015}1$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "473", "queId": "d323e163e77e444c844488ed70c6f84e", "competition_source_list": ["2016年第20届四川成都华杯赛小学中年级竞赛B卷第1~6题60分", "2015年第20届全国华杯赛小学中年级四年级竞赛初赛A卷"], "difficulty": "1", "qtype": "single_choice", "problem": "森林里举行比赛，要派出狮子、老虎、豹子和大象中的两个动物去参加．如果派狮子去，那么也要派老虎去；如果不派豹子去，那么也不能派老虎去；要是豹子参加的话，大象可不愿意去．那么，最后能去参加比赛的是． ", "answer_option_list": [[{"aoVal": "A", "content": "狮子、老虎 "}], [{"aoVal": "B", "content": "老虎、豹子 "}], [{"aoVal": "C", "content": "狮子、豹子 "}], [{"aoVal": "D", "content": "老虎、大象 "}], [{"aoVal": "E", "content": "狮子、大象 "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->逻辑推理->条件型逻辑推理->连线法"], "answer_analysis": ["在逻辑推理中，原命题成立，则逆否命题也成立. 从题意出发： （1）狮子去则老虎去，逆否命题；老虎不去则狮子也不去 （2）不派豹子则不派老虎，逆否命题：派老虎则要派豹子 （3）派豹子则大象不愿意去，逆否命题；大象去则不能派豹子 从（2）出发可以看出答案为$$B$$ 题目要求有两个动物去，可以使用假设法，若狮子去，则老虎去，老虎去则豹子也去，三个动物去，矛盾，所以狮子不去，若豹子不去则老虎不去，那么只有大象去，矛盾，所以豹子去，豹子去则大象不去，由两骄气去得到结论，老虎要去，所以答案是$$B$$，豹子和老虎去. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "293", "queId": "ff8080814518d5240145201ad07c0a74", "competition_source_list": ["2014年全国迎春杯三年级竞赛复赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$6$$、$$7$$、$$8$$、$$9$$填入右边算式的方格中：``$$\\square\\times\\square＋\\square \\square$$''那么这个算式的结果最大为（~~~~~~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$152$$ "}], [{"aoVal": "B", "content": "$$145$$ "}], [{"aoVal": "C", "content": "$$140$$ "}], [{"aoVal": "D", "content": "$$154$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->横式数字谜->横式数字谜的最值"], "answer_analysis": ["结果最大应该是$$7\\times 8+96=152$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1847", "queId": "fbb468d8160249d894a75c5f0ef8dec3", "competition_source_list": ["2012年IMAS小学高年级竞赛第二轮测试真题第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "某购物网站每间隔$$500$$个小时就自动向它的客户的电子邮箱发出一份购物宣传���，小明在上个星期的早上$$9$$点收到一份该网站的购物宣传单，请问小明下次收到该网站的购物宣传单是星期几？ ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}], [{"aoVal": "E", "content": "五 "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$500\\div 24=20$$（天）$$\\cdots$$（$$20$$小时） $$20$$（天）$$\\div 7=2$$（个）$$\\cdots$$$$6$$（天） 星期二$$+2$$个星期$$6$$天$$=$$星期一 $$9$$点$$+20$$个小时$$=29$$个小时$$=$$一天$$5$$个小时． ∴星期二凌晨$$5$$点． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1552", "queId": "4dd507c54a0449d1a5e2802b782d8951", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一串数：$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$34$$，$$55$$，$$89$$$$\\cdots \\cdots $$，其中第一个数是$$2$$，第二个数是$$3$$，从第三个数起，每个数恰好是前两个数的和．那么在这串数中，第$$2019$$个数被$$3$$除后所得余数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["将这一串数写成除以$$3$$的余数，则为：$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0$$，$$1$$，$$1$$，$$2$$，$$0$$，$$2\\cdots \\cdots $$ 所以重复的为：``$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0$$，$$1$$，$$1$$''， 故周期为$$8$$．$$2019\\div 8=252$$（组）$$\\cdots \\cdots 3$$（个），则答案为$$2$$，故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "530", "queId": "fdbf88c6401d45f5b6dac5c626b07456", "competition_source_list": ["2014年迎春杯三年级竞赛初赛", "2014年迎春杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在下列算式的空格中填入互不相同的数字 $$\\square \\times \\left( \\square +\\square \\square \\right)\\times \\left( \\square +\\square +\\square +\\square \\square \\right)\\text{=}2014$$， 其中五个一位数的和最大是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$35$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜"], "answer_analysis": ["$$2014=2\\times 19\\times 53$$，五个一位数之和最大，则两位数应最小 $$2\\times \\left( a+\\overline{1b} \\right)\\times \\left( c+d+e+\\overline{3f} \\right)=2014$$可得$$a+b=9=9+0$$，$$c+d+e+f=23=8+6+5+4$$，则五个一位数的和的最大值为$$2+a+c+d+e=2+9+8+6+5=30$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1145", "queId": "388f4d795d9649eab39fbd1742d5c342", "competition_source_list": ["2017年新希望杯小学高年级六年级竞赛训练题（三）第1题", "2018年湖北武汉新希望杯小学高年级六年级竞赛训练题（三）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "仓库运来含水量为$$96 \\%$$的水果$$1000$$千克，一周后再测发现含水量降低为$$92 \\%$$，现在这批水果的重量是千克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$400$$ "}], [{"aoVal": "B", "content": "$$500$$ "}], [{"aoVal": "C", "content": "$$600$$ "}], [{"aoVal": "D", "content": "$$700$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->浓度问题->抓不变量"], "answer_analysis": ["$$1000\\times (1-96 \\%)\\div (1-92 \\%)=500$$（千克）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "890", "queId": "78adc126a03f43b9883a0366b64e9c50", "competition_source_list": ["2018年IMAS小学高年级竞赛（第一轮）第15题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "将一个正整数的各位数码以相反的顺序排列后，若所得的数与原来的数相同﹐则称这个数为回文数（例如$$909$$与$$1221$$都是回文数）．请问能被$$9$$整除的三位回文数有多少个？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$20$$ "}], [{"aoVal": "E", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["设能被$$9$$整除的回文数为 $$\\overline{aba}$$ ﹐其中$$ 1\\leqslant a\\leqslant 9$$ ，$$ 0\\leqslant b\\leqslant 9$$能被$$9$$整除的数的各位数码之和也能被$$9$$整除﹐反之亦然．因此$$a+b+a=2a+b$$能被$$9$$整除． 当$$2a+b=27$$时，只有$$999$$一个数， 当$$2a+b=18$$时，有$$585$$、$$666$$、$$747$$、$$828$$、$$909$$等五个数， 当$$2a+b=9$$时，有$$171$$、$$252$$、$$333$$、$$414$$等四个数， 故符合要求的回文数共有$$1+5+4=10$$个，故选（$$\\text{A}$$）． ", "设能被$$9$$整除的回文数为 $$\\overline{aba}$$ ﹐其中$$ 1\\leqslant a\\leqslant 9$$ ，$$ 0\\leqslant b\\leqslant 9$$．能被$$9$$整除的数的各位数码之和也能被$$9$$整除﹐反之亦然．
\n当$$a=1$$时，只有$$171$$一个数，
\n当$$a=2$$时，只有$$252$$一个数，
\n当$$a=3$$时，只有$$333$$一个数，
\n当$$a=4$$时，只有$$414$$一个数，
\n当$$a=5$$时﹐只有$$585$$一个数，
\n当$$a=6$$时，只有$$666$$一个数，
\n当$$a=7$$时，只有$$747$$一个数，
\n当$$a=8$$时，只有$$828$$一个数，
\n当$$a=9$$时，只有$$909$$、$$999$$二个数，
\n故符合要求的回文数共有$$10$$个．故选（$$\\text{A}$$）．
"], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3191", "queId": "8afb4948117543279d54c1cda75a0a0c", "competition_source_list": ["2016年创新杯五年级竞赛训练题（三）第7题", "小学高年级六年级其它2015年数学思维能力等级测试复试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$1$ $12$$这$$12$$个数中，最少选出（~ ）个数，使得在选出的数中，一定有一个数是另一个数的$$2$$倍． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["分组，（$$1$$、$$2$$、$$4$$、$$8$$），（$$3$$、$$6$$、$$12$$），（$$5$$、$$10$$），$$7$$、$$9$$、$$11$$，最少$$9$$个． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3419", "queId": "d2e26c96794242808209db2e4e3b2f56", "competition_source_list": ["2017年河南郑州小升初豫才杯第13题3分", "2017年安徽合肥庐江县小升初第18题1分", "2017年河南郑州豫才杯竞赛第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "小红和小军玩``石头、剪刀、布''游戏，两人获胜的可能性都是（~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$50 \\%$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->概率基本概念"], "answer_analysis": ["玩``石头、剪刀、布''游戏，只能出现胜、负、平三种情况，所以两人获胜的可能性都是$$1\\div 3=\\frac{1}{3}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2055", "queId": "ef845155f3ff4170a8e2cbdd33e20584", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第13题", "2018年四川成都锦江区四川师范大学附属第一实验中学小升初（四）第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "同学们一起去划船，但公园船不够多，如果每船坐$$4$$人，会多出$$10$$人；如果每船坐$$5$$人，还会多出$$1$$人．共有只船． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$46$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->盈亏问题->盈亏基本类型->盈亏基本类型盈亏问题"], "answer_analysis": ["盈盈类问题：共有$$(10-1)\\div (5-4)=9$$（只）船． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "668", "queId": "991c5d7186c5412aba09ae994bb90559", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（三）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "房间里有一盏灯亮着，突然停电了，第$$1$$个人进来后按了一下开关，第$$2$$个人进来后按了$$2$$下开关，第$$3$$个人进来后按了$$3$$下开关$$\\cdots \\cdots $$，第$$10$$个人进来后按了$$10$$下开关．那么，来电后，这盏灯是着的． ", "answer_option_list": [[{"aoVal": "A", "content": "开 "}], [{"aoVal": "B", "content": "关 "}], [{"aoVal": "C", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->七大能力->逻辑分析"], "answer_analysis": ["第$$10$$个人进来后，一共拉了开关$$\\left( 1+2+3+\\ldots \\ldots +10 \\right)$$下，这$$10$$个加数中有$$5$$个偶数，任意数量的偶数之和一定是偶数；有$$5$$个奇数，奇数个奇数的和一定是奇数．因为偶数$$+$$奇数$$=$$奇数，而这盏灯开始是亮着的，则拉奇数下开关，灯一定是关着的． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3466", "queId": "eb3870c8868446a3baa88b905da327f3", "competition_source_list": ["2017年新希望杯六年级竞赛训练题（三）第6题", "2018年湖北武汉新希望杯六年级竞赛训练题（三）第6题"], "difficulty": "3", "qtype": "single_choice", "problem": "从$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$16$$中选出$$m$$个数，若选出的数中，每一个数都不是另一个数的$$2$$倍，则$$m$$的最大值为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["先选择$$1$$、$$3$$、$$5$$、$$7$$、$$9$$、$$11$$、$$13$$、$$15$$，再选择$$4$$、$$12$$、$$16$$，共$$11$$个数． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "856", "queId": "8573fbe05d5b4276acbb563726df9165", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛B卷第3题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "有一个六位数$$\\overline{32A34B}$$能被$$88$$整除，这个数除以$$8$$所得的商是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40392$$ "}], [{"aoVal": "B", "content": "$$14382$$ "}], [{"aoVal": "C", "content": "$$40293$$ "}], [{"aoVal": "D", "content": "$$28438$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["被$$88$$整除即同时被$$11$$和$$8$$整除，若一个整数的未尾三位数能被$$8$$整除，则这个数能被$$8$$整除，$$34B$$能被$$8$$整除，得最后一位$$B$$为$$4$$，若一个整数的奇位数字之和与偶位数字之和的差能被$$11$$整除，则这个数能被$$11$$整除. 奇数位数字之和为$$2+3+4=9$$，偶数位数字之和为$$3+A+4=7+A$$，得第三位$$A$$为$$2$$，所以这个数是$$322344$$，这个数除以$$8$$所得的商是$$322344\\div8=40293$$，因此，答案选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3205", "queId": "c2b0d8995a714fcea760b3e0edaca496", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "两个自然数的和是$$25$$，那么这两个自然数的乘积不可能是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$114$$ "}], [{"aoVal": "C", "content": "$$132$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["$$25=1+24=2+23=3+22=4+21=5+20=6+19$$ $$=7+18=8+17=9+16=10+15=11+14=12+13$$． 所以：$$1\\times 24=24$$，$$2\\times 23=46$$，$$3\\times 22=66$$，$$4\\times 21=84$$，$$5\\times 20=100$$，$$6\\times 19=114$$，$$7\\times 18=126$$，$$8\\times 17=136$$，$$9\\times 16=144$$，$$10\\times 15=150$$，$$11\\times 14=154$$，$$12\\times 13=156$$， 故$$132$$是不可能的． 故选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2121", "queId": "fe6da58234774be6a813998a309fc7db", "competition_source_list": ["2003年第1届创新杯六年级竞赛复赛第22题"], "difficulty": "2", "qtype": "single_choice", "problem": "经测算，地球上的资源可供$$100$$亿人生活$$100$$年，或可供$$80$$亿人生活$$300$$年．假设地球上新生资源的生长速度是一定的，那么为了使人类有不断发展的潜力，地球上最多能养活多少亿人? ", "answer_option_list": [[{"aoVal": "A", "content": "$$24000$$ "}], [{"aoVal": "B", "content": "$$10000$$ "}], [{"aoVal": "C", "content": "$$70$$ "}], [{"aoVal": "D", "content": "$$80$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->牛吃草问题->牛吃草转化型->生活中的牛吃草->其他问题"], "answer_analysis": ["设每亿人每年消耗资源量为$$1$$份， 每年新生资源量：$$(80\\times 300-100\\times 100)\\div (300-100)=70$$（份） 即为保证不断发展，地球上最多养活$$70$$亿人． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1151", "queId": "3403cdb6ef2e4a4985ddc1af1aca3121", "competition_source_list": ["2020年新希望杯二年级竞赛初赛（个人战）第6题", "2020年新希望杯二年级竞赛决赛（8月）第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "李叔叔排队买票，他前面有$$5$$人，后面有$$6$$人．一共有人排队买票． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["加上李叔叔自己，一共：$$5+6-1=10$$（人）， 故选择：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "365", "queId": "6a608bd109894904b3214ffd0b59fc49", "competition_source_list": ["2004年第2届创新杯五年级竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$10$$个小数：$$0.6$$，$$0.66$$，$$0.666$$，$$\\cdots \\cdots \\underbrace{0.666\\cdots 666}_{10{个}6}$$如果要从这些小数中取出若干个使取出的数的总和大于$$5$$，那么所取出的数的个数至少是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$个 "}], [{"aoVal": "B", "content": "$$7$$个 "}], [{"aoVal": "C", "content": "$$8$$个 "}], [{"aoVal": "D", "content": "$$9$$个 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["此题应采取求最值的方法解决，看看取最小数与最大数时应取几个，因为$$5\\div 0.6\\approx 8.3$$，$$5\\div \\underbrace{0.666\\cdots 666}_{10{个}6}\\approx 7.5$$，故至少是$$8$$个． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2029", "queId": "c65538f8553c4aaf81a54caee42ec516", "competition_source_list": ["2019年第24届YMO三年级竞赛决赛第7题3分", "2020年第24届YMO三年级竞赛决赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小刚用一根绳子测量一口枯井的深度．他把绳子放入井里，当绳子到达井底后，井外还留有$$12$$米．小刚又把这根绳子对折后再放入井里，绳子到达井底后，井外还留有$$2$$米，这口井深米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->还原问题->单一变量还原问题"], "answer_analysis": ["第$$1$$次井外$$12$$米， 第$$2$$次井外共有$$2\\times 2=4$$（米）， 故井深$$12-4=8$$（米）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2547", "queId": "877dceea60dc4868a9dda0b81d5cccb0", "competition_source_list": ["2004年第2届创新杯六年级竞赛复赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "在下面四个算式中，得数最大的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( \\frac{1}{17}+\\frac{1}{19} \\right)\\times 20$$ "}], [{"aoVal": "B", "content": "$$\\left( \\frac{1}{24}+\\frac{1}{29} \\right)\\times30$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{31}+\\frac{1}{17} \\right)\\times40$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{1}{41}+\\frac{1}{47} \\right)\\times50$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$\\left( \\frac{1}{17}-\\frac{1}{19} \\right)\\div 20=\\frac{2}{17\\times 19}\\div 20=\\frac{1}{17\\times 19}\\times \\frac{1}{10}$$;$$\\left( \\frac{1}{15}-\\frac{1}{21} \\right)\\div 60=\\frac{6}{15\\times 21}\\div 60=\\frac{1}{15\\times 21}\\times \\frac{1}{10}$$; $$\\left( \\frac{1}{13}-\\frac{1}{23} \\right)\\div 100=\\frac{10}{13\\times 23}\\div 100=\\frac{1}{13\\times 23}\\times \\frac{1}{10};$$$$\\left( \\frac{1}{11}-\\frac{1}{25} \\right)\\div 140=\\frac{14}{11\\times 25}\\div 140=\\frac{1}{11\\times 25}\\times \\frac{1}{10};$$ 只需比较$$\\frac{1}{17\\times 19}$$，$$\\frac{1}{15\\times 21}$$，$$\\frac{1}{13\\times 23}$$，$$\\frac{1}{11\\times 25}$$的大小，根据和一定，两数越接近乘 积越大，则$$11\\times 25 \\textless{} 13\\times 23 \\textless{} 15\\times 21 \\textless{} 17\\times 19$$，那么 $$\\frac{1}{11\\times 25}\\textgreater\\frac{1}{13\\times 23}\\textgreater\\frac{1}{15\\times 21}\\textgreater\\frac{1}{17\\times 19}$$，所以答案为$$D$$ "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "471", "queId": "b376903265b943e9a7ead79acfecae12", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一把钥匙只能开一把锁，现有$$4$$把钥匙和$$4$$把锁搞乱了，最多试开次就能确定哪把钥匙开哪把锁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$10$$ "}], [{"aoVal": "E", "content": "$$12$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->组合模块->抽屉原理->最不利原则", "拓展思维->思想->对应思想"], "answer_analysis": ["第一把钥匙最坏的情况要试$$3$$次，把这把钥匙和这把锁拿出；剩下的$$3$$把锁和$$3$$把钥匙，最坏的情况要试$$2$$次，把这把钥匙和这把锁拿出；剩下的$$2$$把锁和$$2$$把钥匙，最坏的情况要试$$1$$次，把这把钥匙和这把锁拿出；剩下的$$1$$把锁和$$1$$把钥匙就不用试了；$$3+2+1=6$$（次）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "243", "queId": "ac251501dfa244578d186e10e2af5171", "competition_source_list": ["2017年河南郑州联合杯竞赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "有红、黄、蓝、白珠子各$$10$$个，装在一只袋子里，为了保证摸出的珠子至少有两个颜色相同，最少需要摸出（~ ）个珠子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$个 "}], [{"aoVal": "B", "content": "$$5$$个 "}], [{"aoVal": "C", "content": "$$4$$个 "}], [{"aoVal": "D", "content": "$$11$$个 "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["最不利原则；$$4+1=5$$个，则最少需要摸出$$5$$个珠子． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3052", "queId": "fc9e8e3b9043479d8bca7d3281b9b398", "competition_source_list": ["2020年广东广州羊排赛四年级竞赛第17题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "简易方程． $$5x+12=52-3x$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$x=4$$． "}], [{"aoVal": "B", "content": "$$x=3$$． "}], [{"aoVal": "C", "content": "$$x=2$$． "}], [{"aoVal": "D", "content": "$$x=5$$． "}]], "knowledge_point_routes": ["课内体系->知识点->式与方程->简易方程->解简易方程->整数简易方程->解方程：ax＋b＝c这种类型", "拓展思维->拓展思维->计算模块->方程基础->一元一次方程"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde5x+12=52-3x$$ 解：$$5x+3x=52-12$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde8x=40$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x=40\\div8$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x=5$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2499", "queId": "1a20ae8d98304af3899f29bbd9cb1aa0", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "先找出下列数排列的规律，然后在括号里填上适当的数．$$1$$，$$4$$，$$9$$，$$16$$，$$25$$，，$$49$$，$$64$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$34$$ "}], [{"aoVal": "B", "content": "$$35$$ "}], [{"aoVal": "C", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->能力->抽象概括"], "answer_analysis": ["这组数是平方数列， 所以第$$6$$个空为$$6^{2}=36$$， 故选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "75", "queId": "1cd7376c603f4eeca133d37f2de3478b", "competition_source_list": ["2008年第6届创新杯六年级竞赛初赛B卷第9题5分", "2008年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "电视台要播放一部30集电视连续剧，若要求每天安排播出的集数互不相等，则该电视连续剧最多可以播( ). ", "answer_option_list": [[{"aoVal": "A", "content": "7天 "}], [{"aoVal": "B", "content": "8天 "}], [{"aoVal": "C", "content": "9天 "}], [{"aoVal": "D", "content": "10天 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->和一定型最值问题->多数之积的最值->拆分数的数目不确定"], "answer_analysis": ["由于希望播出的天数要尽可能的多，所以，在每天播出的集数互不相等的条件下，每天播放的集数应可能的少，又$$1+2+3+4+5+6+7=28$$，如果各天播出的集数分别为1、2、3、4、5、6、7时，那么七天共可播出28集，还剩2集未播出，由于已有过一天播出2集的情况，因此，这余下2集不能再单独一天播出，而只好把它们分到以前的日子里播出，例如，各天播出的集数安排为1、2、3、4、5、7、8或1、2、3、4、5、6、9均可，所以最多可以播7天. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "448", "queId": "9c685a9df90848688bfc5b76a90a99cc", "competition_source_list": ["2018年湖北武汉创新杯小学高年级六年级竞赛初赛数学思维能力等级测试第5题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知两位数与三位数之积$$\\overline{3x}·\\overline{yz6}=7788$$，那么$$x+y+z=$$（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->横式数字谜->与数论的结合"], "answer_analysis": ["由个位分析，$$x=3$$，$$x= 8$$，但是$$7788$$不能被$$38$$整除，则$$\\overline{yz6}=7788\\div 33=236$$．则$$x+y+z=8$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1091", "queId": "3cdd2c9a02b947eabd4f5d72bb47f3a9", "competition_source_list": ["2012年第13届上海中环杯小学中年级三年级竞赛初赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "学校买了$$2$$张桌子和$$3$$把椅子，共付了$$99$$元．一张桌子的价钱和$$4$$把椅子的价钱相等，一张桌子~\\uline{~~~~~~~~~~}~元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$27$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["设一张桌子$$x$$元，一把椅子$$y$$元． 则可得出方程：$$\\begin{cases}2x+3y=99 x=4y \\end{cases}$$，代入消元法，解得方程：$$x=36$$，$$y=9$$． 所以一张桌子$$36$$元，一把椅子$$9$$元． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "780", "queId": "a35b90d09a5b411ca0f1eadaa31cb08f", "competition_source_list": ["2016年第3届广东深圳鹏程杯六年级竞赛集训材料第十八章综合训练题（八）第13题"], "difficulty": "3", "qtype": "single_choice", "problem": "一次数学竞赛，某校$$200$$多人报名，其中有$$\\frac{1}{16}$$的人成绩不足$$60$$分，$$\\frac{1}{15}$$的人成绩不足$$70$$分，$$20 \\%$$的人不足$$80$$分，$$90$$分以上的占总数的$$\\frac{1}{6}$$．那么成绩在$$80$$～$$89$$分的有人. ", "answer_option_list": [[{"aoVal": "A", "content": "$$150$$ "}], [{"aoVal": "B", "content": "$$151$$ "}], [{"aoVal": "C", "content": "$$152$$ "}], [{"aoVal": "D", "content": "$$153$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->因倍应用题->倍数应用题"], "answer_analysis": ["人数必须是整数，总人数是$$16$$，$$15$$，$$5$$，$$6$$的公倍数。求公倍数可以由最小公倍数来求，所以可以求出这次竞赛共有$$240$$（$$16$$、$$15$$、$$5$$、$$6$$的最小公倍数）人参加。 $$80$$～$$90$$分的人数占$$\\left( 1-20 \\% -\\frac{1}{6} \\right)=\\frac{19}{30}$$． 由此可求出$$80$$～$$90$$分的人数． $$240\\times \\frac{19}{30}=152$$（人） 答：成绩在$$80$$～$$90$$分的有$$152$$人． 拓展：$$3$$个数短除法，百分数（六年级上）也叫百分比、百分率 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1863", "queId": "db8feccd7eeb426cab9b29ac9d93d217", "competition_source_list": ["2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（二）"], "difficulty": "1", "qtype": "single_choice", "problem": "$$10$$名同学参加数学竞赛，前$$4$$名同学平均得分$$150$$分，后$$6$$名同学平均得分比$$10$$人的平均分少$$20$$分，这$$10$$名同学的平均分是（~ ）分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$140$$ "}], [{"aoVal": "C", "content": "$$160$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["设这$$10$$名同学的平均得分是$$x$$分，那么$$10$$名同学的总得分是$$10x$$分，则： $$10x6\\times \\left( x20 \\right)=4\\times 150$$，$$10x-6x=480$$，$$\\begin{matrix}4x=480 \\end{matrix}$$，$$x=120$$， 答 ：这$$10$$名同学的平均得分是$$120$$分. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2229", "queId": "5effc8da6cb84618ace8600641c2492e", "competition_source_list": ["竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙三人行路，甲每分钟走$$60$$米，乙每分钟走$$65$$米，丙每分钟走$$70$$米，甲、乙从东镇去西镇，丙从西镇去东镇，三人同时出发，丙与乙相遇后，又经过$$1$$分钟与甲相遇，则东、西两镇间的路程有（ ）米。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$3510$$ "}], [{"aoVal": "B", "content": "$$3750$$ "}], [{"aoVal": "C", "content": "$$2510$$ "}], [{"aoVal": "D", "content": "$$5210$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->多人相遇与追及问题->多人相遇追及问题"], "answer_analysis": ["那$$1$$分钟是甲和丙相遇，所以距离是$$\\left( 60+70 \\right)\\times 1=130$$（米），这距离是乙、丙相遇时间里甲、乙的路程差，所以乙、丙相遇时间$$=130\\div (65-60)=26$$（分），所以路程$$=26\\times (65+70)=3510$$（米）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2209", "queId": "28fb7cbdc42e4a20abf7647181fcb925", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "某人开车往返于深圳和广州之间，从深圳去广州每小时行$$30$$千米，从广州返回深圳每小时行$$60$$千米．那么他往返深圳和广州的平均速度是千米$$/$$时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$50$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->平均速度->公式法"], "answer_analysis": ["深圳和广州的路程为$$60$$千米，则从深圳到广州用时$$60\\div30=2$$小时，从广州到深圳用时$$60\\div60=1$$小时，总共用时$$1+2=3$$小时，平均速度$$=60\\times2\\div3=40$$千米$$/$$小时． 答案：$$B$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "985", "queId": "1297622d3512448e8039ed53d6f84de2", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第4题2分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$、$$B$$、$$C$$、$$D$$、$$E5$$人在一次满分为$$100$$分的考试中，得分都是大于$$91$$的整数．如果$$A$$，$$B$$，$$C$$的平均分是$$95$$分；$$B$$，$$C$$，$$D$$的平均分是$$94$$；$$A$$是第一名；$$E$$是第三名得$$96$$分，问$$D$$得了分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$94$$ "}], [{"aoVal": "B", "content": "$$95$$ "}], [{"aoVal": "C", "content": "$$96$$ "}], [{"aoVal": "D", "content": "$$97$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["$$A$$，$$B$$，$$C$$的总分是：$$95\\times 3=285$$（分）， $$B$$，$$C$$，$$D$$的总分是：$$94\\times 3=282$$（分）， $$A$$比$$D$$就多考了︰$$285-282=3$$（分）， 因为$$E$$是第三名考了$$96$$分，所以，$$D$$有两种可能： 一是$$D$$比$$E$$考得少，鉴于$$A$$是第一名，又比$$D$$多三分，$$A$$只能是$$98$$分，而$$D$$是$$95$$分，$$B$$，$$C$$中有一人考$$97$$分，这样的话，$$B$$，$$C$$中的另一个考得分数就是：$$285-98-97=90$$，这与所有人得分都大于$$91$$是矛盾的，所以，$$D$$的名次一定在$$E$$的前面；即$$D$$是第二名；$$D$$是第二名，得分就要多于$$96$$分，结合$$A$$比$$D$$多$$3$$分，可知$$D$$的得分是$$97$$分，$$A$$的得分是：$$100$$分． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1839", "queId": "8a94ad76e8dd464980593bd6e8879911", "competition_source_list": ["2021年新希望杯三年级竞赛初赛第30题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某个闰年的元旦是星期日，那么这一年的$$2$$月有$$5$$个． ", "answer_option_list": [[{"aoVal": "A", "content": "星期一 "}], [{"aoVal": "B", "content": "星期二 "}], [{"aoVal": "C", "content": "星期三 "}], [{"aoVal": "D", "content": "星期四 "}], [{"aoVal": "E", "content": "星期五 "}], [{"aoVal": "F", "content": "星期六 "}], [{"aoVal": "G", "content": "星期日 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["周期问题： 元旦$$1$$月$$1$$日是星期日，则周期：星期日、星期一、星期二、星期三、星期四、星期五、星期六$$\\cdots\\cdots$$ 从$$1$$月$$1$$日到$$2$$月$$1$$日一共有$$31+1=32$$（天）， $$32\\div7=4$$（周）$$\\cdots\\cdots4$$（天）， 所以$$2$$月$$1$$日是星期三，则$$2$$月份的周期：星期三、星期四、星期五、星期六、星期日、星期一、星期二$$\\cdots\\cdots$$ 闰年的$$2$$月有$$29$$天，$$29\\div7=4$$（周）$$\\cdots\\cdots1$$（天）， 所以$$2$$月$$29$$日是星期三， 因此这一年的$$2$$月有$$5$$个星期三． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "285", "queId": "ff8080814502fa2401450bdf764d178f", "competition_source_list": ["2014年全国迎春杯五年级竞赛初赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "在下列算式的空格中填入互不相同的数字：$$\\square \\times \\left( \\square +\\square \\square \\right)\\times \\left( \\square +\\square +\\square +\\square \\square \\right)=2014$$．其中五个一位数的和最大是（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$35$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$2014=2\\times 19\\times 53$$，五个一位数之和最大，则两位数应最小 $$2\\times (a+\\overline{1b})\\times(c+d+e+\\overline{3f})=2014$$，$$\\Rightarrow \\left { \\begin{align}\\& a+b=9=9+0 \\& c+d+e+f=23=8+6+5+4 \\end{align} \\right.$$，$$\\Rightarrow{{(2+a+c+d+e)}_{\\max }}=2+9+8+6+5=30$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "773", "queId": "9587efdace244e8e950ee5d56bbf9701", "competition_source_list": ["2018年美国数学大联盟杯六年级竞赛初赛第35题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "平方为$$6-$$位数，满足条件的最小的$$3-$$位数的位数之和是多少？ What is the sum of the digits of the least $$3-$$digit integer whose square is a $$6-$$digit integer？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->枚举型最值问题", "Overseas Competition->知识点->数论模块->完全平方数"], "answer_analysis": ["平方为六位数整数的最小三位数整数的位数之和是多少？ 可以先进行尝试，确定这个三位数的大小范围，$$100$$的平方为$$10000$$，$$200$$的平方为$$40000$$，$$300$$的平方为$$90000$$，$$400$$的平方为$$160000$$， 所以这个数位于$$300$$与$$400$$之间，经试验，$$316$$的平方为$$99856$$，$$317$$的平方为$$100489$$，所以这个三位数最小是$$317$$，$$3+1+7=11$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1735", "queId": "8995adcb75244f4fb75809b207c4693f", "competition_source_list": ["2020年新希望杯二年级竞赛初赛（个人战）第6题", "2020年新希望杯二年级竞赛决赛（8月）第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "李叔叔排队买票，他前面有$$11$$人，后面有$$12$$人．一共有人排队买票． ", "answer_option_list": [[{"aoVal": "A", "content": "$$22$$ "}], [{"aoVal": "B", "content": "$$23$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["加上李叔叔自己，一共：$$11+1+12=24$$（人）， 故选择：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "8", "queId": "0162d08da2b8459b9fc4313d182ed5ab", "competition_source_list": ["2015年全国中环杯二年级竞赛决赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个盒子里有一些大小相同的球，其中白的有$$8$$个，黑的有$$9$$个。不许看球，每次拿一个，至少拿~\\uline{~~~~~~~~~~}~次才能保证拿到黑球。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["要保证拿出的球三种颜色都有，考虑倒霉的情���．最倒霉的是拿出的全部是黑色的球，那么这是$$9$$次，接着拿出的全部是白色的球，这是$$8$$次，最后随便拿一个肯定是黄色的，所以至少拿$$9+8+1=18$$（次）才能保证三种颜色的球都有． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1790", "queId": "a8f0b97719bc45d1abbfc1b938b797fc", "competition_source_list": ["2014年走美杯二年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一根$$30$$厘米长的木条，要锯成$$5$$厘米的小段，需要锯（ ）次。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->封闭型->封闭型植树问题->封闭植树类型问题(段数小于等于10)"], "answer_analysis": ["解：$$30\\div 5-1$$ $$=6-1$$ $$=5$$（次） 答：需要锯$$5$$次。 故答案为：$$5$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1223", "queId": "2725ffa7611b473dada43b0588fd73e9", "competition_source_list": ["2013年第11届创新杯四年级竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "一对双胞胎和一组三胞胎$$5$$个人年龄的总和是$$84$$．如果把双胞胎的年龄同三胞胎的年龄互换，那么这$$5$$人年龄的总和将是$$76$$．那么双胞胎的年龄是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["根据题意，设双胞胎的年龄是$$x$$岁，三胞胎的年龄是$$y$$岁，由一对双胞胎和一组三胞胎$$5$$个人年龄的总和是$$84$$可得$$2x+3y=84$$；把双胞胎的年龄同三胞胎的年龄互换，那么这$$5$$人年龄的总和是$$76$$岁，可得$$2y+3x=76$$，联立方程组然后再进一步解答； 设双胞胎的年龄是$$x$$岁，三胞胎的年龄是$$y$$岁； 由题意可得：$$\\begin{cases} 2x+3y=84 2y+3x=76 \\end{cases}$$， 解得：$$x=12$$，$$y=20$$． 答：双胞胎的年龄是$$12$$岁． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2085", "queId": "cb9c1805cc424af0890d10c5022d97c7", "competition_source_list": ["2014年全国迎春杯六年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个半径为$$20$$厘米的蛋糕可以让$$4$$个人吃饱，如果半径增加了$$150 \\%$$，同样高的蛋糕可以让个人吃饱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->几何模块->曲线型->圆与扇形->圆的基本公式"], "answer_analysis": ["由条件，面积变为原来的$${{\\left( 1+150 \\% \\right)}^{2}}$$，所以可供$$4\\times {{\\left( 1+150 \\% \\right)}^{2}}=25$$（个）人吃饱． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1314", "queId": "30eec3435e814efaa30c119e0b1d2e26", "competition_source_list": ["2017年河南郑州豫才杯小学高年级五年级竞赛初赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "游览景区期间，爸爸将车放在停车场，收费标准为：不超过$$1$$小时收费$$3$$元，每多停半小时加收$$1.5$$元．爸爸最终一共交了$$13.5$$元的停车费，他们的车在停车场最多停了（~ ）小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$小时 ~~~ "}], [{"aoVal": "B", "content": "$$4.5$$小时~~ "}], [{"aoVal": "C", "content": "$$5$$小时 "}], [{"aoVal": "D", "content": "$$8$$小时 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->分段计价问题"], "answer_analysis": ["设最多在停车场停了$$x$$小时，$$3+\\left( x-1 \\right)\\times 2\\times 1.5=13.5$$，解得$$x=4.5$$，所以他们的车在停车场最多停$$4.5$$小时． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1935", "queId": "c9ec58ba993743429a22f7375e3b616f", "competition_source_list": ["2022年第9届广东深圳鹏程杯四年级竞赛初赛第21题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "2022年$$2$$月$$22$$日被广大网民称为``世界最爱日'',因为这个日期里面包含六个$$2.$$与它包含相同多$$2$$的日期是$$2022$$年$$12$$月$$22$$日，比它包含更多$$2$$的日期则是$$200$$��后的$$2222$$年$$2$$月$$22$$日. 今年$$2$$月$$22$$日又恰好是星期二，而$$12$$月$$22$$日是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}], [{"aoVal": "E", "content": "五 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->求某日期是周几问题"], "answer_analysis": ["无 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2577", "queId": "2c9ca1cb197a495a9844ff17a1579eef", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（一）第4题", "2017年新希望杯六年级竞赛训练题（一）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "（1）将四个分数按从小到大的顺序排列，正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ "}], [{"aoVal": "C", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ "}], [{"aoVal": "D", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["通分子$$\\frac{5}{14}=\\frac{60}{168}$$， $$\\frac{10}{27}=\\frac{60}{162}$$， $$\\frac{12}{31}=\\frac{60}{155}$$， $$\\frac{20}{53}=\\frac{60}{159}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3104", "queId": "d4993b875a2d4da4a444abcdc686529c", "competition_source_list": ["2004年五年级竞赛创新杯", "2004年第2届创新杯五年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个数把小数点向左移动一位后，就比原数少2.844，这个数原来是（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "0.316 "}], [{"aoVal": "B", "content": "3.16 "}], [{"aoVal": "C", "content": "31.6 "}], [{"aoVal": "D", "content": "316 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数基础->小数点的移动规律"], "answer_analysis": ["设新数为$$x$$，则原数为$$10x$$，$$10x-x=2.844$$，$$x=0.316$$.原数$$=0.316\\times 10=3.16$$. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2233", "queId": "3b8e0c5fb6ad4b5bb00af4d1c63c5b7e", "competition_source_list": ["2008年第8届希望杯六年级竞赛初赛第11题6分", "2019年四川绵阳涪城区绵阳东辰国际学校小升初（十三）第21题2分", "小学高年级五年级下学期其它时钟问题"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$16$$点$$16$$分这个时刻，钟表盘面上时针和分针的夹角是~\\uline{~~~~~~~~~~}~度． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$36$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->时钟问题->时钟上的追及问题->已知时间求角度"], "answer_analysis": ["$$16$$点的时候夹角为$$120$$度，每分钟分针转$$6$$度，时针转$$0.5$$度，$$16$$：$$16$$的时候夹角为$$120-6\\times 16+0.5\\times 16=32$$度． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "732", "queId": "51acb6c975324d839f1e3e594e3a0563", "competition_source_list": ["2014年全国迎春杯五年级竞赛初赛第8题", "2014年全国迎春杯六年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$4$$个质数的积是它们和的$$11$$倍，则它们的和为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$46$$ "}], [{"aoVal": "B", "content": "$$47$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["知识标签->数学思想->逐步调整思想"], "answer_analysis": ["由已知条件，$$4$$个质数中一定有$$11$$，那么则满足$$a\\times b\\times c=a+b+c+11$$，其中$$a$$、$$b$$、$$c$$都是质数．若$$a$$、$$b$$、$$c$$都是奇数，那么等式左边是奇数，右边为偶数，矛盾．若$$a$$、$$b$$、$$c$$中有$$1 $$个偶数，那么一定是$$2$$．即$$a\\times b\\times 2=a+b+2+11$$，此时，根据奇偶性，$$a$$、$$b$$中也必有一个偶数为$$2$$，解得$$a$$、$$b$$、$$c$$、$$d$$为$$2$$、$$2$$、$$5$$、$$11$$和为$$20$$．选项中$$ABC$$均不符合条件，���选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "883", "queId": "bb4f86143025459d9aef59a7ceb26bad", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛B卷第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$120$$有~\\uline{~~~~~~~~~~}~个因数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->数论模块->因数与倍数->因数个数定理"], "answer_analysis": ["因为$$120$$的因数有：$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$8$$、$$10$$、$$12$$、$$15$$、$$20$$、$$24$$、$$30$$、$$40$$、$$60$$、$$120$$，所以一共有$$16$$个因数． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1363", "queId": "62f74eaee41f43fbb25d72070b916c56", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2015$$减去它的$$\\frac{1}{2}$$，再减去余下的$$\\frac{1}{3}$$，再减去余下的$$\\frac{1}{4}$$，$$\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot $$，以此类推，直到减去余下的$$\\frac{1}{2015}$$，最后的结果是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2015$$ "}], [{"aoVal": "B", "content": "$$1042$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->量率对应已知单位1"], "answer_analysis": ["$$2015\\times \\frac{1}{2}\\times \\frac{2}{3}\\times \\frac{3}{4}\\times \\cdot \\cdot \\cdot \\times \\frac{2014}{2015}=1$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1004", "queId": "45cf29af50dc4d13b48d9c7a4cf9e3c5", "competition_source_list": ["2014年迎春杯三年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在一道没有余数的除法中，被除数、除数与商三个数的和是$$103$$，商是$$3$$。被除数是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$75$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->和倍问题->二量和倍问题->两量和倍"], "answer_analysis": ["因为除数$$=$$被除数$\\div3$ 所以被除数是： $100\\div (3+1)\\times3$ $=100\\div4\\times3$ $=75$ 故选：$$\\text{C}$$． ", "除数：$$(103-3)\\div (3+1)=25$$
\n被除数：$$25\\times3=75$$
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "284", "queId": "beddf9a04dc443d38eec0d91fa135a9d", "competition_source_list": ["2020年希望杯二年级竞赛模拟第17题"], "difficulty": "1", "qtype": "single_choice", "problem": "小皮、小舒和小贝是同班同学，他们中一个是班长，一个是学习委员，一个是体育委员． （2020年希望杯二年级） 现在知道： ①小皮的年龄比体育委员的年龄大； ②小舒比学习委员的年龄大； ③小贝和学习委员年龄不同． 那么小皮、小舒和小贝分别担任． ", "answer_option_list": [[{"aoVal": "A", "content": "班长，学习委员，体育委员 "}], [{"aoVal": "B", "content": "学习委员，体育委员，班长 "}], [{"aoVal": "C", "content": "学习委员，班长，体育委员 "}], [{"aoVal": "D", "content": "班长，体育委员，学习委员 "}], [{"aoVal": "E", "content": "体育委员，班长，学习委员 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["由②小舒比学习委员的年龄大；和③小贝和学习委员年龄不同．可知小舒和小贝都不是学习委员，所以学习委员是小皮； 已知小舒比小皮大，并且小皮的年龄比体育委员的年龄大，所以小舒不是体育委员，所以小舒是班长；那么小贝就是体育委员． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2693", "queId": "5f5b21bf34c447dc9ea620f733ad3197", "competition_source_list": ["2008年第6届创新杯四年级竞赛初赛B卷第1题5分", "2008年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$$10001\\times 3333+10001\\times 6666=$$( ). ", "answer_option_list": [[{"aoVal": "A", "content": "999999 "}], [{"aoVal": "B", "content": "9999999 "}], [{"aoVal": "C", "content": "99999999 "}], [{"aoVal": "D", "content": "999999999 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数乘法巧算之提取公因数（普通型）"], "answer_analysis": ["原式$$=10001\\times \\left( 3333+6666 \\right)=10001\\times 9999=99999999$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2506", "queId": "272b4950e854488c9118c3ceacbad15d", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$\\text{a}$$， $$\\text{b}$$是非$$0$$的自然数，并且$$\\text{a}$$\\textless$$\\text{b}$$，则$$\\frac{\\text{a+b}}{\\text{b}}$$的值是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "是$$0$$和$$1$$之间的数 "}], [{"aoVal": "B", "content": "是$$1$$和$$2$$之间的数 "}], [{"aoVal": "C", "content": "可以是$$2$$ "}], [{"aoVal": "D", "content": "可以大于$$2$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["由$$\\frac{a+b}{b}=\\frac{a}{b}+1$$及$$\\text{a}$$\\textless$$\\text{b}$$，知$$1\\textless\\frac{a+b}{b}\\textless2$$，故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "294", "queId": "ff8080814518d5240145201ad74d0a76", "competition_source_list": ["2014年全国迎春杯三年级竞赛复赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙、丁和戊参加$$100$$米比赛，比赛结束后丁说：``我比乙跑得快．''丙说：``戊在我前面冲过终点线．''甲说：``我的名次排在丁的前面，丙的后面．''请根据他们的话排出名次：（~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "戊＞丙＞丁＞甲＞乙 "}], [{"aoVal": "B", "content": "甲＞乙＞丙＞丁＞戊 "}], [{"aoVal": "C", "content": "乙＞丁＞甲＞丙＞戊 "}], [{"aoVal": "D", "content": "戊＞丙＞甲＞丁＞乙 "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["根据丁说的可得``丁＞乙''，根据丙说的可得``戊＞丙''，根据甲说的可得``丙＞甲＞丁''，综合可得``戊＞丙＞甲＞丁＞乙''． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1438", "queId": "b0af3f23879a45c9884e47c1990e9ef4", "competition_source_list": ["2013年IMAS小学中年级竞赛第一轮检测试题第12题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "四位小朋友共有课外读物$$240$$本，甲给了乙$$3$$本，乙给了丙$$4$$本，丙给了丁$$5$$本，丁给了甲$$6$$本，这时他们$$4$$人课外读物的本数都相同．请问他们之中原来课外读物最少的人有多少本？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$57$$ "}], [{"aoVal": "B", "content": "$$58$$ "}], [{"aoVal": "C", "content": "$$59$$ "}], [{"aoVal": "D", "content": "$$60$$ "}], [{"aoVal": "E", "content": "$$61$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["可知这四位小朋友共有$$240$$本课外读物，所以当他们的书都相等时，每人各有$$240\\div 4=60$$（本）． 而题意可知此时甲的书多了$$6-3=3$$（本）、乙的书少了$$4-3=1$$（本）、丙的书少了$$5-4=1$$（本）、丁的书少了$$6-5=1$$（本），因此甲原有$$57$$本书，而乙、丙、丁各原有$$61$$本书．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "277", "queId": "6433a0f865ae48c6919b806a75fb560d", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "（1）一把钥匙只能开一把锁，现有$$4$$把钥匙和$$4$$把锁搞乱了，最多试开次就能确定哪把钥匙开哪把锁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["第一把钥匙最坏的情况要试$$3$$次，把这把钥匙和这把锁拿出；剩下的$$3$$把锁和$$3$$把钥匙，最坏的情况要试$$2$$次，把这把钥匙和这把锁拿出；剩下的$$2$$把锁和$$2$$把钥匙，最坏的情况要试$$1$$次，把这把钥匙和这把锁拿出；剩下的$$1$$把锁和$$1$$把钥匙就不用试了；$$3+2+1=6$$（次）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "241", "queId": "a788a820e8df4dd4924a4c18915c8f0a", "competition_source_list": ["2017年第17届湖北武汉世奥赛五年级竞赛决赛第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "保尔特、罗特、卢特三个人进行赛车比赛，最初的排名：第一是卢特，第二是罗特，第三是保尔特；在比赛过程中，保尔特和卢特三次相互超越，罗特和卢特五次相互超越，保尔特和罗特四次相互超越，经过这样的激战终于到达了终点线．三个人最终的排名，从高到低是（~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "卢特一罗特一保尔特 "}], [{"aoVal": "B", "content": "罗特一保尔特一卢特 "}], [{"aoVal": "C", "content": "保尔特一罗特一卢特 "}], [{"aoVal": "D", "content": "罗特一卢特一保尔特 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["相互超越的次数若是偶数，那么两人的顺序不发生变化；两人互相超越的次数若奇数，则两人先后顺序互换. 保尔特和卢特$$3$$次相互超越，那么顺序互换：保尔特$$\\textgreater$$卢特 罗特和卢特的$$5$$次相互超越，先后顺序互换：罗特$$\\textgreater$$卢特 保尔特和罗特四次相互超越，先后顺序不发生变化：罗特$$\\textgreater$$保尔特 所以最后的排名是：罗特$$\\textgreater$$保尔特$$\\textgreater$$卢特 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2714", "queId": "4504a4fcd58a4c0e9317c22193dd954f", "competition_source_list": ["2020年第1届广东深圳超常思维竞赛六年级竞赛复赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一列数，第一个数是$$105$$，第二个数是$$85$$．从第三个数开始，每个数都是它前面两个数的平均数，那么这列数中第$$2022$$个数的整数部分是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$90$$ "}], [{"aoVal": "B", "content": "$$91$$ "}], [{"aoVal": "C", "content": "$$92$$ "}], [{"aoVal": "D", "content": "$$93$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->公式类->直接求平均数"], "answer_analysis": ["因为$$\\frac{105+85}{2}=95$$，$$\\frac{85+95}{2}=90$$，$$\\frac{95+90}{2}=92.5$$， $$\\frac{90+92.5}{2}=91.25$$，$$\\frac{92.5+91.25}{2}=91.875$$，$$\\cdots $$， 因为$$91.25$$与$$91.875$$的整数部分相同，而两个数的平均数一定介于两个数之间， 所以，后面各数的整数部分均为$$91$$． 当然第$$2022$$个数的整数部分也为$$91$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2189", "queId": "d54b16aadebf4e3d8a89d591b07632e6", "competition_source_list": ["2012年IMAS小学高年级竞赛第一轮检测试题第7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁各有一只手表． （$$1$$）甲的手表快了$$10$$分钟，但他以为慢了$$5$$分钟； （$$2$$）乙的手表慢了$$5$$分钟，但他以为快了$$10$$分钟； （$$3$$）丙的手表快了$$5$$分钟，但他以为快了$$3$$分钟； （$$4$$）丁的手表慢了$$5$$分钟，但他以为慢了$$10$$分钟． 用他们的手表，每个人都认为自己恰好能准时到达学校，请问谁会迟到？ ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "丁 "}], [{"aoVal": "E", "content": "都不会 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["甲早到$$5-(-10)=15$$分钟，丙早到$$5-3=2$$分钟，丁早到$$(-5)-(-10)=5$$分钟，乙迟到$$10-(-5)=15$$分钟． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2307", "queId": "a604d3f3cd5d4f0495e96cc2e5cf0dc5", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "（过程完整）某人骑自行车过一座桥，上桥速度为每小时$$12$$千米，下桥速度为每小时$$24$$千米．而且上桥与下桥所经过的路程相等，中间也没有停顿．问这个人骑车过这座桥，往返的平均速度是每小时千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->平均速度->公式法"], "answer_analysis": ["设上桥的路程为$$24$$千米，则下桥的路程也为$$24$$千米． 所以上桥的时间为：$$24\\div12=2$$（小时）， 下桥的时间为：$$24\\div24=1$$（小时）， 所以全程的平均速度为：$$(24+24)\\div (1+2)=16$$（千米/小时）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2450", "queId": "aff9e96cff944af9b578fc58e779c978", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "（2018 IMAS，Question\\#2） 若$$\\left( \\Delta \\times 2-1 \\right)\\times 2=2018$$，请问$$\\Delta $$代表的数是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$502$$ "}], [{"aoVal": "B", "content": "$$503$$ "}], [{"aoVal": "C", "content": "$$504$$ "}], [{"aoVal": "D", "content": "$$505$$ "}], [{"aoVal": "E", "content": "$$506$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "Overseas Competition->知识点->计算模块->方程基础->一元一次方程"], "answer_analysis": ["$$\\left( \\Delta \\times 2-1 \\right)\\times 2=2018$$， $$\\Delta \\times 2-1=2018\\div 2=1009$$， $$\\Delta \\times 2=1009+1=1010$$， $$\\Delta =1010\\div 2=505$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "742", "queId": "3bcbfde0246548698cf21bad88cda332", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "四位数$$\\overline{A37B}$$能被$$88$$整除，则$$A$$表示的数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["由于$$88=8\\times 11$$，而且$$8\\textbar\\overline{A37B}$$，所以$$\\overline{37B}$$是$$8$$的倍数，则$$B=6$$；$$11\\textbar\\overline{A376}$$，所以$$(A+7)-(3+6)=0$$，则$$A=2$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "275", "queId": "3c3e4efd259840b685a96fd1005117db", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$4$$辆汽车进行了$$4$$场比赛，每场比赛结果如下： （$$1$$）$$1$$号汽车比$$2$$号汽车跑得快； （$$2$$）$$2$$号汽车比$$3$$号汽车跑得快； （$$3$$）$$3$$号汽车比$$4$$号汽车跑得慢； （$$4$$）$$4$$号汽车比$$1$$号汽车跑得快． 汽车跑得最快． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$号 "}], [{"aoVal": "B", "content": "$$2$$号 "}], [{"aoVal": "C", "content": "$$3$$号 "}], [{"aoVal": "D", "content": "$$4$$号 "}]], "knowledge_point_routes": ["拓展思维->能力->推理推导->言语逻辑推理"], "answer_analysis": ["根据（$$1$$）可知，$$1$$号比$$2$$号快．根据（$$2$$）可知，$$2$$号比$$3$$号快．根据（$$3$$）可知，$$4$$号比$$3$$号快．根据（$$4$$）可知，$$4$$号比$$1$$号快．所以$$4$$号快于$$1$$号，$$1$$号快于$$2$$号，$$2$$号快于$$3$$号．故最快的是$$4$$号．故$$\\text{ABC}$$错误，$$\\text{D}$$正确． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2330", "queId": "1be35facd5284184a64ce9669061cec4", "competition_source_list": ["2017年陕西西安碑林区西北工业大学附属中学小升初（二）第3题3分", "2018年湖南长沙雨花区中雅培粹中学小升初第3题3分", "2017年陕西西安碑林区西北工业大学附属中学小升初(十)第3题3分", "2017年陕西西安小升初某工大附中", "2016年创新杯六年级竞赛训练题（四）第4题", "2018年陕西西安小升初分类卷15第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "盒子中原来有$$7$$个小球，魔术师从中任取几个小球，把每一个小球都变成$$7$$个小球放回盒中；他又从中任取一些小球，把每一个小球又都变成$$7$$个小球放回盒中；如此进行，到某一时刻魔术师停止取球变魔术，此时盒中球的总数可能是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2018$$~~~~~~ "}], [{"aoVal": "B", "content": "$$2017$$~~~~~~ "}], [{"aoVal": "C", "content": "$$2016$$~~~~~~ "}], [{"aoVal": "D", "content": "$$2015$$~~~~~~ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["每取出$$1$$个，盒中就增加$$6$$个，有$$7+6n=2017$$，∴$$n=335$$． 而$$\\rm A$$中，$$7+6n=2018$$，$$\\rm C$$中$$7+6n=2016$$，$$\\text{D}$$中$$7+6n=2015$$，$$n$$都不为整数． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "387", "queId": "bb2036e3a63848399f619d7d06cce45d", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$3$$个质数的和是$$20$$，那么这$$3$$个质数乘积的最大值是（~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$130$$ "}], [{"aoVal": "B", "content": "$$154$$ "}], [{"aoVal": "C", "content": "$$182$$ "}], [{"aoVal": "D", "content": "$$312$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->和一定型最值问题->多数之积的最值->拆分数的数目确定"], "answer_analysis": ["其中一个一定为$$2$$，有$$2+7+11=20$$，$$2\\times 7\\times 11=154$$最大． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "866", "queId": "e4665cafd72b4a48a93e992594db9b17", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛B卷第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "下面四个数中可以写成$$3$$个连续两位数的乘积的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1321$$ "}], [{"aoVal": "B", "content": "$$12144$$ "}], [{"aoVal": "C", "content": "$$5812$$ "}], [{"aoVal": "D", "content": "$$44568$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数->分解质因数的应用->已知乘积求因数"], "answer_analysis": ["此题考查了乘积的个位数，明确三个连续两位数中必有一个偶数，一个数能被$$3$$整除，即这三个连续数的积是偶数，而且能被$$3$$整除，是解答此题的关键． 解：三个连续两位数中必有一个偶数，一个数能被$$3$$整除； 三个连续数的积是偶数，而且能被$$3$$整除，故排除（$$\\text{A}$$）、（$$\\text{C}$$）；$$12144=2\\times 2\\times 2\\times 2\\times 3\\times 11\\times 23=22\\times 23\\times 24$$， $$44568=2\\times 2\\times 2\\times 3\\times 3\\times 619$$， 由此可知：$$12144$$正好是三个连续两位数的积，其余两个数不是三个连续两位数的积，故$$\\text{B}$$正确． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "331", "queId": "89451a7024d94b24a0db0b1836609ce7", "competition_source_list": ["2012年第10届创新杯四年级竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "显示在电子钟上的时间是$$5:55$$，下一次电子钟上显示的时间又是全部相同的数字，还要过分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$71$$ "}], [{"aoVal": "B", "content": "$$255$$ "}], [{"aoVal": "C", "content": "$$316$$ "}], [{"aoVal": "D", "content": "$$377$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->时间问题->时间计算"], "answer_analysis": ["因为分钟的十位最大为$$5$$，所以下一次数字相同的时刻为$$11:11$$，$$11:11$$距$$5:55$$有$$5$$小时$$16$$分，即$$316$$分钟．所以选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2765", "queId": "ac9e34e32d514a2f99676b7a434a5264", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（四）第6题"], "difficulty": "3", "qtype": "single_choice", "problem": "从$$0$$、$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$这十个数字中选出九个数字，组成一个两位数、一个三位数和一个四位数，使这三个数的和等于$$2017$$，那么未被选中的数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "$$8$$ "}], [{"aoVal": "F", "content": "$$9$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->整数->整数乘除->整数除法运算->带余除法"], "answer_analysis": ["$$2017$$除以$$9$$余$$1$$，$$0+1+2+3+\\cdots +9=45$$，$$45$$是$$9$$的倍数，$$9-1=8$$，所以未被选中的数字是$$8$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "326", "queId": "4ee313d027824491bd514fb7139f642d", "competition_source_list": ["小学高年级六年级其它2015年数学思维能力等级测试初试第5题4分", "2015年第13届全国创新杯六年级竞赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "某校买来$$36$$套单坐课桌椅，不料发票给墨水弄污了，单价只剩下两个数字：$$\\square 23.\\square \\square $$元，总价只剩下四个数字：$$4\\square 44.2\\square $$元，那么总价应是（~ ~ ~ ）元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4944.24$$ "}], [{"aoVal": "B", "content": "$$4444.20$$ "}], [{"aoVal": "C", "content": "$$4544.28$$ "}], [{"aoVal": "D", "content": "$$4644.20$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$4\\square 44.2\\square =\\square 23.\\square \\square \\times 36$$，只有$$4444.20$$满足． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1086", "queId": "afe51d8c846847c5b0b5f6f2901883b8", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明与小红去摘桃，小明摘下$$13$$个桃，当小明将自己的桃分$$2$$个给小红时，两人的桃就一样多，小红摘了个桃． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["小红接受小明的$$2$$个桃后有：$$13-2=11$$（个）， 所以之前小红有桃：$$11-2=9$$（个）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1852", "queId": "adf90c1c6436468cafcaf17c58ae9ab7", "competition_source_list": ["2013年IMAS小学中年级竞赛第一轮检测试题第16题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有三个旅行者结伴穿越沙漠，途中小明的水喝完了，小东还剩$$5$$瓶矿泉水，小杰剩$$4$$瓶矿泉水，三个人商量将水平分，由小明向小东、小杰总共支付矿泉水费用$$36$$元，请问小明向小东支付了矿泉水费用多少元？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$20$$ "}], [{"aoVal": "E", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["可知每人可分得$$(5+4)\\div 3=3$$瓶水，因此小明需为每瓶水付出$$36\\div 3=12$$元，且可判断出小东给小明$$5-3=2$$瓶矿泉水、小杰给小明$$4-3=1$$瓶矿泉水，故小明应该向小东支付矿泉水费用$$2\\times 12=24$$元． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "575", "queId": "4144f286bf6043dd89b6ec0469c6b251", "competition_source_list": ["2020年希望杯五年级竞赛模拟第29题", "2020年新希望杯五年级竞赛第29题"], "difficulty": "0", "qtype": "single_choice", "problem": "梅森质数是形如「$${{2}^{p}}-1$$」的质数，这里$${{2}^{p}}$$表示$$p$$个$$2$$相乘的积，而且$$p$$也是一个质数。目前用超级计算机已找到了$$51$$个梅森质数，其中第二大的梅森质数是$${{2}^{77232917}}-1$$，那麼以下各数中，哪個數能被$$15$$整除？ ", "answer_option_list": [[{"aoVal": "A", "content": "$${{2}^{77232917}}-2$$ "}], [{"aoVal": "B", "content": "$${{2}^{77232917}}+1$$ "}], [{"aoVal": "C", "content": "$${{2}^{77232917}}+2$$ "}], [{"aoVal": "D", "content": "$${{2}^{77232917}}+3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "海外竞赛体系->Knowledge Point->Calculation Modules"], "answer_analysis": ["$${{2}^{n}}\\div 3$$的余数周期为$$2$$，$$1$$， $$77232917\\div 2\\cdots \\cdots 1$$，$${{2}^{77231917}}\\div 3\\cdots \\cdots 2$$， $${{2}^{n}}\\div 5$$的余数周期为$$2$$，$$4$$，$$3$$，$$1$$． $$77232917\\div 4\\cdots \\cdots 1$$，$${{2}^{77232917}}\\div 5\\cdots \\cdots 2$$， $$3\\times 5+2=17$$，$${{2}^{77232917}}\\equiv 17\\equiv 2\\left( \\bmod 15 \\right)$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2644", "queId": "a2d1a591c57b49c5b04273af70cdc0d4", "competition_source_list": ["2018~2019学年浙江杭州西湖区杭州市行知小学五年级上学期期中期中竞赛第13题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列算式中，结果小于$$1$$的算式是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.99\\div 0.1$$ "}], [{"aoVal": "B", "content": "$$1\\div 0.99$$ "}], [{"aoVal": "C", "content": "$$0.99\\div 0.99$$ "}], [{"aoVal": "D", "content": "$$0.99\\times 0.99$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数与运算->除法"], "answer_analysis": ["$$0.99\\times 0.99=0.9801$$，结果小于$$1$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1135", "queId": "3d16e81a5b234934ba3ccfeeac8b5932", "competition_source_list": ["2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$12$$枚硬币的总值是$$9$$角，其中只有$$5$$分和$$1$$角两种，那么每种硬币各（ ）枚。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题"], "answer_analysis": ["解：$$5$$分的数量： $$(12\\times 1-9)\\div \\left( 1-0.5 \\right)$$ $$=3\\div 0.5$$ $$=6$$（枚）； $$1$$角的硬币数量为：$$12-6=6$$（枚）。 答：每种硬币各$$6$$枚。 故选:C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2692", "queId": "379269f608ca4aedb7cca5cf05dc6b9a", "competition_source_list": ["2017年第1届重庆华杯赛竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$1007\\times \\dfrac{1\\dfrac{3}{4}\\div \\dfrac{3}{4}+3\\div 2\\dfrac{1}{4}+\\dfrac{1}{3}}{\\left( 1+2+3+4+5+6 \\right)\\times 6-73}\\div 19$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->繁分数->繁分数计算"], "answer_analysis": ["$$\\begin{matrix}=1007\\div 19\\times \\dfrac{1\\dfrac{3}{4}\\div \\dfrac{3}{4}+3\\div 2\\dfrac{1}{4}+\\dfrac{1}{3}}{\\left( 1+2+3+4+5+6 \\right)\\times 6-73} =53\\times \\dfrac{\\dfrac{7}{4}\\times \\dfrac{4}{3}+3\\times \\dfrac{4}{9}+\\dfrac{1}{3}}{6\\times 21-73} =53\\times \\dfrac{\\dfrac{7}{3}+\\dfrac{4}{3}+\\dfrac{1}{3}}{126-73} =53\\times \\dfrac{4}{53} =4 \\end{matrix}$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2763", "queId": "ba5bed0c677c4ab0af50d0725d0d7bcc", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛A卷第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "用四舍五入法求$$9.9999$$的近似值，精确到百分位时，结果为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$10.0$$ "}], [{"aoVal": "C", "content": "$$10.00$$ "}], [{"aoVal": "D", "content": "$$10.000$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为$$9.9999$$的千分位是$$9$$，根据四舍五入原则，$$9$$需要进位，所以精确到百分位时，结果为$$10.00$$，故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "194", "queId": "deb031b3a5604895ab6cca363b2393cb", "competition_source_list": ["2013年全国学而思杯三年级竞赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "小何、小琳和小俊参加了一次数独比赛，赛后，他们对比赛结果进行了预测． 小何说：``我是第$$1$$，小琳是第$$2$$，小俊是第$$3$$．'' 小琳说：``我是第$$1$$，小何是第$$2$$，小俊是第$$3$$．'' 小俊说：``我是第$$1$$，小琳是第$$2$$，小何是第$$3$$．'' 如果他们$$3$$人中，有$$1$$人$$3$$句话都预测正确，其余两人都只预测正确了$$1$$句话，那么，$$3$$人的名次按小何、小琳、小俊的顺序组成的$$3$$位数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "1,2,3 "}], [{"aoVal": "B", "content": "1,3,2 "}], [{"aoVal": "C", "content": "2,1,3 "}], [{"aoVal": "D", "content": "2,3,1 "}], [{"aoVal": "E", "content": "3,1,2 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾", "海外竞赛体系->Knowledge Point->Calculation Modules"], "answer_analysis": ["方法一：由于$$3$$人中有且仅有$$1$$人全预测正确，故可枚举全正确的人的情况： 若小何全正确，则小琳只有第$$3$$句正确，小俊只有第$$2$$句正确，符合要求； 若小琳全正确，则小俊全错，不符合要求； 若小俊全正确，则小琳全错，不符合要求； 方法二：注意到小琳和小俊的预测全都不同，故知全正确的人不可能在这两人之中（否则另一个人就全错，不符合要求） 综上，小何全正确，答案为$$123$$. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2886", "queId": "96bbdd44310f482887cf8ea3be01b82f", "competition_source_list": ["2004年第2届创新杯五年级竞赛复赛第7题", "2004年五年级竞赛创新杯", "2005年第3届创新杯五年级竞赛初赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "某月中有三个星期一的日期都是偶数，则该月的$$18$$日一定是( )。 ", "answer_option_list": [[{"aoVal": "A", "content": "星期一 "}], [{"aoVal": "B", "content": "星期三 "}], [{"aoVal": "C", "content": "星期五 "}], [{"aoVal": "D", "content": "星期日 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数表规律->常见数表规律"], "answer_analysis": ["因为此月有三个星期一为偶数，那么该月必有两个星期一为奇数，所以此月有五个星期一，那么第一个星期一只能为该月的$$2$$日(否则这个月没有五个星期一)，故$$18$$日为星期三，故选B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "396", "queId": "8a3b41329e694f11bd9bd42c59e6592b", "competition_source_list": ["2017年全国华杯赛小学中年级竞赛初赛模拟第5题", "其它改编题"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$至$$11$$这$$11$$个自然数中至少选出~\\uline{~~~~~~~~~~}~个不同的数，才能保证其中一定有两个数的和为$$12$$? ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想", "课内体系->能力->逻辑分析"], "answer_analysis": ["把和为$$12$$的两个数分成一组，这样就把这$$11$$个数分成$$6$$组：$$(1,11)$$，$$\\left( 2,10 \\right)$$，$$\\left( 3,9 \\right)$$，$$(4,8)$$，$$(5,7)$$，$$(6)$$．要保证一定有两个数的和为$$12$$，就要保证至少有两个数属于同一组．由抽屉原理可知，从这$$12$$个数中选出$$7$$个数，就一定有两个数属于同一组．此时这两个数的和就是$$12$$．如果从$$6$$组中各取一个数，则取出的这$$6$$个数中，没有两个数的和是$$12$$，因此本题的答案就是至少选出$$7$$个不同的数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1919", "queId": "c534fbb083874f5e94873400faadaf99", "competition_source_list": ["2014年全国华杯赛小学中年级竞赛决赛第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "王老师共买了$$53$$支铅笔，分给了$$A$$，$$B$$，$$C$$，$$D$$四个同学，分到最多的与最少的铅笔数相差不到$$5$$支，如果$$B$$把分到的铅笔全都给$$A$$，那么$$A$$的铅笔数是$$C$$的$$2$$倍；如果$$B$$把分到的铅笔全都给$$C$$，那么$$C$$的铅笔数是$$D$$的$$2$$倍，由此可知，$$B$$分到~\\uline{~~~~~~~~~~}~支铅笔． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["设$$A$$，$$B$$，$$C$$，$$D$$分到的铅笔数分别是$$A$$，$$B$$，$$C$$，$$D$$，由$$B+C=2D$$，知$$C$$、$$D$$、$$B$$依次成等差数列，设公差为$$K$$；由$$A+B=2C$$，知$$A$$、$$C$$、$$B$$依次成等差数列，则公差为$$2K$$；由$$4$$人铅笔数相差不会超过$$4$$，所以$$K=0$$或$$1$$；若$$K=0$$，则$$4\\times B=53$$，但$$53$$不是$$4$$的整数倍； 若$$K=1$$，$$A\\textgreater C\\textgreater D\\textgreater B$$，则$$4\\times C-1=53$$，但$$54$$不是$$4$$的整数倍． 综上所述，$$B$$分到$$15$$支铅笔． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3079", "queId": "bd7b8396c5c14e348a2b6530148fb6c6", "competition_source_list": ["2013年第11届全国创新杯小学高年级五年级竞赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一列数，第$$1$$个数是$$22$$，第$$2$$个数是$$12$$，从第$$3$$个开始，每个数是它前面两个数的平均数，这列数的第$$10$$个数的整数部分（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["第$$3$$个数为$$17$$，第$$4$$个数为$$14.5$$，第$$5$$个数为$$15.75$$，第$$6$$个数为$$15.125\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot $$整数部分一直都是$$15$$，所以第$$10$$个数的整数部分为$$15$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3330", "queId": "f1be9adc45a443599a736ef0a818c717", "competition_source_list": ["2020年第24届YMO四年级竞赛决赛第3题3分", "2019年第24届YMO四年级竞赛决赛第3题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第3题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第3题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$1$$至$$2020$$这$$2020$$个自然数中，每次取两个不同的数相加，和大于$$2020$$的取法共有． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1010000$$ "}], [{"aoVal": "B", "content": "$$1020000$$ "}], [{"aoVal": "C", "content": "$$1020100$$ "}], [{"aoVal": "D", "content": "$$1020200$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["若取了$$2020$$，则另一个可取$$1\\sim 2019$$，共$$2019$$种， 若取了$$2019$$，则另一个可取$$2\\sim 2018$$，共$$2017$$种， 若取了$$2018$$，则另一个可取$$3\\sim 2017$$，共$$2015$$种， $$\\vdots $$ 若取了$$1011$$，则另一个可取$$1010$$，共$$1$$种， 共有$$1+3+5+7+\\cdots +2019={{1010}^{2}}=1020100$$种． 选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2622", "queId": "6330c0fc4e924c049ee18c1b889cecb9", "competition_source_list": ["五年级下学期其它第16讲浓度问题", "2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一容器内装有$$10$$升纯酒精，倒出$$2.5$$升后，用水加满，再倒出$$5$$升，再用水加满，这时容器内的溶液的浓度是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "37.5\\% "}], [{"aoVal": "B", "content": "40\\% "}], [{"aoVal": "C", "content": "62.5\\% "}], [{"aoVal": "D", "content": "50\\% "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->公式类运算->连续自然数非从1的平方需要补全再计算", "课内体系->能力->逻辑分析"], "answer_analysis": ["第二次倒出的纯酒精为$$5\\times \\frac{10-2.5}{10}=3.75$$ 所以两次共倒出纯酒精为$$2.5+3.75=6.25$$（升）， 此时容器内的溶液浓度为$$\\left( 10-6.25 \\right)\\div 10\\times 100 \\%=37.5 \\%$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2409", "queId": "0fc24db754a1473eb650ce557623dc7a", "competition_source_list": ["2018年福建福州河仁杯五年级竞赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "（单选题）如果$$2\\oplus 3=2+22+222$$，$$3\\oplus 2=3+33$$，那么$$4\\oplus 5$$等于． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5555$$ "}], [{"aoVal": "B", "content": "$$6170$$ "}], [{"aoVal": "C", "content": "$$44444$$ "}], [{"aoVal": "D", "content": "$$49380$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["根据题意，已知$$2\\oplus 3=2+22+222$$，$$3\\oplus 2=3+33$$，即用$$\\oplus $$前面的数加上十位个位都是这个数的两位数加上千位十位个位都是这个数的三位数，$$\\oplus $$后面是几就加几个数；据此求出$$4\\oplus 5$$． $$4\\oplus 5=4+44+444+4444+44444=49380$$． 故答案为：$$49380$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1450", "queId": "75cd72f3a4944a46974c9f7785833ee4", "competition_source_list": ["2012年IMAS小学高年级竞赛第一轮检测试题第17题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "姐姐今年$$12$$岁，妹妹今年$$8$$岁，弟弟今年$$3$$岁，他们的生日恰好是同一天．当三人的年龄和为$$50$$岁时，妹妹为多少岁？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$17$$ "}], [{"aoVal": "E", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["今年他们三人的年龄之和为$$23$$岁，当他们的年龄之和为$$50$$岁时，已过了$$(50-23)\\div 3=9$$（年），所以这时妹妹为$$8+9=17$$（岁）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1204", "queId": "6b7d26127e7e48d687ca1771204c5f54", "competition_source_list": ["2020年第24届YMO三年级竞赛决赛第5题3分", "2019年第24届YMO三年级竞赛决赛第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个正方形的操场，在它的外面一圈插上小红旗，四个角上都插一面小红旗，每边都插$$16$$面，一共插了面小红旗． ", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$58$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$62$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["根据题意分析可知，四个角都插上一面小红旗，每边都插$$16$$面， 所以除了端点，每一边有小红旗：$$16-2=14$$（面）， 因为是正方形，每边都有$$14$$面，再加上四个端点，那么一共插了：$$14\\times4+4=60$$（面）红旗， 故答案为：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "944", "queId": "c664095a04dd437ca5c1ce67d4506b99", "competition_source_list": ["2013年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一个四位数，各位数字互不相同，所有数字之和等于$$6$$，并且这个数是$$11$$的倍数，则满足这种要求的四位数共有( )个。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->位值原理运用->位值原理的综合应用"], "answer_analysis": ["解：由题意，组成四位数的四个数字分别为$$0$$、$$1$$、$$2$$、$$3$$，并且这个数是$$11$$的倍数，则奇数位上的数字和等于偶数位上的数字和，等于$$3$$。符合条件的四位数有$$3102$$、$$3201$$、$$1320$$、$$1023$$、$$2310$$、$$2013$$，共$$6$$个。 故选：A。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1441", "queId": "3f849783c7be4a07b0663d5a531635b6", "competition_source_list": ["2020年春蕾杯六年级竞赛第9题2分", "2021年春蕾杯六年级竞赛第4题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一杯$$100$$克盐水，盐和水的比是$$1:4$$，商家为了降低成本，往里面加了水，现在盐和盐水的比为$$1:10$$．加了克的水． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$90$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["$$100$$克盐水，盐比水为$$1:4$$，则盐的含量为：$$100\\times \\frac{1}{5} =20$$ （克）． 加水后，盐的含量不变，而盐比盐水为$$1:10$$，则盐水的总质量为$$20\\times 10=200$$（克），增加的水为：$$200-100=100$$（克）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1103", "queId": "0c5d3e6d7c454f129b24251586ffa23d", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（四）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "某年的二月份有$$4$$个星期一和$$5$$个星期二，则$$2$$月$$15$$号是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["因为这年的二月份有$$5$$个星期二，说明这是一闰年，且$$2$$月$$1$$号是星期二，$$15\\div 7=2$$（周）$$\\ldots \\ldots 1$$（天），所以$$2$$月$$15$$号也是星期二． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1225", "queId": "5917ed654f034251818421fa0f1ea64d", "competition_source_list": ["2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "同学们一起去划船，但公园船不够多，如果每船坐$$4$$人，会多出$$10$$人；如果每船坐$$5$$人，还会多出$$1$$人，共有（ ）人去划船。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$46$$ "}], [{"aoVal": "C", "content": "$$51$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->盈亏问题->盈亏基本类型->盈盈问题"], "answer_analysis": ["解：$$\\left( 10-1 \\right)\\div \\left( 5-4 \\right)$$ $$=9\\div 1$$ $$=9$$（条） $$4\\times 9+10$$ $$=36+10$$ $$=46$$（人） 答：共有$$46$$人去划船。 故选:B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1816", "queId": "7870de5c94144334bb19d26fc6f6408c", "competition_source_list": ["2020年长江杯五年级竞赛复赛B卷第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "有一篮苹果，第一天吃了总数的一半又半个，第二天吃了余下的一半又半个，第三天吃了第二天余下的一半又半个，这时篮子里只剩下一个苹果了．原来篮子里的苹果有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$21$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$16.5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据最后篮内的苹果个数是$$1$$，那第二天吃完后余下的苹果的个数是$$2\\times (1+0.5)$$，第一天吃完后余下的苹果的个数是$$2\\times [2\\times (1+0.5)+0.5]$$，同样道理可以求出原来苹果的个数， 第二天吃完后余下的苹果的个数是��$$2\\times (1+0.5)=3$$（个）， 第一天吃完后余下的苹果的个数是：$$2\\times (3+0.5)=7$$（个）， 原有苹果的个数是：$$2\\times (7+0.5)=2\\times 7.5=15$$（个）． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "626", "queId": "7de53ed0f9864ebe819a873efb8bd5fc", "competition_source_list": ["2017年江苏南京小升初三十六计之26第3题", "2011年北京五年级竞赛"], "difficulty": "4", "qtype": "single_choice", "problem": "有一类各位数字各不相同的五位数$$M$$，它的千位数字比左右两个数字大，十位数字也比左右两位数字大．另有一类各位数字各不相同的五位数$$W$$，它的千位数字比左右两个数字小，十位数字也比左右两位数字小．请问符合要求的数$$M$$与$$W$$，哪一类的个数多:~\\uline{~~~~~~~~~~}~,多~\\uline{~~~~~~~~~~}~个 ", "answer_option_list": [[{"aoVal": "A", "content": "M；630 "}], [{"aoVal": "B", "content": "M；126 "}], [{"aoVal": "C", "content": "W；630 "}], [{"aoVal": "D", "content": "W；126 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$M$$与$$W$$都是五位数，都有千位和十位与其它数位的大小关系，所以两类数有一定的对应关系．比如有一个符合要求的五位数$$M=\\overline{ABCDE}$$($$A$$不为$$0)$$，那么就有一个与之相反并对应的五位数$$\\overline{(9-A)(9-B)(9-C)(9-D)(9-E)}$$必属于$$4$$类，比如$$13254$$为$$M$$类，则与之对应的$$86754$$为$$W$$类． 所以对于$$M$$类的每一个数，$$n-1$$类都有一个数与之对应．但是两类数的个数不是一样多，因为$$M$$类中$$0$$不能做首位，而$$W$$类中$$9$$可以做首位．所以$$W$$类的数比$$M$$类的数要多，多的就是就是首位为$$\\underbrace{{{a}_{n-1}}+{{a}_{n}}=3\\times 3\\times \\ldots \\times3}_{(n-1)3}={{3}^{n-1}}$$的符合要求的数． 计算首位为$${{a}_{1}}=0$$的$$W$$类的数的个数，首先要确定另外四个数，因为要求各不相同，从除$$9$$外的其它$$9$$个数字中选出$$4$$个，有$$C_{9}^{4}=126$$种选法． 对于每一种选法选出来的$$4$$个数，假设其大小关系为$$5$$，由于其中最小的数只能在千位和十位上，最大的数只能在百位和个位上，所以符合要求的数有$$60$$类：①千位、十位排$${{A}_{1}}$$、$${{A}_{2}}$$，有两种方法，百位、十位排$${{A}_{3}}$$、$${{A}_{4}}$$，也有两种方法，故此时共有$$3$$种；②千位、十位排$$0$$、$${{A}_{3}}$$，只能是千位$${{A}_{3}}$$，百位$${{A}_{4}}$$，十位$$3$$，个位$$6$$，只有$$3$$种方法． 根据乘法原理，首位为$$9$$的$$W$$类的数有$$126\\times \\left( 4+1 \\right)=630$$个． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3021", "queId": "e5456531473c462c8368276ee8ba86a2", "competition_source_list": ["2018年IMAS小学高年级竞赛（第二轮）第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$、$$b$$、$$c$$、$$d$$是不为$$0$$且互不相同的数码﹐如果 $$\\overline{ab}+ \\overline{cd}= \\overline{dc}+ \\overline{ba}$$ ﹐则称这个等式为回文式﹐而能写成回文式的两个数则称为一对回文数﹐例如∶$$54+12=21+45=66$$﹐一对回文数的和称为回文和．请问最小的回文和是什么？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$22$$ "}], [{"aoVal": "B", "content": "$$33$$ "}], [{"aoVal": "C", "content": "$$44$$ "}], [{"aoVal": "D", "content": "$$55$$ "}], [{"aoVal": "E", "content": "$$99$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["由$$\\overline{ab}+\\overline{cd}=\\overline{dc}+\\overline{ba}$$知$$10(a+c)+(b+d)=10(b+d)+(a+c)$$， 故$$a+c=b+d$$． 由于能用两种方式表示成两个不同正整数之和的最小数为$$5=1+4=2+3$$﹐ 故回文和的最小值为$$55$$﹐其中一个例子为$$12+43=34+21=55$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1380", "queId": "35ee4e061d1d4688bf1e6f6efd4a4dd9", "competition_source_list": ["2020年亚洲国际数学奥林匹克公开赛（AIMO）四年级竞赛决赛第29题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$4$$ years ago, the sum of the ages of Amy and Bella is $$13$$ years old. The age of Bella after $$17$$ years is $$21$$ years older than the age of Chris $$19$$ years before. The age of Chris now is $$5$$ times the age of Amy. How old is Bella now? $$4$$年前，小艾和小贝的年龄加起来是$$13$$岁．小贝$$17$$年后的年龄比小克$$19$$年前的年龄大$$21$$岁．小克现在的年龄是小艾的$$5$$倍．小贝现在~\\uline{~~~~~~~~~~}~岁. ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$15$$ "}], [{"aoVal": "E", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "Overseas Competition->知识点->应用题模块->年龄问题->年龄问题之差倍型"], "answer_analysis": ["小艾和小贝现在的年龄加起来是$$13+4\\times 2=21$$（岁）， 小贝现在的年龄比小克小$$17+19-21=15$$（岁）， 所以小克和小艾现在的年龄加起来是$$21+15=36$$（岁）， 小艾现在$$36\\div \\left( 5+1 \\right)=6$$（岁），小贝现在$$21-6=15$$（岁）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2094", "queId": "dde1e86a4979445582e940394e5aa3ee", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第13题", "2018年四川成都锦江区四川师范大学附属第一实验中学小升初（四）第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "同学们一起去划船，但公园船不够多，如果每船坐$$4$$人，会多出$$10$$人；如果每船坐$$5$$人，还会多出$$1$$人，共有人去划船． ", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$46$$ "}], [{"aoVal": "C", "content": "$$51$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->盈亏问题->盈亏基本类型->盈亏基本类型盈亏问题"], "answer_analysis": ["盈盈类问题：共有$$(10-1)\\div (5-4)=9$$（只）船，共有$$4\\times 9+10=46$$（人）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3072", "queId": "e61efffae310499f955b976a41e61bed", "competition_source_list": ["2006年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "一个数的小数点向右移动一位，比原数大59.94，这个数是( ) ", "answer_option_list": [[{"aoVal": "A", "content": "6.66 "}], [{"aoVal": "B", "content": "11.66 "}], [{"aoVal": "C", "content": "66.6 "}], [{"aoVal": "D", "content": "116.6 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->差倍->两量差倍->暗倍型二量差倍问题"], "answer_analysis": ["小数点向右移动1位，则新数比原数扩大10倍，即增加9倍，所以原数为$$59.94\\div \\left( 10-1 \\right)=6.66$$，选A. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "352", "queId": "898de766850d4c81a7a5520b08e2b319", "competition_source_list": ["2015年全国中环杯二年级竞赛决赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个盒子里有一些大小相同的球，其中白的有$$8$$个，黑的有$$9$$个。不许看球，每次拿一个不放回，至少拿~\\uline{~~~~~~~~~~}~次才能保证拿到黑球。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["要保证拿到黑球，考虑倒霉的情况．最倒霉的是拿出的全部是白色的球，那么这是$$8$$次，接着只剩黑球时，下一次一定能拿出黑球，所以至少拿$$8+1=9$$（次）才能保证拿到黑球． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "42", "queId": "0ab3832a26a24e43b92b6f8e9d94a2bb", "competition_source_list": ["2017年北京学而思杯四年级竞赛年度教学质量测评第20题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "有一条长度为$$8$$的绳子，将它对折$$3$$次，用剪刀从正中间剪开，得到一些短绳子．那么长度为$$1$$的绳子有（~ ）段． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$~~~~~ "}], [{"aoVal": "B", "content": "$$6$$~~~~~ "}], [{"aoVal": "C", "content": "$$7$$~~~~~ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->归纳递推"], "answer_analysis": ["略 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2100", "queId": "d4f635d4d0de45f091ba7eac5ecb8d58", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（一）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "商店以每副$$30$$元的价格购进一批乒乓球拍，又以每副$$40$$元的价格售出，当剩下$$80$$副时，除已收回购进这批乒乓球拍所用的钱之外，还赚了$$100$$元．这批乒乓球拍共有副． ", "answer_option_list": [[{"aoVal": "A", "content": "$$320$$ "}], [{"aoVal": "B", "content": "$$330$$ "}], [{"aoVal": "C", "content": "$$300$$ "}], [{"aoVal": "D", "content": "$$350$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->盈亏问题->盈亏基本类型->盈亏基本类型盈亏问题"], "answer_analysis": ["假设全部卖出，总盈利为$$40\\times 80+100=3300$$（元），又每副盈利$$40-30=10$$（元），因此这批兵乓球拍共有$$3300\\div 10=330$$（副）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2895", "queId": "6f5abb86d44c410a8b341de1abfea71a", "competition_source_list": ["2016年创新杯五年级竞赛训练题（二）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "循环小数$$0.\\dot{1}99251\\dot{7}$$与$$0.\\dot{3}456\\dot{7}$$．这两个循环小数在小数点后第（~ ）位，首次同时出现在该位中的数字都是$$7$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$35$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->循环小数->循环小数的周期问题"], "answer_analysis": ["$$0.\\dot{1}99251\\dot{7}$$的小数部分$$7$$出现在小数点后的第$$7$$位、第$$14$$位、第$$21$$位$$\\cdots $$ $$0.\\dot{3}456\\dot{7}$$的小数部分$$7$$出现在小数点后的第$$5$$位、第$$10$$位、第$$15$$位$$\\cdots $$ 由于$$\\left[ 5,7 \\right]=35$$，故两个循环小数在小数点后第$$35$$位，首次同时出现在该位中的数字都是$$7$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2612", "queId": "deba86d3d0584d7e9361c3610e342a5c", "competition_source_list": ["2011年全国世奥赛五年级竞赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "在循环小数$$9.617628\\dot{1}$$的某一位上再添一个循环点，使所产生的循环小数尽可能大，则得到的最大的新循环小数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9.61762{8}\\dot{1}$$ "}], [{"aoVal": "B", "content": "$$9.617\\dot62{8}\\dot{1}$$ "}], [{"aoVal": "C", "content": "$$9.6176\\dot2{8}\\dot{1}$$ "}], [{"aoVal": "D", "content": "$$9.61762\\dot{8}\\dot{1}$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->小数->循环小数->循环小数的概念"], "answer_analysis": ["无论在何处添循环节，前八位不变，为$$9.6176281$$ 从下一位开始最大即可，因此在$$8$$上方添循环点． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "480", "queId": "cec8977b09ea4551bc1007bc6c1411d1", "competition_source_list": ["2011年第7届全国新希望杯五年级竞赛A卷第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "杨老师、龙老师、郭老师各带$$1$$名学生参加希望之星数学夏令营，活动期间，六人进行了乒乓球双打友谊赛，规定师生不能搭档． 第一局：杨老师和小刚对龙老师和小王；第二局：龙老师带小雨对杨老师和郭老师的学生．那么郭老师的学生是（~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "小刚 "}], [{"aoVal": "B", "content": "小王~ "}], [{"aoVal": "C", "content": "小雨 "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["龙老师与小王和小丽都搭档过，所以龙老师的学生是小刚．从第二局比赛可以看出郭老师的学生不是小丽，所以郭老师的学生是小王． ~ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "356", "queId": "7322c463fce047889e25a618f14216f6", "competition_source_list": ["2014年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在下列四个算式中：$$\\overline{AB}\\div \\overline{CD}=2$$，$$E\\times F=0$$，$$G-H=1$$，$$I+J=4$$，$$A\\sim J$$代表$$0\\sim 9$$中的不同数字，那么两位数$$\\overline{AB}$$不可能是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$54$$ "}], [{"aoVal": "B", "content": "$$58$$ "}], [{"aoVal": "C", "content": "$$92$$ "}], [{"aoVal": "D", "content": "$$96$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜"], "answer_analysis": ["解：由条件可知：$$E$$、$$F$$中至少有一个为$$0$$，假设$$E$$为$$0$$，另一个可以是任何数。$$I$$和$$J$$有一个是$$3$$，有一个是$$1$$，那么$$0\\sim 9$$中的数字还剩下$$2$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$。 因为$$G-H=1$$， $$G$$、$$H$$是$$9$$，$$8$$时，则$$54\\div 27=2$$，此时$$F=6$$； $$G$$、$$H$$是$$8$$，$$7$$时，则$$92\\div 46=2$$，此时$$F=5$$； $$G$$、$$H$$是$$7$$，$$6$$时，则$$58\\div 29=2$$，此时$$F=4$$； $$G$$、$$H$$是$$6$$，$$5$$时，此时不满足条件； $$G$$、$$H$$是$$5$$，$$4$$时，此时不满足条件。 所以两位数$$\\overline{AB}$$可能是$$54$$、$$58$$、$$92$$，不可能是$$96$$。 故选：D "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2486", "queId": "7e007883bd5741da9519731eed95d847", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第12题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "请问算式$$20\\times 30\\times 40\\times 50$$的乘积末尾有多少个连续的零？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}], [{"aoVal": "E", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数乘除->整数乘法运算->表外乘法计算"], "answer_analysis": ["解法一：$$20\\times 30\\times 40\\times 50=1200000$$．乘积末尾共$$5$$个零． 解法二：$$20$$、$$30$$、$$40$$、$$50$$末尾各有一个零，而$$2$$乘$$5$$又会产生一个零，故末尾共有$$5$$个零． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "983", "queId": "008e465193634d95b732a805020dda3a", "competition_source_list": ["2007年四年级竞赛创新杯", "2007年第5届创新杯四年级竞赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "四年级（1）班50名同学帮助老师把20捆教科书搬到200米外的图书馆，两人抬一捆，大家轮流休息，平均每人抬( )米. ", "answer_option_list": [[{"aoVal": "A", "content": "80 "}], [{"aoVal": "B", "content": "160 "}], [{"aoVal": "C", "content": "180 "}], [{"aoVal": "D", "content": "200 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->归总问题"], "answer_analysis": ["$$200\\times 20\\div 25\\text{=}160$$米 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2321", "queId": "dd4032d908dc482cb0f84cf90879a7aa", "competition_source_list": ["2011年六年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "钟表上$$12$$点$$15$$分，时针与分针夹角为（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$${{90}^{{}^{}\\circ }}$$ "}], [{"aoVal": "B", "content": "$${{82.5}^{{}^{}\\circ }}$$ "}], [{"aoVal": "C", "content": "$${{67.5}^{{}^{}\\circ }}$$ "}], [{"aoVal": "D", "content": "$${{60}^{{}^{}\\circ }}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->时钟问题->时钟上的追及问题->已知时间求角度"], "answer_analysis": ["$$12$$点时，夹角为$${{0}^{{}^{}\\circ }}$$；$$15$$分钟后，分针往前走了$${{90}^{{}^{}\\circ }}$$，时针走了$${{7.5}^{{}^{}\\circ }}$$，此时夹角为$${{82.5}^{{}^{}\\circ }}$$。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3131", "queId": "fa1b19e129cf475c86e750f68fd57454", "competition_source_list": ["2018年福建福州河仁杯五年级竞赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "（单选题）如果$$2\\oplus 3=2+22+222$$，$$3\\oplus 2=3+33$$，那么$$4\\oplus 5$$等于．改编单选题 ", "answer_option_list": [[{"aoVal": "A", "content": "$$5555$$ "}], [{"aoVal": "B", "content": "$$6170$$ "}], [{"aoVal": "C", "content": "$$44444$$ "}], [{"aoVal": "D", "content": "$$49380$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["根据题意，已知$$2\\oplus 3=2+22+222$$，$$3\\oplus 2=3+33$$，即用$$\\oplus $$前面的数加上十位个位都是这个数的两位数加上千位十位个位都是这个数的三位数，$$\\oplus $$后面是几就加几个数；据此求出$$4\\oplus 5$$． $$4\\oplus 5=4+44+444+4444+44444=49380$$． 故答案为：$$49380$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "633", "queId": "41ee5493dcf24aa88c8e0ebc4b126da4", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛B卷第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "某创客小组突发奇想，以每$$12$$摄氏度为$$1$$个温度单位，零下$$24$$摄氏度记作$$0$$，那么$$96$$摄氏度应记为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["这是新定义的问题，主要是要读懂��意， 对于这道题，以每$$12$$摄氏度为$$1$$个温度单位，零下$$24$$摄氏度记作$$0$$， 那么$$96$$摄氏度到零下$$24$$摄氏度一共有： $$96-\\left( -24 \\right)=120$$（摄氏度）， $$120\\div 12=10$$， 那么$$96$$摄氏度应记为$$10$$， 也就是选择$$\\text{B}$$选项． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1596", "queId": "4e14dba856b04f52905f13c04443355b", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第5题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "阳光小学派一些学生去搬树苗，如果每人搬$$6$$棵，则差$$4$$棵，如果每人搬$$8$$棵，则差$$18$$棵，这批树苗有棵． ", "answer_option_list": [[{"aoVal": "A", "content": "$$26$$ "}], [{"aoVal": "B", "content": "$$38$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$56$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["从每人差$$8-6=2$$棵，就导致总数从差$$4$$棵到差$$18$$棵，总共差了$$18-4=14$$棵，所以$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde(18-4)\\div (8-6)$$， $$=14\\div 2$$， $$=7$$（人）， $$6\\times 7-4=38$$（棵）， 答：这批树苗有$$38$$棵． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "332", "queId": "ff80808148880257014888b117d50a1b", "competition_source_list": ["2014年全国华杯赛小学高年级竞赛初赛B卷第1题", "2014年全国华杯赛小学高年级竞赛初赛A卷第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "平面上的四条直线将平面分割成八个部分，则这四条直线中至多有（~ ~ ~ ~）条直线互相平行． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->能力->创新思维"], "answer_analysis": ["这是一道考前公开题．当四条直线相互平行的时候把平面分成五个部分，当三条直线平行，另一条直线与它们相交的时候四条直线恰好把平面分成八个部分．所以选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1127", "queId": "6b3b9936bb46492e8dbc0d0a5ecdb150", "competition_source_list": ["2020年新希望杯五年级竞赛第11题", "2020年希望杯五年级竞赛模拟第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "【$$2020$$五年级卷第$$11$$题】大头儿子给小头爸爸计算每天的交通费用．大头儿子说：``您的车百公里油耗为$$8$$升，每天上下班共行驶$$30$$公里，油价按每升$$7.2$$元计算，您每天上下班需要支付的油费为元．'' ", "answer_option_list": [[{"aoVal": "A", "content": "$$27$$ "}], [{"aoVal": "B", "content": "$$17.28$$ "}], [{"aoVal": "C", "content": "$$1.92$$ "}], [{"aoVal": "D", "content": "$$19.2$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->先归一再归总"], "answer_analysis": ["因为由题干可知，车百公里油耗为$$8$$升，每天上下班共行驶$$30$$公里，则上下班共耗油$$30\\div 100\\times 8=2.4$$（升），油价按每升$$7.2$$元计算，所以每天上下班需要支付的油费为$$7.2\\times 2.4=17.28$$（元）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1695", "queId": "91857126ecd94685995eb1f85b0cc1a9", "competition_source_list": ["2016年北京学而思杯六年级竞赛第14题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$4$$个小于$$60$$的合数，且两两互质，那么这$$4$$个合数之和的最大值为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$217$$ "}], [{"aoVal": "B", "content": "$$219$$ "}], [{"aoVal": "C", "content": "$$221$$ "}], [{"aoVal": "D", "content": "$$223$$ "}], [{"aoVal": "E", "content": "$$225$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["$$4$$个合数之和最大，需要每个合数都尽量大，小于$$60$$的合数最大是$$58$$， 且$$58=2\\times 29$$，因为$$4$$个合数要两两互质， 所以接下来找的合数不能含有质因数$$2$$和$$29$$， 可以依次找到符合要求且最大的合数分别为$$58=2\\times 29$$，$$57=3\\times 19$$，$$55=5\\times 11$$，$$49=7\\times 7$$， 所以这$$4$$个合数之和最大为$$58+57+55+49=219$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1313", "queId": "1fcc4932976945b5b37244d854b76dd4", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小东看叔叔锯木头，锯一次要$$3$$分钟，最后叔叔把木头锯成$$4$$段，叔叔请小东算算，他一共锯了分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["锯成$$4$$段，则需要锯$$3$$次，则所花时间为：$$3+3+3=9$$（分）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "744", "queId": "37a42e492cbb4c6d81d8d4f942be0375", "competition_source_list": ["2020年长江杯五年级竞赛复赛A卷", "2020年长江杯五年级竞赛复赛A卷第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个$$200$$位的自然数，每个数位上的数字都是$$8$$，这个数除以$$7$$后的余数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$7$$的整除特征是：从右往左三位一段，奇数段减偶数段的差能够被$$7$$整除， 则这个数就能够被$$7$$整除． 因此我们可以得到$$888888$$一定可以被$$7$$整除．（$$888888\\div 7=126984$$）这个自然数有$$200$$个$$8$$， $$200\\div 6=33$$（组）$$\\cdots \\cdots 2$$（个）． $$88\\div 7=12\\cdots \\cdots 4$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3232", "queId": "507005d5a3d24c698d23dfa9abe12afa", "competition_source_list": ["2007年第5届创新杯四年级竞赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "五位老朋友$$A$$，$$B$$，$$C$$，$$D$$，$$E$$在公园聚会，见面时候握手致意间侯，已知：$$A$$握了$$4$$次，$$B$$握了$$1$$次，$$C$$握了$$3$$次，$$D$$握了$$2$$次，到此为止，$$E$$握了次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$A$$握了$$4$$次手，即分别与$$B$$、$$C$$、$$D$$、$$E$$各握了一次，此时$$E$$握了$$1$$次， 由于$$B$$握了$$1$$次手，只能和$$A$$握过，和$$E$$没有握手， $$C$$握了$$3$$次手，和$$B$$没握，所以$$C$$是与$$A$$、$$D$$、$$E$$握的；此时$$E$$又握了$$1$$次， 此时$$D$$已与$$A$$、$$C$$握了$$2$$次手，和$$E$$没有握手， 所以此时$$E$$也握了$$2$$次，即是与$$A$$、$$C$$握的手． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2511", "queId": "30252a1dde2b4ce08974a962b8c0e024", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第2题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "在算式$$ +5=13-2$$中，括号中应填入什么数才能使算式成立？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数减法运算->减法横式"], "answer_analysis": ["13-2=11；11-5=6 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1391", "queId": "87d77837ef244186b92c8ab3ec9a6407", "competition_source_list": ["2016年全国世奥赛竞赛A卷第1题", "2016年第16届世奥赛六年级竞赛决赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "熊大和熊二分一块蛋糕，熊二因为分的太少而哭闹，于是熊大把自己蛋糕的$$\\frac{1}{3}$$给了熊二，使熊二的蛋糕增加到原来的$$3$$倍．那么，最终熊二分得整块蛋糕的（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{7}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{7}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["找对数量之间的关系是本题的关键，熊大把自己蛋糕的$$\\frac{1}{3}$$给了熊二，使得熊二的蛋糕增加到原来的三倍，即熊大给熊二的相当于熊二原来的$$2$$倍．假设熊二原来分得了$$1$$份，那么熊大原来分得的$$\\frac{1}{3}$$就是$$2$$份，即熊大原来分得$$6$$份．所以熊二原来分得整块蛋糕的$$\\frac{1}{7}$$，最终熊二分的蛋糕的$$\\frac{3}{7}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2848", "queId": "7770e130039f42afab97cec8a3aefd70", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第2题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "在算式$$（）+5=13-2$$中，括号中应填入什么数才能使算式成立？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数减法运算->减法横式"], "answer_analysis": ["13-2=11；11-5=6 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2440", "queId": "5413bc5a8f8e4ef2b041908b21fa10fb", "competition_source_list": ["2011年第7届全国新希望杯小学高年级六年级竞赛第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "下列选择中，正确的是（~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$-7\\textgreater-3.27$$ "}], [{"aoVal": "B", "content": "$$0\\textless{}-11$$ "}], [{"aoVal": "C", "content": "$$3\\textgreater-27$$ "}], [{"aoVal": "D", "content": "$$-7=7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["正数大于负数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "35", "queId": "4a7dfea5e1404eb8aeb1d73c9db9e8a5", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第10题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "明明、亮亮和刚刚三个好朋友的爸爸，一位是工人，一位是医生，一位是解放军战士．请你根据下面三句话，猜一猜的爸爸是解放军． （$$1$$）明明的爸爸不是工人． （$$2$$）亮亮的爸爸不是医生． （$$3$$）明明的爸爸和亮亮的爸爸正听一位当解放军的爸爸讲战斗故事． ", "answer_option_list": [[{"aoVal": "A", "content": "明明 "}], [{"aoVal": "B", "content": "亮亮 "}], [{"aoVal": "C", "content": "刚刚 "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据第（$$3$$）句话可知，明明的爸爸和亮亮的爸爸都不是解放军，所以刚刚的爸爸是解放军． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2567", "queId": "23eda4fa773d446ba5344eea41a64f2b", "competition_source_list": ["2008年第6届创新杯五年级竞赛初赛B卷第7题5分", "2008年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "小红、小军和小明三人参加一次数学竞赛，一共有$$100$$道题，每个人各解出其中的$$60$$道题，有些题目三人全都解出来了，我们称之为``容易题''，有些题目只有两人解出来我们称之为``中等题''，有些题目只有一人解出来，我们称之为``难题''，每个题都至少被他们中的一人解出，则难题比容易题多道题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->不定方程->不定方程组"], "answer_analysis": ["设容易题、中等题和难题分别有$$x$$，$$y$$，$$z$$道题，则 $$\\begin{cases} x+y+z=100\\textasciitilde\\textasciitilde\\textasciitilde① 3x+2y+z=180\\textasciitilde\\textasciitilde\\textasciitilde② \\end{cases}$$， $$2\\times ①-②$$可以得到$$z-x=20$$，所以难题比容易题多$$20$$道题． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1378", "queId": "3ec35bf983a748eb9d918bf94f6819b2", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "小刚、小强、小明一起买文具，小刚买$$3$$枝铅笔和$$5$$枝圆珠笔共用$$5.4$$元，小强买$$5$$枝铅笔和$$7$$枝圆珠笔共用$$7.8$$元，小明买$$4$$枝铅笔和$$4$$枝圆珠笔共用元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3.4$$ "}], [{"aoVal": "B", "content": "$$6.6$$ "}], [{"aoVal": "C", "content": "$$6.0$$ "}], [{"aoVal": "D", "content": "$$4.8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->多元一次方程解应用题->整数系数二元一次方程组解应用题"], "answer_analysis": ["设每支铅笔$$x$$元，每支圆珠笔$$y$$元，则$$3x+5y=5.4$$，$$5x+7y=7.8$$，得$$2x+2y=2.4$$，所以$$4x+4y=4.8$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "499", "queId": "afc71f39920c4726a6e9af79aa45fc5e", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个长方形的长和宽长度都是整数，如果它的周长是$$22$$，那么，这长方形面积的最大值是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->和一定型最值问题->两数之积的最值"], "answer_analysis": ["长$$+$$宽$$=22\\div 2=11$$．由于长和宽都是整数，欲求面积的最大值，则长和宽越接近越好，即长$$=6$$，宽$$=5$$，故面积为：$$5\\times 6=30$$．所以选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "641", "queId": "38f86964f61845b385ed1a0c6ac0c1c5", "competition_source_list": ["2011年五年级竞赛明心奥数挑战赛"], "difficulty": "1", "qtype": "single_choice", "problem": "$$42$$的因数共有（ ）个。 ", "answer_option_list": [[{"aoVal": "A", "content": "3 "}], [{"aoVal": "B", "content": "6 "}], [{"aoVal": "C", "content": "8 "}], [{"aoVal": "D", "content": "10 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理"], "answer_analysis": ["方法一：枚举法。$$42$$的因数有$$1$$、$$2$$、$$3$$、$$6$$、$$7$$、$$14$$、$$21$$、$$42$$。共$$8$$个。 方法二：$$42=2\\times 3\\times 7$$，因数个数等于指数加$$1$$再连乘，$$\\left( 1+1\\right) \\times\\left(1+1\\right) \\times\\left(1+1\\right) =8$$个。故选C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3424", "queId": "9cedf28367094132b54d3658a92c82d1", "competition_source_list": ["2011年五年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "将各位数字的和为$$4$$的所有自然数由小到大排成一列，$$2011$$是第( )个。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$23$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$25$$ "}], [{"aoVal": "D", "content": "$$27$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["枚举法。各位数字和为$$4$$的自然数依次为：$$4\\text{，}13\\text{，}22\\text{，}31\\text{，}40\\text{，}103\\text{，}112\\text{，}121\\text{，}130\\text{，}202\\text{，}211\\text{，}220\\text{，}301\\text{，}310\\text{，}400\\text{，}1003\\text{，}1012\\text{，}1021\\text{，}1030\\text{，}1102\\text{，}1111\\text{，}1120\\text{，}1201\\text{，}1210\\text{，}1300\\text{，}2002\\text{，}2011$$，即$$2011$$是第$$27$$个。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "556", "queId": "0e26e3e575ba460d96651a1532c2ebf5", "competition_source_list": ["2011年六年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "符号$$\\left[ x \\right]$$表示不大于$$x$$的最大整数，例如$$\\left[ 5 \\right]=5$$、$$\\left[ 6.31 \\right]=6$$。如果$$\\left[ \\frac{3x+7}{7} \\right]=4$$，这样的正整数$$x$$有( ) ", "answer_option_list": [[{"aoVal": "A", "content": "3个 "}], [{"aoVal": "B", "content": "4个 "}], [{"aoVal": "C", "content": "5个 "}], [{"aoVal": "D", "content": "2个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->高斯记号->高斯记号的基本运算"], "answer_analysis": ["根据高斯函数的性质，可得$$4 \\leqslant \\frac{3x+7}{7} \\textless{} 5$$，解得$$ 7 \\leqslant x \\textless{} \\frac{28}{3}$$，对应的整数共$$3$$个数． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "518", "queId": "f8ba1c04c7a747848dca0711ba94c134", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "$$2$$个人同时吃$$2$$个馒头用$$2$$分钟，$$10$$个人同时吃$$10$$个馒头用分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$20$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数学广角->优化->解决问题策略"], "answer_analysis": ["两人同时吃两个馒头的时间与一个人吃一个馒头时间相同，与$$10$$个人同时吃$$10$$个馒头时间不同，都是两分钟． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3307", "queId": "ccac6ecf35254740b3e280eb420f1f43", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第14题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "将一个正整数的各位数码以相反的顺序排列后，若所得的数与原来的数相同，则称这个数为回文数（例如$$909$$与$$1221$$都是回文数）．请问在$$10$$与$$1000$$之间总共有多少个回文数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$90$$ "}], [{"aoVal": "B", "content": "$$99$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$106$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "Overseas Competition->知识点->计数模块->枚举法综合->字典排序法->组数->回文数"], "answer_analysis": ["如果回文数为两位数，则它的个位数码与十位数码必须相同，但十位数码不能为$$0$$， 即共有$$9$$个两位数的回文数． 如果回文数为三位数，则它的个位数码与百位数码必须相同，但百位数码不能为$$0$$，而十位数码可自由选择数码，即共有$$9\\times 10=90$$个三位数的回文数． 而$$1000$$不是回文数，故在$$1$$与$$1000$$之间总共有$$9+90=99$$个回文数． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3438", "queId": "d7de8f193557475085c056751de16bc1", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$10$$个玻璃球分成数量不同的$$3$$堆，共有种不同的分法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["根据题意可得： $$15=1+2+3+9$$；$$15=1+2+4+8$$； $$15=1+2+5+7$$；$$15=1+3+4+7$$； $$15=1+3+5+6$$；$$15=2+3+4+6$$； 一共有$$6$$种． 答：共有$$6$$种不同的分法． 故选：$$\\text{C}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1257", "queId": "5dd8260d1b704801a88c7ec17b59e801", "competition_source_list": ["2017年全国华杯赛小学中年级竞赛初赛模拟第1题", "小学中年级三年级上学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两人在春节一共得$$210$$元压岁钱，甲给乙$$40$$元，还比乙多$$10$$元，那么，原来甲得了~\\uline{~~~~~~~~~~}~元压岁钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$150$$ "}], [{"aoVal": "B", "content": "$$140$$ "}], [{"aoVal": "C", "content": "$$125$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->和差倍问题->和差问题->二量和差问题->明和暗差"], "answer_analysis": ["因为甲给乙$$40$$元，还比乙多$$10$$元，所以甲比乙多$$40\\times 2+10=90$$元， 所以甲：$$(210+90)\\div 2=150$$ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1300", "queId": "fa6fc59a946e4856b6a621ee73c3158e", "competition_source_list": ["2014年全国迎春杯五年级竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "一辆大卡车一次可以装煤$$2.5$$吨，现在要一次运走$$48$$吨煤，那么至少需要辆这样的大卡车． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$48\\div 2.5=19.2$$，所以$$19$$辆不够，至少需要$$20$$辆．选$$\\text{C}$$. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2815", "queId": "c85b9c1f6b5d4e87b057ef688a5c9f8e", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列数中，最接近$$80$$的数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$75$$ "}], [{"aoVal": "B", "content": "$$76$$ "}], [{"aoVal": "C", "content": "$$78$$ "}], [{"aoVal": "D", "content": "$$83$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数减法运算->减法横式"], "answer_analysis": ["对于$$\\text{A}$$选项：$$80-75=5$$；也就是$$75$$与$$80$$的距离是$$5$$； 对于$$\\text{B}$$选项：$$80-76=4$$；也就是$$76$$与$$80$$的距离是$$4$$； 对于$$\\text{C}$$选项：$$80-78=2$$；也就是$$78$$与$$80$$的距离是$$2$$； 对于$$\\text{D}$$选项：$$83-80=3$$；也就是$$83$$与$$80$$的距离是$$3$$． 那么距离$$80$$最近的也就是$$\\text{C}$$选项$$78$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "971", "queId": "f08f6b8fbd03400797d2874cf07589b9", "competition_source_list": ["2019年四川成都青白江区外国语小学小升初（一）第15题2分", "2016年河南郑州联合杯六年级竞赛初赛第14题3分", "2018年陕西西安雁塔区西安交通大学第二附属中学小升初第10题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果一个数恰好等于它的所有因数（本身除外）相加之和，那么这个数就是``完美数''．例如：$$6$$有四个因数$$1$$,$$2$$,$$3$$,$$6$$，除本身$$6$$以外，还有$$1$$,$$2$$,$$3$$三个因数．$$6=1+2+3$$，恰好是所有因数之和，所以$$6$$就是``完美数''．下面的数中是``完美数''的是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$~~~~ "}], [{"aoVal": "B", "content": "$$12$$~~~~ "}], [{"aoVal": "C", "content": "$$15$$~~~~ "}], [{"aoVal": "D", "content": "$$28$$~~~~ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$9\\ne 1+3$$，排除$$9$$；$$12\\ne 1+2+3+4+6$$，排除$$12$$；$$15\\ne 1+3+5$$，排除$$15$$；$$28=1+2+4+7+14$$，故选D． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1569", "queId": "ff8080814518d524014519081d2002f2", "competition_source_list": ["2014年全国迎春杯六年级竞赛初赛第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$在$$B$$地西边$$60$$千米处．甲乙从$$A$$地，丙丁从$$B$$地同时出发．甲、乙、丁都向东行驶，丙向西行驶．已知甲乙丙丁的速度依次成为一个等差数列，甲的速度最快．出发后经过$$n$$小时乙丙相遇，再过$$n$$小时甲在$$C$$地追上丁．则$$B$$、$$C$$两地相距（~~~~~~）千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$90$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["由$n$小时乙丙相遇，知$$n$$小时内$${{S}_{乙}}+{{S}_{丙}}=60$$千米，因此在$$2n$$小时内$${{S}_{乙}}+{{S}_{丙}}=120$$千米．由$$2n$$小时甲追上丁，知$$2n$$小时内$${{S}_{甲}}-{{S}_{丁}}=60$$．由于甲乙丙丁的速度成等差数列，因此甲乙丙丁在$$2n$$小时内的路程也成等差数列，于是由$${{S}_{甲}}-{{S}_{丁}}=60$$知路程的公差为$$60\\div3=20$$千米．再由$${{S}_{乙}}+{{S}_{丙}}=120$$容易解出$${{S}_{乙}}=70$$，$${{S}_{丙}}=50$$千米，进而求出$${{S}_{丁}}=30$$千米．而$${{S}_{丁}}$$恰为$$BC$$两地之间的距离． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "941", "queId": "cad90299340348b2ad86bf7a794ecb38", "competition_source_list": ["2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "前$$10$$个正整数的积，末尾有几个$$0$$？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["求积末尾$$0$$的个数，需要看$$2$$和$$5$$的因数个数．$$2$$的个数多余$$5$$的个数，因此找出前$$10$$个数里有多少个$$5$$，$$5$$本身一个，$$10=2\\times 5$$，一共两个$$5$$，因此，末尾两个$$0$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2964", "queId": "d2746124907e4161bfa8a2bcd456aea3", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "求三个分数$$\\frac{20122012}{20132013}$$，$$\\frac{20132013}{20142014}$$，$$\\frac{20142014}{20152015}$$中值最大的． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{20122012}{20132013}$$ "}], [{"aoVal": "B", "content": "$$\\frac{20132013}{20142014}$$ "}], [{"aoVal": "C", "content": "$$\\frac{20142014}{20152015}$$ "}], [{"aoVal": "D", "content": "三个一样大 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比���与估算->比较大小综合->分数基准数法"], "answer_analysis": ["因为$$\\frac{20122012}{20132013}=\\frac{2012\\times 1001}{2013\\times 1001}=\\frac{2012}{2013}$$， $$\\frac{20132013}{20142014}=\\frac{2013\\times 1001}{2014\\times 1001}=\\frac{2013}{2014}$$，$$\\frac{20142014}{20152015}=\\frac{2014\\times 1001}{2015\\times 1001}=\\frac{2014}{2015}$$，$$1-\\frac{2012}{2013}=\\frac{1}{2013}$$，$$1-\\frac{2013}{2014}=\\frac{1}{2014}$$， $$1-\\frac{2014}{2015}=\\frac{1}{2015}$$． 因为$$\\frac{1}{2015}\\textless{}\\frac{1}{2014}\\textless{}\\frac{1}{2013}$$，所以$$\\frac{2014}{2015}\\textgreater\\frac{2013}{2014}\\textgreater\\frac{2012}{2013}$$．因此，三个分数中，最大的是$$\\frac{20142014}{20152015}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1218", "queId": "2b890396ad7b495f84d01392ca03a1d1", "competition_source_list": ["2015年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "足球友谊比赛的票价是$$50$$元，赛前一小时还有余票，于是决定降价，结果售出的票增加了三分之一，而票房收入增加了四分之一，那么每张票售价降了（ ）元。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$\\frac{25}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{50}{3}$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["解：设原来卖出的票数为$$1$$，则每张票售价降了： $$50-50\\times \\frac{1}{4}\\div \\frac{1}{3}$$ $$=50-\\frac{75}{2}$$ $$=\\frac{25}{2}$$（元） 答：每张票售价降了$$\\frac{25}{2}$$元。 故选：$$B$$。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2319", "queId": "f3ed52289aca45e1ba6d78f3a213191b", "competition_source_list": ["2016年创新杯小学高年级六年级竞赛训练题（一）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "汽车从甲地到乙地，先行上坡，后行下坡，共用$$9.4$$小时．如果甲、乙两地相距$$450$$千米，上坡车速为每小时$$45$$千米，下坡车速为每小时$$50$$千米，且返回时上坡和下坡速度保持不变，那么原路返回要小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9.4$$ "}], [{"aoVal": "B", "content": "$$9.5$$ "}], [{"aoVal": "C", "content": "$$9.6$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题"], "answer_analysis": ["方法一: 从甲地到乙地共有$$450$$千米，共用$$9.4$$小时，上坡车速为每小时$$45$$千米，下坡车速为每小时 50千米，可得上坡所用的时间为$$ (50 \\times 9.4-450)\\div(50-45)=4(个, $$),那么上坡路为 $$ 45 \\times 4=180(千米 $$),下坡路为$$ 450-180=270（千米） $$原路返回时，原来的上坡路变成下 坡路，原来的下坡路变成上坡路，所用时间为$$ 270 \\div 45+180+50=6+3.6=9.6（小时） $$ 方法二: 本题也可以从整体上进行考虑，由于原路返回时，原来的上坡路变成下坡路，原来的下坡路变 成了上坡路，所以往返一次相当于共走了$$450$$千米的上坡路和$$450$$千米的下坡路，那么所用的 总时间为$$ 450 \\div 45+450 \\div 50=19(小+1+) $$,其中从甲地到乙地用了$$9.4$$小时，所以原路返回 要用$$ 19-9.4=9.6(.1,1= $$ 方法三：设原来上坡用了$$x$$小时，$$45x+50\\left( 9.4-x \\right)=450$$，解得$$x=4$$，那么原来上坡是$$45\\times 4=180$$千米，下坡就是$$450-180=270$$千米，返回时下坡变为上坡，上坡变为下坡，所以返回时时间是$$\\frac{270}{45}+\\frac{180}{50}=9.6$$（小时）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3271", "queId": "e37301806df44e349d4a9c876d6a757c", "competition_source_list": ["2017年全国小学生数学学习能力测评六年级竞赛复赛第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知一个三位数的百位、十位和个位分别是：$$a$$，$$b$$，$$c$$，且$$a\\times b\\times c=a+b+c$$，那么满足上述条件的所有三位数的和为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1332$$ "}], [{"aoVal": "B", "content": "$$1032$$ "}], [{"aoVal": "C", "content": "$$1000$$ "}], [{"aoVal": "D", "content": "$$998$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["满足$$a\\times b\\times c=a+b+c$$的只有$$1$$，$$2$$，$$3$$，即$$1\\times 2\\times 3=1+2+3=6$$， 所以这些三位数是$$123$$，$$132$$，$$213$$，$$231$$，$$312$$，$$321$$； 和为$$123+132+213+231+312+321=1332$$． 故$$\\text{BCD}$$错误，$$\\text{A}$$正确． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1157", "queId": "111de4c85bc84fb5aac581ce5cd2e2ec", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（五）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "一队学生排成$$9$$行$$9$$列的方阵，如果去掉最外层一圈，要减少人 ", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$ "}], [{"aoVal": "B", "content": "$$34$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$38$$ "}]], "knowledge_point_routes": ["拓展思维->思想->数形结合思想", "课内体系->思想->数形结合思想"], "answer_analysis": ["$$9\\times 9-7\\times 7=32$$（人）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "244", "queId": "32c1a39e313c45f9a2caf4151a0bd714", "competition_source_list": ["2021年第8届鹏程杯四年级竞赛初赛第15题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁$$4$$ 个人过桥，分别需要$$1$$分钟，$$2$$ 分钟，$$5$$分钟，$$10$$分钟．因为天黑，必须借助手电筒过桥，可是他们只有一个手电筒，并且桥的载重能力有限，最多只能承受两个人的重量，也就是说，每次最多过两人．那么他们过桥的最短时间是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$18$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["甲乙先过去，甲回来，用时$$2+1=3$$分钟，丙丁过去乙回来，用时$$10+2=12$$分钟，甲乙过去用时$$2$$分钟，总共用时$$3+12+2=17$$分钟， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "444", "queId": "e04caf762432431d9e8d8591d1b982c6", "competition_source_list": ["2019年广东深圳全国小学生数学学习能力测评六年级竞赛初赛第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "周长都相等的圆，正方形和长方形，它们的面积相比． ", "answer_option_list": [[{"aoVal": "A", "content": "圆最大 "}], [{"aoVal": "B", "content": "正方形最大 "}], [{"aoVal": "C", "content": "长方形最大 "}], [{"aoVal": "D", "content": "一样大 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["假设正方形、长方形、圆的周长都是$$16$$厘米，则： （$$1$$）正方形的边长：$$16\\div 4=4$$（厘米）， 面积：$$4\\times 4=16$$（平方厘米）； （$$2$$）假设长方形的长为$$6$$厘米，宽为$$2$$厘米， 则面积：$$2\\times 6=12$$（平方厘米）； （$$3$$）圆的半径：$$16\\div 3.14\\div 2=\\frac{800}{314}$$（厘米）， 面积：$$3.14\\times \\left( \\frac{800}{314} \\right)\\times \\left( \\frac{800}{314} \\right)$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=3.14\\times \\frac{800}{314}\\times \\frac{800}{314}$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\frac{6400}{314}$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=20\\frac{60}{157}$$（平方厘米）； 所以，$$12$$平方厘米$$\\textless{}16$$平方厘米$$\\textless{}20\\frac{60}{157}$$平方厘米， 所以$$\\text{A}$$选项是正确的． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2965", "queId": "a08c0edebf39465c8f10e5cdfe497cbb", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "给小数$$0.0123456789$$添加表示循环节的两个圆点，得到一个循环小数．要使得这个循环小数的小数点后第$$100$$位数字是$$8$$，应该在数字和$$9$$上添加圆点． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["A~ 如果循环节是$$123456789$$，循环周期为$$9$$，除了已经有的$$10$$位小数以外，后面还有$$100-10=90$$（位）小数， $90\\div9=10\\cdots0$，所以最后一位是$$9$$，排除 A； B~ 如果循环节是$$23456789$$，循环周期为$$8$$，除了已经有的$$10$$位小数以外，后面还有$$100-10=90$$（位）小数， $90\\div8=11\\cdots2$，所以最后一位是$$3$$，排除 B； C 如果循环节是$$3456789$$，循环周期为$$7$$，除���已经有的$$10$$位小数以外，后面还有$$100-10=90$$（位）小数， $90\\div7=12\\cdots6$，所以最后一位是$$8$$，选择 C； D~ 如果循环节是$$456789$$，循环周期为$$6$$，除了已经有的$$10$$位小数以外，后面还有$$100-10=90$$（位）小数， $90\\div6=15\\cdots0$，所以最后一位是$$9$$，排除 D； 所以，选择 C． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3185", "queId": "38bf3d9c9dbf48ed8b8078109b84be9f", "competition_source_list": ["2017年北京学而思杯小学中年级三年级竞赛年度教学质量测评第17题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小胖和爸爸一起去超市购物．爸爸钱包里有$$2$$张$$100$$元和$$1$$张$$10$$元，他买了一些物品．当小胖四处张望的时候，只听到收银员说：``谢谢，这是找您的$$50$$元．''他看到爸爸的手里只有$$1$$张找回来的$$50$$元钱，请问爸爸买的物品可能是（~ ）元钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$80$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["因为找回的是$$50$$元，则有可能是$$100-50=50$$元，$$100+10-50=60$$元，$$100+100-50=150$$元，$$100+100+10-50=160$$元，则选项中只有$$B$$符合． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1277", "queId": "7e43cecdcd764ed7b9d5d77ffcedd35c", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第8题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "一只兔子$$4$$条腿，一只鸡有$$2$$条腿，已知鸡兔共有$$4$$个头，$$12$$条腿，请你枚举一下，有只兔． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->原型题"], "answer_analysis": ["枚举法： 当有$$1$$只兔，$$3$$只鸡时，共$$1\\times4+3\\times2=10$$（条）腿． 当有$$2$$只兔，$$2$$只鸡时，共$$2\\times4+2\\times2=12$$（条）腿． 所以有$$2$$只兔． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "212", "queId": "47f4da42d6df4c9793b21e5eaeb7fef9", "competition_source_list": ["2014年全国迎春杯六年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "大力、大鹏、大生、大明、大林是天津某相声茶馆里的相声演员，今天，他们每两人要演一场相声．到目前为止，大鹏演了$$1$$场，大生演了$$3$$场，大林演了$$3$$场，大明演了$$4$$场，那么大力演了场 ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->测量->面积->圆的面积->圆的面积公式"], "answer_analysis": ["$$C$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "15", "queId": "01c7c35ea418481f84b998642b3f19bc", "competition_source_list": ["2015年全国中环杯二年级竞赛决赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙三人中有一人是牧师，有一人是骗子，还有一人是赌徒．牧师从不说谎，骗子总是说谎，赌徒有时说真话有时说谎话．甲说:``我是牧师．''乙说:``我是骗子．''丙说:``我是赌徒．''那么，三人中，~\\uline{~~~~~~~~~~}~是牧师． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找一致（同伙）"], "answer_analysis": ["根据条件，牧师不可能说假话，所以牧师一定会说自己是牧师，此题只有一个说自己是牧师．所以甲就是牧师． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1525", "queId": "b0f1f7eb99d44835bfacd5fc9c7e2b2b", "competition_source_list": ["2017年河南郑州K6联赛竞赛模拟第七套第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "梨比苹果多$$50$$个．下面的关系式中，（ ~ ~）是正确的． ", "answer_option_list": [[{"aoVal": "A", "content": "梨的个数$$+50=$$苹果的个数 "}], [{"aoVal": "B", "content": "梨的个数$$=50+$$苹果的个数 "}], [{"aoVal": "C", "content": "梨的个数$$+$$苹果的个数$$=50$$ "}], [{"aoVal": "D", "content": "梨的个数$$=50$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["梨多，所以梨的个数$$=50+$$苹果的个数． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "625", "queId": "fee6d2fd25c746c787fbc922639eeccf", "competition_source_list": ["2018年迎春杯五年级竞赛初赛C卷第2题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个各位数字互不相同的四位数$$\\overline{ABCD}$$，$$\\overline{AB}$$、$$\\overline{BC}$$、$$\\overline{CD}$$都是质数，这个四位数最大是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9874$$ "}], [{"aoVal": "B", "content": "$$9797$$ "}], [{"aoVal": "C", "content": "$$9753$$ "}], [{"aoVal": "D", "content": "$$9371$$ "}], [{"aoVal": "E", "content": "$$9731$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理", "海外竞赛体系->知识点->数论模块->质数与合数"], "answer_analysis": ["要使这个四位数最大，高位要尽可能大，$$\\overline{AB}$$为$$100$$以内最大的质数，即$$97$$，$$\\overline{BC}$$为质数，只能是$$79$$，$$73$$，$$71$$，因为各位数字互不相同，且尽可能大，$$\\overline{BC}=73$$，$$\\overline{CD}$$为质数，只能是$$37$$，$$31$$，因为数字互不相同，$$\\overline{CD}=31$$，所以$$\\overline{ABCD}$$最大是$$9731$$． 故答案为：$$9731$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2981", "queId": "f2a65d87f78341348807ae3f01e398f0", "competition_source_list": ["2020年新希望杯五年级竞赛第14题", "2020年希望杯五年级竞赛模拟第14题"], "difficulty": "1", "qtype": "single_choice", "problem": "【$$2020$$年五年级卷第$$14$$题】猪猪侠在闯关游戏中遇到一个计算题：$$0.004186\\times 8812345.321$$，下列选项中最接近计算结果的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3200$$ "}], [{"aoVal": "B", "content": "$$3600$$ "}], [{"aoVal": "C", "content": "$$32000$$ "}], [{"aoVal": "D", "content": "$$36000$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$0.004186\\times 8812345.321\\approx 0.004\\times 9000000=36000$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1208", "queId": "2b69d57da6a84ac6943a130dbd356f2f", "competition_source_list": ["2010年第10届全国小机灵杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "由两个$$4$$和一个$$5$$组成的所有不同的三位数的平均数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$481$$ "}], [{"aoVal": "B", "content": "$$471$$ "}], [{"aoVal": "C", "content": "$$454$$ "}], [{"aoVal": "D", "content": "$$544$$ "}], [{"aoVal": "E", "content": "$$445$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->公式类->直接求平均数", "Overseas Competition->知识点->应用题模块->平均数问题"], "answer_analysis": ["组成的三位数分别为$$445$$，$$454$$，$$544$$，平均数为$$\\left( 445 + 454 + 544 \\right) \\div 3 = 481$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3278", "queId": "3b3f591301474bccaae55da5f5636ab2", "competition_source_list": ["2009年第7届创新杯四年级竞赛初赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "师徒两人加工同一种零件，每人都把自己的产品装入自己的箩筐中，结果师傅产量是徒弟的两倍，现在装了$$6$$只箩筐，每只箩筐都标了零件的只数：$$78$$只，$$94$$只，$$86$$只，$$87$$只，$$82$$只，$$80$$只．那么( )这两筐是徒弟加工的. ", "answer_option_list": [[{"aoVal": "A", "content": "$$87$$只与$$86$$只 "}], [{"aoVal": "B", "content": "$$87$$只与$$82$$只 "}], [{"aoVal": "C", "content": "$$80$$只与$$87$$只 "}], [{"aoVal": "D", "content": "$$94$$只与$$80$$只 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->整数分拆->整数拆分应用->加法拆数（应用）"], "answer_analysis": ["因为$$\\left( 78+94+86+87+82+80 \\right)\\div \\left( 1+2 \\right)=169$$，又$$87+82=169$$，所以$$87$$只与$$82$$只这两筐是徒弟加工的． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2023", "queId": "ef042f46e0c24ccb86590422b7b69840", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（二）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "小乐前几次数学测验的平均分为$$88$$分，这次要是能考$$100$$分，就能把平均分提高到$$91$$分，这是第次考试． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["根据``以多初少''可知，小乐之前考试的次数为$$\\left( 100-91 \\right)\\div \\left( 91-88 \\right)=3$$（次），这是第$$3+1=4$$（次）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2553", "queId": "504d1514990442babfd9eef9515c87f0", "competition_source_list": ["2015年六年级其它", "2010年第8届希望杯六年级竞赛复赛第2题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知$$1-\\dfrac{1}{6+\\dfrac{1}{6+\\dfrac{1}{6}}}=\\dfrac{1}{A+\\dfrac{1}{B+\\dfrac{1}{C+\\dfrac{1}{C}}}}$$其中$$A$$、$$B$$、$$C$$都是大于$$0$$且互不相同的自然数，则$$\\left( A+B \\right)\\div C=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["根据题意，容易解出$$1-\\dfrac{1}{6+\\dfrac{1}{6+\\dfrac{1}{6}}}=\\dfrac{191}{228}$$， 所以$$A+\\dfrac{1}{B+\\dfrac{1}{C+\\dfrac{1}{C}}}=1+\\dfrac{37}{191}$$， 而$$B+\\dfrac{1}{C+\\dfrac{1}{C}}$$大于$$1$$， 所以$$A=1$$，同理可知，$$B=5$$，$$C=6$$，则$$\\left( A+B \\right)\\div C=1$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1957", "queId": "d7ad140b2d7e4ae8ab9d4501f6d2b39a", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第11题2分"], "difficulty": "2", "qtype": "single_choice", "problem": "张阿姨给幼儿园两个班的孩子分水果，大班每人分得$$5$$个橘子和$$2$$个苹果，小班每人分得$$3$$个橘子和$$2$$个苹果．张阿姨一共分出了$$135$$个橘子和$$70$$个苹果，那么小班有个孩子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$25$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->变型题"], "answer_analysis": ["两班共有$$70\\div 2=35$$（个）孩子， 假设所有人都分得$$5$$个橘子，需要$$35\\times 5=175$$（个）橘子． 则小班有$$\\left( 175-135 \\right)\\div \\left( 5-3 \\right)=20$$（个）孩子． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2602", "queId": "35dd55b4d2b542acbf4d429fca3a57dd", "competition_source_list": ["2016年全国AMC六年级竞赛8第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "$${{13}^{4}}-{{11}^{4}}$$的因数中，只含有质因数$$2$$的最大的因数是（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$64$$ "}], [{"aoVal": "E", "content": "$$128$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->数论模块->分解质因数", "拓展思维->拓展思维->数论模块->分解质因数->分解质因数的应用->已知乘积求因数"], "answer_analysis": ["$${{13}^{4}}-{{11}^{4}}$$ $$=({{13}^{2}}-{{11}^{2}})\\times ({{13}^{2}}+{{11}^{2}})$$ $$=(13-11)\\times (13\\times 11)\\times (169+121)$$ $$=2\\times 24\\times 290$$ $$=2\\times {{2}^{3}}\\times 3\\times 2\\times 145$$ $$={{2}^{5}}\\times 3\\times 145$$ $${{2}^{5}}=32$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "44", "queId": "6b060bff71964305ab702406f2428b3a", "competition_source_list": ["2017年安徽合肥庐江县小升初第19题1分", "2017年河南郑州豫才杯竞赛第14题", "2017年河南郑州小升初豫才杯第14题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列说法正确的是（~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "角的两边是两条直线 "}], [{"aoVal": "B", "content": "知道了物体的方向，就能确定物体的位置 "}], [{"aoVal": "C", "content": "平行四边形有一条对称轴 "}], [{"aoVal": "D", "content": "长方体、正方体、圆柱体的体积都能用``底面积✖️ 高''来计算 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->方向与��标->方向"], "answer_analysis": ["$$A$$不对，角的两边是两条射线；$$B$$不对，缺少观察点无法确定物体的位置；$$C$$不对，平行四边形不是轴对称图形，所以没有对称轴；故选$$D$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3268", "queId": "da2faefd92994e158815e84b841ed19e", "competition_source_list": ["2017年全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "茶商城推销某种产品，有如下优惠：每次第一件全价，第二件$$\\frac{1}{2}$$价，第三件$$\\frac{1}{3}$$价，\\ldots，第十件$$\\frac{1}{10}$$价．甲同学第一次购$$10$$件；乙同学第一次购$$5$$件，第二次购$$5$$件；丙同学第一次购$$4$$件，第二次购$$6$$件；问同样购$$10$$件，谁花的钱最多，谁花的钱最少？（~ ~) ", "answer_option_list": [[{"aoVal": "A", "content": "甲同学花的钱最多，丙同学花的钱最少 "}], [{"aoVal": "B", "content": "乙同学花的钱最多，甲同学花的钱最少 "}], [{"aoVal": "C", "content": "乙同学花的钱最多，丙同学花的钱最少 "}], [{"aoVal": "D", "content": "丙同学花的钱最多，甲同学花的钱最少 "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["设商品单价为$$1$$，则甲同学花的钱为$$1+\\frac{1}{2}+\\frac{1}{3}+...+\\frac{1}{9}+\\frac{1}{10}=\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5} \\right)+\\left( \\frac{1}{6}+\\frac{1}{7}+\\frac{1}{8}+\\frac{1}{9}+\\frac{1}{10} \\right)$$；乙同学花的钱为$$\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5} \\right)\\times 2=\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5} \\right)+\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5} \\right)$$；丙同学花的钱为$$\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4} \\right)+\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6} \\right)=\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5} \\right)+\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{6} \\right)$$． 因 为$$1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}\\textgreater1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{6}\\textgreater\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{8}+\\frac{1}{9}+\\frac{1}{10}$$，所以乙同学花的钱最多，甲同学花的钱最少． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3155", "queId": "6664bc0327c849078d89962fd8d6be1d", "competition_source_list": ["2016年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$1,2,3,4,5,6,7,8$$这$$8$$个数排成一行，使得$$8$$的两边各数之和相等，那么共有（ ）种不同的排法。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$1152$$ "}], [{"aoVal": "B", "content": "$$864$$ "}], [{"aoVal": "C", "content": "$$576$$ "}], [{"aoVal": "D", "content": "$$288$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->乘法原理"], "answer_analysis": ["首先求出$$1,2,3,4,5,6,7$$的和是$$28$$，判断出$$8$$的两边各数之和都是$$14$$，然后分为$$4$$种情况： （$$1$$）$$8$$的一边是$$1,6,7$$，另一边是$$2,3,4,5$$时； （$$2$$）$$8$$的一边是$$2,5,7$$，另一边是$$1,3,4,6$$时； （$$3$$）$$8$$的一边是$$3,4,7$$，另一边是$$1,2,5,6$$时； （$$4$$）$$8$$的一边是$$1,2,4,7$$，另一边是$$3,5,6$$时。 求出每种情况下各有多少种不同的排法，即可求出共有多少种不同的排法。 解：$$1+2+3+4+5+6+7=28$$ $$8$$的两边各数之和是：$$28\\div 2=14$$ （$$1$$）$$8$$的一边是$$1,6,7$$，另一边是$$2,3,4,5$$时， 不同的排法一共有： $$\\left( 3\\times 2\\times 1 \\right)\\times \\left( 4\\times 3\\times 2\\times 1 \\right)\\times 2$$ $$=6\\times 24\\times 2$$ $$=288$$（种） （$$2$$）$$8$$的一边是$$2,5,7$$，另一边是$$1,3,4,6$$时， 不同的排法一共有$$288$$种。 （$$3$$）$$8$$的一边是$$3,4,7,$$另一边是$$1,2,5,6$$时， 不同的排法一共有$$288$$种。 （$$4$$）$$8$$的一边是$$1,2,4,7,$$另一边是$$3,5,6$$时， 不同的排法一共有$$288$$种。 因为$$288\\times 4=1152$$（种）， 所以共有$$1152$$种不同的排法。 答：共有$$1152$$种不同的排法。 故选：A "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "705", "queId": "2929ca71e0314e6cbd32aa08ce434424", "competition_source_list": ["2021年新希望杯五年级竞赛模拟（考前培训100题）第61题"], "difficulty": "1", "qtype": "single_choice", "problem": "自然数$$M$$乘$$13$$的积的末三位数是$$123$$，$$M$$最小是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$471$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["$$M$$取最小时乘积应为$$4$$位数，根据$$13$$的整除特征，三位截断，积的末$$3$$位$$123$$与最高位数字的差必为$$13$$的倍数，而$$123 \\div13$$$\\cdots\\cdots$$$6$$， 故取最高位数字为$$6$$，则$$M=6123\\div13=471$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2520", "queId": "e7c7c8a3ffba4bf1aeafc6ea5199737d", "competition_source_list": ["2008年第6届创新杯五年级竞赛初赛B卷第9题5分", "2008年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "某人沿$$1$$路电车路线行走，每$$12$$分钟有一辆$$1$$路电车从后边追上，每$$4$$分钟有一辆$$1$$路电车从对面同他相遇，假定此人与电车均为匀速，则$$1$$路电车每隔分钟发出一辆. ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->发车问题"], "answer_analysis": ["设人，电车的速度分别为$${{v}_{1}}，{{v}_{2}}$$，两辆相邻电车之间的距离为$$s$$，则$$\\frac{s}{{{v}_{2}}-{{v}_{1}}}=12$$，$$\\frac{s}{{{v}_{1}}+{{v}_{2}}}=4$$，从而$$\\frac{{{v}_{2}}-{{v}_{1}}}{s}=\\frac{1}{12}$$，$$\\frac{{{v}_{2}}+{{v}_{1}}}{s}=\\frac{1}{4}$$，两等式相加得$$\\frac{2{{v}_{2}}}{s}=\\frac{1}{12}+\\frac{1}{4}=\\frac{1}{3}$$，所以$$\\frac{s}{{v}_{2}}=6$$，即电车每隔$$6$$分钟发出一辆． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "174", "queId": "cc1ba1d6d1bf4fdb9ff102fa4a21df19", "competition_source_list": ["2018年IMAS小学高年级竞赛（第一轮）第6题3分", "2018年IMAS小学中年级竞赛（第一轮）第17题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "列车在某天早上$$8$$点$$30$$分从$$A$$地开出，第二天凌晨$$1$$点$$50$$分抵达$$B$$地．已知$$A$$地与$$B$$地没有时差﹐请问该列车全程共用了多少时间？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$小时$$20$$分钟 "}], [{"aoVal": "B", "content": "$$10$$小时$$20$$分钟 "}], [{"aoVal": "C", "content": "$$15$$小时$$20$$分钟 "}], [{"aoVal": "D", "content": "$$16$$小时$$20$$分钟 "}], [{"aoVal": "E", "content": "$$17$$小时$$20$$分钟 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用->度量单位认知"], "answer_analysis": ["可知列车在出发当日共用了$$24$$小时减去$$8$$小时$$30$$分钟，即$$15$$小时$$30$$分钟；在第二日用了$$1$$小时$$50$$分钟，故全程所用的时间是$$16$$小时又$$80$$分钟，即$$17$$小时$$20$$分钟． 故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1694", "queId": "57dc299d8dbe4c32ac99fca7f8d1a8a8", "competition_source_list": ["2013年第25届广东广州五羊杯六年级竞赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "敏华发现徐明在看一本故事书，想借来看．徐明说这本书很厚，页码共有$$741$$个数字．则这本书有页． ", "answer_option_list": [[{"aoVal": "A", "content": "$$280$$ "}], [{"aoVal": "B", "content": "$$281$$ "}], [{"aoVal": "C", "content": "$$282$$ "}], [{"aoVal": "D", "content": "$$283$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["这本书有$$741$$个数字，故页码在$$100\\sim 999$$之间， $$\\frac{741-9-90 \\times 2}{3}=184$$，$$9+90+184=283$$（页）． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2707", "queId": "4945e89891fe403084bfd5dd6a4d97ee", "competition_source_list": ["2005年六年级竞赛创新杯", "2005年第3届创新杯六年级竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "长方体的体积是12，有两对侧面的面积分别是3和12，那么第3对侧面的面积是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "12 "}], [{"aoVal": "B", "content": "6 "}], [{"aoVal": "C", "content": "4 "}], [{"aoVal": "D", "content": "3 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->一元一次方程->整数系数方程"], "answer_analysis": ["设这个长方体的长宽高分别为$$a，b，c$$，依题意有$$abc=12，ab=3，ac=12$$，那么$$b=1，a=3，c=4$$，所以另一对侧面的面积为$$bc=4$$，选$$C$$. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2557", "queId": "473d091da328486984d49e00ee66db8b", "competition_source_list": ["2020年北京迎春杯六年级竞赛模拟初赛第2题3分", "2017年四川成都锦江区四川师范大学附属第一实验中学小升初（六）第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "脱式计算： ~$\\dfrac{9}{13}\\div7+\\dfrac{1}{7}\\times\\dfrac{4}{13}=$~\\uline{~~~~~~~~~~}~ ", "answer_option_list": [[{"aoVal": "A", "content": "$\\dfrac{1}{5}$ "}], [{"aoVal": "B", "content": "$\\dfrac{1}{6}$ "}], [{"aoVal": "C", "content": "$\\dfrac{1}{7}$ "}], [{"aoVal": "D", "content": "$\\dfrac{2}{7}$ "}]], "knowledge_point_routes": ["课内体系->能力->逻辑分析", "拓展思维->拓展思维->计算模块->分数->分数运算->分小四则混合运算"], "answer_analysis": ["$\\dfrac{1}{7}$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "701", "queId": "d0d76eaf186643199ea985ef9657a6ae", "competition_source_list": ["2013年全国华杯赛小学高年级竞赛初赛C卷第4题"], "difficulty": "3", "qtype": "single_choice", "problem": "已知正整数$$A$$分解质因数可以写成$$A={{2}^{\\alpha }}\\times {{3}^{\\beta }}\\times {{5}^{\\gamma }}$$，其中$$\\alpha $$、$$\\beta $$、$$\\gamma $$是自然数．如果$$A$$除以$$2$$的商是完全平方数，$$A$$除以$$3$$的结果是完全立方数，$$A$$除以$$5$$后是某个自然数的五次方，那么$$\\alpha +\\beta +\\gamma $$的最小值是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$31$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->逻辑分析"], "answer_analysis": ["根据``$$A$$的二分之一是完全平方数''可以知道，$$\\alpha -1$$、$$\\beta $$、$$\\gamma $$都是$$2$$的倍数． 根据``$$A$$的三分之一是完全立方数''可以知道，$$\\alpha $$、$$\\beta -1$$、$$\\gamma $$都是$$3$$的倍数． 根据``$$A$$的五分之一是某个自然数的五次方''可以知道，$$\\alpha $$、$$\\beta $$、$$\\gamma -1$$都是$$5$$的倍数． 同时满足三个条件的$$\\alpha $$的最小值恰好是$$\\left[ 3,5 \\right]=15$$；$$\\beta $$的最小值恰好是$$\\left[2,5 \\right]=10$$；$$\\gamma $$的最小值恰好是$$\\left[ 2,3 \\right]=6$$． 所以，$$\\alpha +\\beta +\\gamma $$的最小值是$$15+10+6=31$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1968", "queId": "f323609fbb37457397af0a0e9f2d33eb", "competition_source_list": ["2013年河南郑州中原网杯六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "某班统计数学考试成绩，平均成绩$$84.1$$分，后来发现小红的成绩是$$96$$分，被错写成了$$69$$分，重新计算后，平均成绩是$$84.7$$分，那么这个班有（~ ）名学生． ~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$41$$ "}], [{"aoVal": "B", "content": "$$43$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$47$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["第二次平均分比第一次平均分多$$84.7-84.1=0.6$$，其中小红第二次分数比第一次多$$96-69=27$$，显然，平均分的差$$\\times $$班级人数$$=$$小红少加的分数．故有$$27\\div 0.6=45$$人． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "611", "queId": "5d543549447842de859245275e8f0567", "competition_source_list": ["2009年希望杯四年级竞赛复赛", "2009年希望杯三年级竞赛复赛"], "difficulty": "0", "qtype": "single_choice", "problem": "某数被$$13$$除，商是$$9$$，余数是$$8$$，则某数等于（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$115$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$125$$ "}], [{"aoVal": "D", "content": "$$130$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法"], "answer_analysis": ["被除数$$\\div 13=9\\cdots\\cdots 8$$，则被除数$$=13\\times 9+8=125$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "203", "queId": "87d44c560cad43bc9cf4250742293227", "competition_source_list": ["2012年世界少年奥林匹克数学竞赛三年级竞赛初赛A卷第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "爸爸、妈妈、哥哥、姐姐和我站成一排照相，我挨着爸爸，也挨着妈妈；姐姐挨着妈妈，也挨着哥哥．那么，站在正中间的人是谁？ ", "answer_option_list": [[{"aoVal": "A", "content": "爸爸 "}], [{"aoVal": "B", "content": "妈妈 "}], [{"aoVal": "C", "content": "哥哥 "}], [{"aoVal": "D", "content": "姐姐 "}], [{"aoVal": "E", "content": "我 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["由题干可推拍照顺序可能为爸爸~ 我~ 妈妈~ 姐姐~ 哥哥； 或者 哥哥~ 姐姐~ 妈妈~ 我~ 爸爸，因此妈妈在中间． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1415", "queId": "3ad334339e354684a6d5f90ec108d227", "competition_source_list": ["小学中年级三年级上学期其它", "2017年全国华杯赛小学中年级竞赛初赛模拟第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两人在春节一共得$$200$$元压岁钱，甲给乙$$40$$元，还比乙多$$10$$元，那么，原来甲得了~\\uline{~~~~~~~~~~}~元压岁钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$145$$ "}], [{"aoVal": "B", "content": "$$140$$ "}], [{"aoVal": "C", "content": "$$125$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->和差问题->二量和差问题->明和暗差"], "answer_analysis": ["因为甲给乙$$40$$元，还比乙多$$10$$元，所以甲比乙多$$40\\times 2+10=90$$元， 所以甲：$$(200+90)\\div 2=145$$元． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1669", "queId": "7b887d7572ef4014aa92e1a5914bdba9", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第5题", "2017年全国小升初八中入学备考课程"], "difficulty": "1", "qtype": "single_choice", "problem": "$$12$$枚硬币的总值是$$9$$角，其中只有$$5$$分和$$1$$角的两种，那么每种硬币各（~~~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->变型题"], "answer_analysis": ["鸡兔同笼，假设$$12$$枚都是$$5$$分硬币，则1角硬币共$$(90-12\\times 5)\\div (10-5)=6$$枚，每种硬币各$$6$$枚． ", "$$5$$分的数量：
\n$$(12\\times 1-9)\\div \\left( 1-0.5 \\right)$$
\n$$=3\\div 0.5$$
\n$$=6$$（枚）；
\n$$1$$角的硬币数量为：$$12-6=6$$（枚）．
\n答：每种硬币各$$6$$枚．
\n故选：$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "195", "queId": "59ce02acca194da38aba534c33856dc4", "competition_source_list": ["2015年第14届春蕾杯一年级竞赛初赛第5题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "小王说：小胖比我高；小胖说：我比园园矮；小朱说：我还没有小王高 呢；问四个小朋友哪个最高？答． ", "answer_option_list": [[{"aoVal": "A", "content": "小王 "}], [{"aoVal": "B", "content": "小胖 "}], [{"aoVal": "C", "content": "小朱 "}], [{"aoVal": "D", "content": "园园 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意分析可知，小胖比小王高，所以小胖的身高高于小王，小胖说比园园矮，所以园园的身高高于小胖，小朱没有小王高，所以小王的身高高于小朱，由此可知，四个人的身高从高到低位：园园$$\\textgreater$$小胖$$\\textgreater$$小王$$\\textgreater$$小朱． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1635", "queId": "913cee09003847e480302c818c29850f", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第1题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "一根$$20\\text{m}$$的木棍，要把它锯成每根$$4\\text{m}$$的小木棍，每次只能锯一根木棍．请问一共要锯多少次？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["木棍被锯成了$$20\\div 4=5$$（根），因此只需要锯$$5-1=4$$（次）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "969", "queId": "fe58f813b42343be89d33e985b490215", "competition_source_list": ["2018年IMAS小学高年级竞赛（第二轮）第1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "请问算式$$\\left( 2019-2018 \\right)\\times \\left( 2019-2017 \\right)\\times \\cdots \\times \\left( 2019-2012 \\right)\\times \\left( 2019-2011 \\right)$$的值之质因数分解包含多少个不同的质因数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}], [{"aoVal": "E", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left( 2019-2018 \\right)\\times \\left( 2019-2017 \\right)\\times \\cdots \\times \\left( 2019-2012 \\right)\\times \\left( 2019-2011 \\right)$$ $$=1\\times 2\\times 3\\times 4\\times 5\\times 6\\times 7\\times 8$$ $$=2\\times 3\\times \\left( 2\\times 2 \\right)\\times 5\\times \\left( 2\\times 3 \\right)\\times 7\\times \\left( 2\\times 2\\times 2\\times 2 \\right)$$． 包含不同的质因数有$$2$$、$$3$$、$$5$$、$$7$$，共有$$4$$个． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1150", "queId": "19168869c06743ffbb63a55b6cfe6023", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级三年级竞赛初赛第40题"], "difficulty": "1", "qtype": "single_choice", "problem": "约翰的钱比吉尔多$$20$$美元，他们两个总共有$$40$$美元，约翰有美元 . ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$(40+20)\\div 2=30$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "494", "queId": "dcaf214d85d445f9a91898570f615a57", "competition_source_list": ["2015年北京华杯赛小学高年级竞赛初赛A卷第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "现在从甲、乙、丙、丁四个人中选出两个人参加一项活动．规定：如果甲去，那么乙也去；如果丙不去，那么乙也不去；如果丙去，那么丁不去．最后去参加活动的两个人是谁？ ", "answer_option_list": [[{"aoVal": "A", "content": "甲、乙 "}], [{"aoVal": "B", "content": "乙、丙 "}], [{"aoVal": "C", "content": "甲、丙 "}], [{"aoVal": "D", "content": "乙、丁 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾", "Overseas Competition->知识点->组合模块->逻辑推理"], "answer_analysis": ["题目要求有两个人去，可以使用假设法，若甲去，则乙去，乙去则丙也去．三个人去，矛盾，所以甲不去．若丙不去则乙不去，那么只有丁去，矛盾，所以丙去．丙去则丁不去，由两个人去得到结论，乙要去． 所以答案是$$\\text{B}$$，丙和乙去． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1617", "queId": "ff808081465a848401466a66a6cb1296", "competition_source_list": ["2017年全国小升初八中入学备考课程", "2014年全国华杯赛小学高年级竞赛初赛B卷第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙丙丁四个人今年的年龄之和是$$72$$岁．几年前（至少一年）甲是$$22$$岁时，乙是$$16$$岁．又知道，当甲是$$19$$岁的时候，丙的年龄是丁的$$3$$倍（此时丁至少$$1$$岁）．如果甲乙丙丁四个人的年龄互不相同，那么今年甲的年龄可以有（~ ~ ~ ~）种情况． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型"], "answer_analysis": ["甲乙的年龄差是$$22-16=6$$（岁）；当甲$$19$$岁时， $$13$$岁；至少一年前甲$$22$$岁，所以当甲$$19$$岁的时候，此时至少是$$4$$年前的年龄，那么甲今年至少是$$23$$岁；甲$$19$$岁时，丙的年龄是丁的$$3$$倍，假设丁为$$1$$岁，丙为$$3$$岁，此时四人的年龄和至少是$$19+13+1+3=36$$（岁）；且甲今年的年龄至多为$$19+\\left(72-36 \\right)\\div 4=28$$（岁）；所以甲今年的年龄可能是$$23$$，$$24$$，$$25$$，$$26$$，$$27$$，$$28$$；共$$6$$种，所以选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2597", "queId": "832240c01ab847e5b1b72efa27e538f0", "competition_source_list": ["2017年IMAS小学中年级竞赛（第一轮）第13题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "规定$$*$$是一种运算符号，若$$4*2=86$$、$$6*3=189$$、$$8*4=3212$$、$$9*3=2712$$，请问$$10*2$$是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$128$$ "}], [{"aoVal": "B", "content": "$$208$$ "}], [{"aoVal": "C", "content": "$$2008$$ "}], [{"aoVal": "D", "content": "$$2012$$ "}], [{"aoVal": "E", "content": "$$2020$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["由题意与范例可判断出运算结果为先写出两个数的乘积后紧接着写下两个数的和．因$$10\\times2=20$$、$$10+2=12$$，所以$$10*2=2012$$．故选$$\\text{D}$$． 答案：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3142", "queId": "029aa2ff2bdb4a6181c30a000622c0d6", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第2题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "由数字$$1$$，$$2$$，$$3$$，$$4$$组成没有重复数字的四位数，所有这些四位数的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$666$$ "}], [{"aoVal": "B", "content": "$$22200$$ "}], [{"aoVal": "C", "content": "$$6660$$ "}], [{"aoVal": "D", "content": "$$66660$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["每一个数字在个位、十位、百位、千位均会出现$$3\\times2\\times1=6$$次． 故总和：$$(1+2+3+4)\\times6666=66660$$． 故选择$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2892", "queId": "8dc4f872bdbd44a89cf8b54a74a71a37", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第2题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "在算式$$\\square +5=13-2$$中，$$\\square $$中应填入什么数才能使算式成立？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数减法运算->减法横式"], "answer_analysis": ["13-2=11；11-5=6 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "731", "queId": "90980869635a49c8b78224121af32335", "competition_source_list": ["2013年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "已知正整数$$A$$分解质因数可以写成$$A={{2}^{\\alpha }}\\times {{3}^{\\beta }}\\times {{5}^{\\gamma }}$$， 其中$$\\alpha $$、$$\\beta $$、$$\\gamma $$ 是自然数。 如果$$A$$的二分之一是完全平方数，$$A$$的三分之一是完全立方数，$$A$$的五分之一是某个自然数的五次方，那么$$\\alpha +\\beta +\\gamma $$的最小值是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$31$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->完全平方数->平方数的简单应用"], "answer_analysis": ["根据``$$A$$的二分之一是完全平方数''可以知道，$$\\left( \\alpha -1 \\right)$$、$$\\beta $$、$$\\gamma $$都是$$2$$的倍数。根据``$$A$$的三分之一是完全立方数''可以知道，$$\\alpha $$、$$\\left( \\beta -1 \\right)$$、$$\\gamma $$都是$$3$$的倍数。根据``$$A$$的五分之一是某个自然数的五次方''可以知道，$$\\alpha $$、$$\\beta $$、$$\\left( \\gamma -1 \\right)$$都是$$5$$的倍数。同时满足三个条件的$$\\alpha $$的最小值恰好是$$\\left[ 3 \\right.$$，$$\\left. 5 \\right]=15$$；$$\\beta $$的最小值恰好是$$\\left[ 2 \\right.$$，$$\\left. 5 \\right]=10$$；$$\\gamma $$的最小值恰好是$$\\left[ 2 \\right.$$，$$\\left. 3 \\right]=6$$．所以，$$\\alpha +\\beta +\\gamma $$的最小值是$$15+10+6=31$$。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1092", "queId": "3cdd445910d34db49a8a95fe6b5c76b7", "competition_source_list": ["2005年五年级竞赛创新杯", "2005年第3届创新杯五年级竞赛复赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "某部83集的电视连续剧，从星期三开始在凤凰电视中文台播出，计划除星期六、星期日停播外，每天播出一集，那么最后一集将在星期（ ）播出. ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "四 "}], [{"aoVal": "D", "content": "五 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题"], "answer_analysis": ["从星期三开始，每7天会播出5集，$$83\\div 5=16(周)\\ldots 3（集）$$那么16周后再播出三集就全部播放完毕，那么16周后从星期三开始，播放这最后3集中的第一集，星期四为第二集，星期五为最后一集，选D. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "764", "queId": "40c57df2ee7846aea568637e3a019c67", "competition_source_list": ["2016年全国华杯赛小学高年级竞赛在线模拟第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "（2）在一个七位数中，任何三个连续排列的数字都构成一个能被$$11$$或$$13$$整除的三位数，那么这个七位数最大是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9981733$$ "}], [{"aoVal": "B", "content": "$$9884737$$ "}], [{"aoVal": "C", "content": "$$9978137$$ "}], [{"aoVal": "D", "content": "$$9871773$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->运算求解"], "answer_analysis": ["任何三个连续排列的数字都构成一个三位数，说明没有数字$$0$$，$$\\left. 11 \\right\\textbar990$$，$$\\left. 13 \\right\\textbar988$$则首三位为$$988$$，选择题这是已经得到答案了！接着往下分析$$88$$，$$13\\left\\textbar{} 884 \\right.$$，$$11\\left\\textbar{} 880 \\right.$$，所以第四位最大为$$4$$，依次类推，得到答案，$$9884737$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "828", "queId": "72b7183d1c4e4f35baae05b7262bca26", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级四年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{(2017 Math League, Priamry 4, Question \\#40)} A group of kids aged $$5$$ to $$12$$ went to the movies. The product of the ages of these children is $$3080$$. What is the sum of the ages of these children? 一群$$5$$至$$12$$岁的孩子去看电影．这些孩子的年龄的乘积是$$3080$$．请问这些孩子的年龄的和是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->数论模块->分解质因数", "拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$3080=2^{3}\\times5\\times7\\times11$$, $$5+7+8+11=31$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1888", "queId": "ee0c3f2f59e84112af147cc8409e83b9", "competition_source_list": ["2016年创新杯六年级竞赛训练题（一）第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "一件商品，如果将价格先降低$$20 \\%$$，再提升$$30 \\%$$，那么现在的价格比原来价格提升了（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4 \\%$$ "}], [{"aoVal": "B", "content": "$$6 \\%$$ "}], [{"aoVal": "C", "content": "$$10 \\%$$ "}], [{"aoVal": "D", "content": "$$30 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$\\left( 1-20 \\% \\right)\\times \\left( 1+30 \\% \\right)-1=0.04=4 \\%$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "122", "queId": "1a4479292ae14dc5995dff0377a739ad", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第12题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "小熊、小马、小牛、和小鹿各拿一只水桶同时到一个水龙头前接水，它们只能一个接一个地接水．~ 小熊接一桶水要$$5$$分钟，小马要$$3$$分钟，小牛要$$7$$分钟，小鹿要$$2$$分钟．它们等候时间的总和最少是分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$32$$ "}], [{"aoVal": "C", "content": "$$34$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->统筹规划->简单时间统筹问题->时间总和问题"], "answer_analysis": ["要使它们等候时间（等候时间包括接水时间）的总和最少，应该让接水用时少的先接水，即接水顺序是：小鹿、小马、小熊、小牛． 等候时间总和最少是： $$2\\times 4+3\\times 3+5\\times 2+7=8+9+10+7=34$$（分钟）， 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "837", "queId": "808342c77f504195bdc002483d7bfc38", "competition_source_list": ["其它改编自2015年全国希望杯六年级竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "定义：符号$$\\left { x \\right }$$表示$$x$$的小数部分，如$$\\left { 3.14 \\right }=0.14$$，$$\\left { 0.5 \\right }=0.5$$．那么，$$\\left { \\frac{2015}{3} \\right }+\\left { \\frac{315}{4} \\right }+\\left { \\frac{412}{5} \\right }=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1371}{30}$$ "}], [{"aoVal": "B", "content": "$$\\frac{322}{15}$$ "}], [{"aoVal": "C", "content": "$$\\frac{211}{60}$$ "}], [{"aoVal": "D", "content": "$$\\frac{109}{60}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$\\left { \\frac{2015}{3} \\right }+\\left { \\frac{315}{4} \\right }+\\left { \\frac{412}{5} \\right }=\\frac{2}{3}+\\frac{3}{4}+\\frac{2}{5}=\\frac{109}{60}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1519", "queId": "7624955dc94a44d88966ffaf11da3f36", "competition_source_list": ["走美杯三年级竞赛", "走美杯四年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "幼儿园买了$$8$$辆玩具车，每辆玩具车需要$$92$$元，李老师带了$$720$$元够吗？下面的解答比较合理的是 ", "answer_option_list": [[{"aoVal": "A", "content": "因为$$92\\times 8\\approx$$ 720（元），$$92\\times$$ 8\\textgreater720，所以$$92\\times$$ 8\\textgreater720，够 "}], [{"aoVal": "B", "content": "因为$$92\\times 8\\approx$$ 800（元），$$92\\times$$ 8\\textless800，所以$$92\\times$$ 8\\textless720，够 "}], [{"aoVal": "C", "content": "因为$$92\\times 8\\approx$$ 720（元），$$92\\times$$ 8\\textgreater720，所以$$92\\times$$ 8\\textgreater720，不够 "}], [{"aoVal": "D", "content": "因为$$92\\times 8\\approx$$ 800（元），$$92\\times$$ 8\\textgreater800，所以$$92\\times$$ 8\\textgreater720，不够 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型"], "answer_analysis": ["92\\textgreater90，总价大于$$720$$元，所以不够 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1886", "queId": "bbe00ddc459d4bf4970c706be094c618", "competition_source_list": ["2006年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "张师傅加工一批零件，原计划每天加工$$80$$个，$$5$$天加工完，实际张师傅只用$$4$$天就加工完了，实际每天比原计划每天多加工零件（ ）个。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->归总问题"], "answer_analysis": ["实际每天加工零件的个数为$$80\\times 5\\div 4=100$$（个）；实际每天比原计划每天多加工的个数是$$100-80=20$$（个），故选$$\\text{C}$$．错选$$\\text{A}$$选项是不能根据条件先算出零件的总数量，进而解决问题；错选$$\\text{B}$$、$$\\text{D}$$选项是计算出错． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1143", "queId": "14d411094ee44edcadf9b41fd981ec74", "competition_source_list": ["2009年第8届全国新希望杯三年级竞赛"], "difficulty": "0", "qtype": "single_choice", "problem": "~周末，妈妈去商场买了一些家庭用品，分别是洗衣机（$$1258$$元）、台灯（$$59$$元）、电饭煲（$$365$$元）、电吹风（$$230$$元），这些用品一共需要花费（~）元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2012$$ "}], [{"aoVal": "B", "content": "$$2013$$ "}], [{"aoVal": "C", "content": "$$2014$$ "}], [{"aoVal": "D", "content": "$$2022$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["为了简化计算，我们常把一些数拆开来使它与其他数字之和是整十、整百、整千$$ \\cdots ~\\cdots $$这种方法叫作借数凑整法，我们发现如果将$$1258$$和$$159$$分别拆成$$1250 + 8$$和$$150 + 9$$时，便可以凑出整百，$$365$$可以拆成$$360 + 5$$，也可以拆成$$370 - 5$$，但是由于后面要加上$$230$$，所以拆成$$370 - 5$$时会与$$230$$凑成整百，所以总的花费 $$ = \\left( 1250 + 150 \\right) + \\left( 370 + 230 \\right) + \\left( 8 + 9 - 5 \\right) = 1400 + 600 + 12 = 2012$$（元）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3301", "queId": "5f7df6e3e7b54eb591dfc6d16cda4816", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级六年级竞赛邀请赛训练题（二）"], "difficulty": "1", "qtype": "single_choice", "problem": "用$$1$$、$$2$$、$$3$$、$$4$$、$$5$$这五个数字可组成（~ ~ ~ ）个比$$20000$$大且百位数字不是$$3$$的无重复数字的五位数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$78$$ "}], [{"aoVal": "B", "content": "$$62$$ "}], [{"aoVal": "C", "content": "$$84$$ "}], [{"aoVal": "D", "content": "$$124$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["根据题意，要求这个五位数比$$20000$$大，则首位必须是$$2$$、$$3$$、$$4$$、$$5$$这四个数字，当首位是$$3$$时，百位数不会是数字$$3$$，共有$$A44=24$$种情况，当首位是$$2$$、$$4$$、$$5$$时，由于百位数不能是数字$$3$$，有$$3\\left( A44A33 \\right)=54$$种情况，有分步计数原理，共有$$54+24=78$$个数字． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "751", "queId": "7633be4aa1444bcb85bc9d90f4c7bafb", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛B卷第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$120$$有个偶因数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["因为$$120$$的因数有：$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$8$$、$$10$$、$$12$$、$$15$$、$$20$$、$$24$$、$$30$$、$$40$$、$$60$$、$$120$$，所以一共有$$16$$个因数． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "525", "queId": "d481474d4fa44909ba63765608e28828", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在算式$$2+8+3=2\\bigcirc 8\\square 3$$中，请问应该分别在$$\\bigcirc $$、$$\\square $$填上什么运算符号才可以使得算式正确？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\bigcirc $$填入$$+$$，$$\\square $$填入$$\\times $$ "}], [{"aoVal": "B", "content": "$$\\bigcirc $$填入$$\\times $$，$$\\square $$填入$$-$$ "}], [{"aoVal": "C", "content": "$$\\bigcirc $$填入$$+$$~ ，$$\\square $$填入$$\\div $$ "}], [{"aoVal": "D", "content": "$$\\bigcirc $$填入$$\\times $$，$$\\square $$填入$$\\div $$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["等式左边等于$$13$$，当算式右边依次填入$$\\times $$、$$-$$时，2\\times 8-3=13，等式成立．经验算，其余选项均不正确. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1750", "queId": "b1c9b26ec8334de0a18f27b8b7a856f5", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛决赛第4题", "2018年浙江杭州西湖区小学高年级六年级上学期单元测试《第三单元》第8题3分", "2018年四川成都锦江区四川师范大学附属第一实验中学小升初（四）第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一种商品的利润率为$$20 \\%$$，进价提高$$25 \\% $$后，保持利润不变，那么，进价提价后的利润率为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25 \\%$$ "}], [{"aoVal": "B", "content": "$$20 \\% $$ "}], [{"aoVal": "C", "content": "$$16 \\% $$ "}], [{"aoVal": "D", "content": "$$12.5 \\%$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数的运算->估算->百分数的简单实际问题->折扣、成数、税率、利率", "拓展思维->知识点->应用题模块->经济问题->基本经济概念->利润基本公式->已知利润成本求利润率"], "answer_analysis": ["利润问题．假设成本为$$100$$元，则利润为$$20$$元，现在成本是$$100\\times (1+25 \\%)=125$$元，$$20\\div 125\\times 100 \\%=16 \\%$$． 以进价为单位``1''，利润为20\\%，第二次进价为1+25\\%=125\\%，则第二次利润率$$20 \\%\\div 125 \\%\\times 100 \\%=16 \\%$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "190", "queId": "c788b6742f934097bfebdccfe53bf203", "competition_source_list": ["2017年湖北武汉中环杯四年级竞赛初赛第14题"], "difficulty": "2", "qtype": "single_choice", "problem": "小明和小红有如下对话： 小明说：``我没有超过$$40$$岁．'' 小红说：``我$$38$$岁，你至少比我大$$5$$岁．'' 小明说：``你至少$$39$$岁．'' 已知这三句话都是假话，那么小红~\\uline{~~~~~~~~~~}~岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$37$$ "}], [{"aoVal": "B", "content": "$$38$$ "}], [{"aoVal": "C", "content": "$$39$$ "}], [{"aoVal": "D", "content": "$$41$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾"], "answer_analysis": ["逻辑推理．由第一句话可知，小明超过$$40$$岁； 由第二句话可知，小红不是$$38$$岁，且两人年龄相差不超过$$4$$岁，即小红的年龄大于$$36$$岁； 由第三句话可知，小红的年龄小于$$39$$岁． 则小红的年龄为$$37$$岁． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2158", "queId": "4f879a9761b74a7e9ba5639b72e66dd2", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小$$A$$、小$$B$$、小$$C$$、小$$D$$跑$$50$$米分别用时$$12$$秒、$$14$$秒、$$11$$秒、$$10$$秒，是第二名． ", "answer_option_list": [[{"aoVal": "A", "content": ".小$$A$$ "}], [{"aoVal": "B", "content": "小$$B$$ "}], [{"aoVal": "C", "content": "小$$C$$ "}], [{"aoVal": "D", "content": "小$$D$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["路程一样，则用时越少，跑的越快．根据时间排序，从少到多为$$10$$秒，$$11$$秒，$$12$$秒，$$14$$秒，故第二名为$$11$$秒的，即小$$C$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1025", "queId": "2ea39d8a4f494098bc248ab886b55ea3", "competition_source_list": ["2015年上海走美杯三年级竞赛初赛", "2015年走美杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "小明说：``我妈妈比我大$$24$$岁，两年前妈妈的年龄是我的$$4$$倍。''那么小明今年（ ）岁。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型"], "answer_analysis": ["$$24\\div \\left( 4-1 \\right)+2=24\\div 3+2=8+2=10$$ （岁） "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "925", "queId": "ceb4d02a24314f638c78d8d2fadacb1a", "competition_source_list": ["2016年IMAS小学高年级竞赛第一轮检测试题第15题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$304$$除以一个两位数，余数为$$24$$，请问这样的两位数共有多少个？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法->除法中四量关系"], "answer_analysis": ["被除数$$=$$商$$\\times $$除数$$+$$余数，用$$304-24=280$$，$$280$$一定是该两位数的倍数， 将$$280$$分解因数：$$280=2\\times 2\\times 2\\times 5\\times 7$$， 则符合条件的两位数有$$70$$、$$56$$、$$40$$、$$35$$、$$28$$，共$$5$$种． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2109", "queId": "e74a54f02e91402680069dd6e0d4802b", "competition_source_list": ["2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$3$$盒同样重的苹果，如果从每盒中都取出$$4$$千克，那么盒子里剩下的苹果的重量正好等于原来$$1$$盒苹果的重量，原来每盒苹果重（ ）千克。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->列方程解应用题等量代换"], "answer_analysis": ["解：$$3\\times 4\\div 2$$ $$=12\\div 2$$ $$=6$$（千克） 答：每盒苹果重$$6$$千克。 故选:B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1599", "queId": "5294e33d48634577aaf9f2c7ed8b68df", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两瓶盐水，甲瓶盐水的浓度是乙瓶盐水的$$3$$倍．将$$100$$克甲瓶盐水与$$300$$克乙瓶盐水混合后得到浓度为$$15 \\% $$的新盐水，那么甲瓶盐水的浓度是~\\uline{~~~~~~~~~~}~. ", "answer_option_list": [[{"aoVal": "A", "content": "30\\% "}], [{"aoVal": "B", "content": "25\\% "}], [{"aoVal": "C", "content": "20\\% "}], [{"aoVal": "D", "content": "15\\% "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["设乙瓶盐水的浓度是$$x \\% $$，甲瓶盐水的浓度是$$3x \\% $$，有$$100 \\times 3x \\%~ + 300\\times x \\%~ = (100 + 300) \\times 15 \\% $$，解得$$x = 10$$，即甲瓶盐水的浓度是$$30 \\% $$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1403", "queId": "abf3a45cadef4ced9c42f54c33313bf4", "competition_source_list": ["2021年春蕾杯六年级竞赛第2题2分", "2020年春蕾杯六年级竞赛第7题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "父亲的年龄是女儿现在的年龄时，女儿刚$$4$$岁；当父亲$$79$$岁时，女儿的年龄恰好是父亲现在的年龄，则父亲现在的年龄是~\\uline{~~~~~~~~~~}~岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$54$$ "}], [{"aoVal": "B", "content": "$$64$$ "}], [{"aoVal": "C", "content": "$$52$$ "}], [{"aoVal": "D", "content": "$$56$$ "}], [{"aoVal": "E", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$\\left( 79-4 \\right)\\div 3=75\\div 3=25$$（岁）， $$79-25=54$$（岁）， 答：父亲现在的年龄是$$54$$岁． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1049", "queId": "4a8d3aaab35a4a8290d52e5707ab0eb4", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（六）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "将一个分数的分母减去$$7$$得$$\\frac{7}{9}$$，分母加上$$3$$得$$\\frac{3}{4}$$，这个分数的分母比分子多（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$53$$ "}], [{"aoVal": "B", "content": "$$57$$ "}], [{"aoVal": "C", "content": "$$63$$ "}], [{"aoVal": "D", "content": "$$67$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["这个分数的分母比分子的$$\\frac{9}{7}$$倍多$$7$$，比分子的$$\\frac{4}{3}$$倍少$$3$$． 设分子为$$x$$，$$\\frac{9}{7}x+7=\\frac{4}{3}x-3$$，解得$$x=210$$，分母$$\\frac{4}{3}x-3=277$$． 分母比分子多$$277-210=67$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1027", "queId": "1781528084144b3abcb64a15533fd2d1", "competition_source_list": ["2011年其它", "2011年北京五年级竞赛"], "difficulty": "3", "qtype": "single_choice", "problem": "一项工程，甲单独做要$$12$$小时完成，乙单独做要$$18$$小时完成．若甲先做$$1$$小时，然后乙接替甲做$$1$$小时，再由甲接替乙做$$1$$小时，$$\\cdots$$，两人如此交替工作，请问：完成任务时，共用了多少小时？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$14\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$14\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$7\\frac{1}{5}$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->整体思想"], "answer_analysis": ["① 若甲、乙两人合作共需多少小时？ ~~ $$1 \\div \\left( {\\frac{1}{{12}} + \\frac{1}{{18}}} \\right) = 1 \\div \\frac{5}{{36}} = 7\\frac{1}{5}$$（小时）． ~②甲、乙两人各单独做$$7$$小时后，还剩多少？ ~~ $$1 - 7 \\times \\left( {\\frac{1}{{12}} + \\frac{1}{{18}}} \\right) = 1 - \\frac{{35}}{{36}} = \\frac{1}{{36}}$$． ~③余下的$$\\frac{1}{{36}}$$由甲独做需要多少小时？ ~~ $$\\frac{1}{{36}} \\div \\frac{1}{{12}} = \\frac{1}{3}$$（小时）． ~④共用了多少小时？ ~~ $$7 \\times 2 + \\frac{1}{3} = 14\\frac{1}{3}$$（小时）． 在工程问题中，转换条件是常用手法． 本题中，甲做$$1$$小时，乙做$$1$$小时，相当于他们合作$$1$$小时，也就是每$$2$$小时，相当于两人合做$$1$$小时． 这样先算一下一共进行了多少个这样的$$2$$小时，余下部分问题就好解决了． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "117", "queId": "12742aee661e46c49eaf1c28f40ce06a", "competition_source_list": ["2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（三）"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁四名同学进行象棋比赛，每两人都比赛$$1$$场，规定胜者得$$2$$分，平局各得$$1$$分，输者得$$0$$分．如果最后结果甲得第一，乙，丙并列第二，丁是最后一名，那么乙得分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->逻辑推理->体育比赛->2-1-0 积分制"], "answer_analysis": ["无论每场比赛的结果如何，每场比赛两队的总得分都是$$2$$分，一共有$$4$$个队，会有$$6$$场比赛，共$$2\\times 6=12$$分．每个人打$$3$$场比赛，第一名甲最多全胜得$$6$$分，乙$$+$$丙$$+$$丁$$=6$$分，所以乙和丙大于$$6\\div 3=2$$分．如果乙$$=$$丙$$=4$$分，那么甲$$4$$分，此时总分数大于$$12$$分，不可能，所以乙丙只能得$$3$$分． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "897", "queId": "78e7a3da6c5d40f59e1c77edd90e6864", "competition_source_list": ["2011年五年级竞赛明心奥数挑战赛"], "difficulty": "1", "qtype": "single_choice", "problem": "小丽用一排地砖创造了一种跳跃游戏.她将地砖标上1，2，3，4，\\ldots 并沿这一排地砖跳跃，每两块地砖着地一次，第一步落在第2块地砖上，最后停在倒数第2块地砖上.转身后她从倒数第2块地砖开始向回跳跃，这一次是每三块地砖着地一次，最后停在第1块地砖上.最后她又转身从第1块地砖开始跳跃，每五块地砖着地一次.这一次她又停在倒数第2块地砖上.那么这一排地砖共有( )块(从下列选项中选出符合条件的答案). ", "answer_option_list": [[{"aoVal": "A", "content": "39 "}], [{"aoVal": "B", "content": "40 "}], [{"aoVal": "C", "content": "47 "}], [{"aoVal": "D", "content": "49 "}], [{"aoVal": "E", "content": "53 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->中国剩余定理->逐级满足法"], "answer_analysis": ["这个数除以2余1，除以3余2，除以5余2，是$$17+30k\\left( k=0，1，2\\cdots \\right)$$ 类型的数，选项中47符合，所以选C. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1827", "queId": "adc148dfc8114969a2dac1b97782b1fd", "competition_source_list": ["2020年第1届广东深圳超常思维竞赛五年级竞赛初赛第21题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$1914$$年至$$1918$$年的战争期间，在意大利波河山谷（$$\\text{PoValley}$$）发现了一具骸骨，一件损坏的制服和一支戟（不长于$$10$$英尺的一种武器，$$1$$英尺=$$0.3048\\text{m}$$）．考古学家发现它们是属于一个法国上尉的．该戟的长度（整数）乘该法国上尉被杀死时的那个月份的天数，乘该上尉的死期到其骸骨被发现之间的年数的一半，再乘该上尉死时年龄的一半，等于$$451066$$．该上尉死于战役． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\text{Torino}$$（$$1522$$年$$2$$月） "}], [{"aoVal": "B", "content": "$$\\text{Cremona}$$（$$1712$$年$$3$$月） "}], [{"aoVal": "C", "content": "$$\\text{Pavia}$$（$$1512$$年$$2$$月） "}], [{"aoVal": "D", "content": "$$\\text{Marengo}$$（$$1800$$年$$6$$月） "}], [{"aoVal": "E", "content": "$$\\text{Castiglione}$$（$$1796$$年$$8$$月） "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设$$h=$$载的长度； $$n=$$该战役的月份中的天数； $$p=$$死期与发现日期之间的年数； 且$$a=$$该上尉死时的年龄． 则 $$h\\times n\\times \\frac{p}{2}\\times \\frac{a}{2}=451066$$，或$$h\\times n\\times p\\times a={{2}^{3}}\\times 7\\times 11\\times 29\\times 101$$，对$$n$$仅有的可能性是$$n=28$$或$$n=29$$，因此该月必是二月，首先设$$n=28$$，$$n={{2}^{2}}\\times 7$$． 我们注意到$$p$$不可能是$$2$$，$$29$$，$$11$$，$$22$$，$$58$$也不能是$$29\\times 101$$，仅有的其他可能性是： （$$1$$）$$p=2\\times 101$$，这种情况下$$a\\times h=29\\times 11$$， 但$$a\\ne 11$$（太年轻）且$$h\\ne 11$$（太长）． （$$2$$）$$p=29\\times 11$$，这种情况下$$a=2$$或$$101$$，这两者均不可能， 所以$$n\\ne 28$$． 因此$$n$$必须是$$29$$，且这年是闰年．故选$$\\text{C}$$． 注：尽管上述理由已足以回答此问题，以下的理由也说明该年是$$1512$$年．猜测$$p$$必须接近于$$400$$，唯一的可能性是$$p={{2}^{2}}\\times 101=404$$， 所以该上尉死于$$1914-404$$和$$1918-404$$之间，即$$1510$$年和$$1514$$年之间，仅有的闰年是$$1512$$年．注意，这也给出$$a\\times h=2\\times 7\\times 11$$，得出$$a=22$$，$$h=7$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1501", "queId": "56526885079349619f8c78d92095f777", "competition_source_list": ["2017年第13届湖北武汉新希望杯六年级竞赛决赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "《小学计算秘籍》的正文共$$193$$页，页码是从$$1$$到$$3$$位的连续自然数，这本书正文的页码共有个数码``$$1$$''． ", "answer_option_list": [[{"aoVal": "A", "content": "$$131$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$133$$ "}], [{"aoVal": "D", "content": "$$134$$ "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["百位上是$$1$$：$$100$ $193$$共$$94$$个; 十位上是$$1$$：$$10$ $19$$，$$110$ $119$$共$$20$$个; 个位上是$$1$$：$$1,11,21,31,\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 191$$共$$20$$个; 总共$$94+20+20=134$$（个）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1285", "queId": "f121863db17a4d638db29aa3a57d5ea2", "competition_source_list": ["2020年新希望杯五年级竞赛第11题", "2020年希望杯五年级竞赛模拟第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "【2020五年级卷第$$11$$题】大头儿子给小头爸爸计算每天的交通费用．大头儿子说：``您的车百公里油耗为$$8$$升，每天上下班共行驶$$30$$公里，油价按每升$$7.2$$元计算，您每天上下班需要支付的油费为元．'' ", "answer_option_list": [[{"aoVal": "A", "content": "$$27$$ "}], [{"aoVal": "B", "content": "$$17.28$$ "}], [{"aoVal": "C", "content": "$$1.92$$ "}], [{"aoVal": "D", "content": "$$19.2$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["因为由题干可知，车百公里油耗为$$8$$升，每天上下班共行驶$$30$$公里，则上下班共耗油$$30\\div 100\\times 8=2.4$$（升），油价按每升$$7.2$$元计算，所以每天上下班需要支付的油费为$$7.2\\times 2.4=17.28$$（元）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3406", "queId": "f735ed99132c4db69972e7cb0439c0a2", "competition_source_list": ["2016年创新杯五年级竞赛训练题（一）第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "用$$0$$、$$1$$、$$2$$三个数可以组成很多的自然数，将其从小到大依次排列起来，分别是：$$0$$，$$1$$，$$2$$，$$10$$，$$11$$，$$\\cdots $$，则$$2012$$是其中的第（~ ~ ）个数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$57$$ "}], [{"aoVal": "B", "content": "$$58$$ "}], [{"aoVal": "C", "content": "$$59$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["我们首先考虑在$$2012$$之前有多少个数，在$$2012$$之前的数考虑其各个数位可以选择的数字，我们把$$2012$$前面的所有数当做是四位数，没有的数位拿$$0$$占位置，此时千位数字：$$0$$或$$1$$；百位、十位、个位数字为$$0$$，$$1$$，$$2$$；共$$2\\times 3\\times 3\\times 3=54$$（个）．当千位数字为$$2$$时：即自第$$55$$个开始：$$2000$$，$$2001$$，$$2002$$，$$2010$$，$$2011$$，$$2012$$为第$$60$$个数． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1287", "queId": "4bbba07e4071471f8caa97e4bb1d3805", "competition_source_list": ["2021年世界少年奥林匹克数学竞赛六年级竞赛初赛第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "在外玩耍的奇奇回到家从冰箱里拿出一瓶$$100 \\%$$的纯果汁，一口气喝了$$\\frac{1}{5}$$后又放回了冰箱．第二天妈妈拿出来喝了剩下的 $$\\frac{1}{5}$$ ，觉得太浓，于是就加水兑满，摇匀之后打算明天再喝．第三天奇奇拿出这瓶果汁，一口气喝得只剩一半了．妈妈担心他喝得太多，于是就加了些水把果汁兑满．这时果汁的浓度是~\\uline{~~~~~~~~~~~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$\\% "}], [{"aoVal": "B", "content": "$$30$$\\% "}], [{"aoVal": "C", "content": "$$32$$\\% "}], [{"aoVal": "D", "content": "$$40$$\\% "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["第一天剩下纯果汁：$$1-\\frac{1}{5}=\\frac{4}{5}$$， 第二天剩下纯果汁：$$:\\frac{4}{5}\\times \\left( 1-\\frac{1}{5} \\right)=\\frac{16}{25}$$， 第三天剩下纯果汁：$$\\frac{16}{25}\\times \\left( 1-\\frac{1}{2} \\right)=\\frac{8}{25}$$， 这时果汁的浓度是：$$\\frac{8}{25}\\div 1\\times 100 \\%=32 \\%$$． 答：这时果汁的浓度是$$32 \\%$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1606", "queId": "9a38e3acfb1342d1be9e0fb09f7853ae", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（一）第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "某校学生到郊外植树，已知老师是学生人数的$$\\frac{1}{3}$$．若每位男学生种$$13$$棵树，每位女学生种$$10$$棵树，每位老师种$$15$$棵树，他们共种了$$186$$棵树，那么老师有人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["设男学生有$$a$$人，女学生有$$b$$人，则老师人数为$$\\frac{a+b}{3}$$． 可得不定方程$$13a+10b+15\\times \\frac{a+b}{3}=186$$，化简得到$$6a+5b=62$$，考虑到$$a+b$$为$$3$$的倍数，只有一组解$$a=2$$，$$b=10$$．那么老师有$$\\left( 2+10 \\right)\\div 3=4$$人． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2338", "queId": "209919afa44b4ab691ce9bb80607a2c3", "competition_source_list": ["2017年��20届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "求等差数列：$$1$$，$$6$$，$$11$$，$$16$$，$$\\cdots $$的第$$61$$项是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$296$$ "}], [{"aoVal": "B", "content": "$$301$$ "}], [{"aoVal": "C", "content": "$$306$$ "}], [{"aoVal": "D", "content": "$$311$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列求通项"], "answer_analysis": ["根据题意，这个等差数列的首项是$$1$$，公差是$$6-1=5$$，项数是$$61$$， 根据末项$$=$$首项$$+$$公差$$\\times $$（项数$$-1$$）来计算：$$1+5\\times (61-1)=301$$， 所以第$$61$$项是$$301$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1834", "queId": "cdab317522a2406e9fc68f6d149a3bd9", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛A卷第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "商店有$$A$$，$$B$$两件衣服，都获得$$200$$元的收入．$$A$$赚了$$10 \\%$$，$$B$$亏了$$10 \\%$$，总体上是． ", "answer_option_list": [[{"aoVal": "A", "content": "赚了 "}], [{"aoVal": "B", "content": "亏了 "}], [{"aoVal": "C", "content": "不亏不赚 "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["售价为$$200$$元，$$A$$赚了$$10 \\%$$， $$A$$的进价为： $$200\\div (1+10 \\%)=\\frac{2000}{11}$$（元）， $$B$$亏了$$10 \\%$$，$$B$$的进价为： $$200\\div (1-10 \\%)=\\frac{2000}{9}$$（元）， $$\\frac{2000}{11}+\\frac{2000}{9}=\\frac{40000}{99}$$（元）$$\\textgreater400$$元． 进价$$\\textgreater$$售价，亏了．选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3164", "queId": "1414343afe034b0ba73b9c966f2c0f2a", "competition_source_list": ["2017年第22届全国华杯赛小学高年级竞赛初赛第1题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "两个有限小数的整数部分分别是$$7$$和$$10$$，那么这两个有限小数的积的整数部分有（~ ~ ）种可能的取值. ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$19$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["设这两个数为$$a$$、$$b$$，则有$$7\\times 10\\textless{}ab\\textless{}8\\times 11$$，所以$$70\\textless{}ab \\textless{}88$$，其中的整数部分有$$70$$，$$71$$，$$72$$，$$\\ldots $$，$$87$$，共有$$18$$种． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1515", "queId": "449cf9911bc540918b21865658db5f59", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛决赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$A$$地植树$$1000$$棵，$$B$$地植树$$1250$$棵，甲、乙、丙每天分别能植树$$28$$、$$32$$、$$30$$棵，甲在$$A$$地、乙在$$B$$地、丙跨$$A$$与$$B$$两地，同时开始、同时结束，整个过程所用的时间是天． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->知识点->应用题模块->工程问题->合作工程问题->双工程问题"], "answer_analysis": ["工程问题．从开始到结束三人一共用了$$(1000+1250)\\div (28+32+30)=2250\\div 90=25$$天． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "135", "queId": "16c2083f466845c2b5a5cb00bde5d723", "competition_source_list": ["2019年新加坡高级学府数学竞赛（SASMO）五年级竞赛第14题"], "difficulty": "1", "qtype": "single_choice", "problem": "四个好朋友发言如下: Kieran:我不是最矮的． Franklin:我不是最高的也不是最矮的． Mabel:我是最矮的． Jeff:我是最高的． 如果他们中只有一个人说谎了，那谁是最高的? ", "answer_option_list": [[{"aoVal": "A", "content": "$$Kieran$$ "}], [{"aoVal": "B", "content": "$$Franklin$$ "}], [{"aoVal": "C", "content": "$$Mabel$$ "}], [{"aoVal": "D", "content": "$$Jeff$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->组合模块->逻辑推理->条件型逻辑推理->表格法", "拓展思维->拓展思维->组合模块->逻辑推理"], "answer_analysis": ["如果$$Kieran$$说谎，那么他最矮，则和$$Mabel$$矛盾，同理$$Mabel$$也不能说谎，$$Franklin$$如果说谎的话则和$$Jeff$$或者$$Kieran$$矛盾，所以只能$$Jeff$$说谎，那么$$Jeff$$和$$Franklin$$ 不是最高，$$Mabel$$最矮，所以$$Kieran$$最高. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3026", "queId": "9d29382b4ef549d0adfcdae100750f10", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "找出规律，将你认为合适的数填入（ ），$$2$$、$$4$$、$$3$$、$$9$$、$$4$$、$$16$$、$$5$$、（ ）、（ ）、$$36$$、$$7$$、$$\\cdots $$ 那么正确的数是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$、$$6$$ "}], [{"aoVal": "B", "content": "$$22$$、$$6$$ "}], [{"aoVal": "C", "content": "$$25$$、$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数"], "answer_analysis": ["解：注意到： $$4$$是$$2$$的平方， $$9$$是$$3$$的平方， $$16$$是$$4$$的平方， $$25$$是$$5$$的平方， $$36$$是$$6$$的平方， $$\\cdots $$ 根据这个规律，可知中间两个括号分别应填$$25$$和$$6$$。 故选:C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1205", "queId": "15f86cf9cb0e435dbbaf1a0968b068aa", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "祖玛游戏中，龙嘴里不断吐出很多颜色的龙珠，先$$4$$颗红珠，接着$$3$$颗黄珠，再$$2$$颗绿珠，最后$$1$$颗白珠，按此方式不断重复，从龙嘴里吐出的第$$2021$$颗龙珠是． ", "answer_option_list": [[{"aoVal": "A", "content": "红珠 "}], [{"aoVal": "B", "content": "黄珠 "}], [{"aoVal": "C", "content": "绿珠 "}], [{"aoVal": "D", "content": "白珠 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["$$2021\\div (4+3+2+1)=201（组）\\ldots\\ldots1（个）$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3290", "queId": "ba04d534e6e7442ea88fa8cc88b9e3e8", "competition_source_list": ["2018年河南郑州K6联赛竞赛初赛第23题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "给正方体的六个面分别涂上红黄蓝三种颜色，任意抛$$30$$次，要想红色朝上次数最少，蓝色朝上次数最多，~\\uline{~~~~~~~~~~}~面涂红色，~\\uline{~~~~~~~~~~}~涂黄色． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2,2$$ "}], [{"aoVal": "B", "content": "$$1,2$$ "}], [{"aoVal": "C", "content": "$$1,3$$ "}], [{"aoVal": "D", "content": "$$2,3$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->基本概率->可能的情况"], "answer_analysis": ["可能性；要想红色朝上次数最少，则$$1$$面涂红色，$$2$$面染黄色． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2700", "queId": "762a0d3d4e5541c4b3ce99fceebf2ecd", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛B卷第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "算式$$\\frac{1}{\\dfrac{1}{20}+\\dfrac{1}{21}+\\dfrac{1}{22}+\\cdots +\\dfrac{1}{29}}$$ 的整数部分是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$\\frac{1}{29}\\times 10\\textless{}\\frac{1}{20}+\\frac{1}{21}+\\cdots +\\frac{1}{29}\\textless{}\\frac{1}{20}\\times 10$$， $$\\frac{10}{29}\\textless{}\\frac{1}{20}+\\frac{1}{21}+\\cdots +\\frac{1}{29}\\textless{}\\frac{1}{2}$$， $$\\frac{2}{1}\\textless{}\\frac{1}{\\dfrac{1}{20}+\\dfrac{1}{21}+\\cdots +\\dfrac{1}{29}}\\textless{}\\frac{29}{10}$$， 所以它的整数部分应是$$2$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3347", "queId": "7293b9904328449788b991b276797446", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$、$$B$$、$$C$$、$$D$$、$$E$$、$$F$$六人抽签推选代表，公证人一共制作了六枚外表一模一样的签，其中只有一枚刻着``中''，六人按照字母顺序先后抽取签，抽完不放回，谁抽到``中''字，即被推选为代表，那么这六人中，谁被抽中的概率谁最大？ ", "answer_option_list": [[{"aoVal": "A", "content": "一样大 "}], [{"aoVal": "B", "content": "$$A$$ "}], [{"aoVal": "C", "content": "$$D$$ "}], [{"aoVal": "D", "content": "$$E$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["抽中的概率为$$\\frac{1}{6}$$，没抽到的概率为$$\\frac{5}{6}$$，如果$$A$$没抽中，那么$$B$$有$$\\frac{1}{5}$$的概率抽中，如果$$A$$抽中，那么$$B$$抽中的概率为$$0$$，所以$$B$$抽中的概率为$$\\frac{5}{6}\\times \\frac{1}{5}=\\frac{1}{6}$$． 同理，$$C$$抽中的概率为$$\\frac{5}{6}\\times \\frac{4}{5}\\times \\frac{1}{4}=\\frac{1}{6}$$，$$D$$抽中的概率为$$\\frac{5}{6}\\times\\frac{4}{5}\\times \\frac{3}{4}\\times \\frac{1}{3}=\\frac{1}{6}$$， $$E$$抽中的概率为$$\\frac{5}{6}\\times\\frac{4}{5}\\times \\frac{3}{4}\\times \\frac{2}{3}\\times \\frac{1}{2}=\\frac{1}{6}$$，$$F$$抽中的概率为$$\\frac{5}{6}\\times\\frac{4}{5}\\times \\frac{3}{4}\\times \\frac{2}{3}\\times \\frac{1}{2}\\times1=\\frac{1}{6}$$． 由此可见六人抽中的概率相等，与抽签的先后顺序无关． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "186", "queId": "3e7b6c18a3774f0d86eafe0e308110d3", "competition_source_list": ["2018年湖北武汉新希望杯小学高年级五年级竞赛训练题（四）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "把$$19$$分成几个自然数的和，这几个自然数的乘积最大是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$512$$ "}], [{"aoVal": "B", "content": "$$729$$ "}], [{"aoVal": "C", "content": "$$768$$ "}], [{"aoVal": "D", "content": "$$972$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$3\\times 3=9$$，$$2\\times 2\\times 2=8$$，$$9\\textgreater8$$，所以分成$$2$$个$$3$$比分成$$3$$个$$2$$乘积大； $$3\\times 1=3$$，$$2\\times 2=4$$，$$3\\textless{}4$$，所以分成两个$$2$$比分成$$1$$个$$3$$和$$1$$个$$1$$乘积大，$$19=3+3+3+3+3+2+2$$，乘积为$$3\\times 3\\times 3\\times 3\\times 3\\times 2\\times 2=972$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "527", "queId": "eb482cd1c4f9488e87d5d156fc6d23d8", "competition_source_list": ["2013年IMAS小学中年级竞赛第一轮检测试题第6题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "小王从家到学校只有两种方式可供选择：（$$a$$）步行$$8$$分钟到离家较近的公车站，然后乘$$15$$分钟的公车可导学校；（$$b$$）步行$$10$$分钟到离家较近的地铁站，然后乘$$10$$分钟的地铁可到学校．如果不计等车的时间，请问小王从家到学校至少需要多少分钟？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$25$$ "}], [{"aoVal": "E", "content": "$$33$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["从家到学校，若小王选择第一种方式需要$$8+15=23$$分钟，选择第二种方式则需要$$10+10=20$$分钟，所以小王从家到学校至少需要$$20$$分钟．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2771", "queId": "b13f290d7d9b4de7809702c154b58ad6", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$1\\div (3\\div 5)\\div (5\\div 7)\\div (7\\div 9)\\div (9\\div 13)\\div (13\\div 15)=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["原式$$=1\\div 3\\times 5\\div 5\\times 7\\div 7\\times 9\\div 9\\times 13\\div 13\\times 15$$$$=1\\div 3\\times 15$$$$=5$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1987", "queId": "dc8297b650ad49e5af60027f34ff0a3e", "competition_source_list": ["2021年新希望杯一年级竞赛初赛第14题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$5$$ animals in a row, dogs and cats are adjacent to chickens, and cats and rats are adjacent to rabbits, so are adjacent to cats. $$5$$只动物排成一排，狗和猫都与鸡相邻，猫和鼠都与兔相邻，那么都与猫相邻． ", "answer_option_list": [[{"aoVal": "A", "content": "Rat and Rooster鼠和鸡 "}], [{"aoVal": "B", "content": "Chicken and Rabbit鸡和兔 "}], [{"aoVal": "C", "content": "Rabbit and dog兔和狗 "}], [{"aoVal": "D", "content": "Rabbit and Rat兔和鼠 "}], [{"aoVal": "E", "content": "Rat and dog鼠和狗 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["根据题意分析可知，狗和猫都与鸡相邻，即鸡的左右两边是猫和狗，又因为猫与老鼠都与兔相邻，所以猫的旁边是兔子和老鼠，兔子和老鼠在猫的同一侧，由此可知，与猫相邻的是鸡和兔． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1580", "queId": "bee4316c365949798e2baad788e77b6d", "competition_source_list": ["2008年第6届创新杯四年级竞赛初赛A卷第8题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在一条长$$2008$$米公路的两侧栽树，每隔$$8$$米栽一棵，一共要栽． ", "answer_option_list": [[{"aoVal": "A", "content": "$$251$$棵 "}], [{"aoVal": "B", "content": "$$252$$棵 "}], [{"aoVal": "C", "content": "$$502$$棵 "}], [{"aoVal": "D", "content": "$$504$$棵 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde(2008\\div 8+1)\\times 2$$ $$=(251+1)\\times 2$$ $$=252\\times 2$$ $$=504$$（棵） 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3113", "queId": "e256fe01f40745dbb6b338e0d8b8e00b", "competition_source_list": ["2015年第13届全国创新杯六年级竞赛第1题", "小学高年级六年级其它2015年数学思维能力等级测试初试第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "五楼的王老师病了，小孙帮王老师送早点，从一楼到二楼用了$$\\frac{3}{4}$$分钟，用同样的速度从一楼走到五楼王老师家要用（~ ）分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{15}{4}$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$\\frac{20}{3}$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数运算->分数乘法运算"], "answer_analysis": ["$$\\frac{3}{4}\\times 4=3$$分钟． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1995", "queId": "b873333b2e5248cc86cffaa460b1a8e5", "competition_source_list": ["2013年IMAS小学高年级竞赛第一轮检测试题第15题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有甲、乙两个水桶，甲桶中装了$$\\frac{1}{6}$$满的水，乙桶中装了$$60$$升的水，如果把乙桶中的水全部倒到甲桶中，那么甲桶将装有$$\\frac{1}{2}$$满的水；如果现在把甲桶中的水全部倒到乙桶中，那么乙桶刚好装满水．请问乙桶的容量为多少升？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$70$$ "}], [{"aoVal": "B", "content": "$$80$$ "}], [{"aoVal": "C", "content": "$$90$$ "}], [{"aoVal": "D", "content": "$$100$$ "}], [{"aoVal": "E", "content": "$$180$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["由题意可知甲桶的容量的$$\\frac{1}{2}-\\frac{1}{6}=\\frac{1}{3}$$为$$60$$升，因此甲桶的容量为$$60\\div \\frac{1}{3}=180$$升，故乙桶的容量为$$180\\times \\frac{1}{2}=90$$升． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2822", "queId": "651034d3d9214d6590f6747a6803d197", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "以下哪一个选项中的数小于$$2010000$$？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$201$$万 "}], [{"aoVal": "B", "content": "$$2100000$$ "}], [{"aoVal": "C", "content": "$$102$$万 "}], [{"aoVal": "D", "content": "$$20100000$$ "}], [{"aoVal": "E", "content": "$$210$$万 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\text{A}$$等于数$$2010000$$、$$\\text{B}$$大于数$$2010000$$、$$\\text{C}$$小于数$$2010000$$、$$\\text{D}$$大于数$$2010000$$、$$\\text{E}$$大于数$$2010000$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2348", "queId": "09e9d058f6e74700ab8017c1412cc278", "competition_source_list": ["2017年新希望杯六年级竞赛训练题（一）第4题", "2018年湖北武汉新希望杯六年级竞赛训练题（一）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "将四个分数按从小到大的顺序排列，正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ "}], [{"aoVal": "C", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ "}], [{"aoVal": "D", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数通分法->通分母", "课内体系->能力->运算求解"], "answer_analysis": ["通分子$$\\frac{5}{14}=\\frac{60}{168}$$， $$\\frac{10}{27}=\\frac{60}{162}$$， $$\\frac{12}{31}=\\frac{60}{155}$$， $$\\frac{20}{53}=\\frac{60}{159}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "557", "queId": "056b4b73cc964b44b2ff4d4c3debe1c9", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知一个质数的三倍与另一个质数的五倍的和是$$301$$，则这两个质数的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$61$$ "}], [{"aoVal": "B", "content": "$$85$$ "}], [{"aoVal": "C", "content": "$$99$$ "}], [{"aoVal": "D", "content": "$$61$$或$$99$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["质数是指在大于$$1$$的自然数中，除了$$1$$和它本身以外不再有其他因数的自然数． 在所有质数中，只有$$2$$这一个偶数；其他质数都是奇数； 但是奇数的$$3$$倍或者$$5$$倍都是奇数；两个奇数之和不可能是奇数； 所以这两个质数必然有$$2$$； 若第一个质数是$$2$$，$$2\\times 3+$$质数$$\\times 5=301$$，第二个质数$$=\\left( 301-6 \\right)\\div 5=59$$； 若第二个质数是$$2$$，质数$$\\times 3+2\\times 5=301$$，第一个质数$$=\\left( 301-10 \\right)\\div 3=97$$； 所以这两个质数的和$$61$$或$$99$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1764", "queId": "77d7d5191eb8443b9dc4ba483e085119", "competition_source_list": ["2017年第20届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "有一幢$$10$$层的大楼，由于停电，电梯停开，某人从$$1$$层走到$$3$$层需要$$30$$秒，照这样计算，他从$$3$$层走到$$10$$层需要秒． ", "answer_option_list": [[{"aoVal": "A", "content": "$$70$$秒 "}], [{"aoVal": "B", "content": "$$80$$秒 "}], [{"aoVal": "C", "content": "$$105$$秒 "}], [{"aoVal": "D", "content": "$$120$$秒 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都有->爬楼梯问题"], "answer_analysis": ["从$$1$$层走到$$3$$层走的楼梯层数是：$$3-1=2$$个，走一个楼层用时为：$$30\\div2=15$$秒，那么从$$3$$层走到$$10$$层走的楼梯层数是：$$10-3=7$$个，要用时为：$$15\\times7=105$$秒，据此解答． $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde30\\div (3-1)\\times (10-3)$$ $$=15\\times7$$ $$=105$$（秒）， 答：他从$$3$$层走到$$10$$层需要$$105$$秒． 故答案为：$$105$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2678", "queId": "75f97c9c69e24fcbb973a7351c81f465", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知等差数列$$13$$，$$18$$，$$23$$，$$28$$，$$\\cdots $$，$$1003$$．这个等差数列共有项． ", "answer_option_list": [[{"aoVal": "A", "content": "$$198$$ "}], [{"aoVal": "B", "content": "$$199$$ "}], [{"aoVal": "C", "content": "$$200$$ "}], [{"aoVal": "D", "content": "$$201$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意：公差为$$5$$， 所以项数：$$\\left( 1003-13 \\right)\\div 5+1=199$$． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "884", "queId": "970acb6c2fb8488685ae45b7744170e1", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "用两个质数之和来表示$$100$$有许多种方法，在这些方法中，两个质数中大数减小数的差最小是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["欲求两个质数中大数减小数的差的最小值即这两个质数越接近越好． 则这两��质数分别为$$47$$和$$53$$． 则差：$$53-47=6$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1176", "queId": "ec5538740df44af893c14ebd4bc47549", "competition_source_list": ["2014年迎春杯四年级竞赛初赛", "2014年迎春杯三年级竞赛初赛", "2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "动物园的饲养员把一堆桃子分给若干只猴子，如果每只猴子分$$6$$个，剩$$57$$个桃子；如果每只猴子分$$9$$个，就有$$5$$只猴子一个也分不到，还有一只猴子只分到$$3$$个。那么，共有( )个桃子。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$216$$ "}], [{"aoVal": "B", "content": "$$324$$ "}], [{"aoVal": "C", "content": "$$273$$ "}], [{"aoVal": "D", "content": "$$301$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->盈亏问题->盈亏基本类型->盈亏基本类型盈亏问题"], "answer_analysis": ["每只猴子多分了$$3$$个，分$$5\\times 9+\\left( 9-3 \\right)+57=108$$（个），那么共$$108\\div 3=36$$（只）猴子，共$$36\\times 6+57=273$$（个）桃子。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "213", "queId": "a2b1aa37363b444cb0f92597797304bc", "competition_source_list": ["2006年四年级竞赛创新杯", "2006年第4届创新杯四年级竞赛复赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "桌子上并排放着三张扑克牌，A右边的两张中至少有一张K，而K的左边两张中至少有一张K，黑桃左边的两张至少有一张红桃，而红桃右边的两张中也至少有一张红桃，中间的那张牌是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "红桃A "}], [{"aoVal": "B", "content": "红桃K "}], [{"aoVal": "C", "content": "黑桃A "}], [{"aoVal": "D", "content": "黑桃K "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["从左至右，顺次有三张牌，编号为①②③由第一、二个条件得①为A，②为K，③为K.因为，黑桃左边有两张可得③为黑桃，即可得出③为黑桃K，且①、②中至少有一张是红桃.因为红桃右边有两张，可得①为红桃，即得出①为红桃A，且②、③中至少有一张红桃.因为，③为黑桃K，所以②为红桃.因此得出②为红桃K，即中间的一张为红桃K.故选B "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1994", "queId": "d382c1142f6f4efda3d8f286346db729", "competition_source_list": ["2011年四年级其它", "2016年第3届广东深圳鹏程杯六年级竞赛集训材料第十五章综合训练题（五）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "食品店上午卖出每千克为$$20$$元、$$25$$元、$$30$$元的$$3$$种糖果共$$100$$千克，共收入$$2570$$元．已知其中售出每千克$$25$$元和每千克$$30$$元的糖果共收入了$$1970$$元，那么，每千克$$25$$元的糖果售出了多少千克？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$23$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$25$$ "}], [{"aoVal": "D", "content": "$$26$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->变型题"], "answer_analysis": ["每千克$$25$$元和每千克$$30$$元的糖果共收入了$$1970$$元，则每千克$$20$$元的收入：$$2570-1970=600$$元， 所以卖出：$$600\\div 20=30$$千克， 所以卖出每千克$$25$$元和每千克$$30$$克的糖果共$$100-30=70$$千克， 相当于将题目转换成： 卖出每千克$$25$$元和每千克$$30$$克的糖果共$$70$$千克，收入$$1970$$元，问：每千克$$25$$元的糖果售出了多少千克？ 转换成了最基本的鸡兔同笼问题． 假设全是每千克$$25$$元的，$$\\left( 1970-25\\times 70 \\right)\\div \\left( 30-25\\right)=44$$（千克）， 所以$$30$$元的是$$44$$千克，所以$$25$$元的有：$$70-44=26$$（千克）． 关键：将三种以及更多的动物/东西，转化为两种最基本模型，即：抓住转化后的``头''与``脚''． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "6", "queId": "01209832d21f43598c8e6147842abaad", "competition_source_list": ["1993年四年级竞赛创新杯", "2006年第4届创新杯四年级竞赛初赛B卷第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙、丁四位同学的运动衫上印有不同的号码。 赵说：``甲是$$2$$号，乙是$$3$$号。'' 钱说：``丙是$$4$$号，乙是$$2$$号。'' 孙说：``丁是$$2$$号，丙是$$3$$号。'' 李说：``丁是$$4$$号，甲是$$1$$号。'' 又知道赵、钱、孙、李每人都只说对了一半，那么丙的号码是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["无论赵的哪半句对了，钱的后半句一定错了，那前半句一定对了，丙是$$4$$号。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2389", "queId": "d502b640a5ef4c999b1be584d18dc564", "competition_source_list": ["2020年广东广州羊排赛六年级竞赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$\\frac{5}{7}$$，$$ \\frac{2}{13}$$， $$\\frac{3}{4}$$，$$\\frac{10}{17}$$ 几个分数中，按从大到小排列，排在第二位的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{7}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{13}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{10}{17}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数化小数法"], "answer_analysis": ["$$\\frac{5}{7}=0.714285714285\\cdots \\cdots $$，$$\\frac{2}{13}=0.153846153846\\cdots \\cdots $$， $$\\frac{3}{4}=0.75$$，$$\\frac{10}{17}=0.588235294\\cdots \\cdots $$， $$0.75\\textgreater0.714285714285\\cdots \\cdots \\textgreater0.588235294\\cdots \\cdots \\textgreater$$ $$0.153846153846\\cdots \\cdots $$，所以$$\\frac{3}{4}\\textgreater\\frac{5}{7}\\textgreater\\frac{10}{17}\\textgreater\\frac{2}{13}$$， 则从大到小排列排在第二位的是$$\\frac{5}{7}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2489", "queId": "6243720213274932ad5613d3de10637f", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "把一个自然数$$n$$的数位上的偶数数字相加所得的和记为$$E(n)$$，例如：$$E\\left( 1999 \\right)=0$$，$$E\\left( 2000 \\right)=2$$，$$E\\left( 2021 \\right)=2+2=4$$．则$$E\\left( 1 \\right)+E\\left( 2 \\right)+E\\left( 3 \\right)+\\cdots +E\\left( 100 \\right)=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$400$$ "}], [{"aoVal": "D", "content": "$$2020$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$2$$，$$4$$，$$6$$，$$8$$这$$4$$个数字，每个在个位出现$$10$$次，在十位出现$$10$$次， 所以$$E(1)+E(2)+···+E(100)=(2+4+6+8)\\times (10+10)=400$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "914", "queId": "fc003f9051084739a4223683a1f05a85", "competition_source_list": ["2011年全国美国数学大联盟杯小学高年级五年级竞赛初赛第34题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$180$$的因数中最大的完全平方数比最小的完全平方数大多少? ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$27$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$35$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->因数与倍数基础"], "answer_analysis": ["最大完全平方数是$$36$$，最小完全平方数是$$1$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1667", "queId": "892528168a2e4be5a1936caffece498c", "competition_source_list": ["2021年新希望杯六年级竞赛初赛第19题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "对角巷的魔药店进了一批曼德拉草，按$$100 \\%$$的利润率来定价，结果只售出$$30 \\%$$的曼德拉草．为尽早售出剩下的曼德拉草，魔药店决定打六折销售，结果剩余的曼德拉草销售一空．这批曼德拉草的利润率是~\\uline{~~~~~~~~~~}~$$ \\%$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$44$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$28$$ "}], [{"aoVal": "D", "content": "$$40$$ "}], [{"aoVal": "E", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["假设这一批产品每件成本为$$1$$元，一共有$$100$$件，总成本为$$1\\times 100=100$$（元）， 原价：$$1\\times (1+100 \\%)=2$$（元）， 打折后：$$2\\times 0.6=1.2$$（元）， 原价售出$$30 \\%$$：$$2\\times 100\\times 30 \\%=60$$（元）， 打折后全部售出：$$1.2\\times 100\\times(1-30 \\%)=84$$（元）， 总售价：$$60+84=144$$（元）， 利润率：$$\\frac{144-100}{100}=44 \\%$$． 答：实际利润率为$$44 \\%$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1974", "queId": "c13b52ba0d684e4c83f53c69ea5d183b", "competition_source_list": ["2013年第25届广东广州五羊杯六年级竞赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小红比她爸爸少$$26$$岁，已知三年后她爸爸的年龄是她的三倍，则今年爸爸和小红的岁数的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$43$$ "}], [{"aoVal": "B", "content": "$$46$$ "}], [{"aoVal": "C", "content": "$$49$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型"], "answer_analysis": ["设小红今年$$x$$岁．则能列出方程：$$3(x+3)=x+26+3$$， 解出$$x=10$$，即小红今年$$10$$岁，父亲今年$$36$$岁，所以年龄和为$$46$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "339", "queId": "7741650d7d8b4cd2b911605b443a3d65", "competition_source_list": ["2012年全国学而思杯三年级竞赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "房间里有$$3$$种小动物：小白鼠、小花猫、小黄狗．房间里如果猫的数量不超过狗，狗就会欺负猫；如果鼠的数量不超过猫，猫就会欺负鼠；如果猫、狗数量之和不超过鼠，鼠就会偷吃东西．现在小白鼠、小花猫、小黄狗三种小动物在房间里相安无事，但是再进来任意一只，都会打破平衡．那么，原来房间里有多少只小动物？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}], [{"aoVal": "E", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->比较型逻辑推理", "Overseas Competition->知识点->组合模块->逻辑推理"], "answer_analysis": ["不超过表示等于或小于均可．现在相安无事，说明猫比狗多，鼠比猫多，猫与狗之和比鼠多，从再进来任意$$1$$只，都会打破平衡可知，猫比狗多$$1$$只，鼠比猫多$$1$$只，猫与狗之和比鼠多$$1$$只，经尝试知有$$4$$只鼠，$$3$$只猫，$$2$$只狗． 再进$$1$$只猫，会打破平衡，所以鼠比猫多$$1$$只； 再进$$1$$只狗，会打破平衡，所以猫比狗多$$1$$只； 再进$$1$$只鼠，会打破平衡，所以猫与狗的和比鼠多$$1$$只； 比较第一个条件和第三个条件可知，猫有$$3$$只，进而求得鼠有$$4$$只，狗有$$2$$只． 共有$$4+3+2=9$$（只）小动物． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1511", "queId": "51ebb2ff78fc476b9ca7bde2063b04ca", "competition_source_list": ["2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "某次小明坐缆车上山，从他坐上车开始，一路上他一直在数对面交错而过的缆车数量，一共是$$60$$辆，则在这一整条索道上（双向）共有辆缆车？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$辆 "}], [{"aoVal": "B", "content": "$$60$$辆 "}], [{"aoVal": "C", "content": "$$61$$辆 "}], [{"aoVal": "D", "content": "$$120$$辆 "}]], "knowledge_point_routes": ["拓展思维->七大能力->实践应用"], "answer_analysis": ["缆车是循环的，它遇到的第一辆就是紧随其后的，那么他遇到的最后一辆就是它前面的，所以除了它自己这辆，还有$$60$$辆，一共是$$60+1=61$$辆． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1440", "queId": "e81d669db1e44391be06151742fd663b", "competition_source_list": ["2021年春蕾杯六年级竞赛第2题2分", "2020年春蕾杯六年级竞赛第7题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "父亲的年龄是女儿现在的年龄时，女儿刚$$4$$岁；当父亲$$79$$岁时，女儿的年龄恰好是父亲现在的年龄，则父亲现在的年龄是岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$54$$ "}], [{"aoVal": "B", "content": "$$64$$ "}], [{"aoVal": "C", "content": "$$52$$ "}], [{"aoVal": "D", "content": "$$56$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\left( 79-4 \\right)\\div 3=75\\div 3=25$$（岁）， $$79-25=54$$（岁）， 答：父亲现在的年龄是$$54$$岁． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "596", "queId": "0fcc3db86db240ec86eb7c52f32d138b", "competition_source_list": ["2014年迎春杯三年级竞赛初赛", "2014年迎春杯四年级竞赛初赛", "2014年迎春杯三年级竞赛初赛", "2014年迎春杯四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "你能根据以下的线索找出百宝箱的密码吗？ （$$1$$）密码是一个八位数； （$$2$$）密码既是$$3$$的倍数又是$$25$$的倍数； （$$3$$）这个密码在$$20000000$$到$$30000000$$之间； （$$4$$）百万位与十万位上的数字相同； （$$5$$）百位数字比万位数字小$$2$$； （$$6$$）十万位、万位、千位上数字组成的三位数除以千万位、百万位上数字组成的两位数，商是$$25$$。 依据上面的条件，推理出这个密码应该是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$25526250$$ "}], [{"aoVal": "B", "content": "$$26650350$$ "}], [{"aoVal": "C", "content": "$$27775250$$ "}], [{"aoVal": "D", "content": "$$28870350$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法"], "answer_analysis": ["解：（$$1$$）四个选项都是$$8$$位数； （$$2$$）四选项都是$$25$$的倍数，C的数字和是$$35$$不是$$3$$的倍数，排除C； （$$3$$）都满足条件； （$$4$$）都满足条件； （$$5$$）A百位数字和万位数字相等，D百位数字比万位数字少$$4$$，不满足条件； （$$6$$）B满足条件。 故选：B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1382", "queId": "b9b4adc037374c2db107d0d54829656d", "competition_source_list": ["2004年六年级竞赛创新杯", "2004年第2届创新杯六年级竞赛初赛第5题", "2021年广东广州番禺区执信中学附属小学小升初（分班考）第17题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两根同样长的绳子，甲绳先剪去$$\\frac{1}{3}$$，再剪去$$\\frac{1}{3}$$米；乙绳先剪去$$\\frac{1}{3}$$米，再剪去剩下部分的$$\\frac{1}{3}$$．两根绳子剩下部分的长度相比较是． ", "answer_option_list": [[{"aoVal": "A", "content": "甲绳剩下的部分长 "}], [{"aoVal": "B", "content": "乙绳剩下的部分长 "}], [{"aoVal": "C", "content": "甲绳与乙绳剩下的部分同样长 "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->量率对应已知单位1"], "answer_analysis": ["选$$\\text{B}$$ 假设甲、乙原来长度都是$$a$$米， 甲剩下($$\\frac{2}{3}a-\\frac{1}{3}$$)米； 乙剩下$$\\frac{2}{3}（a-\\frac{1}{3})$$=$$\\frac{2}{3}a-\\frac{2}{9}$$米 所以乙剩下的部分长. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2327", "queId": "f9a8b9dc7d954e768ba17fd1e4147db7", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）"], "difficulty": "1", "qtype": "single_choice", "problem": "一列火车通过一座长$$320$$米的桥用了$$105$$秒，当它通过$$860$$米的隧道时，速度是过桥速度的$$2$$倍，结果用了$$120$$秒，火车通过大桥时的速度是每秒（~ ）米；火车的车身长度为（~ ）米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$；$$100$$ "}], [{"aoVal": "B", "content": "$$4$$；$$100$$ "}], [{"aoVal": "C", "content": "$$8$$；$$100$$ "}], [{"aoVal": "D", "content": "$$8$$；$$200$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["若通过$$860$$米隧道时速度不变则需要$$120\\times 2=240$$（秒），火车过桥速度：$$\\left( 860-320 \\right)\\div \\left( 240-105 \\right)=4$$（米/秒）：火车车身长：$$105\\times 4-320=100$$（米）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1420", "queId": "7a3c52c432404b18aab5b6165cca75ad", "competition_source_list": ["2014年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁四个人今年的年龄之和是$$72$$岁，几年前（至少一年）甲是$$22$$岁时，乙是$$16$$岁，又知道，当甲是$$19$$岁的时候，丙的年龄是丁的$$3$$倍（此时丁至少$$1$$岁）。如果甲、乙、丙、丁四个人的年龄互不相同，那么今年甲的年龄可以有（ ）种情况。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之和倍型"], "answer_analysis": ["已知四个人今年的年龄之和是$$72$$岁，几年前（至少一��）甲是$$22$$岁时，乙是$$16$$岁；当甲是$$19$$岁时，则乙$$13$$岁，丙的年龄是丁的$$3$$倍（此时丁至少$$1$$岁），因为此时是至少$$4$$年前，所以$$4$$个人的年龄和不超过$$72-4\\times 4=56$$（岁）。 设此时丁的年龄是$$x$$岁，丙的年龄是$$3x$$岁，则$$19+13+x+3x\\leqslant 56$$，解得$$x\\leqslant 6$$，又知道$$x\\geqslant 1$$，所以丁的年龄有$$6$$种情况。 如果甲、乙、丙、丁四个人的年龄互不相同，所以今年甲的年龄可以有$$6$$种情况。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1871", "queId": "978a7d82a8cb49c399f08630dbc2070a", "competition_source_list": ["2016年第3届广东深圳鹏程杯六年级竞赛集训材料第十九章综合训练题（九）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "小红从一本书的第$$54$$页阅读到第$$67$$页，小明从第$$95$$页阅读到第$$135$$页，小莉从第$$180$$页阅读到第$$273$$页，他们总共阅读了多少页？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$147$$ "}], [{"aoVal": "B", "content": "$$148$$ "}], [{"aoVal": "C", "content": "$$149$$ "}], [{"aoVal": "D", "content": "$$150$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["解：阿里读的页数是$$67-54+1=14$$， 同理苏明读的页数是$$135-95+1=41$$， 梅莉阅读的页数是$$273-180+1=94$$， 他们共读$$14+41+94=149$$页． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1493", "queId": "447832eddfca452aa873c251cb58becf", "competition_source_list": ["2010年第11届上海中环杯小学中年级三年级竞赛第11题12分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个四口之家，由爸爸、妈妈、大儿子和小儿子组成，他们的年龄之和是$$68$$岁，爸爸比妈妈大$$2$$岁．$$3$$年前，这个家庭的成员的年龄之和是$$57$$岁；$$5$$年前，这个家庭的成员的年龄之和是$$52$$岁．请问这个家庭每个成员现在的年龄各是多少岁？ There are four members in a family, dad, mom, elder brother, and little brother. Their sum of the ages this year is $68$, and dad is $2$ years older than mom. $3$ years ago from now, their sum of the ages was $57$; $5$ years ago from now, their sum of ages was $52$. How old is dad this year? ", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$ "}], [{"aoVal": "B", "content": "$$32$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$27$$ "}], [{"aoVal": "E", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析", "Overseas Competition->知识点->应用题模块->年龄问题"], "answer_analysis": ["$$3$$年前，四人年龄之和是$$68-3\\times 4=56$$（岁），但实际上是$$57$$岁，这说明$$3$$ 年前小儿子还没有出生，是$$2$$年前出生的，所以小儿子今年是$$2$$岁．则其余$$3$$人年龄之和是$$66$$岁．$$5$$年前，$$3$$人的年龄和应该是$$66-5\\times 3=51$$（岁），但实际上是$$52$$岁，这说明$$5$$年前大儿子还没有出生，是$$4$$年前出生的，所以大儿子今年是$$4$$岁．则爸爸妈妈的年龄和是$$62$$岁，而爸爸比妈妈大两岁，所以爸爸今年$$32$$岁，妈妈今年$$30$$岁． $3$ years ago from now, their sum of ages should be $$68-3\\times 4=56$$ but actually it was $57$, which means the little brother was not born $3$ years ago. He was born $2$ years ago. Thus, the little brother this year is $2$ years old and the sum of the other three members this year is $68-2=66$ years old. $5$ years ago from now, the sum of the three members ages should be $66-5\\times 3=51$ but actually it was $52$, whcih means the elder brother was not born $5$ years ago. He was born $4$ years ago. Thus, the elder brother is $4$ years old this year. Thus, the sum of ages of mom and dad this year is $66-4=62$. Dad is $2$ years older than mom, which means dad is $32$ years old and mom is $30$ years old. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1846", "queId": "e00f5265f3fe4eb4871f6ab41a80cb58", "competition_source_list": ["2016年创新杯六年级竞赛训练题（三）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "加工一批零件，甲、乙两人的工效比为$$5:4$$，两人同时加工$$2$$小时以后，甲的机器发生了故障停工修理一小时，（乙按原工效继续工作），然后甲以原工效的$$60 \\%$$与乙合作完了剩下的零件．因此比原计划推迟$$1\\frac{4}{7}$$小时完成全部任务，求实际用（~ ）小时完成全部任务． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4\\frac{4}{7}$$小时 "}], [{"aoVal": "B", "content": "$$6\\frac{4}{7}$$小时 "}], [{"aoVal": "C", "content": "$$7\\frac{4}{7}$$小时 "}], [{"aoVal": "D", "content": "$$8\\frac{4}{7}$$小时 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["略 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1359", "queId": "5539cdc83659420194561a34cc286f14", "competition_source_list": ["2019年美国数学大联盟杯竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "$$8$$千克的香蕉是$$12$$美元．$$12$$千克的香蕉是多少钱？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->先归一再归总"], "answer_analysis": ["$$8$$千克的香蕉是$$12$$美元．$$12$$千克的香蕉是多少钱？ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde12\\div 8\\times 12$$ $$=12\\times 12\\div 8$$ $$=144\\div 8$$ $$=18$$（美元）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2565", "queId": "282b0f37511b456ea00f21f867f8e350", "competition_source_list": ["2017年第17届湖北武汉世奥赛五年级竞赛决赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "有一位疯狂的艺术家为了寻找灵感，把一张厚$$0.1$$毫米的足够大的纸对半撕开，重叠起来，然后再对半撕开重叠起来，假设他如此重复这一过程$$14$$次，并且纸没有倒塌，这叠纸的高度会（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "像山一样高 "}], [{"aoVal": "B", "content": "像平房(一层楼)一样高 "}], [{"aoVal": "C", "content": "像成人一样高 "}], [{"aoVal": "D", "content": "不足$$1$$米 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->乘方->乘方的认识"], "answer_analysis": ["叠纸高$$0.1\\times {{2}^{14}}=1638.4$$（毫米）．像成人一样高． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "609", "queId": "0d08eab315fd45219e91f4f69dae6d1b", "competition_source_list": ["2019年广东深圳全国小学生数学学习能力测评四年级竞赛初赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$9$$和$$5$$之间添上个$$0$$，读作九百万零五． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["最后读作：九百万零五，所以$$9$$在百万位，$$5$$在个位．故中间需要添$$5$$个$$0$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "891", "queId": "cdaa9a7f04d54e678b93a4c8656ce6fc", "competition_source_list": ["2007年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "从1000到2007的自然数中有奇数个因数的整数有( )个. ", "answer_option_list": [[{"aoVal": "A", "content": "10 "}], [{"aoVal": "B", "content": "11 "}], [{"aoVal": "C", "content": "12 "}], [{"aoVal": "D", "content": "13 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->完全平方数->偶指奇因->奇因"], "answer_analysis": ["若为$$n\\textgreater1$$的整数，则$$n=p_{1}^{{{\\alpha }_{1}}}p_{2}^{{{\\alpha }_{2}}}\\cdots p_{r}^{{{\\alpha }_{r}}}$$为质因数分解式，其中$${{p}_{1}} \\textless{} {{p}_{2}} \\textless{} \\cdots \\textless{} {{p}_{r}}$$为质数，所以$$n$$的因数个数为$$d\\left( n \\right)=\\left( {{\\alpha }_{1}}+1 \\right)\\cdots \\left( {{\\alpha }_{r}}+1 \\right)$$，因此$$d\\left( n \\right)$$为奇数时，$${{\\alpha }_{1}}\\text{，}\\alpha {}_{2}\\text{，}\\cdots \\text{，}{{\\alpha }_{r}}\\text{，}$$均为偶数，$$n$$为完全平方数，反过来也成立. $$1000$$到$$2007$$中完全平方数有$$13$$个，即$${{32}^{2}}=1024\\text{，}{{33}^{2}}\\text{，}{{34}^{2}}\\text{，}\\cdots \\text{，}{{44}^{2}}=1936\\left( {{45}^{2}}=2025\\textgreater2007 \\right)$$，从而$$1000$$到$$2007$$中有奇数个因数的整数有$$13$$个. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1924", "queId": "984077c7a7314eadb8f80d45e8f4c266", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级三年级竞赛初赛第40题"], "difficulty": "1", "qtype": "single_choice", "problem": "约翰的钱比吉尔多$$20$$美元，他们两个总共有$$40$$美元．请问约翰有多少美元？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->和差问题->二量和差问题->明和明差"], "answer_analysis": ["$$(40+20)\\div 2=30$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3135", "queId": "129728b71137438c948bd0243da469aa", "competition_source_list": ["2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛"], "difficulty": "3", "qtype": "single_choice", "problem": "在小镇大学里，学生们的专业是音乐、美术，或两者．如果有$$500$$个学生是音乐专业，$$600$$个学生是美术专业，$$300$$个学生是音乐和美术专业，那么大学里有多少个学生？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$500$$ "}], [{"aoVal": "B", "content": "$$800$$ "}], [{"aoVal": "C", "content": "$$1000$$ "}], [{"aoVal": "D", "content": "$$1400$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->逆向思想"], "answer_analysis": ["$$500+600-300=800$$个． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "815", "queId": "77039fdd8845426cb8418fba4f8de31d", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛决赛"], "difficulty": "2", "qtype": "single_choice", "problem": "不超过$$100$$的所有质数的乘积，减去不超过$$100$$的所有个位数字为$$3$$和$$7$$的质数的乘积，所得差的个位数字为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["不超过$$100$$的所有质数包含$$2$$、$$5$$， 所以乘积个位一定为$$0$$． 不超过$$100$$的所有个位数字为$$3$$或$$7$$的质数： $$3$$、$$7$$、$$13$$、$$17$$、$$23$$、$$37$$、$$43$$、 $$47$$、$$53$$、$$67$$、$$73$$、$$83$$、$$97$$，$$3\\times7$$的个位为$$1$$， 那么不超过$$100$$的所有个位数字为$$3$$或$$7$$的质数的乘积个位是$$3$$， 差的个位是$$7$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2642", "queId": "3b0251d5540f42d3bffe820bd1ee6b9e", "competition_source_list": ["2017年第17届湖北武汉世奥赛五年级竞赛决赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "任选$$3$$个不同的数字，按从大到小的顺序排成一个数，再按从小到大的顺序排成一个数，用大数减去小数（如$$1$$，$$2$$，$$0$$，就用$$210-12=198$$），用所得结果的三位数重复上述过程，最后陷入的``数字黑洞''是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$123$$ "}], [{"aoVal": "B", "content": "$$495$$ "}], [{"aoVal": "C", "content": "$$594$$ "}], [{"aoVal": "D", "content": "$$954$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$123$$、$$198$$、$$792$$、$$693$$、$$594$$、$$495$$、$$495$$、$$495$$，可以发现选项的$$4$$个数字都会陷入$$495$$的数字黑洞，所以数字黑洞是$$495$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1384", "queId": "fa8dbd7052d7484b8a0483d64ba1b29c", "competition_source_list": ["2006年第4届创新杯四年级竞赛初赛B卷第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\underbrace{3\\times 3\\times \\cdots \\times 3}_{2006个3}$$减去$$\\underbrace{7\\times 7\\times \\cdots \\times 7}_{100个7}$$，得数的个位数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为$${{3}^{n}}$$的个位数字是$$3$$，$$9$$，$$7$$，$$1$$四个一循环，$$2006\\div 4=501$$（组）$$\\cdots \\cdots 2$$（个）， 所以$${{3}^{2006}}$$个位数字和$${{3}^{2}}$$的个位数字是相同的，即为$$9$$； 因为$${{7}^{n}}$$的个位数字是$$7$$，$$9$$，$$3$$，$$1$$四个一循环，$$100\\div 4=25$$， 所以$${{7}^{100}}$$个位数字和$${{7}^{4}}$$的个位数字是相同的，即为$$1$$； 所以$$\\underbrace{3\\times 3\\times \\cdots \\times 3}_{2006个3}$$减去$$\\underbrace{7\\times 7\\times \\cdots \\times 7}_{100个7}$$，得数的个位数字是$$8$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1862", "queId": "d6f77137933d429dbb1f9262b75bd41f", "competition_source_list": ["2017年希望杯四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "今年，小军$$5$$岁，爸爸$$31$$岁，再过（ ）年，爸爸的年龄是小军的$$3$$倍。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->差倍->两量差倍->整数差倍"], "answer_analysis": ["解：父子年龄差是：$$31-5=26$$（岁）， 爸爸的年龄是小军的$$3$$倍时， 小军的年龄是：$$26\\div \\left( 3-1 \\right)$$ $$=26\\div 2$$ $$=13$$（岁）， $$13-5=8$$（年）， 则再过$$8$$年，爸爸的年龄是小军的$$3$$倍。 故答案为：$$8$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2532", "queId": "1ad716bfab1d43f59d320639344be56f", "competition_source_list": ["2004年五年级竞赛创新杯", "2004年第2届创新杯五年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个数除以一个两位小数，如果把除数去掉小数点，把被除数缩小$$10$$倍，那么商的变化情况是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "扩大$$10$$倍 "}], [{"aoVal": "B", "content": "缩小$$10$$倍 "}], [{"aoVal": "C", "content": "扩大$$1000$$倍 "}], [{"aoVal": "D", "content": "缩小$$1000$$倍 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数加减->小数加减巧算之位值原理"], "answer_analysis": ["把除数为两位小数的小数点去掉，那么除数扩大了$$100$$倍，商就会缩小$$100$$倍.又把被除数缩小$$10$$倍，那么商又缩小了$$10$$倍，一共缩小了$$1000$$倍，选D. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1", "queId": "1be042552e18453d8fcc5111d611feca", "competition_source_list": ["2013年中环杯三年级竞赛决赛"], "difficulty": "0", "qtype": "single_choice", "problem": "星期天，小军帮助妈妈做一些家务。各项家务花的时间为：叠被子$$3$$分钟，洗碗$$8$$分钟，用洗衣机洗衣服$$30$$分钟，晾衣服$$5$$分钟，拖地板$$10$$分钟，削土豆皮$$12$$分钟。 想一想：以上事件中哪些是必须要按顺序完成的？ ", "answer_option_list": [[{"aoVal": "A", "content": "叠被子$$-$$洗碗 "}], [{"aoVal": "B", "content": "用洗衣机洗衣服$$-$$晾衣服 "}], [{"aoVal": "C", "content": "拖地板$$-$$削土豆皮 "}], [{"aoVal": "D", "content": "洗碗$$-$$晾衣服 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->统筹规划->简单时间统筹问题->时间总和问题"], "answer_analysis": ["解： 先用洗衣机洗衣服，然后晾衣服。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1397", "queId": "3a9262421a564fbfa124e139ef8e3c10", "competition_source_list": ["2018年IMAS小学高年级竞赛（第一轮）第7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "百货商店运来$$300$$双球鞋，分别装在$$2$$个木箱、$$9$$个纸箱里，且每个木箱所装的球鞋数量都相同、每个纸箱所装的球鞋数量也都相同．若$$3$$个纸箱所装的球鞋数量与$$1$$个木箱所装的球鞋数量一样，请问每个木箱装多少双球鞋． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$60$$ "}], [{"aoVal": "E", "content": "$$100$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->双归一问题"], "answer_analysis": ["因为$$3$$个纸箱所装的球鞋数量与$$1$$个木箱所装的球鞋数量一样．所以$$9$$个纸箱所装的球鞋数量与$$9\\div 3=3$$个木箱所装的球鞋数量一样．因此$$300$$双球鞋相当于装在$$2+3=5$$个木箱中．故每个木箱装$$300\\div5=60$$双球鞋． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1321", "queId": "47611c3394d74d208cbc8b2f7d4653fe", "competition_source_list": ["2012年IMAS小学高年级竞赛第一轮检测试题第5题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "小明沿着马路的一侧以匀速跑步，马路上相邻两根电线杆之间的距离相等，他从第$$1$$根电线杆开始跑到第$$5$$根电线杆用了$$2$$分钟，他继续以此速度向前跑，当他跑到第$$n$$根电线杆时就理科折返原出发处．若小明全程共用了$$12$$分钟，请问$$n$$之值为何？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$12$$ "}], [{"aoVal": "E", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为小明以$$2$$分钟从第$$1$$根电线杆开始跑到第$$5$$根电线杆共跑了$$4$$个间隔，所以$$4$$分钟可以跑到第$$9$$根电线杆，$$6$$分钟可以跑到第$$13$$根电线杆． 故$$n=13$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3374", "queId": "a4649d06e9034dada9ccd29418bcde19", "competition_source_list": ["2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第6题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第6题3分", "2020年第24届YMO四年级竞赛决赛第6题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第6题3分", "2019年第24届YMO四年级竞赛决赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$1986$$的数字和是$$1+9+8+6=24$$，小于$$2020$$的四位数中数字和等于$$24$$的数共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["小于$$2000$$的四位数千位数字是$$1$$，要它数字和为$$24$$，只需其余三位数字和是$$23$$，因为十位、个位数字和最多为$$9+9=18$$，因此，百位数字至少是$$5$$，于是： 百位为$$5$$时，只有$$1599$$一个； 百位为$$6$$时，只有$$1689$$、$$1698$$两个； 百位为$$7$$时，只有$$1779$$、$$1788$$、$$1797$$三个； 百位为$$8$$时，只有$$1869$$、$$1878$$、$$1887$$、$$1896$$四个； 百位为$$9$$时，只有$$1959$$、$$1968$$、$$1977$$、$$1986$$、$$1995$$五个； 总计共$$1+2+3+4+5=15$$ （个）． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2978", "queId": "c06aa0e5e2fe4e12aac2579bcfaadc94", "competition_source_list": ["2014年全国迎春杯四年级竞赛复赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个$$12$$个数的等差数串，相邻两数的差是$$2$$，且前$$8$$个数的和等于后$$4$$个数的和，那么，这个数串的第二个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->能力->抽象概括"], "answer_analysis": ["根据题意得$$({{a}_{1}}+{{a}_{8}})\\times 8\\div 2=({{a}_{9}}+{{a}_{12}})\\times4\\div 2$$，因为$${{a}_{8}}={{a}_{1}}+14$$，$${{a}_{9}}={{a}_{1}}+16$$，$${{a}_{12}}={{a}_{1}}+22$$， 所以$$({{a}_{1}}+{{a}_{1}}+14)\\times 8\\div2=({{a}_{1}}+16+{{a}_{1}}+22)\\times 4\\div 2$$，解得$${{a}_{1}}=5$$，因此$${{a}_{2}}=5+2=7$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2697", "queId": "3bf37d80a08a489a828df1c93dbb5283", "competition_source_list": ["2011年第7届全国新希望杯小学高年级五年级竞赛A卷第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$x=1+12+123+1234+\\cdots +123456789$$，$$x$$的后三位数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$105$$ "}], [{"aoVal": "B", "content": "$$205$$ "}], [{"aoVal": "C", "content": "$$305$$ "}], [{"aoVal": "D", "content": "$$405$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数加法运算->加法横式"], "answer_analysis": ["只用计算末三位数字和为$$1+12+123+234+345+456+567+678+789=3205$$，故答案为$$205$$，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3245", "queId": "554a5f964b724f51a0f4cc9a738300ae", "competition_source_list": ["2013年IMAS小学中年级竞赛第二轮检测试题第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "礼品店出售三种不同价格的礼品，售价分别是$$30$$元、$$60$$元、$$90$$元；另外还有三种不同款式的礼盒，售价分别是$$20$$元、$$50$$元、$$80$$元．小英想要购买一份礼品与一个礼盒，请问她支付的款项共有多少种不同的可能金额？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}], [{"aoVal": "E", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["依次枚举，$$30+20=50$$（元），$$30+50=80$$（元），$$30+80=110$$（元），$$60+80=140$$（元），$$90+80=170$$（元），共$$5$$种． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1476", "queId": "c33e19e0f74c480cafd9c60f0a7403a9", "competition_source_list": ["2012年第10届全国创新杯小学高年级六年级竞赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "某次知识竞赛共$$5$$道题，全部$$52$$人，答对一道得$$1$$分，已知全部共得$$181$$分，每人至少得$$1$$分，且得$$1$$分的有$$7$$人，得$$2$$分的和得$$3$$分的人一样多，得$$5$$分的人有$$6$$人，则得$$4$$分的有（~ ~ ~ ）人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$31$$ "}], [{"aoVal": "D", "content": "$$35$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->不定方程解应用题"], "answer_analysis": ["错误的题共有$$260-181=79$$题，假设得$$2$$分的和得$$3$$分的为$$a$$人，得$$4$$分的为$$b$$人，可以得到 $$79=28+5a+b$$ 有$$5a+b=51$$，只有$$31$$满足． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2132", "queId": "06691232fb3c4b8c95a2421e5e1a6581", "competition_source_list": ["2015年第20届全国华杯赛小学中年级四年级竞赛初赛A卷第5题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "一只旧钟的分针和时针重合一次，需要经过标准时间$$66$$分钟，那么，这只旧钟的$$24$$小时比标准时间的$$24$$小时（ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "快$$12$$分 "}], [{"aoVal": "B", "content": "快$$6$$分 "}], [{"aoVal": "C", "content": "慢$$6$$分 "}], [{"aoVal": "D", "content": "慢$$12$$分 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->时钟问题->坏钟问题"], "answer_analysis": ["时针速度为每分钟$$0.5$$度，分针速度为每分钟$$6$$度．分钟每比时针多跑一圈，即多跑$$360$$度，时针分针重合一次．经过$$\\frac{360}{6-0.5}=\\frac{720}{11}$$分钟，旧钟时针分针重合一次，需要经过标准时间$$66$$分钟；则旧钟的$$24$$小时，相当于标准时间的$$\\frac{24\\times 60}{\\frac{720}{11}}\\times 66=1452$$分钟，所以比标准时间$$24$$小时对应的$$24\\times 60=1440$$分钟多了$$1452-1440=12$$分钟，即慢了$$12$$分钟． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3365", "queId": "660c2b576d964e37ac7988bc824a99d8", "competition_source_list": ["2021年第8届鹏程杯四年级竞赛初赛第23题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "按英国人的记法，$$2021$$年$$5$$月$$15$$日记作$$5-15-2021$$；按美国人的记法，$$2021$$年$$5$$月$$15$$日记作$$15-5-2021$$．那么，$$2021$$年全年中共有天会让英、美两国人在记法上产生误会． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$66$$ "}], [{"aoVal": "C", "content": "$$132$$ "}], [{"aoVal": "D", "content": "$$144$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["日期$$\\leqslant 12$$的都会产生误会，$$12\\times12=144$$天，但日期和月份相同时不会产生误会，这样的时间共有$$12$$天，所以共有$$144-12=132$$天． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2025", "queId": "cac914d997734af691f9088746126770", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$42$$个小朋友排成一队去秋游，从排头往后数，小$$Y$$是第$$22$$个，从排尾往前数，小$$M$$是第$$22$$个，小$$Y$$和小$$M$$中间有个人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["根据题意分析可知，小$$Y$$后面有$$42-22=20$$（人），$$22-20-1=1$$（人）．那么小$$M$$应该在小$$Y$$的正前方，所以小$$Y$$和小$$M$$之间有$$0$$个人． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3139", "queId": "055e25a19b7a44c2be01bd59d38281e7", "competition_source_list": ["1993年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$7$$个相同的小球放入$$4$$个不同的盒子中，每个盒子中至少放一个球，则共有~\\uline{~~~~~~~~}~种不同的放法。 ", "answer_option_list": [[{"aoVal": "A", "content": "15 "}], [{"aoVal": "B", "content": "18 "}], [{"aoVal": "C", "content": "20 "}], [{"aoVal": "D", "content": "24 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->排列组合->组合->插板法->至少1个"], "answer_analysis": ["插板法，$$\\text{C}_{6}^{3}=20$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2346", "queId": "12d1164202db451a9afdcd9f8f3f5156", "competition_source_list": ["2020年希望杯六年级竞赛（个人赛）第4题", "2020年新希望杯六年级竞赛初赛（个人战）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "【2020新希望杯六年级竞赛初赛】 如果$$x:y=4:7$$，$$z:x=3:5$$，那么$$\\left( x+y \\right):(z+x)=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11:8$$ "}], [{"aoVal": "B", "content": "$$33:55$$ "}], [{"aoVal": "C", "content": "$$32:55$$ "}], [{"aoVal": "D", "content": "$$41:32$$ "}], [{"aoVal": "E", "content": "$$55:32$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["已知$$x:y=4:7$$，$$z:x=3:5$$，那么$$y:x=7:4$$，$$x:z=5:3$$， $$y:x=\\left( 7\\times 5 \\right):\\left( 4\\times 5 \\right)=35:20$$，$$x:z=\\left( 5\\times 4 \\right):\\left( 3\\times 4 \\right)=20:12$$， $$y:x:z=35:20:12$$． 假设$$y=35a$$，$$x=20a$$，$$z=12a$$，其中$$a\\ne 0$$． $$x+y=20a+35a=55a$$，$$z+x=12a+20a=32a$$， 那么$$\\left( x+y \\right):\\left( z+x \\right)=55a:32a=55:32$$． 故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2172", "queId": "2fcd0d323b3f45f8821cca1129b9f648", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛六年级竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明家距学校，乘地铁需要$$30$$分钟，乘公交车需要$$50$$分钟．某天小明因故先乘地铁，再换乘公交车．用了$$40$$分钟到达学校，其中换乘过程用了$$6$$分钟，那么这天小明乘坐公交车用了分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$12$$ "}], [{"aoVal": "E", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["乘车时间是$$40-6=34$$分，假设全是地铁是$$30$$分钟，时间差是$$34-30=4$$ 分钟，需要调整到公交推迟$$4$$分钟，地铁和公交的时间比是$$3:5$$，设地铁时间是多份，公交是$$5$$份时间，$$4\\div \\left( 5-3 \\right)=2$$，公交时间为$$5\\times 2=10$$分钟． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "796", "queId": "95b6ab233a95422eb5436ca98952b58d", "competition_source_list": ["2005年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "两个整数的最大公因数是15，最小公倍数是360，那么这两个整数的和是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "135 "}], [{"aoVal": "B", "content": "165 "}], [{"aoVal": "C", "content": "180 "}], [{"aoVal": "D", "content": "195 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->两数的最小公倍数"], "answer_analysis": ["这两个整数的最大公因数为15，可设这两个数分别为15a，15b($$a \\textless{} b$$且$$a\\text{，}b$$互质)，.那么$$15\\times a\\times b=360$$，即$$a\\times b=24$$，又$$a\\text{，}b$$互质，则$$\\left( a\\text{，}b \\right)=\\left( 1\\text{，}24 \\right)$$或$$\\left( 3\\text{，}8 \\right)$$.当$$\\left( a\\text{，}b \\right)=\\left( 1\\text{，}24 \\right)$$时，这两个整数的和为$$15\\times 1+15\\times 24=15\\times 25=375$$;当$$\\left( a\\text{，}b \\right)=\\left( 3\\text{，}8 \\right)$$时，这两个整数的和为$$15\\times 3+15\\times 8=15\\times 11=165$$.四个选项中只有B符合结论，故选B "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "409", "queId": "fb9d6af8865744b5bef8338039700b97", "competition_source_list": ["2021年新希望杯一年级竞赛初赛第14题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "5、$$5$$只动物排成一排，狗和猫都与鸡相邻，猫和鼠都与兔相邻，那么都与猫相邻． ", "answer_option_list": [[{"aoVal": "A", "content": "鼠和鸡 "}], [{"aoVal": "B", "content": "鸡和兔 "}], [{"aoVal": "C", "content": "兔和狗 "}], [{"aoVal": "D", "content": "兔和鼠 "}], [{"aoVal": "E", "content": "鼠和狗 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据题意分析可知，狗和猫都与鸡相邻，即鸡的左右两边是猫和狗，又因为猫与老鼠都与兔相邻，所以猫的旁边是兔子和老鼠，兔子和老鼠在猫的同一侧，由此可知，与猫相邻的是鸡和兔． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1308", "queId": "2c6b6805a61d47ae855ff7d47d61602f", "competition_source_list": ["2019年全国小学生数学学习能力测评六年级竞赛复赛第9题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "$$0.5$$千克盐溶解在$$20$$千克水中，盐的重量占盐水重量的． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{4}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{41}$$ "}], [{"aoVal": "D", "content": "无选项 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["盐水：$$0.5+20=20.5\\left( \\text{kg} \\right)$$， 盐占盐水：$$0.5\\div 20.5=\\frac{1}{41}$$． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2368", "queId": "06e50d6771364cae92db23b1b9085966", "competition_source_list": ["六年级其它", "2017年河南郑州豫才杯六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "我们定义一种新运算``$$\\oplus $$''如下：当$$a\\textgreater b$$时，$$a\\oplus b=2b-1$$；当$$a\\leqslant b$$时，$$a\\oplus b=a+1$$．则$$\\left( 2\\oplus 3 \\right)\\oplus \\left( 5\\oplus 4 \\right)=$$（ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\left( 2\\oplus 3 \\right)=2+1=3$$，$$\\left( 5\\oplus 4 \\right)=2\\times 4-1=7$$，$$\\left( 2\\oplus 3 \\right)\\oplus \\left( 5\\oplus 4 \\right)=3\\oplus 7=3+1=4$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1531", "queId": "5b09e3f4e9ca49c483d39cfec553d21a", "competition_source_list": ["2016年第12届全国新希望杯小学高年级六年级竞赛复赛第4题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "某实验小学去年参加$$15$$届``新希望杯''书法大赛的学生中，女生占总数的$$\\frac{1}{5}$$，今年参加第$$16$$届``新希望杯''书法大赛的学生比去年增加了$$20 \\%$$，其中女生占总数的$$\\frac{1}{4}$$，那么今年参加书法大赛的女生总数比去年增加了百分之． ", "answer_option_list": [[{"aoVal": "A", "content": "$$49$$ "}], [{"aoVal": "B", "content": "$$48$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$51$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["设去年参加比赛的为单位``$$1$$''，女生占$$\\dfrac{1}{5}$$，即女生人数为$$\\dfrac{1}{5}\\times 1=\\dfrac{1}{5}$$人，今年参加比赛的人数增加$$20 \\%$$，则今年参加比赛人数为$$1\\times \\left( 1+20 \\% \\right)=1.2$$，女生占$$\\dfrac{1}{4}$$，则女生人数为$$1.2\\times \\dfrac{1}{4}=0.3$$人．今年比去年增加了$$\\dfrac{\\left( 0.3-\\dfrac{1}{5} \\right)}{\\dfrac{1}{5}}=50 \\%$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3145", "queId": "8aac50a7519fa10a01519fde4010005d", "competition_source_list": ["2016年全国华杯赛小学高年级竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "从自然数$$1$$，$$2$$，$$3$$，\\ldots，$$2015$$，$$2016$$．任意取$$n$$个不同的数，要求总能在这$$n$$个不同的数中找到$$5$$个数，它们的数字和相等．那么$$n$$的最小值等于． ", "answer_option_list": [[{"aoVal": "A", "content": "$$109$$ "}], [{"aoVal": "B", "content": "$$110$$ "}], [{"aoVal": "C", "content": "$$111$$ "}], [{"aoVal": "D", "content": "$$112$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["$$1$$到$$2016$$中，数字和最大$$28$$， 数字和$$28$$的数只有$$1999$$， 最坏情况：取数字和$$1$$到$$27$$的数各$$4$$个，以及$$1999$$，共$$109$$个数， 再多取一个数就保证有$$5$$个数字和相等，$$n=110$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1085", "queId": "3cd12ade990e4182af38f6e0e5969ac4", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（一）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "小李以每股$$10$$元的相同价格买入$$A$$和$$B$$两只股票共$$1000$$股．此后$$A$$股票先跌$$5 \\% $$，$$B$$股票先涨$$5 \\% $$再跌$$5 \\% $$．若在此期间小李没有再买卖过这两只股票，则现在这$$1000$$股股票的市值是元．~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10250$$ "}], [{"aoVal": "B", "content": "$$9975$$ "}], [{"aoVal": "C", "content": "$$10000$$ "}], [{"aoVal": "D", "content": "$$9750$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->已知成本利润求售价"], "answer_analysis": ["$$10\\times 1000\\times \\left( 1-5 \\% \\right)\\times \\left( 1+5 \\% \\right)=9975$$（元）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1425", "queId": "7a440e0b1596478fab2214e3249eb5c0", "competition_source_list": ["2014年迎春杯三年级竞赛初赛", "2014年迎春杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一辆大卡车一次可以装煤$$2.5$$吨，现在要一次性运走$$48$$吨煤，那么至少需要（ ）辆这样的大卡车。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->乘除法应用->除法应用"], "answer_analysis": ["解：$$48\\div 2.5=19.2\\approx 20$$（辆），由实际情况可知需要$$20$$辆。 答：至少需要$$20$$辆这样的大卡车。 故选：C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3227", "queId": "2c6a712cca3448ee9eb2c7e0b570a3b0", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级六年级竞赛邀请赛训练题（四）"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$0\\sim 8$$这九个数字中选四个数字组成没有重复的四位数，然后将这些四位数由小到大排列，其中第$$2017$$个四位数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2017$$ "}], [{"aoVal": "B", "content": "$$6012$$ "}], [{"aoVal": "C", "content": "$$7012$$ "}], [{"aoVal": "D", "content": "$$8120$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->组数问题->有特殊要求的组数问题"], "answer_analysis": ["组成的四位数一共有$$8\\times 8\\times 7\\times 6=2688$$（个），其中千位为$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$的各有$$336$$个．$$2017336\\times 6=1$$（个），即千位为$$7$$的第$$1$$个数为$$7012$$，选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "922", "queId": "dc206ee05da946a1bc30fb8f13ec072b", "competition_source_list": ["2017年迎春杯五年级竞赛决赛二试第2题10分", "2017年迎春杯六年级竞赛决赛二试第3题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果两个正整数$$A$$和$$B$$满足以下条件： ① $$A(A+1)$$是$$B(B+1)$$的倍数； ②$$ A$$和$$(A+1)$$都不是$$B$$或者$$(B+1)$$的倍数； 那么，$$A+B$$的最小值是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$34$$ "}], [{"aoVal": "B", "content": "$$43$$ "}], [{"aoVal": "C", "content": "$$44$$ "}], [{"aoVal": "D", "content": "$$49$$ "}], [{"aoVal": "E", "content": "$$55$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析", "Overseas Competition->知识点->数论模块"], "answer_analysis": ["$$A$$、$$A+1$$、$$B$$、$$B+1$$均不为质数；也不能是质数的$$n$$次方．所以，$$B$$只能是$$14$$．（$$B$$为$$6$$、$$10$$时，$$B+1$$都是质数），此时$$B+1$$为$$15$$，$$B(B+1)$$含有质因数$$2$$、$$3$$、$$5$$、$$7$$；最小符合条件的$$A$$为$$20$$，所以，$$A+B$$最小值为$$34$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "697", "queId": "2d3025f443d34f3da1ad9e442821c410", "competition_source_list": ["2013年第12届上海小机灵杯小学高年级五年级竞赛初赛第5题1分"], "difficulty": "0", "qtype": "single_choice", "problem": "古时候的原始人捕猎，捕到一只野兽对应一根手指．等到$$10$$根手指用完，就在绳子上打一个结，这就是运用现在的数学中的． ", "answer_option_list": [[{"aoVal": "A", "content": "二进制计数法 "}], [{"aoVal": "B", "content": "五进制计数法 "}], [{"aoVal": "C", "content": "十进制计数法 "}]], "knowledge_point_routes": ["拓展思维->七大能力->实践应用", "课内体系->七大能力->实践应用"], "answer_analysis": ["古时候的原始人捕猎运用现在的数学中的十进制计数法． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "121", "queId": "b01fce4bd8934e8291af3a9a8b89f3c5", "competition_source_list": ["2017年第23届浙江杭州华杯赛小学高年级竞赛卓学堂高端备考活动第7题", "2014年全国中环杯五年级竞赛决赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "有 $$15$$位选手参加一个围棋锦标赛，每两个人之间需要比赛一场． 赢一场得 $$2$$分，平一场各得$$1$$分，输一场得 $$0$$分． 如果一位选手的得分不少于 $$20$$分，他就能获得一份奖品． 那么，最多有位选手能够获得奖品． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->体育比赛->2-1-0 积分制"], "answer_analysis": ["容易知道，这 $$15$$ 位选手一共需要比赛$$C_{15}^{2}=15\\times 7$$ （场），产生$$15\\times 7\\times 2=210$$ （分）， 所以理论上最多有$$\\left[ \\frac{210}{20} \\right]=10$$ （人）能够拿到奖品．接下来，我们要证明，不可能有 $$10$$ 人同时拿到 $$20$$ 分或以上． 假设$${{A}_{1}}.{{A}_{2}}.\\cdots {{A}_{15}}$$ 同时拿到$$20$$ 分或以上，也就意味着$${{A}_{11}}{{A}_{12}}\\cdots {{A}_{15}}$$ 一共只能拿到$$10$$ 分或以下．考虑到$${{A}_{11}}{{A}_{12}}\\cdots {{A}_{15}}$$之间有$$C_{5}^{2}=10$$ （场）比赛，产生了$$20$$ 分的总分，所以不可能只得到$$10$$ 分或以下． 至此，理论上最多只能有 $$9$$ 个人，接下来举例：$${{A}_{1}}.{{A}_{2}}.\\cdots {{A}_{9}}$$互相之间全部打平，然后$${{A}_{1}}.{{A}_{2}}.\\cdots {{A}_{9}}$$每个人都战胜了$${{A}_{11}}{{A}_{12}}\\cdots {{A}_{15}}$$中的所有人，这样$${{A}_{1}}.{{A}_{2}}.\\cdots {{A}_{9}}$$每个人都可以得$$6\\times 2+8=20$$ （分）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "224", "queId": "635e4b4ce28b48fc946665dd738e5666", "competition_source_list": ["2021年新希望杯一年级竞赛初赛第14题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$5$$只动物排成一排，狗和猫都与鸡相邻，猫和鼠都与兔相邻，那么都与猫相邻． ", "answer_option_list": [[{"aoVal": "A", "content": "鼠和鸡 "}], [{"aoVal": "B", "content": "鸡和兔 "}], [{"aoVal": "C", "content": "兔和狗 "}], [{"aoVal": "D", "content": "兔和鼠 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据题意分析可知，狗和猫都与鸡相邻，即鸡的左右两边是猫和狗，又因为猫与老鼠都与兔相邻，所以猫的旁边是兔子和老鼠，兔子和老鼠在猫的同一侧，由此可知，与猫相邻的是鸡和兔． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3371", "queId": "8db0248f21304005bcf6fbc8bebe60aa", "competition_source_list": ["2012年IMAS小学高年级竞赛第二轮测试真题第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "有六对夫妇共$$12$$个人参加宴会，在宴会上，除了自己的妻子之外，每位男宾都与其他每人握手一次，女宾与女宾之间不握手，请问这$$12$$个人之间总共握手了多少次？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$51$$ "}], [{"aoVal": "E", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->排列组合->组合->组合的基本应用"], "answer_analysis": ["用排除法，$$12$$个人共握手$$C_{12}^{2}=12\\times 11\\div 2=66$$（次） 男宾与妻子握手$$6$$次， 女宾与女宾握手$$\\text{C}_{6}^{2}=6\\times 5\\div 2=15$$（次） ∴$$66-6-15=45$$（次）． ", "用排除法，$$12$$个人共握手$$12\\times 11\\div 2=66$$（次）
\n男宾与妻子握手$$6$$次，
\n女宾与女宾握手$$6\\times 5\\div 2=15$$（次）
\n∴$$66-6-15=45$$（次）．
"], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3252", "queId": "24eb63876d7744cbbfebb2a7c0c2de00", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "不妨称各位数字之和为$$7$$的整数为``魔力数''，如$$115$$，$$1312$$等就是魔力数．那么随手写出一个三位数，恰好是``魔力数''的可能性是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{45}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{75}$$ "}], [{"aoVal": "C", "content": "$$\\frac{7}{225}$$ "}], [{"aoVal": "D", "content": "$$\\frac{8}{225}$$ "}]], "knowledge_point_routes": ["拓��思维->思想->枚举思想"], "answer_analysis": ["三位数一共有$$900$$个，满足条件的魔力数有$$\\text{C}_{7+2-1}^{2}=28$$个，则$$\\frac{28}{900}=\\frac{7}{225}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "354", "queId": "5ccc08994f484a97b33b51f893a05820", "competition_source_list": ["2015年第13届全国创新杯小学高年级五年级竞赛初赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "将``$$OPQRST$$''连续接下去可得到；``$$OPQRSTOPQRST\\cdot \\cdot \\cdot $$''，从左至右第$$2015$$个字母应该是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$S$$ "}], [{"aoVal": "B", "content": "$$Q$$ "}], [{"aoVal": "C", "content": "$$O$$ "}], [{"aoVal": "D", "content": "$$T$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->字母规律->数字与字母结合"], "answer_analysis": ["$$2015\\div 6\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 5$$，则$$2015$$个数应为$$S$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2087", "queId": "e6d14d29515f4a639f199531722abb37", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（五）第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "自来水厂有一个水净化池，在某一净化过程中，需同时打开$$3$$根进水管和$$2$$根出水管，一根进水管每小时能进$$50$$立方米的水，一根出水管每小时能出$$80$$立方米的水．水池中原本有$$200$$立方米的水，当水池中的水全部净化完停止进水，在这个过程中国，水池一共出水（~ ）立方米．~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1600$$ "}], [{"aoVal": "B", "content": "$$1800$$ "}], [{"aoVal": "C", "content": "$$3200$$ "}], [{"aoVal": "D", "content": "$$3600$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->进水与排水问题"], "answer_analysis": ["每小时水池的水减少$$2\\times 80-3\\times 50=10$$立方米，出完水需要的时间是$$200\\div 10=20$$小时，共出水$$20\\times 2\\times 80=3200$$立方米． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1898", "queId": "938ce8f5f77f463e92152cca04c36976", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第4题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "云云做了$$18$$朵花，芳芳做了$$8$$朵，云云给芳芳朵花，两人的花就同样多了． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["云云比芳芳多：$$18-8=10$$（朵）， 所以要使两人一样多，应将多余的$$10$$朵花一半分， 即每人$$5$$朵，所以云云给芳芳$$5$$朵． 故选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2674", "queId": "683e2de76d25496aa07892c7c46f00a9", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "【数学人教版】下列算式结果为$$500$$的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5\\times99+1$$ "}], [{"aoVal": "B", "content": "$$100+25\\times4$$ "}], [{"aoVal": "C", "content": "$$88\\times4+37\\times4$$ "}], [{"aoVal": "D", "content": "$$100\\times0\\times5$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数与运算->混合运算->整数四则混合运算"], "answer_analysis": ["A等于$$5\\times(100-1)+1=500-5+1=496$$，$$B$$等于$$100+100=200$$ ，$$C$$等于$$(88+37)\\times4=125\\times 4=500$$ ，$$D$$等于$$0$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "902", "queId": "822ac40360674c1b8ef66114d4f15b7e", "competition_source_list": ["2014年第15届上海中环杯小学高年级五年级竞赛初赛第18题"], "difficulty": "3", "qtype": "single_choice", "problem": "一个五位数$$\\overline{ABCDE}$$是$$2014$$的倍数，并且$$\\overline{CDE}$$恰好有$$16$$个因数，则$$\\overline{ABCDE}$$的最小值是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10070$$ "}], [{"aoVal": "B", "content": "$$16112$$ "}], [{"aoVal": "C", "content": "$$22154$$ "}], [{"aoVal": "D", "content": "$$24168$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->枚举型最值问题"], "answer_analysis": ["最值问题从极端情况出发，既是五位数又是$$2014$$的倍数，最小为$$10070$$；所以$$\\overline{ABCDE}=2014\\times k$$，且$$\\overline{CDE}$$有$$16$$个因数，$$2014\\times k=2014\\times 5+2014\\times n=10070+2014n$$，因此$$\\overline{CDE}=70+14n$$．要求当$$\\overline{CDE}$$有$$16$$个因数时，$$n$$的最小值是几，$$\\overline{CDE}=14\\left( n+5 \\right)=2\\times 7\\times \\left( n+5 \\right)$$从小到大发现$$n=7$$时满足条件，此时$$\\overline{CDE}={{2}^{3}}\\times 3\\times 7$$一共有$$16$$个因数，此时最小值为$$2014\\times \\left( 5+7 \\right)=24168$$． 最值问题从极端情况出发，既是五位数又是$$2014$$的倍数，最小为$$10070$$； 约数个数逆应用，$$16=16=8\\times 2=4\\times 4=4\\times 2\\times 2=2\\times 2\\times 2\\times 2$$，分解质因数后指数可能是（ $$15 $$），（$$7$$，$$1$$），（$$3$$，$$3$$），（$$3$$，$$1$$，$$1$$ ），（$$1$$，$$1$$，$$1$$，$$1$$ ）这几组． $$10070$$，$$70=2\\times 5\\times 7$$，舍 $$12084$$，$$84={{2}^{2}}\\times 3\\times 7$$，舍 $$14098$$，$$98={{2}^{{}}}\\times {{7}^{2}}$$，舍 $$16112$$，$$112={{2}^{4}}\\times 7$$，舍 $$18126$$，$$126=2\\times {{3}^{2}}\\times 7$$，舍 $$20140$$，$$140={{2}^{2}}\\times 5\\times 7$$，舍 $$22154$$，$$154=2\\times 7\\times 11$$，舍 $$24168$$，$$168={{2}^{3}}\\times 3\\times 7$$，符合 $$\\overline{ABCDE}$$最小为$$24168$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1758", "queId": "6186c87759f946ed8872fbade64a9e93", "competition_source_list": ["2014年IMAS小学中年级竞赛第一轮检测试题第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "体育课上，$$30$$位同学排成一排，从$$1$$开始依序报数后，老师说：``报数为$$1$$号到$$10$$号的同学向前走一步，报数为$$20$$号至$$30$$号的同学向后退一步．''请问还有多少位同学原地不动？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$20$$ "}], [{"aoVal": "E", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["因为$$11$$到$$19$$号同学原地不动，所以还有$$19-11+1=9$$位同学原地不动．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2268", "queId": "65d7b0141474413884c22285467119bd", "competition_source_list": ["2016年创新杯小学高年级六年级竞赛训练题（一）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "汽车从甲地到乙地，先行上坡，后行下坡，共用$$9.4$$小时．如果甲、乙两地相距$$450$$千米，上坡车速为每小时$$45$$千米，下坡车速为每小时$$50$$千米，且返回时上坡和下坡速度保持不变，那么原路返回要（~ ~ ~ ）小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9.4$$ "}], [{"aoVal": "B", "content": "$$9.5$$ "}], [{"aoVal": "C", "content": "$$9.6$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题"], "answer_analysis": ["设原来上坡用了$$x$$小时，$$45x+50\\left( 9.4-x \\right)=450$$，解得$$x=4$$，那么原来上坡是$$45\\times 4=180$$千米，下坡就是$$450-180=270$$千米，返回时下坡变为上坡，上坡变为下坡，所以返回时时间是$$\\frac{270}{45}+\\frac{180}{50}=9.6$$（小时）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1833", "queId": "9ba5b7bf91d74f4180ce8c2aff8e8b17", "competition_source_list": ["2020年广东广州羊排赛四年级竞赛第18题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "薇儿参加古诗词大赛，共$$12$$道题．答对一题得$$5$$分，不答或答错一题倒扣$$2$$分．如果薇儿最终分数是$$39$$分，那么她答对了~\\uline{~~~~~~~~~~}~题？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题->方程法解鸡兔同笼"], "answer_analysis": ["根据题目设她答对了$$x$$道题， 因为共$$12$$道题，所以答错了$$(12-x)$$道题，答对一题得$$5$$分，不答或答错一题倒扣$$2$$分． 薇儿最终分数是$$39$$分，可以列方程为：$$5x-(12-x)\\times 2=39$$， 解得$$x=9$$，所以她答对了$$9$$题． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1126", "queId": "18b70ec3135a475da60039e7e22b4d09", "competition_source_list": ["2015年第2届广东深圳鹏���杯四年级竞赛第6题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "爸爸今年的岁数是儿子小明今年岁数的$$3$$倍还大$$4$$岁；再过六年，爸爸的岁数是小明岁数的$$2$$倍还大$$10$$岁．则小明今年是~\\uline{~~~~~~~~~~}~岁． ", "answer_option_list": [[{"aoVal": "A", "content": "在答题卡上提交 "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题->方程法解倍数问题"], "answer_analysis": ["设小明今年是$$x$$岁，则爸爸今年是$$(3x+4)$$岁，再过$$6$$年爸爸就是$$[(3x+4)+6]$$岁， 而六年后，小明长大了$$6$$岁，爸爸也长大了$$6$$岁， 因此，六年后，爸爸年龄是$$[2(x+6)+10]$$岁． 据题意得到$$(3x+4)+6=2(x+6)+10$$，解得$$x=12$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "698", "queId": "28ff3d82e4c541ffbc5e26b3a7201841", "competition_source_list": ["2014年第10届全国新希望杯小学高年级五年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$1.5$$，$$1.6$$，$$2.1$$，$$2.3$$这组数据中加一个数，使这组数据的中位数是$$1.9$$，则下列选项中，满足条件的是（~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1.9$$ "}], [{"aoVal": "B", "content": "$$1.6$$ "}], [{"aoVal": "C", "content": "$$1.7$$ "}], [{"aoVal": "D", "content": "$$2.1$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的认识"], "answer_analysis": ["添加一个数后这列数有奇数个，中位数是最中间的数（由小到大排），则添加的数为$$1.9$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "24", "queId": "02bc475624ea46789fe45d609d85da27", "competition_source_list": ["竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "理发店里同时来了$$A$$、$$B$$、$$C$$三个顾客，只有一个理发师，$$A$$理板寸需要$$7$$分钟，$$B$$理光头需要$$10$$分钟，$$C$$烫卷发需要$$40$$分钟。合理安排这三个人的理发顺序，他们三人所花的时间总和会有最小值，这个最短的时间是（ ）分钟。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$80$$ "}], [{"aoVal": "B", "content": "$$85$$ "}], [{"aoVal": "C", "content": "$$82$$ "}], [{"aoVal": "D", "content": "$$81$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->统筹规划->简单时间统筹问题->时间总和问题"], "answer_analysis": ["按照$$A$$、$$B$$、$$C$$的顺序理发，最短时间为$$7\\times 3+10\\times 2+40=81$$（分）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "516", "queId": "cb37d8dcd80746b88558f3d596cb6d01", "competition_source_list": ["2018年迎春杯四年级竞赛决赛一试第3题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列$$5$$道判断题，这$$5$$道题的题干与它们的答案相关，请判断正误： （$$1$$）第$$2$$题的答案是√． （$$2$$）第$$4$$题的答案是$$\\times $$． （$$3$$）有$$2$$道题的答案是√． （$$4$$）本题和上一题中至少有$$1$$个√． （$$5$$）第$$2$$、$$3$$题答案一致． 如果这$$5$$道题的答案可以使得题干正确，互不矛盾，那么，请在上面的$$5$$个括号内写出这$$5$$道题的答案．请问：正确的题干序号为：~\\uline{~~~~~~~~~~}~（多选题） ", "answer_option_list": [[{"aoVal": "A", "content": "（1）（2） "}], [{"aoVal": "B", "content": "（3）（4） "}], [{"aoVal": "C", "content": "（2）（3）（5） "}], [{"aoVal": "D", "content": "（1）（2）（5） "}], [{"aoVal": "E", "content": "（5） "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["假设第一句描述是正确的，则根据第一句内容``第$$2$$题的答案是对''可知，第二句的描述也是正确的；同理根据第二句的描述``第$$4$$题的答案是错''可以得到，第四句的描述是错误的；再根据第四句的内容``本题和上一题中至少有$$1$$个对''可知，第三句应该选择错；再结合第二句的结果对和第三句的结果错可知，第五句的描述是错的，所以目前我们的结果是``对对错错错''，那么根据这个结果，第三句的描述``有$$2$$道题的答案是对''这句话就应该是对的，出现了矛盾，则我们一开始的假设是错误的，所以第一句的结果应该是错． 第一句的结果是错，则第一句描述的内容``第$$2$$题的答案是对''是不对的，进而可知第二句描述的内容``第$$4$$题的答案是错''也是错的，也就是说第四句话是对的，然后结合第三句和第五句，如果第三句���对的，那么第五句的描述``第$$2$$题、$$3$$题答案一致''就是错的，所以此时的答案就是``错错对对错''，满足题意；如果第三句是错的，则第五句描述``第$$2$$题、$$3$$题答案一致''就是对的，此时的答案就是``错错错对对''，有两个对的，那就和第三句的描述``有$$2$$道题的答案是对''一致，所以出现了矛盾． 所以最后的答案就是错错对对错． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1947", "queId": "bc713ea5f64148bf8eaae5d38a26a84c", "competition_source_list": ["2012年河南郑州二七区郑州实验外国语中学小升初第25题5分", "2011年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "一项工程，甲单独做需要$$30$$天时间，甲、乙合作需要$$12$$天时间，如果乙单独做需要多少时间？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$42$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["将整个工程的工作量看作``$$1$$''个单位，那么甲每天完成总量的$$\\frac{1}{30}$$，甲、乙合作每天完成总量的$$\\frac{1}{12}$$，乙单独做每天能完成总量的$$\\frac{1}{12}-\\frac{1}{30}=\\frac{1}{20}$$，所以乙单独做$$1\\div\\frac{1}{20}=20$$天能完成． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1010", "queId": "12d41a64a30b406894a05f0960ba2eb7", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛A卷第6题3分", "2020年湖北宜昌小升初第15题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个圆的周长增加$$30 \\%$$，这个圆的面积增加$$ \\%$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$69$$ "}], [{"aoVal": "B", "content": "$$90$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["因为圆的周长增加$$30 \\%$$，所以圆的半径也增加$$30 \\%$$．设圆的半径为$$r$$，则增加后的半径为$$(1+30 \\%)r=1.3r$$，所以原来的圆的面积为$$\\pi {{r}^{2}}$$，半径增加后的圆的面积为$$\\pi {{(1.3r)}^{2}}=1.69\\pi {{r}^{2}}$$，故面积增加$$(1.69\\pi {{r}^{2}}-\\pi {{r}^{2}})\\div \\pi {{r}^{2}}=69 \\%$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "823", "queId": "acea09fd29724bb4a830ff305f78b3e1", "competition_source_list": ["2018年湖北武汉新希望杯小学高年级六年级竞赛训练题（四）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "张老师把$$165$$苹果、$$220$$个梨和$$398$$个桔子平均分给小朋友们，最后剩下$$21$$个苹果、$$4$$个梨和$$38$$个桔子没有分出去．每个小朋友分到了个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$165-21=144$$，$$220-4=216$$，$$398-38=360$$，（$$144$$，$$216$$，$$260$$）$$=72$$，所以小朋友的人数应该是$$72$$的因数，又$$38\\textgreater36$$，所以人数只能是$$72$$人，$$144\\div 72+216\\div 72+360\\div 72=10$$（个）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2381", "queId": "137e3827f24741e8a7634ed89996b222", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第7题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$5$$个不同的奇自然数的和为$$85$$，则这五个数中最大数$$M$$的取值范围是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$23\\leqslant M\\leqslant 67$$ "}], [{"aoVal": "B", "content": "$$19\\leqslant M\\leqslant 67$$ "}], [{"aoVal": "C", "content": "$$21\\leqslant M\\leqslant 69$$ "}], [{"aoVal": "D", "content": "$$17\\leqslant M\\leqslant 69$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等差数列->中项定理应用"], "answer_analysis": ["我们尝试考虑最大数$$M$$的最极端情况，当数$$M$$要去的最大值时，其他四个数必须最小，那么分别取$$1$$，$$3$$，$$5$$，$$7$$，此时$$M$$最大值为$$69$$，排除$$A$$、$$B$$选项，为了满足$$M$$是五个数最大的数且最小，那么最好极端的情况应该是这五个数恰好是一个等差的奇数列，根据中项定理易得$$M$$的最小值应为：$$85\\div 5+2+2=21$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2129", "queId": "02311de791cb45a5aa14946611de9998", "competition_source_list": ["2021年第8届鹏程杯四年级竞赛初赛第19题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一列快车和一列慢车相向而行，快车的车长是$$270$$米，慢车的车长是$$360$$米，坐在快车上的人看见慢车驶过的时间是$$12$$秒，那么坐在慢车上的人看见快车驶过的时间是秒． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$12$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["速度和$$=360\\div 12=30$$，$$270\\div 30=9$$（秒）． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3111", "queId": "e24d4a9d03c9443fbe67949ad522a0bb", "competition_source_list": ["2013年IMAS小学中年级竞赛第二轮检测试题第1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$86$$除以一个数，商是一位数，余数是$$6$$．请问``除数''不可能是下列哪个选项的数值？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$40$$ "}], [{"aoVal": "E", "content": "$$80$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数乘除->整数除法运算->带余除法"], "answer_analysis": ["$$86\\div 10=8\\cdots \\cdots 6$$，$$\\text{A}$$符合； $$86\\div 20=4\\cdots \\cdots 6$$，$$\\text{B}$$符合； $$86\\div 30=2\\cdots \\cdots 26$$，$$\\text{C}$$不符合； $$86\\div 80=1\\cdots \\cdots 6$$，$$\\text{D}$$符合． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2584", "queId": "3e7c51268f0e4f708441d991ebff29f1", "competition_source_list": ["2012年全国美国数学大联盟杯小学高年级竞赛初赛第33题", "2013年美国数学大联盟杯小学高年级竞赛初赛第33题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "现有$$6$$个连续的整数，其中最大的数为$$30$$，若另有$$10$$个连续整数的和恰好等于上述$$6$$个连续整数之和，请求出这$$10$$个连续整数中最大的数? ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$26$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$6$$个连续的整数的和$$(25+30)\\times6 \\div2=165$$， $$10$$个连续整数最中间$$165\\div5=33=16+17$$， $$17+4=21$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1148", "queId": "0dba4e79ef67478a881c07e24225b898", "competition_source_list": ["2018年全国小学生数学学习能力测评五年级竞赛初赛第7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "某班共买来$$66$$本课外书，把它们分别放在书架上，每次摆放都是下面一层比上面一层多放$$1$$本书，则至多要放的层数为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["根据``每次摆放都是下面一层比上面一层多放$$1$$本书''得知，每层摆放书的数目成等差数列，要求``至多''摆放几层，需要第一层是$$1$$本，第二层是$$2$$本，第三层是$$3$$本$$\\cdots \\cdots $$， 假设摆放$$n$$层，则：$$1+2+3+\\cdots \\cdots +n=\\frac{n(1+n)}{2}$$， $$66=\\frac{n(1+n)}{2}$$． 得$$n=11$$，所以选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "581", "queId": "0807671281324674a8d28c62ea90ca60", "competition_source_list": ["2018年美国数学大联盟杯五年级竞赛初赛第13题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "\\textbf{(2018 Math league, Primary 5, Question \\#13)} Of the following, which is the sum of two consecutive integers? 下列哪个选项是$$2$$个连续整数的和？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$111111$$ "}], [{"aoVal": "B", "content": "$$222222$$ "}], [{"aoVal": "C", "content": "$$444444$$ "}], [{"aoVal": "D", "content": "$$888888$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["在两个连续整数中，其中一个是双数，一个是单数．一个双数和一个单数的和一定是一个单数，所以两个连续整数的和也定是单数． 在$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$，$$4$$个选项中，只有$$\\text{A}$$是单数，其余$$3$$个都是双数． 故答案选：$$\\text{A}$$ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2091", "queId": "fdd247543f0547ba8232b0e5b695617f", "competition_source_list": ["2020年广东广州羊排赛三年级竞赛第14题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "某部$$87$$集的电视连续剧，每周日、周一、周四、周五、周六都要播出一集，周二、周三停播．这部剧周一大结局，请问它是周几开播的？ ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "四 "}], [{"aoVal": "C", "content": "五 "}], [{"aoVal": "D", "content": "六 "}], [{"aoVal": "E", "content": "日 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$87$$集的电视连续剧在某个星期日开播，每周日、周一、周四、周五、周六都要播一集，五天一周期，$$87\\div 5=17$$（周）$$\\cdots \\cdots 2$$（天），所以最后一集在周一播出，选项$$\\text{A}$$正确． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "637", "queId": "d9cafef543634b12b74e80f5620670dd", "competition_source_list": ["2016年全国世奥赛竞赛A卷第3题", "2016年第16届世奥赛六年级竞赛决赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "美国人口普查局报告说：美国东部时间$$2012$$年$$8$$月$$14$$日下午$$2$$时$$20$$分，美国的人口数字升至$$314159265$$（三亿一千四百一十五万九千两百六十五）人，恰好相当于圆周率（$$ \\pi $$）的一亿倍，更为神奇的时，全世界的河流源头到河口之间曲曲折折的总长度平均约是其源头到河口之间直线距离的$$ \\pi $$倍．圆周率$$ \\pi $$是一个无限不循环小数$$3.1415926\\cdot\\cdot\\cdot\\cdot\\cdot\\cdot$$，我国古代数学家对圆周率的研究做出了巨大的贡献．古算书《周髀算经》记载了``周三径一''；东汉科学家张衡进一步估算的$$ \\pi $$约为$$3.16$$；三国时期的数学家刘徽用``割圆术''求得$$ \\pi $$约为$$3.14$$；魏晋南北朝时期的数学家（~~ ）对圆周率进行了深入的研究，他算出$$ \\pi $$的值介于$$3.1415926$$与$$3.1415927$$之间，并取``约率''为$$\\frac{（~ ）}{7}$$、``密率''为$$\\frac{（~ ）}{113}$$（填分子）作为圆周率的近似值． ", "answer_option_list": [[{"aoVal": "A", "content": "杨辉，$$22$$，$$335$$ "}], [{"aoVal": "B", "content": "祖冲之，$$22$$，$$355$$ "}], [{"aoVal": "C", "content": "韩信，$$11$$，$$355$$ "}], [{"aoVal": "D", "content": "赵爽，$$12$$，$$335$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->进制的性质与应用->进制的认识"], "answer_analysis": ["第一空考察的是数学常识．根据题干中的信息可以知道约率和密率都是接近$$3.14$$的，所以$$3.14\\times 7=21.98\\approx 22$$，$$3.14\\times 113=354.82\\approx 355$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2131", "queId": "03b864df174940e0888573ed05728ba2", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第13题2分"], "difficulty": "2", "qtype": "single_choice", "problem": "一列货车早晨$$6$$时从甲地开往乙地，平均每小时行$$45$$千米，一列客车从乙地开往甲地，平均每小时比货车快$$15$$千米，已知客车比货车迟发$$2$$小时，中午$$12$$时两车同时经过途中某站，然后仍继续前进，问：当客车到达甲地时，货车离乙地还有千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$47.2$$ "}], [{"aoVal": "B", "content": "$$37.5$$ "}], [{"aoVal": "C", "content": "$$24.5$$ "}], [{"aoVal": "D", "content": "$$10.5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["本题考查相遇问题； 客车比货车迟发$$2$$小时，所以客车是$$8$$时出发，到$$12$$时，经过$$4$$小时到达相遇点，而货车到达相遇点共用了$$4+2=6$$（小时），那么甲、乙两地之间的距离就可求出，用货车 速度乘相遇时货车总共用的时间加上客车速度乘相遇时客车所用时间，即$$45\\times \\left( 4+2 \\right)+60\\times 4=510$$（千米）；客车行完全程所需的时间： $$510\\div \\left( 45+15 \\right)=8.5$$（小时）；货车行完全程所需的时间：$$8.5+2=10.5$$（小时）；客车到达甲地时，货车距乙地的距离，用全程减去货车已经行的路程，即$$510-45\\times 10.5=37.5$$��千米）． 客车速度： $$45+15=60$$（千米$$/$$时） 两地距离： $$45\\times \\left( 12-6 \\right)+60\\times \\left( 12-6-2 \\right)=510$$（千米） 客车行完全程所需时间： $$510\\div 60=8.5$$（小时） 客车到达甲地时，货车距乙地的距离： $$510-45\\times \\left( 8.5+2 \\right)=37.5$$（千米） 答：客车到达甲地时，货车离乙地还有$$37.5$$千米． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3107", "queId": "f47f09fec798440b8b4cab02b1c10830", "competition_source_list": ["2018年美国数学大联盟杯五年级竞赛初赛第14题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "三幕戏剧的第二幕占整个戏剧长度的$$\\frac{1}{3}$$，如果第一幕是第三幕的两倍，那么第三幕占整个戏剧的几分之几？ The $$2$$nd act of a $$3$$-act play is $$\\frac{1}{3}$$ the length of the entire play. If the $$1$$ st act is twice as long as the $$3$$rd, what fraction of the play is the $$3$$rd act? ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{9}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{9}$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{9}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用", "Overseas Competition->知识点->计算模块->分数"], "answer_analysis": ["本题题意为``三幕戏剧的第二幕占整个戏剧长度的$$\\frac{1}{3}$$，如果第一幕是第三幕的两倍，那么第三幕占戏剧的几分？'' 根据题意可得，第二幕占整个戏剧长度的$$\\frac{1}{3}$$， 因此第一幕和第三幕占整个长度的$$\\frac{2}{3}$$，第一幕是第三幕的两倍， 因此假设第三幕时长为$$N$$，第一幕的长度为$$2N$$， 因此$$N+2N=\\frac{2}{3}$$， 因此$$N=\\frac{2}{9}$$， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2592", "queId": "a72f07ff6e7545a0b6ba3d95a5ebba7a", "competition_source_list": ["2017年江苏南京小升初工程问题第27题", "2016年北京西城区小升初四中八中分班", "2012年世界少年奥林匹克数学竞赛六年级竞赛初赛A卷第16题10分", "2016年第3届广东深圳鹏程杯六年级竞赛集训材料第一章工程问题第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙加工一批零件，甲、乙合作$$24$$天可以完成．现在甲先做$$16$$天，然后乙再做$$12$$天，还剩下这批零件的$$\\frac{2}{5}$$，已知甲每天比乙多加工$$3$$个零件，求这批零件共多少个？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$240$$ "}], [{"aoVal": "B", "content": "$$280$$ "}], [{"aoVal": "C", "content": "$$300$$ "}], [{"aoVal": "D", "content": "$$360$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算求解", "拓展思维->拓展思维->计算模块->分数->分数基础->分数的认识"], "answer_analysis": ["解题关键在于①甲先做$$16$$天然后乙再做$$12$$天，还剩这批零件的$$\\frac{2}{5}$$，想成甲乙合干$$12$$天，甲独干$$4$$天完成全工程的$$1-\\frac{2}{5}=\\frac{3}{5}$$，从而分别求出甲、乙工作效率．②找出甲比乙每天多加工$$3$$个零件对应的甲乙工效差，即可使问题得解． 解：甲乙合作$$12$$天完成总工程几分之几． $$\\frac{1}{24}\\times 12=\\frac{1}{2}$$ 甲工作效率 $$\\left( 1-\\frac{2}{5}-\\frac{1}{2} \\right)\\div (16-12)=\\frac{1}{40}$$ 乙工作效率 $$\\frac{1}{24}-\\frac{1}{40}=\\frac{1}{60}$$ 这批零件共多少个？ $$3\\div \\left( \\frac{1}{40}-\\frac{1}{60} \\right)=360$$（个） 答：这批零件共$$360$$个． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2766", "queId": "691619244b38471793c09d5cecfeccbc", "competition_source_list": ["2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（二）"], "difficulty": "1", "qtype": "single_choice", "problem": "如果规定$$a*b=13\\times a-b\\div 8$$，那么$$17*24$$的最后结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$218$$ "}], [{"aoVal": "B", "content": "$$208$$ "}], [{"aoVal": "C", "content": "$$198$$ "}], [{"aoVal": "D", "content": "$$200$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由题意知：$$a*b=13\\times a-b\\div 8$$， 则$$17*24=13\\times 17-24\\div 8=221-3=218$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1454", "queId": "9e474070cf7f4c21b31990f7be7feee0", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "姐姐比妹妹的年龄大$$4$$岁，当两人的年龄和为$$50$$岁时，请问妹妹为多少岁？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->逻辑分析"], "answer_analysis": ["姐姐与妹妹的年龄差是$$4$$岁，当两人年龄和为$$50$$岁时，妹妹的年龄等于$$(50-4)\\div 2=23$$（岁）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2026", "queId": "f3ad2ace7b88455989f808776636113d", "competition_source_list": ["2014年全国华杯赛小学中年级竞赛决赛第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "老师共买了$$53$$支铅笔，分给了$$A$$，$$B$$，$$C$$，$$D$$四个同学，分到最多的与最少的铅笔数相差不到$$5$$支，如果$$B$$把分到的铅笔全都给$$A$$，那么$$A$$的铅笔数是$$C$$的$$2$$倍；如果$$B$$把分到的铅笔全都给$$C$$，那么$$C$$的铅笔数是$$D$$的$$2$$倍，由此可知，$$B$$分到~\\uline{~~~~~~~~~~}~支铅笔． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["设$$A$$，$$B$$，$$C$$，$$D$$分到的铅笔数分别是$$A$$，$$B$$，$$C$$，$$D$$，由$$B+C=2D$$，知$$C$$、$$D$$、$$B$$依次成等差数列，设公差为$$K$$；由$$A+B=2C$$，知$$A$$、$$C$$、$$B$$依次成等差数列，则公差为$$2K$$；由$$4$$人铅笔数相差不会超过$$4$$，所以$$K=0$$或$$1$$；若$$K=0$$，则$$4\\times B=53$$，但$$53$$不是$$4$$的整数倍； 若$$K=1$$，$$A\\textgreater C\\textgreater D\\textgreater B$$，则$$4\\times C-1=53$$，但$$54$$不是$$4$$的整数倍． 综上所述，$$B$$分到$$15$$支铅笔． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2316", "queId": "eec6381edd0840e9951ade470b6ce1f2", "competition_source_list": ["2004年第2届创新杯六年级竞赛初赛第6题", "2004年六年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "一辆客车和一辆货车同时从甲、乙两城的中点向相反方向行驶，$$3$$小时后客车到达甲城；货车离乙城还有$$45$$千米，已知货车的速度是客车的$$\\frac{3}{4}$$；甲、乙两城之间的路程是( ) ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$千米 "}], [{"aoVal": "B", "content": "$$180$$千米 "}], [{"aoVal": "C", "content": "$$315$$千米 "}], [{"aoVal": "D", "content": "$$360$$千米 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->两人相遇与追及问题->相遇问题->同时出发相向而行"], "answer_analysis": ["当客车到达甲城时，货车离乙城的距离为$$45$$千米，这说明在相同的时间里，客车比货车多行了$$45$$千米。又已知货车的速度是客车的$$\\frac{3}{4}$$，那么在相同的时间里，货车所走的路程也是客车所走路程的$$\\frac{3}{4}$$，所以客车所走的路程为$$45\\div (1-\\frac{3}{4})=180$$千米，那么甲、乙两城之间的距离为$$180\\times 2=360$$千米，选D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1953", "queId": "fc402a95c0f74317a1bab3c06e10c42b", "competition_source_list": ["2020年新希望杯二年级竞赛初赛（团战）第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$10$$名同学参加$$50$$米赛跑．跑到一半的时候，小明后面有$$5$$人，前面有$$4$$人，之后，没有人超过他，而小明又超过了$$3$$人到达终点．这次比赛没有并列名次，那么小明是． ", "answer_option_list": [[{"aoVal": "A", "content": "第一名 "}], [{"aoVal": "B", "content": "第二名 "}], [{"aoVal": "C", "content": "第三名 "}], [{"aoVal": "D", "content": "第四名 "}], [{"aoVal": "E", "content": "第五名 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["刚开始小明在的名次是第$$5$$名，再超过三个人，因此$$5-3=2$$，小明变成了第$$2$$名. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2857", "queId": "c899e93d4c4f40b3b979930294a436f4", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第8题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "把$$\\frac{3}{21}$$化成小数，小数点后面第$$2019$$个数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$\\frac{3}{21}=\\frac{1}{7}=0.142857$$，则$$2019\\div6\\cdots\\cdots3$$， 所以小数点后第$$2019$$个数字是$$2$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2043", "queId": "cb1b39ac92e6484cae9221c015880272", "competition_source_list": ["2019年四川绵阳涪城区绵阳东辰国际学校小升初第2题3分", "2018年第6届湖北长江杯六年级竞赛初赛B卷第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个三角形的三个内角的度数比是$$5:3:2$$，这个三角形是． ", "answer_option_list": [[{"aoVal": "A", "content": "锐角三角形 "}], [{"aoVal": "B", "content": "直角三角形 "}], [{"aoVal": "C", "content": "钝角三角形 "}], [{"aoVal": "D", "content": "等腰三角形 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->几何模块->直线型->图形认知->图形认知角->三角形内外角度"], "answer_analysis": ["$$2+3=5$$，则为直角三角形． 两个较小角之和大于最大角时为锐角三角形，等于时为直角三角形，小于时为钝角三角形，也可求出最大角，再判断． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "750", "queId": "99f1c4804c984a8db3897d8497c852ad", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛决赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "某四位数减去它的各位数字之和后得到$$200\\square $$，所有这样的四位数从大到小排列，第三个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2016$$ "}], [{"aoVal": "B", "content": "$$2017$$ "}], [{"aoVal": "C", "content": "$$2018$$ "}], [{"aoVal": "D", "content": "$$2019$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["位值原理．假设四位数是$$\\overline{ABCD}$$，那么$$\\overline{ABCD}-(A+B+C+D)=999A+99B+9C=200\\square $$，$$200\\square $$一定为$$9$$的倍数即$$2007$$，$$111A+11B+C=223$$，$$A=2$$，$$B=0$$，$$C=1$$，$$D$$从$$9$$开始从大到小排列第三个是$$7$$，即第三个数是$$2017$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2178", "queId": "46cf507f3eeb4f5d8f52cf61d9ef7387", "competition_source_list": ["2016年创新杯六年级竞赛训练题（三）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "小明从$$A$$地到$$B$$地去，去时走路每小时行$$4$$千米，返回时骑电动车每小时行$$20$$千米，求小明往返一趟平均速度为每小时（~ ）千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$千米 "}], [{"aoVal": "B", "content": "$$6\\frac{2}{3}$$千米 "}], [{"aoVal": "C", "content": "$$8$$千米 "}], [{"aoVal": "D", "content": "$$12$$千米 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["略 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2061", "queId": "e682cf1190244aba8e9c2685f8505093", "competition_source_list": ["2020年第25届YMO五年级竞赛决赛第18题"], "difficulty": "1", "qtype": "single_choice", "problem": "由于天气逐渐冷起来，牧场上的草不仅不长大，反而以固定的速度在减少．已知某块草地上的草可供$$20$$头牛吃$$5$$天，或可供$$15$$头牛吃$$6$$天．照此计算，可供~\\uline{~~~~~~~~~~}~头牛吃$$10$$天． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->牛吃草问题->牛吃草转化型->草衰减牛吃草"], "answer_analysis": ["设每头牛每天吃$$1$$份，则草在$$6-5=1$$天内减少$$20\\times 5-15\\times 6=10$$份，原草量为$$\\left( 20+10 \\right)\\times 5=150$$份．若吃$$10$$天，可有$$150\\div 10-10=5$$头牛． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1196", "queId": "12214b967a234fa0b750734a5b8c3195", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "四棵树上共有$$40$$只鸟，若第一棵树上有$$3$$只鸟飞到第二棵树，第二棵树上有$$4$$只鸟飞到第三棵树，第三棵树有$$5$$只鸟飞到第四棵树，这样四棵树上的鸟就一样多了．第三棵树原来有只鸟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["最后四棵树上的鸟一样多， 则每棵树有$$40\\div 4=10$$（只）鸟． 则第三棵树上的鸟飞到第四棵树之前有$$10+5=15$$（只）鸟， 则第三棵树一开始有$$15-4=11$$（只）鸟． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1139", "queId": "0d7143d22d5f4a86bce356767b011ec8", "competition_source_list": ["2021年第4届山东青岛市南区京山杯六年级竞赛决赛A卷第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两人给一片花园浇水，甲单独做需要$$4$$小时完成浇水任务，乙单独做需要$$6$$小时完成浇水任务．现由甲、乙两人合作，完成浇水任务需要． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2.4$$小时 "}], [{"aoVal": "B", "content": "$$3.2$$小时 "}], [{"aoVal": "C", "content": "$$5$$小时 "}], [{"aoVal": "D", "content": "$$10$$小时 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["设完成浇水任务需要$$x$$小时，依题意有： $$\\left( \\frac{1}{4}+\\frac{1}{6} \\right)x=1$$， 解得$$x=2.4$$． 故完成浇水任务需要$$2.4$$小时． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1279", "queId": "991db21eabaa4e5cbafe9ad0898dc04f", "competition_source_list": ["2014年全国迎春杯六年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个半径为$$30$$厘米的蛋糕可以让$$9$$个食量完全相同的人吃饱，如果半径增加了$$200 \\%$$，同样高的蛋糕可以让个这种人吃饱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$81$$ "}], [{"aoVal": "D", "content": "$$144$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->分百应用题->量率对应已知单位1"], "answer_analysis": ["由条件，面积变为原来的$${{\\left( 1+200 \\% \\right)}^{2}}$$，所以可供$$9\\times {{\\left( 1+200 \\% \\right)}^{2}}=81$$（个）人吃饱． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2396", "queId": "4607d92f4a5240bca87d69536f7f28e5", "competition_source_list": ["2020年新希望杯六年级竞赛（2月）第25题"], "difficulty": "1", "qtype": "single_choice", "problem": "【$$2020$$年六年级卷第$$25$$题】比较大小：$$1+ \\frac{1}{2^{2}}+ \\frac{1}{3^{2}}+ \\frac{1}{4^{2}} \\cdots + \\frac{1}{2020^{2}}$$~\\uline{~~~~~~~~~~}~$$2$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\textgreater$$ "}], [{"aoVal": "B", "content": "$$\\textless{}$$ "}], [{"aoVal": "C", "content": "$$=$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}} ~\\textless{} ~\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{1999\\times 2020}$$ $$=1-\\frac{1}{2020}=\\frac{2019}{2020}$$． ∴$$1+\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}} ~\\textless{} ~1+\\frac{2019}{2020} ~\\textless{} ~2$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2224", "queId": "3af94345dafa44a988e2219b1b69b9d7", "competition_source_list": ["2009年第7届创新杯四年级竞赛初赛第6题4分", "2009年四年级竞赛创新杯"], "difficulty": "3", "qtype": "single_choice", "problem": "小红上山时每走$$30$$分钟休息$$10$$分钟，下山时每走$$30$$分钟休息$$5$$分钟。已知小红下山的速度是上山速度的$$1.5$$倍，如果上山用了$$3$$小时$$50$$分，那么下山用了（ ）小时。 ", "answer_option_list": [[{"aoVal": "A", "content": "1.25 "}], [{"aoVal": "B", "content": "2.0 "}], [{"aoVal": "C", "content": "2.5 "}], [{"aoVal": "D", "content": "2.25 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题"], "answer_analysis": ["上山用了$$60\\times 3+50=230$$（分），由于$$230\\div \\left( 30+10 \\right)=5\\cdots \\cdots 30$$，所以上山途中休息了$$5$$次，因此走了$$230-10\\times 5=180$$（分）。由于下山速度是上山速度的$$1.5$$倍，所以下山共走了$$180\\div 1.5=120$$（分），又由于下山每走$$30$$分钟休息$$5$$分钟，而在走完最后一个$$30$$分钟后已到达山下，不需要休息，所以下山所用总时间为$$\\left( 30+5 \\right)\\times 3+30=135$$（分），即$$2.25$$小时，选$$D$$。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1105", "queId": "186caafcff2c40218b6077a84b0199af", "competition_source_list": ["2014年第3届广东广州羊排赛六年级竞赛第8题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "一根绳子剪成两段，第一段长$$\\frac{7}{13}$$米，第二段长占全长的$$\\frac{7}{13}$$，那么这两段绳子， ． ", "answer_option_list": [[{"aoVal": "A", "content": "第一段长 "}], [{"aoVal": "B", "content": "第二段长 "}], [{"aoVal": "C", "content": "两段一样长 "}], [{"aoVal": "D", "content": "无法比较 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["第二段占全长的$$\\frac{7}{13}$$，则第一段占全长的$$1-\\frac{7}{13}=\\frac{6}{13}$$，显然第二段长． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1266", "queId": "6bbbe28fda824d80a21b41b7c133fc4f", "competition_source_list": ["小学中年级四年级其它第3讲用比例思想解应用题", "2014年全国华杯赛小学高年级竞赛初赛A卷第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "小明所在班级的人数不足$$40$$人，但比$$30$$人多，那么这个班男、女生人数的比不可能是（~ ~ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "$$2:3$$~ "}], [{"aoVal": "B", "content": "$$3:4$$ "}], [{"aoVal": "C", "content": "$$4:5$$ "}], [{"aoVal": "D", "content": "$$3:7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$\\left( 1 \\right)$$如果男女比为$$3:7$$，则总人数是$$3+7=10$$的倍数，而总人数是大于$$30$$小于$$40$$，不能是$$10$$的倍数. $$\\left( 2 \\right)$$当总人数为$$35$$人时，男生有$$14$$人，女生有$$21$$人，比为$$2:3$$. $$\\left( 3 \\right)$$当总人数为$$35$$人时，男生有$$15$$人，女生有$$20$$人，比为$$3:4$$. $$\\left( 4 \\right)$$当总人数为$$36$$人时，男生有$$16$$人，女生有$$20$$人，比为$$4:5$$. 所以选$$\\text{D}$$. ", "解：$$\\rm A$$：$$\\left({2+3}\\right)=5$$
\n大于$$30$$小于$$40$$的数中$$35$$是$$5$$的倍数，所以这个班男、女生人数的比可能是$$2:3$$；
\n$$\\rm B$$：$$3+4=7$$
\n大于$$30$$小于$$40$$的数中$$35$$是$$7$$的倍数，所以这个班男、女生人数的比可能是$$3:4$$；
\n$$\\rm C$$：$$4+5=9$$
\n大于$$30$$小于$$40$$的数中$$36$$是$$9$$的倍数，所以这个班男、女生人数的比可能是$$4:5$$；
\n$$\\rm D$$：$$3+7=10$$
\n大于$$30$$小于$$40$$的数中没有数是$$10$$的倍数，所以这个班男、女生人数的比不可能是$$3:7$$；
\n故选：$$\\rm D$$．
"], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2012", "queId": "e5cfc9a40d1148718b5bf8f0e921c60a", "competition_source_list": ["2006年第4届创新杯四年级竞赛初赛A卷第5题", "2006年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "某月有三个星期日的日期都是偶数，这个月的15号是星期( ). ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "四 "}], [{"aoVal": "C", "content": "五 "}], [{"aoVal": "D", "content": "六 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["在同一个月的日历牌中，左右相邻的日期号码相差1，而上下相邻的日期号码相差7，或者说，在日历牌中，一列的日期是公差为7的等差数列. 这样，1号、8号、15号、22号、29号为一列，共5天，3奇2偶; 2号、9号、16号、23号、30号为一列，共5天，3偶2奇; 3号、10号、17号、24号、31号为一列，共5天，3奇2偶. 对于本题，2号、9号、16号、23号、30号应为星期天才符合3个偶数的要求，故本题中16号为星期天，则15号为星期六，选D "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "374", "queId": "66219bb394c34ab4a941b52dcea65cef", "competition_source_list": ["2020年希望杯二年级竞赛模拟第19题"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$3$$、$$3$$、$$5$$、$$9$$分别放入方格中，和最小是． $$\\square \\square +\\square \\square $$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$74$$ "}], [{"aoVal": "C", "content": "$$92$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["要使得和最小，那么两个因数应该尽可能的小； 如果要求两个因数尽可能的小，那么两个因数十位应该填入较小数， 即这两个因数应该是：$$35+39=74$$， 所以和最小是$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1316", "queId": "94968f7a81ad4c78ba0c87f5c3fdd7c5", "competition_source_list": ["2006年第4届创新杯五年级竞赛初赛B卷第1题", "2006年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "一根长2米的木棍，锯成每段长0.4米的短木棍需要20分钟，那么锯成每段长0.5米的木棍需要( ). ", "answer_option_list": [[{"aoVal": "A", "content": "15分钟 "}], [{"aoVal": "B", "content": "12分钟 "}], [{"aoVal": "C", "content": "10分钟 "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->单归一问题"], "answer_analysis": ["锯一次需要:$$20\\div \\left( 2\\div 0.4-1 \\right)=5$$(分钟)，锯成长0.5米的木棍需要锯$$2\\div 0.5-1=3$$(次)，需要$$3\\times 5=15$$(分钟) "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1083", "queId": "0f87229594744c8eb56d342c0afbb670", "competition_source_list": ["2019年世奥赛一年级竞赛决赛第12题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "第一个筐里有$$14$$个梨，第二个筐里有$$35$$个梨，从第一个筐里拿$$1$$个放到第二个筐里，现在一共有多少个梨? ", "answer_option_list": [[{"aoVal": "A", "content": "$$47$$ "}], [{"aoVal": "B", "content": "$$48$$ "}], [{"aoVal": "C", "content": "$$49$$ "}], [{"aoVal": "D", "content": "$$50$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->加减法应用->加减法应用顺口溜"], "answer_analysis": ["从第$$1$$个盘子里拿$$1$$个梨到第$$2$$个盘子里，但是梨的总数量不变，还是$$14+35=49$$（个）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1253", "queId": "39464b27b034493db260ea389b4a41c9", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小$$Y$$和小$$M$$两人比赛爬楼梯，小$$Y$$跑到$$4$$楼时，小$$M$$恰好跑到$$3$$楼，照这样计算，小$$Y$$跑到$$16$$楼时，小$$M$$跑到了楼． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$4-1=3$$（层楼梯）， $$3-1=2$$（层楼梯）， $$16-1=15$$（层楼梯）， $$15\\times \\frac{2}{3}=10$$（层楼梯）， $$10+1=11$$（层）， 答：小$$M$$跑到了$$11$$层楼． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1535", "queId": "ff75f93ed1ce4b3d888aea71ed00259f", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛三年级竞赛初赛第4题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "有一列数：$$3$$，$$3$$，$$1$$，$$1$$，$$2$$，$$3$$，$$3$$，$$1$$，$$1$$，$$2$$，$$\\cdots \\cdots $$那么第$$53$$个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["这列数按照``$$3$$，$$1$$，$$2$$''的周期排列，则，$$53\\div 3=17\\cdots \\cdots 2$$， 所以第$$53$$个数位这个周期的第$$2$$个数，即为$$1$$． 故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1465", "queId": "9e53db3ee81247c2aa57046448824dd2", "competition_source_list": ["2018年环亚太杯六年级竞赛复赛第15题"], "difficulty": "4", "qtype": "single_choice", "problem": "$$A$$, $$B$$ and $$C$$ are three consecutive numbers that are divisible by $$5$$, $$8$$ and $$9$$ respectively. The smallest possible sum of these three numbers is ． $$A$$、$$B$$、$$C$$是三个连续数，分别可被$$5$$、$$8$$及$$9$$整除．此三个数最小的和是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$555$$ "}], [{"aoVal": "B", "content": "$$666$$ "}], [{"aoVal": "C", "content": "$$777$$ "}], [{"aoVal": "D", "content": "$$888$$ "}], [{"aoVal": "E", "content": "$$999$$ "}]], "knowledge_point_routes": ["海外竞赛体系->知识点->数论模块->整除->整除特征", "拓展思维->思想->分类讨论思想"], "answer_analysis": ["$$A$$能被$$5$$整除，个位就是$$0$$或$$5$$，$$A$$加$$1$$得$$B$$ 是$$8$$的倍数，个位不能是$$1$$，故只能是$$6$$，$$B$$再加$$1$$得$$C$$ 是$$9$$的倍数，个位数字是$$7$$，$$C$$的个位只能由$$3\\times 9=27$$得来． 假设$$C=3\\times 9=27$$，则$$B=26$$，$$26$$不能被$$8$$整除；故排除； 假设$$C=13\\times 9=127$$，则$$B=126$$，$$126$$不能被$$8$$整除；故排除； 假设$$C=23\\times 9=207$$，则$$B=206$$，$$206$$不能被$$8$$整除；故排除； 假设$$C=33\\times 9=297$$，则$$B=296$$，$$296\\div 8=37$$，且$$295\\div 5=59$$；因此，最小$$A=295$$，$$B=296$$，$$C=297$$，三个数的和为：$$295+296+297=888$$． 故选$$888$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2497", "queId": "46b564d86dc34e2d8c258aab6114f4b0", "competition_source_list": ["2014年迎春杯四年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "对于任何自然数，定义$$ni=1\\times 2\\times 3\\times \\cdots \\times n$$，如$$8i=1\\times 2\\times 3\\times \\cdots \\times 8$$。那么，算式：$$2014i+2013i-2012i+2011i+\\cdots -4i+3i-2i+1i$$，计算结果的个位数字是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->定义新运算->直接运算型->普通型"], "answer_analysis": ["解：由新定义$$ni=1\\times 2\\times 3\\times \\cdots \\times n$$可知： $$2014i=1\\times 2\\times 3\\times 4\\times 5\\times 6\\times \\cdots \\times 2012\\times 2013\\times 2014$$， $$2013i=1\\times 2\\times 3\\times 4\\times 5\\times 6\\times \\cdots \\times 2012\\times 2013$$， $$2012i=1\\times 2\\times 3\\times 4\\times 5\\times 6\\times \\cdots \\times 2012$$， $$\\cdots $$ $$5i=1\\times 2\\times 3\\times 4\\times 5$$， 由观察很容易知道，$$2014i$$，$$2013i$$，$$2012i$$，$$\\cdots $$，$$6i$$，$$5i$$的因式中均含有$$2\\times 5$$，所以他们的个位数都为$$0$$。 又因为： $$4i=1\\times 2\\times 3\\times 4=24$$， $$3i=1\\times 2\\times 3=6$$， $$2i=1\\times 2=2$$， $$1i=1$$， 所以$$2014i+2013i-2012i+2011i+\\cdots -4i+3i-2i+1i$$的个位数为：$$0-4+6-2+1=1$$。 故选：B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "390", "queId": "812a74cf0b014d0cac905915c398d52c", "competition_source_list": ["2014年第25届亚太杯四年级竞赛决赛第18题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$$Ellie$$懂英文那么她也懂法语． 如果上述断定为真，那么以下哪一项一定也为真？ ", "answer_option_list": [[{"aoVal": "A", "content": "Ellie懂英文但不懂法语． "}], [{"aoVal": "B", "content": "Ellie懂法语但不懂英文． "}], [{"aoVal": "C", "content": "Ellie既懂英文又懂法语． "}], [{"aoVal": "D", "content": "如果$$Ellie$$不懂法语，那么他一定懂英文． "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理", "Overseas Competition->知识点->组合模块->逻辑推理"], "answer_analysis": ["懂英文$\\to$懂法语 C项说明肯定充分条件也肯定必要条件. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3326", "queId": "4e5b2fc3da6d41a19ac354135f35e0b6", "competition_source_list": ["2017年第15届全国希望杯小学高年级六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$20$$个苹果分给$$3$$个小朋友，每个小朋友至少分$$1$$个，共有（~ ~ ~）种分法$$.$$如果可以有小朋友没分到苹果，共有（~ ~ ~）种分法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$190$$，$$253$$ "}], [{"aoVal": "B", "content": "$$171$$，$$253$$ "}], [{"aoVal": "C", "content": "$$171$$，$$231$$ "}], [{"aoVal": "D", "content": "$$190$$，$$231$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["（$$1$$）当每个小朋友至少分$$1$$个时，可以把$$20$$个苹果排成一排，在$$20$$个苹果的$$19$$个空中插 人$$2$$个板．一个空最多插一个板，把苹果分成$$3$$份，分给$$3$$个小朋友，因此有$$\\text{C}_{19}^{2}=171$$（种）方法． （$$2$$）当可以有小朋友没有分到苹果时，可以先给每个小朋友$$1$$个苹果，就变成了前面的问题， $$23$$个苹果．每个小朋友至少分$$1$$个，所以有$$\\text{C}_{22}^{2}=231$$（种）方法． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "712", "queId": "31e80ab41dde4d7f9bcd81c5f86e277b", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "定义$$n!=1\\times 2\\times 3\\times \\cdots \\times (n-1)\\times n$$，读作$$N$$的阶乘$$2019!$$能被$$7$$整除，如果把这个乘积去反复除以$$7$$，直到不能被$$7$$整除为止，从第一次除以$$7$$开始算，共可除以$$7$$次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$288$$ "}], [{"aoVal": "B", "content": "$$329$$ "}], [{"aoVal": "C", "content": "$$334$$ "}], [{"aoVal": "D", "content": "$$335$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数的性质->余数性质综合"], "answer_analysis": ["求$$2019$$！能被$$7$$整除的次数，只需要计算出$$2019$$！中所有能被$$7$$，$${{7}^{2}}$$，$${{7}^{3}}$$整除的个数，最后相加即可． $$1-2019$$中是$$7$$的倍数共有$$2019\\div 7=288\\ldots \\ldots 3$$，共有$$288$$个．是$${{7}^{2}}$$的倍数共有$$2019\\div 49=41\\ldots \\ldots 10$$，共有$$41$$个，是$${{7}^{3}}$$的倍数共有$$2019\\div 343=5\\ldots \\ldots 304$$，共有$$5$$个，因此共有$$288+41+5=334$$（个），即共可除以$$7$$有$$334$$次，故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1067", "queId": "8f5bd4faf340411da4e57b67b5e98663", "competition_source_list": ["2008年三年级竞赛明心奥数挑战赛"], "difficulty": "0", "qtype": "single_choice", "problem": "一桶油连桶重$$9$$千克，用去一半后，连桶重$$5$$千克，原来桶里的油重（ ）千克。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->加减法应用->同增同减应用题"], "answer_analysis": ["一半重量为$$9-5=4$$（千克），则原来有油$$4\\times 2=8$$（千克）。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1753", "queId": "c419d34436834a259b3d0588066f8dbe", "competition_source_list": ["2007年第5届创新杯四年级竞赛第3题5分", "2007年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "做一批零件，原计划每天生产40个，实际上每天比原计划多生产10个，结果提前5天完成任务，那么这批零件的个数是( )个． ", "answer_option_list": [[{"aoVal": "A", "content": "500 "}], [{"aoVal": "B", "content": "1000 "}], [{"aoVal": "C", "content": "1500 "}], [{"aoVal": "D", "content": "2000 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->变速工程问题"], "answer_analysis": ["$$40\\times 5\\text{=}200$$(个) ，$$200\\div 10\\text{=}20$$(天)，$$40\\times \\left( 20\\text{+}5 \\right)\\text{=}1000$$(个)． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "721", "queId": "2e0326ae090846bca3e93987fd5f85dc", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级五年级竞赛邀请赛训练题（四）"], "difficulty": "1", "qtype": "single_choice", "problem": "两个数最大公因数为$$6$$，最小公倍数为$$180$$，满足条件的数一共有（~ ）组． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$组 "}], [{"aoVal": "B", "content": "$$3$$组 "}], [{"aoVal": "C", "content": "$$4$$组 "}], [{"aoVal": "D", "content": "$$5$$组 "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["依题意不妨假设这两个数分别为$$6a$$和$$6b$$（$$a$$与$$b$$互质），则$$6ab=180$$，$$ab=30$$，$$a$$与$$b$$互质的正整数解有以下$$4$$组$$\\begin{cases} a=1 b=30 \\end{cases}$$，$$\\begin{cases} a=2 b=15 \\end{cases}$$，$$\\begin{cases} a=3 b=10 \\end{cases}$$，$$\\begin{cases} a=5 b=6 \\end{cases}$$. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "987", "queId": "29dab88e227f4e9aaaf1ad954ccfceb8", "competition_source_list": ["2008年四年级竞赛创新杯", "2008年第6届创新杯四年级竞赛初赛A卷第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "传说中的九头鸟每只有9个头，1条尾巴;而九尾鸟每只9条尾巴，1个头.有一些九头鸟和九尾鸟在一起，数它们的头共有580个，数它们的尾巴共有900条.那么九头鸟和九尾鸟共有( )只. ", "answer_option_list": [[{"aoVal": "A", "content": "138 "}], [{"aoVal": "B", "content": "148 "}], [{"aoVal": "C", "content": "158 "}], [{"aoVal": "D", "content": "168 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题"], "answer_analysis": ["将所有九头鸟和九尾鸟的头数和尾巴数加起来，应该是它们只数和的10倍，所以九头鸟和九尾鸟共有$$\\left( 580+900 \\right)\\div 10=148$$只 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "311", "queId": "5757353e531148d5a5a01ec51c215850", "competition_source_list": ["2019年第24届YMO六年级竞��决赛第7题3分"], "difficulty": "3", "qtype": "single_choice", "problem": "有一个$$12$$级的楼梯．某人每次能登上$$1$$级或$$2$$级或$$3$$级，现在他要从地面登上第$$10$$级，有种不同的方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$149$$ "}], [{"aoVal": "B", "content": "$$244$$ "}], [{"aoVal": "C", "content": "$$264$$ "}], [{"aoVal": "D", "content": "$$274$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["设第$$n$$级有$${{a}_{n}}$$种登的方式， 则第$$n-3$$，$$n-2$$，$$n-1$$级再分别登$$3$$，$$2$$，$$1$$级可到第$$n$$级， 则$${{a}_{n}}={{a}_{n-3}}+{{a}_{n-2}}+{{a}_{n-1}}$$， 而$${{a}_{1}}=1$$，$${{a}_{2}}=1+1=2$$，$${{a}_{3}}=1+1+2=4$$， 故$${{a}_{4}}=1+2+4=7$$， $${{a}_{5}}=2+4+7=13$$， $${{a}_{6}}=4+7+13=24$$， $${{a}_{7}}=7+13+24=44$$， $${{a}_{8}}=13+24+44=81$$， $${{a}_{9}}=24+44+81=149$$， $${{a}_{10}}=44+81+149=274$$． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1900", "queId": "c088a3409bcf40c8b2b4db5b3fa94278", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一种特殊的计算器，当输入一个$$10$ $49$$的自然数$$A$$后，计算器会先算$$A+A$$，然后将所得结果的十位和个位顺序颠倒，再加$$2$$后显示出最后的结果．那么，下列四个选项中，可能是最后显示的结果． ", "answer_option_list": [[{"aoVal": "A", "content": "$$41$$ "}], [{"aoVal": "B", "content": "$$42$$ "}], [{"aoVal": "C", "content": "$$43$$ "}], [{"aoVal": "D", "content": "$$44$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["倒推．$$44$$ 对应的是$$44-2=42$$，颠倒后是$$24$$，除以$$2$$ 为$$12$$．符合条件．其他的均不符合条件． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1868", "queId": "b29d9f3f75754639b02f6359c72ca3cb", "competition_source_list": ["2015年华杯赛五年级竞赛初赛", "2015年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "六位同学考试的平均成绩是$$92.5$$分，他们的成绩是互不相同的整数，最高的$$99$$分，最低的$$76$$分。那么，按分数从高到低居第三位的同学的分数至少是分。22西川 ", "answer_option_list": [[{"aoVal": "A", "content": "$$94$$ "}], [{"aoVal": "B", "content": "$$95$$ "}], [{"aoVal": "C", "content": "$$96$$ "}], [{"aoVal": "D", "content": "$$97$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题"], "answer_analysis": ["六位同学的平均成绩为$$92.5$$分，因此这六人的总分为$$92.5\\times 6=555$$（分）。又最高分为$$99$$，最低分为$$76$$，因此剩下四人的分数和为$$380$$分。要使第三名的同学分数尽可能小，那么得使另外三人的分数尽可能大。而第二名最高为$$98$$分，而第四名第五名分数小于第三名，所以第三名至少为：$$(380-98)\\div3+1=95$$（分），选$$\\text{B}$$。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1514", "queId": "83d1ae23327743feb4394a498709c53e", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "某学校准备购买$$30$$个篮球，三家商店每个篮球的售价都是$$25$$元，但优惠方法不同，甲店``买九赠一''，乙店``打八八折''；丙店``满$$100$$元减现金$$10$$元''，为节约资金，应该到店购买． A school is going to buy 30 basketballs. The price of each basketball in the three stores is 25 yuan, but the preferential methods are different. Store A \"buy nine and get one free\", store B \"give a 20\\% discount\";~Store C ~\"full 100 yuan minus 10 yuan in cash\", in order to save money, should go toshop to buy. ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "任意一个店皆可 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["甲店：只需付$$27$$个篮球的钱即可买到$$30$$个篮球，要$$27\\times 25=675$$元； 乙店：$$30\\times 25\\times 88 \\% =660$$元； 丙店：$$25\\times 30=750$$元，减$$7\\times 10=70$$元，花费$$750-70=680$$元． 因为$$660\\textless{}675\\textless{}680$$，所以应到乙店购买． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2508", "queId": "1e8457573b924ce0b194107476792f61", "competition_source_list": ["2018年湖北武汉创新杯小学高年级五年级竞赛初赛数��思维能力等级测试第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "小数乘法：$$0.025\\times 0.04$$的结果的小数位数有位． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数与运算->乘法->小数乘法->小数乘小数"], "answer_analysis": ["$$0.025\\times 0.04=0.001$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2136", "queId": "33627dbeadff4a9d846e22511680f607", "competition_source_list": ["2018年全国华夏杯四年级竞赛数学奥林匹克邀请赛全国总决赛第14题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小奥、小匹两人相距$$400$$米，他们向相同的方向行走，小奥在小匹的后面，小奥以每秒$$6$$米的速度追向小匹，小匹以每秒$$2$$米的速度向前行，那么小奥会在~\\uline{~~~~~~~~~~}~秒后追上小匹． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->两人相遇与追及问题->追及问题->同时出发"], "answer_analysis": ["$$400\\div (6-2)=100$$（秒）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1904", "queId": "d74ceeef5c2b484c8269c68bc8346e55", "competition_source_list": ["2016年北京学而思杯五年级竞赛冲刺讲义", "六年级其它导引"], "difficulty": "3", "qtype": "single_choice", "problem": "水果店进了一批水果，希望卖出去之后得到$$50 \\%$$的利润．当售出六成数量的水果时，由于天气原因水果无法保存，于是商店决定打折处理，结果还是有一成数量的水果烂了，最终只得到了所期望利润的$$34 \\%$$．请问：商店打折处理时打了几折？ ", "answer_option_list": [[{"aoVal": "A", "content": "五折 "}], [{"aoVal": "B", "content": "六折 "}], [{"aoVal": "C", "content": "七折 "}], [{"aoVal": "D", "content": "八折 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["方法一：设水果店打了$$x$$折，把所有的水果看成$$10$$份，设总成本为$$10$$元，即$$1$$份水果的成本是$$1$$元，则原期望利润为$$5$$元．依题意，有六成是按原定价卖的，得到的利润为： $$6\\times 50 \\%=3$$（元）； 三成打折卖掉，所得利润为： $$3\\times \\left( 1+50 \\% \\right)\\times x-3=4.5x-3$$； 还有一成坏掉了，不仅没有利润，还亏了$$1$$元．所以总共利润为： $$3+4.5x-3-1=4.5x-1$$． 实际利润为期望利润的$$34 \\%$$，由此列方程： $$4.5x-1=5\\times 34 \\%$$， 解得：$$x=0.6$$，即打六折． 方法二：设水果的进价为$$a$$，打折为$$x$$，则有：$$ \\left(1.5a\\times 0.6+1.5ax\\times 0.3 \\right)-a=0.34\\times 0.5a$$， 约掉等式两边的$$a$$得： $$1.5\\times 0.6+1.5x\\times 0.3-1=0.34\\times 0.5$$， 解得，$$x=0.6$$，即打了六折． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "301", "queId": "ff80808145ff6bf40146032a1134035a", "competition_source_list": ["2011年全国华杯赛竞赛初赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "老师问$$5$$名学生：``昨天你们有几个人复习数学了？'' 张：``没有人．''李：``一个人．''王：``二个人．''赵：``三个人．''刘：``四个人．'' 老师知道，他们昨天下午有人复习，也有人没复习，复习了的人说的都是真话，没复习的人说的都是假话．那么，昨天这$$5$$个人中复习数学的有（~ ~ ~ ~）个人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾"], "answer_analysis": ["$$5$$名学生说的话相互矛盾，只能有$$1$$人说的是真话，则只有李复习了，说的是真话． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "956", "queId": "f8ee359f577c421f9260a486aa48c716", "competition_source_list": ["2012年第8届全国新希望杯五年级竞赛复赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "二进制数$${{(101)}_{2}}$$，可用十进制表示为$$1\\times {{2}^{2}}+0\\times 2+1=5$$，二进制数$${{(1011)}_{2}}$$可用十进制表示为$${{(1011)}_{2}}=1\\times {{2}^{3}}+0\\times {{2}^{2}}+1\\times 2+1=11$$，那么三��制数$${{(11011)}_{2}}$$，用十进制表示为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$27$$ "}], [{"aoVal": "C", "content": "$$29$$ "}], [{"aoVal": "D", "content": "$$31$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$${{(11011)}_{2}}=1\\times {{2}^{4}}+1\\times {{2}^{3}}+0\\times {{2}^{2}}+1\\times 2+1=27$$，因此选$$B$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1922", "queId": "ce4fd87703c1411ab13f5d6a69dc4469", "competition_source_list": ["2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "奶奶折一个纸鹤用$$3$$分钟，每折好一个需要休息$$1$$分钟，奶奶从$$2$$时$$30$$分开始折，她折好第$$5$$个纸鹤时已经到了（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$时$$45$$分 "}], [{"aoVal": "B", "content": "$$2$$时$$49$$分 "}], [{"aoVal": "C", "content": "$$2$$时$$50$$分 "}], [{"aoVal": "D", "content": "$$2$$时$$53$$分 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题"], "answer_analysis": ["解：$$1\\times \\left( 5-1 \\right)=4$$（分钟） $$3\\times 5=15$$（分钟） $$2$$时$$30$$分$$+4$$分钟$$+15$$分钟$$=2$$时$$49$$分 答：她折好第$$5$$个纸鹤时已经到了$$2$$时$$49$$分。 故选:B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1785", "queId": "a8e81ac69bce4f3db7c2c26765811fc7", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛决赛第10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有两个同样的仓库，搬运完其中一个仓库的货物，甲需要$$6$$小时，乙需要$$7$$小时，丙需要$$14$$小时．甲、乙同时开始各搬运一个仓库的货物，开始时，丙先帮甲搬运，后来又去帮乙搬运，最后两个仓库的货物同时搬完．丙从一个仓库到另一个仓库的时间忽略不计．则丙帮甲搬了小时．【本题欣赏题】答案：B ", "answer_option_list": [[{"aoVal": "A", "content": "$$1.5$$ "}], [{"aoVal": "B", "content": "$$1\\frac{3}{4}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["本题主要考查分数的应用， 首先计算出三个人搬仓库用时，再分别计算甲、乙完成了多少， 然后可知丙完成的为$$1$$减去甲的完成量加上$$1$$减去乙的完成量， 再计算丙帮甲、乙分别用时即可， 三个人都搬了同样长的时间， 甲每小时搬$$\\frac{1}{6}$$，乙每小时搬$$\\frac{1}{7}$$，丙每小时搬$$\\frac{1}{14}$$， 三人每小时共可般$$\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{14}=\\frac{8}{21}$$， 则搬完两个仓库共要$$2\\div \\frac{8}{21}=\\frac{21}{4}$$时， 即三人都同样工作了$$\\frac{21}{4}$$小时， 而$$\\frac{21}{4}$$小时内甲完成$$\\frac{21}{4}\\times \\frac{1}{6}=\\frac{7}{8}$$， 乙完成$$\\frac{21}{4}\\times \\frac{1}{7}=\\frac{3}{4}$$， 即甲有$$\\frac{1}{8}$$是丙完成的，乙有$$\\frac{1}{4}$$是丙帮忙完成的， 丙帮甲、乙分别用时$$\\frac{1}{8}\\div \\frac{1}{14}=\\frac{7}{4}$$时，$$\\frac{1}{4}\\div \\frac{1}{14}=\\frac{7}{2}$$小时． 答：丙帮甲、乙分别用时$$\\frac{7}{4}$$时、$$\\frac{7}{2}小时$$， $$\\frac{7}{4}=1\\frac{3}{4}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3110", "queId": "ddb76f8cd53a4dd781d4ef232d455e94", "competition_source_list": ["2008年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$$0.1+0.2+0.3+\\cdots +9.8+9.9+10+9.9+9.8+\\cdots +0.2+0.1=$$( ) . ", "answer_option_list": [[{"aoVal": "A", "content": "990 "}], [{"aoVal": "B", "content": "1000 "}], [{"aoVal": "C", "content": "1010 "}], [{"aoVal": "D", "content": "1100 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列求和"], "answer_analysis": ["原式$$=2\\times \\left( 0.1+0.2+\\cdots +10 \\right)-10=2\\times \\frac{0.1+10}{2}\\times 100-10=1000$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "99", "queId": "b94b4958579747678cc031ca0663288c", "competition_source_list": ["2016年第21届全国华杯赛小学高年级竞赛初赛B卷第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "``凑$$24$$点''游戏规则是：从一副扑克牌中抽去大小王剩下$$52$$张（如果初练也可以只用$$1-10$$这$$40$$张牌），任意抽去$$4$$张牌（称牌组），用加、减、乘、除（可加括号）���牌面上的数算成$$24$$．每张牌必须用一次且只能用一次，并不能用几张牌组成一个多位数，如果抽出的牌是$$3$$，$$8$$，$$8$$，$$9$$，那么算式为$$(9-8)\\times 8\\times 3$$或$$(9-8\\div 8)\\times 3$$等．在下面$$4$$个选项中，唯一无法凑出$$24$$点的是（ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$，$$2$$，$$2$$，$$3$$ "}], [{"aoVal": "B", "content": "$$1$$，$$4$$，$$6$$，$$7$$ "}], [{"aoVal": "C", "content": "$$1$$，$$5$$，$$5$$，$$5$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["选项$$B$$：$$(7+1-4)\\times 6=24$$； 选项$$C$$：$$(5-1\\div 5)\\times 5=24$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2675", "queId": "3fe37abf96a1458e8588421fb8f29f43", "competition_source_list": ["2017年环亚太杯一年级竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "找规律： $$6$$，$$12$$，$$18$$，$$24$$，~\\uline{~~~~~~~~~~}~，$$36$$，~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "28，40 "}], [{"aoVal": "B", "content": "28，42 "}], [{"aoVal": "C", "content": "30，40 "}], [{"aoVal": "D", "content": "30，42 "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->数据处理"], "answer_analysis": ["每个数之间差了一个6 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1742", "queId": "d1d095663ee74915b3547b2546a270b7", "competition_source_list": ["2020年新希望杯六年级竞赛初赛（团战）第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "火星叔叔马丁从火星来到地球．马丁发现地球上的一个昼夜比火星少$$40$$分钟，而火星上的一年包含$$668$$个昼夜．那么，一个火星年大约是一个地球年的倍． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3.24$$ "}], [{"aoVal": "B", "content": "$$2.88$$ "}], [{"aoVal": "C", "content": "$$2.26$$ "}], [{"aoVal": "D", "content": "$$1.88$$ "}], [{"aoVal": "E", "content": "$$1.78$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->倍的基本计算->多量倍"], "answer_analysis": ["火星一个昼夜有$$24$$小时$$40$$分，火星一年有$$668$$个昼夜， 则火星一年有（$$24$$小时$$40$$分）$$\\times 668\\approx 16477.3$$（小时）， 地球一共有$$24\\times 365=8760$$（小时）， 一个火星年相当于地球$$16477.3\\div 8760\\approx 1.88$$（倍）． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2190", "queId": "2c4b8de22bac448487cb5a016e483d8e", "competition_source_list": ["2012年IMAS小学高年级竞赛第一轮检测试题第7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁各有一只手表． （$$1$$）甲的手表快了$$10$$分钟，但他以为慢了$$5$$分钟； （$$2$$）乙的手表慢了$$5$$分钟，但他以为快了$$10$$分钟； （$$3$$）丙的手表快了$$5$$分钟，但他以为快了$$3$$分钟； （$$4$$）丁的手表慢了$$5$$分红总，但他以为慢了$$10$$分钟． 用他们的手表，每个人都认为自己恰好能准时到达学校，请问谁会迟到？ ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "丁 "}], [{"aoVal": "E", "content": "都不会 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["甲早到$$5-(-10)=15$$分钟，丙早到$$5-3=2$$分钟，丁早到$$(-5)-(-10)=5$$分钟，乙迟到$$10-(-5)=15$$分钟． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "26", "queId": "12de7b1b675847f89116f392cf1f67e4", "competition_source_list": ["2012年第10届创新杯四年级竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "显示在电子钟上的时间是$$5:55$$，下一次电子钟上显示的时间又是全部相同的数字，还要过分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$71$$ "}], [{"aoVal": "B", "content": "$$255$$ "}], [{"aoVal": "C", "content": "$$316$$ "}], [{"aoVal": "D", "content": "$$377$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["因为分钟的十位最大为$$5$$，故下一次数字相同的时刻为$$11:11$$，$$11:11$$距$$5:55$$有$$5$$小时$$16$$分，即$$316$$分钟．所以选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1608", "queId": "ba579af7b16542fe83a96005aceadb50", "competition_source_list": ["2017���全国美国数学大联盟杯五年级竞赛初赛第37题"], "difficulty": "1", "qtype": "single_choice", "problem": "苏珊花了一天中的$$\\frac{1}{2}$$的时间做英语作业，$$\\frac{1}{3}$$的时间做数学作业，请问她这一天还剩余多少个小时？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["$$24\\times \\left(1-\\frac{1}{2}-\\frac{1}{3}\\right)=4$$（小时）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1375", "queId": "79f8a60c99024a13ae9d98b06b318cf4", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "祖玛游戏中，龙嘴里不断吐出很多颜色的龙珠，先$$4$$颗红珠，接着$$3$$颗黄珠，再$$2$$颗绿珠，最后$$1$$颗白珠，按此方式不断重复，从龙嘴里吐出的第$$200$$颗龙珠是． ", "answer_option_list": [[{"aoVal": "A", "content": "红珠 "}], [{"aoVal": "B", "content": "黄珠 "}], [{"aoVal": "C", "content": "绿珠 "}], [{"aoVal": "D", "content": "白珠 "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["$$200\\div (4+3+2+1)=20$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1787", "queId": "ad7170f33df9463ab04bbc585c10e4a9", "competition_source_list": ["2018年湖北武汉创新杯小学高年级六年级竞赛初赛数学思维能力等级测试第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两人出售同一种商品，甲按$$20 \\% $$的利润率定价，售出了$$15$$个；乙按照$$15 \\% $$的利润率定价，售出了$$24$$个，比较甲乙所获利润的多少，你的结论为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "甲获利润多 "}], [{"aoVal": "B", "content": "乙获利润多 "}], [{"aoVal": "C", "content": "两人利润相同 "}], [{"aoVal": "D", "content": "无法比较谁多 "}]], "knowledge_point_routes": ["拓展思维->思想->赋值思想"], "answer_analysis": ["假设成本为$$100$$元，则甲的利润$$ =100\\times 20{}^{0}/{}_{0}\\times 15 =300$$元，乙的利润$$=100\\times 15{}^{0}/{}_{0}\\times 24=360$$元． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1918", "queId": "aa24e2970e574c249d7fbb7095a3d0e7", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明原来有$$20$$个金币，如果小刚给小明$$5$$个金币，小刚就比小明少$$1$$个金币，小刚原来有个金币． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$27$$ "}], [{"aoVal": "C", "content": "$$29$$ "}], [{"aoVal": "D", "content": "$$31$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->移多补少->不等变不等（给完还少）"], "answer_analysis": ["小刚给小明$$5$$个金币， 小明有$$20+5=25$$（个）金币． 而此时小刚比小明少$$1$$个金币， 此时小刚有$$25-1=24$$（个）金币， 则原来小刚有$$24+5=29$$（个）金币， 选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2488", "queId": "1256778036884c6ca2b19aa1de26f1ad", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "三个分数$$\\frac{20122012}{20132013}$$，$$\\frac{20132013}{20142014}$$，$$\\frac{20142014}{20152015}$$中值最大的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{20122012}{20132013}$$ "}], [{"aoVal": "B", "content": "$$\\frac{20132013}{20142014}$$ "}], [{"aoVal": "C", "content": "$$\\frac{20142014}{20152015}$$ "}], [{"aoVal": "D", "content": "三个一样大 "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->分数->分数基础->分数的约分"], "answer_analysis": ["因为$$\\frac{20122012}{20132013}=\\frac{2012\\times 1001}{2013\\times 1001}=\\frac{2012}{2013}$$， $$\\frac{20132013}{20142014}=\\frac{2013\\times 1001}{2014\\times 1001}=\\frac{2013}{2014}$$，$$\\frac{20142014}{20152015}=\\frac{2014\\times 1001}{2015\\times 1001}=\\frac{2014}{2015}$$，$$1-\\frac{2012}{2013}=\\frac{1}{2013}$$，$$1-\\frac{2013}{2014}=\\frac{1}{2014}$$， $$1-\\frac{2014}{2015}=\\frac{1}{2015}$$． 因为$$\\frac{1}{2015}\\textless{}\\frac{1}{2014}\\textless{}\\frac{1}{2013}$$，所以$$\\frac{2014}{2015}\\textgreater\\frac{2013}{2014}\\textgreater\\frac{2012}{2013}$$．因此，三个分数中，最大的是$$\\frac{20142014}{20152015}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2524", "queId": "b02ef4bea36443d19fc86a820e8f5d85", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知等差数列$$13$$，$$18$$，$$23$$，$$28$$，$$\\cdots $$，$$1003$$．这个等差数列共有项． ", "answer_option_list": [[{"aoVal": "A", "content": "$$198$$ "}], [{"aoVal": "B", "content": "$$199$$ "}], [{"aoVal": "C", "content": "$$200$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["根据题意：公差为$$5$$， 所以项数：$$\\left( 1003-13 \\right)\\div 5+1=199$$． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1017", "queId": "2ea0bbea40ee45ed9ebc3d6c9a7f5a0f", "competition_source_list": ["小学中年级三年级上学期其它", "2017年全国华杯赛小学中年级竞赛初赛模拟第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两人在春节一共得了$$200$$元压岁钱，甲给乙$$40$$元，还比乙多$$10$$元，那么，原来甲得了元压岁钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$150$$ "}], [{"aoVal": "B", "content": "$$145$$ "}], [{"aoVal": "C", "content": "$$140$$ "}], [{"aoVal": "D", "content": "$$125$$ "}], [{"aoVal": "E", "content": "$$120$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->应用题模块->和差倍问题", "拓展思维->拓展思维->应用题模块->和差倍问题->和差问题->二量和差问题->明和暗差"], "answer_analysis": ["因为甲给乙$$40$$元，还比乙多$$10$$元，所以甲比乙多$$40\\times 2+10=90$$（元）， 所以甲：$$(200+90)\\div 2=145$$（元）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2240", "queId": "37b59df9fe384fb2a6ced65443c0cf18", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级五年级竞赛邀请赛训练题（四）"], "difficulty": "2", "qtype": "single_choice", "problem": "战士小王从$$A$$地前往$$B$$地，他每走$$40$$分钟就休息$$10$$分钟，到达$$B$$地共用$$4$$小时$$20$$分，从$$B$$地原路返回的速度是去时的$$2$$倍．如果他每走$$35$$分钟就休息$$5$$分钟，从$$B$$地返回$$A$$地用分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$109$$ "}], [{"aoVal": "B", "content": "$$110$$ "}], [{"aoVal": "C", "content": "$$115$$ "}], [{"aoVal": "D", "content": "$$140$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["从$$A$$至$$B$$一共历时$$4$$时$$20$$分$$=260$$分，其中$$40+10=50$$（分）一个周期，由$$260\\div 50=5\\cdots \\cdots 10$$，可知从$$A$$至$$B$$的步行时间$$40\\times 5+10=210$$（分 ），根据返回时速度是去时的$$2$$倍，可得从$$B$$至$$A$$的步行时间为$$210\\div 2=105$$（分），其中$$35$$分为一个周期中的步行时间，得$$105\\div 35=3$$（个），所以从$$B$$返回至$$A$$的总时间为$$35\\times 3+5\\times 2=115$$（分）. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3327", "queId": "9edf642268714579b365ea6ef835c4a1", "competition_source_list": ["2007年五年级竞赛创新杯", "2007年第5届创新杯五年级竞赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "我们称数$$543$$和$$531$$具有位数值递减性，而$$322$$则没有，因为它的个位数等于十位数。在$$100$$与$$599$$之间有( )个具有位数值递减性的数。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->字典排序法"], "answer_analysis": ["$$100\\sim 199$$之间没有，$$200\\sim 299$$之间有$$1$$个：$$210$$;$$300\\sim 399$$之间有$$3$$个：$$310\\text{，}320\\text{，}321$$；$$400\\sim 499$$之间有$$6$$个：$$410\\text{，}420\\text{，}421\\text{，}430\\text{，}431\\text{，}432$$，$$500\\sim 599$$之间有$$10$$个：$$510\\text{，}520\\text{，}521\\text{，}530\\text{，}531\\text{，}532\\text{，}540\\text{，}541\\text{，}542\\text{，}543$$，共有$$1+3+6+10=20$$（个）。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1259", "queId": "305f1dd02664430db79c70cff51ea07e", "competition_source_list": ["2019年陕西延安宝塔区北大培文学校小升初入学真卷第2题3分", "2006年第11届全国华杯赛竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "妈妈告诉馨文同学：``$$2006$$年共有$$53$$个星期日''．聪敏的馨文立到告诉妈妈：$$2007$$年的元旦一定是． ", "answer_option_list": [[{"aoVal": "A", "content": "星期一 "}], [{"aoVal": "B", "content": "星期二 "}], [{"aoVal": "C", "content": "星期六 "}], [{"aoVal": "D", "content": "星期日 "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$2006$$年有$$365$$天，而$$365=7\\times 52+1$$，又已知$$2006$$年有$$53$$个星期天，只能元旦是星期天，且$$12$$月$$31$$日也是星期日，所以，$$2007$$年月的元旦是星期一． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1993", "queId": "c1740269c7ab40d9895347bf871a053d", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "毛毛用围棋子摆成一个四层空心方阵，最外一层每边有围棋子$$10$$个．摆这个方阵共用围棋子个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$ "}], [{"aoVal": "B", "content": "$$108$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$132$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["由题意知，用围棋子摆成一个三层空心方阵，最外一层每边有围棋子$$12$$个，由于相邻两层每边相差$$2$$个，则由外向里的两层每边分别是$$(12-2)$$个、$$(12-2\\times 2)$$个，根据``四周的个数$$=$$（每边的个数$$-1$$）$$\\times 4$$''可分别求得这三层棋子的个数，再相加就是所用的总个数，据此解答． 本题关键是求出每层的个数；方阵问题相关的知识点是：四周的个数$$=$$（每边的个数$$-1$$）$$\\times 4$$，每边的个数$$=$$四周的个数$$\\div 4+1$$，中实方阵的总个数$$=$$每边的个数$$\\times $$每边的个数，空心方阵的总个数$$=$$（最外层每边的个数$$-$$空心方阵的层数）$$\\times $$空心方阵的层数$$\\times 4$$，外层边长数$$^{2}-$$中空边长数$$^{2}=$$实面积数． 最外边一层棋子个数：$$(12-1)\\times 4=44$$（个）， 第二层棋子个数：$$(12-2-1)\\times 4=36$$（个）， 第三层棋子个数：$$(12-2\\times 2-1)\\times 4=28$$（个）， 摆这个方阵共用棋子：$$44+36+28=108$$（个）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3141", "queId": "023f631f80954470b02a694ba84b946b", "competition_source_list": ["2006年第4届创新杯六年级竞赛初赛A卷第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "为了让人们感受随地丢弃废电池对环境造成的影响，某班环保小组的$$6$$名同学记录了自己一学期内用完的电池数量，结果如下（单位：节）：$$33$$，$$25$$，$$28$$，$$26$$，$$25$$，$$31$$．如果该班有$$45$$名学生，那么根据提供的数据，可以估计一学期内全班同学总共用完的电池数量约为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$900$$节 "}], [{"aoVal": "B", "content": "$$1080$$节 "}], [{"aoVal": "C", "content": "$$1260$$节 "}], [{"aoVal": "D", "content": "$$1800$$节 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率"], "answer_analysis": ["$$6$$名同学平均用完电池$$\\left( 33+25+28+26+25+31 \\right)\\div 6=28$$ 节， 全班共用完$$28\\times 45=1260$$ 节. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "141", "queId": "a6f12bb4be034fc793cd66af830d08ab", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{(2018 Youth Mathematics Olympics, Primary 4, Question \\#5)~} A certain school district holds the \"Xinmiao Cup\" primary school football tournament, with a total of $$10$$ football teams participating. The tournament adopts a round-robin system, and each team has to play against all other teams. According to the points ranking, these matches are arranged to be played on the football fields of $$3$$ schools. On average, how many matches should be arranged for each school?（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$） 某学区举行``新苗杯''小学生足球赛，共有$$10$$个足球队比赛，比赛采取循环制，每个队都要和其他各队赛一场，根据积分排名次，这些比赛分别安排在$$3$$个学校的球场上进行，平均每个学校要安排（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$）场比赛． ", "answer_option_list": [[{"aoVal": "A", "content": "$$45$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->组合模块->逻辑推理->体育比赛"], "answer_analysis": ["一共要进行的比赛场次：$$1+2+3+4+\\cdots \\cdots +9=45$$（场）， 则平均每个学校安排的场次：$$45\\div 3=15$$（场）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1782", "queId": "e451facf2fbe4ea98640d3211103e34c", "competition_source_list": ["2006年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "一个数加上$$8$$的和，再乘以$$8$$的积，再减去$$8$$的差，再除以$$8$$的商，等于$$80$$，那么，这个数是( )。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$37$$ "}], [{"aoVal": "B", "content": "$$59$$ "}], [{"aoVal": "C", "content": "$$73$$ "}], [{"aoVal": "D", "content": "$$86$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->还原问题->逆运算"], "answer_analysis": ["一个数$$\\xrightarrow{+8}\\triangle \\xrightarrow{\\times8 }\\triangle \\xrightarrow{-8}\\triangle \\xrightarrow{\\div8 }80$$， 还原就是$$80\\xrightarrow{\\times8 }640\\xrightarrow{+8}648\\xrightarrow{\\div 8}81\\xrightarrow{-8}73$$，算式:$$\\left( 80\\times 8+8 \\right)\\div 8-8=73$$，所以选C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1536", "queId": "a7ce7d0ca1494a5a82d8670f8ea79b75", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（四）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "一项工程，甲、乙两队合作需$$12$$天完成，现在甲队先做$$4$$天，然后乙队接着做$$6$$天，共完成这项工程的$$\\frac{3}{7}$$，那么甲队和乙队的工作效率之比为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4:3$$ "}], [{"aoVal": "B", "content": "$$3:4$$ "}], [{"aoVal": "C", "content": "$$3:2$$ "}], [{"aoVal": "D", "content": "$$2:3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["乙队：$$\\left( \\frac{3}{7}-\\frac{4}{12} \\right)\\div (6-4)=\\frac{1}{21}$$， 甲队：$$\\frac{1}{12}-\\frac{1}{21}=\\frac{1}{28}$$， $$\\frac{1}{28}:\\frac{1}{21}=3:4$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1855", "queId": "fbbac8ad81cc43ec8e6dea42bdbb05c8", "competition_source_list": ["2020年长江杯五年级竞赛复赛A卷", "2020年长江杯五年级竞赛复赛A卷第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2015$$年$$11$$月$$1$$日是星期日，问$$2016$$年$$11$$月$$1$$日是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "三 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "一 "}], [{"aoVal": "D", "content": "日 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["根据年份的判断，我们会发现$$2016$$年是闰年． 所以从$$2015$$年$$11$$月$$1$$日$$\\sim 2016$$年$$11$$月$$1$$日，一共是过了$$366$$天． 若算上$$2015$$年$$11$$月$$1$$日则是$$367$$天． 所以$$367\\div 7=52$$（周）$$\\cdots \\cdots 3$$（天）周期是：周日------周六， 所以$$2016$$年$$11$$月$$1$$日是周二． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3248", "queId": "3eb9d44ba0de4bfabec36c43fdb2e7b6", "competition_source_list": ["2014年迎春杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "今天是$$2013$$年$$12$$月$$21$$日，七位数$$\\overline{ABCDEFG}$$恰好满足：前五位数字组成的五位数$$\\overline{ABCDE}$$是$$2013$$的倍数，后五位数字组成的五位数$$\\overline{CDEFG}$$是$$1221$$的倍数。那么四位数$$\\overline{ABFG}$$的最小值是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$1034$$ "}], [{"aoVal": "B", "content": "$$2021$$ "}], [{"aoVal": "C", "content": "$$2815$$ "}], [{"aoVal": "D", "content": "$$3036$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->多数的最小公倍数"], "answer_analysis": ["解：依题意可知，$$\\overline{ABFG}$$最小，$$\\overline{ABCDE}$$就尽量小，还是$$2013$$的倍数，这个倍数是大于$$10$$的倍数。$$2013\\times $$$$\\overline{mn}$$$$=2000\\times $$$$\\overline{mn}$$$$+13\\times $$$$\\overline{mn}$$，同时发现$$\\overline{CDE}$$是$$13$$的倍数。 $$\\overline{CDEFG}$$因为是五位数还是$$1221$$的倍数最小从$$10$$倍开���枚举。 $$1221\\times 10=12210$$，前三位$$122$$不是$$13$$的倍数， $$1221\\times 11=13431$$，前三位$$134$$不是$$13$$的倍数， $$1221\\times 12=14652$$，前三位$$146$$不是$$13$$的倍数， $$1221\\times 13=15873$$，前三位$$158$$不是$$13$$的倍数， $$1221\\times 14=17094$$，前三位$$170$$不是$$13$$的倍数， $$1221\\times 15=18315$$，前三位$$183$$不是$$13$$的倍数， $$1221\\times 16=19536$$，前三位数$$195\\div 13=15$$，满足条件。$$2013\\times 15=30195$$，$$\\overline{ABFG}$$$$=3036$$。 故选：D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2143", "queId": "138e438338a84a28b23209cbc7ccbb36", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第13题2分"], "difficulty": "2", "qtype": "single_choice", "problem": "一列货车早晨$$6$$时从甲地开往乙地，平均每小时行$$45$$千米，一列客车从乙地开往甲地，平均每小时比货车快$$15$$千米，已知客车比货车迟发$$2$$小时，中午$$12$$时两车同时经过途中某站，然后仍继续前进，问：当客车到达甲地时，货车离乙地还有千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$47.2$$ "}], [{"aoVal": "B", "content": "$$37.5$$ "}], [{"aoVal": "C", "content": "$$24.5$$ "}], [{"aoVal": "D", "content": "$$10.5$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["本题考查相遇问题； 客车比货车迟发$$2$$小时，所以客车是$$8$$时出发，到$$12$$时，经过$$4$$小时到达相遇点，而货车到达相遇点共用了$$4+2=6$$（小时），那么甲、乙两地之间的距离就可求出，用货车 速度乘相遇时货车总共用的时间加上客车速度乘相遇时客车所用时间，即$$45\\times \\left( 4+2 \\right)+60\\times 4=510$$（千米）；客车行完全程所需的时间： $$510\\div \\left( 45+15 \\right)=8.5$$（小时）；货车行完全程所需的时间：$$8.5+2=10.5$$（小时）；客车到达甲地时，货车距乙地的距离，用全程减去货车已经行的路程，即$$510-45\\times 10.5=37.5$$（千米）． 客车速度： $$45+15=60$$（千米$$/$$时） 两地距离： $$45\\times \\left( 12-6 \\right)+60\\times \\left( 12-6-2 \\right)=510$$（千米） 客车行完全程所需时间： $$510\\div 60=8.5$$（小时） 客车到达甲地时，货车距乙地的距离： $$510-45\\times \\left( 8.5+2 \\right)=37.5$$（千米） 答：客车到达甲地时，货车离乙地还有$$37.5$$千米． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2977", "queId": "a9ce0f6214b047ff92dace2379bed966", "competition_source_list": ["2014年IMAS小学高年级竞赛第一轮检测试题第4题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "请问下列哪一项的值最小？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1-\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}-\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{3}-\\frac{1}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{4}-\\frac{1}{5}$$ "}], [{"aoVal": "E", "content": "$$\\frac{1}{5}-\\frac{1}{6}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$1-\\frac{1}{2}=\\frac{1}{2}\\textgreater\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}\\textgreater\\frac{1}{3}-\\frac{1}{4}=\\frac{1}{12}\\textgreater\\frac{1}{4}-\\frac{1}{5}=\\frac{1}{20}\\textgreater\\frac{1}{5}-\\frac{1}{6}=\\frac{1}{30}$$． 故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3372", "queId": "6f360731186d4ca48add7d17227896d0", "competition_source_list": ["2017年全国美国数学大联盟杯小学高年级六年级竞赛初赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$101$$到$$199$$的整数中，有多少个十位数字和个位数字都是偶数的数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$49$$ "}], [{"aoVal": "D", "content": "$$50$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->计数模块->加乘原理"], "answer_analysis": ["在$$101$$到$$199$$的整数中，有多少个十位数字和个位数字都是偶数的数？ whole numbers 整数；$$an$$ even tens digit 一个十位数是偶数的数； an even ones digit 一个个位数是偶数的数． 十位可以选择$$0$$、$$2$$、$$4$$、$$6$$、$$8$$，共$$5$$个数，个位可以选择$$0$$、$$2$$、$$4$$、$$6$$、$$8$$，共$$5$$个数，除去$$100$$这个数, 一共有$$5\\times 5-1=24$$个． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "131", "queId": "46d5f6ba0a8a4562aec7b4829e89da2a", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在垒球比赛中，若赢$$1$$场得$$3$$分，平$$1$$场得$$1$$分，输$$1$$场不得分．每个队都与其他队交锋$$4$$场，这时四个参赛队的总积分为：$$A$$队$$22$$分，$$B$$队$$19$$分，$$C$$队$$14$$分，$$D$$队$$12$$分．那么有（~ ）场比赛为平局． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["对于赛况分析试题，尤其对于与分数相关的试题，最重要的是思维方式，本题如果从整体上来考虑比赛所产生的总分值，问题将迎刃而解，依题意可知比赛总场次为$$24$$场比赛之中，若平局则将会让所有队伍的总分增加$$2$$分（比赛双方均得$$1$$分），若出现了胜败，则所有队伍的总分增加$$3$$分，而现在所有队伍获得的总分值为：$$22+19+14+12=67$$（分），$$24$$场比赛，总分最多24\\times 3=72分，平$$1$$场总分少$$1$$分，$$72-67=5$$分，所以平局$$5$$场. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1806", "queId": "8dd5a33a27db4129a32b529de67a92eb", "competition_source_list": ["2004年第2届创新杯六年级竞赛复赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "某商场的营业额$$2000$$年和$$2001$$年连续两年平均每年比上一年上升$$10 \\%$$，而$$2002$$年和$$2003$$年连续两年平均每年比上一年下降$$10 \\%$$．那么$$2003$$年的营业额与$$1999$$年的营业额相比较． ", "answer_option_list": [[{"aoVal": "A", "content": "下降了$$2 \\%$$ "}], [{"aoVal": "B", "content": "下降了$$1-99 \\%$$ "}], [{"aoVal": "C", "content": "上升了$$2 \\%$$ "}], [{"aoVal": "D", "content": "没有变化 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["可设$$1999$$年的营业额为$$1$$，那么$$2003$$年的营业额可表示$${{(1+10 \\%)}^{2}}{{(1-10 \\%)}^{2}}$$，即可求得． $$2003$$年的营业额为 $${{(1+10 \\%)}^{2}}{{(1-10 \\%)}^{2}}\\approx 98 \\%$$， 即$$2003$$年的营业额是$$1998$$年的$$98 \\%$$． 即$$2003$$年的营业额比$$1998$$年的营业额降低了$$1-98 \\%=2 \\%$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2618", "queId": "a2ac4f9c464c48f69101e73baec27dfc", "competition_source_list": ["2008年全国迎春杯三年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算． $$24+63+52+17+49+81+74+38=$$~\\uline{~~~~~~~~~~}~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$398$$ "}], [{"aoVal": "B", "content": "$$408$$ "}], [{"aoVal": "C", "content": "$$393$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$$$$24+63+52+17+49+81+74+38=$$ $$=(38+52)+(63+17)+(49+81)+74+24$$ $$= 90+80+130+98$$ $$=398$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "133", "queId": "22e78288ac9446d2a68cc0a20071a308", "competition_source_list": ["2006年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "如果一个整数，与1，2，3这三个数，通过加减乘除运算(可以添加括号)组成算式，结果等于24，那么这个整数称为可用的，那么，在4，5，6，7，8，9，10这七个数中，可用的整数有( )个. ", "answer_option_list": [[{"aoVal": "A", "content": "7 "}], [{"aoVal": "B", "content": "5 "}], [{"aoVal": "C", "content": "2 "}], [{"aoVal": "D", "content": "4 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->巧填算符"], "answer_analysis": ["4，5，6，7，8，9，10这7个数中，每个数都能与1，2，3这三个数通过加减乘除组成算式，结果都可以为24，所以可用的数有7个 ，故选A "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "405", "queId": "85e263ab76fe4981944f67c1999de56d", "competition_source_list": ["2006年四年级竞赛创新杯", "2006年第4届创新杯四年级竞赛初赛B卷第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "在下列算式中加一对括号后，算式的最大值是． $$7\\times 9+12\\div 3-2$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$65$$ "}], [{"aoVal": "B", "content": "$$77$$ "}], [{"aoVal": "C", "content": "$$89$$ "}], [{"aoVal": "D", "content": "$$90$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["$$7\\times 9+12\\div 3-2$$加上括号最大是： $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde7\\times (9+12\\div 3)-2$$ $$=7\\times 13-2$$ $$=91-2$$ $$=89$$，加上一个括号后算式的最大值是$$89$$， 故$$\\text{C}$$正确． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1961", "queId": "cea688f33ff44640871cbd28e0fcae76", "competition_source_list": ["2003年第1届创新杯六年级竞赛复赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "四个同学进行计算比赛，比赛内容是：在$$9$$、$$10$$、$$11$$、$$\\cdots \\cdots $$、$$67$$、$$68$$这$$60$$个自然数的相邻两数之间任意添加符号``$$+$$''或``$$-$$''，然后进行计算．四个同学得到的结果分别是$$2274$$、$$2003$$、$$2300$$、$$2320$$，老师看后指出．这四个结果中只有一个是正确的．这个正确的结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2274$$ "}], [{"aoVal": "B", "content": "$$2003$$ "}], [{"aoVal": "C", "content": "$$2300$$ "}], [{"aoVal": "D", "content": "$$2320$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["由于$$9+10+11+\\cdots 68=2310$$，由此可知$$2320$$是错误的．由于$$2274$$、$$2003$$、$$2300$$都小于小于$$2310$$，所以减的数较多，由于减一个数，总和里面就要少这个数的$$2$$倍，如减$$2$$，则是$$2310-2\\times 2=2306$$，所以只要是小于$$2310$$．据此分析即 $$9+10+11+\\cdots 68=2310$$，$$2320\\textgreater2310$$，故$$\\text{D}$$错误； $$\\left( 2310-2274 \\right)\\div 2=18$$，$$18\\div 2=9$$，所以在$$9$$前是减号即可，符合题意． $$\\left( 2310-2003 \\right)=307\\textgreater68$$，错误 $$\\left( 2310-2000 \\right)\\div 2=155\\textgreater68$$，错误． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "647", "queId": "8b10405e15c94145849a2fdb62aed1e4", "competition_source_list": ["2014年全国迎春杯六年级竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "式子$$\\frac{2019}{x-5}$$为整数，则正整数$$x$$有种取值． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数的认识->数的特征->因数->分解质因数"], "answer_analysis": ["因为$$2014=2\\times 19\\times 53$$，$$x+1$$可能的取值为：$$2$$、$$19$$、$$53$$、$$38$$、$$106$$、$$1007$$、$$2014$$共七种． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2118", "queId": "f9be0bc26a2446ecad6235b1aad2c1cb", "competition_source_list": ["2020年第3届山东青岛市南区京山杯六年级竞赛决赛A卷第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "某商店卖出两件衣服，每件$$60$$元，其中一件赚$$25 \\%$$，另一件亏$$25 \\%$$，那么这两件衣服卖出后，商店是． ", "answer_option_list": [[{"aoVal": "A", "content": "不赚不亏 "}], [{"aoVal": "B", "content": "赚$$8$$元 "}], [{"aoVal": "C", "content": "亏$$8$$元 "}], [{"aoVal": "D", "content": "赚$$15$$元 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["设盈利的进价是$$x$$元，亏损的进价是$$y$$元，根据每件$$60$$元， 其中一件赚$$25 \\%$$，另一件亏$$25 \\%$$，可列出方程求解． 本题考查一元一次方程的应用，关键知道利润$$=$$售价$$-$$进价， 根据此可列方程求解． 设盈利的进价是$$x$$元，则 $$x+25 \\%x=60$$， $$x=48$$． 设亏损的进价是$$y$$元，则 $$y-25 \\%y=60$$， $$y=80$$． $$60+60-48-80=-8$$， ∴亏了$$8$$元． 所以$$\\text{C}$$选项是正确的． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "849", "queId": "c41b939c273f4ed584e17ec01c217271", "competition_source_list": ["2016年全国世奥赛小学高年级五年级竞赛A卷第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在一个天平的两边分别放上以下重量的物体，唯一平衡的一组是（~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "左边$$312\\times 2598$$克，右边$$820576$$克 "}], [{"aoVal": "B", "content": "左边$$137\\times 4725$$克，右边$$647335$$克 "}], [{"aoVal": "C", "content": "左边$$110\\times 3457$$克，右边$$380270$$克 "}], [{"aoVal": "D", "content": "左边$$261\\times 1231$$克��右边$$300291$$克 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$\\text{A}$$选项左边是$$3$$的倍数，右边数字之和不是$$3$$的倍数； $$\\text{B}$$选项左边是$$25$$的倍数，右边末位两位不是$$25$$的倍数； $$\\text{D}$$选项左边是$$9$$的倍数，右边数字之和不是$$9$$的倍数； 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2798", "queId": "531b496dbad1451dbcf846d4f4ecd158", "competition_source_list": ["2014年迎春杯五年级竞赛初赛", "2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在所有分母小于$$10$$的最简分数中，最接近$$20.14$$的分数是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{101}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{141}{7}$$ "}], [{"aoVal": "C", "content": "$$\\frac{181}{9}$$ "}], [{"aoVal": "D", "content": "$$\\frac{161}{8}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数化小数法"], "answer_analysis": ["解：A.$$\\frac{101}{5}\\textasciitilde=20.2$$，$$20.2-20.14=0.06$$ B.$$\\frac{141}{7}\\approx 20.14$$，$$20.14-20.14=0$$ C.$$\\frac{181}{9}\\approx 20.11$$，$$20.14-20.11=0.03$$ D.$$\\frac{161}{8}=20.125$$，$$20.14-20.125=0.015$$ 故选：B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1201", "queId": "420dd458d2bc45d786f285a5cd0281ae", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第9题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "有一筐梨，它的一半的一半是$$4$$，这筐梨有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->运算求解"], "answer_analysis": ["一个数的一半的一半是$$4$$，则这个数的一半为$$4\\times 2=8$$，则这个数为$$8\\times 2=16$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2862", "queId": "8d4c02897b09466d94da4261570b0b64", "competition_source_list": ["2014年IMAS小学中年级竞赛第一轮检测试题第1题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "请问算式$$2+0+1+4+2\\times 0\\times 1\\times 4$$的值是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$9$$ "}], [{"aoVal": "E", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$2+0+1+4+2\\times 0\\times 1\\times 4=7$$．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1497", "queId": "564e65d0cfc24e149de776141f5b4d85", "competition_source_list": ["2020年希望杯二年级竞赛模拟第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "学校买回来$$10$$个足球和一些排球．过了一段时间丢失了$$3$$个足球，排球数量不变，这时排球比足球多$$4$$个．学校原来买了个排球． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$14$$ "}], [{"aoVal": "E", "content": "$$17$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["学校买回来$$10$$个足球和一些排球．过了一段时间丢失了$$3$$个足球，此时还剩足球$$10-3=7$$（个），这时排球比足球多$$4$$个，排球有$$7+4=11$$（个），故$$\\text{C}$$正确． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "976", "queId": "04c29a8d1e064ad580144ede86b10b87", "competition_source_list": ["2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第9题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2018$$年$$9$$月$$10$$日教师节是星期一，$$2018$$年$$10$$月$$1$$日国庆节是． ", "answer_option_list": [[{"aoVal": "A", "content": "星期一 "}], [{"aoVal": "B", "content": "星期三 "}], [{"aoVal": "C", "content": "星期四 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["$$9$$月$$10$$日到$$10$$月$$1$$日过了$$30-10+1=21$$（天）， 而星期是$$7$$天一周期， 则$$21\\div 7$$余$$0$$，故仍为星期一，选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2084", "queId": "fda7dacca69042199789ef9560eb4023", "competition_source_list": ["2016年IMAS小学高年级竞赛第一轮检测试题第14题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "图书馆典藏的书籍中有$$12.1 \\%$$是小说，已知有$$1800$$本小说与$$2400$$本其它书籍被借走，这时留在图书馆的书籍有$$12 \\%$$是小说．请问这家图书馆原来典藏的书籍总共有多少本？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1296000$$ "}], [{"aoVal": "B", "content": "$$1582200$$ "}], [{"aoVal": "C", "content": "$$1800000$$ "}], [{"aoVal": "D", "content": "$$1586400$$ "}], [{"aoVal": "E", "content": "$$1291800$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设图书馆原有藏书共$$x$$本，则可列方程$$(12.1 \\%x-1800)\\div (x-1800-2400)=12 \\%$$，解得$$x=1296000$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1495", "queId": "7a9f3365412c48d68f3e5dc1eb9b8bcf", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$${{2019}^{2018}}$$的个位数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$${{9}^{1}}$$的个位为$$9$$，$${{9}^{2}}$$的个位为$$1$$，$${{9}^{3}}$$的个位为$$9$$，$${{9}^{4}}$$的个位为$$1\\cdots \\cdots $$，所以$${{9}^{2018}}$$的个位为$$1$$，即$${{2019}^{2018}}$$的个位为$$1$$． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1542", "queId": "8c4133fe55074be69d07aed23f630e6f", "competition_source_list": ["2016年第15届春蕾杯四年级竞赛决赛第11题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某人想用长绳吊一重物来测量井深，当他将绳子$$2$$折时，露出井口的绳比井深还长$$6$$米，当他把绳子$$4$$折时，露出井口的绳比井深长出$$1$$米，井深多少米？下面方程正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2x+6=4x+1$$ "}], [{"aoVal": "B", "content": "$$2x+2\\times 6=4x+1$$ "}], [{"aoVal": "C", "content": "$$2x+2\\times 6=4x+4$$ "}], [{"aoVal": "D", "content": "$$2x+2\\times 6=4x-1$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["把绳子折成$$2$$段时绳比井深长出$$6$$米，这说明，绳长比井深的$$2$$倍多$$12$$米；把绳子折成$$4$$段时，则绳比井深长$$1$$米，这说明绳长比井深的$$4$$倍多$$4$$米， 所以井深是$$\\left( 12-4 \\right)\\div \\left( 4-2 \\right)=4$$（米）. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2696", "queId": "955009e1379149b980ae385551b265e0", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\left( \\Delta \\times 2-1 \\right)\\times 2=2018$$，请问$$\\Delta $$代表的数是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$502$$ "}], [{"aoVal": "B", "content": "$$503$$ "}], [{"aoVal": "C", "content": "$$504$$ "}], [{"aoVal": "D", "content": "$$505$$ "}], [{"aoVal": "E", "content": "$$506$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\left( \\Delta \\times 2-1 \\right)\\times 2=2018$$， $$\\Delta \\times 2-1=2018\\div 2=1009$$， $$\\Delta \\times 2=1009+1=1010$$， $$\\Delta =1010\\div 2=505$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3446", "queId": "ea53c7b0267d483f9e2f694edd2b81f8", "competition_source_list": ["2015年全国AMC小学高年级六年级竞赛8第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "2015年全国$$AMC$$小学高年级六年级竞赛$$8$$第$$10$$题 How many integers between $$1000$$ and $$9999$$ have four distinct digits? ", "answer_option_list": [[{"aoVal": "A", "content": "$$3024$$ "}], [{"aoVal": "B", "content": "$$4536$$ "}], [{"aoVal": "C", "content": "$$5040$$ "}], [{"aoVal": "D", "content": "$$6480$$ "}], [{"aoVal": "E", "content": "$$6561$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想", "Overseas Competition->知识点->计数模块->枚举法综合->字典排序法"], "answer_analysis": ["翻译：$$1000$$至$$9999$$中有多少个位数字互不相同的四位数？ 组数问题：千位数字有$$1\\sim 9$$共$$9$$种选法，百、十、个位分别有$$9$$、$$8$$、$$7$$种选法，一共有$$9\\times 9\\times 8\\times 7=4536$$（个）数字互不相同的四位数． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1928", "queId": "e51d914817c549788928a8d2749836c6", "competition_source_list": ["2013年第11届全国小机灵杯三年级竞赛决赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "某年的三月份正好有$$4$$个星期三和$$4$$个星期六，那么这年$$3$$月$$1$$日是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}], [{"aoVal": "E", "content": "五 "}], [{"aoVal": "F", "content": "六 "}], [{"aoVal": "G", "content": "日 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数学广角->简单的周期->简单的周期计算"], "answer_analysis": ["$$3$$月有$$31$$天，从中任取连续$$28$$天，正好是四周，恰好有$$4$$个周三和$$4$$个周六， 因此，剩下的$$3$$天中不能有周三和周六 不妨取$$3$$月$$4$$日到$$3$$月$$31$$日，这$$28$$天中恰有$$4$$个周三和$$4$$个周六 剩下的$$1$$日到$$3$$日是连续的三天，其中有不能有周三和周六，发现这三天只能是周日、周一、周二 因此，$$3$$月$$1$$日是周日． "], "answer_value": "G"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1162", "queId": "151b4677e0074284937a0b48cbb6efe8", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2016$$个$$2017$$连乘，积的个位数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->数列操作周期问题->数的周期"], "answer_analysis": ["找规律，$${{7}^{1}}$$个位为$$7$$，$${{7}^{2}}$$个位为$$9$$，$${{7}^{3}}$$个位为$$3$$，$${{7}^{4}}$$个位为$$1$$，$${{7}^{5}}$$个位为$$7$$，$${{7}^{6}}$$个位为$$9$$，即四个数为一周期，$$2016\\div 4=504$$（组）$$\\cdots\\cdots0$$（个），即个位为$$1$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3082", "queId": "d896d1a82cbf4282bb8706c1a23e1b80", "competition_source_list": ["2013年第12届全国小机灵杯小学中年级三年级竞赛初赛第2题1分"], "difficulty": "0", "qtype": "single_choice", "problem": "家中电度表上的一度电表示的耗电量为（ ~~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.1$$千瓦小时 "}], [{"aoVal": "B", "content": "$$1$$千瓦小时 "}], [{"aoVal": "C", "content": "$$100$$瓦小时 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["家中电度表上的一度电表示的耗电量为$$1$$千瓦小时． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2157", "queId": "e7a6c79e6c9b4db1a3c88b552fa99d0c", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "亮亮早上$$8:00$$从甲地出发去乙地，速度是每小时$$8$$千米．他在中间休息了$$1$$小时，结果中午$$12:00$$到达乙地．那么，甲、乙两地之间的距离是（ ）千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人变速问题"], "answer_analysis": ["解：$$12$$时$$-8$$时$$=4$$小时 $$8\\times \\left( 4-1 \\right)$$ $$=8\\times 3$$ $$=24$$（千米） 答：甲、乙两地之间的距离是$$24$$千米． 故选:B． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3270", "queId": "55af4879c2be4b46b812772af3a799cf", "competition_source_list": ["2016年全国华杯赛小学高年级竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$，$$8$$这$$8$$个数排成一行，使得$$8$$的两边各数之和相等，那么共有（ ）种不同的排法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1152$$ "}], [{"aoVal": "B", "content": "$$864$$ "}], [{"aoVal": "C", "content": "$$576$$ "}], [{"aoVal": "D", "content": "$$288$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["首先求出$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$的和是$$28$$，判断出$$8$$的两边各数之和都是$$14$$，然后分为$$4$$种情况： （$$1$$）$$8$$的一边是$$1$$，$$6$$，$$7$$，另一边是$$2$$，$$3$$，$$4$$，$$5$$时； （$$2$$）$$8$$的一边是$$2$$，$$5$$，$$7$$，另一边是$$1$$，$$3$$，$$4$$，$$6$$时； （$$3$$）$$8$$的一边是$$3$$，$$4$$，$$7$$，另一边是$$1$$，$$2$$，$$5$$，$$6$$时； （$$4$$）$$8$$的一边是$$1$$，$$2$$，$$4$$，$$7$$，另一边是$$3$$，$$5$$，$$6$$时． 求出每种情况下各有多少种不同的排法，即可求出共有多少种不同的排法． 解：$$1+2+3+4+5+6+7=28$$ $$8$$的两边各数之和是：$$28\\div 2=14$$ （$$1$$）$$8$$的一边是$$1$$，$$6$$，$$7$$，另一边是$$2$$，$$3$$，$$4$$，$$5$$时， 不同的排法一共有： $$\\left( 3\\times 2\\times 1 \\right)\\times \\left( 4\\times 3\\times 2\\times 1 \\right)\\times 2$$ $$=6\\times 24\\times 2$$ $$=288$$（种） （$$2$$）$$8$$的一边是$$2$$，$$5$$，$$7$$，另一边是$$1$$，$$3$$，$$4$$，$$6$$时， 不同的排法一共有$$288$$种． （$$3$$）$$8$$的一边是$$3$$，$$4$$，$$7$$另一边是$$1$$，$$2$$，$$5$$，$$6$$时， 不同的排法一共有$$288$$种． （$$4$$）$$8$$的一边是$$1$$，$$2$$，$$4$$，$$7$$另一边是$$3$$，$$5$$，$$6$$时， 不同的排法一共有$$288$$种． 因为$$288\\times 4=1152$$（种）， 所以共有$$1152$$种不同的排法． 答：共有$$1152$$种不同的排法． 故选：$$\\rm A$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3443", "queId": "c5fb11324cf04685bc55b62830c70bc2", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "从小华家到学校有$$3$$条路可走，从学校到红领巾公园有$$2$$条路可走，从小华家经过学校到红领巾公园，有种不同的走法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->加法原理->加法原理里的其他类型"], "answer_analysis": ["小华从家到学校有$$3$$条路可选，从学校到公园有$$2$$条路可选，所以小华从家到学校再到公园就有$$3$$个$$2$$种情况可选，即$$2+2+2=6$$（种）． 故选择$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "250", "queId": "a794d511a59b4a46b2f7e47f3572eac4", "competition_source_list": ["2020年长江杯五年级竞赛复赛A卷", "2020年长江杯五年级竞赛复赛A卷第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "现有$$1$$克、$$2$$克、$$4$$克、$$8$$克、$$16$$克的砝码各一个和一个天平，最多能称出种不同的质量． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$31$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["（$$1$$）$$1$$个砝码可以称出的重量有$$5$$种：$$1$$克，$$2$$克，$$4$$克，$$8$$克，$$16$$克； （$$2$$）$$2$$个砝码可以称出的重量有$$10$$种： $$1+2=3$$（克），$$1+4=5$$（克），$$1+8=9$$（克）， $$1+16=17$$（克）； $$2+4=6$$（克），$$2+8=10$$（克），$$2+16=18$$（克）； $$4+8=12$$（克），$$4+16=20$$（克）， $$8+16=24$$（克）； （$$3$$）$$3$$个砝码可以称出的重量有$$10$$种： $$1+2+4=7$$（克），$$1+2+8=11$$（克）， $$1+2+16=19$$（克）， $$1+4+8=13$$（克），$$1+4+16=21$$（克）， $$1+8+16=25$$（克）， $$2+4+8=14$$（克），$$2+4+16=22$$（克）， $$2+8+16=26$$（克）， $$4+8+16=28$$（克）； （$$4$$）$$4$$个砝码可以称出的重量有$$5$$种： $$1+2+4+8=15$$（克），$$1+2+4+16=23$$（克）， $$1+2+8+16=27$$（克）， $$1+4+8+16=29$$（克），$$2+4+8+16=30$$（克）； （$$5$$）$$5$$个砝码可以称出的重量有$$1$$种： $$1+2+4+8+16=31$$（克）． 因为$$5+10+10+5+1=31$$（种）， 所以最多可以称出$$31$$种不同的重量，它们是$$1$$克$$-31$$克． 答：最多能称出$$31$$种不同重量的物体． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1601", "queId": "a7fe009e6d5e42ba82a412ff2418b5dc", "competition_source_list": ["2006年第4届创新杯五年级竞赛初赛A卷第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "一根长木棍上刻有三种刻度，第一种刻度将木棍十等分，第二种刻度将木棍十二等分，第三种刻度将木棍十五等分，如果沿每条刻度线将木棍锯开，木棍总共被锯成． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$段 "}], [{"aoVal": "B", "content": "$$24$$段 "}], [{"aoVal": "C", "content": "$$28$$段 "}], [{"aoVal": "D", "content": "$$30$$段 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["由于$$10$$、$$12$$、$$15$$的最小公倍数是$$60$$， 假定这根木棍的长为$$60$$， 于是，各等分的刻度线的标记处是： 十等分：$$6$$、$$12$$、$$18$$、$$24$$、$$30$$、$$36$$、$$42$$、$$48$$、$$54$$、$$60$$， 十二等分：$$5$$、$$10$$、$$15$$、$$20$$、$$25$$、$$30$$、$$35$$、$$40$$、$$45$$、$$50$$、$$55$$、$$60$$， 十五等分：$$4$$、$$8$$、$$12$$、$$16$$、$$20$$、$$24$$、$$28$$、$$32$$、$$36$$、$$40$$、$$44$$、$$48$$、$$52$$、$$56$$、$$60$$， 这样，把有三个刻度线标记处重合的（$$60$$）去掉， 把有两个刻度线标记处的（$$12$$、$$24$$、$$36$$、$$48$$、$$20$$、$$30$$、40）只算一个， 然后在$$4$$、$$5$$、$$6$$、$$8$$、$$10$$、$$12$$、$$15$$、$$16$$、$$18$$、$$20$$、$$24$$、$$25$$、$$28$$、$$30$$、$$32$$、$$35$$、$$36$$、$$40$$、$$42$$、$$44$$、$$45$$、$$48$$、$$50$$、$$52$$、$$54$$、$$55$$、$$56$$处将木棍锯断，共锯了$$27$$次． 根据植树问题的原理可知： 这根木棍共锯成$$27+1=28$$（段）． 所以$$\\text{C}$$选项是正确的． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1032", "queId": "06bc890f40e8498b952cbec5ee234c7d", "competition_source_list": ["2013年第9届全国新希望杯小学高年级六年级竞赛复赛第1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$16$$个小朋友，其中$$9$$岁的有$$11$$人，$$11$$岁的有$$2$$人，$$13$$岁的有$$3$$人，那么这$$16$$个小朋友的平均年龄是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$岁 "}], [{"aoVal": "B", "content": "$$10.5$$岁 "}], [{"aoVal": "C", "content": "$$11$$岁 "}], [{"aoVal": "D", "content": "$$11.5$$岁 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->公式类->加权平均数"], "answer_analysis": ["$$\\left( 9\\times 11+11\\times 2+13\\times 3 \\right)\\div 16=10$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3045", "queId": "bcd22aa627b84c479008dfba18b52add", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "五个连续正整数的和总是可以被下面哪个数整除？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["连续五个正整数一定是分别除以$$5$$余$$1$$,除以$$5$$余$$2$$,除以$$5$$余$$3$$,除以$$5$$余$$4$$,除以$$5$$余$$0$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "196", "queId": "50bf22cadf16416e9bc25410f0b1f25d", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "盒中有形状、大小、质料相同的红、白、黑颜色的球各$$10$$个，摸出若干个，要保证摸出的球中至少有$$3$$个球同色，摸出球的个数至少为个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->知识点->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["最不利原则．要保证拿到一种颜色至少有$$3$$个，则根据最不利原则，可先取每种颜色$$2$$个，最后取一个不论取哪种颜色，都一定可以满足条件，即需要$$2\\times 3+1=7$$个． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "152", "queId": "d9e95eec88e649b78334f802d72a04ab", "competition_source_list": ["2016年全国华杯赛小学中年级竞赛在线模拟第3题", "2013年全国华杯赛小学中年级竞赛初赛A卷第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "小东、小西、小南、小北四个小朋友在一起做游戏时，捡到了一条红领巾，交给了老师．老师问是谁捡到的？ 小东说：``不是小西．'' 小西说：``是小南.'' 小南说：``小东说的不对．'' 小北说:~~``小南说的也不对．'' 他们之中只有一个人说对了，这个人是． ", "answer_option_list": [[{"aoVal": "A", "content": "小东 "}], [{"aoVal": "B", "content": "小西 "}], [{"aoVal": "C", "content": "小南 "}], [{"aoVal": "D", "content": "小北 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾"], "answer_analysis": ["由于只有一个人说对了，而小北支持小东，那么他们俩都错了，所以反对小东的小南说对了． ", "根据题干分析可得，小南与小北说的话是相互矛盾的，所以两人中一定有一个人说的是正确的，假设小北说的是正确的，则小南说“小东说的不对”是错，可得，小东说的对，这样与已知只有一个人说对了相矛盾，所以此假设不成立，故小南说的是正确的．
\n故选：$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2901", "queId": "7cde35930bdd4c5b84f836bea0bb067a", "competition_source_list": ["2013年河南郑州中原网杯六年级竞赛初赛", "2017年河北邯郸邯山区凌云中学小升初（县城卷）第15题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一堆钢管最上层有$$14$$根，最下层有$$26$$根，每层相差$$1$$根，共有$$13$$层，这堆钢管共有（~ ）根． ", "answer_option_list": [[{"aoVal": "A", "content": "$$210$$ "}], [{"aoVal": "B", "content": "$$220$$ "}], [{"aoVal": "C", "content": "$$240$$ "}], [{"aoVal": "D", "content": "$$260$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["运用等差数列求和公式：$$\\left( 14+26 \\right)\\times 13\\div 2=260$$根． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "923", "queId": "b369149727f44f99a8ac00652377b49b", "competition_source_list": ["2015年第13届全国创新杯小学高年级五年级竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$A$$、$$B$$两个整数，$$A$$的各位数字之和为$$36$$，$$B$$的各位数字之和为$$25$$，且两数相加时进位三次，那么$$A+B$$的各位数字之和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$33$$ "}], [{"aoVal": "B", "content": "$$34$$ "}], [{"aoVal": "C", "content": "$$35$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->逐步调整思想"], "answer_analysis": ["进三次位数字和少$$3\\times 9=27$$，则$$36+25-27=34$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "102", "queId": "9439d12f0eff488cb9a312e5fddec0d3", "competition_source_list": ["2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第6题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一部电影共放映了$$85$$分钟，结束时正好是$$20:40$$分．这部电影是开始放映的． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18:15$$ "}], [{"aoVal": "B", "content": "$$19:15$$ "}], [{"aoVal": "C", "content": "$$20:15$$ "}], [{"aoVal": "D", "content": "$$21:15$$ "}], [{"aoVal": "E", "content": "$$22:05$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->组合模块->时间问题", "拓展思维->拓展思维->组合模块->时间问题->时间计算"], "answer_analysis": ["根据题意分析可知，$$60$$分钟$$=1$$小时，$$85$$分钟$$=1$$小时$$25$$分钟， 用结束的时间减去放映的时长即可得到开始放映的时间， 故列式为：$$20:40-1:25=19:15$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "720", "queId": "43c80ff7fd09450aa604c3dc120139fb", "competition_source_list": ["2006年第4届创新杯六年级竞赛初赛B卷第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "三个质数$$p$$，$$q$$，$$r$$满足$$p+q=r$$，且$$p\\textless9$$，那么$$p$$等于． ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数->特殊质数运用->特殊质数2"], "answer_analysis": ["依题意可知：质数的只有$$2$$是偶质数． 其余的就是奇质数， 所以$$p+q=r$$中含有一个偶质数． $$2$$是最小的质数．且$$p\\textless{}q$$． 所以$$p$$是唯一确定是偶质数$$2$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3068", "queId": "c665adacea3d42579645f883a733cce6", "competition_source_list": ["2020年新希望杯三年级竞赛初赛（团战）第35题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面哪个算式的计算结果是偶数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 784-455 \\right)\\times 39+44\\times 11$$ "}], [{"aoVal": "B", "content": "$$11\\times 1+22\\times 2+33\\times 3+44\\times 4+55\\times 5+66\\times 6+77\\times 7+88\\times 8+99\\times 9$$ "}], [{"aoVal": "C", "content": "$$11\\times 1+22\\times 2+33\\times 3+44\\times 4+55\\times 5+66\\times 6+77\\times 7+88\\times 8+99\\times 9$$ "}], [{"aoVal": "D", "content": "$$123\\times 456+789$$ "}], [{"aoVal": "E", "content": "$$2\\times 4\\times 6\\times \\cdots \\times 2018\\times 2020-1\\times 3\\times 5\\times \\cdots \\times 2017\\times 2019$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["暂无 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "604", "queId": "1d07e18777a3486786e49079a638ad29", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（三）第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "学校举行团体操表演，有$$2310$$名学生参加，分成人数相等的若干队，要求每队人数在$$100$$至$$200$$之间，共有种分法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$2310=2\\times 3\\times 5\\times 7\\times 11$$，$$2310$$在$$100$$至$$200$$之间的因数有：$$2\\times 5\\times 11$$、$$2\\times 7\\times 11$$、$$3\\times 5\\times 7$$、$$3\\times 5\\times 11$$共$$4$$个，所以共有$$4$$种分法． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2173", "queId": "38e39d8d0a604b09ad8eb52c16ee249d", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）"], "difficulty": "1", "qtype": "single_choice", "problem": "一列火车通过一座长$$320$$米的桥用了$$21$$秒，当它通过$$860$$米的隧道时，速度是过桥速度的$$2$$倍，结果用了$$24$$秒，火车通过大桥时的速度是每秒米；火车的车身长度为米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$；$$100$$ "}], [{"aoVal": "B", "content": "$$20$$；$$100$$ "}], [{"aoVal": "C", "content": "$$40$$；$$100$$ "}], [{"aoVal": "D", "content": "$$80$$；$$200$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->火车问题->火车过桥->连续过两桥"], "answer_analysis": ["若通过$$860$$米隧道时速度不变则需要$$24\\times 2=48$$（秒），火车过桥速度：$$\\left( 860-320 \\right)\\div \\left( 48-21 \\right)=20$$（米/秒）：火车车身长：$$21\\times 20-320=100$$（米）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "552", "queId": "05314ef75b83424eb2e172305c765c2b", "competition_source_list": ["2016年华杯赛六年级竞赛初赛", "2016年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在一个七位整数中，任何三个连续排列的数字都能构成一个被$$11$$或$$13$$整除的三位数，那么这个七位数最大是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$9981733$$ "}], [{"aoVal": "B", "content": "$$9884737$$ "}], [{"aoVal": "C", "content": "$$9978197$$ "}], [{"aoVal": "D", "content": "$$9871733$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->整除->整除特征->基础复合数字的整除特征"], "answer_analysis": ["解：在$$7$$位数中，首先分析前三位数字，最大的$$11$$的倍数是$$990$$，最大的$$13$$的倍数是$$988$$，因为$$0$$不能做首位。所以$$7$$位数中不能含有数字$$0$$，$$11$$的倍数的第二大数字是$$979$$，小于$$988$$。所以前三位数字是$$988$$。 第$$4$$位，根据如果是$$11$$的倍数，数字就是$$880$$；如果是$$13$$的倍数就是$$884$$。最大是$$884$$。 第$$5$$位，根据如果是$$11$$的倍数，数字就是$$847$$；如果是$$13$$的倍数就是$$845$$。最大是$$847$$。 第$$6$$位，根据如果是$$11$$的倍数，数字就是$$473$$，如果是$$13$$的倍数，在$$470\\sim479$$之间没有$$13$$的倍数。所以是$$473$$。 第$$7$$位，根据如果是$$11$$的倍数，数字就是$$737$$；如果是$$13$$的倍数，没有符合的数字。 所以这个$$7$$位数是$$9884737$$。 故选：B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2617", "queId": "67b7219d39014a228fef920b53503083", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛B卷第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$21$$个数���成一排：$$0$$，$$1$$，$$3$$，$$8$$，$$21$$，$$\\cdots \\cdots $$，除了两头的两个数外，每个数的$$3$$倍都等于它两边的两个数之和，那么这$$21$$个数中有个奇数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["这道题主要考查学生对排列规律的知识点的掌握情况．我们要根据题目中的数字的排列规律首先确定这组数是偶奇奇的规律，偶数两边肯定都是奇数，奇数两边必然一个奇一个偶．根据规律即可解答． $$1$$、这道题主要考查学生奇数与偶数的知识，我们根据排列规律来解答； $$2$$、根据能被$$2$$整除的数叫做偶数，不能被$$2$$整除的叫做奇数．$$0$$也是偶数．我们发现排列规律是偶奇奇的规律，偶数两边肯定都是奇数，奇数两边必然一个奇一个偶； $$3$$、根据一共有$$21$$个数，所以$$21\\div 3=7$$，$$7\\times 2$$即可解答． $$21\\div 3=7$$，$$7\\times 2=14$$（个）． 所以这$$21$$个数中有$$14$$个奇数． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1005", "queId": "09d3cd89203e48c5900b7e19c3ea5eba", "competition_source_list": ["2015年天津陈省身杯四年级竞赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "冬至又称``冬节''、``贺冬''，是华夏二十四节气之一，与二十四节气中的``夏至''相对．$$2014$$的``冬至''是$$12$$月$$22$$日星期一，那么$$2015$$年``冬至''（也在$$12$$月$$22$$日）是星期~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["从$$2014$$年``冬至''到$$2015$$年``冬至''共$$365$$天，而$$365\\div 7=52\\cdots \\cdots 1$$，因为$$2014$$年``冬至''是星期一，所以$$2015$$年的``冬至''是星期二． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2844", "queId": "6e681ea975ec4ae5953c822bbf3588f8", "competition_source_list": ["2007年走美杯五年级竞赛初赛", "2007年走美杯六年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "$$173\\times 173\\times 173-162\\times 162\\times 162$$的计算结果为（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$926183$$ "}], [{"aoVal": "B", "content": "$$926185$$ "}], [{"aoVal": "C", "content": "$$926187$$ "}], [{"aoVal": "D", "content": "$$926189$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数的性质->余数性质综合"], "answer_analysis": ["$$173\\times 173\\times 173$$ 的个位数是$$7$$，$$162\\times 162\\times 162$$的个位数是$$8$$，所以结果的个位数是$$9$$，只有D符合。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1182", "queId": "11932257cd884e439f86036291c12193", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛A卷第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙三个数的平均数是$$6$$，甲、乙两个数的平均数是$$4$$，乙、丙两个数的平均数是$$5.3$$，乙数是，甲、丙两个数的平均数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.6$$，$$8.7$$ "}], [{"aoVal": "B", "content": "$$0.6$$，$$8.3$$ "}], [{"aoVal": "C", "content": "$$0.4$$，$$8.7$$ "}], [{"aoVal": "D", "content": "$$0.4$$，$$8.3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["甲、乙、丙三个数的平均数是$$6$$，那么甲、乙、丙三个数的和等于$$6\\times 3=18$$，甲、乙两个数的平均数是$$4$$，甲、乙两数的和是$$2\\times 4=8$$，那么丙数等于$$18-8=10$$，乙、丙两数的平均数是$$5.3$$，乙、丙两数的和是$$5.3\\times 2=10.6$$，那么甲数等于$$18-10.6=7.4$$，丙数等于$$10$$，甲数等于$$7.4$$．那么乙数等于$$18-10-7.4=0.6$$，甲、丙的平均数等于$$(7.4+10)\\div 2=8.7$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1691", "queId": "729a905981f24b97a96fac6942846872", "competition_source_list": ["2014年全国华杯赛小学高年级竞赛初赛A卷第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "某次考试有$$50$$道试题，答对一道题得$$3$$分，答错一道题或不答题��扣$$1$$分，小龙得分$$118$$分，则小龙答对了道试题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$42$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$50$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->变型题"], "answer_analysis": ["假设小龙所有的题目都做对了，则小龙的得分为：$$50\\times 3=150$$（分）；实际上小龙的得分为$$118$$分，假设与实际相差：$$150-118=32$$（分）；小龙一道题由对变错会损失：$$3+1=4$$（分），所以小龙错了：$$32\\div 4=8$$（道）题，则小龙答对了：$$50-8=42$$（道）题． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "788", "queId": "72162667f19943b29f1f26fb93a6dbbf", "competition_source_list": ["2014年IMAS小学中年级竞赛第一轮检测试题第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$25$$位小朋友看电视，一条长凳最多可坐$$3$$个小朋友，请问最少需要多少条长凳？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}], [{"aoVal": "E", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$25=8\\times 3+1$$，如果有$$8$$条长凳，最多可坐$$24$$人，还有$$1$$位小朋友没有位置可坐． 所以最少需要$$9$$条长凳．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3407", "queId": "b74ac0626c454193827f57cfdb247691", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛B卷第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "从分别写有数字$$1$$、$$2$$、$$3$$、$$4$$、$$5$$的$$5$$张卡片中，任取两张，把第一张卡片上的数字作为十位数，第二张卡片上的数字作为个位数，组成一个两位数，则组成的数是偶数的概率． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{10}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->计数求概率"], "answer_analysis": ["两位数有$$12$$，$$13$$，$$14$$，$$15$$，$$21$$，$$23$$，$$24$$，$$25$$，$$31$$，$$32$$，$$33$$，$$35$$，$$41$$，$$42$$，$$43$$，$$45$$，$$51$$，$$52$$，$$53$$，$$54$$，共$$20$$种， 偶数有$$12$$，$$14$$，$$24$$，$$32$$，$$34$$，$$42$$，$$52$$，$$54$$，共有$$8$$种， $$8\\div 20=\\frac{4}{10}=\\frac{2}{5}$$， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "769", "queId": "ff8080814518d52401451925caa804fc", "competition_source_list": ["2014年全国迎春杯五年级竞赛初赛第12题"], "difficulty": "2", "qtype": "single_choice", "problem": "今天是$$2013$$年$$12$$月$$21$$日，七位数$$\\overline{ABCDEFG}$$恰好满足：前五位数字组成的五位数$$\\overline{ABCDE}$$是$$2013$$的倍数，后五位数字组成的五位数$$\\overline{CDEFG}$$是$$1221$$的倍数．那么四位数$$\\overline{ABFG}$$ 的最小值是（~~~~~~ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "$$1034$$ "}], [{"aoVal": "B", "content": "$$2021$$ "}], [{"aoVal": "C", "content": "$$2815$$ "}], [{"aoVal": "D", "content": "$$3036$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["要求$$\\overline{ABFG}$$最小也就是要求$$\\overline{ABCDE}$$尽量小，$$\\overline{ABCDE}$$是$$2013$$的倍数， 把$$\\overline{ABCDE}$$看成$$\\overline{AB}\\left\\textbar{} \\overline{CDE} \\right.$$，则$$\\overline{CDE}$$是$$13$$的倍数，而$$\\overline{CDEFG}$$是$$1221$$的倍数， 把$$\\overline{CDEFG}$$看成$$\\overline{CDE}\\left\\textbar\\overline{FG} \\right.$$，从$$1$$倍、$$2$$倍、\\ldots\\ldots 变化的话，前段加$$12$$，后段加$$21$$， 则每$$5$$次，$$\\overline{FG}$$向$$\\overline{CDE}$$进$$1$$， $$12\\times n+1\\times \\left[ \\frac{n}{5} \\right]$$ 是$$13$$的倍数， $$12\\times n+1\\times \\left[ \\frac{n}{5} \\right]=13n-n+\\left[ \\frac{n}{5} \\right]$$， 不难看出$$n$$最小得$$16$$， $$1221\\times16=19536$$，$$195\\div 13=15$$，$$2013\\times 15=30195$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1428", "queId": "55cc97d210094ea784b168a119f92ec1", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "把一根木料截成$$4$$段用$$12$$分钟．照这样的速度，如果把同样的木料截成$$8$$段，要用分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$26$$ "}], [{"aoVal": "C", "content": "$$28$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["从题干可以知道，把一根木料截成$$4$$段用$$12$$分钟； 我们要知道，锯一次可以把一根木料截成$$2$$段；锯二次可以把一根木料截成$$3$$段； 所以把一根木料截成$$4$$段需要锯三次，那么每次用时：$$12\\div3=4$$（分钟）； 照这样的速度，如果把同样的木料截成$$8$$段，即需要锯$$8-1=7$$（次）； 一共用时：$$7\\times4=28$$（分钟）；所以选择$$\\text{C}$$选项． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2944", "queId": "bb9158cadac44c9b97aaa7eafeab1d50", "competition_source_list": ["2017年第17届湖北武汉世奥赛五年级竞赛决赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "``黑洞''原指非常奇怪的天体，它的体积小，密度大，吸引力强，任何物体到了它那里都别想再出来．数字中也有类似的``黑洞''，将自然数经过某种数学运算之后陷入了一种循环的境况．例如，任选$$3$$个不同的数字，按从大到小的顺序排成一个数，再按从小到大的顺序排成一个数，用大数减去小数（如$$1$$，$$2$$，$$0$$，就用$$210-12=198$$），用所得结果的三位数重复上述过程，最后陷入的``数字黑洞''是（~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$123$$ "}], [{"aoVal": "B", "content": "$$495$$ "}], [{"aoVal": "C", "content": "$$594$$ "}], [{"aoVal": "D", "content": "$$954$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["$$123$$、$$198$$、$$792$$、$$693$$、$$594$$、$$495$$、$$495$$、$$495$$，可以发现选项的$$4$$个数字都会陷入$$495$$的数字黑洞，所以数字黑洞是$$495$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "126", "queId": "2b95568bb171467dacd1b8542d723b59", "competition_source_list": ["2016年第14届全国创新杯小学高年级五年级竞赛复赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个盒子里装有标号为$$1 \\tilde{ } 24$$的$$24$$张卡片，要从盒子中任意抽取卡片，至少要抽出（~ ）张卡片，才能保证抽出的卡片标号之差为$$4$$，（大标号减去小标号，卡片$$9$$只看作$$9$$，不能看成$$6$$只看作$$6$$，不能看成$$9$$）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据最不利原则，刚开始可以抽到$$1$$，$$2$$，$$3$$，$$4$$；$$9$$，$$10$$，$$11$$，$$12$$；$$17$$，$$18$$，$$19$$，$$20$$；此时一共抽了$$12$$张，如果下一张不管抽到谁，必然会两个数相差$$4$$，则至少要$$13$$张． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1431", "queId": "ff47a9f34a4d43ebb4aa769e75512540", "competition_source_list": ["2004年第2届创新杯六年级竞赛初赛第2题", "2004年六年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "把$$100$$千克的盐溶于$$1$$吨的水中，盐与盐水的质量之比是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{9}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{10}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{11}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{12}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->应用题->比例应用题", "拓展思维->拓展思维->应用题模块->浓度问题->浓度基本题型->已知溶质溶液求浓度"], "answer_analysis": ["盐的质量为$$100$$千克，盐水的质量为盐与水的质量之和，为$$100$$千克$$+1$$吨$$=100$$千克$$+1000$$千克$$=1100$$（千克），所以盐与盐水的质量比为$$100:1100=1:11$$，选C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1720", "queId": "df781301f134484db994adfdedbc7e08", "competition_source_list": ["2013年IMAS小学中年级竞赛第一轮检测试题第12题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "四位小朋友共有课外读物$$24$$本，甲给了乙$$3$$本，乙给了丙$$4$$本，丙给了丁$$5$$本，丁给了甲$$6$$本，这时他们$$4$$���课外读物的本数都相同．请问他们之中原来课外读物最少的人有多少本？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$57$$ "}], [{"aoVal": "B", "content": "$$58$$ "}], [{"aoVal": "C", "content": "$$59$$ "}], [{"aoVal": "D", "content": "$$60$$ "}], [{"aoVal": "E", "content": "$$61$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["可知这四位小朋友共有$$240$$本课外读物，所以当他们的书都相等时，每人各有$$24\\div 4=6$$（本）． 而题意可知此时甲的书多了$$6-3=3$$（本）、乙的书少了$$4-3=1$$（本）、丙的书少了$$5-4=1$$（本）、丁的书少了$$6-5=1$$（本），因此甲原有$$57$$本书，而乙、丙、丁各原有$$61$$本书．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1819", "queId": "a925c323d54e4a018507e845a6074a50", "competition_source_list": ["2016年全国华夏杯小学低年级一年级竞赛初赛香港賽區第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "参考附图，观察图中规律，从左至右数，第$$13$$个图形是甚麽？ $$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\cdots$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\blacksquare$$ "}], [{"aoVal": "B", "content": "$$\\blacktriangle$$ "}], [{"aoVal": "C", "content": "$$\\bigstar$$ "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["根據規律，「$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$」為一個週期 他的順序是： $$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$ 故答案為：$$\\blacksquare$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "919", "queId": "e088c6efe05548f0ac5dc3b5e2f6ffe2", "competition_source_list": ["2016年IMAS小学高年级竞赛第二轮检测试题第14题20分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知$$n$$，$$k$$为正整数使得$${{n}^{2}}\\textless{}4k\\textless{}{{n}^{2}}+\\frac{2016}{{{n}^{2}}}$$，请问$$n$$最大可能值是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "请将该小题的答案写在答题卡的指定答题区域，考试结束后上传至``本讲巩固'' "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->组合模块最值问题->枚举型最值问题"], "answer_analysis": ["该题为解答题，请将该小题的答案写在答题卡的指定答题区域，考试结束后上传至``本讲巩固'' "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "368", "queId": "8545e9e6ba8541fd815a0f4a2a50cede", "competition_source_list": ["2013年河南郑州中原网杯六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁$$4$$人坐在$$1$$、$$2$$、$$3$$、$$4$$号椅子上，有人说：乙坐在丙旁边，甲坐在乙、丙中间，乙没有坐在$$3$$号椅子上．已知此人说的都是错的，则丁坐在（~ ）号椅子上． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->连线法"], "answer_analysis": ["首先，``乙没有坐在$$3$$号椅子上''是错的，就可以判断出乙在$$3$$号椅子上；再看，``乙坐在丙旁边''是错的，可以判断出丙不在$$2$$，$$4$$号椅子上，所以在$$1$$号椅子；最后，``甲坐在乙、丙中间''是错的，可以判断出甲在$$4$$号，故得出丁在$$2$$号椅子上． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1121", "queId": "4ad1dd1cb29444d7a57081a28242cb2b", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（三）第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "刘老师买了一些苹果和梨，其中梨的数量是苹果的$$2$$倍，如果每名小朋友分$$2$$个苹果和$$3$$个梨，则苹果正好分完，而梨剩余$$23$$个，那么小朋友一共有~\\uline{~~~~~~~~~~}~名． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->变型题"], "answer_analysis": ["假设按照$$2$$个苹果和$$4$$个梨进行分配，当苹果分完时，梨也正好分完，但实际梨还剩余$$23$$个，这是因为每名小朋友实际比假设少分$$43=1$$（个）梨，所以小朋友的人数为$$23\\div 1=23$$（人）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "490", "queId": "e1212d6044bb46d59d9e510da3586f44", "competition_source_list": ["2017年全国美国数学大联盟杯五年级竞赛初赛第40题"], "difficulty": "1", "qtype": "single_choice", "problem": "汤姆建了两个塔．塔$$A$$发射$$5$$次能量波需要正好$$5$$秒．塔$$B$$发射$$10$$次能量波需要正好$$10$$秒．能量波是间隔发送的，并且任何两个连续的能量波之间的时间间隔是相同的．假设塔$$A$$发射$$12$$次能量波需要$${t}_{1}$$秒，假设塔$$B$$发射$$12$$次能量波需要$${t}_{2}$$秒．那么下面哪个判断是正确的？（~ ~) ", "answer_option_list": [[{"aoVal": "A", "content": "$${t}_{1}\\textless{t}_{2}$$ "}], [{"aoVal": "B", "content": "$${t}_{1}={t}_{2}$$ "}], [{"aoVal": "C", "content": "$${t}_{1}\\textgreater{t}_{2}$$ "}], [{"aoVal": "D", "content": "无法判断 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->时间问题->时间计算"], "answer_analysis": ["两个塔发射每一个能量波的时间是相同的，但是他们发送能量波的间隔并不相同．塔$$A$$发射$$5$$次能量波需要正好$$5$$秒，即是$$5$$次发送能量波的时间加上$$4$$次塔$$A$$的发射时间间隔，塔$$B$$发射$$10$$次能量波需要正好$$10$$秒，即是$$10$$次发送能量波的时间加上$$9$$次塔$$B$$的发射时间间隔，而$$10$$秒的时间等于$$5$$次发送能量波的时间加上$$4$$次塔$$A$$的发射时间间隔的两倍即$$10$$次发送能量波的时间加上$$8$$次塔$$A$$的发射时间间隔，故塔$$A$$的发射时间间隔大于塔$$B$$的发射时间间隔．它们发射$$12$$次能量波的时间均为$$12$$次发射能量波的时间加上各自的$$11$$次的发射时间间隔，由于塔$$A$$的发射时间间隔大于塔$$B$$的发射时间间隔，故$${t}_{1}\\textgreater{t}_{2}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1387", "queId": "5e7ff52da02943c38d1cf8af10c91f8e", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛决赛4分"], "difficulty": "1", "qtype": "single_choice", "problem": "六一儿童节用彩色小灯泡布置教室，按``一蓝、三红、二黄、二绿''的规律连接起来，第$$99$$个小灯泡是（~ ）色． ", "answer_option_list": [[{"aoVal": "A", "content": "红 "}], [{"aoVal": "B", "content": "黄 "}], [{"aoVal": "C", "content": "绿 "}], [{"aoVal": "D", "content": "蓝 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$1+3+2+2=8$$（个）灯泡一个循环周期，$$99\\div 8=12$$（组）$$\\cdot \\cdot \\cdot 3$$（个），所以第$$99$$个小灯泡是红色． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3364", "queId": "dfa2d6305a984fb68c5e1f1702d6c723", "competition_source_list": ["2018年第21届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$1\\sim 50$$各数中，数字$$4$$出现了次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->能力->推理推导->言语逻辑推理"], "answer_analysis": ["在$$1\\sim 39$$里，有$$4$$个数字$$4$$，分别是$$4$$、$$14$$、$$24$$、$$34$$， 在$$40$$到$$50$$中，有十个数字$$4$$，分别是$$40$$、$$41$$、$$42$$、$$43$$、$$44$$、$$45$$、$$46$$、$$47$$、$$48$$、$$49$$， 一共出现了$$15$$次． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2627", "queId": "29808b359ac64ef5960a3a9f5a69d46a", "competition_source_list": ["2019年广东广州羊排赛六年级竞赛第6题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "甲、乙都是非零的自然数，已知甲的$$\\frac{2}{3}$$等于乙的$$\\frac{6}{7}$$，则甲$$:$$乙$$=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7:9$$ "}], [{"aoVal": "B", "content": "$$9:7$$ "}], [{"aoVal": "C", "content": "$$1:7$$ "}], [{"aoVal": "D", "content": "$$7:1$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["根据题意分析可知，求一个数的几分之几，用乘法所以甲乙两数之间的关系式为： 甲$$\\times \\frac{2}{3}=$$乙$$\\times \\frac{6}{7}$$， 则甲$$:$$乙$$=\\frac{6}{7}:\\frac{2}{3}$$ $$=\\frac{6}{7}\\times \\frac{3}{2}$$ $$=\\frac{9}{7}$$ $$=9:7$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "200", "queId": "24d76b5a3e204121a1db0bc4859bec5d", "competition_source_list": ["其它改编题", "2017年全国华杯赛小学中年级竞赛初赛模拟第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$至$$11$$这$$11$$个自然数中至少选出~\\uline{~~~~~~~~~~}~个不同的数，才能保证其中一定有两个数的和为$$12$$? ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["把和为$$12$$的两个数分成一组，这样就把这$$11$$个数分成$$6$$组：$$(1,11)$$，$$\\left( 2,10 \\right)$$，$$\\left( 3,9 \\right)$$，$$(4,8)$$，$$(5,7)$$，$$(6)$$．要保证一定有两个数的和为$$12$$，就要保证至少有两个数属于同一组． 根据最不利原则，如果我们从$$6$$组中各取一个数，则取出的这$$6$$个数中，没有两个数的和是$$12$$，因此再选一个数，一定会与前面的某数在同一组，即和为$$12$$．本题的答案就是至少选出$$7$$个不同的数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1587", "queId": "a35492cd5d2e45b0a611bb80f0197ee8", "competition_source_list": ["2020年第23届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$10 $$年前父亲的年龄是儿子的$$7$$倍，$$15$$年后父亲的年龄是儿子的$$2$$倍，今年父亲的年龄是儿子的倍。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$2.5$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->应用题模块->列方程解应用题", "拓展思维->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题->方程法解倍数问题"], "answer_analysis": ["设$$10$$年前儿子的年龄是$$x$$岁，那么父亲的年龄是$$7x$$； 根据题意可得： $$7x+10+15=\\left( x+10+15 \\right)\\times 2$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde7x+25=2x+50$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde5x=25$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x=5$$ 现在儿子的年龄是：$$x+10=5+10=15$$（岁） 父亲的年龄是：$$7x+10=7\\times 5+10=45$$（岁） $$45\\div 15=3$$倍 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2459", "queId": "193bc2e64e404efb9789e0843ad5a951", "competition_source_list": ["2015年第27届广东广州五羊杯小学高年级竞赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$1$$至$$2015$$个连续自然数分别加上$$0.1$$，$$0.2$$、$$0.3$$、$$\\cdots $$、$$201.4$$，$$201.5$$，得到$$2015$$个新数．那么这$$2015$$个新数的平均数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1108.8$$ "}], [{"aoVal": "B", "content": "$$1008.8$$ "}], [{"aoVal": "C", "content": "$$1018.8$$ "}], [{"aoVal": "D", "content": "$$1118.8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\frac{1}{2}\\times \\left( 1+2015+0.1+201.5 \\right)$$ $$=1008+100.8$$ $$=1108.8$$， 本题主要考察等差数列、平均数的概念及小数的基本计算方法和简单应用． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3331", "queId": "4e77459d60fb444f80a66853afd08aaf", "competition_source_list": ["2014年全国创新杯小学高年级五年级竞赛5分"], "difficulty": "1", "qtype": "single_choice", "problem": "国际数学奥林匹克每天考$$3$$道题，每题的评分���$$0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$或$$7$$．有一群学生每人得分的乘积都是$$36$$，而且任意两人各题得分不完全相同，那么这群学生最多有人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->整数分拆->整数拆分应用->乘法拆数（应用）"], "answer_analysis": ["$$36=2\\times 3\\times 6$$，$$\\text{A}_{3}^{3}=6$$，有$$6$$种 $$36=3\\times 3\\times 4$$，有$$3$$种 $$36=1\\times 6\\times 6$$，有$$3$$种 共$$6+3+3=12$$种不同的得分，最多有$$12$$人． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1699", "queId": "65369155e75c4b2ba35360e38bddbf17", "competition_source_list": ["2017年河南郑州豫才杯小学高年级六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "小红每$$7$$天值日一次，小梅每$$9$$天值日一次，$$4$$月$$5$$日这一天两人都是值日生，则下次两人一起值日是（~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$月$$5$$日 "}], [{"aoVal": "B", "content": "$$6$$月$$6$$日 "}], [{"aoVal": "C", "content": "$$6$$月$$7$$日 "}], [{"aoVal": "D", "content": "$$6$$月$$8$$日 "}]], "knowledge_point_routes": ["拓展思维->能力->数感认知->数学概念理解（数）"], "answer_analysis": ["再过$$\\left[ 7,9 \\right]=63$$天两人又一起值日，$$4$$月有$$30$$天，$$5$$月有$$31$$天，所以是$$6$$月$$7$$日． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "334", "queId": "57ed0b9139244c77b9582438caf2f919", "competition_source_list": ["2021年新希望杯三年级竞赛初赛第13题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "【$$2021$$三年级卷第$$13$$题】五盘水果排成一排．苹果和橘子相邻，橘子和草莓相邻，苹果和香蕉不相邻，香蕉和芒果不相邻．那么一定和芒果相邻的是． ", "answer_option_list": [[{"aoVal": "A", "content": "只有苹果 "}], [{"aoVal": "B", "content": "只有橘子 "}], [{"aoVal": "C", "content": "只有草莓 "}], [{"aoVal": "D", "content": "香蕉和草莓 "}], [{"aoVal": "E", "content": "橘子和香蕉 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["根据题意分析可知，苹果和草莓都与橘子相邻， 所以可以确定橘子在苹果和草莓的中间， 假设苹果、橘子、草莓这样排列， 则香蕉在草莓的右边，芒果在苹果的左边， 即苹果与芒果一定相邻， 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3049", "queId": "e5a41db6946146c6b1dab0d68bf61d7d", "competition_source_list": ["2004年第2届创新杯六年级竞赛复赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "在下面四个算式中，得数最大的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( \\frac{1}{17}+\\frac{1}{19} \\right)\\times 20$$ "}], [{"aoVal": "B", "content": "$$\\left( \\frac{1}{24}+\\frac{1}{29} \\right)\\times30$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{31}+\\frac{1}{17} \\right)\\times40$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{1}{41}+\\frac{1}{47} \\right)\\times50$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$\\left( \\frac{1}{17}-\\frac{1}{19} \\right)\\div 20=\\frac{2}{17\\times 19}\\div 20=\\frac{1}{17\\times 19}\\times \\frac{1}{10}$$;$$\\left( \\frac{1}{15}-\\frac{1}{21} \\right)\\div 60=\\frac{6}{15\\times 21}\\div 60=\\frac{1}{15\\times 21}\\times \\frac{1}{10}$$; $$\\left( \\frac{1}{13}-\\frac{1}{23} \\right)\\div 100=\\frac{10}{13\\times 23}\\div 100=\\frac{1}{13\\times 23}\\times \\frac{1}{10};$$$$\\left( \\frac{1}{11}-\\frac{1}{25} \\right)\\div 140=\\frac{14}{11\\times 25}\\div 140=\\frac{1}{11\\times 25}\\times \\frac{1}{10};$$ 只需比较$$\\frac{1}{17\\times 19}$$，$$\\frac{1}{15\\times 21}$$，$$\\frac{1}{13\\times 23}$$，$$\\frac{1}{11\\times 25}$$的大小，根据和一定，两数越接近乘 积越大，则$$11\\times 25 \\textless{} 13\\times 23 \\textless{} 15\\times 21 \\textless{} 17\\times 19$$，那么 $$\\frac{1}{11\\times 25}\\textgreater\\frac{1}{13\\times 23}\\textgreater\\frac{1}{15\\times 21}\\textgreater\\frac{1}{17\\times 19}$$，所以答案为$$D$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1561", "queId": "ff8080814502fa2401450bc687b31598", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "有一种特殊���计算器，当输入一个$$10$ $49$$的自然数后，计算器会先将这个数乘$$2$$，然后将所得结果的十位和个位顺序颠倒，再加$$2$$后显示出最后的结果．那么，下列四个选项中，可能是最后显示的结果． ", "answer_option_list": [[{"aoVal": "A", "content": "$$44$$ "}], [{"aoVal": "B", "content": "$$43$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$41$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["倒推．$$44$$ 对应的是$$44-2=42$$，颠倒后是$$24$$，除以$$2$$ 为$$12$$．符合条件．其他的均不符合条件． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "61", "queId": "6676c0c8079846989aa46779c916fcff", "competition_source_list": ["2022年第9届广东深圳鹏程杯四年级竞赛初赛第30题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在某次田径比赛中，每个项目都只有甲、乙、丙三名运动员参加．比赛结果采用积分 制:第一名得$$A$$分，第二名得$$B$$分，第三名得$$C$$分(这里$$ A\\textgreater B\\textgreater C\\textgreater0 $$,A,B,C都是正整数)。已知所有项目的比赛都没有出现名次并列的情况，且甲积分为$$22$$分，乙、丙积分都为$$9$$分．如果丙是短跑第一名，那么铅球第二名是． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "甲或丙 "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->体育比赛"], "answer_analysis": ["无 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3447", "queId": "f37c3cee00f14e4a85f38581e103bcf3", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（二）第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "老师从小明和其他$$9$$名学生中随意抽取$$2$$人表演节目，小明被抽中的可能性是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{9}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{10}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->基本概率->可能性"], "answer_analysis": ["假设分两次抽，第一次抽中小明的可能性是$$\\frac{1}{10}$$，第二次才抽中小明的可能性是$$\\frac{1}{10}$$，$$\\frac{1}{10}+\\frac{1}{10}=\\frac{1}{5}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2249", "queId": "d5ed70f3b68d4bce919ca38d329f1ad9", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "同学甲、同学乙两人在游乐场的直行道上进行$$200$$米赛跑，当同学甲跑到终点时．同学乙还差$$40$$米，现在两人重新跑，而且速度和原来一样，要使两人同时到达终点，那么同学甲的起跑线应往后退（~ ）米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$55$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["比例解行程；相等的时间内，甲跑了$$200$$米时，乙跑了$$160$$米，甲乙的速度比是$$200:160=5:4$$，要使两人同时到达终点，乙跑了$$200$$米的时间内甲要跑$$200\\div 4\\times 5=250$$米，所以同学甲的起跑线应后退$$250-200=50$$米． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "508", "queId": "c20ab40a557e46b8b966439cc476a7c9", "competition_source_list": ["2020年春蕾杯六年级竞赛第12题2分", "2021年春蕾杯六年级竞赛第7题2分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$，$$B$$，$$C$$，$$D$$四个队举行足球循环赛（即每两个队都要赛一场），胜一场得$$3$$分，平一场得$$1$$分，负一场得$$0$$分．已知： （$$1$$）比赛结束后四个队的得分都是奇数； （$$2$$）$$A$$队总分第一； （$$3$$）$$B$$队恰有两场平局，并且其中一场是与$$C$$队平局． 问：$$D$$队得分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$B$$队两场平局，得$$2$$分，因得分为奇数，所以还有一场应胜，得$$3$$分，共得分$$5$$分． 那么$$A$$队虽是第一，也不可能全胜，则$$A$$队得$$7$$分， 其中平局只能是与$$B$$队了，则$$A$$队胜了$$C$$队和$$D$$队． 因此，$$B$$队胜的一场一定是与$$D$$队，与$$A$$和$$C$$队打平． 因为$$C$$队也必须是奇数，所以$$C$$队一定负于$$D$$队，积$$1$$分，$$D$$队得分$$3$$分． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1983", "queId": "f7e89cf2aa714172a76dfd59b19ad9dd", "competition_source_list": ["2018年第21届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "把一根长$$20$$米的彩带剪成两段，要求较长的比较短的长$$4$$米，较短的长米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["因为通过题干可知，把这根长$$20$$米的彩带剪成两段，设较长的一段长$$x$$米，则较短的一段长$$(20-x)$$米，较长的比较短的长$$4$$米，则可列方程为：$$x-(20-x)=4$$，解得：$$x=12$$，所以较长的一段是$$12$$米，较短的一段是：$$20-12=8$$（米）． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1658", "queId": "64e1aba2d0ad46258c16af661cb2735d", "competition_source_list": ["2007年华杯赛六年级竞赛初赛", "2007年华杯赛五年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "折叠一批纸鹤，甲同学单独折叠需要半小时，乙同学单独折叠需要$$45$$分钟，则甲、乙两名同学共同折叠需要（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$分钟 "}], [{"aoVal": "B", "content": "$$15$$分钟 "}], [{"aoVal": "C", "content": "$$18$$分钟 "}], [{"aoVal": "D", "content": "$$20$$分钟 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->已知工时反推->多人合作"], "answer_analysis": ["解：$$\\frac{1}{\\frac{1}{30}+\\frac{1}{45}}=\\frac{1}{\\frac{5}{6\\times 15}}=18$$（分钟） "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2107", "queId": "ebcac4d6338944dfa6047a70f0448f7f", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（四）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "某年的二月份有$$4$$个星期一和$$5$$个星期二，则$$2$$月$$17$$号是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["因为这年的二月份有$$5$$个星期二，说明这是一闰年，且$$2$$月$$1$$号是星期二，$$17\\div 7=2$$（周）$$\\ldots \\ldots 3$$（天），所以$$2$$月$$17$$号也是星期四． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3280", "queId": "a2e78bb3720e4c0eb4a52e1973d92fed", "competition_source_list": ["2012年全国创新杯五年级竞赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "玛丽有$$6$$张卡片，每张卡片上都写有一个大于$$0$$的自然数．她选取了$$3$$张卡片后，算出了它的总和，他又选另外$$3$$张卡片上的总和，$$\\cdot \\cdot \\cdot $$，她进行了可能的$$20$$种$$3$$张卡片的选择，然后计算，发现有$$10$$种总和等于$$16$$，另外$$10$$种总和等于$$18$$．那么这些卡片中最小的数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->容斥原理->多量容斥的最值问题"], "answer_analysis": ["每个数字用$$10$$次，故所有数字和为$$(16\\times 10+18\\times 10)\\div 10=34$$，而出现了$$10$$ 次$$16$$和$$10$$次$$18$$，故除了最小的，其他$$5$$个都相同，所以只能为$$4$$、$$6$$、$$6$$、$$6$$、$$6$$、$$6$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1991", "queId": "f7f5fe5e76b94288920260d33424efa6", "competition_source_list": ["2009年全国迎春杯小学中年级四年级竞赛初赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "老师买了同样数目的田格本���横线本和练习本．他发给每个同学$$1$$个田格本、$$3$$个横线本和$$5$$个练习本．这时横线本还剩$$24$$个，那么田格本和练习本共剩了~\\uline{~~~~~~~~~~}~个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$54$$ "}], [{"aoVal": "D", "content": "$$56$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["同倍数变化问题．把$$1$$个田格本和$$5$$个练习本捆绑成一组，那么每发$$3$$个横格本，就发一组田格本和练习本，田格本和练习本的总量是横线本的$$2$$倍，每次发的数量也是$$2$$倍，所以剩下的也是$$2$$倍，即$$48$$本． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "824", "queId": "537dc2f02e254540a4cab6d2480956e0", "competition_source_list": ["2015年全国华杯赛小学高年级竞赛复赛A卷第9题", "2015年北京华杯赛小学高年级竞赛复赛A卷第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "两个自然数之和为$$667$$，它们的最小公倍数除以最大公约数所得的商等于$$120$$．求这两个数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$552$$，$$115$$ "}], [{"aoVal": "B", "content": "$$232$$，$$435$$ "}], [{"aoVal": "C", "content": "$$552$$，$$115$$或$$232$$，$$435$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["$$120={{2}^{3}}\\times 3\\times 5$$，$$667=23\\times 29$$．（*） 设这两个自然数是$$da$$和$$db$$，其中$$a$$和$$b$$是互质，$$d$$是这两个自然数的最大公约数，$$d$$是$$667$$的因子，且大于$$1$$，小于$$667$$．无妨设$$a\\textgreater b$$，则有： （1）当$$d=23$$时，由（*），$$a+b=29$$，$$ab={{2}^{3}}\\times 3\\times 5$$，此时易得：$$a=24$$，$$b=5$$，这两个自然数是$$435$$，$$232$$． （2）当$$d=29$$时，由（*），$$a+b=23$$，$$ab={{2}^{3}}\\times 3\\times 5$$，此时易得：$$a=15$$，$$b=8$$，这两个自然数是$$552$$，$$115$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2401", "queId": "25c764a762284971b0654189b50b2486", "competition_source_list": ["2003年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2003+2002-2001-2000+1999+1998-1997-1996+ \\cdots +7+6-5-4+3+2-1$$的计算结果是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$2002$$ "}], [{"aoVal": "B", "content": "$$2003$$ "}], [{"aoVal": "C", "content": "$$2004$$ "}], [{"aoVal": "D", "content": "$$4005$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之分组法"], "answer_analysis": ["注意到$$2003+2002-2001-2000=4$$，$$1999+1998-1997-1996=4$$，$$7+6-5-4=4$$，如果把$$2003$$到$$4$$的$$2000$$个数依次$$4$$个分为一组共有$$(2000\\div 4)$$组，且每组的计算结果都为$$4$$，那么这$$2000\\div 4$$组的和是$$2000\\div 4\\times 4=2000$$，再加上最后的$$3$$个数，结果是：$$2000+3+2-1=2004$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1247", "queId": "9da73342c2944461913254823d326e55", "competition_source_list": ["2021年陕西西安六年级下学期小升初模拟分类专题五十七 （年龄问题 平均数问题）第26题", "2021年陕西西安雁塔区西安高新第一中学小升初（五）第6题3分", "2016年北京走美杯四年级竞赛冲刺讲义", "2019年重庆九龙坡区重庆市育才中学小升初（八）第1题3分", "2018年重庆九龙坡区重庆市育才中学小升初（三）第3题3分", "2014年陕西西安小升初高新一中入学真卷（二）第6题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "六个自然数的平均数是$$7$$，其中前四个数的平均数是$$8$$，第$$4$$个数是$$11$$，求后三个数的平均数~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}], [{"aoVal": "E", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["$$(7\\times6+11-8\\times4)\\div3=7$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2724", "queId": "ff8080814502fa2401450bc65c4a1591", "competition_source_list": ["2014年全国迎春杯四年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$2014\\div (2\\times 2+2\\times 3+3\\times 3)=$$（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$53$$ "}], [{"aoVal": "B", "content": "$$56$$ "}], [{"aoVal": "C", "content": "$$103$$ "}], [{"aoVal": "D", "content": "$$106$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$2014\\div 19=106$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2218", "queId": "4ca61b1457be420d87f8717aab557bc4", "competition_source_list": ["2016年第12届全国新希望杯小学高年级六年级竞赛复赛第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "战士小王从$$A$$地前往$$B$$地送信，他每走$$40$$分钟就休息$$10$$分钟，到达$$B$$地共需$$4$$小时$$20$$分，从$$B$$地原路返回的速度是去时的$$2$$倍，若他每走$$35$$分钟就休息$$15$$分钟．从$$B$$地返回到$$A$$地共需（~ ）分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$125$$ "}], [{"aoVal": "B", "content": "$$130$$ "}], [{"aoVal": "C", "content": "$$135$$ "}], [{"aoVal": "D", "content": "$$140$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["去的过程中一共花了$$4\\times 60+20=260$$分钟，$$260\\div 50=5\\ldots \\ldots 10$$说明走路时间一共是$$5\\times 40+10=210$$分．回来过程中速度为原来的$$2$$倍，走路的时间则为原来的$$\\frac{1}{2}$$，即$$210\\times \\frac{1}{2}=105$$分钟$$105\\div 35=3$$，即一共走了$$3$$次，第$$3$$次走完之后不用休息，则一共花了$$\\left( 35+15 \\right)\\times 3-15=135$$分钟． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2658", "queId": "99a96a4fd9fe41cb8e19d5b069ac4ab9", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知等差数列$$13$$，$$18$$，$$23$$，$$28$$，$$\\cdots $$，$$1003$$。这个等差数列共有项。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$198$$ "}], [{"aoVal": "B", "content": "$$199$$ "}], [{"aoVal": "C", "content": "$$200$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理", "Overseas Competition->知识点->计算模块->数列与数表->等差数列"], "answer_analysis": ["根据题意：公差为$$5$$， 所以项数：$$\\left( 1003-13 \\right)\\div 5+1=199$$。 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1992", "queId": "af7788f4e49b455496761de2dbcba851", "competition_source_list": ["2015年湖北武汉世奥赛小学高年级六年级竞赛模拟训练题（三）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "一台新上市的手机，如果按原价的八五折出售可获利$$268$$元，如果按原价九五折出售可获利$$1276$$元，那么这台苹果$$\\text{iphone6}$$的原价是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8300$$ "}], [{"aoVal": "B", "content": "$$9300$$ "}], [{"aoVal": "C", "content": "$$10080$$ "}], [{"aoVal": "D", "content": "$$12080$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->折扣问题"], "answer_analysis": ["$$\\left( 1276-268 \\right)\\div \\left( 95 \\%- 85 \\% \\right)=10080$$（元）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1240", "queId": "5492fef93b2442eca3624c33283c9c0d", "competition_source_list": ["2018年第21届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一筐水果中，恰好有一半数量是苹果，如果吃掉苹果数量的一半，筐中还剩下$$90$$个．那么这时候筐中还有个苹果． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["$$90$$除以$$1+2$$的和，结果是$$30$$，从后向前推算，剩下的是$$1$$份，那么苹果就是$$2$$份， 其他水果也是$$2$$份，所以可以得到是$$30$$个苹果．故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1477", "queId": "d5b08fba90ee4c81aa0beeb25d16be77", "competition_source_list": ["2016年创新杯五年级竞赛训练题（二）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "``创新号''游轮在长江上航行时遇到龙卷风，船底破损开始进水，发现漏水时，已经进了一些水，水迅速进入船内．如果$$10$$台抽水机抽水，$$20$$分钟抽完，如果$$5$$台抽水机抽水，$$45$$分钟抽完．如果要求$$30$$分钟抽完，要安排（~ ）台抽水机抽水． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["设一台抽水机$$1$$分钟抽$$1$$份水，则每分钟的抽水量为$$\\left( 5\\times 45-10\\times 20 \\right)\\div \\left( 45-20 \\right)=1$$（份），船内原有水量为$$5\\times 45-1\\times 45=180$$（份），所以$$30$$分钟抽完需要$$180\\div 30+1=7$$（台）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1132", "queId": "10a7399e7a9744569aef4c3d31875665", "competition_source_list": ["2020年新希望杯六年级竞赛（2月）第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "海尔兄弟被困在一个无人岛上，他们要做一个独木舟逃出这个无人岛，哥哥单独做要$$6$$小时完成，弟弟单独做要$$9$$小时完成．如果按照哥哥、弟弟、哥哥、弟弟$$\\cdots \\cdots $$的顺序交替工作，每人工作$$1$$小时后交换，那么需要~\\uline{~~~~~~~~~~}~小时能做好独木舟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->接力施工问题"], "answer_analysis": ["哥哥工效：$$\\frac{1}{6}$$，弟弟工效：$$\\frac{1}{9}$$， 则$$1\\div \\left( \\frac{1}{6}+\\frac{1}{9} \\right)=\\frac{18}{5}=3$$（周期）$$\\cdots \\cdots \\frac{3}{5}$$， $$1-3\\times \\left( \\frac{1}{6}+\\frac{1}{9} \\right)=\\frac{1}{6}$$， $$\\frac{1}{6}\\div\\frac{1}{6}=1$$（小时）， 共$$3\\times 2+1=7$$（小时）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "417", "queId": "81ad929ae007494e876679ca333d4f63", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（二）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "有帽子三顶、衣服三件、鞋三双，各自分别是红、黑、蓝三种颜色．$$3$$名同学一起去郊游，他们每人戴的帽子，穿的衣服和鞋都是三种不同的颜色，如果戴黑色帽子的同学穿的不是红色的鞋，那么戴蓝色帽子的同学穿色的衣服． ", "answer_option_list": [[{"aoVal": "A", "content": "红 "}], [{"aoVal": "B", "content": "黑 "}], [{"aoVal": "C", "content": "蓝 "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->表格法"], "answer_analysis": ["戴黑帽的同学穿的不是红鞋，只能穿蓝鞋，红衣服，所以戴蓝帽的同学不可能穿红衣服，只能穿黑衣服． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1286", "queId": "239dab6e31f24ca58f94fa5bce0a6752", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛B卷第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "两件衣服都按$$120$$元出售，其中一件赚了$$20 \\%$$，另一件亏了$$20 \\%$$，那么两件衣服合在一起算，结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "亏了 "}], [{"aoVal": "B", "content": "赚了 "}], [{"aoVal": "C", "content": "不赚不亏 "}], [{"aoVal": "D", "content": "无法比较 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["从题目中我们知道，两件衣服都按$$120$$元出售， 也就是一共入账：$$120\\times2=240$$ （元）； 其中一件赚了$$20 \\%$$，那么成本是：$$120\\div (1+20 \\%)=100$$（元）， 另一件亏了$$20 \\%$$，那么成本是：$$120\\div (1-20 \\%)=150$$（元）； 总成本是：$$100+150=250$$（元），也就是实际上还是亏了． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2064", "queId": "f8de4bdedcb44091bc0814ac41da130d", "competition_source_list": ["2017年河南郑州豫才杯小学高年级五年级竞赛初赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "游览景区期间，爸爸将车放在停车场，收费标准为：不超过$$1$$小时收费$$3$$元，超过$$1$$小时的部分每小时收费$$3$$元．爸爸最终一共交了$$13.5$$元的停车费，他们的车在停车场最多停了小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$小时 "}], [{"aoVal": "B", "content": "$$4.5$$小时 "}], [{"aoVal": "C", "content": "$$5$$小时 "}], [{"aoVal": "D", "content": "$$8$$小时 "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->实践应用"], "answer_analysis": ["设最多在停车场停了$$x$$小时，$$3+\\left( x-1 \\right)\\times 3=13.5$$，解得$$x=4.5$$，所以他们的车在停车场最多停$$4.5$$小时． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1125", "queId": "18b332d2291c4392b95643d47e7c44bd", "competition_source_list": ["2015年第11届全国新希望杯小学高年级六年级竞赛复赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙三名化妆师单独为一名新娘化妆所需时间之比为$$2$$:$$3$$:$$4$$（假设给所有新娘的化妆量都相同），某天他们同时为两名新娘化妆，其中甲负责为新娘$$A$$化妆，乙负责为新娘$$B$$化妆，丙先帮甲为新娘$$A$$化妆，然后紧接着又帮乙为新娘$$B$$化妆，$$72$$分钟后，两名新娘同时化妆完毕，其中丙为新娘$$A$$化妆所用的时间为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$分钟 "}], [{"aoVal": "B", "content": "$$12$$分钟 "}], [{"aoVal": "C", "content": "$$9$$分钟 "}], [{"aoVal": "D", "content": "$$6$$分钟 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题"], "answer_analysis": ["设三人的时间分别为$$2x$$，$$3x$$，$$4x$$，那么有$$2\\div \\left( \\frac{1}{2x}+\\frac{1}{3x}+\\frac{1}{4x} \\right)=72$$，解得$$x=39$$，那么甲的效率是$$\\frac{1}{78}$$，丙的效率是$$\\frac{1}{156}$$，那么丙帮$$A$$的时间是$$\\left( 1-\\frac{72}{78} \\right)\\div \\frac{1}{156}=12$$分钟． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2790", "queId": "604688042b864dddab0b8d7f1fec905c", "competition_source_list": ["2006年第4届创新杯六年级竞赛初赛A卷第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "在下面四个算式中，最大的数应是( )． (1) $$(\\frac{1}{17}-\\frac{1}{19})\\times 20$$ (2) $$(\\frac{1}{24}+\\frac{1}{29})\\times 30$$ (3)$$(\\frac{1}{31}+\\frac{1}{37})\\times 40$$ (4) $$(\\frac{1}{41}+\\frac{1}{47})\\times 50$$ ", "answer_option_list": [[{"aoVal": "A", "content": "($$1$$) "}], [{"aoVal": "B", "content": "($$2$$) "}], [{"aoVal": "C", "content": "($$3$$) "}], [{"aoVal": "D", "content": "($$4$$) "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->算式比较大小"], "answer_analysis": ["$$(1)1\\frac{3}{17}-1\\frac{1}{19}，(2)1\\frac{1}{4}+1\\frac{1}{29}，(3)1\\frac{9}{31}+1\\frac{3}{37}，(4)1\\frac{9}{41}+1\\frac{3}{47}$$，显然($$3$$)最大． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2689", "queId": "51d483c6a7064e8f8a933c520a609612", "competition_source_list": ["2003年第1届创新杯五年级竞赛复赛第1题", "2003年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "将一个数的小数点向右移动三位，再向左移动两位，这时所得到的数是$$4.13$$，那么原来的数是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.413$$ "}], [{"aoVal": "B", "content": "$$4.13$$ "}], [{"aoVal": "C", "content": "$$41.3$$ "}], [{"aoVal": "D", "content": "$$413$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数基础->小数点的移动规律"], "answer_analysis": ["小数点向右移动三位再向左移动两位，实际上是向右移动了一位，向右移动一位后得到的数是$$4.13$$，那么把小数点往左移动一位即得原数，故原数是$$0.413$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1537", "queId": "3c3cfe635ce84575b9578d3059f75a0a", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果把一根木料锯成$$3$$段要用$$6$$分钟，那么用同样的速度把这根木料锯成$$6$$段要用分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["把一根木料锯成$$3$$段，需要锯$$3-1=2$$（次），要用$$6$$分钟，每次需要$$6\\div 2=3$$（分钟），用同样的速度把这根木料锯成$$6$$段，需要锯$$6-1=5$$（次），需要$$5\\times 3=15$$（分钟）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "140", "queId": "501279c398054846a2edeb144cb27111", "competition_source_list": ["2014年中环杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "黑箱中有$$60$$块大小、形状都相同的木块，每$$15$$块涂上相同的颜色，一次至少取出（ ）块才能保证其中至少有$$2$$块木块颜色相同。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["$$60\\div 15=4$$（种） $$4+1=5$$（块） 答：一次至少取出$$5$$块才能保证其中至少有$$2$$块木块颜色相同。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2812", "queId": "6974730a76bb4568a48836ba9f103760", "competition_source_list": ["2013年全国希望杯五年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$26$$个连续的自然数，如果前$$13$$个数的和是$$247$$，那么，后$$13$$个数的和是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$392$$ "}], [{"aoVal": "B", "content": "$$416$$ "}], [{"aoVal": "C", "content": "$$420$$ "}], [{"aoVal": "D", "content": "$$445$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列截断求和"], "answer_analysis": ["方法一：求前$$13$$个数中间数：$$247\\div13=19$$，则后$$13$$个数的第一个数为$$19+6+1=26$$，最后一个数为$$26+12=38$$，所以后$$13$$个数的和是$$(26+38)\\times13\\div2=416$$． 方法二：后$$13$$个数比前$$13$$个数按顺序一一对应，后$$13$$个数的每个数比前$$13$$个数的每个数都多$$13$$． 所以后$$13$$个数的和是$$247+13\\times13=416$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1488", "queId": "90a38f5ece81474282ead00bc5287d01", "competition_source_list": ["2019年广东广州学而思综合能力诊断五年级竞赛第11题12分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A+B+C=2021$$，$$A$$、$$B$$、$$C$$分别有$$8$$、$$10$$、$$11$$个因数，并且$$B$$与$$C$$互质，那么$$A$$是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$360$$ "}], [{"aoVal": "B", "content": "$$380$$ "}], [{"aoVal": "C", "content": "$$390$$ "}], [{"aoVal": "D", "content": "$$420$$ "}], [{"aoVal": "E", "content": "$$430$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理逆应用", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["本题是因数个数定理的逆应用，先从$$A$$、$$B$$、$$C$$三个的因数个数开始讨论：$$C$$有$$11$$个因数，$$11$$是质数，所以$$C$$一定是一个质数的$$10$$次方，又因为三个数的和为$$2021$$，小于$$3$$的$$10$$次方，所以$$C$$只能是$$2$$的$$10$$次方，等于$$1024$$，则$$A+B=2021-1024=997$$，$$A$$有$$8$$个因数，则$$A$$的分解质因数形式可以是$$a$$的$$7$$次方、$$a$$的$$1$$次方乘$$b$$的$$3$$次方、$$a\\times b\\times c$$这$$3$$种可能，$$B$$的分解质因数形式可以是$$c$$的$$9$$次方、$$c$$的$$1$$次方乘$$d$$的$$4$$次方，因为$$B$$和$$C$$互质，所以$$B$$不含质因数$$2$$，根据和的大小可以确定$$B$$不等于$$c$$的$$9$$次方，又$$5$$的$$4$$次方等于$$625$$，$$625$$乘$$3$$等于$$1275$$大于$$997$$，所以$$d$$只能等于$$3$$，$$3$$的$$4$$次方等于$$81$$，$$81$$乘$$5$$等于$$405$$，$$997-405=592$$不符合$$A$$的分解质因数条件，$$81$$乘$$7$$等于$$567$$，$$997-567=430$$符合条件，则$$A$$等于$$430$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2699", "queId": "3bff0c002ee543e4af321dfe28392028", "competition_source_list": ["2006年第4届创新杯六年级竞赛初赛B卷第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$\\frac{1}{2}$$，$$\\frac{1}{3}$$，$$\\frac{1}{4}$$，$$\\frac{1}{5}$$，$$\\frac{1}{6}$$中去掉两个分数，使得剩下的三个分数的和与$$\\frac{6}{7}$$最接近，那么去掉的两个分数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$，$$\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$，$$\\frac{1}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{3}$$，$$\\frac{1}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{3}$$，$$\\frac{1}{5}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}=\\frac{5}{6}+\\frac{9}{20}+\\frac{1}{6}=1\\frac{9}{20}=1.45$$， $$\\frac{6}{7}\\approx 0.857$$， $$1.45-0.857=0.593$$， 所以题目即需��从五个分数中选出两个，使他们的和最接近$$0.593$$，比较后可得应选$$\\frac{1}{3}$$和$$\\frac{1}{4}$$，故$$\\text{C}$$正确． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2238", "queId": "6cf17e89d33f42e2ac460d6c68eed87f", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）"], "difficulty": "1", "qtype": "single_choice", "problem": "【拓4】一列火车通过一座长$$320$$米的桥用了$$21$$秒，当它通过$$860$$米的隧道时，速度是过桥速度的$$2$$倍，结果用了$$24$$秒，火车通过大桥时的速度是每秒米；火车的车身长度为米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$；$$100$$ "}], [{"aoVal": "B", "content": "$$20$$；$$100$$ "}], [{"aoVal": "C", "content": "$$40$$；$$100$$ "}], [{"aoVal": "D", "content": "$$80$$；$$200$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["若通过$$860$$米隧道时速度不变则需要$$24\\times 2=48$$（秒），火车过桥速度：$$\\left( 860-320 \\right)\\div \\left( 48-21 \\right)=20$$（米/秒）：火车车身长：$$21\\times 20-320=100$$（米）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "978", "queId": "0df1ea89e25d4b4ebec1251a51744b5b", "competition_source_list": ["2007年四年级竞赛创新杯", "2007年第5届创新杯四年级竞赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一条绳子，折成相等的3段后，再折成相等的两段，然后从中间剪开，一共可以剪成~\\uline{~~~~~~~~}~段. ", "answer_option_list": [[{"aoVal": "A", "content": "7 "}], [{"aoVal": "B", "content": "8 "}], [{"aoVal": "C", "content": "9 "}], [{"aoVal": "D", "content": "10 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都没有->剪绳子"], "answer_analysis": ["将绳折成3段再对折，相当于折成6段，一刀与这6段共有6个交点，所以将绳剪成7段 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1622", "queId": "bf05389469a54b98a997dcc309ad6e02", "competition_source_list": ["2006年第4届创新杯四年级竞赛初赛A卷第3题", "2006年四年级竞赛创新杯", "2006年三年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "大家都知道，一只正常的猫有$$18$$个爪，每条前腿有$$5$$个爪，每条后腿有$$4$$个爪。在某市的``保护伤残动物之家''收养了$$4$$只伤残猫，每只猫都失去了一条腿，但是每只猫失去的腿都不相同，这$$4$$只伤残猫共有（ ）个爪。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$68$$ "}], [{"aoVal": "B", "content": "$$64$$ "}], [{"aoVal": "C", "content": "$$54$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题"], "answer_analysis": ["$$4$$只正常猫一共有$$18\\times 4=72$$（个）爪，$$4$$只伤残猫失去的腿都不同，一共失去$$5\\times 2+4\\times 2=18$$（个）爪，现在$$4$$只伤残猫一共有$$72-18=54$$（个）爪。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1193", "queId": "15b9feba11a2484f9e15baa074a7c534", "competition_source_list": ["2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第10题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "下面说法正确的有． ①陈小强是$$2009$$年$$2$$月$$29$$日出生的． ②$$21$$时用普通计时法表示是晚上$$10$$时． ③东西相对，南北相对． ④小数$$1.6$$和$$1.8$$之间只有$$1.7$$这一个数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$个 "}], [{"aoVal": "B", "content": "$$2$$个 "}], [{"aoVal": "C", "content": "$$3$$个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["①$$2009$$年是平年，平年$$2$$月只有$$28$$天． ②$$21$$时用普通计时法表示是晚上$$9$$点． ③东西相对南北相对，正确． ④小数$$1.6$$和$$1.8$$之间有无数个数． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1558", "queId": "ff8080814502fa24014507b662ee0b9a", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "过年的时候，球球给客人倒啤酒，一瓶啤酒可以倒满$$4$$杯，球球倒酒的时候总是每杯中有半杯泡沫，啤酒倒成泡沫的体积会涨成原来的$$3$$倍，那么球球倒���酒时，一瓶酒可以倒（~~ ）杯． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["根据题意可知，$$1$$份的啤酒可以变成$$3$$份的泡沫．球球倒的啤酒一半是泡沫，那么我们可以把球球倒的每杯酒分成$$6$$份，那么每倒一杯酒只有$$4$$份．而一瓶啤酒可以倒$$4$$杯共有$$4\\times6＝24$$份．球球倒的每杯酒为$$4$$份，她共可以倒的杯数为：$$24\\div4＝6$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1289", "queId": "1f77d69d07704ed89bf1190388e0304e", "competition_source_list": ["2004年希望杯三年级竞赛复赛", "2004年第2届希望杯五年级竞赛复赛第3题", "2004年希望杯四年级竞赛复赛", "2004年希望杯六年级竞赛复赛", "2004年希望杯二年级竞赛复赛", "2004年希望杯一年级竞赛复赛", "2004年希望杯五年级竞赛复赛"], "difficulty": "1", "qtype": "single_choice", "problem": "在一列数$$8$$、$$6$$、$$4$$、$$2$$、$$6$$、\\ldots 中，从第$$3$$个数开始，每个数都是它前面两个数的和的个位数字．在这串数中，有个数在重复循环. ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["$$2$$、$$2$$、$$4$$、$$8$$、$$2$$、$$6$$、$$2$$、$$2$$、$$4$$、$$8$$\\ldots 发现$$6$$个一循环，$$2004\\div 6=334$$（组），所以第$$2004$$个数和第$$6$$个相同为$$6$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1275", "queId": "1f511cabba434c53af9c2f14a65385ae", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "按规律画图：$$\\bigcirc \\triangle \\square \\square $$☆$$\\bigcirc \\triangle \\square \\square $$☆$$\\bigcirc \\triangle \\square \\square $$☆$$\\cdots \\cdots $$第$$30$$个是图形． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\bigcirc $$ "}], [{"aoVal": "B", "content": "$$\\triangle $$ "}], [{"aoVal": "C", "content": "$$\\square $$ "}], [{"aoVal": "D", "content": "☆ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["从图中可知五个一循环，$$30\\div 5=6\\cdots \\cdots 0$$，所以第三十个与第五个相同． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3127", "queId": "e789f89b75ab4b61aa1da941365a3a56", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第24题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "（5分）分数$$A=1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots +\\frac{1}{16}$$的整数部分是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->放缩法->首尾放缩法"], "answer_analysis": ["$$A\\textgreater1+\\frac{1}{2}+\\frac{1}{4}\\times 2+\\frac{1}{8}\\times 4+\\frac{1}{16}\\times 8=3$$， $$A\\textless{}1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}\\times 4+\\frac{1}{8}\\times 8+\\frac{1}{16}\\textless{}4$$， ∴$$\\left[ A \\right]=3$$． 答：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "48", "queId": "07f2e89632cb4258aae0367923b1de2d", "competition_source_list": ["2011年全国学而思杯五年级竞赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "在足球场上，甲、乙、丙每人都擅长踢以下六个位置中的两个：``前锋''、``前卫''、``前腰''、``后腰''、``后卫''、``守门员''，而且每个位置都只有一个人擅长．此外： （1）前锋夸前腰传球传的好 （2）前腰和后卫常去与甲一起去看电影 （3）前卫请后腰喝过汽水 （4）前锋和后腰很要好 （5）乙不擅长后卫 （6）丙在场上的作用常强于乙和后腰 那么甲擅长的位置是~\\uline{~~~~~~~~~~}~和~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "前锋、前卫 "}], [{"aoVal": "B", "content": "前卫、后腰 "}], [{"aoVal": "C", "content": "守门员、前锋 "}], [{"aoVal": "D", "content": "后腰、守门员 "}]], "knowledge_point_routes": ["课内体系->能力->实践应用", "拓展思维->能力->实践应用"], "answer_analysis": ["甲不是前腰和后卫，前腰和后卫是两个人，乙不是后卫，那么丙是后卫，根据$$6$$可知道甲是后腰，那么根据$$2$$可知道，乙是前腰，根据$$1$$，则乙不是前锋，根据$$4$$，甲也不能是前锋，所以丙是前锋，根据$$3$$，前卫只能是乙，那么剩下的守门员只能是甲，所以甲的位置是后腰和守门员． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2373", "queId": "1c400e6190b44bc090224384c9b260f9", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（一）第4题", "2017年新希望杯六年级竞赛训练题（一）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "1.将四个分数按从小到大的顺序排列，正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ "}], [{"aoVal": "C", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ "}], [{"aoVal": "D", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["通分子$$\\frac{5}{14}=\\frac{60}{168}$$， $$\\frac{10}{27}=\\frac{60}{162}$$， $$\\frac{12}{31}=\\frac{60}{155}$$， $$\\frac{20}{53}=\\frac{60}{159}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "888", "queId": "cda5764b9bce420fa323864c97695496", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$7$$进制中有三位数$$\\overline{abc}$$，化为$$9$$进制为$$\\overline{cba}$$，求这个三位数在十进制中为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$148$$ "}], [{"aoVal": "B", "content": "$$248$$ "}], [{"aoVal": "C", "content": "$$348$$ "}], [{"aoVal": "D", "content": "$$648$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["首先还原为十进制： $${{(\\overline{abc})}_{7}}=a\\times{{7}^{2}}+b\\times {{7}^{1}}+c\\times {{7}^{0}}=49a+7b+c$$；$${{(\\overline{cba})}_{9}}=c\\times{{9}^{2}}+b\\times {{9}^{1}}+a\\times {{9}^{0}}=81c+9b+a$$． 于是$$49a+7b+c=81c+9b+a$$；得到$$48a=80c+2b$$，即$$24a=40c+b$$． 因为$$24a$$是$$8$$的倍数，$$40c$$也是$$8$$的倍数，所以$$b$$也应该是$$8$$的倍数，于是$$b=0$$或$$8$$． 但是在$$7$$进制下，不可能有$$8$$这个数字．于是$$b=0$$，$$24a=40c$$，则$$3a=5c$$． 所以$$a$$为$$5$$的倍数，$$c$$为$$3$$的倍数． 所以，$$a=0$$或$$5$$，但是，首位不可以是$$0$$，于是$$a=5$$，$$c=3$$； 所以$${{(\\overline{abc})}_{7}}={{(503)}_{7}}=5\\times 49+3=248$$ ． 于是，这个三位数在十进制中为$$248$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3260", "queId": "2daa93c673454fd690475d88045e7d2a", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级五年级竞赛邀请赛训练题（一）"], "difficulty": "1", "qtype": "single_choice", "problem": "由$$1$$.$$2$$.$$3$$.$$4$$.$$5$$五个数字组成没有重复数字的三位数，各个数字之和为偶数的有． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$个 "}], [{"aoVal": "B", "content": "$$30$$个 "}], [{"aoVal": "C", "content": "$$36$$个 "}], [{"aoVal": "D", "content": "$$42$$个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->组数问题->有特殊要求的组数问题"], "answer_analysis": ["数字之和为偶数的数有$$123$$、$$125$$、$$134$$、$$145$$、$$235$$、$$345$$，而每种选法又可以有五种不同三位数，所以$$6\\times 6=36$$个不同的三位数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1115", "queId": "0cb3c9d98ece42c3a9bf637a0da03234", "competition_source_list": ["2004年六年级竞赛创新杯", "2004年第2届创新杯六年级竞赛初赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "两个相同的瓶子装满酒精溶液，一个瓶子中酒精与水的体积比是$$3:1$$，另一个瓶子中酒精与水的体积比是$$4:1$$。如果把这两瓶酒精溶液混合，那么混合溶液中酒精与水的体积比是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$7:1$$ "}], [{"aoVal": "B", "content": "$$7:2$$ "}], [{"aoVal": "C", "content": "$$24:7$$ "}], [{"aoVal": "D", "content": "$$31:9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题->按比分配"], "answer_analysis": ["两个相同的瓶子装满酒精溶液，设一个瓶子装溶液的体积是单位``$$1$$''，那么第一个瓶子中酒精体积是$$3\\div (1+3)=\\frac{3}{4}$$，第二个瓶子中酒精体积是$$4\\div (1+4)=\\frac{4}{5}$$，把两瓶酒精溶液混合，两个瓶子的酒精体积和是$$\\frac{3}{4}+\\frac{4}{5}=\\frac{31}{20}$$， 水体积和$$=1+1-\\frac{31}{20}=\\frac{9}{20}$$， 那么酒精与水的体积比是$$\\frac{31}{20}:\\frac{9}{20}=31:9$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2962", "queId": "9c03594ba02a4efd9e1a1633b92dd574", "competition_source_list": ["2020年广东广州羊排赛四年级竞赛第16题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "简易方程． $$10+2x+3=19$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$x=4$$． "}], [{"aoVal": "B", "content": "$$x=3$$． "}], [{"aoVal": "C", "content": "$$x=2$$． "}], [{"aoVal": "D", "content": "$$x=5$$． "}]], "knowledge_point_routes": ["课内体系->思想->方程思想", "拓展思维->拓展思维->计算模块->方程基础->一元一次方程"], "answer_analysis": ["$$\\begin{eqnarray}10+2x+3\\&=\\&19 2x\\&=\\&19-3-10 2x\\&=\\&6 x\\&=\\&3．\\end{eqnarray}$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3011", "queId": "dbfe7dc8fd074e1caefab22504de71dc", "competition_source_list": ["2022年第九届鹏程杯四年级竞赛初赛第20题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "若1※4=1+2+3+4，6※5=6+7+8+9+10，按照这种规则，(4※4)$\\times$5-(5※5)=（）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$55$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "\\[70\\] "}], [{"aoVal": "E", "content": "\\[75\\] "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->定义新运算"], "answer_analysis": ["无 "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1111", "queId": "afee1c54792d4e8f9df06db4b552c4b7", "competition_source_list": ["2014年迎春杯三年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "长方形的周长是$$48$$厘米，已知长是宽的$$2$$倍，那么长方形的长是。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$厘米 "}], [{"aoVal": "B", "content": "$$16$$厘米 "}], [{"aoVal": "C", "content": "$$24$$厘米 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->和倍问题->二量和倍问题->两量和倍"], "answer_analysis": ["暗和型： 周长是$$48$$厘米，那么长$$+$$宽的和：$$48\\div$$ 2=24（厘米） 宽：$$24\\div$$ （2+1）=8（厘米） 长：$$8\\times$$ 2=16（厘米） "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2186", "queId": "2be293f2ad30406fb6c5bfd0feba8e4a", "competition_source_list": ["2019年广东深圳全国小学生数学学习能力测评四年级竞赛初赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$9$$点至$$10$$点之间的某个时刻，$$5$$分钟前分针的位置和$$5$$分钟后时针的位置相同，此时刻是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9:05$$ "}], [{"aoVal": "B", "content": "$$9:35$$ "}], [{"aoVal": "C", "content": "$$9:55$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用->度量单位认知"], "answer_analysis": ["根据选项$$\\text{C}$$，现在是$$9:55$$，那么$$5$$分钟前分针的位置是指向$$10$$，$$5$$分钟后时针也是指向$$10$$．所以$$\\text{C}$$选项答案是满足题目要求的，所以选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2252", "queId": "722e110a727e43d3956c3ec8ec53f368", "competition_source_list": ["2018年全国小学生数学学习能力测评五年级竞赛初赛第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "科学家进行一项实验，每隔$$5$$小时做一次记录，做第十二次记录时，钟表时针恰好指向$$9$$，做第一次记录时，时针指向． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->时间问题->认识钟表"], "answer_analysis": ["先算出从第一次记录到第十二次记录中间经过的时间是多少，第$$1$$次到第$$12$$次之���有个$$11$$间隔：$$5\\times11=55$$（小时）． 时针每过$$12$$小时就会转一圈回到原来的状态，$$55\\div12=4$$（圈）$$\\cdots \\cdots 7$$（小时）． 即时针转了$$4$$圈以后，又经过了$$7$$个小时，时针指向$$9$$， 所以原来第一次时针指向：$$9-7=2$$． 因此，选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2053", "queId": "cb31acc21b824ba99ddd01eaa20f2c0d", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（一）第6题", "2017年新希望杯六年级竞赛训练题（一）第6题"], "difficulty": "3", "qtype": "single_choice", "problem": "将四位数$$12472$$重复写$$100$$次组成一个$$500$$位数$$1247212472\\cdots 12472$$，先删去这个数中所有位于奇数位（从左往右数）上的数字，剩下的数字（顺序不变)组成一个新数：再删去新数中所有位于奇数位上的数字，剩下的数字组成一个新数：$$\\cdots \\cdots $$按上述方法一直删除下去，直到剩下一个数字为止，则最后剩下的数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["将这个$$500$$位数从左到右依次编号第$$1$$位、第$$2$$位、第$$3$$位、$$\\cdots $$、第$$500$$位，且编号不变． 第$$1$$次删数后剩下$$2$$的倍数，第$$2$$次删数后剩下$$22$$的倍数，第$$3$$次删数后剩下$$23$$的倍数，$$\\cdots \\cdots $$，因为$$256\\textless{}500\\textless{}512$$，所以最后剩下的数字在第$$256$$位，$$256\\div 5=51$$（组）$$\\cdots \\cdots 1$$（个），第$$1$$位上的数字是$$1$$，所以最后剩下的数字是$$1$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "87", "queId": "58c06befe3f04c48ac51c1ea4bc3fb9c", "competition_source_list": ["2017年第22届全国华杯赛小学中年级竞赛初赛第2题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$至$$10$$这$$10$$个整数中．至少取个数，才能保证其中有两个数的和等于$$10$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->枚举思想"], "answer_analysis": ["将和为$$10$$的两个数作为一组，有$$\\left( 1,9 \\right)$$，$$\\left( 2,8 \\right)$$，$$\\left( 3,7 \\right)$$，$$\\left( 4,6 \\right)$$，另外$$\\left( 5\\right)$$，$$\\left(10 \\right)$$单独一组，考虑极端情况，若在$$6$$组中每组各取一个数字，也没有两数和为$$10$$，然后在剩下的数中取任何一个，都会有两数能凑成$$10$$．所以至少需要取$$7$$个数． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "447", "queId": "8f04cbdca66748ffaf762af7cd75b86c", "competition_source_list": ["2012年第10届创新杯四年级竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "显示在电子钟上的时间是$$5:55$$，下一次电子钟上显示的时间又是全部相同的数字，还要过分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$71$$ "}], [{"aoVal": "B", "content": "$$255$$ "}], [{"aoVal": "C", "content": "$$316$$ "}], [{"aoVal": "D", "content": "$$377$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为分钟的十位最大为$$5$$，所以下一次数字相同的时刻为$$11:11$$，$$11:11$$距$$5:55$$有$$5$$小时$$16$$分，即$$316$$分钟．所以选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1352", "queId": "5e525e19afa041a6b5942939088b4837", "competition_source_list": ["2006年第4届创新杯四年级竞赛初赛B卷第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个数加上$$8$$的和，再乘$$8$$的积，再减去$$8$$的差，再除以$$8$$的商，等于$$80$$，那么，这个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$37$$ "}], [{"aoVal": "B", "content": "$$59$$ "}], [{"aoVal": "C", "content": "$$73$$ "}], [{"aoVal": "D", "content": "$$86$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["从最后一步推起，``除以$$8$$，其结果等于$$80$$''可以求出被除数：$$80\\times 8=640$$；再看倒数第$$2$$步，``减去$$8$$''得$$640$$，可以求出被减数：$$640+8=648$$；然后看倒数第$$3$$步，``乘$$8$$''得$$648$$，可以求出另一个因数：$$648\\div 8=81$$；最后看���$$1$$步，``某数加上$$8$$''得$$81$$，某数为$$81-8=73$$，由此即可解决问题． $$(80\\times 8+8)\\div 8-8=648\\div 8-8=81-8=73$$． 答：这个数是$$73$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1021", "queId": "1c1784982f274501901989a50424d39c", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛决赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小帅第一天写$$5$$个大字，以后每一天都比前一天多写$$2$$个大字，$$6$$天后小帅一共写了个大字． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$66$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["第一天$$5$$个大字， 第二天$$5+2=7$$个大字， 第三天$$7+2=9$$个大字， 第四天$$9+2=11$$个大字， 第五天$$11+2=13$$个大字， 第六天$$13+2=15$$个大字， 六天共写了$$5+7+9+11+13+15=60$$个大字． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3080", "queId": "ef4669edaa5c4ba6a9251ef3ec1ab7c0", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\left( \\Delta \\times 2-1 \\right)\\times 2=2018$$，请问$$\\Delta $$代表的数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$503$$ "}], [{"aoVal": "B", "content": "$$504$$ "}], [{"aoVal": "C", "content": "$$505$$ "}], [{"aoVal": "D", "content": "$$506$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->一元一次方程->整数系数方程"], "answer_analysis": ["$$\\left( \\Delta \\times 2-1 \\right)\\times 2=2018$$， $$\\Delta \\times 2-1=2018\\div 2=1009$$， $$\\Delta \\times 2=1009+1=1010$$， $$\\Delta =1010\\div 2=505$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1123", "queId": "6b376f4c8d3c4b7eb49fd6b293fcfab6", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（二）第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$A$$、$$B$$瓶浓度不同的盐水，小泉从两瓶中各取$$1$$升混合在一起，得到一份浓度为$$30 \\% $$的盐水；他又将这份盐水与$$1$$升$$A$$瓶盐水混合在一起，最终浓度为$$32 \\% $$，那么$$B$$瓶盐水的浓度是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24 \\% $$ "}], [{"aoVal": "B", "content": "$$36 \\% $$ "}], [{"aoVal": "C", "content": "$$42 \\% $$ "}], [{"aoVal": "D", "content": "$$44 \\% $$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["设$$A$$瓶盐水的浓度是$$x \\%$$，$$B$$瓶盐水的浓度是$$y \\%$$． $$1\\centerdot x \\%+1\\centerdot y \\%=2\\centerdot 30 \\%$$ ① $$2\\centerdot 30 \\% +1\\centerdot x \\%=3\\centerdot 32 \\%$$ ② 解得$$x=36$$，$$y=24$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1407", "queId": "abf5c232a853439481d1dc98f1b1140d", "competition_source_list": ["2019年第24届YMO三年级竞赛决赛第5题3分", "2020年第24届YMO三年级竞赛决赛第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个正方形的操场，在它的外面一圈插上小红旗，四个角上都插一面小红旗，每边都插$$16$$面，一共插了面小红旗． ", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$58$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$62$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->封闭型->封闭型植树问题->封闭植树类型问题(段数大于10)"], "answer_analysis": ["根据题意分析可知，四个角都插上一面小红旗，每边都插$$26$$面， 所以每一边有小红旗：$$26-1=25$$（面）， 因为是正方形，每边都有$$25$$面，那么一共插了：$$25\\times4=100$$（面）红旗， 故答案为：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1693", "queId": "536ebf5e511e40a5b2fafe4fea833f85", "competition_source_list": ["2016年陕西西安小升初工大附中入学真题2第3题", "2017年全国小学生数学学习能力测评五年级竞赛初赛第10题3分", "2016年陕西西安小升初某工大附中", "2018年陕西西安小升初分类卷13第14题"], "difficulty": "1", "qtype": "single_choice", "problem": "把一条细绳先对折，再���它折成相等的三折，接着再对折，然后用剪刀在折过三次的绳中间剪一刀，那么这条绳被剪成（~ ）段． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["对折一次剪一刀是三段$$(2+1)$$， 再把它折成相等的三折，是$$(2\\times 3+1)$$， 接着再对折剪一刀是$$2\\times 3\\times 2+1=13$$（段）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2836", "queId": "5c6ec2c951854df58174ee65981e169b", "competition_source_list": ["2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第2题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "爸爸带了$$269$$元钱，要买$$8$$元一个的笔记本，最多能买个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$ "}], [{"aoVal": "B", "content": "$$33$$ "}], [{"aoVal": "C", "content": "$$34$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["根据题意分析可知，利用关系式：数量$$=$$总价$$\\div $$单价，用爸爸带的钱除以每个笔记本的单价即可得到最多能买多少个笔记本，列式为：$$269\\div 8=33$$（个）$$\\cdots \\cdots 5$$（元），剩下的$$5$$元不足以再买一个笔记本， 所以最多能买$$33$$个， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2449", "queId": "3880d235991640788f6c7d078645d524", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "把一个自然数$$n$$的数位上的偶数数字相加所得的和记为$$E(n)$$，例如：$$E\\left( 1999 \\right)=0$$，$$E\\left( 2000 \\right)=2$$，$$E\\left( 2021 \\right)=2+2=4$$．则$$E\\left( 1 \\right)+E\\left( 2 \\right)+E\\left( 3 \\right)+\\cdots +E\\left( 100 \\right)=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$400$$ "}], [{"aoVal": "D", "content": "$$2020$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->符号代换->代数运算"], "answer_analysis": ["$$2$$，$$4$$，$$6$$，$$8$$这$$4$$个数字，每个在个位出现$$10$$次，在十位出现$$10$$次， 所以$$E(1)+E(2)+···+E(100)=(2+4+6+8)\\times (10+10)=400$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3426", "queId": "e0ad94ad5dd743aca886531b06a6c1a0", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第14题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "由数字$$0$$，$$1$$，$$2$$，$$3$$，$$4$$组成三位数，可以组成个不同的三位数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$80$$ "}], [{"aoVal": "D", "content": "$$100$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["暂无 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1948", "queId": "d79cf47016534dfd8d6d941542133537", "competition_source_list": ["2016年第6届广东广州羊排赛六年级竞赛第9题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "琦琦跳绳$$3$$次，平均每次跳$$156$$下，要想跳$$4$$次后达到``平均每次跳$$160$$下''，他第$$4$$次要跳下． ", "answer_option_list": [[{"aoVal": "A", "content": "$$164$$ "}], [{"aoVal": "B", "content": "$$168$$ "}], [{"aoVal": "C", "content": "$$172$$ "}], [{"aoVal": "D", "content": "$$176$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->容斥求平均数"], "answer_analysis": ["琦琦$$3$$次共跳$$156\\times 3=468$$（下）， 要想跳$$4$$次后达到平均每次跳$$160$$下，$$4$$次一共跳$$4\\times 160=640$$（下）， 他第$$4$$次要跳$$640-468=172$$（下）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3152", "queId": "3ca27c0ca9524b95a5848bd957764e93", "competition_source_list": ["2003年第1届创新杯六年级竞赛初赛第5题", "六年级竞赛创新杯", "2003年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个骰子(小正方体)的六个面上分别写有数字1、2、2、3、3、3，当投掷这个骰子时，数字``2''朝上的可能性是��� ）. ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{6}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->基本概率"], "answer_analysis": ["投掷这个骰子时，6个面都有可能朝上，每个面朝上的可能性是$$\\frac{1}{6}$$，有两个面上写着数字2，数字2朝上的可能性为$$2\\times \\frac{1}{6}=\\frac{1}{3}$$. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1513", "queId": "83d1987763c94bb7b7409bc637365aa9", "competition_source_list": ["2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（一）"], "difficulty": "1", "qtype": "single_choice", "problem": "甲用$$1000$$元购买了一些股票，随即他将这些股票转卖给了乙，获利$$10 \\% $$，而后乙又将这些股票反 卖给了甲，但乙损失了$$10 \\% $$，最后甲按乙卖给甲价格的九折将这些股票卖给了乙，如果在上述的交易过程中其他费用忽略不计，那么甲（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "亏欠了 $$3$$元 "}], [{"aoVal": "B", "content": "赠了$$3$$元 "}], [{"aoVal": "C", "content": "亏了$$1$$元 "}], [{"aoVal": "D", "content": "赚了$$1$$元 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->折扣问题"], "answer_analysis": ["甲第一次卖给乙价格为：$$1000\\times \\left( 1+10 \\% ~\\right)=1100$$元，甲盈利了$$100$$元； 乙转卖给甲价格：$$1100\\times \\left( 110 \\% ~\\right)=99$$元 综上，甲在此次交易过程中盈利了$$1$$元． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1609", "queId": "95a7b48dae824034806bc203ca1f8aa3", "competition_source_list": ["2008年第6届创新杯四年级竞赛初赛B卷第2题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "两根电线杆之间相隔$$115$$米，在它们之间等距离增加$$22$$根电线杆后，第$$2$$根和第$$16$$根电线杆之间相隔． ", "answer_option_list": [[{"aoVal": "A", "content": "$$68$$米 "}], [{"aoVal": "B", "content": "$$70$$米 "}], [{"aoVal": "C", "content": "$$71$$米 "}], [{"aoVal": "D", "content": "$$72$$米 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["在两根电线杆中间等距加$$22$$根电线杆， 有$$22+1=23$$（个）间隔， 那么$$115\\div 23=5$$（米）一个间隔， 从第$$2$$根到第$$16$$根共有$$16-2=14$$（个）间隔， $$14\\times 5=70$$（米）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3294", "queId": "48f07999e6f14e9696eaba9a89727e01", "competition_source_list": ["2020年第24届YMO四年级竞赛决赛第4题3分", "2019年第24届YMO四年级竞赛决赛第4题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "现在有六种不同颜色的笔，把``$$\\text{YMO}$$''三个字母写成三种不同颜色，共有种写法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$216$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["依次填涂， $$\\text{Y}$$有$$6$$中颜色可选， 则$$\\text{M}$$有$$6-1=5$$（种）颜色可选， 则$$\\text{O}$$有$$6-1-1=4$$（种）颜色可选， 则有$$6\\times 5\\times 4=120$$（种）涂法． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1250", "queId": "d099fd921cfd4e09b623888d49526f87", "competition_source_list": ["2018年全国小学生数学学习能力测评五年级竞赛复赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有四个不同的整数，它们的平均数是$$13.75$$，三个大数的平均数是$$15$$，三个小数的平均数是$$12$$，如果第二大的数是奇数，那么它是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型"], "answer_analysis": ["四个数之和为$$13.75\\times 4=55$$，三个大数之和为$$15\\times 3=45$$，最小数为$$55-45=10$$，三个小数之和为$$12\\times 3=36$$，最大数为$$55-36=19$$．中间两数之和为$$45-19=26$$，中间两数平均数为$$26\\div 2=13$$．第二大数是奇数且比$$13$$大比$$19$$小，只有$$15$$和$$17$$两种可能．若第二大数为$$17$$，则第三大数为$$26-17=9$$，比最小数$$10$$小，不成立；所以第二大数为$$15$$，此时第三大数为$$26-15=11$$，成立． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1408", "queId": "296c470ee8ef4ff1b9b5ec932fe6ef68", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛A卷第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明爷爷今年的年龄减去$$15$$后，缩小到原来的$$\\frac{1}{4}$$，再减去$$6$$之后扩大到原来的$$10$$倍，恰好是$$100$$岁，请你算一算，小明的爷爷今年是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$77$$ "}], [{"aoVal": "B", "content": "$$80$$ "}], [{"aoVal": "C", "content": "$$83$$ "}], [{"aoVal": "D", "content": "$$79$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["设爷爷今年$$x$$岁， $$\\left[ \\left( x-15 \\right)\\times \\frac{1}{4}-6 \\right]\\times 10=100$$ $$\\left( x-15 \\right)\\times \\frac{1}{4}-6=100\\div 10$$ $$\\left( x-15 \\right)\\times \\frac{1}{4}-6=10$$ $$\\frac{1}{4}\\left( x-15 \\right)=10+6$$ $$\\frac{1}{4}\\left( x-15 \\right)=16$$ $$x-15=16\\div \\frac{1}{4}$$ $$x-15=64$$ $$x=64+15$$ $$x=79$$． 所以小明的爷爷今年$$79$$岁，故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3308", "queId": "4dd922b2b2b34ce99d2259edb9b7eedb", "competition_source_list": ["2010年第8届创新杯五年级竞赛初赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "十位数字等于百位数字与个位数字之和的三位数共有． ", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$个 "}], [{"aoVal": "B", "content": "$$36$$个 "}], [{"aoVal": "C", "content": "$$45$$个 "}], [{"aoVal": "D", "content": "$$55$$个 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["设三位数为$$\\overline{abc}$$，则$$b=a+c$$， $$a=1$$时，$$c=0$$，$$1$$，$$2$$，$$\\ldots 8$$，共$$9$$个； $$a=2$$时，$$c=0$$，$$1$$，$$2$$，$$\\ldots 7$$，共$$8$$个； $$a=3$$时，$$c=0$$，$$1$$，$$2$$，$$\\ldots 6$$，共$$7$$个； $$a=4$$时，$$c=0$$，$$1$$，$$2$$，$$\\ldots 5$$，共$$6$$个； $$a=5$$时，$$c=0$$，$$1$$，$$2$$，$$3$$，$$4$$，共$$5$$个； $$a=6$$时，$$c=0$$，$$1$$，$$2$$，$$3$$，共$$4$$个； $$a=7$$时，$$c=0$$，$$1$$，$$2$$，共$$3$$个； $$a=8$$时，$$c=0$$，$$1$$，共$$2$$个； $$a=9$$时，$$c=0$$，共$$1$$个； 一共有$$1+ 2+ \\ldots 9= \\frac{9 \\times(1+9)}{2}=45$$个， 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1063", "queId": "082ced4b92e4473691fa7693c9365ba3", "competition_source_list": ["2018年美国数学大联盟杯五年级竞赛初赛第34题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2$$ liter of $$2 \\%$$ fat milk $$+3$$ liter of $$3 \\%$$ fat milk$$=5$$ liter of fat milk． $$2$$升$$2 \\%$$浓度的脂肪乳$$+3$$升$$3 \\%$$浓度的脂肪乳$$=5$$升浓度为的脂肪乳． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2.5 \\%$$ "}], [{"aoVal": "B", "content": "$$2.6 \\%$$ "}], [{"aoVal": "C", "content": "$$5 \\%$$ "}], [{"aoVal": "D", "content": "$$6 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$2$$升$$2 \\%$$浓度的脂肪乳$$+3$$升$$3 \\%$$浓度的脂肪乳$$=5$$升浓度的脂肪乳． $$\\frac{总含量}{总体积}\\times 100 \\%$$， 即$$\\frac{2\\times 2 \\%+3\\times 3 \\%}{2+3}\\times 100 \\%=2.6 \\%$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "578", "queId": "74521a22ce334b41809d59bd985bfbcb", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（五）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "十进制数$$25$$转换成二进制数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11101$$ "}], [{"aoVal": "B", "content": "$$1011$$ "}], [{"aoVal": "C", "content": "$$10101$$ "}], [{"aoVal": "D", "content": "$$11001$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["25$\\div$2=12\\ldots\\ldots1 12$\\div$2=6\\ldots\\ldots0 6$\\div$2=3\\ldots\\ldots0 3$\\div$2=1\\ldots\\ldots1 1$\\div$2=0\\ldots\\ldots1 倒取余数，结果为（11001）\\textsubscript{2} 知识来源------四年级春季第二讲《机器人的聊天记录》 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "894", "queId": "b6e769b14d1346dd9ec9373ae0fd9b27", "competition_source_list": ["2017年全国希望杯小学高年级五年级竞赛初赛考前100题"], "difficulty": "3", "qtype": "single_choice", "problem": "三个质数的平方和是$$390$$，这三个质数的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$21$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$26$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->整体思想"], "answer_analysis": ["因为奇数的平方还是奇数，所以这三个数中有一个是$$2$$．因为$$390-{{2}^{2}}=386$$， 所以，通过质数平方的个位特点，判断得还有一个质数应该是$$5$$． 又$$386-{{5}^{2}}=361$$，且$$361={{19}^{2}}$$． 所以这三个质数分别是$$2$$，$$5$$，$$19$$，和是$$26$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3356", "queId": "b64bafe77a3e4ae485f343a5c7caad6b", "competition_source_list": ["2020年第24届YMO三年级竞赛决赛第3题3分", "2019年第24届YMO三年级竞赛决赛第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$10$$本相同的笔记本分给$$3$$个人，每人至少一本，共有种不同的分配方案． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$45$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["根据题意，$$10$$个相同的小球分给$$3$$个人，每人至少$$1$$个，就是将$$10$$个球分成$$3$$组， 一个人最多分：$$10-1-1=8$$（支）， $$8+7+6+5+4+3+2+1=36$$（种）， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "573", "queId": "073546842f084705b3a8c9fcdd2dcd72", "competition_source_list": ["2007年第5届创新杯五年级竞赛第2题5分", "2007年第5届创新杯五年级竞赛第2题5分", "2007年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "有分别写着$$1$$、$$2$$、$$3$$、$$\\cdots$$、$$13$$的卡片各两张，任意抽出两张，计算这两张卡片上的数的积。这样得到许多不相等的积中，最多有( )个能被$$6$$整除。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$26$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["不超过$${{13}^{2}}=169$$的$$6$$的倍数是$$1\\times 6$$，$$2\\times 6$$，$$\\cdots $$，$$28\\times 6$$有$$28$$个，其中不能写成$$a\\times b\\left( 1\\leqslant a\\leqslant b\\leqslant 13 \\right)$$形式的数是：$$17\\times 6$$，$$19\\times 6$$，$$21\\times 6$$，$$23\\times 6$$，$$25\\times 6$$，$$27\\times 6$$，$$28\\times 6$$，共有$$7$$个。因此，符合要求的整数共有$$28-7=21$$（个）。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2476", "queId": "19a090a80fe34d779854ef4f3520f6a5", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛A卷第4题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "算式$$1\\div \\left( 2\\div 3 \\right)\\div \\left( 3\\div 4 \\right)\\div \\left( 4\\div 5 \\right)\\div \\left( 5\\div 6 \\right)$$的结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$55$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["$$1\\div \\left( 2\\div 3 \\right)\\div \\left( 3\\div 4 \\right)\\div \\left( 4\\div 5 \\right)\\div \\left( 5\\div 6 \\right)$$ $$=1\\div 2\\times 3\\div 3\\times 4\\div 4\\times 5\\div 5\\times 6$$ $$=1\\div 2\\times 6$$ $$=3$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "863", "queId": "a8f35f0ccdef45f5992c447c99d64e2f", "competition_source_list": ["2004年四年级竞赛创新杯", "2004年第2届创新杯六年级竞赛初赛第8题", "2004年五年级竞赛创新杯", "2004年六年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "在大于$$2004$$的自然数中，有些数除以$$69$$的商和余数相等，这样的自然数的个数是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$个 "}], [{"aoVal": "B", "content": "$$40$$个 "}], [{"aoVal": "C", "content": "$$68$$个 "}], [{"aoVal": "D", "content": "无穷多个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->同余->同余定理"], "answer_analysis": ["设符合要求的自然数为$$n$$，则$$n\\textgreater2004$$，且$$n=69q+q$$，$$0\\leqslant q\\leqslant 68$$。故而$$n=70q\\textgreater2004$$得$$q\\geqslant 29$$，因此$$29\\leqslant q\\leqslant 68$$，即$$q$$有$$68-29+1=40$$个，符合要求的自然数有$$40$$个，选B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3166", "queId": "2173c507beb74f588af2e88fcfe71cd7", "competition_source_list": ["2006年第11届全国华杯赛竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "五位同学扮成奥运会吉祥物福娃贝贝、晶晶、欢欢、迎迎和妮妮，排成一排表演节目．如果贝贝和妮妮不相邻，共有(~~~~~~~ )种不同的排法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$72$$ "}], [{"aoVal": "C", "content": "$$96$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["贝贝在左、妮妮在右相邻的排法有$$4\\times 3\\times 2\\times 1=24$$（种），贝贝在右、妮妮在左相邻的排法也有$$4\\times 3\\times 2\\times 1=24$$（种），总的排法$$5\\times 4\\times 3\\times 2\\times 1=120$$（种）．所以贝贝和妮妮不相邻的排法是$$120-2\\times 24=72$$（种）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "998", "queId": "45c7d7b6c08747c19be185cb569e5f7b", "competition_source_list": ["2015年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "新生入校后，合唱队、田径队和舞蹈队共招收学员$$100$$人。如果合唱队招收的人数比田径队多一倍，舞蹈队比合唱队多$$10$$人。那么，舞蹈队招收（ ）人。（注：每人限加入一个队） ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$42$$ "}], [{"aoVal": "C", "content": "$$46$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->乘除法应用->除法应用"], "answer_analysis": ["$$\\left( 100-10 \\right)\\div \\left( 2+2+1 \\right)$$ $$=90\\div 5$$ $$=18$$（人） $$18\\times 2+10$$ $$=36+10$$ $$=46$$（人） 答：舞蹈队招收$$46$$人。 故选：$$C$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2223", "queId": "e3764053ef43442a9d7ec7e432afce55", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（三）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "欧欧从$$A$$地前往$$B$$地办事，他每走$$40$$分钟就休息$$10$$分钟，到达$$B$$地共需$$4$$小时$$20$$分钟，从$$B$$地原路返回的速度是去时的$$2$$倍，若他每走$$35$$分钟就休息$$15$$分钟，则到达$$A$$地共需． ", "answer_option_list": [[{"aoVal": "A", "content": "$$105$$分钟 "}], [{"aoVal": "B", "content": "$$130$$分钟 "}], [{"aoVal": "C", "content": "$$135$$分钟 "}], [{"aoVal": "D", "content": "$$150$$分钟 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题"], "answer_analysis": ["$$60\\times 4+20=50\\times 5+10$$，去时共走了$$40\\times 5+10=210$$（分钟），则返回时共需走$$210\\div 2=105$$（分钟），$$105\\div 35=3$$，加上休息的时间共需要$$105+15\\times \\left( 31 \\right)=135$$（分钟）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1276", "queId": "6bc8ca27e7cb46d4b06770a05282b23c", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "某校开展评选``优秀少先队员''和``好公民''活动，``好公民''占评上人数的$$\\frac{3}{4}$$，``优秀少先队员''占评上人数的$$\\frac{9}{25}$$，同时获得两种称号的有$$44$$人，只获得``优秀少先队员''称号的有（~ ）人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$70$$ "}], [{"aoVal": "B", "content": "$$80$$ "}], [{"aoVal": "C", "content": "$$90$$ "}], [{"aoVal": "D", "content": "$$100$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["把评上的总人数看做单位``$$1$$''，则评上的人数为：$$44\\div \\left( \\frac{3}{4}+\\frac{9}{25}-1 \\right)=400$$人，只得``优秀少先队员''有$$400\\times \\frac{9}{25}-44=100$$人． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2421", "queId": "0c47e60f85254e06bfd18b46ceae3a20", "competition_source_list": ["2005年六年级竞赛创新杯", "2005年第3届创新杯六年级竞赛复赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "规定一种运算``\\textasciitilde''： $$a\\sim b$$表示求$$a$$，$$b$$两个数的差，即$$a$$，$$b$$中较大的数减较小的数，例如，$$5\\sim 4=5-4=1$$，$$1\\sim 4=4-1=3$$ ， $$6\\sim 6=6-6=0$$，那么化简$$\\left( \\frac{355}{118} \\sim 2 \\right)+\\left( \\frac{355}{118} \\sim 3 \\right)+\\left( \\frac{355}{118} \\sim 4 \\right)+\\left( \\frac{355}{118} \\sim 5 \\right)$$得（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->定义新运算->观察规律型->找规律型定义新运算"], "answer_analysis": ["$$\\frac{355}{118}=3\\frac{1}{118}$$，那么$$ \\left( \\frac{355}{118} \\sim 2 \\right)+\\left( \\frac{355}{118} \\sim 3 \\right)+\\left( \\frac{355}{118} \\sim 4 \\right)+\\left( \\frac{355}{118} \\sim 5 \\right)$$ $$=3\\frac{1}{118}-2+3\\frac{1}{118}-3+4-3\\frac{1}{118}+5-3\\frac{1}{118}=4+5-2-3=4$$，选D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1305", "queId": "707fa755351e4acbae085d5d29896d45", "competition_source_list": ["2020年新希望杯五年级竞赛决赛（8月）第17题", "2020年新希望杯五年级竞赛初赛（个人战）第17题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2019$$年国庆节是星期二，则$$2020$$年国庆节是． ", "answer_option_list": [[{"aoVal": "A", "content": "星期一 "}], [{"aoVal": "B", "content": "星期二 "}], [{"aoVal": "C", "content": "星期三 "}], [{"aoVal": "D", "content": "星期四 "}], [{"aoVal": "E", "content": "星期五 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$2020\\div 4=505$$，$$2020$$年是闰年． 从$$2019$$年国庆节经过$$366$$天是$$2020$$年国庆节． $$366\\div 7=52$$（周）$$\\cdots \\cdots 2$$（天）， 故$$2020$$年的国庆节是星期四． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "482", "queId": "ea1590b0eb6e4fdfaec54de6e5c42aef", "competition_source_list": ["2017年第20届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "下面两个算式中的每个汉字代表一个数字，不同的汉字代表不同的数字，那么正确的是． $$\\begin{cases}年年\\times 岁岁 = 花相似 岁岁\\div 年年 = 人\\div 不同 \\end{cases}$$ ", "answer_option_list": [[{"aoVal": "A", "content": "年$$=2$$ "}], [{"aoVal": "B", "content": "岁$$=4$$ "}], [{"aoVal": "C", "content": "花$$=6$$ "}], [{"aoVal": "D", "content": "不$$=1$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["根据分析可得， ``年年''、``岁岁''十位数字和个位数字相同， 所以都是$$11$$的倍数， 即，年$$\\times $$岁$$\\times 121=$$花相似， 又因为不同的汉字代表不同的数字，所以``年 年''或``岁岁''不能是$$11$$，年$$\\times $$岁的积只能是$$6$$或$$8$$， 由于$$22\\times 33=726$$，有相同数字，不合要求， 所以，年$$\\times $$岁的积只能是$$8$$，$$8=2\\times 4$$， 岁岁$$\\div $$年年$$=$$人$$\\div $$不同， 所以``岁''$$ ~\\textless{} ~$$ ``年''， 所以，``年年''$$=44$$、``岁岁''$$=22$$，则$$44\\times 22=968$$， 所以，岁岁$$\\div $$年年$$=22\\div 44=\\frac{1}{2}$$， $$0\\sim 9$$十个数字还剩下$$0$$、$$1$$、$$3$$、$$5$$、$$7$$，那么人$$\\div $$不同$$=5\\div 10=\\frac{1}{2}$$； 即``人''$$=5$$，``不同''$$=10$$； 综上所述，``年''$$=4$$、``岁''$$=2$$、``花''$$=9$$、``相''$$=6$$、``似''$$=8$$、``人''$$=5$$、``不''$$=1$$、``同''$$=0$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "901", "queId": "ae1e79e9e157402b9e6f9e5f728ceed9", "competition_source_list": ["2015年全国美国数学大联盟杯小学高年级五年级竞赛初赛"], "difficulty": "0", "qtype": "single_choice", "problem": "两个连续的非负整数的和不可能是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "质数 "}], [{"aoVal": "B", "content": "完全平方数 "}], [{"aoVal": "C", "content": "$$13$$ 的倍数 "}], [{"aoVal": "D", "content": "偶数 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["两个连续的非负整数的和不可能是偶数 ． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1361", "queId": "316172bc10764f568d9033f2c54b4ef8", "competition_source_list": ["2014年世界少年奥林匹克数学竞赛六年级竞赛初赛B卷第10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "李建和张雷做仰卧起坐，李建先做了$$3$$分钟，然后两人各做了$$5$$分钟，一共做仰卧起坐$$136$$次．已知每分钟李建比张雷平均多做$$4$$次，那么李建比张雷多做了~\\uline{~~~~~~~~~~}~次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$56$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->变型题"], "answer_analysis": ["由题意，假设小建每分钟做仰卧起坐的次数与小雷一样多，这样两人做仰卧起坐的总次数就减少了$$4\\times \\left( 3+5 \\right)=32$$ （次），由此可知小雷每分钟做了$$\\left( 136-32 \\right)\\div \\left( 3+5+5 \\right)=8$$（次），进而可以分别求出小建每分钟做的次数以及两人分别做仰卧起坐的总次数之差；据此解答．解答此题关键是求得小建每分钟做的次数以及两人分别做仰卧起坐的总次数之差． 假设小建每分钟做仰卧起坐的次数与小雷一样多，这样两人做仰卧起坐的总次数就减少了$$4\\times \\left( 3+5 \\right)=32$$（次），由此可知小雷每分钟做了$$\\left( 136-32 \\right)\\div \\left( 3+5+5 \\right)=8$$（次），进而可以分别求出小建每分钟做的次数以及两人分别做仰卧起坐的总次数之差．假设小建每分钟做仰卧起坐的次数与小雷一样多，两人做仰卧起坐的总次数就减少：$$4\\times \\left( 3+5 \\right)=32$$（次）小雷每分钟做：$$\\left( 136-32 \\right)\\div \\left( 3+5+5 \\right)=8$$（次）；小建每分钟做：$$8+4=12$$（次），小建一共做：$$12\\times \\left( 3+5 \\right)=96$$（次）；小雷一共做：$$8\\times 5=40$$（次），小建比小雷多做：$$96-40=56$$（次）． 答：小建比小雷多做$$56$$次． 故答案为：$$56$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1048", "queId": "0abf5bfd5961457c8de69ea56f3bbbba", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛初赛第17题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "搬家公司要搬运$$100$$只花瓶，规定完整运送$$1$$只花瓶得$$3$$元，打破$$1$$只要赔偿$$2$$元．全部搬完后搬家公司共得$$260$$元，若假设他们完整运送了$$x$$只花瓶，则$$x$$表示只． ", "answer_option_list": [[{"aoVal": "A", "content": "$$92$$ "}], [{"aoVal": "B", "content": "$$90$$ "}], [{"aoVal": "C", "content": "$$88$$ "}], [{"aoVal": "D", "content": "$$80$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->运算求解"], "answer_analysis": ["他们完整运送了$$x$$只花瓶，则打破的有$$\\left( 100-x \\right)$$只， 根据题意可列方程：$$3x-2\\times \\left( 100-x \\right)=260$$， 解得$$x=92$$， 即完整运送了$$92$$只花瓶． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1652", "queId": "7ff5c890b8ca46f9ac0734b6218f2d7a", "competition_source_list": ["2020年长江杯五年级竞赛复赛B卷第18题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "在第七届长江杯数学邀请赛中，甲、乙、丙、丁四名学生的成绩如下：甲、乙、丙三人的平均分是$$92$$分，乙、丙、丁三人的平均分为$$90$$分，甲、丁两人的平均分是$$96$$分．问甲多少分？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$99$$ "}], [{"aoVal": "C", "content": "$$96$$ "}], [{"aoVal": "D", "content": "$$93$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["甲$$+$$乙$$+$$丙$$=92\\times 3=276$$（分）， 乙$$+$$丙$$+$$丁$$=90\\times 3=270$$（分）， 故甲$$-$$丁$$=276-270=6$$（分）， 甲$$+$$丁$$=96\\times 2=192$$（分）， 故甲有$$192\\div 2+6\\div 2=99$$（分）， 丁有$$192-99=93$$（分）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2952", "queId": "a4fa296916cf475da433c76dc622ce59", "competition_source_list": ["2017年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "小明家到学校，乘地铁需要$$30$$分钟，乘公交车需要$$50$$分钟。某天小明因故先乘地铁，再换乘公交车，用了$$40$$分钟到达学校，其中换乘过程用了$$6$$分钟，那么这天小明乘坐公交车用了（ ）分钟。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题"], "answer_analysis": ["设从家到学校距离共有$$[30$$，$$50]=150$$（份），那么地铁的速度是$$150\\div 30=5$$（份$$/$$分钟），公交车的速度是$$150\\div 50=3$$（份$$/$$分钟）。设这天小明乘公交用了$$x$$分钟，根据题意列出方程：$$5\\times \\left( 40-6-x \\right)+3x=150$$，解得$$x=10$$。因此小明乘公交用了$$10$$分钟。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1390", "queId": "2916d81df53a4a2fb663ee9eca2f5c3e", "competition_source_list": ["2017年IMAS小学中年级竞赛（第一轮）第11题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$(\\Delta \\div 2-2)\\times 2+2=222$$，请问$$\\Delta $$代表的数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$224$$ "}], [{"aoVal": "C", "content": "$$228$$ "}], [{"aoVal": "D", "content": "$$876$$ "}], [{"aoVal": "E", "content": "$$884$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\Delta \\div 2-2=(222-2)\\div 2$$ $$=220\\div 2=110$$， 因此$$\\Delta =(110+2)\\times 2=112\\times 2=224$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "72", "queId": "33a5905cb6a644da8ba9c0b10e7b08fb", "competition_source_list": ["2011年第7届全国新希望杯小学高年级六年级竞赛第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "盒子里装有分别写着$$1$$，$$2$$，$$3$$，$$4$$，\\ldots，$$100$$的黄色卡片各一张，我们称如下操作为一次操作；从盒子里取出$$m(7\\leqslant m\\leqslant 10)$$张卡片，算出这$$m$$张卡片上各数之和减去$$27$$的差，将其写在一张红色卡片上（不放回）．若干次操作之后，盒子里的卡片全部被取出，若所有红色卡片上的数字之和为$$n$$，那么$$n$$的最大可能值减去最小可能值等于． ", "answer_option_list": [[{"aoVal": "A", "content": "$$108$$ "}], [{"aoVal": "B", "content": "$$96$$ "}], [{"aoVal": "C", "content": "$$88$$ "}], [{"aoVal": "D", "content": "$$81$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->整体思想"], "answer_analysis": ["先算卡片的总值． 最小值$$100\\div 10=10$$次，共减$$10\\times 27$$； 最多$$14$$次，共减去$$14\\times 27$$， 因此差为$$(14-10)\\times 27=108$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2708", "queId": "494bea3026c245d59a119bc044a289b0", "competition_source_list": ["2015年全国AMC六年级竞赛8第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "在她工作的第一天，贾纳贝尔卖出了一个小部件．第二天，她卖了三个小部件．第三天，她卖出了五个小部件，接下来的每一天，她都比前一天多卖出两个小部件．工作$20$天后，贾纳贝尔总共卖出了多少个小部件？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$39$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$210$$ "}], [{"aoVal": "D", "content": "$$400$$ "}], [{"aoVal": "E", "content": "$$401$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列求和", "Overseas Competition->知识点->计算模块->数列与数表->等差数列"], "answer_analysis": ["翻译：贾娜贝尔第一天上班买了$$1$$个小装饰品，第二天买了$$3$$个装饰品，第三天卖出了$$5$$个，以后每一天比前一天多卖出$$2$$个装饰品，那么他工作了$$20$$天，一共卖出个装饰品？ $$1+3+5+\\cdots +\\left[ \\left( 20-1 \\right)\\times 2+1 \\right]$$ $$=1+3+5+\\cdots +39$$ $$=\\left( 1+39 \\right)\\times 20\\div 2$$ $$=400$$． The sum of $$1$$，$$3$$，$$5$$，$$\\cdots \\cdots $$$$39$$ is $$\\frac{\\left( 1+39 \\right)\\left( 20 \\right)}{2}=$$$$\\text{D}$$ $$400$$． The sum is just the sum of the first $$20$$ old integers， which is $${{20}^{2}}=$$$$\\text{D}$$ $$400$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "605", "queId": "105632d525094ee7a22b46016761382e", "competition_source_list": ["2015年全国美国数学大联盟杯小学高年级五年级竞赛初赛第24题"], "difficulty": "0", "qtype": "single_choice", "problem": "$$24$$ 和 $$30$$ 的最小公倍数减去$$24$$ 和$$30$$的最大公因数等于多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$114$$ "}], [{"aoVal": "B", "content": "$$117$$ "}], [{"aoVal": "C", "content": "$$234$$ "}], [{"aoVal": "D", "content": "$$237$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$24$$ 和 $$30$$ 的最小公倍数减去$$24$$ 和$$30$$的最大公因数等于多少？ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1265", "queId": "3df5ab98d1f649ce95a40915ebcee48c", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第13题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "一群学生排成长方形队形．小明所站的位置，从前面数是第$$4$$列，从后面数是第$$7$$列，从左边数是第$$3$$行，从右边数是第$$9$$行．请问这群学生共有多少名？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$90$$名 "}], [{"aoVal": "B", "content": "$$100$$名 "}], [{"aoVal": "C", "content": "$$110$$名 "}], [{"aoVal": "D", "content": "$$120$$名 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->方阵问题->实心方阵->实心方阵基本问题"], "answer_analysis": ["根据题意可知长方形队形中每列有$$3+9-1=11$$（名），每行有$$4+7-1=10$$（名），所以学生总数为$$11\\times 10=110$$（名）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "220", "queId": "b5364f861ce84edd86501127abe5282b", "competition_source_list": ["2015年华杯赛六年级竞赛初赛", "2015年华杯赛四年级竞赛初赛", "2015年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一只旧钟的分针和时针每重合一次，需要经过标准时间$$66$$分。那么，这只旧钟的$$24$$小时比标准时间的$$24$$小时（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "快$$12$$分 "}], [{"aoVal": "B", "content": "快$$6$$分 "}], [{"aoVal": "C", "content": "慢$$6$$分 "}], [{"aoVal": "D", "content": "慢$$12$$分 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->时间问题->时间计算"], "answer_analysis": ["正常时钟分针和时针重合一次需要$$\\frac{360}{360-30}\\times 60=\\frac{720}{11}$$（分），而旧钟需要$$66$$分，因此旧钟比正常时钟慢，且正常时钟和旧钟的时间比为$$\\frac{720}{11}:66=\\frac{720}{726}=\\frac{120}{121}$$，所以正常时钟走$$24$$小时，旧钟需要走$$24$$小时多$$12$$分钟，因此这只旧钟比标准时间慢$$12$$分钟。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "262", "queId": "7f3b608ce64c4431bc72ac96ef351109", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "盒中有形状、大小、质料相同的红、白、黑颜色的球各$$10$$个，摸出若干个，要保证摸出的球中至少有$$3$$个球同色，摸出球的个数至少为个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["最不利原则．要保证拿到一种颜色至少有$$3$$个，则根据最不利原则，可先取每种颜色$$2$$个，最后取一个不论取哪种颜色，都一定可以满足条件，即需要$$2\\times 3+1=7$$个． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1343", "queId": "70ac5311a30c4c629ced142180bf5d32", "competition_source_list": ["其它改编题", "2017年全国华杯赛小学中年级竞赛初赛模拟第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "西点店里卖的面包都是$$5$$个一袋或$$3$$个一袋的，不拆开零售．已知$$5$$个一袋的售价是$$10$$元，$$3$$个一袋的售价是$$7$$元．要给$$48$$位同学每人发$$1$$个面包最少要花~\\uline{~~~~~~~~~~}~元钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$ "}], [{"aoVal": "B", "content": "$$97$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$112$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->归总问题"], "answer_analysis": ["因为买$$15$$个面包时，$$3$$个大袋比$$5$$个小袋划算，所以大袋的单价比小袋的便宜，那么应该尽量多买大袋的． 若买$$9$$个大袋和$$1$$个小袋，$$9\\times 5+1\\times 3=48$$个，总价是$$10\\times 9+7\\times 1=97$$元． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2664", "queId": "883262a912a946c48c9bb81fc25283cf", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个三位小数，精确到��分位是$$10.0$$．这个三位小数最大是，最小是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10.045$$，$$10.001$$ "}], [{"aoVal": "B", "content": "$$10.049$$，$$10.001$$ "}], [{"aoVal": "C", "content": "$$10.049$$，$$9.950$$ "}], [{"aoVal": "D", "content": "$$10.049$$，$$9.999$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["一个三位小数，四舍五入到十分位得到$$10.0$$， 原三位小数通过``四舍''获得最大值，通过``五入''获得最小值． 那么通过四舍的原小数百分位数字可能是：$$1$$、$$2$$、$$3$$、$$4$$， 那么通过五入的原小数百分位数字可能是：$$5$$、$$6$$、$$7$$、$$8$$、$$9$$， 所以最大是$$10.049$$，最小是$$9.950$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2866", "queId": "a4195a30f9bb47a1880a339c03c23328", "competition_source_list": ["2020年新希望杯六年级竞赛（2月）第25题"], "difficulty": "1", "qtype": "single_choice", "problem": "比较大小：$$1+ \\frac{1}{2^{2}}+ \\frac{1}{3^{2}}+ \\frac{1}{4^{2}} \\cdots + \\frac{1}{2020^{2}}$$~\\uline{~~~~~~~~~~}~$$2$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\textgreater$$ "}], [{"aoVal": "B", "content": "$$\\textless{}$$ "}], [{"aoVal": "C", "content": "$$=$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}} ~\\textless{} ~\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{1999\\times 2020}$$ $$=1-\\frac{1}{2020}=\\frac{2019}{2020}$$． ∴$$1+\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}} ~\\textless{} ~1+\\frac{2019}{2020} ~\\textless{} ~2$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "493", "queId": "d3985ecb10cb4d3ba3828281e24f20f5", "competition_source_list": ["2014年全国学而思杯五年级竞赛第20题"], "difficulty": "4", "qtype": "single_choice", "problem": "\\hspace{0pt}老师在黑板上随机写了$$8$$个数，每个数都是$$1$$、$$2$$、$$4$$中的某一个．学生们每次擦去两个相同的数，并把这两个相同的数的和写在黑板上．如果某位同学在黑板上写出了``$$2048$$''，则过程``成功结束''，否则老师就再随机写一个数上去（$$1$$或$$2$$或$$4$$），以保证黑板上仍有$$8$$个数．学生每次成功写数都会得与此数相同的分数，例如：擦去两个$$2$$，写上$$4$$，得到$$4$$分．如果并没有写出$$2048$$，但已没有相同的数可以同时擦去，则过程``失败结束''．请问： (2)若一个过程结束后恰好得到了$$18000$$分，能否是一次``成功结束''？为什么？（请写出具体解题过程） ", "answer_option_list": [[{"aoVal": "A", "content": "能 "}], [{"aoVal": "B", "content": "不能 "}]], "knowledge_point_routes": ["知识标签->数学思想->逐步调整思想"], "answer_analysis": ["（2）同上题，出现了$$2048$$，至少要得如下的分数：$$2048+1024\\times 2+512\\times 4+256\\times 8+128\\times 16+64\\times 32+32\\times64+16\\times 128+8\\times 256$$$$=2048\\times 9=18432$$ 由于$$18000\\textless18432$$，因此$$18000$$分是不可能成功结束的． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "59", "queId": "6b1619ee88d644fcb1ff04491f8e0984", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（一）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "现在有$$21$$本故事书分给$$5$$个人阅读，如果每个人得到的数量均不相同，那么得到故事书最多的人至少可得到本． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->简单抽屉原理"], "answer_analysis": ["五人分别是$$2$$、$$3$$、$$4$$、$$5$$、$$7$$本．（不止一种情况．） "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1310", "queId": "1bcb6aa08fe841d58f612105a27c703f", "competition_source_list": ["2020年新希望杯二年级竞赛决赛（8月）第6题", "2020年新希望杯二年级竞赛初赛（个人战）第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "李叔叔排队买票，他前面有$$5$$人，后面有$$6$$人．一共有人排队买票． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->��想->对应思想"], "answer_analysis": ["加上李叔叔自己，一共：$$5+6+1=12$$（人）， 故选择：$$\\text{C}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2151", "queId": "6691fb20e97644e8a0d10d74d53adfa7", "competition_source_list": ["2011年第9届全国创新杯小学高年级六年级竞赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "钟表上$$12$$点$$15$$分，时针与分针夹角为（~ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "$$90{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$82.5{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$67.5{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$60{}^{}\\circ $$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->时钟问题->时钟上的追及问题->已知时间求角度"], "answer_analysis": ["$$12$$点时，夹角为$$0{}^{}\\circ $$；$$15$$分钟后，分钟往前走了$$90{}^{}\\circ $$，时针走了$$7.5{}^{}\\circ $$，此时夹角为$$82.5{}^{}\\circ $$. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2198", "queId": "9945729ee07e4988a210ec08ea6f85f0", "competition_source_list": ["2018年第23届华杯赛小学中年级竞赛初赛第6题", "2018年华杯赛小学中年级竞赛初赛第6题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$I$$型和$$II$$型电子玩具车各一辆，沿相同的两个圆形轨道跑动，$$I$$型每$$5$$分钟跑$$1$$圈，$$II$$型每$$3$$分钟跑$$1$$圈，某同一时刻，$$I$$型和$$II$$型恰好都开始跑第$$19$$圈，则$$I$$型比$$II$$型提前（~ ）分钟开始跑动． ", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$38$$ "}], [{"aoVal": "D", "content": "$$54$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->环形跑道->相遇与追及结合"], "answer_analysis": ["开始跑第$$19$$圈，说明跑了$$18$$圈： $$I$$型跑$$18$$圈用时：$$18\\times 5=90$$（分钟）； $$II$$型$$18$$圈用时：$$18\\times 3=54$$（分钟）； 故相差$$90-54=36$$（分钟），则提前$$36$$分钟． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2016", "queId": "e1474e433cfc4b51bb7372aed88089bd", "competition_source_list": ["2004年第2届创新杯五年级竞赛初赛第9题", "2004年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "兴农农机厂某车间共有61个工人，已知每个工人平均每天可加工甲种部件5个，或者乙种部件4个，或者丙种部件3个，但加工4个甲种部件，1个乙种部件和6个丙种部件才能配成一套，为了使加工出来的甲、乙、丙三种部件恰好都能配成套，那么，安排加工甲种部件的人数应是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "5人 "}], [{"aoVal": "B", "content": "12人 "}], [{"aoVal": "C", "content": "16人 "}], [{"aoVal": "D", "content": "20人 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->统筹规划->调运中的统筹"], "answer_analysis": ["称1个工人一天的工作量为1个工，则加工一个甲、乙、丙种部件分别需要$$\\frac{1}{5}，\\frac{1}{4}，\\frac{1}{3}$$个工，从而知道，每生产1套，甲部件需要$$\\frac{1}{5}\\times 4=0.8$$个工，乙部件需要$$\\frac{1}{4}\\times 1=0.25$$个工，丙部件需要$$\\frac{1}{3}\\times 6=2$$个工，因此，每天可生产$$61\\div \\left( 0.8+0.25+2 \\right)=20$$套.其中生产甲部件需要$$20\\times 0.8$$个工，即需要16个工人工作1天 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1559", "queId": "ff8080814502fa24014507b668e20b9e", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第14题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙两人合作打一份材料．开始甲每分钟打$$100$$ 个字，乙每分钟打$$200$$个字．合作到完成总量的一半时，甲速度变为原来的$$3$$倍，而乙休息了$$5$$分钟后继续按原速度打字．最后当材料完成时，甲、乙打字数相等．那么，这份材料共（~~~~~~~ ）个字． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3000$$ "}], [{"aoVal": "B", "content": "$$6000$$ "}], [{"aoVal": "C", "content": "$$12000$$ "}], [{"aoVal": "D", "content": "$$18000$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["前一半时乙的工作量是甲的$$2$$倍，所以后一半甲应是乙的$$2$$倍．把后一半工作量分为$$6$$份，甲应为$$4$$份，乙应为$$2$$份，说明乙休息时甲打了$$1$$份，这一份的量是$$100\\times 3\\times 5=1500$$字，故总工作量是$$1500\\times6\\times 2=18000$$字． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "860", "queId": "96b3a1f50ed94ff6ac17c3b77ca08fc2", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（五）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "十进制数$$25$$转换成二进制数是$_{2}$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11101$$ "}], [{"aoVal": "B", "content": "$$1011$$ "}], [{"aoVal": "C", "content": "$$10101$$ "}], [{"aoVal": "D", "content": "$$11001$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->数据处理"], "answer_analysis": ["$$25=16+8+1$$，则$$25$$转换成二进制数是$$11001$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "247", "queId": "682e9d91470d4461ab4cf4e4c9def511", "competition_source_list": ["2020年第24届YMO三年级竞赛决赛第8题3分", "2017年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$至$$10$$这$$10$$个整数中，至少取（ ）个数，才能保证其中有两个数的和等于$$10$$。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->简单抽屉原理"], "answer_analysis": ["先找到和为$$10$$的组合：$$(1$$，$$9)$$、$$(2$$，$$8)$$、$$(3$$，$$7)$$、$$(4$$，$$6)$$，剩下$$5$$、$$10$$，若前面四组数中各取一个，后面$$5$$、$$10$$全取，这样还不能满足题目要求的两数和为$$10$$，如果再取剩下四个中的任意一个，必定会凑出和为$$10$$的组合，故最少要取$$6+1=7$$（个）。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1163", "queId": "1d92185d72af4654be1d24986db1ccdf", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（四）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "（学而思拓展，难度★★☆☆☆） 水果店规定：如果购买芒果不超过$$10$$千克，那么每千克售价$$4$$元；如果超过$$10$$千克，那么超过的部分每千克$$3.5$$元．某人买了$$24$$千克芒果，他应付元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$84$$ "}], [{"aoVal": "B", "content": "$$89$$ "}], [{"aoVal": "C", "content": "$$90$$ "}], [{"aoVal": "D", "content": "$$96$$ "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["$$4\\times 10+3.5\\times (24-10)=89$$（元）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2193", "queId": "2819f7b026304ecf8f928918c48ef46c", "competition_source_list": ["其它改编题", "2018年第8届北京学而思综合能力诊断五年级竞赛年度教学质量监测第10题"], "difficulty": "3", "qtype": "single_choice", "problem": "一辆汽车从甲地出发到$$300$$千米外的乙地去，前$$120$$千米的平均速度为$$40$$千米/时，要想使这辆汽车从甲地到乙地的平均速度为$$50$$千米/时，剩下的路程应以~\\uline{~~~~~~~~~~}~千米/时行驶． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$45$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->行程模块->直线型行程问题->路程速度时间->单人变速问题"], "answer_analysis": ["剩下的路程为：$$300-120=180$$（千米），计划总时间为：$$300\\div 50=6$$（小时），前$$120$$千米已用去$$120\\div 40=3（$$小时），所以剩下路程的速度为： $$（300-120）\\div （6-3）=60$$（千米/时）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "551", "queId": "011a9db0d50b431ea6554b90892d290b", "competition_source_list": ["2008年陈省身杯小学高年级六年级竞赛第12题"], "difficulty": "3", "qtype": "single_choice", "problem": "$$ABCD$$都是小于$$100$$的合数，并且$$ABCD$$两两互质，则$$A+B+C+D$$的最大值为~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "答案请写在答题卡上 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数的认识->数的特征->质数与合数->质数的特征"], "answer_analysis": ["答案请写在答题卡上 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "315", "queId": "4a491ed0990841cca68e362e2c21d337", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛A卷第22��10分"], "difficulty": "2", "qtype": "single_choice", "problem": "在一次数学竞赛中，$$A$$，$$B$$，$$C$$，$$D$$，$$E$$五位同学分别得了前五名（没有并列同一名次的），关于各人的名次大家作出了下面的猜测： $$A$$说：``第二名是$$D$$，第三名是$$B$$.'' $$B$$说：``第二名是$$C$$，第四名是$$E$$.'' $$C$$说：``第一名是$$E$$，第五名是$$A$$.'' $$D$$说：``第三名是$$C$$，第四名是$$A$$.'' $$E$$说：``第二名是$$B$$，第五名是$$D$$.'' 结果每人都只猜对了一半，他们的名次从高到低排列是． ", "answer_option_list": [[{"aoVal": "A", "content": "E、$$C$$、$$B$$、$$A$$、D "}], [{"aoVal": "B", "content": "C、$$B$$、$$E$$、$$A$$、D "}], [{"aoVal": "C", "content": "A、$$D$$、$$B$$、$$C$$、E "}], [{"aoVal": "D", "content": "D、$$A$$、$$C$$、$$B$$、E "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->半真半假"], "answer_analysis": ["假设$$A$$猜的第一句是真的，那么$$B$$猜的第二句是真的，即第四名是$$E$$，那么$$C$$猜的``E是第一名''是错的，$$A$$是第五名，那么$$D$$猜的$$C$$是第三名是对的，那么$$B$$就是第一名，从而$$E$$说的全是错的，所以假设不成立．所以$$A$$猜的第二句是真的，即$$B$$是第三名，那么$$D$$猜的第一句是错的，从而$$A$$是第四名，所以$$C$$猜的第二句是错的，$$E$$是第一名，从而$$B$$猜的$$C$$是第二名是对的，$$E$$猜的第五名是$$\\text{D}$$正确，所以，第一名是$$E$$，第二名是$$C$$，第三名是$$B$$，第四名是$$A$$，第五名是$$D$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1838", "queId": "9bb012723ef44d16af7cf1c74ded7071", "competition_source_list": ["2007年六年级竞赛创新杯", "2007年第5届创新杯六年级竞赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙三人出同样多的钱买同样的笔记本，最后甲、乙都比丙多得3本，甲、乙都给了丙2.4元，那么每本笔记本的价格是( )元 ", "answer_option_list": [[{"aoVal": "A", "content": "0.8 "}], [{"aoVal": "B", "content": "1.2 "}], [{"aoVal": "C", "content": "2.4 "}], [{"aoVal": "D", "content": "4.8 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->设而不求"], "answer_analysis": ["设丙得了$$x$$本，则甲.乙二人各得了$$x+3$$本，总共$$3x+6$$本，平均每人应得$$x+2$$本，而甲比平均数多拿了1本，所以每本笔记本的价格是2.4元. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2257", "queId": "fb1a4c36116f4631a67c63acc4bd2416", "competition_source_list": ["2014年全国华杯赛小学高年级竞赛初赛A卷第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "某学校组织一次远足活动，计划$$10$$点$$10$$分从甲地出发，$$13$$点$$10$$分到达乙地，但出发晚了$$5$$分钟，却早到达了$$4$$分钟．甲乙两地之间的丙地恰好是按照计划时间到达的，那么到达丙地的时间是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$点$$40$$分 "}], [{"aoVal": "B", "content": "$$11$$点$$50$$分 "}], [{"aoVal": "C", "content": "$$12$$点 "}], [{"aoVal": "D", "content": "$$12$$点$$10$$分 "}]], "knowledge_point_routes": ["拓展思维->七大能力->运算求解"], "answer_analysis": ["相当于两辆车$$A$$，$$B$$在同一条路上，一辆按原速行驶，一辆按新的速度行驶．那么，晚出发五分钟相当于$$B$$追已经行驶了$$5$$分钟的$$A$$在丙地追上，当$$B$$到达乙地的时候$$A$$还有$$4$$分钟的路程，这样就是两次追击问题，速度差不变追击路程的比是$$5:4$$所以，（甲地到丙地的距离）:（丙地到乙地的距离）$$=5:4$$，所以时间比为$$5:4$$，所以选择$$B$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "359", "queId": "baeb6e4a2a01451f8bef26cc5f4b1fbf", "competition_source_list": ["2015年第27届广东广州五羊杯小学高年级竞赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在下面的括号中分别填入$$+$$、$$-$$、$$\\times $$、$$\\div $$符号，每个符号只能用一次，使得$$a$$，$$b$$，$$c$$，$$d$$之和为最大． $$\\frac{1}{2}( )\\frac{1}{9}=a$$，$$\\frac{1}{9}( )\\frac{1}{8}=b$$，$$\\frac{1}{4}( )\\frac{1}{7}=c$$，$$\\frac{1}{5}( )\\frac{1}{6}=d$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$+$$，$$-$$，$$\\times $$，$$\\div $$ "}], [{"aoVal": "B", "content": "$$\\div $$，$$-$$，$$+$$，$$\\times $$ "}], [{"aoVal": "C", "content": "$$\\div $$，$$-$$，$$\\times $$，$$+$$ "}], [{"aoVal": "D", "content": "$$\\times $$，$$\\div $$，$$=$$，$$+$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["四个式子中所含的$$8$$个分数都小于二分之一， 所以四个式子的和、差、积都小于$$1$$． 那么我们可以先考虑商最大的情况． $$\\frac{1}{2}\\div \\frac{1}{9}=a$$，$$\\frac{1}{3}-\\frac{1}{8}=b$$，$$\\frac{1}{4}+\\frac{1}{7}=c$$，$$\\frac{1}{5}\\times \\frac{1}{6}=d$$． 和为$$5\\frac{113}{840}$$，但不必求出． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "257", "queId": "ff5e7db85ab84d469da0787a3f72ecf6", "competition_source_list": ["2017年IMAS小学中年级竞赛（第二轮）第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "将算式$$1\\Delta 2\\Delta 3\\Delta 8\\Delta 15$$的$$\\Delta $$处分别填入「$$+$$」或「$$-$$」号后计算此算式的值． 请问总共可以得到多少个不同的正整数值？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$14$$ "}], [{"aoVal": "E", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["当$$15$$前面的符号是「$$-$$」号时， $$1+2+3+8=14\\textless{}15$$，故得不到正整数； 当$$15$$前面的符号是「$$+$$」号时，由于$$1$$前面没有空格，相当于有一个加号， 所以无论后面怎么填，这个算式的结果都是大于$$1$$的， 而$$2$$、$$3$$、$$8$$之间的差各不相等， 因此，这三个数前面任意一个符号不同时得到的结果都不相同， 因此总共可以得到$$2\\times2\\times2=8$$个不同的正整数值． 故选$$\\text{A}$$． ", "可知共有$$2\\times2\\times2\\times2=16$$种填入「$$+$$」或「$$-$$」号的方式：
\n①$$1+2+3+8+15=29$$；
\n②$$1+2+3+8-15=14-15<{}0$$，不为正整数；
\n③$$1+2+3-8+15=13$$；
\n④$$1+2-3+8+15=23$$；
\n⑤$$1-2+3+8+15=25$$；
\n⑥$$1+2+3-8-15=6-23<{}0$$，不为正整数；
\n⑦$$1+2-3+8-15=11-18<{}0$$，不为正整数；
\n⑧$$1-2+3+8-15=10-15<{}0$$，不为正整数；
\n⑨$$1+2-3-8+15=7$$；
\n⑩$$1-2+3-8+15=9$$；
\n⑪$$1-2-3+8+15=19$$；
\n⑫$$1+2-3-8-15=3-26<{}0$$，不为正整数；
\n⑬$$1-2+3-8-15=4-25<{}0$$，不为正整数；
\n⑭$$1-2-3+8-15=9-20<{}0$$，不为正整数；
\n⑮$$1-2-3-8+15=3$$；
\n⑯$$1-2-3-8-15=1-28<{}0$$，不为正整数．
\n其中总共有$$29$$、$$13$$、$$23$$、$$25$$、$$7$$、$$9$$、$$19$$、$$3$$共$$8$$个不同的正整数值．
\n故选$$\\text{A}$$．
"], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "717", "queId": "481bb4542186431988cd6b83f17d903f", "competition_source_list": ["2016年全国华杯赛小学高年级竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "在一个七位整数中，任何三个连续排列的数字都构成一个能被$$11$$或$$13$$整除的三位数，则这个七位数最大是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9981733$$ "}], [{"aoVal": "B", "content": "$$9884737$$ "}], [{"aoVal": "C", "content": "$$9978137$$ "}], [{"aoVal": "D", "content": "$$9871773$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["注意到由于任意三个连续排列的数字都能构成三位数，所以这个七位数的前五个数字不能是$$0$$，逐步极端分析，得$$988=13\\times 76$$，$$884=13\\times 68$$，$$847=11\\times 77$$，$$473=11\\times 43$$，$$737=11\\times 67$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "706", "queId": "b9bd72266e5046e1b05d8bf16f92d16d", "competition_source_list": ["2017年河南郑州K6联赛竞赛模拟第六套第2题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "一个两位数，十位上的数字是$$5$$，个位上的数字是$$a$$，表示这个两位数的式子是（ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50+a$$ "}], [{"aoVal": "B", "content": "$$5+a$$ "}], [{"aoVal": "C", "content": "$$50a$$ "}], [{"aoVal": "D", "content": "$$5a$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["十位上的数字是$$5$$，也就是表示$$5$$个十，即$$50$$；个位上的数字是$$a$$，也就是表示$$a$$个一，即$$a$$．所以这个两位数就是$$50+a$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2852", "queId": "6594c3aee99648ad921473a07d788115", "competition_source_list": ["2018年第8届北京学而思综合能力诊断二年级竞赛年度教学质量第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2018$$俄罗斯世界杯刚刚落下帷幕，其中世界杯英文单词``$$WORLD CUP$$''中不同字母代表 着大于$$1$$的不同数字，请问$$W+O+R+L+D+C+U+P=$$（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$43$$ "}], [{"aoVal": "B", "content": "$$44$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$54$$~ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["大于$$1$$的不同数字为：$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$，所以$$W+O+R+L+D+C+U+P=1+2+3+4+5+6+7+8+9=44$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "410", "queId": "8605aa3c6ca6486c841a9d8ff3430b62", "competition_source_list": ["2008年四年级竞赛创新杯", "2008年第6届创新杯四年级竞赛初赛B卷第10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$16$$个外表一样的球，有$$10$$克和$$9$$克两种重量．先取两个球，天平两边各放一个使天平不平衡．就拿这两个球作为标准，将余下的$$14$$个球分成$$7$$对，用天平与这对标准球逐一比较，结果是$$3$$对较重，$$2$$对较轻，$$2$$对与标准球一样重，那么这$$16$$个球的总重量是克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$134$$ "}], [{"aoVal": "B", "content": "$$135$$ "}], [{"aoVal": "C", "content": "$$143$$ "}], [{"aoVal": "D", "content": "$$153$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["第一次称的两个球，天平不平衡，所以$$1$$个球是$$10$$克，$$1$$个球是$$9$$克，作为标准的这对球重$$10+9=19$$（克）。比标准重的一对球重$$20$$克，比标准轻的一对球重$$18$$克，和标准重一样的一对球重$$19$$克，因此，这$$16$$个球的总重量为：$$20\\times 3+19\\times 3+18\\times 2=153$$（克）。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3216", "queId": "2c09d6cd2ff44cf08f5264af032a3701", "competition_source_list": ["2017年全国美国数学大联盟杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$n$$是一任意自然数．若将$$n$$的各位数字反向排列所得自然数$${n}_{1}$$与$$n$$相等，则称$$n$$为回文数．例如，若$$n=1234321$$，则称$$n$$为一回文数；但若$$n=1234567$$，则$$n$$不是回文数．请问在$$10000$$和$$100000$$之间有多少个回文数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$900$$ "}], [{"aoVal": "B", "content": "$$1000$$ "}], [{"aoVal": "C", "content": "$$1100$$ "}], [{"aoVal": "D", "content": "$$1200$$ "}], [{"aoVal": "E", "content": "$$1500$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["在$$10000$$和$$100000$$之间的数都是五位数，不妨设其为$$\\overline{abcba}$$，$$a$$不为$$0$$有$$9$$种选择，$$b$$、$$c$$均有$$10$$种选择，故共有$$9\\times 10\\times 10=900$$个回文数． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "569", "queId": "04399a8724da4e958a6ddd5ba9c8e2d2", "competition_source_list": ["2017年河南郑州小升初豫才杯第二场第11题", "2017年河南郑州豫才杯竞赛第11题"], "difficulty": "0", "qtype": "single_choice", "problem": "$$100$$以内，能同时被$$9$$和$$5$$整除的最大偶数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$75$$ "}], [{"aoVal": "B", "content": "$$85$$ "}], [{"aoVal": "C", "content": "$$90$$ "}], [{"aoVal": "D", "content": "$$95$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的认识"], "answer_analysis": ["能同时被$$9$$和$$5$$整除的数一定能被$$45$$整除，所以$$100$$以内只有$$90$$符合题意． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1905", "queId": "b78cd33347254d0988bbe586706cf656", "competition_source_list": ["2020年第23届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙丙丁四人参加了一次考试．甲乙的成绩和比丙丁的成绩和高$$17$$分，甲比乙低$$4$$分，丙比丁高$$5$$分．四人中最高分比最低分高分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->能力->符号代换->代数运算"], "answer_analysis": ["首先根据题意，可得乙比甲的成绩高，丙比丁的成绩高，然后根据题意，设乙得了$$x$$分，则甲得了$$x-4$$分，丙得了$$y$$分，则丁得了$$y-5$$分，再根据：甲、乙的成绩和$$-$$丙、丁的成绩和$$=17$$，求出$$x$$、$$y$$的关系，判断出四人中最高分、最低分各是多少，即可求出四人中最高分比最低分高多少． 设乙得了$$x$$分，则甲得了$$x-4$$分，丙得了$$y$$分，则丁得了$$y-5$$分， 所以$$(x+x-4)-(y+y-5)=17$$， 整理，可得：$$2x-2y+1=17$$， 所以$$2x-2y=16$$， 所以$$x-y=8$$， 所以乙比丙得分高； 因为$$x-y=8$$， 所以$$(x-4)-(y-5)=9$$， 所以甲比丁得分高， 所以乙得分最高，丁得分最低， 所以四人中最高分比最低分高： $$x-(y-5)=x-y+5=8+5=13$$（分）， 答：四人中最高分比最低分高$$13$$分． 故答案为：$$13$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1040", "queId": "45e0d9cb707045a3b50d6ba9c6025a60", "competition_source_list": ["2017年四川成都六年级竞赛“全能明星”选拔赛第7题2分"], "difficulty": "2", "qtype": "single_choice", "problem": "某学校合唱队与舞蹈队的人数之比为$$3:2$$，如果将合唱队队员调$$10$$人到舞蹈队，则人数之比为$$7:8$$．合唱队原有． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$人 "}], [{"aoVal": "B", "content": "$$48$$人 "}], [{"aoVal": "C", "content": "$$44$$人 "}], [{"aoVal": "D", "content": "$$45$$人 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["根据合唱队与舞蹈队调人前后人数之比可知，调出$$10$$人前，合唱队人数占全体人数的$$\\frac{3}{3+2}$$，调出$$10$$人后，占全体人数的$$\\frac{7}{7+8}$$，则全体人数为$$10\\div \\left( \\frac{3}{3+2}-\\frac{7}{7+8} \\right)$$人，求出全体人数后，就能根据合唱队人数在全体人数中的占比求出合唱队原来有多少人．列式计算为$$\\left[ 10\\div \\left( \\frac{3}{3+2}-\\frac{7}{7+8} \\right) \\right]\\times \\frac{3}{3+2}=45$$（人）. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2406", "queId": "0fa72b15204a4b69b3f40249f2c0a809", "competition_source_list": ["2018年第8届北京学而思综合能力诊断二年级竞赛年度教学质量第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "算式$$\\textasciitilde7+8\\times 9$$的正确结果是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$65$$ "}], [{"aoVal": "B", "content": "$$72$$ "}], [{"aoVal": "C", "content": "$$79$$ "}], [{"aoVal": "D", "content": "$$135$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->整数->四则混合运算"], "answer_analysis": ["先算乘除，再算加减，$$7+8\\times 9=7+72=79$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2733", "queId": "ff8080814502fa2401450bdf56e11785", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个最大的三位数除以一个整数，得到的商四舍五入保留一位小数后是$$2.5$$，除数最小是（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$400$$ "}], [{"aoVal": "B", "content": "$$396$$ "}], [{"aoVal": "C", "content": "$$392$$ "}], [{"aoVal": "D", "content": "$$388$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["要使得除数最小，那么商就尽可能的大，因此商无限接近于$$2.54\\cdots $$；$$999$$除以$$2.54$$符合条件的结果是$$392$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2063", "queId": "f8d61a430fae4b4b94a15ad2926fb62c", "competition_source_list": ["2017年河南郑州东风杯竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "一家商店将某种服装按成本价提高$$40 \\% $$后标价，又以$$8$$折（即按标价的$$80 \\% $$）优惠卖出，结果每件服装仍可获利$$15$$元，则这种服装每件的成本是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$元 "}], [{"aoVal": "B", "content": "$$125$$元 "}], [{"aoVal": "C", "content": "$$135$$元~~~~~ "}], [{"aoVal": "D", "content": "$$140$$元 "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设这种服装每件的成本是$$x$$元．则有$$\\left( 1+40 \\% ~\\right)x\\times 80 \\%=x+15$$，解得$$x=125$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2027", "queId": "b44f55e4c069493ab535accc2642a889", "competition_source_list": ["走美杯三年级竞赛", "走美杯四年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "幼儿园买了$$8$$辆玩具车，每辆玩具车需要$$92$$元，李老师带了$$720$$元够吗？下面的解答比较合理的是． ", "answer_option_list": [[{"aoVal": "A", "content": "因为$$92\\times 8\\approx$$ 720（元），$$92\\times$$ 8\\textgreater720，所以$$92\\times$$ 8\\textgreater720，够 "}], [{"aoVal": "B", "content": "因为$$92\\times 8\\approx$$ 800（元），$$92\\times$$ 8\\textless800，所以$$92\\times$$ 8\\textless720，够 "}], [{"aoVal": "C", "content": "因为$$92\\times 8\\approx$$ 720（元），$$92\\times$$ 8\\textgreater720，所以$$92\\times$$ 8\\textgreater720，不够 "}], [{"aoVal": "D", "content": "因为$$92\\times 8\\approx$$ 800（元），$$92\\times$$ 8\\textgreater800，所以$$92\\times$$ 8\\textgreater720，不够 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型"], "answer_analysis": ["92\\textgreater90，总价大于$$720$$元，所以不够 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1366", "queId": "fa86a48716fa4139aba8d18e03a1a18b", "competition_source_list": ["2013年华杯赛五年级竞赛初赛", "2020年第24届YMO五年级竞赛决赛第8题3分", "2013年华杯赛六年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "一个盒子里有黑棋子和白棋子若干粒，若取出一粒黑子，则余下的黑子数与白子数之比为$$9:7$$，若放回黑子，再取出一粒白子，则余下的黑子数与白子数之比为$$7:5$$，那么盒子里原有的黑子数比白子数多（ ）个。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题->按比分配"], "answer_analysis": ["取出一粒黑子后，盒子内棋子的总数与取出一粒白子后，盒子内棋子的总数相等，$$\\left[ 9+7,7+5 \\right]=\\left[ 16,12 \\right]=48$$，因此，取出一粒黑子后，黑白子之比为$$9:7=27:21$$,取出一粒白子后，黑白子之比为$$7:5=28:20$$，黑子增加了$$1$$粒，从$$27$$份增加到$$28$$份，因此$$1$$份就是$$1$$粒，于是，盒子里原有$$28$$粒黑子、$$21$$粒白子，一开始黑子比白子多$$7$$个。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1962", "queId": "ceaafad7c61d4c54a4a30b56a0fa4122", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "毛毛用围棋子摆成一个三层空心方阵，最内一层每边有围棋子$$8$$个．毛毛摆这个方阵共用围棋子个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$ "}], [{"aoVal": "B", "content": "$$108$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$132$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->方阵问题->空心方阵->空心方阵的增减"], "answer_analysis": ["$$1$$、这是一道关于方阵的问题，考查的是方阵的知识； $$2$$、由题意知，用围棋子摆成一个三层空心方阵，最内一层每边有围棋子$$8$$个，由于相邻两层每边相差$$2$$个，则由里向外的两层每边分别是$$\\left( 8+2 \\right)$$个、$$(8+2\\times 2)$$个； $$3$$、根据``四周的个数$$=\\left( {} \\right.$$每边的个数$$\\left. -1 \\right)\\times 4$$''可分别求得这三层棋子的个数，再相加就是所用的总个数，据此解答． 根据题意分析可知：最内一层棋子个数为：$$\\left( 8-1 \\right)\\times 4=7\\times 4=28$$（个）， 第二层棋子有：$$\\left( 8+2-1 \\right)\\times 4=9\\times 4=36$$（个）， 第三层棋子有：$$\\left( 8+2\\times 2-1 \\right)\\times 4=11\\times 4=44$$（个）， 所以三层一共有$$28+36+44=108$$（个）． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2930", "queId": "81c25d38ba2447f988dbf2b56cc1d18f", "competition_source_list": ["2018年湖北武汉新希望杯小学高年级五年级竞赛训练题（三）第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "有数组{$$1$$，$$2$$，$$3$$}，{$$2$$，$$4$$，$$6$$}，{$$3$$，$$6$$，$$9$$}，$$\\cdots \\cdots $$根据规律，第$$100$$个数组的三个数的和是（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$600$$ "}], [{"aoVal": "B", "content": "$$603$$ "}], [{"aoVal": "C", "content": "$$606$$ "}], [{"aoVal": "D", "content": "$$609$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["第$$100$$个数组是{$$100$$，$$200$$，$$300$$}，$$100+200+300=600$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "291", "queId": "ff8080814518d52401451925eda00502", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第15题"], "difficulty": "3", "qtype": "single_choice", "problem": "老师把某两位数的六个不同因数分别告诉了$$A\\sim F$$六个聪明诚实的同学． $$A$$和$$B$$同时说：我知道这个数是多少了． $$C$$和$$D$$同时说：听了他们的话，我也知道这个数是多少了． $$E$$：听了他们的话，我知道我的数一定比$$F$$的大． $$F$$：我拿的数的大小在$$C$$和$$D$$之间． 那么六个人拿的数之和是（ ~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$141$$ "}], [{"aoVal": "B", "content": "$$152$$ "}], [{"aoVal": "C", "content": "$$171$$ "}], [{"aoVal": "D", "content": "$$175$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["（1）这个数的因数个数肯定不低于$$6$$个（假定这个数为$$N$$，且拿到的$$6$$个数从大到小分别是$$ABCDEF$$~） （2）有两个人同时第一时间知道结果，这说明以下几个问题： 第一种情况：有一个人知道了最后的结果，这个结果是怎么知道的呢？很简单，他拿到的因数在$$50\\sim 99$$之间（也就是说$$A$$ 的$$2$$倍是$$3$$位数，所以$$A$$其实就是$$N$$） 第二种情况：有一个人拿到的不是最后结果，但是具备以下条件： 这个数的约数少于$$6$$个，比如：有人拿到$$36$$，单他不能断定$$N$$究竟是$$36$$还是$$72$$． 这个数小于$$50$$，不然这个数就只能也是$$N$$了． 这个数大于$$33$$，比如：有人拿到$$29$$，那么他不能断定$$N$$ 是$$58$$还是$$87$$；(这里有个特例是$$27$$，因为$$27\\times 2=54$$，因数个数不少于$$6$$个；$$27\\times 3=81$$，因数个数少于$$6$$个，所以如果拿到$$27$$可以判断$$N$$只能为$$54$$） 这个数还不能是是质数，不然不存在含有这个因数的两位数． 最关键的是，这两人的数是$$2$$倍关系 但是上述内容并不完全正确，需要注意还有一些``奇葩''数：$$17$$、$$19$$、$$23$$也能顺利通过第一轮． 因此，这两个人拿到的数有如下可能： （$$54$$，$$27$$）（$$68$$，$$34$$）（$$70$$，$$35$$）（$$76$$，$$38$$）（$$78$$，$$39$$）（$$92$$，$$46$$）（$$98$$，$$49$$） （3）为了对比清晰，我们再来把上面所有的情况的因数都列举出来： （$$54$$，$$27$$，$$18$$，$$9$$，$$6$$，$$3$$，$$2$$，$$1$$） （$$68$$，$$34$$，$$17$$，$$4$$，$$2$$，$$1$$）（$$\\times$$） （$$70$$，$$35$$，$$14$$，$$10$$，$$7$$，$$5$$，$$2$$，$$1$$） （$$76$$，$$38$$，$$19$$，$$4$$，$$2$$，$$1$$）（$$\\times$$） （$$78$$，$$39$$，$$26$$，$$13$$，$$6$$，$$3$$，$$2$$，$$1$$） （$$92$$，$$46$$，$$23$$，$$4$$，$$2$$，$$1$$）（$$\\times$$） （$$98$$，$$49$$，$$14$$，$$7$$，$$2$$，$$1$$） 对于第一轮通过的数，我们用红色标注，所以$$N$$不能是$$68$$、$$76$$、$$92$$中的任意一个． 之后在考虑第二轮需要通过的两个数． 用紫色标注的$$6$$、$$3$$、$$2$$、$$1$$，因为重复使用，如果出现了也不能判断$$N$$是多少，所以不能作为第二轮通过的数． 用绿色标注的$$14$$和$$7$$也不能作为第二轮通过的数，这样$$N$$也不是$$98$$． 那么通过第二轮的数只有黑色的数． 所以$$N$$ 只能是$$54$$、$$70$$、$$78$$中的一个． 我们再来观察可能满足$$E$$和$$F$$所说的内容： （$$54$$，$$27$$，$$18$$，$$9$$，$$6$$，$$3$$，$$2$$，$$1$$） （$$70$$，$$35$$，$$14$$，$$10$$，$$7$$，$$5$$，$$2$$，$$1$$） （$$78$$，$$39$$，$$26$$，$$13$$，$$6$$，$$3$$，$$2$$，$$1$$） 因为$$F$$说他的数在$$C$$和$$D$$之间，我们发现上面的数据只有当$$N=70$$的时候，$$F=7$$，在$$CD$$$$(10$$和$$5)$$之间，是唯一满足条件的一种情况． 又因为$$E$$ 确定自己比$$F$$的大，那么他拿到的数一定是该组中剩余数里最大的．所以$$E$$拿到的是$$14$$（$$N=70$$~）． 所以$$N=70$$，六个人拿的数之和为：$$70+35+14+10+7+5=141$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "476", "queId": "b80ea67c12bb4ae4bd7d761822b5ea8b", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（一）"], "difficulty": "1", "qtype": "single_choice", "problem": "警察厅坐着的$$5$$位偷盗嫌疑人正在接受警察的盘问，其中只有一名是真正小偷，在这$$5$$人的供述中，只有$$3$$句是正确的，请问小偷是~\\uline{~~~~~~~~~~}~． A说：丁是小偷； B说：我不是小偷，我是无辜的； C说：我非常肯定戊不是小偷； D说：甲在撒谎； E说：乙说的是实话． ", "answer_option_list": [[{"aoVal": "A", "content": "$$A$$ "}], [{"aoVal": "B", "content": "$$B$$ "}], [{"aoVal": "C", "content": "$$C$$ "}], [{"aoVal": "D", "content": "$$D$$ "}], [{"aoVal": "E", "content": "$$E$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找一致（同伙）"], "answer_analysis": ["简化这$$5$$人的供述为： 甲：是丁； 乙：不是乙； 丙：不是戊； 丁：不是丁； 戊：不是乙． 其中乙戊的判断是一样的，可能都对或都都错，经过分析知道乙戊是正确的，那么甲丙丁三人中还有一人是正确的，依次假设推理得甲错丙错丁正确，那么丙的供述''不是戊''就是错误的，小偷是戊． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "960", "queId": "cba7b6f810d64771b91a7f338967e4e8", "competition_source_list": ["2010年五年级竞赛明心奥数挑战赛"], "difficulty": "1", "qtype": "single_choice", "problem": "因为$$2003$$是一个质数，所以$$2003$$年是一个质数年。在$$2003$$年以后的十年中还有一个质数年，这个质数年的年份是下列选项中的~\\uline{~~~~~~~~}~。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$2005$$ "}], [{"aoVal": "B", "content": "$$2007$$ "}], [{"aoVal": "C", "content": "$$2009$$ "}], [{"aoVal": "D", "content": "$$2011$$ "}], [{"aoVal": "E", "content": "$$2013$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数->质数与合数判定"], "answer_analysis": ["$$2011$$是质数，所以选D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3351", "queId": "72cac4c3bd424244af3f6a375eff8b94", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "十位数字和个位数字相加的和是$$12$$，这样的两位数一共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["两位数的十位数字与个位数字之和为$$12$$， 而数字均为$$0\\sim 9$$，故枚举得： $$93$$，$$84$$，$$75$$，$$66$$，$$57$$，$$48$$，$$39$$共$$7$$个． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3027", "queId": "f7aef31ff44c4f63aa0a52a73727cef9", "competition_source_list": ["2004年第2届创新杯六年级竞赛复赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "在下面四个算式中，得数最大的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( \\frac{1}{17}+\\frac{1}{19} \\right)\\times 20$$ "}], [{"aoVal": "B", "content": "$$\\left( \\frac{1}{24}+\\frac{1}{29} \\right)\\times30$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{31}+\\frac{1}{37} \\right)\\times40$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{1}{41}+\\frac{1}{47} \\right)\\times50$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->裂项与通项归纳->分数裂项->分数裂差->两分数间接裂差"], "answer_analysis": ["$$\\left( \\frac{1}{17}+\\frac{1}{19} \\right)\\times 20=(17+3)\\times\\frac{1}{17}+(19+1)\\times\\frac{1}{19}=2+\\frac{3}{17}+\\frac{1}{19}=2+\\frac{9}{51}+\\frac{3}{57}$$ $$\\left( \\frac{1}{24}+\\frac{1}{29} \\right)\\times30=(24+6)\\times\\frac{1}{24}+(29+1)\\times\\frac{1}{29}=2+\\frac{1}{4}+\\frac{1}{29}=2+\\frac{9}{36}+\\frac{3}{87}$$ $$\\left( \\frac{1}{31}+\\frac{1}{37} \\right)\\times40=(31+9)\\times\\frac{1}{31}+(37+3)\\times\\frac{1}{37}=2+\\frac{9}{31}+\\frac{3}{37}$$ $$\\left( \\frac{1}{41}+\\frac{1}{47} \\right)\\times50=(41+9)\\times \\frac{1}{41}+(47+3)\\times \\frac{1}{47}=2+ \\frac{9}{41}+ \\frac{3}{47}$$ 分子相同，分母越小，分数越大。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "392", "queId": "a4796eace85240b4b0dfc9d8e2d0b497", "competition_source_list": ["2013年华杯赛三年级竞赛决赛", "2013年华杯赛六年级竞赛决赛"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$7$$中选择若干个不同的数，使得其中偶数之和等于奇数之和，则符合条件的选法共有（ ）种。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->和一定型最值问题->多数之积的最值->拆分数的数目确定"], "answer_analysis": ["偶数：$$2$$、$$4$$、$$6$$；奇数$$1$$、$$3$$、$$5$$、$$7$$；偶数少，所以看偶数和； 偶数和有：$$2$$（舍掉）、$$4$$、$$6$$、$$8$$、$$10$$、$$12$$； $$4$$：$$4=1+3$$，$$1$$组 $$6$$：$$6=2+4=1+5$$，$$2$$组 $$8$$：$$2+6=1+7=3+5$$，$$2$$组 $$10$$：$$4+6=3+7$$，$$1$$组 $$12$$：$$2+4+6=5+7$$，$$1$$组 所以共有$$1+2+2+1+1=7$$（组）。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "741", "queId": "4d5953eb740a4e4fbf02ae88d3ccbd2b", "competition_source_list": ["2014年希望杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$20140316\\div 5$$的余数是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法"], "answer_analysis": ["$$20140316\\div 5=4028063\\cdots\\cdots 1$$； 答：余数是$$1$$． 故答案为：A． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1109", "queId": "1ceac561ad4f44348a1769657239f7ad", "competition_source_list": ["2015年第13届全国创新杯小学高年级五年级竞赛复赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "三部同样的抽水机同时抽水，抽干一池水需用$$15$$小时，五部这样的抽水机抽干这一池子水需用小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$12$$ "}], [{"aoVal": "E", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->简单工程问题->基本工程问题", "Overseas Competition->知识点->应用题模块->工程问题"], "answer_analysis": ["一台功效$$\\frac{1}{45}$$，五台需要$$9$$小时． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "814", "queId": "6973a2eda3424d82b6da253b09844581", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "定义$$n!=1\\times 2\\times 3\\times \\cdots \\times \\left( n-1 \\right)\\times n$$，在$$1!$$、$$2!$$、$$3!\\cdots32!$$中划去一个，剩余的数的乘积是一个完全平方数，划掉的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14!$$ "}], [{"aoVal": "B", "content": "$$15!$$ "}], [{"aoVal": "C", "content": "$$16!$$ "}], [{"aoVal": "D", "content": "$$15!$$或$$16!$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->完全平方数->平方数的综合应用"], "answer_analysis": ["$$1!\\times 2!\\times 3!\\times \\cdots 32!={{1}^{32}}\\times {{2}^{31}}\\times {{3}^{30}}\\times {{4}^{29}}\\times \\cdots \\times {{32}^{1}}$$， 则所有奇数均为偶次方， 偶数均为奇次方， 则$${{2}^{31}}\\times {{4}^{29}}\\times {{6}^{27}}\\times \\cdots {{32}^{1}}$$均去掉对应的最大偶数次方． 变成$$2\\times 4\\times 6\\times \\cdots \\times 32={{2}^{16}}\\times \\left( 1\\times 2\\times 3\\times \\cdots \\times 16 \\right)={{2}^{16}}\\times 16!={{2}^{16}}\\times 15!\\times16$$， 故去掉$$15!$$或$$16!$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "364", "queId": "c41f6b0df3674fc991fbf759bc1ca242", "competition_source_list": ["2014年世界少年奥林匹克数学竞赛六年级竞赛初赛B卷第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "新学期开学，凯奥斯问艾迪坐在教室里的位置．艾迪神秘地说：``我们班有$$56$$人，从左向右数共$$8$$列，从前向后数共$$7$$行，每两列并成一组．由于我的个子比较高，我的位置可以用一种特殊的符号$$(3,6)$$表示 .''艾迪又说：``我的同桌前面隔一个人是薇儿 .''那么，薇儿的位置表示为~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "（2，4） "}], [{"aoVal": "B", "content": "（2，6） "}], [{"aoVal": "C", "content": "（4，4） "}], [{"aoVal": "D", "content": "（4，6） "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->方向与坐标->坐标"], "answer_analysis": ["（4，4） "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "739", "queId": "5641119262844a56b6472cdc7aa3d8ba", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级三年级竞赛初赛第37题"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$1$$到$$20$$的$$20$$个整数，每个整数除以$$6$$，所得的余数的总和是��少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$47$$ "}], [{"aoVal": "C", "content": "$$46$$ "}], [{"aoVal": "D", "content": "$$45$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$1$ $5$$除以$$6$$的余数分别是$$1$$，$$2$$，$$3$$，$$4$$，$$5$$； $$7$ $11$$除以$$6$$的余数分别是$$1$$，$$2$$，$$3$$，$$4$$，$$5$$； $$13$ $17$$除以$$6$$的余数分别是$$1$$，$$2$$，$$3$$，$$4$$，$$5$$； $$19$$，$$20$$除以$$6$$的余数分别是$$1$$，$$2$$； 所以余数和为$$\\left(1+2+3+4+5\\right)\\times 3+3=48$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "10", "queId": "37d96d620c744685be1ab31930c66bfd", "competition_source_list": ["2003年第1届创新杯五年级竞赛复赛第9题", "2003年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "有一组自然数(数可以重复)，其中包含2003，但不包含数0，这组自然数的平均数是572，如果把2003去掉，那么剩下的平均数就变为413，这组数中出现的数最大可以是（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "2003 "}], [{"aoVal": "B", "content": "3708 "}], [{"aoVal": "C", "content": "3709 "}], [{"aoVal": "D", "content": "3717 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->竖式数字谜->竖式数字谜的最值"], "answer_analysis": ["设这组数中共有$$n$$个数，则$$572n-2003=413\\times \\left( n-1 \\right)$$，所以$$n=10$$，去掉2003 后，九个数的和为$$572\\times 10-2003=3717$$，故九个数中的八个数为1时，另一个数的值最大，故最大的数就可以是$$3717-8=3709$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1872", "queId": "8e9c99e537f54151ae65931c0cf68ec8", "competition_source_list": ["2004年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "某商场的营业额2000年和2001年连续两年平均每年比上一年上升10\\%，而2002年和2003年连续两年平均每年比上一年下降10\\%．那么2003年的营业额与1999年的营业额相比较，( ) ", "answer_option_list": [[{"aoVal": "A", "content": "下降了$$2 \\%$$ "}], [{"aoVal": "B", "content": "下降了$$1.99 \\%$$ "}], [{"aoVal": "C", "content": "上升了$$2 \\%$$ "}], [{"aoVal": "D", "content": "没有变化 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["$$\\left( 1+10 \\% \\right)\\times \\left( 1+10 \\% \\right)\\times \\left( 1-10 \\% \\right)\\times \\left( 1-10 \\% \\right)=98.01 \\%$$，下降$$1.99 \\%$$，选B "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2475", "queId": "b9512ca7bc10490c80d04a262938f531", "competition_source_list": ["2011年全国世奥赛五年级竞赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "在循环小数$$9.617628\\dot{1}$$的某一位上再添上一个循环点，使所产生的循环小数尽可能大，新的循环小数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9.617\\dot{6}28\\dot{1}$$ "}], [{"aoVal": "B", "content": "$$9.61\\dot{7}628\\dot{1}$$ "}], [{"aoVal": "C", "content": "$$9.61762\\dot{8}\\dot{1}$$ "}], [{"aoVal": "D", "content": "$$9.\\dot{6}17628\\dot{1}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->循环小数->循环小数的概念"], "answer_analysis": ["无论在何处添循环节，前八位不变，为$$9.6176281$$. 从下一位开始最大即可，因此在$$8$$上方添循环点． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2154", "queId": "d071d7d80b404df3882e935fceb1653d", "competition_source_list": ["2005年第3届创新杯五年级竞赛初赛第6题", "2005年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "在12时50分，时钟的分针与时针所形成的夹角的大小为( ). ", "answer_option_list": [[{"aoVal": "A", "content": "$${{85}^{\\circ }}$$ "}], [{"aoVal": "B", "content": "$${{90}^{\\circ }}$$ "}], [{"aoVal": "C", "content": "$${{105}^{\\circ }}$$ "}], [{"aoVal": "D", "content": "$${{115}^{\\circ }}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->时钟问题->时钟上的追及问题->已知时间求角度"], "answer_analysis": ["时针的速度为每分钟$${{0.5}^{\\circ }}$$，分针的速度为每分钟$${{6}^{\\circ }}$$.在12点整时，时针和分针重合，那么50分钟后时针和分针之间顺时针的路程差为$$50\\times \\left( {{6}^{\\circ }}-{{0.5}^{\\circ }} \\right)={{275}^{\\circ }}$$，即此时时针和分针的夹角为$${{360}^{\\circ }}-{{275}^{\\circ }}={{85}^{\\circ }}$$，选A. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1339", "queId": "24554023339d463198a753d1251002b6", "competition_source_list": ["2014年广东广州羊排赛六年级竞赛第6题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个三角形的三个内角度数比为$$2:4:3$$，这个三角形是． ", "answer_option_list": [[{"aoVal": "A", "content": "钝角三角形 "}], [{"aoVal": "B", "content": "锐角三角形 "}], [{"aoVal": "C", "content": "直角三角形 "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题->按比分配"], "answer_analysis": ["三角形内角和为$$180$$度，这个三角形最大的角为$$180\\times \\frac{4}{2+4+3}=80$$（度），为锐角三角形． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "160", "queId": "2c34d35e6e08400b987861cf258bd011", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级五年级竞赛邀请赛训练题（二）"], "difficulty": "1", "qtype": "single_choice", "problem": "天气炎热，四$$\\left( 1 \\right)$$班的班长小明要去给全班$$23$$人（包括自己）买饮料，商店老板告诉他现在只有可乐，雪碧和王老吉三种饮料了，那么其中买得最多的饮料于少需要买（~ ）瓶． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["$$23÷3=7\\ldots \\ldots 2$$，$$7+1=8$$（瓶）. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1068", "queId": "0861a7b3cc2541d6a12b22d91afa656e", "competition_source_list": ["2011年第9届全国创新杯小学高年级六年级竞赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "某人年初买了一种股票，该股票当年下跌了$$20 \\%$$，第二年应上涨（~ ）$$ \\%$$才能保持原值． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$35$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->求分率"], "answer_analysis": ["$$1\\div \\left( 1-20 \\% \\right)-1=25 \\%$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1326", "queId": "f5d6ce75f0f2469399c08396f8d1e7e2", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "两人共有钱$$200$$元，如果甲借给乙$$30$$元，那么甲、乙两人的钱数相等．那么甲原来有元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$110$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$130$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["从题干可以知道，两人共有钱$$200$$元，如果甲借给乙$$30$$元，那么甲、乙两人的钱数相等； 即此时每个人有钱：$$200\\div 2=100$$（元）， 那么甲原来有：$$100+30=130$$（元）， 那么乙原来有：$$100-30=70$$（元）． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1304", "queId": "505af15f476d4d269f3ba1da895838c9", "competition_source_list": ["2016年IMAS小学高年级竞赛第一轮检测试题第18题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "食米分三种包装出售：$$5\\operatorname{kg}$$装的，每包售价$$48$$元；$$10\\operatorname{kg}$$装的，每包售价为$$92$$元；$$25\\operatorname{kg}$$装的，每包售价为$$210$$元．若要使得每$$\\operatorname{kg}$$食米的平均售价恰好为$$9$$元，请问至少需购买食米多少包？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}], [{"aoVal": "E", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->多元一次方程解应用题->小数系数方程（组）解题"], "answer_analysis": ["设$$5\\text{kg}$$装的$$a$$包，$$10\\text{kg}$$装的$$b$$包，$$25\\text{kg}$$装的$$c$$包，使平均价格为$$9$$元，可列方程：$$4.8a+92b+210c=9(5a+10b+25c)$$，化简$$3a+2b=15c$$，讨论，当$$c=1$$时，有$$a=3$$，$$b=3$$；$$a=5$$，$$b=0$$；$$a=1$$，$$b=6$$，要最少包，则$$a=5$$，$$b=0$$，$$c=1$$，共$$6$$包． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2080", "queId": "cb87d0a590204e9b9b264574e8b5fe63", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（二）"], "difficulty": "0", "qtype": "single_choice", "problem": "一个数，它减去$$2$$，然后除以$$2$$，再加上$$2$$，最后乘$$2$$，得到的结果是$$2014$$．这个数原来是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2011$$ "}], [{"aoVal": "B", "content": "$$2012$$ "}], [{"aoVal": "C", "content": "$$2013$$ "}], [{"aoVal": "D", "content": "$$2014$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$( 2014\\div 2-2 )\\times 2+2=2012$$. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "344", "queId": "8962c5fdc79e481b9a96ae25e6a56a7e", "competition_source_list": ["2014年第26届广东广州五羊杯六年级竞赛第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙三人在讨论他们的数学成绩，丙给甲和乙看了他自己的数学成绩，但是甲、乙没给任何人看他俩的数学成绩．乙说：``我不是我们班的最低分''，甲补充说：``我不是最高分''．这三名学生成绩从高到低的顺序是． ", "answer_option_list": [[{"aoVal": "A", "content": "丙乙甲 "}], [{"aoVal": "B", "content": "丙甲乙 "}], [{"aoVal": "C", "content": "乙甲丙 "}], [{"aoVal": "D", "content": "乙丙甲 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->比较型逻辑推理"], "answer_analysis": ["由乙说的话，得乙$$\\textgreater$$丙；由甲说的话，得丙$$\\textgreater$$甲． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1079", "queId": "0f73bc2ac3f14560ba177088f6a19a45", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "某学校准备购买$$30$$个篮球，三家商店每个篮球的售价都是$$25$$元，但优惠方法不同，$$A$$店``买九赠一''，$$B$$店``打八八折''；$$C$$店``满$$100$$元减现金$$10$$元''，为节约资金，应该到店购买． A school is going to buy 30 basketballs. The price of each basketball in the three stores is 25 yuan, but the preferential methods are different. Store A \"buy nine and get one free\", store B \"give a 20\\% discount\";~Store C ~\"full 100 yuan minus 10 yuan in cash\", in order to save money, should go to store to buy. ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "任意一个店皆可 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["甲店：只需付$$27$$个篮球的钱即可买到$$30$$个篮球，要$$27\\times 25=675$$元； 乙店：$$30\\times 25\\times 88 \\% =660$$元； 丙店：$$25\\times 30=750$$元，减$$7\\times 10=70$$元，花费$$750-70=680$$元． 因为$$660\\textless{}675\\textless{}680$$，所以应到乙店购买． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2461", "queId": "2af3fe6a13974a0e9e890b7e05dc2a6f", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（一）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "观察下列数表： $$\\frac{1}{1}$$； $$\\frac{2}{1}$$，$$\\frac{1}{2}$$； $$\\frac{3}{1}$$，$$\\frac{2}{2}$$，$$\\frac{1}{3}$$； $$\\frac{4}{1}$$，$$\\frac{3}{2}$$，$$\\frac{2}{3}$$，$$\\frac{1}{4}$$； $$\\cdots\\cdots$$ $$\\frac{2014}{2015}$$这个数位于第行第列． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2014$$，$$2015$$ "}], [{"aoVal": "B", "content": "$$2015$$，$$2014$$ "}], [{"aoVal": "C", "content": "$$4028$$，$$2015$$ "}], [{"aoVal": "D", "content": "$$2015$$，$$4028$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["观察发现，这些分数的分母等于它们所在的列数．每行中各分数的分子与分母的和分别相等，且这个和与$$1$$的差等于该分数所在的行数． 因为$$\\frac{2014}{2015}$$的分母是$$2015$$，且$$2014+20151=4028$$，所以这个数位于第$$4028$$行，第$$2015$$列． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "917", "queId": "93d8504afaec45e5b6a9ecb5e39e5b98", "competition_source_list": ["2017年全国美国数学大联盟杯小学高年级六年级竞赛初赛第40题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "取出$$100$$至$$999$$之间所有这样的整数，这些整数的个位数、十位数、百位数各不相同，而且都不是$$0$$．这些整数的和是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$226800$$ "}], [{"aoVal": "B", "content": "$$251748$$ "}], [{"aoVal": "C", "content": "$$279720$$ "}], [{"aoVal": "D", "content": "$$282840$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["百位数字的和是$$(100+200+...+900)\\times 56=252000$$，十位和个位上的和是$$(1+2+...+9)\\times 8\\times 77=27720$$，所以总和是$$279720$$，选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2805", "queId": "f673d0ae5c9643f78e31fbd2db01179d", "competition_source_list": ["2017年河南郑州K6联赛竞赛模拟第6题", "2017年河南郑州模拟考试k6考试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$20$$米长的铁丝平均截成$$5$$段，在下面的各种说法中，错误的是． ", "answer_option_list": [[{"aoVal": "A", "content": "每段长$$4$$米 "}], [{"aoVal": "B", "content": "每段长度是全长的$$4$$倍 "}], [{"aoVal": "C", "content": "每段长度是全长的$$\\frac{1}{5}$$ "}], [{"aoVal": "D", "content": "两段长$$8$$米 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数的认识->认、读、写数->分数->分数的意义"], "answer_analysis": ["每段长$$20\\div 5=4\\text{m}$$，两段长$$2\\times 4=8\\text{m}$$，每段长度是全长的$$\\frac{1}{5}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "554", "queId": "29e48de3257a4e858d745d760b1495bd", "competition_source_list": ["2016年北京华杯赛小学中年级竞赛复赛A卷第11题15分"], "difficulty": "3", "qtype": "single_choice", "problem": "在$$1$$到$$200$$这$$200$$个自然数中任意选数，至少要选出~\\uline{~~~~~~~~~~}~个才能确保其中必有$$2$$个数的乘积等于$$238$$. ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$100$$ "}], [{"aoVal": "E", "content": "$$198$$ "}], [{"aoVal": "F", "content": "$$199$$ "}], [{"aoVal": "G", "content": "$$235$$ "}], [{"aoVal": "H", "content": "$$237$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$238=2\\times 7\\times 17$$，如果两个小于等于$$200$$的整数的乘积为$$238$$，那么只有可能为$$2\\times 119$$、$$7\\times 34$$、$$14\\times 17$$，那么$$(2,119)$$、$$(7,34)$$、$$(14,17)$$这三组数中至少要有一组全取，至少要取出$$200-6+3+1=198$$（个）． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2826", "queId": "c3d4ab061a4f408385146463218175d6", "competition_source_list": ["2019年广东广州羊排赛六年级竞赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$a=\\frac{19}{27}$$，$$b=\\frac{13}{21}$$，$$c=\\frac{17}{25}$$，则$$a$$、$$b$$、$$c$$的大小顺序是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$a ~\\textless{} ~b ~\\textless{} ~c$$ "}], [{"aoVal": "B", "content": "$$a ~\\textgreater{} ~b ~\\textgreater{} ~c$$ "}], [{"aoVal": "C", "content": "$$b ~\\textless{} ~c ~\\textless{} ~a$$ "}], [{"aoVal": "D", "content": "$$b~ \\textgreater~ c ~\\textgreater{} ~a$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数基准数法"], "answer_analysis": ["三个分数的分子都比分母少$$8$$，$$\\frac{19}{27}=1-\\frac{8}{27}$$，$$\\frac{13}{21}=1-\\frac{8}{21}$$，$$\\frac{17}{25}=1-\\frac{8}{25}$$， 分子相同，分母越大，分数值越小，$$\\frac{8}{27} ~\\textless{} ~\\frac{8}{25} ~\\textless{} ~\\frac{8}{21}$$，所以$$\\frac{19}{27}\\textgreater\\frac{17}{25}\\textgreater\\frac{13}{21}$$，即$$a\\textgreater c\\textgreater b$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2986", "queId": "9383875647aa4bdca61863d4de467d23", "competition_source_list": ["2018年福建福州河仁杯五年级竞赛第4题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若三角形的三边长为$$a$$、$$b$$、$$c$$，且满足$$\\frac{1}{a}+\\frac{1}{b}-\\frac{1}{c}=\\frac{1}{a+b-c}$$，则该三角形一定是． ", "answer_option_list": [[{"aoVal": "A", "content": "直角三角形 "}], [{"aoVal": "B", "content": "等腰三角形 "}], [{"aoVal": "C", "content": "等边三角形 "}], [{"aoVal": "D", "content": "等腰直角三角形 "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["因为$$\\frac{1}{a}+\\frac{1}{b}-\\frac{1}{c}=\\frac{1}{a+b-c}$$ ， 所以$$\\frac{1}{a}-\\frac{1}{a+b-c}=\\frac{1}{c}-\\frac{1}{b}$$ 所以$$\\frac{b-c}{a\\left( a+b-c \\right)}=\\frac{b-c}{bc}$$ ， 所以$$\\left( b-c \\right)\\frac{bc-a\\left( a+b-c \\right)}{abc\\left( a+b-c \\right)}=0$$， 由于分母恒大于$$0$$， 所以$$b-c=0$$ 或 $$bc-a\\left( a+b-c \\right)=0$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde bc+ac={{a}^{2}}+ab$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left( b+a \\right)c=a\\left( a+b \\right)$$ 所以$$b=c$$ 或$$a=c$$ ， 所以这个三角形一定是等腰三角形． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1227", "queId": "1a55f9974ac540a793f4ae74cf60ffe9", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛六年级竞赛决赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明带着一些钱去买钢笔，如果钢笔降价$$10 \\%$$，则可以比原来多买$$30$$支．那么降价$$10 \\%$$后，小明带的钱可以买支钢笔． ", "answer_option_list": [[{"aoVal": "A", "content": "$$260$$ "}], [{"aoVal": "B", "content": "$$300$$ "}], [{"aoVal": "C", "content": "$$320$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["设钢笔价格为$$a$$元/支， 原来可以买$$x$$支， 则总钱数为$$ax$$， 所以有：$$ax=(1-10 \\%)a\\times (x+30)$$ $$x=270$$， 所以降价$$10 \\%$$后，可买：$$270+30=300$$（支）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "676", "queId": "24136235039c4e24a27a61919f865dd1", "competition_source_list": ["2018年IMAS小学高年级竞赛（第一轮）第19题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知正整数$$n$$与$$24$$的最大公因数为$$2$$，且$$n+1$$与$$24$$的最大公因数为$$3$$．请问$$n$$不能取下面哪个值？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$38$$ "}], [{"aoVal": "E", "content": "$$50$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公因数与最大公因数->两数的最大公因数"], "answer_analysis": ["由题意可知，$$n$$可被$$2$$整数，但不可被$$4$$整数，只有选项$$\\text{C}$$符合； 由$$n+1$$有因数$$3$$知$$n$$被$$3$$之后的余数为$$2$$，知其余的选项均符合此条件． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3376", "queId": "b215955a042543a0b05371c46e7eb8c9", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$15$$个玻璃球分成数量不同的$$4$$堆，共有种不同的分法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["根据题意可得： $$15=1+2+3+9$$；$$15=1+2+4+8$$； $$15=1+2+5+7$$；$$15=1+3+4+7$$； $$15=1+3+5+6$$；$$15=2+3+4+6$$； 一共有$$6$$种． 答：共有$$6$$种不同的分法． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "289", "queId": "ff8080814518d52401451919f1a60385", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "一次考试有三道题，四个好朋友考完后互相交流了成绩．发现四人各对了$$3$$、$$2$$、$$1$$、$$0$$题．这时一个路人问：你们考的怎么样啊？ 甲：``我对了两道题，而且比乙对的多，丙考的不如丁．'' 乙：``我全对了，丙全错了，甲考的不如丁．'' 丙：``我对了一道，丁对了两道，乙考的不如甲．'' 丁：``我全对了，丙考的不如我，甲考的不如乙．'' 已知大家都是对了几道题就说几句真话，那么对了$$2$$题的人是（~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "丁 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["全对的人不会说自己对的题少于$$3$$，故只有乙、丁可能全对．若乙全对，则排名是乙、丁、甲、丙，与丙所说的``丁对了$$2$$道''是假话相矛盾；若丁全对，则丙的后两句是假话，不可能是第二名，又由丁的``甲考得不如乙''能知道第二名是乙，故丙全错，甲只有``丙考得不如丁''是真话，排名是丁、乙、甲、丙且$$4$$人的话没有矛盾．综上，答案是B． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "778", "queId": "ccbbb86cfcea4e44807f1f3eaae40b25", "competition_source_list": ["2011年全国美国数学大联盟杯小学高年级五年级竞赛初赛第27题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$12$$和$$45$$的最小公倍数是？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$57$$ "}], [{"aoVal": "B", "content": "$$90$$ "}], [{"aoVal": "C", "content": "$$180$$ "}], [{"aoVal": "D", "content": "$$540$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$12=2^{2}\\times3$$ $$45=3^{2}\\times5$$ $$[12,45]=2^{2}\\times3^{2}\\times5=180$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3266", "queId": "55a71b1b0f5449c4b0a61ac26a302d95", "competition_source_list": ["2011年国奥赛竞赛决赛第6题"], "difficulty": "3", "qtype": "single_choice", "problem": "已知$$a$$、$$b$$的最小公倍数是$$60$$；$$b$$、$$c$$的最小公倍数是$$36$$；$$c$$、$$a$$的最小公倍数是$$90$$，那么满足这些条件的不同数组（$$a$$，$$b$$，$$c$$）共有~\\uline{~~~~~~~~~~}~个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举", "Overseas Competition->知识点->计数模块"], "answer_analysis": ["$$60=2\\times 2\\times 3\\times 5$$，$$36=2\\times 2\\times 3\\times 3$$，$$90=2\\times 3\\times 3\\times 5$$． 因为$$36$$中不含有因数$$5$$，所以$$b$$，$$c$$中不含有因数$$5$$，则$$a$$中必含有$$1$$个$$5$$．$$a$$，$$b$$中有且只有一个能被$$3$$整除且不能被$$9$$整除． $$a$$，$$c$$不能被$$4$$整除，$$b$$可被$$4$$整除，$$c$$能被$$9$$整除． 然后枚举：①$$a=5$$，$$b=12$$，$$c=18$$．②$$a=10$$，$$b=12$$，$$c=9$$．③$$a=10$$，$$b=12$$，$$c=18$$．④$$a=15$$，$$b=4$$，$$c=18$$．⑤$$a=15$$，$$b=12$$，$$c=18$$．⑥$$a=30$$，$$b=4$$，$$c=9$$．⑦$$a=30$$，$$b=4$$，$$c=18$$．⑧$$a=30$$，$$b=12$$，$$c=9$$．⑨$$a=30$$，$$b=12$$，$$c=18$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1328", "queId": "39edc3cb27c34661b5c62abbe665fe8b", "competition_source_list": ["2016年新希望杯小学高年级六年级竞赛训练题（三）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$1$$千克浓度为$$x$$的酒精，与$$2$$千克浓度为$$20 \\% $$的酒精混合后，浓度变为$$0.6x$$，则$$x$$的值为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50 \\% $$ "}], [{"aoVal": "B", "content": "$$48 \\% $$ "}], [{"aoVal": "C", "content": "$$45 \\% $$ "}], [{"aoVal": "D", "content": "$$40 \\% $$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["$$\\frac{x\\times 1+2\\times 20 \\% }{1+2}=0.6x$$，$$x=0.5$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2009", "queId": "f37c3382c00d4665ac1e61eb1897dc72", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "小刚进阅览室看书，当天所看内容占连续$$5$$个整页，页码和为$$600$$，他下次来接着看，起始的页码是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$118$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$122$$ "}], [{"aoVal": "D", "content": "$$123$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["起始为$$x$$，有$$x+(x+1)+(x+2)+(x+3)+(x+4)=600$$．$$5x=590$$，$$x=118$$．下次看起始为$$118+5=123$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "595", "queId": "588fc36eea9e4186937f37a9b2a0eccb", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级六年级竞赛邀请赛训练题（四）"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$n$$是正整数，记$$1\\times 2\\times 3\\times \\cdots \\times n=n!$$，比如$$1!=1$$，$$4=1\\times 2\\times 3\\times 4=24$$等，如果$$M=1!\\times 2!\\times 3!\\times 4!\\times 5!\\times 6!\\times 7!\\times 8!\\times 9!$$，则$$M$$的所有因数中，是完全平方数一共有（~ ~ ~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$112$$ "}], [{"aoVal": "B", "content": "$$336$$ "}], [{"aoVal": "C", "content": "$$448$$ "}], [{"aoVal": "D", "content": "$$672$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数->分解质因数（式）"], "answer_analysis": ["将$$M$$分解质因数后，$$M={{2}^{31}}\\times {{3}^{13}}\\times {{5}^{5}}\\times {{7}^{3}}$$，因为完全平方数中质因数成对出现，按乘法原理，其中完全平方数的因数可以搭配出： 一共有：$$16\\times 7\\times 3\\times 2=672$$（个）． 质因数$$2$$可以取$$0$$、$$2$$、$$4$$、$$6$$、$$\\ldots \\ldots $$、$$30$$一共$$16$$种，同理质因数$$3$$、$$5$$、$$7$$依次有$$7$$、$$3$$、$$2$$种选择法．选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2148", "queId": "13e3bfbcb6ad41c3b6d5c05c2c1e9552", "competition_source_list": ["2020年第1届广东深圳超常思维竞赛四年级竞赛初赛第13题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "摩托车骑手甲和自行车骑手乙同时由$$A$$地去$$B$$地．走过全程的$$\\frac{1}{3}$$时，乙停下来休息，当他休息完准备继续出发时，发现甲距$$B$$地还有全程的 $$\\frac{1}{3}$$．甲到达$$B$$地后一刻也不停留，马上向$$A$$地返回．究竟是甲先到达$$A$$地，还是乙先到达$$B$$地，以下正确的说法是． ", "answer_option_list": [[{"aoVal": "A", "content": "谁先到取决于两人的速度 "}], [{"aoVal": "B", "content": "谁先到取决于$$A$$，$$B$$之间的距离 "}], [{"aoVal": "C", "content": "两人同时到达 "}], [{"aoVal": "D", "content": "甲先到$$A$$地 "}], [{"aoVal": "E", "content": "乙先到$$B$$地 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["乙先到达$$B$$地，选$$\\text{E}$$． 这是因为乙在行完全程的$$\\frac{1}{3}$$时停下来休息，而当他休息后离开时，发现甲距$$B$$地还有全程的$$\\frac{1}{3}$$，这时甲已行驶了全程的$$\\frac{2}{3}$$．也就是说，乙走完全程的$$\\frac{1}{3}$$比甲走完全程的$$\\frac{2}{3}$$要快， 因为甲行驶全程的$$\\frac{2}{3}$$所花的时间是乙行驶全程$$\\frac{1}{3}$$所花的时间再加上乙停下来休息的时间．接下来乙将继续向前行驶的路程是全程的$$\\frac{2}{3}$$，而甲则要行驶全程的$$\\frac{1}{3}+\\frac{3}{3}=\\frac{4}{3}$$．由于乙行驶全程的$$\\frac{1}{3}$$比甲行驶全程的$$\\frac{2}{3}$$快一些，那么乙行驶全程的$$\\frac{2}{3}$$就比甲行驶全程的$$\\frac{4}{3}$$要快一些．故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1398", "queId": "632c47cdf0f048b78c8a43db775249d2", "competition_source_list": ["2018年四川成都锦江区四川师范大学附属第一实验中学小升初（五）第5~0题3分", "2014年全国迎春杯四年级竞赛初赛第5题", "2019年四川成都锦江区四川师范大学附属第一实验中学小升初（八）第6题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "皮皮去动物园参观，动物园的饲养员把一堆桃子分给若干只猴子，如果每只猴子分$$6$$个，剩$$57$$个桃子；如果每只猴子分$$9$$个，就有$$5$$只猴子一个也分不到，还有一只猴子只分到$$3$$个．那么，有个桃子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$216$$ "}], [{"aoVal": "B", "content": "$$324$$ "}], [{"aoVal": "C", "content": "$$273$$ "}], [{"aoVal": "D", "content": "$$301$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["每只猴子多分了$$3$$个，分了$$5\\times 9+(9-3)+57=108$$ （个），那么共$$108\\div 3=36$$（只）猴子．共$$36\\times6+57=273$$（个）桃子． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2868", "queId": "80b230d06a78427ab826aa6c4f43c0e7", "competition_source_list": ["2020年希望杯二年级竞赛模拟第29题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$79-67+121-33=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$200$$ "}], [{"aoVal": "C", "content": "$$300$$ "}], [{"aoVal": "D", "content": "$$400$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之凑整法->加减法凑整综合", "Overseas Competition->知识点->���用题模块->加减法应用"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde79-67+121-33$$ $$=79+121-\\left( 67+33 \\right)$$ $$=200-100$$ $$=100$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3022", "queId": "b7f83fc9ea96422aa39cd1ae5c8476a7", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛初赛第10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\triangle +\\triangle =\\bigcirc +\\bigcirc +\\bigcirc $$，$$\\bigcirc +\\bigcirc +\\bigcirc =\\square +\\square +\\square $$，$$\\bigcirc +\\square +\\triangle +\\triangle =100$$，则$$\\triangle =$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->等量代换->不可直接计算的算式代换"], "answer_analysis": ["因为$$\\triangle +\\triangle =\\bigcirc +\\bigcirc +\\bigcirc $$即$$2\\triangle =3\\bigcirc $$； $$\\bigcirc +\\bigcirc +\\bigcirc =\\square +\\square +\\square $$即$$3\\bigcirc =3\\square $$，则$$\\bigcirc =\\square $$； $$\\bigcirc +\\square +\\triangle +\\triangle =\\bigcirc +\\bigcirc +3\\bigcirc =5\\bigcirc =100$$， 所以$$\\bigcirc =20$$，则$$\\square =20$$，$$\\triangle =30$$． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2295", "queId": "9799b1ae43db4f2da1f2c51e8ca13eae", "competition_source_list": ["2003年第1届创新杯五年级竞赛复赛第6题", "2003年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "一艘轮船从甲港开往乙港，由于顺水，每小时可以航行28千米，3小时到达，这艘轮船从乙港返回甲港时，由于逆水，每小时只能航行21千米.这艘轮船往返一次每小时的平均速度是（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "12千米 "}], [{"aoVal": "B", "content": "24千米 "}], [{"aoVal": "C", "content": "24.5千米 "}], [{"aoVal": "D", "content": "25千米 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->平均速度->公式法"], "answer_analysis": ["甲乙两港之间的距离为:$$28\\times 3=84$$千米，从乙返回甲逆水所用的时间为$$84\\div 21=4$$小时，往返一次的平均速度为:$$84\\times 2\\div \\left( 3+4 \\right)=24$$千米/小时 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2266", "queId": "a8883f7f3f7b451297012ec44d8440b6", "competition_source_list": ["2015年第11届全国新希望杯小学高年级六年级竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "小宇的手表每小时慢$$3$$分钟，如果他在早晨$$6$$:$$30$$将手表与准确时间对准，那么当天小宇手表显示$$12$$:$$50$$的时候，准确时间是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$:$$00$$ "}], [{"aoVal": "B", "content": "$$13$$:$$09$$ "}], [{"aoVal": "C", "content": "$$13$$:$$10$$ "}], [{"aoVal": "D", "content": "$$13$$:$$15$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->时钟问题->坏钟问题"], "answer_analysis": ["手表走$$57$$分钟，正常钟走$$60$$分钟，现在手表走$$6$$小时$$20$$分钟，为$$380$$分钟，那么正常钟应该走$$380\\div 57\\times 60=400$$．为$$6$$小时$$40$$分，此时为$$13$$:$$10$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2001", "queId": "b885fcdf3aa144f6b4bab55ec295920a", "competition_source_list": ["2020年春蕾杯六年级竞赛第9题2分", "2021年春蕾杯六年级竞赛第4题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一杯$$100$$克盐水，盐和水的比是$$1:4$$，商家为了降低成本，往里面加了水，现在盐和盐水的比为$$1:10$$．加了~\\uline{~~~~~~~~~~}~克的水． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$90$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题->单量不变"], "answer_analysis": ["$$100$$克盐水，盐比水为$$1:4$$，则盐的含量为：$$100\\times \\frac{1}{5} =20$$ （克）． 加水后，盐的含量不变，而盐比盐水为$$1:10$$，则盐水的总质量为$$20\\times 10=200$$（克），增加的水为：$$200-100=100$$（克）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "389", "queId": "73adad28be26461199da819b83eb91b8", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）"], "difficulty": "2", "qtype": "single_choice", "problem": "口袋中有红、黑、白、黄四种球，个数分别为$$7$$，$$9$$，$$11$$，$$13$$个，它们的外形和重量都一样，至少摸出（~ ）个球才能保证有$$9$$个颜色相同的球． ", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$ "}], [{"aoVal": "B", "content": "$$29$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->简单抽屉原理"], "answer_analysis": ["从最倒霉的情况着手，把红色取完也到不了$$9$$个同色，所以先把红色取完．然后最倒霉是另外三种颜色每种取$$8$$个，最后加$$1$$，得到最不利情况$$7+8\\times 3+1=32$$（个）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2972", "queId": "a52bc95987e042ab8d8d75968cc70464", "competition_source_list": ["2004年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "古代埃及时，人们最喜欢的是分子为$$1$$的分数，如$$\\frac{1}{2}$$，$$\\frac{1}{3}$$，$$\\frac{1}{4}$$，$$\\cdots $$，$$\\frac{1}{n}$$等，我们不妨称这些分数为单位分数，其他的分数，只有它能写成若干个不同的单位分数之和时，人们才承认它是分数，例如，由于$$\\frac{3}{4}=\\frac{1}{2}+\\frac{1}{4}$$，所以他们承认$$\\frac{3}{4}$$是分数。如果当时只有四个单位分数：$$\\frac{1}{2}$$，$$\\frac{1}{3}$$，$$\\frac{1}{4}$$，$$\\frac{1}{5}$$，那么下列四个分数中，不能承认的分数是( ) ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{7}{12}$$ "}], [{"aoVal": "C", "content": "$$\\frac{9}{20}$$ "}], [{"aoVal": "D", "content": "$$\\frac{9}{10}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->裂项与通项归纳->分数裂项->分数裂和之基础"], "answer_analysis": ["选项A的$$\\frac{5}{6}=\\frac{1}{2}+\\frac{1}{3}$$，选项B的$$\\frac{7}{12}=\\frac{1}{3}+\\frac{1}{4}$$，选项C的$$\\frac{9}{20}=\\frac{1}{5}+\\frac{1}{4}$$，选D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1977", "queId": "d35eaeeb6e4740639ca9d2383b393d9a", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛三年级竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一列数：$$3$$，$$1$$，$$2$$，$$3$$，$$1$$，$$2$$，$$\\cdots \\cdots $$那么第$$53$$个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}]], "knowledge_point_routes": ["拓展思维->能力->归纳总结->归纳推理"], "answer_analysis": ["这列数按照``$$3$$，$$1$$，$$2$$''的周期排列，则，$$53\\div 3=17$$（组）$$\\cdots \\cdots 2$$（个）， 所以第$$53$$个数位这个周期的第$$2$$个数，即为$$1$$． 故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2605", "queId": "24d29e37cd4f45c6857a6eca1d23d368", "competition_source_list": ["2020年北京迎春杯六年级竞赛模拟初赛第1题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$\\left( 7\\frac{4}{25}+8.6 \\right)\\div \\left[ \\left( 4\\frac{5}{7}-0.005\\times 900 \\right)\\div \\frac{6}{7} \\right]=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$63.04$$ "}], [{"aoVal": "B", "content": "$$634$$ "}], [{"aoVal": "C", "content": "$$620$$ "}], [{"aoVal": "D", "content": "$$6304$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["$$\\left( 7\\frac{4}{25}+8.6 \\right)\\div \\left[ \\left( 4\\frac{5}{7}-0.005\\times 900 \\right)\\div \\frac{6}{7} \\right]$$ $$=\\left( 7.16+8.6 \\right)\\div \\left[ \\left( 4\\frac{5}{7}-4.5 \\right)\\div \\frac{6}{7} \\right]$$ $$=15.76\\div \\left( \\frac{3}{14}\\times \\frac{7}{6} \\right)$$ $$=15.76\\div \\frac{1}{4}$$ $$=15.76\\times 4$$ $$=63.04$$ 故答案为：$$63.04$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1154", "queId": "14f1f610a70b44edb4d131a7a877e143", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛公开题"], "difficulty": "1", "qtype": "single_choice", "problem": "小明从家出发，乘地铁到学校需要$$30$$分钟，乘公交车到学校需要$$50$$分钟，某天小明因故先乘地铁，再换乘公交车，用了$$40$$分钟到达学校，其中换乘用了$$6$$分钟，那么这天小明乘坐公交车用了~\\uline{~~~~~~~~~~}~分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题->方程法解其他问题"], "answer_analysis": ["设乘坐公交车$$x$$分钟，乘地铁$$\\left(40-6-x\\right)$$， $$\\frac{1}{50}x+\\frac{1}{30}\\left(40-6-x\\right)=1$$ 解得$$x=10$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2648", "queId": "8bcc26302a7f405d967a090e2a19a2ee", "competition_source_list": ["2017年安徽合肥庐江县小升初第16题1分", "2019年湖南郴州六年级下学期小升初模拟（6）第11题", "2017年河南郑州小升初豫才杯第11题3分", "2019~2020学年北京四年级上学期期末模拟（北京版）（四）第9题", "2017年河南郑州豫才杯竞赛第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$34\\div 6=5\\cdots \\cdots 4$$，如果被除数和除数同时扩大$$100$$倍，余数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$400$$ "}], [{"aoVal": "C", "content": "$$4000$$ "}], [{"aoVal": "D", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数乘除->整数除法运算->带余除法"], "answer_analysis": ["$$3400\\div 600=5\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 400$$，故选$$B$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "440", "queId": "dbadfe1349ce4387bf98325f1815e160", "competition_source_list": ["2010年六年级竞赛创新杯", "2010年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "六名同学参加围棋比赛，每两个人都要比赛一场，胜者得$$2$$分，负者得$$0$$分，比赛结果有两个并列第二名，两个并列第五名，则第一名得了（ ）分。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->体育比赛->单循环赛"], "answer_analysis": ["$$6$$个人，每两个人都要赛一场，一共有$$6\\times 5\\div 2=15$$（场），总分$$30$$分，第一名最多胜$$5$$场得$$10$$分。若第一名不为$$10$$分，则由条件$$6$$人得分最多分别为$$8,6,6,4,2,2$$分，和为$$28\\textless30$$，矛盾，故第一名得$$10$$分。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3214", "queId": "5de9ad4a72074e03b21c68a8ef9ef8fb", "competition_source_list": ["2017年IMAS小学中年级竞赛（第一轮）第17题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个二位数的个位与十位数码之和为$$9$$．将这个二位数乘$$5$$，得到乘积的各位数码之和仍为$$9$$．请问满足上述条件的二位数共有多少个？（） ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["个位与十位数码之和为$$9$$的二位数有$$18$$、$$27$$、$$36$$、$$45$$、$$54$$、$$63$$、$$72$$、$$81$$、$$90$$共$$9$$个数，将这些二位数乘以$$5$$，得到的乘积分别为$$90$$、$$135$$、$$180$$、$$225$$、$$270$$、$$315$$、$$360$$、$$405$$、$$450$$，它们的数码之和仍为$$9$$，所有数都符合题目条件﹐因此总共有$$9$$个． 故逐$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "103", "queId": "198c568b5b53426dab358d5616b07ed9", "competition_source_list": ["2017年全国华杯赛竞赛初赛模拟3第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一种计时方法将一天分为十二个时辰，在$$1729$$人中，至少有~\\uline{~~~~~~~~~~}~人出生在同一个月、同一个时辰，且有相同的生肖． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["因为$$1729=12\\times12\\times12+1$$，所以，至少$$2$$人出生在同一个月、同一个时辰，且有相同的生肖． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1244", "queId": "8fbf3d3ae4f9439ba89199548da37c4f", "competition_source_list": ["2014年全国学而思杯一年级竞赛第5题"], "difficulty": "0", "qtype": "single_choice", "problem": "艾迪、薇儿、加加和减减玩游戏，每人写一个数，然后判断这$$4$$个数相加后的和是单数还是双数．其中一局他们分别写的是：$$9$$、$$13$$、$$471$$、$$1236$$，那么你来判断一下，这$$4$$个数的和是~\\uline{~~~~~~~~~~}~数（填``单''或``双''）． ", "answer_option_list": [[{"aoVal": "A", "content": "单 "}], [{"aoVal": "B", "content": "双 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析", "海外竞赛体系->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["四个数中三个单数一个双数，三个单数和是单数，再加一个双数和是单数． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2673", "queId": "4d2d4095e79d4f9f8d96373cf3c1a552", "competition_source_list": ["2017年河南郑州模拟考试k6考试第6题", "2017年河南郑州K6联赛竞赛模拟第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$3$$米长的铁丝平均截成$$5$$段，在下面的各种说法中，错误的是． ", "answer_option_list": [[{"aoVal": "A", "content": "每段长$$\\frac{3}{5}$$米 "}], [{"aoVal": "B", "content": "每段长度是全长的$$\\frac{3}{5}$$ "}], [{"aoVal": "C", "content": "每段长度是全长的$$\\frac{1}{5}$$ "}], [{"aoVal": "D", "content": "每段长度是$$1$$米的$$\\frac{3}{5}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->应用题->分数百分数应用题->已知单位一和部分量，求分率", "拓展思维->拓展思维->计算模块->分数->分数基础"], "answer_analysis": ["每段长$$3\\div 5=\\frac{3}{5}（\\text{m}）$$，每段长度是全长的$$1\\div 5=\\frac{1}{5}$$，每段长度是$$1$$米的$$\\frac{3}{5}\\div 1=\\frac{3}{5}$$，所以选项$$\\text{B}$$不正确． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "820", "queId": "652200c212ec43c4b07177e8deeb7d8e", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛B卷第5题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$1$$路、$$2$$路和$$5$$路公交车都从东站发车，$$1$$路车每隔$$10$$分钟发一辆，$$2$$路车每隔$$15$$分钟发一辆，而$$5$$路车每隔$$20$$分钟发一辆，当三种路线的车同时发车后，至少要经过分钟这三种路线的车再次同时发车． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$90$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->两数的最小公倍数"], "answer_analysis": ["$$10=2\\times 5$$，$$15=3\\times 5$$，$$20=2\\times 2\\times 5$$， 所以$$10$$，$$15$$，$$20$$的最小公倍数为：$$2\\times 2\\times 3\\times 5=60$$， 所以至少经过$$60$$分钟这三种路线再次同时发车， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2940", "queId": "ff99586bded041d5a9e7a91125180c22", "competition_source_list": ["2012年IMAS小学高年级竞赛第一轮检测试题第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "请问算式$$101010-10101$$之值为何？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$90909$$ "}], [{"aoVal": "B", "content": "$$10101$$ "}], [{"aoVal": "C", "content": "$$100000$$ "}], [{"aoVal": "D", "content": "$$900000$$ "}], [{"aoVal": "E", "content": "$$10000$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数减法运算->减法横式"], "answer_analysis": ["$$101010-10101=90909$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2385", "queId": "210e05b151ea4a8bbc6424834bc0dc01", "competition_source_list": ["2016年IMAS小学高年级竞赛第一轮检测试题第2题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "请问下列哪一项表达式是正确的？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1.2\\times 3.4=12\\times 3.4$$ "}], [{"aoVal": "B", "content": "$$0.98\\times 0.99\\textgreater0.99$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}-\\frac{1}{3}\\textless{}\\frac{1}{3}-\\frac{1}{4}$$ "}], [{"aoVal": "D", "content": "$$10.4\\times 0.1\\textless{}1.04$$ "}], [{"aoVal": "E", "content": "$$1.1\\times 1.1\\textgreater1.1$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->两数相减法"], "answer_analysis": ["$$\\text{A}$$：$$1.2\\times 3.4\\textless{}12\\times 3.4$$~~~~~~~~~~~~ $$\\text{B}$$：$$0.98\\times 0.99\\textless{}0.99$$ $$\\text{C}$$：$$\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$$，$$\\frac{1}{3}-\\frac{1}{4}=\\frac{1}{12}$$，$$\\frac{1}{2}-\\frac{1}{3}\\textgreater\\frac{1}{3}-\\frac{1}{4}$$ $$\\text{D}$$：$$10.4\\times 0.1=1.04$$ "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2841", "queId": "6e5732de09b4408eb9c3ddc89964fb11", "competition_source_list": ["2004年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "有10个小数:0.6，0.66，$$\\cdots ，\\underbrace{0.66\\cdots 66}_{10个6}$$，如果要从这些小数中取出若干个使取出的数的总和大于5，那么所取出的数的个数至少是（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "6个 "}], [{"aoVal": "B", "content": "7个 "}], [{"aoVal": "C", "content": "8个 "}], [{"aoVal": "D", "content": "9个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合"], "answer_analysis": ["$$\\frac{2}{3}=0.\\dot{6}$$，那么这10个小数都小于$$\\frac{2}{3}$$，又$$\\frac{2}{3}\\times 7=\\frac{14}{3}$$小于5，所以至少需要7个以上的小数它们的和才有可能大于5，当取8个小数时，其和$$0.6+0.66+0.666+\\ldots +0.66666666\\textgreater0.66\\times 7+0.6=5.22$$.故选C. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2688", "queId": "48e66c11c5664f128368035db9b2b82b", "competition_source_list": ["2010年四年级竞赛明心奥数挑战赛"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$P\\text{↑}$$表示$$P+1$$，$$P\\text{↓}$$表示$$P-1$$，则$$4\\text{↑}\\times3\\text{↓}$$等于（ ）。 A、$$9\\text{↓}$$ B、$$10\\text{↓}$$ C、$$11\\text{↓}$$ D、$$12\\text{↑}$$ E、$$13\\text{↓}$$ ", "answer_option_list": [[{"aoVal": "A", "content": "A "}], [{"aoVal": "B", "content": "B "}], [{"aoVal": "C", "content": "C "}], [{"aoVal": "D", "content": "D "}], [{"aoVal": "E", "content": "E "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->定义新运算->直接运算型->普通型"], "answer_analysis": ["$$5\\times 2=10=11-1$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "848", "queId": "b1d2180c09b14b68bec5ebdcbf9e19cf", "competition_source_list": ["2008年全国迎春杯三年级竞赛初赛第10题"], "difficulty": "3", "qtype": "single_choice", "problem": "有$$125$$个同样大小的正方体木块，木块的每个面的面积均为$$1$$平方厘米，其中$$63$$个表面涂上白色，还有$$62$$个表面涂上蓝色．将这$$125$$个正方体木块粘在一起，形成一个棱长为$$5$$厘米大正方体木块．这个大正方体木块的表面上，蓝色的面积最多是~\\uline{~~~~~~~~~~}~平方厘米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$110$$ "}], [{"aoVal": "B", "content": "$$112$$ "}], [{"aoVal": "C", "content": "$$114$$ "}], [{"aoVal": "D", "content": "$$116$$ "}], [{"aoVal": "E", "content": "$$118$$ "}]], "knowledge_point_routes": ["海外竞赛体系->知识点->数论模块->质数与合数->特殊质数运用->特殊质数2", "拓展思维->能力->空间想象->立体几何加工"], "answer_analysis": ["显然大正方体表面不能全为蓝色，$$8$$个顶点上的正方体木块表面积是$$3$$平方厘米，棱上的正方体木块表面积是$$2$$平方厘米，面上的正方体木块表面积是$$1$$平方厘米，所以要先在顶点和棱上放蓝色的正方体木块，剩下的放在面上，不放在内部．所以让正方体顶点和棱均为蓝色方块时蓝色表面积最大．此时蓝色的面积为：$$8\\times3+2\\times（5-2）\\times12+（62-8-36）=114$$（平方厘米）. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2718", "queId": "ff8080814502fa24014503a7714901a3", "competition_source_list": ["2014年全国迎春杯六年级竞赛初赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "在算式$$2014\\times(\\frac{1}{19}-\\frac{1}{53})$$的计算结果是（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$34$$ "}], [{"aoVal": "B", "content": "$$68$$ "}], [{"aoVal": "C", "content": "$$144$$ "}], [{"aoVal": "D", "content": "$$72$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["原式=$$2014\\times \\frac{1}{19}-2014\\times \\frac{1}{53}=106-38=68$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2490", "queId": "4b3e0b79fede4eb89219658946cd1372", "competition_source_list": ["2012年IMAS小学高年级竞赛第一轮检测试题第12题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$5$$个苹果与$$6$$根香蕉的重量相同，$$3$$根香蕉与$$4$$个橘子的重量相同，请问$$16$$个橘子与几个苹果的重量相同？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["由题意可知$$6$$根香蕉与$$8$$个橘子的重量相同，从而$$5$$个苹果与$$8$$个橘子的重量相同．所以$$16$$个橘子与$$10$$个苹果的重量相同． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "439", "queId": "86aeef366c8147b6bb5b7c6d9a473801", "competition_source_list": ["2016年创新杯六年级竞赛训练题（四）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一个算式$$12\\times 11\\times 10\\times 9\\times 8\\times 7\\times 6\\times 5\\times 4\\times 3\\times 2\\times 1$$，二娃在上式中把一些``$$\\times $$''换成``$$\\div $$''，计算结果还是自然数，那么这个自然数最小是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$77$$ "}], [{"aoVal": "B", "content": "$$154$$ "}], [{"aoVal": "C", "content": "$$231$$ "}], [{"aoVal": "D", "content": "$$286$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->横式数字谜->与数论的结合"], "answer_analysis": ["将原式中的数分解质因数，数出每个质数的个数，发现$$3$$，$$7$$，$$11$$为奇数个，故提出来． $$\\quad 12\\times 11\\times 10\\times 9\\times 8\\times 7\\times 6\\times 5\\times 4\\times 3\\times 2\\times 1=2\\times 2\\times 3\\times 11\\times 2\\times 5\\times 3\\times 3\\times 2\\times 2\\times 2\\times 7\\times 2\\times 3\\times 5\\times 2\\times 2\\times 3\\times 2\\times 1$$ $$=\\left( 2\\times 2\\times 2\\times 2\\times 2\\times 3\\times 3\\times 5 \\right)\\div \\left( 2\\times 2\\times 2\\times 2\\times 2\\times 3\\times 3\\times 5 \\right)\\times 3\\times 7\\times 11$$ $$=\\left( 12\\times 10\\times 9\\times 4 \\right)\\div \\left( 8\\times 6\\times 5\\times 3\\times 2 \\right)\\times 3\\times 7\\times 11$$ $$=3\\times 7\\times 21$$ $$=231$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2156", "queId": "2f572b6ef2014c9d9e549ea1fff9d290", "competition_source_list": ["2019年广东深圳全国小学生数学学习能力测评四年级竞赛初赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "（2019年活动）在$$9$$点至$$10$$点之间的某个时刻，$$5$$分钟前分针的位置和$$5$$分钟后时针的位置相同，此时刻是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9:05$$ "}], [{"aoVal": "B", "content": "$$9:35$$ "}], [{"aoVal": "C", "content": "$$9:55$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->时间问题->认识钟表"], "answer_analysis": ["根据选项$$\\text{C}$$，现在是$$9:55$$，那么$$5$$分钟前分针的位置是指向$$10$$，$$5$$分钟后时针也是指向$$10$$．所以$$\\text{C}$$选项答案是满足题目要求的． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1591", "queId": "8c6c451f4f354715af6b667eb3a78d01", "competition_source_list": ["2011年北京小学高年级五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙两人原有的钱数之比为$$6:5$$，后来甲又得到$$180$$元，乙又得到$$30$$元，这时甲、乙钱数之比为$$18:11$$．那么原来两人的钱数之和为~\\uline{~~~~~~~~~~}~元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$660$$ "}], [{"aoVal": "B", "content": "$$580$$ "}], [{"aoVal": "C", "content": "$$750$$ "}], [{"aoVal": "D", "content": "$$800$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想", "课内体系->能力->运算求解"], "answer_analysis": ["两人原有钱数之比为$$6$$:$$5$$，如果甲得到$$180$$元，乙得到$$150$$元，那么两人的钱数之比仍为$$6$$:$$5$$，现在甲得到$$180$$元，乙只得到$$30$$元，相当于少得到了$$120$$元，现在两人钱数之比为$$18$$:$$11$$，可以理解为：两人的钱数分别增加$$180$$元和$$150$$元之后，钱数之比为$$18$$：$$15$$，然后乙的钱数减少$$120$$元，两人的钱数之比变为$$18$$:$$11$$，所以$$120$$元相当于$$4$$份，$$1$$份为$$30$$元，后来两人的钱数之和为$$30 \\times (18 + 15) = 990$$元，所以原来两人的总钱数之和为$$990 - 180 - 150 = 660$$元． 设甲、乙原来各有$$6x$$、$$5x$$元，那么有$$(6x+180):(5x+30)=18:11$$，解比例方程得$$x=60$$，原来共有$$60\\times (6+5)=660$$（元）． 由于甲、乙得到的钱数不同，所以差并非不变．为了让得到的钱数���同，可以考虑将乙原有的钱数和得到的钱数都翻$$6$$倍，那么甲、乙原有钱数之比变为$$1:5$$，现有钱数之比变为$$3:11$$，统一差，变为$$2:10$$以及$$3:11$$，一份就是$$180$$元，所以原有钱数之和为$$(2+10\\div 6)\\times 180=660$$． 故答案为：$$660$$元． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2870", "queId": "6171e0535d05441bbe1c66aae8a8853f", "competition_source_list": ["2014年IMAS小学中年级竞赛第一轮检测试题第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "请问以下哪一个数最小？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$298$$ "}], [{"aoVal": "B", "content": "$$312$$ "}], [{"aoVal": "C", "content": "$$231$$ "}], [{"aoVal": "D", "content": "$$357$$ "}], [{"aoVal": "E", "content": "$$101$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数感认知->整数数字加工"], "answer_analysis": ["因为$$101\\textless{}231\\textless{}298\\textless{}312\\textless{}357$$，所以最小的数为$$101$$．故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1524", "queId": "44b856783e844afe81c938c0dea60c74", "competition_source_list": ["2017年第13届湖北武汉新希望杯小学高年级六年级竞赛决赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "若六年级一班的及格率是$$96 \\%$$，则不及格人数与总人数的比是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1:23$$ "}], [{"aoVal": "B", "content": "$$1:24$$ "}], [{"aoVal": "C", "content": "$$1:25$$ "}], [{"aoVal": "D", "content": "$$1:26$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["不及格率$$1-96 \\%=4 \\%$$，不及格人数与总人数的比是$$4 \\%:100 \\%=1:25$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1472", "queId": "63af167b0e6a48b9856efb7b34b3fca1", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（一）第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "小马虎在计算$$8.56$$加上一个一位小数时，由于错误地把数的末尾对齐，结果得到了$$10.21$$，正确的结果应该是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1.65$$ "}], [{"aoVal": "B", "content": "$$16.5$$ "}], [{"aoVal": "C", "content": "$$25.06$$ "}], [{"aoVal": "D", "content": "$$28.45$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["把这个一位小数的末尾与$$8.56$$对齐，相当于把这个一位小数缩小到原数的$$\\frac{1}{10}$$，又因为$$10.21-8.56=1.65$$，所以原来的一位小数是$$16.5$$．正确的结果应是$$16.5+8.56=25.06$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "504", "queId": "b460448896ca42fbbf71acea53243424", "competition_source_list": ["2018年四川成都锦江区四川师范大学附属第一实验中学小升初模拟11第9题3分", "2015年湖北武汉世奥赛五年级竞赛模拟训练题（四）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "一把钥匙只能开一把锁．现有$$10$$把钥匙和$$10$$把锁，但不知道哪把钥匙开哪把锁，最少要试次才能保证配好全部的锁的钥匙． ", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$55$$ "}], [{"aoVal": "D", "content": "$$66$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["从最不利的情况考虑，第$$1$$把锁需要试$$9$$次；第$$2$$把锁需要试$$8$$次$$\\ldots \\ldots $$；第$$9$$把锁需要试$$1$$次．一共需要试$$1+2+3+\\cdots \\cdots +9=45$$（次）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "757", "queId": "3c642d4e9dc64c4292f367f78e031511", "competition_source_list": ["2018年湖北武汉新希望杯小学高年级五年级竞赛训练题（三）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$42$$的因数共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理正应用->总个数"], "answer_analysis": ["$$42=2\\times 3\\times 7$$，$$(1+1)\\times (1+1)\\times (1+1)=8$$（个）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1508", "queId": "7aaee706a02748478ad7d733376ca52b", "competition_source_list": ["2019年第24届YMO三年级竞赛决赛第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "某年的$$2$$月份有$$5$$个星期一，这年的$$2$$月$$22$$号是． ", "answer_option_list": [[{"aoVal": "A", "content": "星期日 "}], [{"aoVal": "B", "content": "星期一 "}], [{"aoVal": "C", "content": "星期二 "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["根据题意分析可知，某年的$$2$$月份有$$5$$个星期一， 因为$$2$$月有$$28$$天则只有$$4$$个周一， 所以这一年的$$2$$月有$$29$$天，且$$2$$月$$1$$日为星期一， 则$$2$$月$$22$$日为星期一． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2948", "queId": "ae02fd4956414084a8dda8e69ff0b0cd", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛决赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$$\\left( 1+0.21+0.32 \\right)\\times \\left( 0.21+0.32+0.43 \\right)-\\left( 1+0.21+0.32+0.43 \\right)\\times \\left( 0.21+0.32 \\right)=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$0.21$$ "}], [{"aoVal": "C", "content": "$$0.32$$ "}], [{"aoVal": "D", "content": "$$0.43$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["原式$$=\\left( 1+0.21+0.32 \\right)\\times \\left( 0.21+0.32 \\right)+\\left( 1+0.21+0.32 \\right)\\times 0.43-\\left( 1+0.21+0.32 \\right)\\times \\left( 0.21+0.32 \\right)-0.43\\times \\left( 0.21+0.32 \\right)$$ $$=\\left[ \\left( 1+0.21+0.32 \\right)\\times \\left( 0.21+0.32 \\right)-\\left( 1+0.21+0.32 \\right)\\times \\left( 0.21+0.32 \\right) \\right]+\\left[ \\left( 1+0.21+0.32 \\right)\\times 0.43-\\left( 0.21+0.32 \\right)\\times 0.43 \\right]$$ $$=0+0.43\\times \\left[ \\left( 1+0.21+0.32 \\right)-\\left( 0.21+0.32 \\right) \\right]$$ $$=0.43\\times 1$$ $$=0.43$$． 故答案选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2273", "queId": "617c2d547f3e4d6ca217ec1a969e5624", "competition_source_list": ["2013年华杯赛六年级竞赛决赛", "2013年华杯赛五年级竞赛决赛"], "difficulty": "2", "qtype": "single_choice", "problem": "两个骑车人在不同的赛道上训练。骑车人$$A$$用圆形赛道，其直径是$$1$$千米，骑车人$$B$$用直线赛道，其长度为$$5$$千米。骑车人$$A$$用$$10$$分钟完成$$3$$圈，而骑车人$$B$$用$$5$$分钟行进了两个来回。那么骑车人$$A$$与骑车人$$B$$的速度比是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$1:1.6\\pi$$ "}], [{"aoVal": "B", "content": "$$\\pi :10$$ "}], [{"aoVal": "C", "content": "$$3:4$$ "}], [{"aoVal": "D", "content": "$$3\\pi :40$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题"], "answer_analysis": ["骑车人$$A$$：$$10$$分钟共走$$3\\pi $$千米，骑车人$$B$$：$$5$$分钟走$$20$$千米，即$$10$$分钟走$$40$$千米，速度比等于路程比是$$3\\pi :40$$。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2445", "queId": "94245526c71643d3aaaa5e621537c1e8", "competition_source_list": ["2017年第15届全国希望杯小学高年级六年级竞赛"], "difficulty": "3", "qtype": "single_choice", "problem": "若$$p$$，$$q$$是非$$0$$的自然数，并且$$p\\textless{}q$$，则四个式子：$$\\frac{p}{q}$$，$$\\frac{q-p}{q}$$，$$\\frac{p+q}{p}$$，$$\\frac{p+q}{q}$$中，值在$$1$$和$$2$$之间的是哪一个？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{p}{q}$$ "}], [{"aoVal": "B", "content": "$$\\frac{q-p}{q}$$ "}], [{"aoVal": "C", "content": "$$\\frac{p+q}{p}$$ "}], [{"aoVal": "D", "content": "$$\\frac{p+q}{q}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->等量代换->代数式"], "answer_analysis": ["由$$p$$，$$q$$是非$$0$$的自然数，并且$$p\\textless{}q$$，可知$$0\\textless{}\\frac{p}{q}\\textless{}1$$，于是有$$\\frac{q-p}{q}=1-\\frac{p}{q}\\textless{}1$$，$$\\frac{p+q}{p}=1+\\frac{q}{p}\\textgreater1+1=2$$，$$1\\textless{}\\frac{p+q}{q}=1+\\frac{p}{q}\\textless{}2$$，故四个式子：$$\\frac{p}{q}$$，$$\\frac{q-p}{q}$$，$$\\frac{p+q}{p}$$，$$\\frac{p+q}{q}$$中，值在$$1$$和 $$2$$之间的是$$\\frac{p+q}{q}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1837", "queId": "d244332c02a146adbd811baf70f9c2d2", "competition_source_list": ["2018年美国数学大联盟杯四年级竞赛初赛第29题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一项工程，$Mike$单独做需要$2$小时完成，$Tom$单独做需要$4$小时完成$$.$$ $Mike$和$Tom$合作$1$小时后$Mike$退出，剩下的工作由$Tom$单独完成，还需要小时． It takes Mike $$2$$ hours to finish a task． It takes $$4$$ hours for Tom to finish the same task． Mike and Tom worked together on this task for one hour before Mike had to leave． How long will it take Tom to finish the rest of the task? ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "Overseas Competition->知识点->应用题模块->工程问题"], "answer_analysis": ["根据内容分析可知：工作时间$$=$$工作总量$$\\div $$工作效率． 假设工作总量为$$1$$，用工作总量减去共同完成的工作量，再除以汤姆的工作效率即可得到剩下的工作汤姆单独完成需要多长时间．故列式计算如下： $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left[ 1-\\left( \\frac{1}{2}+\\frac{1}{4} \\right)\\times 1 \\right]\\div \\frac{1}{4}$$ $$=\\left[ 1-\\left( \\frac{2}{4}+\\frac{1}{4} \\right)\\times 1 \\right]\\div \\frac{1}{4}$$ $$=\\left( 1-\\frac{3}{4}\\times 1 \\right)\\div \\frac{1}{4}$$ $$=\\frac{1}{4}\\times 4$$ $$=1$$（小时）． 故选答案$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "179", "queId": "357d1d4f3324483b8ff4c0b22465966e", "competition_source_list": ["2020年长江杯五年级竞赛复赛A卷", "2020年长江杯五年级竞赛复赛A卷第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "现有$$1$$克、$$2$$克、$$4$$克、$$8$$克、$$16$$克的砝码各一个和一个天平，砝码只能放在一边，最多能称出种不同的质量． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$31$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->智巧趣题->数学趣题->砝码问题"], "answer_analysis": ["（$$1$$）$$1$$个砝码可以称出的重量有$$5$$种：$$1$$克，$$2$$克，$$4$$克，$$8$$克，$$16$$克； （$$2$$）$$2$$个砝码可以称出的重量有$$10$$种： $$1+2=3$$（克），$$1+4=5$$（克），$$1+8=9$$（克）， $$1+16=17$$（克）； $$2+4=6$$（克），$$2+8=10$$（克），$$2+16=18$$（克）； $$4+8=12$$（克），$$4+16=20$$（克）， $$8+16=24$$（克）； （$$3$$）$$3$$个砝码可以称出的重量有$$10$$种： $$1+2+4=7$$（克），$$1+2+8=11$$（克）， $$1+2+16=19$$（克）， $$1+4+8=13$$（克），$$1+4+16=21$$（克）， $$1+8+16=25$$（克）， $$2+4+8=14$$（克），$$2+4+16=22$$（克）， $$2+8+16=26$$（克）， $$4+8+16=28$$（克）； （$$4$$）$$4$$个砝码可以称出的重量有$$5$$种： $$1+2+4+8=15$$（克），$$1+2+4+16=23$$（克）， $$1+2+8+16=27$$（克）， $$1+4+8+16=29$$（克），$$2+4+8+16=30$$（克）； （$$5$$）$$5$$个砝码可以称出的重量有$$1$$种： $$1+2+4+8+16=31$$（克）． 因为$$5+10+10+5+1=31$$（种）， 所以最多可以称出$$31$$种不同的重量，它们是$$1$$克$$-31$$克． 答：最多能称出$$31$$种不同重量的物体． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3188", "queId": "2b23392531014f25a3b7ef0422b00a6d", "competition_source_list": ["2011年北京学而思杯五年级竞赛"], "difficulty": "3", "qtype": "single_choice", "problem": "$$5$$卷百科全书按从第$$1$$卷到第$$5$$卷的递增序号排列，今要将它们变为反序排列，即从第$$5$$卷到第$$1$$卷．如果每次只能调换相邻的两卷，那么最少要调换多少次？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["因为必须是调换相邻的两卷，将第$$5$$卷调至原来第$$1$$卷的位置最少需$$4$$次，得到的顺序为$$51234$$； 现在将第$$4$$卷调至此时第$$1$$卷的位置最少需$$3$$次，得到的顺序为$$54123$$； 现在将第$$3$$卷调至此时第$$1$$卷的位置最少需$$2$$次，得到的顺序为$$54312$$； 最后将第$$1$$卷和第$$2$$卷对调即可． 所以，共需调换$$4+3+2+1=10$$（次）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "424", "queId": "a4e98f8d4bf0405e8ab2e4caed69673d", "competition_source_list": ["2019年全国小学生数学学习能力测评四年级竞赛复赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "现有$$9$$个外观一模一样的乒乓球，其中$$8$$个一样重，另外$$1$$个轻一些，是次品．请你想一想，用天平至少称次，就保证一定能把这个次品找出来． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$3$$个球一组，可分为$$3$$组．分别标号第①组，第②组，第③组． 第$$1$$步：将第①组与第②组称$$1$$次．若平衡，则次品在第③组，第$$2$$步：在第③组的$$3$$个球中任选两个称第$$2$$次，即可判断．若第$$1$$次不平衡，哪组轻，就在哪一组进行第$$2$$次称重．故至少称$$2$$次，即可保证一定将这个次品找出来． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3349", "queId": "8d0350346bc3424ab2a15f04f4b1a06b", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个五位数，由$$1$$，$$2$$，$$3$$三个数码组成，对于其中任何一个数码，如果这个数码是$$1$$，则他后面的数只能写$$2$$；如果这个数码是$$2$$，他后面只能写$$3$$；如果这个数码是$$3$$，他后面可以写$$1$$，也可以写$$3$$．这样的五位数有（~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["首位为$$1$$，有 $$12312$$，$$12331$$，$$12333$$这三种；首位为$$2$$，有$$23123$$，$$23312$$，$$23331$$，$$23333$$这四种；首位为$$3$$，有$$31231$$，$$31233$$，$$33123$$，$$33312$$，$$33331$$，$$33333$$这六种， 所以共$$(4+100)\\div 2=52$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2683", "queId": "b5726f84fe4e4ae0af5fb2738a7aff26", "competition_source_list": ["2008年第6届创新杯四年级竞赛初赛B卷第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$101个$$数之和为$$2008$$；把第一个数减$$1$$，第$$2$$个数加$$2$$，第$$3$$个数减$$3$$，\\ldots，第$$100$$个数加$$100$$，第$$101$$个数减$$101$$，所得的$$101$$个新数之和为( )． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1957$$ "}], [{"aoVal": "B", "content": "$$1959$$ "}], [{"aoVal": "C", "content": "$$2008$$ "}], [{"aoVal": "D", "content": "$$2009$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之分组法"], "answer_analysis": ["所得的$$101$$个新数之和为：$$2008-1+2-3+4-\\cdots -99+100-101=2008+\\underbrace{1+1+1+\\cdots +1}_{50个1}-101=1957$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1481", "queId": "562e7c18c50b4f9199d7b4493cf6188f", "competition_source_list": ["2017年IMAS小学中年级竞赛（第一轮）第14题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明前几次数学考试的平均分是$$88$$分，这次数学考试结束后，小明发现如果他想把平均分提高到$$90$$分，则这次考试必须考到$$98$$分，请问小明总共考了多少次数学考试？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["可知这次考试比所欲达到的平均分数多的$$98-90=8$$分要分配到前几次的考试才能使平均分数增加$$90-88=2$$分，因此小明之前共考了$$8\\div 2=4$$次数学考试，所以连同这一次他总共考了$$4+1=5$$次数学考试．故选$$\\text{C}$$． ", "可知最后一次考试要比所欲达到的平均分数多$$98-88=10$$分才能使平均分数从$$88$$分增加到$$90$$分，即须增加$$2$$分，因此连同这一次小明总共考了$$10\\div 2=5$$次数学考试．故选$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "558", "queId": "66514b59aebd43be92395ef7f85e3f04", "competition_source_list": ["2016年创新杯五年级竞赛训练题（一）第6题"], "difficulty": "3", "qtype": "single_choice", "problem": "黑板上有$$1\\sim 2013$$共$$2013$$个数，每次可以擦掉其中两个数，并且写上这两数之和的数字和．已知黑板上剩下四个数，其乘积为$$27$$，那么这四位数的和是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["一个数除以$$9$$的余数等于这个数各个数字之和除以$$9$$的余数，每次操作将数的和变为数字和，不改变除以$$9$$的余数，$$1+2+3+\\cdots +2013=2014\\times 2013\\div 2=1007\\times 2013$$，$$1007$$除以$$9$$的余数相同，余数为$$3$$，那么最后剩下的四个数字的和除以$$9$$也是余$$3$$的；将$$27$$拆分成四个数的乘积；$$27=3\\times 3\\times 3\\times 1=3\\times 9\\times 1\\times 1=27\\times 1\\times 1\\times 1$$，和分别为$$10$$，$$14$$，$$30$$，只有$$30$$除以$$9$$余数为$$3$$，所以四个数的和为$$30$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1854", "queId": "a06ae88eb0934d54afcd38c112f1a17b", "competition_source_list": ["1990年全国华杯赛竞赛复赛", "1990年第1届全国华杯赛小学高年级竞赛复赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "有三堆棋子，每堆棋子数一样多，并且都只有黑、白两色棋子．第一堆里的黑子和第二堆里的白子一样多，第三堆里的黑子占全部黑子的五分之二，把这三堆棋子集中在一起，问白子占全部的几分之几？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\dfrac{2}{9}$$ "}], [{"aoVal": "B", "content": "$$\\dfrac{4}{9}$$ "}], [{"aoVal": "C", "content": "$$\\dfrac{5}{9}$$ "}], [{"aoVal": "D", "content": "$$\\dfrac{1}{3}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["不妨认为第二堆全是黑子，第一堆全是白子，（即将第一堆黑子与第二堆白子互换） 第二堆黑子是全部棋子的$$\\dfrac{1}{3}$$， 同时，又是黑子的$$1-\\dfrac{2}{5}$$， 所以黑子占全部棋子的：$$\\dfrac{1}{3}\\div~ (1-\\dfrac{2}{5})=\\dfrac{5}{9}$$， 白子占全部棋子的：$$1-\\dfrac{5}{9}=\\dfrac{4}{9}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "4", "queId": "00dc5f1edc0e48668717e9f714f28d59", "competition_source_list": ["2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第6题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一部电影共放映了$$85$$分钟，结束时正好是$$20:40$$分．这部电影是什么时辰开始放映的． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18:15$$ "}], [{"aoVal": "B", "content": "$$19:15$$ "}], [{"aoVal": "C", "content": "$$22:05$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->时间问题->时间计算"], "answer_analysis": ["根据题意分析可知，$$60$$分钟$$=1$$小时，$$85$$分钟$$=1$$小时$$25$$分钟， 用结束的时间减去放映的时长即可得到开始放映的时间， 故列式为：$$20:40-1:25=19:15$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2119", "queId": "fe5c6c97664f43a08f6bab52ed03c69e", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级三年级竞赛初赛第38题"], "difficulty": "1", "qtype": "single_choice", "problem": "一次考试中，总共有$$40$$道题，学生答对一道题获得$$4$$分，答错一道题或者未作答扣$$1$$分．如果小明在考试中获得$$95$$分，那么他答对了多少道题？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$26$$ "}], [{"aoVal": "C", "content": "$$27$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$40-(40\\times 4-95)\\div (4+1)=27$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1892", "queId": "8aac50a74ff4b16201500e825aa92ac3", "competition_source_list": ["2009年全国迎春杯小学中年级四年级竞赛初赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "小明在桌上将若干个红球排成一排，然后在每相邻的$$2$$个球之间放$$2$$个黄球，最后在每相邻的$$2$$个球之间再放$$2$$个蓝球，这时桌上共有$$2008$$个球，那么其中黄球有~\\uline{~~~~~~~~~~}~个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$223$$ "}], [{"aoVal": "B", "content": "$$251$$ "}], [{"aoVal": "C", "content": "$$252$$ "}], [{"aoVal": "D", "content": "$$446$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题"], "answer_analysis": ["假设原来有红球$$x$$个，那么第一次操作放入了$$2（x-1）$$个黄球，球数增加到$$3x-2$$个，第二次操作放入了$$2（3x-2-1）=6x-6$$个篮球，球数增加到$$3x-2+6x-6=9x-8$$个．所以$$9x-8=2008$$，解得：$$x=224$$,所以黄球有$$2（x-1）=446$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "260", "queId": "fac59503ddf442b4ba7aec33da03abf6", "competition_source_list": ["2012年全国希望杯六年级竞赛初赛第17题"], "difficulty": "3", "qtype": "single_choice", "problem": "从$$1$$，$$2$$，$$3$$，$$4$$，$$\\cdots$$，$$15$$，$$16$$这十六个自然数中，任取出$$n$$个数，其中必有这样的两个数：一个是另一个的$$3$$倍，则$$n$$最小是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["有$$3$$倍关系的数放入一组：（$$1$$，$$3$$，$$9$$）、（$$2$$，$$6$$）、（$$4$$，$$12$$）、（$$5$$，$$15$$）， 其余$$7$$个数每个数放入一组，第一组最多取$$2$$个（$$1$$和$$9$$），其余每组最多只能取$$1$$个， 因此最多取$$12$$个保证没有$$3$$倍关系，再多取一个就可以保证有一个数是另一个数的$$3$$倍． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "873", "queId": "96e115f4fb9446e88619866067fc6d6a", "competition_source_list": ["2008年第6届创新杯六年级竞赛初赛A卷第8题5分", "2008年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "把三个最小的连续的奇质数，分别写在三张纸片上，每张纸片上写且仅写一个奇质数，如果随意从其中取出至少一张组成一个数，其中质数有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$个 "}], [{"aoVal": "B", "content": "$$7$$个 "}], [{"aoVal": "C", "content": "$$8$$个 "}], [{"aoVal": "D", "content": "$$9$$个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数->特殊质数运用->常用质数"], "answer_analysis": ["显然，三个最小的连续的奇质数是$$3$$、$$5$$、$$7$$，如果取出$$1$$张，$$3$$、$$5$$、$$7$$都为质数； 如果取出两张，可以组成的两位数有$$35$$、$$53$$、$$37$$、$$73$$、$$57$$、$$75$$， $$35=5\\times 7$$，$$57=3\\times 19$$，$$75=3\\times 5\\times 5$$，所以$$3$$、$$5$$、$$7$$组成的两位数质数有$$53$$、$$37$$、$$73$$； 如果取出三张，可以组成三位数，由于三个数字的和是$$3+5+7=15$$是$$3$$的倍数，所以组成的任何一个三位数都不是质数． 综上，所有的质数有：$$3$$、$$5$$、$$7$$、$$53$$、$$37$$、$$73$$，一共有$$6$$个． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3410", "queId": "ee061a9d0ba643dcb6b506905845b9b3", "competition_source_list": ["2017年北京学而思杯六年级竞赛年度教学质量监测第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "有一个推理游戏叫做``狼人杀''，$$10$$位同学参与游戏，通过抽牌决定所扮演的角色，事先做好$$10$$张卡牌（除所写文字不同，其余均相同），其中有法官牌$$1$$张，预言家牌$$1$$张，女巫牌$$1$$张，猎人牌$$1$$张，狼人牌$$3$$张．艾迪参与游戏，如果只随机抽取一张，那么艾迪抽到狼人牌的概率是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10 \\% $$ "}], [{"aoVal": "B", "content": "$$20 \\%$$ "}], [{"aoVal": "C", "content": "$$30 \\%$$ "}], [{"aoVal": "D", "content": "$$40 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->计数求概率"], "answer_analysis": ["狼人牌有$$3$$张，一共有$$10$$张卡牌．所以抽到狼人牌的概率是$$\\frac{3}{10}=30 \\% $$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2286", "queId": "7429d49cd8d74e42a48268d879c92693", "competition_source_list": ["2017年四川成都六年级竞赛“全能明星”选拔赛第8题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "下午$$2$$点$$10$$分，时钟上的时针和分针所成的锐角是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$10{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$15{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$20{}^{}\\circ $$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["在下午$$2$$点$$10$$分，分针从数字$$12$$开始转了$$10\\times 6{}^{}\\circ =60{}^{}\\circ $$，时针从数字$$2$$开始转了$$10\\times 0.5{}^{}\\circ =5{}^{}\\circ $$，这时时针与分针所成的锐角为$$60{}^{}\\circ +10\\times 0.5{}^{}\\circ -60{}^{}\\circ =5{}^{}\\circ $$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2024", "queId": "e15ae62ab2b04101a70efd0d790df12b", "competition_source_list": ["2016年北京学而思杯小学高年级五年级竞赛冲刺讲义"], "difficulty": "3", "qtype": "single_choice", "problem": "一个足球的售价是$$30$$元，学校现在计划买$$100$$个足球，甲乙丙三家店提供了不同的优惠方案： 甲店：所有足球一律打七五折出售． 乙店：每买$$5$$个足球，就赠送$$2$$个，不够$$5$$个不送． 丙店：$$40$$个以内（包括$$40$$个）不打折，$$40$$个以上到$$70$$个（包括$$70$$个）的部分打七折，$$70$$个以上的部分打五折． 哪一家店花费最多？ ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->经济问题"], "answer_analysis": ["甲：$$30\\times 0.75\\times 100=2250$$（元）； 乙：$$100\\div 7=14\\cdots\\cdots 2$$，$$(14\\times 5+2)\\times 30=2160$$（元）； 丙：$$40\\times 30+30\\times 30\\times 0.7+30\\times 30\\times 0.5=76\\times 30=2280$$（元）． 丙\\textgreater 甲\\textgreater 乙． 丙花费最多． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "248", "queId": "b5628b0faed64eb0aa23b1136c4905ef", "competition_source_list": ["2018年第23届华杯赛小学高年级竞赛初赛第5题", "2019年广东深圳福田区深圳市高级中学小升初第5题", "2018年华杯赛小学高年级竞赛初赛第5题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1\\sim 20$$这$$20$$个整数中任意取$$11$$个数，其中必有两个数的和等于（~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$22$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["将$$1\\sim 20$$分成如下$$10$$组： $$\\left( 1,20 \\right)$$，$$\\left( 2,19 \\right)$$，$$\\left( 3,18 \\right)$$，$$\\cdots \\cdots $$，$$\\left( 10,11 \\right)$$，每组和均为$$21$$，根据抽屉原理，$$10$$组取$$11$$个数，至少有$$2$$个在同一组和为$$21$$，所以选 $$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3016", "queId": "a5e7504e7bca4eebbcb31179b244835a", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛复赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙二人共带$$100$$千克行李乘火车，甲超重部分交款$$5.6$$元，乙找找那个部分交款$$4.4$$元；若二人行李由一人携带乘车，超重部分交款$$15$$元，乘火车时，每人可免费携带行李的重量是（~ ）千克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["假设免费携带$$x$$千克 根据两次付费可得比例关系$$\\frac{100-2x}{100-x}=\\frac{10}{15}$$，解得$$x=25$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3384", "queId": "cd9ac50a23314be28b13951a00e3ebb3", "competition_source_list": ["2016年创新杯小学高年级五年级竞赛训练题（四）第4题", "2012年全国创新杯五年级竞赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "【闯关3】某班共有$$48$$人，其中$$27$$人会游泳，$$33$$人会骑自行车，$$40$$人会乒乓球．那么，这个班至少有个学生这三项运动都会． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->容斥原理->三量容斥"], "answer_analysis": ["让三项都会的人最少，即让所有人都只会两项，则一共会$$48\\times 2=96$$项，二十几会$$27+33+40=100$$项，即至少有$$100-96=4$$人会三项． 构造如下：$$8$$人会游泳$$+$$自行车；$$15$$人会游泳$$+$$乒乓球；$$21$$人会自行车$$+$$乒乓球；$$4$$人全会． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3064", "queId": "bd44bcfb30a143b9b3ad48e312aee725", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第24题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "分数$$A=1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\cdots +\\frac{1}{16}$$的整数部分是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$A\\textgreater1+\\frac{1}{2}+\\frac{1}{4}\\times 2+\\frac{1}{8}\\times 4+\\frac{1}{16}\\times 8=3$$， $$A\\textless{}1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}\\times 4+\\frac{1}{8}\\times 8+\\frac{1}{16}\\textless{}4$$， ∴$$\\left[ A \\right]=3$$． 答：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1325", "queId": "7095ae3210d24b349eacf43461fcbbab", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2013$$年$$12$$月$$21$$日是星期六，那么$$2014$$年的春节，即$$2014$$年$$1$$月$$31$$日是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "四 "}], [{"aoVal": "C", "content": "五 "}], [{"aoVal": "D", "content": "六 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["方法一：星期六有：$$21\\to 28\\to 4(35)\\to 11\\to 18\\to 25$$，所以 $$31$$日是星期五． 方法二：$$10+31=41$$（天），$$41\\div7=5\\cdots 6$$ ，差一天是星期六，所以$$31$$日是星期五． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3038", "queId": "e0e710bfe0a6411589fd40d56f94124d", "competition_source_list": ["2013年IMAS小学高年级竞赛第一轮检测试题第17题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "定义新运算$$a\\Theta b=\\frac{a+b}{2}$$，如果$$\\frac{3}{4}\\Theta \\left( \\frac{1}{6}\\Theta \\square \\right)=\\frac{1}{2}$$，请问$$\\square $$里应填入什么？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{13}{24}$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "$$\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->定义新运算->反解未知数型"], "answer_analysis": ["可观察出新运算即计算两数之平均值，首先把括弧看成一个整体，则括弧里的数与$$\\frac{3}{4}$$之平均等于$$\\frac{1}{2}$$，即$$\\frac{1}{6}\\Theta \\square =1-\\frac{3}{4}=\\frac{1}{4}$$，从而$$\\square $$与$$\\frac{1}{6}$$之平均等于$$\\frac{1}{4}$$，所以可得$$\\square =\\frac{1}{4}\\times 2-\\frac{1}{6}=\\frac{1}{3}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2225", "queId": "55d4eb9719924c42a1f66616a00cbe27", "competition_source_list": ["2006年五年级竞赛创新杯", "2006年第4届创新杯五年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "某人上山速度为a，沿原路下山速度为2a，那么他的平均速度为( ). ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{2}a$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{3}a$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{4}a$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->比例解行程问题->行程中的比例"], "answer_analysis": ["不妨假设这段路程为$$6a$$，则$$6a\\times 2\\div \\left( 6a\\div a+6a\\div 2a \\right)=\\frac{4}{3}a$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2794", "queId": "69455f2f8c4d42e4b94d09f3849da9e6", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "从小到大排列$$99$$个数，每两个相邻数的差都相等，第$$7$$个与第$$93$$个的和为$$262$$，则这列数的第$$50$$个数为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$51$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$131$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数"], "answer_analysis": ["等差数列中，$${{a}_{2}}+{{a}_{93}}=2{{a}_{50}}$$，则$${{a}_{50}}=262\\div 2=131$$．（$${{a}_{n}}$$表示这列数中第$$n$$个数）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2716", "queId": "524b0ee1e8db42ba96aedcfba67b2728", "competition_source_list": ["2008年四年级竞赛创新杯", "2008年第6届创新杯四年级竞赛复赛第7题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "数$$a$$四舍五入后得到的近似值为$$5.3$$.那么$$a$$的取值范围是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5.25 \\textless{} a \\leqslant 5.35$$ "}], [{"aoVal": "B", "content": "$$5.25 \\textless{} a \\textless{} 5.35$$ "}], [{"aoVal": "C", "content": "$$5.25\\leqslant a\\leqslant 5.35$$ "}], [{"aoVal": "D", "content": "$$5.25\\leqslant a \\textless{} 5.35$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数基础->取近似值"], "answer_analysis": ["当$$a=5.35$$时，$$a$$四舍五入得到的近似值为$$5.4$$，不合题意，排除$$\\text{A}$$、$$\\text{C}$$，当$$a=5.25$$时，$$a$$四舍五入得到的近似值为$$5.3$$，符合题意，排除$$\\text{B}$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3265", "queId": "2990035304064b209352af824857e279", "competition_source_list": ["2009年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "小红的爸爸计划这个星期天带小红到东湖或黄鹤楼或中山公园或古琴台去玩，小军的爸爸也有同样的打算，如果到四个地方去玩的可能性相同，那么小红和小军星期天到同一地方玩的机会有（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.25$$ "}], [{"aoVal": "B", "content": "$$0.125$$ "}], [{"aoVal": "C", "content": "$$0.50$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->计数求概率"], "answer_analysis": ["小红、小军星期天到四个地方（东湖、黄鹤楼、中山公园或古琴台）去玩，共有$$4\\times 4=16$$种情况，他们碰巧到同一地方玩，只有4种情况，因此，小红与小军星期天到同一地方玩的机会有$$4\\div 16=0.25$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "746", "queId": "d5c68f9eadb343e7bde3ed15d1ceea55", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$1$$开始的连续若干个自然数，其奇数之和比偶数之和多$$20$$，那么共有个自然数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$39$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$41$$ "}], [{"aoVal": "D", "content": "$$43$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["将从$$0$$开始的连续自然数倒序书写，则相邻的奇数与偶数之差为$$1$$，共计$$20$$组． 所以会得到$$20$$个奇数，$$20$$个偶数．共计$$40$$个数，所以从$$0\\sim 39$$会有$$40$$个数． 故选择$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1426", "queId": "43d16904aacc46d898aca8e0c644212e", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛A卷第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在一次数学竞赛中，甲答错了所有题目的四分之一，乙答错了$$3$$道，甲乙都错的题占题目总数的六分之一，那么甲乙都对的题目共有． 道． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->量率对应求单位1"], "answer_analysis": ["如果甲、乙都错的题，都是$$3$$道， 则题目总数为：$$3\\div \\frac{1}{6}=18$$（道）， 则甲错题数不是整数，故不可以； 甲、乙都错的题是一道， 则题目总数为$$6$$道， 则甲错题数也不是整数，故不可以； 甲、乙都错的题目是$$2$$道， 则题目总数为：$$2\\div \\frac{1}{6}=12$$（道）， 则甲错题数为：$$12\\times \\frac{1}{4}=3$$（道）， 所以甲、乙答错的题目数量为：$$2+1+1=4$$（道）， 所以剩下的就是甲乙都答对的题目数：$$12-4=8$$（道）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "71", "queId": "0fbbb1f5cbbe4b8ea6eec2dc80b5ec75", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在垒球比赛中，若赢$$1$$场得$$3$$分，平$$1$$场得$$1$$分，输$$1$$场不得分．总共$$4$$个队，每个队都与其他队交锋$$4$$场，最终四个参赛队的总积分为：$$A$$队$$22$$分，$$B$$队$$19$$分，$$C$$队$$14$$分，$$D$$队$$12$$分．那么有场比赛为平局． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->逻辑分析"], "answer_analysis": ["依题意可知比赛总场次为$$24$$场 现在所有队伍获得的总分值为：$$22+19+14+12=67$$（分），$$24$$场比赛，最高有$24\\times3=72$分，所以平局有$$72-67=5$$场． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1870", "queId": "bba8bafea29d4506899f790af98a7e0d", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级六年级竞赛邀请赛训练题（四）"], "difficulty": "1", "qtype": "single_choice", "problem": "~ 一项工程，甲、乙两人合做$$12$$天可以完成，乙、丙合做$$15$$天可以完成，甲、丙合做$$20$$天完成，求三队合做（~ ~ ）天可以完成． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$天 "}], [{"aoVal": "B", "content": "$$10$$天 "}], [{"aoVal": "C", "content": "$$12$$天 "}], [{"aoVal": "D", "content": "$$15$$天 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->已知工时反推->多人合作"], "answer_analysis": ["求三队工效和：$$\\left( \\frac{1}{12}+\\frac{1}{15}+\\frac{1}{20} \\right)\\div 2=\\frac{1}{10}$$，$$1\\div \\frac{1}{10}=10$$．选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2562", "queId": "992bed417802493c86aab940d568bb55", "competition_source_list": ["2012年第10届全国创新杯小学高年级六年级竞赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "~$$A$$、$$B$$、$$C$$正整数，且$$A+\\frac{1}{B+\\frac{1}{C+1}}=\\frac{24}{5}$$,则$$A+2B+3C=$$（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->繁分数->繁分数方程"], "answer_analysis": ["$$A+\\frac{1}{B+\\frac{1}{C+1}}=4+\\frac{4}{5}=4+\\frac{1}{1+\\frac{1}{4}}=4+\\frac{1}{1+\\frac{1}{3+1}}$$，那么$$A=4$$，$$B=1$$，$$C=3$$，则$$A+2B+3C=4+2+9=15$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3246", "queId": "9020973ff2e74aa7b26c9e36d5d2add8", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（二）第1题", "五年级其它"], "difficulty": "2", "qtype": "single_choice", "problem": "五年级某班有$$47$$名小朋友，他们中每人至少订了《少年报》、《学习报》、《儿童报》中的一种报刊，则其中至少有名小朋友订的报刊相同． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["每人至少订一种，共有$$3+3+1=7$$种不同的订法，又因为$$47\\div 7=6\\cdots \\cdots 5$$，所以至少有$$6+1=7$$名小朋友订的报刊相同． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "548", "queId": "128b16fb44a946d0886f15e3a17e9ab1", "competition_source_list": ["2004年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "【2004年五年级竞赛创新杯】 有两个两位数，它们的最大公因数是$$8$$，最小公倍数是$$96$$，这两个数的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$78$$ "}], [{"aoVal": "C", "content": "$$84$$ "}], [{"aoVal": "D", "content": "$$96$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公因数与最大公因数->两数的最大公因数"], "answer_analysis": ["因为两个两位数的最大公因数为$$8$$，可设这两个数分别为$$8a，8b$$($$a，b$$互质)，那么$$8\\times a\\times b=96$$，即$$a\\times b=12$$，则$$a，b$$为$$\\left( 3,4 \\right),\\left( 1,12 \\right)$$，又因为这两个数皆为两位数，所以这两个数只能为$$24$$和$$32$$，$$24+32=56$$，选 A. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2117", "queId": "fe56c4b2a0964c45b1563c35f76db5a4", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "【补3】小明参加$$\\text{YMO}$$竞赛，共有$$20$$道赛题，答对一题给$$5$$分，答错一题或不答扣$$1$$分．小明得了$$76$$分．小明答对了题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["根据题意分析可知，假设$$20$$道题全部做对， 则得：$$20\\times5=100$$（分），现在比满分少了$$100-76=24$$（分）， 又因为答错或不答一题比答对一题少$$1+5=6$$（分）， 也就是做错了：$$24\\div6=4$$（题）， 所以做对了：$$20-4=16$$（题）， 故选答案$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "565", "queId": "036f000ead7e4cd78c4df8ddaaca8c58", "competition_source_list": ["2011年全国学而思杯五年级竞赛第5题", "2011年北京学而思综合能力诊断六年级竞赛第3题", "2011年全国学而思杯四年级竞赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知地震级数每升$$1$$级，地震释放能量大约扩大到原来的$$30$$倍，那么$$8$$级地震释放能量$$5$$级地震的多少倍？． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30\\times 3$$ "}], [{"aoVal": "B", "content": "$$30\\times 2$$ "}], [{"aoVal": "C", "content": "$$30\\times 30$$ "}], [{"aoVal": "D", "content": "$$30\\times 30\\times 30$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$8$$级和$$5$$级差了$$3$$级，所以$$8$$级是$$5$$级的$$30\\times 30\\times 30$$倍. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "727", "queId": "55fe28f2522f40d5b0fbb29a590f30a3", "competition_source_list": ["2014年IMAS小学高年级竞赛第二轮检测试题第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个两位数是完全平方数，它的两个数码之和也恰好是完全平方数，请问所有这样的完全平方数之和为多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$110$$ "}], [{"aoVal": "C", "content": "$$117$$ "}], [{"aoVal": "D", "content": "$$181$$ "}], [{"aoVal": "E", "content": "$$271$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["如果一个两位数是完全平方数，那么这个两位数的所有可能值是： $$16$$，$$25$$，$$36$$，$$49$$，$$64$$，$$81$$， 而$$3+6={{3}^{2}}$$，$$8+1={{3}^{2}}$$， 所以符合条件的两位数为$$36$$，$$81$$， $$36+81=117$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2352", "queId": "8aac50a7519fa10a01519fed04c2009e", "competition_source_list": ["2016年全国华杯赛小学中年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$124+129+106+141+237-500+113=$$（~~~~~~~~~ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$350$$ "}], [{"aoVal": "B", "content": "$$360$$ "}], [{"aoVal": "C", "content": "$$37$$ "}], [{"aoVal": "D", "content": "$$380$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["原式$$=\\left( 124+106 \\right)+\\left( 129+141 \\right)+\\left( 237+113 \\right)-500$$ ~~~~~~~~$$=230+270+350-500$$ ~ ~ ~ ~ $$=500-500+350$$ ~ ~ ~ ~ $$=350$$ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1342", "queId": "b9a06b0f40e746f5bc2d70392b3c3bf5", "competition_source_list": ["2020年新希望杯二年级竞赛决赛（8月）第6题", "2020年新希望杯二年级竞赛初赛（个人战）第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "李叔叔排队买票，他前面有$$11$$人，后面有$$12$$人．一共有人排队买票． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$24$$ "}], [{"aoVal": "E", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->应用题模块排队问题->单主角求总数", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["加上李叔叔自己，一共：$$11+1+12=24$$（人）， 故选择：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1261", "queId": "b0366a4d31e44da58fda9701f9255288", "competition_source_list": ["2017年第13届湖北武汉新希望杯六年级竞赛决赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "水池有甲、乙两根进水管，单独打开甲进水管$$6$$小时可注满空水池，单独打开乙进水管$$4$$小时可注满空水池．如果按照甲、乙、甲、乙$$\\ldots\\ldots$$的顺序轮流打开$$1$$小时，注满空水池需小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->工程问题->合作工程问题->接力施工问题"], "answer_analysis": ["甲、乙$$2$$小时一个周期：$$\\frac{1}{4}+\\frac{1}{6}=\\frac{5}{12}$$，两个周期后还剩下：$$1-2\\times \\frac{5}{12}=\\frac{1}{6}$$，剩下的甲做：$$\\frac{1}{6}\\div \\frac{1}{6}=1$$小时，共需$$2\\times 2+1=5$$小时． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1902", "queId": "ae77fc2e337e4aed8ad7f35ccbff16ae", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$42$$个小朋友排成一队去秋游，从排头往后数，小$$Y$$是第$$22$$个，从排尾往前数，小$$M$$是第$$22$$个，小$$Y$$和小$$M$$中间有个人． 42 students are standing in a line. Counting from the front, Y is the 22nd. Counting from the back, M is the 22nd. There are~\\uline{~~~~~~~~~~}~people between Y and M. ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->应用题模块->应用题模块排队问题"], "answer_analysis": ["根据题意分析可知，小$$Y$$后面有$$42-22=20$$（人），$$22-20-1=1$$（人）．那么小$$M$$应该在小$$Y$$的正前方，所以小$$Y$$和小$$M$$之间有$$0$$个人． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2214", "queId": "4808c9e1cd924160bb34fd818261ab19", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第22题"], "difficulty": "2", "qtype": "single_choice", "problem": "某天$$110$$指挥中心接到报警电话，得知有几位驴友被困在深山里的某个角落，马上确定最优解救方案，随后消防官兵乘车立即出发．如果行驶$$2$$个小时后，将车速提高 $$\\frac{1}{5}$$，就可比预定时间提前$$30$$分钟赶到；如果先按原速度行驶$$80$$千米，再将车速提高$$\\frac{1}{4}$$，就可比预定时间提前$$40$$分钟赶到，则消防官兵一共需要乘车行驶的总路程是千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$210$$ "}], [{"aoVal": "B", "content": "$$220$$ "}], [{"aoVal": "C", "content": "$$230$$ "}], [{"aoVal": "D", "content": "$$240$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设原来的车速为$$v$$千米/小时，预定时间为$$t$$小时， 由第一个条件可知$$2v+\\left( 1+\\frac{1}{5} \\right)v\\times \\left( t-2-\\frac{30}{60} \\right)=vt$$， 两边同时除以$$v$$，$$2+\\left( 1+\\frac{1}{5} \\right)\\left( t-2-\\frac{30}{60} \\right)=t$$， 解得$$t=5$$， 由第二个条件可得$$80+\\left( 1+\\frac{1}{4} \\right)v\\times \\left( t-\\frac{80}{v}-\\frac{40}{60} \\right)=vt$$， 把$$t=5$$代入，解得$$v=48$$， 所以总路程$$=48\\times 5=240$$（千米）． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2168", "queId": "220e5d46da7041d6ab6a7c2c91220547", "competition_source_list": ["2021年世界少年奥林匹克数学竞赛五年级竞赛复赛第9题12分"], "difficulty": "1", "qtype": "single_choice", "problem": "两只龙舟同时从江上的东西两岸相对驶出．第一次距东岸$$250$$米处相遇．相遇后继续前进，到达对岸后立即返回，第二次相遇在离西岸$$160$$米处．如果两只龙舟在行驶中速度不变，江面东西两岸的距离是~\\uline{~~~~~~~~~~}~米？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$660$$ "}], [{"aoVal": "B", "content": "$$590$$ "}], [{"aoVal": "C", "content": "$$340$$ "}], [{"aoVal": "D", "content": "$$560$$ "}], [{"aoVal": "E", "content": "$$230$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["根据题意则相距：$$250\\times 3-160=750-160=590$$米． 答：相距$$590$$米． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2600", "queId": "b9ae1bbe156246a2914bff405abcd558", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（二）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "下列说法正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "两个小数相乘，积一定是小数 "}], [{"aoVal": "B", "content": "一个小数四舍五入后是$$2.34$$，这个小数最大可能是$$2.344$$ "}], [{"aoVal": "C", "content": "小数都比整数小 "}], [{"aoVal": "D", "content": "循环小数都是无限小数 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数基础->取近似值"], "answer_analysis": ["$$A$$项，$$2.5\\times 0.4=1$$，不是小数； $$B$$项，这个小数没有最大值，例如：$$2.3499\\cdots $$ $$C$$项，不一定，例如：$$3.6\\textgreater2$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2092", "queId": "f00fd084651640ecab3916629622026d", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（三）第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "``枫叶新希望杯''知识竞赛，共$$50$$道选择题，评分标准是：答对$$1$$题得$$3$$分，答错$$1$$题扣$$1$$分，不答的题得$$0$$分．若陈同学最终得了$$95$$分，则他答错的题最多有 道． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设答错$$x$$道，则答对$$\\frac{95+x}{3}$$，要使$$x$$最大，则不答的题应当最少． 当不答的题为$$0$$道时，$$x+\\frac{95+x}{3}=50$$，$$x=\\frac{55}{4}$$（舍），当不答的题为$$1$$道时，$$x+\\frac{95+x}{3}+1=50$$，$$x=13$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "686", "queId": "8b70fb4aeb5f4a5f96f4bc6a39f622ee", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（三）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2$$、$$4$$、$$6$$、$$8\\cdots \\cdots $$、$$98$$,$$100$$这$$50$$个偶数的各位上的数字之和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$425$$ "}], [{"aoVal": "B", "content": "$$426$$ "}], [{"aoVal": "C", "content": "$$427$$ "}], [{"aoVal": "D", "content": "$$428$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$\\left( 0+2+4+6+8 \\right)\\times 10+\\left( 1+2+\\ldots \\ldots +9 \\right)\\times 5+1=426$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1757", "queId": "80c3faab4b724eee88a19e256d993552", "competition_source_list": ["2017年全国小学生数学学习能力测评六年级竞赛初赛第8题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "一种商品先降价$$\\frac{1}{8}$$又提价$$\\frac{1}{8}$$，现价与原价相比． ", "answer_option_list": [[{"aoVal": "A", "content": "现价高 "}], [{"aoVal": "B", "content": "原价高 "}], [{"aoVal": "C", "content": "相等 "}], [{"aoVal": "D", "content": "无法比较 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->折扣问题"], "answer_analysis": ["设商品的原价为$$x$$， ∴现价为：$$x\\cdot \\left( 1-\\frac{1}{8} \\right)\\cdot \\left( 1+\\frac{1}{8} \\right)=\\frac{63}{64}x$$， ∴$$x\\textgreater\\frac{63}{64}x$$ ， ∴原价$$\\textgreater$$现价． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "886", "queId": "7422f6e22b2c4e0eb600b5831750a5cf", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（五）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "在斐波那契数列：$$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$34$$，$$55$$，$$\\cdots $$中，到第$$2017$$个数止，共有个奇数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1343$$ "}], [{"aoVal": "B", "content": "$$1344$$ "}], [{"aoVal": "C", "content": "$$1345$$ "}], [{"aoVal": "D", "content": "$$1346$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的认识"], "answer_analysis": ["这串数的奇偶性：奇，奇，偶，奇，奇，偶，奇，奇，偶，$$\\cdots \\cdots $$周期为$$3$$，$$2017\\div 3=762\\cdots \\cdots 1$$，奇数共有$$2\\times 672+1=1345$$（个）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "652", "queId": "2733571f0be146d1aece11ef4d2b86cd", "competition_source_list": ["2017年第15届湖北武汉创新杯小学高年级五年级竞赛��赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "课堂上，老师说：\"请打开课本\"，学生问：\"多少页？\"老师说：\"你一眼看见两页页码的乘积是$$930$$\"．学生说：\"知道了．\"则这两页的页码依次是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$，$$31$$ "}], [{"aoVal": "B", "content": "$$6$$，$$155$$ "}], [{"aoVal": "C", "content": "$$5$$，$$186$$ "}], [{"aoVal": "D", "content": "$$1$$，$$930$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数->分解质因数（式）"], "answer_analysis": ["$$930=30\\times 31$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "338", "queId": "539bf589564145ada0a1a262afbe3c14", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "盒中有形状、大小、质料相同的红、白、黑颜色的球各$$10$$个，摸出若干个，要保证摸出的球中至少有$$3$$个球同色，摸出球的个数至少为个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["最不利原则．要保证拿到一种颜色至少有$$3$$个，则根据最不利原则，可先取每种颜色$$2$$个，最后不论取哪种颜色，都一定可以满足条件，即需要$$2\\times 3+1=7$$个． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3363", "queId": "6ecbafe0001e470aabe9adb0002ac3b4", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$600$$的有很多因数，这些因数中有个是$$6$$的倍数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["方法一：枚举法（不推荐）$$600$$的因数有$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$8$$，$$10$$，$$12$$，$$15$$，$$20$$，$$24$$，$$25$$，$$30$$，$$40$$，$$50$$，$$60$$，$$75$$，$$100$$，$$120$$，$$150$$，$$200$$，$$300$$，$$600$$，其中是$$6$$的倍数的有$$6$$，$$12$$，$$24$$，$$30$$，$$60$$，$$120$$，$$150$$，$$300$$，$$600$$这$$9$$个． 故选$$\\text{B}$$． 方法二：因数个数定理。 600=2\\textsuperscript{3}$$\\times $$3$$\\times $$5\\textsuperscript{2}。 思路1：根据加乘原理，$$2$$与$$3$$必须至少选一个，故$$2$$的选法有$$3$$种，$$3$$的选法有$$1$$种，$$5$$的选法有$$3$$种（不选$$5$$、选$$1$$个$$5$$、选$$2$$个5），故$$3\\times 1\\times$$ （2+1）=9（个）； 思路2: $$600\\div 6=100$$， 100=2\\textsuperscript{2}$$\\times $$5\\textsuperscript{2~}$$，再根据因数个数定理：$$ （2+1）$$\\times （2+1）=9$$（个） "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2747", "queId": "88b40b13de9f4e29b3aaf2aa4a7700fc", "competition_source_list": ["2017年第17届全国中环杯三年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$325\\times 337+650\\times 330+975=$$ ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$33700$$ "}], [{"aoVal": "B", "content": "$$65000$$ "}], [{"aoVal": "C", "content": "$$32500$$ "}], [{"aoVal": "D", "content": "$$325000$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->运算求解", "课内体系->七大能力->运算求解"], "answer_analysis": ["原式$$=325\\times 337+325\\times 2\\times 330+325\\times 3$$ $$=325\\times (337+2\\times 330+3)$$ $$=325\\times 1000$$ $$=325000$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1556", "queId": "40b943dcfb6249339b9877dc2d06683b", "competition_source_list": ["2017年全国美国数学大联盟杯小学高年级六年级竞赛第17题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "（这是一道挑战题！）There are only white cars and black cars．$$If$$ the ratio of white cars to black cars is $$3$$ to $$8$$ and there are a total of $$242$$ cars， how many black cars and white cars are there？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$66$$ "}], [{"aoVal": "B", "content": "$$110$$ "}], [{"aoVal": "C", "content": "$$129$$ "}], [{"aoVal": "D", "content": "$$176$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->比例应用题->按比分配"], "answer_analysis": ["停车场只有黑白两种车，白车与黑车的比率是$$3:8$$，黑白车的总数是$$242$$辆．请问黑车比白车多多少辆？ ratio比率； a total of 总数； more．．than 比．．多． 白车数量：$$\\frac{3}{11}\\times 242=66$$辆，黑车数量：$$\\frac{8}{11}\\times 242=176$$辆．黑车比白车多$$176-66=110$$辆． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1233", "queId": "273464817e5d475894d3c84dd47cd07f", "competition_source_list": ["2003年第1届创新杯六年级竞赛复赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "四个同学进行计算比赛，比赛内容是：在$$9$$、$$10$$、$$11$$、$$\\cdots \\cdots $$、$$67$$、$$68$$这$$60$$个自然数的相邻两数之间任意添加符号``$$+$$''或``$$-$$''，然后进行计算．四个同学得到的结果分别是$$2274$$、$$2003$$、$$2300$$、$$2820$$，老师看后指出．这四个结果中只有一个是正确的．这个正确的结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2274$$ "}], [{"aoVal": "B", "content": "$$2003$$ "}], [{"aoVal": "C", "content": "$$2300$$ "}], [{"aoVal": "D", "content": "$$2320$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["由于$$9+10+11+\\cdots 68=2310$$，由此可知$$2320$$是错误的．由于$$2274$$、$$2003$$、$$2300$$都小于小于$$2310$$，所以减的数较多，由于减一个数，总和里面就要少这个数的$$2$$倍，如减$$2$$，则是$$2310-2\\times 2=2306$$，所以只要是小于$$2310$$．据此分析即 $$9+10+11+\\cdots 68=2310$$，$$2320\\textgreater2310$$，故$$\\text{D}$$错误； $$\\left( 2310-2274 \\right)\\div 2=18$$，$$18\\div 2=9$$，所以在$$9$$前是减号即可，符合题意． $$\\left( 2310-2003 \\right)=307\\textgreater68$$，错误 $$\\left( 2310-2000 \\right)\\div 2=155\\textgreater68$$，错误． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2453", "queId": "6208d66c3bec454a8b2319071d908a67", "competition_source_list": ["2017年湖北武汉中环杯五年级竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个等差数列，首项是$$5$$，第$$10$$项为$$32$$，则第$$20$$项为~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$61$$ "}], [{"aoVal": "C", "content": "$$62$$ "}], [{"aoVal": "D", "content": "$$63$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "Overseas Competition->知识点->计算模块->数列与数表->等差数列->等差数列求公差"], "answer_analysis": ["先求公差：$$(32-5)\\div (10-1)=3$$，于是第$$20$$项是$$3\\times (20-1)+5=62$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2444", "queId": "109d9a35e38d4e54a3b302a190926a4b", "competition_source_list": ["2017年IMAS小学高年级竞赛（第一轮）第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$2.\\overline{718}$$、$$2.7\\overline{18}$$、$$2.71\\overline{8}$$、$$2.71828$$按从小到大的顺序排列，请问下列哪一项的式子是正碓的？（注：在小数点后面的数码上方添加横线代表循环小数） ", "answer_option_list": [[{"aoVal": "A", "content": "$$2.7\\overline{18}\\textless2.\\overline{718}\\textless2.71828\\textless2.71\\overline{8}$$ "}], [{"aoVal": "B", "content": "$$2.71828\\textless2.7 \\overline{18}\\textless2.\\overline{718}\\textless2.71 \\overline{8}$$ "}], [{"aoVal": "C", "content": "$$2.7 \\overline{18}\\textless2.71828\\textless2.\\overline{718}\\textless2.71 \\overline{8}$$ "}], [{"aoVal": "D", "content": "$$2.71828\\textless2.71\\overline{8}\\textless2.7\\overline{18}\\textless2.\\overline{718}$$ "}], [{"aoVal": "E", "content": "$$2.7 \\overline{18}\\textless2.\\overline{718}\\textless2.71\\overline{8}\\textless2.71828$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$2.\\overline{718}=2.718718\\cdots $$、$$2.7\\overline{18}=2.71818\\cdots $$、$$2.71\\overline{8}=2.71888\\cdots $$与$$2.71828$$，比较它们小数点后第四位可得$$2.7 \\overline{18}\\textless2.71828\\textless2.\\overline{718}\\textless2.71 \\overline{8}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3286", "queId": "deeb727c6ac24f7a8fa7c1744e739107", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "我们定义像：$$31024$$、$$98567$$这样的五位数为位``神马数''，``神马数''是中间的数字最小，从中间往两边越来越大，且各位数字均不相同，那么，这样的五位数有个 ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1512$$ "}], [{"aoVal": "B", "content": "$$3024$$ "}], [{"aoVal": "C", "content": "$$1510$$ "}], [{"aoVal": "D", "content": "$$3020$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "课内体系->思想->对应思想"], "answer_analysis": ["考察是计数问题中的排列组合． $$0～9$$这$$10$$个数中任意挑选$$5$$个都可以组成``神马数''，$$\\text{C}_{10}^{5}=\\frac{10\\times9\\times 8\\times 7\\times 6}{5\\times 4\\times 3\\times 2\\times 1}=252$$种；在被挑选的$$5$$个数中，最小的放中间，剩下的$$4$$个数进行组合，从中任意挑选$$2$$个可以放在左边或者右边，$$\\text{C}_{4}^{2}=6$$种； 在此一定要注意：$$4$$个数中任选$$2$$个放在左边然后再放到右边数的顺序改变了． 所以共有``神马数''$$252\\times 6=1512$$个． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2247", "queId": "ff8080814502fa2401450b0db35911b5", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第11题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙两车分别从$$A$$、$$B$$两地同时出发，相向而行，甲车的速度大于乙车．甲行驶了$$60$$千米后和乙车在$$C$$点相遇．此后甲车继续向前行驶，乙车掉头与甲车同向行驶．那么当甲车到达$$B$$地时，甲乙两车最远相距（~~~~~~~ ）千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$25$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["假设甲走$$60$$千米时，乙走了$$a$$千米，甲到达$$B$$地时，乙车应走$$\\frac{a}{60}\\times a=\\frac{{{a}^{2}}}{60}$$千米，此时甲、乙相差最远为$$a-\\frac{a^{2}}{60}=\\frac{1}{60}\\times(60-a)\\times a$$，和一定，差小积大，$$60-a=a$$，$$a=30$$．甲、乙最远相差$$30-\\frac{900}{60}=15$$（千米）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "703", "queId": "f5ec2bb01aa94010a94e9a19da69652a", "competition_source_list": ["2011年北京学而思综合能力诊断六年级竞赛第3题", "2011年全国学而思杯四年级竞赛第5题", "2011年全国学而思杯五年级竞赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知地震级数每升$$1$$级，地震释放能量大约扩大到原来的$$30$$倍，那么$$7$$级地震释放能量$$4$$级地震的多少倍？． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30\\times 3$$ "}], [{"aoVal": "B", "content": "$$30\\times 2$$ "}], [{"aoVal": "C", "content": "$$30\\times 30$$ "}], [{"aoVal": "D", "content": "$$30\\times 30\\times 30$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$7$$级和$$4$$级差了$$3$$级，所以$$7$$级是$$4$$级的$$30\\times 30\\times 30$$倍. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2047", "queId": "ef6a9286835e4cefbff20ab9bfac3b18", "competition_source_list": ["2018年湖北武汉创新杯小学高年级五年级竞赛初赛数学思维能力等级测试第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明从学校步行回家，当走了全程的一半时下雨了，他继续走；当雨停了时，剩下路程是他在雨中步行路程的$$\\frac{5}{7}$$，那么，小明在雨中步行的路程是全程的（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{14}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{24}$$ "}], [{"aoVal": "C", "content": "$$\\frac{7}{24}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{7}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["设雨中步行了$$7$$份，还剩下$$5$$份，全程$$\\left( 7+5 \\right)\\times 2=24$$份，小明在雨中步行的路程是全程的$$\\frac{7}{24}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2528", "queId": "4265583ddad14c6d905fa6242e3ad3fd", "competition_source_list": ["2017年北京学而思杯小学中年级三年级竞赛年度教学质量测评第19题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "按规律填数，那么横线上填入的数是． $$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，~\\uline{~~~~~~~~~~}~， $$34$$，$$55$$，$\\cdots\\cdots$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$23$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["斐波那契数串，从第$$3$$个数开始，每��数等于前两个数的和． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1578", "queId": "ff8080814518d524014519096ede031e", "competition_source_list": ["2019年四川成都锦江区四川师范大学附属第一实验中学小升初（八）第6题3分", "2018年四川成都锦江区四川师范大学附属第一实验中学小升初（五）第5~0题3分", "2014年全国迎春杯四年级竞赛初赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "动物园的饲养员把一堆桃子分给若干只猴子，如果每只猴子分$$6$$个，剩$$57$$个桃子；如果每只猴子分$$9$$个，就有$$5$$只猴子一个也分不到，还有一只猴子只分到$$3$$个．那么，有（~~~~ ）个桃子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$216$$ "}], [{"aoVal": "B", "content": "$$324$$ "}], [{"aoVal": "C", "content": "$$273$$ "}], [{"aoVal": "D", "content": "$$301$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["每只猴子多分了$$3$$个，分了$$5\\times 9+(9-3)+57=108$$ （个），那么共$$108\\div 3=36$$（只）猴子．共$$36\\times6+57=273$$（个）桃子． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1520", "queId": "c7f472ce5e6c47edabed6bc1c99e08de", "competition_source_list": ["2007年四年级竞赛创新杯"], "difficulty": "3", "qtype": "single_choice", "problem": "一片匀速生长的草场长满牧草，可供$$5$$头牛吃$$30$$天，或者可供$$4$$头牛吃$$40$$天。已知每头牛每天吃$$1$$份草，那么草场原有份草，每天新长出份草。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$240$$ $$2$$ "}], [{"aoVal": "B", "content": "$$120$$ $$2$$ "}], [{"aoVal": "C", "content": "$$240$$ $$1$$ "}], [{"aoVal": "D", "content": "$$120$$ $$1$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->牛吃草问题->牛吃草基本型->求头数"], "answer_analysis": ["$$5$$头牛$$30$$天共吃$$30\\times5=150$$（份）草，$$4$$头牛$$40$$天共吃$$4\\times40=160$$（份）草，总草量相差部分是因为生长天数不同，故草的生长速度为$$(160-150)\\div(40-30)=1$$（份/天），草场原有$$150-1\\times30=120$$（份）草。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1728", "queId": "ad148df683c046f483a5b7b08904ccdf", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "五个小学生的年龄和是$$54$$岁，三个大学生的年龄和是$$72$$岁，若要这五个小学生的年龄和与这三个大学生的年龄和相等，还需要经过年． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$9$$~ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题基本关系->年龄和"], "answer_analysis": ["$$(72-54)\\div (5-3)=18\\div 2=9$$（年）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2817", "queId": "6dfaf1b673dd4af9b56dea89cf348136", "competition_source_list": ["2017年全国希望杯小学高年级五年级竞赛初赛考前100题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一串数，最前面的$$4$$个数是$$2，0，1，6，$$从第$$5$$个数起，每一个数是它前面相邻$$4$$个数之和的个位数字，问在这一串数中，会依次出现$$2，0，1，7$$这$$4$$个数吗？ ", "answer_option_list": [[{"aoVal": "A", "content": "会 "}], [{"aoVal": "B", "content": "不会 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据规律，这串数是$$2，0，1，6，9，6，2，3，0，1，6，0，7，4，7，8，6，5，6，5，... $$呈现的规律是：偶偶奇偶奇，而$$2，0，1，7$$是：偶偶奇奇 按照上述的规律两个奇数不可能相邻，所以不可能出现$$2，0，1，7$$这$$4$$个数． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1475", "queId": "b5676fedc81c49a8a233818e129c5b09", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "求$$17$$个自然数的平均数，结果保留两位小数，甲得到$$11.28$$，这个数百分位上的数字错了，求正确答案是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11.24$$ "}], [{"aoVal": "B", "content": "$$11.28$$ "}], [{"aoVal": "C", "content": "$$11.29$$ "}], [{"aoVal": "D", "content": "$$11.24$$ 或$$11.29$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["因为$$11.195\\times 17=190.315$$，$$11.295\\times 17=192.015$$，所以$$17$$个自然数的和是$$191$$或$$192$$，$$191\\div 17\\approx 11.24$$，$$192\\div 17\\approx 11.29$$，故正确答案是$$11.24$$ 或$$11.29$$ ． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3296", "queId": "99df8ae3ddfd46858fa8e5c51e3227ee", "competition_source_list": ["2014年IMAS小学中年级竞赛第一轮检测试题第19题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "妈妈给小明一个红色大盒子，里面装着$$6$$个蓝色盒子，每个蓝色盒子里又装着$$4$$个绿色盒子，请问小明总共有多少个盒子？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$25$$ "}], [{"aoVal": "E", "content": "$$31$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->加乘原理综合"], "answer_analysis": ["由题意可知，绿色盒子有$$4\\times 6=24$$个，蓝色盒子有$$6$$个，红色盒子有$$1$$个，所以小明总共 有$$24+6+1=31$$个盒子，故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3190", "queId": "15ac506d2bbb47acb9ae5fd70629a787", "competition_source_list": ["2019年第24届YMO五年级竞赛决赛第10题3分", "2020年第24届YMO五年级竞赛决赛第10题3分"], "difficulty": "3", "qtype": "single_choice", "problem": "有一个密码箱的密码由六位数码组成，结果程序出现了问题，只要和原来的密码数字相同而排列顺序不同也能打开这个密码箱．现在知道有$$180$$个不同的六位数码都能打开，而且这六个数码里没有$$3$$，那么原来这个密码箱的密码数字的和最小是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->组数问题->有特殊要求的组数问题"], "answer_analysis": ["①$$6$$个数字互不相同， 有$$6\\times 5\\times 4\\times 3\\times 2\\times 1=720$$个不同六位数码舍去； ②只有$$2$$个数字相同， 有$$6\\times 5\\times 4\\times 3=360$$个不同六位数码，舍去； ③只有$$3$$个数字相同， 有$$6\\times 5\\times 4=120$$个不同六位数码，舍去； ④有$$2$$对数字分别相同， 有$$6\\times 5\\times \\left( 3+2+1 \\right)=180$$个不同六位数码，符合， 且$$6$$个数字里没有$$3$$，则最小为$$0$$，$$0$$，$$1$$，$$1$$，$$2$$，$$4$$， 故和最小为$$0+0+1+1+2+4=8$$； ⑤其他情况均小于$$180$$个六位数码； 综上，故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2250", "queId": "ff808081466b745501467017af8a07a6", "competition_source_list": ["2014年全国华杯赛小学高年级竞赛初赛B卷第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "小华下午$$2$$点要到少年宫参加活动，但他的手表每个小时快了$$4$$分钟，他特意在上午$$10$$点时对好了表．当小华按照自己的表于下午$$2$$点到少年宫时，实际早到了（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["小华所带的``快表''每小时快了$$4$$分钟，说明准确时间走$$60$$分钟的时候，``快表''已经走了$$64$$分钟了，这样我们就可以得到$$\\frac{快表}{准表}=\\frac{64}{60}=\\frac{16}{15}$$；现在快表走了$$4\\times60=240$$，那么标准表走了$$240\\times 15\\div 16=225$$；所以实际上早到了$$240-225=15$$，选$$B$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1595", "queId": "68e69549a1b34df38d5cb11180b0b45c", "competition_source_list": ["其它改编自2015年全国希望杯六年级竞赛初赛第17题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一项工程，甲单独做需$$6$$小时，乙单独做需$$8$$小时，丙单独做需$$10$$小时，上午$$8$$时三人同时开始，中间甲有事离开，结果到中午$$12$$点工程才完成，则甲离开的时间是上午~\\uline{~~~~~~~~~~}~时~\\uline{~~~~~~~~~~}~分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$；$$30$$ "}], [{"aoVal": "B", "content": "$$8$$；$$36$$ "}], [{"aoVal": "C", "content": "$$9$$；$$30$$ "}], [{"aoVal": "D", "content": "$$9$$；$$36$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["甲效率$$\\frac{1}{6}$$，乙效率$$\\frac{1}{8}$$，丙效率$$\\frac{1}{10}$$， 乙，丙都工作$$12-8=4$$（小时）， 甲工作：$$\\left( 1-\\frac{1}{8}\\times 4-\\frac{1}{10}\\times 4 \\right)\\div \\frac{1}{6}=\\frac{3}{5}$$小时$$=36$$（分）， 甲离开时间为：$$8:36$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2533", "queId": "50181f209d84453ca71458b7827347c0", "competition_source_list": ["四年级其它小学奥数优秀生培养教程7级", "2005年第6届中环杯四年级竞赛复赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "规定：$$a*b=2a+3b$$，$$a\\triangle b=2ab$$，如果$$\\left( 2*x \\right)\\triangle 2=64$$，那么$$x=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["按照$$*$$运算法则为前数$$\\times 2-$$后数$$\\times 3$$，按照$$\\triangle $$运算法则为$$2\\times $$两数之积． 在解方程时，由于括号内含有未知数，所以要由外向内化简方程，一步一步展开． 也可以将$$2*x$$看作一个整体，先求结果，再求$$x$$． 有括号时先算括号里的，$$2*x=2\\times 2+3x=4+3x$$，$$\\left( 2*x \\right)\\triangle 2=\\left( 4+3x \\right)\\triangle 2=2\\times \\left( 4+3x \\right)\\times 2=16+12x$$，得一元一次方程$$16+12x=64$$，解得$$x=4$$． 设$$2*x=A$$，$$A\\Delta 2=2\\times 2\\times A=64,A=16.$$所以，$$2*x=2\\times 2+3x=16$$，解得$$x=4$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1317", "queId": "30f7b0868b05420ba833318adb01e357", "competition_source_list": ["2017年全国华杯赛竞赛初赛模拟试卷2第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "一列数：$$a_{1}，a_{2}，a_{3}，a_{4}，\\cdots，a_{n}，\\cdots，$$其中$$a_{1}=2016，a_{2}=21， a_{n}=（a_{n-1}）+（a_{n-2}）$$这里$$（a_{n-1}）$$表示$$a_{n-1}$$的所有数字之和，那么$$a_{2016}$$=（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->数列操作周期问题->数的周期"], "answer_analysis": ["这列数为$$2016，$$21$$，$$12$$，$$6$$，$$9$$，$$15$$，$$15$$，$$12$$，$$9$$，$$12$$，$$12$$，$$6$$，$$9$$，$$15$$，$$15$$，$$12$$，$$9$$，\\cdots，$$从$$a_{4}$$开始每$$8$$个数一循环．$$2016=3+8\\times251+5$$，所以，$$a_{2016}=a_{8}=12$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2530", "queId": "cbff3d67bae14b1e9c0a953da214fd4d", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "童童在计算有余数的除法时，把被除数$$472$$错看成了$$427$$，结果商比原来小$$5$$，但余数恰好相同，那么这个余数是（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["知识标签->课内题型->综合与实践->应用题->还原问题->抄错数得出结果，求正确的"], "answer_analysis": ["除数$$=(472-427)\\div 5=9$$，$$472\\div 9=52\\cdots \\cdots 4$$，所以余数是$$4$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "562", "queId": "8aac50a7519fa10a01519fd85983003c", "competition_source_list": ["2016年全国华杯赛小学高年级竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "在一个七位整数中，任何三个连续排列的数字都构成一个能被$$11$$或$$13$$整除的三位数，则这个七位数最大是（~~~~~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9981733$$ "}], [{"aoVal": "B", "content": "$$9884737$$ "}], [{"aoVal": "C", "content": "$$9978137$$ "}], [{"aoVal": "D", "content": "$$9871773$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["注意到由于任意三个连续排列的数字都能构成三位数，所以这个七位数的前五个数字不能是$$0$$，逐步极端分析，得$$988=13\\times 76$$，$$884=13\\times 68$$，$$847=11\\times 77$$，$$473=11\\times 43$$，$$737=11\\times 67$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1959", "queId": "aaa0a7abbd074c6c9874a46525b2351a", "competition_source_list": ["2013年第11届创新杯四年级竞赛初赛第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "某便民点销售矿泉水，进货，$$5$$元钱$$4$$瓶，售出时，$$5$$元钱$$3$$瓶，要获利$$300$$元，那么需售瓶． ", "answer_option_list": [[{"aoVal": "A", "content": "$$480$$ "}], [{"aoVal": "B", "content": "$$360$$ "}], [{"aoVal": "C", "content": "$$240$$ "}], [{"aoVal": "D", "content": "$$720$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["已知进货$$5$$元钱$$4$$瓶，每瓶是$$5$$ $$\\div4= \\frac{5}{4}$$ （元）， 售出时，$$5$$元钱$$3$$瓶，每瓶$$5 \\div 3= \\frac{5}{3}$$ 元，每瓶获和 $$\\frac{5}{3}- \\frac{5}{4}= \\frac{5}{12}$$（元），要获利$$300$$元，那么需售多少瓶，用 $$300 \\div \\frac{5}{12}=720$$（元），即可得解． $$5\\div 4=\\frac{5}{4}$$， $$5\\div 3=\\frac{5}{3}$$， $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde300\\div \\left( \\frac{5}{3}-\\frac{5}{4} \\right)$$ $$=300\\div \\left( \\frac{20}{12}-\\frac{15}{12} \\right)$$ $$=300\\div \\frac{5}{12}$$ $$=300\\times \\frac{12}{5}$$ $$=720$$（瓶）， 答：要获利$$300$$元，那么需售$$720$$瓶． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1778", "queId": "e44ff1e3c4a14851ab2b76c0a30bc80e", "competition_source_list": ["2016年全国小学生数学学习能力测评四年级竞赛复赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "马大哈在计算有余数的除法时，把被除数$$113$$错写成$$131$$，结果商比原来的商多$$3$$，但余数恰好相同．该题的余数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["错把被除数$$113$$写成$$131$$，被除数增加了$$131-113=18$$， 因为余数相同，且商多$$3$$，除数增加了$$3$$倍，即$$18$$是除数的$$3$$倍， 所以除数是$$18\\div3=6$$， 那么该题的余数是$$113$$除以$$6$$所得的余数，据此解答． $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde(131-113)\\div3$$ $$=18\\div3$$ $$=6$$ $$113\\div6=18\\cdots\\cdots5$$． 答：原来的除数是$$6$$，余数是$$5$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "949", "queId": "fd2ec54fb71748878b1221681ab8da65", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第20题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "仅用$$1$$和$$2$$两个数码组成的，能被$$9$$整除的八位数有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$32$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->数感认知->数学概念理解（数）"], "answer_analysis": ["能被$$9$$整除的八位数，则这个八位数的数字和必须是$$9$$的倍数， 又因为只能由数字$$1$$和$$2$$构成， 所以数字和只能$$9$$，则能是七个$$1$$和一个$$2$$， 所以只用给$$2$$选一个位置就可以，一共是八位数，所以有八个． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "109", "queId": "66d3c371db2c4eba8dc0afcb68b7783a", "competition_source_list": ["2016年第3届河南K6联赛小学高年级竞赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "四名同学（甲乙丙丁）猜测他们之中谁被评为三好学生． 甲说：``如果我被评上，那么乙也被评上；'' 乙说：``如果我被评上，那么丙也被评上；'' 丙说：``如果丁没评上，那么我也没评上；'' 实际上他们之中只有一个人没评上，并且甲、乙、丙说的都是正确的，则没有被评上三好学生的是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "丁 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数学广角->推理->解决简单逻辑推理问题"], "answer_analysis": ["假设找矛盾 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2776", "queId": "9a51ef7c24aa44bb8d75713139f90ade", "competition_source_list": ["2011年全国美国数学大联盟杯小学高年级五年级竞赛初���第30题"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$10A$$.$$M$$.起，过$$1000$$分钟后是什么时间? ", "answer_option_list": [[{"aoVal": "A", "content": "$$4:40 P.M.$$ "}], [{"aoVal": "B", "content": "$$10:00 P.M.$$ "}], [{"aoVal": "C", "content": "$$2:40 A.M.$$ "}], [{"aoVal": "D", "content": "$$1:00 A.M.$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$1000\\div60=16\\cdots40$$ 过了$$16$$个小时$$40$$分钟． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1708", "queId": "804cb9c19abd4bb3a4cdee2f772fa743", "competition_source_list": ["2011年第7届全国新希望杯五年级竞赛A卷第3题", "2016年创新杯五年级竞赛训练题（四）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "原计划安排若干人进行某项任务，如果增加$$10$$人，$$6$$天可以完成：如果增加$$15$$人，$$5$$天可以完成．那么原计划（~~~ ）天可以完成（每人工作效率相同）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->基本合作问题"], "answer_analysis": ["算出两种方案的差$$5\\times 15-6\\times 10=15$$；即原有$$15$$人，然后，$$(15+10)\\times 6\\div 15=10$$故选$$C$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3144", "queId": "02f3ad7d4d7b4d0b8725ebf827afa3f6", "competition_source_list": ["2017年全国美国数学大联盟杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$n$$是一任意自然数．若将$$n$$的各位数字反向排列所得自然数$${n}_{1}$$与$$n$$相等，则称$$n$$为回文数．例如，若$$n=1234321$$，则称$$n$$为一回文数；但若$$n=1234567$$，则$$n$$不是回文数．请问在$$10000$$和$$100000$$之间有多少个回文数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$900$$ "}], [{"aoVal": "B", "content": "$$1000$$ "}], [{"aoVal": "C", "content": "$$1100$$ "}], [{"aoVal": "D", "content": "$$1200$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["在$$10000$$和$$100000$$之间的数都是五位数，不妨设其为$$\\overline{abcba}$$，$$a$$不为$$0$$有$$9$$种选择，$$b$$、$$c$$均有$$10$$种选择，故共有$$9\\times 10\\times 10=900$$个回文数． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1612", "queId": "b5cc3967a628488aa624d8890c3648ca", "competition_source_list": ["1995年六年级竞赛创新杯", "2007年第5届创新杯六年级竞赛第1题5分", "2007年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "与30以内的奇质数的平均数最接近的整数是( )． ", "answer_option_list": [[{"aoVal": "A", "content": "12 "}], [{"aoVal": "B", "content": "13 "}], [{"aoVal": "C", "content": "14 "}], [{"aoVal": "D", "content": "15 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数->特殊质数运用->常用质数"], "answer_analysis": ["由于30以内奇质数的平均数为$$\\left( 3+5+7+11+13+17+19+23+29 \\right)\\div 9=14.111\\cdots $$，所以与此平均数最接近的整数是14 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2317", "queId": "e5adc4fd74fa4e309b732206b4a62d32", "competition_source_list": ["2017年华杯赛四年级竞赛初赛", "2017年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\text{I}$$型和$$\\text{II}$$型电子玩具车各一辆，沿相同的两个圆形轨道跑动，$$\\text{I}$$型每$$5$$分钟跑一圈，$$\\text{II}$$型每$$3$$分钟跑一圈。某一时刻，$$\\text{I}$$型和$$\\text{II}$$型恰好都开始跑第$$19$$圈，则$$\\text{I}$$型比$$\\text{II}$$型提前（ ）分钟开始跑动。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$38$$ "}], [{"aoVal": "D", "content": "$$54$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->多次相遇和追及"], "answer_analysis": ["解：$$5-3=2$$（分钟），$$18\\times 2=36$$（分钟） 故选：B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1860", "queId": "93014ec0fce644539f2e9e3a40a4284e", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛B卷第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "小亮在期末考试��，道德与法治、语文、数学、英语、科学五科的平均成绩是$$89$$分，道德与法治、数学两科平均成绩是$$91.5$$分，道德与法治、英语两科平均成绩是$$86$$，英语与语文多$$10$$分．下面选项中正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "语文$$79$$分~ ~数学$$94$$分 "}], [{"aoVal": "B", "content": "语文$$79$$分~ ~道德与法治$$83$$分 "}], [{"aoVal": "C", "content": "数学$$100$$分~ ~英语$$79$$分 "}], [{"aoVal": "D", "content": "科学$$94$$分~ ~道德与法治$$89$$分 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["五科的成绩之和是$$89\\times5=445$$（分）， 政治和数学的成绩之和是：$$91．5\\times2=183$$（分）， 语文和英语的成绩之和是：$$84\\times2=168$$（分）， 政治和英语成绩之和是：$$86\\times2=172$$（分）， 所以生物成绩是： 五科成绩之和$$-$$政治和数学成绩之和$$-$$语文和英语成绩之和， 语文成绩是（语文和英语成绩之和$$-10$$）$$\\div2$$， 用语文成绩减去$$10$$分就是英语成绩， 政治成绩$$=$$政治和英语成绩之和$$-$$英语成绩， 数学成绩$$=$$政治和数学成绩之和$$-$$政治成绩， 据此解答即可． 五科成绩之和：$$89\\times5=445$$（分）， 政治和数学成绩之和是：$$91．5\\times2=183$$（分）， 语文和英语的成绩之和是：$$84\\times2=168$$（分）， 政治和英语成绩之和是：$$86\\times2=172$$（分）， 生物成绩：$$445-183-168=94$$（分）， 语文成绩：$$(168-10)\\div2=158\\div2=79$$（分）， 英语成绩：$$79+10=89$$（分）， 政治成绩：$$172-89=83$$（分）， 数学成绩：$$183-83=100$$（分）． 答：五科成绩分别为生物$$94$$分，语文$$79$$分， 英语$$89$$分，政治$$83$$分，数学$$100$$分． 故答案为：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2314", "queId": "eec069f97b6048b48634a7dc96b55cda", "competition_source_list": ["2012年第10届全国创新杯小学高年级六年级竞赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "李军有一个闹钟，但它走时不准，这天下午$$6:00$$把它对准北京时间，可到晚上$$9:00$$时，它才走到$$8:45$$，这天早上李军看闹钟走到$$6:17$$的时候赶去上学，这时候北京时间为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7:15$$ "}], [{"aoVal": "B", "content": "$$7:24$$ "}], [{"aoVal": "C", "content": "$$7:30$$ "}], [{"aoVal": "D", "content": "$$7:35$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["坏钟和标准钟的速度比为：$$165:180=11:12$$，现知道闹钟共走$$12\\times 60+17=737$$格，而标准钟可以转$$737\\times \\frac{12}{11}=804$$格，合计$$13$$个小时$$24$$分钟． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "896", "queId": "d24a111e9c7f40e1bf536a4dfc249255", "competition_source_list": ["2012年第8届全国新希望杯小学高年级六年级竞赛复赛第6题4分"], "difficulty": "4", "qtype": "single_choice", "problem": "$$\\overline{**45}$$，$$\\overline{19*8}$$，$$\\overline{23*1}$$，$$\\overline{3*49}$$是四个四位数，其中$$*$$代表不能辨认的数字，若其中有一个数是完全平方数，那么这个数可能是．（6分） ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\overline{**45}$$ "}], [{"aoVal": "B", "content": "$$\\overline{19*8}$$ "}], [{"aoVal": "C", "content": "$$\\overline{23*1}$$ "}], [{"aoVal": "D", "content": "$$\\overline{3*49}$$ "}]], "knowledge_point_routes": ["拓展思维->知识点->数论模块->完全平方数->平方数的尾数特征"], "answer_analysis": ["完全平方数的个位只能为$$0$$、$$1$$、$$4$$、$$5$$、$$6$$、$$9$$，因此排除$$\\text{B}$$选项．如果一个完全平方数是 $$5$$的倍数，那么它至少是$$25$$的倍数，因此排除$$\\text{A}$$选项．估算$${{48}^{2}}=2304$$，$${{49}^{2}}=2401$$，排除 $$\\text{C}$$选项．经检验$${{57}^{2}}=3249$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2594", "queId": "94b279419d8f43069a2ed99c322e8c1f", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（五）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙三个人合租了一套房子，甲付的钱数等于乙付的钱数的$$1.5$$倍．乙付的钱数是丙付的钱数的$$\\frac{3}{4}$$，已知甲比丙多付了$$100$$元，这套房子的租金是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2000$$元 "}], [{"aoVal": "B", "content": "$$2100$$元 "}], [{"aoVal": "C", "content": "$$2200$$元 "}], [{"aoVal": "D", "content": "$$2300$$元 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比和比例->比例->化连比"], "answer_analysis": ["甲：乙$$=3:2$$，乙：丙$$=3:4$$，甲：乙：丙$$=9:6:8$$，$$100\\div (9-8)\\times (9+6+8)=2300$$（元）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1518", "queId": "490fb6518ef2433c958c233801e4c642", "competition_source_list": ["2014年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "两个正整数的和小于$$100$$，其中一个是另一个的两倍，则这两个正整数的和的最大值是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$83$$ "}], [{"aoVal": "B", "content": "$$99$$ "}], [{"aoVal": "C", "content": "$$96$$ "}], [{"aoVal": "D", "content": "$$98$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->和倍问题->二量和倍问题->两量和倍"], "answer_analysis": ["解：根据$$3$$的倍数特征，不难判断$$83$$和$$98$$都不是$$3$$的倍数，$$99$$和$$96$$都是，但$$99\\textgreater96$$，所以这两个数的最大值是$$99$$。 故选：B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1817", "queId": "a9215a37e2b64a259f45c0b1a4ef6f95", "competition_source_list": ["2020年第23届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$10 $$年前父亲的年龄是儿子的$$7$$倍，$$15$$年后父亲的年龄是儿子的$$2$$倍，今年父亲的年龄是儿子的倍． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$2.5$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设$$10$$年前儿子的年龄是$$x$$岁，那么父亲的年龄是$$7x$$； 根据题意可得： $$7x+10+15=\\left( x+10+15 \\right)\\times 2$$， $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde7x+25=2x+50$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde5x=25$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x=5$$， 现在儿子的年龄是：$$x+10=5+10=15$$（岁）， 父亲的年龄是：$$7x+10=7\\times 5+10=45$$（岁）． $$45\\div 15=3$$（倍）． 故选$$\\text C$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1879", "queId": "fbdb7b81f0e245e1bd2e5cfb64dff707", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "祖玛游戏中，龙嘴里不断吐出很多颜色的龙珠，先$$4$$颗红珠，接着$$3$$颗黄珠，再$$2$$颗绿珠，最后$$1$$颗白珠，按此方式不断重复，从龙嘴里吐出的第$$252$$颗龙珠是． ", "answer_option_list": [[{"aoVal": "A", "content": "红珠 "}], [{"aoVal": "B", "content": "黄珠 "}], [{"aoVal": "C", "content": "绿珠 "}], [{"aoVal": "D", "content": "白珠 "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["$$252\\div (4+3+2+1)=25······2$$．红色 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1492", "queId": "3772c637cf22432cb9b5a0ae1cf6254c", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "兄弟俩今年的年龄和是$$35$$岁，当哥哥像弟弟现在这样大时，弟弟的年龄恰好是哥哥年龄的一半，弟弟今年岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题->方程法解其他问题"], "answer_analysis": ["题目中有$$2$$个时间线，一是现在的年龄和$$35$$岁的现在时，二是当哥哥像弟弟现在这样大时的过去时． 在过去时中：弟弟的年龄是$$1$$���，则哥哥为$$2$$份，所以哥哥比弟弟多$$1$$份． 到现在时的时候：弟弟的年龄$$2$$份，则哥哥为$$3$$份．差仍然是$$1$$份． 所以现在：$$35\\div$$ （2+3）=7（岁/份） 弟弟：$$2\\times$$ 7=14（岁） 哥哥：$$3\\times$$ 7=21（岁） "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3346", "queId": "ba99b7d03a6b484c81204fe0471fa5d5", "competition_source_list": ["2019年美国数学大联盟杯五年级竞赛初赛第34题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "大于$$99$$且小于$$1000$$的整数既不包含数字$$1$$也不包含数字$$2$$有多少个？． ", "answer_option_list": [[{"aoVal": "A", "content": "$$392$$ "}], [{"aoVal": "B", "content": "$$448$$ "}], [{"aoVal": "C", "content": "$$512$$ "}], [{"aoVal": "D", "content": "$$620$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->加乘原理综合", "Overseas Competition->知识点->计数模块->加乘原理"], "answer_analysis": ["本题题意为``大于$$99$$且小于$$1000$$的整数既不包含数字$$1$$也不包含数字$$2$$有多少个？''本题相当于问三位数中即不包括数字$$1$$也不包括数字$$2$$的有多少个，本题考查排列组合，先考虑个位上可能出现的数字有$$0$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$总共$$8$$个，同理十位也是$$8$$个，百位上去掉不能为$$0$$的情况有$$7$$个数字，因此一共有$$7\\times 8\\times 8=448$$（个）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2860", "queId": "a409220b2d0e4bdab5137b0e144fe61c", "competition_source_list": ["2013年全国走美杯五年级竞赛初赛B卷第11题", "2013年全国走美杯六年级竞赛初赛第11题"], "difficulty": "2", "qtype": "single_choice", "problem": "定义$$a \\And b=(a+2)(b+2)-2$$，算式$$1\\times 3\\times 5\\times7\\times 9\\times 11\\times 13-(1 \\And3 \\And5 \\And7 \\And9 \\And11)$$的计算结果是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->数据处理"], "answer_analysis": ["本题可以算出答案，但用递推推出答案会使过程更简单，由定义式可推得：$$1 \\And3=3\\times 5-2$$，$$1 \\And3 \\And5=(1\\times3\\times 5-2+2)\\times 7-2=1\\times 3\\times 5\\times 7-2$$$$\\cdots$$以此类推，$$1 \\And3 \\And5 \\And 7 \\And 9 \\And 11=1\\times 3\\times 5\\times 7\\times 9\\times11\\times 13-2$$，故原式答案为$$2$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1141", "queId": "41b92b668c80479cb94c6b203f9f4873", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级六年级竞赛邀请赛训练题（二）"], "difficulty": "1", "qtype": "single_choice", "problem": "一班和二班的人数之比是$$8:7$$，如果将一班的$$8$$名同学调到二班去，则一班和二班的人数比变为$$4:5$$．原来二班的人数为（~ ）人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["一班原来的人数占两班总人数的$$\\frac{8}{15}$$，$$4+5=9$$，如果将一班的$$8$$名同学调到二班去，那么一班的人数占两班总人数的$$\\frac{4}{9}$$．原来两班总人数是:$$8\\div \\left( 8\\div154\\div9 \\right)=90$$，因此，原来一班有：$$90\\times \\frac{8}{15}=48$$人，二班有：$$90\\times \\frac{7}{15}=42$$人． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2213", "queId": "9970125548804ea186ed6110531d253d", "competition_source_list": ["2018年第12届北京学而思杯四年级竞赛初赛线上第13题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "一列火车全长$$240$$米，每秒行驶$$20$$米，要想通过全长为$$360$$米的大桥，需要多少秒？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$50$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->式与方程->数量关系->路程=速度×时间"], "answer_analysis": ["火车的路程为$$240+360=600$$（米），所以需要$$\\left( 240+360 \\right)\\div 20=30$$（秒）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "92", "queId": "14f69994cbde48d4b22ab76cec919e49", "competition_source_list": ["2019年全国小学生数学学习能力测评五年级竞赛复赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一把钥匙只能开一把锁，现有$$4$$把钥匙$$4$$把锁，但不知哪把钥匙开哪把锁，问最多试次能将所有的锁都找到相对应的钥匙． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["无需将锁打开，只需匹配成功，故最多：$$3+2+1=6$$（次）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "304", "queId": "60165654edbb4418ab26b969c88a45f1", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$20$$个苹果分成数量各不相同的五堆，其中数量最多的一堆最多有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["五堆共有$$20$$个苹果，要使数量最多的那堆最多， 则其它$$4$$堆尽量小， 而其它$$4$$堆最小为$$1$$，$$2$$，$$3$$，$$4$$个苹果， 则数量最多那堆最多， $$20-1-2-3-4=10$$（个）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3010", "queId": "e9af498e2be14bd0a6a0e14f0e487dd1", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$\\frac{0.2\\dot{3}\\dot{4}+\\frac{84}{495}}{0.56\\dot{8}-\\frac{56}{450}}$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{11}{10}$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$\\frac{10}{11}$$ "}], [{"aoVal": "D", "content": "$$\\frac{9}{10}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["原式$$=\\frac{\\frac{234-2}{990}+\\frac{84}{495}}{\\frac{568-56}{900}-\\frac{56}{450}}$$ $$=\\frac{\\frac{232}{990}+\\frac{168}{990}}{\\frac{512}{900}-\\frac{112}{900}}$$ $$=\\frac{\\frac{400}{990}}{\\frac{400}{900}}=\\frac{10}{11}$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2215", "queId": "481bbdd453a14ab7aecf9faa969a6cd0", "competition_source_list": ["2021年河南郑州二七区京广实验学校小升初第5题2分", "2017年江苏镇江句容市句容市崇明小学小升初第16题1分", "2017年河南郑州联合杯竞赛决赛第5题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$9:30$$时，钟面上时针和分针所组成的角是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "直角 "}], [{"aoVal": "B", "content": "锐角 "}], [{"aoVal": "C", "content": "钝角 "}], [{"aoVal": "D", "content": "平角 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["当$$9:30$$时，时针和分针所组成的角是钝角． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1356", "queId": "28a8a1b0ad5644608e42a1061700e5e1", "competition_source_list": ["2020年希望杯二年级竞赛模拟第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "鸭妈妈带着小鸭们在池塘里游玩，黄小鸭发现：有$$2$$只小鸭在它的前面，$$3$$只小鸭在它的后面．池塘里共有几只鸭？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}], [{"aoVal": "E", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->应用题模块排队问题->单主角求总数", "Overseas Competition->知识点->应用题模块->应用题模块排队问题"], "answer_analysis": ["根据题意分析可知，用黄小鸭前面的数量加上后面的数量再加上自己所以一共有： $$2+3+1=6$$（只）小鸭，再加上鸭妈妈 故答案为：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2428", "queId": "21683aeeb00f4e82be6bc7814d66cdd0", "competition_source_list": ["2018年IMAS小学高年级竞赛（第一轮）第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "请问满足下面这个不等式的$$\\square $$中能填入的最大整数是多少？ $$9\\times \\square \\lt 2018$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$202$$ "}], [{"aoVal": "B", "content": "$$212$$ "}], [{"aoVal": "C", "content": "$$218$$ "}], [{"aoVal": "D", "content": "$$224$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->算式比较大小"], "answer_analysis": ["$$\\square ~~\\textless{} ~\\frac{2018}{9}=224\\frac{2}{9}$$，最大的整数为$$224$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1989", "queId": "bce6477f641b4cb2a00d37797138bf45", "competition_source_list": ["2014年第10届全国新希望杯小学高年级六年级竞赛复赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "某大桥由于桥面多处破损进行全面检修，修了一个星期之后，已修和未修的比是$$1:7$$，第二个星期又修了$$500$$米，这时已修和未修的比是$$9:23$$，则该大桥全长是米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3200$$ "}], [{"aoVal": "B", "content": "$$3100$$ "}], [{"aoVal": "C", "content": "$$2012.5$$ "}], [{"aoVal": "D", "content": "$$3500$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题->和不变"], "answer_analysis": ["和不变：$$1:7=4:28$$和$$8$$份$$\\times 4$$ $$9:23$$和$$32$$份 故$$1$$份为$$500\\div \\left( 9-4 \\right)=100$$（米） 全长为$$32\\times 100=3200$$（米） 选$$A$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2698", "queId": "887143bab3b44cefaf3ca1841193bdf1", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛初赛第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "根据``三角形任意两边之和大于第三边''的知识，解答本题： 有不同长度的七条线段，其长度均为整数厘米，最短的是$$1$$厘米，最长的是$$21$$厘米，其中以任何三条线段作``边''都不能组成一个三角形，那么这七条线段中第二长的线段长厘米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["斐波那契数列从第三项开始，等于前两项的和，即$$1$$、$$2$$、$$3$$、$$5$$、$$8$$、$$13$$、$$21$$． 即第二长的为$$13$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1998", "queId": "ab198bc2dd0a4b88be8296da318e8243", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛A卷第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个三位数的各位数字之和是$$17$$，其中十位数字比个位数字大$$1$$，如果把这个三位数的百位数字与个位数字对调，得到一个新的三位数，则新的三位数比原三位数大$$198$$，原数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$476$$ "}], [{"aoVal": "B", "content": "$$400$$ "}], [{"aoVal": "C", "content": "$$470$$ "}], [{"aoVal": "D", "content": "$$486$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设个位是$$a$$，十位是$$a+1$$，百位$$17-a-a-1=16-2a$$． 根据题意列出方程：$$100a+10(a+1)+16-2a-100(16-2a)-(10a+1)-a=198$$， 解这个方程，求出个位数字，然后再求十位与百位数字，解决问题． 设原数个位为$$a$$，则十位为$$a+1$$，百位为$$16-2a$$， 根据题意列方程 $$100a+10(a+1)+16-2a-100(16-2a)-(10a+1)$$， 解得$$a=6$$，则$$a+1=7$$，$$16-2a=4$$； 答：原数为$$476$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1933", "queId": "b3447dc75da7429fbcb9d42b529375c0", "competition_source_list": ["2018年第8届北京学而思综合能力诊断一年级竞赛年度教学质量第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "植树节到了，李老师带着同学们去种树．大家要一起合影，站成了两排，两排人数一样 多．李老师站在后一排，从左数是第$$5$$个，从右数是第$$3$$个，一共有（~ ）个同学去种树． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$20$$~ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["一排的人数：$$5+3-1=7$$（人） 两排的人数：$$7+7=14$$（人） 同学的人数：$$14-1=13$$（人）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1015", "queId": "030235d15d51442e87d3581d59b1cdd9", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个数，它的一半的一半是$$4$$，这个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["一个数的一半的一半是$$4$$，则这个数的一半为$$4+4=8$$，则这个数为$$8+8=16$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "240", "queId": "d5a39b8eed74439dbf67cfac1453b902", "competition_source_list": ["1989年华杯赛六年级竞赛初赛"], "difficulty": "3", "qtype": "single_choice", "problem": "一副扑克牌有四种花色，每种花色有$$13$$张，此外还有两张王牌，从中任意抽牌。那么，最少要抽 张牌，才能保证有$$4$$张牌是同一花色。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["如果在最不利的情况下能完成目标，则保证在任何情况下都能完成。最不利的情况是：抽的前$$12$$张是$$4$$种花色各$$3$$张，再抽两张王牌，这时抽第$$15$$张，无论是哪种花色，都能保证凑成$$4$$张牌同一花色。所以至少要抽$$15$$张牌，才能保证有四张牌是同一花色的。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1214", "queId": "5910049a8cc74c99a9d94fc25953de48", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（四）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "某月所有星期天的日期数加起来是$$70$$，这个月星期三的天数比星期四的天数多．这个月一共有天． ", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$ "}], [{"aoVal": "B", "content": "$$29$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$31$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["经推算此月星期天的日期分别为$$7$$号、$$14$$号、$$21$$号、$$28$$号，又星期三的天数比星期四的天数多，那么最后一天应是星期三，$$31$$号． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2086", "queId": "d0269d3f3d7c4b9892a517aa6eaca1a2", "competition_source_list": ["2014年世界少年奥林匹克数学竞赛三年级竞赛初赛B卷第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "今年爸爸是$$32$$ 岁，儿子是$$4$$ 岁，当父子俩年龄之和是$$50$$ 岁时，应该是~\\uline{~~~~~~~~~~}~年之后的事． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$46$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题基本关系->年龄和"], "answer_analysis": ["今年年龄和：$$32+4=36$$（岁），两人共长：$$50-36=14$$（岁），经过的年头：$$14\\div 2=7$$（年）． 答：当父子年龄和是$$50$$时，应该是$$7$$年以后的事． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1053", "queId": "2a23710ba9a043c492717d77454c4753", "competition_source_list": ["2016年环亚太杯四年级竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "一条路上相邻两棵树的距离都相等，甲、乙二人同时从一端的某棵树出发，同向而行．当甲走到第$$21$$棵树时，回头看见乙到的那棵树与自己正隔着$$3$$棵树．已知乙每分钟走$$64$$米．甲每分钟走~\\uline{~~~~~~~~~~}~米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$51.2$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$70$$ "}], [{"aoVal": "D", "content": "$$80$$ "}], [{"aoVal": "E", "content": "$$90$$ "}], [{"aoVal": "F", "content": "$$100$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["依题意得，这是一道有关路程的题，列式如下： $$64\\times (21-1)\\div (21-4-1)$$ $$=64\\times 20\\div 16$$ $$=80$$． 故答案为：$$80$$．选$$\\text{D}$$. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "34", "queId": "03e675bee5fe4f2d9bac5a56267ba07d", "competition_source_list": ["2017年河南郑州��才杯六年级竞赛决赛4分", "2013年四川成都小升初某师大一中第18题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个口袋装有红、黄、蓝三种不同颜色的小球各$$10$$个，一次摸一个且不放回，要保证能摸出$$10$$个相同颜色的小球，至少要摸出个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->简单抽屉原理"], "answer_analysis": ["把三种颜色看做三个抽屉，从极端考虑：先摸出的是红色球、黄色球和蓝色球各$$9$$个，共$$27$$个球，再摸第$$28$$个球则一定有一种球是同色的，因此至少要摸出$$28$$个球． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1653", "queId": "64d95288a4994a91a9249d759d347823", "competition_source_list": ["2020年长江杯五年级竞赛复赛B卷第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$7$$个人的年龄和是$$190$$岁，其中最小的$$17$$岁，且$$7$$个人中只有两个人的年龄相同，那么，年龄最大的一个人最多岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$73$$ "}], [{"aoVal": "B", "content": "$$77$$ "}], [{"aoVal": "C", "content": "$$90$$ "}], [{"aoVal": "D", "content": "$$78$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["要使年龄最大，那么要使其它的年龄最小，最小的$$17$$岁，只有两个人年龄相同， 那么其它$$5$$个人年龄是$$17$$、$$18$$、$$19$$、$$20$$、$$21$$； 最大的一个人是$$190-17-17-18-19-20-21=78$$（岁）． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "348", "queId": "5c95058df9354bbfad9022e984d850fa", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$4$$个互不相同的自然数，它们的平均数是$$10$$．其中最大的数至少是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->和一定型最值问题->构造和一定最值原理"], "answer_analysis": ["欲求最大的数至少为多少，则这$$4$$个不同的自然数越接近越好．则为$$8$$，$$9$$，$$11$$，$$12$$．故最大的数至少为$$12$$，所以选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "947", "queId": "f3d2f99a04ae4a36a8742ae3a7d4d7d8", "competition_source_list": ["2015年第27届广东广州五羊杯小学高年级竞赛第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$2000$$，$$2001$$，$$2002$$，$$2003$$、$$\\cdots $$、$$2015$$这$$16$$个数中，不能表示成两个完全平方数之差的数有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["显然其中的奇数都可以，这可以利用平方差公式与和差问题说明；偶数必须是$$4$$的倍数，所以有$$2000$$，$$2004$$，$$2008$$，$$2012$$可以做得到．因而共有$$2002$$，$$2006$$，$$2010$$，$$2014$$四个不能．答案为$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2615", "queId": "31bb23e43e2943b882300657437ad3a0", "competition_source_list": ["2020年希望杯二年级竞赛模拟第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "请你根据数串的规律再横线上填上正确的答案：3,6,9,12~\\uline{~~~~~~~~~~}~，15. ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->思想->数形结合思想"], "answer_analysis": ["后一个数等于前一个数$$\\times$$ 2 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2695", "queId": "5f5c0985b9a04ca881f83a420c310085", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\square $$和$$\\bigcirc $$各代表一个数， $$\\square =\\bigcirc +\\bigcirc $$，$$\\bigcirc +\\square +\\square =30$$， $$\\square $$-$$\\bigcirc $$=． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->计算模块->方程基础->等量代换", "拓展思维->能力->逻辑分析"], "answer_analysis": ["分析可知：$$\\square =\\bigcirc +\\bigcirc =2\\bigcirc $$， 把$$\\square =2\\bigcirc $$代入$$\\bigcirc +\\square +\\square =30$$中得： $$\\bigcirc +2\\bigcirc +2\\bigcirc =5\\bigcirc $$，$$5\\bigcirc =30$$， 所以$$\\bigcirc =30\\div 5=6$$， 那么$$\\square =2\\bigcirc =2\\times 6=12$$， 由此可知，$$\\square $$与$$\\bigcirc $$的差为$$12-6=6$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2226", "queId": "e37dce3db39d451090c06d6ed2fa4a8a", "competition_source_list": ["2017年全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "学校到图书馆的路一半上坡、一半下坡．学生$$A$$从学校到图书馆的过程中，下坡的速度是他走全程平均速度的$$2$$倍，那么上坡的速度是他走全程平均速度~\\uline{~~~~~~~~~~}~倍． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->平均速度->设数法"], "answer_analysis": ["设全程的路程为$$\\text{s}$$，时间为$$\\text{t}$$，则学生$$A$$走全程的平均速度为$$\\text{v=}\\frac{\\text{s}}{\\text{t}}$$，学生$$A$$下坡的时间为 $$\\frac{\\frac{\\text{s}}{2}}{2\\text{v}}=\\frac{\\text{s}}{4\\text{v}}=\\frac{\\text{s}}{4\\times \\frac{\\text{s}}{\\text{t}}}=\\frac{\\text{t}}{4}$$，所以，学生$$A$$上坡的时间为$$\\text{t-}\\frac{\\text{t}}{4}=\\frac{3}{4}\\text{t}$$，故学生$$A$$上坡的速度是$$\\frac{\\frac{\\text{s}}{2}}{\\frac{3\\text{t}}{4}}=\\frac{2\\text{s}}{3\\text{t}}=\\frac{2}{3}\\text{v}$$， 即上坡的速度是他走全程平均速度$$\\frac{2}{3}$$倍． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "373", "queId": "b1ed294ff7e048848a8dc1bae7e5ce5b", "competition_source_list": ["2013年华杯赛四年级竞赛初赛", "2013年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有一个人捡到一条红领巾，老师问是谁捡到的？小东说：不是小西；小西说：是小南；小南说：小东说的不对；小北说：小南说的也不对。他们之中只有一个人说对了，这个人是。 ", "answer_option_list": [[{"aoVal": "A", "content": "小东 "}], [{"aoVal": "B", "content": "小西 "}], [{"aoVal": "C", "content": "小南 "}], [{"aoVal": "D", "content": "小北 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理"], "answer_analysis": ["解：根据题干分析可得，小南与小北说的话是相互矛盾的，所以两人中一定有一个人说的是正确的。 假设小北说的是正确的，则小南说``小东说的不对''是错，可得：小东说的对，这样与已知只有一个人说对了相矛盾，所以此假设不成立，故小南说的是正确的。 故选：$$\\text{C}$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "183", "queId": "9de4fa1dea434c2cb2c8eb270ec12faa", "competition_source_list": ["2021年第24届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第17题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙三人中只有$$1$$人会英语．甲说：``我会英语．''乙说：``我不会英语．丙说：``甲不会英语．''三人的话只有一句是真话．会英语的是． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾"], "answer_analysis": ["假设甲会英语， 那么甲和乙说的是真话， 所以和已知矛盾， 所以甲不会英语； 假设乙会英语， 那么甲和乙说的是假话， 丙说的是真话，符合题意； 假设丙会英语， 那么乙和丙说的是真话，和题意矛盾， 所以丙不会英语， 所以会英语的是乙． 选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1084", "queId": "3395ec12e0614a589ee007bdfdf90809", "competition_source_list": ["2004年第2届创新杯六年级竞赛初赛第4题", "2004年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "新村小学有男生$$480$$人，比全校学生人数的$$\\frac{3}{5}$$少$$60$$人，这个学校女生的人数是( ) ", "answer_option_list": [[{"aoVal": "A", "content": "$$220$$人 "}], [{"aoVal": "B", "content": "$$280$$人 "}], [{"aoVal": "C", "content": "$$360$$人 "}], [{"aoVal": "D", "content": "$$420$$人 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["由题意可知，$$480$$人加上$$60$$人正好是全校学生人数的$$\\frac{3}{5}$$，因此全校学生人数为：$$\\left( 480+60 \\right)\\div \\frac{3}{5}$$，然后减去男生人数，就是女生人数．据此解答． $$\\left( 480+60 \\right)\\div \\frac{3}{5}-480=540\\times \\frac{5}{3}-480=900-480=420$$（人）． 答：这个学校女生的人数是$$420$$人． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1791", "queId": "8a124d778c2f48d59e8fcb232ec182f1", "competition_source_list": ["2013年第11届全国小机灵杯三年级竞赛决赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "某年的三月份正好有$$4$$个星期三和$$4$$个星期六，那么这年$$3$$月$$1$$日是星期~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "四 "}], [{"aoVal": "B", "content": "五 "}], [{"aoVal": "C", "content": "六 "}], [{"aoVal": "D", "content": "日 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数学广角->简单的周期->简单的周期计算"], "answer_analysis": ["$$3$$月有$$31$$天，从中任取连续$$28$$天，正好是四周，恰好有$$4$$个周三和$$4$$个周六， 因此，剩下的$$3$$天中不能有周三和周六 不妨取$$3$$月$$4$$日到$$3$$月$$31$$日，这$$28$$天中恰有$$4$$个周三和$$4$$个周六 剩下的$$1$$日到$$3$$日是连续的三天，其中有不能有周三和周六，发现这三天只能是周日、周一、周二 因此，$$3$$月$$1$$日是周日． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3311", "queId": "4517ef8de1de43ae9ece960abc1b81fe", "competition_source_list": ["2016年全国小学生数学学习能力测评五年级竞赛初赛第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$1$$角硬币分正面与反面．拿三个$$1$$角硬币一起投掷一次，得到一个正面和一个反面的可能性为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{8}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["全部情况有：$$2\\times 2=4$$（种）， 最终得到一个正面一个反面可有以下$$2$$种情况： 正反、反正， 故所求概率为$$\\frac{2}{4}$$． 故答案为：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3047", "queId": "d3733a7a7ab248b3a4ce8273f76d79d1", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第19题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "王老师在黑板上写了五个数，代号分别为$$A$$、$$B$$、$$C$$、$$D$$、$$E$$，她告诉大家：$$A$$比$$B$$大；$$C$$比$$D$$大，$$C$$比$$E$$小；$$D$$比$$B$$大；$$E$$比$$A$$小．哪一个数第三大？（~ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$A$$ "}], [{"aoVal": "B", "content": "$$B$$ "}], [{"aoVal": "C", "content": "$$C$$ "}], [{"aoVal": "D", "content": "$$D$$ "}], [{"aoVal": "E", "content": "$$E$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["因为$$A\\textgreater E\\textgreater C\\textgreater D\\textgreater B$$，所以第三大的数为$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "794", "queId": "9eda96c0b0ce4f1ca27f624f801e6673", "competition_source_list": ["小学高年级其它华杯教程", "2019年华杯赛小学高年级竞赛第9题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个八位整数，由$$8$$个不同的数字组成，其中任何两个相邻数字构成的两位整数能被$$13$$或$$17$$整除，这个八位数的数字和等于（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$41$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$38$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["用$$13$$，$$26$$，$$39$$，$$52$$，$$65$$，$$78$$，$$91$$和$$17$$，$$34$$，$$51$$，$$68$$，$$85$$拼出一个八位整数$$39178526$$．除此，无其他解答． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3238", "queId": "43092790f5fb41da86c35b602151dda1", "competition_source_list": ["2020年长江杯六年级竞赛复赛A卷第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$1000$$至$$1999$$这些自然数中，个位数不小于百位数的有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$650$$ "}], [{"aoVal": "B", "content": "$$550$$ "}], [{"aoVal": "C", "content": "$$450$$ "}], [{"aoVal": "D", "content": "$$400$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["百位数字和个位数字相等的数有∶ $$1\\times 10\\times 10\\times 1=100$$（个）， 因为个位数字比百位数字大的和比百位数字小的各占一半， 所以满足条件的数有$$\\left( 1000-100 \\right)\\div 2=450$$（个）． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1583", "queId": "ff8080814518d5240145201acb9d0a72", "competition_source_list": ["2017年全国小升初八中入学备考课程", "2014年全国迎春杯三年级竞赛复赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "过年了，小明家买了很多瓶果汁．年三十喝了总量的一半少$$1$$瓶；初一又喝了剩下的一半；初二又喝了剩下的一半多$$1$$瓶，这时还剩$$2$$瓶没有喝，那么小明家一共买了（~~~~~ ）瓶果汁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$26$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->还原问题->多量还原问题"], "answer_analysis": ["还原法，初二没喝之前有$$(2+1)\\times 2=6$$（瓶），初一没喝之前有$$6\\times 2=12$$（瓶），开始有$$(12-1)\\times 2=22$$（瓶）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3281", "queId": "2e7ce9aeb9a8468796a2ef56f32ebd88", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "用$$26$$厘米长的铁丝围成长方形，长和宽都是整厘米数，可以有种不同的围法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["从题干可以知道，用$$26$$厘米长的铁丝围成长方形，那么长$$+$$宽$$=26\\div 2=13$$（厘米）， 由于长和宽都是整厘米数， $$13=1+12=2+11=3+10=4+9=5+8=6+7$$， 所以不同的围法有$$6$$种． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1487", "queId": "75ff42630cc74dbc9fa6d4c41faa091c", "competition_source_list": ["2021年第24届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第13题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "五个数的平均数是$$12$$，如果把其中一个数改为$$1$$，这时五个数的平均数是$$11$$，这个被改动的数原来是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型"], "answer_analysis": ["从题干可以知道，五个数的平均数是$$12$$， $$12\\times5=60$$， 那么五个数之和是$$60$$； 如果把其中一个数改为$$1$$， 这时五个数的平均数是$$11$$， $$11\\times5=55$$， 此时五个数之和是$$55$$； $$60-55+1=6$$； 那么这个被改动的数原来是$$6$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2394", "queId": "08caf5f192a24f62ba842fb67408fcfc", "competition_source_list": ["2014年迎春杯四年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$2014\\div (2\\times 2+2\\times 3+3\\times 3)=$$（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$53$$ "}], [{"aoVal": "B", "content": "$$56$$ "}], [{"aoVal": "C", "content": "$$103$$ "}], [{"aoVal": "D", "content": "$$106$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->四则���合运算"], "answer_analysis": ["解：$$2014\\div (2\\times 2+2\\times 3+3\\times 3)$$ $$=2014\\div (4+6+9)$$ $$=2014\\div 19$$ $$=106$$ 故选：D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1950", "queId": "d3181eb91a844e5c8aa2ba920c02936c", "competition_source_list": ["2020年新希望杯二年级竞赛初赛（个人战）第4题", "2020年新希望杯二年级竞赛决赛（8月）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$3$$斤菠萝可以做$$6$$个菠萝蛋糕，那么要做$$10$$个菠萝蛋糕，需要斤菠萝． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$20$$ "}], [{"aoVal": "E", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据$$1$$斤菠萝可以做$$2$$个菠萝蛋糕，那么要做$$10$$个菠萝蛋糕，需要$$10\\div2=5$$（斤）菠萝．故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "771", "queId": "ff8080814518d5240145192b5cdb0570", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "老师在黑板上从$$1$$开始将奇数连续地写下去，写了一长串数后，擦去了其中的两个数，将这些奇数隔成了$$3$$串，已知第二串比第一串多$$1$$个数，第三串比第二串多$$1$$个数，且第三串奇数和为$$4147$$，那么被划去的两个奇数的和是（~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$188$$ "}], [{"aoVal": "B", "content": "$$178$$ "}], [{"aoVal": "C", "content": "$$168$$ "}], [{"aoVal": "D", "content": "$$158$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["设第$$1$$段有$$n$$个，则第$$2$$段有$$n+1$$个，第一个擦的奇数是$$2n+1$$，第二个擦的奇数是$$4n+5$$，和为$$6n+6$$，是$$6$$的倍数．只有$$168$$符合． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1951", "queId": "aefb399b9b4545ec8f81b2372ab9d680", "competition_source_list": ["2009年华杯赛三年级竞赛初赛", "2009年华杯赛四年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "开学前6天，小明还没做寒假数学作业，而小强已完成了60道题，开学时，两人都完成了数学作业. 在这6天中，小明做的题的数目是小强的3倍，他平均每天做（ ）道题. ", "answer_option_list": [[{"aoVal": "A", "content": "6 "}], [{"aoVal": "B", "content": "9 "}], [{"aoVal": "C", "content": "12 "}], [{"aoVal": "D", "content": "15 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->差倍->两量差倍->暗差型二量差倍问题"], "answer_analysis": ["由于开学前6天时小强比小明多做了60道题，而开学时两人做的题一样多，所以这6天中小明比小强多做了60道题，而这6天中小明做的题的数目是小强的3倍，所以这6天小明做了$$60\\div \\left( 3-1 \\right)\\times 3=90$$(道) 题，他平均每天做$$90\\div 6=15$$题.正确答案为 D. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1541", "queId": "5f9c17c5300b4bd9aaaf15a7cf3498c0", "competition_source_list": ["2013年第12届全国小机灵杯小学中年级三年级竞赛初赛第18题15分"], "difficulty": "3", "qtype": "single_choice", "problem": "有一串数，第一个数是$$6$$，第二个数是$$3$$，从第二个数起，每个数都比它前面的那个数与后面的那个数的和小$$5$$，那么这串数的前$$200$$个数之和是~\\uline{~~~~~~~~~~}~. ", "answer_option_list": [[{"aoVal": "A", "content": "$$990$$ "}], [{"aoVal": "B", "content": "$$996$$ "}], [{"aoVal": "C", "content": "$$998$$ "}], [{"aoVal": "D", "content": "$$999$$ "}], [{"aoVal": "E", "content": "$$1009$$ "}], [{"aoVal": "F", "content": "$$1010$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->数列操作周期问题->数的周期"], "answer_analysis": ["第一个数为$$6$$，第二个数为$$3$$， 根据``从第二个数起，每个数都比它前面的那个数与后面的那个数的和小$$5$$''，可以得到第三个数为$$3+5-6=2$$； 同理，后面的数依次得到为:$$6$$、$$3$$、$$2$$、$$4$$、$$7$$、$$8$$、$$6$$、$$3$$、$$2$$、$$4$$\\ldots\\ldots． 易知，每$$6$$个数为一个周期循环．$$200\\div 6=33$$（组）$$\\cdots \\cdots 2$$（个）． 所以前$$200$$个数的和为$$33\\times \\left( 6+3+2+4+7+8 \\right)+6+3=999$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1967", "queId": "b821eea1cc454bf19fa5fce9b534f0c4", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛决赛4分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$3$$辆$$A$$种汽车$$4$$小时可运送$$144$$吨货物，$$5$$辆$$A$$种汽车和$$8$$辆$$B$$种汽车$$2$$小时可运送$$296$$吨货物，$$7$$辆$$B$$种汽车$$4$$小时可运送（~ ）吨货物． ", "answer_option_list": [[{"aoVal": "A", "content": "$$196$$ "}], [{"aoVal": "B", "content": "$$285$$ "}], [{"aoVal": "C", "content": "$$273$$ "}], [{"aoVal": "D", "content": "$$308$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["每辆$$A$$种汽车$$1$$小时可运$$144\\div 3\\div 4=12$$吨，则$$5$$辆$$A$$种汽车$$2$$小时可运$$12\\times 5\\times 2=120$$吨，每辆$$B$$种汽车$$1$$小时可运$$\\left( 296-120 \\right)\\div 8\\div 2=11$$吨，所以$$7$$辆$$B$$种汽车$$4$$小时可运$$11\\times 7\\times 4=308$$吨． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2946", "queId": "8e71738457b84083bb93d7b0008ddfb9", "competition_source_list": ["2015年第13届全国创新杯六年级竞赛第3题", "小学高年级六年级其它2015年数学思维能力等级测试初试第3题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "计算：$$\\frac{1}{3\\times 4}+\\frac{1}{4\\times 5}+\\frac{1}{5\\times 6}+\\cdots +\\frac{1}{9\\times 10}+\\frac{1}{10\\times 11}=$$（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{4}{33}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{33}$$ "}], [{"aoVal": "C", "content": "$$\\frac{8}{33}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{11}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["原式$$=\\frac{1}{3}-\\frac{1}{11}=\\frac{8}{33}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2890", "queId": "db2f954751a54dca8ea2423501f9035e", "competition_source_list": ["2013年走美杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "下面算式中结果最大的是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{8}{9}+\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{8}{9}-\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{8}{9}\\times \\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{8}{9}\\div \\frac{1}{2}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数基准数法"], "answer_analysis": ["解：$$\\frac{8}{9}+\\frac{1}{2}\\textgreater\\frac{8}{9}$$， $$\\frac{8}{9}-\\frac{1}{2} \\textless{} \\frac{8}{9}$$， $$\\frac{8}{9}\\times \\frac{1}{2} \\textless{} \\frac{8}{9}$$ $$\\frac{8}{9}\\div \\frac{1}{2}\\textgreater\\frac{8}{9}$$， 又因为，$$\\frac{8}{9}+\\frac{1}{2} \\textless{} \\frac{8}{9}\\div \\frac{1}{2}$$， 所以，结果最大的是$$\\frac{8}{9}\\div \\frac{1}{2}$$。 故选：D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2191", "queId": "c7704744c5b04ad1b02fdcea0882bcb0", "competition_source_list": ["2016年创新杯五年级竞赛训练题（四）第6题", "2015年第11届全国新希望杯五年级竞赛复赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙两船在静水中的速度相同，它们分别从相距$$60$$千米的两港同时出发相向而行，$$2$$小时后相遇，如果两船的速度各增加$$5$$千米/小时，再次从两港同时出发相向而行，那么，它们再次相遇的地点就与前一次的相遇地点相距$$0.45$$千米，则水流的速度是（~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.7$$千米/小时 "}], [{"aoVal": "B", "content": "$$1.4$$千米/小时 "}], [{"aoVal": "C", "content": "$$0.9$$千米/小时 "}], [{"aoVal": "D", "content": "$$1.8$$千米/小时 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["原来的速度和是船速和$$60\\div 2=30$$，那么船速是$$30\\div 2=15$$千米/时，第二次时间为$$60\\div 40=1.5$$小时，那么水速是$$0.45\\div (2-1.5)=0.9$$千米/时． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2337", "queId": "0990668766054838b6175ca8c0b02113", "competition_source_list": ["2014年IMAS小学高年级竞赛第一轮检测试题第2题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "请问下列哪一组数的乘积等于$$2014$$？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$、$$17$$、$$59$$ "}], [{"aoVal": "B", "content": "$$4$$、$$17$$、$$53$$ "}], [{"aoVal": "C", "content": "$$2$$、$$13$$、$$59$$ "}], [{"aoVal": "D", "content": "$$2$$、$$19$$、$$53$$ "}], [{"aoVal": "E", "content": "$$2$$、$$23$$、$$29$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["选项$$\\text{A}$$乘积的末位数为$$8$$，故不合； 选项$$\\text{B}$$的乘积$$4\\times 17\\times 53\\textgreater60\\times 50=3000$$，故不合； 选项$$\\text{C}$$的乘积$$2\\times 13\\times 59\\textless{}30\\times 60=1800$$，故不合； 选项$$\\text{E}$$的乘积$$2\\times 23\\times 29\\textless{}50\\times 30=1500$$，故不合； 只有选项$$\\text{D}$$的乘积$$2\\times 19\\times 53=2014$$，符合所求． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1437", "queId": "5a5b4649a1d14877953dbb71b06ef6a4", "competition_source_list": ["2016年第16届世奥赛六年级竞赛决赛第8题", "2016年全国世奥赛竞赛A卷第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "国际的形势不稳定，对金融市场也造成了巨大的冲击．某月月初，每盎司黄金价格与每桶原油价格之比为$$47:5$$．月末，它们的单价都跌到了$$70$$美元，每盎司黄金价格与每桶原油的价格之比变为$$96:5$$．则月初每盎司黄金价格是（~ ~ ~ ）美元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1262$$ "}], [{"aoVal": "B", "content": "$$1222$$ "}], [{"aoVal": "C", "content": "$$1300$$ "}], [{"aoVal": "D", "content": "$$2496$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["找到石油月初与月末单价的等量关系列出方程，设月初每盎司黄金价格是$$x$$美元，$$x\\div \\frac{47}{5}-70=\\left( x-70 \\right)\\div \\frac{96}{5}$$，解得$$x=1222$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1643", "queId": "ff808081472482f5014724fa9a1200c7", "competition_source_list": ["2013年全国华杯赛小学中年级竞赛初赛A卷第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "张老师每周的周一、周六和周日都跑步锻炼$$20$$分钟，而其余日期每日都跳绳$$20$$分钟．某月他总共跑步$$5$$小时，那么这个月的第$$10$$天是（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "周日 "}], [{"aoVal": "B", "content": "周六 "}], [{"aoVal": "C", "content": "周二 "}], [{"aoVal": "D", "content": "周一 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["一般这种题目问这个月的第几天是周几时，我们要弄清楚这个月的周几被重复出现$$5$$次．计算$$5\\times 60\\div 20=15$$（天）说明这个月周一、周六、周日都是$$5$$次，说明这个月$$31$$天，且第一天是周六，所以第$$10$$天是$$10\\div 7=1\\cdots3$$，周六、周日、周一这个月的第$$10$$天是周一． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2031", "queId": "e17bb6656b4e4a21832e0a5eeae93527", "competition_source_list": ["2016年全国世奥赛五年级竞赛A卷第11题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "素数是指只能被自身和$$1$$整除的正整数，``孪生素数''是指两个相差为$$2$$的素数，例如$$3$$和$$5$$，$$17$$和$$19$$等．而随着素数的增大，下一个素数离上一个素数应该越来越远，古希腊数学家欧几里得猜想，存在无穷多对素数，它们只相差$$2$$，例如$$3$$和$$5$$，$$5$$和$$7$$，$$2003663613\\times 2195000-1$$和$$2003663613\\times 2195000+1$$等等．这就是著名的孪生素数猜想，它与黎曼猜想、哥德巴赫猜想一样让无数数论者为之着迷．但数学中有这样一些孪生自然数，它们相差$$1$$，例如$$24$$和$$25$$；其中$$24$$是$$6$$的倍数，$$25$$是$$5$$的倍数，$$24$$和$$25$$是相邻的自然数；$$35$$是$$5$$的倍数，$$36$$是$$6$$的倍数。$$35$$和$$36$$是相邻的自然数．如果将$$24$$，$$25$$或$$35$$，$$36$$看做一组，那么$$1\\tilde{ }200$$中共有（~ ）组相邻的自然数，其中一个是$$5$$的倍数，另一个是$$6$$的倍数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$1\\sim 30$$个数中有两组这样相邻自然数：$$5$$和$$6$$；$$24$$和$$25$$．$$\\left[ 5,6 \\right]=30$$，同时考虑$$30$$个数为周期． $$5$$和$$6$$一组，下组分别为$$35$$和$$36$$；$$65$$和$$66$$；$$95$$和$$96$$；$$125$$和$$126$$；$$155$$和$$156$$；$$185$$和$$186$$，共$$7$$组． $$24$$和$$25$$为一组，下组分别为$$54$$和$$55$$；$$84$$和$$85$$；$$114$$和$$115$$；$$144$$和$$145$$；$$174$$和$$175$$，共$$6$$组． 故满足条件的数组共有$$7+6=13$$（组）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1739", "queId": "5cd6718edd6b4c60b70de9134ca8fc2d", "competition_source_list": ["2004年六年级竞赛创新杯", "2004年第2届创新杯六年级竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "某市$$2002$$年底统计近$$10$$年平均降雨量为$$631$$毫米，一年后再统计近$$10$$年平均降雨量为$$601$$毫米，如果$$2003$$年的降雨量是$$450$$毫米，那么，$$1993$$年降雨量是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$1232$$毫米 "}], [{"aoVal": "B", "content": "$$616$$毫米 "}], [{"aoVal": "C", "content": "$$750$$毫米 "}], [{"aoVal": "D", "content": "$$480$$毫米 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题"], "answer_analysis": ["从$$1993$$年初到$$2002$$年底这$$10$$年的平均降雨量为$$631$$毫米，那么这$$10$$年的总降雨量为$$631\\times 10=6310$$毫米，一年后，再统计近$$10$$年的平均降雨量为$$601$$毫米，为$$1994$$年到$$2003$$年的平均降雨量，那么这$$10$$年的总降雨量为$$601\\times 10=6010$$毫米，又$$2003$$年的降雨量为$$450$$毫米，那么$$1994$$到$$2002$$年这$$9$$年的总降雨量为$$6010-450=5560$$毫米，所以$$1993$$年的降雨量为$$6310-5560=750$$毫米，选C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3425", "queId": "aa6e26e9ebb14c3186f7b901cafe2271", "competition_source_list": ["2016年第16届世奥赛六年级竞赛决赛第15题", "2016年全国世奥赛竞赛A卷第15题"], "difficulty": "3", "qtype": "single_choice", "problem": "十字绣是用专用的绣线和十字格布，利用经纬交织的搭十字的方法，对照专用的坐标团进行刺绣，任何人都可以秀出同样效果的一种刺绣方法．十字绣是一种古老的民族刺绣，具有悠久的历史．在我们日常的生活中，一直以来就普遍存在你自制的十字绣赫尔工艺品．在纸上，遵照十字绣的刺绣规律，按照$$2$$条横线和$$1$$条竖线这样的顺序一直画．当画完第$$47$$条直线之后，，将四边形染成红色或黄色 （注：相邻的两个四边形颜色不同）欧欧的作品完成时，黄色四边形共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$132$$ "}], [{"aoVal": "B", "content": "$$158$$ "}], [{"aoVal": "C", "content": "$$224$$ "}], [{"aoVal": "D", "content": "$$264$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据题目要求，我们先补充完这个作品．观察图例，第$$1$$条线为横线，第$$2$$条线为横线，第$$3$$条线为竖线，第$$4$$条线为横线，第$$5$$条线为横线，第$$6$$条线为竖线，$$\\cdot \\cdot \\cdot $$．可以推出规律，每话$$3$$条线，是先画两条横线，再画一条竖线，所以可以画到$$45$$条线时，横线有$$30$$条，竖线有$$15$$条．画到$$47$$条，则需要再画两条横线，也就是说$$47$$条线中，共有$$32$$条横线，$$15$$条竖线．纸上面共有$$\\left( 32+1 \\right)\\times \\left( 15+1 \\right)=528$$个四边形．相邻的两个四边形颜色不能相同的，可以参考国际象棋的棋盘，所以红色和黄色的四边形应该各占一半，也就是说黄色四边形的个数为$$528\\div 2=264$$个． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2887", "queId": "f6d1b7b2e8b5402f83445ebe458964a8", "competition_source_list": ["2012年美国数学大联盟杯六年级竞赛初赛第12题5分(每题5分)"], "difficulty": "1", "qtype": "single_choice", "problem": "（The reciprocal of $$\\frac{3}{4}$$）$$\\div $$（the reciprocal of $$\\frac{4}{3}$$）$$=$$．注：$$reciprocal$$是倒数的意思 ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\frac{16}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{9}{16}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{4}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["（$$\\frac{3}{4}$$的倒数）$$\\div $$（$$\\frac{4}{3}$$的倒数）$$=$$乘积是$$1$$的两个数互为倒数，所以$$\\frac{3}{4}$$的倒数是$$1\\div \\frac{3}{4}=\\frac{4}{3}$$，$$\\frac{4}{3}$$的倒数是$$1\\div \\frac{4}{3}=\\frac{3}{4}$$，$$\\frac{4}{3}\\div \\frac{3}{4}=\\frac{16}{9}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1249", "queId": "1ebf3e9fbe7c4db697c81520ab03b217", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "草地上有许多兔子．数一数黑兔与白兔一共$$16$$只，黑兔与灰兔一共$$17$$只，白兔与灰兔一共$$15$$只．问草地上三种兔子一共有只． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$26$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->列方程解应用题等量代换->等量代换替换型"], "answer_analysis": ["草地上三种兔子一共有： $$(16+17+15)\\div 2$$ $$=48\\div 2$$ $$=24$$（只）． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "90", "queId": "2655098ea2d142328f9fdfa922761206", "competition_source_list": ["2017年第15届全国希望杯五年级竞赛第1试试题第18题", "其它改编题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$，$$B$$，$$C$$，$$D$$，$$E$$五人一同参加飞镖大赛，其中只有一人射中飞镖盘的中心，但不知是谁所射． $$A$$说：``不是我射中的，就是$$C$$射中的．'' $$B$$说：``不是$$E$$射中的．'' $$C$$说：``如果不是$$D$$射中的，那么一定是$$B$$射中的．'' $$D$$说：``既不是我射中的，也不是$$B$$射中的．'' $$E$$说：``既不是$$C$$射中的，也不是$$A$$射中的．'' 其中五人中只有两个人说的是对的，由此可以判断射中飞镖盘中心的人是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$A$$ "}], [{"aoVal": "B", "content": "$$B$$ "}], [{"aoVal": "C", "content": "$$C$$ "}], [{"aoVal": "D", "content": "$$D$$ "}], [{"aoVal": "E", "content": "$$E$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾"], "answer_analysis": ["$$A$$和$$E$$说的话对立，$$C$$和$$D$$说的话对立，必有两对两错，故$$B$$说的是错的，则是$$E$$射中的． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1418", "queId": "8bb96d6c905740da8cb669d9fbab8266", "competition_source_list": ["2013年第9届全国新希望杯小学高年级六年级竞赛复赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$16$$个小朋友，其中$$9$$岁的有$$11$$人，$$11$$岁的有$$2$$人，$$13$$岁的有$$3$$人，那么这$$16$$个小朋友的平均年龄是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$岁 "}], [{"aoVal": "B", "content": "$$10.5$$岁 "}], [{"aoVal": "C", "content": "$$11$$岁 "}], [{"aoVal": "D", "content": "$$11.5$$岁 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->公式类->加权平均数"], "answer_analysis": ["$$\\left( 9\\times 11+11\\times 2+13\\times 3 \\right)\\div 16=10$$（岁）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "167", "queId": "1b8f3c3c402845fda15b5ad6771782d8", "competition_source_list": ["2020年新希望杯三年级竞赛初赛（团战）第60题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$6$$人围成一圈玩狼人游戏，他们中有狼人和平民，平民说真话，狼人说假话．在回答``你左边相邻的人是狼人吗？''这个问题时，有$$2$$人回答``是''，有$$4$$人回答``不是''．这$$6$$人中最多有个狼人？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->推理推导->言语逻辑推理"], "answer_analysis": ["暂无 "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "564", "queId": "03388d396a5643298fcb9ac0f6cfedfb", "competition_source_list": ["2018年美国数学大联盟杯三年级竞赛初赛第31题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "下面结果哪个是奇数． ", "answer_option_list": [[{"aoVal": "A", "content": "even $$\\times$$ odd "}], [{"aoVal": "B", "content": "odd $$\\times$$ even "}], [{"aoVal": "C", "content": "odd $$+$$ even "}], [{"aoVal": "D", "content": "odd $$+$$ odd "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["本题题意为``下面结果哪个是奇数''， $$\\text{A}$$选项是偶数$$\\times $$奇数$$=$$偶数， $$\\text{B}$$选项是奇数$$\\times $$偶数$$=$$偶数， $$\\text{C}$$选项是奇数$$+$$偶数$$=$$奇数， $$\\text{D}$$选项是奇数$$+$$奇数$$=$$偶数， 综上结果为奇数的只有$$\\text{C}$$选项． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2535", "queId": "2beef333b33f42ba9030a64e216b8858", "competition_source_list": ["2016年第7届广东广州羊排赛六年级竞赛第6题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列数的正确读法中，能读出三个零的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2002.0$$ "}], [{"aoVal": "B", "content": "$$201602016$$ "}], [{"aoVal": "C", "content": "$$20160201$$ "}], [{"aoVal": "D", "content": "$$1001.0001$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数基础->小数的认识"], "answer_analysis": ["选项$$\\text{A}$$读法为二千零二点零，选项$$\\text{B}$$读法为二亿零一百六十万二千零一十六，选项$$\\text{C}$$读法为二千零一十六万零二百零一，选项$$\\text{D}$$读法为一千零一点零零零一． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1565", "queId": "8c584ddaa66d412698cd7b765bd63d44", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（四）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "三位小朋友一起跳绳，平均每人跳了$$130$$个，小明跳的个数是小亮跳的个数的$$\\frac{1}{2}$$，小亮跳的个数是小强跳的个数的$$\\frac{3}{2}$$，则小亮共跳了（~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$90$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$130$$ "}], [{"aoVal": "D", "content": "$$180$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题"], "answer_analysis": ["设小强跳了$$x$$个，则小亮跳了$$\\frac{3}{2}x$$个，小明跳了$$\\frac{3}{4}x$$个．$$x+\\frac{3}{2}x+\\frac{3}{4}x=130\\times 3$$，解得：$$x=120$$，$$\\frac{3}{2}x=180$$，小亮跳了$$180$$个． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2112", "queId": "f99e0c818d0942b7a93edfa7e51211b6", "competition_source_list": ["2011年世界少年奥林匹克数学竞赛六年级竞赛初赛第3题5分", "2021年陕西西安六年级下学期小升初模拟分类专题五十六（和差倍问题 盈亏问题）第6题", "2018年海南海口琼山区华中师范大学海南附属中学小升初华师杯第10题3分", "2018年浙江金华六年级下学期小升初模拟第8题2分", "2019年陕西西安灞桥区铁一中滨河中学小升初入学真卷2第12题3分", "2016年陕西西安小升初某西北大附中"], "difficulty": "1", "qtype": "single_choice", "problem": "两个桶里共盛水$$40$$斤，若把第一个桶里的水倒$$6$$斤到第$$2$$个桶里，两个桶里的水就一样多，则第一桶有~\\uline{~~~~~~~~~~}~斤水． ", "answer_option_list": [[{"aoVal": "A", "content": "$$23$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$26$$ "}], [{"aoVal": "D", "content": "$$14$$ "}], [{"aoVal": "E", "content": "都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想", "Overseas Competition->知识点->应用题模块->和差倍问题->和差问题"], "answer_analysis": ["由题意的第一个桶比第二个桶多$$12$$斤，所以利用和差关系： 第一桶水：$$(40+12)\\div 2=26$$（斤）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "682", "queId": "2cca264bfb004f3bb44b0190ab7757e0", "competition_source_list": ["2021年新希望杯五年级竞赛初赛第39题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "$$2021$$年，疯狂动物城警察局为了表彰表现优异的警察，给他们授予特殊的警号，这些警号是形如$$\\overline{\\square 2021\\square }$$的六位数，并且都能被$$21$$整除，这样的警号有~\\uline{~~~~~~~~~~}~个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->整除->整除特征->和系整除特征应用"], "answer_analysis": ["$$21=3\\times 7$$，故能被$$21$$整除必能被$$3$$和$$7$$整除，满足被$$3$$整除为各位数字之和为$$3$$的倍数， 故$$2+0+2+1=5$$故余下$$2$$位数字和最大为$$9+9=18$$， 故可取$$1$$，$$4$$，$$7$$，$$10$$，$$13$$，$$16$$， $$1$$：$$1+0$$，$$4$$：$$4+0$$，$$3+1$$，$$2+2$$，$$7$$：$$7+0$$，$$6+1$$，$$5+2$$，$$4+3$$， $$10$$：$$9+1$$，$$8+2$$，$$7+3$$，$$6+4$$，$$5+5$$， $$13$$：$$9+4$$，$$8+5$$，$$7+6$$， $$16$$：$$9+7$$，$$8+8$$， 上述除$$0$$不能在首位外共有：$$1$$：$$1$$种；$$4$$：$$1+2+1=4$$种，$$7$$：$$1+2+2+2=7$$种；$$10$$：$$2+2+2+2+1=9$$种；$$13$$：$$2+2+2=6$$种；$$16$$：$$2+1=3$$种， 能被$$7$$整除特点为（除去个位数字后的数$$-2\\times $$个位）能被$$7$$整除， 故一共有：$$420210$$，$$520212$$，$$620214$$，$$720216$$，$$820218$$共$$5$$个． 故答案为：$$5$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "549", "queId": "096312cc2bb446f3810751d690c6f53e", "competition_source_list": ["2017年全国美国数学大联盟杯小学高年级六年级竞赛初赛第20题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{(2017 Math League, Priamry 6, Question \\#20)} How many whole-number factors does $$8640$$ have？ $$8640$$有多少个因数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$38$$ "}], [{"aoVal": "D", "content": "$$56$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析", "Overseas Competition->知识点->数论模块->因数与倍数->公因数与公倍数"], "answer_analysis": ["$$8640$$有多少个因数？ whole-number ~整数；$$factor$$ ~因数． 它的因数有$$\\left( 6+1 \\right)\\times \\left( 3+1 \\right)\\times \\left( 1+1 \\right)=56$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3234", "queId": "39de7b89d09c4b558bf55e2264699072", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛六年级竞赛决赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "掷一大一小两个骰子，每次掷出的两个点数均为质数的概率是$$ \\%$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$75$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["骰子的点数为$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$． 其中质数有$$2$$，$$3$$，$$5$$， 故$$1$$个骰子掷出质数的概率为$$\\frac{1}{2}$$， 则两次均为质数的概率为$$\\frac{1}{2}\\times\\frac{1}{2}=\\frac{1}{4}=25 \\%$$， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "217", "queId": "5109689684284b2b988540cc3b390cd6", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第15题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在一次数学考试中，共有$$10$$道选择题，评分标准是：基础分为$$10$$分，答对一题得$$3$$分，答错一题倒扣$$1$$分，不答得$$0$$分．已知参赛的学生中，至少有$$3$$人得分相同，则参加考试的学生至少有人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$75$$ "}], [{"aoVal": "B", "content": "$$76$$ "}], [{"aoVal": "C", "content": "$$77$$ "}], [{"aoVal": "D", "content": "$$78$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["总分最高是$$10+10\\times3=40$$分，最低是$$10-10\\times1=0$$，$$0\\sim 40$$共$$41$$种得分都有可能取到，从两端极值考虑，最小端，$$1=10-9\\times1-0\\times1$$，$$2=10-8\\times1-0\\times2$$，$$3=10-7\\times1-0\\times3$$，则后面的分数都能取到，再考虑最大端，若做对$$9$$题，还有一题可以错，也可以不做，分别得到$$36$$分和$$37$$分，则$$38$$、$$39$$取不到，若做对$$8$$题，剩余两题可能全错，全不做，错一题不做一题，得分分别是$$32$$、$$34$$、$$33$$，则$$35$$分取不到，依次类推可以得到$$0\\sim 40$$中$$35$$、$$38$$、$$39$$三种得分取不到，所以得分情况只有$$41-3=38$$种，要想保证有$$3$$人分一样，则每种得分至少有$$2$$人，再多一人就可以满足至少有$$3$$人得分一样，所以是$$38\\times2+1=77$$人． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1000", "queId": "01e9d1e733ed4a249da2845abb844775", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛决赛第4题", "2018年浙江杭州西湖区小学高年级六年级上学期单元测试《第三单元》第8题3分", "2018年四川成都锦江区四川师范大学附属第一实验中学小升初（四）第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一种商品的利润率为$$20 \\%$$，进价提高$$25 \\% $$后，保持利润不变，那么，进价提价后的利润率为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25 \\%$$ "}], [{"aoVal": "B", "content": "$$20 \\% $$ "}], [{"aoVal": "C", "content": "$$16 \\% $$ "}], [{"aoVal": "D", "content": "$$12.5 \\%$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数的运算->估算->百分数的简单实际问题->折扣、成数、税率、利���", "拓展思维->知识点->应用题模块->经济问题->基本经济概念->利润基本公式->已知利润成本求利润率"], "answer_analysis": ["方法一：利润问题．假设成本为$$100$$元，则利润为$$20$$元，现在成本是$$100\\times (1+25 \\%)=125$$元，$$20\\div 125\\times 100 \\%=16 \\%$$． 方法二：$$20\\%\\div （1+25\\%）=16\\%$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "236", "queId": "487a03b15abe481fb146841d770fde68", "competition_source_list": ["2015年湖北武汉世奥赛小学高年级六年级竞赛模拟训练题（三）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "写有$$0$$、$$1$$、$$2$$、$$3\\cdots \\cdots $$、$$9$$的卡片各一张．$$A$$、$$B$$、$$C$$、$$D$$、$$E$$分别拿走$$2$$张，然后报出自己所拿两张卡片上的数的和．已知$$A$$报$$5$$，$$B$$报$$12$$，$$C$$报$$10$$，$$D$$报$$12$$， $$E$$拿的． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$和$$6$$ "}], [{"aoVal": "B", "content": "$$1$$和$$6$$ "}], [{"aoVal": "C", "content": "$$0$$和$$7$$ "}], [{"aoVal": "D", "content": "$$1$$和$$5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$E$$拿的两张卡片的和为$$\\left( 0+1+\\ldots \\ldots +9 \\right)\\left( 5+12+10+12 \\right)=6$$，则$$E$$拿的卡片有三种情况：①$$2+4$$；②$$1+5$$； ③$$0+6$$．而$$5=0+5=1+4=2+3$$，$$12=3+9=4+8=5+7$$，依次假设①②③种情况，进行推理得出矛盾，最后满足条件的是③． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1741", "queId": "d66234c7a48d4b37b18658d54188658e", "competition_source_list": ["2019年第24届YMO四年级竞赛决赛第1题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第1题3分", "2020年第24届YMO四年级竞赛决赛第1题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第1题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲有桌子若干张，乙有椅子若干把．如果乙用全部椅子换回相同数量的桌子，那么需要补给甲$$640$$元；如果乙不补钱，就会少换回$$5$$张桌子．已知$$3$$张桌子比$$5$$把椅子的价钱少$$96$$元．乙原来有椅子把． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["桌子的价钱：$$640\\div 5=128$$（元）， 椅子的价钱：$$(128\\times 3+96)\\div 5=480\\div 5=96$$（元）， 所以乙有椅子的数量为$$640\\div (128-96)=640\\div 32=20$$（把）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1452", "queId": "b0b8bc2251df40798ed633b8b8dc4fca", "competition_source_list": ["2008年第6届创新杯五年级竞赛初赛A卷第4题5分", "2008年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "小华玩射击游戏，打中一枪得$$5$$分，打不中倒扣$$2$$分，他打了$$20$$枪，共得$$51$$分，小华共打中了（ ）枪。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题"], "answer_analysis": ["当全打中时得$$20\\times 5=100$$（分），没打中$$\\left(100-51\\right) \\div \\left(5+2\\right) =7$$（枪），故打中$$20-7=13$$（枪）。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2550", "queId": "8fdd61d34f4d4130a1872f2a0b06e43a", "competition_source_list": ["2014年迎春杯六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "算式$$2014\\times \\left( \\frac{1}{19}-\\frac{1}{53} \\right)$$的计算结果是( )． ", "answer_option_list": [[{"aoVal": "A", "content": "$$34$$ "}], [{"aoVal": "B", "content": "$$68$$ "}], [{"aoVal": "C", "content": "$$144$$ "}], [{"aoVal": "D", "content": "$$72$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数巧算->乘法分配律"], "answer_analysis": ["原式$$=2014\\times \\frac{1}{19}-2014\\times \\frac{1}{53}=106-38=68$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2897", "queId": "8a41bc86a34647249e52fe7cabd971e3", "competition_source_list": ["2017年IMAS小学中年级竞赛（第一轮）第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "请问算式$$25\\times 99+55\\times 5$$的值等于什么？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2750$$ "}], [{"aoVal": "B", "content": "$$2850$$ "}], [{"aoVal": "C", "content": "$$2900$$ "}], [{"aoVal": "D", "content": "$$2950$$ "}], [{"aoVal": "E", "content": "$$3000$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$25\\times 99+55\\times 5=25\\times (99+11)=25\\times 110=2750$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2507", "queId": "46cb1d4d2c954550a1de6dd0246d4fc1", "competition_source_list": ["2018~2019学年浙江杭州西湖区杭州市行知小学五年级上学期期中期中竞赛第17题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "小机灵在用计算器计算``$$6.9\\times 7$$''时，发现计算器的键``$$6$$''坏了，小机灵想到了四种不同的输入方法．请你判断一下，下面选项中的方法是错误的 ", "answer_option_list": [[{"aoVal": "A", "content": "$$2.3\\times 3\\times 7$$ "}], [{"aoVal": "B", "content": "$$13.8\\times 7\\div 2$$ "}], [{"aoVal": "C", "content": "$$2\\times 3\\times 7+0.9\\times 7$$ "}], [{"aoVal": "D", "content": "$$7\\times 7-7$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->小数->小数乘除->小数乘法运算"], "answer_analysis": ["$$7\\times 7-7=7\\times \\left( 7-1 \\right)=7\\times 6=42$$，提取公因数后发现$$\\text{D}$$式变为$$7\\times 6$$，而非原式的$$6.9\\times 7$$，所以错误． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1022", "queId": "45d607aad5a548e0b17bdfb8dc488d57", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "石灰水是把石灰和水按照$$1∶100$$配制而成，要配制$$4545$$千克石灰水，需要石灰多少千克？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$千克 "}], [{"aoVal": "B", "content": "$$45$$千克 "}], [{"aoVal": "C", "content": "$$55$$千克 "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["相当于把石灰水分成了$$101$$份，石灰占$$1$$份．所以石灰有$$4545\\times \\frac{1}{{1{ + }100}}{ = }45$$（千克）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "402", "queId": "a0006f0cc1c64f2a9a9e328114af2d15", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙两人玩拿火柴棍游戏，桌上共有$$10$$根火柴棍，谁取走最后一根谁胜。甲每次可以取走$$1$$根、$$3$$根或$$4$$根（只能取恰好的数量，如果最后剩$$2$$根火柴棍，甲只能取$$1$$根），乙每次可以取$$1$$根或$$2$$根。如果甲先取，那么甲为了取胜，第一次应（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "取$$1$$根 "}], [{"aoVal": "B", "content": "取$$3$$根 "}], [{"aoVal": "C", "content": "取$$4$$根 "}], [{"aoVal": "D", "content": "无论怎么取都无法获胜 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->智巧趣题->数学趣题->火柴棒问题"], "answer_analysis": ["解：无论甲怎么取，乙只要让最后火柴棍剩两根，甲这时只能取$$1$$根，乙胜； 在这之前只要保证火柴剩下$$5$$根，甲取$$1$$根，则乙取$$2$$根，剩$$2$$根，乙胜； 或者甲取$$3$$根，乙取$$2$$根，乙胜；或者甲取$$4$$根，乙取$$1$$根，乙胜。 所以甲无论怎么取都无法获胜。 故选:D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1586", "queId": "ff8080814526d2f401452f9088cd2443", "competition_source_list": ["2014年全国迎春杯五年级竞赛初赛第4题", "2017年全国小升初八中入学备考课程"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$、$$b$$、$$c$$、$$d$$四个数的平均数是$$12.345$$，$$a\\textgreater b\\textgreater c\\textgreater d$$，那么$$b$$（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "大于$$12.345$$ "}], [{"aoVal": "B", "content": "小于$$12.345$$ "}], [{"aoVal": "C", "content": "等于$$12.345$$ "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["排除法，A、B、C三个选项均可找到反例，故无法确定． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "708", "queId": "3a8a78451e924f19951be2d791168446", "competition_source_list": ["2020年春蕾杯六年级竞赛第10题2分", "2021年春蕾杯六年级竞赛第5题2分"], "difficulty": "2", "qtype": "single_choice", "problem": "某班有一个小图书馆，共有$$300$$多本书，从$$1$$开始，图书按自然数的顺序编号，即$$1$$，$$2$$，$$3\\cdots $$．小光看了这图书馆里都被$$2$$，$$3$$和$$8$$整除的书号，共$$16$$本，这个图书馆里至少有 本图书． ", "answer_option_list": [[{"aoVal": "A", "content": "$$381$$ "}], [{"aoVal": "B", "content": "$$382$$ "}], [{"aoVal": "C", "content": "$$383$$ "}], [{"aoVal": "D", "content": "$$384$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->多数的最小公倍数"], "answer_analysis": ["$$2$$、$$3$$、$$8$$的最小公倍数是$$24$$，所以这些书号都是$$24$$的倍数， $$24\\times 16=384$$（本）． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2068", "queId": "d46cfc9a0fe04b46b473df5f8f63d61e", "competition_source_list": ["2013年第12届上海小机灵杯小学中年级四年级竞赛初赛第4题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "一只青蛙掉到了$$20$$米深的井里．每天白天它可以沿着湿滑的井壁向上爬$$3$$米，但它晚上休息时会掉下$$2$$米．青蛙第天才能爬出这口井． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["蜗牛爬井问题． 若最后一天青蛙向上爬$$3$$米到达井口，就不会向下掉了，则之前至少爬$$20-3=17$$（米）； 每天向上爬$$3$$米，晚上休息时会掉下$$2$$米，实际每天向上$$1$$米； $$17\\div 1+1=18$$（天）； 青蛙第$$18$$天才能爬出这口井； 选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "74", "queId": "d9aed025a8f64c03adef18393280886d", "competition_source_list": ["2014年第10届全国新希望杯小学高年级六年级竞赛复赛第6题4分"], "difficulty": "4", "qtype": "single_choice", "problem": "某足球联赛积分规则如下：每队赛$$30$$场，胜一场得$$3$$分，平一场得$$1$$分，负一场得$$0$$分．在$$2013$$赛季中，希望队以总积分$$77$$分位居积分榜首位，且负的场数比平的场数少，则这个赛季该队共胜了（~ ）场比赛． ", "answer_option_list": [[{"aoVal": "A", "content": "$$22$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->体育比赛->3-1-0 积分制"], "answer_analysis": ["设胜了$$x$$场，平$$y$$场，负$$\\left( 30-x-y \\right)$$ 有$$3x+y=77$$ $$30-\\left( x+y \\right)\\textless{}y$$，则$$x=24$$ 选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "115", "queId": "46a85ef9d6634b19b7616e8e69f11c1c", "competition_source_list": ["2015年第20届全国华杯赛小学中年级四年级竞赛初赛A卷", "2016年第20届四川成都华杯赛小学中年级竞赛B卷第1~6题60分"], "difficulty": "1", "qtype": "single_choice", "problem": "一次考试共有$$6$$道选择题，评分规则如下：每人先给$$6$$分，答对一题加$$4$$分，答错一题减$$1$$分，不答得$$0$$分．现有$$51$$名同学参加考试，那么，至少有（~ ~ ~ ~ ~）人得分相同． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["设答对了$$x$$题，答错$$y$$题，$$x+y$$≤$$6$$； 当$$x=6$$时，得分$$30$$分； 当$$x=5$$时，$$y=0$$，$$1$$，对应得分$$26$$，$$25$$； 当$$x=4$$时，$$y=0$$，$$1$$，$$2$$，对就得分$$22$$，$$21$$，$$20$$； 当$$x=3$$时，$$y=0$$，$$1$$，$$2$$，$$3$$，对应得分$$18$$，$$17$$，$$16$$，$$15$$； 当$$x=2$$时，$$y=0$$，$$1$$，$$2$$，$$3$$，$$4$$，对应得$$14$$，$$13$$，$$12$$，$$11$$，$$10$$； 当$$x=1$$时，$$y=0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，对应得$$10$$，$$9$$，$$8$$，$$7$$，$$6$$，$$5$$； 当$$x=0$$时，$$y=0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，对应得分$$6$$，$$5$$，$$4$$，$$3$$，$$2$$，$$1$$，$$0$$； 共计$$25$$种得分，$$51\\div 25=2\\cdots 1$$，则至少$$2+1=3$$人得分相同. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "763", "queId": "6d4923af93604af58417e7dea91c601e", "competition_source_list": ["2005年第3届创新杯六年级竞赛初赛第9题", "2005年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "盒中原有7个小球，魔术师从中取出若干个小球，把每个小球都变成7个小球，将其放回盒中，他又由其中取出若干个小球，把每个小球变成7个小球，再将其放回盒中$$\\cdots \\cdots $$，如此进行到某一时刻，当魔术师停止变魔术时，盒中球的总数可能是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "2003个 "}], [{"aoVal": "B", "content": "2004个 "}], [{"aoVal": "C", "content": "2005个 "}], [{"aoVal": "D", "content": "2006个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法"], "answer_analysis": ["把1个球变成7个球，那么球的数量就增加了6个，则无论这个魔术师怎么变，球增加的数量为6的倍数，又原有7个球，那么球的总数被6除的余数总是1，而$$A，B，C，D$$这4个选项中，只有$$C$$选项的2005除以6的余数为1，所以选$$C$$. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1825", "queId": "d23559f76679428e9da53416a9f03634", "competition_source_list": ["2013年第25届广东广州五羊杯六年级竞赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某学校派发儿童节礼物，其中一种是魔方，全校$$300$$个学生共派发$$510$$个魔方，已知男生每人两个魔方，女生每人一个魔方，则该学校的女生人数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$90$$ "}], [{"aoVal": "B", "content": "$$105$$ "}], [{"aoVal": "C", "content": "$$150$$ "}], [{"aoVal": "D", "content": "$$180$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["设全校有男生有$$x$$人，则$$2x+300-x=510$$， 可解得$$x=210$$， 所以女生有$$90$$人． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "137", "queId": "82c2237cfa954124a0b0b1472d1638cc", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "算式$$a+\\frac{1}{b+\\dfrac{1}{c}}+d+\\frac{1}{e+\\dfrac{1}{f}}+g+\\frac{1}{h+\\dfrac{1}{i}}$$的最大值为，其中每个不同的字母代表不同的非零数码． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25\\frac{1003}{1008}$$ "}], [{"aoVal": "B", "content": "$$25\\frac{611}{1014}$$ "}], [{"aoVal": "C", "content": "$$25\\frac{609}{1026}$$ "}], [{"aoVal": "D", "content": "$$25\\frac{597}{1040}$$ "}], [{"aoVal": "E", "content": "$$25\\frac{620}{1001}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["要使算式的值尽可能大，则$$a$$，$$d$$，$$g$$应尽可能大，可取$$a=9$$，$$d=8$$，$$g=7$$．而$$b$$，$$e$$，$$h$$则应尽可能小，所以可取$$b=1$$，$$e=2$$，$$A=3$$， 由于$$\\frac{1}{m+\\dfrac{1}{n+1}}\\textgreater\\frac{1}{(m+1)+\\dfrac{1}{n}}$$，故取$$c=6$$，而$$f$$，$$i$$只剩下$$4$$和$$5$$可供选择，故可取$$f=5$$，$$i=4$$，故最大值为$$9+\\frac{1}{1+\\dfrac{1}{6}}+8+\\frac{1}{2+\\dfrac{1}{5}}+7+\\frac{1}{3+\\dfrac{1}{4}}=25\\frac{620}{1001}$$． 故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "694", "queId": "5551e3413b4f4270a5e4f900365de7f5", "competition_source_list": ["2006年第11届全国华杯赛竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2008006$$共有( )个质因数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["因为$$2008006=2006\\times 1000+2006=2006\\times 1001=（2\\times 17\\times 59）\\times （7\\times 11\\times 13）$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2517", "queId": "ab8fa36c4fc84023a278d9fcf6d99fed", "competition_source_list": ["2018年湖北武汉创新杯小学高年级六年级竞赛初赛数学思维能力等级测试第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知一班和二班的人数比是$$8:7$$，从一班调$$8$$名同学到二班去，一班和二班的人数之比变为$$4:5$$，那么二班原来的人数为（~ ）人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$ "}], [{"aoVal": "B", "content": "$$35$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["由于前后总和不变，则变比为原��一$$:$$二$$=8:7=24:21$$，现在一$$:$$二$$=4:5=20:25$$，易知$$4$$份$$=8$$人，$$1$$份$$=2$$人，则二班原有$$2\\times 21=42$$人． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "872", "queId": "a90ea7eb17f9454dbd046dae0efccf2e", "competition_source_list": ["2013年第12届上海小机灵杯小学高年级五年级竞赛初赛第5题1分"], "difficulty": "0", "qtype": "single_choice", "problem": "古时候的原始人捕猎，捕到一只野兽对应一根手指．等到$$10$$根手指用完，就在绳子上打一个结，这就是运用现在的数学中的． ", "answer_option_list": [[{"aoVal": "A", "content": "出入相补原理 "}], [{"aoVal": "B", "content": "等差数列求和 "}], [{"aoVal": "C", "content": "十进制计数法 "}], [{"aoVal": "D", "content": "四舍五入 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->进制的性质与应用->进制的认识"], "answer_analysis": ["古时候的原始人捕猎运用现在的数学中的十进制计数法． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "994", "queId": "017eb05bc5b94ddbbc76e27a56396fd3", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "五个小朋友的岁数和为$$47$$岁，三个成人的岁数和为$$73$$岁，到五个小朋友的岁数和与三个成人的岁数和相等时，需经过年． ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["每过一年，两组的年龄差减少$$5-3=2$$（岁），所以需要经过$$(73-47)\\div 2=13$$（年）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2818", "queId": "5345059d7e034de98c9d9c42c9007bfb", "competition_source_list": ["2019年第24届YMO三年级竞赛决赛第4题3分", "2020年第24届YMO三年级竞赛决赛第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "按规律写算式：$$100+2$$，$$98-5$$，$$96+8$$，$$94-11$$，$$\\cdots $$第$$10$$个算式是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$80-27$$ "}], [{"aoVal": "B", "content": "$$82-29$$ "}], [{"aoVal": "C", "content": "$$82+29$$ "}], [{"aoVal": "D", "content": "$$82-27$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["规律：每个算式第一个数是从$$100$$开始，依次少$$2$$， 第二个数是从$$2$$开始，依次多$$3$$， 算式中间的符号是``$$+-$$''循环， 则第$$10$$个算式，中间符号为``$$-$$''， 第$$1$$个数为$$100-9\\times 2=82$$， 第$$2$$个数为$$2+3\\times 9=29$$， 则第$$10$$个算式为``$$82-29$$''． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "585", "queId": "0b5db628be1549f691704e8fd6023303", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "分母是$$2016$$的所有最简真分数的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$288$$ "}], [{"aoVal": "B", "content": "$$576$$ "}], [{"aoVal": "C", "content": "$$144$$ "}], [{"aoVal": "D", "content": "$$200$$ "}], [{"aoVal": "E", "content": "$$300$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["先把$$2016$$分解质因数，可得分子是$$2$$的倍数、$$3$$的倍数、$$7$$的倍数时，都不是最简分数，$$1\\sim 42$$中，不是$$2$$，$$3$$，$$7$$倍数的数有$$1$$，$$5$$，$$11$$，$$13$$，$$17$$，$$19$$，$$23$$，$$25$$，$$29$$，$$31$$，$$37$$，$$41$$总共$$12$$个，和为$$252$$，把这$$12$$个数分别加$$42$$，就得到了$$43\\sim 84$$中不是$$2$$，$$3$$，$$7$$倍数的数，和为$$252+12\\times 42$$，以此类推，$$1\\sim 2016$$中不是$$2$$，$$3$$，$$7$$倍数的数的和是$$252+(252+12\\times 42)+(252+12\\times 42\\times 2)+(252+12\\times 42\\times 3)+\\cdots +(252+12\\times 42\\times 42)$$，求出和再除以$$2016$$即可． $$2016=2\\times 2\\times 2\\times 2\\times 2\\times 3\\times 3\\times 7$$， 所以分子是$$2$$的倍数、$$3$$的倍数、$$7$$的倍数时，都不是最简分数， $$1\\sim 42$$中，不是$$2$$，$$3$$，$$7$$倍数的数有$$1$$，$$5$$，$$11$$，$$13$$，$$17$$，$$19$$，$$23$$，$$25$$，$$29$$，$$31$$，$$37$$，$$41$$总共$$12$$个，和为$$252$$． 把这$$12$$个数分别加$$42$$，就得到了$$43\\sim 84$$中不是$$2$$，$$3$$，$$7$$倍数的数， 和为$$252+12\\times 42$$ $$=252+504$$ $$=756$$， 以此类推，$$1\\sim 2016$$中不是$$2$$，$$3$$，$$7$$倍数的数的和是 $$252+(252+12\\times 42)+(252+12\\times 42\\times 2)+(252+12\\times 42\\times 3)+\\cdots +(252+12\\times 42\\times 42)$$ $$=252+(252+504)+(252+1008)+(252+1512)+\\cdots +(252+21168)$$ $$=252+756+1260+1764+\\cdots +21420$$ $$=580608$$， 分母是$$2016$$的所有最简真分数的和是$$\\frac{580608}{2016}=288$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2906", "queId": "73fe62bbcd374e61b799428e9ae13120", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "算式$$19+28+37+46+55+64+73+82+91$$的计算结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$450$$ "}], [{"aoVal": "B", "content": "$$485$$ "}], [{"aoVal": "C", "content": "$$495$$ "}], [{"aoVal": "D", "content": "$$505$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$19+28+37+46+55+64+73+82+91$$ $$=(19+91)+(28+82)+(37+73)+(46+64)+55$$ $$=110+110+110+110+55$$ $$=495$$． 所以选择$$\\text C$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "951", "queId": "e65a82a801b446ca85d133c38787d9ba", "competition_source_list": ["2013年IMAS小学高年级竞赛第二轮检测试题第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "请问下面哪个数写在$$2013$$后面组成的六位数能同时被$$3$$、$$4$$、$$7$$整除？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$56$$ "}], [{"aoVal": "E", "content": "$$78$$~ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["同时被$$3$$、$$4$$、$$7$$整除意味着能分别整除$$3$$、$$4$$、$$7$$． 整除$$3$$各项数之和为$$3$$的倍数$$2013\\to 2+1+3=6$$，排除$$\\text{B}$$项和$$\\text{D}$$项． 整除$$4$$后两项能整除$$4$$，排除$$\\text{E}$$． 剩余部分代入$$\\text{A}$$项不能整除，故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2525", "queId": "1ed8bf38dc4747b3986938384467ae07", "competition_source_list": ["2017年第8届广东广州羊排赛六年级竞赛第2题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "小伦有$$a$$个苹果，肥罗的苹果比小伦多$$b$$个，两人一共个苹果． ", "answer_option_list": [[{"aoVal": "A", "content": "$$a+b$$ "}], [{"aoVal": "B", "content": "$$2a+b$$ "}], [{"aoVal": "C", "content": "$$a+2b$$ "}], [{"aoVal": "D", "content": "$$2a+2b$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["肥罗有苹果$$(a+b)$$个，两人共有$$a+(a+b)=(2a+b)$$个． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1560", "queId": "5fb96f6ccbc1486b8bcddabf2570f7f0", "competition_source_list": ["2017年第13届湖北武汉新希望杯六年级竞赛决赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "水池有甲、乙两根进水管，单独打开甲进水管$$6$$小时可注满空水池，单独打开乙进水管$$4$$小时可注满空水池．如果按照甲、乙、甲、乙$$\\ldots\\ldots$$的顺序轮流打开$$1$$小时，注满空水池需小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2.4$$~~ "}], [{"aoVal": "B", "content": "$$3$$~~~~~ "}], [{"aoVal": "C", "content": "$$4.4$$~~ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->接力施工问题"], "answer_analysis": ["甲、乙$$2$$小时一个周期：$$\\frac{1}{4}+\\frac{1}{6}=\\frac{5}{12}$$，两个周期后还剩下：$$1-2\\times \\frac{5}{12}=\\frac{1}{6}$$，剩下的甲做：$$\\frac{1}{6}\\div \\frac{1}{6}=1$$小时，共需$$2\\times 2+1=5$$小时． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1818", "queId": "b6c33801f35540aa8493c00fcfbb575f", "competition_source_list": ["2013年第11届创新杯四年级竞赛初赛第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙两名鱼贩，卖同一种鱼，两人卖法如下： 甲卖$$10$$元一斤；乙把鱼分成鱼头和鱼身两部分卖，鱼头$$9.5$$元一斤、鱼身$$10.5$$元一斤．照这样的卖法，甲、乙卖同样的一条鱼，甲、乙所得的钱比较． （提示：鱼身重量大于鱼头重量）． ", "answer_option_list": [[{"aoVal": "A", "content": "乙卖的多 "}], [{"aoVal": "B", "content": "甲卖的多 "}], [{"aoVal": "C", "content": "甲、乙同样多 "}], [{"aoVal": "D", "content": "无法确定谁多 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->不定方程解应用题"], "answer_analysis": ["设鱼头重量为$$a$$，鱼身重量为$$b$$． 甲卖鱼所得钱为：$$10a+10b$$； 乙卖鱼所得钱为∶$$9.5a+10.5b$$， 乙卖鱼所得钱$$-$$甲卖鱼所得钱得： $$\\left( 9.5a+10.5b \\right)-\\left( 10a+10b \\right)$$ $$=0.5b-0.5a$$． 因为$$b\\textgreater a$$， 所以$$0.5b-0.5a\\textgreater0$$． 即乙卖鱼所得钱比甲卖鱼所得钱多． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "433", "queId": "932cd703d4394fbdb4ac975234945c98", "competition_source_list": ["2013年小机灵杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "现有甲、乙、丙三人同时说了以下三句话。甲说：``乙正在说谎。''乙说：``丙正在说谎。''丙说：``他俩正在说谎。''根据三人的对话情况，请你分析，（ ）说的是真话。 ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理"], "answer_analysis": ["解：若甲说的是真话，则乙说谎，丙说真话，但丙说甲在说谎，所以矛盾。因此，甲说谎。 甲说谎，则乙说的是真话，丙是说谎的，此时符合。 所以，三人中，甲说谎，乙说真话，丙说谎。 故答案为：B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2484", "queId": "54657b24f18949d7b3b8a5709f56fcc5", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "根据下面两个算式，则$$\\triangle$$代表． $$\\triangle +\\triangle +\\triangle -\\square -\\square =12$$~ ~ ~ ~ ~ ~ ~ $$\\square +\\square +\\square -\\triangle -\\triangle =2$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["依题意有：$$3\\triangle -2\\square =12$$①，$$3\\square -2\\triangle =2$$②， ①$$\\times 3$$，②$$\\times 2$$得： $$9\\triangle -6\\square =36$$③， $$6\\square -4\\triangle =4$$④， ③$$+$$④得：$$5\\triangle =40$$， 所以$$\\triangle =8$$． 故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "422", "queId": "9bd9b9f3500b43a8972e4de3e65420cb", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个两位数，它$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$、$$10$$倍的数字和都与原两位数的数字和相等，这样的两位数中，最大数减去最小数的差是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$99$$ "}], [{"aoVal": "B", "content": "$$81$$~ "}], [{"aoVal": "C", "content": "$$27$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["记两位数的数字和为$$M$$，并且进了$$k$$次位，则可得$$2M-9k=M$$，得$$M=9k$$，即$$M$$是$$9$$的倍数，即两位数最大为$$99$$，$$M$$最大为$$99$$，$$M$$最小为$$18$$，则差为$$99-18=81$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "193", "queId": "24a3c6acbb454ffdb9cd9adbd9706d08", "competition_source_list": ["2022年新加坡高级学府数学竞赛（SASMO）三年级竞赛第11题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "妈妈回到家，发现她的一个孩子把家里的巧克力都吃光了．她问了她的每个孩子是谁做的，他们的回答如下所示． 奥斯汀:卡洛斯吃了巧克力． 本尼:奥斯汀吃了巧克力． 卡洛斯:本尼在撒谎． 丹妮拉:吃巧克力的不是卡洛斯． 艾拉:我没有吃巧克力． 如果只有两个孩子说的是真话，那么是谁吃了巧克力? ", "answer_option_list": [[{"aoVal": "A", "content": "Austin "}], [{"aoVal": "B", "content": "Carlos "}], [{"aoVal": "C", "content": "Daniela "}], [{"aoVal": "D", "content": "Ella "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找一致（同伙）", "Overseas Competition->知识点->组合模块->逻辑推理->假设型逻辑推理->真假话"], "answer_analysis": ["因为$$Austin$$和$$Daniela$$说的话是矛盾的，$$Benny$$和$$Carlos$$说的话是矛盾的，而五个孩子中有两个孩子说的真话，$$Ella$$说的一定是假话，真相是巧克力是$$Ella$$吃的. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1401", "queId": "31e16d1aa6504a0f9215af0e624f5b64", "competition_source_list": ["走美杯三年级竞赛", "走美杯四年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "幼儿园买了$$8$$辆玩具车，每辆玩具车需要$$92$$元，李老师买这些玩具车，带了$$720$$元够吗？下面的解答比较合理的是． ", "answer_option_list": [[{"aoVal": "A", "content": "因为$$92\\times 8\\approx$$ 720（元），$$92\\times$$ 8\\textgreater720，所以$$92\\times$$ 8\\textgreater720，够 "}], [{"aoVal": "B", "content": "因为$$92\\times 8\\approx$$ 800（元），$$92\\times$$ 8\\textless800，所以$$92\\times$$ 8\\textless720，够 "}], [{"aoVal": "C", "content": "因为$$92\\times 8\\approx$$ 720（元），$$92\\times$$ 8\\textgreater720，所以$$92\\times$$ 8\\textgreater720，不够 "}], [{"aoVal": "D", "content": "因为$$92\\times 8\\approx$$ 800（元），$$92\\times$$ 8\\textgreater800，所以$$92\\times$$ 8\\textgreater720，不够 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型"], "answer_analysis": ["92\\textgreater90，总价大于$$720$$元，所以不够 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2239", "queId": "83d6850c89e549a6a4da9eebc6f76ea8", "competition_source_list": ["2016年河南郑州联合杯小学高年级五年级竞赛复赛改编"], "difficulty": "1", "qtype": "single_choice", "problem": "猎犬发现在它$$10$$米远的前方有一只奔跑着的野兔，猎犬马上紧追上去．猎犬的步子大，它跑$$5$$步的路程，兔子要跑$$9$$步；但兔子动作快，猎犬跑$$2$$步的时间，兔子却能跑$$3$$步．猎犬跑多少米能追上兔子？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$． "}], [{"aoVal": "B", "content": "$$60$$． "}], [{"aoVal": "C", "content": "$$90$$． "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["得到猎犬和野兔的速度比是$$6:5$$，即相同时间路程比是$$6:5$$，得到猎犬跑了$$60$$米． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "889", "queId": "742e8b1b49d2469ba281a34d3c7f6de1", "competition_source_list": ["2017年第13届湖北武汉新希望杯五年级竞赛决赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列说法正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "互质的两个数没有公因数 "}], [{"aoVal": "B", "content": "$$12$$和$$24$$的最大公因数是$$24$$ "}], [{"aoVal": "C", "content": "$$24$$和$$30$$的最大公因数是$$6$$ "}], [{"aoVal": "D", "content": "互质的两个数都是质数 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数"], "answer_analysis": ["$$\\text{A}$$选项：互质的两个数公因数为$$1$$，$$\\text{B}$$选项：$$12$$和$$24$$的最大公因数是$$12$$，$$\\text{D}$$选项：互质的两个数不一定都是质数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1423", "queId": "a76e21635d9f459e8934fc701b951999", "competition_source_list": ["2011年五年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "一台计算机感染了病毒，在计算机的存储器里，从$$2$$到$$9$$的每一个数$$x$$都被$$1+2+\\cdots +x$$这个和代替，例如$$2$$被$$3\\left( 3=1+2 \\right)$$代替，$$5$$被$$15\\left( 15=1+2+3+4+5 \\right)$$代替，计算机的其他功能都正常，如果你计算$$1+3+5$$，计算机显示的结果是( )。 ", "answer_option_list": [[{"aoVal": "A", "content": "9 "}], [{"aoVal": "B", "content": "15 "}], [{"aoVal": "C", "content": "22 "}], [{"aoVal": "D", "content": "25 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->加减法应用"], "answer_analysis": ["$$3$$被$$1+2+3=6$$代替，$$5$$被$$1+2+3+4+5=15$$代替，所以$$1+3+5$$显示为$$1+6+15=22$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1563", "queId": "68c45ade239241758cb17f657e127910", "competition_source_list": ["小学高年级六年级其它2014年数学思维能力等级测试第3题", "2014年第12届全国创新杯六年级竞赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "~猎狗发现一只狐狸在它前$$90$$米处，于是直接扑上去追捕，而狐狸马上闻风逃逸，当狐狸前逃$$1$$米时，猎狗赶上了$$10$$米，如果猎狗和狐狸前进路线相同，当猎狗抓到狐狸时，猎狗总共走了（~ ）米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$118$$ "}], [{"aoVal": "C", "content": "$$115$$ "}], [{"aoVal": "D", "content": "$$110$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$${{V}_{}}+{{V}_{}}=1:10$$，追上时狗比狐狸多跑$$90$$米，即狗跑了$$90\\times \\frac{10}{10-1}=100$$米． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "577", "queId": "4148558bfc27481e85b84cf676360241", "competition_source_list": ["2017年第22届湖北武汉华杯赛小学高年级竞赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知自然数$$n$$有$$10$$个因数，$$2n$$有$$20$$个因数，$$3n$$有$$15$$个因数．那么$$6n$$有~\\uline{~~~~~~~~~~}~个因数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$35$$ "}], [{"aoVal": "E", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理逆应用"], "answer_analysis": ["因为$$\\frac{20}{10}=\\frac{1+1}{0+1}$$，所以$$n$$的质因数分解式中含有$$0$$个$$2$$． 因为$$\\frac{15}{10}=\\frac{2+1}{1+1}$$，所以$$n$$的质因数分解式中含有$$1$$个$$3$$，另一个质因数有$$4$$个． 所以$$6n$$的因数有$$(1+1)(2+1)(4+1)=30$$个． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1146", "queId": "a215a72a85a542e0a6a6debeb02b4d7e", "competition_source_list": ["2015年第11届全国新希望杯小学高年级六年级竞赛复赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "某公司仓库里原有一批存货，以后每天陆续有货入库，且每天进的货一样多，用同样的汽车运货出库，如果每天用$$24$$辆汽车，$$5$$天刚好运完；如果每天用$$18$$辆汽车，$$8$$天刚好运完．现在用若干辆这样的汽车运货出库，运$$4$$天后，仓库每天的进货量是原来每天进货量的$$1.5$$倍，如果要求用$$10$$天时间运完仓库里的货，那么至少需要（~ ）辆这样的汽车（不准超载）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设一辆汽车一天运走$$1$$份货物，那么进货的速度是每天$$\\left( 18\\times 8-24\\times 5 \\right)\\div \\left( 8-5 \\right)=8$$份，原有货物$$18\\times 8-8\\times 8=80$$份，设需要$$x$$辆车，有$$10x\\geqslant 80+4\\times 8+6\\times 1.5\\times 8=184$$，那么$$x$$至少是19． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2793", "queId": "ff8080814724846801472db5f62907d6", "competition_source_list": ["2016年全国华杯赛小学中年级竞赛在线模拟第1题", "2013年全国华杯赛小学中年级竞赛初赛A卷第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$45$$与$$40$$的积的数字和是（~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$45\\times 40=45\\times 2\\times 20=1800$$，数字和$$1+8+0+0=9$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1341", "queId": "70a96e6a813b417098db7e482db0e32d", "competition_source_list": ["2020年长江杯六年级竞赛复赛B卷第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "某淘宝店的一种商品按原定价出售，每件利润为成本的$$30 \\%$$ ，``双$$11$$''这一天搞促销，按原定价打九折出售，结果当天出售的件数是原来每天出售件数的$$6$$倍，该店``双$$11$$''这天经营这种商品的总利润比以前每天增加$$ \\%$$ ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$90$$ "}], [{"aoVal": "D", "content": "$$240$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["后来的售价为原来的： $$(1+30 \\%)\\times 90 \\%=1.3\\times 90 \\%=1.17$$（倍）， 利润为：$$(1.17-1)\\times 6=1.02$$； 增加了： $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde(1.02-0.3)\\div 0.3\\times 100 \\%$$ $$=0.72\\div 0.3\\times 100 \\%$$ $$=2.4\\times 100 \\%$$ $$=240 \\%$$． 故答案为：$$240$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "346", "queId": "8065e52716e34cd69d6de99bd54d6708", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）"], "difficulty": "3", "qtype": "single_choice", "problem": "有一天，彭老师和陈老师约好去打乒乓球，结果彭老师以$$4:0$$完虐陈老师．乒乓球比赛为$$11$$分制，即每局$$11$$分，$$7$$局$$4$$胜制，打成$$10:10$$后必须净胜而且只能净胜$$2$$分．经计算，彭老师四局的总得分为$$48$$分，陈老师总得分为$$39$$分，且每一局比赛分差不超过$$3$$分，则一共有种情况．（不考虑这四局比分之间的顺序） ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数学广角->推理->解决简单逻辑推理问题"], "answer_analysis": ["每周比赛要分出胜负分差必须在$$2$$分或以上，题中又给出每局比赛分差不超过$$3$$分，故每局比赛的分差只有两种可能：差$$2$$分或$$3$$分．且分差为$$3$$分的那局彭老师得分为$$11$$分，总分差为$$48-39=9$$分，故必有$$3$$场分差为$$2$$分，另一场分差为$$3$$分；即有一场的比分为$$118$$，另两场的总比分为$$3731$$，有以下四种情况：①$$11:9$$，$$11:9$$，$$15:13$$②$$11:9$$，$$12:10$$，$$14:12$$③$$11:9$$，$$13:11$$，$$13:11$$④$$12:10$$，$$12:10$$，$$13:11$$．故一共有$$4$$种情况． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "559", "queId": "02e796557c3540ba8231e98c5064227a", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛A卷第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "相邻的两个奇数 ． ", "answer_option_list": [[{"aoVal": "A", "content": "一定是质数 "}], [{"aoVal": "B", "content": "不一定是互质数 "}], [{"aoVal": "C", "content": "没有公因数 "}], [{"aoVal": "D", "content": "最小公倍数是他们的积 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数->质数与合数判定->质数与合数的认识"], "answer_analysis": ["奇数（$$\\text{odd}$$）指不能被$$2$$整除的整数，数学表达形式为：$$2k+1$$，奇数可以分为正奇数和负奇数． 偶数是能够被$$2$$所整除的整数．正偶数也称双数．若某数是$$2$$的倍数，它就是偶数，可表示为$$2n$$； $$\\text{A}$$．质数是指在大于$$1$$的自然数中，除了$$1$$和它本身以外不再有其他因数的自然数．$$7$$和$$9$$是相邻的奇数，但是$$9$$是合数，不是质数；故$$\\text{A}$$错误； $$\\text{B}$$．互质数为数学中的一种概念，即两个或多个整数的公因数只有$$1$$的非零自然数．公因数只有$$1$$的两个非零自然数，叫做互质数．相邻的两个奇数一定是互质数，故$$\\text{B}$$错误； $$\\text{C}$$．$$7$$和$$9$$是相邻的奇数，它们有公因数$$1$$，故$$\\text{C}$$错误； $$\\text{D}$$．相邻的两个奇数最小公倍数是他们的积，故$$\\text{D}$$正确． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2539", "queId": "1f23f23692474b6b894b68dc9e448df2", "competition_source_list": ["2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第4题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "用$$416$$连续减$$8$$，减次后等于$$0$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$52$$ "}], [{"aoVal": "C", "content": "$$54$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意分析可知，求用$$416$$连续减多少个$$8$$后等于$$0$$， 也就是求$$416$$中有多少个$$8$$， 用除法计算，列式为：$$416\\div8=52$$， 所以减$$52$$次后等于$$0$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1462", "queId": "5605a89fe9b04537a517b239f3f12a1d", "competition_source_list": ["2016年新希望杯小学高年级六年级竞赛训练题（六）第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "兰兰和星星去林子里摘桃子，兰兰摘的桃子数是星星的$$\\frac{11}{12}$$，星星又摘了$$40$$个桃子后，兰兰摘得桃子是星星的$$\\frac{3}{4}$$，星星一共比兰兰多摘了（~ ~ ~）个桃子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$45$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$55$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["设星星一开始摘的桃子数是$$12x$$，兰兰摘的桃子数是$$11x$$．$$\\frac{11x}{12x+40}=\\frac{3}{4}$$，解得$$x=15$$． 星星摘的桃子比兰兰多$$12x+40-11x=55$$个． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "608", "queId": "21946cb3545f408187910b35f84620db", "competition_source_list": ["2005年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$P$$和$${{P}^{3}}+5$$都是质数，那么$${{P}^{5}}+5$$（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "一定是质数 "}], [{"aoVal": "B", "content": "一定是合数 "}], [{"aoVal": "C", "content": "可为质数，也可为合数 "}], [{"aoVal": "D", "content": "既不是质数也不是合数 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数->质数与合数判定"], "answer_analysis": ["若$$P$$为奇数，那么$$({{P}^{3}}+5)$$为偶数，又大于$$2$$的偶数都是合数，那么$$P$$是偶数，又$$P$$是质数，所以$$P=2$$，则$${{P}^{5}}+5={{2}^{5}}+5=37$$，选A。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1060", "queId": "1c653a6c39064c869a3c0c73413fc883", "competition_source_list": ["2018年四川成都锦江区四川师范大学附属第一实验中学小升初（四）第5题3分", "2014年全国迎春杯三年级竞赛初赛第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "青青老师给一些同学发臭豆腐，，如果每人发$$4$$块，还剩下$$10$$块；如果每人发$$5$$块，还会剩下$$1$$块，青青老师共有多少块臭豆腐？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$46$$ "}], [{"aoVal": "C", "content": "$$51$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->盈亏问题->盈亏基本类型->盈亏基本类型盈亏问题"], "answer_analysis": ["盈盈类问题：共有$$(10-1)\\div (5-4)=9$$人，共有$$4\\times 9+10=46$$块臭豆腐． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2709", "queId": "6424647b50f04b56adaab3059c068943", "competition_source_list": ["2017年IMAS小学中年级竞赛（第一轮）第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "四个连续的奇数之和是$$72$$，请问这四个数中最大的数是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$21$$ "}], [{"aoVal": "E", "content": "$$23$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由题意可知这四个连续奇数的平均值为$$72\\div 4=18$$， 因此这四个连续奇数为$$15$$、$$17$$、$$19$$、$$21$$， 即最大奇数为$$21$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2379", "queId": "e2f05f1faed24b1d99b401f38ff78e2b", "competition_source_list": ["六年级其它", "2004年第2届创新杯六年级竞赛复赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "古埃及时代，人们最喜欢的是分子为$$1$$的分数，如$$\\frac{1}{2}$$，$$\\frac{1}{3}$$，$$\\frac{1}{4}\\cdots$$，$$\\frac{1}{n}$$，\\ldots 等，我们不妨称这些分数为单位分数，其他的分数，只有它能写成若干个不同的单位分数之和时，人们才承认它是分数，例如，由于$$\\frac{3}{4}=\\frac{1}{2}+\\frac{1}{4}$$，所以他们承认$$\\frac{3}{4}$$是分数．如果当时只知有四个单位分数：$$\\frac{1}{2}$$，$$\\frac{1}{3}$$，$$\\frac{1}{4}$$，$$\\frac{1}{5}$$，那么下列四个分数中，不被承认的分数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{7}{12}$$ "}], [{"aoVal": "C", "content": "$$\\frac{19}{20}$$ "}], [{"aoVal": "D", "content": "$$\\frac{9}{10}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->裂项与通项归纳->分数裂项->分数拆分"], "answer_analysis": ["$$\\frac{5}{6}=\\frac{1}{2}+\\frac{1}{3}$$，$$\\frac{7}{12}=\\frac{1}{2}+\\frac{1}{12}$$，$$\\frac{19}{20}=\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{5}$$．$$D$$不能被拆分． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "21", "queId": "05bef68bce6346598893ff2f788c943f", "competition_source_list": ["2021年新希望杯五年级竞赛初赛第18题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "``水仙花数''是指这样一类数：将各位数字的立方相加，得到的和正好是原来的数，比如$$370$$，$$3^{3}+7^{3}+0^{3}=27+343+0=370$$．将一个数的各位数字的立方相加，得到一个新的数，这称为一次操作．从$$645$$开始不断重复操作，最后得到的水仙花数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$135$$ "}], [{"aoVal": "B", "content": "$$153$$ "}], [{"aoVal": "C", "content": "$$513$$ "}], [{"aoVal": "D", "content": "$$189$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->操作问题->数字操作"], "answer_analysis": ["$$645$$是$$6^{3}+4^{3}+5^{3}=216+64+125=280+125=405$$， $$405$$是$$4^{3}+0^{3}+5^{3}=64+0+125=64+125=189$$， $$189$$是$$1^{3}+8^{3}+9^{3}=1+512+729=513+729=1242$$， $$1242$$是$$1^{3}+2^{3}+4^{3}+2^{3}=1+8+64+8=9+72=81$$， $$81$$是$$8^{3}+1^{3}=512+1=513$$， $$513$$是$$5^{3}+1^{3}+3^{3}=125+1+27=126+27=153$$， $$153$$是$$1^{3}+5^{3}+3^{3}=1+125+27=126+27=153$$， 则最后是$$153$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2058", "queId": "fd5e6aad41fb4d95b817d8690694585f", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（六）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "李家村李家兄弟合作在门口挖一口井，李凡单独挖需要$$10$$天才能挖成，李超单独挖需要$$30$$天才能挖完．兄弟两人同时挖需要（~ ）天．~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$7.5$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["李凡一天挖$$\\frac{1}{10}$$，李超一天挖$$\\frac{1}{30}$$，兄弟合挖一天能完成$$\\frac{1}{30}+\\frac{1}{10}=\\frac{2}{15}$$，需要$$1\\div \\frac{2}{15}=7.5$$天． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1845", "queId": "8e5bd9b4572a4272a757469e3d11139f", "competition_source_list": ["2009年四年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "黑板上写有一个数，男同学从黑板前走过时，把它乘以3减去14，擦去原数，换上答案;女同学从黑板前走过时，把它乘以2减去7，擦去原数，换上答案.全班25名男生和15名女生都走过去以后，老师把最后的数乘以5，减去5，结果是30.那么，黑板上最初的数是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "5 "}], [{"aoVal": "B", "content": "6 "}], [{"aoVal": "C", "content": "7 "}], [{"aoVal": "D", "content": "8 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->操作问题"], "answer_analysis": ["全体同学走过后，黑板上的数是$$\\left( 30+5 \\right)\\div 5=7$$，最后一名学生走过之前，黑板上的数是$$\\left( 7+7 \\right)\\div 2=7$$，或$$\\left( 7+14 \\right)\\div 3=7$$，总之，最后一名学生(即第40名学生)走过之前，黑板上的数还是7，同理，第39名学生走过之前，黑板上的数还是7，\\ldots，因此可知，第1名学生走过之前，黑板上的数还是7，即黑板上最初的数是7. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3461", "queId": "c6dceee8994d476aa60f969cb712e5eb", "competition_source_list": ["2014年广东广州羊排赛六年级竞赛第8题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个袋子里装有$$9$$个白球和$$3$$个红球（球的大小和形状一样），从中任意摸出一个，摸到红球的可能性是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20 \\%$$ "}], [{"aoVal": "B", "content": "$$25 \\%$$ "}], [{"aoVal": "C", "content": "$$30 \\%$$ "}], [{"aoVal": "D", "content": "$$37.5 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->典型问题->摸小球"], "answer_analysis": ["摸到红球可能性为$$3\\div (9+3)\\times 100 \\%=25 \\%$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2789", "queId": "ff808081472482f50147296100e10400", "competition_source_list": ["2013年全国华杯赛小学高年级竞赛初赛C卷第1题", "2017年全国小升初建华入学备考第16题"], "difficulty": "3", "qtype": "single_choice", "problem": "如果$$\\frac{2013\\times 2013}{2014\\times 2014+2012}=\\frac{n}{m}$$（其中$$m$$与$$n$$为互质的自然数），那么$$m+n$$的值是（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1243$$ "}], [{"aoVal": "B", "content": "$$1343$$ "}], [{"aoVal": "C", "content": "$$4025$$ "}], [{"aoVal": "D", "content": "$$4029$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["方法一：找规律 $$\\frac{{{2}^{2}}}{{{3}^{2}}+1}=\\frac{2}{5}$$、$$\\frac{{{3}^{2}}}{{{4}^{2}}+2}=\\frac{1}{2}=\\frac{3}{6}$$、$$\\frac{{{4}^{2}}}{{{5}^{2}}+3}=\\frac{4}{7}$$、$$\\frac{{{5}^{2}}}{{{6}^{2}}+4}=\\frac{5}{8}$$$$\\cdots$$ 规律找到了，$$\\frac{{{n}^{2}}}{{{(n+1)}^{2}}+(n-1)}=\\frac{n}{n+3}$$． $$\\frac{2013\\times 2013}{2014\\times 2014+2012}=\\frac{n}{m}$$=$$\\frac{2013}{2016}=\\frac{671}{672}$$． 方法二：原式= $$\\frac{2013\\times 2013}{(2013+1)\\times (2013+1)+2013-1}=\\frac{2013\\times2013}{2013\\times 2013+2\\times 2013+1+2013-1}=\\frac{2013}{2016}=\\frac{671}{672}$$． 方法三：原式=$$\\frac{2013\\times 2013}{2014\\times (2013+1)+2012}=\\frac{2013\\times2013}{2013\\times 2014+2014+2012}=\\frac{2013}{2016}=\\frac{671}{672}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "683", "queId": "5e4696bc9d474b8fa6901682a11f878c", "competition_source_list": ["2016年第16届世奥赛六年级竞赛决赛第11题", "2016年全国世奥赛竞赛A卷第11题"], "difficulty": "3", "qtype": "single_choice", "problem": "大互联网梅森素数分布式计算（GIMPS）项目又建奇功，据外媒今年$$1$$月$$20$$月日报道，美国州立中密苏里大学阿迪斯库珀CurtisCooper通过GIMPS项目发现了第$$49$$个梅森素数（被称为M74207281），为GIMPS项目诞生$$20$$周免献礼．M74207281这个超大素数$$22338618$$位，是目前已知的最大素数，诞生一台Inter I7-4790 CPU电脑．这是库珀教授第四次通过S项目发现的梅森素数，刷新了他的记录．他上次发现第$$48$$个梅森素数$${{2}^{57885161}}-1$$是在$$2013$$年$$1$$月，由$$17425170$$位．GIMPS项目集合了$$20$$多万台计算机的计算能力，主要任务是不断筛选，寻找更大的梅森素数，尽管一些素数已经被用于加密和其他实际应用任务，但寻找最大的素数仍然出于对学术方面的兴趣．近年来发现的最大素数都是梅森素数．这一命名是为了纪念法国神学家、数学家、音乐理论家马林·梅森（Marin Nersenne，$$1588 \\tilde{ } 1648$$）．下面该同学显显身手了，请从一个$$1 \\tilde{ } 9$$（缺$$8$$）这八个自然数中不重复地用这些数字构造出四个两位质数，并求出它们的和是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$190$$ "}], [{"aoVal": "B", "content": "$$217$$ "}], [{"aoVal": "C", "content": "$$127$$ "}], [{"aoVal": "D", "content": "无法构造 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["我们知道所有的偶数都是合数，除了$$5$$以外，个位为$$5$$的数也都是合数，这$$8$$个数中只有$$1$$、$$3$$、$$5$$、$$7$$、$$9$$放在个位才有可能是质数，所以十位上只能是$$2$$、$$4$$、$$5$$、$$6$$．这四个两位数的和是$$\\left( 2+4+5+6 \\right)\\times 10+1+3+7+9=190$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1550", "queId": "7ae8c4ad7c434d0d9b822b7de2576669", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛二年级竞赛决赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "熊奶奶家的水缸可以装$$8$$桶水，小熊每天早晨提$$4$$桶水，熊奶奶全家每天用$$3$$桶水．熊奶奶问小熊：``这样下去，到第天，我们家的水缸会装满水了． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["小熊都是早晨提$$4$$桶水，则最后一天会提$$4$$桶水到水缸，所以$$8-4=4$$（桶），而前面的每天只装水：$$4-3=1$$（桶）．所以天数为：$$4\\div1+1=5$$（天）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2323", "queId": "d8f35a0f05344c8099d2858c36b22742", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一列快车和一列慢车相向而行，快车的车长是$$280$$米，慢车的车长是$$455$$米，坐在快车上的人看见慢车驶过的时间是$$13$$秒，那么坐在慢车上的人看见快车驶过的时间是秒． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["坐在快车上的人和慢车进��相遇运动，则速度和$$=455\\div 13=35$$（米/秒），坐在慢车上的人和快车进行相遇运动时，则相遇时间$$=280\\div 35=8$$（秒）． 故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "382", "queId": "8a0887ca6a6743038bd6f2a351f1c038", "competition_source_list": ["2016年河南郑州K6联赛小学高年级六年级竞赛第16题2分"], "difficulty": "2", "qtype": "single_choice", "problem": "用两块同样大小的铁皮制成一个长方体和一个正方体铁桶，它们容积相比，（~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "长方体大 "}], [{"aoVal": "B", "content": "正方体大 "}], [{"aoVal": "C", "content": "同样大 "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->最值原理在几何中的应用"], "answer_analysis": ["同样大小的铁皮说明两个铁桶的表面积相等，可以先反过来考虑容积相等的长方体和正方体铁桶，它们的表面积哪一个大．假设两者体积都是$$27$$，正方体表面积是$$3\\times 3\\times 6=54$$；可以拼成一个长$$9$$，宽$$3$$，高$$1$$的长方体，表面积为$$\\left( 9\\times 3+3\\times 1+9\\times 1 \\right)\\times 2=78$$．由此可看出容积相等，正方体的表面积小一些．所以表面积相等，正方体的容积大． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "180", "queId": "8b61b68975b44fb8b4a4720fc7b1d302", "competition_source_list": ["2017年第17届湖北武汉世奥赛五年级竞赛决赛第14题"], "difficulty": "1", "qtype": "single_choice", "problem": "在桌子上有三个盖着盖子的盒子，其中一个盒子内有两粒绿豆，第二个盒子内有两粒红豆，另一个盒子内有一粒绿豆和一粒红豆，三个盒子盖子上分别写着``红豆''，``红、绿豆''和``绿豆''，但是所有标签都标错了．最少（~ ）就能判断出所有盒子内各装着什么豆子． ", "answer_option_list": [[{"aoVal": "A", "content": "从一个盒子取一粒豆子 "}], [{"aoVal": "B", "content": "从两个盒子各取一粒豆子 "}], [{"aoVal": "C", "content": "从一个盒子取两粒豆子 "}], [{"aoVal": "D", "content": "从三个盒子各取一粒豆子 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["因为都错了．从标有``红、绿豆''的盒子中随便取出一个即可，假如取出来的是一颗红豆，则``红、绿''中都是红，剩下的盒子中，标有``红豆''的盒子肯定是绿豆，标有``绿豆''的肯定是``红绿豆''，所以最少从$$1$$个盒子从取出一粒豆子就可以判断出来． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1018", "queId": "0a0bf9dd6a4943e3939ddc37bfb748a3", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（五）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "一根绳子剪去全长的$$\\frac{1}{5}$$后，又接上$$20$$米，接上后的长度是接上前的$$\\frac{5}{4} $$，那么原来的绳子长米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$44$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$125$$ "}], [{"aoVal": "D", "content": "$$150$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数与运算->数的运算的实际应用（应用题）->百分数的简单实际问题->折扣、成数、税率、利率"], "answer_analysis": ["$$20\\div \\left( \\frac{5}{4} -1 \\right)=80$$（米），$$80\\div \\left( 1-\\frac{1}{5} \\right)=100$$（米）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "810", "queId": "76e8f5c4f4904b2e8807119adb13f2fd", "competition_source_list": ["2013年全国学而思杯五年级竞赛A卷第20题"], "difficulty": "3", "qtype": "single_choice", "problem": "思思编了一个计算机程序，在屏幕上显示所有由$$0$$、$$1$$、$$2$$、$$3$$组成的四位编码（数字可以重复使用），每个四位编码都是红、黄、蓝、绿四种颜色中的一种．并且，如果两个编码的每一位数字均不相同，那么这两个编码的颜色也不相同．如果，$$0000$$是红色的、$$1000$$是黄色的、$$2000$$是蓝色的，那么： （1）下列编码中，一定不是红色的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0102$$ "}], [{"aoVal": "B", "content": "$$0312$$ "}], [{"aoVal": "C", "content": "$$2222$$ "}], [{"aoVal": "D", "content": "$$0123$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算求解", "拓展思维->能力->实践应用"], "answer_analysis": ["$$2222$$与$$0000$$的��一位数字均不相同，故$$2222$$一定不是红色，选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1350", "queId": "2ce00020733e48ab8247c597be1dcaf3", "competition_source_list": ["2005年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "小王的年龄的3倍与小谢的年龄的5倍相等，他们的年龄相差4岁，那么10年后( ). ", "answer_option_list": [[{"aoVal": "A", "content": "小王的年龄的5倍与小谢年龄的3倍相等 "}], [{"aoVal": "B", "content": "小王的年龄的4倍与小谢年龄的5倍相等 "}], [{"aoVal": "C", "content": "小王的年龄的5倍与小谢年龄的4倍相等 "}], [{"aoVal": "D", "content": "小王的年龄的3倍与小谢年龄的5倍相等 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题"], "answer_analysis": ["小王的年龄的3倍与小谢的年龄的5倍相等，那么小王的年龄就等于小谢的年龄的$$\\frac{5}{3}$$相等.又他们的年龄相差4岁，所以小谢的年龄为$$4\\div \\left( \\frac{5}{3}-1 \\right)=6$$(岁)，那么小王的年龄为$$6\\times \\frac{5}{3}=10$$(岁).10年后，小谢的年龄为16岁，小王的年龄为20岁，那么小谢的年龄的5倍和小王年龄的4倍相等，选$$B$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "707", "queId": "c301a93843934f7aa45b25ed21998587", "competition_source_list": ["2022年第九届鹏程杯四年级竞赛初赛第29题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$80$$名同学面向老师站成一排，老师先让大家从左到右按$$1$$至$$3$$报数，再让报数$$3$$的倍数的同学向后转，接着又从头开始$$1$$至$$5$$报数，再让报数是$$5$$的倍数的同学向后转，这样做过之后，现在面向老师的同学还有~\\uline{~~~~~~~~~~}~名． ", "answer_option_list": [[{"aoVal": "A", "content": "$$43$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$46$$ "}], [{"aoVal": "D", "content": "$$48$$ "}], [{"aoVal": "E", "content": "$$50$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["无 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1464", "queId": "370eb1beab784dd1aa2ed870951efa0d", "competition_source_list": ["2011年全国美国数学大联盟杯小学高年级五年级竞赛初赛第29题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$400$$个人被堵在上班路上，其中$$160$$人上班要迟到，问：迟到的人占百分之多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$20 \\%$$ "}], [{"aoVal": "B", "content": "$$40 \\%$$ "}], [{"aoVal": "C", "content": "$$60 \\%$$ "}], [{"aoVal": "D", "content": "$$80 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->求分率"], "answer_analysis": ["$$160\\div400=40 \\%$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2587", "queId": "9947183bf8d74d25a81b7b3476b61c93", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$100$$以内的$$25$$个质数中任取两个构成分数的分子和分母，这样的真分数有几个？假分数有几个？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$150$$；$$150$$ "}], [{"aoVal": "B", "content": "$$300$$；$$150$$ "}], [{"aoVal": "C", "content": "$$150$$；$$300$$ "}], [{"aoVal": "D", "content": "$$300$$；$$300$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["$$100$$以内的$$25$$个质数分别为：$$2$$，$$3$$，$$5$$，$$7$$，$$11$$，$$13$$，$$17$$，$$19$$，$$23$$，$$29$$，$$31$$，$$37$$，$$41$$，$$43$$，$$47$$，$$53$$，$$59$$，$$61$$，$$67$$，$$71$$，$$73$$，$$79$$，$$83$$，$$89$$，$$97$$．用$$97$$做分母时，真分数有$$24$$个；用$$89$$做分母时，真分数有$$23$$个；\\ldots；用$$3$$做分母时，真分数有$$1$$个，用$$2$$做分母时，真分数有$$0$$个．所以，满足题意的真分数有$$1+2+3+...+23+24=300$$（个）．因此．满足条件的真分数有$$300$$个．把上面的满足条件真分数的分子、分母颠倒就成了假分数，所以符合条件的假分数也是$$300$$个． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2297", "queId": "97a25c2cc0e94943bebb95136242bb21", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙三人同时从东村去西村，甲骑自行车每小时比乙快$$12$$千米，比丙快$$15$$千米．甲行$$3.5$$小时到达西村，然后立刻返回，在距西村$$30$$千米处和乙相遇，丙出发小时和甲相遇． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$5.6$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->思想->数形结合思想"], "answer_analysis": ["此题主要考查了相遇问题，要熟练掌握，注意速度、时间和路程的关系：速度$$\\times $$时间$$=$$路程，路程$$\\div $$时间$$=$$速度，路程$$\\div $$速度$$=$$时间，解答此题的关键是求出两村之间的距离是多少． 首先根据题意，可得甲乙相遇时，甲比乙多行的路程是$$30\\times2=60$$（千米），再根据路程$$\\div $$速度$$=$$时间，用甲乙相遇时行的路程之差除以他们的速度的差，求出甲乙相遇时用的时间是多少，进而求出甲行$$30$$千米用的时间是多少；然后求出两村之间的距离是多少，再根据路程$$\\div $$时间$$=$$速度，用两村之间的距离的$$2$$倍除以甲丙的速度之和，求出丙行了多长时间和甲相遇即可． 甲的速度是每小时行： $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde30\\div (30\\times2\\div12-3.5)$$ $$=30\\div1.5$$ $$=20$$（千米）， 丙和甲相遇用的时间是： $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde20\\times3.5\\times2\\div (20-15+20)$$ $$=70\\times2\\div25$$ $$=140\\div25$$ $$=5.6$$（小时）． 答：丙行了$$5.6$$小时和甲相遇． 故答案为：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "599", "queId": "0c39c2e36f7e47c88275e551aa87ef39", "competition_source_list": ["2016年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在除法算式中，被除数为$$2016$$，余数为$$7$$，则满足算式的除数共有( )个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法"], "answer_analysis": ["某个数除$$2016$$余$$7$$，于是这个数整除$$2016-7=2009$$，$$2009={{7}^{2}}\\times 41$$，所以$$2009$$共有$$3\\times 2=6$$个约数，其中比$$7$$大的约数有$$4$$个(除了$$1$$和$$7$$)．所以满足要求的除数共有$$4$$个． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2245", "queId": "e3ae4cb4b5c74386886ac50a7fea9932", "competition_source_list": ["2020年新希望杯四年级竞赛决赛（8月）第7题", "2020年新希望杯四年级竞赛初赛（个人战）第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "光头强去赶集，去时步行，速度是$$15$$千米$$/$$时；回来时骑车，速度是$$30$$千米$$/$$时．光头强往返的平均速度是~\\uline{~~~~~~~~~~}~千米$$/$$时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$22.5$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$17.5$$ "}], [{"aoVal": "D", "content": "$$25$$ "}], [{"aoVal": "E", "content": "$$27.5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["路程$$=$$时间$$\\times$$速度， 设路程为$$30$$千米，那么去时的时间是$$30\\div15=2$$（时）， 回来的时间为$$30\\div30=1$$（时）， 因此往返的平均速度是$$(30+30)\\div(2+1)=60\\div3=20$$（千米$$/$$时）． 故答案为$$20$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3442", "queId": "f7f76d1760f8418aaad7d06b1a61b175", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（二）"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$1$$、$$2$$、$$3$$、$$4$$这$$4$$个数字中任选两个不同的数字组成两位数，那么所有两位数中偶数有（~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["先确定个位，本题中个位只能是$$2$$、$$4$$，$$2$$种选择；再确定十位，$$3$$种选择，$$2\\times 3=6$$（个）. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "713", "queId": "295ee02edd604e39a2264cb0a34676e7", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级四年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$1$$开始，鲍勃一共喊了$$2017$$个数，从第一个数之后的每个数都比前一个数大$$4$$．请问以下哪一个数是鲍勃喊过的？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$137$$ "}], [{"aoVal": "B", "content": "$$138$$ "}], [{"aoVal": "C", "content": "$$139$$ "}], [{"aoVal": "D", "content": "$$140$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["除以$$4$$余$$1$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "477", "queId": "f31d5dc7ffc3440db801c9735d733fc5", "competition_source_list": ["2013年全国世奥赛五年级竞赛第17题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "多思希望小学语文组、数学组、英语组的老师参加趣味游戏比赛，比赛项目有吹气球、吃橘子、投篮球．每项比赛各取前三名，第一名得$$5$$分，第二名得$$3$$分，第三名得$$1$$分（单项比赛中名次没有并列情况）．已知语文组进入前三名的老师人数最少，数学组进入前三名的老师人数是语文组的$$2$$倍，并且这两个组所得总分相等并列第一．英语组的老师得了多少分？ ", "answer_option_list": [[{"aoVal": "A", "content": "答案请写在答题卡上 "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["答案请写在答题卡上 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3408", "queId": "c4ec7240d48541c491c0c4dd745d4c49", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "有四张数字卡片，分别为$$9$$、$$8$$、$$5$$、$$0$$．从中挑出两张排成一个两位数，一共可以排成个两位数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据加乘原理：十位上的数有$$3$$种可选，然后个位只有$$3$$种可选，故$$3\\times 3=9$$（种），故$$\\text{A}$$正确． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2780", "queId": "76c34139969245d1957155ee1d7a0624", "competition_source_list": ["2013年全国美国数学大联盟杯小学高年级竞赛初赛第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "【本讲闯关3】 $${{2}^{3}}\\times {{2}^{5}}\\times {{2}^{7}}\\times {{2}^{11}}=$$ ． ", "answer_option_list": [[{"aoVal": "A", "content": "$${{2}^{13}}$$ "}], [{"aoVal": "B", "content": "$${{4}^{26}}$$ "}], [{"aoVal": "C", "content": "$${{2}^{26}}$$ "}], [{"aoVal": "D", "content": "$${{16}^{26}}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->乘方->乘方的运算->乘方的幂运算"], "answer_analysis": ["$$2^{3}\\times2^{5}\\times2^{7}\\times2^{11}=2^{3+5+7+11}=2^{26}=4^{13}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1116", "queId": "4accc8783aab41c7a6605f733ef50e5c", "competition_source_list": ["2012年IMAS小学高年级竞赛第一轮检测试题第17题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "姐姐今年$$12$$岁，妹妹今年$$8$$岁，弟弟今年$$3$$岁，他们的生日恰好是同一天．当三人的年龄和为$$50$$岁时，妹妹为多少岁？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->年龄问题->年龄问题基本关系->年龄和"], "answer_analysis": ["今年他们三人的年龄之和为$$23$$岁，当他们的年龄之和为$$50$$岁时，已过了$$(50-23)\\div 3=9$$年，所以这时妹妹为$$8+9=17$$岁． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "255", "queId": "83b46556807747b5bcb8fcd890ddf03a", "competition_source_list": ["2020年第24届YMO五年级竞赛决赛第6题3分", "2019年第24届YMO五年级竞赛决赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$Y$$、$$M$$、$$O$$、$$T$$四人中只有$$1$$人会开车． $$Y$$说：``我会开''． $$M$$说：``我不会开''． $$O$$说：``$$Y$$不会开''． $$T$$什么也没说． 已知$$Y$$、$$M$$、$$O$$三人只有一人说了真话．会开车的是谁？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$Y$$ "}], [{"aoVal": "B", "content": "$$M$$ "}], [{"aoVal": "C", "content": "$$O$$ "}], [{"aoVal": "D", "content": "$$T$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->组合模块->逻辑推理->假设型逻辑推理->真假话", "拓展思维->能力->逻辑分析"], "answer_analysis": ["对比发现，$$O$$与$$Y$$说的矛盾，相互对立， 则$$O$$与$$Y$$必一对一错， 则$$M$$说假话，则$$M$$会开车，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2481", "queId": "1e03b3c23d00419d9f2018cd2acbc052", "competition_source_list": ["2020年广东广州羊排赛六年级竞赛第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$\\frac{2}{5}a=\\frac{3}{7}b$$，则$$a$$与$$b$$的大小关系为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b$$ "}], [{"aoVal": "B", "content": "$$a=b$$ "}], [{"aoVal": "C", "content": "$$a\\textless{}b$$ "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["当积一定时，一个因数大，则另一个因数就小， $$\\frac{2}{5}=\\frac{14}{35}$$，$$\\frac{3}{7}=\\frac{15}{35}$$，$$\\frac{2}{5}\\textless{}\\frac{3}{7}$$， 所以$$a\\textgreater b$$， 但是当$$a=b=0$$时，$$\\frac{2}{5}a=0=\\frac{3}{7}b$$， 所以$$a$$和$$b$$的大小无法确定． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "887", "queId": "a02477afc1de442289289975658cfd6f", "competition_source_list": ["2015年美国数学大联盟杯六年级竞赛初赛（中国赛区）第34题5分", "2016年全国美国数学大联盟杯小学高年级六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "\\textbf{N} is a two-digit number. When \\textbf{N} is divided by its ones digit， the quotient is $$8$$ and the remainder is $$1$$. When \\textbf{N} is divided by its tens digit， the quotient is $$11$$ and the remainder is $$2$$. What is the value of \\textbf{N}? $$\\text{N}$$是一个两位数．当$$\\text{N}$$被其个位数除的时候，商是$$8$$余数是$$1$$．当$$\\text{N}$$被其十位数除的时候商是$$11$$余数是$$2$$．那么$$\\text{N}$$是几？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$46$$ "}], [{"aoVal": "B", "content": "$$57$$ "}], [{"aoVal": "C", "content": "$$68$$ "}], [{"aoVal": "D", "content": "$$79$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["$$\\text{N}$$是一个两位数．当$$\\text{N}$$被其个位数除的时候，商是$$8$$余数是$$1$$．当$$\\text{N}$$被其十位数除的时候商是$$11$$余数是$$2$$．那么$$\\text{N}$$是几？ 此题最直接的方式，就是四个数试除，答案是$$\\text{B}$$. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2325", "queId": "e6e3876265e5428b9afd8f51bdc3ddd0", "competition_source_list": ["2020年长江杯六年级竞赛复赛A卷第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一辆汽车从荆州出发送一批货物到沙市用了$$50$$分钟，到达目的地后沿原路返回荆州用了$$40$$分钟．这辆汽车从沙市返回荆州时速度提高了$$ \\%$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.5$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["设全程$$a$$千米， 从荆州到长沙的速度是$$a\\div 50=\\frac{a}{50}$$（千米/分钟）， 返回的速度是$$a\\div 40=\\frac{a}{40}$$（千米/分钟）， 返回时速度提高了$$\\left( \\frac{a}{40}-\\frac{a}{50} \\right)\\div \\frac{a}{50}\\times 100 \\%=25 \\%$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "838", "queId": "ff80808149990d5e0149b43ef9ef380a", "competition_source_list": ["2014年北京海淀区五年级其它", "2014年全国迎春杯四年级竞赛复赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "有两根同样长的绳子，第一根平均剪成$$4$$段，第二根平均剪成$$6$$段，已知第一根剪成的每段长度与第二根剪成的每段长度相差$$2$$米．那么，原来两段绳子长度之和是（ ~ ~ ）米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$48$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$\\left[ 4,6 \\right]=12$$，$$12\\div 6=2$$，$$12\\div 4=3$$，$$12\\div 3-2=2$$，$$12\\times ~2\\times ~2=48$$（米）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1323", "queId": "a278d80dbfac401fb7a7720b7efc4b89", "competition_source_list": ["2010年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "$$1$$号混合液中水、油、醋的比例为$$1:2:3$$，$$2$$号混合液中水、油、醋的比例为$$3:4:5$$，将两种混合液倒在一起后，得到的混合液中水、油、醋的比例可能为（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$1:3:5$$ "}], [{"aoVal": "B", "content": "$$2:3:5$$ "}], [{"aoVal": "C", "content": "$$3:5:7$$ "}], [{"aoVal": "D", "content": "$$6:10:13$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->浓度问题->浓度基本题型"], "answer_analysis": ["由于$$2$$种溶液质量不定，混合后的比例应该是有无数种可能。关键点是第二个物质两种溶液比例全是$$\\frac{1}{3}$$，所以新溶液中的比例也是$$\\frac{1}{3}$$。这样排除$$B$$,$$D$$。再分析另两种物质的最大比例$$\\frac{1}{2}$$和最小比例$$\\frac{1}{6}$$。新溶剂的比例不能超过极值。$$A$$中的水比例$$\\frac{1}{9}$$小于最小比例，舍去。所以取$$C$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "700", "queId": "abe2a25851a84a97b815a3938e4ba69f", "competition_source_list": ["2019年广东广州羊排赛六年级竞赛第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "有一个数列$$1$$、$$1$$、$$2$$、$$3$$、$$5$$、$$8$$、$$13$$、$$21$$、$$\\cdots $$，从第三项开始，每一项都是前两项之和，这个数列第$$2019$$项除以$$4$$的余数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["根据余数的性质，和的余数等于余数的和， $$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$\\cdots $$，$$\\div 4$$余$$1$$，$$1$$，$$2$$，$$3$$，$$1$$，$$0$$，$$1$$，$$1$$，$$2$$，$$3$$，$$1$$，$$0$$， $$1$$，$$1$$，$$\\cdots $$，$$6$$个为一周期，$$2019\\div 6=336$$（组）$$\\cdots \\cdots $$$$3$$（个）， 为周期的第$$3$$个，余数为$$2$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3413", "queId": "e0588b6e3aba45aeba75a0dd33971a9a", "competition_source_list": ["2010年中环杯四年级竞赛决赛", "2011年中环杯四年级竞赛初赛", "2011年中环杯四年级竞赛决赛"], "difficulty": "1", "qtype": "single_choice", "problem": "康康到麦当劳买套餐，一份套餐包含了一个汉堡、一份小吃和一杯饮料。服务员告诉他店里有$$8$$种汉堡、$$4$$种小吃、$$5$$种饮料可供选择，那么康康一共可以搭配出（ ）种套餐。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$160$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->乘法原理"], "answer_analysis": ["康康先选择汉堡，共有$$8$$种选择，然后选小吃，有$$4$$种选择，最后选饮料，有$$5$$种选择，根据乘法原理，共有：$$8\\times 4\\times 5=160$$（种）搭配方法。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2809", "queId": "a3a4cbecef9b477d96479b41b7f6ec27", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\square $$和$$\\bigcirc $$各代表一个数，$$\\square =\\bigcirc +\\bigcirc $$，$$\\bigcirc +\\square +\\square =30$$，求$$\\square $$与$$\\bigcirc $$的差是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["分析可知：$$\\square =\\bigcirc +\\bigcirc =2\\bigcirc $$， 把$$\\square =2\\bigcirc $$代入$$\\bigcirc +\\square +\\square =30$$中得： $$\\bigcirc +2\\bigcirc +2\\bigcirc =5\\bigcirc $$，$$5\\bigcirc =30$$， 所以$$\\bigcirc =30\\div 5=6$$， 那么$$\\square =2\\bigcirc =2\\times 6=12$$， 由此可知，$$\\square $$与$$\\bigcirc $$的差为$$12-6=6$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "52", "queId": "0b2c0af4249d4de5ad60a9783116bb67", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛初赛第6题"], "difficulty": "3", "qtype": "single_choice", "problem": "整数$$N=12345678910111213\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 201020112012201320142015$$是由$$1 \\tilde{ } 2015$$这$$2015$$个整数，由小到大的顺序依次写出得到的，那么$$N$$是（~ ）位数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5678$$ "}], [{"aoVal": "B", "content": "$$6947$$ "}], [{"aoVal": "C", "content": "$$6950$$ "}], [{"aoVal": "D", "content": "$$6953$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->图形规律->数量变化"], "answer_analysis": ["$$1\\times 9+2\\times 90+3\\times 900+4\\times 1016=6953$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2310", "queId": "e573d9cf202b4892a0dc23b12be8add1", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛B卷第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两人各走一段路，所行路程的比是$$5:4$$，所用的时间比是$$3:5$$，则甲、乙两人的速度比是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4:3$$ "}], [{"aoVal": "B", "content": "$$3:4$$ "}], [{"aoVal": "C", "content": "$$12:25$$ "}], [{"aoVal": "D", "content": "$$25:12$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->比例解行程问题->行程中的比例"], "answer_analysis": ["本题是一道有关比例问题的求解，熟知路程、速度和时间之间的关系是解题的关键；分析题意，把甲的路程和乙的路程分别看作$$5$$份和$$4$$份，甲和乙的时间分别看作$$3$$份和$$5$$份；然后根据``速度$$=$$路程$$\\div $$时间''分别求出甲和乙的速度，进而再求出最简比即可． 甲的速度为：$$5\\div 3=\\frac{5}{3}$$， 乙的速度为：$$4\\div 5=\\frac{4}{5}$$， 甲乙的速度比为：$$\\frac{5}{3}:\\frac{4}{5}=25:12$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1506", "queId": "df0083bec6b2458b89c2b5d29e589cbb", "competition_source_list": ["2019年美国数学大联盟杯竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "$$8$$千克的香蕉是$$12$$美元．$$12$$千克的香蕉是多少钱？The cost of $$8$$ kilograms of bananas is $$$12$$. What is the cost of $$12$$ kilograms of bananas?． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$8$$千克的香蕉是$$12$$美元．$$12$$千克的香蕉是多少钱？ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde12\\div 8\\times 12$$ $$=12\\times 12\\div 8$$ $$=144\\div 8$$ $$=18$$（美元）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2438", "queId": "6b37c5ff99b0493a8621ec8676816b36", "competition_source_list": ["其它改编题", "2015年湖北武汉世奥赛五年级竞赛模拟训练题（二）第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$(1+0.23+0.34)\\times (0.23+0.34+0.45)$$$$-(1+0.23+0.34+0.45)\\times (0.23+0.34)=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.35$$ "}], [{"aoVal": "B", "content": "$$0.45$$ "}], [{"aoVal": "C", "content": "$$0.55$$ "}], [{"aoVal": "D", "content": "$$0.65$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数换元法"], "answer_analysis": ["设$$a=0.23+0.34$$，$$b=0.23+0.34+0.45$$，原式$$=(1+a)b-(1+b)a=b+ab-a-ab=b-a=0.45$$. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "655", "queId": "22e3872e66c34ca09e697f5cfd2e6b72", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（三）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2$$、$$4$$、$$6$$、$$8\\cdots \\cdots $$、$$98$$,$$100$$这$$50$$个偶数的所有数字之和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$425$$ "}], [{"aoVal": "B", "content": "$$426$$ "}], [{"aoVal": "C", "content": "$$427$$ "}], [{"aoVal": "D", "content": "$$428$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$\\left( 0+2+4+6+8 \\right)\\times 10+\\left( 1+2+\\ldots \\ldots +9 \\right)\\times 5+1=426$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "18", "queId": "05b202ca2ce644158b1579bc0f94cfbb", "competition_source_list": ["2007年第12届全国华杯赛竞赛初赛第6题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "从和为$$55$$的$$10$$个不同的自然数中，取出$$3$$��数后，余下的数之和是$$55$$的$$\\frac{7}{11}$$，则取出的三个数的积最大等于（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$280$$ "}], [{"aoVal": "B", "content": "$$270$$ "}], [{"aoVal": "C", "content": "$$252$$ "}], [{"aoVal": "D", "content": "$$216$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["余下的数之和为：$$55\\times \\frac{7}{11}=35$$，取出的数之和为：$$55-35=20$$， 要使取出的三个数之积尽量大，则取出的三个数应尽量接近， 我们知$$6+7+8=21$$，所以取$$5\\times 7\\times 8=280$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2162", "queId": "2f777d8353894d4495155f94f89095db", "competition_source_list": ["2014年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "新生开学后去远郊步行拉练，到达$$A$$地时比原计划时间$$10$$点$$10$$分晚了$$6$$分钟，到达$$C$$地时比原计划时间$$13$$点$$10$$分早了$$6$$分钟，$$A$$、$$C$$之间恰有一点$$B$$是按照原计划时间到达的，那么到达$$B$$点的时间是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$点$$35$$分 "}], [{"aoVal": "B", "content": "$$12$$点$$5$$分 "}], [{"aoVal": "C", "content": "$$11$$点$$40$$分 "}], [{"aoVal": "D", "content": "$$12$$点$$20$$分 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题"], "answer_analysis": ["由于全程是匀速运动，所以从晚$$6$$分追到早$$6$$分，前半程和后半程所需时间是一样的，所以经过中点的时间应该是不变的，就是$$10$$点$$10$$分和$$13$$点$$10$$分的中点$$11$$点$$40$$分。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1979", "queId": "e0f95d22b7304255a61663bdd82121a3", "competition_source_list": ["2017年河南郑州豫才杯四年级竞赛初赛第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "聪聪所在的光明小学实施了阅读评级制度，级别最高的``状元''，需要达到$$300$$万字的课外阅读量．聪聪已经达到了$$200$$万字的``榜眼''阅读量，他保证在$$100$$天里冲刺到``状元''级别，每天保证阅读时间$$20$$分钟，那么他需要将阅读速度至少提升到（~ ）的水平． ", "answer_option_list": [[{"aoVal": "A", "content": "每分钟$$300$$字 "}], [{"aoVal": "B", "content": "每分钟$$400$$字 "}], [{"aoVal": "C", "content": "每分钟$$500$$字 "}], [{"aoVal": "D", "content": "每分钟$$600$$字 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["$$1000000\\div 100\\div 20=500$$（字），所以阅读速度至少提升到每分钟$$500$$字的水平． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3017", "queId": "bc5a057814404cd086848075152f1231", "competition_source_list": ["2020年希望杯二年级竞赛模拟第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "请你根据数串的规律再横线上填上正确的答案：3,6,12,~\\uline{~~~~~~~~~~}~，48. ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["后一个数等于前一个数$$\\times$$ 2 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1783", "queId": "a8defd58b0684ce78c241341e4072213", "competition_source_list": ["2020年新希望杯六年级竞赛（2月）第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "海尔兄弟被困在一个无人岛上，他们要做一个独木舟逃出这个无人岛，哥哥单独做要$$6$$小时完成，弟弟单独做要$$9$$小时完成．如果按照哥哥、弟弟、哥哥、弟弟$$\\cdots \\cdots $$的顺序交替工作，每人工作$$1$$小时后交换，那么需要 小时能做好独木舟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["哥哥工效：$$\\frac{1}{6}$$，弟弟工效：$$\\frac{1}{9}$$， 则$$1\\div \\left( \\frac{1}{6}+\\frac{1}{9} \\right)=\\frac{18}{5}=3$$（周期）$$\\cdots \\cdots \\frac{3}{5}$$， $$1-3\\times \\left( \\frac{1}{6}+\\frac{1}{9} \\right)=\\frac{1}{6}$$， $$\\frac{1}{6}\\div\\frac{1}{6}=1$$（小时）， 共$$3\\times 2+1=7$$（小时）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2277", "queId": "782e71a984634b92bc0036f4f9d3c3e5", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（三）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "欧欧从$$A$$地前往$$B$$地办事，他每走$$40$$分钟就休息$$10$$分钟，到达$$B$$地共需$$4$$小时$$20$$分钟，从$$B$$地原路返回的速度是去时的$$2$$倍，若他每走$$35$$分钟就休息$$15$$分钟，则到达$$A$$地共需． ", "answer_option_list": [[{"aoVal": "A", "content": "$$105$$分钟 "}], [{"aoVal": "B", "content": "$$130$$分钟 "}], [{"aoVal": "C", "content": "$$135$$分钟 "}], [{"aoVal": "D", "content": "$$150$$分钟 "}]], "knowledge_point_routes": ["知识标签->课内知识点->式与方程->数量关系->路程=速度×时间"], "answer_analysis": ["$$60\\times 4+20=50\\times 5+10$$，去时共走了$$40\\times 5+10=210$$（分钟），则返回时共需走$$210\\div 2=105$$（分钟），$$105\\div 35=3$$，加上休息的时间共需要$$105+15\\times \\left( 3-1 \\right)=135$$（分钟）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2846", "queId": "897eb34a230a4035a024211957a1c845", "competition_source_list": ["2017年环亚太杯一年级竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "找规律填数： $$6$$，$$12$$，$$18$$，$$24$$，~\\uline{~~~~~~~~~~}~，$$36$$，~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "28，40 "}], [{"aoVal": "B", "content": "28，42 "}], [{"aoVal": "C", "content": "30，40 "}], [{"aoVal": "D", "content": "30，42 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数学广角->找规律->数字排列规律"], "answer_analysis": ["每个数之间差了一个6 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1394", "queId": "7a11f92c29374b6da53bcc04f57b90d2", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（一）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "学校秋游共用了$$10$$辆客车，已知大客车每辆坐$$100$$人，小客车每辆坐$$60$$人，大客车比小客车一共多坐$$520$$人．大客车有辆． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->原型题"], "answer_analysis": ["假设这$$10$$辆客车都是大客车，则大客车比小客车多坐$$10\\times 100=1000$$（人），比实际多了$$1000-520=480$$（人），这是因为每把一辆小客车看作大客车，就会多算$$100+60=160$$（人），所以小客车一共有$$480\\div 160=3$$（辆），大客车有$$10-3=7$$（辆）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1709", "queId": "c3e1d86a733448f49c53681b2defc5fd", "competition_source_list": ["2003年五年级竞赛创新杯", "2003年第1届创新杯五年级竞赛复赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$10\\div 7$$的商是循环小数，那么这个循环小数的小数部分第2003位上的数字是（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "7 "}], [{"aoVal": "B", "content": "8 "}], [{"aoVal": "C", "content": "5 "}], [{"aoVal": "D", "content": "1 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题"], "answer_analysis": ["若a为1到6的整数，则$$\\frac{a}{7}$$化成小数时，其小数部分均为1、4、2、8、5、7 这6个数码组成.由于$$10\\div 7=1\\frac{3}{7}=1+0.\\dot{4}2857\\dot{1}$$，而$$2003\\div 6=333\\cdots \\cdots 5$$， 那么小数部分第2003个数为一个周期数码428571中的第5个为7 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1144", "queId": "82792e87f654406bbef71c105171a33e", "competition_source_list": ["2014年全国学而思杯一年级竞赛第5题"], "difficulty": "0", "qtype": "single_choice", "problem": "ài艾dí迪、wēi薇er儿、jiā加jiā加hé和jiǎn减jiǎn减wán玩yóu游xì戏，měi每rén人xiě写yī一gè个shù数，ran然hòu后pàn判duàn断zhè这$$4$$gè个shù数xiàng相jiā加hòu后de的hé和shì是dān单shù数hái还shì是shuāng双shù数．qí其zhōng中yī一jú局tā他men们fēn分bié别xiě写de的shì是：$$9$$、$$13$$、$$471$$、$$1236$$，nà那me么nǐ你lái来pàn判duàn断yī一xià下，zhè这$$4$$gè个shù数de的hé和shì是~\\uline{~~~~~~~~~~}~shù数（tián填``dān单''huò或``shuāng双''）． ", "answer_option_list": [[{"aoVal": "A", "content": "单 "}], [{"aoVal": "B", "content": "双 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["四个数中三个单数一个双数，三个单数和是单数，再加一个双数和是单数． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1906", "queId": "c9b2a554d98a4e779ba68443415190f1", "competition_source_list": ["2018年湖北武汉创新杯小学高年级六年级竞赛初赛数学思维能力等级测试第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙两人加工一批零件，每人加工零件总数的一半，他们同时开始，甲完成任务的$$\\frac{1}{3}$$时乙加工了$$50$$个零件；甲完成任务的$$\\frac{3}{5}$$时乙完成了任务的一半，这批零件共有（~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$600$$ "}], [{"aoVal": "B", "content": "$$420$$ "}], [{"aoVal": "C", "content": "$$350$$ "}], [{"aoVal": "D", "content": "$$360$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->量率对应求单位1"], "answer_analysis": ["由于甲完成任务的$$\\frac{3}{5}$$等于易完成任务的$$\\frac{1}{2}$$，则甲、乙的效率比$$:$$为甲乙$$=\\frac{3}{5}:\\frac{1}{2}=6:5$$，则当乙完成$$50$$个小时，甲完成$$60$$个，则甲的任务量为$$60\\div \\frac{1}{3}=180$$个，则共有$$180\\times 2=360$$个． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2069", "queId": "d8f455275f9047c98f50c4a7c2c2fe8f", "competition_source_list": ["2019年广东广州学而思综合能力诊断五年级竞赛第11题12分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A+B+C=2021$$, $$A$$, $$B$$ and $$C$$ have $$8$$, $$10$$ and $$11$$ factors respectively, and $$B$$ and $$C$$ are mutually prime, then $$A$$ is ． $$A+B+C=2021$$，$$A$$、$$B$$、$$C$$分别有$$8$$、$$10$$、$$11$$个因数，并且$$B$$与$$C$$互质，那么$$A$$是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$360$$ "}], [{"aoVal": "B", "content": "$$380$$ "}], [{"aoVal": "C", "content": "$$390$$ "}], [{"aoVal": "D", "content": "$$420$$ "}], [{"aoVal": "E", "content": "$$430$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理逆应用", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["本题是因数个数定理的逆应用，先从$$A$$、$$B$$、$$C$$三个的因数个数开始讨论：$$C$$有$$11$$个因数，$$11$$是质数，所以$$C$$一定是一个质数的$$10$$次方，又因为三个数的和为$$2021$$，小于$$3$$的$$10$$次方，所以$$C$$只能是$$2$$的$$10$$次方，等于$$1024$$，则$$A+B=2021-1024=997$$，$$A$$有$$8$$个因数，则$$A$$的分解质因数形式可以是$$a$$的$$7$$次方、$$a$$的$$1$$次方乘$$b$$的$$3$$次方、$$a\\times b\\times c$$这$$3$$种可能，$$B$$的分解质因数形式可以是$$c$$的$$9$$次方、$$c$$的$$1$$次方乘$$d$$的$$4$$次方，因为$$B$$和$$C$$互质，所以$$B$$不含质因数$$2$$，根据和的大小可以确定$$B$$不等于$$c$$的$$9$$次方，又$$5$$的$$4$$次方等于$$625$$，$$625$$乘$$3$$等于$$1275$$大于$$997$$，所以$$d$$只能等于$$3$$，$$3$$的$$4$$次方等于$$81$$，$$81$$乘$$5$$等于$$405$$，$$997-405=592$$不符合$$A$$的分解质因数条件，$$81$$乘$$7$$等于$$567$$，$$997-567=430$$符合条件，则$$A$$等于$$430$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3230", "queId": "2c7b81c912b74000bd2c13eac96d1c18", "competition_source_list": ["2012年全国创新杯五年级竞赛第1题5分", "五年级其它", "2016年创新杯小学高年级五年级竞赛训练题（四）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "某班共有$$48$$人，其中$$27$$人会游泳，$$33$$人会骑自行车，$$40$$人会乒乓球．那么，这个班至少有（~ ）个学生这三项运动都会． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->容斥原理->三量容斥"], "answer_analysis": ["让三项都会的人最少，即让所有人都只会两项， 则一共会$$48\\times 2=96$$项，$$27+33+40=100$$项， 即至少有$$100-96=4$$人会三项． 构造如下：$$8$$人会游泳$$+$$自行车； $$15$$人会游泳$$+$$乒乓球；$$21$$人会自行车$$+$$乒乓球；$$4$$人全会． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "680", "queId": "7e734930ab144b07ba09066a2f1610a0", "competition_source_list": ["2017年河南郑州豫才杯竞赛第15题", "2016年河南郑州K6联赛六年级竞赛第17题2分", "2017年河南郑州小升初豫才杯第二场第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "假设$$A=9876543\\times 23456789$$，$$B=9876544\\times 23456788$$，则（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$A\\textgreater B$$ "}], [{"aoVal": "B", "content": "$$A=B$$ "}], [{"aoVal": "C", "content": "$$A\\textless{}B$$ "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["知识标签->拓展思维->数论模块->位值原理与进制->数与数字->比较大小"], "answer_analysis": ["$$A-B=9876543\\times 23456789-9876544\\times 23456788$$ $$ ~=9876543\\times \\left( 23456788+1 \\right)-9876544\\times 23456788 $$ $$ ~=9876543\\times 23456788+9876543-9876544\\times 23456788 $$ ~$$=9876543-23456788\\textless{}0 $$ 所以$$A\\textless{}B$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2281", "queId": "a004c557379549b7be86761ab8333bfa", "competition_source_list": ["2019年广东深圳全国小学生数学学习能力测评四年级竞赛初赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$9$$点至$$10$$点之间的某个时刻，$$5$$分钟前分针的位置和$$5$$分钟后时针的位置相同，此时刻是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9:05$$ "}], [{"aoVal": "B", "content": "$$9:35$$ "}], [{"aoVal": "C", "content": "$$9:55$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->时间问题->认识钟表"], "answer_analysis": ["根据选项$$\\text{C}$$，现在是$$9:55$$，那么$$5$$分钟前分针的位置是指向$$10$$，$$5$$分钟后时针也是指向$$10$$．所以$$\\text{C}$$选项答案是满足题目要求的，所以选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2811", "queId": "4ed6343268eb4b97b34420985970b71d", "competition_source_list": ["2017年全国小升初八中入学备考课程", "2012年全国华杯赛小学高年级竞赛初赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "计算$\\left[ \\left( 80 \\%+\\dfrac{1}{5}\\right)\\times24+6.6\\right]\\div\\dfrac{9}{14}-7.6=$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数与运算->混合运算->分、小数四则混合运算"], "answer_analysis": ["原式= $$\\left[ \\left( 0.8+\\frac{1}{5} \\right)\\times 24+6.6 \\right]\\div \\frac{9}{14}-7.6$$ $$=\\left[ (0.8+0.2)\\times 24+6.6 \\right]\\div \\frac{9}{14}-7.6$$ $$=\\left[ 1\\times 24+6.6 \\right]\\div \\frac{9}{14}-7.6$$ $$=30.6\\div \\frac{9}{14}-7.6$$ $$=30.6\\times \\frac{14}{9}-7.6$$ $$=47.6-7.6$$ $$=40$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "555", "queId": "53b39b9373ed4e3481ae03ca3998221f", "competition_source_list": ["小学高年级六年级其它2014年数学思维能力等级测试第2题", "2014年第12届全国创新杯小学高年级六年级竞赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "~张敏最近搬进了新居，房号是一个三位数，这个数加上各位数上的数字和是$$429$$，那么他的房号三个位数上的数字得到乘积是（~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$\\overline{abc}+a+b+c=429$$，$$101a+11b+2c=429$$， 所以$$a=4$$，$$11b+2c=25$$，$$b=1$$，$$c=7$$， $$4\\times 1\\times 7=28$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1396", "queId": "834337d764b54340ba98dbaebffae14f", "competition_source_list": ["2018年第23届华杯赛小学中年级竞赛初赛第2题", "2018年华杯赛小学中年级竞赛初赛第2题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明把$$6$$个数分别写在三张卡片的正面和反面，每一面写一个数，每张卡片上的$$2$$个数的和相等，然后他将卡片放在桌子上，发现正面上写着$$28$$，$$40$$，$$49$$，反面上的数都只能被$$1$$和它自己整除，那么，反面上的三个数的平均数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$39$$ "}], [{"aoVal": "D", "content": "$$40$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["由题目条件可知，反面的数为质数．又知反面数字和相等，故反面三个数依然相差$$12$$，$$9$$． 数学质数有$$2$$，$$3$$，$$5$$，$$7$$，$$11$$，$$13$$，$$17$$，$$23$$． $$2$$，$$11$$，$$23$$依次相差$$9$$和$$12$$，故确定其平均数为$$\\left( 2+11+23 \\right)\\div 3=12$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3204", "queId": "27376e7ef2eb4636904ba6a8b464d2eb", "competition_source_list": ["2015年全国AMC小学高年级六年级竞赛8第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "\\textbf{2015年 AMC8 第 10 题} 1000至$$9999$$中有多少个位数字互不相同的四位数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3024$$ "}], [{"aoVal": "B", "content": "$$4536$$ "}], [{"aoVal": "C", "content": "$$5040$$ "}], [{"aoVal": "D", "content": "$$6480$$ "}], [{"aoVal": "E", "content": "$$6561$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->计数模块->加乘原理", "拓展思维->能力->运算求解"], "answer_analysis": ["翻译：$$1000$$至$$9999$$中有多少个位数字互不相同的四位数？ 组数问题：千位数字有$$1\\sim 9$$共$$9$$种选法，百、十、个位分别有$$9$$、$$8$$、$$7$$种选法，一共有$$9\\times 9\\times 8\\times 7=4536$$（个）数字互不相同的四位数． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2614", "queId": "be5e7229bd0b499594ff5af0c1a33592", "competition_source_list": ["2015年华杯赛六年级竞赛初赛", "2015年华杯赛五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$(\\frac{9}{20}-\\frac{11}{30}+\\frac{13}{42}-\\frac{15}{56}+\\frac{17}{72})\\times 120-\\frac{1}{3}\\div \\frac{1}{4}=$$( ) ", "answer_option_list": [[{"aoVal": "A", "content": "$$42$$ "}], [{"aoVal": "B", "content": "$$43$$ "}], [{"aoVal": "C", "content": "$$15\\frac{1}{3}$$ "}], [{"aoVal": "D", "content": "$$16\\frac{2}{3}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数运算->分数四则混合运算"], "answer_analysis": ["原式$$=\\left[ \\left( \\frac{1}{4}+\\frac{1}{5} \\right)-\\left( \\frac{1}{5}+\\frac{1}{6} \\right)+\\left( \\frac{1}{6}+\\frac{1}{7} \\right)-\\left( \\frac{1}{7}+\\frac{1}{8} \\right)+\\left( \\frac{1}{8}+\\frac{1}{9} \\right) \\right]\\times 120-\\frac{1}{3}\\div \\frac{1}{4}$$ $$=\\left( \\frac{1}{4}+\\frac{1}{9} \\right)\\times 120-\\frac{1}{3}\\div \\frac{1}{4}$$ $$=\\frac{13}{36}\\times 120-\\frac{4}{3}$$ $$=\\frac{130}{3}-\\frac{4}{3}$$ $$=42$$ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "531", "queId": "fde83dad34084579bb103458fd0a0f4b", "competition_source_list": ["2016年全国华杯赛小学中年级竞赛在线模拟第3题", "2013年全国华杯赛小学中年级竞赛初赛A卷第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "丽丽、莱克西、$$Ellie$$、$$Michelle$$四个小朋友在一起做游戏时，捡到了一条围巾．失主问是谁捡到的？ 丽丽说不是莱克西； 莱克西说是$$Ellie$$； $$Ellie$$说丽丽说的不对； Michelle说$$Ellie$$说的不对． 他们之中只有一个人说对了，这个人是谁？ ", "answer_option_list": [[{"aoVal": "A", "content": "丽丽 "}], [{"aoVal": "B", "content": "莱克西 "}], [{"aoVal": "C", "content": "$$Ellie$$ "}], [{"aoVal": "D", "content": "Michelle "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾", "Overseas Competition->知识点->组合模块->逻辑推理->假设型逻辑推理->真假话"], "answer_analysis": ["由于只有一个人说对了，而$$Michelle$$支持丽丽，那么他们俩都错了，所以反对丽丽的$$Ellie$$说对了． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1458", "queId": "5a787c73bef44d1aa908ec50ccd897a2", "competition_source_list": ["2020年长江杯六年级竞赛复赛A卷第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "李阿姨的淘宝店今天卖出甲乙两件衣服，每件各卖$$120$$元，但甲赚了$$\\frac{1}{4}$$，乙亏本$$\\frac{1}{4}$$．问：李阿姨今天卖出这两件衣服，是赚了还是亏了？ ", "answer_option_list": [[{"aoVal": "A", "content": "赚了 "}], [{"aoVal": "B", "content": "亏了 "}], [{"aoVal": "C", "content": "不赚不亏 "}], [{"aoVal": "D", "content": "无法判断 "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设甲成本为$$x$$元， $$x\\times \\left( 1+\\frac{1}{4} \\right)=120$$， 解得$$x=96$$， 设乙成本为$$x$$元， $$x\\times \\left( 1-\\frac{1}{4} \\right)=120$$， 解得$$x=160$$， 总成本$$96+160=256$$（元）， 总收入为$$120\\times 2=240$$（元）， $$256\\textgreater240$$， 故成本大于收入，亏了． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "456", "queId": "ce3896d59e0143f699c36e487bffd1a5", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "$$2$$个人同时吃$$2$$个馒头用$$2$$分钟，$$10$$个人同时吃$$10$$个馒头用分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->统筹规划->简单时间统筹问题->同时进行问题"], "answer_analysis": ["两人同时吃两个馒头的时间与一个人吃一个馒头时间相同，与$$10$$个人同时吃$$10$$个馒头时间相同，都是两分钟． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2827", "queId": "60bb8e240d2a41c9b6cd38daaf6603dc", "competition_source_list": ["其它改编自2012年全国希望杯六年级竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "在小数$$3.1415926$$的其中两个数字上方加$$2$$个循环点，得到一个循环小数，这样的循环小数中，最小的~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3.\\dot{1}41592\\dot{6}$$ "}], [{"aoVal": "B", "content": "$$3.1\\dot{4}1592\\dot{6}$$ "}], [{"aoVal": "C", "content": "$$3.14\\dot{1}592\\dot{6}$$ "}], [{"aoVal": "D", "content": "$$3.14159\\dot{2}\\dot{6}$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数的认识->比较大小->小数比较大小->多位小数比较大小"], "answer_analysis": ["要使循环小数最小，则循环节开始的几位尽量小，因此从$$1$$开始循环，下一位为$$4$$，依次往下，最小的为$$3.\\dot{1}41592\\dot{6}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "600", "queId": "0ca0c7161d9249c59cd7c9b62d9e1cbe", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛B卷第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$120$$有个因数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为$$120$$的因数有：$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$8$$、$$10$$、$$12$$、$$15$$、$$20$$、$$24$$、$$30$$、$$40$$、$$60$$、$$120$$，所以一共有$$16$$个因数． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1551", "queId": "71cf18462ae94398856f1ebb3b97ee00", "competition_source_list": ["2018年全国小学生数学学习能力测评四年级竞赛初赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙二人比赛爬楼梯，甲跑到第$$4$$层时，乙恰好跑到第$$3$$层．以这样的速度，甲跑到第$$16$$层时，乙跑到第层． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["本来甲乙所处的位置就在$$1$$层， 即甲从$$1$$层到$$4$$层的时间，其实是跑了$$3$$层楼梯的时间， $$\\left( 16-1 \\right)\\div \\left( 4-1 \\right)\\times \\left( 3-1 \\right)=10$$层，$$10+1=11$$层． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1730", "queId": "9ad695d973374524967b72bdf05346fe", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$1$$只小猪的重量等于$$6$$只鸡的重量；$$3$$只鸡的重量等于$$4$$只鸭的重量；$$2$$只鸭的重量等于$$6$$条鱼的重量，那么$$2$$只小猪的重量等于（ ）条鱼的重量。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->列方程解应用题等量代换"], "answer_analysis": ["解：$$6\\div 3\\times 4$$ $$=2\\times 4$$ $$=8$$（只） $$8\\div 2\\times 6$$ $$=4\\times 6$$ $$=24$$（条） $$24\\times 2=48$$（条） 答：$$2$$只小猪的重量等于$$48$$条鱼的重量。 故选:A。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1648", "queId": "ff80808147342b7e014735453a14023d", "competition_source_list": ["2013年全国华杯赛小学高年级竞赛初赛B卷第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙两仓的稻谷数量一样，爸爸，妈妈和阳阳单独运完一仓稻谷分别需要$$10$$天，$$12$$天和$$15$$天．爸爸妈妈同时开始分别运甲、乙两仓的稻谷，阳阳先帮妈妈，后帮爸爸，结果同时运完两仓稻谷，那么阳阳帮妈妈运了天． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["设一个仓库的稻谷量为``$$1$$''，爸爸、妈妈、阳阳的效率分别是$$\\frac{1}{10}$$、$$\\frac{1}{12}$$、$$\\frac{1}{15}$$， 三人同时运完两仓，需要的时间：$$(1+1)\\div \\left( \\frac{1}{10}+\\frac{1}{12}+\\frac{1}{15} \\right)=8$$（天）；妈妈$$8$$天共搬运了：$$8\\times \\frac{1}{12}=\\frac{2}{3}$$（仓）；妈妈剩下的就是阳阳帮妈妈运的，所以，阳阳帮妈妈运了$$\\left( 1-\\frac{2}{3} \\right)\\div \\frac{1}{15}=5$$（天）． 三人一共搬了： $$(1+1)\\div \\left( \\frac{1}{10}+\\frac{1}{12}+\\frac{1}{15} \\right)$$， $$=2\\div \\frac{1}{4}$$ $$=8$$（天）； 阳阳帮妈妈运的天数： $$\\left( 1-\\frac{1}{12}\\times 8 \\right)\\div \\frac{1}{15}$$ $$=\\frac{1}{3}\\times 15$$ $$=5$$（天）； 答：阳阳帮妈妈运了$$5$$天． 故答案为：$$\\rm C$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2315", "queId": "fc9472137ed642cb8cf2a5a511a5f46c", "competition_source_list": ["2016年创新杯五年级竞赛训练题（一）第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "一条船往返于甲乙两港之间，由于甲至乙是顺水行驶，由乙至甲是逆水行驶．已知船在静水中的速度为$$8$$千米/小时，平时逆水航行所用时间是顺水航行所用的时间的$$1.5$$倍．某天恰逢暴雨，水流速度是原来的三倍，这条船往返共用了$$10$$小时．问：甲、乙两地相距千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$25$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->行程模块->流水行船问题->基本流水行船问题->水速变化"], "answer_analysis": ["平时逆水航行所用时间是顺水航行所用时间的$$2$$倍，所以顺水航行速度是逆水航行的$$2$$倍，即$${{V}_{水}}+8=2\\times \\left( 8-{{V}_{水}} \\right)$$，解得：$${{V}_{}}=\\frac{8}{3}$$，所以水速为$$\\frac{8}{3}$$千米/小时，变为原来的$$2$$倍后变为$$\\frac{16}{3}$$千米/小时．设甲、乙两地相距$$S$$千米，则：$$\\frac{S}{\\frac{16}{3}+8}+\\frac{S}{8-\\frac{16}{3}}=9$$ 解得：$$S=20$$，所以甲、乙两地相距$$20$$千米． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2579", "queId": "67663801e9e143a0afe4eb550459430a", "competition_source_list": ["2018年IMAS小学中年级竞赛（第二轮）第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "将一个两位数乘$$3$$再加上$$10$$，然后交换它的个位与十位，最后得到的是$$95$$、$$96$$、$$97$$、$$98$$、$$99$$之中的一个整数，请问原来的两位数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$24$$ "}], [{"aoVal": "E", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->位值原理运用->位值原理的综合应用"], "answer_analysis": ["逆推回去，交换$$95$$、$$96$$、$$97$$、$$98$$、$$99$$的固位与十位，再减$$10$$，得到的数分别为$$49$$、$$59$$、$$69$$、$$79$$、$$89$$，这些数中只有$$69$$是$$3$$的倍数，故原来的两位数是$$69\\div3=23$$． 故选$$\\text{C}$$． ", "将一个两位数乘以$$3$$再加上$$10$$，则所得的数除以$$3$$所得的余数为$$1$$．将这个数的个位与十位交换后所得的数除以$$3$$所得的余数仍然为$$1$$．在$$95$$、$$96$$、$$97$$、$$98$$、$$99$$这些数中只有$$97$$除以$$3$$所得的余数为$$1$$，故原来的两位数是$$\\left(79-10\\right)\\div3=23$$．
\n故选$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2797", "queId": "95dd0a6757a745b093528cd870108fe3", "competition_source_list": ["2016年全国学而思杯一年级竞赛样题第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "这个算式要按照什么顺序计算呢． $$1+2-3+1$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3-3$$ "}], [{"aoVal": "B", "content": "$$1-3$$ "}], [{"aoVal": "C", "content": "$$2-3$$ "}], [{"aoVal": "D", "content": "$$1+1$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["按照从左往右的顺序计算． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2725", "queId": "ff8080814502fa2401450bc65f921593", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "算式$$826446281\\times11\\times 11$$的计算结果是（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9090909091$$ "}], [{"aoVal": "B", "content": "$$909090909091$$ "}], [{"aoVal": "C", "content": "$$10000000001$$ "}], [{"aoVal": "D", "content": "$$100000000001$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数乘除->整数乘法运算->表外乘法计算"], "answer_analysis": ["根据$$11$$乘法的特征``两边一拉，中间相加''可得到结果D "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3097", "queId": "eb16598ef88948e8a7c062edadddaea0", "competition_source_list": ["2018年陕西西安雁塔区西安铁一中小升初（二十六）第20题5分", "2015年世界少年奥林匹克数学竞赛六年级竞赛复赛A卷第10题10分", "六年级其它导引"], "difficulty": "3", "qtype": "single_choice", "problem": "计算：$$\\frac{{{1}^{2}}+{{2}^{2}}}{1\\times 2}+\\frac{{{2}^{2}}+{{3}^{2}}}{2\\times 3}+\\cdots +\\frac{{{18}^{2}}+{{19}^{2}}}{18\\times 19}+\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}$$=~\\uline{~~~~~~~~~~}~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$37\\frac{19}{20}$$ "}], [{"aoVal": "B", "content": "$$28\\frac{19}{20}$$ "}], [{"aoVal": "C", "content": "$$38\\frac{19}{20}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->裂项与通项归纳->分数裂项->分数裂差->两分数间接裂差", "课内体系->思想->转化与化归的思想"], "answer_analysis": ["算式中的分母是裂项计算的最基本形式，但分子比较复杂，我们可以从前几项找找规律： $$\\frac{{{1}^{2}}+{{2}^{2}}}{1\\times 2}=\\frac{5}{2}=2\\frac{1}{2}$$，$$\\frac{{{2}^{2}}+{{3}^{2}}}{2\\times 3}=\\frac{13}{6}=2\\frac{1}{6}$$，$$\\frac{{{3}^{2}}+{{4}^{2}}}{3\\times 4}=\\frac{25}{12}=2\\frac{1}{12}$$． 我们发现一规律：每一项减去$$2$$后，分子就变成了$$1$$． 再来试试最后一项：$$\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}=\\frac{761}{380}=2\\frac{1}{380}$$， 也满足这个规律，这是为什么呢？ 观察每一项的分子和分母，我们发现分子的每个加数都与分母大小接近，可以做如下变形： $$\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}=\\frac{19\\times \\left( 20-1 \\right)+20\\times \\left( 19+1 \\right)}{19\\times 20}$$ $$=\\frac{19\\times 20\\times 2+\\left( 20-19 \\right)}{19\\times 20}$$ $$=2+\\frac{1}{19\\times 20}$$． 算式中的每一项都能像上面一样进行变形，再结合分数裂差，所以： 原式$$=2\\frac{1}{1\\times 2}+2\\frac{1}{2\\times 3}+\\cdots +2\\frac{1}{19\\times 20}$$ $$=2\\times 19+\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{19\\times 20}$$ $$=38+1-\\frac{1}{2}+\\frac{1}{2}-\\frac{1}{3}+\\frac{1}{3}-\\frac{1}{4}+\\cdots +\\frac{1}{18}-\\frac{1}{19}+\\frac{1}{19}-\\frac{1}{20}$$ $$=38+1-\\frac{1}{20}$$ $$=38\\frac{19}{20}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3069", "queId": "e18568104e1b4a60a075b06d0a8e5041", "competition_source_list": ["2006年四年级竞赛创新杯", "2006年第4届创新杯四年级竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "小王老师外出开会，离开的时候撕下了当天的日历，之后$$9$$天没有回家$$.$$ 回来后一次撕下这$$9$$张日历，发现这$$9$$张日期数相加得$$99$$，那么，王老师回家这天是号. ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$15$$ "}], [{"aoVal": "E", "content": "$$16$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->应用题模块->周期问题", "拓展思维->拓展思维->计算模块->数列与数表->等差数列->中项和首项、末项的关系"], "answer_analysis": ["因为$$99\\div 9=11$$，所以第五天也就是中间一天是$$11$$号，故这$$9$$天的日期为：$$7$$、$$8$$、$$9$$、$$10$$���$$11$$、$$12$$、$$13$$、$$14$$、$$15$$，因此，回家是$$15$$号 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "233", "queId": "6383691f57e247de987b7cd927ea3ed9", "competition_source_list": ["其它改编题", "2017年第15届全国希望杯五年级竞赛第1试试题第18题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$，$$B$$，$$C$$，$$D$$，$$E$$五人一同参加飞镖大赛，其中只有一人射中飞镖盘的中心，但不知是谁所射． $$A$$说：``不是我射中的，就是$$C$$射中的．'' $$B$$说：``不是$$E$$射中的．'' $$C$$说：``如果不是$$D$$射中的，那么一定是$$B$$射中的．'' $$D$$说：``既不是我射中的，也不是$$B$$射中的．'' $$E$$说：``既不是$$C$$射中的，也不是$$A$$射中的．'' 其中五人中只有两个人说的是对的，由此可以判断射中飞镖盘中心的人是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$A$$ "}], [{"aoVal": "B", "content": "$$B$$ "}], [{"aoVal": "C", "content": "$$C$$ "}], [{"aoVal": "D", "content": "$$D$$ "}], [{"aoVal": "E", "content": "$$E$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$A$$和$$E$$说的矛盾，$$C$$和$$D$$说的矛盾，必有两对两错，故$$B$$说的是错的，则是$$E$$射中的． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2876", "queId": "cd4e53d7390f4f05a6beae81e0669d7e", "competition_source_list": ["2013年第12届春蕾杯二年级竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一串数按照$$1$$，$$0$$，$$2$$，$$4$$，$$1$$，$$0$$，$$2$$，$$4$$$$\\cdots \\cdots $$的顺序排列，一共写了$$21$$个数，那么这$$21$$个数的和是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$37$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["周期是$$4$$，$$21\\div$$ 4=5（组）$$\\cdots \\cdots $$1（个），每组的和：$$1+0+2+4=7$$，共：$$5\\times$$ 7+1=36. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1480", "queId": "75f478fdef314415b5e525bc90b00702", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛决赛第4题", "2018年四川成都锦江区四川师范大学附属第一实验中学小升初（四）第4题3分", "2018年浙江杭州西湖区小学高年级六年级上学期单元测试《第三单元》第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一种商品的利润率为$$20 \\%$$，进价提高$$25 \\% $$后，保持利润不变，那么，进价提价后的利润率为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25 \\%$$ "}], [{"aoVal": "B", "content": "$$20 \\% $$ "}], [{"aoVal": "C", "content": "$$16 \\% $$ "}], [{"aoVal": "D", "content": "$$12.5 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->利润基本公式->已知利润成本求利润率"], "answer_analysis": ["利润问题．假设成本为$$100$$元，则利润为$$20$$元，现在成本是$$100\\times (1+25 \\%)=125$$（元），$$20\\div 125\\times 100 \\%=16 \\%$$． ", "设这种商品原来的进价为$$x$$，利润为$$m$$，则原售价为$$\\left( \\frac{1}{20\\%} \\right)=\\frac{x}{m}$$，解得$$x=5m$$．因为这种商品的进价提高$$25\\%$$，所以新的进价为$$(1+25\\%)5m=6.25m$$．设提价后的利润率为$$y$$，则有$$6.25m\\times (1+y)=6.25m+m$$，解得$$y=16\\%$$．
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3220", "queId": "1f75e9610e8f4f21bab99d1bc1b36542", "competition_source_list": ["2016年第12届全国新希望杯五年级竞赛决赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$至$$60$$中选出$$6$$个连续自然数，它们的乘积末尾恰有两个$$0$$的取法共有（~~~ ）种． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["由于是连续的$$6$$个数，分解质因数只能合有$${{5}^{2}}$$，但是$$25、50$$中本身分解质因数为$${{5}^{2}}$$，所以共有以下几种情况： $$5-10$$、$$10-15$$、$$15-20$$、$$21-26$$、$$22-27$$、$$23-28$$、$$24-29$$、$$30-35$$、$$35-40$$、$$40-45$$、$$46-51$$、$$47-52$$、$$48-53$$、$$49-54$$、$$55-60$$ 共有$$11$$种． ~ "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "730", "queId": "63b149c46ec1402aa7eae1f0de196d91", "competition_source_list": ["2018年第12届湖北武汉学而思综合能力诊断小学高年级六年级竞赛第7题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "一箱苹果有$$40$$多个，如果把这箱苹果每$$8$$个装一盒，还剩余$$3$$个；如果每$$10$$个装一盒，也剩余$$3$$个．这箱苹果有~\\uline{~~~~~~~~~~}~个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$43$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$42$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["去掉$$3$$个之后，剩余的是$$8$$和$$10$$的公倍数，由于只有$$40$$多个，那么去掉$$3$$个后就是$$40$$个，一共有$$43$$个． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1808", "queId": "a47a71d4c792480f83133cc6d5018a30", "competition_source_list": ["2014年第10届全国新希望杯五年级竞赛复赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "小刚问：``小珍，你爸爸的年龄是多少？小珍风趣地说：``我爸爸现在的年龄等于四年后的岁数的$$3$$倍减去八年前的岁数的$$3$$倍．``小刚很决就回答：``你爸爸现在（~~~ ）岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$38$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["四年后和八年前相隔$$12$$年，爸爸现在的年龄是$$12\\times 3=36$$（岁）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3219", "queId": "2c27e7c7f4b241a9b1e9c220211f2980", "competition_source_list": ["2021年第8届鹏程杯四年级竞赛初赛第20题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$3\\times 3$$方格中的每一个格中填入$$1$$或者$$2$$，如果要求每个$$2\\times 2$$田字方格中$$4$$个数之和都是偶数，那么共有种不同的填数方法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$33$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["先从中间入手，中间的数可以放奇数或者偶数，有两种放法，假设放奇数， 则它的上下左右都可以放奇数或者偶数，共有$$2\\times 2\\times 2\\times 2=16$$种， 此时四个角都只有唯一一种放法，因为每个$$2\\times 2$$的田字格中都已经填了三个数， 和要么是奇数要么是偶数，和是奇数，则剩下的一个角放奇数， 和是偶数则剩下的一个角放偶数，都只有一种，所以一共有$$2\\times 16=32$$种． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "85", "queId": "0d472d32d2a74a5682a25c1402580253", "competition_source_list": ["2013年第9届全国新希望杯五年级竞赛复赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "布袋中的球大小相同，其中黄球和红球各有$$9$$个，蓝球有$$4$$个，绿球有$$3$$个，从布袋中摸出若干个球，为保证摸出的球中至少有$$5$$个球颜色相同，至少要摸出（~~~ ）个球． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$3+4+4+4+1=16$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1333", "queId": "2437cb42797443d58bf2418cf997059e", "competition_source_list": ["2016年创新杯小学高年级六年级竞赛训练题（四）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "四娃抄写一份报告，如果每分钟抄写$$30$$个字，则用若干小时可以抄完．当抄完$$\\frac{2}{5}$$时，将工作效率提高$$20 \\% $$，结果比原计划提前$$20$$分钟完成．问这份报告共有（~ ）字． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5000$$ "}], [{"aoVal": "B", "content": "$$6000$$ "}], [{"aoVal": "C", "content": "$$7500$$ "}], [{"aoVal": "D", "content": "$$9000$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->变速工程问题"], "answer_analysis": ["抄完$$\\frac{2}{5}$$后，前后工作效率之比为$$1:\\left( 1+20 \\% \\right)=5:6$$，所以前后工作时���之比为$$6:5$$．时间相差$$20$$分钟，则按原计划做需要：$$20\\div \\left( 6-5 \\right)\\times 6=120$$（分）．$$120\\div \\left( 1-\\frac{2}{5} \\right)=200$$（分钟）$$30\\times 200=6000$$（字）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "852", "queId": "d1e9a828676e4d45b39a82d4bfc992b2", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$7$$进制中有三位数$$\\overline{abc}$$，化为$$9$$进制为$$\\overline{cba}$$，求这个三位数在十进制中为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$148$$ "}], [{"aoVal": "B", "content": "$$248$$ "}], [{"aoVal": "C", "content": "$$348$$ "}], [{"aoVal": "D", "content": "$$648$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["首先还原为十进制： $${{(\\overline{abc})}_{7}}=a\\times{{7}^{2}}+b\\times {{7}^{1}}+c\\times {{7}^{0}}=49a+7b+c$$；$${{(\\overline{cba})}_{9}}=c\\times{{9}^{2}}+b\\times {{9}^{1}}+a\\times {{9}^{0}}=81c+9b+a$$． 于是$$49a+7b+c=81c+9b+a$$；得到$$48a=80c+2b$$，即$$24a=40c+b$$． 因为$$24a$$是$$8$$的倍数，$$40c$$也是$$8$$的倍数，所以$$b$$也应该是$$8$$的倍数，于是$$b=0$$或$$8$$． 但是在$$7$$进制下，不可能有$$8$$这个数字．于是$$b=0$$，$$24a=40c$$，则$$3a=5c$$． 所以$$a$$为$$5$$的倍数，$$c$$为$$3$$的倍数． 所以，$$a=0$$或$$5$$，但是，首位不可以是$$0$$，于是$$a=5$$，$$c=3$$； 所以$${{(\\overline{abc})}_{7}}={{(503)}_{7}}=5\\times 49+3=248$$ ． 于是，这个三位数在十进制中为$$248$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1675", "queId": "6e010259f8384fe7a2ce0290529e64b5", "competition_source_list": ["2022年第9届广东深圳鹏程杯四年级竞赛初赛第21题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "2022年$$2$$月$$22$$日被广大网民称为``世界最爱日'',因为这个日期里面包含六个$$2.$$与它包含相同多$$2$$的日期是$$2022$$年$$12$$月$$22$$日，比它包含更多$$2$$的日期则是$$200$$年后的$$2222$$年$$2$$月$$22$$日. 今年$$2$$月$$22$$日又恰好是星期二，而$$12$$月$$22$$日是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}], [{"aoVal": "E", "content": "五 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->求某日期是周几问题"], "answer_analysis": ["无 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "317", "queId": "7b6e6513fd054d61a2e2078ff46c055a", "competition_source_list": ["2021年新希望杯三年级竞赛初赛第13题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "五盘水果排成一排．苹果和橘子相邻，橘子和草莓相邻，苹果和香蕉不相邻，香蕉和芒果不相邻．那么一定和芒果相邻的是． ", "answer_option_list": [[{"aoVal": "A", "content": "只有苹果 "}], [{"aoVal": "B", "content": "只有橘子 "}], [{"aoVal": "C", "content": "只有草莓 "}], [{"aoVal": "D", "content": "香蕉和草莓 "}], [{"aoVal": "E", "content": "橘子和香蕉 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意分析可知，苹果和草莓都与橘子相邻， 所以可以确定橘子在苹果和草莓的中间， 假设苹果、橘子、草莓这样排列， 则香蕉在草莓的右边，芒果在苹果的左边， 即苹果与芒果一定相邻， 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "216", "queId": "4c864fb304a0488e9d1546a2c9a381c2", "competition_source_list": ["2021年新希望杯三年级竞赛初赛第13题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "五盘水果排成一排．苹果和橘子相邻，橘子和草莓相邻，苹果和香蕉不相邻，香蕉和芒果不相邻．那么一定和芒果相邻的是谁？ ", "answer_option_list": [[{"aoVal": "A", "content": "只有苹果 "}], [{"aoVal": "B", "content": "只有橘子 "}], [{"aoVal": "C", "content": "香蕉和草莓 "}], [{"aoVal": "D", "content": "橘子和香蕉 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析", "Overseas Competition->知识点->组合模块->逻辑推理"], "answer_analysis": ["根据题意分析可知，苹果和草莓都与橘子相邻， 所以可以确定橘子在苹果和草莓的中间， 假设苹果、橘子、草莓这样排列， 则香蕉在草莓的右边，芒果在苹果的左边， 即苹果与芒果一定相邻， 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2097", "queId": "f02eea4451b748db87a6a7026ffbb4bd", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在``神庙大逃亡''游戏中，吃一个黄色钱币可以得$$1$$元钱；吃一个红色钱币可以得$$3$$元钱；吃一个蓝色钱币可以得$$5$$元钱。已知阿奇在一次游戏中一共吃了$$2800$$个钱币，共获得$$7800$$元，并且吃到蓝色钱币比红色钱币多$$200$$个，那么阿奇吃到了（ ）个红色钱币。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$700$$ "}], [{"aoVal": "B", "content": "$$900$$ "}], [{"aoVal": "C", "content": "$$1200$$ "}], [{"aoVal": "D", "content": "$$1500$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题"], "answer_analysis": ["解：根据分析，把蓝色钱币比红色钱币多的$$200$$个在总数上减去，可以得到他一共吃了：$$2800-200=260$$个钱币， 共获得：$$7800-5\\times 200=6800$$（元），由于红色蓝色一样多后可以看做有两种钱币，一种1元的黄色钱币， 一种是：$$(3+5)\\div 2=4$$元的红蓝钱币，假设$$2600$$个钱币全部是一元的， 那么可得红蓝钱币一共有：$$(6800-2600\\times 1)\\div \\left( 4-1 \\right)=1400$$（个）， 则红色钱币有：$$1400\\div 2=700$$（个）。 故选:A。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3369", "queId": "737e5582d8c34ca6bce286ce2e8d0609", "competition_source_list": ["2012年第10届创新杯四年级竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "艾丽丝的房间里有三条腿的凳子和四条腿的椅子．它们共有$$17$$条腿，那么艾丽丝的房间里有张三条腿的凳子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["根据整数进行枚举尝试，代入枚举法． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1713", "queId": "6e496b31fa1542a0b52aa13ffbf71a5d", "competition_source_list": ["2007年华杯赛六年级竞赛初赛", "2007年华杯赛五年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "地球表面的陆地面积和海洋面积之比是$$29:71$$，其中陆地的四分之三在北半球，那么南、北半球海洋面积之比是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$284:29$$ "}], [{"aoVal": "B", "content": "$$284:87$$ "}], [{"aoVal": "C", "content": "$$87:29$$ "}], [{"aoVal": "D", "content": "$$171:113$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题->按比分配"], "answer_analysis": ["设地球表面积为$$1$$， 则北半球海洋面积为：$$0.5-0.29\\times \\frac{3}{4}=\\frac{2-0.87}{4}=\\frac{1.13}{4}$$ 南半球海洋面积为：$$0.71-\\frac{1.13}{4}=\\frac{2.84-1.13}{4}=\\frac{1.71}{4}$$ 南北半球海洋面积之比为：$$\\frac{1.71}{4}:\\frac{1.13}{4}=171:113$$ "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "959", "queId": "efeac11d3a3c436fa5736ba7ad3a262d", "competition_source_list": ["2008年四年级竞赛创新杯", "2008年第6届创新杯四年级竞赛复赛第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "有一个77位数，它的各位数字都是1，这个数除以7，余数是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "6 "}], [{"aoVal": "B", "content": "4 "}], [{"aoVal": "C", "content": "2 "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->整除->整除特征->差系整除特征"], "answer_analysis": ["因为$$111111\\div 7=15873$$，所以由六个数字1组成的六位数必定是7的倍数，又77被6除余5从而$$\\underbrace{11\\cdots 1}_{77个1}\\text{=}\\underbrace{1\\cdots 1}_{6个1}\\underbrace{0\\cdots 0}_{71个0}\\text{+}\\underbrace{1\\cdots 1}_{6个1}\\underbrace{0\\cdots 0}_{65个0}\\text{+}\\cdots \\underbrace{1\\cdots 1}_{6个1}\\underbrace{0\\cdots 0}_{5个0}\\text{+}\\underbrace{1\\cdots 1}_{5个1}$$所以$$\\underbrace{11\\cdots 1}_{77个1}$$和$$\\underbrace{1\\cdots 1}_{5个1}$$被7除所得余数相同，而$$\\underbrace{1\\cdots 1}_{5个1}\\div 7\\text{=}1587$$余2，所以$$\\underbrace{11\\cdots 1}_{77个1}$$被7除，余数是2. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "423", "queId": "92ebfb615a98433eb3ac539d1d7691ad", "competition_source_list": ["2016年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "``凑$$24$$点''游戏规则是：从一副扑克牌中抽去大小王剩下$$52$$张（如果初练也可以只用$$1\\sim10$$共$$40$$张牌），任意抽取$$4$$张牌（称牌组），用加、减、乘、除（可加括号）把牌面上的数算成$$24$$，每张牌必须用一次且只能用一次，并不能用几张牌组成一个多位数。如果抽出的牌是$$3$$，$$8$$，$$8$$，$$9$$，那么算式为$$\\left( 9-8 \\right)\\times 8\\times 3$$或$$\\left( 9-8\\div 8 \\right)\\times 3$$等。在下面$$4$$个选项中，唯一无法凑出$$24$$点的是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$，$$2$$，$$2$$，$$3$$ "}], [{"aoVal": "B", "content": "$$1$$，$$4$$，$$6$$，$$7$$ "}], [{"aoVal": "C", "content": "$$1$$，$$5$$，$$5$$，$$5$$ "}], [{"aoVal": "D", "content": "$$3$$，$$3$$，$$7$$，$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->巧填算符"], "answer_analysis": ["解：选项B：$$\\left( 7+1-4 \\right)\\times 6=24$$ 选项C：$$\\left( 5-1\\div 5 \\right)\\times 5=24$$ 选项D：$$\\left( 3+3\\div 7 \\right)\\times 7=24$$ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2723", "queId": "5b3ace009b6d4f48aceb7b6ab6fe2756", "competition_source_list": ["2014年IMAS小学高年级竞赛第一轮检测试题第6题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知每袋面粉售价为$$800$$元、每袋白米售价为$$500$$元，小安花$$3400$$元买了几袋面粉和几袋白米，请问小安买了几袋面粉？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->不定方程->加法不定方程", "Overseas Competition->知识点->计算模块->方程基础->不定方程"], "answer_analysis": ["因为$$800\\times 5=4000\\textgreater3400$$，所以小安买面粉的袋数少于$$5$$袋，并且$$3400$$减去面粉的总价后必须是$$500$$的整倍数，所以小安买的面粉只能是$$3$$袋，购买的白米是$$2$$袋：$$800\\times 3+500\\times 2=3400$$元．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "318", "queId": "606f40394d4b484cb0a3f76b5690214e", "competition_source_list": ["2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（一）"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$\\overline{7x}\\times \\overline{yz6}=41808$$，其中$$x$$、$$y$$、$$z$$ 代表非 0 数字，则$$x+y+z=$$（~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$19$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["我们首先通过乘积的个位判断，乘积的个位取决于因数个位的乘积，那么必有$$6\\times x$$的个位数字是$$8$$，那么$$x=3$$或$$8$$；当$$x=3$$时，那么原式为$$73\\times \\overline{yz6}=41808$$，求得$$\\overline{yz6}\\approx 572.71$$，不符合题意，舍去；当$$x=8$$时，那么原式为$$78\\times \\overline{yz6}=41808$$，求得$$\\overline{yz6}=536$$，符合题意；综上所述，有且仅有$$78\\times 536=41808$$符合题意，那么$$x+y+z=8+5+3=16$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1802", "queId": "85a9cc72f1be4ec3859fe127fa8cb73a", "competition_source_list": ["其它改编自2012年全国希望杯六年级竞赛初赛第19题"], "difficulty": "3", "qtype": "single_choice", "problem": "王老师在黑板上写了若干个从$$1$$开始的连续自然数：$$1$$，$$2$$，$$3$$，$$4$$，$$\\cdots$$，然后擦去三个数（其中有两个质数），如果剩下的数的平均数是$$19\\frac{8}{9}$$，那么王老师在黑板上共写了~\\uline{~~~~~~~~~~}~个数，擦去的两个质数的和最大是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$39$$；$$40$$ "}], [{"aoVal": "B", "content": "$$48$$；$$40$$ "}], [{"aoVal": "C", "content": "$$39$$；$$60$$ "}], [{"aoVal": "D", "content": "$$48$$；$$60$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$1+2+3+\\ldots +n=\\frac{(1+n)n}{2}$$，这几个数的平均数是$$\\frac{(1+n)n}{2}\\div n=\\frac{1+n}{2}$$，因此平均数为$$20$$左右，$$n$$应为$$40$$左右，擦掉$$3$$个数后剩下的个数应为$$9$$的倍数，故$$n=39$$，$$1+2+3+\\ldots +39=780$$．$$39$$个数擦掉$$3$$个数后���$$36$$个，和为$$19\\frac{8}{9}\\times36=716$$，$$780-716=64$$，$$39$$以下质数中两个质数和最大为$$37+23=60$$或$$31+29=60$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1245", "queId": "22f091be73524e44b97d6fa4c3fabad4", "competition_source_list": ["2021年第8届鹏程杯四年级竞赛初赛第16题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "爷爷比爸爸大$$27$$岁，妈妈比小鹏也大$$27$$岁，小鹏一家$$4$$口今年的年龄之和为$$132$$岁，而$$5$$年前是$$113$$岁．则爸爸今年的年龄是岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$29$$ "}], [{"aoVal": "B", "content": "$$32$$ "}], [{"aoVal": "C", "content": "$$35$$ "}], [{"aoVal": "D", "content": "$$38$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["五年前的年龄和是$$113$$，则今年应该是$$113+5\\times4=133$$岁，但是今年的年龄和为$$132$$岁，说明五年前，小鹏还没有出生，今年小鹏$$=5-(133-132)=4$$岁，妈妈的年龄$$=27+4=31$$岁，爷爷和爸爸一共$$132-4-31=97$$岁，根据和差公式，爸爸年龄$$=(97-27)\\div2=35$$岁． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2343", "queId": "743fc60116b54a759afb46fefb97a0a9", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（四）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$3\\odot 4=3\\times 4\\times 5\\times 6=360$$，$$2\\odot 5=2\\times 3\\times 4\\times 5\\times 6=720$$，则$$(3\\odot 5)\\div (2\\odot 4)=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$35$$ "}], [{"aoVal": "D", "content": "$$42$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->定义新运算->直接运算型->普通型"], "answer_analysis": ["$$(3\\odot 5)\\div (2\\odot 4)=(3\\times 4\\times 5\\times 6\\times 7)\\div (2\\times 3\\times 4\\times 5)=6\\times 7\\div 2=21$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "854", "queId": "bb02a21fa16442dca3cb47be8ddf58be", "competition_source_list": ["2013年第14届上海中环杯小学高年级五年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$51$$个连续奇数$$1$$、$$3$$、$$5\\cdots 101$$中选取$$k$$个数，使得他们的和为$$2013$$，那么$$k$$的最大值是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$41$$ "}], [{"aoVal": "B", "content": "$$42$$ "}], [{"aoVal": "C", "content": "$$43$$ "}], [{"aoVal": "D", "content": "$$44$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["要使$$k$$最大，则所选的数最好小．$$1+3+\\cdots +89={{45}^{2}}=2025\\textgreater2013$$，所以所选的数必须少于$$45$$个，而$$44$$个奇数的和为偶数，所以$$k$$的最大值理论上为$$43$$．下面开始构造$$1+3+\\cdots +83={{42}^{2}}=1764$$，$$2013-1764=249$$，$$249-101=148$$，将$$83$$换成$$99$$，和增大$$16$$，$$81$$换成$$97$$，和增大$$16$$，$$148\\div 16=9\\cdots 4$$，所以要替换$$9$$个数，再替换$$1$$个数使其大$$4$$即可．所以，可以选$$1$$至$$63$$、$$69$$，以及$$83$$至$$101$$这$$43$$个数． 所以，$$k$$最大为$$43$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2167", "queId": "1534cd40e0194ff5a4164cc96e6c04ce", "competition_source_list": ["2014年四川成都小升初七中嘉祥外国语学校第26题", "2016年河南郑州联合杯六年级竞赛复赛第15题2分", "六年级上学期其它北师大版53天天练"], "difficulty": "1", "qtype": "single_choice", "problem": "小明从$$A$$地到$$B$$地的平均速度为$$4$$米/秒，然后又从$$B$$地按原路以$$6$$米/秒的速度返回$$A$$地，那么小明在$$A$$地与$$B$$地之间行一个来回的平均速度应为米/秒． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4.2$$ "}], [{"aoVal": "B", "content": "$$4.8$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$5.5$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["平均速度；$$1500\\times 2\\div \\left( 2+2.5 \\right)=3000\\div 4.5=\\frac{2000}{3}\\text{km/h}=666\\frac{2}{3}\\text{km/h}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1156", "queId": "4afa2fde865a44f4a60460e66c2e6a19", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛决赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "红花有$$22$$朵，蓝花比红花多$$16$$朵，蓝花比黄花少$$9$$朵．黄花有朵． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$27$$ "}], [{"aoVal": "C", "content": "$$29$$ "}], [{"aoVal": "D", "content": "$$47$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意分析可知，已知红花的数量是$$22$$朵，蓝花比红花多$$16$$朵，即是比$$22$$多$$16$$，所以蓝花的数量是：$$22+16=38$$（朵）；蓝花比黄花少$$9$$朵也就是黄花比蓝花多$$9$$朵，所以黄花的数量是：$$38+9=47$$（朵）． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "651", "queId": "d53392143e9f40388edba3700315a108", "competition_source_list": ["2005年六年级竞赛创新杯", "2005年第3届创新杯六年级竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "一条公路由$$A$$经$$B$$到$$C$$，已知$$A$$.$$B$$相距$$280$$米，$$B$$.$$C$$相距$$315$$米，现要在路边植树．要求相邻两树间的距离相等，并在$$B$$点及$$AB$$.$$BC$$的中点上都要植一棵，那么两树间距离最多有． ", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$米 "}], [{"aoVal": "B", "content": "$$36$$米 "}], [{"aoVal": "C", "content": "$$17.5$$米 "}], [{"aoVal": "D", "content": "$$18$$米 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公因数与最大公因数->多数的最大公因数"], "answer_analysis": ["依题，在$$140$$米、$$280$$米和$$437$$.$$5$$米处都要植上树，那么两树之间的距离能被$$140$$、$$280$$和$$437.5$$整除，又上述三个数的$$2$$倍的最大公约数为$$\\left( 280\\text{，}560\\text{，}875 \\right)=35$$，那么$$140$$、$$280$$和$$437.5$$这三个数的最大公约数为$$17.5$$，即两树之间的距离最大为$$17.5$$米，选$$C$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1901", "queId": "a9f7e5c08bb24404a2636a06dd01cc0f", "competition_source_list": ["2018~2019学年江西南昌湾里区六年级上学期期中第13题1分", "2008年陈省身杯小学高年级六年级竞赛", "2018~2019学年广东广州黄埔区六年级上学期期末第2题2分", "天津六年级上学期单元测试《第一单元3》"], "difficulty": "1", "qtype": "single_choice", "problem": "两根$$2$$米长的铁丝，一根用去了$$\\frac{1}{3}$$，另一根用去$$\\frac{1}{3}$$米，剩下的铁丝． ", "answer_option_list": [[{"aoVal": "A", "content": "第一根长 "}], [{"aoVal": "B", "content": "第二根长 "}], [{"aoVal": "C", "content": "同样长 "}], [{"aoVal": "D", "content": "无法比较哪根长 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数与运算->数的运算的实际应用（应用题）->分数的简单实际问题"], "answer_analysis": ["第一根减去$\\dfrac{2}{3}m$，剩下$\\dfrac{4}{3}m$；第二根减去$\\dfrac{1}{3}m$，剩下$\\dfrac{5}{3}m$．所以选择$B$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "398", "queId": "b6b60925568e45fead1b8992d780dc4b", "competition_source_list": ["2007年华杯赛五年级竞赛初赛", "2007年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "从和为$$55$$的$$10$$个不同的自然数中，取出$$3$$个数后，余下的数之和是$$55$$的$$\\frac{7}{11}$$，则取出的三个数的积最大等于（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$280$$ "}], [{"aoVal": "B", "content": "$$270$$ "}], [{"aoVal": "C", "content": "$$252$$ "}], [{"aoVal": "D", "content": "$$216$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->和一定型最值问题->多数之积的最值"], "answer_analysis": ["余下的数之和为：$$55\\times \\frac{7}{11}=35$$，取出的数之和为:$$55-35=20$$， 要使取出的三个数之积尽量大，则取出的三个数应尽量接近， 我们知道$$6+7+8=21$$，所以取$$5\\times 7\\times 8=280$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "615", "queId": "5d5e5c616b3d4b0ab6513a36f125e1c9", "competition_source_list": ["2015年美国数学大联盟杯六年级竞赛初赛（中国赛区）第39题5分", "2016年全国美国数学大联盟杯小学高年级六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "做家务是迪的兴趣．迪每$$8$$天进行一次大扫除，每$$11$$天洗一次衣服．如果星期天迪做了两件家务，那么下一次她做两件家务会是哪一天？ ", "answer_option_list": [[{"aoVal": "A", "content": "��期三 "}], [{"aoVal": "B", "content": "星期四 "}], [{"aoVal": "C", "content": "星期五 "}], [{"aoVal": "D", "content": "星期六 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$8\\times 11=88$$，$$88$$除以$$7$$余$$4$$，所以正好是周四． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2954", "queId": "c94f40530b5343b58bb75c395938afd8", "competition_source_list": ["2008年华杯赛六年级竞赛初赛", "2008年华杯赛五年级竞赛初赛", "2008年华杯赛四年级竞赛初赛"], "difficulty": "3", "qtype": "single_choice", "problem": "若$$a=\\underbrace{1515\\cdots 15}_{1004个15}\\times \\underbrace{333\\cdots 3}_{2008个3}$$，则整数a的所有位数上的数字和等于( ). ", "answer_option_list": [[{"aoVal": "A", "content": "18063 "}], [{"aoVal": "B", "content": "18072 "}], [{"aoVal": "C", "content": "18079 "}], [{"aoVal": "D", "content": "18054 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["法一:递推: $$15\\times 33=495$$;$$1515\\times 3333=5049495$$;$$151515\\times 333333=50504949495$$;$$15151515\\times 33333333=505050494949495$$，$$\\cdots \\cdots $$ 我们发现$$\\underbrace{1515\\cdots 15}_{n个1\\text{5}}\\times \\underbrace{\\text{33}\\cdots \\text{3}}_{\\text{2}n个\\text{3}}\\text{=}\\underbrace{\\text{50}\\cdots \\text{50}}_{n-1个50}\\underbrace{4949\\cdots 49}_{n个49}5$$ 则$$\\underbrace{1515\\cdots 15}_{1004个1\\text{5}}\\times \\underbrace{\\text{33}\\cdots \\text{3}}_{2008个\\text{3}}\\text{=}\\underbrace{\\text{50}\\cdots \\text{50}}_{1003个50}\\underbrace{4949\\cdots 49}_{1004个49}5$$ 则数字和为$$\\left( 5+0 \\right)\\times 1003+\\left( 4+9 \\right)\\times 1004+5=18072$$ 法二:凑整 $$\\underbrace{1515\\cdots 15}_{1004个15}\\times \\underbrace{333\\cdots 3}_{2008个3}$$ $$\\text{=}\\underbrace{5050\\cdots 50}_{1003个50}5\\times 3\\times \\underbrace{333\\cdots 3}_{2008个3}$$ $$\\text{=}\\underbrace{5050\\cdots 50}_{1003个50}5\\times \\underbrace{999\\cdots 9}_{2008个9}$$ $$=\\underbrace{5050\\cdots 50}_{1003个50}5\\times \\left( 1\\underbrace{000\\cdots 0}_{2008个0}-1 \\right)$$ $$\\text{=}\\underbrace{5050\\cdots 50}_{1003个50}5\\underbrace{000\\cdots 0}_{2008个0}-\\underbrace{5050\\cdots 50}_{1003个50}5$$ $$\\text{=}\\underbrace{5050\\cdots 50}_{1003个50}\\underbrace{4949\\cdots 49}_{1004个49}5$$ 则数字和为$$\\left( 5+0 \\right)\\times 1003+\\left( 4+9 \\right)\\times 1004+5=18072$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3263", "queId": "b535c95982d84bb4bd28e2bfc9840ac6", "competition_source_list": ["2018年第8届北京学而思综合能力诊断一年级竞赛年度教学质量第14题"], "difficulty": "1", "qtype": "single_choice", "problem": "【例4】薇儿今天要去公园玩，妈妈给薇儿准备了$$3$$件上衣、$$2$$条裙子，还有$$2$$顶帽子，薇儿出门必须挑选一件上衣和一条裙子，帽子可以戴也可以不戴．请问，薇儿一共有种不同的搭配方法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$种 "}], [{"aoVal": "B", "content": "$$8$$种 "}], [{"aoVal": "C", "content": "$$12$$种 "}], [{"aoVal": "D", "content": "$$18$$种 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->乘法原理->物品搭配（有特殊要求）"], "answer_analysis": ["帽子可以戴也可以不带，那么在帽子上面就有$$3$$种选择，所以共有$$3\\times 2\\times 3=18$$（种）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1850", "queId": "b2807a2eb6c44470981e8e48c233dbcf", "competition_source_list": ["2017年全国小学生数学学习能力测评六年级竞赛初赛第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "开学前$$6$$天，小明还没做寒假数学作业，而小强已完成了$$60$$道题．开学时，两人都完成了数学作业，在这$$6$$天中，小明做的题量是小强的$$3$$倍，小明平均每天做了道题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为由题干可知， 开学前$$6$$天，小明还没做寒假数学作业， 小强已完成了$$60$$道题，在这$$6$$天中， 小明做的题量是小强的$$3$$倍， 设小明平均每天做了$$x$$道题， 则小强平均每天做$$\\frac{x}{3}$$道题，可列方程为：$$60+\\frac{6x}{3}=6x$$， 解得$$x=15$$， 所以小明平均每天做了$$15$$道题， 故本题答案为$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2035", "queId": "caefc83b4bfe4a1c9fd7eab8577a69ca", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第5题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "小红今年$$7$$岁，她比爸爸小$$29$$岁．去年她比爸爸小岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$ "}], [{"aoVal": "B", "content": "$$29$$ "}], [{"aoVal": "C", "content": "$$22$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["在年龄问题中，小红与爸爸的年龄差不会变．所以去年小红仍然比爸爸小$$29$$岁． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1711", "queId": "7bc7d34a1f66472ea7e2de4f40c87d18", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小黑兔有$$10$$个萝卜，如果小白兔给小黑兔$$4$$个萝卜，它俩的萝卜就一样多，小白兔原来有个萝卜． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["小白兔给小黑兔四个萝卜，小白兔的萝卜减四个，小黑兔的萝卜加四个，它俩的萝卜一样多，说明小白兔开始的时候比小黑兔多八个萝卜，小黑兔有$$10$$个萝卜，则小白兔有$$10+8=18$$（个）萝卜． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1990", "queId": "ab0982ec3a6548188836e8be28e6c319", "competition_source_list": ["2017年IMAS小学高年级竞赛（第一轮）第17题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在一次考试中，某班的平均分数为$$70$$分，其中有两位学生缺考得了$$0$$分．若这两位学生成绩不计，则该班上其他学生的平均分数为$$74$$分，请问这个班上总共有位学生． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$35$$ "}], [{"aoVal": "E", "content": "$$37$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型"], "answer_analysis": ["由于把两位得$$0$$分的学生算进来时，平均分数由$$74$$分变为$$70$$分， 故可视为其他每一位学生都拿出$$74-70=4$$（分）给这两位学生， 此时共拿出$$70\\times 2=140$$（分）， 即可判断出其他学生共有$$140\\div 4=35$$（位）， 因此这个班上共有$$35+2=37$$（位）学生． 故选$$\\text{E}$$． ", "设这个班上共有$$x$$位学生，
\n由题意可得$$74\\left( x-2 \\right)=70x$$，
\n解得$$x=37$$．
\n故选$$\\text{E}$$．
"], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3333", "queId": "ff80808147248448014724de17e80129", "competition_source_list": ["2012年全国华杯赛小学高年级竞赛初赛网络版第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "在由$$1$$，$$3$$，$$4$$，$$7$$，$$9$$组成的没有重复数字的数中，是$$9$$的倍数的有（~ ~ ~ ~）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["数字和是$$9$$的倍数才可以，只有$$9$$．所以只有$$1$$个． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2133", "queId": "a68cb77b84de461eb1e3909320fbc217", "competition_source_list": ["2008年第6届创新杯五年级竞赛初赛B卷第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两辆汽车同时从$$A$$地开往$$B$$地，速度分别是每小时走$$42$$千米和$$38$$千米，甲车到达$$B$$地后立即返回，在距$$B$$地$$20$$千米的地方与乙车相遇，$$A$$、$$B$$两地相距千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$360$$ "}], [{"aoVal": "B", "content": "$$390$$ "}], [{"aoVal": "C", "content": "$$385$$ "}], [{"aoVal": "D", "content": "$$400$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["设$$A$$、$$B$$两地相距$$x$$千米． $$\\begin{eqnarray} \\left( x+20 \\right)\\div 42\\&=\\&\\left( x-20 \\right)\\div 38 \\left( x+20 \\right)\\div 42\\times 38\\&=\\&x-20 \\left( x+20 \\right)\\times 38\\&=\\&\\left( x-20 \\right)\\times 42 38x+760\\&=\\&42x-840 38x+760+840\\&=\\&42x 38x+1600\\&=\\&42x 4x\\&=\\&1600 x\\&=\\&400．\\end{eqnarray}$$ 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "279", "queId": "90eb70ff536d4e5cad2bc3bc4be8bf4a", "competition_source_list": ["2014年IMAS小学高年级竞赛第一轮检测试题第12题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若一个非负整数的五分之一与另一个非负整数的三分之一的和是$$1$$，请问这两个数之和最大可能值是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["设这两个非负整数分别为$$a$$和$$b$$，即$$\\frac{a}{5}+\\frac{b}{3}=1$$，因此，$$b$$只能取$$0$$、$$1$$、$$2$$、$$3$$． 若$$b=0$$，则$$a=5$$；若$$b=1$$，则$$a=\\frac{10}{3}$$不是整数；若$$b=2$$，则$$a=\\frac{5}{3}$$不是整数； 若$$b=3$$，则$$a=0$$．所以$$a+b$$只有两个可能的值：$$5+0=5$$、$$0+3=3$$，最大值为$$5$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1082", "queId": "212a7d0e3e874dafb014c4e420688c65", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（四）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "某公司准备进行促销，奖品总金额为$$2$$万元，中奖率为$$10 \\% $$，每$$200$$元发奖券一张，获奖的总人次为$$500$$，如此促销则公司的销售总额至少为元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$万 "}], [{"aoVal": "B", "content": "$$12$$万 "}], [{"aoVal": "C", "content": "$$80$$万 "}], [{"aoVal": "D", "content": "$$100$$万 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->量率对应求单位1"], "answer_analysis": ["$$500\\div 10 \\% \\times 200=100$$（万元） "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "184", "queId": "9942b2a9aefc46759d72e8523121f325", "competition_source_list": ["2021年新希望杯三年级竞赛初赛第13题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "五盘水果排成一排．苹果和橘子相邻，橘子和草莓相邻，苹果和香蕉不相邻，香蕉和芒果不相邻．那么一定和芒果相邻的是． ", "answer_option_list": [[{"aoVal": "A", "content": "只有苹果 "}], [{"aoVal": "B", "content": "只有橘子 "}], [{"aoVal": "C", "content": "只有草莓 "}], [{"aoVal": "D", "content": "香蕉和草莓 "}], [{"aoVal": "E", "content": "橘子和香蕉 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据题意分析可知，苹果和草莓都与橘子相邻， 所以可以确定橘子在苹果和草莓的中间， 假设苹果、橘子、草莓这样排列， 则香蕉在草莓的右边，芒果在苹果的左边， 即苹果与芒果一定相邻， 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "341", "queId": "ed4a49f5bda442a4bf050d9c2bf1ba38", "competition_source_list": ["2016年全国华杯赛小学中年级竞赛在线模拟第3题", "2013年全国华杯赛小学中年级竞赛初赛A卷第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "小东、小西、小南、小北四个小朋友在一起做游戏时，捡到了一条红领巾，交给了老师．老师问是谁捡到的？小东说不是小西；小西说是小南；小南说小东说的不对；小北说小南说的也不对．他们之中只有一个人说对了，这个人是． ", "answer_option_list": [[{"aoVal": "A", "content": "小东 "}], [{"aoVal": "B", "content": "小西 "}], [{"aoVal": "C", "content": "小南 "}], [{"aoVal": "D", "content": "小北 "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾"], "answer_analysis": ["首先发现矛盾，小东和小南的话存在矛盾，如果小南对，则小东不对因此小南和小东一真一假．其次，小北支持小东，如果小北对，则小南不对，小东对，因此小北和小东同真同假．而只有一个人说对了，那么小北和小东都为假，则小南为真，小南说对了，那么由小东说错可知小西捡到红领巾． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1727", "queId": "ff808081498992ec01498ee74d2b0ba4", "competition_source_list": ["2014年全国华杯赛小学高年级竞赛初赛A卷第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "某次考试有$$50$$道试题，答对一道题得$$3$$分，答错一道题扣$$1$$分，不答题不得分，小龙得分$$120$$分，那么小龙最多答对了（~ ~ ~ ~）道试题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$42$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$50$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["因为要求小龙最多答对了几道题，又因为小龙最后得$$120$$分每对一道得$$3$$分，所以小龙错的题目数是$$3$$的倍数，所以答对与答错题目总数也是$$3$$的倍数且最大为$$48$$道， 所以 $$\\left { \\begin{matrix}\\& 3x-y=120 \\&x+y=48 \\end{matrix} \\right.$$ 解得， $$\\left { \\begin{matrix}\\& x=42 \\& y=6 \\end{matrix} \\right.$$ 所以选择B． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "546", "queId": "feb42727cd2b4103827e146107325319", "competition_source_list": ["2020年第24届YMO六年级竞赛决赛第9题3分", "2019年第24届YMO六年级竞赛决赛第9题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "在下面的四个算式中，最大的得数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2017\\times 2022+2022$$ "}], [{"aoVal": "B", "content": "$$2018\\times 2021+2021$$ "}], [{"aoVal": "C", "content": "$$2019\\times 2020+2020$$ "}], [{"aoVal": "D", "content": "$$2020\\times 2019+2019$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$2017\\times 2022+2022=2022\\times 2018$$， $$2018\\times 2021+2021=2021\\times 2019$$， $$2019\\times 2020+2020=2020\\times 2020$$， $$2020\\times 2019+2019=2019\\times 2021$$， 发现每组两个数之和都为$$4040$$，由最值原理，和一定，差小积大，可知$$\\text{C}$$最大． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1969", "queId": "ceb8312330b44da2b31e06b53b38e725", "competition_source_list": ["2009年第7届创新杯六年级竞赛初赛第7题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "五个互不相等的奇数之和等于$$85$$，其中最大的一个为$$M$$，则$$M$$的取值范围是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$23\\leqslant M\\leqslant 67$$ "}], [{"aoVal": "B", "content": "$$19\\leqslant M\\leqslant 67$$ "}], [{"aoVal": "C", "content": "$$21\\leqslant M\\leqslant 69$$ "}], [{"aoVal": "D", "content": "$$17\\leqslant M\\leqslant 69$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["如果这$$5$$个数比较接近，那么根据平均数$$85\\div5=17$$， 可以知道最大数的最小值是$$17+2+2=21$$， 当其他四个数较小时，最大数的最大值就是$$85-1-3-5+7=69$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1548", "queId": "9573b5c8c78440849f5254affce5f534", "competition_source_list": ["2017年IMAS小学中年级竞赛（第一轮）第14题4分", "2017年IMAS小学中年级竞赛（第一轮）第14题4分", "2017年IMAS小学中年级竞赛（第一轮）第14题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明前几次数学考试的平均分是$$88$$分，这次数学考试结束后，小明努力学习，想把平均分提高，已知小明这次考试成绩为$$98$$分，平均分恰好达到$90$分，请问小明总共考了次数学考试. ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->容斥求平均数"], "answer_analysis": ["可知这次考试比所欲达到的平均分数多的$$98-90=8$$分要分配到前几次的考试才能使平均分数增加$$90-88=2$$分，因此小明之前共考了$$8\\div 2=4$$次数学考试，所以连同这一次他总共考了$$4+1=5$$次数学考试．故选$$\\text{C}$$． 可知最后一次考试要比所欲达到的平均分数多$$98-88=10$$分才能使平均分数从$$88$$分增加到$$90$$分，即须增加$$2$$分，因此连同这一次小明总共考了$$10\\div 2=5$$次数学考试．故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2619", "queId": "87e3148c09464389aff9d6c64ba794a2", "competition_source_list": ["2017年第4届广东深圳鹏程杯小学高年级竞赛第14题15分"], "difficulty": "3", "qtype": "single_choice", "problem": "$$1977$$除本身外有三个因数���$$1$$，$$3$$，$$659$$（注意：$$659$$是个质数，也叫素数）．$$1977$$的数字和为$$1+9+7+7=24$$，它除本身外所有因数的数字和也是$$1+3+(6+5+9)=24$$，我们把具有这种特点的数叫鹏程数，以下是$$8000$$以内的鹏程数： $$6$$，$$33$$，$$87$$，~\\uline{~~~~~~~~~~}~，$$303$$，$$519$$，$$573$$，$$681$$，$$843$$，$$951$$，$$1059$$，$$1329$$，$$1383$$，$$1923$$，$$1977$$，$$2463$$，$$2733$$，$$2787$$，$$2949$$，$$3057$$，$$3273$$，$$3327$$，$$3543$$，$$3651$$，$$3867$$，$$3921$$，$$4083$$，$$4353$$，$$4677$$，$$5163$$，$$5433$$，$$5703$$，$$5919$$，$$6081$$，$$6243$$，$$6297$$，$$6621$$，$$6891$$，$$7053$$，$$7323$$，$$7377$$，$$7647$$，$$7971$$．那么上面的鹏程数中缺少的数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$178$$ "}], [{"aoVal": "B", "content": "$$198$$ "}], [{"aoVal": "C", "content": "$$218$$ "}], [{"aoVal": "D", "content": "$$249$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数"], "answer_analysis": ["249的因数：1,3,83,249．$$1+3+8+3=15$$；$$2+4+9=15$$．所以$$249$$也是鹏程数． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2609", "queId": "8b8c15951689440fa04fa237541621d1", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "求三个分数$$\\frac{20122012}{20132013}$$，$$\\frac{20132013}{20142014}$$，$$\\frac{20142014}{20152015}$$中值最大的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{20122012}{20132013}$$ "}], [{"aoVal": "B", "content": "$$\\frac{20132013}{20142014}$$ "}], [{"aoVal": "C", "content": "$$\\frac{20142014}{20152015}$$ "}], [{"aoVal": "D", "content": "三个一样大 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["因为$$\\frac{20122012}{20132013}=\\frac{2012\\times 1001}{2013\\times 1001}=\\frac{2012}{2013}$$， $$\\frac{20132013}{20142014}=\\frac{2013\\times 1001}{2014\\times 1001}=\\frac{2013}{2014}$$，$$\\frac{20142014}{20152015}=\\frac{2014\\times 1001}{2015\\times 1001}=\\frac{2014}{2015}$$，$$1-\\frac{2012}{2013}=\\frac{1}{2013}$$，$$1-\\frac{2013}{2014}=\\frac{1}{2014}$$， $$1-\\frac{2014}{2015}=\\frac{1}{2015}$$． 因为$$\\frac{1}{2015}\\textless{}\\frac{1}{2014}\\textless{}\\frac{1}{2013}$$，所以$$\\frac{2014}{2015}\\textgreater\\frac{2013}{2014}\\textgreater\\frac{2012}{2013}$$．因此，三个分数中，最大的是$$\\frac{20142014}{20152015}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "920", "queId": "bc351b0111ba443982d5c813ac8afe9a", "competition_source_list": ["2017年第13届湖北武汉新希望杯五年级竞赛决赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "下列说法正确的是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "互质的两个数没有公因数~~~~~~~~~~~~~~~~~~~~ "}], [{"aoVal": "B", "content": "$$12$$和$$18$$的最小公倍数是$$72$$ "}], [{"aoVal": "C", "content": "$$24$$和$$30$$的最大公因数是$$6$$~~~~ "}], [{"aoVal": "D", "content": "两个质数的和一定是偶数 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$\\text{A}$$选项：互质的两个数公因数为$$1$$，$$\\text{B}$$选项：$$12$$和$$18$$的最小公倍数是$$36$$，$$\\text{D}$$选项：$$2+3=5$$为奇数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3128", "queId": "ec24207534c94c7c8719f8a2329fceb4", "competition_source_list": ["2017年北京学而思杯六年级竞赛年度教学质量监测第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "艾迪有一本书，他原计划每天看$$10$$页，$$10$$天看完．现在他已经看了$$5$$天，平均每天看$$8$$页．如果艾迪想按原计划$$10$$天把书看完，那么从现在开始他应该平均每天看~\\uline{~~~~~~~~~~}~页． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->公式类->直接求平均数"], "answer_analysis": ["这一本书一共有$$10\\times 10=100$$页，还剩下$$100-5\\times 8=60$$页要看，所以之后平均每天看$$60\\div 5=12$$页才能按原计划． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2006", "queId": "cf1deff52bd44aa6bf30e03dbdf35f15", "competition_source_list": ["2008年第6届创新杯四年级竞赛初赛A卷第2题5分", "2008年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "小华买1支钢笔和2支圆珠笔共用15元钱，小红买同样的钢笔2支和圆珠笔1支共用21元钱.那么，每支钢笔的价格是( )元. ", "answer_option_list": [[{"aoVal": "A", "content": "8 "}], [{"aoVal": "B", "content": "9 "}], [{"aoVal": "C", "content": "10 "}], [{"aoVal": "D", "content": "11 "}]], "knowledge_point_routes": ["拓展思维->能力->符号代换->代数运算"], "answer_analysis": ["3支钢笔和3支圆珠笔共用$$15+21=36$$元，从而，每支钢笔和每支圆珠笔共用$$36\\div 3=12$$元，所以，每支钢笔的价格是$$21-12=9$$元 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2920", "queId": "861170ebe9cd4f7fb7c3135d2defe865", "competition_source_list": ["2018年第12届北京学而思综合能力诊断小学中年级三年级竞赛第14题"], "difficulty": "3", "qtype": "single_choice", "problem": "某个等差数列第$$8$$项是$$50$$，前$$10$$项的和为$$325$$．那么，在这个数列前$$99$$项中有~\\uline{~~~~~~~~~~}~个奇数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$33$$ "}], [{"aoVal": "C", "content": "$$49$$ "}], [{"aoVal": "D", "content": "$$50$$ "}], [{"aoVal": "E", "content": "$$99$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->计数模块->几何计数->分类枚举法数图形->常规图形枚举计数", "拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列与估算"], "answer_analysis": ["根据题意： $$\\begin{cases}{{a}_{1}}+7d=50 ({{a}_{1}}+{{a}_{1}}+9d)\\times 5=325 \\end{cases}$$ 整理得：$$\\begin{cases}{{a}_{1}}+7d=50 2{{a}_{1}}+9d=65 \\end{cases}$$ 解得：$$\\begin{cases}{{a}_{1}}=1 d=7 \\end{cases}$$ 因此：该数列首项为奇数，公差为奇数， 故数列中奇数项为奇数，偶数项为偶数． 前$$99$$项中共有奇数$$(99+1)\\div 2=50$$（个）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2885", "queId": "9fce1926296541e79443793bf75e5feb", "competition_source_list": ["2004年第2届创新杯六年级竞赛复赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "\\textbf{2.~} 在下面四个算式中，得数最大的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( \\frac{1}{17}-\\frac{1}{19} \\right)\\div 20$$ "}], [{"aoVal": "B", "content": "$$\\left( \\frac{1}{15}-\\frac{1}{21} \\right)\\div 60$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{13}-\\frac{1}{23} \\right)\\div 100$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{1}{11}-\\frac{1}{25} \\right)\\div 140$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$\\left( \\frac{1}{17}-\\frac{1}{19} \\right)\\div 20=\\frac{2}{17\\times 19}\\div 20=\\frac{1}{17\\times 19}\\times \\frac{1}{10}$$;$$\\left( \\frac{1}{15}-\\frac{1}{21} \\right)\\div 60=\\frac{6}{15\\times 21}\\div 60=\\frac{1}{15\\times 21}\\times \\frac{1}{10}$$; $$\\left( \\frac{1}{13}-\\frac{1}{23} \\right)\\div 100=\\frac{10}{13\\times 23}\\div 100=\\frac{1}{13\\times 23}\\times \\frac{1}{10};$$$$\\left( \\frac{1}{11}-\\frac{1}{25} \\right)\\div 140=\\frac{14}{11\\times 25}\\div 140=\\frac{1}{11\\times 25}\\times \\frac{1}{10};$$ 只需比较$$\\frac{1}{17\\times 19}$$，$$\\frac{1}{15\\times 21}$$，$$\\frac{1}{13\\times 23}$$，$$\\frac{1}{11\\times 25}$$的大小，根据和一定，两数越接近乘 积越大，则$$11\\times 25 \\textless{} 13\\times 23 \\textless{} 15\\times 21 \\textless{} 17\\times 19$$，那么 $$\\frac{1}{11\\times 25}\\textgreater\\frac{1}{13\\times 23}\\textgreater\\frac{1}{15\\times 21}\\textgreater\\frac{1}{17\\times 19}$$，所以答案为$$D$$ "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2036", "queId": "d883f5fb32344c20976ad51577b90dfa", "competition_source_list": ["2017年四川成都六年级竞赛“全能明星”选拔赛第3题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一根$$4$$米长的钢材，先截掉总长的$$\\frac{1}{4}$$，再截掉剩余长度的$$\\frac{1}{4}$$，还剩． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$米 "}], [{"aoVal": "B", "content": "$$3$$米 "}], [{"aoVal": "C", "content": "$$2\\frac{3}{4}$$米 "}], [{"aoVal": "D", "content": "$$2\\frac{1}{4}$$米 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->量率对应已知单位1"], "answer_analysis": ["$$4\\times \\left( 1-\\frac{1}{4} \\right)\\times \\left( 1-\\frac{1}{4} \\right)=2\\frac{1}{4}$$（米）． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1814", "queId": "73ef6c642e754ee4a59bcd70bb4f89d7", "competition_source_list": ["2005年五年级竞��创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "小华期末考试，数学、语文、英语三科的平均分是92分，语文、英语两科的平均分是90分;又英语比语文高3分，那么数学比语文高( ). ", "answer_option_list": [[{"aoVal": "A", "content": "8分 "}], [{"aoVal": "B", "content": "7.5分 "}], [{"aoVal": "C", "content": "7分 "}], [{"aoVal": "D", "content": "6.5分 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->公式类"], "answer_analysis": ["数学、语文、英语三科的总成绩是$$92\\times 3=276$$分，语文、英语两科的总成绩是$$90\\times 2=180$$分，那么数学的得分为$$276-180=96$$分.又英语比语文高3分，那么语文的成绩为$$\\left( 180-3 \\right)\\div 2=88.5$$分，所以数学比语文高$$96-88.5=7.5$$分. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "915", "queId": "8f448b90f8094596bbc6faf9d3106718", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "完全平方数是指可以分解成两个相同自然数乘积的数．在整数$$1$$、$$2$$、$$3$$、$$\\cdots \\cdots 99$$、$$100$$中，合数的个数为$$x$$，偶数的个数为$$y$$，完全平方数的个数为$$z$$．则$$x+y+z$$等于． ", "answer_option_list": [[{"aoVal": "A", "content": "$$85$$ "}], [{"aoVal": "B", "content": "$$130$$ "}], [{"aoVal": "C", "content": "$$134$$ "}], [{"aoVal": "D", "content": "$$135$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["合数是指在大于$$1$$的整数中除了能被$$1$$和本身整除外，还能被其他数（$$0$$除外）整除的数．与之相对的是质数，$$1$$到$$100$$中，合数的个数有：$$74$$个；所以$$x=74$$； 偶数有$$50$$个，所以$$y=50$$； 完全平方数分别是：$$1$$、$$4$$、$$9$$、$$16$$、$$25$$、$$36$$、$$49$$、$$64$$、$$81$$、$$100$$这$$10$$个；即$$z=10$$； $$x+y+z=74+50+10=134$$；所以选择$$\\text{C}$$选项． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "30", "queId": "036895656b96400ab97e9fb5f7f833e7", "competition_source_list": ["2015年第20届全国华杯赛小学中年级四年级竞赛初赛A卷", "2016年第20届四川成都华杯赛小学中年级竞赛B卷第1~6题60分"], "difficulty": "1", "qtype": "single_choice", "problem": "森林里举行比赛，要派出狮子、老虎、豹子和大象中的两个动物去参加．如果派狮子去，那么也要派老虎去；如果不派豹子去，那么也不能派老虎去；要是豹子参加的话，大象可不愿意去．那么，最后能去参加比赛的是． ", "answer_option_list": [[{"aoVal": "A", "content": "狮子、老虎 "}], [{"aoVal": "B", "content": "老虎、豹子 "}], [{"aoVal": "C", "content": "狮子、豹子 "}], [{"aoVal": "D", "content": "老虎、大象 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->连线法"], "answer_analysis": ["在逻辑推理中，原命题成立，则逆否命题也成立. 从题意出发： （1）狮子去则老虎去，逆否命题；老虎不去则狮子也不去 （2）不派豹子则不派老虎，逆否命题：派老虎则要派豹子 （3）派豹子则大象不愿意去，逆否命题；大象去则不能派豹子 从（2）出发可以看出答案为$$B$$ 题目要求有两个动物去，可以使用假设法，若狮子去，则老虎去，老虎去则豹子也去，三个动物去，矛盾，所以狮子不去，若豹子不去则老虎不去，那么只有大象去，矛盾，所以豹子去，豹子去则大象不去，由两骄气去得到结论，老虎要去，所以答案是$$B$$，豹子和老虎去. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3200", "queId": "390b26f1dc2f4f82b8cc05c495dec35c", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第14题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "由数字$$0$$，$$1$$，$$2$$，$$3$$，$$4$$组成三位数，可以组成个不同的三位数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$80$$ "}], [{"aoVal": "D", "content": "$$100$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->组数问题->一般组数问题"], "answer_analysis": ["$4\\times5\\times5=100（个）$ "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1238", "queId": "9da2808a8b494e73a262c6cb6af6e0c3", "competition_source_list": ["其它改编自 2014��全国华杯赛小学高年级竞赛复赛A卷第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一杯子装满了浓度为$$16 \\%$$的盐水．有大、中、小铁球各一个，它们的体积比为$$10:4:3$$．首先将小球沉入盐水杯中，结果盐水溢出$$10 \\%$$，取出小球；其次把中球沉入盐水杯中，又将它取出；接着将大球沉入盐水杯中后取出；最后在杯中倒入纯水至杯满为止．此时杯中盐水的浓度是~\\uline{~~~~~~~~~~}~$$ \\%$$．（保留一位小数） ", "answer_option_list": [[{"aoVal": "A", "content": "$$5.3$$ "}], [{"aoVal": "B", "content": "$$10.7$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$5.5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数的运算->估算->百分数的简单实际问题->百分数浓度问题", "拓展思维->知识点->应用题模块->浓度问题->浓度基本题型->已知溶质溶液求浓度"], "answer_analysis": ["大、中、小球体积比为$$10:4:3$$，盐水的$$10 \\%$$对应小球``$$3$$份''体积，则大球``$$10$$份''体积对应盐水的$$10 \\%\\div 3\\times 10=\\frac{1}{3}$$，因此最终溢出的盐水量为杯子容积的$$\\frac{1}{3}$$，此时杯中盐水的浓度为$$16 \\%\\times (1-\\frac{1}{3})\\div 1\\approx 10.7 \\%$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "298", "queId": "527dbf2eaa2f4621a275ff83233b9952", "competition_source_list": ["2015年第14届春蕾杯一年级竞赛初赛第5题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知学而思有女老师：简简，柚子，嘉嘉；男老师有：小南，橙子，小易。 (1) 喜欢点珍珠奶茶外卖的是所有女老师和小南老师； (2) 学而思喜欢点外卖的老师不喜欢臭豆腐； (3) 小易老师最不喜欢的就是臭豆腐了； 那么，聪明的你知道今天中午的臭豆腐外卖最有可能是谁点的吗？ ", "answer_option_list": [[{"aoVal": "A", "content": "小南 "}], [{"aoVal": "B", "content": "橙子 "}], [{"aoVal": "C", "content": "简简 "}], [{"aoVal": "D", "content": "柚子 "}], [{"aoVal": "E", "content": "嘉嘉 "}], [{"aoVal": "F", "content": "小易 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据题意分析可知，小胖比小王高，所以小胖的身高高于小王，小胖说比园园矮，所以园园的身高高于小胖，小朱没有小王高，所以小王的身高高于小朱，由此可知，四个人的身高从高到低位：园园$$\\textgreater$$小胖$$\\textgreater$$小王$$\\textgreater$$小朱． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3471", "queId": "d929b788a5ae4b40bd4fa473010a6d62", "competition_source_list": ["2008年中环杯六年级竞赛决赛", "2008年中环杯五年级竞赛决赛", "2008年第9届中环杯五年级竞赛决赛第8题", "2008年中环杯四年级竞赛决赛"], "difficulty": "3", "qtype": "single_choice", "problem": "有$$49$$个孩子，每人胸前有一个号码，号码从$$1$$到$$49$$各不相同.~ 请你挑出若干个小孩排成一个圆圈，使任 何相邻两个孩子号码数乘积小于100.~ 你最多能够选出个孩子. ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->排列组合->组合->组合的基本运算"], "answer_analysis": ["任意两位数乘积都小于$$100$$，所以相邻两个数必须有一个$$1$$位数，$$1$$位数共有$$9$$个，所以两位数最多也只能挑$$9$$个，因此最多能挑$$18$$个孩子. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "859", "queId": "b68db2d07b634e9cbdcd79e4d2b347b4", "competition_source_list": ["2022年广东深圳华杯赛小学高年级竞赛（惠州市）第6题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$1\\leqslant n\\leqslant 100$$，且$$8n+1$$为完全平方数，则符合条件的整数$$n$$的个数为~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$13$$ "}], [{"aoVal": "E", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析", "Overseas Competition->知识点->数论模块"], "answer_analysis": ["易知$$8n+1$$只能为奇数的平方，设$$8n+1={{\\left( 2l+1 \\right)}^{2}}$$， 其中$$l$$为非负整数，则$$n=\\frac{l\\left( l+1 \\right)}{2}$$， 所以$$1\\leqslant \\frac{l\\left( l+1 \\right)}{2}\\leqslant 100$$， 故$$1\\leqslant l\\leqslant 13$$，所以，满足条件的整数$$n$$有$$13$$个． ∵$$1\\leqslant n\\leqslant 100$$， ∴$$9\\leqslant 8n+1\\leqslant 801$$， ∴$$n$$为如下几个整数时， $$1$$，$$3$$，$$6$$，$$10$$，$$15$$，$$21$$，$$28$$，$$36$$，$$45$$，$$55$$，$$66$$，$$78$$，$$91$$， $$8n+1$$为完全平方数，整数$$n$$的个数为$$13$$． 故答案为：$$13$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2596", "queId": "432913240d4a4026a699b3890b52052e", "competition_source_list": ["2004年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "在下面四个算式中，得数最大的是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( \\frac{1}{17}-\\frac{1}{19} \\right)\\div 20$$ "}], [{"aoVal": "B", "content": "$$\\left( \\frac{1}{15}-\\frac{1}{21} \\right)\\div 60$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{13}-\\frac{1}{23} \\right)\\div 100$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{1}{11}-\\frac{1}{25} \\right)\\div 140$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->整数比较大小"], "answer_analysis": ["$$A=\\frac{1}{17\\times 19\\times 10}$$，$$B=\\frac{1}{15\\times 21\\times 10}$$，$$C=\\frac{1}{13\\times 23\\times 10}$$，$$D=\\frac{1}{11\\times 25\\times 10}$$，显然最大的为$$D$$，选$$D$$. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1716", "queId": "5812a1fe732e456286483580b67c7a57", "competition_source_list": ["2019年陕西延安宝塔区北大培文学校小升初入学真卷第2题3分", "2006年第11届全国华杯赛竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "奶奶告诉小明$$2006$$年共有$$53$$个星期日，聪敏的小明立到告诉奶奶：$$2007$$年的元旦一定是． ", "answer_option_list": [[{"aoVal": "A", "content": "星期一 "}], [{"aoVal": "B", "content": "星期二 "}], [{"aoVal": "C", "content": "星期六 "}], [{"aoVal": "D", "content": "星期日 "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$2006$$年有$$365$$天，而$$365=7\\times 52+1$$，又已知$$2006$$年有$$53$$个星期天，只能元旦是星期天，且$$12$$月$$31$$日也是星期日，所以，$$2007$$年月的元旦是星期一． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2713", "queId": "3c607ca6222142aaafcbffaf260938c3", "competition_source_list": ["2006年第4届创新杯四年级竞赛复赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "某数的小数点向右移一位，则小数值比原来大$$25.65$$，原小数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2.565$$ "}], [{"aoVal": "B", "content": "$$2.56$$ "}], [{"aoVal": "C", "content": "$$2.855$$ "}], [{"aoVal": "D", "content": "$$2.85$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["某数的小数点向右移动一位，相当于此数扩大了$$10$$倍，原数是$$1$$份数，现在的数就是$$10$$份数，现在的数比原数大$$9$$份数，再根据这个数就比原来大$$25.65$$， $$25.65\\div (10-1)=25.65\\div9=2.85$$． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2345", "queId": "059b7f9fbac64ea3aa75d25d85e0835f", "competition_source_list": ["2007年六年级竞赛创新杯", "2007年第5届创新杯六年级竞赛第8题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "某次数学考试共$$5$$道题，全班$$52$$人参加，共做对$$181$$题，已知每人至少做对$$1$$题，做对$$1$$道题的有$$7$$人，做对$$2$$道题的人和做对$$3$$道题的人一样多，做对$$5$$道题的有$$6$$人，那么做对$$4$$道题的人数是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "$$29$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$33$$ "}], [{"aoVal": "D", "content": "$$35$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->多元一次方程组->整数系数方程组"], "answer_analysis": ["解法一:做对两道、三道、四道题的人共有$$52-7-6=39$$人，他们共做对了$$181-1\\times 7-5\\times 6=144$$题.由于做对两道题和三道题的人一样多，我们将他们都看做是做对$$2.5$$道题的人，若这$$39$$人全部都做对$$2.5$$道题，共做对$$2.5\\times 39=97.5$$题，比$$144$$道题少了$$144-97.5=46.5$$道题，这时将做对四道题的同学不看作做对$$2.5$$道题的结果，于是做对四道题的学生有$$46.5\\div \\left( 4-2.5 \\right)=31$$人. 解法二:设做对四道题的人有$$x$$人，做对二道、三道的人各有$$y$$人，则 $$\\begin{cases}x+2y+7+6=52 4x+\\left( 2+3 \\right)y+7+30=181 \\end{cases}$$ 化简得到$$\\begin{cases}x+2y=39 4x+5y=144 \\end{cases}$$， 解得$$x=31$$. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1344", "queId": "28905983bb934ad897e52d7e4a219b5d", "competition_source_list": ["2005年第3届希望杯四年级竞赛复赛第13题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2005$$年$$4$$月$$10$$日是星期日，则$$2005$$年$$6$$月$$1$$日是星期~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->求某日期是周几问题"], "answer_analysis": ["$$4$$月$$10$$日到$$4$$月$$30$$日经过了$$20$$天，$$5$$月有$$31$$天， 再到$$6$$月$$1$$日又经过$$1$$天； 共经过：$$20+31+1=52$$（天）， $$52\\div 7=7$$（周）$$\\cdots \\cdots3$$（天）； 即$$6$$月$$1$$日是星期三． 故答案为：三． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2272", "queId": "6a551c45fb724bbcbcc9780f4d303b83", "competition_source_list": ["2011年四年级竞赛创新杯", "2011年第9届创新杯四年级竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "一辆小汽车每秒行驶$$20$$米，刚驶入隧道时，发现一辆客车正在前面$$180$$米处向前行驶。如果两车速度保持不变，$$90$$秒后两车同时驶出隧道，那么客车每秒行驶米。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->两人相遇与追及问题->相遇问题->同时出发相向而行"], "answer_analysis": ["速度差：$$180\\div 90=2$$（米/秒）， 客车的速度：$$20-2=18$$（米/秒）。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1884", "queId": "c4f8c6f742a04757b872d54877151cc9", "competition_source_list": ["2020年广东广州海珠区广州为明学校卓越杯六年级竞赛初赛第1题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "丽人服装厂去年产值$$240$$万元，比前年增加$$\\frac{1}{5}$$，丽人服装厂这两年产值共有多少万元？应列式为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$240+240\\times \\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$240+240\\times \\left( 1+\\frac{1}{5} \\right)$$ "}], [{"aoVal": "C", "content": "$$240+240\\div \\left( 1+\\frac{1}{5} \\right)$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["丽人服装厂去年产值$$240$$万元， 比前年增加$$\\frac{1}{5}$$，则这两个年产值共有$$240\\div \\left( 1+\\frac{1}{5} \\right)+240$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1463", "queId": "951cb9be42a046d597f2af252b42cf6f", "competition_source_list": ["2020年希望杯五年级竞赛模拟第11题", "2020年新希望杯五年级竞赛第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "大头儿子给小头爸爸计算每天的交通费用．大头儿子说：``您的车百公里油耗为$$8$$升，每天上下班共行驶$$30$$公里，油价按每升$$7.2$$元计算，您每天上下班需要支付的油费为元．'' ", "answer_option_list": [[{"aoVal": "A", "content": "$$27$$ "}], [{"aoVal": "B", "content": "$$17.28$$ "}], [{"aoVal": "C", "content": "$$1.92$$ "}], [{"aoVal": "D", "content": "$$19.2$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为由题干可知，车百公里油耗为$$8$$升，每天上下班共行驶$$30$$公里，则上下班共耗油$$30\\div 100\\times 8=2.4$$（升），油价按每升$$7.2$$元计算，所以每天上下班需要支付的油费为$$7.2\\times 2.4=17.28$$（元）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "865", "queId": "6f4e09ecca6347a59937d0c217da77a0", "competition_source_list": ["2013年IMAS小学高年级竞赛第一轮检测试题第4题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "请问下列哪一个数可被$$6$$整除？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$98$$ "}], [{"aoVal": "B", "content": "$$163$$ "}], [{"aoVal": "C", "content": "$$192$$ "}], [{"aoVal": "D", "content": "$$212$$ "}], [{"aoVal": "E", "content": "$$254$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["被$$6$$整除的数同时被$$2$$和$$3$$整除． 被$$2$$整除的数的末位是偶数，所以$$\\text{B}$$不合； 被$$3$$整除的数之数码和必须被$$3$$整除，$$9+8=17$$、$$1+9+2=12$$、$$2+1+2=5$$、$$2+5+4=11$$故只有$$\\text{C}$$选项符合要求． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3179", "queId": "41cb700127e54f6b80dac5b35f56ce46", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "十位数字和个位数字相加，和是$$11$$的两位数有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["根据题意分析可知，$$11=2+9=3+8=4+7=5+6$$， 所以这样的两位数有：$$29$$；$$92$$；$$38$$；$$83$$；$$47$$；$$74$$；$$56$$；$$65$$一共$$8$$个． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2690", "queId": "6860a388dae34471a7114cffc3104f36", "competition_source_list": ["2017年四川成都六年级竞赛“全能明星”选拔赛第2题2分", "2019年山东青岛市南区山东省青岛育才中学小升初第22题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$\\frac{3}{5}:4=x:5$$时，$$x$$的值是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "C", "content": "$$2\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$3\\frac{3}{4}$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["$$4x=\\frac{3}{5}\\times 5$$，则$$x=\\frac{3}{4}$$． 利用比例基本性质，内项积等于外项积解方程即可，$$x=\\frac{3}{4}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2284", "queId": "f2522cf04fa84e41bac8282e472dc63f", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第9题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "北京、天津相距$$140$$千米，客车和货车同时从北京出发驶向天津，客车每小时行$$70$$千米，货车每小时行$$50$$千米，客车到达天津后停留$$15$$分钟，又以原速度返回北京．则两车首次相遇的地点距离北京千米．（结果保留整数） ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$122$$ "}], [{"aoVal": "C", "content": "$$123$$ "}], [{"aoVal": "D", "content": "$$124$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["客车从北京到天津需要$$140\\div70=2$$（时）， 当客车正要返回时，货车已开出 $$2+15\\div60=2.25$$（时）， 行驶了$$50\\times2.25=112.5$$（千米）， 此时两车相距$$140-112.5=27.5$$（千米）， 还要$$27.5 \\div(70+50)= \\frac{11}{48}$$（小时）才能相遇． 从而相遇地点与天津的距离是客车行$$\\frac{11}{48}$$小时， 所走过的路程为$$70 \\times \\frac{11}{48}\\approx 16$$（千米）． 所以，相遇地点与北京的距离是 $$140-16=124$$（千米）． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1088", "queId": "3837d62febcf4254b4c6f35cff425528", "competition_source_list": ["2013年全国希望杯六年级竞赛初赛第21题"], "difficulty": "1", "qtype": "single_choice", "problem": "小红整理零钱包时发现，包中有面值为$$1$$分、$$2$$分、$$5$$分的硬币共$$25$$枚，总值为$$0.60$$元，则$$5$$分的硬币最多有枚． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设$$1$$、$$2$$、$$5$$分硬币分别有$$xyz$$枚，由题意可知$$\\left { \\begin{align}\\& x+y+z=25 \\& x+2y+5z=60 \\end{align}\\right.$$，两式相减可得$$y+4z=35$$，由此可得$$z$$最多有$$8$$枚． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2625", "queId": "75863f5a92ff4ec0afa00012ce59a6ca", "competition_source_list": ["2013年IMAS小学高年级竞赛第一轮检测试题第2题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "请问下列哪一项��间最接近一天的时间？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.9$$天 "}], [{"aoVal": "B", "content": "$$1.2$$天 "}], [{"aoVal": "C", "content": "$$23$$小时 "}], [{"aoVal": "D", "content": "$$26$$小时 "}], [{"aoVal": "E", "content": "$$1410$$分钟 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->时间问题->时间计算"], "answer_analysis": ["我们首先把所有时间都化成以分钟为单位．一天$$=1440$$分钟，$$0.9$$天$$=1296$$分钟，$$1.2$$天$$=1728$$分钟，$$23$$小时$$=1380$$分钟，$$26$$小时$$=1560$$分钟．而： $$1440-1296=144$$ $$1728-1440=288$$ $$1440-1380=60$$ $$1560-1440=120$$ $$1440-1410=30$$ 所以$$1410$$分钟最接近于一天的时间．故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1422", "queId": "55bdd385b8474795aef19c4bafa1897b", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一种特殊的计算器，当输入一个$$10$ $49$$的自然数后，计算器会先将这个数乘$$2$$，然后将所得结果的十位和个位顺序颠倒，再加$$2$$后显示出最后的结果．那么，下列四个选项中，可能是最后显示的结果． ", "answer_option_list": [[{"aoVal": "A", "content": "$$44$$ "}], [{"aoVal": "B", "content": "$$43$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$41$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->还原问题->逆运算"], "answer_analysis": ["倒推．$$44$$ 对应的是$$44-2=42$$，颠倒后是$$24$$，除以$$2$$ 为$$12$$．符合条件．其他的均不符合条件． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1293", "queId": "2c45fe1695d64306a2f12a8662713fef", "competition_source_list": ["2017年河南郑州联合杯竞赛附加赛第一场第6题2分"], "difficulty": "0", "qtype": "single_choice", "problem": "如果规定电梯运行时上升为``$$+$$''，那么电梯运行``$$-10$$米''表示（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "电梯上升$$10$$米 "}], [{"aoVal": "B", "content": "电梯下降$$10$$米 "}], [{"aoVal": "C", "content": "电梯上升$$0$$米 "}], [{"aoVal": "D", "content": "电梯没有动 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["用正负数来表示具有相反意义的两种量：上升记为正，则下降记为负，所以``$$-10$$米''表示下降$$10$$米． 故选$$\\rm B$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "770", "queId": "ff8080814518d5240145192b34440567", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙二人进行下面的游戏．二人先约定一个整数$$N$$，然后由甲开始，轮流把$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$，$$8$$，$$9$$这九个数字之一填入下面任一方格中：$$\\square$$$$\\square$$$$\\square$$$$\\square$$$$\\square$$$$\\square$$，每一方格只填入一个数字，形成一个数字可以重复的六位数．若这个六位数能被$$N$$整除，乙胜；否则甲胜．当$$N$$小于$$15$$时，使得乙有必胜策略的$$N$$有（~~~~~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["若$$N$$是偶数，甲只需第一次在个位填个奇数，乙必败只需考虑$$N$$是奇数． $$N=1$$，显然乙必胜． $$N=3 9$$，乙只需配数字和$$1-8$$，$$2-7$$，$$3-6$$，$$4-5$$，$$9-9$$即可． $$N=5$$，甲在个位填不是$$5$$的数，乙必败． $$N=7 11 13$$，乙只需配成$$\\overline{abcabc}=\\overline{abc}\\times1001=\\overline{abc}\\times 7\\times 11\\times 13$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1402", "queId": "4383941a7192484b93d8b7b5dd4747ac", "competition_source_list": ["2014年迎春杯四年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "两根同样长的绳子，第一根平均剪成$$4$$段，第二根平均剪成$$6$$段，已知第一根剪成的每段长度与第二根剪成的每段长度相差$$2$$米，那么，原来两根绳子的长度之和是（ ）米。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$48$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->差倍->两量差倍->暗倍型二量差倍问题"], "answer_analysis": ["解：第二根绳子的长度为： $$(2\\times 4)\\div \\left( 6-4 \\right)\\times 6$$ $$=8\\div 2\\times 6$$ $$=4\\times 6$$ $$=24$$（米） 原来两根绳子的长度之和是： $$24\\times 2=48$$（米） 答：原来两根绳子的长度之和是$$48$$米。 故选：D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "536", "queId": "f9a02f36568e41d0a723e3841c4f2453", "competition_source_list": ["2015年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "现在从甲、乙、丙、丁四个人中选出两个人参加一项活动。规定：如果甲去，那么乙也去；如果丙不去，那么乙也不去；如果丙去，那么丁不去。最后去参加活动的两个人是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "甲、乙 "}], [{"aoVal": "B", "content": "乙、丙 "}], [{"aoVal": "C", "content": "甲、丙 "}], [{"aoVal": "D", "content": "乙、丁 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->比较型逻辑推理", "课内体系->知识模块->综合与实践"], "answer_analysis": ["解：根据``如果甲去，那么乙也去''可得：甲在，乙必然也在； 又根据``如果丙不去，那么乙也不去''可得：如果乙去了，丙也一定去了； ``如果丙去；那么丁不去''可得：如果丁去；那么丙不去，同时乙也不去。则根据``甲去，那么乙也去''可得：甲也不去，这样只有丁去，这与两个人参加一项活动相矛盾。 同时满足条件只能是乙、丙参加了活动。 故选：$$B$$。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1483", "queId": "48c84e7fd19e47bfb111a531a93c6903", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "幼儿园老师给小朋友分糖果，如果每个小朋友分$$4$$颗糖果，则多出$$28$$颗糖果；如果有$$4$$个小朋友每人分$$6$$颗，$$6$$个小朋友每人分$$4$$颗，其余的分$$5$$颗，则正好分完．那么原来有颗糖果． ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$140$$ "}], [{"aoVal": "C", "content": "$$148$$ "}], [{"aoVal": "D", "content": "$$150$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->盈亏问题->盈亏基本类型->盈亏基本类型盈亏问题"], "answer_analysis": ["本题的关键是用假设法求出每人分$$5$$颗糖果时少的糖果数，再根据（盈$$+$$亏）$$\\div $$两次分的糖果差$$=$$人数，求出人数． 假设第二次分糖果时每个小朋友都分$$5$$颗，则少$$(5-4)\\times 6-(6-5)\\times 4=2$$（颗），此题则变为每个小朋友多分$$5-4=1$$（颗）糖，则需要$$28+2=30$$（颗）糖，据此可求出小朋友人数，进而可求出糖果数． $$(5-4)\\times 6-(6-5)\\times 4$$ $$=1\\times 6-1\\times 4$$ $$=2$$（颗）， $$(28+2)\\div (5-4)$$ $$=30\\div 1$$ $$=30$$（个）， $$30\\times 4+28$$ $$=120+28$$ $$=148$$（颗）． 答：一共有$$30$$个小朋友和$$148$$颗糖果． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "154", "queId": "35051a9e6e514d83a4d55a7a77811b10", "competition_source_list": ["2008年第6届创新杯五年级竞赛初赛B卷第8题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$1$$克，$$2$$克、$$4$$克、$$8$$克的砝码各一个，从这四个砝码中每次任选$$2$$个砝码使用，能称种不同的重量（砝码也可以放在天平的两边）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["从$$1$$、$$2$$、$$4$$、$$8$$克中任取两个，其和为$$3$$、$$5$$、$$9$$、$$6$$、$$10$$、$$12$$克，从$$1$$、$$2$$、$$4$$、$$8$$克任取两个，大的减去小的，其差为$$1$$、$$3$$、$$7$$、$$2$$、$$6$$、$$4$$克，又因为和与差中的$$3$$克、$$6$$克重复，所以可称出$$6+6-2=10$$ （种）不同重量． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1456", "queId": "b0b9f617eca9416d933226aa37cc0126", "competition_source_list": ["2018年第21届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "王朋的语文、数学、英语三科成绩的平均分是$$90$$分，其中语文、数学两科的���均是$$95$$分，他的英语考了分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$70$$ "}], [{"aoVal": "B", "content": "$$80$$ "}], [{"aoVal": "C", "content": "$$90$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["（语文$$+$$数学$$+$$英语）$$\\div 3=90$$（分）， 语文$$+$$数学$$+$$英语$$=270$$（分）， （语文$$+$$数学）$$\\div 2=95$$（分）， 语文$$+$$数学$$=190$$（分）， 英语：$$270-190=80$$（分）， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1887", "queId": "9c533e62db1d4cbaa3ac419229258703", "competition_source_list": ["2017年第20届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小芳、小王、小刘三个人共有画片$$90$$张，如果小王向小芳借$$10$$张后，又借给小刘$$8$$张，结果三个人的画片张数正好相等．那么下列表述正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "小芳的画片比小王多$$12$$张 "}], [{"aoVal": "B", "content": "小芳的画片比小刘多$$12$$张 "}], [{"aoVal": "C", "content": "小芳的画片比小王少$$12$$张 "}], [{"aoVal": "D", "content": "小王的画片比小刘少$$6$$张 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["因为$$90\\div 3=30$$（张），可得此时每人手里有$$30$$张画片， $$30+8-10=28$$（张），故小王有$$28$$张画片， $$30+10=40$$（张），故小芳有$$40$$张画片， $$30-8=22$$（张），故小刘有$$22$$张画片． 故$$\\text{A}$$：小芳比小王多$$40-28=12$$张画片，$$\\text{A}$$正确，$$\\text{C}$$错误； $$\\text{B}$$：小芳比小刘多$$40-22=18$$张画片，故$$\\text{B}$$错误； $$\\text{D}$$：小王的画片比小刘多，故$$\\text{D}$$错误． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1070", "queId": "25a67a1716a2425289e11ba9b2a97f31", "competition_source_list": ["2017年河南郑州联合杯小学高年级六年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "一根铜线长$$21$$厘米，一根铝线长$$16$$厘米，把这两根金属线剪掉同样长，使剩下的铜线长度恰好是铝线长度的$$2$$倍，问各剪去多少厘米？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["铜线比铝线长$$21-16=5\\left(\\text{cm} \\right)$$，剪掉同样长的一段后，铜线还是比铝线长$$5\\text{cm}$$，剩下的铝线长度：$$5\\div \\left( 2-1 \\right)=5\\left(\\text{cm} \\right)$$，剪去的长度：$$16-5=11\\left(\\text{cm} \\right)$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1603", "queId": "88c9d0647a4648f3b3ddde57dfccc94e", "competition_source_list": ["2020年长江杯五年级竞赛复赛B卷第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "小明拿来一根$$1$$米长的绳子，他从绳子的一端开始每隔$$4$$厘米染一个红点，然后他又从绳子的另一端开始，每隔$$5$$厘米染一个黑点，最后，小明用剪刀沿染有点的地方剪断．问这根绳子被剪成了段． ", "answer_option_list": [[{"aoVal": "A", "content": "$$45$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$39$$ "}], [{"aoVal": "D", "content": "$$41$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都没有->剪绳子"], "answer_analysis": ["从一端开始每隔$$4$$厘米做一个记号， 有记号：$$\\frac{100}{4}-1=24$$， 从一端开始每隔$$5$$厘米做一个记号， 有记号：$$\\frac{100}{5}-1=19$$ ， $$4$$和$$5$$的最小公倍数为$$20$$， 所以每隔$$20$$厘米处的记号重合，有记号：$$\\frac{100}{20}-1=4$$， 一共有记号：$$24+19-4=39$$， 有记号的地方切断，绳子共剪成$$39+1=40$$段， 答：绳子共剪成$$40$$段． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2114", "queId": "e75ac34e997a45b0a822bd87bda40160", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（二）"], "difficulty": "0", "qtype": "single_choice", "problem": "一个数，它减去$$2$$，然后除以$$2$$，再加上$$2$$，最后乘$$2$$，得到的结果是$$2014$$．这个数原来是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2011$$ "}], [{"aoVal": "B", "content": "$$2012$$ "}], [{"aoVal": "C", "content": "$$2013$$ "}], [{"aoVal": "D", "content": "$$2014$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->还原问题->单一变量还原问题"], "answer_analysis": ["$$( 2014\\div 22\\div2 )\\times 2+2=2012$$. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1486", "queId": "b570f15082564ca39a789cc3bdbd8d63", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第5题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "$$20$$个同学排成一队做操，从左边数小文排在第$$12$$个，从右边数小文排在第个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["小文的左边有人：$$12-1=11$$（人）， 所以从右边数小文排在：$$20-11=9$$（个）． 故选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3070", "queId": "e18c40d8b5b940098204434be7f82fe7", "competition_source_list": ["2019年福建泉州鲤城区泉州师范附属小学三年级竞赛模拟第17题"], "difficulty": "2", "qtype": "single_choice", "problem": "一条大街上原有路灯$$201$$盏，相邻两盏路灯相距$$50$$米；现在换新路灯增加了$$50$$盏，相邻两盏路灯的距离是~\\uline{~~~~~~~~~~}~米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$35$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$50$$ "}], [{"aoVal": "E", "content": "$$30$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->计算模块", "拓展思维->思想->对应思想"], "answer_analysis": ["先利用原有的路灯盏数和间隔长度， 求出这条大街的总长度是$$\\left( 201-1 \\right)\\times 50=10000$$（米），换新路灯后，一共有路灯$$201+50=251$$盏，此时的间隔数是$$251-1=250$$，由此即可求出$$1$$个间隔的长度是$$10000\\div 250=40$$（米）． $$\\left( 201-1 \\right)\\times 5\\div \\left( 201+50-1 \\right)=200\\times 50\\div 250=40$$（米）． 所以相邻的两盏路灯的距离是$$40$$米． 故答案为：$$40$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "282", "queId": "ff8080814502fa24014507b673810ba4", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "一只大熊猫从$$A$$地往$$B$$地运送竹子，他每次可以运送$$50$$根，但是他从$$A$$地走到$$B$$地和从$$B$$地返回$$A$$地都要吃$$5$$根，$$A$$地现在有$$200$$根竹子，那么大熊猫最多可以运到$$B$$地（~~~~~~~ ）根． ", "answer_option_list": [[{"aoVal": "A", "content": "$$150$$ "}], [{"aoVal": "B", "content": "$$155$$ "}], [{"aoVal": "C", "content": "$$160$$ "}], [{"aoVal": "D", "content": "$$165$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["运四次，去四次回三次，吃掉了$$5\\times (4+3)=35$$根，则最多可以运到$$B$$地$$200-35=165$$根． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2878", "queId": "ad605ea9b8e3432cb6e92654391027ed", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛初赛第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "根据``三角形两边之和大于第三边''的知识，解答本题： 有不同长度的七条线段，其长度均为整数厘米，最短的是$$1$$厘米，最长的是$$21$$厘米，其中以任何三条线段作``边''都不能组成一个三角形，那么这七条线段中第二长的线段长厘米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->数列与数表->数列规律"], "answer_analysis": ["斐波那契数串从第三项开始，等于前两项的和，即$$1$$、$$2$$、$$3$$、$$5$$、$$8$$、$$13$$、$$21$$． 即第二长的为$$13$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "113", "queId": "66dae4df3e2e41e1b3e4e42aabc3a525", "competition_source_list": ["2004年五年级竞赛创新杯", "2004年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "在某次足球比赛中，每个球队需要打6场比赛，其得分规则是:赢一场得5分，输一场得0分，平一场得2分.某��共得14分，则该球队( ) ", "answer_option_list": [[{"aoVal": "A", "content": "至多赢一场 "}], [{"aoVal": "B", "content": "至少赢三场 "}], [{"aoVal": "C", "content": "至少平三场 "}], [{"aoVal": "D", "content": "赢两场 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->体育比赛->2-0 积分制"], "answer_analysis": ["该球队6场比赛得分为14分，所以该队不可能一场未胜也不可能胜3场，也不可能胜1场(因为$$14-5=9$$分，剩下的比赛不可能得9分)，那么该队一定是胜两场.$$14-5\\times 2=4$$分，$$4\\div 2=2$$场，那么该队还平两场，输两场，选D. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "93", "queId": "111da7821b194f4a93a0f9c02b11dfe3", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）"], "difficulty": "3", "qtype": "single_choice", "problem": "有一天，彭老师和陈老师约好去打乒乓球，结果彭老师以$$4:0$$完虐陈老师．乒乓球比赛为$$11$$分制，即每局$$11$$分，$$7$$局$$4$$胜制，打成$$10:10$$后必须净胜而且只能净胜$$2$$分．经计算，彭老师四局的总得分为$$48$$分，陈老师总得分为$$39$$分，且每一局比赛分差不超过$$3$$分，则一共有种情况．（不考虑这四局比分之间的顺序） ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->逻辑推理->体育比赛->单循环赛"], "answer_analysis": ["每周比赛要分出胜负分差必须在$$2$$分或以上，题中又给出每局比赛分差不超过$$3$$分，故每局比赛的分差只有两种可能：差$$2$$分或$$3$$分．且分差为$$3$$分的那局彭老师得分为$$11$$分，总分差为$$48-39=9$$分，故必有$$3$$场分差为$$2$$分，另一场分差为$$3$$分；即有一场的比分为$$118$$，另两场的总比分为$$37：31$$，有以下四种情况：①$$11:9$$，$$11:9$$，$$15:13$$②$$11:9$$，$$12:10$$，$$14:12$$③$$11:9$$，$$13:11$$，$$13:11$$④$$12:10$$，$$12:10$$，$$13:11$$．故一共有$$4$$种情况． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "719", "queId": "c316ae1840f34fb0961a94c9270ebbd0", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一个长方体的长、宽、高都是整数厘米，它的体积是$$1998$$立方厘米，那么它的长、宽、高的和的最小可能值厘米？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$66$$ "}], [{"aoVal": "B", "content": "$$48$$ "}], [{"aoVal": "C", "content": "$$58$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数->分解质因数（式）"], "answer_analysis": ["我们知道任意个已确定个数的数的乘积一定时，它们相互越接近，和越小．如$$3$$个数的积为$$18$$，则三个数为$$2$$、$$3$$、$$3$$时和最小，为$$8.1998=2\\times 3\\times 3\\times 3\\times 37$$，$$37$$是质数，不能再分解，所以$$2\\times 3\\times 3\\times 3$$对应的两个数应越接近越好．有$$2\\times 3\\times 3\\times 3=6\\times 9$$时，即$$1998=6\\times 9\\times 37$$时，这三个自然数最接近．它们的和为$$6+9+37=52($$厘米)． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3088", "queId": "f408772f25e546f7a9cb245bf9ed0fde", "competition_source_list": ["2008年四年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "计算$$20082008\\times 2007-20072007\\times 2008=$$（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "2007 "}], [{"aoVal": "B", "content": "2008 "}], [{"aoVal": "C", "content": "0 "}], [{"aoVal": "D", "content": "20072008 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之凑整法->整数基准数法"], "answer_analysis": ["原式$$=2008\\times 10001\\times 2007-2007\\times 10001\\times 2008=0$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3264", "queId": "abfd7c7ed8584123a55e75972a4f24e6", "competition_source_list": ["2010年第8届创新杯六年级竞赛初赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$4$$名同学约定去上网，现只有$$3$$台电脑，只好有两个同学上同一台电脑，则共有种不同的上网方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$64$$ "}], [{"aoVal": "B", "content": "$$81$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$72$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["捆绑法，$$\\text{C}_{4}^{2}\\times \\text{A}_{3}^{3}=6\\times 6=36$$． ", "设三台电脑为①②③，四名同学为$$A$$，$$B$$，$$C$$，$$D$$，
\n因为组合数有如下三种：①$$1+1+2$$②$$1+2+1$$③$$2+1+1$$，每一种不同排列数为：$$3\\times 2\\times 2\\times 1=12$$（种），
\n所以不同上网方式有$$12\\times 3=36$$（种）．
\n故选$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "234", "queId": "c32b199060704e80849469a28af5d433", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级六年级竞赛邀请赛训练题（三）"], "difficulty": "2", "qtype": "single_choice", "problem": "有一个$$80$$人的旅游团，其中男$$50$$人，女$$30$$人，他们住的旅馆有$$11$$人、$$7$$人、$$5$$人的三种房间，男女分住不同的房间（所住房间不允许有空床位），他们至少要住个房间． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["题目求房间的最小值，所以$$11$$人的房间要尽量多，并且不能所房间不能有空位，所以还需要其他两种房间来调整． 男$$50$$人，安排$$3$$个$$11$$人房间，$$1$$个$$7$$人房间，$$2$$个$$5$$人房间或$$2$$个$$11$$人房间，$$4$$个$$7$$人房间，共$$6$$个房间； 女$$30$$人，安排$$1$$个$$11$$人房间，$$2$$个$$7$$人房间，$$1$$个$$5$$人房间，共$$4$$个房间，$$6+4=10$$个房间． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2202", "queId": "28b58b861bf84a3895b6f0a7f8149b06", "competition_source_list": ["2016年第28届广东广州五羊杯小学高年级竞赛初赛第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "钟面上$$3$$时分时，分针与时针所成角最小． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->时钟问题->时钟上的追及问题->已知时间求角度"], "answer_analysis": ["$$3$$点时，分针和时针夹角为$$3\\times 30{}^{}\\circ =90{}^{}\\circ $$， 则$$90{}^{}\\circ \\div \\left( 6{}^{}\\circ -0.5{}^{}\\circ \\right)=\\frac{180}{11}=16\\frac{4}{11}\\min $$时， 夹角为$$0{}^{}\\circ $$，而$$0\\textless{}\\frac{4}{11}\\textless{}\\frac{1}{2}$$， 故$$16\\min $$时，分针和时针夹角最小，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1241", "queId": "50016047c0a8419fb1de2ad1b4cc2db2", "competition_source_list": ["2017年全国亚太杯四年级竞赛初赛第23题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$ and $$B$$ are $$40$$ kilometers apart. Two people, Alice and Bob, start from $$A$$ and $$B$$ at the same time and travel in opposite directions, meeting after $$8$$ hours. If two people leave $${A}$$ at the same time and go to $${B}$$, Alice is $$5$$ kilometers ahead of Bob after $$5$$ hours. Alice travels kilometers per hour. $$A$$，$$B$$两地相距$$40$$千米．甲、乙两人，同时分别由$$A$$，$$B$$两地出发，相向而行，$$8$$小时后相遇．如果两人同时由$${A}$$地出发前往$${B}$$地，$$5$$小时后甲在乙前方$$5$$千米处．则甲每小时行千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["设甲的速度为$${a}$$千米/时，乙的速度为$${b}$$千米/时， 所以$$\\left( {a}+{b} \\right)\\times 8=40$$ 从而推出$${a}+{b=}5$$． $$\\left( {a}-{b} \\right)\\times 5=5$$从而推出$${a}-{b=}1$$． 根据和差公式$${a=}\\left( 5+1 \\right)\\div 2=3$$． 甲每小时行$$3$$千米． 速度和为$$40\\div 8=5$$（千米$$/$$小时）， 速度差为$$5\\div 5=1$$（千米$$/$$小时）， 甲速度为$$\\left( 5+1 \\right)\\div 2=3$$（千米$$/$$小时）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "325", "queId": "b16abe893e0d4eaba11864aff555006c", "competition_source_list": ["2015年全国美国数学大联盟杯小学高年级五年级竞赛初赛第32题"], "difficulty": "2", "qtype": "single_choice", "problem": "赛斯在周一下午 $$4$$ 点吃了第一个苹果，然后间隔了 $$100$$ 个小时吃第二个苹果．请问他吃第二个苹果是在周五的什么时候？ ", "answer_option_list": [[{"aoVal": "A", "content": "中午 "}], [{"aoVal": "B", "content": "下午 $$4$$ 点 "}], [{"aoVal": "C", "content": "下午 $$8$$ 点 "}], [{"aoVal": "D", "content": "午夜 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["赛斯在周一下午 $$4$$ 点吃了第一个苹果，然后间隔了 $$100$$ 个小时吃第二个苹果．请问他吃第二个苹果是在周五的下午 $$8$$ 点． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2523", "queId": "34c5d343f6ab4065aa0ef14302b41139", "competition_source_list": ["2017年北京学而思杯小学中年级三年级竞赛年度教学质量测评第19题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "按规律填数，那么横线上填入的数是． $$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，~\\uline{~~~~~~~~~~}~， $$34$$，$$55$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$23$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->斐波那契数列（兔子数列）"], "answer_analysis": ["斐波那契数串：从第$$3$$个数开始，每个数都等于它前两个数之和． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1046", "queId": "4a8aa5c902144d999110d1f5deb5b2dc", "competition_source_list": ["2017年河南郑州联合杯六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "含盐$$30 \\% $$的盐水中，加入$$5$$克盐和$$10$$克水，此时盐水含盐百分比是：（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "大于$$30 \\% $$ "}], [{"aoVal": "B", "content": "等于$$30 \\% $$ "}], [{"aoVal": "C", "content": "小于$$30 \\% $$ "}], [{"aoVal": "D", "content": "无法比较 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["加入的盐水浓度：$$\\frac{5}{5+10}=\\frac{5}{15}=\\frac{1}{3}$$，$$\\frac{1}{3}\\textgreater30 \\% $$，则盐水含盐率会变大，即大于$$30 \\% $$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "864", "queId": "b20fe77c5e3b44beb6ba49852f227961", "competition_source_list": ["2013年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2+2\\times 3+2\\times 3\\times 3+\\cdots +2\\times \\underbrace{3\\times 3\\times \\cdots \\times 3}_{9\\text{个3}}$$的结果的个位数字是( ) ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->完全平方数->平方数的尾数特征"], "answer_analysis": ["解：依题意可知： $$2\\times 3=6$$ $$2\\times 3\\times 3$$尾数是$$8$$。 $$2\\times 3\\times 3\\times 3$$尾数是$$4$$。 $$2\\times 3\\times 3\\times 3\\times 3$$尾数是$$2$$。 发现尾数的数字规律是$$2,6,8,4,2,6,8,4,2,6$$。 $$2+6+8+4+2+6+8+4+2+6$$尾数是$$2+6=8$$。 故选：B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1533", "queId": "d5d471d0d1f44ffc98a8e6ed18136e50", "competition_source_list": ["2011年世界少年奥林匹克数学竞赛六年级竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小宇参加奥数竞赛抢答赛，抢答试题共有$$10$$道，每答对一题得$$8$$分，答错一题倒扣$$5$$分．小宇最终得$$41$$分，他做对~\\uline{~~~~~~~~~~}~题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$37$$ "}], [{"aoVal": "C", "content": "$$7$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["本题是鸡兔同笼问题的变式应用． 解决此类问题一般都用假设法，做对一题得$$8$$分，相当于``兔''，做错一题倒扣$$5$$分，相当于``鸡''，通过适当的替换进行解答． 假设$$10$$道题全部做对，则得分为$$10\\times 8=80$$（分），而答错一题比答对一题少得$$8+5=13$$（分），从题目中可以求出小宇一共少得$$80-41=39$$（分），由此可以求出他做错了$$39\\div 13=3$$（道）． $$(10\\times 8-41)\\div (48+5)$$ $$=39\\div 13$$ $$=3$$（道）， $$10-3=7$$（道）． 故答案为：$$7$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1703", "queId": "7bbaec27388a42c3ae437ab836cbee12", "competition_source_list": ["2014年全国华杯赛小学中年级竞赛决赛第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "王一老师共买了$$53$$支铅笔，分给了$$A$$，$$B$$，$$C$$，$$D$$四个同学，分到最多的与最少的铅笔数相差不到$$5$$支，如果$$B$$把分到的铅笔全都给$$A$$，那么$$A$$的铅笔数是$$C$$的$$2$$倍；如果$$B$$把分到的铅笔全都给$$C$$，那么$$C$$的铅笔数是$$D$$的$$2$$倍，由此可知，$$B$$分到~\\uline{~~~~~~~~~~}~支铅笔． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["设$$A$$，$$B$$，$$C$$，$$D$$分到的铅笔数分别是$$A$$，$$B$$，$$C$$，$$D$$，由$$B+C=2D$$，知$$C$$、$$D$$、$$B$$依次成等差数列，设公差为$$K$$；由$$A+B=2C$$，知$$A$$、$$C$$、$$B$$依次成等差数列，则公差为$$2K$$；由$$4$$人铅笔数相差不会超过$$4$$，所以$$K=0$$或$$1$$；若$$K=0$$，则$$4\\times B=53$$，但$$53$$不是$$4$$的整数倍； 若$$K=1$$，$$A\\textgreater C\\textgreater D\\textgreater B$$，则$$4\\times C-1=53$$，但$$54$$不是$$4$$的整数倍． 综上所述，$$B$$分到$$15$$支铅笔． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1592", "queId": "beed6949607f438d8f776a6d430ac7a8", "competition_source_list": ["2006年第11届全国华杯赛竞赛初赛第3题", "2019年陕西延安宝塔区北大培文学校小升初入学真卷第2题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "奶奶告诉小明：``$$2006$$年共有$$53$$个星期日，聪敏的小明立到告诉奶奶：$$2007$$年的元旦一定是． ", "answer_option_list": [[{"aoVal": "A", "content": "星期一 "}], [{"aoVal": "B", "content": "星期二 "}], [{"aoVal": "C", "content": "星期六 "}], [{"aoVal": "D", "content": "星期日 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["$$2006$$年有$$365$$天，而$$365=7\\times 52+1$$，又已知$$2006$$年有$$53$$个星期天，只能元旦是星期天，且$$12$$月$$31$$日也是星期日，所以，$$2007$$年月的元旦是星期一． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3054", "queId": "e5bfee1ee88f42a5b0a9bfdbcee93369", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛B卷第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\frac{1}{\\dfrac{1}{20}+\\dfrac{1}{21}+\\dfrac{1}{22}+\\cdots +\\dfrac{1}{29}}$$ 的整数部分是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->放缩法->首尾放缩法"], "answer_analysis": ["$$\\frac{1}{29}\\times 10\\textless{}\\frac{1}{20}+\\frac{1}{21}+\\cdots +\\frac{1}{29}\\textless{}\\frac{1}{20}\\times 10$$， $$\\frac{10}{29}\\textless{}\\frac{1}{20}+\\frac{1}{21}+\\cdots +\\frac{1}{29}\\textless{}\\frac{1}{2}$$， $$\\frac{2}{1}\\textless{}\\frac{1}{\\dfrac{1}{20}+\\dfrac{1}{21}+\\cdots +\\dfrac{1}{29}}\\textless{}\\frac{29}{10}$$， 所以它的整数部分应是$$2$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3386", "queId": "8e0a28c8674643558fcbf8fcc0af6972", "competition_source_list": ["2011年全国迎春杯三年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "超市中的某种汉堡每个$$10$$元，这种汉堡最近推出了``买二送一''的优惠活动，即花钱买两个汉堡，就可以免费获得一个汉堡，已知东东和朋友需要买$$9$$个汉堡，那么他们最少需要花~\\uline{~~~~~~~~~~}~元钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$70$$ "}], [{"aoVal": "D", "content": "$$90$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["相当于每$$3$$个汉堡$$20$$元，所以$$9$$个汉堡需要$$60$$元． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3297", "queId": "4d6bdec1f09c491e91994100a71ed328", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "某射手���百步之外射箭恰好射到靶心的概率为$40 \\%$，如果该射手在百步之外连射三箭，三箭全部射中靶心的概率为~\\uline{~~~~~~~~~~}~. ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.064$$ "}], [{"aoVal": "B", "content": "$$0.072$$ "}], [{"aoVal": "C", "content": "$$0.081$$ "}], [{"aoVal": "D", "content": "$$0.09$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["全部射中靶心的概率为$$0.4\\times 0.4\\times 0.4=0.064$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1702", "queId": "b188573db0d944f599c1f60d81aee82e", "competition_source_list": ["1990年第1届全国华杯赛小学高年级竞赛复赛第10题", "1990年全国华杯赛竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "【提升7】 有三堆棋子，每堆棋子数一样多，并且都只有黑、白两色棋子．第一堆里的黑子和第二堆里的白子一样多，第三堆里的黑子占全部黑子的五分之二，把这三堆棋子集中在一起，问白子占全部的几分之几？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\dfrac{3}{9}$$． "}], [{"aoVal": "B", "content": "$$\\dfrac{4}{9}$$． "}], [{"aoVal": "C", "content": "$$\\dfrac{5}{9}$$． "}], [{"aoVal": "D", "content": "$$\\dfrac{7}{9}$$． "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->转化单位1->统一单位1"], "answer_analysis": ["不妨认为第二堆全是黑子，第一堆全是白子，（即将第一堆黑子与第二堆白子互换） 第二堆黑子是全部棋子的$$\\dfrac{1}{3}$$， 同时，又是黑子的$$1-\\dfrac{2}{5}$$， 所以黑子占全部棋子的：$$\\dfrac{1}{3}\\div~ (1-\\dfrac{2}{5})=\\dfrac{5}{9}$$， 白子占全部棋子的：$$1-\\dfrac{5}{9}=\\dfrac{4}{9}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1663", "queId": "bf27f2395fab445f8d11f131e0194d25", "competition_source_list": ["2012年全国美国数学大联盟杯小学高年级竞赛初赛第31题", "2013年美国数学大联盟杯小学高年级竞赛初赛第31题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "上个月我花费了$$24$$美元（＄$$24$$）购买磁铁，其中每根磁铁的价格为$$80$$美分（$$80$$￠），这个月我再次拿出$$24$$美元（＄$$24$$）来购买磁铁，而此时每根磁铁的价格变为$$1.20$$美元（＄$$1.20$$）问：两次购买磁铁的平均单价是？ ", "answer_option_list": [[{"aoVal": "A", "content": "＄$$0.92$$ "}], [{"aoVal": "B", "content": "＄$$0.96$$ "}], [{"aoVal": "C", "content": "＄$$1.00$$ "}], [{"aoVal": "D", "content": "＄$$1.04$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["$$24\\times 2\\div (24\\div 0.8+24\\div1.2)=0.96$$. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2235", "queId": "f626373ec68e4a5896fa96947fb25727", "competition_source_list": ["2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第8题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第8题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第8题3分", "2020年第24届YMO四年级竞赛决赛第8题3分", "2019年第24届YMO四年级竞赛决赛第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "小$$Y$$和小$$M$$两人同时从$$A$$地出发，其中小$$Y$$每天走$$7$$公里；小$$M$$第一天走$$1$$公里，第二天走$$2$$公里，第三天走$$3$$公里，以后每天都比前一天多走$$1$$公里．请问，二人经过天走的路程相同． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["当小$$Y$$和小$$M$$走过的路程相同时，说明小$$Y$$和小$$M$$此时相遇了，要使小$$Y$$和小$$M$$两人相遇，则小$$Y$$的速度不变为每天$$7$$千米，而小$$M$$的速度逐天增加$$1$$千米，最后与小$$Y$$相遇那天速度最快，因此可推出第一天和最后一天的平均速度与第二天和倒数第二天的平均速度相等，都为每天$$7$$千米，设最后一天走$$x$$千米，得出下列方程：$$\\frac{1+x}{2}=7$$，解得$$x=13$$，小$$M$$每天走的千米数每天数相等，因此二人相遇时走了$$13$$天，即两人经过$$13$$天走的路程相同． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "958", "queId": "eb5090d0cba54b50be06eda329c82ccc", "competition_source_list": ["2012年第8届全国新希望杯小学高年级��年级竞赛复赛第13题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "$$\\overline{abcd}$$是一个四位数，且$$\\overline{abcd}+3b+c+8d=2012$$，则$$d$$是。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["课内体系->能力->逻辑分析", "拓展思维->拓展思维->数论模块->位值原理与进制->位值原理运用->位值原理的综合应用"], "answer_analysis": ["位值原理展开：$$1000a+100b+10c+d+3b+c+8d=2012$$，化简得： $$1000a+103b+11c+9d=2012$$， 若$$a=2$$，则有$$103b+11c+9d=12$$，则$$b=0$$，即$$11c+9d=12$$，显然无整数解； 若$$a=1$$，则有$$103b+11c+9d=1012$$，若$$b=8$$，则$$103b+11c+9d$$最大为 $$103\\times 8+11\\times 9+9\\times 9=1004$$，不可能，所以有$$b=9$$，则有$$11c+9d=85$$，$$85\\equiv 4\\left( \\bmod 9 \\right)$$， 即$$11c\\equiv 4\\left( \\bmod 9 \\right)$$，显然当$$c=2$$，$$11c=22$$时，满足$$11c\\equiv 4\\left( \\bmod 9 \\right)$$，此时$$9d=63$$， 即$$d=7$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1498", "queId": "95421c2a031a4bb2a4cf516f4dd57e88", "competition_source_list": ["2017年IMAS小学中年级竞赛（第二轮）第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$30$$个三角排成一列﹐然后依照下面的规律涂上黑色或白色，请问涂上黑色的三角形总共比涂上白色的三角形多几个？ $$▲▲$$$$\\triangle $$$$▲▲$$$$\\triangle $$$$▲▲$$$$\\triangle $$$$\\cdots \\cdots $$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$14$$ "}], [{"aoVal": "E", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->能力->图形认知"], "answer_analysis": ["由图可知，从第一个三角形开始，以每三个三角形为一个周期，每个周期内有$$2$$个黑色三角形与$$1$$个白色三角形，黑色三角形比白色三角形多$$1$$个．$$30$$个三角形共有$$10$$个周期，所以黑色三角形总共比白色三角形多$$10$$个． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "238", "queId": "4cf4e87557b94c748d77c99bf1d17256", "competition_source_list": ["2008年亚太杯竞赛初赛第25题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "在一座岛上，其岛民不是君子就是骗子．君子所言，句句属实，而骗子则句句假话．一个水手登陆这座岛后，遇到了岛民$$\\underline{A}$$，$$\\underline{B}$$与$$\\underline{C}$$．水手问岛民$$\\underline{A}$$，道：``您是个君子还是个骗子呢？''$$\\underline{A}$$回答了，可是水手却听不清楚．他接着问岛民$$\\underline{B}$$，道：``$$\\underline{A}$$说了什么？''$$\\underline{B}$$答道：``$$\\underline{A}$$说他是个骗子．''就在这时，$$\\underline{C}$$喊道：``$$\\underline{B}$$骗人！'' $$\\text{I}$$：无法知道$$\\underline{A}$$究竟是个君子还是个骗子． $$\\text{II}$$：$$\\underline{B}$$是个君子，$$\\underline{C}$$是个骗子． $$\\text{III}$$：$$\\underline{B}$$是个骗子，$$\\underline{C}$$是个君子． ", "answer_option_list": [[{"aoVal": "A", "content": "只有$$\\text{I}$$是正确的． "}], [{"aoVal": "B", "content": "只有$$\\text{II}$$是正确的． "}], [{"aoVal": "C", "content": "只有$$\\text{III}$$是正确的． "}], [{"aoVal": "D", "content": "$$\\text{I}$$与$$\\text{II}$$是正确的． "}], [{"aoVal": "E", "content": "$$\\text{I}$$与$$\\text{III}$$是正确的． "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["①若$$A$$是君子，他会回答：我是君子， 而$$B$$说：``$$A$$说是骗子''，$$B$$说了假话，因此$$B$$是骗子， $${C}$$正确指出$$B$$是骗子，因此$$C$$是君子． ②若$$A$$是骗子，他会回答：我是君子， 而$$B$$复述$$A$$话说$$A$$是骗子，显然未正确真实复述，$$B$$是骗子， $${C}$$指出$$B$$说了假话，$$C$$是君子． 因此，无从得知$$A$$的身份，但$$B$$是骗子，$$C$$是君子可知． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1908", "queId": "aa129c472748461cbb5b5b4a7e830203", "competition_source_list": ["2015年美国数学大联盟杯六年级竞赛初赛（中国赛区）第38题5分", "2016年全国美国数学大联盟杯小学高年级六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "进步先生在不断提高他的考试成绩．他前两次考试的平均分是$$86$$，前三次考试的平均分是$$87$$，前四次考试的平均分是$$88$$���他第四次考了多少分？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$88$$ "}], [{"aoVal": "B", "content": "$$89$$ "}], [{"aoVal": "C", "content": "$$90$$ "}], [{"aoVal": "D", "content": "$$91$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->容斥求平均数"], "answer_analysis": ["$$88\\times 4-87\\times 3=91$$（分）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2821", "queId": "8cdfe6fef1e343dda49937d08ce43df3", "competition_source_list": ["2018~2019学年浙江杭州西湖区杭州市行知小学五年级上学期期中期中竞赛第13题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列算式中，结果小于$$1$$的算式是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.99\\div 0.1$$ "}], [{"aoVal": "B", "content": "$$1\\div 0.99$$ "}], [{"aoVal": "C", "content": "$$0.99\\div 0.99$$ "}], [{"aoVal": "D", "content": "$$0.99\\times 0.99$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->小数->小数乘除->小数乘法运算"], "answer_analysis": ["$$0.99\\times 0.99=0.9801$$，结果小于$$1$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1413", "queId": "32181ca16db8473aa08464ce95814ed1", "competition_source_list": ["2017年四川成都六年级竞赛“全能明星”选拔赛第16题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "水结成冰时它的体积增加了原来的$$\\frac{1}{11}$$，冰化成水后它的体积减少了冰块的~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{11}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{10}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{12}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{9}$$ "}]], "knowledge_point_routes": ["知识标签->课内题型->综合与实践->应用题->分数百分数应用题->需要变换单位一的问题"], "answer_analysis": ["水结成冰时它的体积增加了原来的$$\\frac{1}{11}$$，可得出水为$$11$$份，结冰后冰为$$12$$份 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "943", "queId": "f859ffc16c34464e8f57a4a91a7039b4", "competition_source_list": ["2018年湖北武汉新希望杯小学高年级五年级竞赛训练题（三）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$56$$的因数共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理正应用->总个数"], "answer_analysis": ["$$8$$个 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "666", "queId": "1f39a872b3a1424a834b3bdfc6418c14", "competition_source_list": ["2010年第8届创新杯六年级竞赛初赛第5题4分", "小学中年级四年级上学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "如果形如$$\\overline{3\\square \\square 4}$$的四位数能被$$9$$整除，那么这样的四位数有（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$个 "}], [{"aoVal": "B", "content": "$$10$$个 "}], [{"aoVal": "C", "content": "$$11$$个 "}], [{"aoVal": "D", "content": "$$12$$个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->整除->整除特征->整除特征综合"], "answer_analysis": ["$$9\\textbar\\overline{3AB4}$$，则$$3+A+B+4=7+A+B$$满足是$$9$$的倍数即可． （1）$$A+B=2$$ （2）$$A+B=11$$ $$1+1=0+2=2+0$$ $$2+9=3+8=4+7=5+6$$ 一共$$3+4\\times 2=11$$（个）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "84", "queId": "3d0051a589834c178770e65587915b36", "competition_source_list": ["2011年北京学而思综合能力诊断小学高年级六年级竞赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "有四个人说话，分别如下： $$A$$：我们中至少有一个人说的是正确的． $$B$$：我们中至少有两个人说的是正确的． $$C$$：我们中至少有一个人说的是错误的． $$D$$：我们中至少有两个人说的是错误的． 请问：说错话的是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$A$$ "}], [{"aoVal": "B", "content": "$$B$$ "}], [{"aoVal": "C", "content": "$$C$$ "}], [{"aoVal": "D", "content": "$$D$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["方法一：若没人说对，则$$CD$$说对，矛盾；若$$1$$人说对，则$$ACD$$说对，矛盾；若$$2$$人说对，则$$ABCD$$说对，矛盾；若$$3$$人说对，则$$ABC$$说对，$$D$$错，成立；若$$4$$人说对，则$$AB$$说对，$$CD$$说错，矛盾，因此只能是$$ABC$$说对，$$D$$说错． 方法二：因为四个人，所以至少有两人说错或两人说对，因此$$AB$$一定是正确的，剩下的就容易知道$$D$$是错的． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2908", "queId": "dfe32fcdecee4b1ebb76200fd4938c72", "competition_source_list": ["2017年第22届全国华杯赛小学中年级竞赛初赛第6题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个数串$$219\\cdots$$，从第$$4$$个数字开始，每个数字都是它前面$$3$$个数字和的个位数．下面有$$4$$个四位数：$$1113$$，$$2226$$，$$2125$$，$$2215$$，其中共有个不出现在该数字串中． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数"], "answer_analysis": ["按照题目规则，写出数串$$21922372215847906512811\\cdots$$发现数字的奇偶性为``偶奇奇偶偶奇奇偶偶奇奇偶偶$$\\cdots \\cdots$$''不可能出现三个偶数相连、三个奇数相连或``奇偶奇偶''间隔的情况，可排除$$1113$$，$$2226$$，$$2125$$．且$$2215$$在数串中出现，则选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2531", "queId": "6710741c25a94044bc5418806e355c55", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$2.3\\div 0.08\\div 1.25=$$（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$230$$ "}], [{"aoVal": "B", "content": "$$23$$ "}], [{"aoVal": "C", "content": "$$2.3$$ "}], [{"aoVal": "D", "content": "$$0.23$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["$$2.3\\div 0.08\\div 1.25=2.3\\div (0.08\\times 1.25)=2.3\\div 0.1=23$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2902", "queId": "f6eba03d53254611afad1b47f0848c20", "competition_source_list": ["2012年全国美国数学大联盟杯小学高年级竞赛初赛第25题", "2013年美国数学大联盟杯小学高年级竞赛初赛第25题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "\\textbf{2013年美国数学大联盟小学高年级竞赛初赛} In a garage, the ratio of red cars to black is $$8:5$$, and the ratio of black cars to white car is $$3:4$$． There cannot be fewer thancars in the garage． 在一个车库，红色汽车与黑色汽车的比例是$$8:5$$，黑色汽车与白色汽车比例是$$3:4$$．有不少于辆的汽车在车库里． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$59$$ "}], [{"aoVal": "C", "content": "$$74$$ "}], [{"aoVal": "D", "content": "$$91$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题->复杂的比例问题"], "answer_analysis": ["黑车至少$$15$$辆，红车至少$$24$$辆，白车至少$$20$$辆，共计$$59$$辆 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2184", "queId": "946bc19b04bb4929ae209779ab809d6c", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（二）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲从$$A$$地，乙从$$B$$地同时以匀速相向而行，第一次相遇离$$A$$地$$8$$千米，到达对方起点后立即返回，在离$$B$$地$$4$$千米处第二次相遇，则$$A$$、$$B$$两地相距千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["课内体系->思想->对应思想", "拓展思维->知识点->行程模块->直线型行程问题->多次相遇和追及->往返相遇"], "answer_analysis": ["$8\\times3-4$=20（千米）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1024", "queId": "12ecd19085484a90a0e1e66d76097234", "competition_source_list": ["2020年北京迎春杯六年级竞赛模拟决赛第4题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "一炉铁水凝成铁块，它的体积缩小了$$\\frac{1}{34}$$；那么，这个铁块又融化成铁水（不计损耗），其体积． ", "answer_option_list": [[{"aoVal": "A", "content": "增加了$$\\frac{1}{33}$$ "}], [{"aoVal": "B", "content": "增加了$$\\frac{1}{34}$$ "}], [{"aoVal": "C", "content": "减少了$$\\frac{1}{33}$$ "}], [{"aoVal": "D", "content": "减少了$$\\frac{1}{34}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["铁块的体积是铁水的：$$1-\\frac{1}{34}=\\frac{33}{34}$$， 铁块融化成铁水体积增加：$$\\left( 1-\\frac{33}{34} \\right)\\div \\frac{33}{34}=\\frac{1}{33}$$． 答：这个铁块又融化成铁水（不计损耗），其体积增加了$$\\frac{1}{33}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "567", "queId": "03853fc3ae954a1a8e68b2702bd26081", "competition_source_list": ["2010年四年级竞赛创新杯", "2010年三年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "在一个带余数的除法算式$$A\\div 6=8\\cdots B$$中，$$A$$、$$B$$都是整数，$$A$$不是$$6$$的倍数，那么$$A$$的最大值和最小值的和等于． ", "answer_option_list": [[{"aoVal": "A", "content": "$$49$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$102$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["最大值：$$6\\times 8+5=53$$，最小值：$$6\\times 8+1=49$$.$$53+49=102$$ "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "847", "queId": "b1cdaa03dcb34da2ba68343498129316", "competition_source_list": ["2016年希望杯四年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$60$$的不同因数（$$1$$除外）的个数是( )。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数"], "answer_analysis": ["$$60$$分解质因数为$$60=2\\times 2\\times 3\\times 5$$，再改为标准式是$${{2}^{2}}\\times 3\\times 5$$，再利用因数个数公式，因数个数$$=$$不同质因数指数加$$1$$然后再相乘。 $$60$$的不同因数（$$1$$除外）的个数是$$(2+1)\\times (1+1)\\times (1+1)-1=11$$个。 答：答案是A。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "239", "queId": "99aea4356a9e4aaea66d66f892d56473", "competition_source_list": ["2017年全国小学生数学学习能力测评六年级竞赛初赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "游泳馆在教学楼北偏东$$60$$度方向上，两幢楼相距$$200$$米，游泳馆在实验楼南偏东$$60$$度方向上，两幢楼也相距$$200$$米，则实验楼位于教学楼． ", "answer_option_list": [[{"aoVal": "A", "content": "正东方向$$400$$米 "}], [{"aoVal": "B", "content": "南偏东$$60$$度方向$$200$$米 "}], [{"aoVal": "C", "content": "正北方向$$200$$米 "}], [{"aoVal": "D", "content": "北偏东$$60$$度方向$$200$$米 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->方向与坐标->方向"], "answer_analysis": ["由题意可得实验楼位于教学楼的正北方向，且距离为$$200\\div 2\\times 2=200$$（米）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2294", "queId": "8690bb882f6346ef908aed1643b7fdb5", "competition_source_list": ["2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "(2016 Math League, Grade3） Running Man is running a certain distance at a constant speed of $$300$$m/min and can finish the entire distance in $$3$$ hours. If he ran at a constant speed of $$360$$m/min, how many minutes less would it take him to run this distance? Running Man是以$300$ m/min的恒定速度跑完一段距离，可以在$3$小时内跑完全程$$.$$ 如果以以$360$ m/min的恒定速度跑完这段距离会少花多少分钟? ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$150$$ "}], [{"aoVal": "C", "content": "$$180$$ "}], [{"aoVal": "D", "content": "$$360$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->行程模块->直线型行程问题->路程速度时间", "拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题"], "answer_analysis": ["某人跑步，每分钟$$300$$米，$$3$$小时跑完．假设他提速到每分钟$$360$$米，跑完全程比之前节省几分钟？ $$300\\times 180\\div 360=150$$分钟，$$180-150=30$$分钟． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "857", "queId": "bfab7f6310d74d179f19bfd556662c0d", "competition_source_list": ["2015年IMAS小学高年级竞赛第一轮检测试题第4题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若正整数$$n$$的平方减去$$200$$后所得的差是一个$$4$$的倍数之三位数，请问这样的正整数$$n$$总共有多少个不同的取值？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$17$$ "}], [{"aoVal": "E", "content": "$$32$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$100\\textless{}{{n}^{2}}-200\\textless{}1000$$并且$$n$$为偶数， ∴$$18\\leqslant n\\leqslant 34$$，期间偶数为$$9$$个． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2419", "queId": "0c3d604d7d8a4d1281e069c43ac710f1", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（四）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列说法中，错误的是． ", "answer_option_list": [[{"aoVal": "A", "content": "百分数不一定比$$1$$小$$B$$ "}], [{"aoVal": "B", "content": "$$20$$分$$=\\frac{1}{3}$$时 "}], [{"aoVal": "C", "content": "面积相等的两个圆，它们的半径一定相等 "}], [{"aoVal": "D", "content": "苹果树的棵数比梨树多$$\\frac{1}{5}$$，也就是梨树的棵数比苹果树少$$\\frac{1}{5}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->单位换算->时间单位换算"], "answer_analysis": ["梨树的棵数比苹果树少$$\\frac{1}{6}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2477", "queId": "873810896bfc41ab9092cda32692a4b6", "competition_source_list": ["2020年新希望杯三年级竞赛初赛（团战）第34题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$63$$不能表示成个连续整数的和． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}], [{"aoVal": "E", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["暂无 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2554", "queId": "4bbfba7d69184fc7924e3eb242152463", "competition_source_list": ["2018年陕西西安小升初分类卷10第33题", "2017年河南郑州豫才杯小学高年级六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "一个整数，四舍五入到万位约是$$6$$万，那么这个数最大是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60999$$ "}], [{"aoVal": "B", "content": "$$64449$$ "}], [{"aoVal": "C", "content": "$$64999$$ "}], [{"aoVal": "D", "content": "$$69999$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->大数的估算"], "answer_analysis": ["四舍五入后的近似值是$$6$$万，最大的情况是$$6$$后面的数字小于$$5$$，直接舍去，即为$$64999$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "148", "queId": "f119a8c51fee4df485b7dc78be8144de", "competition_source_list": ["2015年第20届全国华杯赛小学中年级四年级竞赛初赛A卷", "2016年第20届四川成都华杯赛小学中年级竞赛B卷第1~6题60分"], "difficulty": "1", "qtype": "single_choice", "problem": "森林里举行比赛，要派出狮子、老虎、豹子和大象中的两个动物去参加．如果派狮子去，那么也要派老虎去；如果不派豹子去，那么也不能派老虎去；要是豹子参加的话，大象可不愿意去．那么，最后能去参加比赛的是（~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "狮子、老虎 "}], [{"aoVal": "B", "content": "老虎、豹子 "}], [{"aoVal": "C", "content": "狮子、豹子 "}], [{"aoVal": "D", "content": "老虎、大象 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->连线法"], "answer_analysis": ["在逻辑推理中，原命题成立，则逆否命题也成立. 从题意出发： （1）狮子去则老虎去，逆否命题；老虎不去则狮子也不去 （2）不派豹子则不派老虎，逆否命题：派老虎则要派豹子 （3）派豹子则大象不愿意去，逆否命题；大象去则不能派豹子 从（2）出发可以看出答案为$$B$$ 题目要求有两个动物去，可以使用假设法，若狮子去，则老虎去，老虎去则豹子也去，三个动物去，矛盾，所以狮子不去，若豹子不去则老虎不去，那么只有大象去，矛盾，所以豹子去，豹子去则大象不去，由两骄气去得到结论，老虎要去，所以答案是$$B$$��豹子和老虎去. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2825", "queId": "ba9cc890314d452abba20ed00a824437", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$202\\times 404\\times 505\\times 606\\times 808$$的最后两位数是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$80$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["$$2\\times 4\\times 5\\times 6\\times 8=1920$$ $$1920\\times101\\times 101\\times 101\\times 101\\times 101$$不会改变后两位． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "659", "queId": "8b1e90a5d922437fa6a6485319a340a3", "competition_source_list": ["2014年迎春杯六年级竞赛初赛", "2014年迎春杯五年级竞赛初赛", "2014年迎春杯六年级竞赛初赛", "2014年迎春杯三年级竞赛初赛", "2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$4$$个质数的积是它们和的$$11$$倍，则它们的和为（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$46$$ "}], [{"aoVal": "B", "content": "$$47$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "没有符合条件的数 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的乘法规律"], "answer_analysis": ["设四个质数为$$a$$、$$b$$、$$c$$、$$d$$。由已知条件得，$$4$$个质数中一定有$$11$$，那么满足$$a\\times b\\times c=a+b+c+11$$，其中$$a$$、$$b$$、$$c$$都是质数。若$$a$$、$$b$$、$$c$$都是奇数，则等式左边是奇数，右边是偶数，矛盾。若$$a$$、$$b$$、$$c$$中有一个偶数，那么一定是$$2$$，即$$a\\times b\\times 2=a+b+2+11$$，此时，根据奇偶性，$$a$$、$$b$$中也必有一个偶数为$$2$$，解得$$a$$、$$b$$、$$c$$、$$d$$为$$2$$、$$2$$、$$5$$、$$11$$，和为$$20$$。ABC均不符合条件，故选D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "455", "queId": "9812d4f24482420580c6931d94c5d771", "competition_source_list": ["2013年全国学而思杯二年级竞赛第4题"], "difficulty": "0", "qtype": "single_choice", "problem": "狐狸、灰兔、小熊、小猪和松鼠都参加了跳绳比赛，小猪比狐狸少跳了$$3$$下，小猪比小熊少跳了$$1$$下，灰兔比狐狸多跳了$$3$$下，比松鼠少跳$$3$$下，狐狸是第~\\uline{~~~~~~~~~~}~名． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "$$5$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->组合模块->逻辑推理", "拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["根据题目描述得小猪\\textless 小熊\\textless 狐狸\\textless 灰兔\\textless 松鼠． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "295", "queId": "7672a558fe244d518e9e2273d2c145c9", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "（3分）在垒球比赛中，若赢$$1$$场得$$3$$分，平$$1$$场得$$1$$分，输$$1$$场不得分．每个队都与其他队交锋$$4$$场，这时四个参赛队的总积分为：$$A$$队$$22$$分，$$B$$队$$19$$分，$$C$$队$$14$$分，$$D$$队$$12$$分．那么有场比赛为平局． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->逻辑分析"], "answer_analysis": ["对于赛况分析试题，尤其对于与分数相关的试题，最重要的是思维方式，本题如果从整体上来考虑比赛所产生的总分值，问题将迎刃而解，依题意可知比赛总场次为$$24$$场比赛之中，若平局则将会让所有队伍的总分增加$$2$$分（比赛双方均得$$1$$分），若出现了胜败，则所有队伍的总分增加$$3$$分，而现在所有队伍获得的总分值为：$$22+19+14+12=67$$（分），$$24$$场比赛，有$$3$$分，有$$2$$分，总分为$$67$$分，可当做鸡兔同笼问题解答，易得平局有$$5$$场． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1956", "queId": "c10cdbf5d5f64d09864d189410fb8598", "competition_source_list": ["2005年第3届创新杯六年级竞赛初赛第4题", "2005年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "某商场的营业额$$2001$$年比$$2000$$年上升$$10 \\%$$，$$2002$$年又比$$2001$$年上升了$$10 \\%$$，而$$2003$$年和$$2004$$年连续两年比上一年降低$$10 \\%$$，那么$$2004$$年的营业额比$$2000$$年的营业额． ", "answer_option_list": [[{"aoVal": "A", "content": "降低了$$2 \\%$$ "}], [{"aoVal": "B", "content": "没有变化 "}], [{"aoVal": "C", "content": "上升了$$2 \\%$$ "}], [{"aoVal": "D", "content": "降低了$$1.99 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["设$$2000$$年的营业额为``$$1$$''，那么$$2004$$年的营业额为$$1\\times \\left( 1+10 \\% \\right)\\times \\left( 1+10 \\% \\right)\\times \\left( 1-10 \\% \\right)\\times \\left( 1-10 \\% \\right)=0.9801$$，则$$2004$$年的营业额比$$2000$$年的营业额下降了$$\\left( 1-0.9801 \\right)\\div 1\\times 100 \\%=1.99 \\%$$，选$$\\text{D}$$. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1982", "queId": "b8558e88fdd84aa7a5f3b7d651b407b8", "competition_source_list": ["2014年IMAS小学中年级竞赛第一轮检测试题第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "荧幕上闪动着一行字元``$$2014\\text{IMAS}$$''，每经过$$1$$分钟，最左边的字元就会移到这些字元的最右边．请问从开始到再次出现``$$2014\\text{IMAS}$$''需要多少分钟？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["这一串字元共有$$8$$个字元，所以从开始到再次出现``$$2014\\text{IMAS}$$''需要$$8$$分钟．故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3405", "queId": "ffb97d7103954aa796105155549a6a9d", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "某射手在百步之外射箭恰好射到靶心的概率为$40 \\%$，如果该射手在百步之外连射三箭，有一箭射中靶心的概率为~\\uline{~~~~~~~~~~}~. ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.343$$ "}], [{"aoVal": "B", "content": "$$0.512$$ "}], [{"aoVal": "C", "content": "$$0.432$$ "}], [{"aoVal": "D", "content": "$$0.435$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["第一箭射中，其他两箭射空的概率为$$0.4\\times \\left( 1-0.4 \\right)\\times \\left( 1-0.4 \\right)=0.144$$． 第二箭射中，其他两箭射空的概率为$$0.4\\times \\left( 1-0.4 \\right)\\times \\left( 1-0.4 \\right)=0.144$$． 第三箭射中，其他两箭射空的概率为$$0.4\\times \\left( 1-0.4 \\right)\\times \\left( 1-0.4 \\right)=0.144$$． 有一箭射中的概率为$$0.144+0.144+0.144=0.432$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1516", "queId": "44a0a1d45192474a805de4ed476c7e54", "competition_source_list": ["2011年四川成都小升初", "2016年创新杯六年级竞赛训练题（四）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙两人从底楼（第一层）开始比赛爬楼梯（每两层之间楼梯的级数相同），甲跑到第$$4$$层时，乙恰好跑到第三层，甲跑到第$$16$$层时，乙跑到（~ ）层． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$~~~~~ "}], [{"aoVal": "B", "content": "$$10$$~~~ "}], [{"aoVal": "C", "content": "$$11$$~~~ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题->按比分配"], "answer_analysis": ["此题应注意一点，即从一楼到二楼其实只爬了一层楼梯，所以当甲跑到$$4$$层时，只爬$$3$$层楼梯，乙跑到$$3$$楼也只跑了$$2$$层楼梯．甲、乙速度之比为$$3:2$$．甲到$$16$$层时，只爬了$$15$$层楼梯，故$$3:2=15:10$$，$$10+1=11$$，即乙跑到$$11$$层． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "510", "queId": "c69410efa5f94c8fb061cf66555aa52a", "competition_source_list": ["2020年第24届YMO五年级竞赛决赛第6题3分", "2019年第24届YMO五年级竞赛决赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小$$Y$$、小$$M$$、小$$O$$、小$$T$$四人中只有$$1$$人会开车．小$$Y$$说：``我会开''．小$$M$$说：``我不会开''．小$$O$$说：``小$$Y$$不会开''．小$$T$$什么也没说．已知小$$Y$$、小$$M$$、小$$O$$三人只有一人说了真话．会开车的是谁？ ", "answer_option_list": [[{"aoVal": "A", "content": "小$$Y$$ "}], [{"aoVal": "B", "content": "小$$M$$ "}], [{"aoVal": "C", "content": "小$$O$$ "}], [{"aoVal": "D", "content": "小$$T$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析", "Overseas Competition->知识点->组合模块->逻辑推理"], "answer_analysis": ["对比发现，小$$O$$与小$$Y$$说的矛盾，相互对立， 则小$$O$$与小$$Y$$必一对一错， 则小$$M$$说假话，则小$$M$$会开车，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1471", "queId": "cc781109020f40b6bff13f943903e691", "competition_source_list": ["2015年河南郑州K6联赛小学高年级竞赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "教室里的彩灯按照$$5$$盏红灯、$$4$$盏蓝灯、$$3$$盏黄灯的顺序循环出现，则前$$160$$盏中有盏红灯． ", "answer_option_list": [[{"aoVal": "A", "content": "$$65$$ "}], [{"aoVal": "B", "content": "$$69$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$70$$ "}]], "knowledge_point_routes": ["拓展思维->能力->归纳总结->归纳推理"], "answer_analysis": ["彩灯的排列规律是按照$$12$$盏灯一个循环周期，$$80\\div 12=6\\cdot \\cdot \\cdot 8$$，所以第$$80$$盏灯是第$$6$$个周期的第$$8$$个，是蓝灯．$$160\\div 12=13\\cdot \\cdot \\cdot 4$$，所以前$$160$$盏灯中有红灯$$13\\times 5+4=69$$盏． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2759", "queId": "df353c8e742d41c59b60d97108702d10", "competition_source_list": ["2014年IMAS小学中年级竞赛第一轮检测试题第12题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "小兰今年上小学四年级，请问下列哪项的时间与小兰的年龄最接近？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$小时 "}], [{"aoVal": "B", "content": "$$120$$天 "}], [{"aoVal": "C", "content": "$$120$$个星期 "}], [{"aoVal": "D", "content": "$$120$$个月 "}], [{"aoVal": "E", "content": "$$120$$年 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为$$120$$小时$$\\textless{}120$$天$$\\textless{}120$$个星期$$\\textless{}30$$个月$$\\textless{}3$$年，所以前三项的时间都小于$$3$$年，不符合实际． 选项$$\\text{E}$$显然也不符合实际．而$$120$$个月$$=10$$年，比较符合事实，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2302", "queId": "c52a8c3e8b9f46ba923dd5704fd34682", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "某人开车往返于深圳和广州之间，从深圳去广州每小时行$$30$$千米，从广州返回深圳每小时行$$60$$千米．那么他往返深圳和广州的平均速度是千米$$/$$时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$50$$ "}], [{"aoVal": "E", "content": "$$55$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->平均速度->公式法"], "answer_analysis": ["深圳和广州的路程为$$60$$千米，则从深圳到广州用时$$60\\div30=2$$小时，从广州到深圳用时$$60\\div60=1$$小时，总共用时$$1+2=3$$小时，平均速度$$=60\\times2\\div3=40$$千米$$/$$小时． 答案：$$B$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "50", "queId": "53d87ebda6d24d0d9b8e9821a3767a74", "competition_source_list": ["2006年六年级竞赛创新杯", "2006年六年级竞赛创新杯", "2006年五年级竞赛创新杯", "2006年第4届创新杯四年级竞赛初赛A卷第2题", "2006年五年级竞赛创新杯", "2006年四年级竞赛创新杯", "2006年四年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "$$10$$个学生参加一次考试，满分是$$100$$分，这次考试中十个学生的平均分为$$92$$，成绩最差的学生可能得到的最低分为( ) ", "answer_option_list": [[{"aoVal": "A", "content": "$$92$$ "}], [{"aoVal": "B", "content": "$$90$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->竖式数字谜->竖式数字谜的最值"], "answer_analysis": ["这次考试$$10$$个学生的总分为$$92\\times 10=920$$（分），假设有$$9$$个学生的成绩为满分，即$$100\\times 9=900$$（分），那么最差学生的最低分为$$920-900=20$$（分），所以选D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "124", "queId": "3015e8fdfc244e6888383b9a4e0a9734", "competition_source_list": ["2016年创新杯小学高年级六年级竞赛训练题（一）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "欢欢、迎迎各有$$4$$张卡片，每张卡片上各写有一个自然数．两人各出一张卡片，计算两张卡片上所写数的和，结果发现一共能得到$$16$$个不同的和．那么，两人的卡片上所写的数中最大的数最小是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->操作问题->数字操作"], "answer_analysis": ["为使所写的数最大的最小，直接考虑这$$16$$个不同的和最小是$$0$$，最大是$$15$$，设为$$\\text{A}$$、$$\\text{B}$$两组，由于$$0=0+0$$，则两组必都有$$0$$，而$$1=0+1$$，那有一组必有$$1$$，且两数和为$$15$$，那么其中一个数至少是$$8$$，假设最大是$$8$$，那么另一组一定有$$7$$和$$6$$，不能构造出来，假设最大是$$9$$，可以构造出两组，（$$0,2$$，$$4,6$$），（$$0,1$$，$$8,9$$）注：和是$$0$$到$$15$$，也就是二进制的$$0000$$到$$1111$$，那么，显然了，每人控制两位的开关，两个人就能够控制全部四位的开惯了，为了使得最大的数最小，控制最高位的那个人再控制最低位就行了． 一个人控制最高位和最低位：$$0000$$,$$0001$$,$$1000$$,$$1001$$； 另一个人控制中间两位：$$0000$$,$$0010$$,$$0100$$,$$0110$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1804", "queId": "8dd335589fdf43c8bf06ca09b1dc1fb2", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "三年级二班的同学在上游泳课，男生戴蓝泳帽，女生戴红泳帽。 男体育委员说：``我看见的蓝泳帽比红泳帽的$$4$$倍多$$1$$个。'' 女体育委员说：``我看见的蓝泳帽比红泳帽多$$24$$个。'' 根据两位体育委员的话，算出三年级二班共有（ ）位同学。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$37$$ "}], [{"aoVal": "D", "content": "$$38$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->差倍->两量差倍->暗倍型二量差倍问题"], "answer_analysis": ["解：男体育委员少看到一个蓝色帽子，所以实际蓝泳帽比红泳帽的$$4$$倍多$$2$$个，女体育委员少看到一个红色帽子，所以实际蓝泳帽比红泳帽多$$23$$个，红泳帽有$$\\left( 23-2 \\right)\\div \\left( 4-1 \\right)=7$$个，蓝泳帽有$$7+23=30$$个，共有$$30+7=37$$位同学。 故选:C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "784", "queId": "5b6e875ce9df4451bff6699ed3cd5135", "competition_source_list": ["2019年美国数学大联盟杯四年级竞赛初赛第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "三角形的边长是偶数，其周长不可能是哪个选项 ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$64$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["本题题意为``三角形的边长是偶数，其周长不可能是哪个选项''，三角形的周长即为三条边相加，三条边长是偶数，因此三个偶数相加最终结果也是偶数，因此结果不可能是$$\\text{A}$$选项$$9$$，最终结果选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "583", "queId": "0b45d6c9723f4b6ca7d35a9e3652bcd3", "competition_source_list": ["2010年全国华杯赛竞赛复赛第12题"], "difficulty": "2", "qtype": "single_choice", "problem": "华罗庚爷爷出生于$$1910$$年$$11$$月$$12$$日．将这些数字排成一个整数，并且分解成$$19101112=1163\\times16424$$，请问这两个数$$1163$$和$$16424$$中是否有质数？ ", "answer_option_list": [[{"aoVal": "A", "content": "有 "}], [{"aoVal": "B", "content": "没有 "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["$$16424$$是合数，原因是$$16424$$的约数不止两个，除了有$$1$$和本身外还有$$2$$、$$4$$$$\\cdots\\cdots$$等等． $$1163$$是质数，判断方法是:$$ 35^{2}=1225$$，$$34^{2}=1156$$ 最接近$$1163$$，所以用小于$$34$$的所有质数$$2$$、$$3$$、$$5$$、$$7$$、$$11$$、$$13$$、$$17$$、$$19$$、$$23$$、$$29$$、$$31$$去除$$1163$$都除不尽，所以可以判断$$1163$$是质数． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "303", "queId": "600f11f8cf974590a19b6dffa02b2618", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第11题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "把$$2021$$分拆成若干个自然数的和，那么这些自然数的乘积最大为． ", "answer_option_list": [[{"aoVal": "A", "content": "$${{2}^{1009}}\\times 3$$ "}], [{"aoVal": "B", "content": "$${{3}^{673}}\\times 2$$ "}], [{"aoVal": "C", "content": "$${{4}^{504}}\\times 5$$ "}], [{"aoVal": "D", "content": "$${{2}^{4}}\\times {{3}^{671}}$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->和一定型最值问题->多数之积的最值->拆分数的数目不确定"], "answer_analysis": ["把一个数拆成多个自然数的和，要使得这些数的乘积最大，应该尽量平均， 因为和一定时，差小积大，但并不是越小越好， 因为拆成$$1$$，乘积并没有变化， 所以应该拆成$$2$$，但是$$2+2+2=6=3+3$$， 但是$$3\\times 3=9\\textgreater2\\times 2\\times 2=8$$， 所以应该尽量拆成$$3$$． $$2021\\div 3=673\\cdots \\cdots 2$$， 所以把$$2021$$拆成$$673$$个$$3$$和一个$$2$$时，他们的乘积最大，最大的乘积就是$$3$$的$$673$$次方乘$$2$$. 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2880", "queId": "b1fabdf7caa5451bb3cb50d535675e66", "competition_source_list": ["2008年全国迎春杯三年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算． $$24+63+52+17+49+81+74+38+95=$$~\\uline{~~~~~~~~~~}~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$493$$ "}], [{"aoVal": "B", "content": "$$483$$ "}], [{"aoVal": "C", "content": "$$505$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["凑整法： 原式$$=(38+52)+(63+17)+(49+81)+74+24+95$$ $$= 90+80+130+98+95$$ $$=493$$． 拆分凑整法 原式$$=(20+4+60+3+50+2+10+7+40+9+80+1+70+4+30+8+90+5$$ $$=20+80+60+40+10+90+70+30+50+3+7+1+9+2+8+4+4+5$$ $$=100+100+100+100+50+10+10+10+13$$ $$=493$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1302", "queId": "23c374eaf30a4a28b2f9f0530fe2b05c", "competition_source_list": ["2014年迎春杯三年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "长方形的周长是$$48$$厘米，已知长是宽的$$2$$倍，那么长方形的长是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$厘米 "}], [{"aoVal": "B", "content": "$$16$$厘米 "}], [{"aoVal": "C", "content": "$$24$$厘米 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->和倍问题->二量和倍问题->两量和倍"], "answer_analysis": ["解：$$48\\div 2\\div (1+2)\\times 2$$ $$=24\\div 3\\times 2$$ $$=16$$ （厘米） 答：长方形的长是$$16$$厘米。 故选：B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3116", "queId": "d965a1c989f34a21bd66e38c37d86b8d", "competition_source_list": ["2007年华杯赛六年级竞赛初赛", "2007年华杯赛四年级竞赛初赛", "2007年华杯赛五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$$a*b=ab-a+b$$，则$$5*x=19$$的解是$$x=$$（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->定义新运算->反解未知数型"], "answer_analysis": ["$$5*x=19$$ $$5x-5+x=19$$ $$6x=24$$ $$x=4$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2203", "queId": "b06dfe728e3445bb90c4d8c88622fead", "competition_source_list": ["2016年第12届全国新希望杯小学高年级六年级竞赛复赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "8.战士小王从$$A$$地前往$$B$$地送信，他每走$$40$$分钟就休息$$10$$分钟，到达$$B$$地共需$$4$$小时$$20$$分，从$$B$$地原路返回的速度是去时的$$2$$倍，若他每走$$35$$分钟就休息$$15$$分钟．从$$B$$地返回到$$A$$地共需分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$125$$ "}], [{"aoVal": "B", "content": "$$130$$ "}], [{"aoVal": "C", "content": "$$135$$ "}], [{"aoVal": "D", "content": "$$140$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->行程模块->环形跑道->走走停停"], "answer_analysis": ["去的过程中一共花了$$4\\times 6+20=260$$分钟，$$260\\div 50=5\\ldots \\ldots 10$$说明走路时间一共是$$5\\times 40+10=210$$分．回来过程中速度为原来的$$2$$倍，走路的时间则为原来的$$\\frac{1}{2}$$，即$$210\\times \\frac{1}{2}=105$$分钟$$105\\div 35=3$$，即一共走了$$3$$次，第$$3$$次走完之后不用休息，则一共花了$$\\left( 35+15 \\right)\\times 3-15=135$$分钟． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1752", "queId": "a8b20bb9f26d454080fb879de825eb15", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有大、小两桶油共重$$50$$千克，两个桶都倒出同样多的油后，分别还剩$$14$$千克和$$6$$千克，小桶原来装油千克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["从题干可以知道，有大、小两桶油共重$$50$$千克，两个桶都倒出同样多的油后，分别还剩$$14$$千克和$$6$$千克，不妨设倒出来的油有$$x$$千克，那么大桶有油：$$(x+14)$$千克，小桶有油：$$\\left( x+6 \\right)$$千克． $$\\left( x+14 \\right)+\\left( x+6 \\right)=50$$ $$2x+20=50$$ $$x=15$$． 所以倒出部分是$$15$$千克，那么小桶原来装油$$15+6=21$$（千克），即小桶原来装油$$21$$千克． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3315", "queId": "ff8080814518d5240145201b528a0a79", "competition_source_list": ["2014年全国迎春杯三年级竞赛复赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "春节时，妈妈买了$$3$$个完全一样的福袋，小悦想把$$10$$枚相同的一元硬币放到这三个福袋里，如果每个福袋里至少放$$1$$枚，不考虑福袋的先后顺序的话，共有（~~~~~ ）种放法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["枚举法，$$10$$能被拆成哪三个数相加，$$10=1+1+8$$，$$10=1+2+7$$，$$10=1+3+6$$，$$10=1+4+5$$，$$10=2+2+6$$，$$10=2+3+5$$，$$10=2+4+4$$，$$10=3+3+4$$，共$$8$$种． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3030", "queId": "a1aa0fb9750e461599a0bd81b4594e88", "competition_source_list": ["2018年陕西西安碑林区西安市铁一中学小升初（二十六）第20题5分", "2015年世界少年奥林匹克数学竞赛六年级竞赛复赛A卷第10题10分", "六年级其它导引"], "difficulty": "3", "qtype": "single_choice", "problem": "【提升7】 计算：$$\\frac{{{1}^{2}}+{{2}^{2}}}{1\\times 2}+\\frac{{{2}^{2}}+{{3}^{2}}}{2\\times 3}+\\cdots +\\frac{{{18}^{2}}+{{19}^{2}}}{18\\times 19}+\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$38\\frac{19}{20}$$ "}], [{"aoVal": "B", "content": "$$36\\frac{19}{20}$$ "}], [{"aoVal": "C", "content": "$$38\\frac{11}{20}$$ "}], [{"aoVal": "D", "content": "$$36\\frac{13}{20}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->裂项与通项归纳->分数裂项->分数裂差->两分数间接裂差"], "answer_analysis": ["算式中的分母是裂项计算的最基本形式，但分子比较复杂，我们可以从前几项找找规律： $$\\frac{{{1}^{2}}+{{2}^{2}}}{1\\times 2}=\\frac{5}{2}=2\\frac{1}{2}$$，$$\\frac{{{2}^{2}}+{{3}^{2}}}{2\\times 3}=\\frac{13}{6}=2\\frac{1}{6}$$，$$\\frac{{{3}^{2}}+{{4}^{2}}}{3\\times 4}=\\frac{25}{12}=2\\frac{1}{12}$$． 我们发现一规律：每一项减去$$2$$后，分子就变成了$$1$$． 再来试试最后一项：$$\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}=\\frac{761}{380}=2\\frac{1}{380}$$， 也满足这个规律，这是为什么呢？ 观察每一项的分子和分母，我们发现分子的每个加数都与分母大小接近，可以做如下变形： $$\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}=\\frac{19\\times \\left( 20-1 \\right)+20\\times \\left( 19+1 \\right)}{19\\times 20}$$ $$=\\frac{19\\times 20\\times 2+\\left( 20-19 \\right)}{19\\times 20}$$ $$=2+\\frac{1}{19\\times 20}$$． 算式中的每一项都能像上面一样进行变形，所以： 原式$$=2\\frac{1}{1\\times 2}+2\\frac{1}{2\\times 3}+\\cdots +2\\frac{1}{19\\times 20}$$ $$=2\\times 19+\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{19\\times 20}$$ $$=38+1-\\frac{1}{20}$$ $$=38\\frac{19}{20}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3414", "queId": "d7395b6117b4426ab1ff1b795a25a1f3", "competition_source_list": ["2008年第6届创新杯六年级竞赛初赛A卷第2题5分", "2008年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "我们知道，闰年每隔4年一次，今年(2008年)就是一个闰年.但是在世纪之交时，只有年号是400的倍数时，才是闰年，例如1900年就不是闰年，但是2000年是闰年，那么从2001年到3001年，闰年有( ). ", "answer_option_list": [[{"aoVal": "A", "content": "240个 "}], [{"aoVal": "B", "content": "242个 "}], [{"aoVal": "C", "content": "248个 "}], [{"aoVal": "D", "content": "250个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["位于2001与3001之间的4的倍数有250个，其中2100、2200、2300、2500、2600、2700、2900和3000不是闰年，因此只有$$250-8=242$$(个)闰年. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2649", "queId": "3f780c8d548d4a63a9211da16cde5c2d", "competition_source_list": ["2017年河南郑州豫才杯小学高年级五年级竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "《红楼梦》也成《石头记》，开篇第一回即交代这块顽石的尺寸：女娲所炼五色石，高十二丈，见方二十四丈．欢欢经过百度查询得知：$$1$$丈$$=10$$尺$$=100$$寸，$$1$$米$$=3$$尺$$=30$$寸．请你根据以上资料推算一下，这块顽石法人高度最接近（~ ）米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$436$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$12\\times 100=1200$$寸，$$1200\\div 30=40$$米． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1864", "queId": "977a83f2357e4874bf3199093f3bacee", "competition_source_list": ["2005年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "一项工程，甲单独做需要6小时，乙单独做要10小时，如果按甲、乙、甲、乙\\ldots 顺序交替工作，每次1小时，那么需要( )小时完成. ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$7\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$7\\frac{1}{3}$$ "}], [{"aoVal": "D", "content": "$$7\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->接力施工问题"], "answer_analysis": ["若把甲工作1小时后乙再工作1小时看作一个整体，那么一个整体2小时的工作总量为$$\\frac{1}{6}+\\frac{1}{10}=\\frac{4}{15}$$，又$$1\\div \\frac{4}{15}=3\\frac{3}{4}$$，那么这项工程可以包含3个整体，即$$2\\times 3=6$$(小时)完成这项工程的$$\\frac{4}{15}\\times 3=\\frac{4}{5}$$，还剩$$1-\\frac{4}{5}=\\frac{1}{5}$$.这时又由甲来做，而$$\\frac{1}{5}\\textgreater\\frac{1}{6}$$，那么甲又做1小时后这项工程还剩$$\\frac{1}{5}-\\frac{1}{6}=\\frac{1}{30}$$，这$$\\frac{1}{30}$$又由乙来做，需要$$\\frac{1}{30}\\div \\frac{1}{10}=\\frac{1}{3}$$(小时)可以完成.故完成这项工程需要$$6+1+\\frac{1}{3}=7\\frac{1}{3}$$(小时)，选$$C$$. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "561", "queId": "8f490ca59e75472e981cc914b0097736", "competition_source_list": ["2016年全国美国数学大联盟杯五年级竞赛初赛第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "两个合数的和不能为． ", "answer_option_list": [[{"aoVal": "A", "content": "奇数 "}], [{"aoVal": "B", "content": "偶数 "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数->质数与合数判定->质合判断", "Overseas Competition->知识点->数论模块->质数与合数"], "answer_analysis": ["$$11=1+10=2+9=3+8=4+7=5+6$$ None of the pairs of numbers are both composite numbers. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2221", "queId": "cc58318399064ecfb20ec7ef42f1d1c6", "competition_source_list": ["2006年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "从山下到山上的路程是$$1200$$米，小华上山时平均每分钟走$$60$$米，下山时��均每分钟走$$120$$米，则小华往返行程中的平均速度是每分钟走米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$90$$ "}], [{"aoVal": "B", "content": "$$80$$ "}], [{"aoVal": "C", "content": "$$75$$ "}], [{"aoVal": "D", "content": "$$100$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题"], "answer_analysis": ["路程$$=$$速度$$\\times $$时间，而平均速度$$=$$总路程$$\\div $$总时间，因此上山的时间为$$1200\\div 60=20$$（分钟），下山时间为$$1200\\div 120=10$$（分钟），总时间为$$20+10=30$$（分钟），总路程为$$1200\\times 2=2400$$（米） ，平均速度$$2400\\div 30=80$$（米/分）故选$$B$$。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1774", "queId": "a8d3db742e8043f6bb99817f4b526260", "competition_source_list": ["2014年迎春杯三年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "湖边种着一排柳树，每两棵树之间相距$$6$$米。小明从第一棵树跑到第$$200$$棵，一共跑了（ ）米。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$1200$$米 "}], [{"aoVal": "B", "content": "$$1206$$米 "}], [{"aoVal": "C", "content": "$$1194$$米 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题"], "answer_analysis": ["解：$$\\left( 200-1 \\right)\\times 6$$ $$=199\\times 6$$ $$=1194$$（米） 答：小明一共跑了$$1194$$米。 故选：C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "358", "queId": "65d60b9735e6450e9572805a19935e05", "competition_source_list": ["2016年全国世奥赛竞赛A卷第10题", "2016年第16届世奥赛六年级竞赛决赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "奥奥去爬山，第一天在山下的旅馆住下．为了制定第二天的行程，他在网上搜索了许多的相关资料，其中同一个旅行团的几名游客分别给出了以下信息： 游客$$A$$说：我们$$7:30$$上山，$$2$$小时后，距离山顶还有$$1$$千米． 游客$$B$$说：我们上山的速度与下山的速度之比是$$2:3$$． 游客$$C$$说：我们下山用了$$1$$小时，到山脚下刚好是$$12:00$$． 游客$$D$$说：我们下山时可以抄近路，比上山的路程近$$2$$千米． 同一个旅行图里所有人上山、下山的路线、速度都相同．根据以上信息，他们在山顶游览花了（~ ~ ~ ）小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.5$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$1.4$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["我们需要分析题目中的每一句话．找到我们需要的条件．根据$$A$$的话，我们可以知道出发的时间以及上山途中时间与路程之间的关系；从$$B$$的话中，我们知道了上山速度与下山速度之比；从$$C$$的话中，我们可以知道下山的时间和到达的时间；根据最后$$D$$的话，我们可以知道上山的路程和下山的路程之间的关系．那么下山的时间为$$1$$小时，总时间也很容易求出，从$$7:30$$到$$12:00$$一共有$$4.5$$小时，要想求出游览的时间，就必须知道上山花了多少时间．设上山的路程为$$x$$千米，由$$D$$说的可知下山的路程为$$x-2$$千米，由$$C$$说的可知下山的速度为$$\\frac{x-2}{1}$$千米/小时，由$$B$$说的可知上山的速度为$$\\frac{x-2}{1}\\times \\frac{2}{3}=\\frac{2x-4}{3}$$千米/小时根据$$A$$说的列出方程$$\\frac{2x-4}{3}\\times 2+1=x$$，解得$$x=5$$，所以上山的用的时间为$$5\\div 2=2.5$$小时．他们在山顶游览花了$$4.5-2.5-2=1$$小时． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2299", "queId": "c4ee1028d361431db15f86b224a72fd6", "competition_source_list": ["2020年新希望杯六年级竞赛初赛（团战）第42题"], "difficulty": "2", "qtype": "single_choice", "problem": "在纳美克星，每小时有$$100$$分钟．该星球的时钟，时针转一圈是$$20$$小时，分针转一圈是$$100$$分钟，那么从$$0$$时$$0$$分开始，到下一次时针与分针重合，纳美克星上经过了小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{10}{19}$$ "}], [{"aoVal": "B", "content": "$$\\frac{12}{19}$$ "}], [{"aoVal": "C", "content": "$$\\frac{20}{19}$$ "}], [{"aoVal": "D", "content": "$$\\frac{24}{19}$$ "}], [{"aoVal": "E", "content": "$$\\frac{40}{19}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["暂无 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1371", "queId": "a295fdafe8df42d3807b7a376cc872dd", "competition_source_list": ["2013年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "张老师每周的周一、周六和周日都跑步锻炼$$20$$分钟，而其余日期每日都跳绳$$20$$分钟。某月他总共跑步$$5$$小时，那么这个月的第$$10$$天是( ) ", "answer_option_list": [[{"aoVal": "A", "content": "周日 "}], [{"aoVal": "B", "content": "周六 "}], [{"aoVal": "C", "content": "周二 "}], [{"aoVal": "D", "content": "周一 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["$$5$$小时$$=300$$分钟，$$300\\div 20=15$$，$$15\\div 3=5$$，所以一共经历了$$5$$个周六、日、一，但是一个月最多有四个完整的星期，所以第五个星期一定是只有三天：周六、周日、周一。并且这个月也是从周六开始的，第十天是周一。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "5", "queId": "051672eb13fd42828b94f2908319644e", "competition_source_list": ["2011年全国希望杯四年级竞赛初赛第16题"], "difficulty": "2", "qtype": "single_choice", "problem": "王强步行去公园，回来时坐车，往、返用了一个半小时，如果他来回都步行，则需要$$2$$个半小时，那么，他来回都坐车，则需~\\uline{~~~~~~~~~~}~分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->时间问题->时间计算"], "answer_analysis": ["一个半小时是$$90$$分钟，$$2$$个半小时是$$2\\times 60+30=150$$分钟．易知王强步行单程需要$$150\\div 2=75$$分钟，则坐车单程要$$90-75=15$$分钟，所以来回都坐车要$$15\\times 2=30$$分钟． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "661", "queId": "5017ae2091704e459474d685ebce0f4c", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛B卷第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把所有三位数的质数相乘，它们的积不会是一个． ", "answer_option_list": [[{"aoVal": "A", "content": "合数 "}], [{"aoVal": "B", "content": "奇数 "}], [{"aoVal": "C", "content": "偶数 "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["在质数中$$2$$是唯一的偶数，其他质数都是奇数，因为所有三位数的质数都是奇数，奇数和奇数相乘仍然是奇数，既然是若干个三位数的质数相乘，这些质数自然是他们积的因数，所以，积一定是合数，不会是质数，也不会是偶数． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "351", "queId": "a3eaa6c5acbd4f7fa609c29685c24821", "competition_source_list": ["2014年第3届广东广州羊排赛六年级竞赛第9题1分"], "difficulty": "2", "qtype": "single_choice", "problem": "琦琦身陷神秘房间中．房间大门紧锁，上面刻着一段话：``钥匙上的话只有一句是真的．''往门旁一看，挂着三把钥匙，上面各贴着一句话．金钥匙：``这把钥匙不可以打开大门．''银钥匙：``金钥匙可以打开大门．''铜钥匙：``这把钥匙不可以打开大门．''那么，． ", "answer_option_list": [[{"aoVal": "A", "content": "琦琦应该拿金钥匙打开大门 "}], [{"aoVal": "B", "content": "琦琦应该拿银钥匙打开大门 "}], [{"aoVal": "C", "content": "琦琦应该拿铜钥匙打开大门 "}], [{"aoVal": "D", "content": "三把钥匙都不能打开大门，琦琦应该另想办法 "}], [{"aoVal": "E", "content": "三把钥匙都能打开大门 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["本题的思路其实很简单，就是``找矛盾''； 1、三句话只有一句是真的；------题目条件； 2、只要存在一对矛盾则必然一真一假；------矛盾的定义； 3、金钥匙和银钥匙上写的话相互矛盾，根据以上两条，真的那句话一定是其中一句（具体到底是金还是银不重要）； 4、根据$$3$$，则铜钥匙那句一定是假的，但它上面说的本身是否定，否定为假则肯定为真，所以铜钥匙就可以打开大门． 答案选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "314", "queId": "8cb6ad5c87544a0c9db03f7a7a393c18", "competition_source_list": ["2021年超常思维竞赛小学中年级四年级竞赛初赛"], "difficulty": "3", "qtype": "single_choice", "problem": "数学家维纳是控制论的创始人。在他获得哈佛大学博士学位的授予仪式上，有人看他一脸稚气的样子，好奇地询问他的年龄。维纳的回答很有趣，他说:``我的年龄的立方是一个四位数，年龄的四次方是一个六位数，这两个数刚好把$$0\\sim9$$这$$10$$个数字全都用上了，不重也不漏.''那么，维纳这一年~\\uline{~~~~~~~~~~}~岁。(注:数$$a$$的立方等于$$a\\times a\\times a$$，数$$a$$的四次方等于$$a\\times a\\times a\\times a$$) ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$16$$ "}], [{"aoVal": "E", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["假设维纳今年的年龄为$$a$$岁，根据$$a$$的三次方是一个四位数，可以推出$$a$$不超过$$21$$，根据$$a$$的四次方是一个六位数，可以推出$$a$$不小于$$18$$.所以，$$a$$可能为$$18$$、$$19$$、$$20$$、$$21$$. 当$$a$$取$$20$$或$$21$$时，它的三次方和四次方尾数相同，与题意相违，排除; 当$$a=19$$时，$${{\\text{a}}^{4}}=130321$$，$$1$$和$$3$$重复出现，与题意相违，排除$$.$$ 所以，$$a=18$$.经验算，符合题意，所以维纳这一年$$18$$岁. "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1929", "queId": "a5c1aa6e063447b79ea800884b3fdaa1", "competition_source_list": ["2017年新希望杯六年级竞赛训练题（一）第6题", "2018年湖北武汉新希望杯六年级竞赛训练题（一）第6题"], "difficulty": "3", "qtype": "single_choice", "problem": "将五位数$$12472$$重复写$$100$$次组成一个$$500$$位数$$1247212472\\cdots 12472$$，先删去这个数中所有位于奇数位（从左往右数）上的数字，剩下的数字（顺序不变）组成一个新数：再删去新数中所有位于奇数位上的数字，剩下的数字组成一个新数：$$\\cdots \\cdots $$按上述方法一直删除下去，直到剩下一个数字为止，则最后剩下的数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["将这个$$500$$位数从左到右依次编号第$$1$$位、第$$2$$位、第$$3$$位、$$\\cdots $$、第$$500$$位，且编号不变． 第$$1$$次删数后剩下$$2$$的倍数，第$$2$$次删数后剩下$$22$$的倍数，第$$3$$次删数后剩下$$23$$的倍数，$$\\cdots \\cdots $$，因为$$256\\textless{}500\\textless{}512$$，所以最后剩下的数字在第$$256$$位，$$256\\div 5=51$$（组）$$\\cdots \\cdots 1$$（个），第$$1$$位上的数字是$$1$$，所以最后剩下的数字是$$1$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2062", "queId": "fd6c62e8452547958afb39d93d427169", "competition_source_list": ["2008年六年级竞赛创新杯", "2008年第6届创新杯六年级竞赛初赛A卷第10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "展览会上午9点开门，但早就有人排队等着入场，并且从第一个观众来到之后每分钟来到的人数是一定的，如果开3个入场口，9点9分就不再有人排队了;如果开5个入场口，9点5分就没人排队了，问第一个观众来到的时间是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "8:15 "}], [{"aoVal": "B", "content": "8:30 "}], [{"aoVal": "C", "content": "8:45 "}], [{"aoVal": "D", "content": "8:50 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题"], "answer_analysis": ["设第一个观众来到后每分钟来到的人数都是$$a$$，开门前$$x$$分钟开始有人排队，则每个入场口每分钟进场人数为:$$\\frac{\\left( x+9 \\right)a}{9\\times 3}=\\frac{\\left( x+5 \\right)a}{5\\times 5}\\Rightarrow x=45$$，9时-45分=8时15分 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3329", "queId": "e87d2e6b25934d4e911f4901424b35fa", "competition_source_list": ["2019年第24届YMO三年级竞赛决赛第9题3分", "2020年第24届YMO三年级竞赛决赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一些三位数的各位数字都不是$$0$$，且各位数字之和为$$6$$，这样的三位数共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["���举，第$$1$$位为$$1$$，有$$114$$，$$123$$，$$132$$，$$141$$， 第$$1$$位为$$2$$，有$$213$$，$$222$$，$$231$$， 第$$1$$位为$$3$$，有$$312$$，$$321$$， 第$$1$$位为$$4$$，有$$411$$， 共有：$$1+2+3+4=10$$（个）． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "421", "queId": "a969cf0d91444321aeda5bde9471a670", "competition_source_list": ["2004年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "小强、小明、小军是学校田径运动的好手，在学校运动会上他们包揽了$$60$$米短跑、铅球和跳远三项比赛的前三名（没有并列名次）。学校规定第一名得$$5$$分，第二名得$$3$$分，第三名得$$1$$分。已知他们三人中，小军得分最低，小强和小明得分相同，小强得的第一名比小明得的第一名多，那么小强的得分是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$分 "}], [{"aoVal": "B", "content": "$$12$$分 "}], [{"aoVal": "C", "content": "$$11$$分 "}], [{"aoVal": "D", "content": "$$10$$分 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["他们三人包揽了三个项目的前三名，那么他们每人都拿了三个前三名，所以他们三人的得分都为奇数，又他们三人共得了$$3$$个第一名，$$3$$个第二名，$$3$$个第三名，那么他们三人的得分之和为：$$3\\times (5+3+1)=27$$（分）。由小军得分最低知小军得分小于$$27\\div 3=9$$（分），那么小军最多得$$7$$分。若小军的得分为$$7$$分，那么小强和小明的得分为$$(27-7)\\div 2=10$$（分），不为奇数，那么小军的得分不为$$7$$分（同理小军的得分也不为$$3$$分）。若小军的得分为$$5$$分，那么小强和小明的得分都是$$(27-5)\\div 2=11$$（分），这时小军的得分为：$$3+1+1=5$$（分），小强的得分为：$$5+5+1=11$$（分），小明的得分为：$$5+3+3=11$$（分）正好满足题中所有条件，故小强的得分为$$11$$分，选C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "264", "queId": "4d6b56a7c8ee478e9236551121e47c65", "competition_source_list": ["2015年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "春季开学后，有不少同学将部分压岁钱捐给山区的贫困学生；事后，甲、乙、丙、丁$$4$$位同学有如下的对话： 甲：``丙、丁之中至少有$$1$$人捐了款'' 乙：``丁、甲之中至多有$$1$$人捐了款'' 丙：``你们$$3$$人中至少有$$2$$人捐了款'' 丁：``你们$$3$$人中至多有$$2$$人捐了款'' 已知这$$4$$位同学说的都是真话且其中恰有$$2$$位同学捐了款，那么，这$$2$$位同学是． ", "answer_option_list": [[{"aoVal": "A", "content": "甲、乙 "}], [{"aoVal": "B", "content": "丙、丁 "}], [{"aoVal": "C", "content": "甲、丙 "}], [{"aoVal": "D", "content": "乙、丁 "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["由丙的话可知：丙没有捐款；再由甲的话可知：丁捐了款；再由乙的话可知：丁、甲之中最多有$$1$$人捐款．由此推知，甲没有捐款，乙捐了款，捐款的两人为乙和丁，选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3210", "queId": "4b87950f05b0436aac2bb768d2b6b1b9", "competition_source_list": ["2005年六年级竞赛创新杯", "2005年第3届创新杯六年级竞赛复赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "若自然数$$n$$使得作竖式加法$$n+\\left( n+1 \\right)+\\left( n+2 \\right)$$不产生进位现象，便称$$n$$为``连绵数''，例如$$12$$是连绵数，$$13$$不是``连绵数''，那么不超过$$1000$$的``连绵数''共有（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "46个 "}], [{"aoVal": "B", "content": "47个 "}], [{"aoVal": "C", "content": "48个 "}], [{"aoVal": "D", "content": "49个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->乘法原理"], "answer_analysis": ["连绵数$$n$$的个位只能为$$0$$、$$1$$、$$2$$有三种情况，十位只能为$$0$$、$$1$$、$$2$$、$$3$$有四种情况；百位也只能为$$0$$、$$1$$、$$2$$、$$3$$有四种情况。那么在$$0$$$$\\sim$$$$999$$中，连绵数有$$3\\times 4\\times 4=48$$（个），而$$1000$$也为连绵数，那么在不超过$$1000$$的自然数中，连绵数有$$48+1=49$$（个），选$$D$$。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "62", "queId": "0bb9fc4aec6c4bec937b19e2191f213a", "competition_source_list": ["2018年第21届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "某啤酒厂做促销活动，$$3$$个空瓶可以换$$1$$瓶啤酒．爸爸共买了$$13$$瓶啤酒，他一共可以喝到瓶啤酒． ", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$13\\div3=4$$（瓶）$$\\ldots\\ldots1$$（瓶），（喝$$13$$瓶$$+4$$瓶）， $$（4+1）\\div3=1$$（瓶）$$\\ldots\\ldots2$$（瓶），（喝$$1$$瓶$$+1$$瓶）， 所以一共是$$13+4+1+1=19$$（瓶）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2038", "queId": "ef4c741430774ee9b0352c87c8b515f6", "competition_source_list": ["2021年新希望杯三年级竞赛初赛第30题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "【2021三年级卷第$$30$$题】某个闰年的元旦是星期日，那么这一年的$$2$$月有$$5$$个． ", "answer_option_list": [[{"aoVal": "A", "content": "星期一 "}], [{"aoVal": "B", "content": "星期二 "}], [{"aoVal": "C", "content": "星期三 "}], [{"aoVal": "D", "content": "星期四 "}], [{"aoVal": "E", "content": "星期五 "}], [{"aoVal": "F", "content": "星期六 "}], [{"aoVal": "G", "content": "星期日 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["周期问题： 元旦$$1$$月$$1$$日是星期日，则周期：星期日、星期一、星期二、星期三、星期四、星期五、星期六$$\\cdots\\cdots$$ 从$$1$$月$$1$$日到$$2$$月$$1$$日一共有$$31+1=32$$（天）， $$32\\div7=4$$（周）$$\\cdots\\cdots4$$（天）， 所以$$2$$月$$1$$日是星期三，则$$2$$月份的周期：星期三、星期四、星期五、星期六、星期日、星期一、星期二$$\\cdots\\cdots$$ 闰年的$$2$$月有$$29$$天，$$29\\div7=4$$（周）$$\\cdots\\cdots1$$（天）， 所以$$2$$月$$29$$日是星期三， 因此这一年的$$2$$月有$$5$$个星期三． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "33", "queId": "03d013b2848549b68701c3e38ea369fa", "competition_source_list": ["2016年全国华杯赛小学中年级竞赛在线模拟第3题", "2013年全国华杯赛小学中年级竞赛初赛A卷第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "小东、小西、小南、小北四个小朋友在一起做游戏时，捡到了一条红领巾，交给了老师．老师问是谁捡到的？小东说不是小西；小西说是小南；小南说小东说的不对；小北说小南说的不对．他们之中只有一个人说对了，这个人是． ", "answer_option_list": [[{"aoVal": "A", "content": "小东 "}], [{"aoVal": "B", "content": "小西 "}], [{"aoVal": "C", "content": "小南 "}], [{"aoVal": "D", "content": "小北 "}], [{"aoVal": "E", "content": "不知道 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾", "Overseas Competition->知识点->组合模块->逻辑推理"], "answer_analysis": ["由于只有一个人说对了，而小北支持小东，那么他们俩都错了，所以反对小东的小南说对了． 根据题干分析可得，小南与小北说的话是相互矛盾的，所以两人中一定有一个人说的是正确的，假设小北说的是正确的，则小南说``小东说的不对''是错，可得，小东说的对，这样与已知只有一个人说对了相矛盾，所以此假设不成立，故小南说的是正确的． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "836", "queId": "898c8c07681944bb91885b892c19cebe", "competition_source_list": ["2016年全国世奥赛竞赛A卷第11题", "2016年第16届世奥赛六年级竞赛决赛第11题"], "difficulty": "3", "qtype": "single_choice", "problem": "请从一个$$1 \\sim 9$$（缺$$8$$）这八个自然数中不重复地用这些数字构造出四个两位质数，并求出它们的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$190$$ "}], [{"aoVal": "B", "content": "$$217$$ "}], [{"aoVal": "C", "content": "$$127$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数->质数与合数判定->质数与合数的认识"], "answer_analysis": ["我们知道所有的偶数都是合数，除了$$5$$以外，个位为$$5$$的数也都是合数，这$$8$$个数中只有$$1$$、$$3$$、$$5$$、$$7$$、$$9$$放在个位才有可能是质数，所以十位上只能是$$2$$、$$4$$、$$5$$、$$6$$．这四个两位数的和是$$\\left( 2+4+5+6 \\right)\\times 10+1+3+7+9=190$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1811", "queId": "a917beda5bad426e9ad4cfea1820d0f8", "competition_source_list": ["2017年全国华杯赛小学中年级竞赛初赛模拟第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两人在春节一共得$$200$$元压岁钱，甲给乙$$40$$元，还比乙多$$10$$元，那么，原来甲比乙多~\\uline{~~~~~~~~~~}~元压岁钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$70$$ "}], [{"aoVal": "B", "content": "$$80$$ "}], [{"aoVal": "C", "content": "$$90$$ "}], [{"aoVal": "D", "content": "$$100$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->和差倍问题->和差问题"], "answer_analysis": ["甲给乙$$40$$元后还比乙多$$10$$元，那么原来甲比乙多$$90$$元． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1807", "queId": "c8f13ee524b74f249659ebd1a41924ce", "competition_source_list": ["2014年第26届广东广州五羊杯六年级竞赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明买了一瓶$$1.5$$升汽水，他把汽水倒入一个空杯子．某一时刻，瓶中的汽水和杯中的汽水刚好是各自容积的$$\\frac{3}{4}$$，那么杯子的容积是升． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.25$$ "}], [{"aoVal": "B", "content": "$$0.5$$ "}], [{"aoVal": "C", "content": "$$0.75$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["倒后瓶中还有$$1.5\\times \\frac{3}{4}=1.125$$升汽水， 杯中有$$1.5-1.125=0.375$$升汽水，占杯的$$\\frac{3}{4}$$， 故杯的容积是$$0.375\\div \\frac{3}{4}=0.5$$升． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "571", "queId": "6afeeaa9d7d640fdabbc6f810713c11f", "competition_source_list": ["2016年天津走美杯四年级竞赛初赛", "2016年走美杯四年级竞赛决赛", "2016年走美杯四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "恰好有$$12$$个不同因数的最小自然数为（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$65$$ "}], [{"aoVal": "C", "content": "$$63$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理逆应用"], "answer_analysis": ["$$12=1\\times 12=2\\times 6=3\\times 4=2\\times 2\\times 3$$ $${{2}^{11}}=2048$$，$${{2}^{5}}\\times 3=96$$，$${{2}^{3}}\\times {{3}^{2}}=72$$，$${{2}^{2}}\\times 3\\times 5=60$$，最小为$$60$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3176", "queId": "2653343de55648bebaf31f0bcdfc23b1", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$990$$有很多因数，这些因数的平均数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$110$$ "}], [{"aoVal": "B", "content": "$$115$$ "}], [{"aoVal": "C", "content": "$$117$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["$$990$$的因数有：$$1$$；$$990$$；$$2$$；$$495$$；$$3$$；$$330$$；$$5$$；$$198$$；$$6$$；$$165$$；$$9$$；$$110$$；$$10$$；$$99$$；$$11$$；$$90$$；$$15$$；$$66$$；$$18$$；$$55$$；$$22$$；$$45$$；$$30$$；$$33$$，一共$$24$$个， 平均数为：$$(1+990+2+495+3+330+5+198+6+165+9+110+10+99+11+90+15+66+18+55+22+45+30+33)\\div 24$$ $$=2808\\div 24$$ $$=117$$， 故选答案$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "762", "queId": "90efdbd02fe749ff86b8c748a27823e7", "competition_source_list": ["2012年第8届全国新希望杯小学高年级六年级竞赛复赛第6题4分"], "difficulty": "4", "qtype": "single_choice", "problem": "$$\\overline{**45}$$，$$\\overline{19*8}$$，$$\\overline{2*86}$$，$$\\overline{3*49}$$是四个四位数，其中$$*$$代表不能辨认的数字，若其中有一个数是完全平方数，那么这个数可能是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\overline{**45}$$ "}], [{"aoVal": "B", "content": "$$\\overline{19*8}$$ "}], [{"aoVal": "C", "content": "$$\\overline{2*86}$$ "}], [{"aoVal": "D", "content": "$$\\overline{3*49}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->完全平方数->平方数的简单应用"], "answer_analysis": ["完全平方数的个位只能为$$0$$、$$1$$、$$4$$、$$5$$、$$6$$、$$9$$，因此排除$$\\text{B}$$选项．如果一个完全平方数是 $$5$$的倍数，那么它至少是$$25$$的倍数，因此排除$$\\text{A}$$选项．估算$${{48}^{2}}=2304$$，$${{49}^{2}}=2401$$，排除 $$\\text{C}$$选项．经检验$${{57}^{2}}=3249$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3378", "queId": "fb7f1cb50cad4e8e8802283263280c37", "competition_source_list": ["2017年河南郑州K6联赛竞赛模拟第九套第10题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "下面的信息资料中，最适合用条形统计图表示的是（ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "某学校各学科教师人数情况 "}], [{"aoVal": "B", "content": "$$8$$月份气温变化情况 "}], [{"aoVal": "C", "content": "各种消费情况与家庭总收入的关系 "}], [{"aoVal": "D", "content": "商场$$2015$$年每月销售额的变化情况 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->统计图表->单一对象条形图"], "answer_analysis": ["条形统计图能很容易看出数量的多少；折线统计图不仅容易看出数量的多少，而且能反映数量的增减变化情况；扇形统计图能反映部分与整体的关系．根据各统计图的特点，易知选项$$A$$正确． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3470", "queId": "eb5466040bc0441c92b574c57764c049", "competition_source_list": ["2016年创新杯五年级竞赛训练题（一）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "投掷两颗骰子，出现的两个点数构成的二位数正好是一个完全平方数的可能性是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$~~~~~~ "}], [{"aoVal": "B", "content": "$$\\frac{1}{6}$$~~~~~~~~~~~~~~ "}], [{"aoVal": "C", "content": "$$\\frac{2}{9}$$~~~~~~ "}], [{"aoVal": "D", "content": "$$\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->典型问题->掷骰子"], "answer_analysis": ["二位数是完全平方数的有：$$16$$，$$25$$，$$36$$，$$49$$，$$64$$，$$81$$．显然$$49$$和$$81$$不满足，那么概率为：$$\\frac{4\\times 2}{6\\times 6}=\\frac{2}{9}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "634", "queId": "26c23f461e4b42efa4932d09a1adcf24", "competition_source_list": ["2007年华杯赛五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "植树节到了，某市举行大型植树活动，共有1430人参加植树，要把人分成人数相等的若干队，且每队人数在100至200之间，则有分法（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "3种 "}], [{"aoVal": "B", "content": "7种 "}], [{"aoVal": "C", "content": "11种 "}], [{"aoVal": "D", "content": "13种 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数"], "answer_analysis": ["只要找到100到200之间可以整除1430的数即可.1430可分解成2，5，11，13的乘积，所以可以按每组110人，130人，143人分组，共有3个方案.所以答案为A "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "128", "queId": "de7c930eb8b548669f638cb9afa1a0a0", "competition_source_list": ["2020年新希望杯一年级竞赛初赛（巅峰对决）第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙三人站成一排，甲比乙矮，丙比乙高，他们三人分别穿红、黄、蓝三种不同颜色的衣服。已知：最高的人不穿红色，丙比穿蓝色衣服的人高，最矮的人和穿红色衣服的人相邻。甲、乙、丙的衣服颜色对应是~\\uline{~~~~~~~~~~}~。 ", "answer_option_list": [[{"aoVal": "A", "content": "蓝、黄、红 "}], [{"aoVal": "B", "content": "蓝、红、黄 "}], [{"aoVal": "C", "content": "红、黄、蓝 "}], [{"aoVal": "D", "content": "黄、蓝、红 "}]], "knowledge_point_routes": ["拓展思维->能力->推理推导->言语逻辑推理"], "answer_analysis": ["根据甲比乙矮，丙比乙高，可以推理得：丙最高，然后是乙，甲最矮． 最高的人不穿红色，即丙不穿红色，丙比穿蓝色衣服的人高，说明丙和穿蓝色衣服的人不是同一个人，即丙不穿蓝色，所以丙穿黄色．最矮的人和穿红色衣服的人相邻，即甲和穿红色衣服的人相邻，说明甲和穿红色衣服的人不是同一个人，即甲不穿红色，那么甲只能穿蓝色（丙已经穿黄色了），所以剩下的乙穿红色． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "305", "queId": "459becf14f734557903998e4bc401c9a", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛A卷第22题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "在一次数学竞赛中，$$A$$，$$B$$，$$C$$，$$D$$，$$E$$五位同学分别得了前五名（没有并列同一名次的），关于各人的名次大家作出了下面的猜测： $$A$$说：``第二名是$$D$$，第三名是$$B$$.'' $$B$$说：``第二名是$$C$$，第四名是$$E$$.'' $$C$$说：``第一名是$$E$$，第五名是$$A$$.'' $$D$$说：``第三名是$$C$$，第四名是$$A$$.'' $$E$$说：``第二名是$$B$$，第五名是$$D$$.'' 结果每人都只猜对了一半，下面的排名哪一个是正确的？ ", "answer_option_list": [[{"aoVal": "A", "content": "第一名是$$E$$，第二名是$$C$$，第三名是$$B$$，第四名是$$A$$，第五名是$$D$$． "}], [{"aoVal": "B", "content": "第一名是$$C$$，第二名是$$E$$，第三名是$$B$$，第四名是$$A$$，第五名是$$D$$． "}], [{"aoVal": "C", "content": "第一名是$$D$$，第二名是$$C$$，第三名是$$B$$，第四名是$$A$$，第五名是$$E$$． "}], [{"aoVal": "D", "content": "第一名是$$E$$，第二名是$$B$$，第三名是$$C$$，第四名是$$A$$，第五名是$$D$$． "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->半真半假"], "answer_analysis": ["假设$$A$$猜的第一句是真的，那么$$B$$猜的第二句是真的，即第四名是$$E$$，那么$$C$$猜的``E是第一名''是错的，$$A$$是第五名，那么$$D$$猜的$$C$$是第三名是对的，那么$$B$$就是第一名，从而$$E$$说的全是错的，所以假设不成立．所以$$A$$猜的第二句是真的，即$$B$$是第三名，那么$$D$$猜的第一句是错的，从而$$A$$是第四名，所以$$C$$猜的第二句是错的，$$E$$是第一名，从而$$B$$猜的$$C$$是第二名是对的，$$E$$猜的第五名是$$\\text{D}$$正确，所以，第一名是$$E$$，第二名是$$C$$，第三名是$$B$$，第四名是$$A$$，第五名是$$D$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1701", "queId": "a8638045b9594793a061e85f680524f0", "competition_source_list": ["2006年第7届中环杯五年级竞赛决赛第3题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "一本书的页码是连续的自然数，在将这些页码加起来的时候，某个页码被加了两次，得到的不正确的结果是$$1997$$．则这个被加了两次的页码是第~\\uline{~~~~~~~~~~}~页． ", "answer_option_list": [[{"aoVal": "A", "content": "$$31$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$44$$ "}], [{"aoVal": "E", "content": "$$49$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["设共$$n$$页，被加了两次的页码是$$x$$， 则$$n\\left( n+1 \\right)\\div 2\\leqslant 1997$$， 且$$x\\leqslant n$$， 用特殊值法求得$$n=62$$， 则被加了两次的页码是： $$1997-62\\times \\left( 62+1 \\right)\\div 2=x$$， $$x=1997-63\\times31$$， $$x=1997-1953$$， $$x=44$$； 故答案为：$$44$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3019", "queId": "a5f01bb3646545dba3379294f4cef974", "competition_source_list": ["2011年六年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "有$$8$$个数，$$0.\\dot{5}\\dot{1}$$，$$0.5\\dot{1}$$，$$\\frac{2}{3}$$，$$\\frac{5}{9}$$，$$\\frac{24}{47}$$，$$\\frac{13}{25}$$是其中的$$6$$个，如果按照从小到大的顺序排列，第$$4$$个数是$$0.5\\dot{1}$$，那么从大到小排列时，第$$4$$个数是( ) ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.\\dot{5}\\dot{1}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{24}{47}$$ "}], [{"aoVal": "D", "content": "$$\\frac{13}{25}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合"], "answer_analysis": ["六个数的大小关系为$$\\frac{24}{47} \\textless{} 0.5\\dot{1} \\textless{} 0.\\dot{5}\\dot{1} \\textless{} \\frac{13}{25} \\textless{} \\frac{5}{9} \\textless{} \\frac{2}{3}$$，根据已知条件，$$0.5\\dot{1}$$为从小到大第四个，所以从大到小排列第四个数为$$0.\\dot{5}\\dot{1}$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2301", "queId": "f7556eaf07644e3d8b3548d81a747087", "competition_source_list": ["六年级上学期其它第12讲", "2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙二人以均匀的速度分别从$$A$$，$$B$$两地同时出发，相向而行，他们第一次相遇地点离$$A$$地$$4$$千米，相遇后二人继续前进，走到对方出发点后立即返回，在距$$B$$地$$3$$千米处第二次相遇，求两次相遇地点之间的距离是多少千米？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$4\\times 3=12$$（千米），通过画图，我们发现甲走了一个全程多了回来那一段，就是距$$B$$地的$$3$$千米，所以全程是$$12-3=9$$（千米），两次相遇点相距$$9-\\left( 3+4 \\right)=2$$（千米）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3126", "queId": "fe6cd8dee82a42b8a8988f5268092665", "competition_source_list": ["2006年第4届创新杯五年级竞赛初赛B卷第10题", "2006年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "红星小学礼堂共24排座位，每排30座位，全校650人到礼堂开会，那么，至少有( )排座位上坐的人数相同. ", "answer_option_list": [[{"aoVal": "A", "content": "3 "}], [{"aoVal": "B", "content": "4 "}], [{"aoVal": "C", "content": "5 "}], [{"aoVal": "D", "content": "6 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["每一排坐的人数都不同最多可坐$$\\left( 7+30 \\right)\\div 2\\times 24=444$$人 每一排坐的人数有$$2$$个相同最多可坐$$30\\times 2+29\\times 2+......+19\\times 2=588 \\textless{} 650$$ 每一排坐的人数有$$3$$个相同最多可坐$$30\\times 3+29\\times 3+......+23\\times 3=636 \\textless{} 650$$ 每一排坐的人数有$$4$$个相同最多可坐$$30\\times 4+29\\times 4+......+25\\times 4=660\\textgreater650$$ 故，至少有$$4$$排座位上人数相同 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1710", "queId": "72c0104114e9461b8ad1aae84f165a70", "competition_source_list": ["2006年第4届创新杯五年级竞赛初赛B卷第6题", "2006年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "购买十种货物$${{A}_{1}}$$，$${{A}_{2}}$$，$${{A}_{3}}\\cdots {{A}_{10}}$$，如果购买件数依次为$$1$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$，$$8$$，$$9$$，$$10$$，$$11$$件，共需人民$$1995$$元，如果购买件数依次为$$1$$，$$5$$，$$7$$，$$9$$，$$11$$，$$13$$，$$15$$，$$17$$，$$19$$，$$21$$件，共需人民币$$3000$$元，那么各购买一件共需人民币( ). ", "answer_option_list": [[{"aoVal": "A", "content": "1000元 "}], [{"aoVal": "B", "content": "900元 "}], [{"aoVal": "C", "content": "990元 "}], [{"aoVal": "D", "content": "980元 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型"], "answer_analysis": ["将第一种方案购买件数同时都扩大2倍，则购买货物$${{A}_{1}}$$、$${{A}_{2}}$$、$${{A}_{3}}$$、$$\\cdots$$、$${{A}_{10}}$$的件数依次为$$2$$，$$6$$，$$8$$，$$10$$，$$12$$，$$14$$，$$16$$，$$18$$，$$20$$，$$22$$件，需人民币$$1995\\times 2=3990$$(元)，所以各购买一件共需人民币$$3990-3000=990$$(元). "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3465", "queId": "d47c1e8b52474512b3b8dae7366ee8cf", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "2019年第$$24$$届$$YMO$$一年级竞赛决赛第$$6$$题$$3$$分 The sum of the ten digits and the single digit is $$12$$, and there are a total of such two-digit numbers. 十位数字和个位数字相加的和是$$12$$，这样的两位数一共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->整数分拆->简单拆分->加法拆数（指定个数）", "Overseas Competition->知识点->计数模块->枚举法综合->枚举法"], "answer_analysis": ["两位数的十位数字与个位数字之和为$$12$$， 而数字均为$$0\\sim 9$$，故枚举得： $$93$$，$$84$$，$$75$$，$$66$$，$$57$$，$$48$$，$$39$$共$$7$$个． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1101", "queId": "0c495afbdc884e7ea254ae7ae3820125", "competition_source_list": ["2017年全国亚太杯五年级竞赛初赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "小明、小红、小刚三人拥有的藏书数量之比为$$3:5:8$$，三人一共藏书$$48$$本，那么小刚的藏书数量是多少本？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->比例应用题->按比分配"], "answer_analysis": ["设小刚的藏书数量为$$8$$份，则小明和小红的藏书数量分别为$$3$$份和$$5$$份所以一份为$$48\\div \\left( 3+5+8 \\right)=3$$（本）． 所以小刚的藏书数量是$$8\\times 3=24$$ （本）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3223", "queId": "3539eb83adb24f4cbc475eeb04df638d", "competition_source_list": ["2017年全国华杯赛竞赛初赛模拟试卷2第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "两个小数的整数部分分别是$$5$$和$$9$$，那么这两个小数的乘积的整数部分有（ ~）中可能的值． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["设整数部分为$$5$$的小数为$$a$$，则$$5\\leqslant a\\textless6$$，整数部分为$$9$$的小数为$$b$$，则$$9\\leqslant b\\textless10$$，那么$$5\\times9\\leqslant a\\times b\\textless6\\times10$$，所以$$a\\times b$$的整数部分可以等于$$45$$至$$59$$之间的任意整数，共$$15$$种可能． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2246", "queId": "f1958ef111784958811c3a1c2ab624d1", "competition_source_list": ["2011年第7届全国新希望杯六年级竞赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列时刻中，时针和分针所成的角最接近$$30{}^{}\\circ $$是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3:27$$ "}], [{"aoVal": "B", "content": "$$4:17$$ "}], [{"aoVal": "C", "content": "$$5:14$$ "}], [{"aoVal": "D", "content": "$$6:22$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->行程模块->时钟问题->时钟上的追及问题->已知时间求角度"], "answer_analysis": ["b选项：$4\\times30-（6-0.5）=26.5^{\\circ}$，离$$30$$度最近，故选$$b$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "845", "queId": "9671beee77574902b189ccaddcb00711", "competition_source_list": ["2011年第9届全国创新杯小学高年级六年级竞赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "符号$$\\left[ x \\right]$$表示不大于$$x$$的最大整数，例如$$\\left[ 5 \\right]=5$$、$$\\left[ 6.31 \\right]=6$$，如果$$\\left[ \\frac{3x+7}{7} \\right]=4$$，这样的正整数$$x$$有． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$个 "}], [{"aoVal": "B", "content": "$$4$$个 "}], [{"aoVal": "C", "content": "$$5$$个 "}], [{"aoVal": "D", "content": "$$2$$个 "}]], "knowledge_point_routes": ["知识标签->拓展思维->数论模块->高斯记号->高斯记号的复杂应用"], "answer_analysis": ["根据高斯函数的性质，可得$4\\leqslant\\dfrac{3x+7}{7}\\lt5$，解得$$7\\leqslant x\\leqslant 9$$，对应的整数共$$3$$个数． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1184", "queId": "f0fb91cbb3194287be39d6b9ef325cdb", "competition_source_list": ["2015年第11届全国新希望杯五年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "某班五名同学在一次数学考试中的平均成绩为$$105.4$$分，其中四名同学的成绩分别为$$120$$分、$$90$$分、$$102$$分、$$99$$分，那么另外一名同学的成绩是（~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$106$$分 "}], [{"aoVal": "B", "content": "$$110$$分 "}], [{"aoVal": "C", "content": "$$114$$分 "}], [{"aoVal": "D", "content": "$$116$$分 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型"], "answer_analysis": ["$$105.4\\times 5-120-90-102-99=116$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3121", "queId": "f9ae7cd754574f408aa4a4f9e1262255", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "用$$p$$代表黄金分割数，是无限不循环小数，若取小数点后三位时，$$p=0.618$$；取小数点后十位时，$$p=0.6180339887$$这，则（~ ）是一个错误的判断． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{p}$$代表一个特定的数 "}], [{"aoVal": "B", "content": "$${{p}^{2}}$$代表一个特定的数 "}], [{"aoVal": "C", "content": "$$p$$代表一个尚未确定的数 "}], [{"aoVal": "D", "content": "$$\\frac{1-p}{p}$$代表一个特定的数 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->循环小数->循环小数的概念"], "answer_analysis": ["如果圆周率$$\\pi $$，黄金分割数量个特定的无限不循环小数，则$$\\frac{1}{p}$$，$${{p}^{2}}$$和$$\\frac{1-p}{p}$$都是特定的数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2510", "queId": "8b148c006d904c0c86890dc9bdeaa7c3", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "对于任何自然数，定义$$n!=1\\times 2\\times 3\\times \\cdots \\times n$$．那么算式$$2022!-3!$$的计算结果的个位数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["由新定义：$$n!=1\\times 2\\times 3\\times \\ldots\\times n$$得： $$2014!=1\\times 2\\times 3\\times 4\\times 5\\times \\ldots\\times 2021\\times 2022$$ $$=1\\times 3\\times 4\\times 6\\times 7\\times 8\\times \\ldots\\times 2021\\times 2022\\times 10$$ 所以$$1\\times 3\\times 4\\times 6\\times 7\\times 8\\times \\ldots\\times 2021\\times 2022\\times 10$$是$$10$$的倍数， 所以$$2022!$$的个位数为$$0$$； $$3!=1\\times 2\\times 3=6$$ 所以$$2022!-3!$$的个位数也就为：$$10-6=4$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1041", "queId": "070f7b7b9f0d4a8b92587b575f200be2", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第4题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "``数学趣味题数学趣味题$$\\cdots \\cdots $$''依次重复排列，第$$2019$$个字是． ", "answer_option_list": [[{"aoVal": "A", "content": "趣 "}], [{"aoVal": "B", "content": "味 "}], [{"aoVal": "C", "content": "题 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["观察题干可知，$$5$$个字一个循环周期，分别按照数、学、趣、味、题，依次排列，据此求出第$$2019$$个字是第几个循环周期的第几个字即可解答问题． $$2019\\div 5=403$$（组）$$\\cdots \\cdots 4$$（个）， 所以第$$2019$$个字是第$$403$$循环周期的第$$4$$个字，是``味''． 答：第$$2019$$个字是味． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1160", "queId": "112e904a22e14040a556b9ef625cdc7d", "competition_source_list": ["2009年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "2008年父亲的年龄是兄弟二人年龄之和的2倍，是兄弟两人年龄差的7倍，父子三人年龄之和为84，那么哥哥生于( )年. ", "answer_option_list": [[{"aoVal": "A", "content": "1995 "}], [{"aoVal": "B", "content": "1990 "}], [{"aoVal": "C", "content": "1992 "}], [{"aoVal": "D", "content": "1989 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之当当型->年龄轴"], "answer_analysis": ["设父亲年龄为x岁，兄弟二人的年龄分别为a岁、b岁，$$\\left( a\\textgreater b \\right)$$.则$$x=2\\left( a+b \\right)=7\\left( a-b \\right)$$，且$$x+a+b=84$$.将$$a+b=84-x$$代入$$x=2\\left( a+b \\right)$$得$$x=2\\left( 84-x \\right)$$，$$3x=84\\times 2$$.从而$$x=56$$，所以$$a+b=28$$，$$ab=8$$.因此$$a=18$$，$$b=10$$，哥哥生于1990年. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2631", "queId": "ac001b3dc65a4d47a761b6b355e638fa", "competition_source_list": ["2021年新希望杯六年级竞赛初赛第11题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "小糊涂遇到一个问题：比较$$\\frac{99}{100}$$，$$ \\frac{100}{101}$$，$$\\frac{199}{201}$$的大小．他感到很迷糊，请你帮他找到正确的答案． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{199}{201}$$ "}], [{"aoVal": "B", "content": "$$\\frac{199}{201}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{99}{100}$$ "}], [{"aoVal": "C", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{100}{101}$$ "}], [{"aoVal": "D", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater\\frac{99}{100}$$ "}], [{"aoVal": "E", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{99}{100}\\textgreater{} \\frac{199}{201}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数感认知->分数数字加工"], "answer_analysis": ["$$\\frac{99}{100}=1- \\frac{1}{100}=1- \\frac{2}{200}$$， $$\\frac{100}{101}=1- \\frac{1}{101}=1- \\frac{2}{202}$$， $$\\frac{199}{201}=1- \\frac{2}{201}$$， 因为$$\\frac{2}{202}\\textless{} \\frac{2}{201}\\textless{} \\frac{2}{200}$$， 所以$$1- \\frac{2}{202}\\textgreater1- \\frac{2}{201}\\textgreater1- \\frac{2}{200}$$， 即$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{99}{100}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2427", "queId": "100e6e2a9f6e4fb8b264121451fe6b62", "competition_source_list": ["2020年福建河仁杯六年级竞赛初赛A卷第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$\\frac{3}{{{1}^{2}}+1}-\\frac{5}{{{2}^{2}}+2}+\\frac{7}{{{3}^{2}}+3}-\\frac{9}{{{4}^{2}}+4}+\\cdots +\\frac{4039}{{{2019}^{2}}+2019}-\\frac{4041}{{{2020}^{2}}+2020}$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2018}{2019}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2020}{2021}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2019}{2020}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2021}{2022}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\frac{3}{{{1}^{2}}+1}-\\frac{5}{{{2}^{2}}+2}+\\frac{7}{{{3}^{2}}+3}-\\frac{9}{{{4}^{2}}+4}+\\cdots +\\frac{4039}{{{2019}^{2}}+2019}-\\frac{4041}{{{2020}^{2}}+2021}$$ $$=\\frac{3}{1\\times 2}-\\frac{5}{2\\times 3}+\\frac{7}{3\\times 4}-\\frac{9}{4\\times 5}+\\cdots +\\frac{4039}{2019\\times 2020}-\\frac{4041}{2020\\times 2021}$$ $$=\\left( 1+\\frac{1}{2} \\right)-\\left( \\frac{1}{2}+\\frac{1}{3} \\right)+\\left( \\frac{1}{3}+\\frac{1}{4} \\right)-\\left( \\frac{1}{4}+\\frac{1}{5} \\right)+\\cdots +\\left( \\frac{1}{2019}+\\frac{1}{2020} \\right)-\\left( \\frac{1}{2020}+\\frac{1}{2021} \\right)$$ $$=1+\\frac{1}{2}-\\frac{1}{2}-\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{4}-\\frac{1}{5}+\\cdots +\\frac{1}{2019}+\\frac{1}{2020}-\\frac{1}{2020}-\\frac{1}{2021}$$ $$=1-\\frac{1}{2021}$$ $$=\\frac{2020}{2021}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "108", "queId": "d9cb6c5c89ba4e13b97eed2588f694d8", "competition_source_list": ["2014年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "小华下午$$2$$点要到少年宫参加活动，但他的手表每个小时快了$$4$$分钟，他特意在上午$$10$$点时对好了表。当小华按照自己的表于下午$$2$$点到少年宫时，实际早到了（ ）分钟。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->时间问题->时间计算"], "answer_analysis": ["解：依题意可知： 上午十点对好表，标准钟每小时走$$60$$格，小华的表快$$4$$分钟每小时走$$64$$格。路程比例为$$15:16$$。 当小华的表为下午$$2$$点时，小华的表走了$$4$$圈共$$240$$格。 根据比例关系设标准钟走的路程为$$x$$，则有$$15:16=x:240$$，解得$$x=225$$。 $$240-225=15$$（分钟）。 故选：B "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1664", "queId": "4ece93595c12425ebf3b895cc4982e44", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "$$2019$$年元旦是星期二，当年的国庆节是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["从$$2019$$年的$$1$$月$$1$$日至$$2019$$年的$$10$$月$$1$$日共：$$31+28+31+30+31+30+31+31+30+1=274$$（天）， 根据``二、三、四、五、六、日、一''的周期规律：$$274\\div 7=39$$（周）$$\\cdots \\cdots 1$$（天）． 所以$$10$$月$$1$$日是这个周期的第$$1$$天，即星期二． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1795", "queId": "924aee10a21d474baae0f11132ccc5d4", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "君君、嘉嘉、旭旭、宇宇四位同学这次考试的平均分是$$95$$分，如果去掉宇宇的成绩，则其他三位同学的平均分是$$94$$分，那么宇宇这次考了分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$ "}], [{"aoVal": "B", "content": "$$97$$ "}], [{"aoVal": "C", "content": "$$98$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["四位同学的总分为：$$95\\times4=380$$（分）， 去掉宇宇总分：$$94\\times3=282$$（分）， 故，宇宇的分数为：$$380-282=98$$（分）． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1507", "queId": "686e06716f0d44aba382ad946ebafaea", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "猴子摘桃，第一天摘了树上桃子的一半，第二天又摘了余下桃子的一半多$$1$$个，这时树上还有$$9$$个桃子，原来树上有个桃子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$42$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["根据题目叙述倒推，第二天又摘了余下桃子的一半多$$1$$个，这时树上还有$$9$$个桃子，所以第一天摘完还剩桃子$$\\left(9+1\\right)\\div\\frac{1}{2}=20$$（个），因为第一天摘了树上桃子的一半，所以原来树上有$$20\\div\\frac{1}{2}=40$$（个），所以答案为$$40$$个． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1203", "queId": "34633ce4d27d44eca83a7dbfb7f88d5f", "competition_source_list": ["2017年河南郑州联合杯小学高年级六年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "甲数比乙数少$$20 \\% $$，那么，乙数比甲数多（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20 \\% $$ "}], [{"aoVal": "B", "content": "$$25 \\% $$ "}], [{"aoVal": "C", "content": "$$30 \\% $$ "}], [{"aoVal": "D", "content": "$$22.5 \\% $$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->量率对应已知单位1"], "answer_analysis": ["甲数比乙数少$$\\frac{1}{5}$$，设乙为5份，甲为4份，乙数比甲数多$$\\frac{5-4}{4}=\\frac{1}{4}=25 \\%$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2754", "queId": "768fc09119714d7ca65b3c892dcb2a30", "competition_source_list": ["2016年IMAS小学高年级竞赛第二轮检测试题第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "请问算式$$666+669+699+999$$的值为多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2433$$ "}], [{"aoVal": "B", "content": "$$2970$$ "}], [{"aoVal": "C", "content": "$$2973$$ "}], [{"aoVal": "D", "content": "$$3030$$ "}], [{"aoVal": "E", "content": "$$3033$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["$$666+669+699+999$$ $$=670+670+700+1000-4-1-1-1$$ $$=3040-7$$ $$=3033$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3358", "queId": "6e9ae6622a3040229bb3eea76fa740f7", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{(2019 Youth Mathematics Olympics, Primary 1, Question \\#6)~} The sum of the ten digits and the single digit is $$12$$, and there are a total of such two-digit numbers. 十位数字和个位数字相加的和是$$12$$，这样的两位数一共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->计数模块->枚举法综合->枚举法", "拓展思维->能力->逻辑分析"], "answer_analysis": ["两位数的十位数字与个位数字之和为$$12$$， 而数字均为$$0\\sim 9$$，故枚举得： $$93$$，$$84$$，$$75$$，$$66$$，$$57$$，$$48$$，$$39$$共$$7$$个． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1474", "queId": "561d2b6171c4464eb9e359c17b9bc6ca", "competition_source_list": ["其它改编自2014年全国希望杯六年级竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知三个分数的和是$$\\frac{10}{11}$$ ，并且它们的分母相同，分子的比是$$2：3：4$$ ，那么，这三个分数中最大的是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{20}{99}$$ "}], [{"aoVal": "B", "content": "$$\\frac{30}{99}$$ "}], [{"aoVal": "C", "content": "$$\\frac{40}{99}$$ "}], [{"aoVal": "D", "content": "$$\\frac{50}{99}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["假设分数的分母为$$y$$，分子分别为$$2x$$，$$3x$$，$$4x$$，根据题意$$\\frac{2x}{y}+\\frac{3x}{y}+\\frac{4x}{y}=\\frac{10}{11}$$，所以解出最大分数为$$\\frac{4x}{y}=\\frac{40}{99}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "78", "queId": "98b5657c8d2e47c2a3d135db304cb333", "competition_source_list": ["1989年华杯赛六年级竞赛初赛"], "difficulty": "3", "qtype": "single_choice", "problem": "一副扑克牌有四种花色，每种花色有$$13$$张，此外还有两张王牌，从中任意抽牌。那么，最少要抽张牌，才能保证有$$4$$张牌是同一花色。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["如果在最不利的情况下能完成目标，则保证在任何情况下都能完成。最不利的情况是：抽的前$$12$$张是$$4$$种花色各$$3$$张，再抽两张王牌，这时抽第$$15$$张，无论是哪种花色，都能保证凑成$$4$$张牌同一花色。所以至少要抽$$15$$张牌，才能保证有四张牌是同一花色的。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3061", "queId": "b8b4f20de24b4a0e9a34ae50175a3365", "competition_source_list": ["2020年希望杯五年级竞赛模拟第14题", "2020年新希望杯五年级竞赛第14题"], "difficulty": "1", "qtype": "single_choice", "problem": "艾迪在闯关游戏中遇到一个计算题：$$0.004186\\times 8812345.321$$，下列选项中最接近计算结果的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3200$$ "}], [{"aoVal": "B", "content": "$$3600$$ "}], [{"aoVal": "C", "content": "$$32000$$ "}], [{"aoVal": "D", "content": "$$36000$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$0.004186\\times 8812345.321\\approx 0.004\\times 9000000=36000$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2916", "queId": "adc51b42382d4a719e3829a04bc85277", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（三）第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "小敏参加数学竞赛，题目的类型分为$$3$$种：$$A$$类题目每题$$7$$分，$$B$$类题目每题$$4$$分，$$C$$ 类题目每题$$1$$分．小敏考完后核对答案发现她一共做对了$$20$$道题目，那么小敏可能的得分是分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$51$$ "}], [{"aoVal": "B", "content": "$$67$$ "}], [{"aoVal": "C", "content": "$$86$$ "}], [{"aoVal": "D", "content": "$$100$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为$$1$$，$$4$$，$$7$$都是除以$$3$$余$$1$$的数，小敏答对$$20$$道题，$$\\left( 20\\times 1 \\right)\\div 3=6\\ldots \\ldots 2$$，所以她的分数也一定是除以$$3$$余$$2$$的数，综合比较$$A$$，$$B$$，$$C$$，$$D$$四个选项，不难发现，只有$$C$$选项符合题意，即小敏可能的得分是$$86$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "930", "queId": "b863c4d0799c4e68b1006a39cf17fc09", "competition_source_list": ["2018年迎春杯小学高年级竞赛决赛A卷第9题12分"], "difficulty": "3", "qtype": "single_choice", "problem": "算式$$\\underbrace{20182018\\cdots 2018}_{18个2018}\\times 151514141313\\cdots 0707$$的计算结果中有~\\uline{~~~~~~~~~~}~个奇数数字． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$63$$ "}], [{"aoVal": "C", "content": "$$72$$ "}], [{"aoVal": "D", "content": "$$81$$ "}], [{"aoVal": "E", "content": "$$96$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想", "海外竞赛体系->知识点->数论模块->质数与合数->特殊质数运用->特殊质数2"], "answer_analysis": ["设$$A=151514141313\\cdots 0707$$， （$$1$$）由四位数截断法知$$A$$是$$9999$$的倍数，不妨设$$A=9999B$$，则$$B$$是个$$32$$位数； （$$2$$）原式$$=\\underbrace{20182018\\cdots 2018}_{18个2018}\\times 9999\\times B=\\underbrace{99999999\\cdots 9999}_{18个9999}\\times 2018\\times B=\\underbrace{99\\cdots 9}_{72个9}\\times 2018\\times B$$ 因为$$2018\\times B$$是一个不超过$$72$$位的数，无论它是多少，$$\\underbrace{99\\cdots 9}_{72个9}\\times 2018\\times B$$的计算结果中都是$$72$$个奇数数字． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3050", "queId": "f7fb117acadf4270a58dca23c6eaca94", "competition_source_list": ["2020年广东广州羊排赛六年级竞赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$\\frac{5}{7}$$，$$ \\frac{2}{13}$$， $$\\frac{3}{4}$$，$$\\frac{10}{17}$$ 几个分数中，按从大到小排列，排在第二位的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{7}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{13}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{10}{17}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["通分子，都变成$$30$$去比较大小 则从大到小排列排在第二位的是$$\\frac{5}{7}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1570", "queId": "ff8080814518d5240145190910980309", "competition_source_list": ["2014年全国迎春杯五年级竞赛初赛第3题"], "difficulty": "0", "qtype": "single_choice", "problem": "一辆大卡车一次可以装煤$$2.5$$吨，现在要一次运走$$48$$吨煤，那么至少需要（~~~ ~）辆这样的大卡车． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["$$48\\div 2.5=19\\cdots 0.5$$，$$19+1=20$$(辆)． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1971", "queId": "a1d10a2ebf3741328a06fc9d3cfe85e4", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "下列各组数中，平均数较大的是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$与$$101$$之间的$$2$$倍数 "}], [{"aoVal": "B", "content": "$$1$$与$$101$$之间的$$3$$倍数 "}], [{"aoVal": "C", "content": "$$1$$与$$101$$之间的$$4$$倍数 "}], [{"aoVal": "D", "content": "$$1$$与$$101$$之间的$$6$$倍数 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["根据等差数列相关概念可知，一个等差数列的平均数，其实就是这个数列的首项与末项的平均数，简化计算后易知$$1$$与$$101$$之间$$4$$的倍数的平均数最大，其值为$$(4+100)\\div 2=52$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1131", "queId": "33dde079f5834b28baea4e56ef5873a4", "competition_source_list": ["2014年广东广州羊排赛六年级竞赛第9题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "往$$100$$克浓度为$$20 \\%$$的盐水中加入$$10$$克水和$$10$$克盐，浓度变成． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20 \\%$$ "}], [{"aoVal": "B", "content": "$$25 \\%$$ "}], [{"aoVal": "C", "content": "$$27.5 \\%$$ "}], [{"aoVal": "D", "content": "$$30 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->浓度问题->浓度基本题型->已知溶质溶液求浓度"], "answer_analysis": ["最后浓度变为$$\\frac{100\\times 20 \\%+10}{100+10+10}\\times 100 \\%=25 \\%$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1627", "queId": "bf082ceb76294cbaba52e7082ce871a2", "competition_source_list": ["2015年第4届广东广州羊排赛六年级竞赛第10题1分"], "difficulty": "0", "qtype": "single_choice", "problem": "某款机器人用``$$+$$''\\,``$$-$$''符号分别表示向右走、向左走．如向右走$$3$$步记为``$$+3$$''，向左走$$2$$步记为``$$-2$$''．某天内，机器人的行走记录如下： $$+3$$、$$-5$$、$$+4$$、$$-2$$、$$+3$$ 最终机器人的位置在起点~\\uline{~~~~~~~~~~}~方~\\uline{~~~~~~~~~~}~步的地方． ", "answer_option_list": [[{"aoVal": "A", "content": "左，$$2$$ "}], [{"aoVal": "B", "content": "左，$$3$$ "}], [{"aoVal": "C", "content": "右，$$2$$ "}], [{"aoVal": "D", "content": "右，$$3$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->加减法应用->加减法应用顺口溜"], "answer_analysis": ["相当于共向右走了$$3+4+3=10$$（步），共向左走了$$5+2=7$$（步），$$10-7=3$$（步），相当于在起点右方$$3$$步的地方． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1910", "queId": "aa17b0eeda1044eeab6524b91d0c7619", "competition_source_list": ["2017年全国华杯赛小学中年级竞赛初赛模拟第1题", "小学中年级三年级上学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两人在��节一共得$$200$$元压岁钱，甲给乙$$40$$元，甲和乙钱数相等，那么，原来甲得了~\\uline{~~~~~~~~~~}~元压岁钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$150$$ "}], [{"aoVal": "B", "content": "$$140$$ "}], [{"aoVal": "C", "content": "$$130$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->和差问题->二量和差问题->明和暗差"], "answer_analysis": ["因为甲给乙$$40$$元，甲和乙钱数相等，所以甲$$200+40\\times$$ 2=280（元），$$280\\div$$ 2=140（元） "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2831", "queId": "5c63e8f531764df482c188b6d5f2159c", "competition_source_list": ["2020年新希望杯六年级竞赛（2月）第25题"], "difficulty": "1", "qtype": "single_choice", "problem": "比较大小：$$1+ \\frac{1}{2^{2}}+ \\frac{1}{3^{2}}+ \\frac{1}{4^{2}} \\cdots + \\frac{1}{2020^{2}}$$~\\uline{~~~~~~~~~~}~$$2$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\textgreater$$ "}], [{"aoVal": "B", "content": "$$\\textless{}$$ "}], [{"aoVal": "C", "content": "$$=$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}} ~\\textless{} ~\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{1999\\times 2020}$$ $$=1-\\frac{1}{2020}=\\frac{2019}{2020}$$． ∴$$1+\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}} ~\\textless{} ~1+\\frac{2019}{2020} ~\\textless{} ~2$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "197", "queId": "da135a92e8204ea59f303092b26ecb62", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛初赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "质料、型号相同的红、白、黑色袜子各$$5$$双，拆开后混装在暗箱中，从中摸出若干只袜子，要能配成$$2$$双（只要两只袜子同色，即为一双），至多摸出只． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->抽象概括"], "answer_analysis": ["考虑最不利情况，先摸到一双，两只，再每种颜色各一只，共三只，再任意摸一只就可以配成两双，所以一共至多$$6$$只． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "758", "queId": "450ab68529b348078d09b09d2de28aa9", "competition_source_list": ["2016年全国小学生数学学习能力测评四年级竞赛复赛第7题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "下列各数中，读出的``零''最多． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60006000$$ "}], [{"aoVal": "B", "content": "$$60606060$$ "}], [{"aoVal": "C", "content": "$$6006060$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->数与数字->数、数位、数字的认识"], "answer_analysis": ["$$60006000$$读作六千万六千，没有读出``零''； $$60606060$$读作六千零六十万六千零六十，读了两个``零''； $$6006060$$读作六百万六千零六十，读了一个``零''． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2422", "queId": "25e9f239d53c4a4fbc66e6b8516bf58a", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第1题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "请问算式$$19\\times 1+19\\times 3+19\\times 5+19\\times 7+\\cdots +19\\times 19$$的值等于什么？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1900$$ "}], [{"aoVal": "B", "content": "$$1919$$ "}], [{"aoVal": "C", "content": "$$2900$$ "}], [{"aoVal": "D", "content": "$$2919$$ "}], [{"aoVal": "E", "content": "$$3800$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数乘法巧算之提取公因数（普通型）"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde19\\times 1+19\\times 3+19\\times 5+19\\times 7+\\cdots +19\\times 19$$ $$=19\\times \\left( 1+3+5+7+\\cdots +19 \\right)$$ $$=19\\times \\frac{\\left( 1+19 \\right)\\times 10}{2}$$ $$=19\\times 100$$ $$=1900$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2884", "queId": "6f2a5e83ff2f4d2493e6995377cd9832", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛A卷第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "圆锥的底面圆的周长和它的体积． ", "answer_option_list": [[{"aoVal": "A", "content": "成正比例 "}], [{"aoVal": "B", "content": "成反比例 "}], [{"aoVal": "C", "content": "不成比例 "}], [{"aoVal": "D", "content": "无选项 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->几何模块->立体图形->圆柱与圆锥->圆锥的基本公式"], "answer_analysis": ["圆锥的底面周长表示为$$2\\pi r$$， 圆锥的体积表示为$$\\frac{1}{3}\\pi {{r}^{2}}$$， 两者之间没有任何关系，所以故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "161", "queId": "8784f038e0e8496fa777ed4ee14f5a46", "competition_source_list": ["2011年第7届全国新希望杯小学高年级六年级竞赛第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "盒子里装有分别写着$$1$$，$$2$$，$$3$$，$$4$$，\\ldots，$$100$$的黄色卡片各一张，我们称如下操作为一次操作；从盒子里取出$$m(7\\leqslant m\\leqslant 10)$$张卡片，算出这$$m$$张卡片上各数之和减去$$27$$的差，将写在一张红色卡片上（不放回）．若干次操作之后，盒子里的卡片全部被取出，若所有红色卡片上的数字之和为$$n$$，那么$$n$$的最大可能值减去最小可能值等于． ", "answer_option_list": [[{"aoVal": "A", "content": "$$108$$ "}], [{"aoVal": "B", "content": "$$96$$ "}], [{"aoVal": "C", "content": "$$88$$ "}], [{"aoVal": "D", "content": "$$81$$ "}], [{"aoVal": "E", "content": "$$54$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["先算卡片的总值． 最小值$$100\\div 10=10$$次，共减$$10\\times 27$$； 最多$$14$$次，共减去$$14\\times 27$$， 因此差为$$(14-10)\\times 27=108$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2260", "queId": "57cb09674dcd44a389bc5fbe4fcdab95", "competition_source_list": ["2017年第13届湖北武汉新希望杯六年级竞赛决赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "泸昆高铁最后一段贵阳至昆明于$$2016$$年$$12$$月$$28$$日开通运营，这对我国``一带一路''战略的实施和区域经济发展都有着重大意义．$$G1375$$次高铁$$11:16$$从上海虹桥站出发，当天$$22:54$$到达昆明南站，全程共$$1593$$千米，途中站点共计停车$$56$$分钟，扣除停车时间，$$G1375$$次高铁的平均速度为（~ ）千米/时．（结果保留整数） ", "answer_option_list": [[{"aoVal": "A", "content": "$$148$$~ "}], [{"aoVal": "B", "content": "$$149$$~ "}], [{"aoVal": "C", "content": "$$150$$~ "}], [{"aoVal": "D", "content": "$$151$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->平均速度->公式法"], "answer_analysis": ["$$11:16-22:54$$共$$11$$小时$$38$$分，其中停车$$56$$分钟，运行$$10$$小时$$42$$分钟．$$1593\\div 10.7\\approx 149$$千米/时． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "656", "queId": "1a83b5d03fbe4a6c85a2870c665167a8", "competition_source_list": ["2016年IMAS小学高年级竞赛第一轮检测试题第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "请问$$192$$与$$120$$的公因数共有多少个？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理正应用->总个数"], "answer_analysis": ["$$(192,120)={{2}^{3}}\\times 3$$，所以因数个数有$$(3+1)\\times (1+1)=8$$个． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1432", "queId": "da3ca35220ff4ad596de1e7f94f384a8", "competition_source_list": ["2020年新希望杯二年级竞赛初赛（团战）第54题"], "difficulty": "1", "qtype": "single_choice", "problem": "二年级一班有$$25$$人，其中男生有$$11$$人．全班属兔的有$$8$$人，其余的同学都属龙，那么女生中至少有人属龙． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}], [{"aoVal": "E", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["二年级一班有$$25$$人，其中男生有$$11$$人，则女生有$$25-11=14$$人，$$8$$人属兔，则属龙的有$$25-8=17$$人，女生中至少有$$14-8=6$$人属龙． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1451", "queId": "9512418177684402a37a381a589cb735", "competition_source_list": ["2013年全国迎春杯三年级竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "四个海盗杰克、吉米、汤姆和桑吉共分$$280$$个金币．杰克说：``我分到的金币比吉米少$$11$$ 个，比汤姆多$$15$$个，比桑吉少$$20$$个．''那么，桑吉分到了个金币． ", "answer_option_list": [[{"aoVal": "A", "content": "$$84$$ "}], [{"aoVal": "B", "content": "$$85$$ "}], [{"aoVal": "C", "content": "$$86$$ "}], [{"aoVal": "D", "content": "$$87$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->和差倍问题->和差问题->多量和差问题"], "answer_analysis": ["设杰克分到的金币为$$1$$份量，则有 $$\\left. \\begin{matrix}1+11 1-15 1+20 \\end{matrix} \\right }\\Rightarrow 4+16=280\\Rightarrow 1=66\\Rightarrow 66+20=86$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1345", "queId": "59b51d397ec54f15aa661abe8fe9c9ed", "competition_source_list": ["2017年IMAS小学高年级竞赛（第二轮）第1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "2017年IMAS 将$$80$$个三角排成一列﹐然后依照下面的规律涂上黑色或白色﹐请问涂上黑色的三角形总共比涂上白色的三角形多几个？ ▲▲▲$$\\triangle \\triangle $$▲▲▲$$\\triangle \\triangle $$▲▲▲$$\\triangle \\triangle \\cdots \\cdots $$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}], [{"aoVal": "E", "content": "$$32$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "Overseas Competition->知识点->应用题模块->周期问题"], "answer_analysis": ["由图可知，从第一个三角形开始，以每五个三角形为一个周期，每个周期内有$$3$$个黑色三角形与$$2$$个白色三角形，即黑色三角形比白色三角形多$$1$$个．可知$$80$$个三角形共有$$16$$个周期， 所以黑色三角形比白色三角形总共多$$16$$个． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2512", "queId": "4b669b50cbd44fe6aa0c39bdf4dce495", "competition_source_list": ["2007年第5届创新杯四年级竞赛第1题5分", "2007年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "9个连续自然数的和是2007，其中最小的自然数是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "225 "}], [{"aoVal": "B", "content": "223 "}], [{"aoVal": "C", "content": "221 "}], [{"aoVal": "D", "content": "219 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列求和"], "answer_analysis": ["$$2007\\div 9-4\\text{=}219$$ "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1798", "queId": "812ac99060b64acebc0ef33a16b59656", "competition_source_list": ["2016年第7届广东广州羊排赛六年级竞赛第7题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "琦琦语文考了$$88$$分，英语考了$$85$$分，如果想要平均分达到$$90$$分，那么他的数学成绩至少要达到分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$ "}], [{"aoVal": "B", "content": "$$97$$ "}], [{"aoVal": "C", "content": "$$98$$ "}], [{"aoVal": "D", "content": "$$99$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["数序至少要$$90\\times 3-88-85=97$$（分）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2956", "queId": "bb9fa74e014d412b81e4d786634cd4b8", "competition_source_list": ["2017年河南郑州豫才杯小学高年级六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "小明上网时，电脑提示要输入开机密码，小明并不知道．这时小明爸爸写了一列有规律的数：$$1$$、$$2$$、$$3$$、$$5$$、$$8$$、~\\uline{~~~~~~~~~~}~共$$7$$个数字是密码，并提示说横线上是一个两位数，请聪明的你根据数字规律帮小明写出密码是（~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1235810$$ "}], [{"aoVal": "B", "content": "$$1235811$$ "}], [{"aoVal": "C", "content": "$$1235812$$ "}], [{"aoVal": "D", "content": "$$1235813$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数"], "answer_analysis": ["观察数列可发现：从第$$3$$项开始，每一项都等于前两项之和．所以第$$6$$项为$$5+8=13$$，即$$1235813$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1975", "queId": "d7d0047f84f9461fb46e8d628f179c9b", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（三）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个长方形水库长$$420$$米、宽$$280$$米，水库周围栽有杨树和柳树，相邻两棵杨树之间有$$3$$棵柳树，任意相邻两棵树的距离都是$$7$$米．柳树共有棵． ", "answer_option_list": [[{"aoVal": "A", "content": "$$116$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$136$$ "}], [{"aoVal": "D", "content": "$$150$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$7\\times (1+3)=28$$（米），$$(420+280)\\times 2\\div 28=50$$（段），$$3\\times 50=150$$（棵）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2727", "queId": "ff8080814502fa2401450bc68de1159c", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列算式结果为$$500$$的是（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5\\times99+1$$ "}], [{"aoVal": "B", "content": "$$100+25\\times4$$ "}], [{"aoVal": "C", "content": "$$88\\times4+37\\times4$$ "}], [{"aoVal": "D", "content": "$$100\\times0\\times5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->四则混合运算"], "answer_analysis": ["A等于$$5\\times(100-1)+1=500-5+1=496$$，B等于$$100+100=200$$ ，C等于$$(88+37)\\times4=125\\times 4=500$$ ，D等于$$0$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3048", "queId": "b3e9a417bcf24915b227a985b8d78922", "competition_source_list": ["2017年IMAS小学高年级竞赛（第一轮）第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "请问算式$$\\frac{20\\times 17}{2+0+1+7}$$的值等于什么？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$340$$ "}], [{"aoVal": "B", "content": "$$\\frac{34}{2017}$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$20$$ "}], [{"aoVal": "E", "content": "$$34$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$\\frac{20\\times 17}{2+0+1+7}=\\frac{20\\times 17}{10}=2\\times 17=34$$． 故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2824", "queId": "8938226f4dc94690b6c85ed2c9145ac7", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（五）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "古埃及计算圆的面积的方法：圆的面积等于直径减去直径的$$\\frac{1}{9}$$，然后再平方．由此来看，古埃及人认为圆周率是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3.16$$ "}], [{"aoVal": "B", "content": "$$3.15$$ "}], [{"aoVal": "C", "content": "$$3.14$$ "}], [{"aoVal": "D", "content": "$$3.13$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->一元一次方程->分数、小数系数方程"], "answer_analysis": ["$$\\frac{1}{4}\\pi {{d}^{2}}={{\\left( d-\\frac{1}{9}d \\right)}^{2}}$$，解得$$\\pi \\approx 3.16$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2386", "queId": "0f416f5966c1426cb715ff86ff96d985", "competition_source_list": ["2014年IMAS小学高年级竞赛第一轮检测试题第1题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "请问算式$$2015+1520+5201$$的值等于什么？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$8236$$ "}], [{"aoVal": "B", "content": "$$8506$$ "}], [{"aoVal": "C", "content": "$$8736$$ "}], [{"aoVal": "D", "content": "$$8836$$ "}], [{"aoVal": "E", "content": "$$9716$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["解法$$1$$：$$2015+1520+5201=8736$$，故选$$\\text{C}$$． 解法$$2$$：$$2015+1520+5201=2015+152+1520+5201-152=8888-152=8736$$，故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1164", "queId": "1147219097774076aeae82815ce1079d", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第16题"], "difficulty": "2", "qtype": "single_choice", "problem": "中国正在进入老龄化社会，社区中老年人数量日渐增多．今有五位老人的年龄互不相同，其中年龄最大的比年龄最小的大$$6$$岁，已知他们的平均年龄为$$85$$岁，则其中年龄最大的一位老人为岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$87$$ "}], [{"aoVal": "B", "content": "$$88$$ "}], [{"aoVal": "C", "content": "$$89$$ "}], [{"aoVal": "D", "content": "$$90$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["设这五个老人的年龄为$$a$$，$$b$$，$$c$$，$$d$$，$$e$$，且$$a\\textless{}b\\textless{}c\\textless{}d\\textless{}e$$， 则$$e-a=6$$，$$a+b+c+d+e=85\\times 5=425$$． ①因为$$d\\leqslant e-1$$，$$c\\leqslant e-2$$，$$b\\leqslant e-3$$，$$a=e-6$$， 所以$$(e-6)+(e-3)+(e-2)+(e-1)+e\\geqslant 425$$， $$e\\geqslant 87.4$$； ②因为$$b\\geqslant a+1$$，$$c\\geqslant a+2$$，$$d\\geqslant a+3$$，$$e=a+6$$， $$a+(a+1)+(a+2)+(a+3)+(a+6)\\leqslant 425$$， $$a\\leqslant 82.6$$， 即$$e=a+6\\leqslant 88.6$$， 所以$$87.4\\leqslant e\\leqslant 88.6$$， 所以$$e=88$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3036", "queId": "f32dfa5f79944467855a45fcbd78d5ff", "competition_source_list": ["2022年六年级竞赛（世界少年奥林匹克思维能力测评地方选拔活动）第4题"], "difficulty": "3", "qtype": "single_choice", "problem": "新学期性格活泼开朗的莉莉要竞选文艺委员，按规定需$$\\frac{3}{4}$$的选票才能当选，计算$$\\frac{2}{3}$$的选票后，她得到的选票已达到当选票数的$$\\frac{5}{6}$$，她还要得到剩下选票的~\\uline{~~~~~~~~~~}~才能当选。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac58$$ "}], [{"aoVal": "B", "content": "$$\\frac56$$ "}], [{"aoVal": "C", "content": "$$\\frac14$$ "}], [{"aoVal": "D", "content": "$$\\frac38$$ "}], [{"aoVal": "E", "content": "$$\\frac23$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数"], "answer_analysis": ["【分析】假设全班一共$$120$$人，按规定需$$\\frac34$$的选票才能当选，也就是至少需要得到$$90$$票，现在已经得到当选票数的$$\\frac56$$，求出已获得的选票数量，求出还需要的选票数量，再计算还需要的数量占余下票数的几分之几。【详解】假设全班一共$$60$$人；$$120\\times\\frac34=90$$（张）$$90\\times\\frac56=75$$（张）$$120\\times\\frac23=80$$（张）$$120-80=40$$（张）$$90-75=15$$（张）$$15\\div40=\\frac38$$【点睛】本题考查的是基础的分数乘除法应用题，找准单位``1''是解题的关键。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3436", "queId": "aadf8324cb0e479399705299f5fba97c", "competition_source_list": ["2015年华杯赛四年级竞赛初赛", "2015年华杯赛三年级竞赛初赛", "2015年华杯赛四年级竞赛初赛", "2015年华杯赛三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "小明有多张面额为$$\\text{1}$$元、$$\\text{2}$$元和$$\\text{5}$$元的人民币，他想用其中不多于$$\\text{1}0$$张的人民币购买一只价格为$$\\text{18}$$元的风筝，要求至少用其中两种面额的人民币。那么，不同的付款方式有（ ）种。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\text{3}$$ "}], [{"aoVal": "B", "content": "$$\\text{9}$$ "}], [{"aoVal": "C", "content": "$$\\text{11}$$ "}], [{"aoVal": "D", "content": "$$\\text{8}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->加法原理"], "answer_analysis": ["设所用的$$\\text{1}$$元、$$\\text{2}$$元和$$\\text{5}$$元的人民币张数依次为$$x$$，$$y$$，$$z$$。由已知可得： $$x+2y+5z=18$$，$$x+y+z\\leqslant 10$$ 当$$z=0$$时，$$\\left( x\\text{，}y \\right)$$只能取$$\\left( 2\\text{，}8 \\right)$$，共$$\\text{1}$$种； 当$$z=1$$时，$$\\left( x\\text{，}y \\right)$$只能取$$\\left( 1\\text{，}6 \\right)$$，$$\\left( 3\\text{，}5 \\right)$$，$$\\left( 5\\text{，}4 \\right)$$，共$$\\text{3}$$种； 当$$z=2$$时，$$\\left( x\\text{，}y \\right)$$只能取$$\\left( 0\\text{，}4 \\right)$$，$$\\left( 2\\text{，}3 \\right)$$，$$\\left( 4\\text{，}2 \\right)$$，$$\\left( 6\\text{，}1 \\right)$$，$$\\left( 8\\text{，}0 \\right)$$，共$$\\text{5}$$种； 当$$z=3$$时，$$\\left( x\\text{，}y \\right)$$只能取$$\\left( 1\\text{，}1 \\right)$$，$$\\left( 3\\text{，}0 \\right)$$，共$$\\text{2}$$种。 综上可得，一共有：$$\\text{1}+\\text{3}+\\text{5}+\\text{2}=\\text{11}$$（种）不同的付款方式。 选C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2545", "queId": "50373d75eef643029f81495059936454", "competition_source_list": ["2019年美国数学大联盟杯五年级竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$${{1}^{2019}}+{{1}^{2020}}=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$1$$的任意次方都为$$1$$，所以$$1$$的$$2019$$次方等于$$1$$，$$1$$的$$2020$$次方等于$$1$$，$$1+1=2$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1605", "queId": "e86f2a4c61c94a8fb7f81def197071a1", "competition_source_list": ["走美杯三年级竞赛", "走美杯四年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "下面问题不可以用算式$$50\\times 2\\times 3$$解决的是。 ", "answer_option_list": [[{"aoVal": "A", "content": "惠民超市一天卖出$$3$$盒保温杯，每盒装有$$2$$个杯子，每个杯子$$50$$元，一共卖了多少元 "}], [{"aoVal": "B", "content": "乐乐在长$$50$$米的游泳池里游了三个来回，他游了多少米 "}], [{"aoVal": "C", "content": "2箱蜜蜂一年可以产$$50$$千克的蜂蜜，照这样计算，$$3$$箱蜜蜂一年可以产多少千克的蜂蜜 "}], [{"aoVal": "D", "content": "悠悠一家三口人，每人每天早上各吃一个$$50$$克的鸡蛋，两天吃的鸡蛋共多少克 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型"], "answer_analysis": ["略 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3112", "queId": "f92cfb2fa49d43eea52629f2c5502136", "competition_source_list": ["2013年IMAS小学中年级竞赛第二轮检测试题第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$7$$名选手参加赛跑，选手的编号从$$1$$号开始连续编排．除了小明外，其它$$6$$名选手的号码之总和是$$26$$．那么小明的号码是。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}], [{"aoVal": "E", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列的概念"], "answer_analysis": ["从$$1$$号到$$7$$号选手的编码是从$$1$$到$$7$$的自然数，找到中间数是$$4$$，那么他们的和是$$4\\times 7=28$$，除了小明之外其他选手的编号和是$$26$$，所以小明的编号是$$28-26=2$$，即$$2$$号。选择A "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2590", "queId": "abcdcd6c845646ff900e88d97143c93e", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（二）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "【2016年新希望杯六年级竞赛题】 小张打算购买围巾和手套送给朋友们，预算不超过$$500$$元，已知围巾单价是$$70$$元，手套的单价是$$60$$元，如果小张至少要买$$2$$条围巾和$$3$$双手套，那么有种不同的选购方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["设围巾$$x$$条，手套$$y$$条． $$\\begin{cases}70x+60y\\leqslant 500 x\\geqslant 2y\\geqslant 3 \\end{cases}$$ $$\\begin{cases}x=2 y=3 \\end{cases}$$，$$\\begin{cases}x=2 y=4 \\end{cases}$$，$$\\begin{cases}x=2 y=5 \\end{cases}$$，$$\\begin{cases}x=2 y=6 \\end{cases}$$，$$\\begin{cases}x=3 y=3 \\end{cases}$$，$$\\begin{cases}x=3 y=4 \\end{cases}$$，$$\\begin{cases}x=4 y=3 \\end{cases}$$，$$7$$种． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1490", "queId": "e836f914bc3c4456b47d3e4ce7fef1b1", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（二）"], "difficulty": "1", "qtype": "single_choice", "problem": "一次考试，甲、乙、丙三人的平均分为$$90$$分．其中甲、乙二人的平均分为$$88$$分，那么丙得了（~ ）分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$92$$ "}], [{"aoVal": "B", "content": "$$93$$ "}], [{"aoVal": "C", "content": "$$94$$ "}], [{"aoVal": "D", "content": "$$95$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$90\\times 3=270$$（分），甲，乙，丙总分是$$270$$分， $$88\\times 2=176$$（分），甲，丙总分是$$176$$分， $$270-176=94$$（分），丙的得分是$$94$$分． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1760", "queId": "b1dfd6e70c0840a0890576d2ca4547f9", "competition_source_list": ["2008年第6届创新杯五年级竞赛初赛A卷第5题5分", "2008年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "如果有2008个学生排成一列，从第一个学生开始，按1，2，3，4，5，4，3，2，1，2，3，4，5，4，3，2，1，2，3，$$\\cdots$$ 报数，则第2000个学生所报的数为( ) ", "answer_option_list": [[{"aoVal": "A", "content": "2 "}], [{"aoVal": "B", "content": "3 "}], [{"aoVal": "C", "content": "4 "}], [{"aoVal": "D", "content": "5 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["设$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$4$$，$$3$$，$$2$$为一节，每节$$8$$个数，由于$$2000=8\\times 250$$，所以，第2000个学生所报的数是第$$250$$节的最后一个数，因此，第2000个学生所报的数为$$2$$. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1272", "queId": "2c1a889ea182405fa104a43cd1037568", "competition_source_list": ["走美杯三年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "泡泡比毛毛小$$7$$岁，再过$$4$$年泡泡的年龄将是毛毛年龄的一半，他们今年的年龄总和是（ ）岁。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型"], "answer_analysis": ["再过$$4$$年毛毛的年龄将是泡泡年龄的$$2$$倍，所以那时泡泡的年龄为$$7$$岁，毛毛为$$7\\times 2=14$$(岁)，他们今年的年龄总和为$$7+14-4\\times 2=13$$(岁)。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2059", "queId": "eb09bcc67477471e885429ed6bb2beef", "competition_source_list": ["2008年第6届创新杯六年级竞赛初赛A卷第7题5分", "2008年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "有一些糖，每人分5块多10块;如果现有的人数增加到原来的1.5倍，则每人分4块就少2块，那么这些糖在( )块之间. ", "answer_option_list": [[{"aoVal": "A", "content": "80-\\/-90 "}], [{"aoVal": "B", "content": "68-\\/-78 "}], [{"aoVal": "C", "content": "56-\\/-62 "}], [{"aoVal": "D", "content": "95-\\/-102 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题"], "answer_analysis": ["设现有人数为$$x$$，则糖块数为$$5x+10$$，依题意得$$5x+10=4\\times1.5x-2$$，解得$$x=12$$，所以糖块数为$$5\\times 12+10=70$$.即在这些糖在68-\\/-78块之间. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "522", "queId": "d4675c9c3f264be3b1e7a282c3bc6e52", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "在算式：$$2\\times \\square \\square \\square =\\square \\square \\square $$的六个空格中，分别填入$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$这六个数字，使算式成立，并且算式的积能被$$13$$整除，那么这个积是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$234$$ "}], [{"aoVal": "B", "content": "$$286$$ "}], [{"aoVal": "C", "content": "$$534$$ "}], [{"aoVal": "D", "content": "$$654$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["先从个位数考虑，有$$2\\times 2=4$$，$$2\\times 3=6$$，$$2\\times 6=12$$，$$2\\times 7=14$$．再考虑乘数的百位只能是$$2$$或$$3$$，因此只有三种可能的填法： $$2\\times 273=546$$， $$2\\times 327=654$$， $$2\\times 267=534$$， 其中只有$$546$$能被$$13$$整除，因此这个积是$$546$$． 故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "992", "queId": "054854c072d543119269ba611b3cd0a9", "competition_source_list": ["2012年全国美国数学大联盟杯小学高年级竞赛初赛第32题", "2013年美国数学大联盟杯小学高年级竞赛初赛第32题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在数轴中请求出到$$1.75$$和$$7.25$$距离相同的点？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2.75$$ "}], [{"aoVal": "B", "content": "$$3.25$$ "}], [{"aoVal": "C", "content": "$$3.75$$ "}], [{"aoVal": "D", "content": "$$4.5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->公式类->直接求平均数"], "answer_analysis": ["$$(1.75+7.25)\\div2=4.5$$. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2914", "queId": "7d0786a15396498688d248d7c10f99d4", "competition_source_list": ["2008年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "张扬已经进行了$$20$$场比赛，并且赢了$$95 \\%$$的比赛，如果他以后每一场都获胜，当赢得$$96 \\%$$的比赛时，他又赢了（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$场 "}], [{"aoVal": "B", "content": "$$3$$场 "}], [{"aoVal": "C", "content": "$$4$$场 "}], [{"aoVal": "D", "content": "$$5$$场 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->一元一次方程->分数、小数系数方程"], "answer_analysis": ["解法一：张扬已经赢了$$20\\times 95 \\%=19$$（场）比赛，只输了$$1$$场，为了赢得$$96 \\%$$的比赛，他输的场数只能占总场数的$$4 \\%$$，$$1\\div 4 \\%=25$$（场），所以他至少还要赢$$25-20=5$$（场）。 解法二：在$$20$$场比赛中，张扬已经赢了$$20\\times 95 \\%=19$$（场），如果他以后每一场都赢，设他再赢$$x$$场，则$$19+x=\\left( 20+x \\right)\\times 96 \\%$$，解得$$x=5$$。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2891", "queId": "a46530340ece46eca86a6a3f555b9142", "competition_source_list": ["2008年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "下列各分数中，分数值最小的是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{7}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{15}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{15}{112}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数通分法->通分子"], "answer_analysis": ["由于$$\\frac{1}{7}=\\frac{2}{14}\\textgreater\\frac{2}{15}\\textgreater\\frac{2}{16}=\\frac{1}{8}$$，$$\\frac{15}{112}\\textgreater\\frac{14}{112}=\\frac{1}{8}$$，所以$$\\frac{1}{8} \\textless{} \\frac{1}{7}$$，$$\\frac{1}{8} \\textless{} \\frac{2}{15}$$，$$\\frac{1}{8} \\textless{} \\frac{15}{112}$$因此，这些分数中最小的是$$\\frac{1}{8}$$. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2740", "queId": "ff8080814518d52401451925852804f4", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第11题"], "difficulty": "2", "qtype": "single_choice", "problem": "三位数$$N$$，分别减$$3$$、加$$4$$、除以$$5$$、乘$$6$$，得到四个整数，已知这四个数的数字和恰好是$$4$$个连续的自然数，那么满足条件的三位数$$N$$有（~~~~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["考虑到一定会有进位，退位．设原数数字和为$$a$$，则$$-3$$，$$+4$$定不是差$$7$$，否则无法成为连续$$4$$个自然数．$$\\div 5$$说明末位为$$0$$或$$5$$，当末位为$$5$$时，$$-3$$，$$+4$$均不进位退位．当末位为$$0$$时，$$-3$$退位，符合． 所以$$-3$$相当于数字和多$$6$$，$$a+6$$；$$+4$$相当于数字和多$$4$$，$$a+4$$；$$\\div 5$$ 相当于数字和$$\\times 2$$，$$a\\times 2$$；$$a\\times 2$$、$$a+2$$、$$a+4$$连续，$$a\\times 2$$为$$a+7$$，$$a+5$$，$$a+3$$中的一个． 分类讨论得到$$a\\times 2=a+5$$成立，所以$$a=5$$，数字和为$$5$$，尾数为$$0$$的有，$$500$$（舍弃），$$410$$，$$320$$，$$230$$，$$140$$，共$$4$$个． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "795", "queId": "4e565c26018846368d0c26153e790ca1", "competition_source_list": ["2019~2020学年陕西宝鸡渭滨区五年级上学期期末第22题1分", "2019~2020学年山东济南高新区五年级下学期期末第10题1分", "2020年广东深圳龙岗区亚迪学校迎春杯五年级竞赛模拟第20题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "在下面四组数中，组中的两个数都是质数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$，$$31$$ "}], [{"aoVal": "B", "content": "$$13$$，$$33$$ "}], [{"aoVal": "C", "content": "$$35$$，$$45$$ "}], [{"aoVal": "D", "content": "$$37$$，$$97$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\text{A}$$中$$21$$不是质数，故$$\\text{A}$$错误； $$\\text{B}$$中$$33$$不是质数，故$$\\text{B}$$错误． $$\\text{C}$$中$$35$$，$$45$$都不是质数，故$$\\text{C}$$错误． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1045", "queId": "0742d19203814a3194e6a393f1f5c7b9", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（一）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "五年级（$$1$$）班原有$$39$$名学生，平均体重为$$48$$千克，班里转来新生小林后，平均体重变为$$48.1$$千克，那么小林的体重是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$48.2$$千克 "}], [{"aoVal": "B", "content": "$$49$$千克 "}], [{"aoVal": "C", "content": "$$50$$千克 "}], [{"aoVal": "D", "content": "$$52$$千克 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->公式类->加权平均数"], "answer_analysis": ["$$48+(48.1-48)\\times 40=52$$（千克）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2669", "queId": "32d3b50f3d014ecda9891bf03505f2c2", "competition_source_list": ["2008年第6届创新杯六年级竞赛复赛第2题4分", "2008年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "规定$$\\max \\left( a,b \\right)$$表示$$a$$，$$b$$两个数中较大的一个，$$\\min \\left( a,b \\right)$$表示$$a$$，$$b$$两个数中较小的一个，则$$\\max \\left[ \\min \\left( 2006,2008 \\right),\\min \\left( 2007,2009 \\right) \\right]$$等于． ", "answer_option_list": [[{"aoVal": "A", "content": "2006 "}], [{"aoVal": "B", "content": "2007 "}], [{"aoVal": "C", "content": "2008 "}], [{"aoVal": "D", "content": "2009 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->定义新运算->直接运算型->普通型"], "answer_analysis": ["由$$\\text{min}\\left( 2006\\text{，}2008 \\right)=2006$$，$$\\min \\left( 2007\\text{，}2009 \\right)=2007$$得，原式$$=\\max \\left( 2006\\text{，}2007 \\right)=2007$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2464", "queId": "220495ff8eea42878cef0bdb878ed2ee", "competition_source_list": ["2018年美国数学大联盟杯五年级竞赛初赛第33题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$\\left(2^{2} \\times 2^{4} \\times 2^{6}\\times \\ldots ~\\times 2^{98} \\times 2^{100}\\right) \\div \\left(2^{1} \\times 2^{3} \\times 2^{5}\\times \\ldots ~\\times 2^{97} \\times 2^{99}\\right)=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$${{2}^{49}}$$ "}], [{"aoVal": "C", "content": "$${{2}^{50}}$$ "}], [{"aoVal": "D", "content": "$${{2}^{100}}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left(2^{2} \\times 2^{4} \\times 2^{6}\\times \\ldots ~\\times 2^{98} \\times 2^{100}\\right) \\div \\left(2^{1} \\times 2^{3} \\times 2^{5}\\times \\ldots ~\\times 2^{97} \\times 2^{99}\\right)$$ $$=\\left( {{2}^{2}}\\div {{2}^{1}} \\right)\\times \\left( {{2}^{4}}\\div {{2}^{3}} \\right)\\times \\left( {{2}^{6}}\\div {{2}^{5}} \\right)\\times \\ldots \\times \\left( {{2}^{98}}\\div {{2}^{97}} \\right)\\times \\left( {{2}^{100}}\\div {{2}^{99}} \\right)$$ $$=\\underbrace{2\\times 2\\times 2\\times \\cdots \\times 2}_{50个2}$$ $$={{2}^{50}}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3390", "queId": "9b9fddf4c3884f5fbcf59112ab39d4cb", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛二年级竞赛决赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "二（$$1$$）班的同学算两道口算题，第一道算对的有$$20$$人，第二道算对的有$$31$$人，两道都对的有$$10$$人，这个班共有个同学． ", "answer_option_list": [[{"aoVal": "A", "content": "$$41$$ "}], [{"aoVal": "B", "content": "$$51$$ "}], [{"aoVal": "C", "content": "$$61$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["根据重叠问题，可知这个班上共有同学：$$20+31-10=41$$人． 故选择$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "229", "queId": "43e8f2cccce74a0799a70701b88ce913", "competition_source_list": ["2018年华杯赛小学中年级竞赛初赛第4题10分", "2018年第23届华杯赛小学中年级竞赛初赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$6\\times 6$$网格的所有方格中放入围棋子，每个方格方$$1$$枚棋子，要求每行中的白色棋子的数目互不相等，每列中的白色棋子的数目都相等，那么这个$$6\\times 6$$网格中共有枚黑色围棋子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["可以考虑纵向每列白色棋子数相等，故白棋子数量一定是$$6$$的倍数；考虑横向，每行棋子数不同，且最少为$$0$$，最多为$$6$$，白色棋子数最少为$$0+1+2+3+4+5=15$$，最多是$$1+2+3+4+5+6=21$$，故$$15\\sim 21$$间能被$$6$$整除的数只有$$18$$，故白棋子数量为$$18$$，黑棋子有$$36-18=18$$． ", "首先考虑纵向，每列白色棋子数相等，故白棋子数量一定是$$6$$的倍数，
\n考虑横向每行棋子数不同，且最少为$$0$$，最多为$$6$$，
\n白棋子数量最少为$$0+1+2+3+4+5=15$$，
\n最多为$$1+2+3+4+5+6=21$$，
\n易知$$15-21$$间能被$$6$$整除的只有$$18$$，
\n故白棋子数量为$$18$$，
\n黑棋子数量为$$6\\times 6-18=18$$．
\n故选$$\\text{A}$$．
"], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "738", "queId": "b56fb8e2218a4d789730925b90609fc6", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（二）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一列数：$$1234$$、$$5678$$、$$9101112$$、$$13141516\\cdots \\cdots $$每个数都是由四个连续的自然数组 成．其中只有一个十三位数，它的各数位上的数字之和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$76$$ "}], [{"aoVal": "B", "content": "$$77$$ "}], [{"aoVal": "C", "content": "$$78$$ "}], [{"aoVal": "D", "content": "$$79$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["这个十三位数只能是$$9979989991000$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "984", "queId": "04e8ee87898a47f6bc470ef866643555", "competition_source_list": ["2012年第10届创新杯四年级竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "某运输公司为玻璃厂运来$$1000$$个玻璃镜框到仓库，双方商定每个玻璃镜框运费$$5$$元，如果打碎$$1$$个，这一个玻璃镜框不但不给运费，而且要赔偿$$20$$元，结果到目的地结算时，玻璃厂共付出运费$$4475$$元，那么运输过程中打碎了个玻璃镜框． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$26$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["假设玻璃镜框全部完好，玻璃厂应付$$1000\\times 5=5000$$元，与实际情况相比少付了$$5000-4475=525$$元，每打碎一个镜框少付$$5+20=25$$元，因此一共打碎了$$525\\div 25=21$$个． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1748", "queId": "e8d3430b73f04775b30aa07e351feb2f", "competition_source_list": ["2005年第3届创新杯六年级竞赛初赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "---件工程，甲单独做要$$6$$小时，乙单独做要$$10$$小时，如果按甲、乙、甲、乙$$\\cdots \\cdots $$顺序交替工作，每次$$1$$小时，那么需要小时完成． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$7\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$7\\frac{1}{3}$$ "}], [{"aoVal": "D", "content": "$$7\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->接力施工问题"], "answer_analysis": ["甲乙合作完成需要： $$1\\div \\left( \\frac{1}{6}+\\frac{1}{10} \\right)$$ $$=1\\div \\frac{4}{15}$$ $$=3.75$$（小时）， 每人工作$$3$$小时，还剩下： $$1-\\left( \\frac{1}{6}+\\frac{1}{10} \\right)\\times 3$$ $$=1-\\frac{4}{5}$$ $$=\\frac{1}{5}$$， 甲再工作$$1$$小时，剩下的由乙完成需要： $$\\left( \\frac{1}{5}-\\frac{1}{6} \\right)\\div \\frac{1}{10}$$ $$=\\frac{1}{30}\\div \\frac{1}{10}$$ $$=\\frac{1}{3}$$（小时）， 一共$$3\\times 2+1+\\frac{1}{3}=7\\frac{1}{3}$$（小时）． 答：需要$$7\\frac{1}{3}$$时完成． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1521", "queId": "bebf1a032fdc41ae9357de2ad9477086", "competition_source_list": ["2017年全国希望杯小学高年级五年级竞赛初赛考前100题"], "difficulty": "2", "qtype": "single_choice", "problem": "1.价格相同的一种商品，甲店：买四赠一．乙店：优惠$$\\frac{1}{4}$$．如果只从经济方面考虑，你选择去哪？ ", "answer_option_list": [[{"aoVal": "A", "content": "甲店 "}], [{"aoVal": "B", "content": "乙店 "}], [{"aoVal": "C", "content": "都一样 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["设此商品的单价为$$a$$元，``买四赠一''即用$$4$$件的钱买了$$5$$件商品，购买单价为$$\\frac{4}{5}a=0.8a$$；``优惠$$\\frac{1}{4}$$''表示每件单价为$$\\left( 1-\\frac{1}{4} \\right)a=0.75a$$，故选择乙店． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2137", "queId": "258cfd8d577949b28654bb2b84e9b396", "competition_source_list": ["2016年广东深圳四年级模拟考试鹏程杯集训", "2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有甲、乙、丙$$3$$人，甲每分钟走$$100$$米，乙每分钟走$$80$$米，丙每分钟走$$75$$米．现在甲从东村，乙、丙两人从西村同时出发相向而行，在途中甲与乙相遇$$6$$分钟后，甲又与丙相遇． 那么，东、西两村之间的距离是多少米？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$37800$$ "}], [{"aoVal": "B", "content": "$$38800$$ "}], [{"aoVal": "C", "content": "$$39900$$ "}], [{"aoVal": "D", "content": "太难了，容我在喝一杯$$82$$年的雪碧压压惊 "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["甲、丙$$6$$分钟相遇的路程：$$\\left( {100 + 75} \\right) \\times 6 = 1050$$$$($$米$$)$$； 甲、乙相遇的时间为：$$1050 \\div \\left( {80 - 75} \\right) = 210$$$$($$分钟$$)$$； 东、西两村之间的距离为：$$\\left( {100 + 80} \\right) \\times 210 = 37800$$$$($$米$$)$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "871", "queId": "813f1400bef74ffc8677e8ca0142576f", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛A卷第6题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$5$$路和$$9$$路公共汽车早上$$6$$时$$40$$分同时发车，$$5$$路公共汽车每隔$$10$$分钟发一辆车，$$9$$路公共汽车每隔$$8$$分钟发一辆车，这两路车第$$20$$次同时发车是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8:00$$ "}], [{"aoVal": "B", "content": "$$18:40$$ "}], [{"aoVal": "C", "content": "$$19:20$$ "}], [{"aoVal": "D", "content": "$$20:00$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->因倍应用题->倍数应用题"], "answer_analysis": ["$$10$$和$$8$$的最小公倍数是$$40$$， $$40\\times20=800$$（分）， $$6$$时$$20$$分再过$$800$$分钟为$$20$$时， 所以第$$20$$次同时发车为$$20$$时． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "369", "queId": "f21943325e6442c09f751996f45052bd", "competition_source_list": ["2017年全国美国数学大联盟杯小学高年级五年级竞赛初赛第29题"], "difficulty": "1", "qtype": "single_choice", "problem": "2017年全国美国数学大联盟杯小学高年级五年级竞赛初赛第$$29$$题 The minimum number of people needed in a room so that there are always at least five people in the room born in the same month is~\\uline{~~~~~~~~~~}~． 翻译：一个房间需要的最少多少人，才能使得总是有至少五个人是同一个月份出生的？ the minimum number：最小值；$$at$$ least：最少；$$born$$ in：出生；$$the$$ same month：同一个月份； ", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$49$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$61$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数学广角->鸽巢问题->利用抽屉原理解决实际问题"], "answer_analysis": ["一个房间需要的最少多少人，才能使得总是有至少五个人是同一个月份出生的？ the minimum number：最小值；$$at$$ least：最少；$$born$$ in：出生；$$the$$ same month：同一个月份； $$4\\times 12+1=49$$，所以选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1571", "queId": "ff8080814518d5240145190915fe030b", "competition_source_list": ["2014年全国迎春杯五年级竞赛初赛第6题", "2014年全国迎春杯六年级竞赛初赛第6题", "2016年陕西西安小升初交大附中入学真卷9第11题", "2014年北京六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁四人拿出同样多的钱，一起订购同样规格的若干件新年礼物，礼物买来后，甲、乙、丙分别比丁多拿了$$3$$，$$7$$，$$14$$件礼物，最后结算时，乙付给了丁$$14$$元钱，并且乙没有付给甲钱．那么丙应该再付给丁（~~~ ~）元钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$56$$ "}], [{"aoVal": "D", "content": "$$70$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设丁拿了$$a$$件礼物，则四人花同样的钱，每人可以拿到$$a+\\frac{3+7+14}{4}=a+6$$件礼物， 实际情况：丁少拿了$$6$$件，乙多拿了$$1$$件，给丁$$14$$元，则货物单价$$14$$元， 丙多拿了$$14-6=8$$件，$$3$$件给甲，$$5$$件给丁，$$5\\times14=70$$元． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1094", "queId": "b48e4cb1a9174ae5990848dbd7d0bc5b", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛B卷第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一只手表每小时慢$$5$$分钟，照这样计算，早上$$6$$时对准标准时间后，当手表指示下午$$5$$时整时，标准时间是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16:05$$ "}], [{"aoVal": "B", "content": "$$17:55$$ "}], [{"aoVal": "C", "content": "$$18:00$$ "}], [{"aoVal": "D", "content": "$$18:05$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["本题中的相等关系是： 这只手表慢的时间$$-$$手表每小时比准确时间慢$$5$$分钟$$\\times $$标准时间经过的时间$$=0$$， 设标准时间经过了$$x$$小时，根据等量关系列方程求解即可． $$5+12=17$$时， 设标准时间经过了$$x$$小时，则 $$\\left( 6+x-17 \\right)\\times 60-5x=0$$， $$60\\left( x-11 \\right)-5x=0$$， $$60x-660-5x=0$$， $$55x=660$$， $$x=12$$． $$6:00+12=18:00$$． 所以准确时间应该是$$18:00$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "164", "queId": "42ab9ae47b164e9e943a8956cd860b01", "competition_source_list": ["2020年迎春杯六年级竞赛第10题12分"], "difficulty": "3", "qtype": "single_choice", "problem": "小周老师写了一个两位质数，并将这个质数个位数字告诉了甲，十位数字告诉了乙，十位数字与个位数字之和告诉了丙，十位数字与个位数字之差（大减小）告诉了丁． 丙说：在我说话之前，甲一定认为乙不知道这个质数是多少． 丙说完之后，乙说：在我说话之前，甲一定认为丁不知道这个质数是多少． 那么，这个质数是~\\uline{~~~~~~~~~~}~．（甲、乙、丙、丁四位同学诚实且聪明） ", "answer_option_list": [[{"aoVal": "A", "content": "$$23$$ "}], [{"aoVal": "B", "content": "$$29$$ "}], [{"aoVal": "C", "content": "$$41$$ "}], [{"aoVal": "D", "content": "$$61$$ "}], [{"aoVal": "E", "content": "$$97$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数感认知->数学概念理解（数）", "海外竞赛体系->知识点->组合模块->逻辑推理"], "answer_analysis": ["所有的两位质数：$$11$$、$$31$$、$$41$$、$$61$$、$$71$$； $$13$$、$$23$$、$$43$$、$$53$$、$$73$$、$$83$$； $$17$$、$$37$$、$$47$$、$$67$$、$$97$$； $$19$$、$$29$$、$$59$$、$$79$$、$$89$$． 丙：丙说话之前甲只知道个位，就知道乙猜不出， 所以个位不可能是$$7$$， 丙手上拿的是数字和，通过数字和来确定这个信息的， 因此数字和不可能是$$8$$、$$10$$、$$11$$、$$13$$、$$16$$，将数字和为这些的都排除， 只剩下$$9$$个数：$$11$$、$$13$$、$$23$$、$$31$$、$$41$$、$$43$$、$$59$$、$$61$$、$$89$$． 乙：乙说话之前，丁拿的是数字差，甲一定猜不出， 所以排除$$11$$、$$41$$、$$59$$、$$61$$，那么个位不可能是$$1$$和$$9$$， 继续排除$$31$$、$$59$$、$$89$$． 剩下$$3$$个数：$$13$$、$$23$$、$$43$$， 乙通过十位确定甲一定知道丁猜不出，那么十位不可能是$$1$$和$$4$$， 故这个质数一定是$$23$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3037", "queId": "cec43207b7064e98a2023c039129eb74", "competition_source_list": ["2011年全国世奥赛三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$100 + 99 - 98 + 97 - 96 + ~\\cdots ~+ 3 - 2 + 1$$的结果是（ ~~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$101$$ "}], [{"aoVal": "B", "content": "$$150$$ "}], [{"aoVal": "C", "content": "$$149$$ "}], [{"aoVal": "D", "content": "$$151$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["观察$$100$$和$$1$$之间的数，将$$\\left( 99 - 98\\right)$$、$$\\left( 97 - 96 \\right)$$、$$\\left( 95 - 94 \\right)$$、$$ \\cdots $$、$$\\left( 3 - 2 \\right)$$进行组合（ 除去$${1}$$和$${100}$$后，奇数前面是加号，偶数前面是减号），所以一共有$$\\left( 100 - 2 \\right) \\div 2 = 49$$个$$1$$，所以原式$$ = 100 + 1 + 49 = 150$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "206", "queId": "250202111bab4e7a9477aa80f96208b3", "competition_source_list": ["2015年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "春季开学后，有不少同学都将部分压岁钱捐给山区的贫困学生；事后，甲、乙、丙、丁$$4$$位同学有如下的对话： 甲：``丙、丁之中至少有$$1$$人捐了款'' 乙：``丁、甲之中至多有$$1$$人捐了款'' 丙：``你们$$3$$人中至少有$$2$$人捐了款'' 丁：``你们$$3$$人中至多有$$2$$人捐了款'' 已知这$$4$$位同学说的都是真话且其中恰有$$2$$位同学捐了款，那么，这$$2$$位同学是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "甲、乙 "}], [{"aoVal": "B", "content": "丙、丁 "}], [{"aoVal": "C", "content": "甲、丙 "}], [{"aoVal": "D", "content": "乙、丁 "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["由丙的话可知：丙没有捐款；再由甲的话可知：丁捐了款；再由乙的话可知：丁、甲之中最多有$$1$$人捐款．由此推知，甲没有捐款，乙捐了款，捐款的两人为乙和丁，选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "225", "queId": "8368dc822e354db184a76c0e11073ebe", "competition_source_list": ["2013年全国迎春杯小学中年级竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "小明碰到了三个人，其中一位是牧师、一位是骗子、一位是疯子．牧师只说真话，骗子只说假话，疯子有时说真话，有时说假话．第一位说：``我是疯子．''第二位说：``你胡说，你才不是疯子呢！''第三位说：``别说了，我是疯子．''那么，这三个人中第~\\uline{~~~~~~~~~~}~位是疯子． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["第一个和第三个都说自己是疯子，所以他们不是牧师，只有第二个是牧师．而牧师说的是真话，所以第一个不是疯子．因此第三个才是真正的疯子． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "996", "queId": "0185a543e4a7449faadc0544d2e9d08d", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第1题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一项工程预计$$15$$人每天做$$4$$小时，$$18$$天可以完成，后来增加$$3$$人，并且工作时间增加$$1$$小时，这项工程天完成． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["首先要求出工作总量，即$$15\\times 4\\times 18$$，再用工作总量除以工作人数，再除以每天的工作时间，即为所需要的天数．解决此题的关键是先求出总工作量． $$15\\times 4\\times 18\\div \\left[ \\left( 15+3 \\right)\\times \\left( 4+1 \\right) \\right]$$ $$=1080\\div 90$$ $$=12$$（天）． 答：这项工程$$12$$天完成． 故选 $$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "232", "queId": "7ef1f938a70848d792b217637698a54a", "competition_source_list": ["2017年全国美国数学大联盟杯小学高年级五年级竞赛初赛第33题"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{(2017 US Math League, Priamry 5, Question \\#33)} $$A$$, $$B$$, $$C$$, $$D$$, $$E$$, $$F$$, and $$G$$ participated in a chess tournament. Each player must play each of his six opponents exactly once. So far, $$A$$ has played $$1$$ match. $$B$$ has played $$2$$ matches. $$C$$ has played $$3$$ matches. $$D$$ has played $$4$$ matches. $$E$$ has played $$5$$ matches and $$F$$ has played $$6$$ matches. How many matches has $$G$$ played at this point?（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$） 甲、乙、丙、丁、戊、己、庚一起参加象棋锦标赛，每一位选手必须要跟六个对手分别对赛一次．目前为止，甲比赛了一场，乙两场，丙三场，丁四场，戊五场，己六场，那么目前庚比赛了几场？（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$） ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "Overseas Competition->知识点->组合模块->逻辑推理->体育比赛"], "answer_analysis": ["根据题意，$$F$$已经跟每个人都比赛过，包括$$G$$；同理，$$D$$，$$E$$也跟$$G$$比赛过，所以$$G$$比了$$3$$场 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1707", "queId": "7bc22ee509a44c59860d7ac6597bd0cb", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（四）"], "difficulty": "1", "qtype": "single_choice", "problem": "有一年六月份日历中周五比周四多$$1$$个，那么这年七月一日为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "周四 "}], [{"aoVal": "B", "content": "周五 "}], [{"aoVal": "C", "content": "周六 "}], [{"aoVal": "D", "content": "周日 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$6$$月份有$$30$$天，按照每同$$7$$天计算，$$30\\div 7=4$$（周）$$\\cdots \\cdots 2$$（天），根据这个$$6$$月的周五比周四多$$1$$个，可知多余的$$2$$天必为周五和周六，即这个$$6$$月的最后一天是星期 六，所以$$7$$月$$1$$日是周日． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3043", "queId": "d36359aa8639416fae03040e53301667", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛决赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$a$$、$$b$$互素，且两个最简分数之和为$$\\frac{m}{a}+\\frac{n}{b}=\\frac{31}{35}$$，则$$\\frac{a}{m}+\\frac{b}{n}-\\frac{1}{m\\times n}=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数->质数与合数判定->质数与合数的认识"], "answer_analysis": ["数论．$$a$$、$$b$$互素，且计算结果分母为$$35$$，因此$$a$$、$$b$$为$$5$$和$$7$$，代入原式得$$\\frac{7m+5n}{35}=\\frac{31}{35}$$，则$$m=3$$，$$n=2$$，所以$$\\frac{5}{3}+\\frac{7}{2}-\\frac{1}{2\\times 3}=5$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "868", "queId": "9fe438e6e1b944e7ac303d8d98230707", "competition_source_list": ["2018年美国数学大联盟杯四年级竞赛初赛第32题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "九个连续的正整数的总和总是能被整除 The sum of nine consecutive positive integers is always divisible by． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$9$$个连续正整数的和总是可以被整除． 设$$9$$个数中的中间数为$$a$$，则这$$9$$个数可表示为：$$a-4$$，$$a-3$$，$$a-2$$，$$a-1$$，$$a$$，$$a+1$$，$$a+2$$，$$a+3$$，$$a+4$$， 所以和为：$$\\left(a-4\\right)+\\left(a-3\\right)+\\left(a-2\\right)+\\left(a-1\\right)+a+\\left(a+1\\right)+\\left(a+2\\right)+\\left(a+3\\right)$$ $$+\\left(a+4\\right)=9a$$，即不管$$a$$为什么数时，其和必能被$$9$$整除． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2371", "queId": "2eb6337dc82c442599a65cddb4a07343", "competition_source_list": ["2016年全国中环杯四年级竞赛初赛第11题"], "difficulty": "2", "qtype": "single_choice", "problem": "神庙里有一把古老的秤，对于重量小于$$1000$$克的物体，这把秤会显示其正确的重量；对于重量大于等于$$1000$$克的物体，这把秤会显示出一个大于等于$$1000$$的随机数．艾迪有五个物品，各自的重量都小于$$1000$$克，我们分别用$$P$$、$$Q$$、$$R$$、$$S$$、$$T$$表示它们的重量．将这五个物品两两配对放到秤上进行称重，得到下面的结果：$$Q+S=1200$$（克）、$$R+T=2100$$（克）、$$Q+T=800$$（克）、$$Q+R=900$$（克）、$$P+T=700$$（克）．那么这五个物品的重量从重到轻的顺序为~\\uline{~~~~~~~~~~}~$$\\textgreater$$~\\uline{~~~~~~~~~~}~$$\\textgreater$$~\\uline{~~~~~~~~~~}~$$\\textgreater$$~\\uline{~~~~~~~~~~}~$$\\textgreater$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$R\\textgreater T\\textgreater S\\textgreater Q\\textgreater P$$ "}], [{"aoVal": "B", "content": "$$R\\textgreater S\\textgreater T\\textgreater Q\\textgreater P$$ "}], [{"aoVal": "C", "content": "$$S\\textgreater R\\textgreater Q\\textgreater T\\textgreater P$$ "}], [{"aoVal": "D", "content": "$$S\\textgreater R\\textgreater T\\textgreater Q\\textgreater P$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$Q+T=800$$①； $$Q+R=900$$②；$$P+T=700$$③；$$Q+S\\geqslant 1000$$④；$$R+T\\geqslant 1000$$⑤；由①②得：$$R\\textgreater T$$；由①③得：$$Q\\textgreater P$$；由②④得：$$S\\textgreater R$$；由②⑤得：$$T\\textgreater Q$$；所以：$$S\\textgreater R\\textgreater T\\textgreater Q\\textgreater P$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2388", "queId": "08922c007d594a87b32c2fc8b247caf1", "competition_source_list": ["2017年河南郑州豫才杯五年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "《红楼梦》是一部长篇章回体小说，全书分为$$120$$回，字数总计为$$729636$$个字．章回体小说是中国古典小说的长篇小说的主要形式．章回体的一个特点即是每``回''叙述一个较为完整的故事段落，篇幅长短相当．由此我们可以推算出，此书每回的字数约为（~ ）字左右． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$万~~~~~~~~~~ "}], [{"aoVal": "B", "content": "$$0.06$$万~~~~ "}], [{"aoVal": "C", "content": "$$6000$$~~~~~~~ "}], [{"aoVal": "D", "content": "$$600$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$729636\\div 120\\approx 6000$$字． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3464", "queId": "f8ef63625f1f44b38ea60e3504c991a0", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "某射手在百步之外射箭恰好射到靶心的概率为$40 \\%$，如果该射手在百步之外连射三箭，有两箭射中靶心的概率为~\\uline{~~~~~~~~~~}~. ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.216$$ "}], [{"aoVal": "B", "content": "$$0.432$$ "}], [{"aoVal": "C", "content": "$$0.144$$ "}], [{"aoVal": "D", "content": "$$0.288$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["第一箭射空，其他两箭射中的概率为$$\\left( 1-0.4 \\right)\\times 0.4\\times 0.4=0.096$$． 第二箭射空，其他两箭射中的概率为$$\\left( 1-0.4 \\right)\\times 0.4\\times 0.4=0.096$$． 第三箭射空，其他两箭射中的概率为$$\\left( 1-0.4 \\right)\\times 0.4\\times 0.4=0.096$$． 有两箭射空的概率为$$0.96+0.96+0.96=0.288$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1348", "queId": "7e7d0d73baca424a9dd59e93529d0ed6", "competition_source_list": ["2012年全国迎春杯六年级竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "一辆玩具汽车，第一天按$$100 \\%$$的利润定价，无人来买；第二天降价$$10 \\%$$，还是无人买；第三天再降价$$360$$元，终于卖出．已知卖出的价格是进价的$$1.44$$倍，那么这辆玩具汽车的进价是~\\uline{~~~~~~~~~~}~元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$800$$ "}], [{"aoVal": "B", "content": "$$1000$$ "}], [{"aoVal": "C", "content": "$$1200$$ "}], [{"aoVal": "D", "content": "$$1400$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->经济问题->基本经济概念->折扣问题"], "answer_analysis": ["设进价为``$$1$$''，则第一天定价为``$$2$$''，第二天定价为``$$1.8$$''，最终售价为``$$1.44$$''． ``$$1.8$$''与``$$1.44$$''的差价等于$$360$$元，可知进价``$$1$$''$$=1000$$元． 设进价为$$x$$元． $$2x\\times (1-10 \\%)-360=1.44x$$， 解得：$$x=1000$$， 故答案为：$$1000$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1294", "queId": "2c4957341f4b4e55af308f736b608247", "competition_source_list": ["2013年IMAS小学中年级竞赛第一轮检测试题第15题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "有三只小兔子小白，小花和小黑在田里拔萝卜．已知小白和小花共拔了$$13$$根萝卜；小花和小黑共拔了$$11$$根萝卜；小黑和小白共拔了$$16$$根萝卜．请问这三只小兔子共拔了多少根萝卜？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$16$$ "}], [{"aoVal": "E", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["可发现$$13+11+16=40$$恰为三只小兔子总共拔的萝卜数之$$2$$倍，因此三只小兔子共拔了$$40\\div 2=20$$根萝卜．故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1725", "queId": "84f3bc739ab64c62a29a5a388ea6fc20", "competition_source_list": ["2020年广东广州羊排赛三年级竞赛第14题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "某部$$87$$集的电视连续剧在某个星期日开播，每周日、周一、周四、周五、周六都要播出一集，周二、周三停播．问：最后一集在周几播出？． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "四 "}], [{"aoVal": "C", "content": "五 "}], [{"aoVal": "D", "content": "六 "}], [{"aoVal": "E", "content": "日 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["$$87$$集的电视连续剧在某个星期日开播，每周日、周一、周四、周五、周六都要播一集，五天一周期，$$87\\div 5=17$$（周）$$\\cdots \\cdots 2$$（天），所以最后一集在周一播出，选项$$\\text{A}$$正确． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "774", "queId": "7affcc007ccb48b8abbee3b401d86903", "competition_source_list": ["2012年第8届全国新希望杯五年级竞赛复赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$\\overline{20\\square 12\\square }$$的$$\\square $$内填上合适的数字，使该六位数能同时被$$2、$$3$$、5$$整除，不同的填法有． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$种 "}], [{"aoVal": "B", "content": "$$4$$种 "}], [{"aoVal": "C", "content": "$$5$$种 "}], [{"aoVal": "D", "content": "$$6$$种 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$\\overline{20\\square 12\\square }$$末位肯定为$$0$$，那么$$\\overline{20\\square 120}$$是$$3$$的倍数，中间的$$\\square $$有三种选择$$1、$$4$$、7$$，因此最后只有$$3$$种填法． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2830", "queId": "a3d1b930a24f474597b0bb7ea47e1da1", "competition_source_list": ["2008年六年级竞赛创新杯", "2008年第6届创新杯六年级竞赛初赛B卷第10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\overline{abcd}$$是一个四位的自然数，已知$$\\overline{abcd}-\\overline{abc}-\\overline{ab}-a=2008$$，这个四位数是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "2275 "}], [{"aoVal": "B", "content": "2175 "}], [{"aoVal": "C", "content": "2257 "}], [{"aoVal": "D", "content": "2157 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->大数的估算"], "answer_analysis": ["由已知等式知，其中$$\\left( 1000a+100b+10c+d \\right)-\\left( 100a+10b+c \\right)-\\left( 10a+b \\right)-a=889a+89b+9c+d=2008$$，其中$$1\\leqslant a\\leqslant 9$$，$$0\\leqslant b$$，$$c$$，$$d\\leqslant 9$$.显然$$a=2$$，于是$$89b+9c+d=230$$，从而$$b=2$$，$$9c+d=52$$，所以$$c=5$$，$$d=7$$，因此，这个四位数$$\\overline{abcd}=2257$$. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1896", "queId": "b2e7fc32ab164e01bda10abdd29790b0", "competition_source_list": ["2017年IMAS小学中年级竞赛（第一轮）第11题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$(\\Delta \\div 2-2)\\times 2+2=222$$，请问$$\\Delta $$代表的数是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$224$$ "}], [{"aoVal": "C", "content": "$$228$$ "}], [{"aoVal": "D", "content": "$$876$$ "}], [{"aoVal": "E", "content": "$$884$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["$$\\Delta \\div 2-2=(222-2)\\div 2$$ $$=220\\div 2=110$$， 因此$$\\Delta =(110+2)\\times 2=112\\times 2=224$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2280", "queId": "8de71bbfe1354b6dad41d64e52950d3c", "competition_source_list": ["2019年广东深圳全国小学生数学学习能力测评四年级竞赛初赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$9$$点至$$10$$点之间的某个时刻，$$5$$分钟前分针的位置和$$5$$分钟后时针的位置相同，此时刻是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9:05$$ "}], [{"aoVal": "B", "content": "$$9:35$$ "}], [{"aoVal": "C", "content": "$$9:55$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->时间问题->认识钟表"], "answer_analysis": ["根据选项$$\\text{C}$$，现在是$$9:55$$，那么$$5$$分钟前分针的位置是指向$$10$$，$$5$$分钟后时针也是指向$$10$$．所以$$\\text{C}$$选项答案是满足���目要求的，故选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "827", "queId": "a3cc47094bea48a08f07d2918d7c6ecf", "competition_source_list": ["2006年华杯赛六年级竞赛初赛", "2006年华杯赛五年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "2008006共有( )个质因数. ", "answer_option_list": [[{"aoVal": "A", "content": "4 "}], [{"aoVal": "B", "content": "5 "}], [{"aoVal": "C", "content": "6 "}], [{"aoVal": "D", "content": "7 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->整除->整除特征->差系整除特征"], "answer_analysis": ["因为$$2008006=2006\\times 1000+2006=2006\\times 1001=\\left( 2\\times 17\\times 59 \\right)\\times \\left( 7\\times 11\\times 13 \\right)$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1745", "queId": "f6ab2479421641ddb270a9410d9d01a2", "competition_source_list": ["2016年第21届全国华杯赛小学高年级竞赛初赛B卷第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$为自然数，$$\\dfrac{4}{15}=\\frac{1}{a}+\\frac{1}{b}$$那么$$a+b$$的最小值是（ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->加减法应用->最值问题"], "answer_analysis": ["极端思考$$\\frac{4}{15}=\\frac{1}{6}+\\frac{1}{10}=\\frac{1}{5}+\\dfrac{1}{15}=\\frac{1}{4}+\\frac{1}{60}$$，则$$a+b$$最小为$$16$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3118", "queId": "f99de094d99f4ae19cb61292caf6b871", "competition_source_list": ["2006年五年级竞赛创新杯", "2006年第4届创新杯五年级竞赛初赛B卷第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "某年某月内有三个星期天的日期都是偶数，那么这个月的14日是星期( ). ", "answer_option_list": [[{"aoVal": "A", "content": "二 "}], [{"aoVal": "B", "content": "三 "}], [{"aoVal": "C", "content": "四 "}], [{"aoVal": "D", "content": "五 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数表规律->常见数表规律"], "answer_analysis": ["因为此月有三个星期天的日期都是偶数，所以该月必有两个星期天的日期为奇数，从而该月有5个星期天，由此可知该月的第一个星期天的日期只能是2号(否则，这个月就没有5个星期天)，从而5个星期天的日期为2，9，16，23，30，故14号为星期五 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3313", "queId": "ff8080814518d52401451925774804f0", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "我们定义像：$$31024$$、$$98567$$这样的五位数为位``神马数''，``神马数''是中间的数字最小，从中间往两边越来越大，且各位数字均不相同，那么，这样的五位数有（~ ~ ~ ~~）个 ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1512$$ "}], [{"aoVal": "B", "content": "$$3024$$ "}], [{"aoVal": "C", "content": "$$1510$$ "}], [{"aoVal": "D", "content": "$$3020$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["考察是计数问题中的排列组合． $$0～9$$是个数中任意挑选$$5$$个都可以组成``神马数''，$$\\text{C}_{10}^{5}=\\frac{10\\times9\\times 8\\times 7\\times 6}{5\\times 4\\times 3\\times 2\\times 1}=252$$种；在被挑选的$$5$$个数中，最小的放中间，剩下的$$4$$个数进行组合，从中任意挑选$$2$$个可以放在左边或者右边，$$\\text{C}_{4}^{2}=6$$种； 在此一定要注意：$$4$$个数中任选$$2$$个放在左边然后再放到右边数的顺序改变了． 所以共有``神马数''$$252\\times 6＝1512$$个． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2402", "queId": "3cd28305f2f249dbb2898da5e96087cc", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第9题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "小明在写作业时，不小心把一个三位小数的小数点忘掉了，结果所得的整数是$$60.03$$的$$100$$倍．原来的小数是多少，你知道吗？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$6003$$ "}], [{"aoVal": "B", "content": "$$600.3$$ "}], [{"aoVal": "C", "content": "$$6.03$$ "}], [{"aoVal": "D", "content": "$$6.003$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数基础->小数点的移动规律"], "answer_analysis": ["把一个三位小数的小数点忘掉了， 结果所得的整数是$$60.03$$的$$100$$倍即$$60.03\\times 100=6003$$． 原来的小数是$$60003\\div 1000=6.003$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2800", "queId": "a39e2b3c7b594371a8177ebb347bffc9", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知等差数列$$13$$，$$18$$，$$23$$，$$28$$，$$\\cdots $$，$$1003$$．这个等差数列共有项． ", "answer_option_list": [[{"aoVal": "A", "content": "$$198$$ "}], [{"aoVal": "B", "content": "$$199$$ "}], [{"aoVal": "C", "content": "$$200$$ "}], [{"aoVal": "D", "content": "$$201$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意：公差为$$5$$， 所以项数：$$\\left( 1003-13 \\right)\\div 5+1=199$$． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1381", "queId": "94c33d709ba546a0b08432e7cc6623bb", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "某学校准备购买$$30$$个篮球，三家商店每个篮球的售价都是$$25$$元，但优惠方法不同，甲店``买九赠一''，乙店``打八八折''；丙店``满$$100$$元减现金$$10$$元''，为节约资金，应该到（~ ）店购买． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "任意一个店皆可 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["甲店：只需付$$27$$个篮球的钱即可买到$$30$$个篮球，要$$27\\times 25=675$$元； 乙店：$$30\\times 25\\times 88 \\% =660$$元； 丙店：$$25\\times 30=750$$元，减$$7\\times 10=70$$元，花费$$750-70=680$$元． 因为$$660\\textless{}675\\textless{}680$$，所以应到乙店购买． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "737", "queId": "33154aa4cedb45e2914a1b58fab56bdd", "competition_source_list": ["2016年第16届世奥赛六年级竞赛决赛第11题", "2016年全国世奥赛竞赛A卷第11题"], "difficulty": "3", "qtype": "single_choice", "problem": "请从一个$$1 \\sim 9$$（缺$$8$$）这八个自然数中不重复地用这些数字构造出四个两位质数，并求出它们的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$190$$ "}], [{"aoVal": "B", "content": "$$217$$ "}], [{"aoVal": "C", "content": "$$127$$ "}], [{"aoVal": "D", "content": "无法构造 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数->质数与合数判定->质数与合数的认识"], "answer_analysis": ["我们知道所有的偶数都是合数，除了$$5$$以外，个位为$$5$$的数也都是合数，这$$8$$个数中只有$$1$$、$$3$$、$$5$$、$$7$$、$$9$$放在个位才有可能是质数，所以十位上只能是$$2$$、$$4$$、$$5$$、$$6$$．这四个两位数的和是$$\\left( 2+4+5+6 \\right)\\times 10+1+3+7+9=190$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2803", "queId": "5bf7727f81174be1a92f3ebb1f9c67aa", "competition_source_list": ["2017年河南郑州联合杯竞赛附加赛第一场第5题2分"], "difficulty": "0", "qtype": "single_choice", "problem": "用乘法分配律可以将$$ab+b$$改写成（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$(a+b)b$$ "}], [{"aoVal": "B", "content": "$$a(a+b)$$ "}], [{"aoVal": "C", "content": "$$(a+0)b$$ "}], [{"aoVal": "D", "content": "$$(a+1)b$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数乘法巧算之提取公因数（普通型）"], "answer_analysis": ["$$ab+b=a\\times b+1\\times b=\\left( a+1 \\right)b$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "249", "queId": "7f0e63f0cf7c48ca9d75429a3af9a9dc", "competition_source_list": ["2020年广东广州羊排赛四年级竞赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "六支队伍进行单循环赛，每两队都要赛一场．如果赛平，每队各得$$1$$分，否则胜队得$$2$$分，负队得$$0$$分．那么，打完所有比赛后，六支队伍的总得分是分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$30$$ "}], [{"aoVal": "E", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->体育比赛->2-1-0 积分制"], "answer_analysis": ["在一场比赛中，如果一胜一负， 则胜得$$2$$分，负得$$0$$分， 总分为$$2+0=2$$分， 如果赛平，则总分为$$1+1=2$$分， 即在一场比赛中，无论结果如何，比赛总分是不变的，都是$$2$$分， $$6$$支队伍进行单循环赛，共有比赛：$$5+4+3+2+1=15$$场， 所以六支队伍的总得分是：$$15\\times 2=30$$分， 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1329", "queId": "4bf45b311f7c4546a9e2c659338b1782", "competition_source_list": ["2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "在琼斯食堂里，每份晚餐有$$8$$个肉丸和$$3$$块蒜蓉面包．今天卖出的肉丸数量比卖出的蒜蓉面包数量多$$600$$．今天琼斯卖出了多少份晚餐？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$360$$ "}], [{"aoVal": "D", "content": "$$960$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["每一份晚餐肉丸比蒜蓉多$$5$$，一共多了$$600$$，因此卖了$$600\\div 5=120$$（份）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1168", "queId": "98d3f6753e3647a09286fd888d1d41a9", "competition_source_list": ["2022年第9届广东深圳鹏程杯四年级竞赛初赛第23题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$1$$、$$2$$、$$3$$、\\ldots\\ldots、$$1000$$顺次写下来，形成一个数1234567891011\\ldots9979989991000.这个数从左到右第$$2022$$个数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$6$$ "}], [{"aoVal": "E", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->页码问题"], "answer_analysis": ["无 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3288", "queId": "5a99a54fbdc8470b80c70ef4365db8bb", "competition_source_list": ["2021年第24届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第14题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "用数字$$0$$，$$1$$，$$2$$，$$3$$能组成个数字不重复的三位数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->字典排序法->组数->数字组数（不可重复）"], "answer_analysis": ["因为用数字$$0$$，$$1$$，$$2$$，$$3$$可以组成的三位数有：$$120$$、$$102$$、$$210$$、$$201$$、$$310$$、$$130$$、$$301$$、$$103$$、$$230$$、$$203$$、$$320$$、$$302$$、$$123$$、$$132$$、$$213$$、$$231$$、$$321$$、$$312$$，所以一共有$$18$$个不重复的三位数， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2863", "queId": "a8a5e0d490574a2ca3e702af6e0a4a12", "competition_source_list": ["2018~2019学年浙江杭州西湖区杭州市行知小学五年级上学期期中期中竞赛第17题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "小机灵在用计算器计算``$$6.9\\times 7$$''时，发现计算器的键``$$6$$''坏了，小机灵想到了四种不同的输入方法．请你判断一下，下面选项中的方法是错误的 ", "answer_option_list": [[{"aoVal": "A", "content": "$$2.3\\times 3\\times 7$$ "}], [{"aoVal": "B", "content": "$$13.8\\times 7\\div 2$$ "}], [{"aoVal": "C", "content": "$$2\\times 3\\times 7+0.9\\times 7$$ "}], [{"aoVal": "D", "content": "$$7\\times 7-7$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["$$7\\times 7-7=7\\times \\left( 7-1 \\right)=7\\times 6=42$$，提取公因数后发现$$\\text{D}$$式变为$$7\\times 6$$，而非原式的$$6.9\\times 7$$，所以错误． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2921", "queId": "add2799cd96940f3a1b9449b17fd73f4", "competition_source_list": ["2018年陕西西安雁塔区西安铁一中小升初（二十六）第20题5分", "2015年世界少年奥林匹克数学竞赛六年级竞赛复赛A卷第10题10分", "六年级其它导引"], "difficulty": "3", "qtype": "single_choice", "problem": "计算：$$\\frac{{{1}^{2}}+{{2}^{2}}}{1\\times 2}+\\frac{{{2}^{2}}+{{3}^{2}}}{2\\times 3}+\\cdots +\\frac{{{18}^{2}}+{{19}^{2}}}{18\\times 19}+\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}$$=~\\uline{~~~~~~~~~~}~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$37\\frac{19}{20}$$ "}], [{"aoVal": "B", "content": "$$28\\frac{19}{20}$$ "}], [{"aoVal": "C", "content": "$$38\\frac{19}{20}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "课内体系->能力->运算求解"], "answer_analysis": ["算式中的分母是裂项计算的最基本形式，但分子比较复杂，我们可以从前几项找找规律： $$\\frac{{{1}^{2}}+{{2}^{2}}}{1\\times 2}=\\frac{5}{2}=2\\frac{1}{2}$$，$$\\frac{{{2}^{2}}+{{3}^{2}}}{2\\times 3}=\\frac{13}{6}=2\\frac{1}{6}$$，$$\\frac{{{3}^{2}}+{{4}^{2}}}{3\\times 4}=\\frac{25}{12}=2\\frac{1}{12}$$． 我们发现一规律：每一项减去$$2$$后，分子就变成了$$1$$． 再来试试最后一项：$$\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}=\\frac{761}{380}=2\\frac{1}{380}$$， 也满足这个规律，这是为什么呢？ 观察每一项的分子和分母，我们发现分子的每个加数都与分母大小接近，可以做如下变形： $$\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}=\\frac{19\\times \\left( 20-1 \\right)+20\\times \\left( 19+1 \\right)}{19\\times 20}$$ $$=\\frac{19\\times 20\\times 2+\\left( 20-19 \\right)}{19\\times 20}$$ $$=2+\\frac{1}{19\\times 20}$$． 算式中的每一项都能像上面一样进行变形，再结合分数裂差，所以： 原式$$=2\\frac{1}{1\\times 2}+2\\frac{1}{2\\times 3}+\\cdots +2\\frac{1}{19\\times 20}$$ $$=2\\times 19+\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{19\\times 20}$$ $$=38+1-\\frac{1}{20}$$ $$=38\\frac{19}{20}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1114", "queId": "33c293a238044286871e3de8402ce02e", "competition_source_list": ["2013年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "某日，甲学校买了$$56$$千克水果糖，每千克$$8.06$$元．过了几日，乙学校也需要买同样的$$56$$千克水果糖，不过正好赶上促销活动，每千克水果糖降价$$0.56$$元，而且只要买水果糖都会额外赠送$$5 \\%$$同样的水果糖．那么乙学校将比甲学校少花元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$51.36$$ "}], [{"aoVal": "C", "content": "$$31.36$$ "}], [{"aoVal": "D", "content": "$$10.36$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->折扣问题"], "answer_analysis": ["甲学校花钱： $$56\\times 8.06=451.36$$（元）； 乙学校花钱： $$56\\div \\left( 1+5 \\% \\right)\\times\\left(8.06-0.56 \\right)$$， $$=\\frac{160}{3}\\times 7.5$$， $$=400$$（元）； 乙学校将比甲学校少花： $$451.36-400=51.36$$（元）； 答：乙学校将比甲学校少花$$51.36$$元． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "543", "queId": "f0af3924013b4f289e446aa8d92e92d3", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$63$$分成两个自然数，要使这两个数的乘积最大，这两个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "0,63 "}], [{"aoVal": "B", "content": "30,33 "}], [{"aoVal": "C", "content": "31,32 "}], [{"aoVal": "D", "content": "1,62 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->和一定型最值问题->构造和一定最值原理"], "answer_analysis": ["和一定，差小积大 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1800", "queId": "6f597fc82ae9429bb0164657aee01f3b", "competition_source_list": ["2006年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "$$12345678910111213\\cdots 20052006$$是( )位数. ", "answer_option_list": [[{"aoVal": "A", "content": "6913 "}], [{"aoVal": "B", "content": "6914 "}], [{"aoVal": "C", "content": "6915 "}], [{"aoVal": "D", "content": "6917 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->页码问题->数字与页码的对应问题->数字个数"], "answer_analysis": ["求多位数共有多少个数码，分几个阶段来计算(一位，二位，三位，四位).$$9\\times 1+90\\times 2+900\\times 3+\\left( 2006-999 \\right)\\times 4=6917$$(位) "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2393", "queId": "0b959586ee754baabe674f9e0722b34f", "competition_source_list": ["2020年重庆渝中区重庆市巴蜀中学校小升初（十一）第2题2分", "2018年华杯赛小学中年级竞赛初赛第1题10分", "2018年第23届华杯赛小学中年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$A$$、$$B$$均为小于$$1$$的小数，算式$$A\\times B+0.1$$的结果．（2020BS） ", "answer_option_list": [[{"aoVal": "A", "content": "大于$$1$$ "}], [{"aoVal": "B", "content": "小于$$1$$ "}], [{"aoVal": "C", "content": "等于$$1$$ "}], [{"aoVal": "D", "content": "无法确定和$$1$$的大小 "}]], "knowledge_point_routes": ["拓展思维->思想->赋值思想"], "answer_analysis": ["$$0.2+0.3+0.1=0.16\\textless{}1$$ $$0.99\\times 0.99+0.1=1.081\\textgreater1$$ 又因为小数是连续的，因此可以取到等于$$1$$的$$A$$、$$B$$值，故$$A\\times B+0.1$$的计算结果与$$1$$的大小无法比较． 因为$$A$$，$$B$$都小于$$1$$， 所以无法确定$$A\\times B$$与$$0.9$$的大小关系． ①当$$A\\times B\\textless{}0.9$$，则$$A\\times B+0.1\\textless{}1$$， ②当$$A\\times B=0.9$$，则$$A\\times B+0.1=1$$， ③当$$A\\times B\\textgreater0.9$$，则$$A\\times B+0.1\\textgreater1$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2418", "queId": "2152c1bd3aae4b87b58bb1ce985922ea", "competition_source_list": ["2020年第1届广东深圳超常思维竞赛五年级竞赛初赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "找规律：$$16\\textasciitilde\\textasciitilde20\\textasciitilde\\textasciitilde25\\textasciitilde\\textasciitilde30\\textasciitilde\\textasciitilde36\\textasciitilde\\textasciitilde42\\textasciitilde\\textasciitilde49$$，则空格处的正确答案是 ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$64$$ "}], [{"aoVal": "C", "content": "$$72$$ "}], [{"aoVal": "D", "content": "$$81$$ "}], [{"aoVal": "E", "content": "$$100$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数"], "answer_analysis": ["这组数的规律如下：$$4\\times4$$，$$4\\times5$$，$$5\\times5$$，$$5\\times6$$，$$6\\times6$$，$$6\\times7$$，$$7\\times7$$，$$7\\times8$$，$$7\\times8=56$$．故选$$\\text{A}$$． ", "$$16+4=20$$，$$20+5=25$$，$$25+5=30$$，$$30+6=36$$，$$36+6=42$$，$$42+7=49$$，$$49+7=56$$．故选$$\\text{A}$$．
"], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2742", "queId": "ff8080814518d524014519271d85053a", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "童童在计算有余数的除法时，把被除数$$472$$错看成了$$427$$，结果商比原来小$$5$$，但余数恰好相同，那么这个余数是（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法->除法中四量关系"], "answer_analysis": ["除数$$=(472-427)\\div 5=9$$，$$472\\equiv 4(\\bmod9)$$，所以余数是4． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2005", "queId": "b894bb61ebac4b6abc94a31736e023e5", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$3$$年前爷爷的年龄是小明年龄的$$4$$倍，$$5$$年后爷爷的年龄是小明年龄的$$3$$倍，爷爷今年是岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$67$$ "}], [{"aoVal": "C", "content": "$$70$$ "}], [{"aoVal": "D", "content": "$$76$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["设小明今年$$x$$岁，爷爷今年$$y$$岁， $$y-3=4\\left( x-3 \\right)$$， $$y+5=3\\left( x+5 \\right)$$， 解得$$x=19$$，$$y=67$$， 则爷爷今年$$67$$岁． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "664", "queId": "2bfb8173cb274461a4a28137b1b31d75", "competition_source_list": ["2011年全国学而思杯五年级竞赛第5题", "2011年北京学而思综合能力诊断六年级竞赛第3题", "2011年全国学而思杯四年级竞赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "2018年$$9$$月$$8$$日，云南墨江发生大约$$6$$级地震．$$2008$$年$$5$$月$$12$$日，汶川发生$$8$$级大地震．已知地震级数每升$$1$$级，地震释放能量大约扩大到原来的$$30$$倍，那么汶川大地震释放能量是云南墨江地震的~\\uline{~~~~~~~~~~}~倍． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$90$$ "}], [{"aoVal": "C", "content": "$$900$$ "}], [{"aoVal": "D", "content": "$$27000$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$8$$级和$$6$$级差了$$2$$级，那么30*30=900 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2850", "queId": "658e2a9d0a0e4e1dbe4508836c97a058", "competition_source_list": ["四年级其它", "2012年全国学而思杯三年级竞赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "算式：$$103\\times 107-91\\times 99$$的计算结果是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2022$$ "}], [{"aoVal": "B", "content": "$$2012$$ "}], [{"aoVal": "C", "content": "$$2002$$ "}], [{"aoVal": "D", "content": "$$2000$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->公式类运算->平方差公式->平方差公式逆向应用", "Overseas Competition->知识点->计算模块->公式类运算->平方差公式->平方差公式逆向应用"], "answer_analysis": ["$$\\begin{align}\\& 103\\times 107-91\\times99 \\& =(105-2)\\times (105+2)-(95-4)\\times (95+4) \\&=({{105}^{2}}-{{2}^{2}})-({{95}^{2}}-{{4}^{2}}) \\&={{105}^{2}}-{{2}^{2}}-{{95}^{2}}+{{4}^{2}} \\&=({{105}^{2}}-{{95}^{2}})+({{4}^{2}}-{{2}^{2}}) \\& =(105+95)\\times(105-95)+12 \\& =200\\times 10+12 \\& =2012 \\end{align}$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "547", "queId": "005512b0eb50425a82d25ca8c59f4866", "competition_source_list": ["2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第3题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一道除法题，商和除数都是$$8$$，余数是$$7$$，被除数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$57$$ "}], [{"aoVal": "B", "content": "$$64$$ "}], [{"aoVal": "C", "content": "$$71$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法->除法中四量关系"], "answer_analysis": ["被除数$$\\div $$除数$$=$$商$$+$$余数；被除数$$=$$除数$$\\times $$商$$+$$余数； 从题干可以知道，一道除法题，商和除数都是$$8$$，余数是$$7$$； 所以被除数$$=8\\times 8+7=71$$． 所以选择$$\\text{C}$$选项． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "990", "queId": "2096c9fa1f6244fb9b3520e3a1405543", "competition_source_list": ["走美杯四年级竞赛", "走美杯三年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "小明的妈妈比小明大$$26$$岁，去年妈妈的岁数正好是小明的$$3$$倍，小明今年（ ）岁。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型"], "answer_analysis": ["去年妈妈比小明大$$2$$倍，则去年小明$$26\\div 2=13$$（岁），小明今年$$13+1=14$$（岁）。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1292", "queId": "1b845f70956f4040b02c7e04f2e69a7a", "competition_source_list": ["2017年湖北武汉创新杯六年级竞赛邀请赛训练题（三）"], "difficulty": "2", "qtype": "single_choice", "problem": "金星上有三种奇怪的生物，$$3$$脚$$5$$尾的独头鸟：$$2$$脚$$1$$尾的四头蛇；$$5$$脚$$3$$尾的五头龙．所有生物加起来共$$106$$个头，$$98$$只脚和$$76$$条尾巴，那么这三种动物一共（~ ）只． ", "answer_option_list": [[{"aoVal": "A", "content": "$$26$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["解设独头鸟$$x$$只，四头蛇$$y$$只，五头龙$$\\text{z}$$只，其方程组 $$\\begin{cases} x+4y+5z=106 3x+2y+5z=98 5x+y+3z=76 \\end{cases}$$解得$$\\begin{cases} x=6 y=10 z=12 \\end{cases}$$ $$6+10+12=28$$（只）. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1427", "queId": "8373cebd7bc246ee9c8eb04eb97982ac", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "小林今年$$8$$岁．小林满$$10$$岁时，爸爸正好$$39$$岁，爸爸今年岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$29$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$37$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["爸爸比小林大：$$39-10=29$$（岁），所以今年爸爸的年龄为：$$29+8=37$$（岁）． 故选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1409", "queId": "510c3c2136034398848f809af597d5f4", "competition_source_list": ["2020年第24届YMO三年级竞赛决赛第5题3分", "2019年第24届YMO三年级竞赛决赛第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个正方形的操场，在它的外面一圈插上小红旗，四个角上都插一面小红旗，每边都插$$26$$面，一共插了面小红旗． ", "answer_option_list": [[{"aoVal": "A", "content": "$$92$$ "}], [{"aoVal": "B", "content": "$$96$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$104$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->封闭型->封闭型植树问题->封闭植树类型问题(段数大于10)"], "answer_analysis": ["根据题意分析可知，四个角都插上一面小红旗，每边都插$$26$$面， 所以每一边有小红旗：$$26-1=25$$（面）， 因为是正方形，每边都有$$25$$面，那么一共插了：$$25\\times4=100$$（面）红旗， 故答案为：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1970", "queId": "b8278aa442a64462b7085d7e61afe9f7", "competition_source_list": ["2016年广东深圳华杯赛小学高年级竞赛冬令营二试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "数学竞赛原定一等奖$$10$$人、二等奖$$20$$人，现在将一等奖中得分靠后的$$4$$人调整为二等奖，这样得二等奖的学生的平均分提高了$$1$$分，得一等奖的学生的平均分提高了$$3$$分．那么原来一等奖平均得分比二等奖平均分多多少分？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$9.5$$ "}], [{"aoVal": "B", "content": "$$10.2$$ "}], [{"aoVal": "C", "content": "$$10.5$$ "}], [{"aoVal": "D", "content": "$$11.5$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->列方程解应用题->设而不求"], "answer_analysis": ["设原来一等奖平均分为$$x$$分，二等奖平均分为$$y$$分． $$10x+20y=\\left( 10-4 \\right)\\times \\left( x+3 \\right)+\\left( 20+4 \\right)\\times \\left( y+1 \\right)$$ $$10x+20y=6x+18+24y+24$$ $$4x-4y=42$$ $$x-y=10.5$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1473", "queId": "372ef32705ed4f5b943c5efb0f810696", "competition_source_list": ["2009年第9届全国走美杯四年级竞赛初赛"], "difficulty": "3", "qtype": "single_choice", "problem": "小华问陈老师今年有多少岁，陈老师说:``当我像你这么大时，我的年龄是你年龄的$$10$$倍，当你像我这么大时，我已经$$56$$岁了．''陈老师今年多少岁？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$38$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$28$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["和差倍问题（年龄问题），差不变，设当陈老师与小华一样大时，小华为$$1$$份，则陈老师为$$10$$份，此时年龄差为$$9$$份，所以现在小华为$$10$$份，陈老师为$$19$$份，当小华像陈老师一样大时，小华$$19$$份，陈老师为$$28$$份，此时$$1$$份为$$2$$，所以陈老师今年$$38$$岁． $$（10-1）\\times 3+1=28$$ 学生：$$56\\div 28=2$$（岁） 老师：$$2\\times 10=20$$（岁） $$20+（20-2）=38$$（岁） "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3178", "queId": "2f7830cec449471489686d54e4dbef1d", "competition_source_list": ["2017年河南郑州联合杯竞赛第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "某汽车销售处有$$65$$辆小汽车，其中$$45$$辆有空调，$$30$$辆有高级音响，$$12$$辆有空调也有高级音响，没有空调也没有高级音响的汽车有（~ ）辆． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$~~~~~ "}], [{"aoVal": "B", "content": "$$8$$~~~~~ "}], [{"aoVal": "C", "content": "$$10$$~~~ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->容斥原理->三量容斥"], "answer_analysis": ["容斥原理；没有空调也没有高级音响的有$$65-\\left( 45+30-12 \\right)=2$$辆． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "64", "queId": "2ef54e132c8047b0b130878302d4bc7e", "competition_source_list": ["六年级其它", "2017年华��赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "两个有限小数的整数部分分别是$$7$$和$$10$$，那么这两个有限小数的积的整数部分有种可能的取值． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$19$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["由已知设这两个数分别为$$a$$，$$b$$，可得$$7\\times 10 \\textless{} a\\times b \\textless{} 8\\times 11$$，即$$70 \\textless{} a\\times b \\textless{} 88$$，则乘积的整数部分$$M$$满足$$M\\leqslant a\\times b$$，则$$70\\leqslant M \\textless{} 88$$，因此可得整数部分可以取$$70$$到$$87$$的所有整数，共有$$87-70+1=18$$（个），因此选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1400", "queId": "3a9a26fd902d4f4ca403e217a758bffa", "competition_source_list": ["2005年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "某班学生的达标人数是没有达标人数的$$\\frac{1}{4}$$，如果又有$$2$$人达标，这时达标人数是没有达标人数的$$\\frac{1}{3}$$，那么全班有（ ）人。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题", "课内体系->知识模块->综合与实践"], "answer_analysis": ["由达标人数是没有达标人数的$$\\frac{1}{4}$$知： 达标人数占总人数的$$\\frac{1}{1+4}=\\frac{1}{5}$$。又有$$2$$人达标后， 达标人数是没有达标人数的$$\\frac{1}{3}$$， 那么这时达标人数占总人数的$$\\frac{1}{1+3}=\\frac{1}{4}$$。 所以这个班的总人数为$$2\\div \\left( \\frac{1}{4}-\\frac{1}{5} \\right)=40$$（人）， 故而选B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1504", "queId": "ff676fab3110463cb348de43e1ad0bd0", "competition_source_list": ["2014年IMAS小学中年级竞赛第一轮检测试题第17题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "王师傅搬$$40$$玻璃，每搬一块得$$2$$元，如果打碎一块玻璃不但没有搬运费，还要赔$$8$$元，最后王师傅拿到了$$60$$元，请问王师傅打碎了多少块玻璃？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->差倍->两量差倍->变倍型二量差倍问题"], "answer_analysis": ["假设王师傅没有打碎玻璃，则他应该拿到$$2\\times 40=80$$元，但他只拿到$$60$$元，所以他打碎了$$(80-60)\\div (2+8)=2$$块玻璃． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1585", "queId": "40e4d19b1773495095f7f918478a5f13", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2014$$年$$2$$月$$6$$日是星期四，小胖决定从这天起（含$$2$$月$$6$$日）练习计算，一直练习到$$2$$月$$17$$日，（含$$2$$月$$17$$日）开学为止．但是中间如果遇到周六和周日，小胖还是决定休息一下，不做练习．已知他第一天做$$1$$道题，第二天做$$3$$道题，第三天做$$5$$道题，依此变化做下去，那么小胖这段时间一共做了道计算练习题. ", "answer_option_list": [[{"aoVal": "A", "content": "$$144$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$81$$ "}], [{"aoVal": "D", "content": "$$64$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["解：依题意可知： 从$$2$$月$$6$$日到$$2$$月$$17$$日为止，一共有$$17-6+1=12$$（天）； 其中有$$2$$个星期六，星期日，工作了$$12-4=8$$（天）. 共完成$$1+3+5+7+9+11+13+15=64$$（道）题. 故选：$$\\text{D}$$. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "638", "queId": "2b3f57680d464571a75e08d50dd3fb3c", "competition_source_list": ["2014年迎春杯六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "式子$$\\frac{2014}{x+1}$$为整数，则正整数$$x$$有( )种取值。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数"], "answer_analysis": ["因为$$2014=2\\times 19\\times 53$$，$$x+1$$全部可能的取值为$$2014$$的$$1$$以外的全部因数，根据因数个数公式，故共有$${{\\left( 1+1 \\right)}^{3}}-1=7$$种。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2775", "queId": "5bad78e59d5149ef9c06fbfbe7b92c61", "competition_source_list": ["2006年四年级竞赛创新杯", "2006年第4届创新杯四年级竞赛初赛A卷第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面的算式是按一定的规律排列的:$$4+2$$，$$5+8$$ ，$$6+14$$，$$7+20\\cdots $$那么和为83的算式是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "$$16+67$$ "}], [{"aoVal": "B", "content": "$$15+68$$ "}], [{"aoVal": "C", "content": "$$14+69$$ "}], [{"aoVal": "D", "content": "$$17+66$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列求项数"], "answer_analysis": ["观察这组式子可知两加数分别由两组等差数列组成，且和亦为等差数列 第一组:4，5，6，7，\\ldots 公差1; 第二组:2，8，14，20，\\ldots 公差6; 各组和:6，13，20，27，\\ldots 公差7; 因此，当和为83时，可求得项数为$$(83-6)\\div 7+1=12$$所以第12项的算式为:$$\\left[ 4+\\left( 12-1 \\right)\\times 1 \\right]+\\left[ \\left( 12-1 \\right)\\times 6+2 \\right]=15+68$$，选B "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "434", "queId": "822c9f1ac99b4189b7990922772fd420", "competition_source_list": ["2012年第10届创新杯四年级竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "显示在电子钟上的时间是$$5:55$$，下一次电子钟上显示的时间又是全部相同的数字，还要过分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$71$$ "}], [{"aoVal": "B", "content": "$$255$$ "}], [{"aoVal": "C", "content": "$$316$$ "}], [{"aoVal": "D", "content": "$$377$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["因为分钟的十位最大为$$5$$，故下一次数字相同的时刻为$$11:11$$，$$11:11$$距$$5:55$$有$$5$$小时$$16$$分，即$$316$$分钟． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2313", "queId": "e5a51986a7c54763a8c75881b9e56a7e", "competition_source_list": ["2016年创新杯五年级竞赛训练题（一）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙两人在学校到体育场的路上练习竞走，甲每分比乙多走$$10$$米，上午$$9$$点两人同时从学校出发，上午$$10$$点甲到达体育场后立即返回学校，在距体育场$$310$$米处遇到乙，那么学校到体育馆的距离为（~ ）米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9610$$ "}], [{"aoVal": "B", "content": "$$9300$$ "}], [{"aoVal": "C", "content": "$$8600$$ "}], [{"aoVal": "D", "content": "$$7000$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["根据题意有相遇时甲比乙多走了$$310\\times 2=620$$（米），已知每分钟甲比乙多走$$10$$米，那么从出发到相遇时间为$$620\\div 10=62$$（分），形完全程甲需要$$60$$分钟，那么走$$310$$米需要$$2$$分钟，那么甲的速度为$$310\\div 2=155$$（米）每分钟，所以学校到体育馆的距离为：$$60\\times 155=9300$$（米）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1607", "queId": "4574c364eaa949d8a300d1b1541f73a2", "competition_source_list": ["2018年IMAS小学高年级竞赛（第一轮）第11题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "\\uline{小明}在作除法时，把除数$$45$$写成$$54$$，结果得到的商是$$18$$且余数为$$18$$．请问正确的商应该是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$22$$ "}], [{"aoVal": "D", "content": "$$24$$ "}], [{"aoVal": "E", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["由题意知被除数为$$54\\times 18+18=990$$， 因此正确的商为$$990\\div 45=22$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2048", "queId": "dd376ed13120437498ae8455d620befe", "competition_source_list": ["2017年全国希望杯小学高年级五年级竞赛初赛考前100题"], "difficulty": "3", "qtype": "single_choice", "problem": "小聪赶着一头猪到山外的生猪收购站去卖．过秤知猪重$$150$$斤，他和收购站的工作人员有如下对话： 收购员：你这头猪肚子这么大有这么重，是不是故意让猪吃了很咸的猪食，然后大量喝水造成的？不收！ 小聪：我们家有诚信的家风，绝不会这样！请收购吧，我走了很远的山路猜到这里． 收购员：如果马上收购，猪的重量要打九折，如果你明天早上来，当面再称的重量，收购价提高两成五，两种选择由你来决定！ 请帮助小聪做出选择，并说明原因． ", "answer_option_list": [[{"aoVal": "A", "content": "第一种 "}], [{"aoVal": "B", "content": "第二种 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题->方程法解其他问题"], "answer_analysis": ["设收购价为$$a$$元/斤，则第二天``保本''情况下，猪的重量应为 $$0.9\\times 150a\\div \\left[ a\\left( 1+25 \\% \\right) \\right]=0.9\\times 150a\\div 1.25a=108$$（斤） 就是说，及时猪的重量减轻了$$42$$斤，也能``保本''，也比第一种选择好，根据生活经验可知，一夜之后猪的重量不可能减轻$$42$$斤，所以应该果断选择第二种！ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3375", "queId": "96c41632fe374477a365b3b351deebe1", "competition_source_list": ["2009年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "师徒两人加工同一种零件，每人都把自己的产品装入自己的箩筐中，结果师傅产量是徒弟的两倍，现在装了$$6$$只箩筐，每只箩筐都标了零件的只数：$$78$$只，$$94$$只，$$86$$只，$$87$$只，$$82$$只，$$80$$只。那么( )这两筐是徒弟加工的。 ", "answer_option_list": [[{"aoVal": "A", "content": "87只与86只 "}], [{"aoVal": "B", "content": "87只与82只 "}], [{"aoVal": "C", "content": "80只与87只 "}], [{"aoVal": "D", "content": "94只与80只 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法"], "answer_analysis": ["因为$$\\left( 78+94+86+87+82+80 \\right)\\div \\left( 1+2 \\right)=169$$，又$$87+82=169$$，所以$$87$$只与$$82$$只这两筐是徒弟加工的。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "821", "queId": "9f255ff8cf9e48e596d08e791d5dffa4", "competition_source_list": ["2014年迎春杯四年级竞赛初赛", "2014年迎春杯三年级竞赛初赛", "2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一些糖果，如果每天吃$$3$$个，十多天吃完，最后一天只吃了$$2$$个；如果每天吃$$4$$个，不到$$10$$天就吃完了，最后一天只吃了$$3$$个。那么，这些糖果原来有（ ）个。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$35$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法"], "answer_analysis": ["如果每天吃$$3$$个，十多天吃完，最后一天只吃了$$2$$个，说明糖果至少有$$3\\times 10+2=32$$（个），且糖果数应除以$$3$$余$$2$$；如果每天吃$$4$$个，不到$$10$$天就吃完了，最后一天吃了$$3$$个，说明糖果至多有$$4\\times 8+3=35$$（个），且糖果数应除以$$4$$余$$3$$。综上，糖果有$$35$$个。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "12", "queId": "4a6c95f3b4ad41eb857ce0290604a957", "competition_source_list": ["2017年第16届春蕾杯二年级竞赛决赛第12题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "把一张正方形纸对折$$1$$次，可以得到$$2$$个完全相同的长方形；再对折$$1$$次，可以得到$$4$$个完全相同的长方形；再对折$$1$$次，可以得到$$\\cdots \\cdots $$那么把这张正方形的纸连续对折多少次，可以得到$$128$$个完全相同的长方形？ ", "answer_option_list": [[{"aoVal": "A", "content": "6次 "}], [{"aoVal": "B", "content": "7次 "}], [{"aoVal": "C", "content": "8次 "}], [{"aoVal": "D", "content": "64次 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["（$$1$$）对折$$3$$次，得到$$8$$个完全相同的长方形； （$$2$$）对折$$4$$次，得到$$16$$个完全相同的长方形； （$$3$$）对折$$5$$次，得到$$32$$个完全相同的长方形； （$$4$$）对折$$6$$次，得到$$64$$个完全相同的长方形； （$$5$$）对折$$7$$次，得到$$128$$个完全相同的长方形； 答：把这张正方形的纸连续对折$$7$$次，可以得到$$128$$个完全相同的长方形． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "436", "queId": "a0a56f3611f54d8cb485926409f22c79", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（五）第6题"], "difficulty": "3", "qtype": "single_choice", "problem": "用$$1\\tilde{ }9$$这九个数字组成三个三位数，每个数字只用一次，要求它们的和是奇数，那么这三个三位数的和最大是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2521$$ "}], [{"aoVal": "B", "content": "$$2529$$ "}], [{"aoVal": "C", "content": "$$2539$$ "}], [{"aoVal": "D", "content": "$$2547$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["先考虑最大，$$741+852+963=2556$$，再调整相近的两个数字$$4$$和$$3$$，则$$731+852+964=2547$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2144", "queId": "0f70646727664cb796f44ea932142bf3", "competition_source_list": ["2016年全国小学生数学学习能力测评四年级竞赛复赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "上午$$9$$点$$30$$分时针与分针组成的角是度． ", "answer_option_list": [[{"aoVal": "A", "content": "$$75$$ "}], [{"aoVal": "B", "content": "$$90$$ "}], [{"aoVal": "C", "content": "$$105$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$9$$点$$30$$分时时针与分针构成的角度为$$30\\times 9-30\\times 5.5=105$$（度）． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2089", "queId": "fdc8257132b944dabe5d657d499db487", "competition_source_list": ["2011年广东深圳华杯赛小学中年级竞赛复赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "一列数：$$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，34$\\cdot\\cdot\\cdot$从第二个开始，每个数都是最靠近它前两个数的和．那么第$$100$$个数除以$$3$$的余数是~\\uline{~~~~~~~~~~}~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "无余数 "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["除以$$3$$的余数为$$1$$，$$1$$，$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0$$，$$1$$，$$1$$，$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0.$$八个一周期 100$\\div8=12\\cdot\\cdot\\cdot\\cdot\\cdot\\cdot4$ 第$$100$$个数是除以$$3$$的余数和第$$4$$个数除以$$3$$的余数一样．所以第$$100$$个数除以$$3$$的余数是$$0$$． 故答案是没有余数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1503", "queId": "df005691d6ee4b7aaa1460a7c55b7fa1", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "大李和小李两人一共做了$$68$$个幸运星，大李的数量比小李的$$4$$倍少$$2$$个，如果设小李做了$$x$$个幸运星，则下列方程正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4x-2=68$$ "}], [{"aoVal": "B", "content": "$$(4x-2)+x=68$$ "}], [{"aoVal": "C", "content": "$$(4x+2)+x=68$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["设小李做了$$x$$个幸运星，则大李做了$$(4x-2)$$个，根据题意列方程：$$(4x-2)+x=68$$，故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "928", "queId": "bcbb90873a9b462f844c63fd4acb8a09", "competition_source_list": ["2012年美国数学大联盟杯六年级竞赛初赛第33题5分(每题5分)"], "difficulty": "1", "qtype": "single_choice", "problem": "$$264$$和的最大公因数是$$132$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$66$$ "}], [{"aoVal": "C", "content": "$$528$$ "}], [{"aoVal": "D", "content": "$$660$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["公约数，亦称``公因数''．它是指能同时整除几个整数的数．如果一个整数同时是几个整数的约数，称这个整数为它们的``公约数''；公因数中最大的称为最大公因数． $$264$$和$$528$$的最大公因数为$$264$$； $$264$$和$$660$$的最大公因数为$$132$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "169", "queId": "35411fca5f7b4c4e9f6227ef87dfa378", "competition_source_list": ["2003年第9届华杯赛竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "一副扑克牌有$$54$$张，最少要抽取几张牌，方能使其中至少有$$2$$张牌有相同的点数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$54$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->简单抽屉原理"], "answer_analysis": ["如果不算大、小王，每个花色$$13$$张牌，只需$$14$$张便一定有两张相同点数的牌，加上大、小王，则需要$$16$$张牌． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3379", "queId": "a90c956fc6294c7aa35d2440e62c3531", "competition_source_list": ["2008年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$1$$克、$$2$$克、$$4$$克、$$8$$克的砝码各一个，从这$$4$$个砝码中每次任选$$2$$个使用，能称出（ ）种不同的重量。（砝码也可以放在天平的两边） ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["从$$1\\text{、}2\\text{、}4\\text{、}8$$克中任取两个，其和为$$3\\text{、}5\\text{、}9\\text{、}6\\text{、}10\\text{、}12$$克，从$$1\\text{、}2\\text{、}4\\text{、}8$$克任取两个，大的减去小的，其差为$$1\\text{、}3\\text{、}7\\text{、}2\\text{、}6\\text{、}4$$克，又因为和与差中的$$3$$克$$\\text{、}6$$克重复，所以可称出$$6+6-2=10$$（种）不同的重量。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1786", "queId": "f22edb1683e848a7a49cda2fcc48df1b", "competition_source_list": ["2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "祖玛游戏中，龙嘴里不断吐出很多颜色的龙珠，先$$4$$颗红珠，接着$$3$$颗黄珠，再$$2$$颗绿珠，最后$$1$$颗白珠，按此方式不断重复，从龙嘴里吐出的第$$2000$$颗龙珠是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "红珠 "}], [{"aoVal": "B", "content": "黄珠 "}], [{"aoVal": "C", "content": "绿珠 "}], [{"aoVal": "D", "content": "白珠 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["解：$$2000\\div (4+3+2+1)$$ $$=2000\\div 10$$ $$=200$$（组） 商是$$200$$，没有余数，说明第$$2000$$颗龙珠是$$200$$组的最后一个，是白珠。 答：从龙嘴里吐出的第$$2000$$颗龙珠是白珠。 故选:D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "77", "queId": "6fd4094d371d447886c8f9c317afcc9d", "competition_source_list": ["2003年五年级竞赛创新杯", "2003年六年级竞赛创新杯", "2003年第1届创新杯五年级竞赛复赛第10题", "2003年四年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "一个聪明的海盗，他根据藏宝图的指引来到了传说中的藏宝岛，他上岛后发现岛上确实有三个厚重的石门，而其中之一就是通往宝藏之门. 右边的石门上写着:此非通往宝藏之门. 中间的石门上写着:此是通往宝藏之门. 左边的石门上写着:右边的石门确实不是通往宝藏之门. 根据藏宝图所附的提示得知，这三道石门上的陈述至少有一句是真的，有一句是假的，那么真正通往宝藏之门的是（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "右边的石门 "}], [{"aoVal": "B", "content": "中间的石门 "}], [{"aoVal": "C", "content": "左边的石门 "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理"], "answer_analysis": ["用假设法，假设右边门是，则左中右都是假的，所以假设不符;假设中间门是，则左中右都是真的，不符;假设左边门是，则左右是真的，中间是假的，符合.所以真正通往宝藏之门是左边的石门，选C. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2357", "queId": "037a92247d87499aa63ad80700b2db23", "competition_source_list": ["2004年第2届创新杯六年级竞赛复赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "在下面四个算式中，得数最大的是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( \\frac{1}{17}-\\frac{1}{19} \\right)\\div 20$$ "}], [{"aoVal": "B", "content": "$$\\left( \\frac{1}{15}-\\frac{1}{21} \\right)\\div 60$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{13}-\\frac{1}{23} \\right)\\div 100$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{1}{11}-\\frac{1}{25} \\right)\\div 140$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->和一定型最值问题->构造和一定最值原理"], "answer_analysis": ["$$\\left( \\frac{1}{17}-\\frac{1}{19} \\right)\\div 20=\\frac{2}{17\\times 19}\\div 20=\\frac{1}{17\\times 19}\\times \\frac{1}{10}$$;$$\\left( \\frac{1}{15}-\\frac{1}{21} \\right)\\div 60=\\frac{6}{15\\times 21}\\div 60=\\frac{1}{15\\times 21}\\times \\frac{1}{10}$$; $$\\left( \\frac{1}{13}-\\frac{1}{23} \\right)\\div 100=\\frac{10}{13\\times 23}\\div 100=\\frac{1}{13\\times 23}\\times \\frac{1}{10};$$$$\\left( \\frac{1}{11}-\\frac{1}{25} \\right)\\div 140=\\frac{14}{11\\times 25}\\div 140=\\frac{1}{11\\times 25}\\times \\frac{1}{10};$$ 只需比较$$\\frac{1}{17\\times 19}$$，$$\\frac{1}{15\\times 21}$$，$$\\frac{1}{13\\times 23}$$，$$\\frac{1}{11\\times 25}$$的大小，根据和一定，两数越接近乘 积越大，则$$11\\times 25 \\textless{} 13\\times 23 \\textless{} 15\\times 21 \\textless{} 17\\times 19$$，那么 $$\\frac{1}{11\\times 25}\\textgreater\\frac{1}{13\\times 23}\\textgreater\\frac{1}{15\\times 21}\\textgreater\\frac{1}{17\\times 19}$$，所以答案为$$D$$ "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1779", "queId": "c8cb2f284ba942169d39ad6ebb1e42d3", "competition_source_list": ["2012年IMAS小学高年级竞赛第一轮检测试题第13题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "小英从上星期五开始观察一株风信子，当时有些花已经开了．从此之后，每天新开的花朵数刚好等于前一天已开的花朵数，在这个过程中没有花凋谢．如果风信子的花朵全部开的那一天是星期四，请问花刚好开完一半的那一天是星期几？ ", "answer_option_list": [[{"aoVal": "A", "content": "星期六 "}], [{"aoVal": "B", "content": "星期天 "}], [{"aoVal": "C", "content": "星期一 "}], [{"aoVal": "D", "content": "星期二 "}], [{"aoVal": "E", "content": "星期三 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["因为每天新开的花朵数刚好等于前一天已开的花朵数，所以最后一天新开的花朵数刚好等于总花朵数的一半．这说明花刚好开完一半的那一天是最后一天的前一天，也就是星期三． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1761", "queId": "ed79aaf290ec4c659894c847e5a3d831", "competition_source_list": ["2020年广东广州海珠区广州为明学校卓越杯六年级竞赛初赛第5题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "水结成冰体积增加$$\\frac{1}{10}$$，冰化成水体积减少． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{10}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{11}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "E", "content": "$$\\frac{1}{8}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["设水的体积是``$$1$$''， 那么水结成冰体积是$$1\\times \\left( 1+\\frac{1}{10} \\right)=\\frac{11}{10}$$， 冰化成水体积减少$$\\left( \\frac{11}{10}-1 \\right)\\div \\frac{11}{10}=\\frac{1}{11}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "875", "queId": "96e4a402918d41008b0c69a5e26d15cf", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（二）第16题", "2017年新希望杯六年级竞赛训练题（二）第16题"], "difficulty": "3", "qtype": "single_choice", "problem": "质数$$a$$、$$b$$、$$c$$满足：$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=19350$$，则$$a+b+c=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$128$$ "}], [{"aoVal": "B", "content": "$$146$$ "}], [{"aoVal": "C", "content": "$$185$$ "}], [{"aoVal": "D", "content": "$$149$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$19350=$$奇数$$+$$奇数$$+$$偶数，所以$$3$$个质数中必定有一个是$$2$$； 由于质数中除了$$2$$、$$5$$以外，其他的质数都以$$1$$、$$3$$、$$7$$、$$9$$结尾，因此它们的平方末尾相应为$$1$$、$$9$$、$$9$$、$$1$$，相加后不可能得到$$6$$，而$$5$$的平方个位还是$$5$$，所以其中一个奇数必定为$$5$$．$$19350-4-25=19321$$，$$139$$的平方是$$19321$$，$$2+5+139=146$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "968", "queId": "e2d587eb418f4e21a554fc1ced9bb87f", "competition_source_list": ["2006年四年级竞赛创新杯", "2006年第4届创新杯四年级竞赛初赛B卷第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "0\\textasciitilde4五个数字组成的，各位数字不同的最大五位数与最小五位数相差( ). ", "answer_option_list": [[{"aoVal": "A", "content": "30870 "}], [{"aoVal": "B", "content": "32900 "}], [{"aoVal": "C", "content": "32976 "}], [{"aoVal": "D", "content": "10000 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->位值原理运用->位值原理的综合应用"], "answer_analysis": ["$$43210-10234=32976$$，选C "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1776", "queId": "6628fb6bbcb64907add42c90c42d7c60", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "天气炎热，维维去帮爸爸妈妈排队买饮料，维维发现他的前面有$$3$$人，后面有$$4$$人，聪明的你赶快开动大脑想想这里一共有多少个小朋友在排队买饮料呢？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["除了涛涛前面的人和涛涛本身，剩下的就是排在涛涛后面的人， 所以共有：$$26-10-1=15$$（人）． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "799", "queId": "60342feddb764954864d04f3928ec397", "competition_source_list": ["2006年第4届创新杯五年级竞赛复赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$1234567891011121314\\cdots 20052006$$是位数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6913$$ "}], [{"aoVal": "B", "content": "$$6914$$ "}], [{"aoVal": "C", "content": "$$6915$$ "}], [{"aoVal": "D", "content": "$$6917$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->数与数字->数、数位、数字的认识"], "answer_analysis": ["$$1\\sim 9$$，共有$$9$$个数字组成， $$10\\sim 99$$共有$$2\\times 90=180$$个数字组成， $$100\\sim 999$$，共有$$3\\times 900=2700$$个数字组成， $$1000\\sim 2006$$共有$$4\\times 1007=4028$$个数字组成， 所以$$1234567891011121314\\cdots 20052006$$是由：$$9+180+2700+4028=6917$$个数字组成． 则其是$$6917$$位数． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3261", "queId": "b531f7c275ea4e598c4624ad7328ab7d", "competition_source_list": ["2016年全国小学生数学学习能力测评五年级竞赛初赛第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$1$$角硬币分正面与反面．拿三个$$1$$角硬币一起投掷一次，得到两个正面和一个反面的可能性为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{8}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->典型问题->抛硬币"], "answer_analysis": ["全部情况有：$$2\\times 2\\times 2=8$$（种）， 最终得到两个正面一个反面可有以下三种情况： 正正反、正反正、反正正， 故所求概率为$$\\frac{3}{8}$$． 故答案为：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "862", "queId": "bb1c98913ccd403f9aca01cc6414fe1c", "competition_source_list": ["2018年美国数学大联盟杯五年级竞赛初赛第24题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{(2018 Math League, Priamry 5, Question \\#24)} My favorite number has $$6$$ different factors. If the product of all $$6$$ factors is $${{12}^{3}}$$，what$$$$ is the sum of the factors of my favorite number? 我最喜欢的数有$$6$$个不同的因数．如果所有$$6$$个因数的乘积为$${{12}^{3}}$$，我最喜欢的数的因数之和是多少？（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$） ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->数论模块->因数与倍数->因数个数定理"], "answer_analysis": ["我们知道：$${{12}^{3}}=1728$$； 将$$1728$$分解质因数：$$1728=2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 3\\times 3\\times 3$$； 由于$$6$$个因数各不相同，将质因数进行组合； 所以这六个因数可以是：$$1$$、$$2$$、$$3$$、$$4$$、$$6$$、$$12$$； 那么$$6$$个因数之和是：$$1+2+3+4+6+12=28$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "736", "queId": "374ef944bd874c3dad6bf6941d79c5c6", "competition_source_list": ["2012年美国数学大联盟杯六年级竞赛初赛第29题5分(每题5分)"], "difficulty": "1", "qtype": "single_choice", "problem": "$$12$$的所有因数之和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$28$$ "}], [{"aoVal": "D", "content": "$$56$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["整数$$a$$除以整数$$b(b\\ne 0)$$除得的商正好是整数而没有余数，我们就说$$a$$能被$$b$$整除，或$$b$$能整除$$a$$． $$a$$称为$$b$$的倍数，$$b$$称为$$a$$的因数． $$12$$的因数有：$$1$$、$$2$$、$$3$$、$$4$$、$$6$$、$$12$$， 它们的和是：$$1+2+3+4+6+12=28$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2176", "queId": "d5301253ff7d4d7f8db42628c10c75c1", "competition_source_list": ["2016年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$A,B$$两地相距$$300$$米。甲、乙两人同时分别从$$A,B$$两地出发，相向而行，在距$$A$$地$$140$$米处相遇； 如果乙每秒多行$$1$$米，则两人相遇处距$$B$$地$$180$$米。那么乙原来的速度是每秒（ ）米。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\frac{3}{5}$$ "}], [{"aoVal": "B", "content": "$$2\\frac{4}{5}$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$3\\frac{1}{5}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->比例解行程问题->行程中的比例"], "answer_analysis": ["本题是典型的利用正反比例解行程的问题。首先根据不变量判断是否成正反比例。两次相遇过程中两人的时间相同，路程比等于速度比。两次相遇过程中甲的速度没变，通分比较乙的，即可解决问题。 解：第一次相遇过程中，甲、乙两人的路程之比是$$140:\\left( 300-140 \\right)=7:8$$，时间相同，路程比就是速度比。 第二次相遇过程中，甲、乙两人的路程之比是$$\\left( 300-180 \\right):180=2:3$$，速度比也是$$2:3$$。 在两次相遇问题中，甲的速度是保持不变的。通分得，第一次速度比为$$7:8=14:16$$；第二次速度比为$$2:3=14:21$$。 速度从$$16$$份增加到$$21$$份，速度增加$$1$$米/秒，即$$1\\div \\left( 21-16 \\right)=\\frac{1}{5}$$。 乙原来的速度是$$16\\times \\frac{1}{5}=3.2$$（米/秒）$$=3\\frac{1}{5}$$（米/秒）。 故选：D "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2832", "queId": "b6243428bce244f89fbc77165d32c5c5", "competition_source_list": ["2018~2019学年浙江杭州西湖区杭州市行知小学五年级上学期期中期中竞赛第17题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "小机灵在用计算器计算``$$6.9\\times 7$$''时，发现计算器的键``$$6$$''坏了，小机灵想到了四种不同的输入方法．请你判断一下，下面选项中的方法是错误的 ", "answer_option_list": [[{"aoVal": "A", "content": "$$2.3\\times 3\\times 7$$ "}], [{"aoVal": "B", "content": "$$13.8\\times 7\\div 2$$ "}], [{"aoVal": "C", "content": "$$2\\times 3\\times 7+0.9\\times 7$$ "}], [{"aoVal": "D", "content": "$$7\\times 7-7$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->小数->小数乘除->小数乘法运算"], "answer_analysis": ["$$7\\times 7-7=7\\times \\left( 7-1 \\right)=7\\times 6=42$$，提取公因数后发现$$\\text{D}$$式变为$$7\\times 6$$，而非原式的$$6.9\\times 7$$，所以错误． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1372", "queId": "47b69bf977024b01b79b84a4b804f773", "competition_source_list": ["2006年第4届创新杯四年级竞赛初赛B卷第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "某次数学比赛，分两种方法给分，一种是答对了一题给$$5$$分，不答给$$2$$分，答错不给分；另一种是先给$$40$$分，答对一题给$$3$$分，不答不给分，答错扣$$1$$分．某考生两种判分方法均得$$81$$分，这次比赛共有( )题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$22$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["选项$$\\rm A$$：$$12$$题，按第一种给分方法，最多得$$12\\times 5=60$$分，排除$$\\rm A$$；选项$$\\rm B$$：$$15$$题，按第一种给分方法，最多得$$15\\times 5=75$$分，排除$$\\rm B$$； 选项$$\\rm D$$：$$17$$题，按第一种给分方法，得$$81$$分只有一种方法，$$81=5\\times 15+2\\times 3$$，因此比赛试题最少应该有$$15+3=18$$题，排除$$\\rm D$$．故选$$\\rm C$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "205", "queId": "3eda02c4b9194a45ba1796bd1d992acf", "competition_source_list": ["2021年新希望杯一年级竞赛初赛第14题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$5$$只动物排成一排，狗和猫都与鸡相邻，猫和鼠都与兔相邻，那么都与猫相邻． ", "answer_option_list": [[{"aoVal": "A", "content": "鼠和鸡 "}], [{"aoVal": "B", "content": "鸡和兔 "}], [{"aoVal": "C", "content": "兔和狗 "}], [{"aoVal": "D", "content": "兔和鼠 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意分析可知，狗和猫都与鸡相邻，即鸡的左右两边是猫和狗，又因为猫与老鼠都与兔相邻，所以猫的旁边是兔子和老鼠，兔子和老鼠在猫的同一侧，由此可知，与猫相邻的是鸡和兔． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2179", "queId": "2744bdaf70b74e6bb028b77deee741b7", "competition_source_list": ["2016年新希望杯小学高年级六年级竞赛训练题（一）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$A$$、$$B$$两地相距$$600$$千米．甲、乙两车同时从$$A$$、$$B$$两地相向而行，$$3$$小时后相遇．甲速度是乙的$$1.5$$倍，则甲的速度是千米/时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$80$$ "}], [{"aoVal": "B", "content": "$$90$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设甲速度是$$3x$$千米/时，以速度是$$2x$$千米/时，$$600=\\left( 2x+3x \\right)\\times 3$$，$$3x=120$$，即甲的速度是$$120$$千米/时 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "375", "queId": "a4447bef78ef4ee6bd94e2f3a9ec632b", "competition_source_list": ["2013年河南郑州中原网杯六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁$$4$$人坐在$$1$$、$$2$$、$$3$$、$$4$$号椅子上，有人说：乙坐在丙旁边，甲坐在乙、丙中间，乙没有坐在$$3$$号椅子上．已知此人说的都是错的， 则丁坐在几号椅子上？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->组合模块->逻辑推理->条件型逻辑推理->神推理", "拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->连线法"], "answer_analysis": ["首先，``乙没有坐在$$3$$号椅子上''是错的，就可以判断出乙在$$3$$号椅子上；再看，``乙坐在丙旁边''是错的，可以判断出丙不在$$2$$，$$4$$号椅子上，所以在$$1$$号椅子；最后，``甲坐在乙、丙中间''是错的，可以判断出甲在$$4$$号，故得出丁在$$2$$号椅子上． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1647", "queId": "4a3abc651f774b6daf03e28723631671", "competition_source_list": ["2018年四川成都锦江区四川师范大学附属第一实验中学小升初（四）第5题3分", "2014年全国迎春杯三年级竞赛初赛第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "同学们一起去划船，但公园船不够多，如果每条船坐$$4$$人，会多出$$10$$人；如果每条船坐$$5$$人，还会多出$$1$$人，共有人去划船． ", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$46$$ "}], [{"aoVal": "C", "content": "$$51$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->盈亏问题->盈亏基本类型->盈亏基本类型盈亏问题"], "answer_analysis": ["盈盈类问题：共有$$(10-1)\\div (5-4)=9$$（条）船，共有$$4\\times 9+10=46$$人． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1419", "queId": "51325d32706c4104bbb9a462f109045f", "competition_source_list": ["2010年华杯赛四年级竞赛初赛", "2010年华杯赛三年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "两条纸带，较长的一条为$$23$$cm，较短的一条为$$15$$cm. 把两条纸带剪下同样长的一段后，剩下的两条纸带中，要求较长的纸带的长度不少于较短的纸带长度的两倍，那么剪下的长度至少是（ ）厘米. ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->加减法应用->同增同减应用题"], "answer_analysis": ["考虑到``同增同减差不变''的原则，在剪之后，两条纸带长度的差仍为$$23-15=8$$(厘米). 较长的纸带的长度不少于较短的纸带长度的两倍，故他们长度的差不少于较短的纸带长度的一倍， 较短的纸带不能超过$$8$$厘米，故至少剪下了$$7$$厘米。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2953", "queId": "ffa1b030d1fe4ad2aa6494f9dbcdc188", "competition_source_list": ["2017年河南郑州联合杯竞赛决赛第10题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一列数$$1$$，$$2$$，$$2$$，$$3$$，$$3$$，$$3$$，$$4$$，$$4$$，$$4$$，$$4$$，$$...$$ ．中的第$$35$$个数为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "无答案 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$1+2+\\cdots +7=\\left( 1+7 \\right)\\times 7\\div 2=28$$，$$35-28=7$$，则第$$35$$个数为$$8$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1540", "queId": "3c41e0d9e6774abf836906f67f97a559", "competition_source_list": ["2015年第11届全国新希望杯小学高年级六年级竞赛复赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "星星、希希、望望三人过年期间一共收了$$1800$$元压岁钱，如果星星把自己的钱数的$$\\frac{1}{3}$$给希希，然后希希把自己现有钱数的$$\\frac{1}{4}$$给望望，望望再把自己现有钱数的$$\\frac{1}{5}$$给星星，此时二个人手中的钱就一样多了，那么希希原来的钱数比望望原来的钱数多（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$35$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$125$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["相互给，总和不变，给完后三人的钱数都是$$1800\\div 3=600$$，望望得到希希的钱后是$$600\\div \\frac{4}{5}=750$$，给希希$$150$$后，星星是$$600$$，那么说明星星给出$$\\frac{1}{3}$$后是$$450$$，那么星星原来是$$450\\div \\frac{2}{3}=675$$（元），那么给了希希$$225$$元，希希得到后有$$600\\div \\frac{3}{4}=800$$（元），那么希希原有$$800-225=575$$（元）．那么望望原有$$1800-675-575=550$$（元），所以希希原来的钱比望望原来的钱多$$25$$元． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "199", "queId": "28f3316cabe149fcb68a14853643aa18", "competition_source_list": ["2014年第3届广东广州羊排赛六年级竞赛第9题1分"], "difficulty": "2", "qtype": "single_choice", "problem": "琦琦身陷神秘房间中．房间大门紧锁，上面刻着一段话：``钥匙上的话只有一句是真的．''往门旁一看，挂着三把钥匙，上面各贴着一句话．金钥匙：``这把钥匙不可以打开大门．''银钥匙：``金钥匙可以打开大门．''铜钥匙：``这把钥匙不可以打开大门．''那么，（ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "琦琦应该拿金钥匙打开大门 "}], [{"aoVal": "B", "content": "琦琦应该拿银钥匙打开大门 "}], [{"aoVal": "C", "content": "琦琦应该拿铜钥匙打开大门 "}], [{"aoVal": "D", "content": "三把钥匙都不能打开大门，琦琦应该另想办法 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾"], "answer_analysis": ["本题的思路其实很简单，就是``找矛盾''； 1、三句话只有一句是真的；------题目条件； 2、只要存在一对矛盾则必然一真一假；------矛盾的定义； 3、金钥匙和银钥匙上写的话相互矛盾，根据以上两条，真的那句话一定是其中一句（具体到底是金还是银不重要）； 4、根据$$3$$，则铜钥匙那句一定是假的，但它上面说的本身是否定，否定为假则肯定为真，所以铜钥匙就可以打开大门． 答案选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "650", "queId": "16682faeed844e519bf4848665814fd7", "competition_source_list": ["2006年华杯赛六年级竞赛初赛", "2006年华杯赛五年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "2008006共有个质因数. ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->质数与合数"], "answer_analysis": ["因为$$2008006=2006\\times 1000+2006=2006\\times 1001=\\left( 2\\times 17\\times 59 \\right)\\times \\left( 7\\times 11\\times 13 \\right)$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2560", "queId": "2c66de2f335949e7ac51cdf47f715f03", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级六年级竞赛邀请赛训练题（一）"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$2000$$，$$2001$$，$$2002$$，$$2003$$，$$\\ldots \\ldots $$，$$2017$$这$$18$$个连续自然数中能表示为两个自然数平方之差的数有（~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->公式类运算->完全平方公式"], "answer_analysis": ["根据完全平方公式有$${{\\left( a+1 \\right)}^{2}}={{a}^{2}}+2a+1$$，那么可以得到$${{\\left( a+1 \\right)}^{2}}{{a}^{2}}=2a+1$$，$$2a+1$$是一个奇数，那么所在奇数共有$$9$$个，都可以表示为两个自然数平方之形；若一个偶数能表示为两个平方数之差，都为$${{a}^{2}}{{b}^{2}}=\\left( a+b \\right)\\left( ab \\right)$$，此时$$\\left( a+b \\right)$$与$$\\left( a-b \\right)$$同奇偶性，必同为偶数，那么此时$${{a}^{2}}{{b}^{2}}=\\left( a+b \\right)\\left( ab \\right)$$必为$$4$$的偶数，也就是说$$4$$的偶数才能表示为两个平方数之和，那么$$2000$$，$$2004$$，$$2006$$，$$2012$$，$$2016$$也是可以表示成两个平方数之差的，所以一共有$$9+5=14$$个． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2745", "queId": "ff8080814518d5240145201b60020a7d", "competition_source_list": ["2014年全国迎春杯三年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "找出规律，将你认为合适的数填入(~ ~ ~ ），$$2$$、$$4$$、$$3$$、$$9$$、$$4$$、$$16$$、$$5$$、 ~( ~ ~ ~ ~ ~)、 ~( ~ ~ ~ ~ )、$$36$$、 $$7$$、$$\\cdots$$那么正确的数是（~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$、$$6$$ "}], [{"aoVal": "B", "content": "$$22$$、$$6$$ "}], [{"aoVal": "C", "content": "$$25$$、$$6$$ "}], [{"aoVal": "D", "content": "$$25$$、$$26$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$2$$、$${ 2 }^{ 2 }$$、$$3$$、$${ 3 }^{ 2 }$$、$$4$$、$${ 4 }^{ 2 }$$、$$5$$、（$${5 }^{ 2 }$$）、（$$6$$ ）、$${ 6 }^{ 2 }$$、$$7$$$$\\cdots$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2358", "queId": "06826e60ba0040cc8f092b3c41d18b4a", "competition_source_list": ["2011年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "$$1\\times 2\\times 3\\times 4\\times 5\\times \\cdots \\times 21\\div 343$$，则商的千位上的数字是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之位值原理"], "answer_analysis": ["易知最后的几位为$$0$$，$$0$$的个数由$$5$$和$$2$$的个数决定，在这里显然$$2$$的个数多于$$5$$；我们只需要考虑$$5$$的个数即可。$$1\\times 2\\times 3\\times 4\\times 5\\times \\cdots \\times 21$$中含有$$4$$个$$5$$，而$$343={{7}^{3}}$$，所以商千位上的数字为$$0$$。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2641", "queId": "6c8666acdc714e3280eef1f4f87e0d06", "competition_source_list": ["2019年亚洲国际数学奥林匹克公开赛（AIMO）三年级竞赛决赛第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算． $$13+16+19+22+ 25+ 28+31+34+37+40=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$286$$ "}], [{"aoVal": "B", "content": "$$265$$ "}], [{"aoVal": "C", "content": "$$258$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列求和"], "answer_analysis": ["原式$$=\\left( 13+40 \\right)+\\left( 16+37 \\right)+\\left( 19+34 \\right)+\\left( 22+31 \\right)+\\left( 25+28 \\right)$$ $$=53\\times 5$$ $$=265$$． $$(13+40)\\times 10\\div 2=265$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3392", "queId": "8645d3cb153e4df8a87b929347d0e886", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（四）第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "一袋准备要爆花的玉米，其中$$\\frac{2}{3}$$是白粒的，$$\\frac{1}{3}$$是黄粒的．白粒的$$\\frac{1}{2}$$会爆开，黄粒的$$\\frac{2}{3}$$会爆开．今从该袋中任选一粒放入锅中发生爆花，则是白粒玉米的概率是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{4}{7}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{5}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->加乘原理求概率"], "answer_analysis": ["会爆开的白粒$$\\frac{2}{3}\\times \\frac{1}{2}=\\frac{1}{3}$$，会爆开的黄粒$$\\frac{1}{3}\\times \\frac{2}{3}=\\frac{2}{9}$$，会爆开的白粒是会爆开的玉米总数的$$\\frac{1}{3}\\div \\left( \\frac{1}{3}+\\frac{2}{9} \\right)=\\frac{3}{5}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1019", "queId": "035d0fe6ddd845e19b6d576a7d0da19b", "competition_source_list": ["2015年华杯赛五年级竞赛初赛", "2015年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "六位同学考试的平均成绩是$$92.5$$分，他们的成绩是互不相同的整数，最高的$$99$$分，最低的$$76$$分。那么，按分数从高到低居第三位的同学的分数至少是（ ）分。 ", "answer_option_list": [[{"aoVal": "A", "content": "94 "}], [{"aoVal": "B", "content": "95 "}], [{"aoVal": "C", "content": "96 "}], [{"aoVal": "D", "content": "97 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题"], "answer_analysis": ["六位同学的平均成绩为$$92.5$$分，因此这六人的总分为$$92.5\\times 6=555$$（分）。又最高分为$$99$$，最低分为$$76$$，因此剩下四人的分数和为$$380$$分。要使第三名的同学分数尽可能小，那么得使另外三人的分数尽可能大。而第二名最高为$$98$$分，而第四名第五名分数小于第三名，所以第三名至少为：$$(380-98)\\div3+1=95$$（分），选$$B$$。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "957", "queId": "dd85a44e9f044b7d83ea1a5c97414cef", "competition_source_list": ["2015年第27届广东广州五羊杯小学高年级竞赛第10题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$15$$位同学，每位同学都有编号，他们是从$$1$$号到$$15$$号．$$1$$号同学写了一个自然数，$$2$$号同学说：``这个数能被$$2$$整除．''$$3$$号同学说：``这个数能被$$3$$整除．''$$4$$号同学说：``这个数能被$$4$$整除．''$$\\cdots \\cdots $$，依次下去，每位同学都说，这个数能被他（或她）的编号整除．$$1$$号同学一一验证，发现只有编号相邻的两个同学说得不对．那么，说得不对的同学，他们的编号是两个连续的自然数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$、$$6$$ "}], [{"aoVal": "B", "content": "$$6$$、$$7$$ "}], [{"aoVal": "C", "content": "$$7$$、$$8$$ "}], [{"aoVal": "D", "content": "$$8$$、$$9$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["首先可以判断编号为$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$的同学说的一定都对， 否则的话，其中说的不对的同学的编号乘以$$2$$后所得的编号也将说得不对， 这样就与``只有编号相邻的两个同学说得不对''矛盾． 因而这个数能被$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$整除， 从而也能被$$10$$，$$12$$，$$14$$整除， 即编号为$$10$$，$$12$$，$$14$$的同学说得也对． 于是可以判定编号为$$11$$，$$13$$，$$15$$的同学也对． 现在只剩下$$8$$号和$$9$$号了，所以错的是$$8$$，$$9$$两号， 答案为$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1673", "queId": "d190eb727e824be6ac177c88cd04f020", "competition_source_list": ["2017年全国华杯赛竞赛初赛模拟3第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "养鸡场购进一批种蛋，若能全部孵化成小鸡则有$$2.5$$倍的收益，但是实际上��得的收益仅仅为$$1.25$$倍，这批种蛋的孵化率是~\\uline{~~~~~~~~~~}~$$ \\%$$（百分子保留一位小数）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$64.3$$ "}], [{"aoVal": "B", "content": "$$63.3$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$70$$ "}]], "knowledge_point_routes": ["拓展思维->思想->赋值思想"], "answer_analysis": ["解法一：这一批种蛋的孵化收益损失$$2\\frac{1}{2}-1\\frac{1}{4}=1\\frac{1}{4}$$（倍），若是全部孵化失败，则要损失$$2\\frac{1}{2}+1=3\\frac{1}{2}$$（倍），所以这批种蛋的孵化率是： $$（1-1\\frac{1}{4}\\div3\\frac{1}{2}）\\times100 \\%=\\frac{9}{14}\\times100 \\%=64.3 \\%$$． 解法二：假设每个种蛋的成本价为$$1$$元，$$100$$个种蛋总成本价$$100$$元．又设$$100$$个种种蛋$$S$$个孵化出来，则孵化出后可卖$$S（2\\frac{1}{2}+1）$$元．于是$$S（2\\frac{1}{2}+1）\\div100=1\\frac{1}{4}+1$$， 因此，这批种蛋的孵化率是$$\\frac{9}{14}\\times100 \\%=64.3 \\%$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "524", "queId": "f442abd3b0f2414f96b4d10d52bc0cc3", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第3题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "芳芳煮熟$$2$$个生鸡蛋用$$6$$分钟，煮熟$$10$$个生鸡蛋用分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->统筹规划->简单时间统筹问题->时间总和问题"], "answer_analysis": ["鸡蛋可以一起煮，所以所花的时间是一样的． 故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2657", "queId": "b0b5cb382193412898a153325fbfdb2e", "competition_source_list": ["2014年第10届全国新希望杯小学高年级六年级竞赛复赛第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "对自然数$$n$$进行如下操作：如果$$n$$是偶数，就把它除以$$2$$，如果$$n$$是奇数，就把它加上$$105$$．现在对$$123$$进行有限次操作，得到的结果不可能是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$57$$ "}], [{"aoVal": "B", "content": "$$93$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$114$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由于$$123$$是$$3$$的倍数，操作过程中得到数一定为$$3$$的倍数，故不可能得到$$100$$选$$C$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "276", "queId": "6d2e1e5eb3bb4c8da02fdc700eb660de", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在一个口袋里有$$8$$个黑球、$$3$$个白球、$$9$$个红球，至少取出个球，才能保证其中有黑球． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["根据题意分析可知，考虑最坏的情况，先摸到的都是其他颜色的球，即白球和红球，把$$3+9=12$$（个）球全摸完然后再接下来摸到的球一定是黑球也就是：$$12+1=13$$（个），故选答案$$\\text C$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "809", "queId": "6957bf1cbbda4ad09081c2a77a630087", "competition_source_list": ["2004年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "有两个两位数，它们的最大公因数是$$8$$，最小公倍数是$$96$$，这两个数的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "56 "}], [{"aoVal": "B", "content": "78 "}], [{"aoVal": "C", "content": "84 "}], [{"aoVal": "D", "content": "96 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公因数与最大公因数->两数的最大公因数"], "answer_analysis": ["因为两个两位数的最大公因数为8，可设这两个数分别为$$8a\\text{，}8b$$($$a\\text{，}b$$互质)，那么$$8\\times a\\times b=96$$，即$$a\\times b=12$$，则$$a\\text{，}b$$为$$\\left( 3\\text{，}4 \\right)\\text{，}\\left( 1\\text{，}12 \\right)$$，又因为这两个数皆为两位数，所以这两个数只能为24和32，$$24+32=56$$，选 A. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1765", "queId": "6608dd7f709945a0a879b82d91695c50", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（二）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "某货运公司运送一批货物，原计划安排$$18$$辆小卡车和$$12$$辆大卡车刚好运$$4$$次，已知$$2$$辆大卡车与$$5$$辆小卡车装的重量相同，现在只能派出$$8$$辆小卡车，需运次才能把货物运完． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$12$$辆大卡车装的重量等于$$12\\div 2\\times 5=30$$辆小卡车，则原计划运$$1$$次的货物现在要运$$\\left( 18+30 \\right)\\div 8=6$$（次），所以一共需运$$6\\times 4=24$$次． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "267", "queId": "83d78f31dfcc4276b36d670b7c3e11a8", "competition_source_list": ["2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一次考试有三道题，四个好朋友考完后互相交流了成绩，发现四人各对了$$3$$、$$2$$、$$1$$、$$0$$ 道题。这时一个路人问：``你们考得怎么样啊？'' 甲：``我对了两道题，而且比乙对的多，丙考得不如丁。'' 乙：``我全对了，丙全错了，甲考得不如丁。'' 丙：``我对了一道，丁对了两道，乙考得不如甲。'' 丁：``我全对了，丙考得不如我，甲考得不如乙。'' 已知大家都是对了几道题就说几句真话，那么对了$$2$$题的人是~\\uline{~~~~~~~~}~。 ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "丁 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理"], "answer_analysis": ["全对的人不会说自己对的题少于$$3$$，故只有乙、丁可能全对。若乙全对，则排名是乙、丁、甲、丙，与丙所说的``丁对了$$2$$道''是假话相矛盾；若丁全对，则丙的后两句是假话，不可能是第二名，又由丁的``甲考得不如乙''能知道第二名是乙，故丙全错，甲只有``丙考得不如丁''是真话，排名是丁、乙、甲、丙且$$4$$人的话没有矛盾。综上，答案是B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "674", "queId": "67558def52984186b07fc234da86d942", "competition_source_list": ["2014年IMAS小学高年级竞赛第一轮检测试题第7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "若甲数除以$$5$$余$$2$$、乙数除以$$5$$余$$4$$，请问甲、乙两数之和除以$$5$$的余数是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}], [{"aoVal": "E", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["将甲、乙两数余数之和$$2+4=6$$继续除以$$5$$，可得其余数为$$1$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2774", "queId": "45b15afcf45645bea34f5680f758605b", "competition_source_list": ["2020年新希望杯六年级竞赛初赛（个人战）第4题", "2020年希望杯六年级竞赛（个人赛）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "【2020希望杯六年级竞赛初赛】 如果$$x:y=4:7$$，$$z:x=3:5$$，那么$$\\left( x+y \\right):(z+x)=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11:8$$ "}], [{"aoVal": "B", "content": "$$33:55$$ "}], [{"aoVal": "C", "content": "$$32:55$$ "}], [{"aoVal": "D", "content": "$$41:32$$ "}], [{"aoVal": "E", "content": "$$55:32$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["已知$$x:y=4:7$$，$$z:x=3:5$$，那么$$y:x=7:4$$，$$x:z=5:3$$， $$y:x=\\left( 7\\times 5 \\right):\\left( 4\\times 5 \\right)=35:20$$，$$x:z=\\left( 5\\times 4 \\right):\\left( 3\\times 4 \\right)=20:12$$， $$y:x:z=35:20:12$$． 假设$$y=35a$$，$$x=20a$$，$$z=12a$$，其中$$a\\ne 0$$． $$x+y=20a+35a=55a$$，$$z+x=12a+20a=32a$$， 那么$$\\left( x+y \\right):\\left( z+x \\right)=55a:32a=55:32$$． 故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3171", "queId": "4f82532f1b0945efac7ebbdcaa0831ea", "competition_source_list": ["2015年湖北武汉世奥赛小学高年级六年级竞赛模拟��练题（四）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "四位同学到商店买毛笔或铅笔，每人只买了$$1$$支笔，而且至少有$$1$$人买了铅笔，则可能的买法有（~ ~ ~ ~）种． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->乘法原理->物品搭配（有特殊要求）"], "answer_analysis": ["每位同学有两种买法，排除全部是毛笔，则共有$$2\\times 2\\times 2\\times 2-1=15$$（种）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1362", "queId": "28bee930d52440d999bd63254b1386d0", "competition_source_list": ["其它", "2010年北京学而思综合能力诊断三年级竞赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "某次数学竞赛，试题共有$$10$$道，每做对一题得$$6$$分，每做错一题倒扣$$2$$分，小红最终得$$44$$分，做对的题比做错的题多~\\uline{~~~~~~~~~~}~道． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$8$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["$$\\left( 60-44 \\right)\\div 8=2$$，做错$$2$$道题，做对$$8$$道题，对的比错的多$$6$$道． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2998", "queId": "c09ede4a0b804b71b43d5ef2cb128fcb", "competition_source_list": ["2017年第15届全国希望杯小学高年级六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$9$$个正方形放在一行，第$$1$$个正方形的面积为$$1$$，从第$$2$$个正方形开始，每个正方形的面积都是前一个正方形面积的一半，试比较第$$2$$个到第$$9$$个正方形的面积之和与第$$1$$个正方形面积的大小． ", "answer_option_list": [[{"aoVal": "A", "content": "第$$2$$个到第$$9$$个正方形的面积之和大 "}], [{"aoVal": "B", "content": "第$$1$$个正方形面积大 "}], [{"aoVal": "C", "content": "一样大 "}], [{"aoVal": "D", "content": "不能判断 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["第$$2$$个到第$$9$$个正方形的面积之和为$$\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+\\frac{1}{32}+\\frac{1}{64}+\\frac{1}{128}+\\frac{1}{256}$$ $$=\\frac{1}{256}\\left( 128+64+32+16+8+4+2+1 \\right)$$ $$=\\frac{255}{256}\\textless{}1$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1216", "queId": "1e69375093124e51858de5b1becc4d50", "competition_source_list": ["2014年迎春杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a$$、$$b$$、$$c$$、$$d$$四个数的平均数是$$12.345$$，$$a\\textgreater b\\textgreater c\\textgreater d$$，那么$$b$$（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "大于$$12.345$$ "}], [{"aoVal": "B", "content": "小于$$12.345$$ "}], [{"aoVal": "C", "content": "等于$$12.345$$ "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->公式类"], "answer_analysis": ["解：因为$$a$$、$$b$$、$$c$$、$$d$$四个数的平均数是$$12.345$$，$$a\\textgreater b\\textgreater c\\textgreater d$$， 所以$$a$$一定大于$$12.345$$，$$d$$一定小于$$12.345$$， 但是$$b$$的取值无法确定，$$b$$可能大于$$12.345$$，也有可能小于$$12.345$$或等于$$12.345$$。 故选：D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3000", "queId": "c5266d4023514625a71935532ed5fe6d", "competition_source_list": ["竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "计算:$$0.123\\times 958958+877\\times 613.613-34.5\\times 1231.23=$$( ). ", "answer_option_list": [[{"aoVal": "A", "content": "$$613613$$ "}], [{"aoVal": "B", "content": "$$6136.13$$ "}], [{"aoVal": "C", "content": "61361.3 "}], [{"aoVal": "D", "content": "以上答案都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数提取公因数->小数乘法巧算之提取公因数（普通型）"], "answer_analysis": ["原式 $$=123\\times 958.958-123\\times 1.001\\times 345+877\\times 613.613 $$ $$=123\\times \\left( 958.958-345.345 \\right)+877\\times 613.613 $$ $$=123\\times 613.613+877\\times 613.613 $$ $$=613\\times 1.001\\times \\left( 123+877 \\right) $$ $$=613\\times 1001 $$ $$=613613 $$ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "729", "queId": "32da61ffac2c48928ce9ef6b79f374d1", "competition_source_list": ["2017年新希望杯小学高年级六年级竞赛训练题（二）第5题", "2018年湖北武汉新希望杯六年级竞赛训练题（二）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$1\\tilde{ }200$$的自然数中，如果一个数能被$$2$$或$$7$$整除，就称这个数是``双七数''，那么``双七数''共有（~~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$112$$ "}], [{"aoVal": "B", "content": "$$114$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$128$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$\\left[ \\frac{200}{2} \\right]+\\left[ \\frac{200}{7} \\right]-\\left[ \\frac{200}{2\\times 7} \\right]=100+28-14=114$$（个）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "114", "queId": "38f0899231e84303826ab6b14e7cc8a8", "competition_source_list": ["2017年第15届湖北武汉创新杯小学高年级六年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "若两个三位数的和为$$\\overline{ \\square \\square\\square } +\\overline{\\square \\square \\square } =1949$$，那么$$6$$个$$\\square $$中的数字之和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$23$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$41$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->横式数字谜->与数论的结合"], "answer_analysis": ["和有千位，则百位一定向前进一位，和的百位为$$9$$且向前进位，则十位一定向百位进一位，和的个位为$$9$$，则个位一定不向前进位，所以总共进两位，数字之和减少$$9\\times 2=18$$，则$$\\overline{\\square } $$中的数字之和是$$1+9+4+9+18=41$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "534", "queId": "f4f774fad04d43e69660158a99200518", "competition_source_list": ["2019年第24届YMO六年级竞赛决赛第7题3分"], "difficulty": "3", "qtype": "single_choice", "problem": "有一个$$12$$级的楼梯．某人每次能登上$$1$$级或$$2$$级或$$3$$级，现在他要从地面登上第$$12$$级，有种不同的方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$149$$ "}], [{"aoVal": "B", "content": "$$244$$ "}], [{"aoVal": "C", "content": "$$264$$ "}], [{"aoVal": "D", "content": "$$274$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["设第$$n$$级有$${{a}_{n}}$$种登的方式， 则第$$n-3$$，$$n-2$$，$$n-1$$级再分别登$$3$$，$$2$$，$$1$$级可到第$$n$$级， 则$${{a}_{n}}={{a}_{n-3}}+{{a}_{n-2}}+{{a}_{n-1}}$$， 而$${{a}_{1}}=1$$，$${{a}_{2}}=1+1=2$$，$${{a}_{3}}=1+1+2=4$$， 故$${{a}_{4}}=1+2+4=7$$， $${{a}_{5}}=2+4+7=13$$， $${{a}_{6}}=4+7+13=24$$， $${{a}_{7}}=7+13+24=44$$， $${{a}_{8}}=13+24+44=81$$， $${{a}_{9}}=24+44+81=149$$， $${{a}_{10}}=44+81+149=274$$． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "916", "queId": "d2cf8027528a4881a7e0720da101aa1d", "competition_source_list": ["2016年第21届全国华杯赛小学高年级竞赛初赛B卷第2题"], "difficulty": "3", "qtype": "single_choice", "problem": "有一种数，是以法国数学家梅森的名字命名的，它们就是形如$$2^{n}-1$$($$n$$为质数）的梅森数，当梅森数是质数时就叫梅森质数，是合数时就叫梅森合数，例如：$$2^{2}-1=3$$就是第一个梅森质数．第一个梅森合数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$127$$ "}], [{"aoVal": "D", "content": "$$2047$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["选项$$\\rm A$$：$$2^{n}-1=4$$，$$n$$无整数解； 选项$$\\rm B$$：$$2^{n}-1=15$$，$$n$$为$$4$$，但$$n$$不是质数，故舍去； 选项$$\\rm C$$：$$2^{n}-1=127$$，$$n$$为$$7$$，$$127$$不是合数，故舍去； 选项$$\\rm D$$：$$2^{n}-1=2047$$，$$n$$为$$11$$，$$n$$为质数，且$$2047=23\\times 89$$，是合数，满足条件． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1379", "queId": "5555b5f7b08e49ea90cb331c12a11968", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "祖玛游戏中，龙嘴里不断吐出很多颜色的龙珠，先$$4$$颗红珠，接着$$3$$颗黄珠，再$$2$$颗绿珠，最后$$1$$颗白珠，按此方式不断重复，从龙嘴里吐出的第$$2000$$颗龙珠是． （2014年二年级竞赛初赛第$$8$$题） ", "answer_option_list": [[{"aoVal": "A", "content": "红珠 "}], [{"aoVal": "B", "content": "黄珠 "}], [{"aoVal": "C", "content": "绿珠 "}], [{"aoVal": "D", "content": "白珠 "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["$$2000\\div (4+3+2+1)=200$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "134", "queId": "5496ba24c6e948ada71b1c97569bdb45", "competition_source_list": ["2017年第22届全国华杯赛小学高年级竞赛初赛第6题10分"], "difficulty": "3", "qtype": "single_choice", "problem": "从$$0$$至$$9$$中选择四个不同的数字分别填入下方的四个括号中，共有~\\uline{~~~~~~~~~~}~种填法使得下面这句话是正确的． 这个话里有个数大于$$1$$，有个数大于$$2$$，有个数大于$$3$$，有个数大于$$4$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["设四个空依次为$$a，b，c，d$$，即，``这句话里有$$a$$个数大于$$1$$，有$$b$$个数大于$$2$$，有$$c$$个数大于$$3$$，有$$d$$个数大于$$4$$．''根据包含关系，必有$$a\\textgreater b\\textgreater c\\textgreater d\\geqslant 0$$，所以$$a\\textgreater b\\geqslant 2$$，所以至少有$$5$$个数大于$$1$$($$a，b$$以及给出的$$2，3，4$$)，所以$$a\\geqslant 5$$，说明至少有一个数大于$$4$$(这个数是$$a$$)，所以$$d\\geqslant 1$$．从$$d$$开始考虑: 若$$d=1$$，说明只有$$1$$个数大于$$4$$(只能为$$a$$)，且$$a，b，c$$均大于$$1$$，于是有$$a\\textgreater4\\geqslant b\\textgreater c\\geqslant 2$$，则有$$6$$个数大于$$1$$($$a，b，c$$以及给出的$$2，3，4$$)，所以$$a=6$$．由$$a\\textgreater4\\geqslant b\\textgreater c\\geqslant 2$$可知$$b\\geqslant 3$$说明至少有$$4$$个数大于$$2$$($$a，b$$以及给出的$$3，4$$)，所以$$b\\geqslant 4$$，又因为$$b\\geqslant 4$$，所以有$$b=4$$．此时有$$3$$个数大于$$3$$($$a，b$$以及给出的$$4$$），但此时发现$$8$$个数中有$$5$$个大于$$2$$，和$$b=4$$矛盾； 若$$d=2$$，说明有$$2$$个数大于$$4$$(只能为$$a，b$$)，且$$a，b，c$$均大于$$2$$，于是有$$a\\textgreater b\\textgreater4\\geqslant c\\geqslant 3$$，则有$$7$$个数大于$$1$$($$a，b，c$$以及给出的$$2，3，4$$)，$$5$$个数大于$$2$$($$a，b，c$$以及给出的$$3，4$$)，所以$$a=6$$，$$b=5$$．此时无论$$c=3$$或$$c=4$$均满足条件，这时的两组解为$$a=7$$，$$b=5$$，$$c=3$$，$$d=2$$和$$a=7$$，$$b=5$$，$$c=4$$，$$d=2$$． 若$$d=3$$，说明有$$3$$个数大于$$4$$(为$$a，b，c$$)，那么大于$$3$$的数有$$4$$个($$a，b，c$$以及给出的$$4$$)，所以$$c=4$$，但$$c\\textgreater4$$，矛盾，所以$$d=3$$不存在． 若$$d=4$$，说明有$$4$$个数大于$$4$$(为$$a，b，c，d$$)，但$$d=4$$，与$$d\\textgreater4$$矛盾． 显然$$d=5$$不存在，所以共有$$2$$组解． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "16", "queId": "01d9d2aeb27a4c0c8734f9e5ee0672e8", "competition_source_list": ["2020年希望杯二年级竞赛模拟第19题"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$3$$、$$3$$、$$5$$、$$9$$分别放入方格中，不能重复使用，那么和最小是． $$\\square \\square +\\square \\square $$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$74$$ "}], [{"aoVal": "C", "content": "$$92$$ "}], [{"aoVal": "D", "content": "$$146$$ "}], [{"aoVal": "E", "content": "$$188$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["要使得和最小，那么两个因数应该尽可能的小； 如果要求两个因数尽可能的小，那么两个因数十位应该填入较小数， 即这两个因数应该是：$$35+39=74$$， 所以和最小是$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3433", "queId": "e0e43bba29eb4292bc676c53b6ed7e43", "competition_source_list": ["2015年第27届广东广州五羊杯小学高年级竞赛第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "有一类两位数，只有$$4$$个因数，并且个位和十位上的数字是相邻的自然数，那么这样的两位数共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["解析分类讨论，有序列举． 有$$4$$个因数��整数只有两种情况：立方数和两个质因数的乘积． 不难验证，立方数中没有任何一个满足条件， 所以只能是第二种情况． 1．自然数由大到小：$$10=2\\times5$$；$$21=3\\times7$$；$$65=13\\times5$$；$$87=3\\times29$$； 2．自然数由小到大：$$34=2\\times17$$，所以共有$$5$$个． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1813", "queId": "bb4248def3c446898d2305d1656057bd", "competition_source_list": ["2021年迎春杯小学高年级六年级竞赛决赛C卷第7题10分"], "difficulty": "3", "qtype": "single_choice", "problem": "甲、乙、丙、丁各戴了一顶帽子，帽子上写了一个$$1\\sim9$$中的自然数，且互不相同．每人能看到别人帽子上的数，但看不到自己的．他们有如下对话： 甲说：``我看到的三个数两两互质．'' 乙、丙同时说：``那我知道我帽子上的数了．'' 如果他们都聪明且诚实，那么丁帽子上的数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["本题中，通过甲说出的``另外三个数两两互质''，乙丙就知道了自己的数，说明乙、丙、丁三个人的数字，在$$1\\sim 9$$当中是唯一固定的一组三个数明确的互质数即``$$5$$、$$6$$、$$7$$''三个数，而在这三个数中，只有``$$6$$''在$$1$$$$\\sim 9$$当中有且仅有$$5$$、$$7$$两个互质数，所以乙、丙两个人的数为$$5$$、$$7$$或$$7$$、$$5$$，丁为$$6$$． $$1$$$$\\sim9$$各个数字对应的互质数． $$1$$：$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$， $$2$$：$$3$$、$$5$$、$$7$$、$$9$$， $$3$$：$$2$$、$$4$$、$$5$$、$$7$$、$$8$$， $$4$$：$$3$$、$$5$$、$$7$$、$$9$$， $$5$$：$$2$$、$$3$$、$$4$$、$$6$$、$$7$$、$$8$$、$$9$$， $$6$$：$$5$$、$$7$$， $$8$$：$$3$$、$$5$$、$$7$$、$$9$$， $$9$$：$$2$$、$$4$$、$$5$$、$$7$$、$$8$$． 由此可见，只有$$6$$在$$1\\sim9$$的范围内有且仅有两个互质数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2621", "queId": "50fa2ac9b07041c1ba7dff6d143ca7cf", "competition_source_list": ["2017年四川六年级竞赛排位赛第17题2分", "2020~2021学年福建福州鼓楼区福州教育学院附属第二小学六年级上学期期中第5题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a$$、$$b$$、$$c$$是不同的自然数，而且$$a\\times \\frac{c}{b} ~\\textless{} ~a$$，那么正确的一个结论是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$a ~\\textless{} ~b$$ "}], [{"aoVal": "B", "content": "$$b ~\\textless{} ~c$$ "}], [{"aoVal": "C", "content": "$$a ~\\textless{} ~c$$ "}], [{"aoVal": "D", "content": "$$c ~\\textless{} ~b$$ "}]], "knowledge_point_routes": ["拓展思维->能力->抽象概括"], "answer_analysis": ["因为$$a$$、$$b$$、$$c$$是不同的自然数，$$a\\times \\frac{c}{b} ~\\textless{} ~a$$，则$$ac ~\\textless{} ~ab$$，即$$c ~\\textless{} ~b$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3020", "queId": "b7e4520fee404a7fb2fe34c2911b48d9", "competition_source_list": ["2020年第1届广东深圳超常思维竞赛四年级竞赛初赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$$1+3+5+7+\\cdots +2019+2021=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$${{1011}^{2}}$$ "}], [{"aoVal": "B", "content": "$${{1000}^{2}}$$ "}], [{"aoVal": "C", "content": "$${{999}^{2}}$$ "}], [{"aoVal": "D", "content": "$${{998}^{2}}$$ "}], [{"aoVal": "E", "content": "以上都不正确 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["中项定理：等差数列的和等于中项的平方． 中项：$$\\left( 1+2021 \\right)\\div 2=1011$$， 所以答案为$$1011\\times 1011$$，即选择$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "612", "queId": "18c91d273c6743c58bc10903476ceb82", "competition_source_list": ["2014年IMAS小学高年级竞赛第二轮检测试题第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个两位数是完全平方数，它的两个数字之和也恰好是完全平方数，请问所有这样的完全平方数之和为多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$110$$ "}], [{"aoVal": "C", "content": "$$117$$ "}], [{"aoVal": "D", "content": "$$181$$ "}], [{"aoVal": "E", "content": "$$271$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->完全平方数->平方数的尾数特征"], "answer_analysis": ["如果一个两位数是完全平方数，那么这个两位数的所有可能值是： $$16$$，$$25$$，$$36$$，$$49$$，$$64$$，$$81$$， 而$$3+6={{3}^{2}}$$，$$8+1={{3}^{2}}$$， 所以符合条件的两位数为$$36$$，$$81$$， $$36+81=117$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2781", "queId": "acb1695a28704072a8e6265f4414e1ce", "competition_source_list": ["2017年全国亚太杯四年级竞赛初赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "算式$$42+402+4002+\\cdots \\cdots +\\underbrace{400\\cdots 002}_{5个0}$$ 的计算结果是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4444452$$ "}], [{"aoVal": "B", "content": "$$4444450$$ "}], [{"aoVal": "C", "content": "$$4444450$$ "}], [{"aoVal": "D", "content": "$$4444428$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["根据位值原理，原式$$=4\\times 1111110+2\\times 6=4444452$$ ． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2470", "queId": "5d83a03f98274306a6f64eaa3afe01f3", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛A卷第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$0.00000025\\times 0.000004$$的结果的小数部分有个$$0$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数乘除->小数乘法运算"], "answer_analysis": ["因为$$0.\\underbrace{00000025}_{8位}\\times 0.\\underbrace{000004}_{6位}=0.\\underbrace{000000000001}_{12位}$$， 故小数部分有$$11$$个$$0$$， 故答案是：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "290", "queId": "ff8080814518d5240145192574fe04ee", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "将一个数加上或减去或乘或除以一个一位数（$$0$$不是一位数）视为一次操作，比如$$53$$可以通过加$$3$$，除以$$7$$，除以$$8$$三次操作变成$$1$$．那么$$2014$$至少经过（~~~~~~~ ）次操作可变成$$1$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["$$2014$$要变成$$1$$就需要除以一个数，而除数只能是一位数，那么这个除数显然是越大越好．第一次操作$$2014+2=2016$$；第二次操作$$2016\\div9=224$$；第三次操作$$224\\div 8=28$$；第四次操作$$28\\div 7=4$$；第五次操作$$4\\div 4=1$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2676", "queId": "cc7c2cdb421841b9bce34b35eb52d5b7", "competition_source_list": ["2017年四川六年级竞赛排位赛第16题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$\\frac{3}{8}$$的分子加上$$3$$，要使分数的大小不变，分母应该加上． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["$$3+3=6$$，$$6\\div 3=2$$，分子扩大到原来的$$2$$倍，要使分数大小不变，分母也要扩大到原来的$$2$$倍，即$$8\\times 2=16$$，$$16-8=8$$，分母应加上$$8$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3167", "queId": "6fd8245d4c2447c8a9ddc05f98c467f8", "competition_source_list": ["2020年长江杯五年级竞赛复赛B卷第3题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个盒子里共有$$96$$颗雨花石，明明从这个盒子里把雨花石全部拿出来，每次拿的颗数要相等．如果不能一次全部拿出，也不能一颗一颗地拿出，那么，明明共有种不同的拿法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["不能一颗一颗地拿，不能一次全拿出去，而且每次拿的颗数要相等，最后全部拿完，也就是说次数$$\\times $$每次拿的颗数$$=96$$．$$96=1\\times 96$$，$$96=2\\times 48$$，$$96=3\\times 32$$，$$96=4\\times 24$$，$$96=6\\times 16$$，$$96=8\\times 12$$，那么每次拿的颗数可以是$$2$$、$$48$$、$$3$$、$$32$$、$$4$$、$$24$$、$$6$$、$$16$$、$$8$$、$$12$$，一共有$$10$$种拿法． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1773", "queId": "77ed427eb6304936bd90e4b228d950ea", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "毛毛用围棋子摆成一个三层空心方阵，最内一层每边有围棋子$$8$$个．毛毛摆这个方阵共用围棋子个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$ "}], [{"aoVal": "B", "content": "$$108$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$132$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$1$$、这是一道关于方阵的问题，考查的是方阵的知识； $$2$$、由题意知，用围棋子摆成一个三层空心方阵，最内一层每边有围棋子$$8$$个，由于相邻两层每边相差$$2$$个，相邻两层相差$$8$$个，求出最内一层后依次加$$8$$即可。 根据题意分析可知：最内一层棋子个数为：$$\\left( 8-1 \\right)\\times 4=7\\times 4=28$$（个）， 第二层棋子有：$$28+8=36$$（个）， 第三层棋子有：$$36+8=44$$（个）， 所以三层一共有$$28+36+44=108$$（个）． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2210", "queId": "756967750313406c8e1d745b73fbd9b4", "competition_source_list": ["2015年第11届全国新希望杯小学高年级六年级竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "小明的手表每小时慢$$3$$分钟，如果他在早晨$$6$$:$$30$$将手表与准确时间对准，那么当天小明手表显示$$12$$:$$50$$的时候，准确时间是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$:$$00$$ "}], [{"aoVal": "B", "content": "$$13$$:$$09$$ "}], [{"aoVal": "C", "content": "$$13$$:$$10$$ "}], [{"aoVal": "D", "content": "$$13$$:$$15$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->时钟问题->坏钟问题"], "answer_analysis": ["手表走$$57$$分钟，正常钟走$$60$$分钟，现在手表走$$6$$小时$$20$$分钟，为$$380$$分钟，那么正常钟应该走$$380\\div 57\\times 60=400$$．为$$6$$小时$$40$$分，此时为$$13$$:$$10$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "312", "queId": "ccdcc8cc15134479a41511e22d245f16", "competition_source_list": ["2021年新希望杯三年级竞赛初赛第13题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "【2021三年级卷第$$13$$题】五盘水果排成一排．苹果和橘子相邻，橘子和草莓相邻，苹果和香蕉不相邻，香蕉和芒果不相邻．那么一定和芒果相邻的是． ", "answer_option_list": [[{"aoVal": "A", "content": "只有苹果 "}], [{"aoVal": "B", "content": "只有橘子 "}], [{"aoVal": "C", "content": "只有草莓 "}], [{"aoVal": "D", "content": "香蕉和草莓 "}], [{"aoVal": "E", "content": "橘子和香蕉 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意分析可知，苹果和草莓都与橘子相邻， 所以可以确定橘子在苹果和草莓的中间， 假设苹果、橘子、草莓这样排列， 则香蕉在草莓的右边，芒果在苹果的左边， 即苹果与芒果一定相邻， 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2390", "queId": "1803f467a1164d2b9a4b87d311071aba", "competition_source_list": ["2017年第8届广东广州羊排赛六年级竞赛第4题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "省略下列各数的最后一位数得到的近似数与$$9.24$$大小不同的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9.237$$ "}], [{"aoVal": "B", "content": "$$9.2395$$ "}], [{"aoVal": "C", "content": "$$9.2449$$ "}], [{"aoVal": "D", "content": "$$9.244$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数基础->取近似值"], "answer_analysis": ["$$9.237\\approx 9.24$$，$$9.2395\\approx 9.240$$，$$9.2449\\approx 9.245$$，$$9.244\\approx 9.24$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1229", "queId": "7e14a4a993614e98b354003f0e6b66e4", "competition_source_list": ["2016年创新杯五年级竞赛训练题（二）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$198896$$后面添一长串数，组成一个$$2016$$位数，添加的方法是：将末位数字乘$$7$$，取积的个位数字加在末位的后面，例如：$$198896$$的末位数字是$$6$$，$$6\\times 7=42$$，在$$6$$后面添加数字$$2$$得$$1988962$$，$$2\\times 7=14$$，在$$2$$后面添加$$4$$得$$19889624$$，继续按此规则添加下去组成一个$$2016$$位的数，这个$$2016$$位的数的尾数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["按照题目的要求将这个数向后写$$10$$位，得到$$1988962486248624\\cdots $$，可以看到从第$$6$$位开始每$$4$$个数字一循环，与$$\\left( 2016-5 \\right)\\div 4=502$$（组）$$\\cdots 3$$（个），所以第$$2016$$位的数字为$$4$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2072", "queId": "cff12befed3c4eb3b745eac7ffc55be5", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一串数：$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$34$$，$$55$$，$$89$$$$\\cdots \\cdots $$，其中第一个数$$2$$，第二个数$$3$$，从第三个数起，每个数恰好是前两个数的和．那么在这串数中，第$$2019$$个数被$$3$$除后所得余数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["拓展思维->能力->归纳总结->归纳推理"], "answer_analysis": ["将这一串数写成除以$$3$$的余数，则为：$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0$$，$$1$$，$$1$$，$$2$$，$$0$$，$$2\\cdots \\cdots $$ 所以重复的为：``$$2$$，$$0$$，$$2$$，$$2$$，$$1$$，$$0$$，$$1$$，$$1$$''， 故周期为$$8$$．$$2019\\div 8=252$$（组）$$\\cdots \\cdots 3$$（个），则答案为$$2$$，故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1153", "queId": "827c4c38a5534aaca9136918392fa7d1", "competition_source_list": ["2003年第1届创新杯六年级竞赛复赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "四个同学进行计算比赛，比赛内容是：在$$9$$、$$10$$、$$11$$、$$\\cdots \\cdots $$、$$67$$、$$68$$这$$60$$个自然数的相邻两数之间任意添加符号``$$+$$''或``$$-$$''，然后进行计算．四个同学得到的结果分别是$$2274$$、$$2003$$、$$2300$$、$$2320$$，老师看后指出．这四个结果中只有一个是正确的．这个正确的结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2274$$ "}], [{"aoVal": "B", "content": "$$2003$$ "}], [{"aoVal": "C", "content": "$$2300$$ "}], [{"aoVal": "D", "content": "$$2320$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["由于$$9+10+11+\\cdots 68=2310$$，由此可知$$2320$$是错误的．由于$$2274$$、$$2003$$、$$2300$$都小于小于$$2310$$，所以减的数较多，由于减一个数，总和里面就要少这个数的$$2$$倍，如减$$2$$，则是$$2310-2\\times 2=2306$$，所以只要是小于$$2310$$．据此分析即 $$9+10+11+\\cdots 68=2310$$，$$2320\\textgreater2310$$，故$$\\text{D}$$错误； $$\\left( 2310-2274 \\right)\\div 2=18$$，$$18\\div 2=9$$，所以在$$9$$前是减号即可，符合题意． $$\\left( 2310-2003 \\right)=307\\textgreater68$$，错误 $$\\left( 2310-2000 \\right)\\div 2=155\\textgreater68$$，错误． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1288", "queId": "54dc63565f814fb3a0489aa7127b94e8", "competition_source_list": ["2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一只大熊猫从$$A$$地往$$B$$地运送竹子，它每次可以运送$$50$$根，但是它从$$A$$地走到$$B$$地和从$$B$$地返回$$A$$地都要吃$$5$$根，$$A$$地现在有$$200$$根竹子，那么大熊猫最多可以运到$$B$$地（ ）根。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$150$$ "}], [{"aoVal": "B", "content": "$$155$$ "}], [{"aoVal": "C", "content": "$$160$$ "}], [{"aoVal": "D", "content": "$$165$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->乘除法应用->乘法应用（顺口溜）"], "answer_analysis": ["解：由题意，运四次，去四次回三次，吃掉了$$5\\times (4+3)=35$$根，则最多可以运到$$B$$地$$200-35=165$$根。 故选:D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1527", "queId": "5684aa516e9645cbaece3c8dfb613e56", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个数，它的一半的一半是$$4$$，这个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["一个数的一半的一半是$$4$$，则这个数的一半为$$4\\times 2=8$$，则这个数为$$8\\times 2=16$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2955", "queId": "9bf8b4e1455b409f8de985f1c58c7e4c", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（四）第6题"], "difficulty": "3", "qtype": "single_choice", "problem": "从$$0$$、$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$这十个数字中选出九个数字，组成一个两位数、一个三位数和一个四位数，使这三个数的和等于$$2017$$，那么未被选中的数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$2017$$除以$$9$$余$$1$$，$$0+1+2+3+\\cdots +9=45$$，$$45$$是$$9$$的倍数，$$9-1=8$$，所以未被选中的数字是$$8$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2984", "queId": "f74ac5fe3f6e45e8b00e523a5166b3a9", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$2015\\times 2015-2014\\times 2016=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2014$$ "}], [{"aoVal": "C", "content": "$$2015$$ "}], [{"aoVal": "D", "content": "$$2016$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->公式类运算->平方差公式->平方差公式逆向应用"], "answer_analysis": ["$${{2015}^{2}}-(2015-1)\\times (2015+1)={{2015}^{2}}-{{2015}^{2}}+1=1$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2207", "queId": "3eb4bc2a2825416c9df8fb1c8255d101", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）"], "difficulty": "1", "qtype": "single_choice", "problem": "一列火车通过一座长$$320$$米的桥用了$$105$$秒，当它通过$$860$$米的隧道时，速度是过桥速度的$$2$$倍，结果用了$$120$$秒，火车通过大桥时的速度是每秒米；火车的车身长度为米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$；$$100$$ "}], [{"aoVal": "B", "content": "$$4$$；$$100$$ "}], [{"aoVal": "C", "content": "$$8$$；$$100$$ "}], [{"aoVal": "D", "content": "$$8$$；$$200$$ "}]], "knowledge_point_routes": ["课内体系->知识点->应用题->行程应用题->火车过桥完全过桥", "拓展思维->思想->对应思想"], "answer_analysis": ["若通过$$860$$米隧道时速度不变则需要$$120\\times 2=240$$（秒），火车过桥速度：$$\\left( 860-320 \\right)\\div \\left( 240-105 \\right)=4$$（米/秒）：火车车身长：$$105\\times 4-320=100$$（米）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3224", "queId": "6745abad3b244368bc11cb40689947cb", "competition_source_list": ["2014年第2届广东广州羊排赛六年级竞赛第4题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "袋子中有$$4$$个红球，$$6$$个黄球，$$8$$个蓝球和$$3$$个绿球．从中随机拿出一个，是黄球的概率是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{7}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{5}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["共有$$4+6+8+3=21$$（个）球，拿到黄球概率为$$6\\div 21=\\frac{2}{7}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1006", "queId": "1c065f5617f046ec8c3cf057e06cd16f", "competition_source_list": ["2017年第17届全国中环杯三年级竞赛初赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$100$$个数的平均数为$$1$$，增加一个数$$102$$之后，这$$101$$个数的平均数为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->公式类->加权平均数"], "answer_analysis": ["$$1\\times 100+102\\div（ 100+1）=2$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3177", "queId": "827a73fcd55a469f9f50123b8d10138a", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛初赛模拟"], "difficulty": "1", "qtype": "single_choice", "problem": "六个小朋友排成一排照相，其中有四个男生和两个女生，两个女生必须站在一起而且不能站在边上，则一共有种不同的排列方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$144$$ "}], [{"aoVal": "D", "content": "$$240$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["先捆绑，再插空，$A_{2}^{\\^{2}}\\times C_{3}^{1}\\times A_{4}^{4}=144$种． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2138", "queId": "c716332e8f504c90897ed34a6bdb91f4", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛复赛第8题", "2011年全国创新杯五年级竞赛第8题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "甲、乙两车分别从$$A$$、$$B$$两地同时相向开出，$$4$$小时后两车相遇，然后各自继续行驶$$3$$小时，此时甲车距$$B$$地$$10$$千米，乙车距$$A$$地$$80$$千米，那么$$AB$$两地相距（~ ）千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$350$$ "}], [{"aoVal": "B", "content": "$$360$$ "}], [{"aoVal": "C", "content": "$$370$$ "}], [{"aoVal": "D", "content": "$$380$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->两人相遇与追及问题->相遇问题->同时出发相向而行"], "answer_analysis": ["$$4$$小时甲乙共行了$$AB$$，$$3$$小时甲乙共行了$$\\frac{3}{4}AB$$，$$(10+80)\\div \\left( 1-\\frac{3}{4} \\right)=360$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "191", "queId": "50ab989a58b044af8c05bea54c107139", "competition_source_list": ["2018年第21届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第6题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "一副扑克牌去掉大小王共有$$52$$张，共有$$13$$种点数，最少要抽取张牌，方能保证其中至少有$$2$$张牌有相同的点数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["扑克牌的点数有$$1$ $13$$点，所以要保证$$2$$张牌有相同的点数．最坏的情况就是抽到$$1$ $13$$，这$$13$$张牌，再来一张，就一定和之前的$$13$$张相同，所以是$$13+1=14$$（张）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2454", "queId": "19074618ec8d40a5ada6edb7ddb5965e", "competition_source_list": ["2019年福建泉州鲤城区泉州师范附属小学三年级竞赛模拟第2题2分", "2019年广西防城港小升初模拟第13题", "2015~2016学年内蒙古乌兰察布丰镇市丰镇实验小学四年级上学期期中第25题5分", "2019年广西防城港小升初模拟第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$9+99+999+9999+99999=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$111105$$ "}], [{"aoVal": "B", "content": "$$1111105$$ "}], [{"aoVal": "C", "content": "$$111104$$ "}], [{"aoVal": "D", "content": "$$111005$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算求解", "拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之凑整法"], "answer_analysis": ["根据题意，把$$9$$、$$99$$、$$999$$、$$9999$$、$$99999$$看作$$\\left(10-1\\right)$$、$$\\left(100-1\\right)$$、$$\\left(1000-1\\right)$$、$$\\left(10000-1\\right)$$、$$\\left(100000-1\\right)$$，再进行计算． $$9+99+999+9999+99999$$ $$=\\left(10-1\\right)+\\left(100-1\\right)+\\left(1000-1\\right)+\\left(10000-1\\right)+\\left(100000-1\\right)$$ $$=\\left(100000+10000+1000+100+10\\right)-\\left(1+1+1+1+1\\right)$$ $$=111110-5$$ $$=111105$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2344", "queId": "0203e0f25745468e9e0547dc86ed62fc", "competition_source_list": ["2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2$$个樱桃的价钱与$$3$$个苹果的价钱一样，但是一个苹果的大小却是一个樱桃的$$12$$倍，如果妈妈用买$$1$$箱樱桃的钱买同样大小箱子的苹果，能买( )箱。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$27$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->等量代换"], "answer_analysis": ["$$12$$个樱桃的钱可以买$$18$$个苹果，大小是$$1$$个苹果的大小，所以$$1$$个苹果大小的樱桃可以买到$$18$$个苹果，所以$$1$$箱樱桃的钱可以买$$18$$箱苹果。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "991", "queId": "3c7cd9341ad34ec397f5a0bb02c3faf2", "competition_source_list": ["2020年长江杯六年级竞赛复赛A卷第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "``双$$11$$''前夕，某天猫店购进一批商品，按期望获得$$60 \\%$$的利润定价，在卖出$$70 \\%$$的商品后，进行``双$$11$$''促销，该天猫店决定按定价折扣销售，货销完后，所获得的全部利润只是原来期望利润的$$80 \\%$$，该店``双$$11$$''销售剩余部分商品时打的折扣是． ", "answer_option_list": [[{"aoVal": "A", "content": "六折 "}], [{"aoVal": "B", "content": "七折 "}], [{"aoVal": "C", "content": "七五折 "}], [{"aoVal": "D", "content": "八折 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["假设商品的成本是``$$1$$''，原来希望获得利润$$60 \\%$$，现出售$$70 \\%$$的商品已获得利润$$60 \\%\\times 70 \\%=0.42$$． 剩下的$$30 \\%$$商品将要获得利润：$$0.6\\times 80 \\%-0.42=0.06$$，因此这剩下$$30 \\%$$的商品的售价是：$$1\\times 30 \\%+0.06=0.36$$（元）原来定价是：$$1\\times 30 \\%\\times (1+60 \\%)=0.48$$，$$(0.36\\div 0.48)\\times 100 \\%=75 \\%$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1320", "queId": "1fe9fabb0e7945b5b03e0f1933d9171b", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（六）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "李家村李家兄弟合作在门口挖一口井，李凡单独挖需要$$10$$天才能挖成，李超单独挖需要$$15$$天才能挖完．兄弟两人同时挖需要天． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数与运算->加法->分数加法->异分母分数加法"], "answer_analysis": ["李凡一天挖$$\\frac{1}{10}$$，李超一天挖$$\\frac{1}{30}$$，兄弟合挖一天能完成$$\\frac{1}{30}+\\frac{1}{10}=\\frac{2}{15}$$，需要$$1\\div \\frac{2}{15}=7.5$$天． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "936", "queId": "fca6163f959a461c9246663834221a8e", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "妈妈在家开着灯做饭，突然停电了．爸爸回家按了$$4$$下开关，小林回家又按了$$3$$下开关．当来电的时候，灯泡． ", "answer_option_list": [[{"aoVal": "A", "content": "亮 "}], [{"aoVal": "B", "content": "不亮 "}], [{"aoVal": "C", "content": "不能确定 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["开单数次数，改变原来灯的状态，开双数次，不改变原来的状态．总共开的次数为：$$4+3=7$$（次）， 所以改变了原来的状态，原来是开着的， 所以现在不亮． 故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1467", "queId": "6cc178e6ed1e496db9ac31a4991db464", "competition_source_list": ["2012年全国美国数学大联盟杯小学高年级竞赛初赛第30题", "2013年美国数学大联盟杯小学高年级竞赛初赛第30题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "花园里栽种着雏菊和玫瑰两种花共计$$66$$朵，其中每出现$$5$$朵雏菊便另有$$6$$朵玫瑰．问：花园里有玫瑰（~~ ）朵？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->比例应用题->按比分配"], "answer_analysis": ["$$66\\div(6+5)=6$$， 花园里有玫瑰$$6\\times6=36$$朵． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "964", "queId": "d980a5804d734ed4929bf4be62e0b37c", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$\\square $$里填上适当的数字，使得七位数$$\\overline{2019\\square \\square \\square }$$能同时被$$9$$、$$25$$、$$8$$整除，符合条件的七位数的数字和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$27$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["要使这个数能同时被$$9$$，$$25$$，$$8$$整除， 那么这个数一定是$$9$$，$$25$$，$$8$$的公倍数， 首先算出$$9$$，$$25$$，$$8$$的最小公倍数是$$1800$$， 然后可以使用试除法：$$2019999\\div 1800=1122\\cdots \\cdots 399$$， 所以这个七位数为$$2019999-399=2019600$$， 数字和是$$2+0+1+9+6+0+0=18$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1594", "queId": "faef5314ba114e9cbbfd04296c7a0b10", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛A卷第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "每次考试试卷总分是$$120$$分，小红四次考试的平均成绩是$$105$$分．为了使平均成绩尽快达到$$110$$分，她至少需要再考次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["根据题意分析可知，总分数$$\\div $$考的次数$$=$$平均分数，；可设他至少要考$$x$$次才能尽快达到$$110$$分以上，那么 小红的总分数为：$$120x+105\\times 4$$，小红考试的次数为：$$x+4$$，小红平均分应为$$\\geqslant $$$$100$$，根据公式列方程解答即可． 设小红至少要考$$x$$次才能尽快达到$$110$$分以上， $$\\left( 120x+105\\times 4 \\right)$$$$\\div \\left( x+4 \\right)\\geqslant 110$$， $$120x+420\\geqslant 110x+440$$， $$120x-110x\\geqslant 440-420$$， $$10x\\geqslant 20$$， $$x\\geqslant 2$$． 即再考$$2$$次满分平均分可达到$$110$$分以上． 故选答案$$B$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1697", "queId": "7bb14cd8374d4d38adf6550647aba206", "competition_source_list": ["2017年全国美国数学大联盟杯小学高年级六年级竞赛第17题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "（这是一道挑战题！）There are only white cars and black cars．$$If$$ the ratio of white cars to black cars is $$3$$ to $$8$$ and there are a total of $$242$$ cars， how many black cars and white cars are there？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$66$$ "}], [{"aoVal": "B", "content": "$$110$$ "}], [{"aoVal": "C", "content": "$$129$$ "}], [{"aoVal": "D", "content": "$$176$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->比例应用题->按比分配"], "answer_analysis": ["停车场只有黑白两种车，白车与黑车的比率是$$3:8$$，黑白车的总数是$$242$$辆．请问黑车比白车多多少辆？ ratio比率； a total of 总数； more．．than 比．．多． 白车数量：$$\\frac{3}{11}\\times 242=66$$辆，黑车数量：$$\\frac{8}{11}\\times 242=176$$辆．黑车比白车多$$176-66=110$$辆． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "937", "queId": "d39b76e2ce0948bdb57dd64a25524865", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛B卷第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知一个自然数有$$12$$个不同的因数，那么这个数最小是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$64$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["∵$$12=(2+1)\\times (1+1)\\times (1+1)$$， ∴最小是$$4\\times3\\times5=60$$， $$12$$个因数为：$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$10$$、$$15$$、$$12$$、$$20$$、$$30$$、$$60$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "49", "queId": "080ac401862e49d688247f52722a94dd", "competition_source_list": ["2008年四年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "在下面的数字之间添上四个加号``+''，不同的填法所算出的结果中最小值是( ). 1( )2( )3( )4( )5( )6( )7( )8 ", "answer_option_list": [[{"aoVal": "A", "content": "117 "}], [{"aoVal": "B", "content": "118 "}], [{"aoVal": "C", "content": "119 "}], [{"aoVal": "D", "content": "120 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->竖式数字谜->竖式数字谜的最值"], "answer_analysis": ["在数串(列)1，2，3，4，5，6，7，8之间任意添上四个``+''号，可将它分成5段，因而可分成5个整数求和，所以有以下三种情况: ⑴1个四位数和4个一位数的和，此时和的最小值为$$1234+5+6+7+8=1260$$; ⑵1个三位数，1个两位数与3个一位数的和，此时最小值为$$123+45+6+7+8=189$$; ⑶3个两位数与2个一位数的和，此时最小值为$$12+34+56+7+8=117$$. 因此和的最小值是117 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2736", "queId": "68c936b71c0a431fa4da7b513ce8e70f", "competition_source_list": ["2016年第28届广东广州五羊杯小学高年级竞赛初赛第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$1:9000000$$地图上，$$A$$、$$B$$两地相距$$2$$厘米，早上八点，一辆车从$$A$$地以$$50\\text{km}/\\text{h}$$的速度开往$$B$$，那么在时汽车到达$$B$$地． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$点 "}], [{"aoVal": "B", "content": "$$11$$点$$30$$分 "}], [{"aoVal": "C", "content": "$$11$$点$$36$$分 "}], [{"aoVal": "D", "content": "$$11$$点$$45$$分 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比和比例->比例->比例尺"], "answer_analysis": ["$$AB$$距离为$$2\\times 9000000=18000000\\text{cm=180}$$千米， 则需要$$180\\div 50=3.6\\text{h}=3\\text{h}36\\min $$， 则$$8:00+3:36=11:36$$到达． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "413", "queId": "f703934550544aefa90fb1462d5244c6", "competition_source_list": ["2021年新希望杯五年级竞赛初赛第14题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在比武大会上，熊猫阿宝和金猴两人进行比试，最多七局，谁先获得四局胜利，谁就是胜者，那么一共有~\\uline{~~~~~~~~~~}~种可能的比试情况． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$70$$ "}], [{"aoVal": "C", "content": "$$38$$ "}], [{"aoVal": "D", "content": "$$56$$ "}], [{"aoVal": "E", "content": "$$62$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["假设阿宝胜，有发以下几种获胜情况（用``$$+$$''表示胜``$$-$$''表示，负） ①连胜四局$$1$$种， $$++++$$， ②四胜一负$$4$$种， $$-++++$$；$$+-+++$$；$$++-++$$；$$+++-+$$ ③四胜两负$$10$$种， $$-\\/-++++$$；$$+-\\/-+++$$；$$++-\\/-++$$；$$+++-\\/-+$$；$$-+-+++$$；$$-++-++$$；$$-+++-+$$；$$+-+-++$$；$$+-++-+$$；$$++-+-+$$． ④四胜三负， （1）甲负三局连负$$4$$种， $$-\\/-\\/-++++$$；$$+-\\/-\\/-+++$$；$$++-\\/-\\/-++$$；$$+++-\\/-\\/-+$$． （2）甲连负两局$$12$$种， $$-\\/-+-+++$$；$$-\\/-++-++$$；$$-\\/-+++-+$$，$$+-\\/-+-++$$；$$+-\\/-++-+$$；$$++-\\/-+-+$$；$$-+-\\/-+++$$；$$-++-\\/-++$$；$$-+++-\\/-+$$；$$+-+-\\/-++$$；$$+-++-\\/-+$$；$$++-+-\\/-+$$． （3）甲负三局全都分开$$4$$种， $$-+-+-++$$，$$-+-++-+$$；$$-++-+-+$$；$$+-+-+-+$$． $$1+4+10+4+12+4=35$$（种）$$35+35=70$$种， 当考虑金猴的情况时与上述一致也为$$35$$种，因此一共$$35+35=70$$种可能． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2968", "queId": "932b947ce1fa4b0e9ac790ec811ed31d", "competition_source_list": ["2017年全国小升初八中入学备考课程", "2008年全国华杯赛竞赛复赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "对于大于零的分数，有如下$$4$$个结论： ①两个真分数的和是真分数； ②两个真分数的积是真分数； ③一个真分数与一个假分数的和是一个假分数； ④一个真分数与一个假分数的积是一个假分数． 其中正确的有（~~~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->拓展���维->计算模块->分数->分数运算->分数加减"], "answer_analysis": ["对于这种类型的题目，我们可以采取``反驳''的方法来做，找出每个不成立的案例来，若找不到则正确．①反例：$$\\frac{1}{2}+\\frac{1}{2}=1$$，$$\\frac{4}{5}+\\frac{3}{5}=\\frac{7}{5}$$；④反例：$$\\frac{1}{2}\\times \\frac{3}{2}=\\frac{3}{4}$$，$$\\frac{1}{5}\\times \\frac{8}{5}=\\frac{8}{25}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3221", "queId": "a70973afeffe4f30978d8c5952978f70", "competition_source_list": ["2018年第8届北京学而思综合能力诊断二年级竞赛年度教学质量第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "五一到了，花花想要出去旅游，她有$$2$$顶帽子、$$3$$条项链、$$5$$种头花、$$2$$双鞋、$$3$$条白色裙子、$$4$$条粉色裙子、$$1$$条蓝色裙子，每类物品中各选一样进行搭配．那么，一共有种不同的搭配方法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$460$$ "}], [{"aoVal": "C", "content": "$$480$$ "}], [{"aoVal": "D", "content": "$$720$$ "}]], "knowledge_point_routes": ["课内体系->思想->对应思想", "拓展思维->能力->运算求解"], "answer_analysis": ["$$2\\times 2\\times 3=12$$（种）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1826", "queId": "929be6aa07964550b2b91b1a9bf72b40", "competition_source_list": ["2017年全国小升初八中入学备考课程", "2014年全国华杯赛小学高年级竞赛初赛B卷第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙丙丁四个人今年的年龄之和是$$72$$岁．几年前（至少一年）甲是$$22$$岁时，乙是$$16$$岁．又知道，当甲是$$19$$岁的时候，丙的年龄是丁的$$3$$倍（此时丁至少$$1$$岁）．如果甲乙丙丁四个人的年龄互不相同，那么今年甲的年龄可以有种情况． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->年龄问题->年龄问题之差倍型"], "answer_analysis": ["甲乙的年龄差是$$22-16=6$$岁；当甲$$19$$岁时， $$13$$岁；至少一年前甲$$22$$岁，所以当甲$$19$$岁的时候，此时至少是$$4$$年前的年龄，那么甲今年至少是$$23$$岁；甲$$19$$岁时，丙的年龄是丁的$$3$$倍，假设丁为$$1$$岁，丙为$$3$$岁，此时四人的年龄和至少是$$19+13+1+3=36$$岁；且甲今年的年龄至多为$$19+\\left(72-36 \\right)\\div 4=28$$；所以甲今年的年龄可能是$$23$$，$$24$$，$$25$$，$$26$$，$$27$$，$$28$$；共$$6$$种，所以选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3002", "queId": "bc126778752e4148b3f54f30ab612784", "competition_source_list": ["2014年IMAS小学中年级竞赛第二轮检测试题第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "请问$$32$$个$$1000$$、$$19$$个$$100$$、$$29$$个$$10$$之总和等于多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3219290$$ "}], [{"aoVal": "B", "content": "$$321929$$ "}], [{"aoVal": "C", "content": "$$342190$$ "}], [{"aoVal": "D", "content": "$$34190$$ "}], [{"aoVal": "E", "content": "$$32129$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->四则混合运算"], "answer_analysis": ["$$32\\times 1000+19\\times 100+29\\times 10=34190$$，答案选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1008", "queId": "37e3ad8702034f1489c3d5cf5ec6a4ea", "competition_source_list": ["2017年第16届湖北武汉创新杯小学高年级六年级竞赛邀请赛训练题（四）"], "difficulty": "1", "qtype": "single_choice", "problem": "~春节中甲、乙两人出售同一种产品，甲按$$20 \\% $$的利润定价，一共售出$$15$$个，乙按$$15 \\% $$的利润定价，一共售出$$24$$个甲、乙获利润比较结论为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "甲获利润较多 "}], [{"aoVal": "B", "content": "乙获利润较多 "}], [{"aoVal": "C", "content": "两人利润相同 "}], [{"aoVal": "D", "content": "无法比较谁多 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->已知成本售价求利润"], "answer_analysis": ["设原成本为$$1$$元/个，甲获利$$15\\times 20 \\% =3$$（元），乙获利$$0.15\\times 24=3.6$$（元）．选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "253", "queId": "884bbed70af843ae81f91f75ea0c10a7", "competition_source_list": ["1989年华杯赛六年级竞赛初赛"], "difficulty": "3", "qtype": "single_choice", "problem": "一副扑克牌有四种花色，每种花色有$$13$$张，此外还有两张王牌，从中任意抽牌。那么，最少要抽（ ）张牌，才能保证有$$4$$张牌是同一花色。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["如果在最不利的情况下能完成目标，则保证在任何情况下都能完成。最不利的情况是：抽的前$$12$$张是$$4$$种花色各$$3$$张，再抽两张王牌，这时抽第$$15$$张，无论是哪种花色，都能保证凑成$$4$$张牌同一花色。所以至少要抽$$15$$张牌，才能保证有四张牌是同一花色的。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1882", "queId": "b757297e8faf4bf5a57ee4a6032b1578", "competition_source_list": ["2011年世界少年奥林匹克数学竞赛六年级竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小宇参加奥数竞赛抢答赛，抢答试题共有$$10$$道，每答对一题得$$8$$分，答错一题倒扣$$5$$分．小宇最终得$$41$$分，他做对~\\uline{~~~~~~~~~~}~题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$37$$ "}], [{"aoVal": "C", "content": "$$7$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["本题是鸡兔同笼问题的变式应用． 解决此类问题一般都用假设法，做对一题得$$8$$分，相当于``兔''，做错一题倒扣$$5$$分，相当于``鸡''，通过适当的替换进行解答． 假设$$10$$道题全部做对，则得分为$$10\\times 8=80$$（分），而答错一题比答对一题少得$$8+5=13$$（分），从题目中可以求出小宇一共少得$$80-41=39$$（分），由此可以求出他做错了$$39\\div 13=3$$（道）． $$(10\\times 8-41)\\div (48+5)$$ $$=39\\div 13$$ $$=3$$（道）， $$10-3=7$$（道）． 故答案为：$$7$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2659", "queId": "da44f96b5ab14700a99d0d0714da8b72", "competition_source_list": ["2009年五年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "一列有明显规律的数$$5$$，$$8$$，$$11$$，$$14$$，$$17$$，$$\\cdots$$第（ ）个数为$$2009$$。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$667$$ "}], [{"aoVal": "B", "content": "$$668$$ "}], [{"aoVal": "C", "content": "$$669$$ "}], [{"aoVal": "D", "content": "$$700$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列求项数"], "answer_analysis": ["项数$$=\\left( 2009-5 \\right)\\div 3+1=669$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1353", "queId": "ff21ec4e6de94a59b95fb50cee8a3813", "competition_source_list": ["2018年第12届北京学而思综合能力诊断小学高年级五年级竞赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "取经路上，师徒四人被妖精重重包围．孙悟空一个人，预计$$30$$分钟可以消灭所有妖精，如果猪八戒帮助他，可以比预计提前$$6$$分钟消灭所有妖精；如果猪八戒和沙和尚一起帮助他，可以比预计提前$$10$$分钟消灭所有妖精．现在猪八戒要保护师傅，只有沙和尚帮助孙悟空，那么需要分钟才能消灭所有妖精． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$22$$ "}], [{"aoVal": "E", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["孙悟空打妖精的工作效率是$$\\frac{1}{30}$$； 猪八戒与孙悟空合作打妖精的工作效率是$$\\frac{1}{24}$$， 则有猪八戒打妖精的工作效率是$$\\frac{1}{24}-\\frac{1}{30}=\\frac{1}{120}$$； 沙和尚、猪八戒与孙悟空合作打妖精的工作效率是$$\\frac{1}{20}$$， 所以沙和尚和孙悟空合作打妖精的工作效率为$$\\frac{1}{20}-\\frac{1}{120}=\\frac{1}{24}$$， 需$$1\\div \\frac{1}{24}=24$$分钟． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1662", "queId": "accd290176bf443db05ced82b2b08f5a", "competition_source_list": ["2017年河南郑州联合杯竞赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明买了一本$$513$$页的小说，数字$$1$$在页码中出现了次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$206$$ "}], [{"aoVal": "B", "content": "$$203$$ "}], [{"aoVal": "C", "content": "$$153$$ "}], [{"aoVal": "D", "content": "$$211$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->页码问题->数码综合问题"], "answer_analysis": ["页码问题；$$1\\sim 99$$中数字``$$1$$''出现了$$20$$次，$$100\\sim 199$$中数字``$$1$$''出现了$$120$$个，剩下的$$200\\sim 500$$有$$3\\times 20=60$$（个），$$501\\sim 513$$中数字``$$1$$''出现了$$6$$次，所以一共有$$20+120+60+6=206$$（次）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1209", "queId": "1a1af13a6b4d4bab95364658c582443e", "competition_source_list": ["2012年美国数学大联盟杯六年级竞赛初赛第32题5分(每题5分)"], "difficulty": "1", "qtype": "single_choice", "problem": "我两年前的年龄加上两年后的年龄是$$24$$岁，则我现在几岁？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题基本关系->年龄和"], "answer_analysis": ["根据题意分析可知，假设现在是$$x$$岁，那么两年前的年龄为$$(x-2)$$岁，两年后的年龄为$$(x+2)$$岁，根据等量关系列方程如下： $$(x-2)+(x+2)=24$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x-2+x+2=24$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x+x=24+2-2$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde2x=24$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x=12$$． 即现在$$12$$岁， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2383", "queId": "5d204a536ef848919a01250f0dca240a", "competition_source_list": ["2016年IMAS小学高年级竞赛第一轮检测试题第2题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "课后测 请问下列哪一项表达式是正确的？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1.2\\times 3.4=12\\times 3.4$$ "}], [{"aoVal": "B", "content": "$$0.98\\times 0.99\\textgreater0.99$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}-\\frac{1}{3}\\textless{}\\frac{1}{3}-\\frac{1}{4}$$ "}], [{"aoVal": "D", "content": "$$10.4\\times 0.1\\textless{}1.04$$ "}], [{"aoVal": "E", "content": "$$1.1\\times 1.1\\textgreater1.1$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->数论模块->位值原理与进制->数与数字->比较大小"], "answer_analysis": ["$$\\text{A}$$：$$1.2\\times 3.4\\textless{}12\\times 3.4$$ $$\\text{B}$$：$$0.98\\times 0.99\\textless{}0.99$$ $$\\text{C}$$：$$\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$$，$$\\frac{1}{3}-\\frac{1}{4}=\\frac{1}{12}$$，$$\\frac{1}{2}-\\frac{1}{3}\\textgreater\\frac{1}{3}-\\frac{1}{4}$$ $$\\text{D}$$：$$10.4\\times 0.1=1.04$$ "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2274", "queId": "77d8a02e3fc84712a9ef3e99ab27ab04", "competition_source_list": ["2013年北京迎春杯六年级竞赛", "2013年全国迎春杯六年级竞赛初赛第9题"], "difficulty": "3", "qtype": "single_choice", "problem": "甲、乙二车分别从$$A$$、$$B$$两地同时出发，相向匀速而行，当甲行驶过$$AB$$中点$$12$$千米时，两车相遇．若甲比乙晚出发$$10$$分钟，则两车恰好相遇在$$AB$$中点，且甲到$$B$$地时，乙距离$$A$$地还有$$20$$千米．那么$$AB$$两地间的距离是~\\uline{~~~~~~~~~~}~千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$110$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$130$$ "}], [{"aoVal": "E", "content": "$$140$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->比例解行程问题->行程中的比例", "Overseas Competition->知识点->行程模块"], "answer_analysis": ["方法一：由第一个条件知甲乙速度比为$$（\\frac { S }{ 2 } +12）：（\\frac { S }{ 2 } -12）$$；再由第二条件知甲乙速度比为$$\\frac { S }{ 2 } ：（\\frac { S }{ 2 } -20）$$，两者相等得$$S=120$$千米． 方法二：设全长为$$2S$$千米，则第一次相遇时，甲走了$$S+12$$千米，乙走了$$S-12$$千米；第二次相遇点走到各自目的地时，甲走了$$S$$千米，乙走了$$S-20$$千米．由于两人速度不变，故以速度比作等量关系列方程可得：$$\\frac{S+12}{S-12}=\\frac{S}{S-20}$$，解得$$S=60$$，故全长为$$2S=120$$千米． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "733", "queId": "3fe8fcc043894df48727aef004cc6051", "competition_source_list": ["2014年走美杯六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "十进制的$$1039$$用二进制表示是$$($$~\\uline{~~~~~~~~~~~~~~~~}~$$)_2$$。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$10000001111$$ "}], [{"aoVal": "B", "content": "$$10000001110$$ "}], [{"aoVal": "C", "content": "$$10000000111$$ "}], [{"aoVal": "D", "content": "$$100000011001$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->进制的性质与应用->进制间的互化"], "answer_analysis": ["用除$$2$$倒取余数法，得到$$10000001111$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2943", "queId": "a068f4948b394233abad5596b2185536", "competition_source_list": ["2018年四川成都小升初公立名校初一开学考（一）第5题5分", "2008年全国华杯赛竞赛初赛第6题", "2008年第13届全国迎春杯竞赛初赛第6题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a=\\frac{2005\\times 2006}{2007\\times 2008}$$，$$b=\\frac{2006\\times 2007}{2008\\times 2009}$$，$$c=\\frac{2007\\times 2008}{2009\\times 2010}$$，则有（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b\\textgreater c$$ "}], [{"aoVal": "B", "content": "$$a\\textgreater c\\textgreater b$$ "}], [{"aoVal": "C", "content": "$$a\\textless{}c\\textless{}b$$ "}], [{"aoVal": "D", "content": "$$a\\textless{}b\\textless{}c$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["由糖水原理可知$$a\\textless{}b\\textless{}c$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1820", "queId": "d22ba96b34ff4b01947b2f41ce0f189a", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛第7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "帅帅家购买了一套现价$$12$$万的新房，采用分期付款的形式，购房时，第一年付款$$3$$万元，从第二年起，每年应付房款$$5000$$元加上上一年欠款的利息，若剩余欠款的年利率为$$0.4 \\% $$，第~\\uline{~~~~~~~~~~}~年帅帅家需要交房款$$5200$$元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->利息问题"], "answer_analysis": ["第一年付：$$30000$$元，第二年付：$$5000+90000\\times 0.4 \\% =5360$$（元），第三年付：$$5000+85000\\times 0.4 \\% =5340$$（元），第四年付：$$5000+80000\\times 0.4 \\% =5320$$（元），$$\\cdot \\cdot \\cdot $$，以此类推，第十年付：$$5200$$元． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "783", "queId": "6d74ee57646b4fd48eee5de18a4b8751", "competition_source_list": ["2014年第12届全国小机灵杯小学高年级五年级竞赛决赛第14题"], "difficulty": "3", "qtype": "single_choice", "problem": "幼儿园老师把$$270$$个苹果、$$180$$个梨和$$235$$个橘子平均分给大班小朋友，余下的苹果、梨和橘子的数量之比是$$3:2:1$$．大班有~\\uline{~~~~~~~~~~}~名小朋友． ", "answer_option_list": [[{"aoVal": "A", "content": "请在答题卡上作答 "}]], "knowledge_point_routes": ["课内体系->思想->逐步调整思想", "拓展思维->七大能力->实践应用"], "answer_analysis": ["如果橘子的个数扩大到原来的$$3$$倍，那么余下的橘子数量将与余下苹果的相等； 即$$235\\times 3=705$$与$$270$$分别除以大班小朋友的人数，余数相同； 大班小朋友的人数是$$705-270=435$$的因数； 如果橘子的个数扩大到原来的$$2$$倍，那么余下的橘子数量将与余下的梨的相等； 即$$235\\times 2=470$$与$$180$$分别除以大班小朋友的人数，余数相同； 大班小朋友的人数是$$470-180=290$$的因数； 则大班小朋友的人数是$$435$$和$$290$$的公因数； $$(435,290)=5\\times 29=145$$； 大班小朋友的人数是$$145$$的因数：$$1$$、$$5$$、$$29$$、$$145$$ 当大班有$$1$$名小朋友时，$$270\\div 1=270$$，$$180\\div 1=180$$，$$235\\div 1=235$$； 余下$$0$$个苹果、$$0$$个梨、$$0$$个橘子，不符题意； 当大班有$$5$$名小朋友时，$$270\\div 5=54$$，$$180\\div 5=36$$，$$235\\div 5=47$$； 余下$$0$$个苹果、$$0$$个梨、$$0$$个橘子，不符题意； 当大班有$$29$$名小朋友时，$$270\\div 29=9\\cdots \\cdots 9$$，$$180\\div 29=6\\cdots \\cdots 6$$，$$235\\div 29=8\\cdots \\cdots 3$$； 余下$$9$$个苹果、$$6$$个梨、$$3$$个橘子，$$9:6:3=3:2:1$$，符合题意； 当大班有$$145$$名小朋友时，$$270\\div 145=1\\cdots \\cdots 125$$，$$180\\div 145=1\\cdots \\cdots 35$$，$$235\\div 145=1\\cdots \\cdots 90$$； 余下$$125$$个苹果、$$35$$个梨、$$90$$个橘子，$$125:35:90\\ne 3:2:1$$，不符题意； 经检验，大班有$$29$$名小朋友． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "406", "queId": "8e0a0c9be1e6408cb7eb3e1495a80637", "competition_source_list": ["2017年北京学而思杯四年级竞赛年度教学质量测评第19题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "有黑白卡片各$$6$$张，它们的正面分别写着数字$$1$$到$$6$$．把它们充分打乱顺序后，从中拿走$$2$$张（黑白各$$1$$张），把剩下的$$10$$张按照下列规则进行排列： 规则$$1$$：从左到右，按照数字从小到大的顺序排列； 规则$$2$$：当黑白卡片的数字相同时，黑卡片排在白卡片的左边． 把这$$10$$张卡片按照上述规则排列，保持顺序不变，翻到背面朝上，从左到右卡片颜色为：黑白黑白黑白白黑黑白． 请问：拿走的$$2$$张卡片上的数字分别为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "黑卡是$$3$$，白卡是$$4$$ "}], [{"aoVal": "B", "content": "黑卡是$$4$$，白卡是$$5$$ "}], [{"aoVal": "C", "content": "白卡是$$4$$，黑卡是$$5$$~~~~~~ "}], [{"aoVal": "D", "content": "白卡是$$5$$，黑卡是$$6$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["略 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2903", "queId": "bfdf1c2e769344869160b82ccb72263e", "competition_source_list": ["2014年全国华杯赛小学高年级竞赛复赛C卷第12题"], "difficulty": "2", "qtype": "single_choice", "problem": "某自然数减去$$39$$是一个完全平方数，减去$$144$$也是一个完全平方数，此自然数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$400$$、$$208$$、$$160$$ "}], [{"aoVal": "B", "content": "$$2848$$、$$400$$、$$160$$ "}], [{"aoVal": "C", "content": "$$2848$$、$$400$$、$$208$$、$$160$$ "}], [{"aoVal": "D", "content": "$$2848$$、$$400$$、$$208$$、$$190$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["设这两个完全平方数分别为$${{a}^{2}}$$和$${{b}^{2}}$$（$$a$$、$$b$$均为整数），则$${{a}^{2}}-{{b}^{2}}=144-39=105$$． 由平方差公式得：$$\\left( a+b \\right)\\left( a-b \\right)=105$$． 考虑到$$105=1\\times 105=3\\times 35=5\\times 21=7\\times 15$$，因此有$$4$$种可能： $$\\begin{cases}a+b=105 a-b=1 \\end{cases}$$，$$\\begin{cases}a+b=35 a-b=3 \\end{cases}$$，$$\\begin{cases}a+b=21 a-b=5 \\end{cases}$$，$$\\begin{cases}a+b=15 a-b=7 \\end{cases}$$ 解得：$$\\begin{cases}a=53 b=52 \\end{cases}$$，$$\\begin{cases}a=19 b=16 \\end{cases}$$，$$\\begin{cases}a=13 b=8 \\end{cases}$$，$$\\begin{cases}a=11 b=4 \\end{cases}$$． 原数为$${{a}^{2}}+39$$，对应$$4$$种可能：$$2848$$、$$400$$、$$208$$、$$160$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1178", "queId": "6221023c036e4605a77dd2bda14d0d19", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "毛毛用围棋子摆成一个三层空心方阵，最外一层每边有围棋子$$12$$个．摆这个方阵共用围棋子个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$ "}], [{"aoVal": "B", "content": "$$108$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$132$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["由��意知，用围棋子摆成一个三层空心方阵，最外一层每边有围棋子$$12$$个，由于相邻两层每边相差$$2$$个，则由外向里的两层每边分别是$$(12-2)$$个、$$(12-2\\times 2)$$个，根据``四周的个数$$=$$（每边的个数$$-1$$）$$\\times 4$$''可分别求得这三层棋子的个数，再相加就是所用的总个数，据此解答． 本题关键是求出每层的个数；方阵问题相关的知识点是：四周的个数$$=$$（每边的个数$$-1$$）$$\\times 4$$，每边的个数$$=$$四周的个数$$\\div 4+1$$，中实方阵的总个数$$=$$每边的个数$$\\times $$每边的个数，空心方阵的总个数$$=$$（最外层每边的个数$$-$$空心方阵的层数）$$\\times $$空心方阵的层数$$\\times 4$$，外层边长数$$^{2}-$$中空边长数$$^{2}=$$实面积数． 最外边一层棋子个数：$$(12-1)\\times 4=44$$（个）， 第二层棋子个数：$$(12-2-1)\\times 4=36$$（个）， 第三层棋子个数：$$(12-2\\times 2-1)\\times 4=28$$（个）， 摆这个方阵共用棋子：$$44+36+28=108$$（个）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1659", "queId": "69617415d9c14a4d9932242a43beaedd", "competition_source_list": ["2013年第25届广东广州五羊杯六年级竞赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "今天是星期四，从第二天算起，则第$$2013$$天是． ", "answer_option_list": [[{"aoVal": "A", "content": "星期一 "}], [{"aoVal": "B", "content": "星期二 "}], [{"aoVal": "C", "content": "星期三 "}], [{"aoVal": "D", "content": "星期四 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->求再过几日是周几问题"], "answer_analysis": ["$$2013\\div 7=287$$（组）$$\\cdots \\cdots 4$$（个），星期四后的第四天是星期一． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1619", "queId": "49f642e12cb14504811bdc0a1a6d1653", "competition_source_list": ["2018年华杯赛六年级竞赛模拟卷"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙两队都可以胜任一项工程，甲单独做需要$$20$$天，乙单独做需要$$35$$天。甲队做了几天之后，由于突发状况而退出，停工了$$1$$天，剩下的由乙队完成，竣工时一共用了$$30$$天。那么，乙队做了（ ）天。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$23$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->接力施工问题"], "answer_analysis": ["可以设乙队工作了$$x$$天，列方程得：$$\\frac{30-1-x}{20}+\\frac{x}{35}=1$$，解得：$$x=21$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2353", "queId": "8aac50a7519fa10a01519ff4869700ca", "competition_source_list": ["2016年全国华杯赛小学中年级竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "在除法算式中，被除数为$$2016$$，余数为$$7$$，则满足算式的除数共有（~~ ~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["某个数除$$2016$$余$$7$$，于是这个数整除$$2016-7=2009$$，$$2009={{7}^{2}}\\times {{41}^{1}}$$，所以$$2009$$共有$$6$$个约数，其中比$$7$$大的约数有$$4$$个（除了$$1$$和$$7$$），所以满足要求的除数共有$$4$$个． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1177", "queId": "1180db4484cd4c16b8e9c7f571ffe191", "competition_source_list": ["2020年新希望杯四年级竞赛决赛（8月）第7题", "2020年新希望杯四年级竞赛初赛（个人战）第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "光头强去赶集，去时步行，速度是$$15$$千米$$/$$时；回来时骑车，速度是$$30$$千米$$/$$时．光头强往返的平均速度是~\\uline{~~~~~~~~~~}~千米$$/$$时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$25$$ "}], [{"aoVal": "E", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->平均速度->设数法", "Overseas Competition->知识点->应用题模块"], "answer_analysis": ["路程$$=$$时间$$\\times$$速度， 设路程为$$30$$千米，那么去时的时间是$$30\\div15=2$$（时）， 回来的时间为$$30\\div30=1$$（时）， 因此往返的平均速度是$$(30+30)\\div(2+1)=60\\div3=20$$（千米$$/$$时）． 故答案为$$20$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3189", "queId": "344298b2b3a646f58e8df758ff357ae5", "competition_source_list": ["2017年河南郑州联合杯竞赛决赛第6题2分", "2021年河南郑州二七区京广实验学校小升初第6题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "小红买售价$$4.4$$元的钢笔一支，根据你的生活经验，结合人民币币值的特点，下列付钱方式不合理的是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "付出$$4.5$$元，找回$$0.1$$元 "}], [{"aoVal": "B", "content": "付出$$4.7$$元，找回$$0.3$$元 "}], [{"aoVal": "C", "content": "付出$$5.4$$元，找回$$1.0$$元 "}], [{"aoVal": "D", "content": "付出$$10$$元，找回$$5.6$$元 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["B选项不合理． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "818", "queId": "84a8ad3470834fda84506b305b9e9068", "competition_source_list": ["2019年江苏南京栖霞区南外仙林分校小学部六年级下学期单元测试《正比例和反比例》第29题2分", "2015年全国迎春杯小学高年级竞赛复赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果一个自然数的各位数字能够分成两组，使得每组中的数字之和相等，则称这个数为``均衡数''．例如$$25254$$是``均衡数''，因为$$5+2+2=4+5$$．如果相邻的两个自然数都是``均衡数''，则称这对``均衡数''为``孪生均衡数''．那么最小的一对``孪生均衡数''的和是 ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$999$$ "}], [{"aoVal": "B", "content": "$$1099$$ "}], [{"aoVal": "C", "content": "$$2099$$ "}], [{"aoVal": "D", "content": "$$3099$$ "}]], "knowledge_point_routes": ["课内体系->七大能力->逻辑分析", "拓展思维->拓展思维->数论模块->奇数与偶数"], "answer_analysis": ["两位数没有符合要求的数，$$99$$、$$100$$亦不符合，故知至少为三位数．两个相邻数数字和都是偶数，说明必有进位，且三位数必然只进$$1$$次位（数字和加$$1$$再减$$9$$），即这两个数是$$\\overline{ab9}$$和$$\\overline{a\\left( b+1 \\right)0}$$，必有$$a+b=9$$和$$a=b+1$$，故这两个数为$$549$$和$$550$$．$$549+550=1099$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "822", "queId": "6e1934cdf0904575a44016317763ae7a", "competition_source_list": ["2006年第4届创新杯四年级竞赛初赛A卷第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$1\\times 1+2\\times 2+3\\times 3+\\cdots +2005\\times 2005+2006\\times 2006$$的个位数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["$$1\\times 1=1$$， $$2\\times 2=4$$， $$3\\times 3=9$$， $$4\\times 4=16$$， $$5\\times 5=25$$， $$6\\times 6=36$$， $$7\\times 7=49$$， $$8\\times 8=64$$， $$9\\times 9=81$$， $$10\\times 10=100$$， 我们可以把每$$10$$个数当作一个整体，它们的和的个位数是$$5$$， 从$$1$$到$$2006$$可以看作$$200$$个上面的整体和剩下$$6$$个数的乘积， 所以$$200$$个整体的和的个位数字应该是$$5$$乘$$200$$的个位，即为$$0$$， 显然剩下的$$6$$个数的个位是$$1$$， 所以最终的个位就应该为$$1$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1478", "queId": "51ac51ed624a4198af6ef7b4eb35d22c", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（四）第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "市政府要修建一条新地铁线路，其中一段交给甲、乙两个工程队完成．甲工程队每天能完成整个工程任务的$$\\frac{1}{16}$$，乙工程队每天能完成整个工程任务的$$\\frac{1}{24}$$．现在两队合作$$4$$天后，甲工程队被抽调离开，由乙队独自完成剩下的部分需要（~ ）天．~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["甲、乙两队第一阶段完成了$$\\left( \\frac{1}{16}+\\frac{1}{24} \\right)\\times 4=\\frac{5}{12}$$，乙队第二阶段需要$$\\left( 1-\\frac{5}{12} \\right)\\div \\frac{1}{24}=14$$天． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3192", "queId": "4fc891b2b711463f9ac81d6a11049c49", "competition_source_list": ["2008年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$1$$到$$50$$这五十个自然数中，取两个不同的数相加，要使它们的和大于$$50$$，共有( )种不同的取法。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$560$$ "}], [{"aoVal": "B", "content": "$$605$$ "}], [{"aoVal": "C", "content": "$$625$$ "}], [{"aoVal": "D", "content": "$$725$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->字典排序法"], "answer_analysis": ["两个数中小的数为$$1$$，有$$\\left( 1\\text{，}50 \\right)$$，$$1$$种取法； 同理，小的数为$$2$$，有$$\\left( 2\\text{，}50 \\right)$$，$$\\left( 2\\text{，}49 \\right)$$，$$2$$种取法； 小的数为$$3$$，有$$\\left( 3\\text{，}50 \\right)$$，$$\\left( 3\\text{，}49 \\right)$$，$$\\left( 3\\text{，}48 \\right)$$，$$3$$种取法； $$\\cdots\\cdots$$ 小的数为$$25$$，有$$\\left( 25\\text{，}50 \\right)$$，$$\\left( 26\\text{，}50 \\right)$$，$$\\cdots$$，$$\\left( 25\\text{，}26 \\right)$$，$$25$$种取法； 小的数为$$26$$，有$$\\left( 26\\text{，}50 \\right)$$，$$\\left( 26\\text{，}49 \\right)$$，$$\\cdots$$，$$\\left( 26\\text{，}27 \\right)$$，$$24$$种取法； $$\\cdots\\cdots$$ 小的数为$$49$$，有$$\\left( 49\\text{，}50 \\right)$$，$$1$$种取法。 所以共有$$1+2+3+\\cdots +25+24+\\cdots +3+2+1$$ $$=2\\times \\left( 1+2+3+\\cdots +25 \\right)-25=2\\times \\frac{25\\times 26}{2}-25=25\\times 26-25$$ $$=25\\times 25=625$$（种）不同的取法。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3450", "queId": "f3d356c1be3f4cf8a8d9c7ee36578be9", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（三）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$15$$个相同的悠悠球分装到四个相同的纸盒中，要求每个盒子里至少装一个，且每个盒子装的数量都不相同，一共有种不同的装法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["将$$15$$拆成四个不同的加数，按最小的加数分类数：$$15=1+2+3+9=1+2+4+8=1+2+5+7=1+3+4+7=1+3+5+6=2+3+4+6$$，共$$6$$种不同的分类． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2758", "queId": "457a7909d46545bc8e5dea539577bb87", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面计算结果等于$$9$$的是（ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3\\times 3\\div 3+3$$ "}], [{"aoVal": "B", "content": "$$3\\div 3+3\\times 3$$ "}], [{"aoVal": "C", "content": "$$3\\times 3-3+3$$ "}], [{"aoVal": "D", "content": "$$3\\div 3+3\\div 3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["解：$$\\rm A$$、$$3\\times 3\\div 3+3$$ $$=3+3$$ $$=6$$ $$\\rm B$$、$$3\\div 3+3\\times 3$$ $$=1+9$$ $$=10$$ $$\\rm C$$、$$3\\times 3-3+3$$ $$=9-3+3$$ $$=9$$ $$\\rm D$$、$$3\\div 3+3\\div 3$$ $$=1+1$$ $$=2$$ 故选：$$\\rm C$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "355", "queId": "91d8654287004305899244a2daf03ce3", "competition_source_list": ["2014年全国学而思杯一年级竞赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$5$$名分别来自美国、俄国、中国、日本和韩国的运动员参加冬奥会滑雪决赛，比赛结束后： 美国运动员说：俄国运动员紧跟在我后面； 俄国运动员说：我才不是最后一名； 中国运动员说：我比日本国人和美国人都快； 韩国运动员说：有三个人比我先到达终点． 那么，~\\uline{~~~~~~~~~~}~国运动员是第一名． ", "answer_option_list": [[{"aoVal": "A", "content": "美国 "}], [{"aoVal": "B", "content": "俄国 "}], [{"aoVal": "C", "content": "中国 "}], [{"aoVal": "D", "content": "日本 "}], [{"aoVal": "E", "content": "韩国 "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->逻辑推理->比较型逻辑推理"], "answer_analysis": ["中国比日本、美国快，美国比俄国快，所以日本、美国、俄国都不是第一，而韩国也不是第一个到的，所以中国是第一． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "613", "queId": "0d3414b355da4790ac34efd69b29b070", "competition_source_list": ["竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "记$$A=8\\times 8\\times 8\\times \\cdots\\times 8$$（共$$25$$个$$8$$相乘），$$A\\div 9$$的余数为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数的性质->余数找规律"], "answer_analysis": ["$$8$$的若干次方除以$$9$$的余数是以$$8$$、$$1$$为周期的数，$$25$$为奇数，对应余数为$$8$$。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "474", "queId": "fc4a9333bef447ddb9abbc0514e49ccd", "competition_source_list": ["2012年第10届创新杯四年级竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "显示在电子钟上的时间是$$5:55$$，下一次电子钟上显示的时间又是全部相同的数字，还要过分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$71$$ "}], [{"aoVal": "B", "content": "$$255$$ "}], [{"aoVal": "C", "content": "$$316$$ "}], [{"aoVal": "D", "content": "$$377$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["因为分钟的十位最大为$$5$$，故下一次数字相同的时刻为$$11:11$$，$$11:11$$距$$5:55$$有$$5$$小时$$16$$分，即$$316$$分钟．所以选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "419", "queId": "c9301aa7fe654db6b01ecddf387f7e73", "competition_source_list": ["2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛"], "difficulty": "3", "qtype": "single_choice", "problem": "一个程序员给编号为$$1-40$$的$$40$$个机器人写好程序，让它们来判断某句话的对错．程序员为了提高准确率，在它们都进行过一次判断后，让从编号为$$1$$的机器人开始，修改自己的答案．如果其余$$39$$个机器人回答``对''多，则自己的回答也为``对''，否则回答``错''．如果最开始有$$20$$个机器人回答``对''，$$20$$个机器人回答``错''，并且一号机器人回答``对''．那么这$$40$$个机器人按序修改自己的答案后，有多少人回答``对''？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->答案（数字）正误问题"], "answer_analysis": ["由于第一个机器人回答对，所以说明除了第一个外的机器人，开始时$$19$$个回答对，$$20$$个回答错，第一个机器人根据判定需要改答案，其他机器根据第一个机器人改了答案判断，应该都回答错． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1385", "queId": "4c52eb9d09664c81b4576f0d173771cd", "competition_source_list": ["2006年五年级竞赛创新杯", "2006年第4届创新杯五年级竞赛初赛A卷第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "一本书中间有一张被撕掉了，余下各页码数之和正好等于1000，这本书原有( )页. ", "answer_option_list": [[{"aoVal": "A", "content": "40 "}], [{"aoVal": "B", "content": "45 "}], [{"aoVal": "C", "content": "48 "}], [{"aoVal": "D", "content": "50 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->页码问题"], "answer_analysis": ["设撕掉这张的两个页码的和为x，原书有n页，则所有页码和为$$\\left( 1+2+3+\\cdots +n \\right)-x=\\frac{n\\left( n+1 \\right)}{2}-x=1000$$，只有当$$n=45$$时，其所有页码和为$$45\\times 46\\div 2=1035$$，恰好大于1000且最接近1000，故撕掉这张纸的两个页码的和为$$1035-1000=35$$，撕掉的这张的两个页码为17，18.所以这本书原有45页，故选B. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "370", "queId": "89d16a180f4149a2893ff61e16a2ce08", "competition_source_list": ["2018年第17届春蕾杯一年级竞赛初赛第14题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "四耳穴学校也举办了一场足球比赛，皮皮、蛋君和大强所在的三个队伍分别获得了金、银、铜牌．根据下面三句话，猜一猜他们分别获得什么奖牌？ 皮皮：``我分到的不是金牌．'' 大强：``他们俩的队伍一支获得金牌，一支获得银牌．'' 请问，蛋君所在的队伍获得什么奖牌？ ", "answer_option_list": [[{"aoVal": "A", "content": "金 "}], [{"aoVal": "B", "content": "银 "}], [{"aoVal": "C", "content": "铜 "}]], "knowledge_point_routes": ["拓展思维->能力->推理推导->言语逻辑推理"], "answer_analysis": ["推理题目 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1499", "queId": "6cefec8852384b2b9a5efce51c89d04d", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果某国物价下降$$50 \\%$$，那么原来买$$1$$件东西的钱现在就能买$$2$$件．$$1$$件变为$$2$$件增加了$$100 \\%$$，这就相当于该国居民手中的钱增值了$$100 \\%$$，如果物价上涨了$$25 \\%$$，那么相当于手中的钱贬值了$$ \\%$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$13$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->折扣问题"], "answer_analysis": ["不难列出算式：$$\\left[ 1-\\frac{1}{1+25 \\%} \\right]\\times100 \\%=20 \\%$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2783", "queId": "ccdc9313e6e94306984ad9cbdfe1c13a", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$9$$个数：$$1$$，$$\\frac{2}{5}$$，$$7$$，$$8$$，$$\\frac{5}{4}$$，$$1.\\dot{2}$$，$$15$$，$$3.75$$，$$0.\\dot{7}$$中取一个数作被除数，再取另外两个数，用它们的和作除数，使商为整数，这样的算式存在吗？ ", "answer_option_list": [[{"aoVal": "A", "content": "存在 "}], [{"aoVal": "B", "content": "不存在 "}], [{"aoVal": "C", "content": "不能确定 "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["$$8\\div \\left( 7+1 \\right)=1$$，$$15\\div \\left( 7+8 \\right)=1$$，$$15\\div \\left( \\frac{5}{4}+3.75 \\right)=3$$，$$8\\div \\left( 1.\\dot{2}+0.\\dot{7} \\right)=4$$，$$7\\div \\left( \\frac{2}{5}+1 \\right)=5$$等． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "426", "queId": "7d7d580bf5024f47a8466e6c5e5aaeab", "competition_source_list": ["2014年全国创新杯小学高年级五年级竞赛第4题5分", "2016年创新杯小学高年级五年级竞赛训练题（四）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$1$$，$$2$$，$$\\cdot \\cdot \\cdot $$，$$79$$这$$79$$个数中，选出若干个数来，使得选出的这些数中，任何两个数之差（大减小）都不等于$$1$$，$$2$$，$$4$$．那么至多可以选出（~ ~ ~ ）个数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$26$$ "}], [{"aoVal": "B", "content": "$$27$$ "}], [{"aoVal": "C", "content": "$$28$$ "}], [{"aoVal": "D", "content": "$$29$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["取$$1$$，$$4$$，$$7$$，$$\\cdot \\cdot \\cdot $$，$$79$$，最多有$$(79-1)\\div 3+1=27$$个． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2564", "queId": "674e71ddaecb411086cd03ac2d59b68c", "competition_source_list": ["2008年四年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "101个数之和为2008;把第一个数减1，第2个数加2，第3个数减3，\\ldots，第100个数加100，第101个数减101，所得的101个新数之和为( ). ", "answer_option_list": [[{"aoVal": "A", "content": "1957 "}], [{"aoVal": "B", "content": "1959 "}], [{"aoVal": "C", "content": "2008 "}], [{"aoVal": "D", "content": "2009 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之分组法"], "answer_analysis": ["所得的101个新数之和为: $$2008-1+2-3+4-\\cdots -99+100-101=2008+\\underbrace{1+1+1+\\cdots +1}_{50个1}-101=1957$$ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1138", "queId": "3881252f9d0d4c4a9d1b2928d9baef9c", "competition_source_list": ["2014年全国创新杯小学高年级五年级竞赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某出租车的收费标准是：起步价$$7$$元（$$3$$千米以内包括$$3$$千米），$$3$$千米后每增加$$1$$千米加收$$2.4$$元（不足$$1$$千米按$$1$$千米计），某人乘坐出租车从甲地到乙地付车费$$19$$元，那么下列句子中正确的是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "甲地到乙地的路程为$$8$$千米 "}], [{"aoVal": "B", "content": "甲乙两地路程为$$7$$千米 "}], [{"aoVal": "C", "content": "甲乙两地路程��于$$7$$千米但不超过$$8$$千米 "}], [{"aoVal": "D", "content": "甲乙两地路程最多为$$8$$千米 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$19-7=12$$，$$12\\div 2.4=5$$，故甲、乙两地路程大于$$7$$千米但不超过$$8$$千米． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1955", "queId": "e55364bd3a524623b8cec30c05955744", "competition_source_list": ["2018年第21届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有大、中、小三筐菠萝，小筐装的是中筐的一半，中筐比大筐少装$$24$$千克，大筐装的是小筐的$$5$$倍．大、中、小三筐一共装菠萝千克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$56$$ "}], [{"aoVal": "C", "content": "$$64$$ "}], [{"aoVal": "D", "content": "$$80$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["根据题意分析可知，因为小框装的是中筐的一半且大框装的是小筐的$$5$$倍可知， 大筐装的是中筐的$$5\\div 2=2.5$$倍， 又因为中框比大框少装$$24$$千克， 用差倍问题求出小筐的能装：$$24\\div \\left( 5-2 \\right)=8$$（千克）， 中筐装：$$8\\times 2=16$$（千克）， 大筐装：$$16\\times 2=32$$（千克）， 所以三个筐一共能装：$$8+16+32=56$$（千克）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3063", "queId": "b8bc913057414da49c9a5bbe39260f04", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "西元$$1202$$年，意大利数学家斐波那契（Fibonacci，$$1170\\sim 1250$$），在所著的《算术书（Liber Abaci）》中，提出了以下有趣的问题：一对兔子在它们出生整整两个月以后可以生一对小兔子（雌、雄兔各一只），其后每隔一个月又可再生一对小兔子．现有一对刚生下来的小兔子﹐如果兔子都不死亡，请问经过$$12$$个月后（即在第$$13$$个月初）总共有多少对兔子？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$144$$ "}], [{"aoVal": "B", "content": "$$233$$ "}], [{"aoVal": "C", "content": "$$234$$ "}], [{"aoVal": "D", "content": "$$235$$ "}], [{"aoVal": "E", "content": "$$377$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->斐波那契数列（兔子数列）"], "answer_analysis": ["逐月推算，我们可以得到下面的一列数： $$1$$、$$1$$、$$2$$、$$3$$、$$5$$、$$8$$、$$13$$、$$21$$、$$34$$、$$55$$、$$89$$、$$144$$、$$233$$、$$\\cdots $$ 其中第$$13$$个数为$$233$$，所以经过$$12$$个月后共有$$233$$对兔子． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2778", "queId": "6dae651848b348fb8a9167fa5de9c821", "competition_source_list": ["2017年环亚太杯一年级竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "找规律，选一选： $$6$$，$$12$$，$$18$$，$$24$$，~\\uline{~~~~~~~~~~}~，$$36$$，~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "28，40 "}], [{"aoVal": "B", "content": "28，42 "}], [{"aoVal": "C", "content": "30，40 "}], [{"aoVal": "D", "content": "30，42 "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数"], "answer_analysis": ["每个数之间差了一个6 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1033", "queId": "0a56fcc8142043b1acd341867cfe686c", "competition_source_list": ["2010年全国华杯赛竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "两个水池内有金鱼若干条，数目相同．亮亮和红红进行捞鱼比赛，一共比两次，每次捞一池，第一个水池内的金鱼被捞完时，亮亮和红红所捞到的金鱼数目比是$$3:4$$；捞完第二个水池内的金鱼时，亮亮比第一次多捞$$33$$条，与红红捞到的金鱼数目比是$$5:3$$，每个水池内有金鱼条． ", "answer_option_list": [[{"aoVal": "A", "content": "$$112$$ "}], [{"aoVal": "B", "content": "$$168$$ "}], [{"aoVal": "C", "content": "$$224$$ "}], [{"aoVal": "D", "content": "$$336$$ "}]], "knowledge_point_routes": ["课内体系->七大能力->实践应用", "拓展思维->知识点->应用题模块->比例应用题->按比分配"], "answer_analysis": ["本题原意为两人捞第一个水池内的金鱼，亮亮与红红捞到得金鱼数之比为$$3:4$$，共捞了$$7$$份；这样���第二个水池内涝完后，亮亮和红红所捞到的金鱼数目比是$$5:3$$，共捞了$$8$$份；由于两个水池内的鱼的量是相等的，则找$$\\left[ 7,8 \\right]=56$$． 两个水池内的总份数，均统一为$$56$$份，则在捞第一个水池时，亮亮和红红所捞到的金鱼数目之比为：$$3:4=24:32$$； 捞第二个水池时，亮亮和红红所捞到的金鱼数目之比为：$$5:3=35:21$$． 亮亮第一次捞了$$24$$份，第二次捞了$$35$$份，差了$$11$$份，为$$33$$条，则$$1$$份为$$3$$条． 所以原来每隔水池内的金鱼为：$$3\\times56=168$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1878", "queId": "c05b75ee5df04182b9285c9a8c604f6e", "competition_source_list": ["2008年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "$$A$$、$$B$$、$$C$$、$$D$$四个数，每次去掉一个数，将其余三个数求平均数.这样计算了四次，得到下面四个数:23，26，30，33，$$A$$、$$B$$、$$C$$、$$D$$四个数的平均数是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "24 "}], [{"aoVal": "B", "content": "26 "}], [{"aoVal": "C", "content": "28 "}], [{"aoVal": "D", "content": "30 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型"], "answer_analysis": ["不妨设$$A\\geqslant B\\geqslant C\\geqslant D$$，则 $$B\\text{+}C\\text{+}D=23\\times 3=69$$ $$A\\text{+}C\\text{+}D=26\\times 3=78$$ $$A\\text{+}B\\text{+}D=30\\times 3=90$$ $$A\\text{+}B\\text{+}C=33\\times 3=99$$ 四个等式相加得$$3\\times \\left( A\\text{+}B\\text{+}C\\text{+}D \\right)=336$$，从而$$A\\text{+}B\\text{+}C\\text{+}D\\text{=}112$$，所以$$A$$、$$B$$、$$C$$、$$D$$的平均数为$$112\\div 4\\text{=}28$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1687", "queId": "80297962280f4c76bc3b146c19efe163", "competition_source_list": ["2009年第9届全国新希望杯三年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "一串彩珠按``$$1$$红$$2$$蓝$$4$$黄''的顺序依次排列，第$${32}$$颗彩珠是~\\uline{~~~~~~~~~~}~色． ", "answer_option_list": [[{"aoVal": "A", "content": "红 "}], [{"aoVal": "B", "content": "蓝 "}], [{"aoVal": "C", "content": "黄 "}], [{"aoVal": "D", "content": "白 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["将``$$1$$红$$2$$蓝$$4$$黄''看作一组，共有$$7$$颗，$$32 \\div 7 = 4 \\cdots ~\\cdots 4$$，说明这一组共出现了$$4$$次，余数是$$4$$说明还依次出现了$$1$$颗红珠、$$2$$颗蓝珠、$$1$$颗黄珠，所以第$$32$$颗彩珠是黄色． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3482", "queId": "fec52e32149c45ddb0993819e3d4c890", "competition_source_list": ["2020年希望杯一年级竞赛模拟第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "每两人之间都握一次手（不能重复计数），$$5$$个小朋友一共要握次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["因为由题干可知，一共有$$4$$个人，每两人之间需要握一次手，我们可以将这$$4$$个人分别用$$A$$、$$B$$、$$C$$和$$D$$来指代，则所有的握手情况为：$$AB$$，$$AC$$，$$AD$$，$$BC$$，$$BD$$和$$CD$$，所以一共要握$$6$$次，故本题答案为$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1433", "queId": "43e85924e22c4eea8f7021e970dd139b", "competition_source_list": ["2018年四川成都锦江区四川师范大学附属第一实验中学小升初模拟8第3题3分", "2018年全国小学生数学学习能力测评六年级竞赛初赛第8题3分", "2018年湖南长沙小升初数学入学试卷第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "王师傅加工一批零件，$$\\frac{1}{2}$$小时加工了这批零件的$$\\frac{3}{8}$$，全部加工完还需要( )小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{10}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{6}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["首先根据王师傅加工一批零件，$$\\frac{1}{2}$$小时加工了这批零件的$$\\frac{3}{8}$$，工作效率$$=$$工作量$$\\div $$工作时间，求出每小时加工这批零件的几分之几；求出剩下的工作量，然后根据工作时间$$=$$工��量$$\\div $$工作效率，求出全部加工完还需要多少小时即可． 解；$$\\frac{3}{8}\\div \\frac{1}{2}=\\frac{3}{4}$$ $$(1-\\frac{3}{8})\\div \\frac{3}{4}$$ $$=\\frac{5}{8}\\div \\frac{3}{4}$$ $$=\\frac{5}{6}$$(小时) 答:全部加工完还需要$$\\frac{5}{6}$$小时． 故选$$\\rm D$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2995", "queId": "ce36ce976f3b45339902bed7a3c60882", "competition_source_list": ["2018年第8届北京学而思综合能力诊断二年级竞赛年度教学质量第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "算式$$\\textasciitilde8+9\\times 5$$的正确结果是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$85$$ "}], [{"aoVal": "B", "content": "$$55$$ "}], [{"aoVal": "C", "content": "$$53$$ "}], [{"aoVal": "D", "content": "$$62$$~ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->四则混合运算"], "answer_analysis": ["先算乘除，再算加减，$$8+9\\times 5=8+45=53$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2898", "queId": "8de584514277495c9cb6c6939d0d9f24", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛A卷第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$15-18$$这些数的平方的倒数乘，可约分的数最多． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数基础->分数的约分"], "answer_analysis": ["$${{15}^{2}}=225$$，$${{16}^{2}}=256$$，$${{17}^{2}}=289$$，$${{18}^{2}}=324$$， $$\\text{A}$$. $$\\frac{1}{225}\\times 2=\\frac{2}{225}$$，$$\\frac{1}{256}\\times 2=\\frac{2}{256}=\\frac{1}{128}$$， $$\\frac{1}{289}\\times 2=\\frac{2}{289}$$，$$\\frac{1}{324}\\times 2=\\frac{2}{324}=\\frac{1}{162}$$； $$\\text{B}$$. $$\\frac{1}{225}\\times 3=\\frac{3}{225}=\\frac{1}{75}$$，$$\\frac{1}{256}\\times 2=\\frac{3}{256}$$， $$\\frac{1}{289}\\times 3=\\frac{3}{289}$$，$$\\frac{1}{324}\\times 3=\\frac{3}{324}=\\frac{1}{108}$$； $$\\text{C}$$. $$\\frac{1}{225}\\times 4=\\frac{4}{225}$$，$$\\frac{1}{256}\\times 4=\\frac{4}{256}=\\frac{2}{128}=\\frac{1}{64}$$ $$\\frac{1}{289}\\times 4=\\frac{4}{289}$$，$$\\frac{1}{324}\\times =\\frac{4}{324}=\\frac{2}{162}=\\frac{1}{81}$$； $$\\text{D}$$. $$\\frac{1}{225}\\times 5=\\frac{5}{225}=\\frac{1}{45}$$，$$\\frac{1}{256}\\times 5=\\frac{5}{256}$$， $$\\frac{1}{289}\\times 5=\\frac{5}{289}$$，$$\\frac{1}{324}\\times 5=\\frac{5}{324}$$． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3295", "queId": "48f5125a54f3493ab47a92bc40226b03", "competition_source_list": ["2017年全国小学生数学学习能力测评四年级竞赛初赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "用$$1$$、$$2$$、$$3$$、$$4$$这四个数字可以组成许多四位数，将它们从小到大依次排列，那么$$4123$$这个四位数排在第个位置上． ", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由$$1$$、$$2$$、$$3$$、$$4$$这四个数字组成的四位数有： $$1234$$；$$1243$$；$$1324$$；$$1342$$；$$1423$$；$$1432$$； $$2134$$；$$2143$$；$$2314$$；$$2341$$；$$2413$$；$$2431$$； $$3124$$；$$3142$$；$$3214$$；$$3241$$；$$3412$$；$$3421$$； $$4123$$；$$4132$$；$$4213$$；$$4231$$；$$4312$$；$$4321$$， 所以将它们从小到大的顺序排列，$$4123$$是第$$19$$个． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "366", "queId": "b66a3e33a16e4782a8a2adfdf91221e4", "competition_source_list": ["2011年第7届全国新希望杯小学高年级六年级竞赛第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "盒子里装有分别写着$$1$$，$$2$$，$$3$$，$$4$$，\\ldots，$$100$$的黄色卡片各一张，我们称如下操作为一次操作；从盒子里取出$$m(7\\leqslant m\\leqslant 10)$$张卡片，算出这$$m$$张卡片上各数之和减去$$27$$的差，将写在一张红色卡片上（不放回）．若干次操作之后，盒子里的卡片全部被取出，若所有红色卡片上的数字之和为$$n$$，那么$$n$$的最大可能值减去最小可能值等于（~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$108$$ "}], [{"aoVal": "B", "content": "$$96$$ "}], [{"aoVal": "C", "content": "$$88$$ "}], [{"aoVal": "D", "content": "$$81$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->操作问题->数字操作"], "answer_analysis": ["先算卡片的总值． 最小值$$100\\div 10=10$$次，共减$$10\\times 27$$； 最多$$14$$次，共减去$$14\\times 27$$， 因此差为$$(14-10)\\times 27=108$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1014", "queId": "02e9303452034cba82dda970fcac4752", "competition_source_list": ["2015年全国迎春杯四年级竞赛初赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "王伯伯养了一些鸡、兔和鹤．其中鹤白天双足站立，夜间则单足站立；鸡晚上睡觉时则把头藏起来．细心的悦悦发现：不论白天还是晚上，足数和头数的差都一样，那么，如果白天悦悦可以数出$$56$$条腿，晚上会数出~\\uline{~~~~~~~~~~}~个头． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题->分组法解鸡兔同笼->腿倍型"], "answer_analysis": ["白天比晚上多了一个鸡头，还多了一只鹤脚；无论晚上还是白天，足数和头数的差都一样，所以鹤的数量和鸡的数量是一样的．将鸡和鹤打一个包，则在白天这个包和兔子腿数一样为$$4$$，在晚上这个包和兔子头数一样为$$1$$；则可以得出晚上的头数为$$56\\div 4=14$$（个）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "166", "queId": "a70a1835e06d4b0d80b3b5e910c6a7fb", "competition_source_list": ["2016年全国小学生数学学习能力测评四年级竞赛初赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$100$$个球放到$$36$$个盒子里．已知任何连续的$$5$$个盒子中共有$$14$$个球，问最后一个盒子里有多少个球？ Put a total of $100$ balls into $36$ boxes. Given that any consecutive $5$ boxes have a sum of $14$ balls, the last box has~\\uline{~~~~~~~~~~}~balls. ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->组合模块->操作与策略->统筹规划"], "answer_analysis": ["$36\\div5=7R1$. Thus, for the first $35$ boxes, there is a total of $$14\\times7=98$$ balls. Therefore, there are $100-98=2$ balls. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2762", "queId": "84410d2730cf47b5b53ea6ad91c8bf91", "competition_source_list": ["六年级其它导引", "2015年世界少年奥林匹克数学竞赛六年级竞赛复赛A卷第10题10分", "2018年陕西西安碑林区西安市铁一中学小升初（二十六）第20题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "【补4】 计算：$$\\frac{{{1}^{2}}+{{2}^{2}}}{1\\times 2}+\\frac{{{2}^{2}}+{{3}^{2}}}{2\\times 3}+\\cdots +\\frac{{{18}^{2}}+{{19}^{2}}}{18\\times 19}+\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$38\\frac{19}{20}$$ "}], [{"aoVal": "B", "content": "$$38\\frac{17}{20}$$ "}], [{"aoVal": "C", "content": "$$38\\frac{13}{20}$$ "}], [{"aoVal": "D", "content": "$$38\\frac{1}{20}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["算式中的分母是裂项计算的最基本形式，但分子比较复杂，我们可以从前几项找找规律： $$\\frac{{{1}^{2}}+{{2}^{2}}}{1\\times 2}=\\frac{5}{2}=2\\frac{1}{2}$$，$$\\frac{{{2}^{2}}+{{3}^{2}}}{2\\times 3}=\\frac{13}{6}=2\\frac{1}{6}$$，$$\\frac{{{3}^{2}}+{{4}^{2}}}{3\\times 4}=\\frac{25}{12}=2\\frac{1}{12}$$． 我们发现一规律：每一项减去$$2$$后，分子就变成了$$1$$． 再来试试最后一项：$$\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}=\\frac{761}{380}=2\\frac{1}{380}$$， 也满足这个规律，这是为什么呢？ 观察每一项的分子和分母，我们发现分子的每个加数都与分母大小接近，可以做如下变形： $$\\frac{{{19}^{2}}+{{20}^{2}}}{19\\times 20}=\\frac{19\\times \\left( 20-1 \\right)+20\\times \\left( 19+1 \\right)}{19\\times 20}$$ $$=\\frac{19\\times 20\\times 2+\\left( 20-19 \\right)}{19\\times 20}$$ $$=2+\\frac{1}{19\\times 20}$$． 算式中的每一项都能像上面一样进行变形，所以： 原式$$=2\\frac{1}{1\\times 2}+2\\frac{1}{2\\times 3}+\\cdots +2\\frac{1}{19\\times 20}$$ $$=2\\times 19+\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{19\\times 20}$$ $$=38+1-\\frac{1}{20}$$ $$=38\\frac{19}{20}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3387", "queId": "adba0f59ad464189bcf144eba4e7c5ae", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛初赛第8题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "整数$$146$$和$$234$$的三个数位上数字的乘积都是$$24$$（注：$$1\\times 4\\times6 =24$$，$$2\\times 3\\times 4=24$$）．那么共有个三位数其各位数字的乘积是$$72$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$26$$ "}], [{"aoVal": "D", "content": "$$27$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["$$1\\times 8\\times 9=72$$， ∴各位数字乘积为$$72$$的有 $$189$$，$$198$$，$$819$$，$$891$$，$$918$$，$$981$$，$$6$$个． ∵$$2\\times 4\\times 9=72$$， ∴三位数各位数字乘积为$$72$$的有 $$249$$，$$294$$，$$429$$，$$492$$，$$924$$，$$942$$，$$6$$个． ∵$$3\\times 3\\times 8=72$$， ∴三位数各位数字乘积为$$72$$的有 $$338$$，$$383$$，$$833$$，$$3$$个． ∵$$3\\times 4\\times 6=72$$， ∴三位数各位数字乘积为$$72$$的有 $$346$$，$$364$$，$$463$$，$$436$$，$$643$$，$$634$$，共$$6$$个． ∵$$2\\times 6\\times 6=72$$， ∴三位数各位数字乘积为$$72$$的有 $$266$$，$$626$$，$$662$$，$$3$$个． 综上所述共有$$6+6+3+6+3=24$$个． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1448", "queId": "838ac350937441018f8450369233741c", "competition_source_list": ["2020年广东广州羊排赛五年级竞赛第8题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "$$2020$$年元旦是周三，以这天为第$$1$$天，从这天起的第天是周三． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$105$$ "}], [{"aoVal": "C", "content": "$$205$$ "}], [{"aoVal": "D", "content": "$$366$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["根据周期问题的公式：总数$$\\div $$周期$$=$$周期数$$\\cdots \\cdots $$余数，余数是几，就是周期中的第几个．本题最后这一天是周三，刚好是周期中的第一天，说明余数就是$$1$$，又因为一周有$$7$$天，依次判断选项中的数除以$$7$$的余数． $$\\text{A}$$.$$15\\div 7=2$$（周）$$\\cdots \\cdots 1$$（天）；$$\\text{B}$$.$$105\\div 7=15$$（周）； $$\\text{C}$$.$$205\\div 7=29$$（周）$$\\cdots \\cdots 2$$（天）；$$\\text{D}$$.$$366\\div 7=52$$（周）$$\\cdots \\cdots 2$$（天）． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "716", "queId": "634956ef59c2495a8302d31e49b036c1", "competition_source_list": ["2006年第4届创新杯六年级竞赛复赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "若数$$n=20\\times 30\\times 40\\times 50\\times 60\\times 70\\times 80\\times 90\\times 100\\times 110\\times 120\\times 130$$，则不是$$n$$的约数的最小质数是( )． ", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "非上述答案 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["原算式中有约数$$2$$，$$3$$，$$5$$，$$7$$，$$11$$，$$13$$，没有约数$$17$$，那么不是$$n$$的约数的最小质数是$$17$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3310", "queId": "ff8080814502fa2401450753fb4d09ed", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "在纸上任意写一个自然数，把这张纸旋转$$180$$度，数值不变，如$$0$$、$$11$$、$$96$$、$$888$$等，我们把这样的数称为``神马数''．在所有五位数中共有（~~~~~~~ ）个不同的``神马数''． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->组数问题->有特殊要求的组数问题"], "answer_analysis": ["设这个数为$$\\overline{ABCBA}$$，$$A$$位可以填$$11$$，$$ 88$$，$$ 69$$，$$96$$，$$4$$种情况，$$B$$位可以填$00$$，$$11$$，$$88$$，$$69$$，$$96$，$$5$$种情况，$$C$$位可以填$$0$$，$$1$$，$$8$$，$$3$$种情况，$$4\\times 5\\times 3=60$$（个）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2867", "queId": "80b1c2036446447bb5705abe8df5ac3c", "competition_source_list": ["2017年四川六年级竞赛排位赛第20题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "有甲、乙、丙三个数，甲$$:$$丙$$=4:3$$，丙$$:$$乙$$=5:3$$，如果甲、乙两数的和是$$87$$，则乙、丙两数相差． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$\\frac{174}{7}$$ "}], [{"aoVal": "C", "content": "$$144$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["甲$$:$$丙$$=4:3=20:15$$，丙$$:$$乙$$=5:3=15:9$$， 则甲$$:$$乙$$:$$丙$$=20:9:15$$，每份为$$87\\div (20+9)=3$$， 则丙数$$-$$乙数$$=(15-9)\\times 3=18$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "138", "queId": "42617762cc144ef39a042a1791a388cb", "competition_source_list": ["2017年第15届全国希望杯五年级竞赛第1试试题第18题", "其它改编题"], "difficulty": "2", "qtype": "single_choice", "problem": "【例4】$$A$$，$$B$$，$$C$$，$$D$$，$$E$$五人一同参加飞镖大赛，其中只有一人射中飞镖盘的中心，但不知是谁所射． $$A$$说：``不是我射中的，就是$$C$$射中的．'' $$B$$说：``不是$$E$$射中的．'' $$C$$说：``如果不是$$D$$射中的，那么一定是$$B$$射中的．'' $$D$$说：``既不是我射中的，也不是$$B$$射中的．'' $$E$$说：``既不是$$C$$射中的，也不是$$A$$射中的．'' 其中五人中只有两个人说的是对的，由此可以判断射中飞镖盘中心的人是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$A$$ "}], [{"aoVal": "B", "content": "$$B$$ "}], [{"aoVal": "C", "content": "$$C$$ "}], [{"aoVal": "D", "content": "$$D$$ "}], [{"aoVal": "E", "content": "$$E$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$A$$和$$E$$说的话对立，$$C$$和$$D$$说的话对立，必有两对两错，故$$B$$说的是错的，则是$$E$$射中的． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2011", "queId": "e13cc3e1cf8b453295732f22ade99670", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第8题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "鸡兔同笼，共有$$4$$个头，$$12$$条腿，有只兔． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->原型题"], "answer_analysis": ["枚举法： 当有$$1$$只兔，$$3$$只鸡时，共$$1\\times4+3\\times2=10$$（条）腿． 当有$$2$$只兔，$$2$$只鸡时，共$$2\\times4+2\\times2=12$$（条）腿． 所以有$$2$$只兔． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3130", "queId": "f5809cca601944bbb88fa8a437d631fc", "competition_source_list": ["2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "如果你花$$30$$分钟做了$$35$$道题目，那么用同样的速度，你做$$315$$道题目至少需要多少小时？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->整数->四则混合运算"], "answer_analysis": ["$$35$$个题目需要$$30$$分钟，$$315$$里面有$$9$$个$$35$$，所以需要的时间是$$30\\times 9=270$$分钟，等于$$4.5$$小时． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2125", "queId": "00175456487f4f9da59ce5e0ae64b547", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（五）第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两人一起参加$$3000$$米长跑比赛，甲以匀速$$8$$米/秒跑完了全程，乙先以$$10$$米/秒跑了一段时间，后减速为$$5$$米/秒跑完了全程．当甲跑到终点的时候，乙距离终点还有$$125$$米．乙跑完全程用$$5$$米/秒的速度跑了（~ ~ ）秒． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$150$$ "}], [{"aoVal": "C", "content": "$$200$$ "}], [{"aoVal": "D", "content": "$$250$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人变速问题"], "answer_analysis": ["甲跑完全程共花了$$3000\\div 8=375$$秒，乙在这段时间跑了$$3000-125=2875$$米． 设乙用$$10$$米/秒速度跑了$$x$$秒，则他用$$5$$米/秒的速度跑了$$\\left( 375-x \\right)$$秒．$$10x+\\left( 375-x \\right)\\times 5=2875$$，解得$$x=200$$，用$$5$$米/秒的速度所跑的路程为$$3000-200\\times 10=1000$$米． 所用时间为$$1000\\div 5=200$$秒． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1539", "queId": "71bcefd196c648259f163cf044af2cdc", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第5题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "$$20$$个同学排成一队做操，从左边数小文排在第$$12$$个，从右边数小文排在第个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["小文的左边有人：$$12-1=11$$（人）， 所以从右边数小文排在：$$20-11=9$$（个）． 故选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1064", "queId": "0b35203fc06b44e287e3c28a912fc016", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第4题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "如果把一根木棍锯成$$5$$段需要$$4$$分钟．如果锯成$$10$$段需要分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都没有->锯木头类型问题"], "answer_analysis": ["锯成$$5$$段，需要锯$$4$$次，所以每次$$1$$分钟．锯成$$10$$段，需要锯$$9$$次，所以需要$$9$$分钟，故选择$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1822", "queId": "e4802ef280d9425ab9d8b665407580f7", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "祖玛游戏中，龙嘴里不断吐出很多颜色的龙珠，先$$4$$颗红珠，接着$$3$$颗黄珠，再$$2$$颗绿珠，最后$$1$$颗白珠，若按此方式不断重复，则从龙嘴里吐出的第$$2000$$颗龙珠是． ", "answer_option_list": [[{"aoVal": "A", "content": "红珠 "}], [{"aoVal": "B", "content": "黄珠 "}], [{"aoVal": "C", "content": "绿珠 "}], [{"aoVal": "D", "content": "白珠 "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["$$2000\\div (4+3+2+1)=200$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "94", "queId": "340a6638f6644a8b8dc51363e7d34aa7", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一把钥匙只能开一把锁，现有$$4$$把钥匙和$$4$$把锁搞乱了，最多试开次就能确定哪把钥匙开哪把锁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["第一把钥匙最坏的情况要试$$3$$次，把这把钥匙和这把锁拿出；剩下的$$3$$把锁和$$3$$把钥匙，最坏的情况要试$$2$$次，把这把钥匙和这把锁拿出；剩下的$$2$$把锁和$$2$$把钥匙，最坏的情况要试$$1$$次，把这把钥匙和这把锁拿出；剩下的$$1$$把锁和$$1$$把钥匙就不用试了；$$3+2+1=6$$（次）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "139", "queId": "1ac24ad3cb314aa6a7c52f172177b05f", "competition_source_list": ["2016年IMAS小学高年级竞赛第一轮检测试题第13题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知袋子中装有$$n$$颗小球，依次编号为$$1$$、$$2$$、$$3$$、$$\\cdots$$、$$n$$，每次都从袋子中取出两颗球，把它们的编号相加并记下结果，然后把它们放回袋子内．重复抽取直到袋子中每一对小球都被取到为止，记录中恰好有$$215$$种不同的数值，请问$$n$$的值是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$105$$ "}], [{"aoVal": "C", "content": "$$108$$ "}], [{"aoVal": "D", "content": "$$109$$ "}], [{"aoVal": "E", "content": "$$215$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["每两个数相加，和最小为$$1+2=3$$，最大为$$n+(n-1)=2n-1$$，现在共有$$215$$种不同的数值，即最大的数值为$$215+3-1=217$$，$$217=2n-1$$，$$n=109$$． ", "最大值和最小值之间有多少数字就是多少种可能性．$$2n -1-3=215$$．
"], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2498", "queId": "2295a3df2f52407f877d7cb7587340b9", "competition_source_list": ["2017年第15届湖北武汉创新杯小学高年级五年级竞赛决赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "天安门广场是世界上最大的广场，面积约为$$44$$万平方米．已知$$1$$亩$$=\\frac{2000}{3}$$平方米，则天安门广场的面积约为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$600$$亩 "}], [{"aoVal": "B", "content": "$$630$$亩 "}], [{"aoVal": "C", "content": "$$660$$亩 "}], [{"aoVal": "D", "content": "$$690$$亩 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$440000\\div \\frac{2000}{3}=660$$（亩）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "681", "queId": "676f4591184f44159afb2c82cc76fa1e", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛A卷第6题3分", "2018年上海小学高年级五年级竞赛A卷"], "difficulty": "1", "qtype": "single_choice", "problem": "$$5$$路和$$9$$路公共汽车早上$$6$$时$$40$$分同时发车，$$5$$路公共汽车每隔$$10$$分钟发一辆车，$$9$$路公共汽车每隔$$8$$分钟发一辆车，这两路车第$$20$$次同时发车是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8:00$$ "}], [{"aoVal": "B", "content": "$$18:40$$ "}], [{"aoVal": "C", "content": "$$19:20$$ "}], [{"aoVal": "D", "content": "$$20:00$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->两数的最小公倍数"], "answer_analysis": ["$$10$$和$$8$$的最小公倍数是$$40$$， $$40\\times20=800$$（分钟）， $$6$$时$$20$$分再过$$800$$分钟为$$20$$时， 所以第$$20$$次同时发车为$$20$$时． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "189", "queId": "205ae59dbf97407c86433eade75c920f", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在垒球比赛中，若赢$$1$$场得$$3$$分，平$$1$$场得$$1$$分，输$$1$$场不得分．每个队都与其他队交锋$$4$$场，这时四个参赛队的总积分为：$$A$$队$$22$$分，$$B$$队$$19$$分，$$C$$队$$14$$分，$$D$$队$$12$$分．那么有（~ ）场比赛为平局． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->分类讨论思想"], "answer_analysis": ["对于赛况分析试题，尤其对于与分数相关的试题，最重要的是思维方式，本题如果从整体上来考虑比赛所产生的总分值，问题将迎刃而解，依题意可知比赛总场次为$$24$$场比赛之中，若平局则将会让所有队伍的总分增加$$2$$分（比赛双方均得$$1$$分），若出现了胜败，则所有队伍的总分增加$$3$$分，而现在所有队伍获得的总分值为：$$22+19+14+12=67$$（分），$$24$$场比赛，总分最多$$24\\times 3=72$$分，平$$1$$场总分少$$1$$分，$$72-67=5$$分，所以平局$$5$$场. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2253", "queId": "4a47642dc5a142d88d2ad30d46576e3f", "competition_source_list": ["2017年第22届全国华杯赛小学中年级竞赛初赛第4题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "猎豹跑一步长为$$2$$米，狐狸跑一步长为$$1$$米，猎豹跑$$2$$步的时间狐狸跑$$3$$步．猎豹距离狐狸$$30$$米，则猎豹跑动~\\uline{~~~~~~~~~~}~米可追上狐狸． ", "answer_option_list": [[{"aoVal": "A", "content": "$$90$$ "}], [{"aoVal": "B", "content": "$$105$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$135$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["在同样时间内，猎豹跑$$2$$步可以跑$$4$$米，孤狸跑$$3$$步可以跑$$3$$米，那么$${{V}_{豹}}:{{V}_{狐}}=4:3$$，猎豹跑距离气$${{S}_{豹}}=30\\times \\frac{4}{4-3}=120$$米． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3089", "queId": "dd47d0da8b174aeca8395b1f73a93a3e", "competition_source_list": ["2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（三）"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$和$$b$$均是正整数，并且$$\\frac{a}{7}+\\frac{b}{11}$$的值四舍五入的百分位等于$$1.03$$，那么$$a+b=$$（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\frac{a}{7}+\\frac{b}{11}=\\frac{11a+7b}{77}=1.03$$ $$11a+7b\\approx 79.31~ ， 11a+7b=79$$ 由于$$a$$和$$b$$都是正整数，$$a=4$$，$$b=5$$，$$a+b=9$$. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1365", "queId": "28c81a8a383943128eb00f5b9ba3863e", "competition_source_list": ["2014年全国华杯赛小学高年级竞赛初赛A卷第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "【数学人教版】某次考试有$$50$$道试题，答对一道题得$$3$$分，答错一道题或不答题会扣$$1$$分，小龙得分$$118$$分，则小龙答对了道试题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$42$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$50$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->方程思想"], "answer_analysis": ["假设小龙所有的题目都做对了，则小龙的得分为：$$50\\times 3=150$$（分）；实际上小龙的得分为$$118$$分，假设与实际相差：$$150-118=32$$（分）；小龙一道题由对变错会损失：$$3+1=4$$（分），所以小龙错了：$$32\\div 4=8$$（道）题，则小龙答对了：$$50-8=42$$（道）题． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1167", "queId": "116047fdb2ca4d5f972379887e22a895", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "小刚、小强、小明一起买文具，小刚买$$3$$支铅笔和$$5$$支圆珠笔共用$$5.4$$元，小强买$$5$$支铅笔和$$7$$支圆珠笔共用$$7.8$$元，小明买$$4$$支铅笔和$$4$$支圆珠笔共用元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3.4$$ "}], [{"aoVal": "B", "content": "$$4.8$$ "}], [{"aoVal": "C", "content": "$$6.0$$ "}], [{"aoVal": "D", "content": "$$6.6$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设每支铅笔$$x$$元，每支圆珠笔$$y$$元，则$$3x+5y=5.4$$，$$5x+7y=7.8$$，得$$2x+2y=2.4$$，所以$$4x+4y=4.8$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2933", "queId": "9752800056d54be991d69bed2475b807", "competition_source_list": ["2013年第12届全国小机灵杯小学中年级三年级竞赛初赛第6题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "对于两个数$$a$$和$$b$$，规定一种新运算，$$a\\Delta b=3\\times a+2\\times b$$，$$a\\nabla b=2\\times a+3\\times b$$，那么$$2\\Delta \\left( 3\\nabla 4 \\right)=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$42$$ "}], [{"aoVal": "B", "content": "$$47$$ "}], [{"aoVal": "C", "content": "$$51$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->定义新运算->直接运算型->普通型"], "answer_analysis": ["$$3\\nabla 4=2\\times 3+3\\times 4=18$$，$$2\\Delta \\left( 3\\nabla 4 \\right)=2\\Delta 18=3\\times 2+2\\times 18=42$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2753", "queId": "e8679025ee8d4fb2b63e0e0fa863e758", "competition_source_list": ["2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（一）"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$N=1\\frac{1}{6}+2\\frac{1}{12}+\\ldots \\ldots +9\\frac{1}{110}$$，那么比$$N$$大的自然数最小值为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$45$$ "}], [{"aoVal": "B", "content": "$$46$$ "}], [{"aoVal": "C", "content": "$$47$$ "}], [{"aoVal": "D", "content": "$$48$$ "}]], "knowledge_point_routes": ["拓展思维->能力->归纳总结->归纳推理"], "answer_analysis": ["原式$$=\\left( 1+2+3+\\cdots \\cdots +9 \\right)+\\left( \\frac{1}{6}+\\frac{1}{12}+\\cdots \\cdots +\\frac{1}{110} \\right)$$ $$=45+\\left( \\frac{1}{2\\times 3}+\\frac{1}{3\\times 4}+\\cdots \\cdots +\\frac{1}{10\\times 11} \\right)$$ $$=45+\\left( \\frac{1}{2}\\times \\frac{1}{11} \\right)$$ $$=45\\frac{9}{22}$$ 所以符合题意��最小值为$$46$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1335", "queId": "676b0454b50b4e648c522a55f8ac6218", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（五）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$9$$行$$9$$列的方阵树林，如果去掉最外层$$2$$行$$2$$列，要减少棵树． ", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$ "}], [{"aoVal": "B", "content": "$$34$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$38$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->方阵问题->实心方阵->实心方阵的增减", "课内体系->思想->数形结合思想"], "answer_analysis": ["$$9\\times 9-7\\times 7=32$$（棵）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1874", "queId": "a5223a7746ac42b1802fe3269e3b7a8f", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一本书，小明从第$$10$$页看到第$$25$$页，他看了页． ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["$$25-10+1=16$$． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2769", "queId": "6da179f0313e4b388156dd6d28b74dd8", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第2题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "在算式$$+5=13-2$$中，括号中应填入什么数才能使算式成立？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数减法运算->减法横式"], "answer_analysis": ["13-2=11；11-5=6 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1035", "queId": "0e95cfa449454fabb295f2fe24e87f0e", "competition_source_list": ["2012年IMAS小学高年级竞赛第一轮检测试题第18题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "老王在机场工作，他每上$$8$$天班后，就连续休息两天，如果这个星期六和星期天连休二天他都休息，请问至少再过几个星期后，他才能又在星期天休息？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}], [{"aoVal": "E", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["老王接下来的休息时间为：周二和周三；周五和周六；周一和周二；周四和周五；周日和周一；$$\\cdots\\cdots$$． 所以他再次在星期天休息时已经过了$$10\\times 4+9=49$$天，刚好$$7$$个星期． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3086", "queId": "bd9cac9ff9d344e9aa52d18a55361368", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛初赛第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "根据``三角形任意两边之和大于第三边''的知识，解答本题： 有不同长度的七条线段，其长度均为整数厘米，最短的是$$1$$厘米，最长的是$$21$$厘米，其中以任何三条线段作``边''都不能组成一个三角形，那么这七条线段中第二长的线段长厘米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["斐波那契数列从第三项开始，等于前两项的和，即$$1$$、$$2$$、$$3$$、$$5$$、$$8$$、$$13$$、$$21$$． 即第二长的为$$13$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "452", "queId": "8f0f40848f6a4253be5828b817464fdd", "competition_source_list": ["2020年希望杯二年级竞赛模拟第17题"], "difficulty": "1", "qtype": "single_choice", "problem": "小皮、小舒和小贝是同班同学，他们中一个是班长，一个是学习委员，一个是体育委员．现在知道： ①小皮的年龄比体育委员的年龄大； ②���舒比学习委员的年龄大； ③小贝和学习委员年龄不同． 那么小皮、小舒和小贝分别担任． ", "answer_option_list": [[{"aoVal": "A", "content": "班长，学习委员，体育委员 "}], [{"aoVal": "B", "content": "学习委员，体育委员，班长 "}], [{"aoVal": "C", "content": "学习委员，班长，体育委员 "}], [{"aoVal": "D", "content": "班长，体育委员，学习委员 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["由②小舒比学习委员的年龄大；和③小贝和学习委员年龄不同．可知小舒和小贝都不是学习委员，所以学习委员是小皮； 已知小舒比小皮大，并且小皮的年龄比体育委员的年龄大，所以小舒不是体育委员，所以小舒是班长；那么小贝就是体育委员． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2374", "queId": "336106070fa54f478c6a17045a56ff9c", "competition_source_list": ["2011年第9届全国创新杯小学高年级六年级竞赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "定义两种运算：$$a\\oplus b=a+b-1$$，$$a\\otimes b=ab-1$$．如果$$4\\otimes \\left[ \\left( 6\\oplus x \\right)\\oplus \\left( 3\\otimes 5 \\right) \\right]=79$$，则$$x$$等于( )． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$6\\oplus x=6+x-1=5+x$$；$$3\\otimes 5=3\\times 5-1=14$$；$$\\left( 5+x \\right)\\oplus 14=5+x+14-1=x+18$$； $$4\\otimes \\left( 18+x \\right)=4\\times \\left( 18+x \\right)-1=4x+71$$；$$4x+71=79$$，得$$x=2$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1867", "queId": "fbc4a38924854eb7991c19197acdf40a", "competition_source_list": ["2018年湖北武汉创新杯小学高年级五年级竞赛初赛数学思维能力等级测试第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明、小强各买了单价为$$10$$元、$$15$$元两种价格的书，每人买的书两种价格的都有，各自的书款总额都是$$90$$元，但所买的数的本书不同，那么两人买的书共有（~ ）本． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["设$$10$$元的有$$x$$本，$$15$$元的有$$y$$本，则$$10x+15y=90$$，解得$$\\begin{cases} x=6 y=2 \\end{cases}$$，$$\\begin{cases} x=3 y=4 \\end{cases}$$，共$$6+2+3+4=15$$本． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3206", "queId": "1a7f2e663d5d484abb048f8985fd7efc", "competition_source_list": ["2011年北京五年级竞赛", "2011年北京六年级期中"], "difficulty": "2", "qtype": "single_choice", "problem": "$$4$$个人聚会，每人各带$$2$$件礼品，分赠给其余$$3$$个人中的$$2$$人．则：至少有$$2$$对人，每对人是互赠过礼品的．~\\uline{~~~~~~~~~~}~ ", "answer_option_list": [[{"aoVal": "A", "content": "对 "}], [{"aoVal": "B", "content": "不对 "}]], "knowledge_point_routes": ["知识标签->数学思想->枚举思想"], "answer_analysis": ["将这四个人用$$4$$个点表示，如果两个人之间送过礼，就在两点之间连一条线． 由于每人送出$$2$$件礼物，图中共有$$4\\times 2=8$$条线，由于每人礼品都分赠给$$2$$个人，所以每两点之间至多有$$1+1=2$$条线．四点间，每两点连一条线，一共$$6$$条线，现在有$$8$$条线，说明必有两点之间连了$$2$$条线，还有另外两点（有一点可以与前面的点相同）之间也连了$$2$$条线． 即为所证结论． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1155", "queId": "3d2e8e0f6c834ae9814ee89ed61b9b0e", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2013$$年$$12$$月$$21$$日是星期六，那么$$2014$$年的春节，即$$2014$$年$$1$$月$$31$$日是星期． $ $ $ $ $ $ ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "四 "}], [{"aoVal": "C", "content": "五 "}], [{"aoVal": "D", "content": "六 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["星期六有：$$21\\to 28\\to 4(35)\\to 11\\to 18\\to 25$$，所以 $$31$$日是星期五． $$10+31=41$$（天），$$41\\div7=5\\cdots 6$$ ，差一天是星期六，所以$$31$$日是星期五． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2987", "queId": "b2eef23883ff431c9f50b9b7b5023dd0", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$\\frac{37037037037\\times 73}{2703703703701}=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$27$$ "}], [{"aoVal": "C", "content": "$$37$$ "}], [{"aoVal": "D", "content": "无法计算 "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["原式$$=\\frac{37\\times 1001001001\\times 73}{2701\\times 1001001001}$$ $$=\\frac{2701\\times 1001001001}{2701\\times 1001001001}$$ $$=1$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "151", "queId": "628c3b5ede004a4784d781d2eb619ba0", "competition_source_list": ["2019年第24届YMO六年级竞赛决赛第7题3分"], "difficulty": "3", "qtype": "single_choice", "problem": "有一个$$12$$级的楼梯．某人每次能登上$$1$$级或$$2$$级或$$3$$级，现在他要从地面登上第$$12$$级，有种不同的方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$149$$ "}], [{"aoVal": "B", "content": "$$274$$ "}], [{"aoVal": "C", "content": "$$504$$ "}], [{"aoVal": "D", "content": "$$927$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["设第$$n$$级有$${{a}_{n}}$$种登的方式， 则第$$n-3$$，$$n-2$$，$$n-1$$级再分别登$$3$$，$$2$$，$$1$$级可到第$$n$$级， 则$${{a}_{n}}={{a}_{n-3}}+{{a}_{n-2}}+{{a}_{n-1}}$$， 而$${{a}_{1}}=1$$，$${{a}_{2}}=1+1=2$$，$${{a}_{3}}=1+1+2=4$$， 故$${{a}_{4}}=1+2+4=7$$， $${{a}_{5}}=2+4+7=13$$， $${{a}_{6}}=4+7+13=24$$， $${{a}_{7}}=7+13+24=44$$， $${{a}_{8}}=13+24+44=81$$， $${{a}_{9}}=24+44+81=149$$， $${{a}_{10}}=44+81+149=274$$， $${{a}_{11}}=81+149+274=504$$， $${{a}_{12}}=149+274+504=927$$． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1038", "queId": "06eacdb28421499883db75681b9ff994", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛B卷第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "商店有$$A$$、$$B$$两件衣服，都卖$$200$$元，$$A$$赚了十分之一，$$B$$亏了十分之一，总体上是． ", "answer_option_list": [[{"aoVal": "A", "content": "赚了 "}], [{"aoVal": "B", "content": "亏了 "}], [{"aoVal": "C", "content": "不亏不赚 "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->已知售价利润求成本"], "answer_analysis": ["售价为$$200$$元，$$A$$赚了十分之一，$$A$$的进价为$$200\\div \\left( 1+10 \\% \\right)=\\frac{2000}{11}$$（元）， $$B$$亏了十分之一，$$B$$的进价为$$200\\div \\left( 1-10 \\% \\right)=\\frac{2000}{9}$$（元）， $$\\frac{2000}{11}+\\frac{2000}{9}=\\frac{40000}{99}$$（元）$$\\textgreater400$$元， 进价$$\\textgreater$$售价，亏了． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2350", "queId": "8aac50a7519fa10a01519fd4faa4002b", "competition_source_list": ["2016年全国华杯赛小学高年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "算式$$\\underbrace{999\\cdots 9}_{2016}\\times \\underbrace{999\\cdots 9}_{2016}$$的结果中含有（ ~ ~ ~ ~）个数字$$0$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2017$$ "}], [{"aoVal": "B", "content": "$$2016$$ "}], [{"aoVal": "C", "content": "$$2015$$ "}], [{"aoVal": "D", "content": "$$2014$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$${{\\left( {{10}^{2016}}-1 \\right)}^{2}}=\\left( {{10}^{2016}}-2 \\right)\\times {{10}^{2016}}+1=\\underbrace{999\\cdots 99}_{2015}8\\underbrace{000\\cdots 00}_{2015}1$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "806", "queId": "ff80808147248448014724de1968012d", "competition_source_list": ["2012年全国华杯赛小学高年级竞赛初赛网络版第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "两个数的最大公因数是$$20$$，最小公倍数是$$100$$，下面说法正确的有个． （1）两个数的乘积是$$2000$$． （2）两个数都扩大$$10$$倍，最大公因数扩大$$100$$倍． （3）两个数都扩大$$10$$倍，最小公倍数扩大$$10$$倍． （4）两个数都扩大$$10$$倍，两个数乘积扩大$$100$$倍． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$（1）$$$$（3）$$$$（4）$$正确． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "204", "queId": "902ee41d952d474f886cd0bdb5b37531", "competition_source_list": ["2013年小机灵杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "古时候的原始人捕猎，捕到一只野兽对应一根手指。等到$$10$$根手指都用完，就在绳子上打一个结，这就是运用现在的数学中的（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "出入相补原理 "}], [{"aoVal": "B", "content": "等差数列求和 "}], [{"aoVal": "C", "content": "十进制计数法 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->智巧趣题"], "answer_analysis": ["解：古时候的原始人捕猎，捕到一只野兽对应一根手指。等到$$10$$根手指都用完，就在绳子上打一个结，这就是运用现在的数学中的十进制计数法； 故选：$$C$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2364", "queId": "047c1a6892d44100b42a46f246e7b5d7", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第2题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "在算式$$\\square +5=13-2$$中，请问$$\\square $$中应填入什么数才能使算式正确？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数减法运算->减法横式"], "answer_analysis": ["13-2=11；11-5=6 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "416", "queId": "78c7a7da757b47a5b0c2b40f4c90d2c8", "competition_source_list": ["2017年IMAS小学高年级竞赛（第一轮）第11题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$的数码卡片各一张，小李每次取出$$2$$张，记录下它们的差（大的数减小的数），然后把这两张卡片扔掉．取完这六张卡片后，请问小李记录下的三个差之和最大可能值为多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["小李记录下的三个差之和等于六个数码的其中三个数码减去另外三个数码， 所以最大可能值为$$6+5+4-3-2-1=9$$． 故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2049", "queId": "eae7efabd11a491ea930257b03240fe8", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（四）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "水果店规定：如果购买芒果不超过$$10$$千克，那么每千克售价$$4$$元；如果超过$$10$$千克，那么超过的部分每千克$$3.5$$元．某人买了$$24$$千克芒果，他应付元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$84$$ "}], [{"aoVal": "B", "content": "$$89$$ "}], [{"aoVal": "C", "content": "$$90$$ "}], [{"aoVal": "D", "content": "$$96$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->分段计价问题"], "answer_analysis": ["$$4\\times 10+3.5\\times (24-10)=89$$（元）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "593", "queId": "7467a847fc424ee882285afac2e4cb9f", "competition_source_list": ["2006年华杯赛五年级竞赛初赛", "2006年华杯赛六年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "2007005共有个质因数. ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数"], "answer_analysis": ["因为$$2007005=2005\\times 1000+2005=2005\\times 1001=\\left( 5\\times 401\\times )( 7\\times 11\\times 13 \\right)$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2804", "queId": "8002adff54074710a97a3b13bec2edd3", "competition_source_list": ["2012年全国希望杯五年级竞赛初赛第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "计算：$$21.49+52.37-0.4+5.51-11.37-6.6=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$41$$ "}], [{"aoVal": "B", "content": "$$51$$ "}], [{"aoVal": "C", "content": "$$61$$ "}], [{"aoVal": "D", "content": "$$71$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["$$21.49+52.37-0.4+5.51-11.37-6.6$$$$=(21.49+5.51)+(52.37-11.37)-(0.4+6.6)$$$$=27+41-7$$$$=61$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3362", "queId": "8d509065dcce4cdf9c309b6b0d3e1ec8", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "不妨称各位数字之和为$$7$$的整数为``魔力数''，如$$115$$，$$1312$$等就是魔力数．那么随手写出一个三位数，恰好是``魔力数''的可能性是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{45}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{75}$$ "}], [{"aoVal": "C", "content": "$$\\frac{7}{225}$$ "}], [{"aoVal": "D", "content": "$$\\frac{8}{225}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["三位数一共有$$900$$个满足条件的魔力数有$$\\text{C}_{7+2-1}^{2}=28$$个，则$$\\frac{28}{900}=\\frac{7}{225}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1688", "queId": "917e0a45e24040759f92eab66af8047c", "competition_source_list": ["2007年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "一条绳子，折成相等的$$3$$段后，再对折成相等的两段，然后从中间剪开，一共可以剪成~\\uline{~~~~~~~~}~段。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都没有->剪绳子"], "answer_analysis": ["将绳折成$$3$$段再对折，相当于折成$$6$$段，一刀与这$$6$$段共有$$6$$个交点，所以将绳剪成$$7$$段。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1206", "queId": "4b42ceb7f4144f149e8f945da32ba889", "competition_source_list": ["2018年全国小学生数学学习能力测评四年级竞赛复赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "学校有象棋，跳棋共$$26$$副，$$2$$人下一副象棋，$$6$$人下一副跳棋，恰好可供$$120$$名同学进行活动，象棋有副． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["利用假设的方法来解答， 假设$$26$$副都是跳棋，那么$$26$$副跳棋可供$$6\\times 26=156$$（人）进行活动， 与$$120$$人相差$$156-120=36$$（人）， 每副象棋和跳棋的人数相差$$6-2=4$$（人）， 用一共相差的人数$$36$$除以$$4$$就求出象棋有多少副，$$36\\div 4=9$$（副）． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2992", "queId": "a57bb94584134e5fb6289af83d836e47", "competition_source_list": ["2014年迎春杯三年级竞赛初赛", "2014年迎春杯四年级竞赛初赛", "2014年迎春杯四年级竞赛初赛", "2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "下面计算结果等于$$9$$的是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$3\\times 3\\div 3+3$$ "}], [{"aoVal": "B", "content": "$$3\\div 3+3\\times 3$$ "}], [{"aoVal": "C", "content": "$$3\\times 3-3+3$$ "}], [{"aoVal": "D", "content": "$$3\\div 3+3\\div 3$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->四则混合运算"], "answer_analysis": ["解：A、$$3\\times 3\\div 3+3$$ $$=3+3$$ $$=6$$ B、$$3\\div 3+3\\times 3$$ $$=1+9$$ $$=10$$ C、$$3\\times 3-3+3$$ $$=9-3+3$$ $$=9$$ D、$$3\\div 3+3\\div 3$$ $$=1+1$$ $$=2$$ 故选：C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "384", "queId": "923e9c283f57445da28185a896629e0e", "competition_source_list": ["2021年新希望杯一年级竞赛初赛第14题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$5$$只动物排成一排，狗和猫都与鸡相邻，猫和鼠都与兔相邻，那么都与猫相邻． ", "answer_option_list": [[{"aoVal": "A", "content": "鼠和鸡 "}], [{"aoVal": "B", "content": "鸡和兔 "}], [{"aoVal": "C", "content": "兔和狗 "}], [{"aoVal": "D", "content": "兔和鼠 "}], [{"aoVal": "E", "content": "鼠和狗 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["根据题意分析可知，狗和猫都与鸡相邻，即鸡的左右两边是猫和狗，又因为猫与老鼠都与兔相邻，所以猫的旁边是兔子和老鼠，兔子和老鼠在猫的同一侧，由此可知，与猫相邻的是鸡和兔． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "850", "queId": "89c25eab2fc64218869a27fda8859631", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$20$$到$$100$$这$$100$$个正整数中，不能被$$2$$，$$3$$，$$5$$，$$7$$中任何一个数整除的数有（~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$19$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->整除->整除特征->整除初识"], "answer_analysis": ["在$$20100$$以内不能被$$2$$，$$3$$，$$5$$，$$7$$整除的数其实就是比$$20$$大的质数，只要理解了这一点，就可以直接得出答案．$$100$$以内有$$25$$质数，$$20$$以内有$$8$$个质数，$$258=17$$（个）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2060", "queId": "e1f07df818ed402a8a1d70bd82131f5a", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小$$\\text{Y}$$和小$$\\text{M}$$共同组装$$15$$个机器人玩具．小$$\\text{Y}$$每$$2$$小时组装$$1$$个机器人玩具，小$$\\text{M}$$每$$3$$小时组装一个机器人玩具，两人同时开始组装，小时能完成任务． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["根据题意分析可知，用工作总量除以工作效率就能得到工作时间，小$$\\text{Y}$$的工作效率为：$$1\\div 2=\\frac{1}{2}$$，小$$\\text{M}$$的工作效率为：$$1\\div 3=\\frac{1}{3}$$，所以两人同时组装，完成的时间为： $$15 \\div \\left( \\frac{1}{2}+ \\frac{1}{3}\\right)$$ $$=15 \\div \\frac{5}{6}$$ $$=18$$（小时） 故答案为：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "673", "queId": "1bd6bca400a5427385c61a8a6374c8eb", "competition_source_list": ["2011年北京五年级竞赛", "2011年其它", "2019年陕西西安新城区爱知中学小升初入学真卷2第24题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两车分别从 $$A$$、$$B$$两地同时出发，相向而行．出发时，甲、乙的速度之比是 $$5 ： 4$$，相遇后甲的速度减少$$20 \\%$$，乙的速度增加 $$20 \\%$$．这样当甲到达$$B$$地时，乙离$$A$$地还有$$10$$千米．那么$$A$$、$$B$$两地相距多少千米？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$300$$ "}], [{"aoVal": "B", "content": "$$350$$ "}], [{"aoVal": "C", "content": "$$400$$ "}], [{"aoVal": "D", "content": "$$450$$ "}], [{"aoVal": "E", "content": "$$480$$ "}]], "knowledge_point_routes": ["海外竞赛体系->知识点->数论模块->质数与合数->特殊质数运用->特殊质数2", "拓展思维->思想->整体思想"], "answer_analysis": ["两车相遇时甲走了全程的$$\\frac{5}{9}$$，乙走了全程的$$\\frac{4}{9}$$， 之后甲的速度减少$$20 \\%$$，乙的速度增加$$20 \\%$$， 此时甲、乙的速度比为$$5 \\times (1 - 20 \\% )：$$4$$ \\times (1 + 20 \\% ) = 5：6$$ ， 所以甲到达$$B$$地时，乙又走了$$\\frac{4}{9} \\times \\frac{6}{5} = \\frac{8}{{15}}$$，距离 $$A$$地$$\\frac{5}{9} - \\frac{8}{{15}} = \\frac{1}{{45}}$$， 所以$$A$$、$$B$$两地的距离为$$10 \\div \\frac{1}{{45}} = 450$$ $$($$千米$$)$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "899", "queId": "81e40461001c4fb9a2e650aca64a3add", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$42$$个边长为$$1$$厘米的小方块粘在一起形成一个长方体，若长方体地面的周长为$$18$$厘米，则长方体的高是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$厘米 "}], [{"aoVal": "B", "content": "$$3$$厘米 "}], [{"aoVal": "C", "content": "$$4$$厘米 "}], [{"aoVal": "D", "content": "$$7$$厘米 "}]], "knowledge_point_routes": ["拓展思维->能力->图形认知"], "answer_analysis": ["根据题意我们必须将$$42$$拆分成$$3$$个自然数的乘积，且恰好其中长与高之和应为$$18\\div 2=9cm$$，易得：$$42=7\\times 2\\times 3$$，故长为$$7$$厘米，宽为$$2$$厘米，高为$$3$$厘米． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "945", "queId": "e615099e054d4f078d9bfaeef50456d1", "competition_source_list": ["2019年美国数学大联盟杯五年级竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "下面哪个数不是三个不同质数的乘积（正好三个）？ Which of the following is not the product of exactly three different prime factors? ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$70$$ "}], [{"aoVal": "C", "content": "$$105$$ "}], [{"aoVal": "D", "content": "$$175$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数->分解质因数（式）", "Overseas Competition->知识点->数论模块->分解质因数"], "answer_analysis": ["题目翻译：下面哪一项不是三个完全不同的质数的乘积？ 把每个选项的数进行分解质因数． $$30=2\\times3\\times5$$，是三个不同的质数的乘积． $$70=2\\times5\\times7$$，是三个不同的质数的乘积． $$105=3\\times5\\times7$$，是三个不同的质数的乘积． $$175=5\\times5\\times7$$，不是三个不同的质数的乘积． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "793", "queId": "ff808081465a848401465b54d7b90281", "competition_source_list": ["2014年全国华杯赛小学高年级竞赛初赛B卷第6题", "2016年吉林长春二道区吉大附中力旺实验小学小升初第9题2分"], "difficulty": "2", "qtype": "single_choice", "problem": "有七张卡片，每张卡片上写有一个数字，这七张卡片摆成一排，就组成了七位数$$2014315$$．将这七张卡片全部分给甲、乙、丙、丁四人，每人至多分$$2$$张．他们各说了一句话： 甲：``如果交换我卡片上的$$2$$个数字在七位数中的位置，那么新的七位数就是$$8$$的倍数'' 乙：``如果交换我卡片上的$$2$$个数字在七位数中的位置，那么新的七位数仍不是$$9$$的倍数'' 丙：``如果交换我卡片上的$$2$$个数字在七位数中的位置，那么新的七位数就是$$10$$的倍数'' 丁：``如果交换我卡片上的$$2$$个数字在七位数中的位置，那么新的七位数就是$$11$$的倍数'' 已知四人中恰有一个人说了谎，那么说谎的人是（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "丁 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["可以直接判断乙必定说的是真话，可以直接判断乙必定是说真话的，乙中的数字不管怎么变换都不可能是$$9$$的倍数． 因为七位数的数字之和为$$2+0+1+4+3+1+5=16$$，不是$$9$$的倍数； 如果丙说真话，那么他手中的数字是$$0$$和$$5$$，可以实现对调位置后被$$10$$整除； 如果甲说真话，那么他手中的数字只能是$$5$$和$$2$$，可以实现对调位置后被$$8$$整除； 如果丁说真话，那么他手中的数字只能是$$0$$和$$3$$，这样才能使得奇数位数字之和减去偶数位数字之和的差是$$11$$的倍数$$(2314015$$，$$(5+0+1+2)-(1+4+3)=0$$)． 综上，如果丙说真话，那甲和丁都是说谎话的人，两个人说谎话，不符合题意，所以说谎话的人是丙，选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1376", "queId": "755ebdd376da4b6fb81f528136c93f86", "competition_source_list": ["2017年IMAS小学高年级竞赛（第二轮）第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个正数去掉小数部分后得到一个整数，将这个整数加上原来的正数所得之和，再与$$5$$相乘，最后得到$$22.1$$．请问原来这个正数是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4.42$$ "}], [{"aoVal": "B", "content": "$$0.42$$ "}], [{"aoVal": "C", "content": "$$4.41$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "$$2.42$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["可知原来的正数与其整数部分相加所得之和为$$22.1\\div 5=4.42$$， 因此原来的正数之小数部分为$$0.42$$，且整数部分为$$4\\div 2=2$$， 所以原来这个正数是$$2.42$$． 故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1639", "queId": "4e897ae0cc2e4a799a40fdc721a49a19", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛A卷第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一条公路，甲队单独修需要$$24$$天完成，乙队单独修需要$$30$$天完成．甲乙两队合修几天后，乙队停工休息，甲队继续修了$$6$$天公路修好．乙队修了天． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->休息工程问题"], "answer_analysis": ["把这条公路的工程量看成单位``$$1$$''， 那么甲的工作效率就是$$\\frac{1}{24}$$，乙的工作效率就是$$\\frac{1}{30}$$， 合作的工作效率就是二者的和， 用总工作量减去甲独干的工作量就是合干的工作量， 用合干的工作量除以合作的工作效率就是合干的时间，也就是乙队干的时间． $$\\left( 1-\\frac{1}{24}\\times 6 \\right)\\div \\left( \\frac{1}{24}+\\frac{1}{30} \\right)$$ $$=\\left( 1-\\frac{1}{4} \\right)\\div \\frac{3}{40}$$ $$=\\frac{3}{4}\\div \\frac{3}{40}$$ $$=10$$（天）． 答：乙队修了$$10$$天． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "786", "queId": "d5f61594f99d4b47a3a40e265b9d1cbe", "competition_source_list": ["2016年全国美国数学大联盟杯小学高年级六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "做家务是郝美女老师的兴趣．迪每$$8$$天进行一次大扫除，每$$11$$天洗一次衣服．如果星期天她做了两件家务，那么下一次她做两件家务会是哪一天？ ", "answer_option_list": [[{"aoVal": "A", "content": "星期三 "}], [{"aoVal": "B", "content": "星期四 "}], [{"aoVal": "C", "content": "星期五 "}], [{"aoVal": "D", "content": "星期六 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数的认识->数的特征->倍数->两个数的公倍数"], "answer_analysis": ["$$8\\times 11=88$$，$$88$$除以$$7$$余$$4$$，所以正好是周四. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2232", "queId": "7161cc01cd2b4628b6120e442df34044", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛决赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "十八世纪，某国、某人在浓雾中步行，另一乘马车之人从他身后来到他身边，他问马车的速度是多少，对方答道：每分钟$$176$$米．二人各自继续相同而行，$$5$$分钟后，乘马车之人在他前方$$660$$米处隐于浓雾中看不见了，问步行人每分钟步行． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$米 "}], [{"aoVal": "B", "content": "$$22$$米 "}], [{"aoVal": "C", "content": "$$33$$米 "}], [{"aoVal": "D", "content": "$$44$$米 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$176-660\\div 5=44\\text{m/min}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1723", "queId": "e8b8f108ca294fbb8824747ebff03d9c", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一根细绳对折两次后，从正中间剪开，最短的一段长$$1$$米，这根细绳原来长米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都没有->剪绳子"], "answer_analysis": ["根据题意分析可知，一根细绳对折两次后被分成了$$4$$段，从中间剪开后，分为两个一样长的短段和一个长段，长段的距离是短段的$$2$$倍， 所以细绳原来长：$$2+1+1=4$$（米）． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "228", "queId": "5ee2ff196d324dbcac89176297cf1e4a", "competition_source_list": ["2021年第24届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第15题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "用$$26$$根长度$$1$$厘米的小棒拼长方形，长方形的面积最大是平方厘米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$42$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->组合模块最值问题->最值原理在几何中的应用"], "answer_analysis": ["$$26\\div2=13$$（厘米），要使长方形的面积最大，它的长和宽最接近，$$7+6=13$$（厘米），$$7\\times6=42$$（平方厘米），所以长方形的面积最大是$$42$$平方厘米． 故答案为：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2990", "queId": "ffde55bafb434963ab377db21b63b380", "competition_source_list": ["2017年河南郑州联合杯竞赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a=\\frac{1234567890}{2345678901}$$，$$b=\\frac{1234567890-2017}{2345678901-2016}$$，$$a$$与$$b$$的大小关系是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b$$~ "}], [{"aoVal": "B", "content": "$$a\\textless{}b$$~~~~~~~ "}], [{"aoVal": "C", "content": "$$a=b$$~ "}], [{"aoVal": "D", "content": "无法判断 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->糖水原理法->加糖"], "answer_analysis": ["比较大小；$$\\frac{1234567890}{2345678901}\\textgreater\\frac{1234567890-2016}{2345678901-2016}\\textgreater\\frac{1234567890-2017}{2345678901-2016}$$，则$$a\\textgreater b$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2923", "queId": "92bf04be1ae942e39b9fdad56ed1e2b3", "competition_source_list": ["2003年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "用方砖铺地，当铺地面积一定时，方砖的边长和所需方砖块数的关系是( ) ", "answer_option_list": [[{"aoVal": "A", "content": "正比例关系 "}], [{"aoVal": "B", "content": "反比例关系 "}], [{"aoVal": "C", "content": "不成比例关系 "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比和比例->比例->正比例与反比例"], "answer_analysis": ["$$s=a{{l}^{2}}$$，其中a为块数，$${l}$$为边长，故不成比例 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "675", "queId": "82fbf33fd7744dc3926e1ce9a3fa0de7", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第3题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "下列数中，$$32$$、$$324$$、$$3244$$、$$32444$$、$$324444$$、$$3244444$$、$$32444444$$、$$324444444$$，有个是完全平方数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$32=2\\times2\\times2\\times2\\times2={{4}^{2}}\\times2$$，不是平方数． $$324={{18}^{2}}$$是平方数． $$32444={{2}^{2}}\\times8111$$不是平方数． $$324444={{2}^{2}}\\times3\\times19\\times1423$$不是平方数． $$3244444={{2}^{2}}\\times7\\times115873$$，而$$115873$$不含质因数$$7$$，故不是． $$32444444={{2}^{2}}\\times23\\times352657$$，而$$352657$$不含质因数$$23$$，故不是． $$324444444={{2}^{2}}\\times3\\times27037037$$，而$$27037037$$不含质因数$$37$$，故不是． 所以选择$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "603", "queId": "7dc4d32a01b7498da11562ad9b937e8a", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛决赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "五个数，两两相加，再把所得的和相加，总和为$$2064$$，原来五个数的和为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2064$$ "}], [{"aoVal": "B", "content": "$$2068$$ "}], [{"aoVal": "C", "content": "$$7099$$ "}], [{"aoVal": "D", "content": "$$516$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->位值原理运用->位值原理的综合应用"], "answer_analysis": ["五个数两两相加再求和，每个数计算$$4$$次，则五个数的和为：$$2064\\div 4=516$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1386", "queId": "31a43c81eb1b4b4fb146e1d9e0c849e5", "competition_source_list": ["2020年广东广州羊排赛三年级竞赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个工人$$3$$小时磨了$$60$$千克面粉，照这样计算，这个工人磨完$$200$$千克面粉，一共要 小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["一个工人$$3$$小时磨了$$60$$千克面粉， 所以$$1$$小时磨面粉$$60\\div3=20$$（千克）， 这个工人磨完$$200$$千克面粉，一共要$$200\\div20=10$$（时）， 所以答案为$$10$$小时． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2455", "queId": "66abdab8a1d246548df974c39dd6e831", "competition_source_list": ["2017年全国美国数学大联盟杯小学高年级五年级竞赛初赛第34题"], "difficulty": "3", "qtype": "single_choice", "problem": "从前一百个正整数中，最多能选出多少个不同整数，使任意其中三个都不能作为同一个三角形的三边长？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->几何模块->直线型->图形认知->三角形->三角形三边关系", "拓展思维->能力->逻辑分析"], "answer_analysis": ["To be the lengths of a triangle, the sum of two sides must be longer than the third side. Therefore, we just need to make the third number exactly the sum of the two former numbers, which is also the Fibonacci sequence. $$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$34$$，$$55$$，$$89$$. The last number to be chosen is $$89$$. A total of $$10$$ numbers. Choose $$C$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1978", "queId": "aae8ce8dd24f4aaa80b2115a097fdddb", "competition_source_list": ["2013年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙两仓的稻谷数量一样，爸爸，妈妈和阳阳单独运完一仓稻谷分别需要$$10$$天，$$12$$天和$$15$$天。爸爸妈妈同时开始分别运甲、乙两仓的稻谷，阳阳先帮妈妈，后帮爸爸，结果同时运完两仓稻谷，那么阳阳帮妈妈运了（ ）天。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题"], "answer_analysis": ["解：三人一共搬了： $$\\left( 1+1 \\right)\\div \\left( \\frac{1}{10}+\\frac{1}{12}+\\frac{1}{15} \\right)$$ $$=2\\div \\frac{1}{4}$$ $$=8$$（天） 阳阳帮妈妈运的天数： $$\\left( 1-\\frac{1}{12}\\times 8 \\right)\\div \\frac{1}{15}$$ $$=\\frac{1}{3}\\times 15$$ $$=5$$（天） 答：阳阳帮妈妈运了$$5$$天。 故选：C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "198", "queId": "24bb45fc993c461e80b53dca33060009", "competition_source_list": ["2020年新希望杯五年级竞赛第33题", "2020年希望杯五年级竞赛模拟第33题"], "difficulty": "0", "qtype": "single_choice", "problem": "斯普林特老师在$$3$$个小箱中各放一个有颜色的球，让四只忍者神龟猜箱子中球的颜色． 李奥纳多说：``$$1$$号箱中放红球，$$2$$号箱中放黑球，$$3$$号箱中放黄球．'' 拉斐尔说：``$$1$$号箱中放橙球，$$2$$号箱中放黑球，$$3$$号箱中放绿球．'' 米开朗琪罗说：``$$1$$号箱中放蓝球，$$2$$号箱中放橙球，$$3$$号箱中放紫球．'' 多纳泰罗说：``$$1$$号箱中放橙球，$$2$$号箱中放绿球，$$3$$号箱中放蓝球．'' 斯普林特老师说：``你们中有一个人恰好猜对了两个，其余三人都只猜对一个．'' 那么$$3$$号箱中放的是球． ", "answer_option_list": [[{"aoVal": "A", "content": "黄 "}], [{"aoVal": "B", "content": "黑 "}], [{"aoVal": "C", "content": "红 "}], [{"aoVal": "D", "content": "橙 "}], [{"aoVal": "E", "content": "蓝 "}], [{"aoVal": "F", "content": "紫 "}], [{"aoVal": "G", "content": "绿 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["共猜对$$5$$次，由题意知：$$1$$号箱被猜$$2$$次橙，$$2$$号箱被猜$$2$$次黑，则$$1$$号箱为橙或$$2$$号箱为黑． 假设$$1$$号为橙色，则拉斐尔和多纳泰罗一个猜中$$1$$次，一人猜中两次． 假设拉斐尔猜中$$1$$次，则$$2$$号不为黑，$$3$$号不为绿，则李奥纳多猜中$$3$$号为黄，米开朗琪罗全猜错． 假设有问题，则拉斐尔猜中$$2$$次，$$1$$号为橙，$$2$$号不为黑，$$3$$号为绿，则李奥纳多全猜错，则$$1$$号为橙，$$2$$号为黑，$$3$$号不为绿，拉斐尔猜中$$2$$次，李奥纳多猜中$$2$$号为黑，多纳泰罗猜中$$1$$号为橙，米开朗琪罗猜中$$3$$号为紫． "], "answer_value": "F"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1340", "queId": "2cd0102c3aa04253bbc93625c0290b3c", "competition_source_list": ["2017年全国华杯赛小学中年级竞赛初赛模拟第1题", "小学中年级三年级上学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两人在春节一共得$$210$$元压岁钱，甲给乙$$40$$元，还比乙多$$10$$元，那么，原来甲得了元压岁钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$150$$ "}], [{"aoVal": "B", "content": "$$140$$ "}], [{"aoVal": "C", "content": "$$125$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->和差问题->二量和差问题->明和暗差"], "answer_analysis": ["因为甲给乙$$40$$元，还比乙多$$10$$元，所以甲比乙多$$40\\times 2+10=90$$元， 所以甲：$$(210+90)\\div 2=150$$ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "342", "queId": "7bcd5933858740c5b93e1ca5f8657f20", "competition_source_list": ["2021年新希望杯二年级竞赛初赛第30题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "谎言岛有一半的人只在星期三、星期五、星期六说谎，另一半的人只在星期二、星期四、星期日说谎．某一天，岛上的所有人都说：``我明天说真话．''那么，这一天是． （2021年新希望杯二年级竞赛初赛数学试卷） ", "answer_option_list": [[{"aoVal": "A", "content": "星期二 "}], [{"aoVal": "B", "content": "星期三 "}], [{"aoVal": "C", "content": "星期五 "}], [{"aoVal": "D", "content": "星期六 "}], [{"aoVal": "E", "content": "星期日 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["前一半$$7$$天的情况：真、真、假、真、假、假、真， 后一半$$7$$天的情况：真、假、真、假、真、真、假， $$\\text{A}$$选项：若这一天是星期二，则前一半人在今天说的是真话，那么明天应该说真话才合理，但星期三他们说的是假话，相互矛盾．后一半人在今天说的是假话，那么明天应该说假话才合理，但是星期三他们说的是真话，也相互矛盾，故$$\\text{A}$$错误； $$\\text{B}$$选项：若这一天是星期三，则前一半人在今天说的是假话，那么明天应该说假话才合理，但星期四他们说的是真话，相互矛盾．后一半人在今天说的是真话，那么明天应该说真话才合理，但是星期四他们说的是假话，也相互矛盾，故$$\\text{B}$$错误； $$\\text{C}$$选项：若这一天是星期五，则前一半人在今天说的是假话，那么明天应该说假话才合理，星期六他们说的正好是假话，符合．后一半人在今天说的是真话，那么明天应该说真话才合理，星期六他们说的正好是真话，符合，故$$\\text{C}$$正确； $$\\text{D}$$选项：若这一天是星期六，则前一半人在今天说的是假话，那么明天应该说假话才合理，但星期日他们说的是真话，相互矛盾．后一半人在今天说的是真话，那么明天应该说真话才合理，但是星期日他们说的是假话，也相互矛盾，故$$\\text{D}$$错误； E选项：若这一天是星期日，则前一半人在今天说的是真话，那么明天应该说真话才合理，星期一他们说的是真话，符合．后一半人在今天说的是假话，那么明天应该说假话才合理，但是星期一他们说的是真话，相互矛盾，故$$\\text{E}$$错误． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "819", "queId": "bf39b0079136456fa142deedeea40061", "competition_source_list": ["2006年第4届创新杯五年级竞赛复赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "某加油站有二位员工，从今年$$1$$月$$1$$日起规定：员工甲每工作$$3$$天后休息$$1$$天，员工乙每工作$$5$$天后休息$$2$$天，当遇到二人都休息时，必须另聘一位临时工，则今年共有~\\uline{~~~~~~~~~~}~天要聘$$1$$个时工． ", "answer_option_list": [[{"aoVal": "A", "content": "$$26$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["甲每到$$4$$的倍数就休息，而乙每到$$7$$的倍数和比$$7$$的倍数少一天都休息． 因为$$4$$和$$7$$的最小公倍数是$$28$$， 因为今年是平年， 所以在$$28$$的倍数体息的日子时：$$365\\div28\\approx 15$$（天）， 而在比$$7$$的倍数少一天休息时，甲乙第一次重逢的日子是第二十天，以后每隔$$28$$天就共同休息一天，$$365-20=345$$（天），$$345\\div28\\approx 12$$（天）， 所以甲乙两人共同休息的天数是$$15+12+1=28$$（天��． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2283", "queId": "787d64ec61af406d84bad2edcdd517c8", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（二）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "一艘轮船先顺水航行$$40$$千米，再逆水航行$$24$$千米，共用$$8$$小时．若该船先逆水航行$$20$$千米，再顺水航行$$60$$千米，也要用$$8$$小时，则在静水种这艘船每小时航行（~ ）千米．~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->流水行船问题->基本流水行船问题->四个速度->基本行程"], "answer_analysis": ["该船顺流行$$120$$千米，再逆流行$$72$$千米，共用时$$24$$小时；若逆流行$$40$$千米，再顺流行$$120$$千米，共用时$$16$$小时． 所以，逆流$$72-40=32$$千米，用时$$24-16=8$$小时，逆流船速是$$\\frac{32}{8}=4$$千米/时，则顺流船速$$40\\div \\left( 8-\\frac{24}{4} \\right)=20$$千米/时，静水船速是$$\\left( 4+20 \\right)\\div 2=12$$千米/时． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "909", "queId": "93735f2654344587acb165160a11279b", "competition_source_list": ["2017年全国小学生数学学习能力测评五年级竞赛复赛第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "下面$$4$$个数中，恰有一个数是两个相邻整数的乘积，这个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5096303$$ "}], [{"aoVal": "B", "content": "$$5096304$$ "}], [{"aoVal": "C", "content": "$$5096305$$ "}], [{"aoVal": "D", "content": "$$5096306$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->尾数特征->末一位数"], "answer_analysis": ["因为两个相邻整数相乘，积的个位数字只可能是$$0$$，$$2$$，$$6$$之中的一个，可以直接排除$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$，只留下$$\\text{D}$$． ", "分解质因数后重新搭配：$$5096306=\\left( 2\\times 1129 \\right)\\times \\left( 37\\times 61 \\right)=2258\\times 2257$$．注意：遇到像这种数比较大的时候，一般都会想到尾数规律，进而排除不成立的选项，而不是第二种直接求出具体的两个数．
"], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1677", "queId": "5350f37b43de4bf3a96ec103b41080a0", "competition_source_list": ["2006年四年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "$$3\\times 3\\times \\cdots \\times 3$$(2006个3)减去$$7\\times 7\\times \\cdots \\times 7$$(100个7)，得数的个位数字是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "2 "}], [{"aoVal": "C", "content": "6 "}], [{"aoVal": "D", "content": "8 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["每4个3相乘，尾数是1.因$$2006=4\\times 501+2$$，所以2006个3相乘的尾数为$$3\\times 3=9$$;每4个7相乘，尾数为1，因$$100=4\\times 25$$，所以100个7相乘的尾数为1;故两个积的个位数字相减，结果为$$9-1=8$$，选D "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1369", "queId": "28d01a0636e44d89a372d3d94c73bd1e", "competition_source_list": ["2015年上海走美杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有一筐苹果，第一次取出全部的一半多$$2$$个，第二次取出余下的一半少$$1$$个，筐中还剩$$4$$个，筐中原有苹果（ ）个。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->还原问题->两量还原问题"], "answer_analysis": ["第一次取后还剩下$$\\left( 4-1 \\right)\\times 2=6$$（个），原来有$$\\left( 6+2 \\right)\\times 2=16$$（个）。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1640", "queId": "c844947b4015460bafc1a83debf034a7", "competition_source_list": ["2010年六年级竞赛创新杯", "2010年第8届创新杯六年级竞赛初赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "某工程进行招标，甲、乙两工程队承包$$2\\frac{2}{5}$$天完成，需人民币$$1800$$元；乙、丙两工程队承包$$3\\frac{3}{4}$$天完成，需人民币$$1500$$元；甲、丙两工程队承包$$2\\frac{6}{7}$$天完成，需人民币$$1600$$元．现要求某队单独承包一星期内完成，所需费用最省，则被招标的是工程队. ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "甲或乙 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->列方程解应用题等量代换->等量代换加减型"], "answer_analysis": ["甲和乙一天可以完成$$\\frac{5}{12}$$，乙和丙一天可以完成$$\\frac{4}{15}$$，甲和丙一天可以完成$$\\frac{7}{20}$$，则三个工程队一天一共可以完成$$\\frac{1}{2}\\times \\left( \\frac{5}{12}+\\frac{4}{15}+\\frac{7}{20} \\right)=\\frac{31}{60}$$，则甲工程队的效率为：$$\\frac{31}{60}-\\frac{4}{15}=\\frac{1}{4}$$，乙工程队的效率为：$$\\frac{5}{12}-\\frac{1}{4}=\\frac{1}{6}$$，丙工程队的效率为：$$\\frac{4}{15}-\\frac{1}{6}=\\frac{1}{10}$$．按要求，只有甲、乙工程队能在一星期内完成．甲、乙、丙$$3$$队一天的工程款是：$$\\left( 1800\\div 2\\frac{2}{5}+1500\\div 3\\frac{3}{4}+1600\\div 2\\frac{6}{7} \\right)\\div 2=855$$（元）. 甲工程队一天的工程款是:$$855-1500\\div 3\\frac{3}{4}=455$$（元），甲工程队总的工程款是:$$455\\times 4=1820$$（元）； 乙工程队一天的工程款是：$$855-1600\\div 2\\frac{6}{7}=295$$（元），乙工程队总的工程款是:$$295\\times 6=1770$$（元）．所以乙工程队所需费用最省，乙工程队被招标. ", "甲乙每天$$1800\\div 2\\frac{2}{5}=1800\\div \\frac{12}{5}=750$$（元）；
\n乙丙每天$$1500\\div 3\\frac{3}{4}=1500\\div \\frac{15}{4}=400$$（元）；
\n甲丙每天$$1600\\div 2\\frac{6}{7}=1600\\div \\frac{20}{7}=560$$（元），
\n甲乙丙每天：$$\\left( 750+400+560 \\right)\\div 2=1710\\div 2=855$$（元），
\n则甲每天：$$855-400=455$$（元），
\n乙每天：$$855-560=295$$（元），
\n丙每天：$$855-750=105$$（元）．
\n因为一星期完成任务，所以乙队最省．
\n答：被招标的是乙工程队．
\n故选$$\\text{B}$$．
\n"], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3124", "queId": "fe5654925f83468c8b8aa18b3f2a909b", "competition_source_list": ["2017年新希望杯六年级竞赛训练题（一）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列说法正确的是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "女生人数与全班人数的比是$$4:9$$，男生人数与女生人数的比是$$4:5$$． "}], [{"aoVal": "B", "content": "比的前项和后项同时乘一个相同的数，比值不变． "}], [{"aoVal": "C", "content": "最简单的整数比，就是比的前项和后项都是质数的比． "}], [{"aoVal": "D", "content": "如果$$A:B=C$$，那么是比的前项，$$A$$是比的后项． "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比和比例->比->求比值"], "answer_analysis": ["$$\\text{A}$$项，男生人数与女生人数的比是$$5:4$$．$$\\text{B}$$项，同时乘的数不能是$$0$$．$$\\text{C}$$项，最简比的前后项是互质数，不一定都是质数． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2899", "queId": "8de91c17285744e58176708c281597fa", "competition_source_list": ["2008年第6届创新杯六年级竞赛初赛A卷第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列各分数中，分数值最小的是( )． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{7}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{15}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{15}{112}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数通分法->通分子"], "answer_analysis": ["由于$$\\frac{1}{7}=\\frac{2}{14}\\textgreater\\frac{2}{15}\\textgreater\\frac{2}{16}=\\frac{1}{8}$$，$$\\frac{15}{112}\\textgreater\\frac{14}{112}=\\frac{1}{8}$$，所以$$\\frac{1}{8} \\textless{} \\frac{1}{7}$$，$$\\frac{1}{8} \\textless{} \\frac{2}{15}$$，$$\\frac{1}{8} \\textless{} \\frac{15}{112}$$因此，这些分数中最小的是$$\\frac{1}{8}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "876", "queId": "e471215e1a56414ea5626733a8eff444", "competition_source_list": ["2019年广东广州羊排赛六年级竞赛第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "有一个数列$$1$$、$$1$$、$$2$$、$$3$$、$$5$$、$$8$$、$$13$$、$$21$$、$$\\cdots $$，从第三项开始，每一项都是前两项之和，这个数列第$$2019$$项除以$$4$$的余数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数的性质->余数找规律"], "answer_analysis": ["根据余数的性质，和的余数等于余数的和， $$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$\\cdots $$，$$\\div 4$$余$$1$$，$$1$$，$$2$$，$$3$$，$$1$$，$$0$$，$$1$$，$$1$$，$$2$$，$$3$$，$$1$$，$$0$$， $$1$$，$$1$$，$$\\cdots $$，$$6$$个为一周期，$$2019\\div 6=336$$（组）$$\\cdots \\cdots $$$$3$$（个）， 为周期的第$$3$$个，余数为$$2$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2746", "queId": "ecfed7315ae342d4a183c33dc4cf038b", "competition_source_list": ["2017年河南郑州豫才杯四年级竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "聪聪在课余时间最爱阅读科普类读物，还养成了随手摘抄的好习惯．以下是他在《少儿百科全书》丛书中摘录的几个数据，其中有两个他故意写错了．根据你的常识，你能看出来是哪些么？（~ ） （$$1$$）我国东北三省总面积是$$800$$万平方千米； （$$2$$）$$100$$万次心跳是一个正常人$$9.9$$天心跳的次数； （$$3$$）$$1$$亿张$$A4$$纸摞起来的厚度可以相当于$$3$$层楼的高度； （$$4$$）光的速度大约是每秒$$299800$$米； ", "answer_option_list": [[{"aoVal": "A", "content": "（$$1$$）和（$$4$$）~~~~~~~ "}], [{"aoVal": "B", "content": "（$$2$$）和（$$4$$）~~~~~~~~~~~~~~~ "}], [{"aoVal": "C", "content": "（$$1$$）和（$$3$$）~~~~~~~ "}], [{"aoVal": "D", "content": "（$$2$$）和（$$4$$） "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->单位换算->面积单位换算"], "answer_analysis": ["（$$1$$）我国东北三省总面积是$$78$$万平方千米，所以错了；（$$2$$）$$100$$万次$$1$$心跳是 一个正常人$$9.9$$天心跳的次数；（$$3$$）$$1$$层楼大概有$$3$$米，$$3$$层楼的高度大约有$$9$$米，$$1$$亿张$$A4$$纸摞起来的高度相当于$$9$$米；（$$4$$）光的速度大约是每秒$$3$$亿米每秒，所以错了；综上所述，（$$1$$）和（$$4$$）错了．~~~~ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "308", "queId": "5ba7181f4a3641e1bfd0cd1ded9f743b", "competition_source_list": ["2011年全国希望杯四年级竞赛初赛第20题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙、丁、戊五人猜测全班个人学科总成绩的前五名： 甲：``第一名是$$D$$，第$$5$$名是$$E$$．'' 乙：``第二名是$$A$$，第四名是$$C$$．'' 丙：``第三名是$$D$$，第四名是$$A$$．'' 丁：``第一名是$$C$$，第三名是$$B$$．'' 戊：``第二名是$$C$$，第四名是$$B$$．'' 若每个人都是只猜对一个人的名次，且每个名次只有一个人猜对，则第一、二、三、四、五名分别是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$DEBAC$$ "}], [{"aoVal": "B", "content": "$$CABDE$$ "}], [{"aoVal": "C", "content": "$$DCBAE$$ "}], [{"aoVal": "D", "content": "$$CADBE$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->半真半假", "课内体系->知识点->数学广角->推理->假设法判断真假"], "answer_analysis": ["假设法． 第一步：假设甲说的前半句是真的，那么$$D$$是第$$1$$名， 那么此时丙说的前半句错，后半句对．则$$A$$是第$$4$$名． 同理乙的后半句对，$$C$$是第$$4$$名．矛盾． 由此至甲的后半句对． 第二步：已知$$E$$是第$$5$$名，$$D$$不是第$$1$$名． 和第一名有关的话只剩下丁说的，设$$C$$是第$$1$$名． 则戊：``第$$2$$名是$$C$$，第$$4$$名是$$B$$''．可知前错后对，$$B$$是第$$4$$名． 且有乙：``第二名是$$A$$，第四名是$$C$$''．可知，$$A$$是第$$2$$名． $$D$$是第$$3$$名． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3292", "queId": "8c0b7e68bac5480eb6287362c424f195", "competition_source_list": ["2006年第4届创新杯五年级竞赛复赛第8题", "2006年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "有7个相同的小球放入4个不同的盒子中，每个盒子中至少放一个球，则共有( )种不同的放法. ", "answer_option_list": [[{"aoVal": "A", "content": "15 "}], [{"aoVal": "B", "content": "18 "}], [{"aoVal": "C", "content": "20 "}], [{"aoVal": "D", "content": "24 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->字典排序法->非数字排列"], "answer_analysis": ["$$7=1+1+1+4$$有$$4$$种放法. $$7=1+1+2+3$$有$$12$$种放法. $$7=2+2+2+1$$有$$4$$种放法. 共有$$4+12+4=20$$种不同的放法. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "226", "queId": "36afcb0fc2114826ab7c5acdd0a9f7e9", "competition_source_list": ["2015年第14届春蕾杯一年级竞赛初赛第5题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "小王说：小胖比我高；小胖说：我比园园矮；小朱说：我还没有小王高呢； 那么四个小朋友~\\uline{~~~~~~~~~~}~最高． ", "answer_option_list": [[{"aoVal": "A", "content": "小王 "}], [{"aoVal": "B", "content": "小胖 "}], [{"aoVal": "C", "content": "小朱 "}], [{"aoVal": "D", "content": "园园 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意分析可知，小胖比小王高，所以小胖的身高高于小王，小胖说比园园矮，所以园园的身高高于小胖，小朱没有小王高，所以小王的身高高于小朱，由此可知，四个人的身高从高到低位：园园$$\\textgreater$$小胖$$\\textgreater$$小王$$\\textgreater$$小朱． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3242", "queId": "35bd405001524d868cca42c78eff645f", "competition_source_list": ["2017年全国华杯赛竞赛初赛模拟题1第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "用红色和黄色给正方体的$$6$$个面染色，每个面必须染色，染色后经过旋转和翻转后相同的算同一种，共有种不同染色方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["分类计数 用$$1$$种颜色染色，则有$$2$$种不同的染色方式； 用$$2$$种颜色染色，再分类如下计数： （$$1$$）仅$$1$$个面是红色，有$$1$$种不同的染色方式； （$$2$$）仅$$2$$个面是红色，相对和相邻，有$$2$$种不同的染色方式； （$$3$$）仅$$3$$个面是红色，其中有$$2$$个相对，经适当转动，固定为红色上底面和红色下底面，仅有$$1$$种不同的染色方式：$$3$$个红色面两两相邻，仅有$$1$$种不同的染色方式； （$$4$$）有$$4$$个红色面，即有$$2$$个黄色面，不同染色方式的个数同（$$2$$）； （$$5$$）有$$5$$个红色面，即有$$1$$个黄色面，不同染色方式的个数同（$$1$$）． 共有$$10$$种不同的染色方式． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1173", "queId": "1dac340d4b0045bdb2d89e3072a6efa5", "competition_source_list": ["2018年湖北武汉创新杯小学高年级六年级竞赛初赛数学思维能力等级测试第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两人出售成本相同的同一种商品，甲按$$20 \\% $$的利润率定价，售出了$$15$$个；乙按照$$15 \\% $$的利润率定价，售出了$$24$$个，比较甲乙所获利润的多少，你的结论为． ", "answer_option_list": [[{"aoVal": "A", "content": "甲获利润多 "}], [{"aoVal": "B", "content": "乙获利润多 "}], [{"aoVal": "C", "content": "两人利润相同 "}], [{"aoVal": "D", "content": "无法比较谁多 "}]], "knowledge_point_routes": ["拓展思维->思想->赋值思想"], "answer_analysis": ["假设成本为$$100$$元，则甲的利润$$ =100\\times 20 \\%\\times 15 =300$$（元），乙的利润$$=100\\times 15 \\%\\times 24=360$$（元）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1553", "queId": "450716fcf23f4a9b920ab32a53c9c292", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "有一对姐妹的生日在同一天，但姐姐比妹妹的年龄大$$4$$岁，当两人的年龄和为$$50$$岁时，请问妹妹为多少岁？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$25$$ "}], [{"aoVal": "E", "content": "$$27$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题基本关系->年龄和"], "answer_analysis": ["姐姐与妹妹的年龄差是$$4$$岁，当两人年龄和为$$50$$岁时，妹妹的年龄等于$$(50-4)\\div 2=23$$（岁）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "41", "queId": "0aa7bc5bb73c41fbafc45fb81179e2ba", "competition_source_list": ["2014年迎春杯四年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在下面的每个方框中填入``$$+$$''或``$$-$$''，得到所有不同计算结果的总和是（ ） $$25\\square 9\\square 7\\square 5\\square 3\\square 1$$。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$540$$ "}], [{"aoVal": "B", "content": "$$600$$ "}], [{"aoVal": "C", "content": "$$630$$ "}], [{"aoVal": "D", "content": "$$650$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->巧填算符"], "answer_analysis": ["解：由于$$25+9+7+5+3+1=50$$，所以我们猜测$$0\\sim 50$$之间的所有偶数都有可能得到， $$0\\sim 50$$所有偶数的总和是$$(0+50)\\times 26\\div 2=650$$； 当把$$1$$前面的$$+$$号变成$$-$$号，可得$$25+9+7+5+3-1=48$$，比$$50$$小$$1\\times 2$$， 当把$$3$$前面的$$+$$号变成$$-$$号，可得$$25+9+7+5-3+1=44$$，比$$50$$小$$3\\times 2$$， 当把$$3$$和$$1$$前面的$$+$$号变成$$-$$号，可得$$25+9+7+5-3-1=42$$，比$$50$$小$$4\\times 2$$， 当把$$5$$前面的$$+$$号变成$$-$$号，可得$$25+9+7-5+3+1=40$$，比$$50$$小$$5\\times 2$$， $$\\cdots \\cdots $$ $$22=1+5+7+9$$，因此当把$$1$$，$$5$$，$$7$$，$$9$$前面的$$+$$号变成$$-$$号，可得$$25-9-7-5+3-1=6$$， $$24=3+5+7+9$$，因此当把$$3$$，$$5$$，$$7$$，$$9$$前面的$$+$$号变成$$-$$号，可得$$25-9-7-5-3+1=2$$， $$25=1+3+5+7+9$$，因此当把$$1$$，$$3$$，$$5$$，$$7$$，$$9$$前面的$$+$$号变成$$-$$号，可得$$25-9-7-5-3-1=0$$， 根据上述规律可得，但是数字$$2$$和$$23$$无法凑出来，那么偶数$$4$$和$$46$$无法取到， 所以答案是：$$650-4-46=600$$。 故选：B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1907", "queId": "ce38939d9b3a4495b8939f0b85e9515f", "competition_source_list": ["2011年第9届全国创新杯小学高年级六年级竞赛第18题"], "difficulty": "2", "qtype": "single_choice", "problem": "一次英语竞赛原定一等奖$$10$$人，二等奖$$20$$人，现将一等奖中最后$$4$$人调整为二等奖，这样得一等奖的平均分提高了$$3$$分，得二等奖的平均分提高了$$2$$分，则原一等奖平均分比二等奖平均分高~\\uline{~~~~~~~~~~}~分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16.4$$ "}], [{"aoVal": "B", "content": "$$16.5$$ "}], [{"aoVal": "C", "content": "$$16.6$$ "}], [{"aoVal": "D", "content": "$$17.5$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->逻辑分析"], "answer_analysis": ["可用方程解决，假设原来一等奖的平均分为$$a$$，二等奖的平均分为$$b$$，根据题意可以得到：$$10a+20b=6\\left( a+3 \\right)+24\\left( b+2 \\right)$$，得$$a-b=16.5$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1671", "queId": "53451c56366d45969a78dfa6eb8d5f16", "competition_source_list": ["2016年新希望杯小学高年级六年级竞赛训练题（六）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "政府机关每年花费$$20000$$元采购本子、笔、打印纸这三种办公用品，其中打印纸的费用是其他两种办公用品费用和的$$3$$倍．前半年政府每个月采购的打印纸数量都是一样多，$$7$$月份开始比$$6$$月份多$$10 \\% $$，之后$$8$$月份比$$6$$月份多$$20 \\% $$，$$9$$月份比$$6$$月份多$$30 \\% $$，$$\\cdots\\cdots$$，$$12$$月比$$6$$月多$$60 \\% $$．$$8$$月份政府机关花费（~ ~ ~ ）元采购打印纸．（精确到小数点后一位）~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1000.0$$ "}], [{"aoVal": "B", "content": "$$1063.8$$ "}], [{"aoVal": "C", "content": "$$1276.6$$ "}], [{"aoVal": "D", "content": "$$1333.3$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["全年打印纸总共花了$$20000\\times \\frac{3}{4}=15000$$元，设第一个月打印纸花了$$x$$元，则全年$$12x+\\frac{\\left( 0.1+0.6 \\right)\\times 6}{2}x=15000$$，解得：$$x=\\frac{150000}{141}$$． 那么$$8$$月份用了$$x\\times 120 \\% =\\frac{180000}{141}=1276.6$$元． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1674", "queId": "650731559ea642b68e7e64f3c26b8b4d", "competition_source_list": ["小学高年级六年级其它2014年数学思维能力等级测试第4题", "2014年第12届全国创新杯六年级竞赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "~小明收集了$$10$$册数学题，每册的题目数相同，并且连续编号（例如，第$$2$$册中的第一个题目比第$$1$$册中最后一个题的编号大$$1$$），一天他发现编号为$$351$$的数学题在第$$5$$册上，编号为$$689$$的数学题在第$$8$$册上，那么每册各有（~ ~ ~ ）个数学题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$70$$ "}], [{"aoVal": "B", "content": "$$71$$ "}], [{"aoVal": "C", "content": "$$85$$ "}], [{"aoVal": "D", "content": "$$87$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设每册$$n$$道题，则有 $$\\left { \\begin{matrix}351\\textgreater4n 689\\textless{}8n \\end{matrix} \\right.\\Rightarrow \\left { \\begin{matrix}n\\textless{}87\\frac{3}{4} n\\textgreater86\\frac{1}{8} \\end{matrix} \\right.$$ 故$$n$$只能为$$87$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3025", "queId": "d322b459694649369eb76fa0341949c6", "competition_source_list": ["2016年全国中环杯四年级竞赛初赛第11题"], "difficulty": "2", "qtype": "single_choice", "problem": "神庙里有一把古老的秤，对于重量小于$$1000$$克的物体，这把秤会显示其正确的重量；对于重量大于等于$$1000$$克的物体，这把秤会显示出一个大于等于$$1000$$的随机数．艾迪有五个物品，各自的重量都小于$$1000$$克，我们分别用$$P$$、$$Q$$、$$R$$、$$S$$、$$T$$表示它们的重量．将这五个物品两两配对放到秤上进行称重，得到下面的结果：$$Q+S=1200$$（克）、$$R+T=2100$$（克）、$$Q+T=800$$（克）、$$Q+R=900$$（克）、$$P+T=700$$（克）．那么这五个物品的重量从重到轻的顺序为~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$S\\textgreater R\\textgreater T\\textgreater P\\textgreater Q$$ "}], [{"aoVal": "B", "content": "$$S\\textgreater T\\textgreater R\\textgreater Q\\textgreater P$$ "}], [{"aoVal": "C", "content": "$$S\\textgreater R\\textgreater T\\textgreater Q\\textgreater P$$ "}], [{"aoVal": "D", "content": "$$S\\textgreater T\\textgreater R\\textgreater P\\textgreater Q$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->方程基础->不等式->不等式组求解"], "answer_analysis": ["$$Q+T=800$$①； $$Q+R=900$$②；$$P+T=700$$③；$$Q+S\\geqslant 1000$$④；$$R+T\\geqslant 1000$$⑤；由①②得：$$R\\textgreater T$$；由①③得：$$Q\\textgreater P$$；由②④得：$$S\\textgreater R$$；由②⑤得：$$T\\textgreater Q$$；所以：$$S\\textgreater R\\textgreater T\\textgreater Q\\textgreater P$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "800", "queId": "e880e88ce7054b00b9ccdce0e16b8ad0", "competition_source_list": ["2008年第6届创新杯六年级竞赛复赛第10题4分", "2008年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "在黑板上任意写一个正整数，在不是它的因数的正整数中，找出最小的正整数，擦去原数，写上找到的这个数.这样继续下去，最多只要擦( )次，黑板上就会出现2. ", "answer_option_list": [[{"aoVal": "A", "content": "2 "}], [{"aoVal": "B", "content": "3 "}], [{"aoVal": "C", "content": "4 "}], [{"aoVal": "D", "content": "5 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->操作问题"], "answer_analysis": ["用符号$$a\\to b$$表示``擦去原数$$a$$，改写为$$b$$''的一次操作. 如:$$5\\to 2$$(擦1次就可得到2); $${{2}^{3}}=8\\to 3\\to 2$$(擦2次就可得到2); $${2\\times 2\\times 3\\times 5\\times 7}=420\\to 8={{2}^{3}}\\to 3\\to 2$$(擦3次就可得到2). 再试几次，没有出现需要擦4次或更多次才出现2的情况，因此初步断定选B. 严格证明如下: （1）如果黑板上最初写的是奇数，只要擦1次就可出现2. （2）如果黑板上最初写的是偶数，擦过1次，改写的数如果是奇数，显然再擦1次就可出现2(共擦2次即可). （3）如果黑板上最初写的是偶数，擦过1次，改写的数如果还是偶数，那么可以证明这个偶数不可能有奇质数因子，即这个改写的数为$${{2}^{k}}$$($$k\\geqslant 1$$)，从而第2次改写得3，第3次改写得2. 事实上，设最初写的偶数为$$a$$，改写的偶数为$$b$$，则1，2，3，\\ldots，$$b-1$$都是$$a$$的因数，而$$b$$不能整除$$a$$，假设$$b$$有奇质数因子，则$$b={{2}^{k}}{{p}_{1}}^{{{a}_{1}}}\\cdots {{p}_{s}}^{{{a}_{s}}}$$为质因数分解式，其中$${{p}_{1}} \\textless{} {{p}_{2}}\\cdots \\textless{} {{p}_{s}}$$，均为奇质数，$$s\\geqslant 1$$，$$k\\geqslant 1$$.显然，$$1 \\textless{} {{2}^{k}} \\textless{} b$$，$$1 \\textless{} {{p}_{i}}^{{{a}_{i}}} \\textless{} b$$($$i=1$$，2\\ldots，$$s$$)，所以$${{2}^{k}}\\textbar a$$，$${{p}_{i}}^{{{a}_{i}}}\\textbar a$$($$i=1$$，2，\\ldots，$$s$$)，从而$${{2}^{k}}p_{1}^{{{a}_{1}}}\\cdots {{p}_{s}}^{{{a}_{s}}}\\textbar a$$，即$$b\\textbar a$$，与$$b$$不能整除$$a$$矛盾，故$$b$$没有奇质数因子. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3317", "queId": "fae9d99c035645c299487c37453f1d50", "competition_source_list": ["2009年五年级竞赛创新杯", "2009年四年级竞赛创新杯", "2009年三年级竞赛创新杯", "2009年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "从5名奥运志愿者中选出3名，分别从事翻译、导游、保洁三项不同的工作，每人承担一项，其中只有甲不能从事翻译工作，则不同的选派方案有( ). ", "answer_option_list": [[{"aoVal": "A", "content": "24种 "}], [{"aoVal": "B", "content": "36种 "}], [{"aoVal": "C", "content": "48种 "}], [{"aoVal": "D", "content": "60种 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->排列组合->排列组合综合"], "answer_analysis": ["因为翻译可以从除甲外的4人中选1个担当，有4种方法:又导游、保洁可从余下的4人中选2个担当，有$$4\\times 3=12$$种选法，所以有$$4\\times 12=48$$种选派方案. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "462", "queId": "93e6600e30314041917cc0aabe518cd9", "competition_source_list": ["2017年河南郑州东风杯竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "■◇◇●●●■◇◇●●●■◇◇●●●$$\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot $$，照这样的规律摆，第$$40$$个图形是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "■~~~~~ "}], [{"aoVal": "B", "content": "◇~~~~~~~~~~~~ "}], [{"aoVal": "C", "content": "●~~~~~ "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->能力->图形认知"], "answer_analysis": ["观察可知，■◇◇●●●，$$6$$个图形一个循环周期，$$40\\div 6=6\\cdots \\cdots4$$，所以第$$40$$个图形是一个周期中的第$$4$$个，是●． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2983", "queId": "dbc4913181d74045a9d97aee08998eec", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$2.\\dot{0}1\\dot{6}\\div \\left( 20.1\\dot{6}+2\\frac{19}{90} \\right)$$=~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\frac{10}{11}$$ "}], [{"aoVal": "C", "content": "$$\\frac{10}{22}$$ "}], [{"aoVal": "D", "content": "$$\\frac{10}{111}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->循环小数->循环小数混合运算"], "answer_analysis": ["原式$$=2\\frac{16}{999}\\times \\frac{1}{20\\frac{16-1}{90}+2\\frac{19}{90}}$$ $$=\\frac{2\\times 999+16}{999}\\times \\frac{1}{22\\frac{34}{90}}$$ $$=\\frac{2000+16-2}{999}\\times \\frac{90}{2014}$$ $$=\\frac{10}{111}$$ "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3046", "queId": "c5e73133b52248e9873f4d90ce814487", "competition_source_list": ["2017年第20届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "两数相减，减数增加$$7$$，要使差增加$$7$$，被减数应该． ", "answer_option_list": [[{"aoVal": "A", "content": "减少$$14$$ "}], [{"aoVal": "B", "content": "减少$$3$$ "}], [{"aoVal": "C", "content": "增加$$3$$ "}], [{"aoVal": "D", "content": "增加$$14$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["两数相减，当减数增加$$7$$，被减数不变则差减少了$$7$$． 又因为要使得差增加$$7$$， 所以被减数应增加：$$7+7=14$$， 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2764", "queId": "4e4ba2ebd91643c280c44c266409c628", "competition_source_list": ["2020年亚洲国际数学奥林匹克公开赛（AIMO）二年级竞赛决赛第23题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$199\\times 11-199\\times 2-199\\times 8-199$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$198$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$1990$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数乘法巧算之提取公因数（普通型）"], "answer_analysis": ["原式$$\\begin{eqnarray}\\&=\\&199\\times \\left( 11-2-8-1 \\right)\\end{eqnarray}$$ $$\\begin{eqnarray} \\&=\\&199\\times 0 \\&=\\&0\\end{eqnarray}$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2941", "queId": "e4aca3d96db445b19ace1db06883ba9a", "competition_source_list": ["其它改编自2015年全国希望杯六年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+\\frac{1}{32}=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{31}{32}$$ "}], [{"aoVal": "B", "content": "$$\\frac{15}{16}$$ "}], [{"aoVal": "C", "content": "$$\\frac{7}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{4}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["原式$$=\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{8}+\\frac{1}{16}+\\left( \\frac{1}{32}+\\frac{1}{32} \\right)-\\frac{1}{32}$$$$=1-\\frac{1}{32}=\\frac{31}{32}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2099", "queId": "fe0516ed3bbc486da970edf15976a0ea", "competition_source_list": ["2004年六年级竞赛创新杯", "2004年第2届创新杯六年级竞赛复赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "Ⅰ号混合液由柠檬汁、油、和醋以$$1:2:3$$的比例配成，Ⅱ号混合液由同样这三种液体以$$3:4:5$$的比例配成，将两种混合液倒在一起后，可以调成下面哪一种比例的混合液（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$2:5:8$$ "}], [{"aoVal": "B", "content": "$$4:5:6$$ "}], [{"aoVal": "C", "content": "$$3:5:7$$ "}], [{"aoVal": "D", "content": "$$5:6:7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->浓度问题->浓度基本题型"], "answer_analysis": ["设取Ⅰ号溶液$$6x$$升，那么一号溶液中有柠檬汁$$x$$升，油$$2x$$升，醋$$3x$$升.同样设取Ⅱ号溶液$$12y$$升，那么Ⅱ号溶液中有柠檬汁$$3y$$升，油$$4y$$升，醋$$5y$$升。现在可以用排除法对四个选项进行分析，为了便于描述，现设混合后三种液体的比为$$a:b:c$$.。 我们先观察选项$$A$$，由$$2(x+3y)=2x+6y\\textgreater2x+4y$$得$$2a\\textgreater b$$，而选项$$A$$中的$$2\\times 2 \\textless{} 5$$，所以选项$$A$$不能被调出； 再来观察选项$$B$$，$$(2x+4y)-(x+3y)=x+y$$，由$$4(x+y)=4x+4y\\textgreater2x+4y$$得$$4(b-a)\\textgreater b$$即$$3b\\textgreater4a$$，而选项$$B$$中的$$3\\times 5 \\textless{} 4\\times 4$$，所以选项$$B$$不能被调出；同理选项$$D$$中的$$3\\times 6 \\textless{} 4\\times 5$$，选项$$D$$也不能被调出； 那么能被调出的只有选项$$C$$。实际上，当$$x=3$$，$$y=1$$时，混合后三种液体的比正好为$$3:5:7$$，选$$C$$.。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3236", "queId": "2cc3b0eccf9944a291f35e1053fa09c9", "competition_source_list": ["2018年河南郑州K6联赛竞赛初赛第23题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "给一个正方体的六个面分别涂上红黄蓝三种颜色，任意抛$$50$$次，要想红色朝上次数最多，蓝色朝上次数最少，~\\uline{~~~~~~~~~~}~面涂红色，~\\uline{~~~~~~~~~~}~涂蓝色． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2,2$$ "}], [{"aoVal": "B", "content": "$$1,2$$ "}], [{"aoVal": "C", "content": "$$1,3$$ "}], [{"aoVal": "D", "content": "$$3,1$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计数模块->统计与概率->概率->基本概率->可能的情况"], "answer_analysis": ["可能性；要想红色朝上次数最少，则$$1$$面涂红色，$$2$$面染黄色． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3360", "queId": "a8a2f0bbd52c47dba3ca5ef6950bd071", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "春节时，妈妈买了$$3$$个完全一样的福袋，小悦想把$$10$$枚相同的一元硬币放到这三个福袋里，如果每个福袋里至少放$$1$$枚，不考虑福袋的先后顺序的话，共有（ ）种放法。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["解：枚举法，$$10$$能变成哪三个数相加，$$10=1+1+8=1+2+7=1+3+6=1+4+5=2+2+6=2+3+5=2+4+4=3+3+4$$，共$$8$$种。 故选:C。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3366", "queId": "8558971e5b01433fa23aa13e7b5351c2", "competition_source_list": ["2017年第20届世界少年奥林匹克数学竞赛四年级竞赛初赛（中国区）第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "丽丽有$$1$$角、$$2$$角、$$5$$角的硬币各$$5$$枚，若她想购买$$1$$元钱的商品，可以有种付钱的方法（不找零）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["$2$枚$$5$$角；$$1$$枚$$5$$角$$+2$$枚$$2$$角$$+1$$枚$$1$$角；$$1$$枚$$5$$角$$+1$$枚$$2$$角$$+3$$枚$$1$$角；$$1$$枚$$5$$角$$+5$$枚$$1$$角；$$5$$枚$$2$$角；$$4$$枚$$2$$角$$+2$$枚$$1$$角；$$3$$枚$$2$$角$$+4$$枚$$1$$角；一共是$$7$$种情况． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1406", "queId": "5a11adfd06184a43a2bc71be03024302", "competition_source_list": ["2005年第3届创新杯五年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "小华期末考试，数学、语文、英语三科的平均分是$$92$$分；语文、英语两科的平均分是$$90$$分；又英语比语文高$$3$$分，那么数学比语文高． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$分 "}], [{"aoVal": "B", "content": "$$7.5$$分 "}], [{"aoVal": "C", "content": "$$7$$分 "}], [{"aoVal": "D", "content": "$$6.5$$分 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["数学、语文、英语三科的总成绩是$$92\\times 3=276$$分， 语文、英语两科的总成绩是$$90\\times 2=180$$分， 那么数学的得分为$$276-180=96$$分， 又英语比语文高$$3$$分， 那么语文的成绩为$$\\left( 180-3 \\right)\\div 2=88.5$$分， 所以数学比语文高$$96-88.5=7.5$$分． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2050", "queId": "fd425719f59a4299ba099b7c4a89e9c0", "competition_source_list": ["2020年新希望杯二年级竞赛初赛（团战）第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$10$$名同学参加$$50$$米赛跑．跑到一半的时候，小明后面有$$5$$人，前面有$$4$$人，之后，没有人超过他，而小明又超过了$$3$$人到达终点．这次比赛没有并列名次，那么小明是． ", "answer_option_list": [[{"aoVal": "A", "content": "第一名 "}], [{"aoVal": "B", "content": "第二名 "}], [{"aoVal": "C", "content": "第三名 "}], [{"aoVal": "D", "content": "第四名 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["小明后面有$$5$$人，前面有$$4$$人，则他此时排在第$$5$$，之后，没有人超过他，小明又超过了$$3$$人后，他此时就是第二名． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "790", "queId": "69090f65d312430ebeca201700ef0a1a", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（五）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "十进制数$$25$$转换成二进制数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11101$$ "}], [{"aoVal": "B", "content": "$$1011$$ "}], [{"aoVal": "C", "content": "$$10101$$ "}], [{"aoVal": "D", "content": "$$11001$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->进制的性质与应用->进制间的互化"], "answer_analysis": ["短除法，倒取余数 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "316", "queId": "7ff6a31104e34079a604024863698bf3", "competition_source_list": ["2019年第22届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明早晨起床，要完成这几件事：起床穿衣$$3$$分钟，刷牙洗脸$$5$$分钟，在火炉上烧水煮面要$$10$$分钟，吃面要$$8$$分钟，整理房间$$5$$分钟，为了尽快做完这些事，最少用分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->统筹规划->简单时间统筹问题->同时进行问题"], "answer_analysis": ["从题干可以知道，小明早晨起床，要完成这几件事：起床穿衣$$3$$分钟，刷牙洗脸$$5$$分钟，在火炉上烧水煮面要$$10$$分钟，吃面要$$8$$分钟，整理房间$$5$$分钟． 为了最优化时间，先起床穿衣，再在火炉上烧水煮面，煮面的过程中刷牙洗脸、整理房间；最后吃面； $$3+10+8=21$$（分钟）； 即为了尽快做完这些事，最少用$$21$$分钟． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1780", "queId": "856bfa8ec9f04646985f351a483d2745", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "下列各组数中，平均数较大的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$与$$101$$之间的$$2$$倍数 "}], [{"aoVal": "B", "content": "$$1$$与$$101$$之间的$$3$$倍数 "}], [{"aoVal": "C", "content": "$$1$$与$$101$$之间的$$4$$倍数 "}], [{"aoVal": "D", "content": "$$1$$与$$101$$之间的$$6$$倍数 "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["一个等差数串的平均数，是这个数串首项与末项的平均数，简化计算后易知$$1$$与$$101$$之间$$4$$的倍数的平均数最大，其值为$$(4+100)\\div 2=52$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1689", "queId": "ace45189688d456ea776c9a523e41eac", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一条毛毛虫由幼虫长成成虫，每天长大一倍，$$30$$天能长到$$20$$厘米．问长到$$5$$厘米时要用天． ", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$ "}], [{"aoVal": "B", "content": "$$27$$ "}], [{"aoVal": "C", "content": "$$26$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["已知一条毛毛虫由幼虫长成成虫，每天长大一倍，$$30$$天能长到$$20$$厘米，可根据最后结果出发，逐步向前一步一步推理； 按从后往前推理可知最后一天长的是前一天的$$2$$倍，最后一天是$$20$$厘米，长到$$10$$厘米的时候是$$30$$天$$-1$$天； 根据以上分析可得$$29$$天时是$$10$$厘米，由此可推出长到$$5$$厘米时是几天，进而得出答案． $$30-1-1=28$$（天）， 答：长到$$5$$厘米时要用$$28$$天． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "372", "queId": "9b1b2510a8f2419db6dd8b96ea8838b9", "competition_source_list": ["2019年广东深圳全国小学生数学学习能力测评四年级竞赛初赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一种水草每天长一倍，到第$$3$$天时长到$$32$$平方米，那么到第天时长到$$64$$平方米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["每天长一倍，第$$3$$天是$$32$$平方米，那么第$$4$$天，就应该是$$32\\times 2=64$$平方米． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2460", "queId": "cbdacae97fe2479eb7f65a3f010345b3", "competition_source_list": ["2016年IMAS小学高年级竞赛第二轮检测试题第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "课后测 正整数$$a$$、$$b$$、$$c$$、$$d$$满足$$\\frac{1}{a-2013}=\\frac{1}{b+2014}=\\frac{1}{c-2015}=\\frac{1}{d+2016}$$，请问下列哪一项关于$$a$$、$$b$$、$$c$$、$$d$$的大小顺序是正确的？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$b\\textless{}d\\textless{}a\\textless{}c$$ "}], [{"aoVal": "B", "content": "$$d\\textless{}b\\textless{}a\\textless{}c$$ "}], [{"aoVal": "C", "content": "$$d\\textless{}a\\textless{}b\\textless{}c$$ "}], [{"aoVal": "D", "content": "$$d\\textless{}b\\textless{}c\\textless{}a$$ "}], [{"aoVal": "E", "content": "$$b\\textless{}d\\textless{}c\\textless{}a$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["因为$$\\frac{1}{a-2013}=\\frac{1}{b+2014}=\\frac{1}{c-2015}+\\frac{1}{d+2016}$$， 所以$$a-2013=b+2014=c-2015=d+2016$$， 所以$$d\\textless{}b\\textless{}a\\textless{}c$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "435", "queId": "c96a8864997f4ec7a1a7ce0d3af8ee55", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（一）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁四名白领同在一栋$$55$$层高的写字楼里办公，他们分别来自广西、湖南、四川和江苏．已知：①甲在的层数比丙在的层数高，但比丁在的层数低；②乙在的层数比四川人在的层数低；③广西人与四川人、江苏人相隔的层数一样；④广西人在的层数是湖南人和四川人在的层数的和．根据以上条件可知，甲是（~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "广西人 "}], [{"aoVal": "B", "content": "湖南人 "}], [{"aoVal": "C", "content": "四川人 "}], [{"aoVal": "D", "content": "江苏人 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["从层数看，丁\\textgreater 甲\\textgreater 丙\\textgreater 乙，江苏\\textgreater 广西\\textgreater 四川\\textgreater 湖南，所以甲是广西人． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "401", "queId": "e912de3cd0ca400b9f51918bc8c658b5", "competition_source_list": ["2012年第8届全国新希望杯小学高年级六年级竞赛复赛第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "现有$$A$$、$$B$$、$$C$$、$$D$$四个钢珠，用天平两个两个称，共称了六次，其中最重的是$$B$$和$$C$$，最轻的是$$A$$和$$D$$，第二重的是$$A$$和$$B$$．这四个钢珠按重量从重到轻排列依次是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$B\\textgreater A\\textgreater C\\textgreater D$$ "}], [{"aoVal": "B", "content": "$$B\\textgreater C\\textgreater A\\textgreater D$$ "}], [{"aoVal": "C", "content": "$$C\\textgreater B\\textgreater A\\textgreater D$$ "}], [{"aoVal": "D", "content": "$$A\\textgreater B\\textgreater D\\textgreater C$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据题意有$$B$$和$$C$$最重，第二重的是$$A$$和$$B$$，因此$$B\\textgreater C\\textgreater A\\textgreater D$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2201", "queId": "3149b168d5ff4aba8e7bf69022f4f536", "competition_source_list": ["2017年全国亚太杯五年级竞赛初赛第24题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲和乙两人分别从圆形场地的直径两端点同时开始以匀速按相反的方向绕此圆形路线运动．当乙走了$$250$$米以后，他们第一次相遇．在甲走完一周前$$80$$米处又第二次相遇．此圆形场地的周长为~\\uline{~~~~~~~~~~}~米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1440$$ "}], [{"aoVal": "B", "content": "$$1340$$ "}], [{"aoVal": "C", "content": "$$1250$$ "}], [{"aoVal": "D", "content": "$$1025$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["第一次相遇甲和乙共走了半圈，此时乙走了$$250$$ 米． 而第二次相遇，甲和乙一共走了$$3$$个半圈，所以此时乙走了$$250\\times 3=750$$米． 因为第二次相遇在甲走完一周前$$80$$米处，所以半圈的路线长为$$750-80=670$$米． 所以此圆形场地的周长为$$670\\times 2=1340$$米． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2279", "queId": "8ddba7c40f634002909ac622fc93d032", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "2. 同学甲、同学乙两人在游乐场的直行道上进行$$200$$米赛跑，当同学甲跑到终点时．同学乙还差$$40$$米，现在两人重新跑，而且速度和原来一样，要使两人同时到达终点，那么同学甲的起跑线应往后退米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$55$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["比例解行程；相等的时间内，甲跑了$$200$$米时，乙跑了$$160$$米，甲乙的速度比是$$200:160=5:4$$，要使两人同时到达终点，乙跑了$$200$$米的时间内甲要跑$$200\\div 4\\times 5=250$$米，所以同学甲的起跑线应后退$$250-200=50$$米． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2645", "queId": "999837297178473a8343f3feb3d270eb", "competition_source_list": ["2015年IMAS小学中年级竞赛第二轮检测试题第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$6\\otimes 2=6+66=72$$且$$2\\otimes 3=2+22+222=246$$，请问$$5\\otimes 3$$的值为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3735$$ "}], [{"aoVal": "B", "content": "$$605$$ "}], [{"aoVal": "C", "content": "$$615$$ "}], [{"aoVal": "D", "content": "$$625$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["由题意，发现规律$$a\\otimes b=a+aa+aaa+\\cdots +\\underbrace{aa\\cdots a}_{b}$$，则$$5\\otimes 3=5+55+555=615$$，故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "851", "queId": "7c4b93cfaaa74b3abd4acde315663beb", "competition_source_list": ["2018年美国数学大联盟杯五年级竞赛初赛第13题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "下列哪个选项是$$2$$个连续整数的和？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$111111$$ "}], [{"aoVal": "B", "content": "$$222222$$ "}], [{"aoVal": "C", "content": "$$444444$$ "}], [{"aoVal": "D", "content": "$$888888$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["在两个连续整数中，其中一个是双数，一个是单数．一个双数和一个单数的和一定是一个单数，所以两个连续整数的和也定是单数． 在$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$，$$4$$个选项中，只有$$\\text{A}$$是单数，其余$$3$$个都是双数． 故答案选：$$\\text{A}$$ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "570", "queId": "7d999547509d43aa9869b56a3adf782e", "competition_source_list": ["2017年第4届广东深圳鹏程杯小学高年级竞赛第3题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$69$$、$$90$$和$$125$$被大于$$1$$的整数$$m$$除的余数都相同，那么$$m$$是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数与运算->除法->整数除法->带余除法"], "answer_analysis": ["根据余数的可减性，这个数可以同时整除$$90-69=21$$和$$125-90=35$$，所以这个数能整除$$（21,35）=7$$，即这个数是$$7$$的因数，而$$7$$的因数有$$1$$和$$7$$，又因为这个数大于$$1$$，所以经检验，这个整数$$m$$为$$7$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1023", "queId": "45d6182ee5874ae4962a53b251e7c38a", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第8题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "【破1】鸡兔同笼，共有$$4$$个头，$$12$$条腿，有只兔． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["枚举法： 当有$$1$$只兔，$$3$$只鸡时，共$$1\\times4+3\\times2=10$$（条）腿． 当有$$2$$只兔，$$2$$只鸡时，共$$2\\times4+2\\times2=12$$（条）腿． 所以有$$2$$只兔． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "677", "queId": "2c9a5f667c9d45bf9175356afa9e8237", "competition_source_list": ["2016年全国美国数学大联盟杯小学高年级六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "做家务是郝美女老师的兴趣．她每$$8$$天进行一次大扫除，每$$11$$天洗一次衣服．如果星期天她做了两件家务，那么下一次她做两件家务会是哪一天？ ", "answer_option_list": [[{"aoVal": "A", "content": "星期三 "}], [{"aoVal": "B", "content": "星期四 "}], [{"aoVal": "C", "content": "星期五 "}], [{"aoVal": "D", "content": "星期六 "}]], "knowledge_point_routes": ["知识标签->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->两数的最小公倍数"], "answer_analysis": ["$$8\\times 11=88$$，$$88$$除以$$7$$余$$4$$，所以正好是周四. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1679", "queId": "5c202dc90ca6494a9f5f67bc79ab7354", "competition_source_list": ["2009年第7届创新杯六年级竞赛初赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$P$$和$$Q$$共做一事$$2$$天可完成，$$Q$$和$$R$$共做此事$$4$$天可完成，$$P$$和$$R$$共做此事$$2.4$$天可完成，$$P$$一人做完此事完成的天数是( )天. ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$P$$和$$Q$$一天可以完成$$\\frac{1}{2}$$，$$Q$$和$$R$$一天可以完成$$\\frac{1}{4}$$，$$P$$和$$R$$一天可以完成$$\\frac{1}{2.4}=\\frac{5}{12}$$，则三人一天一共可以完成$$\\frac{1}{2}\\times \\left( \\frac{1}{2}+\\frac{1}{4}+\\frac{5}{12} \\right)=\\frac{7}{12}$$，$$P$$的工作效率为$$\\frac{7}{12}-\\frac{1}{4}=\\frac{1}{3}$$，因此$$P$$单独做需要$$3$$天完成. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "563", "queId": "09fbcedfa00442f4824cdb5741b4486b", "competition_source_list": ["2005年六年级竞赛创新杯", "2005年第3届创新杯六年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "任意两个��数之和（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "一定是偶数 "}], [{"aoVal": "B", "content": "一定是质数 "}], [{"aoVal": "C", "content": "一定是合数 "}], [{"aoVal": "D", "content": "可能是偶数，也可能是质数，也可能是合数 "}]], "knowledge_point_routes": ["课内体系->知识模块->数与代数", "拓展思维->拓展思维->数论模块->质数与合数"], "answer_analysis": ["$$3+5=8$$，$$8$$为偶数也为合数；$$2+3=5$$，$$5$$为质数，故选D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2839", "queId": "8965d51026a04a789cb7d05648ad0758", "competition_source_list": ["2010年五年级竞赛明心奥数挑战赛", "2010年六年级竞赛明心奥数挑战赛"], "difficulty": "0", "qtype": "single_choice", "problem": "下列各选项中的值最接近$$9$$的是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$9.2$$ "}], [{"aoVal": "B", "content": "$$8.17$$ "}], [{"aoVal": "C", "content": "$$8.7$$ "}], [{"aoVal": "D", "content": "$$9.21$$ "}], [{"aoVal": "E", "content": "$$8.71$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->小数比较大小"], "answer_analysis": ["$$9.21\\textgreater9.2\\textgreater9\\textgreater8.71\\textgreater8.7\\textgreater8.17$$，$$9.2-9=0.2 \\textless{} 9-8.71=0.29$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2287", "queId": "d6cc0a9fc6804fbdabe72633115987a3", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）"], "difficulty": "1", "qtype": "single_choice", "problem": "一列火车通过一座长$$320$$米的桥用了$$105$$秒，当它通过$$860$$米的隧道时，速度是过桥速度的$$2$$倍，结果用了$$120$$秒，火车通过大桥时的速度是每秒米；火车的车身长度为米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$；$$100$$ "}], [{"aoVal": "B", "content": "$$4$$；$$100$$ "}], [{"aoVal": "C", "content": "$$8$$；$$100$$ "}], [{"aoVal": "D", "content": "$$8$$；$$200$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["若通过$$860$$米隧道时速度不变则需要$$120\\times 2=240$$（秒），火车过桥速度：$$\\left( 860-320 \\right)\\div \\left( 240-105 \\right)=4$$（米/秒）；火车车身长：$$105\\times 4-320=100$$（米）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2959", "queId": "d270ab5822df4854a00727860c841cde", "competition_source_list": ["2019年全国小学生数学学习能力测评四年级竞赛复赛第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一位疯狂的艺术家为了寻找灵感，把一张厚为$$0.1$$毫米的很大的纸对半撕开，重叠起来，然后再撕成两半叠起来．假设他如此重复这一过程$$25$$次，那么这叠纸会． ", "answer_option_list": [[{"aoVal": "A", "content": "像山一样高 "}], [{"aoVal": "B", "content": "像一栋房子高 "}], [{"aoVal": "C", "content": "像一个人一样高 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$2$$的十次方为$$1024$$，$$2$$的五次方为$$32$$， 所以$$0.1\\times1024\\times1024\\times32=3355443.2$$毫米$$=3355.4432$$米． 故像山一样高． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2869", "queId": "a41d0f94fe8a431f87b4b489050cddc6", "competition_source_list": ["2018年IMAS小学高年级竞赛（第一轮）第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$\\frac{2}{3}$$的分母加上$$6$$，且要使得原分数大小不变﹐请问分子应该加上多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["$$\\frac{2}{3}=\\frac{6}{9}=\\frac{2+4}{3+6}$$． 故选$$\\text{B}$$． ", "可知原分数的分母为分子的$$1.5$$倍．
\n当分母增加$$6$$之后，为了使这个比例不变，分子应该加上$$6\\div 1.5=4$$．
\n故选$$\\text{B}$$．
"], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1579", "queId": "ff8080814518d524014519098ce20320", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "一些糖果，如果每天吃$$3$$个，十多天吃完，最后一天只吃了$$2$$个，如果每天吃$$4$$个，不到$$10$$天就吃完了，最后一天吃了$$3$$个．那么，这些糖果原来有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$35$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->盈亏问题->盈亏基本类型->盈亏基本类型盈亏问题"], "answer_analysis": ["如果每天吃$$3$$个，十多天吃完，最后一天只吃了$$2$$个，说明糖果至少有$$3\\times 10+2=32$$（个），且糖果数应除以$$3$$ 余$$2$$；如果每天吃$$4$$个，不到$$10$$天就吃完了，最后一天吃了$$3$$个，说明糖果至多有$$4\\times 8+3=35$$（个），且除以$$4$$余$$3$$． 综上，糖果有$$35$$个． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "858", "queId": "db23e19dbf7044ff8899d17b5d19ee08", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛复赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个整数去除$$151$$、$$197$$、$$238$$所得$$3$$个余数的和是$$31$$，所得$$3$$个商的和是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["$$151+197+238-31=555$$是除数的倍数，$$555=3\\times 5\\times 37$$，由于余数和是$$31$$，检验$$37$$成立，商分别是$$4$$、$$5$$、$$6$$，所以和是$$15$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3129", "queId": "f0ba74d663054a38beb77c141ef70728", "competition_source_list": ["其它改编自2012年全国希望杯六年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$\\frac{251}{2008\\times 2009}+\\frac{251}{2009\\times 2010}=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2009}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2010}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{8040}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{4020}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->裂项与通项归纳->分数裂项->分数裂差->两分数间接裂差"], "answer_analysis": ["$$\\begin{align}\\& ~ \\frac{251}{2008\\times 2009}+\\frac{251}{2009\\times 2010} \\& =251\\times(\\frac{1}{2008}-\\frac{1}{2009}+\\frac{1}{2009}-\\frac{1}{2010}) \\&=251\\times (\\frac{1}{2008}-\\frac{1}{2010}) \\& =\\frac{1}{8040} \\end{align}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2046", "queId": "ef697593c2224315a3fa7162a329d2c8", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一种特殊的计算器，当输入一个$$10$ $49$$的自然数后，计算器会先将这个数乘$$2$$，然后将所得结果的十位和个位顺序颠倒，再加$$2$$后显示出最后的结果．那么，下列四个选项中，可能是最后显示的结果． ", "answer_option_list": [[{"aoVal": "A", "content": "$$44$$ "}], [{"aoVal": "B", "content": "$$43$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$41$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->运算求解"], "answer_analysis": ["倒推．$$44$$ 对应的是$$44-2=42$$，颠倒后是$$24$$，除以$$2$$ 为$$12$$．符合条件．其他的均不符合条件． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1482", "queId": "562ff3747b174afe858aed87590d1816", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "祖玛游戏中，龙嘴里不断吐出很多颜色的龙珠，先$$4$$颗红珠，接着$$3$$颗黄珠，再$$2$$颗绿珠，最后$$1$$颗白珠，按此方式不断重复，从龙嘴里吐出的第$$2000$$颗龙珠是． （2017年迎春杯二年级竞赛初赛第$$8$$题） ", "answer_option_list": [[{"aoVal": "A", "content": "红珠 "}], [{"aoVal": "B", "content": "黄珠 "}], [{"aoVal": "C", "content": "绿珠 "}], [{"aoVal": "D", "content": "白珠 "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["$$2000\\div (4+3+2+1)=200$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "671", "queId": "30bba94f8bab4cbc81b54a376566b4c8", "competition_source_list": ["2018年全国小学生数学学习能力测评四年级竞赛复赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "用$$8$$个$$3$$和$$1$$个$$0$$组成的九位数有若干个，其中除以$$4$$余$$1$$的数有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["用$$8$$个$$3$$和$$1$$个$$0$$组成的九位数，其中$$1$$个$$3$$必须放在最高位上， 所以这个$$9$$位数最后两位只有三种情况：即$$03$$，$$30$$，$$33$$； $$4$$的倍数的特点是末尾两位数是$$4$$的倍数，整个数就是$$4$$的倍数， 要使除以$$4$$余数是$$1$$，那么末尾两位数是比$$4$$的倍数多$$1$$的数， 只有$$33$$符合要求，所以前$$7$$位是由$$1$$个$$0$$和$$6$$个$$3$$组成，可以是： $$303333333$$， $$330333333$$， $$333033333$$， $$333303333$$， $$333330333$$， $$333333033$$， 一共有$$6$$个． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2578", "queId": "599e32dd60a1495eb5273614e12f9c68", "competition_source_list": ["2013年IMAS小学高年级竞赛第一轮检测试题第18题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "小明在做一道将三个正整数连乘的习题时，错当把这三个正整数相加．令人惊奇的是，他所得的结果竟然与这三个正整数连乘的正确答案相同．请问这三个正整数的总和是什么？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["设这三个数从小到大分别为$$a$$、$$b$$、$$c$$．显然可知$$a$$、$$b$$、$$c$$不可以都等于$$1$$，否则$$abc=1\\ne 1+1+1=3$$，所以$$c\\geqslant 2$$．由题意可知，$$a$$只能是$$1$$，否则三个数的乘积将大于它们之和（因为$$abc\\geqslant 2bc=bc+bc\\textgreater2b+c\\geqslant a+b+c$$），所以$$1+b+c=bc$$． 同样，如果$$b$$大于或等于$$3$$，则$$bc$$大于或等于$$c$$的$$3$$倍，而$$1+b+c$$小于$$c$$的$$3$$倍，所以$$b$$只能是$$2$$． 因此，$$1+2+c=2c$$，得$$c=3$$． 故这个三个正整数之总和为$$1+2+3=6$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2540", "queId": "c2c1b76f86af4a4a8df6c6e89a39282a", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛B卷第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个分数的分母是分子的$$5$$倍，分子和分母同时扩大$$3$$倍，相当于分子增加$$8$$，分母增加． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["分数的分子和分母同时乘或者除以一个相同的数（$$0$$除外），分数的大小不变．这叫做分数的基本性质．一个分数的分母是分子的$$5$$倍，比如分子是$$A$$，分母是$$5A$$；扩大$$3$$倍这个分数变为：$$\\frac{A}{5A}=\\frac{3A}{15A}$$；根据题意我们道，相当于分子增加$$8$$，也就是$$3A-A=8$$，所以$$A=4$$； $$15A-5A=15\\times 4-5\\times 4=40$$， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1217", "queId": "1a3e2f8bbe9c42328e5b6efcbf59d9aa", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "十个数的平均数是$$57$$，从小到大排列，前四个数的平均数是$$37$$，后七个数的平均数是$$67$$，则第四个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$44$$ "}], [{"aoVal": "B", "content": "$$47$$ "}], [{"aoVal": "C", "content": "$$52$$ "}], [{"aoVal": "D", "content": "$$62$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$37\\times 4+67\\times 7-57\\times 10=47$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "966", "queId": "fe2875e124fe46b99b54f8911bbfa8cc", "competition_source_list": ["竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有一个长方体，长、宽、高都是整数厘米，它的相邻三个面的面积分别是96平方厘米、40平方厘米和60平方厘米，这个长方体的体积是( )立方厘米. ", "answer_option_list": [[{"aoVal": "A", "content": "300 "}], [{"aoVal": "B", "content": "360 "}], [{"aoVal": "C", "content": "400 "}], [{"aoVal": "D", "content": "480 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->几何模块->立体图形->长方体与正方体->长方体与正方体基本概念运用->长方体表面积"], "answer_analysis": ["相乘分解，因为$$96\\times 40\\times 60={{\\left( {{2}^{5}}\\times 3\\times 5 \\right)}^{2}}$$，所以长方体体积为$${{2}^{5}}\\times 3\\times 5=480$$（立方厘米） "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "391", "queId": "a90559c673b0449ea56943d1b30e3954", "competition_source_list": ["2011年全国华杯赛竞赛初赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "老师问学生：``昨天你们有几个人复习数学了？'' 张：``没有人．'' 李：``一个人．'' 王：``二个人．'' 赵：``三个人．'' 刘：``四个人．'' 老师知道，他们昨天下午有人复习，也有人不复习，复习了的人说的都是真话，没复习的人说的都是假话．那么，昨天这$$5$$个人中复习数学的有个人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["知识标签->课内题型->综合与实践->智巧趣题->逻辑推理->排除矛盾法"], "answer_analysis": ["$$5$$名学生说的话相互矛盾，只能有$$1$$人说的是真话，则只有李复习了，说的是真话． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "478", "queId": "a1ce585018a64daba0bb9b800d691608", "competition_source_list": ["2019年第24届YMO六年级竞赛决赛第7题3分"], "difficulty": "3", "qtype": "single_choice", "problem": "有一个$$12$$级的楼梯．某人每次能登上$$1$$级或$$2$$级或$$3$$级，现在他要从地面登上第$$10$$级，有种不同的方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$149$$ "}], [{"aoVal": "B", "content": "$$244$$ "}], [{"aoVal": "C", "content": "$$264$$ "}], [{"aoVal": "D", "content": "$$274$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->归纳递推->斐波那契数列递推", "Overseas Competition->知识点->组合模块->操作与策略->归纳递推"], "answer_analysis": ["设第$$n$$级有$${{a}_{n}}$$种登的方式， 则第$$n-3$$，$$n-2$$，$$n-1$$级再分别登$$3$$，$$2$$，$$1$$级可到第$$n$$级， 则$${{a}_{n}}={{a}_{n-3}}+{{a}_{n-2}}+{{a}_{n-1}}$$， 而$${{a}_{1}}=1$$，$${{a}_{2}}=1+1=2$$，$${{a}_{3}}=1+1+2=4$$， 故$${{a}_{4}}=1+2+4=7$$， $${{a}_{5}}=2+4+7=13$$， $${{a}_{6}}=4+7+13=24$$， $${{a}_{7}}=7+13+24=44$$， $${{a}_{8}}=13+24+44=81$$， $${{a}_{9}}=24+44+81=149$$， $${{a}_{10}}=44+81+149=274$$． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3258", "queId": "31d72939d4194d2bb7ac9bf142d112f3", "competition_source_list": ["2016年创新杯六年级竞赛训练题（四）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "三娃的书架只有$$1$$本书，后来又新买$$3$$本不同的书．他想把书放到书架上，那么这$$4$$本书一共有种不同的放法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数学广角->排列组合->排列数"], "answer_analysis": ["$$24$$ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2328", "queId": "f08509fc0fd64e6ba5c8ff4114ad01b8", "competition_source_list": ["2013年第25届广东广州五羊杯六年级竞赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "张涛周日去爬山，他上山的速度为$$0.5\\text{m/s}$$，登顶后立刻下山，下山的速度为$$2\\text{m/s}$$，假设上山下山的路程一样，问小张上下山的平均速度为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.7\\text{m/s}$$ "}], [{"aoVal": "B", "content": "$$0.8\\text{m/s}$$ "}], [{"aoVal": "C", "content": "$$1\\text{m/s}$$ "}], [{"aoVal": "D", "content": "$$1.25\\text{m/s}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->平均速度->设数法"], "answer_analysis": ["平均速度$$=$$总路程$$\\div $$总时间，将上山的路程看作单位``$$1$$''， 则有$$\\dfrac{2}{\\dfrac{1}{0.5}+ \\dfrac{1}{2}}= \\dfrac{4}{5}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "127", "queId": "2734b200480843028a9f844f4a5b8f53", "competition_source_list": ["2013年全国希望杯六年级竞赛初赛第18题"], "difficulty": "0", "qtype": "single_choice", "problem": "某次数学大比拼，甲、乙、丙$$3$$人中只有一人获奖. 甲说：``我获奖了.'' 乙说：``我没获奖.'' 丙说：``甲没有获奖.'' 他们的话中只有一句是真话，则获奖的是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["甲和丙的话相互矛盾，一定是一真一假，所以乙说的是假话，那么乙获奖了． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2570", "queId": "62c13b9671034ba88afe78ddddac4b45", "competition_source_list": ["2014年全国迎春杯竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2014$$年$$2$$月$$6$$日是星期四，小胖决定从这天起（含$$2$$月$$6$$日）练习计算，一直练习到$$2$$月$$17$$日，（含$$2$$月$$17$$日）开学为止．但是中间如果遇到周六和周日，小胖还是决定休息一下，不做练习．已知他第一天做$$1$$道题，第二天做$$3$$道题，第三天做$$5$$道题，依此变化做下去，那么小胖这段时间一共做了道计算练习题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$144$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$81$$ "}], [{"aoVal": "D", "content": "$$64$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["从$$2$$月$$6$$日到$$2$$月$$17$$日为止，一共有$$17-6+1=12$$（天）其中有$$2$$个星期六，星期日．工作了$$12-4=8$$（天），共完成$$1+3+5+7+9+11+13+15=64$$（题）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1972", "queId": "ea06561cbf5d4986b75e0e7d8b51cae8", "competition_source_list": ["2021年第4届山东青岛市南区京山杯六年级竞赛决赛A卷第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一件上衣按成本价提高$$50 \\%$$后，以$$105$$元售出，则这件上衣的利润为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$元 "}], [{"aoVal": "B", "content": "$$25$$元 "}], [{"aoVal": "C", "content": "$$30$$元 "}], [{"aoVal": "D", "content": "$$35$$元 "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设成本为$$x$$元，由题意可得等量关系：$$\\left( 1+50 \\% \\right)\\times $$成本价$$=$$售价，进而得到方程，可算出成本价，再利用售价$$-$$成本价$$=$$利润．此题主要考查了一元一次方程的应用，关键是正确理解题意，找出题目中的等量关系，列出方程． 设成本为$$x$$元，由题意得： $$\\left( 1+50 \\% \\right)x=105$$， 解得：$$x=70$$， $$105-70=35$$（元）， 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1564", "queId": "4de818336d054fe099de86203fb3ee61", "competition_source_list": ["2003年第1届创新杯六年级竞赛初赛第8题", "2003年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "据日本一个联合小组最近在东京宣布，他们使用1秒钟能进行2万亿计算的超级计算机，将圆周率π计算到小数后12411亿位，计算过程耗时601小时56分，假如一个人1秒钟读2个阿拉伯数字，那么他日夜不停地读完这个π的数值要用的时间是（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "1万年 "}], [{"aoVal": "B", "content": "2万年 "}], [{"aoVal": "C", "content": "3万年 "}], [{"aoVal": "D", "content": "4万年 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->归一归总问题->乘除法应用->除法应用"], "answer_analysis": ["这个人1秒钟读2个数字，那么1天可以读$$2\\times 60\\times 60\\times 24=172800$$(个)数字，一年（平年）可以读$$172800\\times 365=6307200$$个数字，那么1万年大约可以读6307亿个数字，现要读12411亿个数字，大约要读$$12411\\div 6307\\approx 2$$(万年). "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1718", "queId": "774d9dd481744fb18a70499810870ed2", "competition_source_list": ["2017年第20届世界少年奥林匹克数学竞赛四年级竞赛初赛（中国区）第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "鸭子和兔子共有$$30$$只，共有脚$$86$$只．问鸭子和兔子分别有只． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$、$$18$$ "}], [{"aoVal": "B", "content": "$$9$$、$$21$$ "}], [{"aoVal": "C", "content": "$$17$$、$$13$$ "}], [{"aoVal": "D", "content": "$$19$$、$$11$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->原型题"], "answer_analysis": ["假设都是鸭子，共有脚：$$2\\times 30=60$$（只）， 所以有兔子$$(86-60)\\div (4-2)=13$$（只）， 鸭子：$$30-13=17$$（只）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1191", "queId": "873b349f01f7498685ba640cb24b1011", "competition_source_list": ["2015年上海走美杯五年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有一筐苹果，第一次取出全部的一半多$$2$$个，第二次取出余下的一半少$$3$$个，筐中还剩$$24$$个，筐中原有苹果（ ）个。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$104$$ "}], [{"aoVal": "B", "content": "$$88$$ "}], [{"aoVal": "C", "content": "$$46$$ "}], [{"aoVal": "D", "content": "$$96$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->还原问题->两量还原问题"], "answer_analysis": ["方法一：第一次取后还剩下$$\\left( 24-3 \\right)\\times 2=42$$（个），原来有$$\\left( 42+2 \\right)\\times 2=88$$（个）； 方法二：设原有苹果$$x$$个，$$(\\frac{x}{2}-2)\\times \\frac{1}{2}+3=24$$，解得$$x=88$$。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2741", "queId": "ff8080814518d524014519271a710538", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "对于任何自然数，定义$$n!=1\\times 2\\times 3\\times \\cdots \\times n$$．那么算式$$2014!-3!$$的计算结果的个位数字是（~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["由新定义：$$n!=1×2×3×\\ldots×n$$得： $$2014!=1×2×3×4×5×\\ldots×2013×2014$$ $$=1×3×4×6×7×8×\\ldots×2013×2014×10$$ 所以$$1×3×4×6×7×8×\\ldots×2013×2014×10$$是$$10$$的倍数， 所以$$2014!$$的个位数为$$0$$； $$3!=1×2×3=6$$ 所以$$2014!-3!$$的个位数也就为：$$10-6=4$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1502", "queId": "3bdf3557003e4dea88801ba54a617bdb", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "$$2017$$年元旦是星期二，$$2019$$的元旦节是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "一 "}], [{"aoVal": "B", "content": "二 "}], [{"aoVal": "C", "content": "三 "}], [{"aoVal": "D", "content": "四 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["从$$2019$$年的$$1$$月$$1$$日至$$2019$$年的$$10$$月$$1$$日共：$$31+28+31+30+31+30+31+31+30+1=274$$（天）， 根据``二、三、四、五、六、日、一''的周期规律：$$274\\div 7=39$$（周）$$\\cdots \\cdots 1$$（天）． 所以$$10$$月$$1$$日是这个周期的第$$1$$天，即星期二． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2976", "queId": "cdfa9467a0674a9b9d6d4110c30697ee", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$2015\\times 2015-2014\\times 2016=$$（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2014$$ "}], [{"aoVal": "C", "content": "$$2015$$ "}], [{"aoVal": "D", "content": "$$2016$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$${{2015}^{2}}-(2015-1)\\times (2015+1)={{2015}^{2}}-{{2015}^{2}}+1=1$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3101", "queId": "d916a0ec7e0d48839e63c7b9786338fd", "competition_source_list": ["2020年第24届YMO三年级竞赛决赛第4题3分", "2019年第24届YMO三年级竞赛决赛第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "按规律写算式：$$100+2$$，$$98-5$$，$$96+8$$，$$94-11$$，$$\\cdots $$第$$10$$个算式是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$80-27$$ "}], [{"aoVal": "B", "content": "$$82-29$$ "}], [{"aoVal": "C", "content": "$$82+29$$ "}], [{"aoVal": "D", "content": "$$82-27$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数", "Overseas Competition->知识点->组合模块->操作与策略->归纳递推"], "answer_analysis": ["规律：每个算式第一个数是从$$100$$开始，依次少$$2$$， 第二个数是从$$2$$开始，依次多$$3$$， 算式中间的符号是``$$+-$$''循环， 则第$$10$$个算式，中间符号为``$$-$$''， 第$$1$$个数为$$100-9\\times 2=82$$， 第$$2$$个数为$$2+3\\times 9=29$$， 则第$$10$$个算式为``$$82-29$$''． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2022", "queId": "b43ec7dc9a7e49ec99e1257a720e5feb", "competition_source_list": ["2015年世界少年奥林匹克数学竞赛三年级竞赛复赛A卷第7题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "一只蜗牛沿着柱子往上爬，白天往上爬$$5$$米，晚上又下滑$$3$$米，已知这只蜗牛是第$$6$$天爬上柱子顶端的，那么这根柱子最长有~\\uline{~~~~~~~~~~}~米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "课内体系->能力->实践应用"], "answer_analysis": ["因为蜗牛白天爬$$5$$米，晚上下滑$$3$$米，所以相当于蜗牛一天往上爬$$5-3=2$$米，这只蜗牛前五天一共爬了$$5\\times 2=10$$米．因为这只蜗牛是第六天爬上柱子顶端的，所以这根柱子最长有$$10+5=15$$米． 故答案为：$$15$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3103", "queId": "e6be16c9fc8c4930bd01497df967cacd", "competition_source_list": ["2019年第24届YMO四年级竞赛决赛第9题3分", "2020年第24届YMO四年级竞赛决赛第9题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$5$$，$$a$$，$$b$$，$$c$$，$$4035$$这五个数成等差数列，则$$b=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2017$$ "}], [{"aoVal": "B", "content": "$$2018$$ "}], [{"aoVal": "C", "content": "$$2019$$ "}], [{"aoVal": "D", "content": "$$2020$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$5$$，$$a$$，$$b$$，$$c$$，$$4035$$是等差数列， 则$$5$$，$$b$$，$$4035$$也是等差数列， 则$$b=\\left( 4035+5 \\right)\\div 2$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=4040\\div 2$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=2020$$． 选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2175", "queId": "1a2c235f999f4acea1b4217e981b1db7", "competition_source_list": ["2018年全国小学生数学学习能力测评五年级竞赛初赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两船在静水中的速度分别为每小时$$36$$千米和每小时$$28$$千米，今从相隔$$192$$千米的两港同时相对行驶，甲船逆水而上，乙船顺水而下，那么小时后两船相遇． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$2.5$$ "}], [{"aoVal": "D", "content": "$$3.5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->流水行船问题->基本流水行船问题->四个速度->基本行程"], "answer_analysis": ["相遇问题中，路程和$$=$$速度和$$\\times $$相遇时间， 所以：相遇时间$$=$$路程和$$\\div $$速度和． 甲船逆水而上速度是甲船静水速度减去水速，即：$$36-$$水速； 乙船顺水而下速度是乙船静水速度加上水速，即：$$28+$$水速； 甲乙速度和为：$$36-$$水速$$+28+$$水速$$=64$$（千米/小时）． 甲乙相遇时间为：$$192\\div64=3$$（小时）． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "357", "queId": "9af469b99dca46a988d1f7f431df4b34", "competition_source_list": ["2018年四川成都锦江区四川师范大学附属第一实验中学小升初模拟11第9题3分", "2015年湖北武汉世奥赛五年级竞赛模拟训练题（四）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "一把钥匙只能开一把锁．现有$$12$$把钥匙和$$11$$把锁，但不知道哪把钥匙开哪把锁，最少要试次才能保证配好全部的锁的钥匙． ", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$55$$ "}], [{"aoVal": "D", "content": "$$65$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["从最不利的情况考虑，第$$1$$把锁需要试$$11$$次；第$$2$$把锁需要试$$10$$次$$\\ldots \\ldots $$；一共需要试$$2+3+\\cdots \\cdots +11=65$$（次）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3462", "queId": "c6e3e897626b4ac3a302cae9b74d8c79", "competition_source_list": ["2019年第24届YMO三年级竞赛决赛第3题3分", "2020年第24届YMO三年级竞赛决赛第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$10$$本相同的笔记本分给$$3$$个人，每人至少一本，共有~\\uline{~~~~~~~~~~}~种不同的分配方案． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$45$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["根据题意，$$10$$个相同的小球分给$$3$$个人，每人至少$$1$$个，就是将$$10$$个球分成$$3$$组， 一个人最多分：$$10-1-1=8$$（支）， $$8+7+6+5+4+3+2+1=36$$（种）， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3328", "queId": "6da3a7d1808640aead3894892c5c6b27", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "用数字$$0$$、$$1$$、$$2$$、$$3$$、$$4$$最多可以组成个无重复数字的四位数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$96$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["由于首位不能为$$0$$，故有$$4$$种选择，其它三位，任意选$$3$$个数即可，根据分步计数原理可得一共有$$4\\times 4\\times 3\\times 2=96$$个． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "537", "queId": "ebe7e65fc6c6436db8ed2dae8b147972", "competition_source_list": ["2016年全国中环杯五年级竞赛中小学生思维能力训练活动决赛"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$E、$$U$$、$$L$$、$$S$$、$$R$$、T$$分别表示$$1、$$2$$、$$3$$、$$4$$、$$5$$、6$$（不同的字母表示不同的数字），且满足： $$（1）E+U+L= 6$$； $$（2）S+R+U+T= 18$$； $$（3）U\\times T=15$$； $$（4）S \\times L = 8$$． 则六位数$$\\overline{EULSRT}$$ =~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$123456$$ "}], [{"aoVal": "B", "content": "$$132465$$ "}], [{"aoVal": "C", "content": "$$145632$$ "}], [{"aoVal": "D", "content": "$$325416$$ "}], [{"aoVal": "E", "content": "$$146325$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["（$$1$$）因为$$E+U+L= 6$$； 而$$1+2+3=6$$； 所以$${E， U ，L}=~ {1，$$2$$，$$3$$ }$$； （$$2$$）因为$$S+R+U+T = 18$$； 而$$6+5+4+3=18$$； 所以$${S， R ，$$U$$ ，$$T$$ }= {6，$$5$$，$$4$$，$$3$$ }$$； （$$3$$）因为$$U\\times T=15$$； 而$$15=1\\times15=3\\times5$$； 所以$${U，T}={3，$$5$$ }$$； （$$4$$）因为$$S\\times L = 8 $$； 而$$8=1\\times8=2\\times4$$； 所以$${S，L}= {2，$$4$$ }$$． 由（$$1$$）和（$$3$$），得$$U= 3$$，则$$T= 5$$； 由（$$1$$）和（$$4$$），得$$L= 2$$，则$$S= 4$$； 最后分别结合（$$1$$）和（$$2$$），得$$E =1$$，$$R = 6$$； 故六位数$$\\overline{EULSRT}=132465$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1039", "queId": "0ea1fb95fca34bbab0eaa86179ee1ae0", "competition_source_list": ["2019年广东深圳全国小学生数学学习能力测评五年级竞赛初赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "五年级数学测评试卷共有$$15$$小题，做对一题得$$10$$分，做错一题扣$$4$$分，不答得$$0$$分，陈莉得了$$88$$分，她有道题未答． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["假设她全做对，可得：$$15\\times 10=150$$（分）， 现在得了$$88$$分，少了：$$150-88=62$$（分）， 做错一题，不但得不到$$10$$分，还扣$$4$$分， 说明错一题，少得$$10+4=14$$（分）， 不答得$$0$$分，说明不答少得$$10$$分， ∵$$62\\div 14=4$$（题）$$\\cdots 6$$（���），$$6$$不是$$10$$的倍数， 不合题意； $$62\\div 14=3$$（题）$$\\cdots 20$$（分），$$20$$是$$10$$的倍数， 符合题意； ∴未答的题有$$20\\div 10=2$$（题）． 选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1526", "queId": "567dbfc84a5c425c9d28d9f6c0606522", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛A卷第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "学校买了$$4$$张桌子，$$9$$把椅子，共用了$$252$$元，$$1$$张桌子和$$3$$把椅子的价钱正好相等，一张桌子元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$38$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["设椅子的单价是$$x$$元，则桌子的价格是$$3x$$元， $$3x\\times4+x\\times9=252$$ $$21x=252$$ $$x=12$$（元）， 桌子的单价：$$3\\times12=36$$（元）． 答：桌子、椅子的单价分别是$$12$$元、$$36$$元． 故答案选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "68", "queId": "ec4053f7f2e64dbd9fd7b595555d8c07", "competition_source_list": ["其它改编题", "2017年全国华杯赛小学中年级竞赛初赛模拟第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$至$$11$$这$$11$$个自然数中至少选出~\\uline{~~~~~~~~~~}~个不同的数，才能保证其中一定有两个数的和为$$12$$? ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["把和为$$12$$的两个数分成一组，这样就把这$$11$$个数分成$$6$$组：$$(1,11)$$，$$\\left( 2,10 \\right)$$，$$\\left( 3,9 \\right)$$，$$(4,8)$$，$$(5,7)$$，$$(6)$$．要保证一定有两个数的和为$$12$$，就要保证至少有两个数属于同一组． 由抽屉原理可知，从这$$12$$个数中选出$$7$$个数，就一定有两个数属于同一组．此时这两个数的和就是$$12$$． 如果我们从$$6$$组中各取一个数，则取出的这$$6$$个数中，没有两个数的和是$$12$$，因此本题的答案就是至少选出$$7$$个不同的数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1923", "queId": "983e3f268e08460c9d9ac0ecd0068e06", "competition_source_list": ["2015年第4届广东广州羊排赛六年级竞赛第8题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个三角形的三个内角角度之比为$$2:3:5$$，这是一个 ． ", "answer_option_list": [[{"aoVal": "A", "content": "锐角三角形 "}], [{"aoVal": "B", "content": "直角三角形 "}], [{"aoVal": "C", "content": "钝角三角形 "}], [{"aoVal": "D", "content": "等腰三角形 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["最大的角为$$180\\div (2+3+5)\\times 5=90$$（度），是直角三角形． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "130", "queId": "168d59abe6f3432a9ddf27c28d2eb70a", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$4$$辆汽车进行了$$4$$场比赛，每场比赛结果如下： （$$1$$）$$1$$号汽车比$$2$$号汽车跑得快； （$$2$$）$$2$$号汽车比$$3$$号汽车跑得快； （$$3$$）$$3$$号汽车比$$4$$号汽车跑得慢； （$$4$$）$$4$$号汽车比$$1$$号汽车跑得快． 汽车跑得最快． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$号 "}], [{"aoVal": "B", "content": "$$2$$号 "}], [{"aoVal": "C", "content": "$$3$$号 "}], [{"aoVal": "D", "content": "$$4$$号 "}]], "knowledge_point_routes": ["拓展思维->能力->推理推导->言语逻辑推理", "Overseas Competition->知识点->组合模块->逻辑推理"], "answer_analysis": ["根据（$$1$$）可知，$$1$$号比$$2$$号快．根据（$$2$$）可知，$$2$$号比$$3$$号快．根据（$$3$$）可知，$$4$$号比$$3$$号快．根据（$$4$$）可知，$$4$$号比$$1$$号快．所以$$4$$号快于$$1$$号，$$1$$号快于$$2$$号，$$2$$号快于$$3$$号．故最快的是$$4$$号．故$$\\text{ABC}$$错误，$$\\text{D}$$正确． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "918", "queId": "c53f24be7ba0487092db1b193d48fa35", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个自然数，用它分别去除$$62$$、$$90$$、$$130$$都有余数，这三个余数的和是$$24$$．这三个余数中最大的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$19$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["假设这个自然数为$$n$$，$$n$$除$$62$$、$$90$$、$$130$$的余数分别为$$a$$、$$b$$、$$c$$， 则$$62-a$$，$$90-b$$，$$130-c$$都是$$n$$的倍数， 可得$$\\left( 62-a \\right)+\\left( 90-b \\right)+\\left( 130-c \\right)=62+90+130-\\left( a+b+c \\right)$$， 得$$282-24=258$$，$$258=2\\times 3\\times 43$$， 则$$n$$可能为$$2$$、$$3$$、$$6$$、$$43$$， 又由于三个余数的和为$$24$$， 则$$abc$$中至少有一个要大于$$8$$， 由于除数大于余数， 因此$$n=43$$， 所以$$a=19$$，$$b=4$$，$$c=1$$， 因此三个余数中最大的是$$19$$， 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2258", "queId": "d18fc67485314198bc66bde5ddf2a702", "competition_source_list": ["2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（一）"], "difficulty": "2", "qtype": "single_choice", "problem": "在一个流穿梭的早晨，$$XRS$$张老师要去赶飞机，早上$$9$$点整出发，到达机场时惊奇的发现已知下午$$2$$点多了，他拿出了自己的量角器发现此时时针和分针刚好关于表盘上$$3$$点时刻上下对称，那么此时是下午$$2$$点（~ ）分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16\\frac{4}{11}$$ "}], [{"aoVal": "B", "content": "$$13\\frac{11}{13}$$ "}], [{"aoVal": "C", "content": "$$18\\frac{6}{13}$$ "}], [{"aoVal": "D", "content": "$$21\\frac{9}{11}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据题意我们可以得到如果以$$2$$点整为初始时刻，那么从初始时刻开始，时针和分针共走了$$120 {}^{}\\circ $$，已知时针、分针速度分别为$$6 {}^{}\\circ $$每分，$$0.5 {}^{}\\circ $$每分，那么此时为下午$$2$$点过$$120\\left( 6 {}^{}\\circ +0.5 {}^{}\\circ ~\\right)=18\\frac{6}{13}$$分． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "870", "queId": "c8ef2561213f406189e50885335b32d7", "competition_source_list": ["2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛"], "difficulty": "0", "qtype": "single_choice", "problem": "一个奇数乘$$3$$以后将会是 （~ ~ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "质数 "}], [{"aoVal": "B", "content": "合数 "}], [{"aoVal": "C", "content": "奇数 "}], [{"aoVal": "D", "content": "偶数 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的乘法规律"], "answer_analysis": ["奇数$$\\times 3$$只能是奇数．最容易做错的是选合数，反例：$$1\\times 3=3$$，$$3$$不是合数． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "566", "queId": "20b6cab5ac544f6b803249684a1fc44f", "competition_source_list": ["2012年美国数学大联盟杯六年级竞赛初赛第33题5分(每题5分)"], "difficulty": "1", "qtype": "single_choice", "problem": "$264$ and have a greatest common factor of $132$. $$264$$和的最大公因数是$$132$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$66$$ "}], [{"aoVal": "C", "content": "$$528$$ "}], [{"aoVal": "D", "content": "$$660$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->数论模块->因数与倍数->公因数与公倍数", "拓展思维->能力->运算求解"], "answer_analysis": ["公约数，亦称``公因数''．它是指能同时整除几个整数的数．如果一个整数同时是几个整数的约数，称这个整数为它们的``公约数''；公因数中最大的称为最大公因数． $$264$$和$$528$$的最大公因数为$$264$$； $$264$$和$$660$$的最大公因数为$$132$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2271", "queId": "db07463d27784e7fb6209ff0dd6e53db", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "小宏$$2015$$年$$10$$月长跑了$$17$$次，每次跑运的路程相同；小海$$10$$月长跑了$$11$$次，每次跑动的路程相同，小宏跑$$5$$次的路程等于小海跑$$3$$次的路程，则在$$10$$月（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "小宏长跑的总路程多于小海长跑的总路程 "}], [{"aoVal": "B", "content": "小宏长跑的总路程少于小海长跑的总路程 "}], [{"aoVal": "C", "content": "小宏长跑的总路程等于小海长跑的总路程 "}], [{"aoVal": "D", "content": "小宏长跑的次数少于小海长跑的次数 "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["小海跑$$1$$次的路程$$=\\frac{5}{3}\\times $$小宏跑$$1$$次的路程，故小海长跑$$11$$次$$=11\\times \\frac{5}{3}\\times $$小宏跑$$1$$次的路程$$=$$小宏长跑$$18\\frac{1}{3}$$次，所以选$$B$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1632", "queId": "7b4c997046db48a184a2327a248d036d", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第23题"], "difficulty": "2", "qtype": "single_choice", "problem": "---项铺路工程，如果甲队单独做$$100$$天可以完成，乙队单独做$$150$$天可以完成．现在两队同时施工，工作效率比单独做提高$$20 \\%$$，当工程完成$$\\frac{2}{5}$$时，正好赶上疫情，影响施工进度，使得每天少铺$$70$$米，结果前后一起共用了$$90$$天完成这项工程．则整个工程铺路米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6125$$ "}], [{"aoVal": "B", "content": "$$6135$$ "}], [{"aoVal": "C", "content": "$$6145$$ "}], [{"aoVal": "D", "content": "$$6155$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->变速工程问题"], "answer_analysis": ["原来甲乙合作工效$$\\left( \\frac{1}{100}+\\frac{1}{150} \\right)\\times (1+20 \\%)=\\frac{1}{50}$$， 完成前面$$\\frac{2}{5}$$的工作量需工时$$\\frac{2}{5}\\div \\frac{1}{50}=20$$（天）， 完成后面$$\\frac{3}{5}$$的工作量需工时$$90-20=70$$（天）， 工作效率是$$\\frac{3}{5}\\div 70=\\frac{3}{350}$$， 工效之差为$$\\frac{1}{50}-\\frac{3}{350}=\\frac{4}{350}=\\frac{2}{175}$$， 总工作量为$$70\\div \\frac{2}{175}=6125$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3305", "queId": "5b14e8f16a1b4366a4d3d681ba1ced3e", "competition_source_list": ["2020年新希望杯二年级竞赛初赛（团战）第52题"], "difficulty": "1", "qtype": "single_choice", "problem": "个位数字与十位数字不同的两位数共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$90$$ "}], [{"aoVal": "B", "content": "$$82$$ "}], [{"aoVal": "C", "content": "$$81$$ "}], [{"aoVal": "D", "content": "$$80$$ "}], [{"aoVal": "E", "content": "$$79$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->字典排序法->组数->数字组数（规定数位大小）"], "answer_analysis": ["根据题意分析可知，个数数字与十位数字相同的两位数有$$10$$个，两位数一共有$$91$$个， 由此可知，个位数字与十位数字不同的两位数有$$91-10=81$$（个）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1926", "queId": "93e1ef277c8744ea85c526cd644ea9e4", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "石灰水是用石灰和水按照$$1∶100$$配制而成，要配制$$4545$$千克的石灰水，需要石灰多少千克？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$千克 "}], [{"aoVal": "B", "content": "$$45$$千克 "}], [{"aoVal": "C", "content": "$$55$$千克 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数与运算->数的运算的实际应用（应用题）->分数的简单实际问题"], "answer_analysis": ["相当于把石灰水分成了$$101$$份，石灰占$$1$$份．所以石灰有$$4545\\times \\frac{1}{{1{ + }100}}{ = }45$$（千克）． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2515", "queId": "27528719f3094a57ac5cfb5df014c7a5", "competition_source_list": ["2018年湖北武汉新希望杯小学高年级六年级竞赛训练题（二）第2题", "2017年新希望杯小学高年级六年级竞赛训练题（二）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$9$$个数从左到右排成一行，从第$$3$$个数开始，每个数恰好等于它前面两个数之和．如果第$$8$$个数和第$$9$$个数分别是$$76$$和$$123$$，那么第$$1$$个数是（ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律"], "answer_analysis": ["$$123-76=47$$，$$76-47=29$$，$$47-29=18$$，$$29-18=11$$，$$18-11=7$$,$$11-7=4$$，$$7-4=3$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2945", "queId": "adfff7bb5b69450d85c1e7f21b34b8de", "competition_source_list": ["2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第2题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "爸爸带了$$269$$元钱，要买$$8$$元一个的笔记本，最多能买个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$ "}], [{"aoVal": "B", "content": "$$33$$ "}], [{"aoVal": "C", "content": "$$34$$ "}], [{"aoVal": "D", "content": "$$35$$ "}], [{"aoVal": "E", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用", "Overseas Competition->知识点->计算模块->整数->整数乘除->整数除法运算"], "answer_analysis": ["根据题意分析可知，利用关系式：数量$$=$$总价$$\\div $$单价，用爸爸带的钱除以每个笔记本的单价即可得到最多能买多少个笔记本，列式为：$$269\\div 8=33$$（个）$$\\cdots \\cdots 5$$（元），剩下的$$5$$元不足以再买一个笔记本， 所以最多能买$$33$$个， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1921", "queId": "aea65772d1e24c2a8cf4360e80d22e50", "competition_source_list": ["2014年北京六年级竞赛", "2016年陕西西安小升初交大附中入学真卷9第11题", "2014年全国迎春杯六年级竞赛初赛第6题", "2014年全国迎春杯五年级竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁四人拿出同样多的钱，一起订购同样规格的若干件新年礼物，礼物买来后，甲、乙、丙分别比丁多拿了$$3$$，$$7$$，$$14$$件礼物，最后结算时，乙付给了丁$$14$$元钱，并且乙没有付给甲钱．那么丙应该再付给丁元钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$56$$ "}], [{"aoVal": "D", "content": "$$70$$ "}], [{"aoVal": "E", "content": "$$75$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设丁拿了$$a$$件礼物，则四人花同样的钱，每人可以拿到$$a+\\frac{3+7+14}{4}=a+6$$件礼物， 实际情况：丁少拿了$$6$$件，乙多拿了$$1$$件，给丁$$14$$元，则货物单价$$14$$元， 丙多拿了$$14-6=8$$件，$$3$$件给甲，$$5$$件给丁，$$5\\times14=70$$元． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "787", "queId": "df33ea6cf7244368aa645a182e396c86", "competition_source_list": ["2018年全国小学生数学学习能力测评六年级竞赛初赛第9题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "把一批书按$$2:3:4$$或$$2:4:5$$两种方案分给甲、乙、丙三个班，都可以将这批书正好分完，这批书可能有本． ", "answer_option_list": [[{"aoVal": "A", "content": "$$90$$ "}], [{"aoVal": "B", "content": "$$99$$ "}], [{"aoVal": "C", "content": "$$110$$ "}], [{"aoVal": "D", "content": "$$180$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->因倍应用题->倍数应用题"], "answer_analysis": ["如果按照$$2:3:4$$发书，相当于将书分成$$2+3+4=9$$份；如果按照$$2:4:5$$发书，相当于将书分成$$2+4+5=11$$份，因此这批书是$$9$$和$$11$$的公倍数，只有$$\\text{B}$$选项符合． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3452", "queId": "cb10162a53c04c87ab813454152b41dc", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{(2019 Youth Mathematics Olympics, Primary 5, Question \\#7)} $990$ has many factors, and the average of these factors is（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$）. $$990$$有很多因数，这些因数的平均数是（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$110$$ "}], [{"aoVal": "B", "content": "$$115$$ "}], [{"aoVal": "C", "content": "$$117$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->数论模块->因数与倍数->因数个数定理", "拓展思维->思想->对应思想"], "answer_analysis": ["$$990$$的因数有：$$1$$；$$990$$；$$2$$；$$495$$；$$3$$；$$330$$；$$5$$；$$198$$；$$6$$；$$165$$；$$9$$；$$110$$；$$10$$；$$99$$；$$11$$；$$90$$；$$15$$；$$66$$；$$18$$；$$55$$；$$22$$；$$45$$；$$30$$；$$33$$，一共$$24$$个， 平均数为：$$(1+990+2+495+3+330+5+198+6+165+9+110+10+99+11+90+15+66+18+55+22+45+30+33)\\div 24$$ $$=2808\\div 24$$ $$=117$$， 故选答案$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1461", "queId": "7f04b52772fa43ecafd4a35b67b03870", "competition_source_list": ["2013年第11届创新杯三年级竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "某班有$$50$$多人上体育课，他们站成一排，老师让他们按$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$循环报数，最后一人报的数是$$4$$，这个班有人上体育课． ", "answer_option_list": [[{"aoVal": "A", "content": "$$51$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$53$$ "}], [{"aoVal": "D", "content": "$$57$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期"], "answer_analysis": ["也就是说，有这样一个两位数，首先他是一个十位是$$5$$的两位数，并且除以$$7$$的余数是$$4$$，那么最终只可能是$$53$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "95", "queId": "38a1ad9d4ab0407f93bd7f6dffc4190b", "competition_source_list": ["2017~2018学年浙江杭州五年级期中", "2017年第22届湖北武汉华杯赛三年级竞赛复赛（华罗庚金杯）第2题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "在下面加法竖式中，八个不同的字母分别代表$$2\\sim 9$$这八个数字，其中相同的字母代表相同的数字，不同的字母代表不同的数字，那么$$\\overline{NINE}=$$~\\uline{~~~~~~~~~~}~． $$\\frac{\\begin{matrix} \\begin{matrix} + \\end{matrix} \\begin{matrix}O T S \\end{matrix} \\begin{matrix}N W I \\end{matrix} \\begin{matrix}E O X \\end{matrix} \\end{matrix}}{\\begin{matrix} N\\textasciitilde I\\textasciitilde\\textasciitilde N\\textasciitilde E \\end{matrix}}$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1986$$ "}], [{"aoVal": "B", "content": "$$2034$$ "}], [{"aoVal": "C", "content": "$$2176$$ "}], [{"aoVal": "D", "content": "$$2458$$ "}], [{"aoVal": "E", "content": "$$2526$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜", "Overseas Competition->知识点->组合模块->数字谜->竖式数字谜"], "answer_analysis": ["因为字母代表$$2\\sim 9$$八个数字，故$$N$$只能为$$2$$． $$\\frac{\\begin{matrix} \\begin{matrix} + \\end{matrix} \\begin{matrix}O T S \\end{matrix} \\begin{matrix}N W I \\end{matrix} \\begin{matrix}E O X \\end{matrix} \\end{matrix}}{\\begin{matrix} N I N E \\end{matrix}}$$， $$O+E+X=E$$，故$$O+X=10$$，可为$$3+7$$，$$4+6$$；$$W+I=9$$，当$$O=3$$，$$I=4$$或$$5$$， 当$$I=4$$，$$W=5$$，或$$I=5$$，$$W=4$$， $$T+S$$需要为$$20$$或$$21$$，不可能； 当$$O=7$$，$$I=4$$或$$5$$，$$T+S=16$$或$$17$$，此时$$E=6$$． $$\\frac{\\begin{matrix} \\begin{matrix} + \\end{matrix} \\begin{matrix}7 8 9 \\end{matrix} \\begin{matrix}2 4 5 \\end{matrix} \\begin{matrix}6 7 3 \\end{matrix} \\end{matrix}}{\\begin{matrix} 2 5 2 6 \\end{matrix}}$$或$$\\frac{\\begin{matrix} \\begin{matrix} + \\end{matrix} \\begin{matrix}7 9 8 \\end{matrix} \\begin{matrix}2 4 5 \\end{matrix} \\begin{matrix}6 7 3 \\end{matrix} \\end{matrix}}{\\begin{matrix} 2 5 2 6 \\end{matrix}}$$ "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2628", "queId": "298f33cd2a7c4e1c8a6df31da2af2c99", "competition_source_list": ["2020年广东广州羊排赛六年级竞赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在一张地图上用$$1$$厘米的长度表示实际$$300$$千米的距离，则这张地图的比例尺是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1:300$$ "}], [{"aoVal": "B", "content": "$$300:1$$ "}], [{"aoVal": "C", "content": "$$1:30000000$$ "}], [{"aoVal": "D", "content": "$$30000000:1$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$300$$千米$$=30000000$$厘米； 比例尺$$=$$图上距离$$:$$实际距离$$=1:300000000$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1090", "queId": "0c05145fc49c4ecf99eaaa4a3193bd54", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一件商品先涨价$$15 \\%$$，再降价$$15 \\%$$，该商品的价格． ", "answer_option_list": [[{"aoVal": "A", "content": "比原价低 "}], [{"aoVal": "B", "content": "比原价高 "}], [{"aoVal": "C", "content": "与原价相同 "}], [{"aoVal": "D", "content": "无法判断 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->折扣问题"], "answer_analysis": ["先把原价看作单位``$$1$$''； 涨价后的价钱是原价的$$\\left( 1+15 \\% \\right)$$； 后又降价$$15 \\%$$，是降低涨价后的价格的$$15 \\%$$， 据此可知现在的价格是原价的百分之几，进而得出结论． $$\\left( 1+15 \\% \\right)\\times \\left( 1-15 \\% \\right)$$ $$=1.15\\times 0.85$$ $$=97.8 \\%$$； $$97.8 \\% ~\\textless{} ~1$$，即现价比原价低． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2919", "queId": "ff8409bd9b9b4096bb7492c250707069", "competition_source_list": ["2020年新希望杯三年级竞赛初赛（团战）第35题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面哪个算式的计算结果是偶数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 784-455 \\right)\\times 39+44\\times 11$$ "}], [{"aoVal": "B", "content": "$$11\\times 1+22\\times 2+33\\times 3+44\\times 4+55\\times 5+66\\times 6+77\\times 7+88\\times 8+99\\times 9$$ "}], [{"aoVal": "C", "content": "$$11\\times 1+22\\times 2+33\\times 3+44\\times 4+55\\times 5+66\\times 6+77\\times 7+88\\times 8+99\\times 9$$ "}], [{"aoVal": "D", "content": "$$123\\times 456+789$$ "}], [{"aoVal": "E", "content": "$$2\\times 4\\times 6\\times \\cdots \\times 2018\\times 2020-1\\times 3\\times 5\\times \\cdots \\times 2017\\times 2019$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的混合计算"], "answer_analysis": ["暂无 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1358", "queId": "28b414e32ee84eab937c336318e812fc", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛决赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一张课桌$$60$$元，比一把椅子多$$20$$元．一套课桌椅元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$80$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意分析可知， 一张桌子比一把椅子多$$20$$元， 那么一把椅子就比一张桌子少$$20$$元， 所以一把椅子的价钱是：$$60-20=40$$（元）， 那么一套桌椅的价钱是：$$60+40=100$$（元）． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1829", "queId": "bff916051f804fba9c2a528e924effb6", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "小红家住五楼，他每上一层楼要走$$20$$级台阶．小红每天回家要走级台阶． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$80$$ "}], [{"aoVal": "C", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都有->爬楼梯问题"], "answer_analysis": ["小红每天回家要走：$$20\\times \\left(5-1\\right)=80$$（级）． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2306", "queId": "aef3d295a3c745ceb5772ff057e4e092", "competition_source_list": ["2019年广东深圳全国小学生数学学习能力测评四年级竞赛初赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$9$$点至$$10$$点之间的某个时刻，$$5$$分钟前分针的位置和$$5$$分钟后时针的位置相同，此时刻是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9:05$$ "}], [{"aoVal": "B", "content": "$$9:35$$ "}], [{"aoVal": "C", "content": "$$9:55$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据选项$$\\text{C}$$，现在是$$9:55$$，那么$$5$$分钟前分针的位置是指向$$10$$，$$5$$分钟后时针也是指向$$10$$．所以$$\\text{C}$$选项答案是满足题目要求的． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3085", "queId": "ead9f72813dc4bb5aeccdeb3b9fd2e33", "competition_source_list": ["2014年第10届全国新希望杯小学高年级六年级竞赛复赛第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "对自然数$$n$$进行如下操作：如果$$n$$是偶数，就把它除以$$2$$，如果$$n$$��奇数，就把它加上$$7$$．现在对$$154$$进行有限次操作，得到的结果不可能是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->数的认识->数的特征->倍数->倍的认识"], "answer_analysis": ["154按照操作依次为$$77$$，$$84$$，$$42$$，$$21$$，$$28$$，$$14$$，$$7$$，$$14$$，7\\ldots\\ldots，不可能得到$$11$$． 154和$$7$$都是$$7$$点倍数，每次操作后最终结果都是$$7$$的倍数，不可能得到11. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2552", "queId": "5969867fec86420198b259fcb7b217d8", "competition_source_list": ["2014年迎春杯四年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一个$$12$$项的等差数列，公差是$$2$$，且前$$8$$项的和等于后$$4$$项的和，那么，这个数列的第二项是~\\uline{~~~~~~~~}~。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列求通项"], "answer_analysis": ["根据题意得$$({{a}_{1}}+{{a}_{8}})\\times 8\\div 2=({{a}_{9}}+{{a}_{12}})\\times 4\\div 2$$，因为$${{a}_{8}}={{a}_{1}}+14$$，$${{a}_{9}}={{a}_{1}}+16$$，$${{a}_{12}}={{a}_{1}}+22$$， 所以$$({{a}_{1}}+{{a}_{1}}+14)\\times 8\\div 2=({{a}_{1}}+16+{{a}_{1}}+22)\\times 4\\div 2$$，解得$${{a}_{1}}=5$$，因此$${{a}_{2}}=5+2=7$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2950", "queId": "cdd4dd731e9a4485a6a2d41edbfec183", "competition_source_list": ["2006年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "一个数的小数点向右移动一位，比原数大$$36$$，这个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$11.$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数基础->小数点的移动规律"], "answer_analysis": ["小数点向右移动$$1$$位，则新数比原数扩大$$10$$倍，即增加$$9$$倍，所以原数为$$59.94\\div \\left( 10-1 \\right)=6.66$$，选$$\\text{A}$$. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1621", "queId": "b5d2b7f267724df49b397c0ad7b59775", "competition_source_list": ["2022年第9届广东深圳鹏程杯四年级竞赛初赛第13题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一根圆柱形木料长$$200$$厘米，每$$40$$厘米锯成一段，每锯一段要$$2$$分钟，且每锯完一段要休息$$3$$分钟，那么这根木料锯完至少需要分钟 ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$20$$ "}], [{"aoVal": "E", "content": "$$17$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都没有->锯木头类型问题"], "answer_analysis": ["无 "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "363", "queId": "89c7a469ed9b492e80a1568d8c5edaac", "competition_source_list": ["2011年全国华杯赛竞赛初赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "老师问学生：``昨天你们有几个人复习数学了？'' 张：``没有人．''李：``一个人．''王：``二个人．''赵：``三个人．''刘：``四个人．'' 老师知道，他们昨天下午有人复习，也有人不复习，复习了的人说的都是真话，没复习的人说的都是假话．那么，昨天这$$5$$个人中复习数学的有个人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["$$5$$名学生说的话相互矛盾，只能有$$1$$人说的是真话，则只有李复习了，说的是真话． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1784", "queId": "dfb959be9820468497565da86d0b5924", "competition_source_list": ["2008年第6届创新杯四年级竞赛初赛B卷第9题5分", "2008年四年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "一串数��成一列，它们的规律是:前两个数都是1，从第三个数开始，每一个数都是它前面两个数的和.如下所示: 1，1，2，3，5，8，13，21，34，55，$$\\cdots$$ 这串数的前100个数(包括第100个数)中，偶数的个数是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "50 "}], [{"aoVal": "B", "content": "49 "}], [{"aoVal": "C", "content": "34 "}], [{"aoVal": "D", "content": "33 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["这串数的规律是``奇奇偶''循环的，在前99个数中，有33个偶数，第100个数是奇数，所以，这串数中前100个数有33个偶数. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1726", "queId": "9640e29674b4472aa9154277b0613391", "competition_source_list": ["2008年四年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "一个车队以$$5$$米/秒的速度完全经过一座长$$200$$米的大桥，共用时$$145$$秒。已知每辆车长$$5$$米，相邻两辆车相隔$$8$$米，那么这个车队共有车（ ）辆。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$39$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$41$$ "}], [{"aoVal": "D", "content": "$$42$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题"], "answer_analysis": ["车队$$145$$秒行的路程为$$5\\times 145=725$$（米） 车队的长度为$$725-200=525$$（米） 车队共有车$$\\left( 525-5 \\right) \\div\\left( 5+8 \\right)+1=41$$（辆） "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1717", "queId": "69ce00277cb5411cab65edc42c6a3395", "competition_source_list": ["2017年河南郑州联合杯竞赛附加赛第一场第3题2分"], "difficulty": "2", "qtype": "single_choice", "problem": "在一个减法算式里，被减数与减数以及差的和是$$160$$，且减数是差的$$3$$倍，差是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$80$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["被减数$$+$$减数$$+$$差$$=160$$，减数$$=3$$差，又被减数$$=$$减数$$+$$差； 则有被减数$$=$$减数$$+$$差$$=160\\div 2=80$$，即$$3$$差$$+$$差$$=80$$，所以差$$=80\\div 4=20$$． 故选A． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1615", "queId": "5724a89be4f44da492b406615039b8c1", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛B卷第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把一杯$$200$$克甜度是$$30 \\%$$的糖水倒入另外一杯$$300$$克甜度是$$15 \\%$$的糖水里，结果得到甜度是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$22$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["第一杯的糖：$$200\\times 30 \\%=60$$（克）， 第二杯的糖：$$300\\times 15 \\%=45$$（克）， 两杯混到一起后浓度为 $$(45+60)\\div (200+300)$$ $$=105\\div 500$$ $$=0.21$$ $$=21 \\%$$， 甜度为$$21$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "684", "queId": "288be384bfd840c3bcbe1524702de36a", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "由$$3$$个不同的自然数组成一等式：$$\\square +\\triangle +\\bigcirc =\\square \\times \\triangle -\\bigcirc $$这三个数中最多有个奇数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$0$$ "}], [{"aoVal": "E", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["如果这三个数中有$$2$$个奇数和$$1$$个偶数，那么等式左边必为偶数，等式右边必为奇数，不可能．如果这三个数均为奇数，那么等式左边必为奇数，而等式右边必为偶数，不可能．因此，这三个数中最多有$$1$$个奇数，例如$$2+4+1=2\\times 4-1$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3163", "queId": "4abb2d7ecd14401ea5b39eb2483c9274", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "袋子里装着只有颜色不同的若干黑、白、红三种颜色的小球．其中红色小球的颗数是白色的$$\\frac{1}{6}$$．黑色小球的颗数是红色小球的一半．如果从盒子中任意摸出一颗小球，摸出黑色小球的可能性是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{12}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{14}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{15}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->统计与概率->概率->典型问题->摸小球"], "answer_analysis": ["由题意可知，红$$:$$白$$=1:6$$，黑$$:$$红$$=1:2$$，则黑$$:$$红$$:$$白$$=1:2:12$$．摸出黑色小球的可能性为$$\\frac{1}{1+2+12}=\\frac{1}{15}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "711", "queId": "363e8cb5277d48d39771d92fb0d65cf2", "competition_source_list": ["2006年第4届创新杯五年级竞赛复赛第5题", "2006年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "在一根长100厘米的木棍上，自左向右每隔6厘米染一个红点，同时自右向左每隔5厘米也染一个红点，然后沿红点处将木棍逐段锯开，那么，长度是1厘米的短木棍有( )条. ", "answer_option_list": [[{"aoVal": "A", "content": "7 "}], [{"aoVal": "B", "content": "8 "}], [{"aoVal": "C", "content": "9 "}], [{"aoVal": "D", "content": "10 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->因倍应用题->倍数应用题"], "answer_analysis": ["不妨将100厘米长的木棍视为有刻度的米尺，那么从左到右每隔6厘米染一个红点，这些红点在米尺上的量数为(单位:厘米): $$6$$，$$12$$，$$18$$，$$24$$，$$30$$，$$36$$，$$\\cdots$$，$$60$$;$$66$$，$$\\cdots$$，$$90$$;$$96$$① 从右到左每隔5厘米染一个红点，这些红点在米尺上的量数为(单位:厘米):$$5$$，$$10$$，$$15$$，$$20$$，$$25$$，$$30$$;$$35$$，$$\\cdots$$ ，$$60$$;$$65$$，$$\\cdots$$ ，$$90$$;$$95$$② 比较上述两个数列①和②，知 在$$0\\sim30$$内的红点，有两组$$\\left( 5,6 \\right)\\left( 24,25 \\right)$$，两个相邻红点相距1厘米; 在$$30\\sim60$$内的红点，有两组$$\\left( 35,36 \\right)\\left( 54,55 \\right)$$，两个相邻红点相距1厘米; 在$$60\\sim90$$内的红点，有两组$$\\left( 65,66 \\right)\\left( 84,85 \\right)$$，两个相邻红点相距1厘米; 在$$90\\sim100$$内的红点，有一组$$\\left( 95,96 \\right)$$，两个相邻红点相距$$1$$厘米; 因此，沿红点锯开，长1厘米得短木棍有7条 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3115", "queId": "ddeb5f5cc4584483aa54bf2b165d74ad", "competition_source_list": ["2006年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "某小数的小数点向右移动一位，则数值比原来大25.65，原小数是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "2.565 "}], [{"aoVal": "B", "content": "2.56 "}], [{"aoVal": "C", "content": "2.855 "}], [{"aoVal": "D", "content": "2.85 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数基础->小数点的移动规律"], "answer_analysis": ["小数点向右移动一位，则扩大10倍，增加9倍.因此原数为$$25.65\\div \\left( 10-1 \\right)=2.85$$，故选D. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "118", "queId": "ec623f79a86d44598fe5adad7fe82edf", "competition_source_list": ["2014年全国华杯赛小学中年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "两个正整数的和小于$100$，其中一个是另一个的两倍，则这两个正整数的和的最大值是（~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$83$ "}], [{"aoVal": "B", "content": "$99$ "}], [{"aoVal": "C", "content": "$96$ "}], [{"aoVal": "D", "content": "$98$~ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["较小数为$$a$$，较大数为$$2a$$，和为$$a+2a=3a$$． 和小于$$100$$，最大为$$99$$， $$3a=99$$ $$a=33$$． 较大数：$$2\\times 33=66$$，符合题意． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "572", "queId": "0a8deb33bb5f4f589dadb123cf288a94", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一个三位数各个数位上的数字都不相同。把$$2$$写在这个三位数的左端得到一个四位数；把$$2$$写在这个三位数的右端得到一个四位数；这两个四位���相差$$945$$，那么这个三位数是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$117$$ "}], [{"aoVal": "B", "content": "$$327$$ "}], [{"aoVal": "C", "content": "$$219$$ "}], [{"aoVal": "D", "content": "$$312$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->位值原理与进制->位值原理运用->位值原理的综合应用"], "answer_analysis": ["解：设这个三位数是$$\\overline{abc}$$，依题意可知$$\\overline{2abc}-\\overline{abc2}=945$$或者$$\\overline{abc2}-\\overline{2abc}=945$$， 按照不完全拆分把$$abc$$看成一组， 当：$$\\overline{2abc}-\\overline{abc2}=945$$， $$2000+\\overline{abc}-10\\overline{abc}-2=945$$， $$1998-9\\overline{abc}=945$$， $$\\overline{abc}=117$$（与题中说互不相同矛盾）， 当：$$\\overline{abc2}-\\overline{2abc}=945$$， $$10\\overline{abc}+2-2000-\\overline{abc}=945$$， $$9\\overline{abc}-1998=945$$， $$\\overline{abc}=327$$（满足条件）。 故选:B。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3451", "queId": "dd2018aed8174598b48141d43543bbf5", "competition_source_list": ["2017年IMAS小学中年级竞赛（第一轮）第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$六张卡片，从中任意取出两张卡片可以组成一个两位数．把组成的所有两位数按从小到大的顺序排列﹐请问第$$21$$个二位数是？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$43$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$46$$ "}], [{"aoVal": "D", "content": "$$51$$ "}], [{"aoVal": "E", "content": "$$61$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["由题意，十位数为$$1$$的两位数有$$12$$、$$13$$、$$14$$、$$15$$、$$16$$共$$5$$个，同理可得十位数为$$2$$、$$3$$、$$4$$的两位数也是各有$$5$$个，因此第$$21$$个两位数是$$51$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1054", "queId": "0ada4e8505be4304bd236e2b10c6b034", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛A卷第9题3分", "六年级其它"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2018$$年春节有个小程序可以发红包，但是每发一次要收$$2 \\%$$的服务费．张紫亦的爸爸、妈妈和他分别发了$$3$$个红包，被收取$$4.5$$元的服务费，爸爸发的最多．请问下面那个数可能是爸爸发的钱数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$66$$ "}], [{"aoVal": "C", "content": "$$75$$ "}], [{"aoVal": "D", "content": "$$88$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型"], "answer_analysis": ["已知服务费占总钱数的$$2 \\%$$， 服务费是$$4.5$$元，总钱数$$=4.5\\div 2 \\%=225$$（元）． $$225$$元是爸爸、妈妈和张紫亦发的总钱数， 平均每个人发$$225\\div 3=75$$（元）， 爸爸发的最多一定大于$$75$$元，选项$$\\text{D}$$正确． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3212", "queId": "5de2f99569c640e485593ec4ebbf7f26", "competition_source_list": ["2008年第6届创新杯六年级竞赛初赛B卷第3题5分", "2008年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$1\\text{、}2\\text{、}3\\text{、}\\cdots\\text{、}100$$这$$100$$个整数中，能被$$2$$或$$3$$整除的数一共有（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$85$$个 "}], [{"aoVal": "B", "content": "$$67$$个 "}], [{"aoVal": "C", "content": "$$34$$个 "}], [{"aoVal": "D", "content": "$$17$$个 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->容斥原理->二量容斥"], "answer_analysis": ["在$$1\\text{、}2\\text{、}3\\text{、}\\cdots\\text{、}100$$这$$100$$个整数中，$$2$$的倍数有$$50$$个，$$3$$的倍数有$$33$$个，$$6$$的倍数有$$16$$个，所以能被$$2$$或$$3$$整除的整数有$$50+33-16=67$$（个）。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1410", "queId": "5eadc4231f3e444aab46ba23a02016ee", "competition_source_list": ["2008年第6届创新杯五年级竞赛初赛B卷第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在植树节时，某班每人应平均植树$$6$$株，如果只由女生完成，每人应植$$15$$株，如果只由男生完成，则每人应植树株． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->多元一次方程解应用题->整数系数二元一次方程组解应用题"], "answer_analysis": ["设全班有$$x$$人，女生有$$y$$人， 可得$$6x=15y$$， $$x:y=15:6$$， 所以女生占全班的$$6\\div 15=\\frac{6}{15}=\\frac{2}{5}$$， 男生占$$1-\\frac{2}{5}=\\frac{3}{5}$$， 总棵树不变，男生人数$$:$$女生人数$$=3:2$$， 故只由男生完成，男生应植树$$15\\times \\frac{2}{3}=10$$（株）． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3156", "queId": "0f032374f5cf4fb4bff2db0c3e573675", "competition_source_list": ["2011年北京学而思杯五年级竞赛"], "difficulty": "3", "qtype": "single_choice", "problem": "有两个整数，它们的和恰好是两个数字相同的两位数，它们的乘积恰好是三个数字相同的三位数．求这两个整数分别是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$37$$和$$18$$ "}], [{"aoVal": "B", "content": "$$74$$和$$3$$ "}], [{"aoVal": "C", "content": "$$74$$和$$3$$或$$37$$和$$18$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["两位数中，数字相同的两位数有$$11$$、$$22$$、$$33$$、$$44$$、$$55$$、$$66$$、$$77$$、$$88$$、$$99$$共九个，它们中的每个数都可以表示成两个整数相加的形式，例如$$33=1+32=2+31=3+30=\\cdots \\cdots =16+17$$，共有$$16$$种形式，如果把每个数都这样分解，再相乘，看哪两个数的乘积是三个数字相同的三位数，显然太繁琐了．可以从乘积入手，因为三个数字相同的三位数有$$111$$、$$222$$、$$333$$、$$444$$、$$555$$、$$666$$、$$777$$、$$888$$、$$999$$，每个数都是$$111$$的倍数，而$$111=37\\times 3$$，因此把这九个数表示成一个两位数与一个一位数或两个两位数相乘时，必有一个因数是$$37$$或$$37$$的倍数，但只能是$$37$$的$$2$$倍（想想为什么？）$$3$$倍就不是两位数了． 把九个三位数分解：$$111=37\\times 3$$、$$222=37\\times 6=74\\times 3$$、$$333=37\\times 9$$、$$444=37\\times12=74\\times 6$$、$$555=37\\times 15$$、$$666=37\\times 18=74\\times 9$$、$$777=37\\times 21$$、$$888=37\\times24=74\\times 12$$、$$999=37\\times 27$$． 把两个因数相加，只有($$74+3$$)$$=77$$和($$37+18$$)$$=55$$的两位数字相同．所以满足题意的答案是$$74$$和$$3$$，$$37$$和$$18$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1237", "queId": "797a8ce6570b4ee8a36b9a6522a0696d", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛B卷第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "两桶食用油，从第一桶倒入第二桶$$18$$千克，两桶重量相等，现在第一桶用$$9$$分之$$7$$，第二桶用$$5$$分之$$3$$，两桶油剩下的重量相等，问两桶油原来各重多少千克？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$66$$，$$30$$ "}], [{"aoVal": "B", "content": "$$72$$，$$36$$ "}], [{"aoVal": "C", "content": "$$81$$，$$45$$ "}], [{"aoVal": "D", "content": "$$90$$，$$54$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为从第一桶倒入第二桶$$18$$千克，两桶重量相等， 所以第一桶重量$$-$$第二桶重量$$=36$$千克，选项$$\\text{ABCD}$$都满足， 且知第一桶重量$$\\times \\frac{2}{9}=$$第二桶重量$$\\times \\frac{2}{5}$$，只有$$\\text{C}$$项满足， 故答案选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "36", "queId": "d05731d5cc01463787d3a4b86d550413", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛复赛第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "一副扑克牌去掉大、小王共有$$52$$张，最上面的一张是红桃$$K$$，如果每次把最上面的$$18$$张牌移到最下面而不改变他们的顺序，至少经过（~ ）次移动，红桃$$K$$才会有出现在最上面． ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$26$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["每次移动$$18$$张，那么移动牌的总数是$$18$$的倍数，要使红桃$$K$$又出现在最上面，那么$$18\\times $$次数的乘积应该是$$52$$的倍数，那么至少是$$26$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1190", "queId": "15a7393a02484bc082df07c2fc8b4901", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛决赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$A$$地植树$$400$$棵，$$B$$地植树$$600$$棵，甲、乙、丙每天分别能植树$$25$$、$$52$$、$$48$$棵，甲在$$A$$地、乙在$$B$$地、丙跨$$A$$与$$B$$两地，同时开始、同时结束，整个过程所用的时间是天． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->双工程问题"], "answer_analysis": ["工程问题．从开始到结束三人一共用了$$(400+600)\\div (25+52+48)=1000\\div 125=8$$天． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2637", "queId": "d5916431e63743e58331a33f7495a992", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "算式$$826446281\\times11\\times 11$$的计算结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9090909091$$ "}], [{"aoVal": "B", "content": "$$909090909091$$ "}], [{"aoVal": "C", "content": "$$10000000001$$ "}], [{"aoVal": "D", "content": "$$100000000001$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["根据$$11$$乘法的特征``两边一拉，中间相加''可得到结果D 本题考察乘$$11$$的速算诀窍. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "347", "queId": "91b17dd1e25e4dd1a1d0d7dcf5b938fb", "competition_source_list": ["2016年全国华杯赛小学中年级竞赛在线模拟第3题", "2013年全国华杯赛小学中年级竞赛初赛A卷第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "小东、小西、小南、小北四个小朋友在一起做游戏时，捡到了一条红领巾，交给了老师．老师问是谁捡到的？小东说不是小西；小西说是小南；小南说小东说的不对；小北说小南说的不对．他们之中只有一个人说对了，这个人是． ", "answer_option_list": [[{"aoVal": "A", "content": "小东 "}], [{"aoVal": "B", "content": "小西 "}], [{"aoVal": "C", "content": "小南 "}], [{"aoVal": "D", "content": "小北 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾"], "answer_analysis": ["由于只有一个人说对了，而小北支持小东，那么他们俩都错了，所以反对小东的小南说对了． ", "根据题干分析可得，小南与小北说的话是相互矛盾的，所以两人中一定有一个人说的是正确的，假设小北说的是正确的，则小南说“小东说的不对”是错，可得，小东说的对，这样与已知只有一个人说对了相矛盾，所以此假设不成立，故小南说的是正确的．
\n故选：$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "690", "queId": "6789a0f74b1649518d8d85a63bb5d556", "competition_source_list": ["2007年第5届创新杯五年级竞赛第1题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1000$$到$$2007$$的自然数中有奇数个因数的数有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["有奇数个因数的数必是完全平方数，如果不是完全平方数，它们的余数都是成对出现的，因此原题等价于求$$1000$$到$$2007$$之间的完全平方数，$${{32}^{2}}=1024$$，$${{33}^{2}}=1089$$，$${{34}^{2}}=1156$$，$$\\cdots \\cdots $$，$${{42}^{2}}=1764$$，$${{43}^{2}}=1849$$，$${{44}^{2}}=1936$$，$${{45}^{2}}=2025\\textgreater2007$$，所以一共有：$$44-32+1=13$$（个）． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1568", "queId": "ff808081451596cc014516b4bff0024b", "competition_source_list": ["2014年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "等腰三角形底边上的高为$$8$$，周长为$$32$$，则三角形的面积为（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$48$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$321$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["底边上的高为$$8$$，高将等腰三角形分成了两个相等的直角三���形，周长为$$32\\div 2+8=24$$，其中一条直角边（高）为$$8$$另一条（底边的一半）记作$$a$$、则斜边（等腰三角形的腰）应为$$16-a$$，根据勾股定理$${{(16-a)}^{2}}={{8}^{2}}+{{a}^{2}}$$，求得$$a=6$$所以三角形的面积为$$48$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3", "queId": "00afae649a0f4efc9f59b9671c1ae118", "competition_source_list": ["2011年第7届全国新希望杯小学高年级六年级竞赛第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "盒子里装有分别写着$$1$$，$$2$$，$$3$$，$$4$$，\\ldots，$$100$$的黄色卡片各一张，我们称如下操作为一次操作；从盒子里取出$$m(7\\leqslant m\\leqslant 10)$$张卡片，算出这$$m$$张卡片上各数之和减去$$27$$的差，将写在一张红色卡片上（不放回）．若干次操作之后，盒子里的卡片全部被取出，若所有红色卡片上的数字之和为$$n$$，那么$$n$$的最大可能值减去最小可能值等于． ", "answer_option_list": [[{"aoVal": "A", "content": "$$108$$ "}], [{"aoVal": "B", "content": "$$96$$ "}], [{"aoVal": "C", "content": "$$88$$ "}], [{"aoVal": "D", "content": "$$81$$ "}], [{"aoVal": "E", "content": "$$75$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["先算卡片的总值． 最小值$$100\\div 10=10$$次，共减$$10\\times 27$$； 最多$$14$$次，共减去$$14\\times 27$$， 因此差为$$(14-10)\\times 27=108$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "825", "queId": "5c4df7410d614de492c7dd2240f31ed3", "competition_source_list": ["2020年第23届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把一堆桃子分给若干只猴子，如果每只猴子分$$6$$个，剩$$57$$个桃子；如果每只猴子分$$9$$个，就有$$5$$只猴子一个也分不到，还有一只猴子只分到$$3$$个．那么，一共有个桃子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$216$$ "}], [{"aoVal": "B", "content": "$$324$$ "}], [{"aoVal": "C", "content": "$$273$$ "}], [{"aoVal": "D", "content": "$$301$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["如果每只猴子分$$6$$个，剩$$57$$个桃子；如果每只猴子分$$9$$个，就有$$5$$只猴子一个也分不到，还有一只猴子只分到$$3$$个，证明少了$$5\\times9+6=51$$（个），猴子共有：$$(57+51)\\div (9-6)=36$$（只）；桃子共有：$$36\\times6+57=273$$（个） 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "576", "queId": "0ab7295a27f042fabbdc344c3bc96cd3", "competition_source_list": ["2017年IMAS小学高年级竞赛（第一轮）第14题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "某公共汽车总站有两条路线，第一条每$$8$$分钟发一辆车、第二条每$$10$$分钟发一辆车，且在早上$$6:00$$两条路线同时发出第一辆车．请问下面哪一项是两条路线同时发车的时刻？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$7:30$$ "}], [{"aoVal": "B", "content": "$$8:20$$ "}], [{"aoVal": "C", "content": "$$9:40$$ "}], [{"aoVal": "D", "content": "$$10:00$$ "}], [{"aoVal": "E", "content": "$$11:00$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["由题意可知，每隔$$\\left[ 8,10 \\right]=40$$分钟两条线路同时发出一辆车． 而$$7:30$$与$$6:00$$相差$$90$$分钟、$$8:20$$与$$6:00$$相差$$140$$分钟、$$9:40$$与$$6:00$$相差$$220$$分钟、$$10:00$$与$$6:00$$相差$$240$$分钟、$$11:00$$与$$6:00$$相差$$300$$分钟，其中仅$$10:00$$与$$6:00$$相差的分钟数是$$40$$的倍数． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "962", "queId": "f4922c07a5634e64942115a5fed24673", "competition_source_list": ["2017年第17届世奥赛六年级竞赛决赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "通过下面算式观察个位上数字是$$5$$的数的平方： $${{15}^{2}}=225$$可写成$$100\\times 1\\times \\left( 1+1 \\right)+25$$ $${{25}^{2}}=625$$可写成$$100\\times 2\\times \\left( 2+1 \\right)+25$$ $${{35}^{2}}=1225$$可写成$$100\\times 3\\times \\left( 3+1 \\right)+25$$ $${{45}^{2}}=2025$$可写成$$100\\times 4\\times \\left( 4+1 \\right)+25$$······ 请你推测$${{\\left( 10n+5 \\right)}^{2}}$$可写成． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100n\\left( n+1 \\right)+25$$ "}], [{"aoVal": "B", "content": "$$100\\left( n+1 \\right)+25$$ "}], [{"aoVal": "C", "content": "$$100{{n}^{2}}+25$$ "}], [{"aoVal": "D", "content": "$$100{{n}^{2}}+100n+5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->完全平方数->平方数的简单应用"], "answer_analysis": ["$${{\\left( 10n+5 \\right)}^{2}}=100n\\left( n+1 \\right)+25=100{{n}^{2}}+100n+25$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3463", "queId": "eb2c244e22764dc5ada2b8f122b7a667", "competition_source_list": ["2017年北京迎春杯小学高年级五年级竞赛模拟"], "difficulty": "3", "qtype": "single_choice", "problem": "小悦与阿奇比赛下象棋，两人水平相当，约定赛$$7$$局，先赢$$4$$局者胜，现在已经比赛了$$3$$局，小悦胜了$$2$$局，阿奇胜了$$1$$局，请问：小悦最后胜利的概率有多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{8}$$ "}], [{"aoVal": "B", "content": "$$\\frac{11}{16}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{16}$$ "}], [{"aoVal": "D", "content": "$$\\frac{7}{16}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$\\rm B$$ "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "207", "queId": "70e30d127b0645fc8ea0277fcec0cad9", "competition_source_list": ["2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）"], "difficulty": "3", "qtype": "single_choice", "problem": "有一天，彭老师和陈老师约好去打乒乓球，结果彭老师以$$4:0$$完虐陈老师．乒乓球比赛为$$11$$分制，即每局$$11$$分，$$7$$局$$4$$胜制，打成$$10:10$$后必须净胜而且只能净胜$$2$$分．经计算，彭老师四局的总得分为$$48$$分，陈老师总得分为$$39$$分，且每一局比赛分差不超过$$3$$分，则一共有（~ ）种情况．（不考虑这四局比分之间的顺序） ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理->体育比赛->单循环赛"], "answer_analysis": ["每周比赛要分出胜负分差必须在$$2$$分或以上，题中又给出每局比赛分差不超过$$3$$分，故每局比赛的分差只有两种可能：差$$2$$分或$$3$$分．且分差为$$3$$分的那局彭老师得分为$$11$$分，总分差为$$4839=9$$分，故必有$$3$$场分差为$$2$$分，另一场分差为$$3$$分；即有一场的比分为$$118$$，另两场的总比分为$$3731$$，有以下四种情况：①$$11:9$$，$$11:9$$，$$15:13$$②$$11:9$$，$$12:10$$，$$14:12$$③$$11:9$$，$$13:11$$，$$13:11$$④$$12:10$$，$$12:10$$，$$13:11$$．故一共有$$4$$种情况． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2997", "queId": "e9980a8868434ee3a5591426ae546c80", "competition_source_list": ["其它改编自2015年全国希望杯六年级竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "观察下面一列数的规律．这列数从左往右第$$100$$个数是~\\uline{~~~~~~~~~~}~． $$\\frac{1}{2}$$，$$\\frac{3}{5}$$，$$\\frac{5}{8}$$，$$\\frac{7}{11}$$，$$\\frac{9}{14}$$，\\ldots\\ldots{} ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{199}{302}$$ "}], [{"aoVal": "B", "content": "$$\\frac{201}{299}$$ "}], [{"aoVal": "C", "content": "$$\\frac{201}{302}$$ "}], [{"aoVal": "D", "content": "$$\\frac{199}{299}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["分子是公差为$$2$$的等差数列，第$$100$$项是：$$1+\\left( 100-1 \\right)\\times 2=199$$， 分母是公差为$$3$$的等差数列，第$$100$$项是$$2+\\left( 100-1 \\right)\\times 3=299$$， 所以第$$100$$个数是$$\\frac{199}{299}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "841", "queId": "a40ae920a7c9428681c5306ceb3d9a0b", "competition_source_list": ["2017年河南郑州联合杯竞赛附加赛第一场第1题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "王小天有若干张$$10$$元、$$5$$元的纸币，这两种纸币的张数相同，那么王小天可能有（~ ）元钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$51$$ "}], [{"aoVal": "C", "content": "$$75$$ "}], [{"aoVal": "D", "content": "$$100$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["因为张数相同，所以先求出$$1$$张$$10$$元和$$5$$元共有$$15$$元，再看一下选项中哪个是$$15$$的倍数即可．容易看出$$C$$正确． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "165", "queId": "23a9582fffa24ced904d5cc5409748bf", "competition_source_list": ["2020年新希望杯二年级竞赛初赛（团战）第24题"], "difficulty": "1", "qtype": "single_choice", "problem": "有甲、乙、丙、丁四盘苹果，乙不是最多的，但比甲、丁多，甲没有丁多．按照苹果从少到多的顺序，四盘苹果分别是． ", "answer_option_list": [[{"aoVal": "A", "content": "甲、丁、乙、丙 "}], [{"aoVal": "B", "content": "丙、甲、乙、丁 "}], [{"aoVal": "C", "content": "丙、乙、丁、甲 "}], [{"aoVal": "D", "content": "乙、丙、丁、甲 "}], [{"aoVal": "E", "content": "丙、甲、丁、乙 "}]], "knowledge_point_routes": ["拓展思维->能力->推理推导->言语逻辑推理"], "answer_analysis": ["暂无 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1835", "queId": "78b29284999e4b7bafb2afbe5d5a29ac", "competition_source_list": ["2017年河南郑州联合杯竞赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明买了一本$$513$$页的小说，数字$$1$$在页码中出现了（~ ）次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$153$$ "}], [{"aoVal": "B", "content": "$$203$$ "}], [{"aoVal": "C", "content": "$$206$$ "}], [{"aoVal": "D", "content": "$$211$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["页码问题；$$1\\sim 99$$中数字``$$1$$''出现了$$20$$次，$$100\\sim 199$$中数字``$$1$$''出现了$$120$$个，剩下的$$200\\sim 500$$有$$3\\times 20=60$$（个），$$501\\sim 513$$中数字``$$1$$''出现了$$6$$次，所以一共有$$20+120+60+6=206$$（次）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1256", "queId": "16e030a8b2984b08ab1356fa66d3baf5", "competition_source_list": ["2014年第2届广东广州羊排赛六年级竞赛第3题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "一班共有$$48$$名同学，那么男女生人数之比可能是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5:4$$ "}], [{"aoVal": "B", "content": "$$6:5$$ "}], [{"aoVal": "C", "content": "$$7:4$$ "}], [{"aoVal": "D", "content": "$$7:5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["人数为$$48$$，则比的总份数必须是$$48$$的因数，只有$$7+5=12$$符合． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1424", "queId": "51402efb96f64ac199f7965ebbad9127", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（五）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "一队学生排成$$9$$行$$9$$列的方阵，如果去掉最外层$$2$$行$$2$$列，要减少人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$ "}], [{"aoVal": "B", "content": "$$34$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$38$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->方阵问题->实心方阵->实心方阵的增减"], "answer_analysis": ["$$9\\times 9-7\\times 7=32$$（人）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1903", "queId": "8f23a2291bc343088c473307c9ed143e", "competition_source_list": ["2018年环亚太杯三年级竞赛初赛第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{(2009 Math kangaroo Problems, Level 1-2, Question \\#7)} After Snow White, the Prince and the 7 Dwarfs ate one apple each, there were 4 apples left in the basket. How many apples were there in the basket before they ate any？ 翻译：在白雪公主之后，王子和七个小矮人也每人吃了$$1$$个苹果，篮子里还剩下$$4$$个苹果．那么他们吃苹果之前篮子里有多少苹果？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["吃掉的苹果加上篮子里剩下的苹果就是原来一共的苹果数量．白雪公主吃了一个苹果，王子吃了一个苹果，再加上七个小矮人各吃了一个苹果，所以一共吃了$$1+1+7=9$$个苹果，再加上篮子里还剩下的$$4$$个苹果，那么原来没吃苹果之前就有$$9+4=13$$个苹果． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2580", "queId": "3583793597664660a2825db4dd025e7e", "competition_source_list": ["2015年IMAS小学高年级竞赛第一轮检测试题第1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "请问算式$$32\\times 37\\times 75$$的值为多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$88075$$ "}], [{"aoVal": "B", "content": "$$88800$$ "}], [{"aoVal": "C", "content": "$$88200$$ "}], [{"aoVal": "D", "content": "$$74000$$ "}], [{"aoVal": "E", "content": "$$80800$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$32\\times 37\\times 75$$ $$=4\\times 8\\times 37\\times 3\\times 25$$ $$=(4\\times 25)\\times (37\\times 3)\\times 8$$ $$=100\\times 111\\times 8$$ $$=88800$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "906", "queId": "9c25d72396c541bd89ae5df62d7e5c5a", "competition_source_list": ["2020年广东深圳龙岗区亚迪学校迎春杯五年级竞赛模拟第20题2分", "2019~2020学年陕西宝鸡渭滨区五年级上学期期末第22题1分", "2019~2020学年山东济南高新区五年级下学期期末第10题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "在下面四组数中，组中的数都是质数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$31$$，$$71$$，$$91$$ "}], [{"aoVal": "B", "content": "$$13$$，$$21$$，$$37$$ "}], [{"aoVal": "C", "content": "$$17$$，$$37$$，$$85$$ "}], [{"aoVal": "D", "content": "$$43$$，$$53$$，$$73$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$\\text{A}$$中$$91$$不是质数，故$$\\text{A}$$错误； $$\\text{B}$$中$$$21$$不是质数，故$$\\text{B}$$错误． $$\\text{C}$$中$$85$$不是质数，故$$\\text{C}$$错误． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2828", "queId": "d1a133cef0b6466e94b5fe8ca873440d", "competition_source_list": ["2016年河南郑州K6联赛六年级竞赛第14题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$20$$克盐放入$$100$$克水中，盐与水的比、盐与盐水的比分别是（~ ）和（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1:6$$和$$1:5$$ "}], [{"aoVal": "B", "content": "$$1:5$$和$$1:6$$ "}], [{"aoVal": "C", "content": "$$1:4$$和$$1:5$$ "}], [{"aoVal": "D", "content": "$$1:4$$和$$1:6$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["盐与水的比为$$20:100=1:5$$，盐与盐水的比是$$20:\\left( 20+100 \\right)=1:6$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3427", "queId": "aeef0597581e4b54ac1e622b9829cd11", "competition_source_list": ["2004年第2届创新杯五年级竞赛复赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$50$$束鲜花中，$$16$$束有月季花，$$15$$束有马蹄莲，$$21$$束有白兰花，有$$7$$束中既有月季花又有马蹄莲，有$$8$$束中既有马蹄莲又有白兰花，有$$10$$束中既有月季花又有白兰花，还有$$5$$束鲜花中，月季花、马蹄莲、白兰花都有．则$$50$$束鲜花中，上述三种花都没有的花束共有． ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$束 "}], [{"aoVal": "B", "content": "$$18$$束 "}], [{"aoVal": "C", "content": "$$19$$束 "}], [{"aoVal": "D", "content": "$$20$$束 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["先求出有月季花、马蹄莲、白兰花的一共有多少束，然后减去有其中$$2$$种花的束数，再加上三种花都有束数，就是三种花至少有一种的有多少束，再用总数$$50$$束，减去至少有一种的束数，就是三种都没有的束数． $$3$$种花至少有一种的有： $$16+15+21-7-8-10+5$$ $$=52-\\left( 7+8+10 \\right)+5$$ $$=52-25+5$$ $$=32$$ （束） $$50-32=18$$ （束） 答：月季花、马蹄莲、白兰花三种花都没有的花束共有$$18$$束． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "898", "queId": "f716d6e72b134ec8a3db13aba6ffbe95", "competition_source_list": ["2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（二）"], "difficulty": "2", "qtype": "single_choice", "problem": "有一两位数$$N$$，在它的两头各添上一个$$1$$以后就变成一个四位的数$$M$$．若$$M$$是$$N$$的$$23$$倍，那么当$$M$$最小时，$$N$$的值是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$77$$ "}], [{"aoVal": "C", "content": "$$160$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由整除关系容易得到$$N$$的最后一位是$$7$$，显然$$N=7$$不成立，即$$K$$大于等于$$2$$，于是当$$K$$大于等于$$2$$时可以设$$N\\text{=X7}$$（表示$$X$$与$$7$$连在一起，$$X$$是一个数，不是$$X*7$$，例如$$N=857$$，则$$\\text{X=85}$$），于是$$\\text{M=1X71}$$（同上，时表示连在一起的数），把$$M$$、$$N$$表示如下：$$M={{10}^{\\wedge }}\\left( K+1 \\right)+X*100+71$$，$$N=10*X+7$$，由条件$$M=23*N$$，展开$${{10}^{\\wedge }}\\left( K+1 \\right)+X*100+71=\\left( 10*7 \\right)*23=230*X+161$$，化简得$${{10}^{\\wedge }}K=13*X+9$$，容易看出，当$$K=2$$时，$$X=7$$符合条件，所以$$K$$最小为$$2$$时，$$N=7$$，$$\\text{M=1771}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "869", "queId": "ad8ad067ffea4aefb8120119eefbb100", "competition_source_list": ["2013年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1\\sim 11$$这$$11$$个整数中任意取出$$6$$个数，则下列结论正确的有（ ）个。 ①其中必有两个数互质； ②其中必有一个数是其中另一个数的倍数； ③其中必有一个数的$$2$$倍是其中另一个数的倍数。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->两数的最小公倍数"], "answer_analysis": ["（$$1$$）根据互质数的意义，公因数只有$$1$$的两个数叫做互质数，在这$$11$$个中，质数有$$2$$、$$3$$、$$5$$、$$7$$、$$11$$，任何两个质数一定是互质数，又因为在这$$11$$个数中偶数有$$2$$、$$4$$、$$6$$、$$8$$、$$10$$五个，奇数有六个，所以任意取出$$6$$个数，其中必有两个数互质。 （$$2$$）比如取$$5$$个偶数一个奇数，在这$$5$$个偶数中$$4$$、$$6$$、$$8$$、$$10$$都是$$2$$的倍数，如果取$$1$$、$$3$$、$$5$$、$$7$$、$$9$$、$$11$$，其中$$9$$是$$3$$的倍数；如果取$$6$$、$$7$$、$$8$$、$$9$$、$$10$$、$$11$$，就没有倍数关系； （$$3$$）比如$$1$$的$$2$$倍是$$2$$，$$2$$是$$2$$的倍数，$$2$$的$$2$$倍是$$4$$，$$4$$是$$4$$的倍数.据此解答。 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "607", "queId": "10612f786d98473682cdaba3ad156342", "competition_source_list": ["2014年全国创新杯五年级竞赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "在整数$$0$$、$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$中，质数的个数为$$x$$，偶数有$$y$$个，完全平方数的个数为$$z$$．则$$x+y+z$$ 等于． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->完全平方数->平方数的简单应用"], "answer_analysis": ["质数有$$4$$个，偶数有$$5$$个，完全平方数有$$4$$个（$$0$$也是）$$4+5+4=13$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2311", "queId": "fc819369ac714decb680f537d9d7fcf7", "competition_source_list": ["2010年第8届创新杯六年级竞赛初赛第1题4分", "2010年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "因武汉白沙洲大桥维修造成堵车，某人上班车速降低了$$\\frac{1}{7}$$，那么他在路上的时间增加了（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{7}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{9}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->比例解行程问题->行程中的比例"], "answer_analysis": ["速度比为$${{v}_{1}}:{{v}_{2}}=1:\\left( 1-\\frac{1}{7} \\right)=7:6$$，路程一定，时间比为$${{t}_{1}}:{{t}_{2}}=6:7$$，则在路上的时间增加了$$\\frac{1}{6}$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2909", "queId": "b6c44fdba1ab41729c9e29eae637c94c", "competition_source_list": ["2014年IMAS小学高年级竞赛第一轮检测试题第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "把数$$38$$、$$79$$、$$17$$、$$43$$、$$74$$、$$96$$、$$87$$重新排成一排，使得从第二个数开始，每个数的十位数码都等于前一个数的个位数码，请问排列后第四个数是什么？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$38$$ "}], [{"aoVal": "B", "content": "$$43$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$96$$ "}], [{"aoVal": "E", "content": "$$87$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数"], "answer_analysis": ["重新排列后，除了最前面与最后面的数码外，每个数码都要出现偶数次，现只有数码$$1$$与$$6$$各出现一次，故$$17$$必须是第一个数，接下来的排列方式为：$$17$$、$$74$$、$$43$$、$$38$$、$$87$$、$$79$$、$$96$$，故第四个数是$$38$$．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2719", "queId": "ff8080814502fa24014503a78d2001ad", "competition_source_list": ["2014年全国迎春杯六年级竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "算式$$2013\\times \\frac{2015}{2014}+2014\\times\\frac{2016}{2015}+\\frac{4029}{2014\\times 2015}$$计算结果是（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4027$$ "}], [{"aoVal": "B", "content": "$$4029$$ "}], [{"aoVal": "C", "content": "$$2013$$ "}], [{"aoVal": "D", "content": "$$2015$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数运算->分数四则混合运算"], "answer_analysis": ["$$2013\\times \\frac{2015}{2014}\\textgreater2013$$，$$2014\\times\\frac{2016}{2015}\\textgreater2014$$结果大于$$4027$$．结果为B． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1618", "queId": "df3bbbfb45764573aebf4551d7cbfe46", "competition_source_list": ["2013年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "三个自然数$$A$$、$$B$$、$$C$$之和是$$111$$，已知$$A$$、$$B$$的平均数是$$31$$，$$A$$、$$C$$的平均数是$$37$$．那么$$B$$、$$C$$的平均数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$34$$ "}], [{"aoVal": "B", "content": "$$37$$ "}], [{"aoVal": "C", "content": "$$43$$ "}], [{"aoVal": "D", "content": "$$68$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型"], "answer_analysis": ["$$C=111-31\\times 2=49$$ $$B=111-37\\times 2=37$$ $$\\left( 49+37 \\right)\\div 2$$ $$=86\\div 2$$ $$=43$$ 答：$$B$$、$$C$$的平均数是$$43$$． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2423", "queId": "0c5127c0f1df4291b69770c82e65dd35", "competition_source_list": ["2009年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "$$4$$吨葡萄在新疆测得含水量为$$55 \\%$$，运抵武昌后测得含水量为$$40 \\%$$，运抵武昌后葡萄剩下（ ）吨。 ", "answer_option_list": [[{"aoVal": "A", "content": "1 "}], [{"aoVal": "B", "content": "2 "}], [{"aoVal": "C", "content": "3 "}], [{"aoVal": "D", "content": "4 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->一元一次方程->分数、小数系数方程"], "answer_analysis": ["不妨称不含水的葡萄为``干葡萄''。设运抵武昌后葡萄还剩$$x$$吨，由于运输途中``干葡萄''的重量不变，所以$$x\\times \\left( 1-40 \\% \\right)=4\\times \\left( 1-55 \\% \\right)$$，解得$$x=3$$，所以运抵武昌后葡萄剩下$$3$$吨。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2073", "queId": "cb77a07e035e4f0ea983bc190a21f345", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛A卷第2题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "在含盐为$$5 \\%$$的$$100$$克盐水中，再分别加入$$10$$克盐和$$40$$克水后，盐与水的比是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20:1$$ "}], [{"aoVal": "B", "content": "$$1:10$$ "}], [{"aoVal": "C", "content": "$$10:9$$ "}], [{"aoVal": "D", "content": "$$1:9$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["原有的盐有$$100$$克$$\\times 5 \\%=5$$克，所以原有水$$95$$克． 加入$$10$$克盐和$$40$$克水后，有盐$$5$$克$$+10$$克$$=15$$克，有水$$95$$克$$+40$$克$$=135$$克． 所以盐$$:$$水$$=15$$克$$:135$$克$$=1:9$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "779", "queId": "9ec053c8620a446cbb3e12199c3d0723", "competition_source_list": ["2017年全国美国数学大联盟杯小学中年级四年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "一群$$5$$至$$12$$岁的孩子去看电影．这些孩子的年龄的乘积是$$3080$$．请问这些孩子的年龄的和是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$3080=2^{3}\\times5\\times7\\times11$$, $$5+7+8+11=31$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "321", "queId": "64ee522b1fce495fadde8a1fb8fea49c", "competition_source_list": ["2022年陕西西安六年级下学期小升初模拟《推理问题》第9题", "2016年陕西西安小升初工大附中入学真题4第10题", "2021年陕西西安碑林区西安铁一中龙岗中学小升初（三）第10题3分", "2016年陕西西安小升初工大附中", "2010年天津陈省身杯三年级竞赛第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙三个同学中有一个人在同学们都不在时把教室打扫干净，事后老师问他们是谁做的好事，甲说：``是乙干的''，乙说：``不是我干的''，丙说：``不是我干的''，如果他们中有两人说了谎话，一个人说的是真话，由此断定是~\\uline{~~~~~~~~~~}~干的． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["这是逻辑推理题，理清思路即可．假设甲说的是真话，那么是乙干的，这时丙说的话是真话，与只有一人说真话产生矛盾，因此甲说的是假话，即不是乙干的，所以乙说的是真话，从而丙说的是假话，故是丙干的． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "110", "queId": "829a014c892748f98cc1749659eabb88", "competition_source_list": ["2016年环亚太杯三年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "现有甲、乙两个水桶，甲桶容量$7$公升，乙桶容量$$3$$公升，甲桶装满了水，甲桶给乙桶~\\uline{~~~~~~~~~~}~公升的水后，两桶的水一样多． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$2.5$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "$$3.5$$ "}], [{"aoVal": "F", "content": "$$4.5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析", "课内体系->知识点->数学广角->优化->解决问题策略"], "answer_analysis": ["$$\\frac{7}{2}=3.5$$公升$$\\textgreater3$$公升，所以只能将$$3$$公升的水桶装满，且$$7$$公升水桶中仍有$$3$$公升，则$$7-3=4$$公升，即给$$4$$公升水给$$3$$公升水桶，则$$1$$公升溢出，两个容器水一样多． 故答案为：$$4$$．选$$\\text{D}$$. "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2076", "queId": "c7022107f51e4dd2b7ff8816327b2421", "competition_source_list": ["2008年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "李平的妈妈买了几瓶饮料。第一天，他们全家喝了全部饮料的一半零半瓶；第二天，李平招待同学，又喝了第一天剩下的饮料的一半零半瓶；第三天，李平又喝了第二天剩下的饮料的一半零半瓶。这三天，正好把妈妈买的全部饮料喝光，则妈妈买的饮料一共有( )。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$瓶 "}], [{"aoVal": "B", "content": "$$6$$瓶 "}], [{"aoVal": "C", "content": "$$7$$瓶 "}], [{"aoVal": "D", "content": "$$8$$瓶 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["对每个选项进行检验，如（$$C$$）：$$7$$瓶，第一天喝了$$3.5+0.5=4$$（瓶），剩下$$3$$瓶；第二天喝了$$1.5+0.5=2$$（瓶），剩下$$1$$瓶；第三天喝了$$0.5+0.5=1$$（瓶），正好把妈妈买的$$7$$瓶饮料喝光。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2425", "queId": "33b58a81569849d58e033b512a8e006b", "competition_source_list": ["2019年美国数学大联盟杯五年级竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "与$$60\\times 120$$结果相同的算式是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$180+240$$ "}], [{"aoVal": "B", "content": "$$180+540$$ "}], [{"aoVal": "C", "content": "$$1800+2400$$ "}], [{"aoVal": "D", "content": "$$1800+5400$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$60\\times 120=7200$$， $$\\text{A}$$选项中，$$180+240=420$$； $$\\text{B}$$选项中，$$180+540=720$$； $$\\text{C}$$选项中，$$1800+2400=4200$$； $$\\text{D}$$选项中，$$1800+5400=7200$$． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2681", "queId": "3318b34321154e14ae48b27ccb416719", "competition_source_list": ["2006年第4届创新杯六年级竞赛初赛B卷第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "完成一件工作，甲要$$\\frac{1}{5}$$小时，乙要$$\\frac{1}{3}$$小时，甲与乙的工作效率比是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2:6$$ "}], [{"aoVal": "B", "content": "$$5:3$$ "}], [{"aoVal": "C", "content": "$$3:5$$ "}], [{"aoVal": "D", "content": "$$6:2$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["把工作总量看作``$$1$$''，根据工作总量$$\\div $$工作时间$$=$$工作效率，分别求出甲、乙的工作效率，再写出对应的比，根据比的基本性质化成最简整数比． $$\\left( 1\\div \\frac{1}{5} \\right):\\left( 1\\div \\frac{1}{3} \\right)=5:3$$． 答：甲与乙的工作效率比是$$5:3$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1071", "queId": "98a40144b7664e388f5d22f8945abaa0", "competition_source_list": ["2017年第13届湖北武汉新希望杯六年级竞赛决赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "《学而思外思班终极秘籍》的正文共$$193$$页，页码是从$$1$$到$$3$$位的连续自然数，这本书正文的页码共有个数码``$$1$$''． ", "answer_option_list": [[{"aoVal": "A", "content": "$$131$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$133$$ "}], [{"aoVal": "D", "content": "$$134$$ "}]], "knowledge_point_routes": ["拓展思维->知识点->应用题模块->页码问题->数码综合问题", "课内体系->七大能力->逻辑分析"], "answer_analysis": ["百位上是$$1$$：$$100$ $193$$共$$94$$个; 十位上是$$1$$：$$10$ $19$$，$$110$ $119$$共$$20$$个; 个位上是$$1$$：$$1,11,21,31,\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 191$$共$$20$$个; 总共$$94+20+20=134$$（个）． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2706", "queId": "7f5fafe0caa44f569755e91c09d7f21b", "competition_source_list": ["2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "下列算式结果为$$500$$的是( ) ", "answer_option_list": [[{"aoVal": "A", "content": "$$5\\times 99+1$$ "}], [{"aoVal": "B", "content": "$$100+25\\times 4$$ "}], [{"aoVal": "C", "content": "$$88\\times 4\\text{ }+37\\times 4$$ "}], [{"aoVal": "D", "content": "$$100\\times 0\\times 5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数乘法巧算之提取公因数（普通型）"], "answer_analysis": ["$$88\\times 4+37\\times 4=\\left( 88+37 \\right)\\times 4=125\\times 4=500$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2384", "queId": "3cbc83dfef144ab7abc8dbf6c232db57", "competition_source_list": ["2017年第22届湖北武汉华杯赛三年级竞赛复赛（华罗庚金杯）第1题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$$(888+777)\\div (666+555+444)=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["前后括号同时除以$$111$$，即$$(8+7)\\div (6+5+4)=1$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3213", "queId": "1f0ba593d63c4492aa3811de7bfafb36", "competition_source_list": ["2008年第6届创新杯四年级竞赛初赛A卷第4题5分", "2008年四年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "如果两个四位数的差等于2008，我们就称这两个四位数组成了一个``创新数对''，那么``创新数对''共有( )个. ", "answer_option_list": [[{"aoVal": "A", "content": "6991 "}], [{"aoVal": "B", "content": "6992 "}], [{"aoVal": "C", "content": "6993 "}], [{"aoVal": "D", "content": "6994 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["在这些``创新数对''中，被减数最大的是9999，此时减数是7991，而减数最小的为1000，所以``创新数对''共有$$7991-1000+1=6992$$个 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1958", "queId": "aa9e8b3810fd40f3b4db0fd0cb7d3bae", "competition_source_list": ["2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小东看叔叔锯木头，锯一次需要用时$$3$$分，最后叔叔把木头锯成$$4$$段，叔叔请小东算算，他一共用了分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都没有->锯木头类型问题"], "answer_analysis": ["锯成$$4$$段，则需要锯$$3$$次，则所花时间为：$$3+3+3=9$$（分）． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2398", "queId": "0f8c10f8623e4e1798c6c1e578b2e0cd", "competition_source_list": ["2017年陕西西安碑林区西北工业大学附属中学小升初（二）第3题3分", "2017年陕西西安碑林区西北工业大学附属中学小升初(十)第3题3分", "2016年创新杯六年级竞赛训练题（四）第4题", "2017年陕西西安小升初某工大附中", "2018年湖南长沙雨花区中雅培粹中学小升初第3题3分", "2018年陕西西安小升初分类卷15第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "（$$2018$$.ZY）盒子中原来有$$7$$个小球，魔术师从中任取几个小球，把每一个小球都变成$$7$$个小球放回盒中；他又从中任取一些小球，把每一个小球又都变成$$7$$个小球放回盒中；如此进行，到某一时刻魔术师停止取球变魔术，此时盒中球的总数可能是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2018$$ "}], [{"aoVal": "B", "content": "$$2017$$ "}], [{"aoVal": "C", "content": "$$2016$$ "}], [{"aoVal": "D", "content": "$$2015$$ "}]], "knowledge_point_routes": ["拓展思维->能力->抽象概括"], "answer_analysis": ["每取出$$1$$个，盒中就增加$$6$$个，有$$7+6n=2017$$，∴$$n=335$$． 而$$\\rm A$$中，$$7+6n=2018$$，$$\\rm C$$中$$7+6n=2016$$，$$\\text{D}$$中$$7+6n=2015$$，$$n$$都不为整数． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2261", "queId": "893b500eb7e34a64a8399bfe582d3899", "competition_source_list": ["2014年全国华杯赛小学高年级竞赛初赛B卷第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "小华下午$$2$$点要到少年宫参加活动，但他的手表每个小时快了$$4$$分钟，他特意在上午$$10$$点时对好了表．当小华按照自己的表于下午$$2$$点到少年宫时，实际早到了分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["小华所带的``快表''每小时快了$$4$$分钟，说明准确时间走$$60$$分钟的时候，``快表''已经走了$$64$$分钟了，这样我们就可以得到$$\\frac{快表}{准表}=\\frac{64}{60}=\\frac{16}{15}$$；现在快表走了$$4\\times60=240$$，那么标准表走了$$240\\times 15\\div 16=225$$；所以实际上早到了$$240-225=15$$，选$$B$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2663", "queId": "681c654c41c3450aaa595f05f4863d94", "competition_source_list": ["2020年希望杯一年级竞赛模拟第29题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面的算式中，相同的图形代表相同的数，不同的图形代表不同的数． $$\\square +\\square +\\square =\\triangle +\\triangle $$ $$\\triangle =\\bigcirc +\\bigcirc +\\bigcirc $$ $$\\square =?$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\bigcirc +\\bigcirc $$ "}], [{"aoVal": "B", "content": "$$\\bigcirc +\\bigcirc+\\bigcirc $$ "}], [{"aoVal": "C", "content": "$$\\bigcirc +\\bigcirc+\\bigcirc +\\bigcirc $$ "}], [{"aoVal": "D", "content": "$$\\bigcirc +\\bigcirc+\\bigcirc +\\bigcirc +\\bigcirc $$ "}], [{"aoVal": "E", "content": "$$\\bigcirc +\\bigcirc+\\bigcirc +\\bigcirc +\\bigcirc +\\bigcirc $$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->等量代换->可直接计算的算式代换"], "answer_analysis": ["看图可知$$3$$个正方形相加$$=2$$个三角形相加，$$1$$个三角形$$=3$$个圆形相加，所以$$2$$个三角形$$=3+3=6$$（个）圆形相加，所以$$3$$个正方形相加$$=6$$个圆形相加，可以得到$$1$$个正方形$$=2$$个圆形相加，故选项$$\\text{A}$$正确． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "483", "queId": "c5d785a46c2840589896d936a8f39ae2", "competition_source_list": ["2012年全国创新杯五年级竞赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$1$$、$$2$$、$$3$$、$$4\\cdot \\cdot \\cdot 28$$、$$29$$、$$30$$这$$30$$个数从左往右依次排列成一个$$51$$位数，这个数被$$9$$除的余数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$123\\cdot \\cdot \\cdot 9$$为$$9$$的倍数；$$1011\\cdot \\cdot \\cdot 27$$为$$9$$的倍数；$$282930$$的数字和为$$24$$，除以$$9$$ 余$$6$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2544", "queId": "236749d1129341aeb6ef06cf64768dbd", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（一）第4题", "2017年新希望杯六年级竞赛训练题（一）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "将四个分数按从小到大的顺序排列，正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ "}], [{"aoVal": "C", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ "}], [{"aoVal": "D", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数感认知->分数数字加工"], "answer_analysis": ["通分子$$\\frac{5}{14}=\\frac{60}{168}$$， $$\\frac{10}{27}=\\frac{60}{162}$$， $$\\frac{12}{31}=\\frac{60}{155}$$， $$\\frac{20}{53}=\\frac{60}{159}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2737", "queId": "452746b988a74128a12efd99e38ac8d3", "competition_source_list": ["2013年河南郑州中原网杯小学高年级六年级竞赛初赛"], "difficulty": "3", "qtype": "single_choice", "problem": "有$$10$$个表面涂红色的正方体，它们的棱长分别是$$1$$，$$2$$，$$3$$，$$\\cdot \\cdot \\cdot $$，$$10$$厘米，如果把这些正方体全部割成棱长为$$1\\text{cm}$$的小正方体，在这些小正方体中至少有一面是红色的块数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1000$$ "}], [{"aoVal": "B", "content": "$$1250$$ "}], [{"aoVal": "C", "content": "$$1729$$ "}], [{"aoVal": "D", "content": "$$2009$$ "}]], "knowledge_point_routes": ["拓展思维->能力->图形认知"], "answer_analysis": ["求至少有一面的块数我们可以先求一面都没有的块数，从第三个正方体开始我们可以去掉外层的正方体得到中间的$$1$$个正方体，以此类推可得到一组新的正方体，$$1$$，$$2$$，$$3$$，$$\\cdot \\cdot \\cdot $$，$$8$$，那我们就可以按照块数得到一组新的数列$${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+\\cdot \\cdot \\cdot +{{8}^{3}}=1296$$，而总块数是$${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+\\cdot \\cdot \\cdot +{{10}^{3}}=3025$$，则至少有一面的块数为$$3025-1296=1729$$个．数列立方和公式：$$1+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+\\cdot \\cdot \\cdot +{{n}^{3}}={{\\left[ \\frac{n\\left( n+1 \\right)}{2} \\right]}^{2}}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "104", "queId": "de6ae5eeea9143b39516551c35658c17", "competition_source_list": ["2019年第24届YMO五年级竞赛决赛第6题3分", "2020年第24届YMO五年级竞赛决赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "2020第$$24$$届$$YMO$$五年级决赛 小$$Y$$、小$$M$$、小$$O$$、小$$T$$四人中只有$$1$$人会开车．小$$Y$$说：``我会开''．小$$M$$说：``我不会开''．小$$O$$说：``小$$Y$$不会开''．小$$T$$什么也没说．已知小$$Y$$、小$$M$$、小$$O$$三人只有一人说了真话．会开车的是． ", "answer_option_list": [[{"aoVal": "A", "content": "小$$Y$$ "}], [{"aoVal": "B", "content": "小$$M$$ "}], [{"aoVal": "C", "content": "小$$O$$ "}], [{"aoVal": "D", "content": "小$$T$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->组合模块->逻辑推理", "拓展思维->能力->创新思维"], "answer_analysis": ["对比发现，小$$O$$与小$$Y$$说的矛盾，相互对立， 则小$$O$$与小$$Y$$必一对一错， 则小$$M$$说假话，则小$$M$$会开车，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3041", "queId": "d7dc6f1e927c4b778a90fef6815e071e", "competition_source_list": ["2016年第3届广东深圳鹏程杯小学高年级六年级竞赛集训材料第十章从算术到代数第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "解下列方程： 已知关于$$x$$的方程$$2(x+1)=3(x-1)$$的解为$$a+2$$，求关于$$y$$方程$$2[2(y+3)-3(y-a)]=3a$$的解． ", "answer_option_list": [[{"aoVal": "A", "content": "$$y=7\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$y=8\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$y=9\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$y=10\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->式与方程->简易方程（一元一次）->解简易方程->整数简易方程"], "answer_analysis": ["解：把$$a+2$$代入方程$$2(x+1)=3(x-1)$$ 得：$$2(a+2+1)=3(a+2-1)$$ $$2(a+3)=3(a+1)$$ $$2a+6=3a+3$$ $$3=3a-2a$$ $$a=3$$ 把$$a=3$$代入方程$$2[2(y+3)-3(y-a)]=3a$$ 得：$$2[2(y+3)-3(y-3)]=3\\times 3$$ $$2(2y+6-3y+9)=9$$ $$2(15-y)=9$$ $$30-2y=9$$ $$30-9=2y$$ $$21=2y$$ $$y=10\\frac{1}{2}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1416", "queId": "5a2dd8297c254dc7b2e9df9688c93dee", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛决赛第10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有两个同样的仓库，搬运完其中一个仓库的货物，甲需要$$6$$小时，乙需要$$7$$小时，丙需要$$14$$小时．甲、乙同时开始各搬运一个仓库的货物，开始时，丙先帮甲搬运，后来又去帮乙搬运，最后两个仓库的货物同时搬完．丙从一个仓库到另一个仓库的时间忽略不计．则丙帮甲搬了小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1.5$$ "}], [{"aoVal": "B", "content": "$$1\\frac{3}{4}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["本题主要考查分数的应用， 首先计算出三个人搬仓库用时，再分别计算甲、乙完成了多少， 然后可知丙完成的为$$1$$减去甲的完成量加上$$1$$减去乙的完成量， 再计算丙帮甲、乙分别用时即可， 三个人都搬了同样长的时间， 甲每小时搬$$\\frac{1}{6}$$，乙每小时搬$$\\frac{1}{7}$$，丙每小时搬$$\\frac{1}{14}$$， 三人每小时共可般$$\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{14}=\\frac{8}{21}$$， 则搬完两个仓库共要$$2\\div \\frac{8}{21}=\\frac{21}{4}$$时， 即三人都同样工作了$$\\frac{21}{4}$$小时， 而$$\\frac{21}{4}$$小时内甲完成$$\\frac{21}{4}\\times \\frac{1}{6}=\\frac{7}{8}$$， 乙完成$$\\frac{21}{4}\\times \\frac{1}{7}=\\frac{3}{4}$$， 即甲有$$\\frac{1}{8}$$是丙完成的，乙有$$\\frac{1}{4}$$是丙帮忙完成的， 丙帮甲、乙分别用时$$\\frac{1}{8}\\div \\frac{1}{14}=\\frac{7}{4}$$时，$$\\frac{1}{4}\\div \\frac{1}{14}=\\frac{7}{2}$$小时． 答：丙帮甲、乙分别用时$$\\frac{7}{4}$$时、$$\\frac{7}{2}小时$$， $$\\frac{7}{4}=1\\frac{3}{4}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1273", "queId": "8b3805873c4c4beca708ca8913000c27", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（二）第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "办公室有甲、乙、丙、丁$$4$$位同事，甲比乙大$$5$$岁，丙比丁大$$2$$岁．丁$$3$$年前参加工作，当时$$22$$岁．他们四人现在的年龄之和为$$127$$岁．那么乙现在的年龄是（~ ）岁．~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$27$$ "}], [{"aoVal": "C", "content": "$$35$$ "}], [{"aoVal": "D", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["丁今年$$22+3=25$$（岁），丙今年$$25+2=27$$（岁），甲、乙年龄和是$$127-25-27=75$$（岁），甲比乙大$$5$$岁，则乙的年龄是$$\\left( 75-5 \\right)\\div 2=35$$（岁）． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1899", "queId": "e05dafa3119b42f6888b353f8fc17df5", "competition_source_list": ["2011年全国美国数学大联盟杯小学高年级五年级竞赛初赛第32题"], "difficulty": "1", "qtype": "single_choice", "problem": "我班上男生和女生的比例是$$3:4$$．请问我班上可能有多少名学生？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$70$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["是$$3+4=7$$的倍数． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2958", "queId": "86749103e58f4161ac2cf263b706100b", "competition_source_list": ["2021年第8届鹏程杯四年级竞赛初赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$2020\\times 20212021-2021\\times 20202020=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2020$$ "}], [{"aoVal": "D", "content": "$$2021$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["原式$$=2020\\times 2021\\times 10001-2021\\times 2020\\times 10001=0$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2513", "queId": "1e9dc6b3183546c89fa079b3e4b4e78e", "competition_source_list": ["2020年希望杯六年级竞赛模拟第25题"], "difficulty": "1", "qtype": "single_choice", "problem": "比较大小： $$1+\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\frac{1}{{{4}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}}$$~\\uline{~~~~~~~~~~}~$$2$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\textgreater$$ "}], [{"aoVal": "B", "content": "$$\\textless{}$$ "}], [{"aoVal": "C", "content": "$$=$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为$$\\frac{1}{{{2}^{2}}}\\textless{}\\frac{1}{1\\times 2}$$，$$\\frac{1}{{{3}^{2}}}\\textless{}\\frac{1}{2\\times 3}$$，$$\\frac{1}{{{4}^{2}}}\\textless{}\\frac{1}{3\\times 4}$$， 所以$$\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\frac{1}{{{4}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}}\\textless{}\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{2019\\times 2020}$$， 因为$$\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\frac{1}{3\\times 4}\\cdots +\\frac{1}{2019\\times 2020}$$ $$=1-\\frac{1}{2}+\\frac{1}{2}-\\frac{1}{3}+\\frac{1}{3}-\\frac{1}{4}+\\cdots +\\frac{1}{2019}-\\frac{1}{2020}$$ $$=1-\\frac{1}{2020}$$ $$=\\frac{2019}{2020}$$， 所以$$\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}}\\textless{}\\frac{2019}{2020}$$， 所以$$1+\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\frac{1}{{{4}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}}\\textless{}2$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "91", "queId": "a21645299bc34646bc42929ae153cf3e", "competition_source_list": ["2006年五年级竞赛创新杯", "2006年第4届创新杯五年级竞赛初赛A卷第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "五名选手在一次数学竞赛中共得404分，每人得分互不相同，并且其中得分最高为90分，那么得分最低的选手至少得( )分. ", "answer_option_list": [[{"aoVal": "A", "content": "45 "}], [{"aoVal": "B", "content": "50 "}], [{"aoVal": "C", "content": "55 "}], [{"aoVal": "D", "content": "60 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["要使某选手得分最低，其他选手得分必须尽可能的高，又知每人得分互不相同，所以前四名选手得分依次为90，89，88，87，因此，得出得分最低的选手至少得$$404-\\left( 90+89+88+87 \\right)=50$$分 "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "459", "queId": "aa34f1fbfb8b4b58b74958d081562271", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛B卷第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$16$$个外形完全相同的小零件，其中$$15$$个是正品，$$1$$个是次品，正品重量都相等，次品比正品稍重一些，一架无砝码的天平至少称次可把次品找出来． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["天平是用来称量物体质量的工具，此题并不是称量物体的质量，而是使用天平来比较物体质量的大小，所以，在调好的天平两盘中分别放上物体，当哪边的托盘上升，则说明这边托盘中的物体质量偏小． 将$$16$$个零件分成$$6$$、$$6$$、$$4$$三组，先称量$$6$$，$$6$$的一组如果重量相等，则将$$4$$的那一组分为$$2$$，$$2$$的两组，接着称量，找出较重的一组，分为$$1$$，$$1$$的两组，进行第三次称量，找出次品；如果$$6$$，$$6$$的重量不相等，找出较重的一组分为$$3$$，$$3$$的两组，进行第二次称量，找出较重的一组，分为$$1$$，$$1$$，$$1$$的三组，任意挑选其中的两个，如果重量一样，那么另一个就是次品，再或者可以直接称量出来．这样只需$$3$$次即可找出次品． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3342", "queId": "ed2a7153c7664f1c87691c6ac0725f74", "competition_source_list": ["2022年袋鼠数学竞赛(Math Kangaroo)小学高年级竞赛第19题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "丽丽有 4 只狗，每只狗的体重都是整数 kg，且重量各不相��．已知它们的总重量为 60kg，第二重的狗有 28kg．请问第三重的狗体重是？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2kg$$ "}], [{"aoVal": "B", "content": "$$3kg$$ "}], [{"aoVal": "C", "content": "$$4kg$$ "}], [{"aoVal": "D", "content": "$$5kg$$ "}], [{"aoVal": "E", "content": "$$6kg$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->整数分拆"], "answer_analysis": ["According to the given information, the 2nd heaviest dog is 28kg, so the sum of the remaining 3 dogs is: 60 - 28 = 32kg. So the heaviest one has at least 29kg, and the sum of the weight of the 2 remaining dogs is 3kg. Since none of them weigh the same, the weight of the 2 remaining dogs is 2kg and 1kg respectively. Choose A. 本题中，第二重的狗体重为$$28$$千克，剩余$$3$$只狗的体重和为：$$60-28=32$$（千克）．此时最重的狗体重至少为$$29$$千克，那么剩下的两只狗体重之和至多只有$$3$$千克，则它们的体重只能是$$1$$千克和$$2$$千克．因此第三重的狗体重为$$2$$千克．选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2474", "queId": "19977b1c84604a1aa83bd13e891ebcd4", "competition_source_list": ["2010年北京学而思综合能力诊断三年级竞赛第3题"], "difficulty": "3", "qtype": "single_choice", "problem": "在下面的一列数中，从左向右数，第$$8$$个数是~\\uline{~~~~~~~~~~}~． $$1$$，$$4$$，$$10$$，$$20$$，$$35$$，$$\\cdots$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$116$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$140$$ "}], [{"aoVal": "D", "content": "$$156$$ "}], [{"aoVal": "E", "content": "$$180$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数"], "answer_analysis": ["每两个数差值为$$3$$，$$6$$，$$10$$，$$15$$，$$\\cdots$$ 它们的差值为$$3$$，$$4$$，$$5$$，$$\\cdots$$ 所以第$$8$$个数是$$120$$． $$1+3=4$$，$$4+6=10$$，$$10+10=20$$，$$20+15=35$$，$$35+21=56$$，$$56+28=84$$，$$84+36=120$$，每次加上的数$$3$$，$$6$$，$$10$$，$$15$$，$$21$$，$$28$$，$$36$$又有规律，$$3+3=6$$，$$6+4=10$$，$$10+5=15$$，$$15+6=21$$，$$21+7=28$$，$$28+8=36$$，所以第$$8$$个数是$$120$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2630", "queId": "5122e0d2d1ef41f0ae20c6a656428f2f", "competition_source_list": ["2008年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "植树节时，某班每人应平均植树6株，如果只有女生来完成，每人应植树15株，如果只有男生来完成，则每人应植树( )株. ", "answer_option_list": [[{"aoVal": "A", "content": "8 "}], [{"aoVal": "B", "content": "10 "}], [{"aoVal": "C", "content": "12 "}], [{"aoVal": "D", "content": "14 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->设而不求"], "answer_analysis": ["设共有[6,15]=90棵数，所以共有90÷6=15位同学，其中女生有90÷15=6，那么男生有15-6=9位，平均植树90÷9=10棵，故选B "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2254", "queId": "64e5b44502f244c5b98cd81af6be389f", "competition_source_list": ["2017年全国华杯赛竞赛初赛模拟试卷3第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲，乙两个小朋友，在一条环形路上面跑步，同时从同地反向跑，已知甲小朋友的速度是每秒$$5$$米，乙小朋友的速度是每秒$$7$$米，在$$14$$分钟内，他们相遇了 $$21$$次，则环形路长为（~ ~ ~）米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$480$$ "}], [{"aoVal": "B", "content": "$$510$$ "}], [{"aoVal": "C", "content": "$$450$$ "}], [{"aoVal": "D", "content": "$$620$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->环形跑道->环形跑道中的相遇->环形相遇同时同地出发"], "answer_analysis": ["设环形路长为$$M$$米，则甲、乙相遇$$1$$次所需时间是： $$\\frac{M}{5+7}=\\frac{M}{12}$$（秒），所以$$14\\times60\\div\\frac{M}{12}=21$$，因此，$$M=480$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2586", "queId": "3e84907eec734be48a5239812d1de420", "competition_source_list": ["2019年浙江杭州滨江区杭州江南实验学校五年级竞赛模拟（江南杯）第11题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "两个因数的积保留两位小数后约是$$9.28$$，准确数可能是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9.2729$$ "}], [{"aoVal": "B", "content": "$$9.276$$ "}], [{"aoVal": "C", "content": "$$9.286$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["要考虑$$9.28$$是一个三位小数的近似数，有两种情况：``四舍''得到的$$9.28$$最大是$$9.284$$，``五入''得到的$$9.28$$最小是$$9.275$$，即该数的取值是大于或等于$$9.275$$并且小于或等于$$9.284$$；由此进行选择即可． ``四舍''得到的$$9.28$$最大是$$9.284$$，``五入''得到的$$9.28$$最小是$$9.275$$，即该数的取值是大于或等于$$9.275$$并且小于或等于$$9.284$$，结合选项可知：准确值可能是$$9.276$$，故选$$\\text{B}$$． 故答案为：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "300", "queId": "52a720834f234a2787cba03af546580b", "competition_source_list": ["2015年第13届全国创新杯小学高年级五年级竞赛复赛第5题", "2016年创新杯小学高年级五年级竞赛训练题（三）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "三个相邻奇数的积是一个六位数$$\\overline{3abcd3}$$，那么这三个奇数的和为（~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$203$$ "}], [{"aoVal": "B", "content": "$$205$$ "}], [{"aoVal": "C", "content": "$$207$$ "}], [{"aoVal": "D", "content": "$$209$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["末尾是$$3$$，三个奇数的末尾只能是$$7$$、$$9$$、$$1$$，考虑$$6$$位数，三个奇数是两位数，且首位在$$6 \\tilde{ } 7$$之间，那么有$$67\\times 69\\times 71=328233$$满足，和是$$207$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2256", "queId": "4ede08deae0a4b1cbe55e8ffd1a6b38d", "competition_source_list": ["2017年第13届湖北武汉新希望杯六年级竞赛决赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "泸昆高铁最后一段贵阳至昆明于$$2016$$年$$12$$月$$28$$日开通运营，这对我国``一带一路''战略的实施和区域经济发展都有着重大意义．$$G1375$$次高铁$$11:06$$从上海虹桥站出发，当天$$22:56$$到达昆明南站，全程共$$1518$$千米，途中站点共计停车$$50$$分钟，扣除停车时间，$$G1375$$次高铁的平均速度为千米/时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$148$$~ "}], [{"aoVal": "B", "content": "$$138$$ "}], [{"aoVal": "C", "content": "$$150$$~ "}], [{"aoVal": "D", "content": "$$151$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["$$11:06至22:56$$共$$11$$小时$$50$$分，其中停车$$50$$分钟，运行$$11$$小时． 平均速度为$$1518\\div11=138\\rm km/h$$. "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1613", "queId": "4e47a7c8129743c5ab4cc74a25b28ab3", "competition_source_list": ["小学中年级三年级上学期其它", "2017年全国华杯赛小学中年级竞赛初赛模拟第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两人在春节一共得$$210$$元压岁钱，甲给乙$$40$$元，还比乙多$$10$$元，那么，原来甲得了元压岁钱． ", "answer_option_list": [[{"aoVal": "A", "content": "$$150$$ "}], [{"aoVal": "B", "content": "$$140$$ "}], [{"aoVal": "C", "content": "$$125$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->和差倍问题->和差问题->二量和差问题->明和暗差"], "answer_analysis": ["因为甲给乙$$40$$元，还比乙多$$10$$元，所以甲比乙多$$40\\times 2+10=90$$元， 所以甲：$$(210+90)\\div 2=150$$ "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "927", "queId": "a64cb54217b745c68bdc5ec72089061e", "competition_source_list": ["2009年第7届创新杯六年级竞赛初赛第10题4分", "2009年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "$${{1949}^{2009}}$$的末两位数字是( ). ", "answer_option_list": [[{"aoVal": "A", "content": "49 "}], [{"aoVal": "B", "content": "81 "}], [{"aoVal": "C", "content": "01 "}], [{"aoVal": "D", "content": "69 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数的性质->余数的可乘性"], "answer_analysis": ["显然$${{1949}^{2009}}$$的末两位数字与$${{49}^{2009}}$$的末两位数字相同.由于$${{49}^{1}}$$的末两位数字为49，$${{49}^{2}}$$的末两位数字为01，$${{49}^{3}}$$的末两位数字为$$49$$，$${{49}^{4}}$$的末两位数字为01，\\ldots，可见$${{49}^{n}}\\left( n=1，2，3，4，... \\right)$$的末两位数字是以2为周期循环出现49和01的，所以$${{49}^{2009}}$$的末两位数字与$${{49}^{1}}$$的末两位数字相同，都是49. "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2691", "queId": "99da7f2d52364633b8d4062970ea28fb", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛A卷第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$0.988\\times a=0.999\\times b$$（$$a$$，$$b$$均不为$$0$$），那么$$a$$与$$b$$的大小关系是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless{}b$$ "}], [{"aoVal": "B", "content": "$$a\\textgreater b$$ "}], [{"aoVal": "C", "content": "$$a=b$$ "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->算式比较大小"], "answer_analysis": ["因为$$0.988\\times a$$的积和$$0.999\\times b$$的积相等， 而$$0.988\\textless{}0.999$$， 即一个因数变大，要使积不变，另一个因数需要变小， 所以$$a\\textgreater b$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1690", "queId": "ff808081488801c601488c270e6f0f18", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "甲从一个鱼摊上买了三条鱼，平均每条$$a$$元，又从另一个鱼摊上买了两条鱼，平均每条$$b$$元，后来他又以每条$$\\frac{a+b}{2}$$元的价格把鱼全部卖给了乙，结果发现赔了钱，原因是（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b$$ "}], [{"aoVal": "B", "content": "$$a\\textless{b}$$ "}], [{"aoVal": "C", "content": "$$a=b$$ "}], [{"aoVal": "D", "content": "与$$a$$和$$b$$的大小无关 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["可根据题意算出买进和卖出鱼所花钱的总数，买进一共花了$$(3a+2b)$$元，一共卖了$$\\frac{a+b}{2}$$$$\\times5$$元，前者大于后者，即可推导出$$a\\textgreater b$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1756", "queId": "9f93dd2bbd8c441c9d88c8ab09515991", "competition_source_list": ["2013年第9届全国新希望杯小学高年级六年级竞赛复赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "其中$$9$$岁的有$$11$$人，$$11$$岁的有$$2$$人，$$13$$岁的有$$3$$人，那么这$$16$$个小朋友的平均年龄是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$岁 "}], [{"aoVal": "B", "content": "$$10.5$$岁 "}], [{"aoVal": "C", "content": "$$11$$岁 "}], [{"aoVal": "D", "content": "$$11.5$$岁 "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->平均数问题->公式类->加权平均数"], "answer_analysis": ["$$\\left( 9\\times 11+11\\times 2+13\\times 3 \\right)\\div 16=10$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2829", "queId": "7bbf1d8816034849bb06e8d0d538d730", "competition_source_list": ["2006年五年级竞赛创新杯", "2006年第4届创新杯五年级竞赛初赛A卷第8题", "2006年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙两人对同一数作除法，甲将其除以8，乙将其除以9，甲所得的商与乙所得的余数之和为13，则甲所得的余数为( )． ", "answer_option_list": [[{"aoVal": "A", "content": "2 "}], [{"aoVal": "B", "content": "3 "}], [{"aoVal": "C", "content": "4 "}], [{"aoVal": "D", "content": "5 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法"], "answer_analysis": ["解法一:设这个相同的数为$$8m+n$$或$$9\\left( m-1 \\right)+a$$，即$$8m+n=9\\left( m-1 \\right)+a$$，又$$m+a=13$$，所以$$n=m+a-9=13-9=4$$，故选C. 解法二:设这个相同的数为a，则$$\\begin{cases}a=8m+n\\left( 0\\leqslant n \\textless{} 8 \\right) a=9{{m}^{'}}+{{n}^{'}} m+{{n}^{'}}=13 \\end{cases}$$① 所以，$$a=9{{m}^{'}}+13-m$$② ①$$-$$②得$$9m-9{{m}^{'}}+n-13=0$$，所以$$9\\left( m-{m}' \\right)=13-n$$，又$$5 \\textless{} 13-n\\leqslant 13$$，所以$$5 \\textless{} 9\\left( m-{m}' \\right)\\leqslant 13$$，因此，$$m-{{m}^{'}}=1$$，从而$$9=13-n$$，故$$n=4$$. "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3302", "queId": "406d64c912ad4148906f3ffcbb6085a8", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$42$$ children are lined up to go on an autumn trip,~ $$Y$$ is the $$22$$th from the front and $$M$$ is the~~$$22$$th from the back of the line. How many children are between $$Y$$ and $$X$$? $$42$$个小朋友排成一队去秋游，从排头往后数，小$$Y$$是第$$22$$个，从排尾往��数，小$$M$$是第$$22$$个，小$$Y$$和小$$M$$中间有个人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}], [{"aoVal": "E", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解", "Overseas Competition->知识点->计数模块->加乘原理->排队问题"], "answer_analysis": ["根据题意分析可知，小$$Y$$后面有$$42-22=20$$（人），$$22-20-1=1$$（人）．那么小$$M$$应该在小$$Y$$的正前方，所以小$$Y$$和小$$M$$之间有$$0$$个人． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1588", "queId": "b5b578f7d3b84ef4a9d8444d03c36e19", "competition_source_list": ["2018年全国小学生数学学习能力测评六年级竞赛复赛第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "球从高处自由落下，每次弹起的高度是前次下落高度的$$\\frac{2}{5}$$，如果球从$$75$$米处落下，第二次弹起的高度是米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["第一次弹起：$$75\\times \\frac{2}{5}=30$$（米）， 第二次弹起：$$30\\times \\frac{2}{5}=12$$（米）． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3388", "queId": "a0200657f5e44be99cd16dd0de716548", "competition_source_list": ["2008年第6届创新杯四年级竞赛初赛B卷第4题5分", "2008年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "由$$1$$，$$3$$，$$5$$，$$7$$，$$9$$五个数组成甲组数，$$2$$，$$4$$，$$6$$，$$8$$，四个数组成乙组数，从甲，乙两组数中各取一个数相加，可以得到不同的和的个数是( )。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["和可以是$$3$$，$$5$$，$$7$$，$$9$$，$$11$$，$$13$$，$$15$$和$$17$$，所以和的个数是$$8$$。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2791", "queId": "9a6185d3482f406c954ba7f4f3c1fb4e", "competition_source_list": ["2021年新希望杯四年级竞赛初赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$2788\\div 4\\div 27+565\\div (27\\times 5)=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$300$$ "}], [{"aoVal": "C", "content": "$$27$$ "}], [{"aoVal": "D", "content": "$$30$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数除法巧算之提取公除数（普通型）"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde2788\\div 4\\div 27+565\\div (27\\times 5)$$ $$=697\\div 27+565\\div 5\\div 27$$ $$=697\\div 27+113\\div 27$$ $$=(697+113)\\div 27$$ $$=810\\div 27$$ $$=30$$． 故答案为：$$30$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "201", "queId": "24d77807fa714bc2a97225317de7da65", "competition_source_list": ["2007年四年级竞赛创新杯", "2007年第5届创新杯四年级竞赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设m和n是选自自然数$$1\\sim 1004$$中的两个不同的数，那么$$\\left( m+n \\right)\\div \\left( m-n \\right)$$的最大值 是( )。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$2000$$ "}], [{"aoVal": "B", "content": "$$2005$$ "}], [{"aoVal": "C", "content": "$$2007$$ "}], [{"aoVal": "D", "content": "$$2009$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["为使$$\\left( m+n \\right)\\div \\left( m-n \\right)$$最大，应使$$m+n$$尽可能大，而$$m-n$$尽可能小，所以$$m=1004$$，$$n=1003$$，$$\\left( m+n \\right)\\div \\left( m-n \\right)=2007$$为最大值。 "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "979", "queId": "04d732ff8554407aa30528b2137215d7", "competition_source_list": ["2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有大、小两桶油共重$$44$$千克，两个桶都倒出同样多的油后，分别还剩$$14$$千克和$$10$$千克，小桶原来装油千克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->加减法应用->同增同减应用题->连筐（桶、瓶等）带物（考虑筐桶瓶等重量）"], "answer_analysis": ["倒了$$44-14-10=20$$千克，倒了相同多，小桶倒了$$20\\div 2=10$$千克，小桶原来$$10+10=20$$千克。 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2187", "queId": "6281a498f02b4c7c8e0c4ce53f0d3993", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "12. 同学甲、同学乙两人在游乐场的直行道上进行$$200$$米赛跑，当同学甲跑到终点时．同学乙还差$$40$$米，现在两人重新跑，而且速度和原来一样，要使两人同时到达终点，那么同学甲的起跑线应往后退米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$55$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->比例解行程问题->行程中的比例"], "answer_analysis": ["比例解行程；相等的时间内，甲跑了$$200$$米时，乙跑了$$160$$米，甲乙的速度比是$$200:160=5:4$$，要使两人同时到达终点，乙跑了$$200$$米的时间内甲要跑$$200\\div 4\\times 5=250$$米，所以同学甲的起跑线应后退$$250-200=50$$米． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2633", "queId": "c7b2bb9c7dc84dd19ba03242de08e251", "competition_source_list": ["2020年希望杯二年级竞赛模拟第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "请你根据数串的规律再横线上填上正确的答案：3,6,9,~\\uline{~~~~~~~~~~}~，15. ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数"], "answer_analysis": ["后一个数等于前一个数$$\\times$$ 2 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1793", "queId": "8593ebc84ec34f79a24c0c8e0082a414", "competition_source_list": ["2020年新希望杯二年级竞赛初赛（团战）第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$10$$名同学参加$$50$$米赛跑．跑到一半的时候，小明后面有$$5$$人，前面有$$4$$人，之后，没有人超过他，而小明又超过了$$3$$人到达终点．这次比赛没有并列名次，那么小明是． ", "answer_option_list": [[{"aoVal": "A", "content": "第一名 "}], [{"aoVal": "B", "content": "第二名 "}], [{"aoVal": "C", "content": "第三名 "}], [{"aoVal": "D", "content": "第四名 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["小明后面有$$5$$人，前面有$$4$$人，则他此时排在第$$5$$，之后，没有人超过他，小明又超过了$$3$$人后，他此时就是第二名． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "695", "queId": "4341126073c04edab976eb89aad96dde", "competition_source_list": ["2016年创新杯小学高年级六年级竞赛训练题（四）第7题"], "difficulty": "3", "qtype": "single_choice", "problem": "在$$\\left[ \\frac{{{1}^{2}}}{2017} \\right]$$，$$\\left[ \\frac{{{3}^{2}}}{2017} \\right]$$，$$\\left[ \\frac{{{5}^{2}}}{2017} \\right]$$，\\ldots，$$\\left[ \\frac{{{2017}^{2}}}{2017} \\right]$$这$$1009$$个数中共有个互不相同的数（$$\\left[ {}\\textasciitilde\\textasciitilde\\textasciitilde{} \\right]$$表示取整数，如$$\\left[ \\frac{13}{5} \\right]=2$$）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$883$$ "}], [{"aoVal": "B", "content": "$$882$$ "}], [{"aoVal": "C", "content": "$$888$$ "}], [{"aoVal": "D", "content": "$$1509$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["假设相邻两数为$$a$$、$$b$$，且$$a$$大$$b$$小，其实我们知道，$$a-b$$的差小于$$1$$时，$$[a]=[b]$$或者$$[a]-[b]=1$$；当$$a-b$$的差大于等于$$1$$时，$$[a]$$和$$[b]$$一定不同． 从小到大看，取整符号中的第$$n$$项是$$\\frac{{{\\left( 2n-1 \\right)}^{2}}}{2017}$$，第$$\\left( n+1 \\right)$$项是$$\\frac{{{\\left( 2n+1 \\right)}^{2}}}{2017}$$， 相减并利用平方差公式为$$\\frac{{{\\left( 2n+1 \\right)}^{2}}}{2017}-\\frac{{{\\left( 2n-1 \\right)}^{2}}}{2017}=\\frac{8n}{2017}$$，要想大于等于$$1$$，则$$n$$至少取$$253$$．所以取整符号中的第$$253$$项之后的数之差均大于$$1$$，即各不相同． 第$$253$$项为$$\\frac{{{505}^{2}}}{2017}=126$$，$$1-253$$项中，取整后的数每个整数都取过$$0-126$$有$$127$$个不一样的数．从第$$254-1009$$项中，每个数都不相同， 有$$1009-253=756$$个数．$$127+756=883$$（个）． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "756", "queId": "68b3ddd539c2475f8b6eec9222f79d21", "competition_source_list": ["2003年第1届创新杯六年级竞赛复赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果形如$$\\overline{2\\square \\square 3}$$的四位数能被$$9$$整除，那么这样的四位数的个数有多少个？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$个 "}], [{"aoVal": "B", "content": "$$9$$个 "}], [{"aoVal": "C", "content": "$$11$$个 "}], [{"aoVal": "D", "content": "$$12$$个 "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["如果$$\\overline{2ab3}$$能被$$9$$整除，那么$$(2+a+b+3)$$一定是$$9$$的倍数，即$$(a+b+5)$$是$$9$$的倍数．所以$$(a+b+5)$$只能等于$$9$$或$$18$$． 经枚举得，当$$a+b+5=9$$时，这个四位数有：$$2043$$、$$2133$$、$$2223$$、$$2313$$、$$2403$$．共$$5$$种情况． 当$$a+b+5=18$$时，这个四位数有：$$2943$$、$$2853$$、$$2763$$、$$2673$$、$$2583$$、$$2493$$．共$$6$$种情况． 综上，这样的四位数的个数有$$5+6=11$$（个）． ", "若$$9\\left| \\overline{2ab3} \\right.$$，则$$9\\left| \\left( 2+a+b+3 \\right) \\right.$$，则$$\\left( a+b \\right)$$除以$$9$$余$$4$$， 所以$$\\overline{ab}$$除以$$9$$余$$4$$，
\n所以$$\\overline{ab}=4,4+9,4+9\\times 2,4+9\\times 3,\\ldots ,4+9\\times 10$$， 所以有$$11$$个．
"], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1338", "queId": "c784572ad04941f8abc70b4fcd921714", "competition_source_list": ["2014年全国迎春杯四年级竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "有四个数，它们的和是$$45$$，把第一个数加$$１$$，第二个数减$$１$$，第三个数乘$$2$$，第四个数除以$$2$$，得到的结果都相同．那么，原来这四个数依次是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$，$$10$$，$$10$$，$$10$$ "}], [{"aoVal": "B", "content": "$$12$$，$$8$$，$$20$$，$$5$$ "}], [{"aoVal": "C", "content": "$$9$$，$$11$$，$$5$$，$$20$$ "}], [{"aoVal": "D", "content": "$$9$$，$$11$$，$$12$$，$$13$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设相同的结果为$$x$$，根据题意有：$$x-１+x+１+2x+0.5x=45$$，$$\\Rightarrow x=10$$ 易知原来的$$4$$ 个数依次是$$9$$，$$11$$，$$5$$，$$20$$． "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2412", "queId": "46197c3db2474327b7ddf7680601c1bc", "competition_source_list": ["2013年走美杯六年级竞赛初赛", "2013年走美杯六年级竞赛", "2013年浙江杭州走美杯六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$183\\times279\\times361-182\\times278\\times360$$的计算结果是（ ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$217017$$ "}], [{"aoVal": "B", "content": "$$207217$$ "}], [{"aoVal": "C", "content": "$$207216$$ "}], [{"aoVal": "D", "content": "$$217016$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之凑整法->整数基准数法"], "answer_analysis": ["解：$$183\\times279\\times361-182\\times278\\times360$$ $$=\\left(182+1\\right)\\times\\left(278+1\\right)\\times\\left(360+1\\right)-182\\times278\\times360$$ $$=182\\times278\\times360+182\\times278+182\\times360+278\\times360+182+278+360+1-182\\times278\\times360$$ $$=182\\times278+182\\times360+278\\times360+182+278+360+1$$ $$=50596+65520+100080+182+278+360+1$$ $$=217017$$ 故答案是$$217017$$。 "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3161", "queId": "3cc58f7eba3f4855bb9d2be78cf24d85", "competition_source_list": ["2009年全国迎春杯小学中年级竞赛复赛第7题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "过年了，妈妈买了$$7$$件不同的礼物，要送给亲朋好友的$$5$$个孩子每人一件．其中姐姐的儿子小强想从智力拼图和遥控汽车中选一个，朋友的女儿小玉想从学习机和遥控汽车中选一件．那么妈妈送出这$$5$$件礼物共有~\\uline{~~~~~~~~~~}~种方��． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$90$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$150$$ "}], [{"aoVal": "E", "content": "$$180$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "Overseas Competition->知识点->计数模块->加乘原理"], "answer_analysis": ["假如给小强的是智力拼图，则有$$2\\times 5\\times 4\\times 3=120$$种方法． 假如给小强的是遥控汽车，则有$$1\\times 5\\times 4\\times 3=60$$种方法． 总共有$$120+60=180$$种方法． 若将遥控汽车给小强，则学习机要给小玉，此时另外$$3$$个孩子在剩余$$5$$件礼物中任选$$3$$件，有$$5\\times 4\\times 3=60$$种方法；若将遥控汽车给小玉，则智力拼图要给小强，此时也有$$60$$种方法；若遥控汽车既不给小强、也不给小玉，则智力拼图要给小强，学习机要给小玉，此时仍然有$$60$$种方法. 所以共有$$60+60+60=180$$种方法. "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1388", "queId": "24f8f99b382d49359c2d7aee14e898dc", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "下式中，$$\\square $$和$$\\triangle $$分别代表（ ） $$\\square +\\square +\\square +\\triangle +\\triangle +\\triangle =27$$ $$\\triangle +\\triangle +\\square =12$$。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$和$$4$$ "}], [{"aoVal": "B", "content": "$$3$$和$$6$$ "}], [{"aoVal": "C", "content": "$$4$$和$$6$$ "}], [{"aoVal": "D", "content": "$$6$$和$$3$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->列方程解应用题->列方程解应用题等量代换"], "answer_analysis": ["解：$$\\square +\\square +\\square +\\triangle +\\triangle +\\triangle =27$$， 得$$\\square +\\triangle =9$$①， $$\\triangle +\\triangle +\\square =12$$得$$2\\triangle +\\square =12$$②， ②$$-$$①得：$$\\triangle =3$$， 把$$\\triangle =3$$代入①得： $$\\square =6$$ 故选：D。 "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "724", "queId": "3b1f33e49f23408dad88e81ff79f6be8", "competition_source_list": ["2013年IMAS小学高年级竞赛第一轮检测试题第12题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "有四个连续的二位数，将每个数的两个数码相乘，依序得到乘积$$24$$、$$28$$、$$32$$、$$36$$，请问这四个连续的二位数之总和是什么？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$136$$ "}], [{"aoVal": "C", "content": "$$160$$ "}], [{"aoVal": "D", "content": "$$172$$ "}], [{"aoVal": "E", "content": "$$190$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数->分解质因数的应用"], "answer_analysis": ["解法$$1$$：将$$24$$、$$28$$、$$32$$、$$36$$分别表示成两个数码之积的行驶，即可得 $$24=3\\times 8=4\\times 6$$ $$28=4\\times 7$$ $$32=4\\times 8$$ $$36=4\\times 9=6\\times 6$$ 因这四个两位数为四个连续的两位数，故此时可判断出这四个连续的正整数为$$46$$、$$47$$、$$48$$、$$49$$，其和为$$190$$．故选$$\\text{E}$$． 解法$$2$$：由题意可知，这些所得到的乘积不包含数码$$0$$，所以这四个两位数的十位数必须相同．设它们的十位数码为$$a$$．由于两个相邻的个位数是互质的，所以$$24$$、$$28$$、$$32$$、$$36$$的最大公因数为$$a$$． 因此，$$a=4$$，且这四个数的个位数分别为$$6$$、$$7$$、$$8$$、$$9$$．即这四个整数之和是$$46+47+48+49=190$$．故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2926", "queId": "92c81fb7e34a4436a5203fadd9aba92d", "competition_source_list": ["2017年第1届重庆华杯赛竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\frac{2013\\times 2013}{2014\\times 2014+2012}=\\frac{n}{m}( m,n$$互质$$ ),m+n=$$~\\uline{~~~~~~~~~~}~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1234$$ "}], [{"aoVal": "B", "content": "$$1243$$ "}], [{"aoVal": "C", "content": "$$1343$$ "}], [{"aoVal": "D", "content": "$$1244$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["~ ~ $$\\frac{2013\\times 2013}{2014\\times 2014+2012}=\\frac{2013\\times 2013}{\\left( 2013+1 \\right)\\times \\left( 2013+1 \\right)+2013-1}$$ $$=\\frac{2013\\times 2013}{2013\\times 2013+2013\\times 2+1+2013-1}$$ $$=\\frac{2013\\times 2013}{2013\\times 2013+3\\times 2013} $$ $$=\\frac{671}{672}$$ $$n+m=671+672=1343$$ "], "answer_value": "C"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "982", "queId": "04d9ec253b1f4a679995ec040c67eb7e", "competition_source_list": ["2016年创新杯六年级竞赛训练题（二）第4题"], "difficulty": "3", "qtype": "single_choice", "problem": "自然保护区中原有一定量的牛，并保持着一定的自然增长率。为进行原始物种的迁出工作，我们将自然保护区的草地按照$$1:2:3$$的面积比分成三块。工作人员先用$$1$$年时间逐步迁出了一号草地的种群，接着又用$$4$$年去暗处了二号种群．此时，仍需要年能够迁出第三块保护区的牛。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$216$$ "}], [{"aoVal": "B", "content": "$$288$$ "}], [{"aoVal": "C", "content": "$$324$$ "}], [{"aoVal": "D", "content": "$$396$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["设$$1$$份草地原有草量为$$x$$，每天生长量为$$y$$，第三块草地需要$$n$$天吃完． 根据牛每天吃的草量相等可得： $$\\frac{x+12y}{12}=\\frac{2x+2\\left( 12y+48y \\right)}{48}$$ 解得$$x=36y$$，这群牛的每天吃草量为$$4y$$． 故第三块草地的等量关系为$$\\frac{3x+3\\left( 12y+48y+ny \\right)}{4y}=n$$，解得$$n=288$$． "], "answer_value": "B"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "403", "queId": "e47999701c3d4119aa932b2c1462b886", "competition_source_list": ["2018年第8届北京学而思综合能力诊断六年级竞赛年度教学质量监测第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$6$$支球队进行足球比赛，每两支队之间都要赛一场，规定胜一场得$$3$$分，平一场各得$$1$$分，负一场不得分．全部比赛结束后，发现共有$$4$$场平局，且其中$$5$$支球队共得了$$31$$分，则第$$6$$支球队得了分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析", "课内体系->能力->逻辑分析"], "answer_analysis": ["每场平局两队共得$$2$$分，如果分出胜负则两队共得$$3$$分．$$6$$支球队共要比$$6\\times 5\\div 2=15$$场比赛， 其中有$$4$$场平局，所以有$$15-4=11$$场分出了胜负，那么$$6$$支球队总得分为$$2\\times 4+3\\times 11=41$$分， 其中$$5$$支球队共得了$$31$$分，所以第$$6$$支球队得了$$41-31=10$$分． "], "answer_value": "A"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2471", "queId": "11b98f33d7754d9b8c195dc637baed0d", "competition_source_list": ["2021年新希望杯六年级竞赛初赛第11题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "【$$2021$$年六年级卷第$$11$$题】小糊涂遇到一个问题：比较$$\\frac{99}{100}$$，$$ \\frac{100}{101}$$，$$\\frac{199}{201}$$的大小．他感到很迷糊，请你帮他找到正确的答案． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{199}{201}$$ "}], [{"aoVal": "B", "content": "$$\\frac{199}{201}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{99}{100}$$ "}], [{"aoVal": "C", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{100}{101}$$ "}], [{"aoVal": "D", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater\\frac{99}{100}$$ "}], [{"aoVal": "E", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{99}{100}\\textgreater{} \\frac{199}{201}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数感认知->分数数字加工"], "answer_analysis": ["$$\\frac{99}{100}=1- \\frac{1}{100}=1- \\frac{2}{200}$$， $$\\frac{100}{101}=1- \\frac{1}{101}=1- \\frac{2}{202}$$， $$\\frac{199}{201}=1- \\frac{2}{201}$$， 因为$$\\frac{2}{202}\\textless{} \\frac{2}{201}\\textless{} \\frac{2}{200}$$， 所以$$1- \\frac{2}{202}\\textgreater1- \\frac{2}{201}\\textgreater1- \\frac{2}{200}$$， 即$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{99}{100}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2728", "queId": "bedca10707d14c16ab90fbf2c898fcdb", "competition_source_list": ["2018年IMAS小学中年级竞赛（第二轮）第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "请问$$100-97+94-91+88-85+\\cdots +4-1$$之值等于多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$45$$ "}], [{"aoVal": "B", "content": "$$48$$ "}], [{"aoVal": "C", "content": "$$51$$ "}], [{"aoVal": "D", "content": "$$54$$ "}], [{"aoVal": "E", "content": "$$57$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律"], "answer_analysis": ["方法一：等差数列的通项公式 第$$n$$个数$$=$$第$$1$$个数+（n-1）\\times 公差 $$100-97+94-91+88-85+\\cdots +4-1$$ $$=\\left( 100-97 \\right)+\\left( 94-91 \\right)+\\left( 88-85 \\right)+\\cdots +\\left( 4-1 \\right)$$ $$=\\underbrace{3+3+3+\\cdots +3}_{17项}$$ $$=3\\times 17$$ $$=51$$． ①先分组再等差数列求有多少组：以$$4$$为首项，公差为6 ②先等差数列求有多少个数再除以$$2$$求有多少组：以$$1$$为首项，公差为3 故选$$\\text{C}$$． 方法二：等差数列求和公式 等差数列的和=（第$$1$$个数$$+$$第$$n$$个数）\\times 项数$$\\div$$ 2 $$100-97+94-91+88-85+\\cdots +4-1$$ $$=\\left( 100+94+88+\\cdots +4 \\right)-\\left( 97+91+85+\\cdots +1 \\right)$$ $$=\\frac{\\left( 100+4 \\right)\\times 17}{2}-\\frac{\\left( 97+1 \\right)\\times 17}{2}$$ $$=\\left( 104-98 \\right)\\times \\frac{17}{2}$$ $$=6\\times \\frac{17}{2}$$ $$=51$$． 故选$$\\text{C}$$ "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "90", "queId": "258acd0ce78d4125ab22c0739b27d907", "competition_source_list": ["1994年第5届希望杯初二竞赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$y=a{{x}^{7}}+b{{x}^{5}}+c{{x}^{3}}+dx+e$$，其中$$a,b,c,d,e$$为常数，当$$x=2$$时，$$y=23$$；当$$x=-2$$时，$$y=-35$$，那么$$e$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$-6$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$-12$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->二元一次方程（组）->含参二元一次方程组->根据解的等量关系解含参的二元一次方程组", "课内体系->能力->运算能力"], "answer_analysis": ["由题设知，当$$x=2$$时， $$23={{2}^{7}}a+{{2}^{5}}b+{{2}^{3}}c+2d+e$$ 当$$x=-2$$时， $$-35={{\\left( -2 \\right)}^{7}}a+{{\\left( -2 \\right)}^{5}}b+{{\\left( -2 \\right)}^{3}}c+\\left( -2 \\right)d+e$$ 由①+②，则得$$2e=-12$$，所以$$e=-6$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "728", "queId": "ac6ec3620e924e609ef65fa41355bcbc", "competition_source_list": ["2017~2018学年5月四川成都天府新区 成都市实验外国语学校（西区）初二下学期月考第10题3分", "2019~2020学年3月浙江杭州滨江区杭二中白马湖学校初一下学期周测C卷第9题", "初二其它", "1996年第7届希望杯初二竞赛第3题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "【SZ】已知$${{x}^{2}}+ax-12$$能分解成两个整系数的一次因式的乘积，则符合条件的整数$$a$$的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$个 "}], [{"aoVal": "B", "content": "$$4$$个 "}], [{"aoVal": "C", "content": "$$6$$个 "}], [{"aoVal": "D", "content": "$$8$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->因式分解的基础->已知因式分解结果求参数"], "answer_analysis": ["设$${{x}^{2}}+ax-12=(x+m)(x+n)$$（$$m$$，$$n$$为整数）， 则$$a=m+n$$，$$mn=-12$$共有$$6$$种结果． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "358", "queId": "7e27544af0044292b5d9bcede847ed9b", "competition_source_list": ["2018~2019学年浙江宁波鄞州区宁波市鄞州蓝青学校初二下学期期末第4题3分", "2019年浙江温州瓯海区浙江省温州中学初三自主招生", "2011年竞赛第1题4分", "2019年浙江温州瓯海区浙江省温州中学初三自主招生第1题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$x=\\frac{\\sqrt{5}-3}{2}$$，则代数式$$x\\left( x+1 \\right)\\left( x+2 \\right)\\left( x+3 \\right)$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式化简求值"], "answer_analysis": ["解：$$\\because x=\\frac{\\sqrt{5}-3}{2}$$， $$\\therefore 2x=\\sqrt{5}-3$$， $$2x+3=\\sqrt{5}$$ $${{\\left( 2x+3 \\right)}^{2}}={{\\left( \\sqrt{5} \\right)}^{2}}$$， $$4{{x}^{2}}+12x+9=5$$， $$\\therefore {{x}^{2}}+3x=-1$$， $$\\therefore $$原式$$=\\left( {{x}^{2}}+3x \\right)\\left( {{x}^{2}}+3x+2 \\right)$$ $$=-1\\times \\left( -1+2 \\right)$$ $$=-1$$； 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1170", "queId": "8aac49074e023206014e201bd95d6526", "competition_source_list": ["1994年第5届全国希望杯初一竞赛初赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$n$$是整数，那么被$$3$$整除并且商恰为$$n$$的那个数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{n}{3}$$ "}], [{"aoVal": "B", "content": "$$n+3$$ "}], [{"aoVal": "C", "content": "$$3n$$ "}], [{"aoVal": "D", "content": "$${{n}^{3}}$$ "}]], "knowledge_point_routes": ["课内体系->能力->分析和解决问题能力", "课内体系->知识点->式->整式的加减->整式有关的概念->代数式->列代数式"], "answer_analysis": ["被$$3$$整除的商恰好为$$n$$的数是$$3n$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "943", "queId": "89c9a2e3de67465cb5c5ccdb62745f3f", "competition_source_list": ["1996年第7届希望杯初二竞赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "暑假里父亲、儿子、女儿准备外出旅行，咨询时了解到，甲旅行社规定：若大人买一张全票，则两个孩子的费用可按全票价的七折优惠；乙旅行社规定：三人旅行可按团体票计价，即按原价的$$80 \\%$$收费，若两家旅行社的原价相同，则当实际收费时．", "answer_option_list": [[{"aoVal": "A", "content": "甲比乙低 "}], [{"aoVal": "B", "content": "乙比甲低 "}], [{"aoVal": "C", "content": "甲、乙相同 "}], [{"aoVal": "D", "content": "是甲低还是乙低，视原价而定 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的经济问题->一元一次方程的经济问题-打折"], "answer_analysis": ["设原价为$$a$$元，则 甲旅行社收费$$=a+2\\cdot70 \\%a=2.4a$$， 乙旅行社收费$$=3\\cdot 80 \\%a=2.4a$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "236", "queId": "4666c24514ff48728513e85c45f7a9b4", "competition_source_list": ["2013年第24届全国希望杯初一竞赛复赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$$\\left\\textbar{} x+1 \\right\\textbar+\\left\\textbar{} 2x-1 \\right\\textbar=1$$的整数解的个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->一元一次方程->解一元一次方程->含绝对值的一元一次方程"], "answer_analysis": ["由$$\\left\\textbar{} x+1 \\right\\textbar$$依次取$$0$$，$$1$$， 可得方程组$$\\left { \\begin{matrix}x+1=0 2x-1=\\pm 1 \\end{matrix} \\right.$$或$$\\left { \\begin{matrix}x+1=1 2x-1=0 \\end{matrix} \\right.$$， 解之，可得两方程均无解． 所以，原方程整数解的个数为$$0$$ "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1141", "queId": "d6cd11f3c2ec40368f8ef9199713456a", "competition_source_list": ["2009年第20届希望杯初二竞赛第2试第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若三角形的三边长$$a$$，$$b$$，$$c$$满足$$a\\textless{}b\\textless{}c$$，且$${{a}^{2}}+bc=t_{1}^{2}$$，$${{b}^{2}}+ca=t_{2}^{2}$$，$${{c}^{2}}+ab=t_{3}^{2}$$，则$$t_{1}^{2}$$、$$t_{2}^{2}$$、$$t_{3}^{2}$$中．", "answer_option_list": [[{"aoVal": "A", "content": "$$t_{1}^{2}$$最大 "}], [{"aoVal": "B", "content": "$$t_{2}^{2}$$最大 "}], [{"aoVal": "C", "content": "$$t_{3}^{2}$$最大 "}], [{"aoVal": "D", "content": "$$t_{3}^{2}$$最小 "}]], "knowledge_point_routes": ["知识标签->学习能力->运算能力", "知识标签->题型->三角形->三角形及多边形->与三角形有关的线段->题型：与三边关系有关的证明", "知识标签->题型->式->整式的乘除->乘法公式->题型：利用平方差公式计算", "知识标签->题型->式->因式分解->提公因式法与公式法->题型：提公因式法", "知识标签->方法->作差法", "知识标签->知识点->式->整式的乘除->乘法公式->平方差公式", "知识标签->知识点->式->因式分解->因式分解：提公因式法", "知识标签->知识点->三角形->三角形及多边形->与三角形有关的线段->三角形的三边关系"], "answer_analysis": ["由$$t_{1}^{2}-t_{2}^{2}=({{a}^{2}}+bc)-({{b}^{2}}+ca)=(a-b)(a+b-c)\\textless{}0$$，得$$t_{1}^{2}\\textless{}t_{2}^{2}$$， 由$$t_{2}^{2}-t_{3}^{2}=({{b}^{2}}+ca)-({{c}^{2}}+ab)=(b-c)(b+c-a)\\textless{}0$$， 得$$t_{2}^{2}\\textless{}t_{3}^{2}$$，所以$$t_{1}^{2}\\textless{}t_{2}^{2}\\textless{}t_{3}^{2}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1504", "queId": "c1cee196695047bfb1bd528eaf1181fd", "competition_source_list": ["2017年湖南长沙天心区湘郡培粹实验中学初二竞赛（觉园杯）第1题4分", "2019年湖南长沙天心区湘郡培粹实验中学初二竞赛初赛(觉园杯)第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$x\\ne-1$$，$$0$$，$$1$$，则$$\\frac{x-1}{\\textbar x-1\\textbar}+\\frac{\\textbar x\\textbar}{x}+\\frac{x+1}{\\textbar x+1\\textbar}$$的值可能是（ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "比$$3$$大的数 "}], [{"aoVal": "B", "content": "比$$-3$$小的数 "}], [{"aoVal": "C", "content": "$$\\pm1$$，$$\\pm3$$ "}], [{"aoVal": "D", "content": "比$$-3$$大，并且比$$3$$小的数 "}]], "knowledge_point_routes": ["课内体系->思想->分类讨论思想", "课内体系->知识点->数->有理数->绝对值->绝对值的非负性", "课内体系->知识点->数->有理数->绝对值->绝对值综合", "课内体系->能力->运算能力"], "answer_analysis": ["当$$x ~\\textless{} ~-1$$时，$$\\frac{x-1}{\\textbar x-1\\textbar}+\\frac{\\textbar x\\textbar}{x}+\\frac{x+1}{\\textbar x+1\\textbar}=-1-1-1=-3$$， 当$$-1 ~\\textless{} ~x ~\\textless{} ~0$$时，$$\\frac{x-1}{\\textbar x-1\\textbar}+\\frac{\\textbar x\\textbar}{x}+\\frac{x+1}{\\textbar x+1\\textbar}=-1-1+1=-1$$， 当$$x\\textgreater1$$时，$$\\frac{x-1}{\\textbar x-1\\textbar}+\\frac{\\textbar x\\textbar}{x}+\\frac{x+1}{\\textbar x+1\\textbar}=1+1+1=3$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1255", "queId": "a0adba9c44d045aabf497eb2fa3a7fbb", "competition_source_list": ["2023年浙江宁波鄞州区初三竞赛联考学校：鄞实、曙光、鄞外、高桥、雅戈尔、集士港第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若关于$$x$$的不等式组$$\\begin{cases} 3-2x \\geq 1 x \\geq m+1 \\end{cases}$$共有$$2$$个整数解，则$$m$$的取值范围是", "answer_option_list": [[{"aoVal": "A", "content": "$\\quad m=-1 $ "}], [{"aoVal": "B", "content": "$-2\\lt m\\leqslant-1$ "}], [{"aoVal": "C", "content": "$-2\\leqslant m\\leqslant-1$ "}], [{"aoVal": "D", "content": "$m\\textless-1$ "}]], "knowledge_point_routes": ["课内体系->思想->方程思想", "课内体系->知识点->方程与不等式->不等式（组）->含参不等式->由不等式（组）的整数解情况求参数范围", "课内体系->方法->代入法", "课内体系->能力->运算能力"], "answer_analysis": ["解：$解不等式3-2x\\geqslant1,得:x\\leqslant1.$ $\\because不等式组共有2个整数解,$ $\\therefore不等式组的整数解为1,0;$ 则$-1\\lt m\\leqslant0,$ 解得$-2\\lt m\\leqslant-1,$ 故选$B.$ "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1533", "queId": "c69c8cf026b14bc1af3e8042e3aa11e2", "competition_source_list": ["1999年第16届全国初中数学联赛竞赛第4题7分"], "difficulty": "3", "qtype": "single_choice", "problem": "若函数$$y=\\frac{1}{2}({{x}^{2}}-100x+196+\\textbar{{x}^{2}}-100x+196\\textbar)$$，则当自变量$$x$$取$$1$$，$$2$$，$$3$$，$$\\ldots$$，$$100$$这$$100$$个自然数时，函数值的和是．", "answer_option_list": [[{"aoVal": "A", "content": "$$540$$ "}], [{"aoVal": "B", "content": "$$390$$ "}], [{"aoVal": "C", "content": "$$194$$ "}], [{"aoVal": "D", "content": "$$195$$ "}], [{"aoVal": "E", "content": "$$197$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数", "竞赛->知识点->函数->二次函数->二次函数应用"], "answer_analysis": ["$${{x}^{2}}-100x+196=(x-2)(x-98)$$， ∴当$$2\\leqslant x\\leqslant 98$$时，$${{x}^{2}}-100x+196\\textless{}0$$，$$\\textbar{{x}^{2}}-100x+196\\textbar=-({{x}^{2}}-100x+196)$$． ∴当自变量$$x$$取$$2$$，$$3$$，$$\\ldots$$，$$98$$时，函数值都为$$0$$． 而当$$x$$取$$1$$，$$99$$，$$100$$时，$$\\textbar{{x}^{2}}-100x+196\\textbar={{x}^{2}}-100x+196$$， 故所求的和为：$$y(1)+y(99)+y(100)=390$$，故选择$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "45", "queId": "4f19f2e6a7c343cd9dd0685f4acefa9a", "competition_source_list": ["1995年第6届希望杯初二竞赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$是非零实数，那么$$\\frac{a}{\\textbar a\\textbar}+\\frac{{{a}^{2}}}{\\textbar{{a}^{2}}\\textbar}+\\frac{{{a}^{3}}}{\\textbar{{a}^{3}}\\textbar}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$或$$-1$$ "}], [{"aoVal": "B", "content": "$$-3$$或$$1$$ "}], [{"aoVal": "C", "content": "$$3$$或$$1$$ "}], [{"aoVal": "D", "content": "$$-3$$或$$-1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->|a|/a的化简", "竞赛->知识点->数与式->绝对值->给定范围绝对值化简"], "answer_analysis": ["当$$a\\textgreater0$$时，$$\\textbar a\\textbar=a$$， ∴原式$$=1+1+1=3$$； 当$$a\\textless{}0$$时，$$\\textbar a\\textbar=-a$$，$$\\textbar{{a}^{3}}\\textbar=-{{a}^{3}}$$， ∴原式$$=-1+1-1=-1$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1294", "queId": "8aac50a74ed448af014efdff86131c36", "competition_source_list": ["初一上学期单元测试《整式的加减》整式及相关概念第29题", "2001年第12届希望杯初一竞赛第2试第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知有如下一组$$x$$，$$y$$和$$z$$的单项式： $$7{{x}^{3}}{{z}^{2}}$$，$$8{{x}^{3}}y$$，$$\\frac{1}{2}{{x}^{2}}yz$$，$$-3x{{y}^{2}}z$$，$$9{{x}^{4}}yz$$，$$z{{y}^{2}}$$，$$-\\frac{1}{5}xyz$$，$$9{{y}^{3}}z$$，$$x{{z}^{2}}y$$，$$0.3{{z}^{3}}$$ 我们用下面的方法确定它们的先后次序：对任意两个单项式，先看$$x$$的幂次，规定$$x$$的幂次高的单项式排在$$x$$的幂次低的单项式的前面；再看$$y$$的幂次．规定$$y$$的幂次高的排在$$y$$的幂次低的前面；再看$$z$$的幂次，规定$$z$$的幂次高的排在$$z$$的幂次低的前面．将这组单项式按上述法则排序，那么$$9{{y}^{3}}z$$应排在（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "第$$2$$位 "}], [{"aoVal": "B", "content": "第$$4$$位 "}], [{"aoVal": "C", "content": "第$$6$$位 "}], [{"aoVal": "D", "content": "第$$8$$位 "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "课内体系->知识点->式->整式的加减->整式有关的概念->单项式->单项式的定义"], "answer_analysis": ["按照题意把这几个单项式排列如下： $$9{{x}^{4}}zy$$，$$8{{x}^{3}}y$$，$$7{{x}^{3}}{{z}^{2}}$$，$$\\frac{1}{2}{{x}^{2}}yz$$，$$-3x{{y}^{2}}z$$，$$-\\frac{1}{5}xyz$$，$$x{{z}^{2}}y$$，$$9{{y}^{3}}z$$，$$z{{y}^{2}}$$，$$3{{z}^{3}}$$，$$0$$． ∴$$9{{y}^{3}}z$$应排在第$$8$$位． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1091", "queId": "ff8080814d9539f1014d9b5a53790a02", "competition_source_list": ["1992年第3届全国希望杯初一竞赛初赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "两个$$10$$次多项式的和是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$次多项式 "}], [{"aoVal": "B", "content": "$$10$$次多项式 "}], [{"aoVal": "C", "content": "$$100$$次多项式 "}], [{"aoVal": "D", "content": "不高于$$10$$次的多项式 "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式有关的概念->多项式->多项式的定义"], "answer_analysis": ["多项式$${{x}^{10}}+x$$与$$-{{x}^{10}}+{{x}^{2}}$$之和为$${{x}^{2}}+x$$是个次数低于$$10$$次的多项式，选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1447", "queId": "b8488fb9a0eb4eabb3ddcb487f79cb1d", "competition_source_list": ["2003年第14届希望杯初二竞赛第1试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$为三角形的三边长，化简$$\\textbar a+b+c\\textbar-\\textbar a-b-c\\textbar-\\textbar a-b+c\\textbar-\\textbar a+b-c\\textbar$$，结果是（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$2a+2b+2c$$ "}], [{"aoVal": "C", "content": "$$4a$$ "}], [{"aoVal": "D", "content": "$$2b-2c$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->利用题设条件推理化简绝对值", "课内体系->知识点->三角形->三角形及多边形->与三角形有关的线段->三角形的三边关系"], "answer_analysis": ["因为三角形的任意两边之和大于第三边， 所以$$a+b+c\\textgreater0$$，$$a-b-c\\textless{}0$$，$$a-b+c\\textgreater0$$，$$a+b-c\\textgreater0$$， 所以原式$$=a+b+c+(a-b-c)-(a-b+c)-(a+b-c)=0$$， 选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "951", "queId": "66178d35010d4d5485e5e13dabaa1087", "competition_source_list": ["2000年第17届全国初中数学联赛竞赛第4题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "正整数$$n$$小于$$100$$，并满足等式$$\\left\\lfloor \\frac{n}{2} \\right\\rfloor +\\left\\lfloor \\frac{n}{3} \\right\\rfloor +\\left\\lfloor \\frac{n}{6} \\right\\rfloor =n$$，其中$$\\left\\lfloor x \\right\\rfloor $$表示不超过$$x$$的最大整数，这样的正整数$$n$$有( )个.", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$个 "}], [{"aoVal": "B", "content": "$$3$$个 "}], [{"aoVal": "C", "content": "$$12$$个 "}], [{"aoVal": "D", "content": "$$16$$个 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->综合与实践->规律探究与程序框图->定义新运算"], "answer_analysis": ["分析：利用$$\\left\\lfloor x \\right\\rfloor $$表示不超过$$x$$的最大整数的性质，求出$$\\left\\lfloor \\frac{n}{2} \\right\\rfloor +\\left\\lfloor \\frac{n}{3} \\right\\rfloor +\\left\\lfloor \\frac{n}{6} \\right\\rfloor $$的范围． $$\\because \\left\\lfloor \\frac{n}{2} \\right\\rfloor \\leqslant \\frac{n}{2}$$，$$\\left\\lfloor \\frac{n}{3} \\right\\rfloor \\leqslant \\frac{n}{3}$$，$$\\left\\lfloor \\frac{n}{6} \\right\\rfloor \\leqslant \\frac{n}{6}$$$$\\therefore \\left\\lfloor \\frac{n}{2} \\right\\rfloor +\\left\\lfloor \\frac{n}{3} \\right\\rfloor +\\left\\lfloor \\frac{n}{6} \\right\\rfloor \\leqslant \\frac{n}{2}+\\frac{n}{3}+\\frac{n}{6}=n$$． 又$$\\because \\left\\lfloor \\frac{n}{2} \\right\\rfloor +\\left\\lfloor \\frac{n}{3} \\right\\rfloor +\\left\\lfloor \\frac{n}{6} \\right\\rfloor =n$$ $$\\therefore \\left\\lfloor \\frac{n}{2} \\right\\rfloor =\\frac{n}{2}$$，$$\\left\\lfloor \\frac{n}{3} \\right\\rfloor =\\frac{n}{3}$$，$$\\left\\lfloor \\frac{n}{6} \\right\\rfloor =\\frac{n}{6}$$ $$\\therefore n$$的个数有$$\\left\\lfloor \\frac{100}{6} \\right\\rfloor =16$$个，选D． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1574", "queId": "fe06f2ad8a844b91af23c94a1248a6ad", "competition_source_list": ["1999年第10届希望杯初二竞赛第1试第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$${{x}^{3}}+3{{x}^{2}}-3x+k$$有一个因式是$$x+1$$，则$$k=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$-3$$ "}], [{"aoVal": "C", "content": "$$-5$$ "}], [{"aoVal": "D", "content": "$$-7$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->方法->待定系数法", "课内体系->知识点->式->因式分解->其他方法->待定系数法"], "answer_analysis": ["依题意，原多项式当$$x=-1$$时，其值等于$$0$$，即$${{\\left( -1 \\right)}^{3}}+3{{\\left( -1 \\right)}^{2}}-3\\left( -1 \\right)+k=0$$，从而$$k=-5$$． 依题意$$x+1$$也是多项式， $${{\\left( x+1 \\right)}^{3}}-\\left( {{x}^{3}}+3{{x}^{2}}-3x+k \\right)=6x+\\left( 1-k \\right)$$的因式，故$$1-k=6$$，即$$k=-5$$． 依题意可设$${{x}^{3}}+3{{x}^{2}}-3x+k=\\left( x+1 \\right)\\left( {{x}^{2}}+ax+b \\right)={{x}^{3}}+\\left( a+1 \\right){{x}^{2}}+\\left( a+b \\right)x+b$$． 比较同次幂系数得$$\\begin{cases}a+1=3 a+b=-3 k=b \\end{cases}$$， ∴$$\\begin{cases}a=2 b=-5 k=-5 \\end{cases}$$． 故$$k=-5$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1370", "queId": "8f42f9e84ff44b629058d08ff53c8551", "competition_source_list": ["初一上学期单元测试《解读绝对值》第6题", "2013年第24届全国希望杯初一竞赛初赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有理数$$a$$，$$b$$，$$c$$，$$d$$满足$$a\\textless{}b\\textless{}0\\textless{}c\\textless{}d$$，并且$$\\left\\textbar{} b \\right\\textbar\\textless{}c\\textless{}\\left\\textbar{} a \\right\\textbar\\textless{}d$$，则$$a+b+c+d$$的值．", "answer_option_list": [[{"aoVal": "A", "content": "大于$$0$$ "}], [{"aoVal": "B", "content": "等于$$0$$ "}], [{"aoVal": "C", "content": "小于$$0$$ "}], [{"aoVal": "D", "content": "与$$0$$的大小关系不确定 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质", "课内体系->能力->运算能力"], "answer_analysis": ["方法一： ∵$$b\\textless{}0\\textless{}c$$，$$\\left\\textbar{} b \\right\\textbar\\textless{}c$$， ∴$$-b\\textless{}c$$， ∴$$b+c\\textgreater0$$． ∵$$a\\textless{}0\\textless{}d$$，$$\\left\\textbar{} a \\right\\textbar\\textless{}d$$， ∴$$-a\\textless{}d$$， ∴$$a+d\\textgreater0$$． ∴$$a+b+c+d\\textgreater0$$． 方法二： 取特殊值，依题设，可假定$$b=-2$$，$$c=3$$，$$a=-4$$，$$d=5$$， 则$$a+b+c+d=2\\textgreater0$$， 所以$$a+b+c+d$$的值大于$$0$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1094", "queId": "ff8080814d9539f1014d9b60d5570a3c", "competition_source_list": ["1992年第3届全国希望杯初一竞赛初赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a\\textless{}0$$，$$b\\textgreater0$$，且$$\\left\\textbar{} a \\right\\textbar\\textless{}\\left\\textbar{} b \\right\\textbar$$，那么下列式子中结果是正数的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$(a-b)(ab+a)$$ "}], [{"aoVal": "B", "content": "$$(a+b)(a-b)$$ "}], [{"aoVal": "C", "content": "$$(a+b)(ab+a)$$ "}], [{"aoVal": "D", "content": "$$(ab-b)(a+b)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数乘法->有理数的乘法法则", "课内体系->知识点->数->有理数->绝对值->利用题设条件推理化简绝对值"], "answer_analysis": ["因为$$a\\textless{}0$$，$$b\\textgreater0$$．所以$$\\left\\textbar{} a \\right\\textbar=-a$$，$$\\left\\textbar{} b \\right\\textbar=b$$． 由于$$\\left\\textbar{} a \\right\\textbar\\textless{}\\left\\textbar{} b \\right\\textbar$$得$$-a\\textless{}b$$，因此$$a+b\\textgreater0$$，$$a-b\\textless{}0$$． $$ab+a\\textless{}0$$，$$ab-b\\textless{}0$$． 所以应有$$(a-b)(ab+a)\\textgreater0$$成立，选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1002", "queId": "8a1b784b6f28450e98f6848b1afa8e96", "competition_source_list": ["2013年第18届华杯赛初一竞赛初赛第6题", "其它"], "difficulty": "3", "qtype": "single_choice", "problem": "满足不等式$$\\frac{2}{3}\\textless{}\\frac{5}{n}\\textless{}\\frac{3}{m}$$的有序整数对$$(m,n)$$的个数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->一元一次不等式组的整数解"], "answer_analysis": ["由$$\\frac{2}{3}\\textless{}\\frac{5}{n}\\textless{}\\frac{3}{m}$$，得到$$\\frac{m}{3}\\textless{}\\frac{n}{5}\\textless{}\\frac{3}{2}$$，即$$\\frac{5}{3}m\\textless{}n\\textless{}\\frac{15}{2}$$，因此$$n\\leqslant 7$$． $$n=7\\Rightarrow m\\textless{}\\frac{3}{5}n=\\frac{21}{5}$$，$$m=1,2,3,4$$， $$n=6\\Rightarrow m\\textless{}\\frac{3}{5}n=\\frac{18}{5}$$，$$m=1,2,3$$， $$n=5\\Rightarrow m\\textless{}\\frac{3}{5}n=3$$，$$m=1,2$$， $$n=4\\Rightarrow m\\textless{}\\frac{3}{5}n=\\frac{12}{5}$$，$$m=1,2$$， $$n=3\\Rightarrow m\\textless{}\\frac{3}{5}n=\\frac{9}{5}$$，$$m=1$$， $$n=2\\Rightarrow m\\textless{}\\frac{3}{5}n=\\frac{6}{5}$$，$$m=1$$， ∴满足不等式的$$(m,n)$$一共有$$4+3+2+2+1+1=13$$对． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "447", "queId": "79cd910b28164674bf72e119496f35fa", "competition_source_list": ["初一上学期单元测试《解读绝对值》第6题", "2013年第24届全国希望杯初一竞赛初赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有理数$$a$$，$$b$$，$$c$$，$$d$$满足$$a\\textless{}b\\textless{}0\\textless{}c\\textless{}d$$，并且$$\\left\\textbar{} b \\right\\textbar\\textless{}c\\textless{}\\left\\textbar{} a \\right\\textbar\\textless{}d$$，则$$a+b+c+d$$的值．", "answer_option_list": [[{"aoVal": "A", "content": "大于$$0$$ "}], [{"aoVal": "B", "content": "等于$$0$$ "}], [{"aoVal": "C", "content": "小于$$0$$ "}], [{"aoVal": "D", "content": "与$$0$$的大小关系不确定 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["方法一： ∵$$b\\textless{}0\\textless{}c$$，$$\\left\\textbar{} b \\right\\textbar\\textless{}c$$， ∴$$-b\\textless{}c$$， ∴$$b+c\\textgreater0$$． ∵$$a\\textless{}0\\textless{}d$$，$$\\left\\textbar{} a \\right\\textbar\\textless{}d$$， ∴$$-a\\textless{}d$$， ∴$$a+d\\textgreater0$$． ∴$$a+b+c+d\\textgreater0$$． 方法二： 取特殊值，依题设，可假定$$b=-2$$，$$c=3$$，$$a=-4$$，$$d=5$$， 则$$a+b+c+d=2\\textgreater0$$， 所以$$a+b+c+d$$的值大于$$0$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "140", "queId": "25ceeec57a544c428b9670a3df11cc4f", "competition_source_list": ["2021年福建竞赛（\"大梦杯”青少年水平测试）第3~3题"], "difficulty": "2", "qtype": "single_choice", "problem": "将形如3\\emph{m}和$${{2}^{n}}$$（\\emph{m，n}为正整数）的正整数从小到大排列，并依次记为$${{a}_{1}},{{a}_{2}},{{a}_{3}}\\cdots $$若第\\emph{k}个数$${{a}_{k}}=2022$$，则\\emph{k}的值为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "682 "}], [{"aoVal": "B", "content": "683 "}], [{"aoVal": "C", "content": "684 "}], [{"aoVal": "D", "content": "685 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 先确定$$3m$$和$${{2}^{n}}$$不等，考虑在从小到大排列的形如$$3m$$($$m$$为正整数)的正整数3，6，9，27，\\ldots 中，从小到大添加形如$${{2}^{n}}$$($$n$$为正整数)的数．再根据$$2022=3\\times 674$$即可．\\\\ 【详解】\\\\ 易知形如$$3m$$和$${{2}^{n}}$$($$m$$，$$n$$为正整数)的正整数不可能相等．\\\\ 考虑在从小到大排列的形如$$3m$$($$m$$为正整数)的正整数3，6，9，27，\\ldots 中，从小到大添加形如$${{2}^{n}}$$($$n$$为正整数)的数．\\\\ 由$$2022=3\\times 674$$知，将形如$$3m$$($$m$$为正整数)的正整数从小到大排列，2022是第674个数．\\\\ 由于$${{2}^{10}}=1024\\textless{} 2022$$，$${{2}^{11}}=2048\\textgreater2022$$，所以有10个形如$${{2}^{n}}$$($$n$$为正整数)的数小于2022，这10个数排在2022前面．\\\\ 所以$$k=674+10=684$$．\\\\ 【点睛】\\\\ 本题考查数字排列规律问题，掌握因数分解方法，有理数大小比较是解题关键． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "267", "queId": "66c6af2c892b45489ce1255366f67b4a", "competition_source_list": ["1998年第9届希望杯初一竞赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "数$${{(-1)}^{1998}}$$是．", "answer_option_list": [[{"aoVal": "A", "content": "最大的负数 "}], [{"aoVal": "B", "content": "最小的非负数 "}], [{"aoVal": "C", "content": "最小的正整数 "}], [{"aoVal": "D", "content": "绝对值最小的整数 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类", "课内体系->能力->运算能力"], "answer_analysis": ["$${{(-1)}^{1998}}=+1$$．排除$$\\text{A}$$．由于最小的非负数是$$0$$，排除$$\\text{B}$$． 绝对值最小的整数也是$$0$$，排除$$\\text{D}$$．显然应选$$\\text{C}$$．事实上$$+1$$是最小的正整数． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "309", "queId": "1a26b825e0a84b25a3df7c409a6bdd79", "competition_source_list": ["初一单元测试《分式的概念、性质及运算》第18题", "2013年第24届全国希望杯初二竞赛复赛第18题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$${{x}^{2}}-x-1=0$$，则$$\\frac{{{x}^{3}}+x+1}{{{x}^{4}}}$$的值为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的运算->分式加减、乘除、乘方混合运算", "课内体系->知识点->式->分式->分式化简求值->分式化简求值-整体代入求值"], "answer_analysis": ["由条件得$${{x}^{2}}=x+1$$，原式$$=\\frac{{{x}^{3}}+{{x}^{2}}}{{{x}^{4}}}=\\frac{{{x}^{2}}\\left( x+1 \\right)}{{{x}^{4}}}=\\frac{x+1}{{{x}^{2}}}=1$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "960", "queId": "92200b71e01648248b9ee0528c859385", "competition_source_list": ["1999年第10届希望杯初一竞赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\left\\textbar{} a \\right\\textbar=1$$，则$${{a}^{4}}$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->方法->整体法", "课内体系->知识点->数->有理数->绝对值->绝对值的非负性", "课内体系->知识点->数->有理数->绝对值->绝对值的代数意义"], "answer_analysis": ["因为$$\\left\\textbar{} a \\right\\textbar=1$$，所以$$a=\\pm 1$$，因此$${{a}^{4}}=1$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "382", "queId": "1b2b75c8e4f643119f82d09ca5536926", "competition_source_list": ["2018年全国竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "解方程：$$\\frac{x+6}{x+1}-\\frac{3{{x}^{2}}+10x+4}{{{x}^{2}}+3x+2}+\\frac{2x+1}{x+2}=0$$的解为．", "answer_option_list": [[{"aoVal": "A", "content": "$$x=9$$ "}], [{"aoVal": "B", "content": "$$x=-9$$ "}], [{"aoVal": "C", "content": "无解 "}], [{"aoVal": "D", "content": "$$x=-\\frac{1}{9}$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->运算能力", "知识标签->知识点->方程与不等式->分式方程->解分式方程", "知识标签->题型->方程与不等式->分式方程->分式方程的解与解分式方程->题型：解可化为一元一次方程的分式方程"], "answer_analysis": ["原方程可变为$$1+\\frac{5}{x+1}-\\left( 3+\\frac{x-2}{{{x}^{2}}+3x+2} \\right)+2-\\frac{3}{x+2}=0$$， 整理得$$\\frac{5}{x+1}-\\frac{3}{x+2}-\\frac{x-2}{{{x}^{2}}+3x+2}=0$$， 去分母、整理得$$x+9=0$$， 计算得出$$x=-9$$． 经检验知，$$x=-9$$是原方程的根． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "235", "queId": "15148aa5e85c4ef89858dcce3d23a7f3", "competition_source_list": ["2016年第33届全国全国初中数学联赛竞赛第6题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "设实数$$x$$，$$y$$，$$z$$，满足$$x+y+z=1$$，则$$M=xy+2yz+3xz$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["知识标签->题型->式->整式加减->整式加减化简求值->题型：整体代入化简求值", "知识标签->知识点->式->整式的加减->整式的定义", "知识标签->知识点->式->整式的乘除->乘法公式->完全平方公式", "知识标签->方法->配方法"], "answer_analysis": ["$$M=xy+(2y+3x)z$$ $$=xy+(2y+3x)(1-x-y)$$ $$=-3{{x}^{2}}-4xy-2{{y}^{2}}+3x+2y$$ $$=-2\\left[ {{y}^{2}}+2\\left( x-\\frac{1}{2} \\right)y+{{\\left( x-\\frac{1}{2} \\right)}^{2}} \\right]-3{{x}^{2}}+3x+2{{\\left( x-\\frac{1}{2} \\right)}^{2}}$$ $$=-2{{\\left( y+x-\\frac{1}{2} \\right)}^{2}}-{{x}^{2}}+x+\\frac{1}{2}$$ $$=-2{{\\left( y+x-\\frac{1}{2} \\right)}^{2}}-{{\\left( x-\\frac{1}{2} \\right)}^{2}}+\\frac{3}{4}$$ $$\\leqslant \\frac{3}{4}$$． 当且仅当$$x=\\frac{1}{2}$$，$$y=0$$时，不等式取等号， 故$${{M}_{\\max }}=\\frac{3}{4}$$． 令$$y=1-x-z$$，代入$$M$$，则： $$M=x\\left( 1-x-z \\right)+2\\left( 1-x-z \\right)z+3xz=x-{{x}^{2}}-xz+2z-2xz-2{{z}^{2}}+3xz$$ $$=-{{x}^{2}}+x-2{{z}^{2}}+2z=-{{\\left( x-\\frac{1}{2} \\right)}^{2}}-2{{\\left( z-\\frac{1}{2} \\right)}^{2}}+\\frac{3}{4}$$，$${{M}_{\\min }}=\\frac{3}{4}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "397", "queId": "ec78c5693d6249b7a0d89578b1b13baa", "competition_source_list": ["1995年第12届全国初中数学联赛竞赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "方程组$$\\begin{cases}xy+yz=63 xz+yz=23 \\end{cases}$$的正整数解的组数是（~ ）.", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->其他方程->多元二次方程（组）", "课内体系->能力->运算能力"], "answer_analysis": ["先由第二个方程确定$$z=1$$，进而原方程组可化为$$\\begin{cases}xy+y=63 x+y=23 \\end{cases}$$， 消去$$y$$可得：$$\\left( 23-x \\right)\\left( x+1 \\right)=63$$，即$${{x}^{2}}-22x+40=0$$． 解得：$${{x}_{1}}=20$$，$${{x}_{2}}=2$$． 所以，原方程组有两组解：（$$2$$，$$21$$ ，$$1$$），（$$20$$，$$3$$ ，$$1$$）． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1291", "queId": "a0c3451673124664ba01ad46618003f6", "competition_source_list": ["初一下学期其它第2题", "2003年竞赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$4x-3y-6z=0$$，$$x+2y-7z=0(xyz\\ne 0)$$，则代数式$$\\frac{5{{x}^{2}}+2{{y}^{2}}-{{z}^{2}}}{2{{x}^{2}}-3{{y}^{2}}-10{{z}^{2}}}$$的值是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{19}{2}$$ "}], [{"aoVal": "C", "content": "$$-15$$ "}], [{"aoVal": "D", "content": "$$-13$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式化简求值->分式化简求值-条件化简求值", "课内体系->能力->运算能力"], "answer_analysis": ["视$$z$$为常数，将两个方程转化为关于$$x$$，$$y$$的二元一次方程组$$\\begin{cases}4x-3y=6z x+2y=7z \\end{cases}$$②$$\\times 4-$$①得的$$11y=22z$$， $$y=2z$$，把$$y=2z$$代入①得$$4x-6z=6z$$，解得$$x=3z$$， 所以原式$$=\\frac{5{{x}^{2}}+2{{y}^{2}}-{{z}^{2}}}{2{{x}^{2}}-3{{y}^{2}}-10{{z}^{2}}}=\\frac{5{{(3z)}^{2}}+2{{(2z)}^{2}}-{{z}^{2}}}{2{{(3z)}^{2}}-3{{(2z)}^{2}}-10{{z}^{2}}}=-\\frac{52}{4}=-13$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1250", "queId": "9798a701e9f147faa7bdfae663281355", "competition_source_list": ["2012年第23届全国希望杯初二竞赛初赛第18题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a+{{x}^{2}}=2011$$，$$b+{{x}^{2}}=2012$$，$$c+{{x}^{2}}=2013$$，且$$abc=24$$，则$$\\frac{a}{bc}+\\frac{c}{ab}+\\frac{b}{ac}-\\frac{1}{a}-\\frac{1}{b}-\\frac{1}{c}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "frac{1}{2}$$ "}], [{"aoVal": "B", "content": "frac{1}{4}$$ "}], [{"aoVal": "C", "content": "frac{1}{8}$$ "}], [{"aoVal": "D", "content": "frac{1}{16}$$ "}]], "knowledge_point_routes": ["课内体系->方法->配方法", "课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式"], "answer_analysis": ["由$$a+{{x}^{2}}=2011$$，$$b+{{x}^{2}}=2012$$，$$c+{{x}^{2}}=2013$$， 得$$b-a=1$$，$$c-b=1$$，$$c-a=2$$． $$\\frac{a}{bc}+\\frac{c}{ab}+\\frac{b}{ac}-\\frac{1}{a}-\\frac{1}{b}-\\frac{1}{c}$$ $$=\\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac}{abc}$$ $$=\\frac{{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(a-c)}^{2}}}{2abc}$$ $$=\\frac{1+1+4}{2\\times 24}$$ $$=\\frac{1}{8}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "318", "queId": "1651703ed1424d9b8770e5484980f1e1", "competition_source_list": ["2019年四川南充初三自主招生", "竞赛2021年宁波小强基第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a$$，$$b$$是方程$${{x}^{2}}+20x+1=0$$的两个根，$$c$$，$$d$$是方程$${{x}^{2}}-17x+1=0$$的两个根，则代数式$$\\left( a+c \\right)\\left( b+c \\right)\\left( a-d \\right)\\left( b-d \\right)$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2017$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$340$$ "}], [{"aoVal": "D", "content": "$$-111$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元二次方程->一元二次方程的基础->一元二次方程的根"], "answer_analysis": ["解：由题意可得$$a+b=-20$$，$$ab=1$$，$$c+d=17$$，$$cd=1$$ $$\\therefore \\left( a+c \\right)\\left( b+c \\right)\\left( a-d \\right)\\left( b-d \\right)$$ $$=\\left[ ab+\\left( a+b \\right)c+{{c}^{2}} \\right]\\left[ ab-\\left( a+b \\right)d+{{d}^{2}} \\right]$$ $$=\\left( 1-20c+{{c}^{2}} \\right)\\left( 1+20d+{{d}^{2}} \\right)$$ $$=1+20d+{{d}^{2}}-20c-400-20d+{{c}^{2}}+20c+1$$ $$={{d}^{2}}+{{c}^{2}}+2-400$$ $$={{\\left( c+d \\right)}^{2}}-400$$ $$={{17}^{2}}-400$$ $$=-111$$， 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "987", "queId": "ff8080814cfa9b24014cff6b19470fb8", "competition_source_list": ["初二上学期单元测试《分解方法的延拓（1）》第21题", "2002年第13届希望杯初二竞赛第2试第2题", "初二其它"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a\\textgreater b\\textgreater c$$，$$M={{a}^{2}}b+{{b}^{2}}c+{{c}^{2}}a$$，$$N=a{{b}^{2}}+b{{c}^{2}}+c{{a}^{2}}$$，则$$M$$与$$N$$的大小关系是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$M\\textless{}N$$ "}], [{"aoVal": "B", "content": "$$M\\textgreater N$$ "}], [{"aoVal": "C", "content": "$$M=N$$ "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->因式分解的应用", "课内体系->能力->运算能力"], "answer_analysis": ["$$M-N={{a}^{2}}b+{{b}^{2}}c+{{c}^{2}}a-(a{{b}^{2}}+b{{c}^{2}}+c{{a}^{2}})$$ $$=(b-c){{a}^{2}}-({{b}^{2}}-{{c}^{2}})a+({{b}^{2}}c-b{{c}^{2}})$$ $$=(b-c){{a}^{2}}-(b+c)(b-c)a+bc(b-c)$$ $$=(b-c)\\left[ {{a}^{2}}-(b+c)a+bc \\right]$$ $$=(b-c)(a-b)(a-c)$$， ∵$$a\\textgreater b\\textgreater c$$， ∴$$b-c\\textgreater0$$，$$a-b\\textgreater0$$，$$a-c\\textgreater0$$， ∴$$M-N=(b-c)(a-b)(a-c)\\textgreater0$$， ∴$$M\\textgreater N$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1501", "queId": "c1c5fc21baae486b9b0df7fdd41986b9", "competition_source_list": ["2018年全美数学竞赛（AMC）竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "计算 $$1+3+5+\\cdots +2021+2023-2-4-6-\\cdots -2020-2022$$？", "answer_option_list": [[{"aoVal": "A", "content": "$$-1012$$ "}], [{"aoVal": "B", "content": "$$-1011$$ "}], [{"aoVal": "C", "content": "$$1010$$ "}], [{"aoVal": "D", "content": "$$1011$$ "}], [{"aoVal": "E", "content": "$$1012$$ "}]], "knowledge_point_routes": ["美国amc8->知识点->数与运算->数列与数表->等差数列"], "answer_analysis": ["计算 $$1+3+5+\\cdots +2021+2023-2-4-6-\\cdots -2020-2022$$？ 我们可以将给定的表达式重写为$$1+\\left( 3-2 \\right)+\\left( 5-4 \\right)+\\cdot \\cdot \\cdot +\\left( 2023-2022 \\right)+\\left( 2021-2020 \\right)=1+1+1+\\cdot \\cdot \\cdot +1$$，$$1\\text{s}$$的数与$$1$$，$$3$$，$$5$$，7$$\\cdot \\cdot \\cdot $$，$$2021$$，$$2023$$中的数是相同，因此答案是$$1012$$． 故选$$\\text{E}$$． "], "answer_value": "E"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1275", "queId": "8ec5611bfd964d6a8a14bfe0d4ae67ed", "competition_source_list": ["2011年竞赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{（难）}设$$a=\\sqrt{7}-1$$，则代数式$$3{{a}^{3}}+12{{a}^{2}}-6a-12$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$4\\sqrt{7}+10$$ "}], [{"aoVal": "D", "content": "$$4\\sqrt{7}+12$$ "}]], "knowledge_point_routes": ["课内体系->方法->代入法", "课内体系->知识点->式->整式的乘除->整式的乘除运算->整式乘除的综合", "课内体系->知识点->式->整式的乘除->整式乘除化简求值", "课内体系->能力->运算能力"], "answer_analysis": ["由题意得$$a+1=\\sqrt{7}$$，即$${{a}^{2}}=6-2a$$， 所以$$3{{a}^{3}}+12{{a}^{2}}-6a-12=3a\\left( 6-2a \\right)+12\\left( 6-2a \\right)-6a-12$$ $$=-6{{a}^{2}}-12a+60$$ $$=-6\\left( 6-2a \\right)-12a+60$$ $$=24$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "693", "queId": "ac57432147ab40fe808124b1453b3ae7", "competition_source_list": ["2007年第12届华杯赛初一竞赛初赛第3题"], "difficulty": "0", "qtype": "single_choice", "problem": "如果一个多项式的各项的次数都相同，则称该多项式为齐次多项式．例如：$${{x}^{3}}+2x{{y}^{2}}+2xyz+{{y}^{3}}$$是$$3$$次齐次多项式．若$${{x}^{m+2}}{{y}^{2}}+3x{{y}^{3}}{{z}^{2}}$$是齐次多项式，则$$m$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->整式的乘除运算"], "answer_analysis": ["因为$${{x}^{m+2}}{{y}^{2}}+3x{{y}^{3}}{{z}^{2}}$$是齐次多项式，并且第二个单项式的次数为$$6$$， 所以$$m+2+2=6$$，即$$m=2$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1444", "queId": "fc712cf520584ea3b6401b1c79cb2584", "competition_source_list": ["2006年第17届希望杯初二竞赛第2试第5题", "初二其它"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$m={{2006}^{2}}+{{2006}^{2}}\\times {{2007}^{2}}+{{2007}^{2}}$$，则$$m$$．", "answer_option_list": [[{"aoVal": "A", "content": "是完全平方数，还是奇数 "}], [{"aoVal": "B", "content": "是完全平方数，还是偶数 "}], [{"aoVal": "C", "content": "不是完全平方数，但是奇数 "}], [{"aoVal": "D", "content": "不是完全平方数，但是偶数 "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->十字相乘法->二次项系数为±1的十字相乘", "课内体系->能力->运算能力"], "answer_analysis": ["$$m={{2006}^{2}}+{{2006}^{2}}\\times {{2007}^{2}}+{{2007}^{2}}$$ $$={{2006}^{2}}+{{2006}^{2}}{{\\left( 2006+1 \\right)}^{2}}+{{\\left( 2006+1 \\right)}^{2}}$$ $$={{2006}^{4}}+2\\times {{2006}^{3}}+3\\times {{2006}^{2}}+2\\times 2006+1$$ $$={{\\left( {{2006}^{2}}+2006+1 \\right)}^{2}}$$ ∴$$m$$是奇数． 所以选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "460", "queId": "2436caeb223041db8fc8aa427fdb0c7f", "competition_source_list": ["2007年第12届华杯赛初一竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$1-{{(-1)}^{2}}+\\frac{-1\\times {{(-1)}^{3}}-2}{2\\times (-1)+1}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["原式$$=1-1+\\frac{-1\\times (-1)-2}{-2+1}$$ $$=\\frac{1-2}{-1}$$ $$=1$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "962", "queId": "dfb31ec6198b4030978d74a2d89f9bcf", "competition_source_list": ["2004年第15届希望杯初二竞赛第1试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "某出版社计划出版一套百科全书，固定成本为$$8$$万元，每印刷一套需增加成本$$20$$元．如果每套书定价$$100$$元，卖出后有$$3$$成收入给承销商，出版社要盈利$$10 \\%$$，那么该书至少应发行．（精确到千位）", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$千套 "}], [{"aoVal": "B", "content": "$$3$$千套 "}], [{"aoVal": "C", "content": "$$4$$千套 "}], [{"aoVal": "D", "content": "$$5$$千套 "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->一次方程->一元一次方程"], "answer_analysis": ["设出版社发行$$x$$套书， 则$$100\\left( 1-0.3 \\right)x\\geqslant \\left( 80000+20x \\right)\\left( 1+10 \\% \\right)$$， $$70x\\geqslant 88000+22x$$， $$x\\geqslant 1833\\frac{1}{3}$$， 至少应发行$$2$$千套． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "157", "queId": "1875d515f00c4e1e9cb4ccc865a1612b", "competition_source_list": ["2012年竞赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$a=-2+\\sqrt{2}$$，那么$$1+\\frac{1}{2+\\dfrac{1}{3+a}}$$的值为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$2\\sqrt{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算"], "answer_analysis": ["$$1+\\frac{1}{2+\\dfrac{1}{3+a}}=1+\\frac{1}{2+\\dfrac{1}{1+\\sqrt{2}}}$$ $$=1+\\frac{1}{2+\\sqrt{2}-1}=1+\\frac{1}{\\sqrt{2}+1}=1+\\sqrt{2}-1=\\sqrt{2}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "114", "queId": "0b5a5995283f4782949dbaca86e6a0ca", "competition_source_list": ["2002年第13届希望杯初二竞赛第1试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列各式分解因式后，可表示为一次因式乘积的是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{x}^{3}}-9{{x}^{2}}+27x-27$$ "}], [{"aoVal": "B", "content": "$${{x}^{3}}-{{x}^{2}}+27x-27$$ "}], [{"aoVal": "C", "content": "$${{x}^{4}}-{{x}^{3}}+27x-27$$ "}], [{"aoVal": "D", "content": "$${{x}^{3}}-3{{x}^{2}}+9x-27$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->因式分解->公式法->提公因式+平方差", "课内体系->知识点->式->因式分解->提公因式法"], "answer_analysis": ["因为$${{x}^{3}}-9{{x}^{2}}+27x-27=({{x}^{3}}-27)+27x-9{{x}^{2}}$$ $$=(x-3)({{x}^{2}}+3x+9)-9x(x-3)$$ $$=(x-3)({{x}^{2}}+3x+9-9x)$$ $$={{(x-3)}^{3}}$$， 故$$\\text{A}$$正确； 又因为$${{x}^{3}}-{{x}^{2}}+27x-27=(x-1)({{x}^{2}}+27)$$ $${{x}^{4}}-{{x}^{3}}+27x-27=(x-1)(x+3)({{x}^{2}}-3x+9)$$ $${{x}^{3}}-3{{x}^{2}}+9x-27=(x-3)({{x}^{2}}+9)$$， 故$$\\text{BCD}$$错误． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1454", "queId": "d7f841ddfd7f4bb78da306f6e608f38b", "competition_source_list": ["2012年第29届全国全国初中数学联赛竞赛第1题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "9.已知$$a=\\sqrt{2}-1$$，$$b=\\sqrt{3}-\\sqrt{2}$$，$$c=\\sqrt{10}-2$$，那么$$a$$、$$b$$、$$c$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless b\\textless c$$ "}], [{"aoVal": "B", "content": "$$a\\textless c\\textless b$$ "}], [{"aoVal": "C", "content": "$$b\\textless a\\textless c$$ "}], [{"aoVal": "D", "content": "$$b\\textless c\\textless a$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->运算能力", "知识标签->题型->数->实数->无理数有关的计算->题型：无理数大小的比较", "知识标签->知识点->式->二次根式->二次根式的运算->二次根式的混合运算"], "answer_analysis": ["$$a=\\sqrt{2}-1=\\frac{\\sqrt{2}-1}{1}=\\frac{1}{\\sqrt{2}+1}$$， $$b=\\sqrt{3}-\\sqrt{2}=\\frac{1}{\\sqrt{3}+\\sqrt{2}}$$， $$c=\\sqrt{6}-2=\\frac{2}{\\sqrt{6}+2}=\\frac{1}{\\sqrt{\\frac{3}{2}}+1}$$， 由$$\\sqrt{\\frac{3}{2}}+1\\textless{}\\sqrt{2}+1\\textless{}\\sqrt{3}+\\sqrt{2}$$，显然：$$b\\textless{}a\\textless{}c$$． 估算或作差 $$a-b=2\\sqrt{2}-\\sqrt{3}-1$$，$$2\\sqrt{2}\\approx 2.828$$，$$\\sqrt{3}+1\\approx 2.732$$，故$$2\\sqrt{2}\\textgreater\\sqrt{3}+1$$∴$$a\\textgreater b$$； $$c-a=\\sqrt{6}-\\sqrt{2}-1$$，$${{\\left( \\sqrt{6} \\right)}^{2}}=6$$，$${{\\left( \\sqrt{2}+1 \\right)}^{2}}\\approx {{2.414}^{2}} ~\\textless{} ~6$$，故$$\\sqrt{6}\\textgreater\\sqrt{2}+1$$，∴$$c\\textgreater a$$； ∴$$b ~\\textless{} ~a ~\\textless{} ~c$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "148", "queId": "5d3adc2a63984a488b0222d7d96fc126", "competition_source_list": ["2011年第28届全国全国初中数学联赛竞赛第1题7分", "初一单元测试《有条件的分式化解与求值》第22题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a+b=2$$，$$\\frac{{{(1-a)}^{2}}}{b}+\\frac{{{(1-b)}^{2}}}{a}=-4$$，则$$ab$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$-\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->其他方程->二元二次方程（组）", "课内体系->能力->运算能力"], "answer_analysis": ["由$$\\frac{{{(1-a)}^{2}}}{b}+\\frac{{{(1-b)}^{2}}}{a}=-4$$可得$$a{{(1-a)}^{2}}+b{{(1-b)}^{2}}=-4ab$$， 即$$(a+b)-2({{a}^{2}}+{{b}^{2}})+{{a}^{3}}+{{b}^{3}}+4ab=0$$， 即$$2-2({{a}^{2}}+{{b}^{2}})+2({{a}^{2}}-ab+{{b}^{2}})+4ab=0$$， 即$$2-2ab+4ab=0$$， 所以$$ab=-1$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "474", "queId": "a72b35b753d945d6b685f835f61b49a8", "competition_source_list": ["2018~2019学年3月河南洛阳西工区洛阳外国语学校初二下学期月考第6题3分", "2018~2019学年3月天津南开区天津市南开翔宇学校初二下学期月考第8题3分", "2019~2020学年山东枣庄峄城区初二上学期期末第2题3分", "2020~2021学年10月四川成都彭州市成都七中嘉祥外国语学校北城分校初二上学期月考第6题3分", "2018~2019学年山东青岛市南区青岛大学附属中学初二上学期期末第3题3分", "2020~2021学年9月四川成都青羊区四川师范大学实验外国语学校初二上学期周测C卷第3题3分", "2020~2021学年3月天津南开区天津美达菲国际学校初二下学期月考第8题3分", "2018~2019学年3月天津河西区天津市新华中学初二下学期月考第3题3分", "2019~2020学年3月江西南昌东湖区南昌市百树学校初二下学期月考第6题3分", "2018~2019学年5月广东深圳罗湖区深圳市罗湖区翠园中学（初中部）初一下学期周测B卷（竞赛班）第2题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列结论中，错误的有． ①在$$\\text{Rt}\\triangle ABC$$中，已知两边长分别为$$3$$和$$4$$，则第三边的长为$$5$$； ②$$\\triangle ABC$$的三边长分别为$$AB$$，$$BC$$，$$AC$$，若$$B{{C}^{2}}+A{{C}^{2}}=A{{B}^{2}}$$，则$$\\angle A=90{}^{}\\circ $$； ③ 在$$\\triangle ABC$$中，若$$\\angle A:\\angle B:\\angle C=1:5:6$$，则$$\\triangle ABC$$是直角三角形； ④若三角形的三边长之比为$$3:4:5$$，则该三角形是直角三角形．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$个 "}], [{"aoVal": "B", "content": "$$1$$个 "}], [{"aoVal": "C", "content": "$$2$$个 "}], [{"aoVal": "D", "content": "$$3$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->勾股定理及应用->勾股定理基础->勾股逆定理的应用", "课内体系->能力->推理论证能力"], "answer_analysis": ["①在$$\\text{Rt}\\triangle ABC$$中，已知两边长分别为$$3$$和$$4$$，则第三边的长为$$5$$或$$\\sqrt{7}$$，错误； ②$$\\triangle ABC$$的三边长分别为$$AB$$，$$BC$$，$$AC$$，若$$B{{C}^{2}}+A{{C}^{2}}=A{{B}^{2}}$$，则$$\\angle A=90{}^{}\\circ $$，错误； ③在$$\\triangle ABC$$中，若若$$\\angle A:\\angle B:\\angle C=1:5:6$$，则$$\\triangle ABC$$是直角三角形，正确； ④若三角形的三边长之比为$$3:4:5$$，则该三角形是直角三角形，正确． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1547", "queId": "e1efb646307a4796b1c72c12425084b4", "competition_source_list": ["2013年第18届华杯赛初一竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$a$$，$$b$$，$$c$$都是大于$$-\\frac{1}{2}$$的负数，那么下列式子成立的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a+c-b ~\\textless{} ~0$$ "}], [{"aoVal": "B", "content": "$${{a}^{2}}-{{b}^{2}}-{{c}^{2}}\\textgreater0$$ "}], [{"aoVal": "C", "content": "$$abc\\textgreater-\\frac{1}{8}$$ "}], [{"aoVal": "D", "content": "$$\\left\\textbar{} abc \\right\\textbar\\textgreater\\frac{1}{8}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["$$\\text{A}$$．取$$a=c=-\\frac{1}{12}$$，$$b=-\\frac{1}{3}$$，则有：$$a+c+b=-\\frac{1}{12}-\\frac{1}{12}+\\frac{1}{3}=\\frac{1}{6}\\textgreater0$$，故$$\\text{A}$$错误； $$\\text{BD}$$．取$$a=b=c=-\\frac{1}{3}$$，则有：$${{a}^{2}}-{{b}^{2}}-{{c}^{2}}=-\\frac{1}{9} ~\\textless{} ~0$$，$$\\left\\textbar{} abc \\right\\textbar=\\frac{1}{27} ~\\textless{} ~\\frac{1}{8}$$，故$$\\text{BD}$$错误； 因为$$a$$，$$b$$为负数，所以$$ab\\textgreater0$$．因为$$c\\textgreater-\\frac{1}{2}$$，所以$$abc\\textgreater-\\frac{1}{2}ab$$． 又因为$$b\\textgreater-\\frac{1}{2}$$，所以$$-\\frac{1}{2}ab ~\\textless{} ~\\frac{1}{4}$$，所以$$-\\frac{1}{2}ab\\textgreater\\frac{1}{4}a\\textgreater-\\frac{1}{8}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "416", "queId": "1bba0daca2e2422cb8a3cb057c131510", "competition_source_list": ["2003年第20届全国初中数学联赛竞赛第10题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知正整数$$a$$，$$b$$之差为$$120$$，它们的最小公倍数是其最大公约数的$$105$$倍，那么$$a$$，$$b$$中较大的数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$125$$ "}], [{"aoVal": "B", "content": "$$175$$ "}], [{"aoVal": "C", "content": "$$225$$ "}], [{"aoVal": "D", "content": "$$275$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->数的特征->整除的条件", "竞赛->知识点->数论->整除->因数与倍数"], "answer_analysis": ["设两个数的最大公约数为$$d$$，大数为$$md$$，小数为$$nd$$，其中$$m$$，$$n$$互质，则最小公倍数为$$mnd$$．由已知得$$mn=105$$，$$(m-n)d=120$$． 由于$$m\\textgreater n$$，所以$$m$$只可能是$$105$$，$$35$$，$$21$$，$$15$$． 对应的$$n$$分别为$$1$$，$$3$$，$$5$$，$$7$$． 只有在$$m=15$$，$$n=7$$时$$d$$为整数，$$d=15$$． 所以大数为$$md=225$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "756", "queId": "9592fb29be7a485790d29679b43b86d6", "competition_source_list": ["2002年第13届希望杯初二竞赛第2试第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a=\\frac{\\sqrt[3]{3}}{\\sqrt{2}}$$，$$b=\\frac{\\sqrt[3]{2}+m}{\\sqrt{3}+m}$$，$$c=\\frac{\\sqrt[3]{3}+m}{\\sqrt{2}+m}$$，其中$$m\\textgreater0$$，那$$a$$，$$b$$，$$c$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b\\textgreater c$$ "}], [{"aoVal": "B", "content": "$$c\\textgreater a\\textgreater b$$ "}], [{"aoVal": "C", "content": "$$a\\textgreater c\\textgreater b$$ "}], [{"aoVal": "D", "content": "$$b\\textgreater c\\textgreater a$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->二次根式->二次根式的性质与运算"], "answer_analysis": ["因为$$\\sqrt[3]{3}=\\sqrt[6]{9}$$，$$\\sqrt{2}=\\sqrt[6]{8}$$． 因而$$\\sqrt[3]{3}\\textgreater\\sqrt{2}$$，$$a=\\frac{\\sqrt[3]{3}}{\\sqrt{2}}\\textgreater1$$，$$c=\\frac{\\sqrt[3]{3}+m}{\\sqrt{2}+m}\\textgreater1$$． 于是$$a-c=\\frac{\\sqrt[3]{3}}{\\sqrt{2}}-\\frac{\\sqrt[3]{3}+m}{\\sqrt{2}+m}$$ $$=\\frac{\\sqrt[3]{3}\\cdot \\sqrt{2}+m\\sqrt[3]{3}-\\sqrt{2}\\cdot \\sqrt[3]{3}-m\\sqrt{2}}{\\sqrt{2}\\left( \\sqrt{2}+m \\right)}\\textgreater0$$， 所以$$a\\textgreater c\\textgreater1$$， 又$$b=\\frac{\\sqrt[3]{2}+m}{\\sqrt{3}+m} ~\\textless{} ~1$$， 故$$a\\textgreater c\\textgreater b$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "702", "queId": "492372173e1d40478568c594a806abd9", "competition_source_list": ["2002年第13届希望杯初一竞赛第2试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "当$$x$$取$$1$$到$$10$$之间的质数时，四个整式：$${{x}^{2}}+2$$，$${{x}^{2}}+4$$，$${{x}^{2}}+6$$和$${{x}^{2}}+8$$的值中，共有质数个．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->素数与合数"], "answer_analysis": ["因为$$1\\sim 10$$之间的质数有$$2$$，$$3$$，$$5$$，$$7$$共$$4$$个． 当$$x=2$$时，四个整式的值都是大于$$2$$的偶数，不是质数； 当$$x=3$$时，四个整式的值分别为$$11$$，$$13$$，$$15$$，$$17$$； 当$$x=5$$时，四个整式的值分别为$$27$$，$$29$$，$$31$$，$$33$$； 当$$x=7$$时，四个整式的值分别为$$51$$，$$53$$，$$55$$，$$57$$． 其中只有$$11$$，$$13$$，$$17$$，$$29$$，$$31$$，$$53$$为质数，共$$6$$个． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "187", "queId": "2619520d5faf413daedbaf2619cde7fc", "competition_source_list": ["2014年第25届全国希望杯初二竞赛初赛第10题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "将$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$，$$8$$这$$8$$个数排成一行，使$$8$$的两边各数的和相等，则不同的排法有．", "answer_option_list": [[{"aoVal": "A", "content": "$$144$$种 "}], [{"aoVal": "B", "content": "$$288$$种 "}], [{"aoVal": "C", "content": "$$576$$种 "}], [{"aoVal": "D", "content": "$$1152$$种 "}]], "knowledge_point_routes": ["课内体系->能力->分析和解决问题能力", "竞赛->知识点->组合->排列与组合"], "answer_analysis": ["因为去掉$$8$$以后，余下$$7$$个数的和是$$1+2+3+4+5+6+7=28$$， 所以$$8$$的两边各数的和分别是$$14$$． 因为$$7+6=13$$， 所以$$14$$至少是由$$3$$个数相加而成的， 所以$$8$$的两边分别有$$3$$个数和$$4$$个数． 因为$$(7,6,1)$$，$$(7,5,2)$$，$$(7,4,3)$$，$$(6,5,3)$$这四组数的和都是$$14$$， 每组中的数的排列方法有$$3\\times 2\\times 1=6$$种， 与其对应的另外$$4$$个数（$$8$$除外）有$$4\\times 3\\times 2\\times 1=24$$种排列方法， 当排成横行时，这$$4$$组数有排在$$8$$的左边和右边两种排列方法， 因此，每组数的排列方法有$$2\\times 6\\times 24=288$$（种）． 所以，满足题意的排列方法有$$4\\times 288=1152$$（种）． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1313", "queId": "c07e939daabc4e19b4d0f95df0ea021d", "competition_source_list": ["2011年第16届华杯赛初一竞赛初赛第3题", "2018~2019学年甘肃兰州城关区兰州市第三十五中学初二下学期期中第12题4分", "初一下学期单元测试《不等式与不等式组》一元一次不等式及其应用第28题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a$$，$$b$$是常数，不等式$$\\frac{x}{a}+\\frac{1}{b}\\textgreater0$$的解集为$$x\\textless{}\\frac{1}{5}$$，则关于$$x$$的不等式$$bx-a\\textgreater0$$的解集是．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textgreater\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$x\\textless{}-\\frac{1}{5}$$ "}], [{"aoVal": "C", "content": "$$x\\textgreater-\\frac{1}{5}$$ "}], [{"aoVal": "D", "content": "$$x\\textless{}\\frac{1}{5}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->解一元一次不等式"], "answer_analysis": ["原不等式变形得：$$\\frac{x}{a}\\textgreater-\\frac{1}{b}$$， ∵$$x\\textless{}\\frac{1}{5}$$， ∴$$a\\textless{}0$$． 解不等式得：$$x\\textless{}-\\frac{a}{b}$$，$$-\\frac{a}{b}=\\frac{1}{5}$$，即$$b=-5a$$． ∴$$-5ax-a\\textgreater0$$， ∴$$x\\textgreater-\\frac{1}{5}$$． 因为不等式等$$\\frac{x}{a}+\\frac{1}{b}\\textgreater0$$的解集为$$x ~\\textless{} ~\\frac{1}{5}$$，故必有$$a ~\\textless{} ~0$$，所以$$\\frac{-a}{b}=\\frac{1}{5}$$，并且$$b\\textgreater0$$，所以由$$bx-a\\textgreater0$$得到$$x\\textgreater-\\frac{1}{5}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "257", "queId": "74a562c360f6460a8b159aa224a3e5e9", "competition_source_list": ["2006年第17届希望杯初一竞赛初赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "对于数$$x$$，符号[$$x$$]表示不大于$$x$$的最大整数．例如$$[3.14]=3$$，$$[-7.59]=-8$$，则满足关系式$$\\left[ \\frac{3x+7}{7} \\right]=4$$的$$x$$的整数值有．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$个 "}], [{"aoVal": "B", "content": "$$5$$个 "}], [{"aoVal": "C", "content": "$$4$$个 "}], [{"aoVal": "D", "content": "$$3$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->综合与实践->规律探究与程序框图->定义新运算", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->解一元一次不等式组", "课内体系->能力->推理论证能力"], "answer_analysis": ["根据题意可得不等式$$4\\leqslant \\frac{3x+7}{7} \\textless{} 5$$，得$$7\\leqslant x \\textless{} \\frac{28}{3}$$因为$$x$$取整数值，解得$$x=7$$，$$8$$，$$9$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1469", "queId": "eecc141d5048458cbbe77817a9d2d1d5", "competition_source_list": ["2012年第23届全国希望杯初一竞赛初赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若两位数$$\\overline{ab}$$是质数，交换数字后得到的两位数$$\\overline{ba}$$也是质数，则称$$\\overline{ab}$$为绝对质数，在大于$$11$$的两位数中绝对质数有（~ ~ ）个．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类"], "answer_analysis": ["大于$$11$$的两位数中绝对质数有$$13$$，$$17$$，$$31$$，$$37$$，$$71$$，$$73$$，$$79$$，$$97$$，共$$8$$个． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1544", "queId": "e667c3e59e894a6992f63ce143382a78", "competition_source_list": ["2018~2019学年浙江绍兴初一上学期期末第8题3分", "2020年湖南长沙天心区湘郡培粹实验中学初一竞赛初赛（9月）第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$m$$辆客车及$$n$$个人，若每辆客车乘$$40$$人，则还有$$25$$人不能上车；若每辆客车乘$$45$$人，则还有$$5$$人不能上车．有下列四个等式：①$$40m+25=45m+5$$；②$$\\frac{n-25}{40}=\\frac{n-5}{45}$$；③$$\\frac{n+25}{40}=\\frac{n+5}{45}$$；④$$40m+25=45m-5$$．其中正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "①③ "}], [{"aoVal": "B", "content": "①② "}], [{"aoVal": "C", "content": "②④ "}], [{"aoVal": "D", "content": "③④ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的调配问题", "课内体系->能力->运算能力"], "answer_analysis": ["根据人数不变，列出方程：$$40m+25=45m+5$$； 根据客车数不变，列出方程：$$\\frac{n-25}{40}=\\frac{n-5}{45}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "228", "queId": "389c594b7cbc4121b03497f3f928d93c", "competition_source_list": ["2016年第27届全国希望杯初一竞赛初赛第15题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$x$$，$$y$$都是正整数，且$${{x}^{2}}-{{y}^{2}}=45$$，这样的$$(x,y)$$共有 组．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->其他方程->不定方程"], "answer_analysis": ["∵$$x$$是$$6$$的倍数， 设$$x=6m$$，则$${{y}^{2}}=36{{m}^{2}}-2016=36({{m}^{2}}-56)$$， ∴$$y$$也能被$$6$$整除， 设$$y=6n$$，其中$$n\\textless{}m$$， ∴$$36({{m}^{2}}-{{n}^{2}})=36\\times 56$$， ∴$$(m+n)(m-n)=56$$． ∵$$m+n$$与$$m-n$$有相同的奇偶性， 且$$56=7\\times 8=14\\times 4=28\\times 2=56\\times 1$$， ∴$$\\begin{cases}m+n=14 m-n=4 \\end{cases}$$或$$\\begin{cases}m+n=28 m-n=2 \\end{cases}$$， ∴$$\\begin{cases}m=9 n=5 \\end{cases}$$或$$\\begin{cases}m=15 n=13 \\end{cases}$$, ∴$$\\begin{cases}x=54 y=30 \\end{cases}$$或$$\\begin{cases}x=90 y=78 \\end{cases}$$． 故这样的$$(x,y)$$共有$$2$$组． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1153", "queId": "81ae35c209914096a2b7303aab8e463f", "competition_source_list": ["2007年第18届希望杯初一竞赛复赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$9$$个数中：$$-5$$，$$-4$$，$$-3$$，$$-2$$，$$-1$$，$$0$$，$$1$$，$$2$$，$$3$$中，能使不等式$$-3{{x}^{2}}\\textless{}-14$$成立的数的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质", "课内体系->能力->运算能力"], "answer_analysis": ["不等式$$-3{{x}^{2}}\\textless{}-14$$等价于$$3{{x}^{2}}\\textgreater14$$，逐一代入检验知：当$$x=-5$$，$$-4$$，$$-3$$，$$3$$时成立，即满足不等式$$-3{{x}^{2}}\\textless{}-14$$的数有$$4$$个． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "845", "queId": "f1dac1e4cbcd4d9d98fdc2df5b2bf911", "competition_source_list": ["2017年第19届浙江宁波余姚市余姚市实验学校初三竞赛（实验杯）第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$x+y=3$$，则点$$\\left( x,y \\right)$$一定不在．", "answer_option_list": [[{"aoVal": "A", "content": "第一象限 "}], [{"aoVal": "B", "content": "第二象限 "}], [{"aoVal": "C", "content": "第三象限 "}], [{"aoVal": "D", "content": "第四象限 "}]], "knowledge_point_routes": ["课内体系->知识点->函数->平面直角坐标系->坐标系基础->象限内坐标的特征"], "answer_analysis": ["因为$$x+y=3$$， 所以$$x$$和$$y$$中至少有一点大于$$0$$， 所以点$$(x,y)$$一定不在第三象限． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "842", "queId": "5c1ae934b1aa4fda834fb2389d44c8e4", "competition_source_list": ["2009年第20届希望杯初一竞赛第1试第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$2005$$，$$2007$$，$$2009$$这三个数中，质数有．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$个 "}], [{"aoVal": "B", "content": "$$1$$个 "}], [{"aoVal": "C", "content": "$$2$$个 "}], [{"aoVal": "D", "content": "$$3$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->素数与合数"], "answer_analysis": ["显然，$$2005$$能被$$5$$整除，$$2007$$能被$$3$$整除，它们都是合数．$$2009={{7}^{2}}\\times 41$$，也是合数． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1330", "queId": "8aac50a751148307015114dce1e903c7", "competition_source_list": ["2015年第26届全国希望杯初二竞赛复赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$y=(k-\\frac{1}{k})x+\\frac{1}{k}(0\\textless{}k\\textless{}1)$$是关于$$x$$的一次函数，当$$1\\leqslant x\\leqslant 2$$时，$$y$$的最大值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$k$$ "}], [{"aoVal": "D", "content": "$$2k-\\frac{1}{k}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->一次函数->一次函数基础->一次函数的增减性"], "answer_analysis": ["当$$0\\textless{}k\\textless{}1$$时，$$k-\\frac{1}{k}\\textless{}0$$，$$y$$随$$x$$的增大而减小，故在$$1\\leqslant x\\leqslant 2$$时，$$y$$在$$x=1$$时最大，为$$k$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1104", "queId": "ff8080814d9efd56014da55b58280779", "competition_source_list": ["1992年第3届全国希望杯初一竞赛复赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "四个互不相等的正数$$a$$，$$b$$，$$c$$，$$d$$中，$$a$$最大，$$d$$最小，且$$\\frac{a}{b}=\\frac{c}{d}$$，则$$a+d$$与$$b+c$$的大小关系是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a+d\\textless{}b+c$$ "}], [{"aoVal": "B", "content": "$$a+d\\textgreater b+c$$ "}], [{"aoVal": "C", "content": "$$a+d=b+c$$ "}], [{"aoVal": "D", "content": "不确定的 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性"], "answer_analysis": ["在四个互不相等的正数$$a$$，$$b$$，$$c$$，$$d$$中，$$a$$最大，$$d$$最小， 因此有$$a\\textgreater b$$，$$a\\textgreater c$$，$$a\\textgreater d$$，$$b\\textgreater d$$，$$c\\textgreater d$$． 由$$\\frac{a}{b}=\\frac{c}{d}\\Leftrightarrow \\frac{a}{b}-1=\\frac{c}{d}-1\\Leftrightarrow \\frac{a-b}{b}=\\frac{c-d}{d}\\Leftrightarrow \\frac{a-b}{c-d}=\\frac{b}{d}$$． 因为$$b\\textgreater d\\textgreater0$$，所以$$\\frac{a-b}{c-d}=\\frac{b}{d}\\textgreater1$$， 所以$$a-b\\textgreater c-d$$， 所以$$a+b\\textgreater b+c$$成立，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "420", "queId": "1bc3f60399c24acab04618331e135c61", "competition_source_list": ["2002年第13届希望杯初二竞赛第1试第21题"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$${{x}^{2}}-xy-5x+5y-1=0$$的整数解有几组？？", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}], [{"aoVal": "E", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->二次方程->二元二次方程组", "课内体系->知识点->式->整式的加减->整式有关的概念->代数式"], "answer_analysis": ["由$${{x}^{2}}-xy-5x+5y-1=0$$可得$$x\\left( x-y \\right)-5\\left( x-y \\right)-1=0$$， 即有$$\\left( x-y \\right)\\left( x-5 \\right)=1$$． 因为$$x$$，$$y$$都是整数， 所以$$\\begin{cases}x-y=1 x-5=1 \\end{cases}$$或$$\\begin{cases}x-y=-1 x-5=-1 \\end{cases}$$． 即$$\\begin{cases}x=6 y=5 \\end{cases}$$或$$\\begin{cases}x=4 y=5 \\end{cases}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "313", "queId": "1e5f973e56024445812adfbfcb56bf96", "competition_source_list": ["2006年第17届希望杯初二竞赛第1试第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "对实数$$a$$，$$b$$，定义运算``$$*$$''如下： $$a*b=\\begin{cases}{{a}^{2}}b\\quad a\\geqslant b a{{b}^{2}}\\quad a\\textless{}b \\end{cases}$$． 现已知$$3*m=36$$，则实数$$m$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$\\pm 2\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$4$$或$$\\pm 2\\sqrt{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->综合与实践->规律探究与程序框图->定义新运算", "课内体系->知识点->式->整式的乘除->整式的乘除运算->单项式乘单项式", "课内体系->思想->分类讨论思想", "课内体系->能力->运算能力"], "answer_analysis": ["当$$3\\geqslant m$$时，有 $$3*m={{3}^{2}}\\cdot m=9m=36$$， 解得$$m=4$$，与$$3\\geqslant m$$予盾，舍去． 当$$3\\textless{}m$$时，有$$3*m=3{{m}^{2}}=36$$， 解得$$m=\\pm 2\\sqrt{3}$$，因为$$3\\textless{}m$$，故舍去$$m=-2\\sqrt{3}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "194", "queId": "147b270734ec4a629835a8e591538cc2", "competition_source_list": ["2008年第19届希望杯初一竞赛第1试第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "正方形内有一点$$A$$，到各边的距离从小到大依次是$$1$$，$$2$$，$$5$$，$$6$$，则正方形的面积是．", "answer_option_list": [[{"aoVal": "A", "content": "$$33$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$49$$ "}]], "knowledge_point_routes": ["竞赛->知识点->四边形->特殊平行四边形"], "answer_analysis": ["由于$$A$$在正方形内， 所以$$A$$到两组对边的距离之和相等， 由于只有$$1+6=2+5$$， 于是，正方形的边长只能为$$7$$， 故面积是$${{7}^{2}}=49$$， 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "816", "queId": "7264a252206a415d80857d335800c768", "competition_source_list": ["2013年竞赛第4题4分", "初三上学期其它"], "difficulty": "2", "qtype": "single_choice", "problem": "设函数$$y=\\left( \\sqrt{4+x}+\\sqrt{4-x}+1 \\right)\\left( \\sqrt{16-{{x}^{2}}}+2 \\right)$$，则$$y$$的取值范围是（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\leqslant y\\leqslant 20$$ "}], [{"aoVal": "B", "content": "$$2\\leqslant y\\leqslant 30$$ "}], [{"aoVal": "C", "content": "$$4\\sqrt{2}+2\\leqslant y\\leqslant 20$$ "}], [{"aoVal": "D", "content": "$$4\\sqrt{2}+2\\leqslant y\\leqslant 30$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的基础->二次根式有意义的条件"], "answer_analysis": ["因为$$\\begin{matrix}y=\\left( \\sqrt{\\left( 4+x \\right)+\\left( 4-x \\right)+2\\sqrt{16-{{x}^{2}}}}+1 \\right)\\left( \\sqrt{16-{{x}^{2}}}+2 \\right) =\\left( \\sqrt{8+2\\sqrt{16-{{x}^{2}}}}+1 \\right)\\left( \\sqrt{16-{{x}^{2}}}+2 \\right)， ~ \\end{matrix}$$ 所以当$$x=\\pm 4$$时，$$y$$的最小值为$$2\\left( \\sqrt{8}+1 \\right)=4\\sqrt{2}+2$$；当$$x=0$$时，$$y$$的最大值为$$30$$．故$$y$$的取值范围为$$4\\sqrt{2}+2\\leqslant y\\leqslant 30$$．故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "116", "queId": "08b6ae32fffe46f2a0bf99aeb5d02fed", "competition_source_list": ["2018年全国初中数学联赛竞赛B卷"], "difficulty": "2", "qtype": "single_choice", "problem": "满足$${{({{x}^{2}}+x-1)}^{x+2}}=1$$的整数$$x$$的个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元二次方程", "课内体系->知识点->式->整式的乘除->幂的运算", "课内体系->能力->运算能力", "课内体系->能力->抽象概括能力"], "answer_analysis": ["当$$x+2=0$$且$${{x}^{2}}+x-1\\ne 0$$时，$$x=-2$$， 当$${{x}^{2}}+x-1=1$$时，$$x=-2$$或$$x=1$$， 当$${{x}^{2}}+x-1=-1$$且$$x+2$$为偶数时，$$x=0$$， 所以，满足条件的整数$$x$$有$$3$$个． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1485", "queId": "dcb5097fc8534d08b96ecfd924ee27f5", "competition_source_list": ["1991年第2届希望杯初二竞赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$y\\textgreater0$$时，$$\\sqrt{-{{x}^{3}}y}$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$-x\\sqrt{xy}$$ "}], [{"aoVal": "B", "content": "$$x\\sqrt{xy}$$ "}], [{"aoVal": "C", "content": "$$-x\\sqrt{-xy}$$ "}], [{"aoVal": "D", "content": "$$x\\sqrt{-xy}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->二次根式->二次根式的性质与运算"], "answer_analysis": ["由$$y\\textgreater0$$，可知$$x\\textless{}0$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "685", "queId": "4910951259f34365a515acf7d3fef2e0", "competition_source_list": ["2019年广东惠州惠城区光正实验学校初二竞赛第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$${{(\\sqrt{7})}^{2}}-\\sqrt{{{(-4)}^{2}}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$-3$$ "}], [{"aoVal": "D", "content": "$$-11$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的加减以及混合运算"], "answer_analysis": ["$${{(\\sqrt{7})}^{2}}-\\sqrt{{{(-4)}^{2}}}=7-4=3$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "939", "queId": "d676b497a2d74d9ca6cfd7c757247e4c", "competition_source_list": ["1995年第12届全国初中数学联赛竞赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果方程$$\\left( x-1 \\right)\\left( {{x}^{2}}-2x+m \\right)=0$$的三根可以作为一个三角形的三边之长，那么实数$$m$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0\\leqslant m\\leqslant 1$$ "}], [{"aoVal": "B", "content": "$$m\\geqslant \\frac{3}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}\\textless{}m\\leqslant 1$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{4}\\leqslant m\\leqslant 1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->二次方程->一元二次方程的根与系数的关系", "竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["显然，方程的一个根为$$1$$，另两根之和为$${{x}_{1}}+{{x}_{2}}=2\\textgreater1$$，三根能作为一个三角形的三边，须且只须$$\\left\\textbar{} {{x}_{1}}-{{x}_{2}} \\right\\textbar\\textless{}1$$， 又$$\\left\\textbar{} {{x}_{1}}-{{x}_{2}} \\right\\textbar=\\frac{\\sqrt{\\Delta }}{\\left\\textbar{} a \\right\\textbar}=\\sqrt{4-4m}\\textless{}1$$， 有$$0\\leqslant 4-4m\\textless{}1$$， 解得$$\\frac{3}{4}\\textless{}m\\leqslant 1$$， 但作为选择题，只须取$$m=\\frac{3}{4}$$代入，得方程的根为$$1$$、$$\\frac{3}{2}$$、$$\\frac{1}{2}$$，不能组成三角形，故包括$$\\frac{3}{4}$$的$$\\text{A}$$、$$\\text{B}$$、$$\\text{D}$$均可否定，选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "472", "queId": "35a551685829403f98430deebff29b0c", "competition_source_list": ["2021年福建竞赛（\"大梦杯”青少年水平测试）第2~2题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知实数\\emph{x，y}满足$$\\frac{26{{x}^{3}}{{y}^{3}}}{{{x}^{6}}-27{{y}^{6}}}=1$$且$${{x}^{2}}\\ne {{y}^{2}}$$，则$$\\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}$$的值为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "2 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 由$$\\frac{26{{x}^{3}}{{y}^{3}}}{{{x}^{6}}-27{{y}^{6}}}=1$$可得$${{x}^{6}}-26{{x}^{3}}{{y}^{3}}-27{{y}^{6}}=0$$，进而可得$${{\\left( \\frac{x}{y} \\right)}^{6}}-26{{\\left( \\frac{x}{y} \\right)}^{3}}-27=0$$，解得$$\\frac{x}{y}=-1$$或$$\\frac{x}{y}=3$$，然后再对$$\\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}$$进行变形即可解答．\\\\ 【详解】\\\\ 解：∵$$\\frac{26{{x}^{3}}{{y}^{3}}}{{{x}^{6}}-27{{y}^{6}}}=1$$，得$${{x}^{6}}-26{{x}^{3}}{{y}^{3}}-27{{y}^{6}}=0$$，\\\\ 即$${{\\left( \\frac{x}{y} \\right)}^{6}}-26{{\\left( \\frac{x}{y} \\right)}^{3}}-27=0$$．\\\\ ∴$${{\\left( \\frac{x}{y} \\right)}^{3}}=-1$$或$${{\\left( \\frac{x}{y} \\right)}^{3}}=27$$．\\\\ 即$$\\frac{x}{y}=-1$$或$$\\frac{x}{y}=3$$．\\\\ ∴$${{x}^{2}}\\ne {{y}^{2}}$$，所以$$\\frac{x}{y}=3$$，$$\\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}=\\frac{{{\\left( \\frac{x}{y} \\right)}^{2}}+1}{{{\\left( \\frac{x}{y} \\right)}^{2}}-1}=\\frac{9+1}{9-1}=\\frac{5}{4}$$．\\\\ 故选：A．\\\\ 【点睛】\\\\ 本题主要考查了分式的化简求值、立方根、解一元二次方程等知识点，解题的关键是灵活应用相关定义和运算法则以及整体法来求解． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "173", "queId": "58a61c0789c3425982fd036681f7a9d7", "competition_source_list": ["2019年安徽蚌埠初三中考一模（局属学校）第9题4分", "2018年湖北黄冈中考真题第6题3分", "2018~2019学年10月河南洛阳洛龙区洛阳地矿双语学校初三上学期月考第8题3分", "2019~2020学年12月河北石家庄新华区石门实验学校初三上学期月考第13题", "2019~2020学年10月安徽合肥包河区合肥市第四十八中学初三上学期月考第10题4分", "2019~2020学年江苏苏州姑苏区苏州市草桥中学校初三下学期单元测试《二次函数》第4题", "2018~2019学年4月广东深圳宝安区深圳市石岩公学初三下学期周测（第12周）第11题3分", "2018~2019学年9月湖北武汉蔡甸区初三上学期月考第9题3分", "2018~2019学年天津滨海新区初三上学期期末第12题3分", "2018年第20届浙江宁波余姚市余姚市实验学校初三竞赛决赛（实验杯）第6题4分", "2018~2019学年福建厦门思明区厦门松柏中学初三上学期期中第9题4分", "2019~2020学年10月江苏苏州姑苏区金阊实验中学初三上学期月考第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$a\\leqslant x\\leqslant a+1$$时，函数$$y={{x}^{2}}-2x-1$$的最小值为$$-1$$，则$$a$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$0$$或$$2$$ "}], [{"aoVal": "D", "content": "$$-1$$或$$2$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力"], "answer_analysis": ["当$$y=1$$时，有$${{x}^{2}}-2x+1=1$$， 解得：$${{x}_{1}}=0$$，$${{x}_{2}}=2$$． ∵当$$a\\leqslant x\\leqslant a+1$$时，函数有最小值$$1$$， ∴$$a=2$$或$$a+1=0$$， ∴$$a=2$$或$$a=-1$$， 故选：$$\\text{D}$$． ∵$$y={{x}^{2}}-2x+1={{\\left( x-1 \\right)}^{2}}$$， ∴①当$$a+1\\textless{}1$$，即$$a\\textless{}0$$时，函数的最小值为$$y={{\\left( a+1-1 \\right)}^{2}}=1$$， ∴$$a=\\pm 1$$， ∴$$a=-1$$． ②当$$a\\textgreater1$$时，函数的最小值为$$y={{\\left( a-1 \\right)}^{2}}=1$$， ∴$$a=2$$或$$a=0$$， ∴$$a=2$$． 综上所述，$$a=-1$$或$$2$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1380", "queId": "c9cecdc6e25e4cd1ad619dd2d6bc960a", "competition_source_list": ["2001年第12届希望杯初一竞赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "当$$x=\\frac{2}{3}$$时，代数式$$1+3x$$的值是$$-\\frac{1}{3}$$的．", "answer_option_list": [[{"aoVal": "A", "content": "绝对值 "}], [{"aoVal": "B", "content": "倒数 "}], [{"aoVal": "C", "content": "相反数 "}], [{"aoVal": "D", "content": "倒数的相反数 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->相反数", "课内体系->能力->运算能力"], "answer_analysis": ["$$x=\\frac{2}{3}$$时，代数式$$1+3x$$的值是$$1+3\\times \\left( \\frac{2}{3} \\right)=3$$，而$$3$$是$$-\\frac{1}{3}$$的倒数的相反数． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "282", "queId": "3450802b62b34139ba9b27fc1e62490a", "competition_source_list": ["2007年第18届希望杯初一竞赛初赛第10题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "对任意四个有理数$$a$$，$$b$$，$$c$$，$$d$$，定义新运算： $$\\left\\textbar{} \\begin{matrix}a b c d \\end{matrix} \\right\\textbar=ad-bc$$ 已知$$\\left\\textbar{} \\begin{matrix}2x -4 x 1 \\end{matrix} \\right\\textbar=18$$，则$$x=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->综合与实践->规律探究与程序框图->定义新运算", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->解一元一次不等式", "课内体系->知识点->式->整式的加减->整式的加减运算"], "answer_analysis": ["由已知新运算的定义，知 $$\\left\\textbar{} \\begin{matrix}2x -4 x 1 \\end{matrix} \\right\\textbar=2x-\\left( -4 \\right)x=18$$， 解得$$x=3$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "758", "queId": "a7f6815b70ea4e149982e6622ddc5f3b", "competition_source_list": ["2001年第12届希望杯初二竞赛第1试第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "某工厂生产的灯泡中有$$\\frac{1}{5}$$是次品，实际在检查时，只发现其中的$$\\frac{4}{5}$$被剔除，另有$$\\frac{1}{20}$$的正品也被误以为是次品而剔除，其余的灯泡全部上市出售，那么该工厂出售的灯泡中次品所占的百分率是．", "answer_option_list": [[{"aoVal": "A", "content": "$$4 \\%$$ "}], [{"aoVal": "B", "content": "$$5 \\%$$ "}], [{"aoVal": "C", "content": "$$6.25 \\%$$ "}], [{"aoVal": "D", "content": "$$7.25 \\%$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->一次方程->一元一次方程"], "answer_analysis": ["设工厂的总产量为$$M$$，则次品为$$\\frac{1}{5}M$$；正品为$$\\frac{4}{5}M$$； 检查时，被剔除的次品数为$$\\frac{1}{5}M\\times \\frac{4}{5}=\\frac{4}{25}M$$， 被剔除的正品数为$$\\frac{4}{5}M\\times \\frac{1}{20}=\\frac{1}{25}M$$． 所以出售的产品中次品数为$$\\frac{1}{5}M-\\frac{4}{25}M=\\frac{1}{25}M$$． 而出售产品的总量为$$\\dfrac{1}{25}M+\\left( \\dfrac{4}{5}M-\\frac{1}{25}M \\right)=\\frac{4}{5}M$$． 所以该工厂出售的灯泡中次品所占的百分率为$$\\frac{\\dfrac{1}{25}M}{\\dfrac{4}{5}M}\\times 100 \\%=5 \\%$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1248", "queId": "8aac490751148307015127b8b7113655", "competition_source_list": ["初三上学期单元测试《概率初步》第22题", "2009年竞赛第3题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "将一枚六个面编号分别为$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$的质地均匀的正方体骰子先后投掷两次，记第一次掷出的点数为$$a$$，第二次掷出的点数为$$b$$，则使关于$$x$$，$$y$$的方程组$$\\begin{cases}ax+by=3 x+2y=2 \\end{cases}$$只有正数解的概率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{12}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{18}$$ "}], [{"aoVal": "D", "content": "$$\\frac{13}{36}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->统计与概率->概率->概率的计算方法"], "answer_analysis": ["当$$2a-b=0$$时，方程组无解． 当$$2a-b\\ne 0$$时，方程组的解为$$\\begin{cases}x=\\dfrac{6-2b}{2a-b} y=\\dfrac{2a-3}{2a-b} \\end{cases}$$， 由已知，得$$\\begin{cases}\\dfrac{6-2b}{2a-b}\\textgreater0 \\dfrac{2a-3}{2a-b}\\textgreater0 \\end{cases}$$ 即$$\\begin{cases}2a-b\\textgreater0 a\\textgreater-\\dfrac{3}{2} b\\textless{}3 \\end{cases}$$， 或$$\\begin{cases}2a-b\\textless{}0 a\\textless{}\\dfrac{3}{2} b\\textgreater3 \\end{cases}$$， 由$$a$$，$$b$$的实际意义为$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，可得$$a=2$$或$$3$$或$$4$$或$$5$$或$$6$$，$$b=1$$或$$2$$， 共有$$5\\times 2=10$$种情况；或$$a=1$$，$$b=4$$或$$5$$或$$6$$，共$$3$$种情况． 又掷两次骰子出现的基本事件共$$6\\times 6=36$$种情况，故所求的概率为$$\\frac{13}{36}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "923", "queId": "ff8080814a39795c014a50b60d584292", "competition_source_list": ["初二下学期其它", "2000年第11届希望杯初一竞赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$是不为$$0$$的整数，并且关于$$x$$的方程$$ax=2{{a}^{3}}-3{{a}^{2}}-5a+4$$有整数解，则$$a$$的值共有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$个 "}], [{"aoVal": "B", "content": "$$3$$个 "}], [{"aoVal": "C", "content": "$$6$$个 "}], [{"aoVal": "D", "content": "$$9$$个 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->一元一次方程->含参一元一次方程->一元一次方程的含参整数解", "课内体系->知识点->方程与不等式->一元一次方程->解一元一次方程->分离法解一元一次方程"], "answer_analysis": ["由原方程可知，$$x=2{{a}^{2}}-3a-5+\\frac{4}{a}$$． 由于$$a$$是不为$$0$$的整数且$$x$$为整数， 所以$$a=1$$，$$-1$$，$$2$$，$$-2$$，$$4$$，$$-4$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1576", "queId": "e29433ead46e43c4bd31b9b91e04b67e", "competition_source_list": ["2002年第13届希望杯初二竞赛第2试第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$${{a}_{i}}=1989+i$$，当$$i$$取$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$100$$时，得到$$100$$个分式$$\\frac{i}{{{a}_{i}}}$$（如$$i=5$$，则$$\\frac{i}{{{a}_{i}}}=\\frac{5}{1989+5}=\\frac{5}{1994}$$），在这$$100$$个分式中，最简分式的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$58$$ "}], [{"aoVal": "C", "content": "$$63$$ "}], [{"aoVal": "D", "content": "$$65$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的运算->最简分式", "课内体系->知识点->式->分式->分式的基础->最简公分母"], "answer_analysis": ["当$$\\frac{i}{{{a}_{i}}}$$为可约分式时，$$i$$与$${{a}_{i}}$$的最大公约数$$(i,{{a}_{i}})=d\\textgreater1$$)， 而$$({{a}_{i}}-i,i)=({{a}_{i}},i)=d\\textgreater1$$． 所以，当$$\\frac{i}{{{a}_{i}}}$$为可约分式时，$$(1989,i)=d\\textgreater1$$． 由$$1989=3\\times 3\\times 13\\times 17$$，知 d的可能取值是$$1989$$的正约数$$3$$，$$9$$，$$13$$，$$17$$，$$39$$，$$51$$，$$117$$，$$153$$，$$221$$，$$663$$，$$1989$$． 又$$d\\textgreater1$$，$$i\\leqslant 100$$． 因此，$$d=3$$，$$9$$，$$13$$，$$17$$，$$39$$，$$51$$． （1）当$$d=3$$时，$$i=3$$，$$6$$，$$9$$，$$\\cdots $$，$$99$$共$$33$$个； （2）当$$d=9$$时，$$i=9$$，$$18$$，$$27$$，$$\\cdots $$，$$99$$共$$11$$个，而这$$11$$个数均与（$$1$$）中的数重合； （3）当$$d=13$$时，$$i=13$$，$$26$$，$$39$$，$$\\cdots $$，$$91$$共$$7$$个，其中$$i=39$$，$$78$$已包含在（$$1$$）中，故符合条件的分数只有$$5$$个； （4）当$$d=17$$时，$$i=17$$，$$34$$，$$51$$，$$68$$，$$85$$，共$$5$$个，其中$$i=51$$已包含在（$$1$$）中，故符合条件的分数只有$$4$$个； （5）当$$d=39$$时，$$i=39$$，$$78$$共$$2$$个，均与（$$1$$）中的数重合； （6）当$$d=51$$时，$$i=51$$已包含在（$$1$$）中． 故可以约分的分数共有$$33+5+4=42$$个，最简分式有$$58$$个． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "97", "queId": "1c6465c07ec14243a9df061efcde60a0", "competition_source_list": ["1997年第8届希望杯初二竞赛第2试第3题"], "difficulty": "0", "qtype": "single_choice", "problem": "下列图形中，不是轴对称图形的是．", "answer_option_list": [[{"aoVal": "A", "content": "直角三角形$$ABC$$ "}], [{"aoVal": "B", "content": "角$$DOE$$ "}], [{"aoVal": "C", "content": "等边三角形$$FGH$$ "}], [{"aoVal": "D", "content": "线段$$MN$$ "}]], "knowledge_point_routes": ["竞赛->知识点->几何变换->对称"], "answer_analysis": ["角$$DOE$$的对称轴是它的角平分线所在的直线； 等边三角形$$FGH$$的对称轴有三条，它们是三边的垂直平分线； 线段$$MN$$的对称轴是$$MN$$的垂直平分线， 只有直角三角形$$ABC$$不是轴对称图形． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1195", "queId": "8aac49074e442d81014e4d1503f0260c", "competition_source_list": ["1995年第6届全国希望杯初一竞赛复赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$P=-\\frac{1}{12345\\times 12346}$$，$$Q=-\\frac{1}{12344\\times 12346}$$，$$R=-\\frac{1}{12344\\times 12345}$$，则$$P$$，$$Q$$，$$R$$的大小关系是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$P\\textgreater Q\\textgreater R$$ "}], [{"aoVal": "B", "content": "$$Q\\textgreater P\\textgreater R$$ "}], [{"aoVal": "C", "content": "$$P\\textgreater R\\textgreater Q$$ "}], [{"aoVal": "D", "content": "$$R\\textgreater Q\\textgreater P$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性"], "answer_analysis": ["因为$$12344\\textless{}12345\\textless{}12346$$， 所以$$12344\\times 12345\\textless{}12344\\times 12346\\textless{}12345\\times 12346$$， 所以$$\\frac{1}{12344\\times 12345}\\textgreater\\frac{1}{12344\\times 12346}\\textgreater\\frac{1}{12345\\times 12346}$$， 所以$$-\\frac{1}{12344\\times 12345}\\textless{}-\\frac{1}{12344\\times 12346}\\textless{}-\\frac{1}{12345\\times 12346}$$， 所以$$R\\textless{}Q\\textless{}P$$．选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1047", "queId": "ff8080814d7978b9014d86d1c88c2566", "competition_source_list": ["1991年第2届全国希望杯初一竞赛初赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "数$$1$$是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "最小整数 "}], [{"aoVal": "B", "content": "最小正数 "}], [{"aoVal": "C", "content": "最小正整数 "}], [{"aoVal": "D", "content": "最小有理数 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类"], "answer_analysis": ["整数无最小数，排除$$\\text{A}$$； 正数无最小数，排除$$\\text{B}$$； $$0$$是最小正整数，选择C； 有理数无最小数，排除$$\\text{D}$$． 考试的时候，当时规定$$0$$不是自然数，后来$$0$$也规定为自然数了． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "202", "queId": "5d59b2db49d54e14862aaea4b331cc76", "competition_source_list": ["初一下学期其它第18题", "2016~2017学年陕西西安高新区西安高新第一中学初二上学期期末第24题7分", "2008年第19届希望杯初二竞赛第2试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "关于$$x$$，$$y$$的方程组$$\\begin{cases}x+ay+1=0 bx-2y+1=0 \\end{cases}$$有无数组解，则$$a$$，$$b$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a=0$$，$$b=0$$ "}], [{"aoVal": "B", "content": "$$a=-2$$，$$b=1$$ "}], [{"aoVal": "C", "content": "$$a=2$$，$$b=-1$$ "}], [{"aoVal": "D", "content": "$$a=2$$，$$b=1$$ ~ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->二元一次方程组的解", "课内体系->知识点->方程与不等式->二元一次方程（组）->含参二元一次方程组->二元一次方程组解的情况", "课内体系->能力->运算能力"], "answer_analysis": ["方程组有无数组解于是有$$\\frac{{{a}_{1}}}{{{a}_{2}}}=\\frac{{{b}_{1}}}{{{b}_{2}}}=\\frac{{{c}_{1}}}{{{c}_{2}}}$$， 于是$$\\frac{1}{b}=\\frac{a}{-2}=\\frac{1}{1}$$， ∴$$a=-2$$，$$b=1$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "915", "queId": "6a1add00bcea47c4ad767000862ea4d5", "competition_source_list": ["1999年第10届希望杯初一竞赛第6题", "2018~2019学年安徽宿州灵璧县尹集中学初一上学期期中第16题3分", "2016~2017学年9月吉林长春农安县农安县第一中学初一上学期月考第6题3分", "2019~2020学年9月江苏连云港海州区连云港市新海实验中学（苍梧校区）初一上学期周测B卷第5题3分", "2017~2018学年广东深圳福田区深圳外国语学校初一上学期期中第3题3分", "2015~2016学年浙江杭州上城区初一上学期期中第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$是最小的自然数，$$b$$是最大的负整数，$$c$$是绝对值最小的有理数，则$$a-b+c$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类", "课内体系->知识点->数->有理数->有理数基础运算->有理数减法->有理数加减混合运算", "课内体系->能力->运算能力"], "answer_analysis": ["因为$$a$$是最小的自然数，所以$$a=0$$， 因为$$b$$是最大的负整数，所以$$b=-1$$， 因为$$c$$是绝对值最小的有理数，所以$$c=0$$， 所以$$a-b+c=0-\\left(-1\\right)+0=1$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "315", "queId": "625364ec59ba44dc858be988151b89bf", "competition_source_list": ["1997年第8届全国希望杯初一竞赛复赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "有四个关于$$x$$的方程： ①$$x-2=-1$$；②$$\\left( x-2 \\right)+\\left( x-1 \\right)=-1+\\left( x-1 \\right)$$；③$$x=0$$；④$$x-2+\\frac{1}{x-1}=-1+\\frac{1}{x-1}$$． 其中同解的两个方程是．", "answer_option_list": [[{"aoVal": "A", "content": "①与② "}], [{"aoVal": "B", "content": "①与③ "}], [{"aoVal": "C", "content": "①与④ "}], [{"aoVal": "D", "content": "②与④ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->分式方程->分式方程的解与解分式方程->分式方程的解"], "answer_analysis": ["方程①的解$$x=1$$，将$$x=1$$代入方程②，方程②成立， ∴$$x=1$$也是方程②的解． 方程①和②是同解方程，而①与③显然不同解； ①的解代入④，④无意义． ∴$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$都不正确，只有$$\\text{A}$$正确． 所以选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "208", "queId": "58bd599fad474e1aa76ef736a98cb1b6", "competition_source_list": ["2007年竞赛第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "方程$${{x}^{3}}+6{{x}^{2}}+5x={{y}^{3}}-y+2$$的整数解$$\\left( x,y \\right)$$的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "无穷多 "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->二次方程->特殊方程的解法"], "answer_analysis": ["原方程可化为$$x\\left( x+1 \\right)\\left( x+2 \\right)+3\\left( {{x}^{2}}+x \\right)=y\\left( y-1 \\right)\\left( y+1 \\right)+2$$，因为三个连续整数的乘积是$$3$$的倍数，所以上式左边是$$3$$的倍数，而右边除以$$3$$余$$2$$，这是不可能的，所以，原方程无整数解． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "383", "queId": "c76596654af749dfbae0743c97e5b2de", "competition_source_list": ["1994年第11届全国初中数学联赛竞赛第6题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "若方程$$\\sqrt{x-p}=x$$有两个不相等的实根，则实数$$p$$的取值范围是（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$p\\leqslant 0$$ "}], [{"aoVal": "B", "content": "$$p\\textless{}\\frac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$0\\leqslant p\\textless{}\\frac{1}{4}$$ "}], [{"aoVal": "D", "content": "$$p\\geqslant \\frac{1}{4}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->一元二次方程根的判别式", "课内体系->知识点->式->二次根式->二次根式的基础->二次根式有意义的条件"], "answer_analysis": ["本题可以用特殊值法，排除不合理的选项： 取$$p=-1$$，代入原方程得$$\\sqrt{x+1}=x$$，即$${{x}^{2}}-x-1=0$$，此时，方程有一个负根，于是可排除$$\\text{A}$$，$$\\text{B}$$， 取$$p=1$$，代入原方程得$${{x}^{2}}-x+1=0$$，无解，故排除$$\\text{D}$$， 因此，应选$$\\text{C}$$． ", "原方程可化为：$${{x}^{2}}+x+p=0$$，$$x\\geqslant 0$$，
\n∵方程有两个不相等的实根，
\n∴$$\\Delta =1-4p>0$$，
\n解得$$p < \\frac{1}{4}$$，
\n设方程两根为$${{x}_{1}}$$，$${{x}_{2}}$$，则必有$${{x}_{1}}\\geqslant 0$$，$${{x}_{2}}\\geqslant 0$$，于是得$$0\\leqslant p < \\frac{1}{4}$$．
\n故选$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1442", "queId": "bcb585d78f0343d2a4fafeb820ca1ce7", "competition_source_list": ["2015年竞赛第30题"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{2011 Think Cup G5} 艾尔、比和卡尔以恒定的速度刷墙，他们正在一起刷一所房子．艾尔和比一起可以在$$12$$小时内完成这项工作；艾尔和卡尔可以在$$15$$分钟内完成，比和卡尔可以在$$20$$分钟内完成．三个人一起粉刷房子要花多少小时？．", "answer_option_list": [[{"aoVal": "A", "content": "$$8.5$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$10.5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->十字相乘法->二次项系数为±1的十字相乘", "美国AMC8->Knowledge Point->Word Problem->Work Word Problems->Cooperative Work Word Problems"], "answer_analysis": ["设$$a$$、$$b$$、$$c$$是每个人$$1$$小时内可��粉刷的房子的比例，则$$a+b=\\frac{1}{12}$$，$$a+c=\\frac{1}{15}$$，$$b+c=\\frac{1}{20}$$．把$$3$$个等式加起来我们得到$$2a+2b+2c=\\frac{1}{5}$$，相等地，$$a+b+c=\\frac{1}{10}$$．因此，他们在$$1$$小时内刷完$$\\frac{1}{10}$$的房子，$$10$$小时内刷完整个房子． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "782", "queId": "45a076d791a44db09c2021fd58222887", "competition_source_list": ["2014年第25届全国希望杯初一竞赛复赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若有理数$$a$$、$$b$$、$$c$$两两不等，则$$\\frac{a-b}{b-c}$$，$$\\frac{b-c}{c-a}$$，$$\\frac{c-a}{a-b}$$中负数的个数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->正数和负数->正数、负数定义", "课内体系->知识点->数->有理数->有理数基础运算->有理数减法->确定几个数和或差的符号", "课内体系->能力->运算能力"], "answer_analysis": ["方法一： 不妨设$$a\\textgreater b\\textgreater c$$， 则$$\\frac{a-b}{b-c}\\textgreater0$$，$$\\frac{b-c}{c-a}\\textless{}0$$，$$\\frac{c-a}{a-b}\\textless{}0$$， 所以有$$2$$个负数． 方法二： 取$$a=0$$，$$b=1$$，$$c=-2$$， 则$$\\frac{a-b}{b-c}=\\frac{-1}{3}=-\\frac{1}{3}$$，$$\\frac{b-c}{c-a}=\\frac{3}{-2}=-\\frac{3}{2}$$，$$\\frac{c-a}{a-b}=\\frac{-2}{-1}=2$$， 所以有$$2$$个负数． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "272", "queId": "2fb443408f034fcb9719bcb21a76c923", "competition_source_list": ["2019~2020学年3月广东深圳福田区深圳明德实验学校初三下学期月考第3题", "2018~2019学年5月广东深圳罗湖区深圳中学初中部初一上学期周测A卷竞赛班初一第23次第1题3分", "2019~2020学年4月四川成都金牛区成都七中万达学校初三下学期周测B卷第3题3分", "2020年广东中山市南朗镇中山纪念中学初三中考一模第3题3分", "2019年四川成都邛崃市初三中考二模第4题3分", "2019年四川成都青羊区初三中考二模第4题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "港珠澳大桥东起香港国际机场附近的香港口岸人工岛，向西横跨伶仃洋海域后连接珠海和澳门人工岛，止于珠海洪湾，全长$$55$$千米，设计时速$$100$$千米/小时，工程项目总投资额$$1269$$亿元，用科学记数法表示$$1269$$亿元为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1269\\times {{10}^{8}}$$ "}], [{"aoVal": "B", "content": "$$1.269\\times {{10}^{8}}$$ "}], [{"aoVal": "C", "content": "$$1.269\\times {{10}^{10}}$$ "}], [{"aoVal": "D", "content": "$$1.269\\times {{10}^{11}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->科学记数法->科学记数法：表示较大的数"], "answer_analysis": ["$$1269$$亿元$$=1269 000 00=1.269\\times {{10}^{11}}$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1343", "queId": "dbd6d50d7327435cb9d06170422bcbea", "competition_source_list": ["2015年第32届全国全国初中数学联赛竞赛A卷第5题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知实数$$x$$，$$y$$满足关系式$$xy-x-y=1$$，则$${{x}^{2}}+{{y}^{2}}$$的最小值为（~ ~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$3-2\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$6-4\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$6+4\\sqrt{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元二次方程->解一元二次方程->换元法求一元二次方程的根", "课内体系->方法->配方法"], "answer_analysis": ["设$$x+y=t$$， 则$$xy=x+y+1=t+1$$， ∴$$x$$，$$y$$是关于$$m$$的一元二次方程$${{m}^{2}}-tm+t+1=0$$的两个实数根， ∴$$\\Delta ={{t}^{2}}-4(t+1)\\geqslant 0$$， 解得$$t\\geqslant 2+2\\sqrt{2}$$或$$t\\leqslant 2-2\\sqrt{2}$$． ∵$${{x}^{2}}+{{y}^{2}}={{(x+y)}^{2}}-2xy={{t}^{2}}-2(t+1)={{(t-1)}^{2}}-3$$， ∴当$$t=2-2\\sqrt{2}$$（即$$x=y=1-\\sqrt{2}$$）时，$${{x}^{2}}+{{y}^{2}}$$取得最小值，最小值为$$6-4\\sqrt{2}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1071", "queId": "ff8080814d7978b9014d897647403530", "competition_source_list": ["2000年第11届希望杯初二竞赛第2试第7题", "初一其它"], "difficulty": "3", "qtype": "single_choice", "problem": "三元方程$$x+y+z=1999$$的非负整数解的个数有（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$20001999$$个 "}], [{"aoVal": "B", "content": "$$19992000$$个 "}], [{"aoVal": "C", "content": "$$2001000$$个 "}], [{"aoVal": "D", "content": "$$2001999$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->其他方程->三元一次方程组->三元一次方程组的定义", "课内体系->知识点->方程与不等式->其他方程->三元一次方程组->解三元一次方程组"], "answer_analysis": ["当$$x=0$$时，$$y+z=1999$$，$$y$$分别取$$0$$，$$1$$，$$2$$，\\ldots，$$1999$$时，$$z$$取$$1999$$，$$1998$$，\\ldots，$$0$$，有$$2000$$个整数解； 当$$x=1$$时，$$y+z=1998$$，有$$1999$$个整数解， 当$$x=2$$时，$$y+z=1997$$，有$$1998$$个整数解； \\ldots 当$$x=1999$$时，$$y+z=1$$，只有$$1$$组整数解．故非负整数解共有$$2000+1999+1998+\\cdot \\cdot \\cdot +3+2+1=2001000$$（个）． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1114", "queId": "ff8080814d9efd56014daa7506d80a99", "competition_source_list": ["1993年第4届全国希望杯初一竞赛初赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲的$$6$$张卡片上分别写有$$-4$$，$$-1$$，$$-2.5$$，$$-0.01$$，$$-3\\frac{3}{4}$$，$$-15$$，乙的$$6$$张卡片上分别写有$$-5$$，$$-1$$，$$0.1$$，$$-0.001$$，$$-8$$，$$-12\\frac{1}{2}$$，则乙的卡片上的最小数$$a$$与甲的卡片上的最大数$$b$$的比$$\\frac{a}{b}$$的值等于（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1250$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$0.1$$ "}], [{"aoVal": "D", "content": "$$800$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数除法运算"], "answer_analysis": ["观察知，乙的卡片上的最小数$$a=-12\\frac{1}{2}$$，甲的卡片上的最大数$$b=-0.01$$． 所以$$\\frac{a}{b}=\\frac{-12\\frac{1}{2}}{-0.01}=1250$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1475", "queId": "b41001ece7a64317bd596b681aa39fae", "competition_source_list": ["2013年四川成都成华区成都石室中学初中学校初三自主招生第4题5分", "2011年竞赛第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$\\angle A$$，$$\\angle B$$是两个锐角，且满足$${{\\sin }^{2}}A+{{\\cos }^{2}}B=\\frac{5}{4}t$$，$${{\\cos }^{2}}A+{{\\sin }^{2}}B=\\frac{3}{4}{{t}^{2}}$$，则实数$$t$$所有可能的值的和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{8}{3}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{5}{3}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$\\frac{11}{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->锐角三角函数及解直角三角形->解直角三角形", "课内体系->知识点->方程与不等式->一元二次方程->一元二次方程的基础->一元二次方程的根", "课内体系->知识点->方程与不等式->一元二次方程->解一元二次方程->因式分解法求一元二次方程的根", "课内体系->能力->推理论证能力", "课内体系->思想->方程思想"], "answer_analysis": ["根据已知得：$${{\\sin }^{2}}A+{{\\cos }^{2}}B+{{\\cos }^{2}}A+{{\\sin }^{2}}B=\\frac{3}{4}{{t}^{2}}+\\frac{5}{4}t$$，即$$2=\\frac{3}{4}{{t}^{2}}+\\frac{5}{4}t$$， ∴$$3{{t}^{2}}+5t-8=0$$，解得$${{t}_{1}}=1$$，$${{t}_{2}}=-\\frac{8}{3}$$，又∵$${{\\sin }^{2}}A+{{\\cos }^{2}}B=\\frac{5}{4}t\\textgreater0$$，即$$t\\textgreater0$$， ∴$$t=1$$， 故$$t$$所有可能的值的和为$$1$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "89", "queId": "7da23e42a2694446a2809d2150664564", "competition_source_list": ["2009年第14届华杯赛初一竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知关于$$x$$，$$y$$的方程组$$\\begin{cases}2x+3y=-5 3x+7y=m \\end{cases}$$，当$$-20 ~\\textless{} ~m ~\\textless{} ~-10$$时有整数解，则$${{x}^{2}}+xy+{{y}^{2}}$$的值等于~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->一次方程->方程组", "课内体系->知识点->式->因式分解->十字相乘法->二次项系数为±1的十字相乘"], "answer_analysis": ["解方程组$$\\begin{cases}2x+3y=-5 3x+7y=m \\end{cases}$$，解得：$$\\begin{cases}x=-\\dfrac{3m+35}{5} y=\\dfrac{2m+15}{5} \\end{cases}$$， 当$$-20\\textless{}m\\textless{}-10$$时有整数解， 即$$x$$，$$y$$的值都是整数，则只有当$$m=-15$$时，$$x$$、$$y$$的值是整数， 当$$m=-15$$时，$$x=2$$，$$y=-3$$， 则$${{x}^{2}}+xy+{{y}^{2}}={{2}^{2}}+2\\times (-3)+{{(-3)}^{2}}=7$$． 故答案为：$$7$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "366", "queId": "34dd04c5b03a48d1a5573349480dd283", "competition_source_list": ["2009年第20届希望杯初二竞赛第2试第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知多项式$${{p}_{1}}\\left( x \\right)=2{{x}^{2}}-5x+1$$和$${{p}_{2}}\\left( x \\right)=3x-4$$，则$${{p}_{1}}\\left( x \\right)\\times {{p}_{2}}\\left( x \\right)$$的最简结果为．", "answer_option_list": [[{"aoVal": "A", "content": "$$6{{x}^{3}}-23{{x}^{2}}+23x-4$$ "}], [{"aoVal": "B", "content": "$$6{{x}^{3}}+23{{x}^{2}}-23x-4$$ "}], [{"aoVal": "C", "content": "$$6{{x}^{3}}-23{{x}^{2}}-23x+4$$ "}], [{"aoVal": "D", "content": "$$6{{x}^{3}}+23{{x}^{2}}+23x+4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->整式的乘除运算"], "answer_analysis": ["$${{p}_{1}}\\left( x \\right)\\times {{p}_{2}}\\left( x \\right)=\\left( 2{{x}^{2}}-5x+1 \\right)\\left( 3x-4 \\right)$$ $$=6{{x}^{3}}-23{{x}^{2}}+23x-4$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "8", "queId": "0dee35eb4e4e43f5b283e594ab96a594", "competition_source_list": ["2019~2020学年9月江苏无锡梁溪区东林中学初一上学期月考第8题3分", "1996年第7届全国希望杯初一竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$$x\\textless-2$$，那么化简$$\\left\\textbar{} 1-\\left\\textbar{} 1+x \\right\\textbar{} \\right\\textbar$$的结果为． If $$x\\textless-2$$, the result of simplifying $$\\left\\textbar{} 1-\\left\\textbar{} 1+x \\right\\textbar{} \\right\\textbar$$ is ．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2-x$$ "}], [{"aoVal": "B", "content": "$$2+x$$ "}], [{"aoVal": "C", "content": "$$x$$ "}], [{"aoVal": "D", "content": "$$-x$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->已知范围化简绝对值", "课内体系->能力->运算能力"], "answer_analysis": ["∵$$x\\textless{}-2$$， ∴$$1+x\\textless{}0$$， ∴$$\\textbar1+x\\textbar=-x-1$$， ∴$$\\textbar1-\\textbar1+x\\textbar\\textbar=\\textbar1-(-x-1)\\textbar$$ $$=\\textbar2+x\\textbar$$ $$=-2-x$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "263", "queId": "38c6675b8e324f4bbae15b43a0b3d036", "competition_source_list": ["2002年第13届希望杯初二竞赛第2试第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个长方体的棱长都是正整数，体积是$$2002$$，若对应棱长相等的长方体算作同一种长方体，那么这样的长方体．", "answer_option_list": [[{"aoVal": "A", "content": "有$$6$$种 "}], [{"aoVal": "B", "content": "有$$12$$种 "}], [{"aoVal": "C", "content": "有$$14$$种 "}], [{"aoVal": "D", "content": "多于$$16$$种 "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->几何图形->立体图形与平面图形->点线面体", "课内体系->能力->运算能力", "竞赛->知识点->数论->整除->因数与倍数"], "answer_analysis": ["不失一般性，可设长方体在同一顶点处的三条棱长分别为$$a\\leqslant b\\leqslant c$$， 又$$2002=1\\times 2\\times 7\\times 11\\times 13$$． ①$$a=b=1$$时，$$c=2002$$，这样的长方体只有$$1$$种； ②当$$a=1$$，$$b$$是质数$$2$$，$$7$$，$$11$$，$$13$$中的一个时，这样的长方体共有$$4$$种； ③当$$a=1$$，$$b$$是$$2$$，$$7$$，$$11$$，$$13$$中的两个数的乘积时，则有$$b=2\\times 7$$，$$b=2\\times 11$$，$$b=2\\times 13$$共$$3$$种可能（b不能是$$7\\times 11$$或$$7\\times 13$$或$$11\\times 13$$）； ④当$$a=2$$时，$$b$$是$$7$$，$$11$$，$$13$$中的一个时，这样的长方体有$$3$$种； ⑤当$$a=7$$时，$$b$$是$$11$$或$$13$$中的一个时，这样的长方体有$$2$$种； ⑥当$$a=11$$时，$$b$$只能取$$13$$，这样的长方体只有$$1$$种． 综上所述，符合条件的长方体有$$14$$种． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "306", "queId": "3d8dd65473a6490d90d7c509f7315610", "competition_source_list": ["2017年安徽芜湖镜湖区芜湖市第一中学初三自主招生第6题6分", "2004年竞赛第1题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知实数$$a\\ne b$$，且满足$${{\\left( a+1 \\right)}^{2}}=3-3\\left( a+1 \\right)$$，$$3\\left( b+1 \\right)=3-{{\\left( b+1 \\right)}^{2}}$$，则$$b\\sqrt{\\frac{b}{a}}+a\\sqrt{\\frac{a}{b}}$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$23$$ "}], [{"aoVal": "B", "content": "$$-23$$ "}], [{"aoVal": "C", "content": "$$-2$$ "}], [{"aoVal": "D", "content": "$$-13$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元二次方程->根与系数的关系->一元二次方程根与系数的关系", "课内体系->知识点->方程与不等式->一元二次方程->一元二次方程的基础->一元二次方程的根"], "answer_analysis": ["容易知道$$a,b$$是方程$${{\\left( x+1 \\right)}^{2}}=3-3\\left( x+1 \\right)$$的两个根， 整理得$${{x}^{2}}+5x+1=0$$， 于是有$$a+b=-5,ab=1$$，于是$$a\\textless{}0,b\\textless{}0$$， $$b\\sqrt{\\frac{b}{a}}+a\\sqrt{\\frac{a}{b}}$$， $$=b\\sqrt{\\frac{{{b}^{2}}}{ab}}+a\\sqrt{\\frac{{{a}^{2}}}{ab}}$$， $$=\\frac{b\\left\\textbar{} b \\right\\textbar+a\\left\\textbar{} a \\right\\textbar}{\\sqrt{ab}}$$， $$=-\\frac{\\left( {{a}^{2}}+{{b}^{2}} \\right)}{\\sqrt{ab}}$$， $$=-\\frac{{{\\left( a+b \\right)}^{2}}-2ab}{\\sqrt{ab}}$$， $$=-23$$， 故答案为$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1523", "queId": "cf81a689f0e6482b96bc6b8fa4f506e9", "competition_source_list": ["2007年竞赛第1题6分", "2017~2018学年江苏无锡滨湖区无锡外国语学校初一下学期期中第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "\\textbf{【七年级新思维$$-93$$页$$-18$$题】} 方程组$$\\begin{cases}\\left\\textbar{} x \\right\\textbar+y=12 x+\\left\\textbar{} y \\right\\textbar=6 \\end{cases}$$的解的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["知识标签->知识点->方程与不等式->二元一次方程（组）->二元一次方程组->二元一次方程组的解", "知识标签->知识点->方程与不等式->二元一次方程（组）->二元一次方程组->加减消元法解二元一次方程组", "知识标签->题型->方程与不等式->二元一次方程（组）->解二元一次方程组->题型：解含绝对值的方程组", "知识标签->数学思想->分类讨论思想", "知识标签->学习能力->运算能力"], "answer_analysis": ["①当$$x\\textgreater0$$，$$y\\textgreater0$$时，原方程组为$$\\begin{cases}x+y=12 x+y=6 \\end{cases}$$显然无解； ②当$$x ~\\textless{} ~0$$，$$y\\textgreater0$$时，原方程组为$$\\begin{cases}-x+y=12 x+y=6 \\end{cases}$$，解得$$\\begin{cases}x=-3 y=9 \\end{cases}$$； ③当$$x\\textgreater0$$，$$y ~\\textless{} ~0$$时，原方程组为$$\\begin{cases}x+y=12 x-y=6 \\end{cases}$$，解得$$\\begin{cases}x=9 y=3 \\end{cases}$$，舍去； ④当$$x ~\\textless{} ~0$$，$$y ~\\textless{} ~0$$时，原方程组为$$\\begin{cases}-x+y=12 x-y=6 \\end{cases}$$显然无解； 综上，只有$$1$$组解．故选$$\\text{A}$$． 若$$x\\geqslant 0$$，则$$\\begin{cases}x+y=12 x+\\left\\textbar{} y \\right\\textbar=6 \\end{cases}$$，于是$$\\left\\textbar{} y \\right\\textbar-y=-6$$，显然不可能． 若$$x ~\\textless{} ~0$$，则$$\\begin{cases}-x+y=12 x+\\left\\textbar{} y \\right\\textbar=6 \\end{cases}$$， 于是$$\\left\\textbar{} y \\right\\textbar+y=18$$，解得$$y=9$$，进而求得$$x=-3$$． 所以，原方程组的解为$$\\begin{cases}x=-3 y=9 \\end{cases}$$，只有$$1$$个解． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "40", "queId": "030080857e2e4df6a00941b9bc290295", "competition_source_list": ["2014年第31届全国全国初中数学联赛竞赛第5题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$[t]$$表示不超过实数$$t$$的最大整数，令$$ {t }=t-[t]$$．已知实数$$x$$满足$${{x}^{3}}+\\frac{1}{{{x}^{3}}}=18$$，则$$ {x }+\\left { \\frac{1}{x} \\right }=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$3-\\sqrt{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}(3-\\sqrt{5})$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->综合与实践->规律探究与程序框图->定义新运算", "课内体系->知识点->式->整式的乘除->乘法公式->和与差的立方公式", "课内体系->能力->运算能力"], "answer_analysis": ["设$$x+\\frac{1}{x}=a$$， 则$${{x}^{3}}+\\frac{1}{{{x}^{3}}}=\\left( x+\\frac{1}{x} \\right)\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}}-1 \\right)=\\left( x+\\frac{1}{x} \\right)\\left[ {{\\left( x+\\frac{1}{x} \\right)}^{2}}-3 \\right]=a({{a}^{2}}-3)$$， 所以$$a({{a}^{2}}-3)=18$$， 因式分解得$$(a-3)({{a}^{2}}+3a+6)=0$$， 所以$$a=3$$． 由$$x+\\frac{1}{x}=3$$解得$$x=\\frac{1}{2}(3\\pm \\sqrt{5})$$， 显然$$0\\textless{} {x }\\textless{}1$$，$$0\\textless{}\\left { \\frac{1}{x} \\right }\\textless{}1$$， 所以$$ {x }+\\left { \\frac{1}{x} \\right }=1$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "863", "queId": "acef7efed8d64a63a6d96aa88bb2f162", "competition_source_list": ["2015年第32届全国全国初中数学联赛竞赛A卷第1题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "设实数$$a$$，$$b$$，$$c$$满足：$$a+b+c=3$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=4$$，则$$\\frac{{{a}^{2}}+{{b}^{2}}}{2-c}+\\frac{{{b}^{2}}+{{c}^{2}}}{2-a}+\\frac{{{c}^{2}}+{{a}^{2}}}{2-b}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->乘法公式->平方差公式的计算", "课内体系->知识点->式->分式->分式的运算->分式加减、乘除、乘方混合运算"], "answer_analysis": ["∵$$a+b+c=3$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=4$$， ∴$$\\frac{{{a}^{2}}+{{b}^{2}}}{2-c}+\\frac{{{b}^{2}}+{{c}^{2}}}{2-a}+\\frac{{{c}^{2}}+{{a}^{2}}}{2-b}$$ $$=\\frac{4-{{c}^{2}}}{2-c}+\\frac{4-{{a}^{2}}}{2-a}+\\frac{4-{{b}^{2}}}{2-b}$$ $$=2+c+2+a+2+b$$ $$=a+b+c+6$$ $$=3+6$$ $$=9$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "292", "queId": "4b360dedf10548b78e64b953189f862e", "competition_source_list": ["2012年第23届全国希望杯初一竞赛复赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "身高两两不同的$$30$$个学生向老师站成一排．其中恰有$$11$$个学生高于自己左侧相邻的同学．那么高于自己右侧相邻学生的学生有（~ ）人．", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$19$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->容斥原理", "课内体系->知识点->数->有理数->有理数比较大小"], "answer_analysis": ["$$30$$个学生，从左向右数有$$29$$个相邻组$$({{a}_{i}},{{a}_{i+1}})$$，$$i=1$$，$$2$$，$$\\cdots $$，$$29$$． 因为已知$${{a}_{i}}\\textless{}{{a}_{i+1}}$$的有$$11$$对， 所以$${{a}_{i}}\\textgreater{{a}_{i+1}}$$的有$$29-11=18$$（对）． ", "$$30$$人有$$29$$对相邻的同学
\n其中有$$11$$对右侧高
\n则有$$18$$对左侧高
"], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1337", "queId": "8aac50a7511483070151193811840fad", "competition_source_list": ["2015年第26届全国希望杯初三竞赛初赛（特）第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知函数$$y=\\frac{3x+m}{1-2x}$$，当$$x\\textless{}\\frac{1}{2}$$时，$$y$$随$$x$$的增大而减小，则实数$$m$$的取值范围是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{3}{2}\\textless{}m\\textless{}0$$ "}], [{"aoVal": "B", "content": "$$m\\textless{}-\\frac{3}{2}$$ "}], [{"aoVal": "C", "content": "$$m\\textgreater0$$ "}], [{"aoVal": "D", "content": "$$m\\textless{}-\\frac{3}{2}$$或$$m\\textgreater0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->函数概念和图象->函数的相关概念->探究函数性质"], "answer_analysis": ["令$${{x}_{2}}\\textless{}{{x}_{1}}\\textless{}\\frac{1}{2}$$， 当$$x={{x}_{1}}$$时，$$y=\\frac{3{{x}_{1}}+m}{1-2{{x}_{1}}}$$， 当$$x={{x}_{2}}$$时，$$y=\\frac{3{{x}_{2}}+m}{1-2{{x}_{2}}}$$， ∴$$\\frac{3{{x}_{1}}+m}{1-2{{x}_{1}}}-\\frac{3{{x}_{2}}+m}{1-2{{x}_{2}}}=\\frac{(3{{x}_{1}}+m)(1-2{{x}_{2}})-(1-2{{x}_{1}})(3{{x}_{2}}+m)}{(1-2{{x}_{2}})(1-2{{x}_{1}})}=\\frac{(3+2m)({{x}_{1}}-{{x}_{2}})}{(1-2{{x}_{2}})(1-2{{x}_{1}})}$$， ∵$$y$$随$$x$$的增大而减小， ∴$$\\frac{3{{x}_{1}}+m}{1-2{{x}_{1}}}-\\frac{3{{x}_{2}}+m}{1-2{{x}_{2}}}\\textless{}0$$， 即$$\\frac{(3+2m)({{x}_{1}}-{{x}_{2}})}{(1-2{{x}_{2}})(1-2{{x}_{1}})}\\textless{}0$$， 解得$$m\\textless{}-\\frac{3}{2}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1586", "queId": "e2c2a61c4e9c481cbd73799778d30a86", "competition_source_list": ["1990年第1届希望杯初二竞赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$x=1$$时，$${{a}_{0}}{{x}^{10}}-{{a}_{1}}{{x}^{9}}+{{a}_{0}}{{x}^{8}}-{{a}_{1}}{{x}^{7}}-{{a}_{1}}{{x}^{6}}+{{a}_{1}}{{x}^{5}}-{{a}_{0}}{{x}^{4}}+{{a}_{1}}{{x}^{3}}-{{a}_{0}}{{x}^{2}}+{{a}_{1}}x$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$${{a}_{0}}$$ "}], [{"aoVal": "C", "content": "$${{a}_{1}}$$ "}], [{"aoVal": "D", "content": "$${{a}_{0}}-{{a}_{1}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->整式的乘除运算"], "answer_analysis": ["以$$x=1$$代入，得 $${{a}_{0}}-{{a}_{1}}+{{a}_{0}}-{{a}_{1}}-{{a}_{1}}+{{a}_{1}}-{{a}_{0}}+{{a}_{1}}-{{a}_{0}}+{{a}_{1}}$$ $$=2{{a}_{0}}-3{{a}_{1}}+3{{a}_{1}}-2{{a}_{0}}$$ $$=0$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "660", "queId": "9e6b3fbf2e5044aaa014e9e0936fa159", "competition_source_list": ["2008年第19届希望杯初二竞赛第2试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "在平面直角坐标系内，有等腰三角形$$AOB$$，$$O$$是坐标原点，点$$A$$的坐标是$$(a,b)$$，底边$$AB$$的中线在第一、三象限的角平分线上，则点$$B$$的坐标是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(b,a)$$ "}], [{"aoVal": "B", "content": "$$(-a,-b)$$ "}], [{"aoVal": "C", "content": "$$(a,-b)$$ "}], [{"aoVal": "D", "content": "$$(-a,b)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->几何变换->对称"], "answer_analysis": ["因为$$\\triangle OAB$$是等腰三角形，$$O$$为顶点， 所以$$OA=OB$$，又$$AB$$为底边， 所以$$AB$$垂直于中线，即垂直于直线$$y=x$$， 不防设$$a=2$$，$$b=1$$，画图可知$$A(2,1)$$关于$$y=x$$的对称点为$$(1,2)$$． 点$$A(a,b)$$关于$$y=x$$的对称点为$$(b,a)$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1150", "queId": "cdb23b6731324ecbaa14c7b3b33cad46", "competition_source_list": ["1999年第10届希望杯初二竞赛第1试第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "两个数$$a$$，$$b$$，且$$a\\textless b$$，把$$a$$到$$b$$的所有数记做$$[a, b]$$，例如$$1$$到$$4$$的所有数记做$$[1,4]$$，如果$$5 \\leqslant m \\leqslant 15$$，$$20 \\leqslant n \\leqslant 30$$，那么$$\\frac{m}{n}$$的一切值包含在内．", "answer_option_list": [[{"aoVal": "A", "content": "$$[5,30]$$ "}], [{"aoVal": "B", "content": "$$\\left[\\frac{1}{4}, \\frac{3}{4}\\right]$$ "}], [{"aoVal": "C", "content": "$$\\left[\\frac{1}{6}, \\frac{2}{3}\\right]$$ "}], [{"aoVal": "D", "content": "$$\\left[\\frac{1}{6}, \\frac{7}{8}\\right]$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性", "课内体系->能力->数据处理能力"], "answer_analysis": ["因为$$5 \\leqslant m \\leqslant 15$$，$$20 \\leqslant n \\leqslant 30$$， 所以当$$m=5$$，$$n=30$$时，$$\\frac{m}{n}=\\frac{1}{6}$$是$$\\frac{m}{n}$$中的最小值， 当$$m=15$$，$$n=20$$时，$$\\frac{m}{n}=\\frac{3}{4}\\left(\\textless\\frac{7}{8}\\right)$$是$$\\frac{m}{n}$$中的最大值， 所以$$\\frac{m}{n}$$中的所有值都在$$\\left[\\frac{1}{6}, \\frac{7}{8}\\right]$$内． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "868", "queId": "cd0c0ae91f9c4efd8f847bc63376cf44", "competition_source_list": ["2018~2019学年5月广东深圳罗湖区深圳中学初中部初一上学期周测A卷竞赛班初一第23次第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知方程组$$\\begin{cases}{{a}_{1}}x+{{b}_{1}}y={{c}_{1}} {{a}_{2}}x+{{b}_{2}}y={{c}_{1}} \\end{cases}$$的解是$$\\begin{cases}x=6 y=8 \\end{cases}$$，则方程组$$\\begin{cases}3{{a}_{1}}x+4{{b}_{1}}y=5{{c}_{1}} 3{{a}_{2}}x+4{{b}_{2}}y=5{{c}_{1}} \\end{cases}$$的解是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\begin{cases}x=6 y=8 \\end{cases}$$ "}], [{"aoVal": "B", "content": "$$\\begin{cases}x=8 y=6 \\end{cases}$$ "}], [{"aoVal": "C", "content": "$$\\begin{cases}x=8 y=10 \\end{cases}$$ "}], [{"aoVal": "D", "content": "$$\\begin{cases}x=10 y=10 \\end{cases}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->二元一次方程组的解", "课内体系->知识点->方程与不等式->二元一次方程（组）->二元一次方程->由二元一次方程的解求参数的值", "课内体系->能力->运算能力"], "answer_analysis": ["方程组$$\\begin{cases}3{{a}_{1}}x+4{{b}_{1}}y=5{{c}_{1}} 3{{a}_{2}}x+4{{b}_{2}}y=5{{c}_{1}} \\end{cases}$$可化为： $$\\begin{cases}{{a}_{1}}\\cdot \\dfrac{3}{5}x+{{b}_{1}}\\cdot \\dfrac{4}{5}y={{c}_{1}} {{a}_{2}}\\cdot \\dfrac{3}{5}x+{{b}_{2}}\\cdot \\dfrac{4}{5}y={{c}_{1}} \\end{cases}$$， ∵方程组$$\\begin{cases}{{a}_{1}}x+{{b}_{1}}y={{c}_{1}} {{a}_{2}}x+{{b}_{2}}y={{c}_{1}} \\end{cases}$$的解是$$\\begin{cases}x=6 y=8 \\end{cases}$$， ∴$$\\begin{cases}\\dfrac{3}{5}x=6 \\dfrac{4}{5}y=8 \\end{cases}$$， ∴$$\\begin{cases}x=10 y=10 \\end{cases}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "984", "queId": "ff8080814cfa9b24014cfe9600870a3b", "competition_source_list": ["上海自主招生近5年（2015-2019）真题题型分类汇编第15题", "2017~2018学年5月四川成都天府新区 成都市实验外国语学校（西区）初二下学期月考第10题3分", "2019~2020学年3月浙江杭州滨江区杭州二中白马湖学校初一下学期周测C卷第9题", "初二其它", "1996年第7届希望杯初二竞赛第3题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$${{x}^{2}}+ax-12$$能分解成两个整数系数的一次因式的乘积，则符合条件的整数$$a$$的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$个 "}], [{"aoVal": "B", "content": "$$4$$个 "}], [{"aoVal": "C", "content": "$$6$$个 "}], [{"aoVal": "D", "content": "$$8$$个 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->因式分解->因式分解的基础->已知因式分解结果求参数"], "answer_analysis": ["设$$x^{2}+ax-12$$能分解成两个整数系数的一次因式的乘积， 即$$x^{2}+ax-12=(x+m)(x+n)$$，$$m$$，$$n$$是整数， $$\\therefore x^{2}+ax-12=x^{2}+(m+n)x+mn$$， $$\\therefore \\begin{cases}mn=-12 \\m+n=a\\end{cases}$$， $$\\because m$$，$$n$$是整数，且$$mn=-12$$， 有$$\\begin{cases}m=12 \\n=-1\\end{cases}$$，$$\\begin{cases}m=-1 \\n=12\\end{cases}$$，$$\\begin{cases}m=6 \\n=-2\\end{cases}$$，$$\\begin{cases}m=-2 \\n=6\\end{cases}$$， $$\\begin{cases}m=4 \\n=-3\\end{cases}$$，$$\\begin{cases}m=-3 \\n=4\\end{cases}$$，$$\\begin{cases}m=3 \\n=-4\\end{cases}$$，$$\\begin{cases}m=-3 \\n=4\\end{cases}$$，$$\\begin{cases}m=2 \\n=-6\\end{cases}$$， $$\\begin{cases}m=-6 \\n=2\\end{cases}$$，$$\\begin{cases}m=1 \\n=-12\\end{cases}$$，$$\\begin{cases}m=-12 \\n=1\\end{cases}$$，共$$12$$种情况． 而$$a=m+n$$，只有$$6$$种结果， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "211", "queId": "748d7324fb5c4841a1cb786ac5b57435", "competition_source_list": ["2007年第18届希望杯初一竞赛初赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$p$$，$$q$$，$$r$$，$$s$$是互不相同的正整数，且满足$$\\frac{p}{q}=\\frac{r}{s}$$，则( )．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{p}{s}=\\frac{r}{q}$$ "}], [{"aoVal": "B", "content": "$$\\frac{p}{r}=\\frac{s}{q}$$ "}], [{"aoVal": "C", "content": "$$\\frac{p}{q}=\\frac{p+r}{q+s}$$ "}], [{"aoVal": "D", "content": "$$\\frac{r}{s}\\ne\\frac{r-p}{s-q}$$ "}]], "knowledge_point_routes": ["课内体系->思想->方程思想", "课内体系->能力->运算能力", "课内体系->知识点->三角形->相似三角形->比例线段->比例的综合应用", "课内体系->知识点->三角形->相似三角形->比例线段->比例线段的性质"], "answer_analysis": ["可用特殊值法．取$$p=1$$，$$q=2$$，$$r=3$$，$$s=6$$时，满足已知等式$$\\frac{p}{q}=\\frac{r}{s}$$，但选项$$\\text{A}$$、$$\\text{B}$$、$$\\text{D}$$均不成立，则选项$$\\text{C}$$一定成立．证明如下： 设$$\\frac{p}{q}=\\frac{r}{s}=k$$，则$$p=kq$$，$$r=ks$$，$$\\frac{p+r}{q+s}=\\frac{kq+ks}{q+s}=k=\\frac{p}{q}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "77", "queId": "076ed741f2cb4477a682cf71b7054cad", "competition_source_list": ["2018~2019学年浙江嘉兴初一上学期期末第7题3分", "2018~2019学年5月广东深圳罗湖区深圳中学初中部初一上学期周测A卷竞赛班初一第23次第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "单项式$$3{{x}^{2m}}{{y}^{n-1}}$$与单项式$$-\\frac{1}{2}{{x}^{2}}y$$是同类项，则$$m-2n$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "$$-3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式有关的概念->整式->由同类项求参数的值", "课内体系->知识点->式->整式的加减->整式有关的概念->整式->同类项的定义", "课内体系->能力->运算能力"], "answer_analysis": ["由题意，得： $$2m=2$$，$$n-1=1$$． ∴$$m=1$$，$$n=2$$． ∴$$m-2n=1-2\\times 1=-1$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "226", "queId": "e3078adea4dc48699222686857ba0603", "competition_source_list": ["2017~2018学年广东深圳南山区南山实验教育集团麒麟中学初一下学期开学考试第9题", "2000年第11届希望杯初一竞赛第10题", "2017~2018学年江苏无锡江阴市江阴市敔山湾实验学校初三下学期期中第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明编制了一个计算程序．当输入任一有理数，显示屏的结果总等于所输入有理数的平方与$$1$$之��．若输入$$-1$$，并将所显示的结果再次输入，这时显示的结果应当是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->数->实数->实数运算->其他实数的运算"], "answer_analysis": ["∵当输入任一有理数，显示屏的结果总等于所输入有理数的平方与$$1$$之和， ∴若输入$$-1$$，则显示屏的结果为$${{(-1)}^{2}}+1=2$$，再将$$2$$输入，则显示屏的结果为$${{2}^{2}}+1=5$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "524", "queId": "6c3e0b09b35d43fbbf507435d9cee0ab", "competition_source_list": ["2002年竞赛第1题5分", "2016~2017学年江西景德镇初二下学期期中景德镇一中1班第1题4分", "2020~2021学年重庆北碚区重庆市西南大学附属中学初二上学期开学考试第10题4分", "初二上学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a\\textless{}b\\textless{}0$$，$${{a}^{2}}+{{b}^{2}}=4ab$$，则$$\\frac{a+b}{a-b}$$的值为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{6}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式化简求值->分式化简求值-条件化简求值", "课内体系->知识点->式->分式->分式的运算->分式加减、乘除、乘方混合运算", "课内体系->能力->运算能力"], "answer_analysis": ["$$a+b=-\\sqrt{6ab}$$，$$a-b=-\\sqrt{2ab}$$． ∵$${{a}^{2}}+{{b}^{2}}=4ab$$， ∴$${{a}^{2}}+{{b}^{2}}+2ab={{\\left( a+b \\right)}^{2}}=6ab$$① ∴$${{a}^{2}}+{{b}^{2}}-2ab={{\\left( a-b \\right)}^{2}}=2ab$$② $$\\frac{①}{②}$$，得~~$$\\frac{{{\\left( a+b \\right)}^{2}}}{{{\\left( a-b \\right)}^{2}}}=\\frac{6ab}{2ab}$$ ∵$$a\\textless{}b\\textless{}0$$， ∴$$ab\\textgreater0$$，$$a+b\\textless{}0$$，$$a-b\\textless{}0$$， ∴$$\\frac{{{\\left( a+b \\right)}^{2}}}{{{\\left( a-b \\right)}^{2}}}={{\\left( \\frac{a+b}{a-b} \\right)}^{2}}=3$$， ∴$$\\frac{a+b}{a-b}=\\sqrt{3}$$． 所以选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1026", "queId": "ff8080814d56502a014d678da1742146", "competition_source_list": ["2015年第26届全国希望杯初一竞赛复赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两人沿同一路线骑车（匀速）从$$A$$站到$$B$$站，甲要用$$30$$分钟，乙要用$$40$$分钟．如果乙比甲早出发$$5$$分钟去$$B$$站，则甲追上乙时，是甲出发后的第（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$分钟 "}], [{"aoVal": "B", "content": "$$13$$分钟 "}], [{"aoVal": "C", "content": "$$14$$分钟 "}], [{"aoVal": "D", "content": "$$15$$分钟 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->能力->分析和解决问题能力", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的行程问题->一元一次方程的行程问题-追及问题"], "answer_analysis": ["甲每分钟行驶路程的$$\\frac{1}{30}$$，乙每分钟行驶路程的$$\\frac{1}{40}$$． 设$$x$$分钟后，甲追上乙， 由题意，得$$\\frac{x}{30}=\\frac{x+5}{40}$$， 解得$$x=15$$． 答：甲出发后的第$$15$$分钟，追上乙． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "492", "queId": "249e209ffada4f1b80ebf095e9d9afb8", "competition_source_list": ["1998年第9届希望杯初二竞赛第2试第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "在方程组$$\\begin{cases}x+y+z=0 {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=-36 \\end{cases}$$中，$$x$$，$$y$$，$$z$$是互不相等的整数，那么此方程组的解的组数为( )．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "多于$$6$$ "}], [{"aoVal": "D", "content": "少于$$3$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->其他方程->高次方程"], "answer_analysis": ["利用$${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\\left( x+y+z \\right)\\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-xz+yz \\right)=0$$，把原方程组转化为解不定方程$$3xyz=-36$$． 因为$${{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=\\left( x+y+z \\right)\\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \\right)=0$$， 所以$${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$$，从而得$$3xyz=-36$$，即$$xyz=-12$$． 因此$$x$$，$$y$$，$$z$$中一定是两正一负，且$$x+y+z=0$$． 又$$12=1\\times 1\\times 12=1\\times 2\\times 6=1\\times 3\\times 4=2\\times 2\\times 3$$， 则上述两种组合中，只有$$12=1\\times 3\\times 4$$符合条件． 所以$$\\begin{cases}x=1 y=3 z=-4 \\end{cases}$$或$$\\begin{cases}x=1 y=-4 z=3 \\end{cases}$$或$$\\begin{cases}x=3 y=1 z=-4 \\end{cases}$$或$$\\begin{cases}x=3 y=-4 z=1 \\end{cases}$$或$$\\begin{cases}x=-4 y=1 z=3 \\end{cases}$$或$$\\begin{cases}x=-4 y=3 z=1 \\end{cases}$$， 共有$$6$$个解． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1445", "queId": "ca4701a6df9a4747b872b655b4fe43c3", "competition_source_list": ["2013年第24届全国希望杯初二竞赛初赛第23题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "若关于$$x$$的方程$$\\frac{2}{x-2}+\\frac{mx}{{{x}^{2}}-4}=\\frac{3}{x+2}$$有增根，则$$m$$的值为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$-4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$-4$$或$$6$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->分式方程->分式方程的解与解分式方程->分式方程的增根问题"], "answer_analysis": ["当$$x\\ne \\pm 2$$时，可在方程两边同乘$${{x}^{2}}-4$$， 得$$2(x+2)+mx=3(x-2)$$， 当$${{x}^{2}}-4=0$$，即$$x=\\pm 2$$时，方程有增根． 将$$x=\\pm 2$$分别代入上式，得$$m=-4$$或$$6$$． 综上，当$$m=-4$$或$$6$$时，方程有增根． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1024", "queId": "ff8080814d56502a014d6781ad62210e", "competition_source_list": ["2015年第26届全国希望杯初一竞赛复赛第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "若$$a+2015=0$$，则$$a-2015$$的值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-4030$$ "}], [{"aoVal": "B", "content": "$$-2015$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$2015$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数基础运算->有理数减法->有理数减法运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数减法->有理数加减混合运算"], "answer_analysis": ["∵$$a+2015=0$$， ∴$$a=-2015$$， ∴$$a-2015=-4030$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "664", "queId": "48da6e491bc14283a5fa413bebc254e2", "competition_source_list": ["2013年竞赛第9题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "实数$$a$$，$$b$$，$$c$$，$$d$$满足：一元二次方程$${{x}^{2}}+cx+d=0$$的两根为$$a$$，$$b$$，一元二次方程$${{x}^{2}}+ax+b=0$$的两根为$$c$$，$$d$$，则所有满足条件的数组$$\\left( a , b , c , d \\right)$$共~\\uline{~~~~~~~~~~}~组．", "answer_option_list": [[{"aoVal": "A", "content": "1组 "}], [{"aoVal": "B", "content": "2组 "}], [{"aoVal": "C", "content": "3组 "}], [{"aoVal": "D", "content": "无数组 "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->二次方程->一元二次方程的根与系数的关系"], "answer_analysis": ["由一元二次方程根与系数的关系得 $$\\begin{cases}a+b=-c ab=d c+d=-a cd=b \\end{cases}$$， 由上式，可知$$b=-a-c=d$$． 若$$b=d\\ne 0$$，则$$a=\\frac{d}{b}=1$$，$$c=\\frac{b}{d}=1$$，进而$$b=d=-a-c=-2$$． 若$$b=d=0$$，则$$c=-a$$，有$$\\left( a , b , c , d \\right)=\\left( t , 0 , -t , 0 \\right)$$（$$t$$为任意实数）． 经检验，数组$$\\left( 1 , -2 , 1 , -2 \\right)$$与$$\\left( t , 0 , -t , 0 \\right)$$（$$t$$为任意实数）满足条件． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1035", "queId": "ff8080814d7978b9014d865320cd23ac", "competition_source_list": ["1990年第1届全国希望杯初一竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面的说法中正确的是（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "单项式与单项式的和是单项式 "}], [{"aoVal": "B", "content": "单项式与单项式的和是多项式 "}], [{"aoVal": "C", "content": "多项式与多项式的和是多项式 "}], [{"aoVal": "D", "content": "整式与整式的和是整式 "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式有关的概念->单项式->单项式的定义", "课内体系->知识点->式->整式的加减->整式有关的概念->多项式->多项式的定义"], "answer_analysis": ["$${{x}^{2}}$$，$$2{{x}^{2}}$$，$${{x}^{3}}$$都是单项式． 两个单项式$${{x}^{3}}$$，$${{x}^{2}}$$之和为$${{x}^{3}}+{{x}^{2}}$$是多项式，排除$$\\text{A}$$． 两个单项式$${{x}^{2}}$$，$$2{{x}^{2}}$$之和为$$3{{x}^{2}}$$是单项式，排除$$\\text{B}$$． 两个多项式$${{x}^{3}}+{{x}^{2}}$$与$${{x}^{3}}-{{x}^{2}}$$之和为$$2{{x}^{3}}$$是个单项式，排除$$\\text{C}$$，因此选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "834", "queId": "c3c16d8dec3547b6b6ce45bd4cc3d008", "competition_source_list": ["2023年浙江宁波初三竞赛鄞州区联考学校：鄞实、曙光、鄞外、高桥、雅戈尔、集士港"], "difficulty": "0", "qtype": "single_choice", "problem": "下列计算正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3{{x}^{3}}\\cdot 2{{x}^{2}} y=6{{x}^{5}}$$ "}], [{"aoVal": "B", "content": "$$2{{a}^{2}}\\cdot 3{{a}^{3}}=6{{a}^{5}}$$ "}], [{"aoVal": "C", "content": "$$\\left( -2x \\right)\\cdot \\left( -5{{x}^{2}} y \\right)= -10 {{x}^{3}}y$$ "}], [{"aoVal": "D", "content": "$$\\left( -2xy \\right)\\cdot \\left( -3{{x}^{2}}y \\right)=6{{x}^{3}}y$$ "}]], "knowledge_point_routes": ["课内体系->方法->整体法", "课内体系->知识点->式->整式的乘除->整式的乘除运算->单项式乘单项式", "课内体系->思想->整体思想", "课内体系->能力->运算能力"], "answer_analysis": ["$A.$$3x^{2}\\times2x^{2}y=6x^{5}y,故此选项错误;$ $B.2a^{2}\\times3a^{3}=6a^{5},故此选项正确;$ $C.(-2x)\\times(-5x^{2}y)=10x^{3}y,故此选项错误;$ $D.(-2xy)\\times(-3x^{2}y)=6x^{3}y^{2},故此选项错误.$ 故选$B.$ "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1012", "queId": "96d5c362907b420d99f4751e5f106fc3", "competition_source_list": ["2020年第22届浙江宁波余姚市余姚市实验学校初三竞赛（实验杯）第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$y=-2x^{2}-4x-1(-3\\leqslant x\\leqslant 0)$$的最大值与最小值之和是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-6$$ "}], [{"aoVal": "B", "content": "$$-3$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$-8$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->二次函数->二次函数y=ax^2+bx+c 的图象和性质->二次函数y=ax^2+bx+c 的最值问题->给定范围求二次函数最值"], "answer_analysis": ["∵$$y=-2x^{2}-4x-1$$ $$=-2(x^{2}+2x+1)+1$$ $$=-2(x+1)^{2}+1$$ ∴二次函数的对称轴为直线$$x=-1$$且$$a=-2\\textless0$$， ∴当$$-3\\leqslant x\\leqslant -1$$时， $$y$$随$$x$$的增大而增大， ∴最大值为：$$1$$， 最小值为：$$-2(-3+1)^{2}+1=-7$$， 当$$-1\\textless x\\leqslant 0$$时， $$y$$随$$x$$的增大而减小， 此时最小值为：$$-2(0+1)^{2}+1=-1$$， ∴综上，最小值为$$-7$$，最大值为$$1$$， ∴他们之和为：$$-7+1=-6$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "884", "queId": "5c7f743a69254c2d82b476e42885e2f8", "competition_source_list": ["全国全国初中数学联赛竞赛C卷"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$\\begin{cases}{{a}^{4}}+9=2b(2c+b) {{b}^{4}}+9=2c(2a+c) {{c}^{4}}+9=2a(2b+a) \\end{cases}$$，则$$a-b+c$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$\\pm 1$$ "}], [{"aoVal": "C", "content": "$$\\pm \\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$\\pm 3$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->乘法公式->配方思想的运用"], "answer_analysis": ["$$\\left. \\begin{matrix}{{a}^{4}}+9=4bc+2{{b}^{2}} {{a}^{4}}+9=4ca+2{{c}^{2}} {{a}^{4}}+9=4ab+2{{a}^{2}} \\end{matrix} \\right }\\Rightarrow {{a}^{4}}+9+{{b}^{4}}+9+{{c}^{4}}+9-2{{a}^{2}}-2{{b}^{2}}-2{{c}^{2}}-4bc-4ca-4ab=0$$ $${{({{a}^{2}}-3)}^{2}}+{{(b^{2}-3)}^{2}}+({{c}^{2}}-3)^{2}+2[{{(a-b)}^{2}}+{{(b-c)}^{2}}+(c-a)^{2}]=0$$． $$\\therefore $$$$a=b=c=\\pm \\sqrt{3}$$，故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "122", "queId": "1c8ff85333564991ae0610f649e9a658", "competition_source_list": ["2003年第20届全国初中数学联赛竞赛第4题7分", "初三上学期其它"], "difficulty": "2", "qtype": "single_choice", "problem": "满足等式$$x\\sqrt{y}+\\sqrt{x}y-\\sqrt{2003x}-\\sqrt{2003y}+\\sqrt{2003xy}=2003$$的正整数对$$\\left( x,y \\right)$$的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->其他方程->无理方程"], "answer_analysis": ["\\textbf{（知识点：无理方程）} 已知等式可化为 $$\\sqrt{y}{{\\left( \\sqrt{x} \\right)}^{2}}+\\left[ {{\\left( \\sqrt{y} \\right)}^{2}}+\\sqrt{2003}\\sqrt{y}-\\sqrt{2003} \\right]\\sqrt{x}$$$$-\\sqrt{2003}\\left( \\sqrt{y}+\\sqrt{2003} \\right)=0$$， 所以$$\\left( \\sqrt{y}\\sqrt{x}-\\sqrt{2003} \\right)\\left( \\sqrt{x}+\\sqrt{y}+\\sqrt{2003} \\right)=0$$． 又由题意，$$\\sqrt{x}+\\sqrt{y}+\\sqrt{2003}\\textgreater0$$， 则$$\\sqrt{y}\\sqrt{x}-\\sqrt{2003}=0\\Rightarrow xy=2003$$． 而$$2003$$为质数，且$$x$$、$$y$$为正整数，所以$$\\left( x,y \\right)=\\left( 2003,1 \\right)$$，$$\\left( 1,2003 \\right)$$．所以选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1514", "queId": "fd0469990a634e099fb60b7bf6ee9cca", "competition_source_list": ["2009年全美数学竞赛（AMC）竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有多少个三位数，其各个数位之积等于$$24$$？．", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$21$$ "}], [{"aoVal": "E", "content": "$$24$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数乘法->有理数乘法运算", "美国AMC8->Knowledge Point->Number Theory->Factors and Multiples->Factors and Multiples Basics", "美国AMC8->Knowledge Point->Number Theory->Place Value and Number Bases->Numbers and Digits"], "answer_analysis": ["有多少个三位数，其各个数位之积等于$$24$$？ 从最小到最大列出的数字，三位整数位$$138$$，$$146$$，$$226$$，$$234$$，$$226$$可按$$\\frac{3！}{2！}=3$$方式排列，其他三种可按$$3！=6$$方式排列，有$$3+6(3)=\\text{D}21$$个三位数的正整数． With the digits listed from least to greatest, the $$3$$-digit integers are $$138$$, $$146$$, $$226$$, $$234$$, $$226$$ can be arranged in $$\\frac{3!}{2!}=3$$ ways, and the other three can be arranged in $$3!= 6 $$ways. There are $$3+6(3)=21$$,$$3$$-digit positive integers. "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1347", "queId": "8aac50a75139269a0151626e686052f8", "competition_source_list": ["初三上学期其它", "2010年竞赛第2题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "若实数$$a$$、$$b$$满足$$\\frac{1}{2}a-ab+{{b}^{2}}+2=0$$，则实数$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\leqslant -2$$ "}], [{"aoVal": "B", "content": "$$a\\geqslant 4$$ "}], [{"aoVal": "C", "content": "$$a\\leqslant -2$$或$$a\\geqslant 4$$ "}], [{"aoVal": "D", "content": "$$-2\\leqslant a\\leqslant 4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->由一元二次方程根的情况确定参数", "课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->一元二次方程根的判别式"], "answer_analysis": ["原方程即为$${{b}^{2}}-ab+\\frac{1}{2}a+2=0$$， $$\\Delta ={{(-a)}^{2}}-4(\\frac{1}{2}a+2)\\geqslant 0$$， 解得$$a\\leqslant -2$$或$$a\\geqslant 4$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "451", "queId": "31159716cf6245d29285293701a16c56", "competition_source_list": ["2009年第20届希望杯初二竞赛第2试第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "面积是$$48$$的矩形的边长和对角线的长都是整数，则它的周长等于( )．", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$40$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->整除的概念与基本性质"], "answer_analysis": ["设矩形的边长分别是$$a$$、$$b$$，对角线的长是$$c$$，则$${{a}^{2}}+{{b}^{2}}={{c}^{2}}$$． 已知矩形的面积是$$ab=48=3\\times {{2}^{4}}$$，$$a$$，$$b$$都是整数，不妨设$$\\alpha \\leqslant b$$，则$$(a,b)$$可能是 $$(1,48)$$，$$(2,24)$$，$$(3,16)$$，$$(4,12)$$，$$(6,8)$$， 分别代入$${{a}^{2}}+{{b}^{2}}={{c}^{2}}$$，只有当$$a=6$$，$$b=8$$时，$$c$$才是整数$$10$$，其他情况得到的$$c$$的值都不是整数． 所以，矩形的边长分别是$$6$$、$$8$$，周长是$$28$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "478", "queId": "314565bad37c41a0a0d2254a9cefd32f", "competition_source_list": ["2016年全国华杯赛初一竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x+y+z=5$$，$$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=5$$，$$xyz=1$$，则$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["知识标签->知识点->式->分式->分式的运算->分式的加减", "知识标签->知识点->式->整式的乘除->乘法公式->完全平方公式", "知识标签->学习能力->运算能力", "知识标签->题型->式->整式的乘除->乘法公式->题型：利用完全平方公式计算"], "answer_analysis": ["∵$$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=5$$，$$xyz=1$$， ∴$$\\frac{yz+xz+xy}{xyz}=5$$，即$$xy+yz+xz=5$$． ∵$$x+y+z=5$$， ∴$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{\\left( x+y+z \\right)}^{2}}-2\\left( xy+yz+xz \\right)=15$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "118", "queId": "33874ebe19b74c32ac21913d7e26a4f1", "competition_source_list": ["2011年第22届全国希望杯初二竞赛复赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$A:B=\\sqrt[3]{2}:\\sqrt{3}$$，$$A=\\sqrt{2}$$，$$C=\\sqrt{\\frac{29}{10}}$$，则$$B$$，$$C$$的大小关系是 .", "answer_option_list": [[{"aoVal": "A", "content": "$$B\\textgreater C$$ "}], [{"aoVal": "B", "content": "$$B=C$$ "}], [{"aoVal": "C", "content": "$$B\\textless{}C$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->无理数有关的计算->无理数的估算", "课内体系->知识点->数->实数->无理数有关的计算->无理数大小的比较"], "answer_analysis": ["因为$$\\sqrt[3]{2}\\textless{}\\sqrt{2}$$， 则由题设得$$B=\\frac{\\sqrt{3}}{\\sqrt[3]{2}}A\\textgreater\\frac{\\sqrt{3}}{\\sqrt{2}}A=\\sqrt{3}$$， 而$$C=\\sqrt{2.9}\\textless{}\\sqrt{3}$$， 所以$$B\\textgreater C$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "594", "queId": "9e354f530dc84fe68cc2cce7669ed7fd", "competition_source_list": ["2001年第12届希望杯初一竞赛第10题", "初一上学期单元测试《一元一次方程》第20题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$k$$为整数，则使得方程$$(k-1999)x=2001-2000x$$的解也是整数的$$k$$值有（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$个 "}], [{"aoVal": "B", "content": "$$8$$个 "}], [{"aoVal": "C", "content": "$$12$$个 "}], [{"aoVal": "D", "content": "$$16$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->含参一元一次方程->一元一次方程的含参整数解"], "answer_analysis": ["$$x=\\frac{2001}{k+1}$$为整数，又$$2001=1\\times 3\\times 23\\times 29$$， $$k+1$$可取$$\\pm 1$$、$$\\pm 3$$、$$\\pm 23$$、$$\\pm 29$$、$$\\pm (3\\times 23)$$、$$\\pm (3\\times 29)$$、$$\\pm (23\\times 29)$$、$$\\pm 2001$$共$$16$$个值， 其对应的$$k$$值也有$$16$$个． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "764", "queId": "ba51df32a06a4d0fb8b1685e2cf0d7c1", "competition_source_list": ["2000年第17届全国初中数学联赛竞赛第2题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$\\frac{x}{3y}=\\frac{y}{2x-5y}=\\frac{6x-15y}{x}$$，则$$\\frac{4{{x}^{2}}-5xy+6{{y}^{2}}}{{{x}^{2}}-2xy+3{{y}^{2}}}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{9}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{9}{4}$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->整式的乘除运算->多项式除以多项式", "课内体系->知识点->式->分式->分式的运算->分式加减、乘除、乘方混合运算", "课内体系->能力->运算能力"], "answer_analysis": ["首先，我们可以利用已知的三个相等的分式，求出$$x$$，$$y$$之间的关系： 由$$\\frac{x}{3y}=\\frac{y}{2x-5y}$$，得到$$2{{x}^{2}}-5xy-3{{y}^{2}}=0$$，将$$2{{x}^{2}}-5xy-3{{y}^{2}}$$进行因式分解得到： $$2{{x}^{2}}-5xy-3{{y}^{2}}=\\left( x-3y \\right)\\left( 2x+y \\right)=0$$， $$\\therefore $$$$x=3y$$或$$y=-2x$$． 若$$y=-2x$$，$$\\frac{6x-15y}{x}=36$$，而$$\\frac{x}{3y}=\\frac{1}{6}$$， $$\\therefore $$$$\\frac{y}{2x-5y}\\ne \\frac{6x-15y}{x}$$． 若$$x=3y$$，则$$\\frac{x}{3y}=\\frac{y}{2x-5y}=\\frac{6x-15y}{x}=1$$， $$\\therefore $$$$x=3y$$． 将$$x=3y$$代入分式中得：$$\\frac{4{{x}^{2}}-5xy+6{{y}^{2}}}{{{x}^{2}}-2xy+3{{y}^{2}}}=\\frac{36{{y}^{2}}-15{{y}^{2}}+6{{y}^{2}}}{9{{y}^{2}}-6{{y}^{2}}+3{{y}^{2}}}=\\frac{9}{2}$$． ", "由已知条件知$$x\\ne 0$$，$$y\\ne 0$$，
\n把已知等式变形并利用等比消去$$y$$，
\n得$$\\frac{25x}{75y}=\\frac{15y}{30x-75y}=\\frac{6x-15y}{x}$$
\n$$=\\frac{25x+15y+\\left( 6x-15y \\right)}{75y+\\left( 30x-75y \\right)+x}$$
\n$$=\\frac{31x}{31x}$$
\n$$=1$$，
\n则$$x=3y$$，
\n故$$\\frac{4{{x}^{2}}-5xy+6{{y}^{2}}}{{{x}^{2}}-2xy+3{{y}^{2}}}$$
\n$$=\\frac{36{{y}^{2}}-15{{y}^{2}}+6{{y}^{2}}}{9{{y}^{2}}-6{{y}^{2}}+3{{y}^{2}}}$$
\n$$=\\frac{27{{y}^{2}}}{6{{y}^{2}}}$$
\n$$=\\frac{9}{2}$$．
"], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "562", "queId": "be742ed808a14b6ca24cf542eb711fb7", "competition_source_list": ["1998年第9届希望杯初一竞赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$19a+98b=0$$，则$$ab$$是（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "正数 "}], [{"aoVal": "B", "content": "非正数 "}], [{"aoVal": "C", "content": "负数 "}], [{"aoVal": "D", "content": "非负数 "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->整式乘除化简求值->整式乘除化简求值-条件化简求值", "课内体系->能力->运算能力"], "answer_analysis": ["∵$$19a+98b=0$$， ∴$$a=-\\frac{98}{19}b$$， ∴$$ab=-\\frac{98}{19}{{b}^{2}}\\leqslant 0$$， 故$$ab$$为非正数. "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "880", "queId": "69cd5e4311bc4895a9b87571fc3b1b65", "competition_source_list": ["1996年第7届希望杯初二竞赛第3题", "上海自主招生近5年（2015-2019）真题题型分类汇编第15题", "2017~2018学年5月四川成都天府新区 成都市实验外国语学校（西区）初二下学期月考第10题3分", "初二其它", "2019~2020学年3月浙江杭州滨江区杭州二中白马湖学校初一下学期周测C卷第9题"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$${{x}^{2}}+ax-12$$能分解成两个整数系数的一次因式的乘积，则符合条件的整数$$a$$的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$个 "}], [{"aoVal": "B", "content": "$$4$$个 "}], [{"aoVal": "C", "content": "$$6$$个 "}], [{"aoVal": "D", "content": "$$8$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->因式分解的基础->已知因式分解结果求参数", "课内体系->能力->运算能力"], "answer_analysis": ["设$$x^{2}+ax-12$$能分解成两个整数系数的一次因式的乘积， 即$$x^{2}+ax-12=(x+m)(x+n)$$，$$m$$，$$n$$是整数， $$\\therefore x^{2}+ax-12=x^{2}+(m+n)x+mn$$， $$\\therefore \\begin{cases}mn=-12 \\m+n=a\\end{cases}$$， $$\\because m$$，$$n$$是整数，且$$mn=-12$$， 有$$\\begin{cases}m=12 \\n=-1\\end{cases}$$，$$\\begin{cases}m=-1 \\n=12\\end{cases}$$，$$\\begin{cases}m=6 \\n=-2\\end{cases}$$，$$\\begin{cases}m=-2 \\n=6\\end{cases}$$， $$\\begin{cases}m=4 \\n=-3\\end{cases}$$，$$\\begin{cases}m=-3 \\n=4\\end{cases}$$，$$\\begin{cases}m=3 \\n=-4\\end{cases}$$，$$\\begin{cases}m=-3 \\n=4\\end{cases}$$，$$\\begin{cases}m=2 \\n=-6\\end{cases}$$， $$\\begin{cases}m=-6 \\n=2\\end{cases}$$，$$\\begin{cases}m=1 \\n=-12\\end{cases}$$，$$\\begin{cases}m=-12 \\n=1\\end{cases}$$，共$$12$$种情况． 而$$a=m+n$$，只有$$6$$种结果， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1554", "queId": "eb2e48c768ee4ababc569070373ab54e", "competition_source_list": ["2014年第25届全国希望杯初一竞赛初赛第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若实数$$x$$，$$y$$，$$z$$满足$$\\left\\textbar{} x+z \\right\\textbar+{{(x-y)}^{2}}=0$$，则$${{\\left( \\frac{x}{z} \\right)}^{2}}+{{\\left( \\frac{y}{x} \\right)}^{2}}$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->平方根->非负性的应用", "课内体系->知识点->数->有理数->绝对值->绝对值的非负性", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->有理数乘方运算", "课内体系->能力->运算能力"], "answer_analysis": ["由绝对值得非负性及完全平方的非负性， 可知$$x+z=0$$，$$x-y=0$$， ∴$$\\frac{x}{z}=-1$$，$$\\frac{y}{x}=1$$， ∴$${{\\left( \\frac{x}{z} \\right)}^{2}}+{{\\left( \\frac{y}{x} \\right)}^{2}}=1+1=2$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "550", "queId": "31f5862a634f4731959595a6d7dce35d", "competition_source_list": ["2002年第13届希望杯初一竞赛第1试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面四个命题中，正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "一切有理数的倒数还是有理数 "}], [{"aoVal": "B", "content": "一切正有理数的相反数必是负有理数 "}], [{"aoVal": "C", "content": "一切有理数的绝对值必是正有理数 "}], [{"aoVal": "D", "content": "一切有理数的���方是正有理数 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->绝对值->绝对值的代数意义"], "answer_analysis": ["因为有理数$$0$$无倒数；$$0$$的绝对值是零，不是正有理数；$$0$$的平方是$$0$$，$$0$$不是正有理数． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "980", "queId": "81091b0f5e7941a08ab5f5ea71ff2394", "competition_source_list": ["2020~2021学年6月重庆渝中区重庆市巴蜀中学校初一下学期周测C卷第10题4分", "2018~2019学年5月广东深圳罗湖区深圳市罗湖区翠园中学（初中部）初一下学期周测B卷（竞赛班）第10题2分", "2018~2019学年河北石家庄新华区石家庄第四十二中学初一下学期期中第16题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知关于$$x$$的方程$$9x-3=kx+14$$有整数解，且关于$$x$$的不等式组$$\\begin{cases}\\dfrac{x+15}{2}\\textgreater x+5 \\dfrac{3x}{2}\\geqslant \\dfrac{k-2}{8}-2x \\end{cases}$$有且只有$$4$$个整数解，则满足条件的整数$$k$$有个．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->解一元一次不等式组", "课内体系->知识点->方程与不等式->不等式（组）->含参不等式->由不等式（组）的整数解情况求参数范围", "课内体系->能力->运算能力"], "answer_analysis": ["解关于$$x$$的方程$$9x-3=kx+14$$得：$$x=\\frac{17}{9-k}$$， ∵方程有整数解， ∴$$9-k=\\pm 1$$或$$9-k=\\pm 17$$， 解得：$$k=8$$或$$10$$或$$-8$$或$$26$$， 解不等式组$$\\begin{cases}\\dfrac{x+15}{2}\\textgreater x+5 \\dfrac{3x}{2}\\geqslant \\dfrac{k-2}{8}-2x \\end{cases}$$，得不等式组的解集为$$\\frac{k-2}{28}\\leqslant x\\textless{}5$$， ∵不等式组有且只有四个整数解， ∴$$0\\textless{}\\frac{k-2}{28}\\leqslant 1$$，解得：$$2\\textless{}k\\leqslant 30$$， 所以满足条件的整数$$k$$的值为$$8$$或$$10$$或$$26$$． 故答案选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "991", "queId": "6f35b6fd46374be4a8591eee27daace2", "competition_source_list": ["2013年第24届全国希望杯初二竞赛复赛第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "在无理数$$\\sqrt{5}$$，$$\\sqrt{6}$$，$$\\sqrt{7}$$，$$\\sqrt{8}$$中，介于$$\\frac{\\sqrt{8}+1}{2}$$与$$\\frac{\\sqrt{26}+1}{2}$$之间的数有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$个 "}], [{"aoVal": "B", "content": "$$2$$个 "}], [{"aoVal": "C", "content": "$$3$$个 "}], [{"aoVal": "D", "content": "$$4$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->无理数有关的计算->无理数大小的比较"], "answer_analysis": ["因为$$\\sqrt{8}\\textless{}\\sqrt{9}=3$$， 所以$$\\frac{\\sqrt{8}+1}{2}\\textless{}\\frac{3+1}{2}=2$$． 因为$$\\sqrt{26}\\textgreater\\sqrt{25}=5$$， 所以$$\\frac{\\sqrt{26}+1}{2}\\textgreater\\frac{5+1}{2}=3$$． 因为无理数$$\\sqrt{5}$$，$$\\sqrt{6}$$，$$\\sqrt{7}$$，$$\\sqrt{8}$$都介于$$2$$与$$3$$之间， 所以在无理数$$\\sqrt{5}$$，$$\\sqrt{6}$$，$$\\sqrt{7}$$，$$\\sqrt{8}$$中，介于$$\\frac{\\sqrt{8}+1}{2}$$与$$\\frac{\\sqrt{26}+1}{2}$$之间的数有$$4$$个． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "568", "queId": "7594601d492e48b18ecd962567ca1886", "competition_source_list": ["2001年第12届希望杯初二竞赛第2试第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知两个不同的质数$$p$$、$$q$$满足下列关系：$$p^{2}-2001p+m=0$$，$$q^{2}-2001q+m=0$$，$$m$$是适当的整数，那么$$p^{2}+q^{2}$$的数值是（ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4004006$$ "}], [{"aoVal": "B", "content": "$$3996005$$ "}], [{"aoVal": "C", "content": "$$3996003$$ "}], [{"aoVal": "D", "content": "$$4004004$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->含乘方的有理数混合运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算"], "answer_analysis": ["两式相减，得$$(p-q)(p+q-2001)=0$$， ∵$$p\\ne q$$， ∴$$p+q=2001$$，而$$p$$、$$q$$为质数， ∴$$p$$、$$q$$中有一个为$$2$$，另一个为$$1999$$． ∴$$p^{2}+q^{2}=2^{2}+1999^{2}=3996005$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "147", "queId": "0fe4c094fff34652999c56637e3e5e4e", "competition_source_list": ["2015年第26届全国希望杯初一竞赛复赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "小明、小红、小华、小彬四人中的一人书包里有苹果，老师问：谁的书包里有苹果？四人回答如下： 小明：苹果不在我这里； 小红：苹果在小彬那里； 小华：苹果在小红那里； 小彬：苹果不在我这里． 若其中只有一人说了假话，则书包里有苹果的是（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "小明 "}], [{"aoVal": "B", "content": "小红 "}], [{"aoVal": "C", "content": "小华 "}], [{"aoVal": "D", "content": "小彬 "}]], "knowledge_point_routes": ["小升初->小升初知识点->组合模块->逻辑推理"], "answer_analysis": ["用假设法： （$$1$$）假设小明说了假话，那么苹果在小明那里，小红和小华就说的是假话，与题设冲突； （$$2$$）假设小红说的是假话，其他人说的都是真话，那么苹果在小红那里； （$$3$$）假设小华说了假话，则小红说``苹果在小彬那里''与小彬说``苹果不在我这里''都是真话，互相矛盾； （$$4$$）假设小彬说的是假话，那么小华说的就是假话，这样小彬、小华都说假话，与``其中只有一人说的是假话''的题设不符． 因此苹果在小红那里． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1157", "queId": "a04c036d3b8a43f4bec67b538bae6e32", "competition_source_list": ["2019年第1届广东深圳罗湖区深圳中学初中部初一竞赛（凤凰木杯）第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$m=1+ \\sqrt{2}$$，$$n=1- \\sqrt{2}$$且$$(7{{m}^{2}}-14m+a)(3{{n}^{2}}-6n-7)=8$$，则$$a$$的值等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$-9$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$-5$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->二次根式->二次根式的性质与运算"], "answer_analysis": ["由 $$m=1+ \\sqrt{2}$$得$$m-1=\\sqrt{2}$$， 两边平方，得$${{m}^{2}}-2m+1=2$$， 即$${{m}^{2}}-2m=1$$，同理得$${{n}^{2}}-2n=1$$， 又$$(7{{m}^{2}}-14m+a)(3{{n}^{2}}-6n-7)=8$$， 所以$$\\left( 7+a \\right)\\left( 3-7 \\right)=8$$， 解得$$a=-9$$， 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1262", "queId": "8aac50a74e023208014e3f5aeca71935", "competition_source_list": ["1995年第6届全国希望杯初一竞赛初赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$A={{a}^{2}}+{{b}^{2}}-{{c}^{2}}$$，$$B=-4{{a}^{2}}+2{{b}^{2}}+3{{c}^{2}}$$，若$$A+B+C=0$$，则$$C=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$5{{a}^{2}}+3{{b}^{2}}+2{{c}^{2}}$$ "}], [{"aoVal": "B", "content": "$$5{{a}^{2}}-3{{b}^{2}}+4{{c}^{2}}$$ "}], [{"aoVal": "C", "content": "$$3{{a}^{2}}-3{{b}^{2}}-2{{c}^{2}}$$ "}], [{"aoVal": "D", "content": "$$3{{a}^{2}}+{{b}^{2}}+4{{c}^{2}}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的加减->整式的加减运算->整式加减", "课内体系->知识点->式->整式的加减->整式有关的概念->整式->整式的定义"], "answer_analysis": ["∵$$A+B+C=0$$， ∴$$C=-A-B=-\\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \\right)-\\left( -4{{a}^{2}}+2{{b}^{2}}+3{{c}^{2}} \\right)$$ $$=-{{a}^{2}}-{{b}^{2}}+{{c}^{2}}+4{{a}^{2}}-2{{b}^{2}}-3{{c}^{2}}=3{{a}^{2}}-3{{b}^{2}}-2{{c}^{2}}$$．选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1585", "queId": "f9a0e198ece44318bf70d1711f07589d", "competition_source_list": ["2007年第18届希望杯初二竞赛第1试第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "若$$x=1$$满足$$2m{{x}^{2}}-{{m}^{2}}x-m=0$$，则$$m$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$0$$或$$1$$ "}], [{"aoVal": "D", "content": "任意实数 "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->含参方程"], "answer_analysis": ["将$$x=1$$代入到题设的等式，得：$$m-{{m}^{2}}=0$$， 解得$$m=0$$或$$1$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1564", "queId": "ddb72402fedc42cbb864a6ef6b6df448", "competition_source_list": ["2011年第22届全国希望杯初二竞赛初赛第17题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "设完全平方数$$A$$是$$11$$个连续整数的平方和，则$$A$$的最小值是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$81$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$121$$ "}], [{"aoVal": "D", "content": "$$144$$ "}]], "knowledge_point_routes": ["课内体系->能力->抽象概括能力", "课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式", "竞赛->知识点->数论->同余->完全平方数"], "answer_analysis": ["设$$11$$个连续整数中，中间的数为$$x$$， 则这$$11$$个数的平方和可以写成 $${{(x-5)}^{2}}+{{(x-4)}^{2}}+\\cdots +{{x}^{2}}+\\cdots +{{(x+4)}^{2}}+{{(x+5)}^{2}}$$ $$=11{{x}^{2}}+2({{5}^{2}}+{{4}^{2}}+{{3}^{2}}+{{2}^{2}}+{{1}^{2}})$$ $$=11({{x}^{2}}+10)$$， 则$$A=11({{x}^{2}}+10)$$， 当$$x=\\pm 1$$时，$$A$$取最小值$$121$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "63", "queId": "8aac50a7519fa10a0151b8248cd34eeb", "competition_source_list": ["2016年初二上学期其它", "2010年第21届希望杯初二竞赛第2试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\frac{x}{2}-\\frac{y}{3}=1$$，则代数式$$\\frac{9x+y-18}{9x-y-18}$$的值．", "answer_option_list": [[{"aoVal": "A", "content": "等于$$\\frac{7}{5}$$ "}], [{"aoVal": "B", "content": "等于$$\\frac{5}{7}$$ "}], [{"aoVal": "C", "content": "等于$$\\frac{5}{7}$$或不存在 "}], [{"aoVal": "D", "content": "等于$$\\frac{7}{5}$$或不存在 "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的运算->分式加减、乘除、乘方混合运算", "课内体系->知识点->式->分式->分式化简求值->分式化简求值-整体代入求值", "课内体系->知识点->式->分式->分式化简求值->分式化简求值-条件化简求值", "课内体系->能力->运算能力"], "answer_analysis": ["∵$$\\frac{x}{2}-\\frac{y}{3}=1$$，∴$$y=\\frac{3}{2}x-3$$， 将其代入代数式，得$$\\frac{9x+y-18}{9x-y-18}=\\frac{9x+\\frac{3}{2}x-3-18}{9x-\\frac{3}{2}x+3-18}=\\frac{21x-42}{15x-30}=\\frac{7\\left( x-2 \\right)}{5\\left( x-2 \\right)}$$， 当$$x\\ne 2$$时，原式$$=\\frac{7}{5}$$；当$$x=2$$时，原式的值不存在． ", "由$$\\frac{x}{2}-\\frac{y}{3}=1$$得$$3x-2y=6$$，
\n∴$$9x=6y+18$$．
\n于是$$\\frac{9x+y-18}{9x-y-18}=\\frac{\\left( 6y+18 \\right)+y-18}{\\left( 6y+18 \\right)-y-18}=\\frac{7y}{5y}$$．
\n当$$y\\ne 0$$（即$$x\\ne 2$$）时，原式$$=\\frac{7}{5}$$；
\n当$$y=0$$（即$$x=2$$）时，原式无意义．
\n故选$$\\text{D}$$．
"], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "56", "queId": "067a24d2698c45bbb391f458c720dd79", "competition_source_list": ["2009年第20届希望杯初二竞赛第1试第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$100\\sim 1000$$的整数中（含$$100$$和$$1000$$），既不是完全平方数，也不是完全立方数的数有．", "answer_option_list": [[{"aoVal": "A", "content": "$$890$$个 "}], [{"aoVal": "B", "content": "$$884$$个 "}], [{"aoVal": "C", "content": "$$874$$个 "}], [{"aoVal": "D", "content": "$$864$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["用$$A(n)$$表示不大于$$n$$的非零自然数中既不是完全平方数，也不是完全立方数的数的个数． 由于$$9 ~\\textless{} ~\\sqrt{99} ~\\textless{} ~10$$，$$4 ~\\textless{} ~\\sqrt[3]{99} ~\\textless{} ~5$$，$$2 ~\\textless{} ~\\sqrt[6]{99} ~\\textless{} ~3$$， 所以$$A(99)=99-(9+4-2)=88$$． 由于$$31 ~\\textless{} ~\\sqrt{1000} ~\\textless{} ~32$$，$$\\sqrt[3]{1000}=10$$，$$3 ~\\textless{} ~\\sqrt[6]{1000} ~\\textless{} ~4$$， 所以$$A(1000)=1000-(31+10-3)=962$$． 故有$$A(1000)-A(99)=962-88=874$$． 即在$$100\\sim 1000$$的整数中（含$$100$$和$$1000$$），既不是完全平方数也不是完全立方数的数有$$874$$个． 故选$$\\text{C}$$． ", "因为$$100={{10}^{2}}$$，$${{31}^{2}} < 1000 < {{32}^{2}}$$，
\n所以，$$100\\sim 1000$$中，完全平方数有$${{10}^{2}}$$，$${{11}^{2}}$$，$${{12}^{2}}$$，$$\\cdots $$，$${{31}^{2}}$$，共$$22$$个．
\n因为$${{4}^{3}} < 100 < {{5}^{3}}$$，$$1000={{10}^{3}}$$，
\n所以，$$100\\sim 1000$$中，完全立方数有$${{5}^{3}}$$，$${{6}^{3}}$$，$$\\cdots $$，$${{9}^{3}}$$，$${{10}^{3}}$$共$$6$$个，
\n其中$${{9}^{3}}={{3}^{6}}={{27}^{2}}$$既是完全平方数，也是完全立方数．
\n所以，在$$100\\sim 1000$$的整数中（含$$100$$和$$1000$$），既不是完全平方数也不是完全立方数的数有$$901-22-6+1=874$$（个）．
\n故选$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1367", "queId": "c9b947fbb512465ab87a5179ccc0a0a4", "competition_source_list": ["2007年第18届希望杯初二竞赛第2试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "古人用天干和地支记次序，其中天干有$$10$$个：甲乙丙丁戊己庚辛壬癸．地支有$$12$$个：子丑寅卯辰巳午未申酉戌亥，将天干的$$10$$个汉字和地支的$$12$$个汉字对应排列成如下两行： 甲乙丙丁戊己庚辛壬癸甲乙丙丁戊己庚辛壬癸甲乙丙丁$$\\cdots \\cdots $$ 子丑寅卯辰巳午未申酉戌亥子丑寅卯辰巳午未申酉戌亥$$\\cdots \\cdots $$ 从左向右数，第$$1$$列是甲子，第$$2$$列是乙丑，第$$3$$列是丙寅$$\\cdots \\cdots $$，我国的农历纪年就是按这个顺序得来的，如公历$$2007$$年是农历丁亥年．那么从该年往后，农历纪年为甲亥年的那一年在．", "answer_option_list": [[{"aoVal": "A", "content": "$$2019$$年 "}], [{"aoVal": "B", "content": "$$2031$$年 "}], [{"aoVal": "C", "content": "$$2043$$年 "}], [{"aoVal": "D", "content": "没有对应的年号 "}]], "knowledge_point_routes": ["竞赛->知识点->组合->排列与组合"], "answer_analysis": ["天干有$$10$$个，地支有$$12$$个，它们对应排列，天干中排在奇数位的甲、丙$$\\cdots \\cdots $$，只能与文中排在奇数位的子、寅$$\\cdots \\cdots $$组合成农历纪年，而亥排在地支中的第$$12$$位，所以不可能甲亥年． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1288", "queId": "a541399264974e6db749b09d01c37245", "competition_source_list": ["2016年第27届全国希望杯初二竞赛复赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知关于$$x$$的不等式组$$\\left { \\begin{matrix}x-3m\\textless{}0 n-2x\\textless{}0 \\end{matrix} \\right.$$的解集是$$-1\\textless{}x\\textless{}3$$，则$${{(m+n)}^{2016}}=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->含参不等式->由不等式（组）的解集求参数的范围"], "answer_analysis": ["解不等式组$$\\left { \\begin{matrix}x-3m\\textless{}0 n-2x\\textless{}0 \\end{matrix} \\right.$$，得$$\\frac{n}{2}\\textless{}x\\textless{}3m$$， 由题意，得$$3m=3$$，$$\\frac{n}{2}=-1$$， 解得$$m=1$$，$$n=-2$$， ∴$${{(m+n)}^{2016}}={{(-1)}^{2016}}=1$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1268", "queId": "8aac50a74e442d83014e4ccdd0ac1d50", "competition_source_list": ["1995年第6届全国希望杯初一竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\left\\textbar{} a \\right\\textbar=-a$$，则化简$$\\left\\textbar{} a-1 \\right\\textbar-\\left\\textbar{} a-2 \\right\\textbar$$所得的结果是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2a-3$$ "}], [{"aoVal": "D", "content": "$$3-2a$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->利用题设条件推理化简绝对值"], "answer_analysis": ["∵$$\\left\\textbar{} a \\right\\textbar=-a$$，∴$$a\\leqslant 0$$． $$\\left\\textbar{} a-1 \\right\\textbar-\\left\\textbar{} a-2 \\right\\textbar=-\\left( a-1 \\right)+\\left( a-2 \\right)=-1$$，选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1273", "queId": "8aac50a74e442d83014e4cd4134a1dc2", "competition_source_list": ["1995年第6届全国希望杯初一竞赛复赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "某同学到集贸市场买苹果，买每公斤$$3$$元的苹果用去所带钱数的一半，而其余的钱都买了每公斤$$2$$元的苹果，则该同学所买的苹果的平均价格是每公斤（~ ）元．", "answer_option_list": [[{"aoVal": "A", "content": "$$2.6$$ "}], [{"aoVal": "B", "content": "$$2.5$$ "}], [{"aoVal": "C", "content": "$$2.4$$ "}], [{"aoVal": "D", "content": "$$2.3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->二元一次方程（组）->二元一次方程组应用题->二元一次方程组的和差倍分"], "answer_analysis": ["设该同学买了$$3$$元一公斤的苹果$$x$$公斤，$$2$$元一公斤的苹果$$y$$公斤， 所用的钱数都是所带钱数的一半，所以$$3x=2y$$，即$$y=\\frac{3}{2}x$$． 因此该同学共买了$$x+y$$公斤苹果，花去了$$3x+2y=6x$$元． 所以所买的苹果平均价格是每公斤$$\\frac{3x+2y}{x+y}=\\frac{6x}{\\frac{5}{2}x}=\\frac{12}{5}=2.4$$元．选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "278", "queId": "79520a3b15bd4352a3e96e1ec24ab67a", "competition_source_list": ["2000年第11届希望杯初二竞赛第2试第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "$${{a}^{2}}-{{b}^{2}}=1+\\sqrt{2}$$，$${{b}^{2}}-{{c}^{2}}=1-\\sqrt{2}$$, find $${{a}^{4}}+{{b}^{4}}+{{c}^{4}}-{{a}^{2}}{{b}^{2}}-{{b}^{2}}{{c}^{2}}-{{c}^{2}}{{a}^{2}}$$. 设$${{a}^{2}}-{{b}^{2}}=1+\\sqrt{2}$$，$${{b}^{2}}-{{c}^{2}}=1-\\sqrt{2}$$，则$${{a}^{4}}+{{b}^{4}}+{{c}^{4}}-{{a}^{2}}{{b}^{2}}-{{b}^{2}}{{c}^{2}}-{{c}^{2}}{{a}^{2}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->乘法公式", "课内体系->知识点->式->整式的加减->整式有关的概念->整式"], "answer_analysis": ["因为$${{a}^{2}}-{{b}^{2}}=1+\\sqrt{2}$$， $${{b}^{2}}-{{c}^{2}}=1-\\sqrt{2}$$， 所以$${{a}^{2}}-{{c}^{2}}=2$$， 所以$${{a}^{4}}+{{b}^{4}}+{{c}^{4}}-{{a}^{2}}{{b}^{2}}-{{b}^{2}}{{c}^{2}}-{{c}^{2}}{{a}^{2}}$$ $$=\\frac{1}{2}[({{a}^{4}}-2{{a}^{2}}{{b}^{2}}+{{b}^{4}})+({{b}^{4}}-2{{b}^{2}}{{c}^{2}}+{{c}^{4}})+({{c}^{4}}-2{{a}^{2}}{{c}^{2}}+{{a}^{4}})]$$ $$=\\frac{1}{2}[{{({{a}^{2}}-{{b}^{2}})}^{2}}+{{({{b}^{2}}-{{c}^{2}})}^{2}}+{{({{c}^{2}}-{{a}^{2}})}^{2}}]$$ $$=\\frac{1}{2}[{{(1+\\sqrt{2})}^{2}}+{{(1-\\sqrt{2})}^{2}}+{{2}^{2}}]$$ $$=5$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1438", "queId": "aac6a6fca6d14d9d8788d4966f565403", "competition_source_list": ["2001年第18届全国初中数学联赛竞赛第1题", "2019~2020学年12月四川资阳雁江区资阳市雁江区第二中学初三上学期周测D卷第10题3分", "2016~2017学年9月湖北武汉武昌区武汉初级中学初二上学期月考第10题3分", "2016~2017学年3月湖北武汉武昌区武汉初级中学初二下学期月考第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a$$、$$b$$、$$c$$、为有理数，且等式$$a+b\\sqrt{2}+c\\sqrt{3}=\\sqrt{5+2\\sqrt{6}}$$成立，则$$2a+999b+1001c$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1999$$ "}], [{"aoVal": "B", "content": "$$2000$$ "}], [{"aoVal": "C", "content": "$$2001$$ "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算", "课内体系->知识点->方程与不等式->等式与方程->等式->等式的性质->等式性质1"], "answer_analysis": ["本题需要比较等式两边的各项，利用有理数部分等于理数部分，无理数部分等于无理数部分来求$$a$$、$$b$$、$$c$$的值，由于： $$5+2\\sqrt{6}={{\\left( \\sqrt{3}+\\sqrt{2} \\right)}^{2}}$$． 所以：$$a+b\\sqrt{2}+c\\sqrt{3}=\\sqrt{2}+\\sqrt{3}$$． 则$$a=0$$，$$b=1$$，$$c=1$$，∴$$2a+999b+1001c=2000$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "671", "queId": "3bc66e9af33b4af5a704f3710947dca0", "competition_source_list": ["2002年第13届希望杯初一竞赛第1试第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$x$$取$$1$$到$$10$$的整数时，整式$$x^{2}+x+11$$所对应的数值中质数的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "竞赛->知识点->数论->整除->素数与合数"], "answer_analysis": ["当$$x$$取$$1$$到$$10$$的整数时， $$x^{2}+x+11$$所对应的数值依次为： $$13$$，$$17$$，$$23$$，$$31$$，$$41$$，$$53$$，$$67$$，$$83$$，$$101$$，$$121$$． 其中质数为： $$13$$，$$17$$，$$23$$，$$31$$，$$41$$，$$53$$，$$67$$，$$83$$，$$101$$，共$$9$$个． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "556", "queId": "7a28358b3f534fa3bb91fd0eacb390ad", "competition_source_list": ["2019年浙江杭州拱墅区杭州市文澜中学初二竞赛第6题3分", "2019~2020学年浙江杭州拱墅区杭州市文澜中学初二上学期单元测试《一次函数》第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "直线$$y=kx+b$$经过点$$(m ,1)$$和点$$(-1,m)$$ ，其中$$0 \\textless{} m \\textless{} 1$$ ， 则$$k$$ ，$$b$$ 应满足的条件是．", "answer_option_list": [[{"aoVal": "A", "content": "$$k\\textgreater0$$且$$b\\textgreater0$$ "}], [{"aoVal": "B", "content": "$$k\\textless0$$且$$b\\textgreater0$$ "}], [{"aoVal": "C", "content": "$$k\\textgreater0$$且$$b\\textless0$$ "}], [{"aoVal": "D", "content": "$$k\\textless0$$且$$b\\textless0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->一次函数->一次函数基础->一次函数的增减性"], "answer_analysis": ["∵$$m\\textgreater1\\textgreater-1$$，$$1 ~\\textless{} ~m$$，即自变量的值越大函数值越小，即函数值随着自变量的增大而减小，∴$$k ~\\textless{} ~0$$． 过第一象限内的点$$(1,m)$$和第四象限内的点$$(m,-1)$$画直线，知此直线与$$y$$轴正半轴相交，所以，$$b\\textgreater0$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "321", "queId": "b4c4608d742a4bd6ad77ef19e43b3420", "competition_source_list": ["2001年第18届全国初中数学联赛竞赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$ab≠1$$，且有$$5{{a}^{2}}+2001a+9=0$$及$$9{{b}^{2}}+2001b+5=0$$，则$$\\frac{a}{b}$$的值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{9}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{9}$$ "}], [{"aoVal": "C", "content": "$$-\\frac{2001}{5}$$ "}], [{"aoVal": "D", "content": "$$-\\frac{2001}{9}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->二次方程->一元二次方程的根与系数的关系"], "answer_analysis": ["显然可以看出方程系数相同，可以利用根与系数关系来求解： $$2{{a}^{2}}+2001a+9=0$$， $$9{{b}^{2}}+2001b+5=0$$（显然$$b=0$$不是方程的解）． ∴$$5\\frac{1}{{{b}^{2}}}+2001\\frac{1}{b}+9=0$$， 故$$a$$与$$\\frac{1}{b}$$都是方程$$5{{x}^{2}}+2001x+9=0$$的根， 但$$a\\ne \\frac{1}{b}$$，由$$\\Delta =0$$， 即$$a$$与$$\\frac{1}{b}$$是此方程的相异实根． 从而$$a\\cdot \\frac{1}{b}=\\frac{9}{5}$$． 故选 A． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1130", "queId": "ff8080814db3e529014dd125654f5317", "competition_source_list": ["2012年第23届全国希望杯初二竞赛初赛第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "一次函数$$y=({{m}^{2}}-4)x+(1-m)$$和$$y=(m+2)x+({{m}^{2}}-3)$$的图象分别与$$y$$轴交于点$$P$$和$$Q$$，这两点关于$$x$$轴对称，则$$m$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$2$$或$$-1$$ "}], [{"aoVal": "C", "content": "$$1$$或$$-1$$ "}], [{"aoVal": "D", "content": "$$-1$$ "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "课内体系->知识点->函数->一次函数->一次函数的几何变换->一次函数图象关于坐标轴对称"], "answer_analysis": ["一次函数$$y=({{m}^{2}}-4)x+(1-m)$$的图象与$$y$$轴的交点$$P$$为$$(0,1-m)$$， 一次函数$$y=(m+2)x+({{m}^{2}}-3)$$的图象与$$y$$轴的交点$$Q$$为$$(0,{{m}^{2}}-3)$$， 因为$$P$$和$$Q$$关于$$x$$轴对称， 所以$$1-m+{{m}^{2}}-3=0$$， 解得$${{m}_{1}}=2$$，$${{m}_{2}}=-1$$． 又因为两个函数都是一次函数， 所以$${{m}^{2}}-4\\ne 0$$且$$m+2\\ne 0$$， 所以$$m=2$$不合题意，舍去， 所以$$m=-1$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "996", "queId": "6f3ea325555a499eae368666ba0155c6", "competition_source_list": ["2011年第22届全国希望杯初二竞赛初赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一辆汽车从$$A$$地匀速驶往$$B$$地，如果汽车行驶的速度增加$$a \\%$$，则所用时间减少$$b \\%$$，则$$a$$，$$b$$的关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$b=\\frac{100a}{1+a \\%}$$ "}], [{"aoVal": "B", "content": "$$b=\\frac{100}{1+a \\%}$$ "}], [{"aoVal": "C", "content": "$$b=\\frac{a}{1+a}$$ "}], [{"aoVal": "D", "content": "$$b=\\frac{100a}{100+a}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的行程问题->一元一次方程的行程问题-变速问题"], "answer_analysis": ["设$$A$$，$$B$$两地之间的距离为$$s$$，汽车行驶的速度为$$v$$，汽车从$$A$$地到$$B$$地所用的时间为$$t$$， 则有$$s=vt=v(1+a \\%)\\cdot t(1-b \\%)$$， 即$$\\frac{100+a}{100}\\cdot \\frac{100-b}{100}=1$$， 解得$$b=\\frac{100a}{100+a}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "384", "queId": "23614d800dcc43bcb1b2131d77afd761", "competition_source_list": ["初一其它", "河北竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$、$$b$$、$$c$$满足$${{a}^{2}}+2b=7$$，$${{b}^{2}}-2c=-1$$，$${{c}^{2}}-6a=-17$$，则$$a+b+c$$的值等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->其他方程->多元二次方程（组）"], "answer_analysis": ["∵$$\\begin{cases}{{a}^{2}}+2b=7① {{b}^{2}}-2c=-1 ② {{c}^{2}}-6a=-17③ \\end{cases}$$ ∴①$$+$$②$$+$$③得，$${{a}^{2}}+2b+{{b}^{2}}-2c+{{c}^{2}}-6a=-11$$ $${{a}^{2}}-6a+9+{{b}^{2}}+2b+1+{{c}^{2}}-2c+1=0$$ $${{(a-3)}^{2}}+{{(b+1)}^{2}}+{{(c-1)}^{2}}=0$$ ∴$$a-3=0$$，$$b+1=0$$，$$c-1=0$$ ∴$$a=3$$，$$b=-1$$，$$c=1$$ ∴$$a+b+c=3$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "635", "queId": "99b98090352f4f77be6d20467bf9916d", "competition_source_list": ["1991年第2届希望杯初二竞赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙二人，从$$M$$地同时出发去$$N$$地．甲用一半时间以每小时$$a$$公里的速度行走，另一半时间以每小时$$b$$公里的速度行走；乙以每小时$$a$$公里的速度行走一半路程，另一半路程以每小时$$b$$公里的速度行走．若$$a\\ne b$$时，则到达$$N$$地．", "answer_option_list": [[{"aoVal": "A", "content": "二人同时 "}], [{"aoVal": "B", "content": "甲先 "}], [{"aoVal": "C", "content": "乙先 "}], [{"aoVal": "D", "content": "若$$a\\textgreater b$$时，甲先到达，若$$a ~\\textless{} ~b$$时，乙先到达 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->整式的乘除运算"], "answer_analysis": ["设$$M$$，$$N$$两地距离$$S$$，甲需时间$${{t}_{1}}$$，乙需时间$${{t}_{2}}$$， 则$$\\frac{a{{t}_{1}}}{2}+\\frac{b{{t}_{1}}}{2}=S$$，$${{t}_{2}}=\\dfrac{\\dfrac{S}{2}}{a}+\\dfrac{\\dfrac{S}{2}}{b}$$， 即$${{t}_{1}}=\\frac{2S}{a+b}$$，$${{t}_{2}}=\\frac{\\left( a+b \\right)S}{2ab}$$， $${{t}_{1}}-{{t}_{2}}=\\frac{2S}{a+b}-\\frac{\\left( a+b \\right)S}{2ab}$$ $$=\\frac{S\\left[ 4ab-{{\\left( a+b \\right)}^{2}} \\right]}{2ab\\left( a+b \\right)}$$ $$=\\frac{-S{{\\left( a-b \\right)}^{2}}}{2ab\\left( a+b \\right)} ~\\textless{} ~0$$． 所以$${{t}_{1}} ~\\textless{} ~{{t}_{2}}$$，即甲先． 另外：设$$a=1$$，$$b=2$$，则甲走了$$6$$小时，共走了$$9$$公里，这时乙走的时间为$$\\frac{4.5}{1}+\\frac{4.5}{2}=\\frac{3}{2}\\times 4.5\\textgreater6$$， 从这个计算中，可以看到，$$a$$，$$b$$的值互换，不影响结果． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "16", "queId": "0977ee96ed5d45b681b5aefd11be7bcd", "competition_source_list": ["2019~2020学年四川眉山东坡区眉山东辰国际学校初三上学期期中第33题5分", "2001年第12届希望杯初二竞赛第2试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "化简代数式$$\\sqrt{3+2\\sqrt{2}}+\\sqrt{3-2\\sqrt{2}}$$的结果是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$1+\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$2+\\sqrt{2}$$ "}], [{"aoVal": "D", "content": "$$2\\sqrt{2}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->二次根式->二次根式的运算->多重二次根式"], "answer_analysis": ["设$$y=\\sqrt{3+2\\sqrt{2}}+\\sqrt{3-2\\sqrt{2}}$$，$$y\\textgreater0$$， 所以$${{y}^{2}}=3+2\\sqrt{2}+2\\sqrt{3+2\\sqrt{2}}\\cdot \\sqrt{3-2\\sqrt{2}}+3-2\\sqrt{2}$$ $$=6+2\\sqrt{9-8}$$ $$=8$$， 所以$$y=2\\sqrt{2}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "270", "queId": "22380d4644834359905105f9548085cd", "competition_source_list": ["2015年第32届全国全国初中数学联赛竞赛B卷第6题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$n$$是小于$$100$$的正整数且使$$2{{n}^{2}}-3n-2$$是$$6$$的倍数，则符合条件的所有正整数$$n$$的和是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$784$$ "}], [{"aoVal": "B", "content": "$$850$$ "}], [{"aoVal": "C", "content": "$$1536$$ "}], [{"aoVal": "D", "content": "$$1634$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->整除的概念与基本性质", "课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算"], "answer_analysis": ["∵$$2{{n}^{2}}-3n-2$$是$$6$$的倍数， ∴$$2\\textbar({{2n}^{2}}-3n-2)$$， ∴$$2\\textbar3n$$， ∴$$2\\textbar n$$． 设$$n=2m$$（$$m$$是正整数）， 则$$2{{n}^{2}}-3n-2=8{{m}^{2}}-6m-2=6{{m}^{2}}-6m+2({{m}^{2}}-1)$$． 又∵$$2{{n}^{2}}-3n-2$$是$$6$$的倍数， ∴$${{m}^{2}}-1$$是$$3$$的倍数， ∴$$m=3k+1$$或$$m=3k+2$$，其中$$k$$是非负整数， ∴$$n=2(3k+1)=6k+2$$或$$n=2(3k+2)=6k+4$$，其中$$k$$是非负整数， ∴符合条件的所有正整数$$n$$的和为 $$(2+8+14+\\cdots +92+98)+(4+10+16+\\cdots +88+94)=1634$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1408", "queId": "a1876fd351224cd68cb7e47d70f4e387", "competition_source_list": ["2016年第33届全国全国初中数学联赛竞赛第1题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "用$$\\left[ x \\right]$$表示不超过$$x$$的最大整数，把$$x-\\left[ x \\right]$$称为$$x$$的小数部分．已知$$t=\\frac{1}{2-\\sqrt{3}}$$，$$a$$是$$t$$的小数部分，$$b$$是$$-t$$的小数部分，则$$\\frac{1}{2b}-\\frac{1}{a}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{3}}{2}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->综合与实践->规律探究与程序框图->定义新运算", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算", "课内体系->知识点->数->实数->无理数有关的计算->无理数大小的比较", "课内体系->能力->运算能力", "课内体系->能力->推理论证能力"], "answer_analysis": ["∵$$t=\\frac{1}{2-\\sqrt{3}}=2+\\sqrt{3}$$，$$1\\textless{}\\sqrt{3}\\textless{}2$$， ∴$$3\\textless{}2+\\sqrt{3}\\textless{}4$$，即$$3\\textless{}t\\textless{}4$$， ∴$$a=t-3=\\sqrt{3}-1$$． 又$$-t=-2-\\sqrt{3}$$，$$-2\\textless{}-\\sqrt{3}\\textless{}-1$$， ∴$$-4\\textless{}-2-\\sqrt{3}\\textless{}-3$$， ∴$$b=-t-(-4)=2-\\sqrt{3}$$， ∴$$\\frac{1}{2b}-\\frac{1}{a}=\\frac{1}{2(2-\\sqrt{3})}-\\frac{1}{\\sqrt{3}-1}=\\frac{2+\\sqrt{3}}{2}-\\frac{\\sqrt{3}+1}{2}=\\frac{1}{2}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1569", "queId": "fdef853b0a4546da8f7ce29cf4c240f8", "competition_source_list": ["初一上学期单元测试《一元一次方程》第19题", "2013年第24届全国希望杯初一竞赛复赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\left( {{y}^{2}}-1 \\right){{x}^{2}}+\\left( y+1 \\right)x+9=0$$是关于$$x$$的一元一次方程，则代数式$$\\left( 4x+y \\right)\\left( 2x-y \\right)+y$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$54$$ "}], [{"aoVal": "B", "content": "$$56$$ "}], [{"aoVal": "C", "content": "$$169$$ "}], [{"aoVal": "D", "content": "$$171$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->一元一次方程的相关概念->一元一次方程的定义", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程的相关概念->一元一次方程的解", "课内体系->知识点->式->整式的加减->整式有关的概念->整式->整式的定义", "课内体系->知识点->式->整式的加减->整式的加减运算", "课内体系->能力->运算能力"], "answer_analysis": ["因为$$\\left( {{y}^{2}}-1 \\right){{x}^{2}}+\\left( y+1 \\right)x+9=0$$是关于$$x$$的一元一次方程， 所以$${{y}^{2}}-1=0$$，$$y+1\\ne 0$$， 所以$$y=1$$． 从而方程$$\\left( {{y}^{2}}-1 \\right){{x}^{2}}+\\left( y+1 \\right)x+9=0$$可变形为$$2x+9=0$$， 解得$$x=-\\frac{9}{2}$$． 所以$$\\left( 4x+y \\right)\\left( 2x-y \\right)+y$$ $$=\\left[ 4\\left( -\\frac{9}{2} \\right)+1 \\right]\\left[ 2\\left( -\\frac{9}{2} \\right)-1 \\right]+1$$ $$=-17\\times (-10)+1$$ $$=170+1$$ $$=171$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1048", "queId": "ff8080814d7978b9014d86d2d0e72570", "competition_source_list": ["1991年第2届全国希望杯初一竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a\\textgreater b$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{a}\\textless\\frac{1}{b}$$ "}], [{"aoVal": "B", "content": "$$-a\\textless-b$$ "}], [{"aoVal": "C", "content": "$$\\left\\textbar{} a \\right\\textbar\\textgreater\\left\\textbar{} b \\right\\textbar$$ "}], [{"aoVal": "D", "content": "$${{a}^{2}}\\textgreater{{b}^{2}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["因为$$3\\textgreater-2$$，有$$\\frac{1}{3}\\textgreater-\\frac{1}{2}$$，排除$$\\text{A}$$； 因为$$2\\textgreater-3$$，有$$\\left\\textbar{} 2 \\right\\textbar\\textless\\left\\textbar{} -3 \\right\\textbar$$，排除$$\\text{C}$$； 因为$$2\\textgreater-3$$，有$${{2}^{2}}\\textless{{(-3)}^{2}}$$，排除$$\\text{D}$$； 事实上，$$a\\textgreater b$$必有$$-a\\textless-b$$．选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "511", "queId": "b5181e9882f64a25811910d66f3ba4fb", "competition_source_list": ["2017~2018学年上海浦东新区上海中学东校初一上学期期中第16题2分", "2006年第17届希望杯初一竞赛初赛第10题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有如下四个叙述： ①当$$0\\textless{}x\\textless{}1$$时，$$\\frac{1}{1+x}\\textless{}1-x+{{x}^{2}}$$； ②当$$0\\textless{}x\\textless{}1$$时，$$\\frac{1}{1+x}\\textgreater1-x+{{x}^{2}}$$； ③当$$-1\\textless{}x\\textless{}0$$时，$$\\frac{1}{1+x}\\textless{}1-x+{{x}^{2}}$$； ④当$$-1\\textless{}x\\textless{}0$$时，$$\\frac{1}{1+x}\\textgreater1-x+{{x}^{2}}$$， 其中正确的叙述是．", "answer_option_list": [[{"aoVal": "A", "content": "①③ "}], [{"aoVal": "B", "content": "②④ "}], [{"aoVal": "C", "content": "①④ "}], [{"aoVal": "D", "content": "②③ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->整式的乘除运算->整式乘除的综合", "课内体系->知识点->式->整式的乘除->乘法公式->和与差的立方公式", "课内体系->知识点->式->整式的乘除->整式乘除化简求值", "课内体系->知识点->式->因式分解->公式法->利用立方和与立方差公式因式分解", "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质", "课内体系->能力->运算能力", "课内体系->能力->推理论证能力", "课内体系->思想->转化与化归思想"], "answer_analysis": ["当$$0\\textless{}x\\textless{}1$$或$$-1\\textless{}x\\textless{}0$$时，$$\\frac{1}{1+x}$$和$$1-x+{{x}^{2}}$$都大于$$0$$，所以两式的比值大于$$0$$， 又$$(1-x+{{x}^{2}})\\div \\frac{1}{1+x}=(1-x+{{x}^{2}})(1+x)=1+{{x}^{3}}$$， 当$$0\\textless{}x\\textless{}1$$时，$$1+{{x}^{3}}\\textgreater1$$，所以①正确；②不正确； 当$$-1\\textless{}x\\textless{}0$$时，$$1+{{x}^{3}}\\textless{}1$$，所以③不正确，④正确． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1552", "queId": "cfeef127272840c38fc4a56f37ef46d0", "competition_source_list": ["2020~2021学年四川内江市中区内江市第六初级中学校初二上学期期中第12题3分", "2014~2015学年北京海淀区科迪实验中学初一下学期期中第10题", "2019~2020学年天津和平区天津市益中学校初一下学期期中第16题2分", "2009年竞赛第1题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知非零实数$$a$$、$$b$$满足$$\\left\\textbar{} 2a-4 \\right\\textbar+\\left\\textbar{} b+2 \\right\\textbar+\\sqrt{\\left( a-3 \\right){{b}^{2}}}+4=2a$$，则$$a+b$$等于 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的基础->二次根式的性质", "课内体系->能力->运算能力"], "answer_analysis": ["方法一：由条件得$$a\\geqslant 3$$，原等式为$$\\left\\textbar{} b+2 \\right\\textbar+\\sqrt{\\left( a-3 \\right){{b}^{2}}}=0$$，$$a+b=1$$． 方法二：由题意$$(a-3){{b}^{2}}\\geqslant 0$$，∵$${{b}^{2}}\\geqslant 0$$，∴$$a-3\\geqslant 0$$，∴$$2a-4\\textgreater0$$， ∴原式可化简为$$2a-4+\\textbar b+2\\textbar+\\sqrt{(a-3){{b}^{2}}}+4=2a$$，$$\\textbar b+2\\textbar+\\sqrt{(a-3){{b}^{2}}}=0$$， ∴$$b+2=(a-3){{b}^{2}}=0$$，∴$$a=3$$，$$b=-2$$，选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1072", "queId": "a4807cf7cac54265904f0c80c5f0a8c5", "competition_source_list": ["2002年第13届希望杯初二竞赛第1试第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$b \\textless{} 0$$，$$0 \\textless{} \\textbar a\\textbar{} \\textless{} \\textbar b\\textbar{} \\textless{} \\textbar c\\textbar$$，且$$\\sqrt{\\frac{a{{b}^{2}}}{c}}=\\frac{b}{c}\\sqrt{ac}$$，则$$abc$$的大小顺序是( )", "answer_option_list": [[{"aoVal": "A", "content": "$$a \\textless{} b \\textless{} c$$ "}], [{"aoVal": "B", "content": "$$c \\textless{} b \\textless{} a$$ "}], [{"aoVal": "C", "content": "$$b \\textless{} a \\textless{} c$$ "}], [{"aoVal": "D", "content": "$$b \\textless{} c \\textless{} a$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->实数运算->实数基础运算", "课内体系->知识点->数->实数->无理数有关的计算->无理数大小的比较"], "answer_analysis": ["由题意$$\\frac{b}{c}\\textgreater0$$，$$b \\textless{} 0$$，所以得到$$c \\textless{} 0$$，又$$ac\\textgreater0$$，$$c \\textless{} 0$$，可以得到$$a \\textless{} 0$$ 所以$$a$$，$$b$$，$$c \\textless{} 0$$而$$0 \\textless{} \\textbar a\\textbar{} \\textless{} \\textbar b\\textbar{} \\textless{} \\textbar c\\textbar$$，故$$c \\textless{} b \\textless{} a$$选B "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1479", "queId": "b8921956449c46709d498b1bc4c3dec1", "competition_source_list": ["2014年第25届全国希望杯初一��赛复赛第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$p={{3}^{70}}$$，$$q={{5}^{56}}$$，$$r={{6}^{42}}$$，$$s={{17}^{28}}$$．这$$4$$个数中，最大的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$p$$ "}], [{"aoVal": "B", "content": "$$q$$ "}], [{"aoVal": "C", "content": "$$r$$ "}], [{"aoVal": "D", "content": "$$s$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->实数运算->其他实数的运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->有理数乘方运算", "课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性"], "answer_analysis": ["观察到，$$70$$，$$56$$，$$42$$，$$28$$的最大公约数为$$14$$． ∴题设中的四个数可化为 $$p={{3}^{70}}={{3}^{5\\times 4}}$$， $$q={{5}^{56}}={{5}^{4\\times 14}}$$， $$r={{6}^{42}}={{6}^{3\\times 14}}$$， $$s={{17}^{28}}={{17}^{2\\times 14}}$$． ∵$${{3}^{5}}=243$$，$${{5}^{4}}=625$$，$${{6}^{3}}=216$$，$${{17}^{2}}=289$$， ∴这$$4$$个数中，最大的是$$q$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "36", "queId": "05ce0a5cd40b4f1ab2dc36c13feee723", "competition_source_list": ["2008年第19届希望杯初二竞赛第2试第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$是方程$${{x}^{3}}+3x-1=0$$的一个实数根，则直线$$y=ax+1-a$$不经过．", "answer_option_list": [[{"aoVal": "A", "content": "第一象限 "}], [{"aoVal": "B", "content": "第二象限 "}], [{"aoVal": "C", "content": "第三象限 "}], [{"aoVal": "D", "content": "第四象限 "}]], "knowledge_point_routes": ["课内体系->知识点->函数->一次函数->一次函数基础", "课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->由一元二次方程根的情况确定参数", "课内体系->知识点->方程与不等式->一元二次方程->根与系数的关系->一元二次方程根与系数的关系", "课内体系->能力->抽象概括能力"], "answer_analysis": ["当$$x\\leqslant 0$$时，$${{x}^{2}}+3x-1 ~\\textless{} ~0$$，此时$$x\\leqslant 0$$时，原方程无解， 当$$x\\geqslant \\frac{1}{3}$$时，$${{x}^{3}}+3x-1\\textgreater0$$时，此时原方程的实数根只能在$$\\left( 0,\\frac{1}{3} \\right)$$之间，因为$$a$$是方程$${{x}^{3}}+3x-1=0$$的一个实数根，所以$$0 ~\\textless{} ~a ~\\textless{} ~\\frac{1}{3}$$，对于相线$$y=ax+1-a$$，由于$$a\\textgreater0$$，$$1-a\\textgreater0$$，所以，直线不经过第四象限． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1545", "queId": "e1e54dfc08aa4357aa3bec838f6c59b5", "competition_source_list": ["2020~2021学年北京北京汇文中学初一上学期期末第10题2分", "2007年竞赛第1题6分", "2017~2018学年江苏无锡滨湖区无锡外国语学校初一下学期期中第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "方程组$$\\begin{cases}\\left\\textbar{} x \\right\\textbar+y=12 x+\\left\\textbar{} y \\right\\textbar=6 \\end{cases}$$的解的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->解含绝对值的方程组"], "answer_analysis": ["①当$$x\\textgreater0$$，$$y\\textgreater0$$时，原方程组为$$\\begin{cases}x+y=12 x+y=6 \\end{cases}$$显然无解； ②当$$x ~\\textless{} ~0$$，$$y\\textgreater0$$时，原方程组为$$\\begin{cases}-x+y=12 x+y=6 \\end{cases}$$，解得$$\\begin{cases}x=-3 y=9 \\end{cases}$$； ③当$$x\\textgreater0$$，$$y ~\\textless{} ~0$$时，原方程组为$$\\begin{cases}x+y=12 x-y=6 \\end{cases}$$，解得$$\\begin{cases}x=9 y=3 \\end{cases}$$，舍去； ④当$$x ~\\textless{} ~0$$，$$y ~\\textless{} ~0$$时，原方程组为$$\\begin{cases}-x+y=12 x-y=6 \\end{cases}$$显然无解； 综上，只有$$1$$组解．故选$$\\text{A}$$． ", "若$$x\\geqslant 0$$，则$$\\begin{cases}x+y=12 \\\\ x+\\left| y \\right|=6 \\\\\\end{cases}$$，于是$$\\left| y \\right|-y=-6$$，显然不可能．
\n若$$x < 0$$，则$$\\begin{cases}-x+y=12 \\\\ x+\\left| y \\right|=6 \\\\\\end{cases}$$，
\n于是$$\\left| y \\right|+y=18$$，解得$$y=9$$，进而求得$$x=-3$$．
\n所以，原方程组的解为$$\\begin{cases}x=-3 \\\\ y=9 \\\\\\end{cases}$$，只有$$1$$个解．
\n故选$$\\text{A}$$．
"], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "961", "queId": "662ceeddb12243959017385e2b552a52", "competition_source_list": ["2019~2020学年10月安徽合肥包河区合肥市第四十八中学初三上学期月考第10题4分", "2019年安徽蚌埠初三中考一模（局属学校）第9题4分", "2018~2019学年9月湖北武汉蔡甸区初三上学期月考第9题3分", "2018年湖北黄冈中考真题第6题3分", "2018~2019学年天津滨海新区初三上学期期末第12题3分", "2019~2020学年12月河北石家庄新华区石门实验学校初三上学期月考第13题", "2018年第20届浙江宁波余姚市余姚市实验学校初三竞赛决赛（实验杯）第6题4分", "2018~2019学年福建厦门思明区厦门松柏中学初三上学期期中第9题4分", "2019~2020学年10月江苏苏州姑苏区金阊实验中学初三上学期月考第9题3分", "2018~2019学年4月广东深圳宝安区深圳市石岩公学初三下学期周测（第12周）第11题3分", "2018~2019学年10月河南洛阳洛龙区洛阳地矿双语学校初三上学期月考第8题3分", "2019~2020学年江苏苏州姑苏区苏州市草桥中学校初三下学期单元测试《二次函数》第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$a\\leqslant x\\leqslant a+1$$时，函数$$y={{x}^{2}}-2x+1$$的最小值为$$1$$，则$$a$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$0$$或$$2$$ "}], [{"aoVal": "D", "content": "$$-1$$或$$2$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力"], "answer_analysis": ["当$$y=1$$时，有$${{x}^{2}}-2x+1=1$$， 解得：$${{x}_{1}}=0$$，$${{x}_{2}}=2$$． ∵当$$a\\leqslant x\\leqslant a+1$$时，函数有最小值$$1$$， ∴$$a=2$$或$$a+1=0$$， ∴$$a=2$$或$$a=-1$$， 故选：$$\\text{D}$$． ", "∵$$y={{x}^{2}}-2x+1={{\\left( x-1 \\right)}^{2}}$$，
\n∴①当$$a+1<{}1$$，即$$a<{}0$$时，函数的最小值为$$y={{\\left( a+1-1 \\right)}^{2}}=1$$，
\n∴$$a=\\pm 1$$，
\n∴$$a=-1$$．
\n②当$$a>1$$时，函数的最小值为$$y={{\\left( a-1 \\right)}^{2}}=1$$，
\n∴$$a=2$$或$$a=0$$，
\n∴$$a=2$$．
\n综上所述，$$a=-1$$或$$2$$．
"], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "288", "queId": "2fd7002459364dbc9711ff5194ec383d", "competition_source_list": ["2007年第18届希望杯初二竞赛第1试第4题", "其它"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a$$是正整数，方程组$$\\begin{cases}ax+4y=8 3x+2y=6 \\end{cases}$$的解满足$$x\\textgreater0$$，$$y\\textless{}0$$，则$$a$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$4$$，$$5$$，$$6$$以外的其他正整数 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->含参不等式->由不等式（组）的解集求参数的范围", "课内体系->能力->推理论证能力"], "answer_analysis": ["$$\\begin{cases}ax+4y=8 ① 3x+2y=6 ②\\end{cases}\\begin{matrix} \\end{matrix}$$， ①$$-$$②$$\\times 2$$：$$ax-bx=8-12$$ ∴$$(a-b)x=-4$$ ∵$$x\\textgreater0$$， ∴$$a-b\\textless{}0$$，$$a\\textless{}b$$． ①$$\\times 3-$$②$$\\times a$$：$$12y-2ay=24-6a$$， ∴$$(6-a)y=12-3a$$． ∵$$y\\textless{}0$$且$$a\\textless{}6$$， ∴$$12-3a\\textless{}0$$， ∴$$a\\textgreater4$$， ∴$$4\\textless{}a\\textless{}6$$， 又∵$$a$$为正整数， ∴$$a=5$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "836", "queId": "b165b5a3e6714a3487f9d4fe386f8c82", "competition_source_list": ["2012年第17届华杯赛初一竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$ab\\textless{}0$$，$$a-b\\textgreater0$$，则$$a$$，$$b$$两数的正负情况为．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater0$$，$$b\\textless{}0$$ "}], [{"aoVal": "B", "content": "$$a\\textgreater0$$，$$b\\textgreater0$$ "}], [{"aoVal": "C", "content": "$$a\\textless{}0$$，$$b\\textgreater0$$ "}], [{"aoVal": "D", "content": "$$a\\textgreater0$$，$$b\\textless{}0$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["因为$$ab\\textless{}0$$， 所以$$a\\textgreater0$$，$$b\\textgreater0$$，或$$a\\textless{}0$$，$$b\\textgreater0$$． 又因为$$a-b\\textgreater0$$， 所以$$a\\textgreater0$$，$$-b\\textgreater0$$， 即$$a\\textgreater0$$，$$b\\textless{}0$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "798", "queId": "91399da06b774840a39a97a142cbab7c", "competition_source_list": ["2017~2018学年浙江金华金东区初二上学期期���第10题3分", "2018~2019学年辽宁大连甘井子区初一下学期期末第10题3分", "2018~2019学年湖北武汉武昌区初一下学期期中（十五校联考）第8题3分", "2017~2018学年9月安徽安庆潜山县潜山第四中学初二上学期月考第6题4分", "2018年湖南长沙初二竞赛长郡教育集团（觉园杯）第1题4分", "2018~2019学年浙江宁波镇海区宁波市镇海蛟川书院初二上学期期中第4题4分", "2017~2018学年浙江宁波镇海区宁波市镇海蛟川书院初二上学期期中第9题4分", "2020~2021学年4月浙江杭州上城区杭州市建兰中学初三下学期周测A卷第1题4分", "2017~2018学年12月陕西西安碑林区工大附中初二上学期月考第7题3分", "2017年广西贵港中考真题第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在平面直角坐标系中，点$$P(m-3,4-2m)$$不可能在．", "answer_option_list": [[{"aoVal": "A", "content": "第一象限 "}], [{"aoVal": "B", "content": "第二象限 "}], [{"aoVal": "C", "content": "第三象限 "}], [{"aoVal": "D", "content": "第四象限 "}]], "knowledge_point_routes": ["课内体系->知识点->函数->平面直角坐标系->坐标系基础->象限内坐标的特征"], "answer_analysis": ["①$$m-3\\textgreater0$$，即$$m\\textgreater3$$时， $$-2m\\textless{}-6$$，$$4-2m\\textless{}-2$$， 所以，点$$P(m-3,4-2m)$$在第四象限，不可能在第一象限． ②$$m-3\\textless{}0$$，即$$m\\textless{}3$$时， $$-2m\\textgreater-6$$，$$4-2m\\textgreater-2$$， 点$$P(m-3,4-2m)$$可以在第二或三象限， 综上所述，点$$P$$不可能在第一象限． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "666", "queId": "b0d913debd544abead1b51bdb70368dc", "competition_source_list": ["2019年浙江杭州拱墅区杭州市文澜中学初二竞赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "若代数式$$\\sqrt{\\frac{x+2}{{{x}^{2}}}}$$在实数范围内有意义，则$$x$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textgreater2$$ "}], [{"aoVal": "B", "content": "$$x\\textgreater-2$$ "}], [{"aoVal": "C", "content": "$$x\\textgreater-2$$且$$x\\ne 0$$ "}], [{"aoVal": "D", "content": "$$x\\geqslant -2$$且$$x\\ne 0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的基础->分式有意义的条件", "课内体系->知识点->式->二次根式->二次根式的基础->二次根式有意义的条件"], "answer_analysis": ["∵$$\\sqrt{\\frac{x+2}{{{x}^{2}}}}$$在实数范围内有意义， ∴$$\\begin{cases}\\dfrac{x+2}{{{x}^{2}}}\\geqslant 0 {{x}^{2}}\\ne 0 \\end{cases}$$， ∴$$x\\geqslant -2$$且$$x\\ne 0$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1349", "queId": "8aac50a75139269a0151679842a2626a", "competition_source_list": ["2008年全国初中数学联赛竞赛", "初三上学期其它"], "difficulty": "2", "qtype": "single_choice", "problem": "设$${{a}^{2}}+1=3a$$，$${{b}^{2}}+1=3b$$，且$$a\\ne b$$，则代数式$$\\frac{1}{{{a}^{2}}}+\\frac{1}{{{b}^{2}}}$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元二次方程->根与系数的关系->一元二次方程根与系数的关系"], "answer_analysis": ["由题设条件可知$${{a}^{2}}-3a+1=0$$，$${{b}^{2}}-3b+1=0$$，且$$a\\ne b$$，所以$$ab$$是一元二次方程$${{x}^{2}}-3x+1=0$$的两根，故$$a+b=3$$，$$ab=1$$，因此$$\\frac{1}{{{a}^{2}}}+\\frac{1}{{{b}^{2}}}=\\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}{{b}^{2}}}=\\frac{{{(a+b)}^{2}}-2ab}{{{(ab)}^{2}}}=\\frac{{{3}^{2}}-2\\times 1}{{{1}^{2}}}=7$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "333", "queId": "d09535c3de96486e9bd2f65ccac84965", "competition_source_list": ["2015年第26届全国希望杯初三竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$4{{x}^{4}}+3{{x}^{3}}-2{{x}^{2}}+3x+4=0$$，则$$x+\\frac{1}{x}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{4}$$ "}], [{"aoVal": "B", "content": "$$-2$$或$$\\frac{5}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$-2$$ "}]], "knowledge_point_routes": ["知识标签->知识点->式->因式分解->因式分解：其他方法", "知识标签->题型->式->整式的乘除->乘法公式->题型：配方思想的运用"], "answer_analysis": ["∵$$4{{x}^{4}}+3{{x}^{3}}-2{{x}^{2}}+3x+4={{(1+x)}^{2}}(4{{x}^{2}}-5x+4)=0$$， ∴$$1+x=0$$或$$4{{x}^{2}}-5x+4=0$$， 当$$1+x=0$$，则$$x=-1$$，∴$$x+\\frac{1}{x}=-2$$． 当$$4{{x}^{2}}-5x+4=0$$，$$\\Delta ={{5}^{2}}-4\\times 4\\times 4=-39\\textless{}0$$，此方程无解． ∴$$x+\\frac{1}{x}=-2$$． 原方程化为$$4\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}} \\right)+3\\left( x+\\frac{1}{x} \\right)-2=0$$． 故选$$\\text{D}$$ "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1357", "queId": "f2c40bed91da4f0a8fb397337e28dd66", "competition_source_list": ["2001年第12届希望杯初一竞赛第6题"], "difficulty": "0", "qtype": "single_choice", "problem": "珠穆朗玛峰峰顶比吐鲁番盆地底部高$$9003$$米，已知，珠穆朗玛峰海拔高度是$$8848$$米，则吐鲁番盆地的海拔高度是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-155$$米 "}], [{"aoVal": "B", "content": "$$155$$米 "}], [{"aoVal": "C", "content": "$$-17851$$米 "}], [{"aoVal": "D", "content": "$$17651$$米 "}]], "knowledge_point_routes": ["课内体系->方法->整体法", "课内体系->知识点->数->有理数->正数和负数->相反意义的量"], "answer_analysis": ["设吐鲁番盆地的海拔高度是$$x$$米，则依题意得方程$$8848-x=9003$$， 所以$$x=8848-9003=--155$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "378", "queId": "6bb76336eb2a4780b4c32ad4f8678484", "competition_source_list": ["2000年第11届希望杯初一竞赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "有如下四个命题： ①有理数的相反数是正数． ②两个同类项的数字系数是相同的． ③两个有理数的和的绝对值大于这两个有理数绝对值的和． ④两个负有理数的比值是正数． 其中真命题有．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$个 "}], [{"aoVal": "B", "content": "$$3$$个 "}], [{"aoVal": "C", "content": "$$2$$个 "}], [{"aoVal": "D", "content": "$$1$$个 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类", "课内体系->知识点->几何图形初步->命题与证明"], "answer_analysis": ["由于有理数$$0$$的相反数是$$0$$，所以``①有理数的相反数是正数''是假命题． 由于同类项只是说字母相同且相同字母的乘方次数也相同的项，所以``②两个同类项的数字系数是相同的''是个假命题．$$\\left\\textbar{} (-2)+(-1) \\right\\textbar=3=\\left\\textbar{} -2 \\right\\textbar+\\left\\textbar{} -1 \\right\\textbar$$，所以``③两个有理数的和的绝对值大于这两个有理数绝对值的和''是假命题．而两个负有理数的比值是正数显然是真命题．所以，$$4$$个命题中只有$$1$$个是真命题． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "617", "queId": "3b41d4ddc8cd4e138a0757f1c174e640", "competition_source_list": ["1992年第9届全国初中数学联赛竞赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$${{x}_{1}}$$，$${{x}_{2}}$$，$${{x}_{3}}$$，$$\\cdots $$，$${{x}_{9}}$$均为正整数，且$${{x}_{1}}\\textless{}{{x}_{2}}\\textless{}\\cdots \\textless{}{{x}_{9}}$$，$${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{9}}=220$$，则当$${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}$$的值最大时，$${{x}_{9}}-{{x}_{1}}$$的最小值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数加法->有理数加法运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数加法->有理数的加法法则", "课内体系->能力->运算能力"], "answer_analysis": ["先证$${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{5}}\\leqslant 110$$． 若不然，设$${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{5}}\\textgreater110$$，则$${{x}_{5}}=25$$，从而$${{x}_{6}}\\geqslant 26$$，$${{x}_{7}}\\geqslant 27$$，$${{x}_{8}}\\geqslant 28$$，$${{x}_{9}}\\geqslant 29$$．于是，与假设矛盾． 若取$${{x}_{1}}=20$$，$${{x}_{2}}=21$$，$${{x}_{3}}=22$$，$${{x}_{4}}=23$$，$${{x}_{5}}=24$$，则所以当$${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{5}}$$取得最大值时，$${{x}_{1}}$$最大的值是$$20$$． 若取$${{x}_{6}}=27$$，$${{x}_{7}}=27$$，$${{x}_{8}}=28$$，$${{x}_{9}}=29$$，则$${{x}_{6}}+{{x}_{7}}+{{x}_{8}}+{{x}_{9}}=110$$． 所以，当$${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{5}}$$取得最大值时，$${{x}_{9}}$$最小的值是$$29$$． 因此$${{x}_{9}}-{{x}_{1}}$$的最小值是$$29-20=9$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "507", "queId": "50c44989eb674fb3813cbaaa78bc2962", "competition_source_list": ["2001年第18届全国初中数学联赛竞赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x$$、$$y$$是正整数，并且$$xy+x+y=23$$，$${{x}^{2}}y+x{{y}^{2}}=120$$，则$${{x}^{2}}+{{y}^{2}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$34$$ "}], [{"aoVal": "C", "content": "$$35$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["知识标签->题型->方程与不等式->一元二次方程->根与系数的关系->题型：韦达定理应用", "知识标签->知识点->方程与不等式->一元二次方程->一元二次方程根的判别式", "知识标签->知识点->方程与不等式->一元二次方程->一元二次方程的根与系数的关系"], "answer_analysis": ["由于$$xy+\\left( x+y \\right)=23$$，$$xy\\cdot \\left( x+y \\right)=120$$， 则$$x$$，$$y$$与$$\\left( x+y \\right)$$为方程$${{t}^{2}}-23t+120=0$$的两个根，得到$${{t}_{1}}=8$$，$${{t}_{2}}=15$$， 即$$xy=8$$，$$x+y=15$$ ①，或者$$xy=15$$，$$x+y=8$$ ②， ①的时候$$x$$，$$y$$为方程$${{u}^{2}}-15u+8=0$$的根，$${{\\Delta }_{1}}={{15}^{2}}-32=193$$，不是完全平方数，$$xy$$不可能为题目中要求的正整数，舍， ②的时候$$x$$，$$y$$为方程$${{u}^{2}}-8u+15=0$$的根，$${{u}_{1}}=3$$，$${{u}_{2}}=5$$， 故$${{x}^{2}}+{{y}^{2}}={{\\left( x+y \\right)}^{2}}-2xy=34$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "603", "queId": "b548f6fa5e6e4896ae9a553e0c2f1d91", "competition_source_list": ["2019~2020学年河南郑州金水区郑州八中初一上学期期中第8题3分", "初一上学期单元测试《解读绝对值》第3题", "2020~2021学年9月湖北宜昌伍家岗区宜昌英杰学校初一上学期月考（英杰教育集团）第11题3分", "2017年湖南长沙天心区初二竞赛长郡教育集团抵达杯第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a$$，$$b$$，$$c$$是非零有理数，且$$a+b+c=0$$，则$$\\frac{a}{\\textbar a\\textbar}+\\frac{b}{\\textbar b\\textbar}+\\frac{c}{\\textbar c\\textbar}+\\frac{abc}{\\textbar abc\\textbar}$$所有可能的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$或$$-1$$ "}], [{"aoVal": "C", "content": "$$2$$或$$-2$$ "}], [{"aoVal": "D", "content": "$$0$$或$$-2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->|a|/a的化简"], "answer_analysis": ["∵$$a+b+c=0$$且$$a$$、$$b$$、$$c$$均$$\\ne 0$$， ∴$$a$$、$$b$$、$$c$$三数符号为两正一负或两负一正， 不妨设， ①$$a$$，$$b$$为正，$$c$$为负，此时$$abc$$为负： $$\\frac{a}{\\left\\textbar{} a \\right\\textbar}+\\frac{b}{\\left\\textbar{} b \\right\\textbar}+\\frac{c}{\\left\\textbar{} c \\right\\textbar}+\\frac{abc}{\\left\\textbar{} abc \\right\\textbar}$$ $$=1+1+(-1)+(-1)$$ $$=0$$． ②$$a$$，$$b$$为负，$$c$$为正，此时$$abc$$为正， $$\\frac{a}{\\left\\textbar{} a \\right\\textbar}+\\frac{b}{\\left\\textbar{} b \\right\\textbar}+\\frac{c}{\\left\\textbar{} c \\right\\textbar}+\\frac{abc}{\\left\\textbar{} abc \\right\\textbar}$$ $$=-1+(-1)+1+1$$ $$=0$$． 综上原式$$=0$$，故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "103", "queId": "6b11147d9ef04b1785092bd258548ef3", "competition_source_list": ["2011年第22届全国希望杯初一竞赛复赛第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\left\\textbar{} x \\right\\textbar\\leqslant 3$$，$$\\left\\textbar{} y \\right\\textbar\\leqslant 1$$，$$\\left\\textbar{} z \\right\\textbar\\leqslant 4$$且$$\\left\\textbar{} x-2y+z \\right\\textbar=9$$，则$${{x}^{2}}{{y}^{2011}}{{z}^{3}}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$432$$ "}], [{"aoVal": "B", "content": "$$576$$ "}], [{"aoVal": "C", "content": "$$-432$$ "}], [{"aoVal": "D", "content": "$$-576$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->幂的运算->幂的综合运算", "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["只要$$\\left\\textbar{} x \\right\\textbar\\textless{}3$$，$$\\left\\textbar{} y \\right\\textbar\\textless{}1$$，$$\\left\\textbar{} z \\right\\textbar\\textless{}4$$中至少有一个成立， 则$$\\left\\textbar{} x-2y+z \\right\\textbar\\leqslant \\left\\textbar{} x \\right\\textbar+2\\left\\textbar{} y \\right\\textbar+\\left\\textbar{} z \\right\\textbar\\textless{}9$$， 这与$$\\left\\textbar{} x-2y+z \\right\\textbar=9$$矛盾， 所以当且仅当$$\\left\\textbar{} x \\right\\textbar=3$$，$$\\left\\textbar{} y \\right\\textbar=1$$，$$\\left\\textbar{} z \\right\\textbar=4$$同时成立时，才能满足$$\\left\\textbar{} x-2y+z \\right\\textbar=9$$， 此时$$x=3$$，$$y=-1$$，$$z=4$$或$$x=-3$$，$$y=1$$，$$z=-4$$， 所以$${{x}^{2}}{{y}^{2011}}{{z}^{3}}={{3}^{2}}\\times {{(-1)}^{2011}}\\times {{4}^{3}}=-576$$或$${{x}^{2}}{{y}^{2011}}{{z}^{3}}={{(-3)}^{2}}\\times {{1}^{2011}}\\times {{(-4)}^{3}}=-576$$， 综上，$${{x}^{2}}{{y}^{2011}}{{z}^{3}}$$的值是$$-576$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "221", "queId": "0db891868d8741afad969ff8d9ba9e7d", "competition_source_list": ["2002年竞赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$，$$b$$，$$c$$为实数，$$x={{a}^{2}}-2b+\\frac{ \\pi }{3}$$，$$y={{b}^{2}}-2c+\\frac{ \\pi }{6}$$，$$z={{c}^{2}}-2a+\\frac{ \\pi }{2}$$，则$$x$$，$$y$$，$$z$$中，至少有一个值（ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "大于$$0$$ "}], [{"aoVal": "B", "content": "等于$$0$$ "}], [{"aoVal": "C", "content": "不大于$$0$$ "}], [{"aoVal": "D", "content": "小于$$0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->含乘方的有理数混合运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->有理数乘方运算", "课内体系->能力->运算能力"], "answer_analysis": ["$$x+y+z$$ $$={{a}^{2}}-2a+{{b}^{2}}-2b+{{c}^{2}}-2c+\\pi $$ $$={{(a-1)}^{2}}+{{(b-1)}^{2}}+{{(c-1)}^{2}}+(\\pi -3)$$ $$\\textgreater0$$， $$x$$，$$y$$，$$z$$至少有$$1$$个值大于$$0$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1516", "queId": "d3f1ef6f47264eb1b46195cf688cce07", "competition_source_list": ["2012年第17届华杯赛初一竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果正整数$$x$$与$$y$$使得$$\\frac{x{{y}^{2}}}{x+y}$$的值为质数，那么$$x+y$$共有种可能的值．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["设$$(x,y)=d$$，则存在正整数$$a$$，$$b$$，使得$$x=da$$，$$y=db$$，不妨设$$a\\geqslant b$$． 于是质数： $$p=\\frac{x{{y}^{2}}}{x+y}=\\frac{da{{(db)}^{2}}}{ad+db}=\\frac{{{d}^{2}}a{{b}^{2}}}{a+b}$$，即$${{d}^{2}}a{{b}^{2}}=p(a+b)$$．① 由于$$(a,a+b)=(b,a+b)=(a,b)=1$$，且$$a{{b}^{2}}\\textbar p(a+b)$$， 所以，$$a{{b}^{2}}\\textbar p$$，进而$$a=p$$，$$b=1$$，或$$a=b=1$$． ①当$$a=p$$，$$b=1$$时，$${{d}^{2}}=p+1$$． 所以$$(d+1)(d-1)=p$$．进而，$$d-1=1$$，$$d+1=p$$，$$d=2$$，$$p=3$$． 此时$$x=6$$，$$y=2$$，$$x+y=8$$． ②当$$a=b=1$$时，$${{d}^{2}}=2p$$．所以$$d=2$$，$$p=2$$． 此时$$x=2$$，$$y=2$$．$$x+y=4$$． 所以$$x+y$$共有两种可能的值． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1208", "queId": "8aac49074e724b45014e87cc50d55157", "competition_source_list": ["1996年第7届全国希望杯初一竞赛复赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "在某浓度的盐水中加入一杯水后，得到新盐水，它的浓度为$$20 \\%$$，又在新盐水中加入与前述一杯水的重量相等的纯盐合，盐水浓度变为$$33\\frac{1}{3} \\%$$，那么原来盐水的浓度是．", "answer_option_list": [[{"aoVal": "A", "content": "$$23 \\%$$ "}], [{"aoVal": "B", "content": "$$25 \\%$$ "}], [{"aoVal": "C", "content": "$$30 \\%$$ "}], [{"aoVal": "D", "content": "$$32 \\%$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->二元一次方程（组）->二元一次方程组应用题->二元一次方程组的数字问题"], "answer_analysis": ["设原盐水溶液为$$a$$克，其中含纯盐$$m$$克，后加入``一杯水''为$$x$$克， 依题意得$$\\begin{cases}(a+x)\\times 20 \\%=m (a+x+x)\\times 33\\dfrac{1}{3} \\%=m+x \\end{cases}$$， 整理得$$\\begin{cases}a+x=5m a-x=3m \\end{cases}$$， 解得$$2a=8m$$，∴$$a=4m$$． 于是原盐水的浓度为$$\\frac{m}{a}=\\frac{m}{4m}=25 \\%$$，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "696", "queId": "5af563f916fd4fd78df0b7d741bf7be8", "competition_source_list": ["2019年浙江杭州拱墅区杭州市文澜中学初二竞赛第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在平面直角坐标系中，若点$$P(m-2,m+1)$$在第二象限，则$$m$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$m ~\\textless~ -1$$ "}], [{"aoVal": "B", "content": "$$m\\textgreater2$$ "}], [{"aoVal": "C", "content": "$$-1 ~\\textless{} ~m ~\\textless{} ~2$$ "}], [{"aoVal": "D", "content": "$$m\\textgreater-1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->平面直角坐标系->坐标系基础->象限内坐标的特征"], "answer_analysis": ["∵$$P(m-2,m+1)$$在第二象限， ∴$$\\begin{cases}m-2 ~\\textless{} ~0 m+1\\textgreater0 \\end{cases}$$． ∴$$-1 ~\\textless{} ~m ~\\textless{} ~2$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "179", "queId": "0cc87f7059f840e99a8c431814252ad1", "competition_source_list": ["2009年第20届希望杯初二竞赛第2试第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "若不等式组$$\\begin{cases}-x+4m\\textless{}x+10 x+1\\textgreater m \\end{cases}$$的解集是$$x\\textgreater4$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$m\\leqslant \\frac{9}{2}$$ "}], [{"aoVal": "B", "content": "$$m\\leqslant 5$$ "}], [{"aoVal": "C", "content": "$$m=\\frac{9}{2}$$ "}], [{"aoVal": "D", "content": "$$m=5$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->不等式->一次含参不等式"], "answer_analysis": ["$$-x+4m\\textless{}x+10$$，得$$x\\textgreater2m-5$$， $$x+1\\textgreater m$$，得$$x\\textgreater m-1$$， 因为不等式组的解集是$$x\\textgreater4$$，所以必须是$$\\begin{cases}2m-5\\leqslant 4 m-1\\leqslant 4 \\end{cases}$$成立，且其中至少有一个等式成立， 若$$2m-5=4$$，则$$m=\\frac{9}{2}$$，且$$m=\\frac{9}{2}$$，使得$$m-1\\leqslant 4$$成立； 若$$m-1=4$$，则$$m=5$$，当$$m=5$$时，$$2m-5\\leqslant 4$$不成立，所以$$m=5$$不符合原不等式组． 综上得，当$$m=\\frac{9}{2}$$时，原不等式组的解集是$$x\\textgreater4$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1215", "queId": "92f332b3318840d88561c07f096342de", "competition_source_list": ["2008年第19届希望杯初二竞赛第1试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "若单项式$$-2{{a}^{\\textbar x\\textbar}}{{b}^{\\textbar5x\\textbar}}$$和$${{3}^{2}}{{a}^{2}}{{b}^{3-x}}$$的次数相同，则$$x$$的整数值等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$\\pm 1$$ "}], [{"aoVal": "D", "content": "$$\\pm 1$$以外的数 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->整式的乘除运算"], "answer_analysis": ["由题意，知$$\\textbar x\\textbar+\\textbar5x\\textbar=2+(3-x)$$， 当$$x\\geqslant 0$$时，则$$6x=5-x$$， 解得$$x=\\frac{5}{7}$$，不合题意，舍去． 当$$x\\textless{}0$$时，则$$-6x=5-x$$， 解得$$x=-1$$，符合题意． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "867", "queId": "9aa71db43efb4b998b06dd453a2bab04", "competition_source_list": ["2006年第17届希望杯初一竞赛初赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有如下四个命题： ①两个符号相反的分数之间至少有一个正整数； ②两个符号相反的分数之间至少有一个负整数； ③两个符号相反的分数之间至少有一个整数； ④两个符号相反的分数之间至少有一个有理数． 其中真命题的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类", "课内体系->知识点->几何图形初步->命题与证明", "课内体系->能力->推理论证能力"], "answer_analysis": ["如$$-\\frac{1}{2}$$和$$\\frac{1}{2}$$之间既没有正整数，也没有负整数， 所以命题①、②不正确．$$0$$介于两个符号相反的分数之间，所以命题③、④正确． 故选($$\\text{B}$$)． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "353", "queId": "2bcac3fe58f64092ba12d22405639b45", "competition_source_list": ["2006年第17届希望杯初二竞赛第2试第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a$$、$$b$$、$$c$$都是大于$$1$$的自然数，且$${{a}^{c}}=252b$$，则$$a$$的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$42$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题", "竞赛->知识点->数论->整除->因数与倍数"], "answer_analysis": ["因为$$252={{2}^{2}}\\times {{3}^{2}}\\times 7$$， 所以，要使$${{a}^{c}}=252b$$成立，$$a$$中一定含有$$2$$，$$3$$，$$7$$这三个因数， 所以$$a$$的最小值是$$2\\times 3\\times 7=42$$， 如取$$a=42$$，$$b=7$$，$$c=2$$， 得$${{42}^{2}}=252\\times 7$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "817", "queId": "acc6571216fb4a57844c12626cf1d845", "competition_source_list": ["2009年第20届希望杯初二竞赛第2试第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "任何一个正整数$$n$$都可以写成两个正整数相乘的形式，对于两个乘数的差的绝对值最小的一种分解$$n=p\\times q$$（$$p\\leqslant q$$)可称为正整数$$n$$的最佳分解，并规定$$F(n)=\\frac{p}{q}$$．如：$$12=1\\times 12=2\\times 6=3\\times 4$$，则$$F(12)=\\frac{3}{4}$$． 则在以下结论 ①$$F(2)=\\frac{1}{2}$$． ②$$F(24)=\\frac{3}{8}$$． ③若$$n$$是一个完全平方数，则$$F(n)=1$$． ④若$$n$$是一个完全立方数，即$$n={{a}^{3}}$$（$$a$$是正整数)，则$$F(n)=\\frac{1}{a}$$中，正确的结论有．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$个 "}], [{"aoVal": "B", "content": "$$3$$个 "}], [{"aoVal": "C", "content": "$$2$$个 "}], [{"aoVal": "D", "content": "$$1$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->数论->同余->完全平方数"], "answer_analysis": ["因为$$2=1\\times 2$$，所以$$F(2)=\\frac{1}{2}$$，①正确； 因为$$24=1\\times 24=2\\times 12=3\\times 8=4\\times 6$$，所以$$F(24)=\\frac{4}{6}=\\frac{2}{3}$$，②错误； 若$$n$$是一个完全平方数，设$$n={{a}^{2}}$$，则$$F(n)=\\frac{a}{a}=1$$，③正确； 若$$n$$是一个完全立方数，$$F(n)$$可能不等于$$\\frac{1}{a}$$，如$$64=43=8\\times 8$$，$$F(64)=1$$，④错误． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "565", "queId": "559d4eb851e8462fb138dcad01350420", "competition_source_list": ["2014年第25届全国希望杯初一竞赛初赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a+b+c=0$$，则$$\\frac{\\left\\textbar{} a \\right\\textbar}{a}+\\frac{\\left\\textbar{} b \\right\\textbar}{b}+\\frac{\\left\\textbar{} c \\right\\textbar}{c}+\\frac{\\left\\textbar{} ab \\right\\textbar}{ab}+\\frac{\\left\\textbar{} ac \\right\\textbar}{ac}+\\frac{\\left\\textbar{} bc \\right\\textbar}{bc}+\\frac{\\left\\textbar{} abc \\right\\textbar}{abc}$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-7$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->绝对值的非负性", "课内体系->知识点->数->有理数->绝对值->|a|/a的化简"], "answer_analysis": ["方法一： 依题意，$$a$$，$$b$$，$$c$$都不为$$0$$， ∵$$a+b+c=0$$， ∴$$a$$，$$b$$，$$c$$中既有正数，又有负数． 若$$a$$，$$b$$，$$c$$中只有一个正数，不妨设$$a\\textgreater0$$，则$$b\\textless{}0$$，$$c\\textless{}0$$， 则$$\\frac{\\left\\textbar{} a \\right\\textbar}{a}+\\frac{\\left\\textbar{} b \\right\\textbar}{b}+\\frac{\\left\\textbar{} c \\right\\textbar}{c}+\\frac{\\left\\textbar{} ab \\right\\textbar}{ab}+\\frac{\\left\\textbar{} ac \\right\\textbar}{ac}+\\frac{\\left\\textbar{} bc \\right\\textbar}{bc}+\\frac{\\left\\textbar{} abc \\right\\textbar}{abc}$$ $$=1-1-1-1-1+1+1$$ $$=-1$$． 若$$a$$，$$b$$，$$c$$中只有一个负数，不妨设$$a\\textless{}0$$，则$$b\\textgreater0$$，$$c\\textgreater0$$， 则$$\\frac{\\left\\textbar{} a \\right\\textbar}{a}+\\frac{\\left\\textbar{} b \\right\\textbar}{b}+\\frac{\\left\\textbar{} c \\right\\textbar}{c}+\\frac{\\left\\textbar{} ab \\right\\textbar}{ab}+\\frac{\\left\\textbar{} ac \\right\\textbar}{ac}+\\frac{\\left\\textbar{} bc \\right\\textbar}{bc}+\\frac{\\left\\textbar{} abc \\right\\textbar}{abc}$$ $$=-1+1+1-1-1+1-1$$ $$=-1$$． 方法二： 取特殊值． ∵$$a+b+c=0$$，不妨设$$a=2$$，$$b=c=-1$$， 则$$\\frac{\\left\\textbar{} a \\right\\textbar}{a}+\\frac{\\left\\textbar{} b \\right\\textbar}{b}+\\frac{\\left\\textbar{} c \\right\\textbar}{c}+\\frac{\\left\\textbar{} ab \\right\\textbar}{ab}+\\frac{\\left\\textbar{} ac \\right\\textbar}{ac}+\\frac{\\left\\textbar{} bc \\right\\textbar}{bc}+\\frac{\\left\\textbar{} abc \\right\\textbar}{abc}$$ $$=1-1-1-1-1+1+1$$ $$=-1$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "25", "queId": "019f34295c3349c7a85f5d4766de4ade", "competition_source_list": ["2007年第18届希望杯初一竞赛复赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "假定未拧紧的水龙头每秒钟渗出$$2$$滴水，每滴水约$$0.05$$毫升．现有一个水龙头未拧紧，$$4$$小时后���才被发现并拧紧，在这段时间内，水龙头共滴水约（ ）（用科学记数法表示，结果保留两位有效数字）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1440$$毫升 "}], [{"aoVal": "B", "content": "$$1.4\\times {{10}^{3}}$$毫升 "}], [{"aoVal": "C", "content": "$$0.14\\times {{10}^{4}}$$毫升 "}], [{"aoVal": "D", "content": "$$14\\times {{10}^{2}}$$毫升 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->科学记数法->科学记数法：表示较大的数"], "answer_analysis": ["$$0.05\\times 2\\times 4\\times 60\\times 60=1440=1.4\\times {{10}^{3}}$$（毫升）． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "526", "queId": "556ac3d0fb5e47d1b23e64c574062c18", "competition_source_list": ["2013年山东青岛黄岛区山东省青岛第九中学自主招生", "2018年浙江宁波余姚市余姚市实验学校初二竞赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$x=\\frac{\\sqrt{5}-3}{2}$$，则代数式$$x(x+1)\\left( x+2 \\right)\\left( x+3 \\right)$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->二次根式->二次根式的性质与运算"], "answer_analysis": ["$$x=\\frac{\\sqrt{5}-3}{2}$$，$$2x=\\sqrt{5}-3$$ ，$$2x+3=\\sqrt{5}$$， 两边同时平方整理得，$${{x}^{2}}+3x=-1$$． $$x(x+1)(x+2)(x+3)=({{x}^{2}}+3x)({{x}^{2}}+3x+2)$$$$=-1\\times (-1+2)=-1$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1199", "queId": "c9395fc2390c452eb1c5f6f490bedc7e", "competition_source_list": ["1993年第10届全国初中数学联赛竞赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\sqrt[3]{3}{{\\left( \\sqrt[3]{\\frac{4}{9}}-\\sqrt[3]{\\frac{2}{9}}+\\sqrt[3]{\\frac{1}{9}} \\right)}^{-1}}$$可以化简成（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt[3]{3}\\left( \\sqrt[3]{2}+1 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\sqrt[3]{3}\\left( \\sqrt[3]{2}-1 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\sqrt[3]{2}-1$$ "}], [{"aoVal": "D", "content": "$$\\sqrt[3]{2}+1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式化简求值", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算", "课内体系->能力->运算能力"], "answer_analysis": ["原式$$={{3}^{\\frac{1}{3}}}{{\\left( \\frac{1}{9} \\right)}^{-\\frac{1}{3}}}{{\\left( {{2}^{\\frac{2}{3}}}-{{2}^{\\frac{1}{3}}}+1 \\right)}^{-1}}$$ $$=3\\left[ \\frac{{{\\left( {{2}^{\\frac{1}{3}}} \\right)}^{3}}+1}{{{2}^{\\frac{1}{3}}}+1} \\right]^{-1}$$ $$={{2}^{\\frac{1}{3}}}+1$$ $$=\\sqrt[3]{2}+1$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "374", "queId": "5de0df3b26874de591aa3f675e42bbc5", "competition_source_list": ["2016年上海浦东新区进才中学自主招生第6题10分", "2016年第33届全国全国初中数学联赛竞赛第2题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "三种图书的单价分别为$$10$$元、$$15$$元和$$20$$元，某学校计划恰好用$$500$$元购买上述图书$$30$$本，那么不同的购书方案有种．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$12$$ "}], [{"aoVal": "E", "content": "$$13$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->其他方程->三元一次方程组->三元一次方程组的实际应用"], "answer_analysis": ["设购买三种图书的数量分别为$$x$$，$$y$$，$$z$$， 则$$\\left { \\begin{array}{*{35}{l}} x+y+z=30 10x+15y+20z=500 \\end{array} \\right.$$， 即$$\\begin{cases}y+z=30-x 3y+4z=100-2x \\end{cases}$$， 解得$$\\begin{cases}y=20-2x z=10+x \\end{cases}$$， 依题意得，$$x$$，$$y$$，$$z$$，为自然数（非负整数）， 故$$0\\leqslant x\\leqslant 10$$，$$x$$有$$11$$种可能的取值（分别为$$0$$，$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$10$$）， 对于每一个$$x$$值，$$y$$和$$z$$都有唯一的值（自然数）相对应， 即不同的购书方案共有$$11$$种． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1594", "queId": "fea48df0cddc4ccda73ea6d0c4040b43", "competition_source_list": ["初一下学期其它", "1994年第11届全国初中数学联赛竞赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$x=\\frac{1+\\sqrt{1994}}{2}$$时，多项式$${{(4{{x}^{3}}-1997x-1994)}^{2001}}$$的��为（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$${{2}^{2001}}$$ "}], [{"aoVal": "D", "content": "$$-{{2}^{2001}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式化简求值->二次根式的化简求值——利用完全平方"], "answer_analysis": ["∵$$x=\\frac{1+\\sqrt{1994}}{2}$$， ∴$${{(2x-1)}^{2}}=1994$$， 即$$4{{x}^{2}}-4x-1993=0$$， ∴$${{(4{{x}^{3}}-1997x-1994)}^{2001}}$$ $$={{[(4{{x}^{2}}-4x-1993)x+(4{{x}^{2}}-4x-1993)-1]}^{2001}}$$ $$={{(-1)}^{2001}}$$ $$=-1$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1118", "queId": "ff8080814d9efd56014daa79ad010ac2", "competition_source_list": ["1993年第4届全国希望杯初一竞赛初赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$a=(\\frac{-3.14}{3.13})\\div 3.12$$，$$b=(\\frac{2.14}{-2.13})\\div 2.12$$，$$c=(\\frac{1.14}{1.13})\\div (-1.12)$$，则$$a$$，$$b$$，$$c$$的大小关系是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b\\textgreater c$$ "}], [{"aoVal": "B", "content": "$$a\\textgreater c\\textgreater b$$ "}], [{"aoVal": "C", "content": "$$b\\textgreater c\\textgreater a$$ "}], [{"aoVal": "D", "content": "$$c\\textgreater b\\textgreater a$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性", "课内体系->能力->运算能力"], "answer_analysis": ["容易看出$$a$$，$$b$$，$$c$$均为负数，我们看$$\\left\\textbar{} a \\right\\textbar$$，$$\\left\\textbar{} b \\right\\textbar$$，$$\\left\\textbar{} c \\right\\textbar$$． ∵$$\\frac{3.14}{3.13}=1+\\frac{1}{313}$$，$$\\frac{2.14}{2.13}=1+\\frac{1}{213}$$，$$\\frac{1.14}{1.13}=1+\\frac{1}{113}$$， 又$$313\\textgreater213\\textgreater113$$，知$$\\frac{3.14}{3.13}\\textless{}\\frac{2.14}{2.13}\\textless{}\\frac{1.14}{1.13}$$， 又$$3.13\\textgreater2.13\\textgreater1.13$$， ∴$$\\frac{3.14}{3.13}\\div 3.13\\textless{}\\frac{2.14}{2.13}\\div 2.13\\textless{}\\frac{1.14}{1.13}\\div 1.13$$， ∴$$\\left\\textbar{} a \\right\\textbar\\textless{}\\left\\textbar{} b \\right\\textbar\\textless{}\\left\\textbar{} c \\right\\textbar$$， ∴$$a\\textgreater b\\textgreater c$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "762", "queId": "45599f2290fc4b9fa45cd6d82c88338d", "competition_source_list": ["1993年第10届全国初中数学联赛竞赛第4题", "初一上学期其它"], "difficulty": "2", "qtype": "single_choice", "problem": "$${{x}_{1}}$$、$${{x}_{2}}$$、$${{x}_{3}}$$、$${{x}_{4}}$$、$${{x}_{5}}$$满足方程组$$\\begin{cases}{{x}_{1}}+{{x}_{2}}+{{x}_{3}}={{a}_{1}} {{x}_{2}}+{{x}_{3}}+{{x}_{4}}={{a}_{2}} {{x}_{3}}+{{x}_{4}}+{{x}_{5}}={{a}_{3}} {{x}_{4}}+{{x}_{5}}+{{x}_{1}}={{a}_{4}} {{x}_{5}}+{{x}_{1}}+{{x}_{2}}={{a}_{5}} \\end{cases}$$， 其中$${{a}_{1}}$$、$${{a}_{2}}$$、$${{a}_{3}}$$、$${{a}_{4}}$$、$${{a}_{5}}$$是常数，且$${{a}_{1}}\\textgreater{{a}_{2}}\\textgreater{{a}_{3}}\\textgreater{{a}_{4}}\\textgreater{{a}_{5}}$$，则$${{x}_{1}}$$、$${{x}_{2}}$$、$${{x}_{3}}$$、$${{x}_{4}}$$、$${{x}_{5}}$$的大小顺序是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{x}_{1}}\\textgreater{{x}_{2}}\\textgreater{{x}_{3}}\\textgreater{{x}_{4}}\\textgreater{{x}_{5}}$$ "}], [{"aoVal": "B", "content": "$${{x}_{4}}\\textgreater{{x}_{2}}\\textgreater{{x}_{1}}\\textgreater{{x}_{3}}\\textgreater{{x}_{5}}$$ "}], [{"aoVal": "C", "content": "$${{x}_{3}}\\textgreater{{x}_{1}}\\textgreater{{x}_{4}}\\textgreater{{x}_{2}}\\textgreater{{x}_{5}}$$ "}], [{"aoVal": "D", "content": "$${{x}_{5}}\\textgreater{{x}_{3}}\\textgreater{{x}_{1}}\\textgreater{{x}_{4}}\\textgreater{{x}_{2}}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->其他方程->三元一次方程组->解多元一次方程组", "课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->加减消元法解二元一次方程组"], "answer_analysis": ["\\textbf{（知识点：解多元一次方程组）} 给定方程组中的方程按顺序两两相减分别得： $${{x}_{1}}-{{x}_{4}}={{a}_{1}}-{{a}_{2}}$$，$${{x}_{2}}-{{x}_{5}}={{a}_{2}}-{{a}_{3}}$$，$${{x}_{3}}-{{x}_{1}}={{a}_{3}}-{{a}_{4}}$$，$${{x}_{4}}-{{x}_{2}}={{a}_{4}}-{{a}_{5}}$$， 由于$${{a}_{1}}\\textgreater{{a}_{2}}\\textgreater{{a}_{3}}\\textgreater{{a}_{4}}\\textgreater{{a}_{5}}$$，有$${{x}_{1}}\\textgreater{{x}_{4}}$$，$${{x}_{2}}\\textgreater{{x}_{5}}$$，$${{x}_{3}}\\textgreater{{x}_{1}}$$，$${{x}_{4}}\\textgreater{{x}_{2}}$$，因此$${{x}_{3}}\\textgreater{{x}_{1}}\\textgreater{{x}_{4}}\\textgreater{{x}_{2}}\\textgreater{{x}_{5}}$$． 所以选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1213", "queId": "c942880b33f942548e9397704b44ea28", "competition_source_list": ["2002年第13届希望杯初二竞赛第2试第3题", "初一上学期单元测试《几何图形初步》角第16题"], "difficulty": "2", "qtype": "single_choice", "problem": "上午九点钟的时候，时针与分针成直角，那么下一次时针与分针成直角的时间是． At 9:00 a.m., the hour hand is at right angles to the minute hand, so the next time the hour hand is at right angles to the minute hand is ．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$:$$30$$ "}], [{"aoVal": "B", "content": "$$10$$:$$5$$ "}], [{"aoVal": "C", "content": "$$10$$:$$5\\frac{5}{11}$$ "}], [{"aoVal": "D", "content": "$$9$$:$$32\\frac{8}{11}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的钟表问题", "课内体系->能力->运算能力", "课内体系->能力->推理论证能力", "课内体系->能力->分析和解决问题能力"], "answer_analysis": ["设再次转成直角的时间间隔为$$x$$，则 $$(6-\\frac{1}{2})x=180$$， ∴$$x=32\\frac{8}{11}$$． 所以下一次时针与分针成直角的时间为$$9$$时$$32\\frac{8}{11}$$分． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1378", "queId": "d76bc00d030149059628a06fe5339748", "competition_source_list": ["其它", "2009年第20届希望杯初二竞赛第2试第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "电影票有$$10$$元、$$15$$元、$$20$$元三种票价，班长用$$500$$元买了$$30$$张电影票，其中票价为$$20$$元的比票价为$$10$$元的多（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$张 "}], [{"aoVal": "B", "content": "$$15$$张 "}], [{"aoVal": "C", "content": "$$10$$张 "}], [{"aoVal": "D", "content": "$$5$$张 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->其他方程->三元一次方程组->解三元一次方程组", "课内体系->知识点->方程与不等式->其他方程->三元一次方程组->三元一次方程组的实际应用"], "answer_analysis": ["设购买$$10$$元、$$15$$元、$$20$$元的电影票分别为$$x$$、$$y$$、$$z$$张， 则根据题意，得$$\\begin{cases}\\textasciitilde x+y+z=10 ① \\textasciitilde10x+15y+20z=500 ②\\end{cases}$$， 方法一：①$$-$$②$$\\times 10$$得， $$5y+10z=200$$， ∴$$z=\\frac{40-y}{2}=20-\\frac{y}{2}$$． ①$$\\times 20-$$②得， $$10x+5y=100$$， ∴$$x=\\frac{20-y}{2}=10-\\frac{y}{2}$$． 方法二：②$$-$$①$$\\times 15$$得， $$-5x+5z=50$$ ∴$$z-x=10$$． ∴票价为$$20$$元的比票价为$$10$$元的多$$10$$张． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "783", "queId": "9a498d26d1df418a9210f5e31a1b9db9", "competition_source_list": ["2002年第13届希望杯初二竞赛第2试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "有理数$$a$$，$$b$$，$$c$$满足下列条件：$$a+b+c=0$$且$$abc ~\\textless{} ~0$$，那么$$\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$$的值．", "answer_option_list": [[{"aoVal": "A", "content": "是正数 "}], [{"aoVal": "B", "content": "是零 "}], [{"aoVal": "C", "content": "是负数 "}], [{"aoVal": "D", "content": "不能确定是正数、负数或$$0$$． "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的运算->分式加减、乘除、乘方混合运算", "课内体系->知识点->式->分式->分式化简求值->分式恒等变形", "课内体系->能力->运算能力"], "answer_analysis": ["由$$abc ~\\textless{} ~0$$知$$a$$、$$b$$、$$c$$均不为$$0$$． $$\\therefore $$ $${{\\left( a+b+c \\right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\\left( ab+bc+ac \\right)=0$$， $$\\therefore ab+bc+ca=-\\frac{1}{2}\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\right) ~\\textless{} ~0$$， $$\\therefore $$$$\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=\\frac{bc+ac+ab}{abc}\\textgreater0$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1228", "queId": "c041b9458f174cf68b3ce824d348d249", "competition_source_list": ["2006年全美数学竞赛（AMC）竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "羽毛球单打比赛一共有$$6$$名选手参赛．每名选手只和其他选手比赛一次，没有平局．如果$$A$$选手赢了$$4$$场，$$B$$选手赢了$$3$$场，$$C$$选手赢了$$3$$场，$$D$$选手赢了$$2$$场，$$E$$选手赢了$$2$$场，那么第六位选手赢了场．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}], [{"aoVal": "E", "content": "$$4$$ "}]], "knowledge_point_routes": ["美国amc8->知识点->组合->逻辑推理->体育比赛"], "answer_analysis": ["由于有$$6$$名选手，总共有$$\\frac{6(6-1)}{2}=15$$场比赛．到目前为止，已经有$$4+3+3+2+2=14$$场比赛结束了（每场比赛一个人赢），所以第六位选手需要赢$$15-14=1$$场． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1079", "queId": "ff8080814d95366b014d988f71d213cf", "competition_source_list": ["2011年第22届全国希望杯初二竞赛复赛第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$${{a}^{2}}-a=7$$，则代数式$$\\frac{a-1}{a+2}\\centerdot \\frac{{{a}^{2}}-4}{{{a}^{2}}-2a+1}\\div \\frac{1}{{{a}^{2}}-1}$$的值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$\\frac{7}{2}$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的运算->分式乘除混合运算", "课内体系->知识点->式->分式->分式的运算->分式加减、乘除、乘方混合运算", "课内体系->能力->运算能力"], "answer_analysis": ["$$\\frac{a-1}{a+2}\\cdot \\frac{{{a}^{2}}-4}{{{a}^{2}}-2a+1}\\div \\frac{1}{{{a}^{2}}-1}$$ $$=\\frac{a-1}{a+2}\\cdot \\frac{(a+2)(a-2)}{{{(a-1)}^{2}}}\\cdot (a+1)(a-1)$$ $$=(a+1)(a-2)$$ $$={{a}^{2}}-a-2$$ $$=7-2$$ $$=5$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1503", "queId": "ef138c99c4824a04882663c839178383", "competition_source_list": ["2016年第27届全国希望杯初二竞赛初赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知不等式$$\\left\\textbar{} x+1 \\right\\textbar\\textgreater2$$与不等式$$\\left\\textbar{} x \\right\\textbar\\leqslant a$$（$$a\\geqslant 0$$）的解集没有公共部分，那么$$a$$的取值范围是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0\\leqslant a\\leqslant 1$$ "}], [{"aoVal": "B", "content": "$$1\\leqslant a\\leqslant 2$$ "}], [{"aoVal": "C", "content": "$$0\\leqslant a\\leqslant 2$$ "}], [{"aoVal": "D", "content": "$$a\\geqslant 1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->解含绝对值的一元一次不等式（组）"], "answer_analysis": ["不等式$$\\left\\textbar{} x+1 \\right\\textbar\\textgreater2$$的解集为$$x\\textless{}-3$$或$$x\\textgreater1$$． $$\\left\\textbar{} x \\right\\textbar\\leqslant a$$（$$a\\geqslant 0$$）的解集为$$-a\\leqslant x\\leqslant a$$． ∵两不等式的解集没有公共部分， ∴$$a\\leqslant 1$$， ∴$$0\\leqslant a\\leqslant 1$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1304", "queId": "97d5f4af6eb440ea94cb1a14f1aa99db", "competition_source_list": ["2001年第12届希望杯初一竞赛第2试第3题"], "difficulty": "0", "qtype": "single_choice", "problem": "我国古代伟大的数学家祖冲之在$$1500$$年以前就已经相当精确地算出圆周率$$\\pi $$是在$$3.1415926$$和$$3.1415927$$之间，并取$$\\frac{355}{113}$$为密率、$$\\frac{22}{7}$$为约率，则（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$3.1415 \\textless{} \\pi \\textless{} \\frac{333}{106}$$ "}], [{"aoVal": "B", "content": "$$\\frac{355}{113} \\textless{} \\pi \\textless{} \\frac{22}{7}$$ "}], [{"aoVal": "C", "content": "$$\\frac{333}{106} \\textless{} \\pi \\textless{} \\frac{355}{113}$$ "}], [{"aoVal": "D", "content": "$$\\frac{22}{7} \\textless{} \\pi \\textless{} 1.429$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->无理数有关的计算->无理数的估算"], "answer_analysis": ["对$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$四个答案进行比较分析即可得到答案$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1513", "queId": "f3c8b87aeb724096b730c5686170897d", "competition_source_list": ["2000年第11届希望杯初一竞赛第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "在某班的新年晚会上，每个同学都写若干字条祝福他人．已知在任意四个人中，每一位都祝福其他三人中的至少一位．那么该班中没有得到其他同学祝福字条的同学最多有位．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$． "}], [{"aoVal": "B", "content": "$$2$$． "}], [{"aoVal": "C", "content": "$$3$$． "}], [{"aoVal": "D", "content": "$$4$$． "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式的加减运算"], "answer_analysis": ["假设有$$A$$、$$B$$两同学没有得到祝福字条，现在任选另外两位同学记为$$C$$和$$D$$，则$$A$$、$$B$$、$$C$$、$$D$$组成一���四人组，由于$$C$$至少祝福了$$A$$、$$B$$、$$D$$中一位，而$$A$$、$$B$$没有得到任何同学的祝福，所以$$D$$得到$$C$$的祝福，同理$$C$$也得到$$D$$的祝福，由$$C$$、$$D$$的任意性可以得出，除$$A$$、$$B$$之外，班上其他同学都得到了祝福字条．以上分析说明，未得到祝福字条的同学不能超过$$2$$人． 在$$A$$、$$B$$给除他俩之外的同学都写了祝福字条，其余同学也都给除$$A$$、$$B$$之外的所有同学写了祝福字条的情况下，题目条件满足，即存在恰有两位同学未得到祝福字条的情况，所以最多有两位同学未得到祝福字条．选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "543", "queId": "3a90801895934115ab29dea816c2ffc3", "competition_source_list": ["2020~2021学年9月湖北宜昌伍家岗区宜昌英杰学校初一上学期月考（英杰教育集团）第11题3分", "初一上学期单元测试《解读绝对值》第3题", "2017年湖南长沙天心区初二竞赛长郡教育集团抵达杯第1题5分", "2019~2020学年河南郑州金水区郑州市第八中学初一上学期期中第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$是非零有理数，且$$a+b+c=0$$，则$$\\frac{a}{\\textbar a\\textbar}+\\frac{b}{\\textbar b\\textbar}+\\frac{c}{\\textbar c\\textbar}+\\frac{abc}{\\textbar abc\\textbar}$$可能的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$或$$-1$$ "}], [{"aoVal": "C", "content": "$$2$$或$$-2$$ "}], [{"aoVal": "D", "content": "$$0$$或$$-2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->绝对值综合", "课内体系->能力->推理论证能力"], "answer_analysis": ["∵$$a+b+c=0$$且$$a$$、$$b$$、$$c$$均$$\\ne 0$$， ∴$$a$$、$$b$$、$$c$$三数符号为两正一负或两负一正， 不妨设， ①$$a$$，$$b$$为正，$$c$$为负，此时$$abc$$为负： $$\\frac{a}{\\left\\textbar{} a \\right\\textbar}+\\frac{b}{\\left\\textbar{} b \\right\\textbar}+\\frac{c}{\\left\\textbar{} c \\right\\textbar}+\\frac{abc}{\\left\\textbar{} abc \\right\\textbar}$$ $$=1+1+(-1)+(-1)$$ $$=0$$． ②$$a$$，$$b$$为负，$$c$$为正，此时$$abc$$为正， $$\\frac{a}{\\left\\textbar{} a \\right\\textbar}+\\frac{b}{\\left\\textbar{} b \\right\\textbar}+\\frac{c}{\\left\\textbar{} c \\right\\textbar}+\\frac{abc}{\\left\\textbar{} abc \\right\\textbar}$$ $$=-1+(-1)+1+1$$ $$=0$$． 综上原式$$=0$$，故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1494", "queId": "c62fbbba557448b0825048e223c8dafe", "competition_source_list": ["2002年第19届全国初中数学联赛竞赛第6题7分"], "difficulty": "3", "qtype": "single_choice", "problem": "如果对于不小于$$8$$的自然数$$n$$，当$$3n+1$$是一个完全平方数时，$$n+1$$都能表示成$$k$$个完全平方数的和，那么$$k$$的最小值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->同余->完全平方数"], "answer_analysis": ["$$3n+1$$是一个完全平方数，那这个完全平方数$$a$$显然不是$$3$$的倍数． 设$$a=3t\\pm 1$$，则$${{a}^{2}}=9{{t}^{2}}\\pm 6t+1$$， $${{a}^{2}}=3n+1=9{{t}^{2}}\\pm 6t+1$$，$$\\textasciitilde\\Rightarrow n=3{{t}^{2}}\\pm 2t$$， $$n+1=3{{t}^{2}}\\pm 2t+1={{(t\\pm 1)}^{2}}+{{t}^{2}}+{{t}^{2}}$$． 而若$$k$$为$$1$$，当$$n=21$$时$$3n+1=64$$是完全平方数，但是$$n+1=22$$不能写成一个或两个完全平方数的和，故$$k$$最小为$$3$$，选择$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "58", "queId": "0465e7938e634c3f87d9dd27426112c5", "competition_source_list": ["初一上学期单元测试《解读绝对值》第3题", "2017年湖南长沙天心区初二竞赛长郡教育集团抵达杯第1题5分", "2020~2021学年9月湖北宜昌伍家岗区宜昌英杰学校初一上学期月考（英杰教育集团）第11题3分", "2019~2020学年河南郑州金水区郑州市第八中学初一上学期期中第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$是非零有理数，且$$a+b+c=0$$，那么$$\\frac{a}{\\textbar a\\textbar}+\\frac{b}{\\textbar b\\textbar}+\\frac{c}{\\textbar c\\textbar}+\\frac{abc}{\\textbar abc\\textbar}$$的所有可能的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$或$$-1$$ "}], [{"aoVal": "C", "content": "$$2$$或$$-2$$ "}], [{"aoVal": "D", "content": "$$0$$或$$-2$$ "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "课内体系->知识点->数->有理数->绝对值->绝对值综合"], "answer_analysis": ["∵$$a+b+c=0$$且$$a$$、$$b$$、$$c$$均$$\\ne 0$$， ∴$$a$$、$$b$$、$$c$$三数符号为两正一负或两负一正， 不妨设， ①$$a$$，$$b$$为正，$$c$$为负，此时$$abc$$为负： $$\\frac{a}{\\left\\textbar{} a \\right\\textbar}+\\frac{b}{\\left\\textbar{} b \\right\\textbar}+\\frac{c}{\\left\\textbar{} c \\right\\textbar}+\\frac{abc}{\\left\\textbar{} abc \\right\\textbar}$$ $$=1+1+(-1)+(-1)$$ $$=0$$． ②$$a$$，$$b$$为负，$$c$$为正，此时$$abc$$为正， $$\\frac{a}{\\left\\textbar{} a \\right\\textbar}+\\frac{b}{\\left\\textbar{} b \\right\\textbar}+\\frac{c}{\\left\\textbar{} c \\right\\textbar}+\\frac{abc}{\\left\\textbar{} abc \\right\\textbar}$$ $$=-1+(-1)+1+1$$ $$=0$$． 综上原式$$=0$$，故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "269", "queId": "26ad98dcef294cf9b741c7d789f2af0a", "competition_source_list": ["1999年第10届希望杯初二竞赛第2试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知实数$$x$$满足条件$$x\\textgreater\\sqrt{2}x+1$$，那么$$\\sqrt{{{(x+2)}^{2}}}+\\sqrt[3]{{{(x-3)}^{3}}}$$的值等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$2x-1$$ "}], [{"aoVal": "B", "content": "$$-2x+1$$ "}], [{"aoVal": "C", "content": "$$-5$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的运算->二次根式的加减以及混合运算"], "answer_analysis": ["由$$x\\textgreater\\sqrt{2}x+1$$所以$$(\\sqrt{2}-1)x\\textless{}-1$$， 所以$$x\\textless{}-\\frac{1}{\\sqrt{2}-1}=-(\\sqrt{2}+1)\\textless{}-2$$， 所以$$x\\textless{}-2$$，$$x+2\\textless{}0$$， 所以原式$$=-(x+2)+(x-3)=-5$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "215", "queId": "265070ef5f124727995d90bdb347e33c", "competition_source_list": ["初二下学期单元测试《二次根式》二次根式的运算第43题", "2002年第19届全国初中数学联赛竞赛第1题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a=\\sqrt{2}-1$$，$$b=\\sqrt{3}-\\sqrt{2}$$，$$c=\\sqrt{6}-2$$，那么$$a$$，$$b$$，$$c$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless{}b\\textless{}c$$ "}], [{"aoVal": "B", "content": "$$a\\textless{}c\\textless{}b$$ "}], [{"aoVal": "C", "content": "$$b\\textless{}a\\textless{}c$$ "}], [{"aoVal": "D", "content": "$$b\\textless{}c\\textless{}a$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的运算->二次根式比较大小"], "answer_analysis": ["∵$$a-b$$ $$=\\sqrt{2}-1-\\left( 2\\sqrt{2}-\\sqrt{6} \\right)$$ $$=\\sqrt{6}-\\left( 1+\\sqrt{2} \\right)$$ $$\\approx 2.449-2.414\\textgreater0$$， ∴$$a\\textgreater b$$； ∵$$a-c$$ $$=\\sqrt{2}-1-\\left( \\sqrt{6}-2 \\right)$$ $$=\\sqrt{2}+1-\\sqrt{6}$$ $$\\approx 2.414-2.449\\textless{}0$$， ∴$$a\\textless{}c$$； 于是$$b\\textless{}a\\textless{}c$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "586", "queId": "b9d8daf27a2f41c080ab0b5e0b64361d", "competition_source_list": ["2000年第11届希望杯初二竞赛第2试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a+b+c=0$$，$$abc\\textgreater0$$，则$$\\frac{b+c}{\\left\\textbar{} a \\right\\textbar}+\\frac{c+a}{\\left\\textbar{} b \\right\\textbar}+\\frac{a+b}{\\left\\textbar{} c \\right\\textbar}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-3$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$3$$或$$-1$$ "}], [{"aoVal": "D", "content": "$$-3$$或$$1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["因为 $$a+b+c=0$$，$$abc\\textgreater0$$， 所以$$a$$，$$b$$，$$c$$中有两个负数． 由$$a+b+c=0$$，可得$$b+c=-a$$，$$c+a=-b$$，$$a+b=-c$$， 所以原式变形为$$\\frac{-a}{\\left\\textbar{} a \\right\\textbar}+\\frac{-b}{\\left\\textbar{} b \\right\\textbar}+\\frac{-c}{\\left\\textbar{} c \\right\\textbar}$$ ． 而$$\\frac{-a}{\\left\\textbar{} a \\right\\textbar}$$ ， $$\\frac{-b}{\\left\\textbar{} b \\right\\textbar}$$ ，$$\\frac{-c}{\\left\\textbar{} c \\right\\textbar}$$ ， 必有两个式子的值为$$1$$，另一个式子的值为$$-1$$， 所以原式的值为$$1$$． ", "\n特殊值法．
\n由$$a+b+c=0$$，$$abc>0$$，
\n可设$$a=2$$，$$b=c=-1$$，
\n将其代入原式，得原式$$=\\frac{-2}{2}+\\frac{1}{1}+\\frac{1}{1}=1$$ ．
\n故选$$\\text{B}$$．
"], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1437", "queId": "bca8f824ea0c47dea1953dfca62fd670", "competition_source_list": ["1998年竞赛（全国初中数学竞赛）第1题6分", "2015~2016学年浙江温州乐清市乐清市育英寄宿学校初三上学期期中实验A班第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$，$$c$$都是实数，并且$$a\\textgreater b\\textgreater c$$，那么下列式子中正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$ab\\textgreater bc$$ "}], [{"aoVal": "B", "content": "$$a+b\\textgreater b+c$$ "}], [{"aoVal": "C", "content": "$$a-b\\textgreater b-c$$ "}], [{"aoVal": "D", "content": "$$\\frac{a}{c}\\textgreater\\frac{b}{c}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["根据不等式性质： ∵$$a\\textgreater b\\textgreater c$$， ∴$$a+b\\textgreater b+c$$，故$$\\text{B}$$正确， ∵$$a$$、$$b$$、$$c$$的符号不确定，∴$$\\text{A}$$、$$\\text{C}$$、$$\\text{D}$$不成立， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "500", "queId": "24acb477fc8743189b1224d2c70292c5", "competition_source_list": ["2005年第16届希望杯初二竞赛复赛第10题4分", "初一下学期其它第17题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知整数$$x$$，$$y$$，$$z$$满足$$x\\leqslant y\\leqslant z$$，且$$\\begin{cases}\\left\\textbar{} x+y \\right\\textbar+\\left\\textbar{} y+z \\right\\textbar+\\left\\textbar{} z+x \\right\\textbar=4 ① \\left\\textbar{} x-y \\right\\textbar+\\left\\textbar{} y-z \\right\\textbar+\\left\\textbar{} z-x \\right\\textbar=2 ②\\end{cases}$$，那么$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}$$的值等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$2$$或$$14$$ "}], [{"aoVal": "D", "content": "$$14$$或$$17$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->解含绝对值的方程组"], "answer_analysis": ["由②可知$$\\left\\textbar{} x-y \\right\\textbar$$、$$\\left\\textbar{} y-z \\right\\textbar$$、$$\\left\\textbar{} z-x \\right\\textbar$$中必须有一个为$$0$$，只有$$x-y=0$$，$$x=y$$，代入②中得$$\\left\\textbar{} z-x \\right\\textbar=1$$ $$z=x+1$$将$$x=y$$，$$z=x+1$$代入①中，得$$\\left\\textbar{} x \\right\\textbar+\\left\\textbar{} 2x+1 \\right\\textbar=2$$，$$\\left\\textbar{} x \\right\\textbar=0$$，$$1$$，$$2$$．经检验$$\\left\\textbar{} x \\right\\textbar=1$$，依据题意得． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1301", "queId": "c072e5e3171e484e9e75a36e418ae0c6", "competition_source_list": ["2005年第16届希望杯初二竞赛初赛第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x=3$$是不等式$$mx+2\\textless{}1-4m$$的一个解，如果$$m$$是整数，那么$$m$$的最大值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$-2$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->一元一次不等式的解集", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->解一元一次不等式", "课内体系->知识点->方程与不等式->不等式（组）->含参不等式->由不等式（组）的解集求参数的范围"], "answer_analysis": ["因为$$x=3$$满足不等式$$mx+2\\textless{}1-4m$$，所以$$3m+2\\textless{}1-4m$$， 解得$$m\\textless{}-\\frac{1}{7}$$， 又$$m$$为整数，所以$$m$$的最大值是$$-1$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "146", "queId": "0c2a69b222d14f25adf5c04a0b98113b", "competition_source_list": ["1997年第14届全国初中数学联赛竞赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "下述四个命题中，假命题的个数是（~ ）． ①)一个数的倒数等于自身，那么这个数是$$1$$； ②对角线互相垂直且相等的四边形是正方形； ③$${{a}^{2}}$$的平方根是$$\\pm \\left\\textbar{} a \\right\\textbar$$； ④大于直角的角一定是钝角．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$个 "}], [{"aoVal": "B", "content": "$$2$$个 "}], [{"aoVal": "C", "content": "$$3$$个 "}], [{"aoVal": "D", "content": "$$4$$个 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->四边形->特殊平行四边形->正方形->正方形的判定"], "answer_analysis": ["($$1$$)$$-1$$的倒数也是他本身，所以这个命题不对． ($$2$$)对角线互要垂直且相等的平行四边形才是正方形，所以这个命题也不对． ($$3$$)$${{a}^{2}}$$的平方根是$$\\pm \\left\\textbar{} a \\right\\textbar$$，这是对的． ($$4$$)大于直角的角一定是钝角，钝角是$$90-180$$度之间的角． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "654", "queId": "4d3afa6f7a8e4b07a4fab7c8168dd442", "competition_source_list": ["2011年第22届全国希望杯初二竞赛复赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个凸四边形的四个内角可以（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "都是锐角 "}], [{"aoVal": "B", "content": "都是直角 "}], [{"aoVal": "C", "content": "都是钝角 "}], [{"aoVal": "D", "content": "有三个是直角，另一个是锐角或钝角 "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->三角形及多边形->多边形->多边形的内角和定理", "课内体系->知识点->三角形->三角形及多边形->多边形->正多边形->求正多边形的内角"], "answer_analysis": ["凸四边形的四个内角的和是$$360{}^{}\\circ $$． 若四个内角都是锐角，则内角和小于$$360{}^{}\\circ $$； 若四个内角都是钝角，则内角和大于$$360{}^{}\\circ $$； 若有三个是直角，另一个是锐角或钝角，则内角和不为$$360{}^{}\\circ $$； 若四个内角都是直角，则内角和等于$$360{}^{}\\circ $$，这时可以构成凸四边形，符合题意． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "429", "queId": "35549ffcaa8f4b07b6d53fee6f9cb8c2", "competition_source_list": ["2020~2021学年四川德阳初一上学期期末第11题7分", "2000年第11届希望杯初一竞赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "若四个有理数$$a$$，$$b$$，$$c$$，$$d$$满足$$\\frac{1}{a-1997}=\\frac{1}{b+1998}=\\frac{1}{c-1999}=\\frac{1}{d+2000}$$，则$$a$$，$$b$$，$$c$$，$$d$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater c\\textgreater b\\textgreater d$$ "}], [{"aoVal": "B", "content": "$$b\\textgreater d\\textgreater a\\textgreater c$$ "}], [{"aoVal": "C", "content": "$$c\\textgreater a\\textgreater b\\textgreater d$$ "}], [{"aoVal": "D", "content": "$$d\\textgreater b\\textgreater a\\textgreater c$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性"], "answer_analysis": ["由$$\\frac{1}{a-1997}=\\frac{1}{b+1998}=\\frac{1}{c-1999}=\\frac{1}{d+2000}$$， 可知$$a-1997=b+1998=c-1999=d+2000$$， 由这个连等式可得：$$a\\textgreater b$$，$$a\\textless{}c$$，$$a\\textgreater d$$；$$b\\textless{}c$$，$$b\\textgreater d$$，$$c\\textless{}d$$， 由此可得$$c\\textgreater a\\textgreater b\\textgreater d$$， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1226", "queId": "db985128eba54f0c86cf02390cb731ed", "competition_source_list": ["2018年第29届希望杯初一竞赛初赛第16题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "将一个棱长是整数厘米的长方体的各表面都刷成红色，然后将这个长方体分割成若干个棱长为$$1$$厘米的小正方体，若任何一面都没有涂色的小正方体有$$11$$个，则原来的长方体的体积是~\\uline{~~~~~~~~~~}~立方厘米． Brush each surface of a box with integral centimeter of edge length into red, and then divide the box into several small cubes with an edge length of $$1$$ cm. If there are $$11$$ cubes without color on any face, the volume of the original box is~\\uline{~~~~~~~~~~}~cubic centimeter.", "answer_option_list": [[{"aoVal": "A", "content": "$$117$$ "}], [{"aoVal": "B", "content": "$$99$$ "}], [{"aoVal": "C", "content": "$$96$$ "}], [{"aoVal": "D", "content": "$$84$$ "}], [{"aoVal": "E", "content": "$$48$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->能力->分析和解决问题能力", "课内体系->知识点->几何图形初步->几何图形->立体图形与平面图形->正方体堆积图形的问题"], "answer_analysis": ["设该长方体的棱长分别为$$a$$厘米，$$b$$厘米，$$c$$厘米，$$(a\\geqslant b\\geqslant c)$$， 则由题意得$$(a-2)(b-2)(c-2)=11$$， ∵$$11$$为质数， ∴$$a-2=11$$，$$b-2=1$$，$$c-2=1$$， ∴$$a=13$$，$$b=3$$，$$c=3$$， ∴长方体的体积$$V=abc=117$$立方厘米． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1168", "queId": "8aac49074e023206014e2017ad9d64fe", "competition_source_list": ["1994年第5届全国希望杯初一竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\frac{\\left\\textbar{} 1-9 \\right\\textbar-\\left\\textbar{} 9-4 \\right\\textbar}{1-9-9+4}$$的值的负倒数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{3}{13}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$-1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->倒数与负倒数->负倒数的定义"], "answer_analysis": ["$$\\frac{\\left\\textbar{} 1-9 \\right\\textbar-\\left\\textbar{} 9-4 \\right\\textbar}{1-9-9+4}=\\frac{8-5}{-13}=-\\frac{3}{13}$$，其负倒数为$$\\frac{13}{3}=4\\frac{1}{3}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1060", "queId": "ff8080814d7978b9014d88b7a7372a25", "competition_source_list": ["1991年第2届全国希望杯初一竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$，$$b$$为正整数$$(a\\textgreater b)$$．$$p$$是$$a$$，$$b$$的最大公约数，$$q$$是$$a$$，$$b$$的最小公倍数．则$$p$$，$$q$$，$$a$$，$$b$$的大小关系是（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$p\\geqslant q\\geqslant a\\textgreater b$$ "}], [{"aoVal": "B", "content": "$$q\\geqslant a\\textgreater b\\geqslant p$$ "}], [{"aoVal": "C", "content": "$$q\\geqslant p\\geqslant a\\textgreater b$$ "}], [{"aoVal": "D", "content": "$$p\\geqslant a\\textgreater b\\geqslant q$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小", "竞赛->知识点->数论->整除->因数与倍数"], "answer_analysis": ["两个自然数的最小公倍数一定不小于两数中较大者． 两个自然数的最大公约数一定不大于两数中较小者． 所以$$q\\geqslant a\\textgreater b\\geqslant p$$． 选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1506", "queId": "d85fc5cc49164212b8236e749866fbb3", "competition_source_list": ["2013年第24届全国希望杯初二竞赛初赛第7题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$-12≤{}x≤{}300$$，且$$m=\\left\\textbar{} \\left\\textbar{} x \\right\\textbar-100 \\right\\textbar$$的值为整数，则$$m$$的值有．", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$个 "}], [{"aoVal": "B", "content": "$$101$$个 "}], [{"aoVal": "C", "content": "$$201$$个 "}], [{"aoVal": "D", "content": "$$203$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->已知范围化简绝对值", "课内体系->能力->运算能力"], "answer_analysis": ["此题暂无解析 "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "158", "queId": "8f6d876a816f4b76ac39bdd0ec63aae5", "competition_source_list": ["2007年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "从写有$$A$$、$$B$$、$$C$$、$$D$$的四张红牌和$$A$$、$$B$$、$$C$$、$$D$$的四张绿牌中抽取两张．获胜对子为两张相同花色或相同数字的．则抽到获胜对子的概率是？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac 27$$ "}], [{"aoVal": "B", "content": "$$\\frac 38$$ "}], [{"aoVal": "C", "content": "$$\\frac 12$$ "}], [{"aoVal": "D", "content": "$$\\frac 47$$ "}], [{"aoVal": "E", "content": "$$\\frac 58$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Counting, Probability and Statistics->Classical Probability", "课内体系->知识点->统计与概率"], "answer_analysis": ["从写有$$A$$、$$B$$、$$C$$、$$D$$的四张红牌和$$A$$、$$B$$、$$C$$、$$D$$的四张绿牌中抽取两张．获胜对子为两张相同花色或相同数字的．则抽到获胜对子的概率是？ 有四种方法可以选择同一个字母的一对，并且$$2\\left( \\left( \\begin{matrix}4 2 \\end{matrix} \\right) \\right)=12$$方法去选择一对相同颜色．一共有$$\\left( \\begin{matrix}8 2 \\end{matrix} \\right)=28$$方法去选择一对，所以概率是$$\\frac{4+12}{28}=\\frac{4}{7}$$． 故选$$\\text{D}$$． There are $$4$$ ways of choosing a winning pair of the same letter, and $$2\\left( \\left( \\begin{matrix}4 2 \\end{matrix} \\right) \\right)=12$$ ways to choose a pair of the same color. There\\textquotesingle s a total of $$\\left( \\begin{matrix}8 2 \\end{matrix} \\right)=28$$ ways to choose a pair, so the probability is $$\\frac{4+12}{28}=\\frac{4}{7}$$. "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "873", "queId": "b620a93731de4354b60909cb45ba5e04", "competition_source_list": ["2019年第1届广东深圳罗湖区深圳中学初中部初一竞赛（凤凰木杯）第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a$$，$$b$$都是正整数，且满足$$a_{{}}^{2}-b_{{}}^{2}+9=2018$$，则$$a+b$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$44$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$46$$ "}], [{"aoVal": "D", "content": "$$49$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->平方差公式的计算", "课内体系->能力->运算能力"], "answer_analysis": ["∵$$a_{{}}^{2}-b_{{}}^{2}+9-2018$$， ∴$$a_{{}}^{2}-b_{{}}^{2}=2009$$， ∴$$(a+b)(a-b)=2009$$， ∵$$2009=7\\times 7\\times 41=41\\times 49$$， ∵$$a\\cdot b$$都是正整数， ∴$$a+b$$为正整数，且$$a+b\\textgreater a-b$$， ∴$$a+b=49$$， 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1427", "queId": "dc4507f3c6e4423bbf1d61c25028ca9c", "competition_source_list": ["2011年第22届全国希望杯初一竞赛初赛第3题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "对有理数$$a$$，$$b$$，有以下四个判断： ①若$$\\left\\textbar{} a \\right\\textbar=b$$，则$$a=b$$； ②若$$\\left\\textbar{} a \\right\\textbar\\textgreater b$$，则$$\\left\\textbar{} a \\right\\textbar\\textgreater\\left\\textbar{} b \\right\\textbar$$； ③若$$a=-b$$，则$${{\\left( -a \\right)}^{2}}={{b}^{2}}$$； ④若$$\\left\\textbar{} a \\right\\textbar\\textless{}\\left\\textbar{} b \\right\\textbar$$，则$$a\\textless{}b$$． 其中正确的判断的个数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->绝对值综合", "课内体系->知识点->数->有理数->绝对值->绝对值的几何意义", "课内体系->知识点->数->有理数->绝对值->绝对值的非负性"], "answer_analysis": ["①令$$a=-2$$，$$b=2$$，则①错误； ②令$$a=2$$，$$b=-3$$，则②错误； ③若$$a=-b$$，则$${{\\left( -a \\right)}^{2}}={{\\left[ -(-b) \\right]}^{2}}={{b}^{2}}$$，则③正确； ④令$$a=-1$$，$$b=-2$$，则④错误． 故正确的判断的个数是$$1$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1113", "queId": "c46ecf0a3cbf4fad9d667cd7b189a900", "competition_source_list": ["2020~2021学年四川成都金牛区成都外国语学校初二上学期期中模拟第8题3分", "2018~2019学年广东深圳罗湖区深圳中学初中部初一下学期期中（竞赛班）第5题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$x=\\frac{4}{\\sqrt{5}+3}$$，$$y=\\sqrt{5}-3$$，则$$x$$，$$y$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textgreater y$$ "}], [{"aoVal": "B", "content": "$$x\\geqslant y$$ "}], [{"aoVal": "C", "content": "$$x\\textless{}y$$ "}], [{"aoVal": "D", "content": "$$x=y$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->无理数有关的计算->无理数大小的比较"], "answer_analysis": ["∵$$x=\\frac{4}{\\sqrt{5}+3}=\\frac{4(\\sqrt{5}-3)}{(\\sqrt{5}+3)(\\sqrt{5}-3)}=-(\\sqrt{5}-3)=3-\\sqrt{5}\\textgreater0$$， $$y=\\sqrt{5}-3\\textless{}0$$， ∴$$x\\textgreater y$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "682", "queId": "4d770b275d5b4c5092e11ea0d3bb814d", "competition_source_list": ["2007年竞赛第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "口袋中有$$20$$个球，其中白球$$9$$个，红球$$5$$个，黑球$$6$$个，现从中任取$$10$$个球，使得白球不少于$$2$$个但不多于$$8$$个，红球不少于$$2$$个，黑球不多于$$3$$个，那么上述取法的种数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->计数问题-枚举法"], "answer_analysis": ["用枚举法： 红球个数 白球个数 黑球个数 种数 $$5$$ $$2$$，$$3$$，$$4$$，$$5$$ $$3$$，$$2$$，$$1$$，$$0$$ $$4$$ $$4$$ $$3$$，$$4$$，$$5$$，$$6$$ $$3$$，$$2$$，$$1$$，$$0$$ $$4$$ $$3$$ $$4$$，$$5$$，$$6$$，$$7$$ $$3$$，$$2$$，$$1$$，$$0$$ $$4$$ $$2$$ $$5$$，$$6$$，$$7$$，$$8$$ $$3$$，$$2$$，$$1$$，$$0$$ $$4$$ 所以共$$16$$种．故选$$B$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1256", "queId": "8aac50a74e023208014e3f487f4a18de", "competition_source_list": ["1995年第6届全国希望杯初一竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "有理数$$-\\frac{95}{a-19}$$的值一定不是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$-19$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式有关的概念->代数式->代数式的定义", "课内体系->知识点->式->整式的加减->整式有关的概念->代数式->列代数式"], "answer_analysis": ["当$$a=14$$时，$$-\\frac{95}{a-19}=19$$，排除$$\\text{A}$$； 当$$a=24$$时，$$-\\frac{95}{a-19}=-19$$，排除$$\\text{B}$$； 当$$a=-76$$时，$$-\\frac{95}{a-19}=1$$，排除$$\\text{D}$$； 因此，选$$\\text{C}$$．事实上，对任意$$a\\ne 19$$，$$-\\frac{95}{a-19}$$一定不是$$0$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "346", "queId": "2bb3c3e7b6854cb8bb2431ae3e19e14f", "competition_source_list": ["1993年第4届希望杯初二竞赛第14题"], "difficulty": "1", "qtype": "single_choice", "problem": "若方程$$9{{x}^{2}}-6(a+1)x+{{a}^{2}}-3=0$$的两根之积等于$$1$$，则$$a$$的值是（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\pm 2\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$2\\sqrt{3}$$ "}], [{"aoVal": "C", "content": "$$\\pm 2\\sqrt{2}$$ "}], [{"aoVal": "D", "content": "$$2\\sqrt{2}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->一元二次方程->根与系数的关系->一元二次方程根与系数的关系", "课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->一元二次方程根的判别式"], "answer_analysis": ["因为$$\\Delta \\geqslant 0$$，所以$$36{{(a+1)}^{2}}-36({{a}^{2}}-3)\\geqslant 0$$，所以$$a\\geqslant -2$$，又$${{x}_{1}}\\cdot {{x}_{2}}=1$$，所以$$\\frac{{{a}^{2}}-3}{9}=1$$，解得$$a=\\pm 2\\sqrt{3}$$．又由$$a\\geqslant -2$$，所以$$a=2\\sqrt{3}$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "284", "queId": "19d3f2be250940fd9fc1b77ea8707f71", "competition_source_list": ["2019~2020学年12月四川资阳雁江区资阳市雁江区第二中学初三上学期周测D卷第10题3分", "2016~2017学年3月湖北武汉武昌区武汉初级中学初二下学期月考第10题3分", "2016~2017学年9月湖北武汉武昌区武汉初级中学初二上学期月考第10题3分", "2001年第18届全国初中数学联赛竞赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a$$、$$b$$、$$c$$、为有理数，且等式$$a+b\\sqrt{2}+c\\sqrt{3}=\\sqrt{5+2\\sqrt{6}}$$成立，则$$2a+999b+1001c$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1999$$ "}], [{"aoVal": "B", "content": "$$2000$$ "}], [{"aoVal": "C", "content": "$$2001$$ "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算", "课内体系->知识点->方程与不等式->等式与方程->等式->等式的性质->等式性质1", "课内体系->能力->运算能力"], "answer_analysis": ["本题需要比较等式两边的各项，利用有理数部分等于理数部分，无理数部分等于无理数部分来求$$a$$、$$b$$、$$c$$的值，由于： $$5+2\\sqrt{6}={{\\left( \\sqrt{3}+\\sqrt{2} \\right)}^{2}}$$． 所以：$$a+b\\sqrt{2}+c\\sqrt{3}=\\sqrt{2}+\\sqrt{3}$$． 则$$a=0$$，$$b=1$$，$$c=1$$，∴$$2a+999b+1001c=2000$$． 所以选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "820", "queId": "4ec153fa24bd4c7585cf399ce3b4490b", "competition_source_list": ["1992年第9届全国初中数学联赛竞赛第3题", "初一上学期其它"], "difficulty": "2", "qtype": "single_choice", "problem": "若$${{x}^{2}}-13x+1=0$$，则$${{x}^{4}}-{{x}^{-4}}$$的个位数字是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式", "课内体系->能力->运算能力"], "answer_analysis": ["由$${{x}^{2}}-13x+1$$可得到$$x+\\dfrac{1}{x}=13$$，所以$${{x}^{2}}+\\dfrac{1}{{{x}^{2}}}={{(x+\\dfrac{1}{x})}^{2}}-2={{13}^{2}}-2=167$$． 同样的$${{x}^{4}}+\\dfrac{1}{{{x}^{4}}}={{({{x}^{2}}+\\dfrac{1}{{{x}^{2}}})}^{2}}-2$$，可以根据个位数字可以直接判断结果的个位数字为$$7$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1497", "queId": "cab476bd1c4f4eae94bc6e2c1d7c9692", "competition_source_list": ["1991年第2届希望杯初二竞赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个两位数，用它的个位、十位上的两个数之和的$$3$$倍减去$$-2$$，仍得原数，这个两位数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$26$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$38$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["设这个两位数为$$\\overline{ab}$$，则$$3(a+b)+2=10a+b$$， 即$$7a=2b+2$$， 可见$$a$$只能为偶数，$$b+1$$是$$7$$的倍数． ∴这个两位数是$$28$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "850", "queId": "4f0830a901d84618a08517b3535cf5e7", "competition_source_list": ["2010年第21届希望杯初二竞赛第1试第6题", "北京初二上学期单元测试《二次根式的运算》第22题"], "difficulty": "3", "qtype": "single_choice", "problem": "设$$p=\\sqrt[3]{7a+1}+\\sqrt[3]{7b+1}+\\sqrt[3]{7c+1}+\\sqrt[3]{7d+1}$$，其中$$a$$、$$b$$、$$c$$、$$d$$是正实数，且$$a+b+c+d=1$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$p\\textgreater5$$ "}], [{"aoVal": "B", "content": "$$p\\textless{}5$$ "}], [{"aoVal": "C", "content": "$$p\\textless{}4$$ "}], [{"aoVal": "D", "content": "$$p=5$$ "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质", "课内体系->知识点->式->因式分解->公式法->利用立方和与立方差公式因式分解", "课内体系->知识点->式->整式的乘除->乘法公式->和与差的立方公式"], "answer_analysis": ["∵$$a$$、$$b$$、$$c$$、$$d$$是正实数，且$$a+b+c+d=1$$， ∴$$0\\textless{}a\\textless{}1$$， ∴$$a\\textgreater{{a}^{2}}\\textgreater{{a}^{3}}$$， ∴$$7a+1=a+3a+3a+1\\textgreater{{a}^{3}}+3{{a}^{2}}+3a+1={{(a+1)}^{3}}$$， ∴$$\\sqrt[3]{7a+1}\\textgreater\\sqrt[3]{{{(a+1)}^{3}}}=a+1$$， 同理$$\\sqrt[3]{7b+1}\\textgreater b+1\\sqrt[3]{7c+1}\\textgreater c+1\\sqrt[3]{7d+1}\\textgreater d+1$$， ∴$$p=\\sqrt[3]{7a+1}+\\sqrt[3]{7b+1}+\\sqrt[3]{7c+1}+\\sqrt[3]{7d+1}$$， $$\\textgreater a+1+b+1+c+1+d+1$$， $$=a+b+c+d+4=5$$， ∴$$p\\textgreater5$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "931", "queId": "65d4c99313444a55a3b7403dabcd11ff", "competition_source_list": ["1994年第11届全国初中数学联赛竞赛第4题6分", "初一下学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{2021-2022年南昌二中期末考试第$$2$$题} 当$$x=\\frac{1+\\sqrt{1994}}{2}$$时，多项式$${{(4{{x}^{3}}-1997x-1994)}^{2001}}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$${{2}^{2001}}$$ "}], [{"aoVal": "D", "content": "$$-{{2}^{2001}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式化简求值->二次根式的化简求值——利用完全平方"], "answer_analysis": ["∵$$x=\\frac{1+\\sqrt{1994}}{2}$$， ∴$${{(2x-1)}^{2}}=1994$$， 即$$4{{x}^{2}}-4x-1993=0$$， ∴$${{(4{{x}^{3}}-1997x-1994)}^{2001}}$$ $$={{[(4{{x}^{2}}-4x-1993)x+(4{{x}^{2}}-4x-1993)-1]}^{2001}}$$ $$={{(-1)}^{2001}}$$ $$=-1$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "251", "queId": "b0060b5d894f42f392614af886ad7646", "competition_source_list": ["2003年第14届希望杯初二竞赛第2试第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "凸$$n$$边形（$$n\\geqslant 4$$）中，不算两个最大的内角，其余内角的和为$$1100{}^{}\\circ $$，则$$n$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$10$$或$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "课内体系->知识点->三角形->三角形及多边形->多边形->多边形的内角和定理", "课内体系->知识点->三角形->三角形及多边形->多边形->求多边形的内角和"], "answer_analysis": ["因为，在四边形中，两个最大的内角的和大于等于$$180{}^{}\\circ $$， 所以，凸$$n$$边形（$$n\\geqslant 4$$）中，两个最大的内角和也大于等于$$180{}^{}\\circ $$，同时小于$$360{}^{}\\circ $$， 设两个最大的内角的和为$$k{}^{}\\circ $$，则$$1100+180\\leqslant 1100+k=(n-2)\\times 180 ~\\textless{} ~1100+360$$， 所以$$7\\frac{1}{9}\\leqslant n-2 ~\\textless{} ~8\\frac{1}{9}$$，$$9\\frac{1}{9}\\leqslant n ~\\textless{} ~10\\frac{1}{9}$$，$$n=10$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "988", "queId": "ff8080814cfa9b24014d02bbcad3135e", "competition_source_list": ["2006年第17届希望杯初二竞赛第2试第5题", "初二其它"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$m={{2006}^{2}}+{{2006}^{2}}\\times {{2007}^{2}}+{{2007}^{2}}$$，则$$m$$（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "是完全平方数，还是奇数 "}], [{"aoVal": "B", "content": "是完全平方数，还是偶数 "}], [{"aoVal": "C", "content": "不是完全平方数，但是奇数 "}], [{"aoVal": "D", "content": "不是完全平方数，但是偶数 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->因式分解->十字相乘法->二次项系数为±1的十字相乘"], "answer_analysis": ["$$m={{2006}^{2}}+{{2006}^{2}}\\times {{2007}^{2}}+{{2007}^{2}}$$ $$={{2006}^{2}}+{{2006}^{2}}{{\\left( 2006+1 \\right)}^{2}}+{{\\left( 2006+1 \\right)}^{2}}$$ $$={{2006}^{4}}+2\\times {{2006}^{3}}+3\\times {{2006}^{2}}+2\\times 2006+1$$ $$={{\\left( {{2006}^{2}}+2006+1 \\right)}^{2}}$$ ∴$$m$$是奇数． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1127", "queId": "d22cc70b85cb44b3a538b90ddeb04ad8", "competition_source_list": ["1990年第1届希望杯初二竞赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "把二次根式$$a\\sqrt{-\\frac{1}{a}}$$化为最简二次根式，结果是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{-a}$$ "}], [{"aoVal": "B", "content": "$$-\\sqrt{a}$$ "}], [{"aoVal": "C", "content": "$$-\\sqrt{-a}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{a}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的基础->最简二次根式"], "answer_analysis": ["因为$$a\\textless{}0$$，所以$$a\\sqrt{-\\frac{1}{a}}=-\\sqrt{{{(-a)}^{2}}\\frac{1}{-a}}=-\\sqrt{-a}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "831", "queId": "607f0eccb7594bba808621281385dccf", "competition_source_list": ["2016年第27届全国希望杯初一竞赛复赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在一家水果店，小明买了$$1$$斤苹果，$$4$$斤西瓜，$$2$$斤橙子，$$1$$斤葡萄，共付$$27.6$$元；小惠买了$$2$$斤苹果，$$6$$斤西瓜，$$2$$斤橙子，$$2$$斤葡萄，共付$$32.2$$元，则买$$1$$斤西瓜和$$1$$斤橙子需付（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$元 "}], [{"aoVal": "B", "content": "$$14.8$$元 "}], [{"aoVal": "C", "content": "$$11.5$$元 "}], [{"aoVal": "D", "content": "$$10.7$$元 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->二元一次方程（组）->二元一次方程组应用题->二元一次方程组的经济问题", "课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->加减消元法解二元一次方程组"], "answer_analysis": ["设$$1$$斤苹果$$x$$元，$$1$$斤西瓜$$y$$元，$$1$$斤橙子$$z$$元，$$1$$斤葡萄$$w$$元． 由题意，得$$\\left { \\begin{matrix}x+4y+2z+w=27.6\\&① 2x+6y+2z+2w=32.2 \\&② \\end{matrix} \\right.\\begin{matrix} \\end{matrix}$$， $$2\\times $$①$$-$$②，得$$2y+2z=23$$， ∴$$y+z=11.5$$． 答：买$$1$$斤西瓜和$$1$$斤橙子需付$$11.5$$元． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "787", "queId": "4e6843c4e15e46249f597d5f2b9f315a", "competition_source_list": ["1993年第4届希望杯初二竞赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知四个命题：①$$-1$$是$$1$$的平方根．②负数没有立方根．③无限小数不一定是无理数．④$$\\sqrt{3a}$$一定没有意义．其中正确的命题的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->无理数有关的计算->无理数的定义", "课内体系->知识点->数->实数->立方根->平方根和立方根综合", "课内体系->知识点->数->实数->平方根", "课内体系->知识点->几何图形初步->命题与证明"], "answer_analysis": ["命题①，③是正确的，②，④不正确． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "534", "queId": "4c64b07e8c934609a6ec061eb34ca8cb", "competition_source_list": ["2001年竞赛（全国初中数学竞赛）第3题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "如果$$a$$，$$b$$是质数，且$${{a}^{2}}-13a+m=0$$，$${{b}^{2}}-13b+m=0$$那么$$\\frac{b}{a}+\\frac{a}{b}$$的值为（ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{123}{22}$$ "}], [{"aoVal": "B", "content": "$$\\frac{125}{22}$$或$$2$$ "}], [{"aoVal": "C", "content": "$$\\frac{125}{22}$$ "}], [{"aoVal": "D", "content": "$$\\frac{123}{22}$$或$$2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式化简求值->分式化简求值-条件化简求值", "课内体系->思想->分类讨论思想"], "answer_analysis": ["有两种情况： ①若$$a=b$$，则$$\\frac{b}{a}+\\frac{a}{b}=2$$； ②若$$a\\ne b$$，根据题意，$$a$$、$$b$$是方程$${{x}^{2}}-13x+m=0$$的根， 则$$a+b=13$$，因为$$a$$，$$b$$是质数且和为奇数，所以两数分别为$$2$$和$$11$$．此时$$\\frac{b}{a}+\\frac{a}{b}=\\frac{2}{11}+\\frac{11}{2}=\\frac{125}{22}$$． 方法二：两式相减，消$$m$$，$${{a}^{2}}-{{b}^{2}}-13a+13b=0$$，$$\\left( a-b \\right)\\left( a+b-13 \\right)=0$$，所以有$$a=b$$或$$a+b=13$$. "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1374", "queId": "a5a4cbf96ce649b58abb2b961d2d5394", "competition_source_list": ["2020~2021学年10月湖南长沙天心区长郡外国语实验中学初一上学期周测A卷", "1996年第7届希望杯初二竞赛第2题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知：$$-1 ~\\textless{} ~b ~\\textless{} ~a ~\\textless{} ~0$$，那么$$a+b$$，$$a-b$$，$$a+1$$，$$a-1$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a+b ~\\textless{} ~a-b ~\\textless{} ~a-1 ~\\textless{} ~a+1$$ "}], [{"aoVal": "B", "content": "$$a+1\\textgreater a+b\\textgreater a-b\\textgreater a-1$$ "}], [{"aoVal": "C", "content": "$$a-1 ~\\textless{} ~a+b ~\\textless{} ~a-b ~\\textless{} ~a+1$$ "}], [{"aoVal": "D", "content": "$$a+b\\textgreater a-b\\textgreater a+1\\textgreater a-1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["因为$$-1 ~\\textless{} ~b ~\\textless{} ~a ~\\textless{} ~0$$，所以$$a+b ~\\textless{} ~a-b$$． 因为$$b\\textgreater-1$$，所以$$a-1 ~\\textless{} ~a+b$$． 又因为$$-b ~\\textless{} ~1$$，所以$$a-b ~\\textless{} ~a+1$$． 综上得$$a-1 ~\\textless{} ~a+b ~\\textless{} ~a-b ~\\textless{} ~a+1$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1171", "queId": "8aac49074e023206014e201d01216530", "competition_source_list": ["1994年第5届全国希望杯初一竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$x:y=3:2$$并且$$x+3y=27$$，则$$x$$，$$y$$中较小的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->等式与方程->等式->等式的定义", "课内体系->知识点->方程与不等式->二元一次方程（组）->二元一次方程->二元一次方程的解"], "answer_analysis": ["由$$x:y=3:2$$得$$x=1.5y$$， 代入$$x+3y=27$$得$$4.5y=27$$， 于是$$y=6$$，$$x=9$$， 所以$$x$$，$$y$$中较小的那个数是$$6$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "461", "queId": "243c5b94cf5a48a6ba60d7d45f1b9f7d", "competition_source_list": ["2007年第12届华杯赛初一竞赛初赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$是整数，则以下四个代数式中，不可能得整数值的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3a+2}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2-a}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3a+1}{6}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5a-2}{7}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数减法->有理数加减混合运算", "课内体系->能力->运算能力"], "answer_analysis": ["取$$a=1$$，$$\\frac{3a+2}{5}$$，否定$$\\text{A}$$． 取$$a=5$$，$$\\frac{2-a}{3}=-1$$，否定$$\\text{B}$$． 取$$a=-1$$，$$\\frac{5a-2}{7}=-1$$，否定$$\\text{D}$$． 又当$$a$$是整数时，$$3a+1$$表示被$$3$$除余$$1$$的整数，故不可能是$$6$$的倍数． 所以$$\\frac{3a+1}{6}$$不可能得整数值． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "957", "queId": "b1edd5f8eaf14806ab2edbf63e458d7f", "competition_source_list": ["2022~2023学年浙江宁波初三月考（六校强基竞赛）第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "二次函数$y=x^{2}+2x+c$的图象与\\emph{x}轴的两个交点为$A\\left(x_{1},0\\right)$，$B\\left(x_{2},0\\right)$，且$x_{1}\\textless{} x_{2}$，点$P\\left(m,n\\right)$是图象上一点，那么下列判断正确的是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "当$n\\textgreater{} 0$时，$m\\textless{} x_{1}$ "}], [{"aoVal": "B", "content": "当$n\\textgreater{} 0$时，$m\\textgreater{} x_{2}$ "}], [{"aoVal": "C", "content": "当$n\\textless{} 0$时，$m\\textless{} 0$ "}], [{"aoVal": "D", "content": "当$n\\textless{} 0$时，$x_{1}\\textless{} m\\textless{} x_{2}$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->二次函数"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据二次函数的图象与性质可进行排除选项．\\\\ 【详解】\\\\ 解：由二次函数$y=x^{2}+2x+c$可知开口向上，对称轴为直线$x=-1$，\\\\ 当$x\\textless{} -1$时，$y$随$x$的增大而减小，当$x\\textgreater{} -1$时，\\emph{y}随\\emph{x}的增大而增大；\\\\ ∵$A\\left(x_{1},0\\right)$，$B\\left(x_{2},0\\right)$是二次函数与\\emph{x}轴的交点，点$P\\left(m,n\\right)$是图象上的一点，\\\\ ∴当$n\\textgreater{} 0$时，则$m\\textless{} x_{1}$或$m\\textgreater{} x_{2}$；故$\\mathrm{A}$、$\\mathrm{B}$选项错误；\\\\ 当$n\\textless{} 0$时，则$x_{1}\\textless{} m\\textless{} x_{2}$，故$\\mathrm{D}$正确；当$x_{2}\\textgreater{} 0$且$n\\textless{} 0$时，此时有可能$m\\textgreater{} 0$，故$\\mathrm{C}$错误；\\\\ 故选$\\mathrm{D}$．\\\\ 【点睛】\\\\ 本题主要考查二次函数的图象与性质，熟练掌握二次函数的图象与性质是解题的关键． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "767", "queId": "456c005c61ec467595ce123d18f888b2", "competition_source_list": ["2016年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "数字$$N$$是个两位数． $$N$$除以$$9$$，余数为$$1$$． $$N$$除以$$10$$，余数为$$3$$． 则$$N$$除以$$11$$时，余数为几？", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->十字相乘法->二次项系数为±1的十字相乘", "美国AMC8->Knowledge Point->Number Theory->Remainder Problems->Remainder Characteristics"], "answer_analysis": ["数字$$N$$是个两位数． $$N$$除以$$9$$，余数为$$1$$． $$N$$除以$$10$$，余数为$$3$$． 则$$N$$除以$$11$$时，余数为几？ 从第二个子弹开始，我们知道第二个数字必须是$$3$$，因为除以$$9$$还有一个余数$$1$$，$$9$$的倍数必须以$$2$$结束，我们找到： $$9\\left( 1 \\right)=9$$ $$9\\left( 2 \\right)=18$$ $$9\\left( 3 \\right)=27$$ $$9\\left( 4 \\right)=36$$ $$9\\left( 5 \\right)=45$$ $$9\\left( 6 \\right)=54$$ $$9\\left( 7 \\right)=63$$ $$9\\left( 8 \\right)=72$$ 这个上数字$$71+1=73$$都满足条件，我们减去$$11$$的小于$$73$$的最大倍数得到剩余的，也就是$$73-11\\left( 6 \\right)=73-66=7$$． 故选$$\\text{E}$$． From the second bullet point, we know that the second digit must be $$3$$. Because there is a remainder of $$1$$ when it is divided by $$9$$ the multiple of $$9$$ must end in a $$2$$. We now look for this one: $$9(1) = 9$$ $$9(2) =18$$ $$9(3) = 27$$ $$9(4) = 36$$ $$9(5) = 45$$ $$9(6) =54$$ $$9(8) = 72$$ The number $$72 +1= 73$$ satisfies both conditions. We subtract the biggest multiple of $$11$$ less than $$73$$ to get the remainder. Thus, $$73- 11(6) = 73 - 66 =\\boxed{(\\rm E) 7 }$$. "], "answer_value": "E"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1095", "queId": "ff8080814d9539f1014d9b61d3560a50", "competition_source_list": ["1992年第3届全国希望杯初一竞赛初赛第8题", "2019~2020学年辽宁沈阳皇姑区沈阳市33中学初一上学期期中第5题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$2a+5b$$减去$$4a-4b$$的一半，应当得到（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4a-b$$ "}], [{"aoVal": "B", "content": "$$b-a$$ "}], [{"aoVal": "C", "content": "$$a-9b$$ "}], [{"aoVal": "D", "content": "$$7b$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式的加减运算->整式加减", "课内体系->能力->运算能力"], "answer_analysis": ["$$(2a+5b)-\\frac{1}{2}(4a-4b)=2a+5b-2a+2b=7b$$，选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "611", "queId": "dedec3331c4b49029aae50b6c3a85fb6", "competition_source_list": ["2005年第16届希望杯初二竞赛复赛第1题4分", "其它"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a$$，$$b$$均为正整数，且$$m=ab(a+b)$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$m$$一定是奇数 "}], [{"aoVal": "B", "content": "$$m$$一定是偶数 "}], [{"aoVal": "C", "content": "只有$$a$$,$$b$$当均为偶数时，$$m$$是偶数 "}], [{"aoVal": "D", "content": "只有$$a$$,$$b$$当一个为偶数、另一个为奇数时，$$m$$是偶数 ~ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->同余->奇数与偶数", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算"], "answer_analysis": ["$$m=ab(a+b)$$ 若$$a$$，$$b$$中有一个为偶数，$$m$$必为偶数． 若$$a$$，$$b$$均为奇数，$$a+b$$为偶数，$$m$$必为偶数． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "750", "queId": "45401202750b4d4b8a3a2a44553606ae", "competition_source_list": ["2016年第27届全国希望杯初一竞赛初赛第5题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若非零自然数$$a$$，$$b$$的最大公约数与最小公倍数之和恰等于$$a$$，$$b$$的乘积，则$${{\\left( \\frac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \\right)}^{10}}=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$1024$$ "}], [{"aoVal": "C", "content": "$$2104$$ "}], [{"aoVal": "D", "content": "$$2016$$ "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算", "竞赛->知识点->数论->整除->因数与倍数"], "answer_analysis": ["设$$a$$，$$b$$的最大公约数是$$c$$， 则它们的最小公倍数为$$c\\cdot \\frac{a}{c}\\cdot \\frac{b}{c}=\\frac{ab}{c}$$， 由题意，得$$c+\\frac{ab}{c}=ab$$， ∴$$ab=\\frac{{{c}^{2}}}{c-1}$$． 若$$c$$为奇数，则$${{c}^{2}}$$为奇数，$$c-1$$为偶数，$$\\frac{{{c}^{2}}}{c-1}$$不是整数，舍去． 若$$c$$为偶数，则$${{c}^{2}}$$为偶数，$$c-1$$为奇数，当且仅当$$c=2$$时，$$\\frac{{{c}^{2}}}{c-1}$$为整数． ∴$$ab=\\frac{{{2}^{2}}}{2-1}=4$$． 又∵$$a$$，$$b$$均为正整数，且它们的最大公约数为$$2$$， ∴$$a=b=2$$， ∴$${{\\left( \\frac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \\right)}^{10}}={{\\left( \\frac{16}{8} \\right)}^{10}}={{2}^{10}}=1024$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "113", "queId": "824eef26e9214a5d9267c8195d5264e7", "competition_source_list": ["2009年第20届希望杯初二竞赛第1试第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知在代数式$$a+bx+c{{x}^{2}}$$中，$$a$$，$$b$$，$$c$$都是整数，当$$x=3$$时，该式的值是$$2008$$；当$$x=7$$时，该式的值是$$2009$$，这样的代数式有．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$个 "}], [{"aoVal": "B", "content": "$$1$$个 "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "无穷多个 "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式的加减运算->整式加减"], "answer_analysis": ["当$$x=3$$时，$$a+3b+9c=2008$$，① 当$$x=7$$时，$$a+7b+49c=2009$$， ② ②$$-$$①得，$$4b+40c=1$$． 当$$b$$，$$c$$都是整数时，上式左边总为偶数，不可能等于$$1$$，所以不存在这样的代数式． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "459", "queId": "358e5a5e75ce425cbe27cf9d551207c1", "competition_source_list": ["2014年第25届全国希望杯初一竞赛复赛第10题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "水池$$A$$和$$B$$都是深$$1.2$$米，底部是$$3$$米$$\\times 2$$米的长方体．$$1$$号阀门$$18$$分钟可将无水的$$A$$池注满，$$2$$号阀门$$24$$分钟可将$$A$$池中满池的水注入$$B$$池．最初$$A$$、$$B$$均为空池，若同时打开$$1$$号，$$2$$号阀门，当$$A$$池水深$$0.4$$米时，同时关闭两个阀门，这时$$B$$池中有水（~ ）立方米．", "answer_option_list": [[{"aoVal": "A", "content": "$$0.9$$ "}], [{"aoVal": "B", "content": "$$1.8$$ "}], [{"aoVal": "C", "content": "$$3.6$$ "}], [{"aoVal": "D", "content": "$$7.2$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->能力->分析和解决问题能力", "课内体系->知识点->数->有理数->有理数与实际问题"], "answer_analysis": ["由已知条件可得，$$1$$号阀门每分钟注入$$\\frac{1}{18}$$池水，而$$2$$号阀门每分钟放出$$\\frac{1}{24}$$池水． 到$$A$$池深$$0.4$$米时，$$A$$池中正好留存了$$\\frac{1}{3}$$池水， 则注水时间为$$\\frac{1}{3}\\div \\left( \\frac{1}{18}-\\frac{1}{24} \\right)=24$$（分钟）， 故同时关闭阀门时，恰好放了$$24$$分钟水，正好把$$B$$池放满， 故$$B$$水池中有水$$3\\times 2\\times 1.2=7.2$$（立方米）． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "770", "queId": "7b21b04006b240a68ce5ad0f29fc41e5", "competition_source_list": ["1999年第10届希望杯初一竞赛第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "定义：一个工厂一年的生产增长率就是： $$\\frac{当年产值 - 前一年产值}{前一年产值}\\times 100 \\%$$ 如果该工厂$$2000$$年的产值要达到$$1998$$年产值的$$1.44$$倍，而且每年的生产增长率都是$$x$$，则$$x$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$5 \\%$$ "}], [{"aoVal": "B", "content": "$$10 \\%$$ "}], [{"aoVal": "C", "content": "$$15 \\%$$ "}], [{"aoVal": "D", "content": "$$20 \\%$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元二次方程->一元二次方程与实际问题->一元二次方程的增长率问题"], "answer_analysis": ["设$$1998$$年的产值为$$1$$，则$$1999$$年产值为$$1+x$$，$$2000$$年产值为$$(1+x)+(1+x)x={{(1+x)}^{2}}$$．依题意，应当$${{(1+x)}^{2}}=1.44$$． 因为$${{(\\pm 1.2)}^{2}}=1.44$$，且$$x\\textgreater0$$， 所以$$1+x=1.2$$，$$x=0.2=20 \\%$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "322", "queId": "5db5e22431124ee0959658d3c9c44596", "competition_source_list": ["2001年第6届华杯赛初一竞赛决赛第19题"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$100$$以内的质数从小到大排成一个数字串，依次完成以下五项工作叫做一次操作： （$$1$$）将左边第一个数码移到数字串的最右边； （$$2$$）从左到右两位一节组成若干个两位数； （$$3$$）从这些两位数中划去合数； （$$4$$）所剩的两位质数中有相同者，保留左边的一个，其余划去； （$$5$$）所余的两位质数保持数码次序又组成一个新的数字串． 问：经过$$1997$$次操作，所得到数字串是什么？", "answer_option_list": [[{"aoVal": "A", "content": "$$1173$$ "}], [{"aoVal": "B", "content": "$$1731$$ "}], [{"aoVal": "C", "content": "$$7311$$ "}], [{"aoVal": "D", "content": "$$3117$$ "}], [{"aoVal": "E", "content": "$$1713$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->操作与游戏", "课内体系->知识点->综合与实践->规律探究与程序框图->规律探究->数列找规律->数列找规律-具有周期规律的数列"], "answer_analysis": ["第$$1$$次操作得数字串$$711131737$$；第$$2$$次操作得数字串$$11133173$$；第$$3$$次操作得数字串$$111731$$；第$$4$$次操作得数字串$$1173$$；第$$5$$次操作得数字串$$1731$$；第$$6$$次操作得数字串$$7311$$；第$$7$$次操作得数字串$$3117$$；第$$8$$次操作得数字串$$1173$$，以下以$$4$$为周期循环，即$$4k$$次操作所得数字串均为$$1173$$．$$1996=4\\times 499$$，所以第$$1996$$次操作得数字串$$1173$$，因此，第$$1997$$次操作得数字串$$1731$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "387", "queId": "3982b468c8bc41cc9f3a23f358a8e0f0", "competition_source_list": ["1998年第9届希望杯初一竞赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "有下面$$4$$个命题： ①两个数的差一定是正数． ②两个整式的和一定是整式． ③两个同类项的数字系数相同． ④若两个角的和等于$$180{}^{}\\circ $$，则这两个角互为邻补角． 其中真命题的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->几何图形初步->角->角的基础问题"], "answer_analysis": ["由$$3-4=-1$$，知命题①不真； $$3a{{b}^{2}}$$与$$5a{{b}^{2}}$$是同类项，但数字系数不同，③不真； 由于两条平行线被第三条直线所截，同旁内角之和为$$180{}^{}\\circ $$，但它们并不互为邻补角．命题④不真． 易知，两个整式的和仍是整式是真命题． 所以只有$$1$$个真命题，选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1344", "queId": "ffe0640a91794ccf943c1a3a585e9d49", "competition_source_list": ["2006年第17届希望杯初二竞赛第2试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列四组根式中，是同类二次根式的一组是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{2.5}$$和$$2\\sqrt{0.5}$$ "}], [{"aoVal": "B", "content": "$$3a\\sqrt{a}$$和$$3b\\sqrt{b}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{{{a}^{2}}b}$$和$$\\sqrt{a{{b}^{2}}}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{a{{b}^{7}}{{c}^{3}}}$$和$$\\sqrt{\\frac{{{c}^{3}}}{ab}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->二次根式->二次根式的性质与运算"], "answer_analysis": ["$$\\sqrt{a{{b}^{7}}{{c}^{3}}}={{b}^{3}}c\\sqrt{abc}$$，$$\\sqrt{\\frac{{{c}^{3}}}{ab}}=\\frac{c}{ab}\\sqrt{abc}$$，它们是同类二次根式． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "347", "queId": "875ed1005b3d465fb10cd609fd6128f7", "competition_source_list": ["2009年第20届希望杯初一竞赛第1试第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，$$2\\angle A=3\\angle B$$，且$$\\angle C-30{}^{}\\circ =\\angle A+\\angle B$$，则$$\\triangle ABC$$是．", "answer_option_list": [[{"aoVal": "A", "content": "锐角三角形 "}], [{"aoVal": "B", "content": "钝角三角形 "}], [{"aoVal": "C", "content": "有一个角是$$30{}^{}\\circ $$的直角三角形 "}], [{"aoVal": "D", "content": "等腰直角三角形 "}]], "knowledge_point_routes": ["课内体系->能力->分析和解决问题能力", "课内体系->知识点->三角形->三角形及多边形->三角形的基础->三角形的分类", "课内体系->知识点->三角形->三角形及多边形->与三角形有关的角->三角形内角和的应用"], "answer_analysis": ["由题意得：$$\\begin{cases}\\angle A=\\dfrac{3}{2}\\angle B \\angle C=\\angle A+\\angle B+30{}^{}\\circ \\angle A+\\angle B+\\angle C=180{}^{}\\circ \\end{cases}$$， 解得$$\\angle A=45{}^{}\\circ $$，$$\\angle B=30{}^{}\\circ $$，$$\\angle C=105{}^{}\\circ $$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "452", "queId": "55155eddb05e4362a11e55f1bde32222", "competition_source_list": ["2001年第12届希望杯初二竞赛第1试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$y={{x}^{4}}-4{{x}^{3}}+8{{x}^{2}}-8x+5$$，其中$$x$$为任意实数，则$$y$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "一切实数 "}], [{"aoVal": "B", "content": "一切正实数 "}], [{"aoVal": "C", "content": "一切大于或等于$$5$$的实数 "}], [{"aoVal": "D", "content": "一切大于或等于$$2$$的实数 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->二次函数->二次函数的最大值与最小值"], "answer_analysis": ["$$y={{x}^{4}}-4{{x}^{3}}+8{{x}^{2}}-8x+5$$ $$={{x}^{4}}+4{{x}^{2}}+4-4{{x}^{3}}+4{{x}^{2}}-8x+1$$ $$={{({{x}^{2}}+2)}^{2}}-4x({{x}^{2}}+2)+{{(2x)}^{2}}+1$$ $$={{[({{x}^{2}}+2)-2x]}^{2}}+1$$ $$={{[{{(x-1)}^{2}}+1]}^{2}}+1$$. 因为$${{(x-1)}^{2}}\\geqslant 0$$，所以$${{(x-1)}^{2}}+1\\geqslant 1$$. 所以 当$$x=1$$时，$$y$$取得最小值$$2$$， 即$$y$$的取值范围是一切大于或等于$$2$$的实数．故选$$\\text{D}$$. "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1280", "queId": "8aac50a74e724b3f014e83231e553c2a", "competition_source_list": ["1996年第7届全国希望杯初一竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果一个方程的解都能满足另一个方程，那么，这两个方程（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "是同解方程 "}], [{"aoVal": "B", "content": "不是同解方程 "}], [{"aoVal": "C", "content": "是同一个方程 "}], [{"aoVal": "D", "content": "可能不是同解方程 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->一元一次方程->解一元一次方程->方程解的关系"], "answer_analysis": ["当另一个方程的解也都满足第一个方程时，这两个方程才是同解方程，因此排除$$\\text{B}$$． 但另一个方程的解不都满足第一个方程时，它们不是同解方程，所以排除$$\\text{A}$$、$$\\text{C}$$， 因此选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "482", "queId": "2481ca16aa1040a59b9de9463d8015a2", "competition_source_list": ["2011年第16届华杯赛初一竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "船在江中顺水航行与逆水航行的速度之比为$$7:2$$，那么它在两港间往返一次的平均速度与顺水速度之比为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{7}{14}$$ "}], [{"aoVal": "B", "content": "$$\\frac{9}{14}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{9}$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{9}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->方程应用"], "answer_analysis": ["由已知条件可设船顺水航行的速度是$$7v$$，逆水航行的速度是$$2v$$，设两港之间的距离为$$s$$，则船往返一次的平均速度为$$\\frac{2s}{\\frac{s}{7v}+\\frac{s}{2v}}=\\frac{28}{9}v$$，从而平均速度与顺水速度之比为$$\\frac{28}{9}v:7v=\\frac{4}{9}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1381", "queId": "984ad06b86174492b6a781f012cdc9d9", "competition_source_list": ["2008年第19届希望杯初一竞赛第2试第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$是前$$3$$个质数，并且$$a\\textless{}b\\textless{}c$$，现给出下列四个判断： ①$${{(a+b)}^{2}}$$不能被$$c$$整除； ②$${{a}^{2}}+{{b}^{2}}$$不能被$$c$$整除； ③$${{(b+c)}^{2}}$$不能被$$a$$整除； ④$${{a}^{2}}+{{c}^{2}}$$不能被$$b$$整除． 其中不正确的判断是．", "answer_option_list": [[{"aoVal": "A", "content": "①、② "}], [{"aoVal": "B", "content": "①、③ "}], [{"aoVal": "C", "content": "②、③ "}], [{"aoVal": "D", "content": "③、④ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->整除的概念与基本性质"], "answer_analysis": ["由题意，得$$a=2$$，$$b=3$$，$$c=5$$， $${{(a+b)}^{2}}=25$$，$$c=5$$，①不正确； $${{a}^{2}}+{{b}^{2}}=13$$，$$c=5$$，②正确； $${{(b+c)}^{2}}=64$$，$$a=2$$，③不正确； $${{a}^{2}}+{{c}^{2}}=29$$，$$b=3$$，④正确； 不正确的判断是①、③． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1259", "queId": "8aac50a74e023208014e3f51e1ed190a", "competition_source_list": ["1995年第6届全国希望杯初一竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果由四舍五入得到的近似数是$$35$$，那么在下列各数中不可能是真值的数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$34.49$$ "}], [{"aoVal": "B", "content": "$$34.51$$ "}], [{"aoVal": "C", "content": "$$34.99$$ "}], [{"aoVal": "D", "content": "$$35.01$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->近似数与有效数字->精确数位"], "answer_analysis": ["由于$$34.51$$，$$34.99$$，$$35.01$$四舍五入的近似值都可能是$$35$$， 而只有$$34.49$$不可能是真值，选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "990", "queId": "ff8080814cfa9b24014d02c616cf13b3", "competition_source_list": ["2012年竞赛第5题3分", "初二其它"], "difficulty": "4", "qtype": "single_choice", "problem": "黑板上写有$$1$$，$$\\frac{1}{2}$$，$$\\cdots$$，$$\\frac{1}{100}$$共$$100$$个数字，每次操作先从黑板上的数中选取两个数$$a$$，$$b$$，然后删去$$a$$，$$b$$，并在黑板上写上数$$a+b+ab$$，则经过$$99$$次操作后黑板上剩下的数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$2012$$ "}], [{"aoVal": "B", "content": "$$101$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$99$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->因式分解的应用", "课内体系->能力->运算能力"], "answer_analysis": ["由$$a+b+ab+1=(a+1)(b+1)$$得，每次操作前和操作后，黑板上的每个数加$$1$$后的乘积不变， 设经过$$99$$次操作后黑板上剩下的数为$$x$$， 则$$x+1=(1+1)( \\frac{1}{2}+1)(\\frac{1}{3}+1)\\cdots (\\frac{1}{100}+1)$$， 解得$$x=100$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "473", "queId": "900d596c8cfe47c98bab74b84dd1c179", "competition_source_list": ["2016年全国华杯赛初一竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x+y+z=5$$，$$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=5$$，$$xyz=1$$，则$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=$$ ．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式", "课内体系->知识点->式->分式->分式的运算->分式加减混合运算"], "answer_analysis": ["∵$$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=5$$，$$xyz=1$$， ∴$$\\frac{yz+xz+xy}{xyz}=5$$，即$$xy+yz+xz=5$$． ∵$$x+y+z=5$$， ∴$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{\\left( x+y+z \\right)}^{2}}-2\\left( xy+yz+xz \\right)=15$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "650", "queId": "8848e375733645e691b00629b1e92ba9", "competition_source_list": ["2006年第17届希望杯初二竞赛第1试第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "对实数$$a$$，$$b$$，定义运算``$$*$$''如下： $$a*b=\\begin{cases}{{a}^{2}}b\\quad a\\geqslant b a{{b}^{2}}\\quad a\\textless{}b \\end{cases}$$． 现已知$$3*m=36$$，则实数$$m$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$\\pm 2\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$4$$或$$\\pm 2\\sqrt{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->整式的乘除运算->单项式乘单项式", "课内体系->知识点->综合与实践->规律探究与程序框图->定义新运算", "课内体系->能力->运算能力", "课内体系->思想->分类讨论思想"], "answer_analysis": ["当$$3\\geqslant m$$时，有 $$3*m={{3}^{2}}\\cdot m=9m=36$$， 解得$$m=4$$，与$$3\\geqslant m$$予盾，舍去． 当$$3\\textless{}m$$时，有$$3*m=3{{m}^{2}}=36$$， 解得$$m=\\pm 2\\sqrt{3}$$，因为$$3\\textless{}m$$，因此舍去$$m=-2\\sqrt{3}$$． 所以选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "265", "queId": "2fac29f13d514cd5903083fe6adb7b67", "competition_source_list": ["2019~2020学年10月江苏无锡梁溪区东林中学初一上学期月考第10题3分", "1995年第6届全国希望杯初一竞赛复赛第1题", "2017~2018学年10月江苏扬州高邮市高邮市南海中学初一上学期月考第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$y$$是正数，且$$x+y\\textless{}0$$，则在下列结论中，错误的一个是． If $$y$$ is a positive number and $$x+y\\textless{}0$$, then which of the following conclusions is wrong ？．", "answer_option_list": [[{"aoVal": "A", "content": "$${{x}^{3}}y\\textgreater0$$ "}], [{"aoVal": "B", "content": "$$x+\\left\\textbar{} y \\right\\textbar\\textless{}0$$ "}], [{"aoVal": "C", "content": "$$\\left\\textbar{} x \\right\\textbar+y\\textgreater0$$ "}], [{"aoVal": "D", "content": "$$x-{{y}^{2}}\\textless{}0$$ "}], [{"aoVal": "E", "content": "Both A and C "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["∵$$y\\textgreater0$$，若$$x\\geqslant 0$$则$$x+y\\textgreater0$$，与$$x+y\\textless{}0$$矛盾． 所以由$$y\\textgreater0$$，$$x+y\\textless{}0$$必有$$x\\textless{}0$$． 因此，$${{x}^{3}}\\textless{}0$$，$${{x}^{3}}y\\textless{}0$$，即$$\\text{A}$$是错误的． 事实上，$$y\\textgreater0$$，$$x+y\\textless{}0$$，即$$x+\\left\\textbar{} y \\right\\textbar\\textless{}0$$，$$\\text{B}$$成立． $$\\left\\textbar{} x \\right\\textbar+y\\textgreater0$$，$$\\text{C}$$成立．$$x\\textless{}0$$，$${{y}^{2}}\\textgreater0$$，$$x-{{y}^{2}}\\textless{}0$$，$$\\text{D}$$成立．因此，选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "925", "queId": "fb4e78e3bd044a47beab37ba5a36cbba", "competition_source_list": ["2000年第17届全国初中数学联赛竞赛第1题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$\\sqrt{14+6\\sqrt{5}}-\\sqrt{14-6\\sqrt{5}}$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{5}$$ "}], [{"aoVal": "C", "content": "$$2\\sqrt{5}$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算", "课内体系->知识点->式->二次根式->二次根式的运算->多重二次根式"], "answer_analysis": ["$$\\sqrt{14+6\\sqrt{5}}-\\sqrt{14-6\\sqrt{5}}=\\sqrt{14+2\\sqrt{45}}-\\sqrt{14-2\\sqrt{45}}=\\sqrt{{{(\\sqrt{9}+\\sqrt{5})}^{2}}}-\\sqrt{{{(\\sqrt{9}-\\sqrt{5})}^{2}}}$$， $$=\\sqrt{9}+\\sqrt{5}-\\sqrt{9}+\\sqrt{5}=2\\sqrt{5}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "330", "queId": "301caada2b9045bdb54d8159a1fe68b1", "competition_source_list": ["2001年第12届希望杯初二竞赛第1试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "代数式$$\\sqrt{x}+\\sqrt{x-1}+\\sqrt{x-2}$$的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1+\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "不存在的 "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的基础->二次根式有意义的条件", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的加减以及混合运算", "课内体系->知识点->式->二次根式->二次根式化简求值"], "answer_analysis": ["要使代数式$$\\sqrt{x}+\\sqrt{x-1}+\\sqrt{x-2}$$有意义，需满足条件： 当$$\\begin{cases}x\\geqslant 0 x-1\\geqslant 0 x-2\\geqslant 0 \\end{cases}$$，所以$$x\\geqslant 2$$， 当$$x\\geqslant 2$$时， 即$$\\sqrt{x}+\\sqrt{x-1}+\\sqrt{x-2}\\geqslant \\sqrt{2}+\\sqrt{2-1}+\\sqrt{2-2}=\\sqrt{2}+1$$， 即$$x=2$$时，原代数式取得最小值$$\\sqrt{2}+1$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1535", "queId": "cb254cd9d4ec483e9a9378591a650cbf", "competition_source_list": ["2014年第25届全国希望杯初二竞赛初赛第10题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "将$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$，$$8$$这$$8$$个数排成一行，要使$$8$$的两边各数的和相等，那么不同的排法一共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$1152$$种 "}], [{"aoVal": "B", "content": "$$576$$种 "}], [{"aoVal": "C", "content": "$$288$$种 "}], [{"aoVal": "D", "content": "$$144$$种 "}]], "knowledge_point_routes": ["课内体系->能力->分析和解决问题能力", "竞赛->知识点->组合->排列与组合"], "answer_analysis": ["因为去掉$$8$$以后，余下$$7$$个数的和是$$1+2+3+4+5+6+7=28$$， 所以$$8$$的两边各数的和分别是$$14$$． 因为$$7+6=13$$， 所以$$14$$至少是由$$3$$个数相加而成的， 所以$$8$$的两边分别有$$3$$个数和$$4$$个数． 因为$$(7,6,1)$$，$$(7,5,2)$$，$$(7,4,3)$$，$$(6,5,3)$$这四组数的和都是$$14$$， 每组中的数的排列方法有$$3\\times 2\\times 1=6$$种， 与其对应的另外$$4$$个数（$$8$$除外）有$$4\\times 3\\times 2\\times 1=24$$种排列方法， 当排成横行时，这$$4$$组数有排在$$8$$的左边和右边两种排列方法， 因此，每组数的排列方法有$$2\\times 6\\times 24=288$$（种）． 所以，满足题意的排列方法有$$4\\times 288=1152$$（种）． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1098", "queId": "e91191e9fbb5487ebe9cbb3b7e4b357a", "competition_source_list": ["初二上学期其它", "2002年竞赛第1题5分", "2015年山东青岛黄岛区山东省青岛第九中学自主招生", "2016~2017学年江西景德镇初二下学期期中景德镇一中1班第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a\\textless{}b\\textless{}0$$，$${{a}^{2}}+{{b}^{2}}=4ab$$，则$$\\frac{a+b}{a-b}$$的值为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{6}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["知识标签->题型->式->分式->分式化简求值->题型：分式条件化简求值", "知识标签->学习能力->运算能力", "知识标签->知识点->式->分式->分式的运算->分式的混合运算"], "answer_analysis": ["$$a+b=-\\sqrt{6ab}$$，$$a-b=-\\sqrt{2ab}$$． ∵$${{a}^{2}}+{{b}^{2}}=4ab$$， ∴$${{a}^{2}}+{{b}^{2}}+2ab={{\\left( a+b \\right)}^{2}}=6ab$$① ∴$${{a}^{2}}+{{b}^{2}}-2ab={{\\left( a-b \\right)}^{2}}=2ab$$② $$\\frac{①}{②}$$，得~~$$\\frac{{{\\left( a+b \\right)}^{2}}}{{{\\left( a-b \\right)}^{2}}}=\\frac{6ab}{2ab}$$ ∵$$a\\textless{}b\\textless{}0$$， ∴$$ab\\textgreater0$$，$$a+b\\textless{}0$$，$$a-b\\textless{}0$$， ∴$$\\frac{{{\\left( a+b \\right)}^{2}}}{{{\\left( a-b \\right)}^{2}}}={{\\left( \\frac{a+b}{a-b} \\right)}^{2}}=3$$， ∴$$\\frac{a+b}{a-b}=\\sqrt{3}$$． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "784", "queId": "4e5e27c12c124757acb8bfd97e1f00ad", "competition_source_list": ["2016年山东青岛竞赛二中杯", "初二其它"], "difficulty": "3", "qtype": "single_choice", "problem": "✩✩✩✩✩黑板上写有$$1$$，$$\\frac{1}{2}$$，$$\\cdots$$，$$\\frac{1}{100}$$共$$100$$个数字，每次操作先从黑板上的数中选取两个数$$a$$，$$b$$，然后删去$$a$$，$$b$$，并在黑板上写上数$$a+b+ab$$，则经过$$99$$次操作后黑板上剩下的数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$2012$$ "}], [{"aoVal": "B", "content": "$$101$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$99$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->其他方法", "课内体系->知识点->式->因式分解->因式分解的应用"], "answer_analysis": ["由$$a+b+ab+1=(a+1)(b+1)$$得，每次操作前和操作后，黑板上的每个数加$$1$$后的乘积不变， 设经过$$99$$次操作后黑板上剩下的数为$$x$$， 则$$x+1=(1+1)( \\frac{1}{2}+1)(\\frac{1}{3}+1)\\cdots (\\frac{1}{100}+1)$$， 解得$$x=100$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1425", "queId": "d7b513c56a9d4544b6cc57352ca6e46e", "competition_source_list": ["2002年第13届希望杯初一竞赛第2试第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "~下列四个命题： ①如果两个角是对顶角，则这两个角相等． ②如果两个角相等，则这两个角是对顶角． ③如果两个角不是对顶角，则这两个角不相等． ④如果两个角不相等，则这两个角不是对顶角． 其中正确的命题有（ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "1个 "}], [{"aoVal": "B", "content": "2个 "}], [{"aoVal": "C", "content": "3个 "}], [{"aoVal": "D", "content": "4个 "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->相交线与平行线->相交线->对顶角的定义与性质", "课内体系->知识点->几何图形初步->相交线与平行线->相交线->计算对顶角、邻补角度数"], "answer_analysis": ["②两个角相等，不一定为对顶角； ③两个角不是对项角，也可以相等． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1282", "queId": "8aac50a74e724b3f014e83287fc93c4a", "competition_source_list": ["1996年第7届全国希望杯初一竞赛初赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "线段$$AB=1996$$厘米，$$P$$、$$Q$$是线段$$AB$$上的两个点，线段$$AQ=1200$$厘米，线段$$BP=1050$$厘米，则线段$$PQ=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$254$$厘米 "}], [{"aoVal": "B", "content": "$$150$$厘米 "}], [{"aoVal": "C", "content": "$$127$$厘米 "}], [{"aoVal": "D", "content": "$$871$$厘米 "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->直线、射线、线段->直线、射线、线段的运算->线段和差的计算问题->线段和差-需要分类讨论"], "answer_analysis": ["$$PQ=AQ+PB-AB=1200+1050-1996=254$$（厘米），选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "233", "queId": "de62b1a47b3948ffbf19ec2db2051e6c", "competition_source_list": ["2013年第18届华杯赛初一竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$-6$$、$$-4$$、$$-3$$、$$-2$$、$$-1$$、$$3$$、$$6$$中选择任意两个数字，然后将它们相乘，得到的乘积的最大值为$$a$$，最小值为$$b$$，则$$\\frac{a}{b}$$的值为。 Pick any two numbers from $$-6$$, $$-4$$, $$-3$$, $$-2$$, $$-1$$, $$3$$, $$6$$ and multiply them, the maximum value of the resulting product is $$a$$, and the minimum value is $$b$$, then the value of $$\\frac{a}{b}$$ is ．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{2}{3}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{3}{4}$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "E", "content": "$$-\\frac{1}{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->整式的乘除运算", "课内体系->知识点->数->有理数->分数"], "answer_analysis": ["由条件可知，$$a=(-6)\\times (-4)=24$$，$$b=(-6)\\times 6=-36$$，所以 $$\\frac{a}{b}=-\\frac{24}{36}=-\\frac{2}{3}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1490", "queId": "bd26af76fd284ff6b163facd5c4725fd", "competition_source_list": ["1998年第9届希望杯初二竞赛第1试第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$${{a}_{1}}$$，$${{a}_{2}}$$，$${{b}_{1}}$$，$${{b}_{2}}$$均为正数，且$${{a}_{1}}\\geqslant {{a}_{2}}$$，$${{a}_{1}}\\leqslant {{b}_{1}}$$，$${{a}_{1}}{{a}_{2}}\\leqslant {{b}_{1}}{{b}_{2}}$$，则$${{a}_{1}}+{{a}_{2}}$$与$${{b}_{1}}+{{b}_{2}}$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{a}_{1}}+{{a}_{2}}\\leqslant {{b}_{1}}+{{b}_{2}}$$ "}], [{"aoVal": "B", "content": "$${{a}_{1}}+{{a}_{2}}\\geqslant {{b}_{1}}+{{b}_{2}}$$ "}], [{"aoVal": "C", "content": "$${{a}_{1}}+{{a}_{2}}={{b}_{1}}+{{b}_{2}}$$ "}], [{"aoVal": "D", "content": "无法确定的 "}]], "knowledge_point_routes": ["课内体系->能力->分析和解决问题能力", "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["因为$${{a}_{1}}$$，$${{a}_{2}}$$，$${{b}_{1}}$$，$${{b}_{2}}$$均为正数，且$${{a}_{1}}{{a}_{2}}\\leqslant {{b}_{1}}{{b}_{2}}$$． 所以$${{b}_{2}}\\geqslant \\frac{{{a}_{1}}{{a}_{2}}}{{{b}_{1}}}$$， 所以$${{b}_{1}}+{{b}_{2}}\\geqslant {{b}_{1}}+\\frac{{{a}_{1}}{{a}_{2}}}{{{b}_{1}}}$$， 所以$$({{b}_{1}}+{{b}_{2}})-({{a}_{1}}+{{a}_{2}})\\geqslant {{b}_{1}}+\\frac{{{a}_{1}}{{a}_{2}}}{{{b}_{1}}}-{{a}_{1}}-{{a}_{2}}$$ $$=({{b}_{1}}-{{a}_{1}})+\\frac{{{a}_{2}}({{a}_{1}}-{{b}_{1}})}{{{b}_{1}}}=({{b}_{1}}-{{a}_{1}})\\left( 1-\\frac{{{a}_{2}}}{{{b}_{1}}} \\right)$$． 因为$${{a}_{1}}\\leqslant {{b}_{1}}$$，$${{a}_{2}}\\leqslant {{a}_{1}}$$， 所以$$0\\textless{}{{a}_{2}}\\leqslant {{b}_{1}}$$，$$0\\textless{}\\frac{{{a}_{2}}}{{{b}_{1}}}\\leqslant 1$$． 所以$${{b}_{1}}-{{a}_{1}}\\geqslant 0$$，$$\\left( 1-\\frac{{{a}_{2}}}{{{b}_{1}}} \\right)\\geqslant 0$$． 所以$${{b}_{1}}+{{b}_{2}}\\geqslant {{a}_{1}}+{{a}_{2}}$$． 又当$${{a}_{1}}$$，$${{a}_{2}}$$，$${{b}_{1}}$$，$${{b}_{2}}$$均相等时，等号成立． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1001", "queId": "811f79ab954a40b0a3ebfe0681de12eb", "competition_source_list": ["1999年竞赛（全国初中数学竞赛）第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某城市按以下规定收取煤气费：（$$1$$）每月所用煤气按整立方米数计算．（$$2$$）若每月用煤气不超过$$60$$立方米，按每立方米$$0.8$$元收费；若超过$$60$$立方米，超过部分按每立方米$$1.2$$元收费．已知某户人家某月的煤气费平均每立方米$$0.88$$元，则这户人家需要交煤气费．", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$元 "}], [{"aoVal": "B", "content": "$$66$$元 "}], [{"aoVal": "C", "content": "$$70$$元 "}], [{"aoVal": "D", "content": "$$75$$元 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->能力->推理论证能力", "课内体系->能力->分析和解决问题能力", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的梯度计价问题"], "answer_analysis": ["由于$$0.88\\textgreater0.8$$，所以这户人家用的煤气超过$$60$$立方米．设用了$$x$$立方米，则 $$60\\cdot 0.8+1.2\\left( x-60 \\right)=0.88x$$， 解得$$x=75$$， $$75\\times 0.88=66$$（元）． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "121", "queId": "08c9a902b82a4712906b52444ae5b23d", "competition_source_list": ["初二下学期单元测试《二次根式》二次根式的运算第43题", "2002年第19届全国初中数学联赛竞赛第1题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a=\\sqrt{2}-1$$，$$b=2\\sqrt{2}-\\sqrt{6}$$，$$c=\\sqrt{6}-2$$，那么$$a$$，$$b$$，$$c$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless{}b\\textless{}c$$ "}], [{"aoVal": "B", "content": "$$a\\textless{}c\\textless{}b$$ "}], [{"aoVal": "C", "content": "$$b\\textless{}a\\textless{}c$$ "}], [{"aoVal": "D", "content": "$$b\\textless{}c\\textless{}a$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的运算->二次根式比较大小"], "answer_analysis": ["∵$$a-b$$ $$=\\sqrt{2}-1-\\left( 2\\sqrt{2}-\\sqrt{6} \\right)$$ $$=\\sqrt{6}-\\left( 1+\\sqrt{2} \\right)$$ $$\\approx 2.449-2.414\\textgreater0$$， ∴$$a\\textgreater b$$； ∵$$a-c$$ $$=\\sqrt{2}-1-\\left( \\sqrt{6}-2 \\right)$$ $$=\\sqrt{2}+1-\\sqrt{6}$$ $$\\approx 2.414-2.449\\textless{}0$$， ∴$$a\\textless{}c$$； 于是$$b\\textless{}a\\textless{}c$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1385", "queId": "e08dfe5a7ea2428ea576d4182352e676", "competition_source_list": ["2010年第21届全国希望杯初一竞赛初赛第10题4分", "2019~2020学年浙江宁波镇海区宁波市镇海蛟川书院初一上学期期中第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a$$和$$b$$是有理数，若$$a+b=0$$，$${{a}^{2}}+{{b}^{2}}\\ne 0$$，则在$$a$$和$$b$$之间一定（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "存在负整数 "}], [{"aoVal": "B", "content": "存在正整数 "}], [{"aoVal": "C", "content": "存在负分数 "}], [{"aoVal": "D", "content": "不存在正分数 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->相反数->相反数的性质"], "answer_analysis": ["由可知$$a\\ne 0$$且$$b\\ne 0$$． 再由$$a+b=0$$，可知$$a=-b$$． 不妨设$$a\\textgreater b$$，则$$a\\textgreater0\\textgreater b$$， 则在$$a$$和$$b$$之间一定存在负分数． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1105", "queId": "ff8080814d9efd56014da55d10ad0783", "competition_source_list": ["1992年第3届全国希望杯初一竞赛复赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$p$$为偶数，$$q$$为奇数，方程组$$\\left { \\begin{matrix}x-1992y=p 1993x+3y=q \\end{matrix} \\right.$$的解是整数，那么（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$x$$是奇数，$$y$$是偶数 "}], [{"aoVal": "B", "content": "$$x$$是偶数，$$y$$是奇数 "}], [{"aoVal": "C", "content": "$$x$$是偶数，$$y$$是偶数 "}], [{"aoVal": "D", "content": "$$x$$是奇数，$$y$$是奇数 "}]], "knowledge_point_routes": ["课内体系->思想->方程思想", "课内体系->思想->整体思想", "课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->二元��次方程组的定义"], "answer_analysis": ["∵$$p$$为偶数，$$q$$为奇数，其解$$x$$，$$y$$又是整数， 由$$x-1992y=p$$可知$$x$$为偶数，由$$1993x+3y=q$$可知$$y$$是奇数，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1159", "queId": "a963448b95b344979d84234fb52c0bcd", "competition_source_list": ["2000年第11届希望杯初一竞赛第6题"], "difficulty": "0", "qtype": "single_choice", "problem": "某件商品，若按标价的八折出售，可获利$$20 \\%$$，若按原价出售，则可获利（ ~）", "answer_option_list": [[{"aoVal": "A", "content": "$$25 \\%$$ "}], [{"aoVal": "B", "content": "$$40 \\%$$ "}], [{"aoVal": "C", "content": "$$50 \\%$$ "}], [{"aoVal": "D", "content": "$$66.7 \\%$$ "}]], "knowledge_point_routes": ["课内体系->能力->分析和解决问题能力", "课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的经济问题->一元一次方程的经济问题-打折"], "answer_analysis": ["设标价为$$x$$，进价为$$a$$，由题意得$$0.8x-a=0.2a$$解得$$x=1.5a$$，按原价出售可获利$$\\frac{1.5a-a}{a}=0.5$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1190", "queId": "8aac49074e023206014e25c1b67272c1", "competition_source_list": ["1997年第8届全国希望杯初一竞赛初赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "有理数$$b$$满足$$\\left\\textbar{} b \\right\\textbar\\textless{}3$$，并且有理数$$a$$使得$$a\\textless{}b$$恒能成立，则$$a$$的取值范围是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "小于或等于$$3$$的有理数 "}], [{"aoVal": "B", "content": "小于$$3$$的有理数 "}], [{"aoVal": "C", "content": "小于或等于$$-3$$的有理数 "}], [{"aoVal": "D", "content": "小于$$-3$$的有理数 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->绝对值的非负性", "课内体系->知识点->数->有理数->绝对值->绝对值的代数意义", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->解一元一次不等式"], "answer_analysis": ["$$\\left\\textbar{} b \\right\\textbar\\textless{}3$$就是$$-3\\textless{}b\\textless{}3$$，只有当$$a\\leqslant -3$$时，$$a\\textless{}b$$恒成立，选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "266", "queId": "2b1dbf618f5d42acb952995257c36471", "competition_source_list": ["2018年第29届希望杯初一竞赛初赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$A$$、 $$B$$ 两家商店的笔记本定价都是$$10$$元一本．已知在$$A$$商店每购$$5$$本赠一本，在$$B$$商店，超过$$5$$本（含$$5$$本），每本八五折．小明需要购买$$32$$本笔记本，则他至少花元．", "answer_option_list": [[{"aoVal": "A", "content": "$$267$$ "}], [{"aoVal": "B", "content": "$$268$$ "}], [{"aoVal": "C", "content": "$$270$$ "}], [{"aoVal": "D", "content": "$$272$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->能力->分析和解决问题能力", "课内体系->知识点->数->有理数->有理数与实际问题->有理数乘除法与实际问题"], "answer_analysis": ["在$$\\text{A}$$商店每购$$5$$本赠一本，$$5\\div6 \\approx0.83$$，低于八五折， 所以可以在$$\\text{A}$$商店购买$$24$$本，在$$\\text{B}$$商店购买$$8$$本； 或者在$$\\text{A}$$商店购买直接购买$$32$$本． 在$$\\text{A}$$商店购买$$24$$本，在$$\\text{B}$$商店购买$$8$$本， 花费$$20\\times10+8\\times10\\times0.85=268$$（元）； 在$$\\text{A}$$商店购买直接购买$$32$$本，花费$$25\\times10+2\\times10=270$$（元）． 综上，小明需要购买$$32$$本笔记本，则他至少花$$268$$元． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "23", "queId": "098ecdea94bf4c8b9a2512aebc91ebdf", "competition_source_list": ["2000年第11届希望杯初二竞赛第1试第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "互不相等的三个正数$$a$$、$$b$$、$$c$$恰为一个三角形的三条边长，则以下列三数为长度的线段一定能做成三角形的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{a}$$，$$\\frac{1}{b}$$，$$\\frac{1}{c}$$ "}], [{"aoVal": "B", "content": "$$a^{2}$$，$$ b^{2}$$，$$c^{2}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{a}$$，$$\\sqrt{b}$$，$$ \\sqrt{c}$$ "}], [{"aoVal": "D", "content": "$$\\textbar a-b\\textbar$$，$$\\textbar b-c\\textbar$$，$$\\textbar c-a\\textbar$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["如$$2$$，$$4$$，$$5$$为边可以组成三角形，而$$\\frac{1}{2}$$，$$ \\frac{1}{4}$$，$$ \\frac{1}{5}$$不能构成三角形，排除$$\\text{A}$$． 如$$3$$，$$4$$，$$5$$为边可以组成三角形，而$$3^{2}$$，$$ 4^{2}$$，$$ 5^{2}$$不能构成三角形，排除$$\\text{B}$$． 如$$2$$，$$3$$，$$4$$为边可以组成三角形，而$$\\textbar2-3\\textbar$$，$$\\textbar3-4\\textbar$$，$$\\textbar4-2\\textbar$$不能构成三角形，排除$$\\text{D}$$． 不妨设$$a\\textgreater b\\textgreater c\\textgreater0$$，满足$$a\\textless b+c$$， 所以$$(\\sqrt{b}+\\sqrt{c})^{2}=b+2 \\sqrt{b c}+c\\textgreater a$$， 所以$$\\sqrt{b}+\\sqrt{c}\\textgreater\\sqrt{a}$$， 所以$$\\sqrt{a}$$，$$ \\sqrt{b}$$，$$ \\sqrt{c}$$可以构成三角形． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1484", "queId": "c61bb6f52c064154bde99c7da1a63c43", "competition_source_list": ["2016年第27届全国希望杯初二竞赛初赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若关于$$x$$的不等式$$\\left { \\begin{matrix}2x-1\\textgreater3(x-2) 2x+m-1\\textgreater0 \\end{matrix} \\right.$$，只有四个整数解，则$$m$$的取值范围是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1\\textless{}m\\textless{}1$$ "}], [{"aoVal": "B", "content": "$$-1\\leqslant m\\textless{}1$$ "}], [{"aoVal": "C", "content": "$$-1\\textless{}m\\leqslant 1$$ "}], [{"aoVal": "D", "content": "$$-1\\leqslant m\\leqslant 1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->解一元一次不等式", "课内体系->知识点->方程与不等式->不等式（组）->含参不等式->由不等式（组）的整数解情况求参数范围"], "answer_analysis": ["解$$2x-1\\textgreater3(x-2)$$，得$$x\\textless{}5$$； 解$$2x+m-1\\textgreater0$$，得$$x\\textgreater\\frac{1-m}{2}$$． ∵不等式组只有四个整数解， ∴$$0\\leqslant \\frac{1-m}{2}\\textless{}1$$， 解得$$-1\\textless{}m\\leqslant 1$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "916", "queId": "bf74d980ac604d3fb07486b5ee780bbe", "competition_source_list": ["初二下学期其它", "1996年第7届希望杯初二竞赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a=\\dfrac{19961995}{1995}$$，$$b=\\dfrac{19951996}{1996}$$，$$c=\\dfrac{19951996}{1995}$$，$$d=\\dfrac{19961995}{1996}$$，则下列不等关系中成立的是（ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b\\textgreater c\\textgreater d$$ "}], [{"aoVal": "B", "content": "$$c\\textgreater a\\textgreater d\\textgreater b$$ "}], [{"aoVal": "C", "content": "$$a\\textgreater d\\textgreater c\\textgreater b$$ "}], [{"aoVal": "D", "content": "$$a\\textgreater c\\textgreater d\\textgreater b$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数除法运算", "课内体系->能力->运算能力"], "answer_analysis": ["$$a=10001+\\dfrac{10000}{1995}$$，$$b=10001-\\dfrac{10000}{1996}$$，$$c=10001+\\dfrac{1}{1995}$$，$$d=10001-\\dfrac{1}{1996}$$． 所以$$a\\textgreater c\\textgreater d\\textgreater b$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "619", "queId": "b5591c917b91452690f4ef563e35cd8a", "competition_source_list": ["1994年第5届希望杯初二竞赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "直角三角形的三条边的长度是正整数，其中一条直角的长度是$$13$$，那么它的周长为．", "answer_option_list": [[{"aoVal": "A", "content": "$$182$$ "}], [{"aoVal": "B", "content": "$$180$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->特殊三角形->直角三角形"], "answer_analysis": ["设另一条直角边的长度为$$x$$，斜边的长度为$$z$$，则$${{z}^{2}}-{{x}^{2}}={{13}^{2}}$$， 由于$$x$$，$$z$$均为正整数，且$$z\\textgreater x$$， 于是由$$\\left( z+x \\right)\\left( z-x \\right)=169\\times 1$$． 可得$$\\begin{cases}z+x=169 z-x=1 \\end{cases}$$． 所以三角形的周长为$$169+13=182$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "613", "queId": "2e56e107a7cc46bb932171a4b3ee3573", "competition_source_list": ["1998年第9届希望杯初二竞赛第1试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "将多项式$${{x}^{2}}-4{{y}^{2}}-9{{z}^{2}}-12yz$$分解成因式的积，结果是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(x+2y-3z)(x-2y-3z)$$ "}], [{"aoVal": "B", "content": "$$(x-2y-3z)(x-2y+3z)$$ "}], [{"aoVal": "C", "content": "$$(x+2y+3z)(x+2y-3z)$$ "}], [{"aoVal": "D", "content": "$$(x+2y+3z)(x-2y-3z)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->因式分解->因式分解：公式法"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde{{x}^{2}}-4{{y}^{2}}-9{{z}^{2}}-12yz$$ $$={{x}^{2}}-(4{{y}^{2}}+12yz+9{{z}^{2}})$$ $$={{x}^{2}}-{{(2y+3z)}^{2}}$$ $$=[x+(2y+3z)][x-(2y+3z)]$$ $$=(x+2y+3z)(x-2y-3z)$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1453", "queId": "a67bdc2b5f054819a6f18dc4fee38cdd", "competition_source_list": ["1991年第2届希望杯初二竞赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "两个正数$$m$$，$$n$$的比是$$t\\left( t~ \\textgreater~ 1 \\right)$$．若$$m+n=s$$，则$$m$$，$$n$$中较小的数可以表示为．", "answer_option_list": [[{"aoVal": "A", "content": "$$ts$$ "}], [{"aoVal": "B", "content": "$$s-ts$$ "}], [{"aoVal": "C", "content": "$$\\frac{ts}{1+s}$$ "}], [{"aoVal": "D", "content": "$$\\frac{s}{1+t}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["设$$m=3$$，$$n=1$$，则$$t=3$$，$$s=4$$，只有$$\\frac{s}{1+t}=\\frac{4}{1+3}=1=n$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1595", "queId": "fea977030c3248a8bf85e46e06663be9", "competition_source_list": ["2014年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "昨天，四个孩子在城市医院出生。假设每个孩子是男孩或女孩的可能性是相等的。以下哪种结果最有可能？ Four children were born at City Hospital yesterday. Assume each child is equally likely to be a boy or a girl. Which of the following outcomes is most likely?．", "answer_option_list": [[{"aoVal": "A", "content": "all $$4$$ are boys 4个都是男孩 "}], [{"aoVal": "B", "content": "all $$4$$ are girls 四个都是女孩 "}], [{"aoVal": "C", "content": "$$2$$ are girls and $$2$$ are boys 2个女孩和$$2$$个男孩 "}], [{"aoVal": "D", "content": "$$3$$ are of one gender and $$1$$ is of the other gender 3个是同一种性别，$$1$$个是另一种性别 "}], [{"aoVal": "E", "content": "all of these outcomes are equally likely 所有这些结果都有同样的可能性 "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Counting, Probability and Statistics->Statistical Distribution", "课内体系->知识点->统计与概率->概率->可能性的大小"], "answer_analysis": ["城市医院昨日诞下四名婴儿．假设每个婴儿各可能有一半几率为男孩或女孩．则下列哪个事件最有可能？ A．$$4$$个男婴 B．$$4$$个女婴 C．$$2$$男婴$$2$$女婴 D．$$3$$个为一种性别，另一个为异性 E．以上事件皆等可能 我们先把例子分析一下．$$A$$发生的概率为$${{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{16}$$，$$B$$发生的概率为$${{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{16}$$．$$C$$发生的概率为$$\\left( \\begin{matrix}4 2 \\end{matrix} \\right)\\cdot {{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{3}{8}$$．因为我们需要从$$4$$个孩子中选择$$2$$个成为女孩． $$D$$方面，有两种可能的情况，即$$3$$个女孩和$$1$$个男孩或$$3$$个男孩和$$1$$个女孩．第一种情况的概率就$$\\left( \\begin{matrix}4 1 \\end{matrix} \\right)\\cdot {{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{4}$$．因为我们需要从四个孩子中选择一个成为男孩．但是，第二种情况也有同样的可能性，因为我们在$$4$$个孩子中选择了$$1$$个是女孩，所以总的可能性是$$\\frac{1}{4}\\cdot 2=\\frac{1}{2}$$． 所以在四个分数中，$$D$$就最大的．所以我们的答案是$$3$$ are of one gender and $$1$$ is of the other gender． 故选$$\\text{D}$$． 可能性列在帕斯卡三角形的第四行，最左边的$$1$$是所有男孩的可能性，最右边的$$1$$是所有女孩的可能性．由于 pascal 的三角形第四行为$$1$$，$$4$$，$$6$$，$$4$$，$$1$$，$$6$$，均为每个性别的两个孩子的可能性，因此共有$$8$$个可能性，其中一个性别的三个孩子和另一个性别的一个孩子．由于总共有$$2^{4}=16$$个儿童性别的可能性． 故选$$\\text{D}$$． We\\textquotesingle ll just start by breaking cases down. The probability of $$A$$ occurring is $${{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{16}$$. The probability of $$B$$ occurring is $${{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{16}$$. The probability of $$C$$ occurring is $$\\left( \\begin{matrix}4 2 \\end{matrix} \\right)\\cdot {{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{3}{8}$$, because we need to choose $$2$$ of the $$4$$ children to be girls. For $$D$$, there are two possible cases, $$3$$ girls and $$1$$ boy or $$3$$ boys and $$1$$ girl. The probability of the first case is $$\\left( \\frac{4}{1} \\right)\\cdot {{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{4}$$ because we need to choose $$1$$ of the $$4$$ children to be a boy. However, the second case has the same probability because we are choosing $$1$$ of the $$4$$ children to be a girl, so the total probability is $$\\frac{1}{4}\\cdot 2=\\frac{1}{2}$$. So out of the four fractions, $$D$$ is the largest. So our answer is $$\\left( \\text{D} \\right)$$ $$3$$ of one gender and $$1$$ of the other. The possibilities are listed out in the fourth row of Pascal\\textquotesingle s triangle, with the leftmost $$1$$ being the possibility of all boys and the rightmost $$1$$ being the possibility of all girls. Since the fourth row of Pascal\\textquotesingle s Triangle goes $$1$$, $$4$$, $$6$$, $$4$$, $$1$$ and $$6$$ are all the possibilities of two children from each gender, there are a total of $$8$$ possibilities of three children from one gender and one from the other. Since there are a total of $${{2}^{4}}=16$$ total possibilities for the gender of the children, $$\\text{D}$$ has the highest probability. "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1593", "queId": "fa070c2944804de882b68bceed8d78f1", "competition_source_list": ["2018~2019学年9月湖南长沙雨花区中雅培粹学校初一上学期月考第12题3分", "1999年第10届希望杯初一竞赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$n$$是奇自然数，$${{a}_{1}}$$，$${{a}_{2}}$$，\\ldots，$${{a}_{n}}$$是$$n$$个互不相同的负整数，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$({{a}_{1}}+1)({{a}_{2}}+2)\\cdots ({{a}_{n}}+n)$$是正整数 "}], [{"aoVal": "B", "content": "$$({{a}_{1}}-1)({{a}_{2}}-2)\\cdots ({{a}_{n}}-n)$$是正整数 "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{{{a}_{1}}}+1 \\right)\\left( \\frac{1}{{{a}_{2}}}+2 \\right)\\cdots \\left( \\frac{1}{{{a}_{n}}}+n \\right)$$是正数 "}], [{"aoVal": "D", "content": "$$\\left( 1-\\frac{1}{{{a}_{1}}} \\right)\\left( 2-\\frac{1}{{{a}_{2}}} \\right)\\cdots \\left( n-\\frac{1}{{{a}_{n}}} \\right)$$是正数 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->能力->抽象概括能力", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->含乘方的有理数混合运算", "课内体系->知识点->综合与实践->规律探究与程序框图->规律探究->算式找规律"], "answer_analysis": ["$${{a}_{1}}$$，$${{a}_{2}}$$，\\ldots，$${{a}_{n}}$$是$$n$$个互不相同的负整数，其中$$n$$是奇自然数． 若$${{a}_{1}}=-1$$，$$({{a}_{1}}+1)=0$$，则$$({{a}_{1}}+1)({{a}_{2}}+2)\\cdots ({{a}_{n}}+n)=0$$，排除$$\\text{A}$$． 若$${{a}_{1}}=-1$$，$${{a}_{2}}=-2$$，$${{a}_{3}}=-3$$，\\ldots，$${{a}_{n}}=-n$$时， $$({{a}_{1}}-1)({{a}_{2}}-2)\\cdots ({{a}_{n}}-n)=(-2)(-4)(-6)\\cdots (-2n)$$$$={{(-1)}^{n}}2\\times 4\\times 6\\times \\cdots \\times (2n) ~\\textless{} ~0$$（因为$$n$$是奇数）．故排除$$\\text{B}$$． 若$${{a}_{1}}=-1$$时，$$\\left( \\frac{1}{{{a}_{1}}}+1 \\right)=0$$， 故$$\\left( \\frac{1}{{{a}_{1}}}+1 \\right)\\left( \\frac{1}{{{a}_{2}}}+2 \\right)\\cdots \\left( \\frac{1}{{{a}_{n}}}+n \\right)=0$$， 排除$$\\text{C}$$．所以应选$$\\text{D}$$． 事实上，若$${{a}_{1}} ~\\textless{} ~0$$，$${{a}_{2}} ~\\textless{} ~0$$，\\ldots，$${{a}_{n}} ~\\textless{} ~0$$，则$$-\\frac{1}{{{a}_{1}}}\\textgreater0$$，$$-\\frac{1}{{{a}_{2}}}\\textgreater0$$，\\ldots，$$-\\frac{1}{{{a}_{n}}}\\textgreater0$$． 所以$$1-\\frac{1}{{{a}_{1}}}\\textgreater0$$，$$2-\\frac{1}{{{a}_{2}}}\\textgreater0$$，\\ldots，$$n-\\frac{1}{{{a}_{n}}}\\textgreater0$$． 所以$$\\left( 1-\\frac{1}{{{a}_{1}}} \\right)\\left( 2-\\frac{1}{{{a}_{2}}} \\right)\\cdots \\left( n-\\frac{1}{{{a}_{n}}} \\right)\\textgreater0$$，选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "127", "queId": "6fc13c40b9114d42a80d7f27649f042b", "competition_source_list": ["2017~2018学年福建泉州丰泽区泉州实验中学（圣湖校区）初一下学期期中第8题4分", "1995年第6届全国希望杯初一竞赛复赛第10题", "2015~2016学年北京海淀区清华大学附属中学初一下学期期中第10题3分"], "difficulty": "3", "qtype": "single_choice", "problem": "某项球类规则达标测验，规定满分$$100$$分，$$60$$分及格，模拟考试与正式考试形式相同，都是$$25$$道选择题，每题答对记$$4$$分，答错或不答记$$0$$分，并规定正式考试中要有$$80$$分的试题就是模拟考试中的原题．假设某人在模拟考试中答对的试题，在正式考试中仍能答对，某人欲在正式考试中确保及格，则他在模拟考试���，至少要得．", "answer_option_list": [[{"aoVal": "A", "content": "$$80$$分 "}], [{"aoVal": "B", "content": "$$76$$分 "}], [{"aoVal": "C", "content": "$$75$$分 "}], [{"aoVal": "D", "content": "$$64$$分 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->不等式组应用->不等式组的其他实际问题"], "answer_analysis": ["设在模拟考试中至少要得$$x$$分，则在模拟考试中至少做对$$\\frac{x}{4}$$道题， 做错或不会做的题至多是$$25-\\frac{x}{4}$$道题． 在正式考试中要出现模拟考试中$$80$$分的试题，即$$\\frac{80}{4}=20$$道题． 如果最坏的可能，即其余$$20$$分题（$$5$$道新题）某人全不会做， 而且模拟考试中$$25-\\frac{x}{4}$$道失分的题又全出现在正式考试试题之中， 并且该生在模拟考试后也没能复习纠错，仍按错误答案在正规考试中失分， 这时该生只能从$$\\frac{80}{4}-(25-\\frac{x}{4})$$道题中取得及格分， 即$$\\left[ \\frac{80}{4}-(25-\\frac{x}{4}) \\right]\\times 4\\geqslant 60$$，解得$$x\\geqslant 80$$． 即某人欲在正式考试中确保及格，则他在模拟考试中至少要得$$80$$分．选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "851", "queId": "5c32c3cb23154fcb821ffac061a3e73b", "competition_source_list": ["2011年第22届全国希望杯初二竞赛初赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在平面直角坐标系中，横、纵坐标都是整数的点称为整点，已知$$k$$为整数，若函数$$y=2x-1$$与$$y=kx+k$$的图象的交点是整点，则$$k$$的值有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$个 "}], [{"aoVal": "B", "content": "$$3$$个 "}], [{"aoVal": "C", "content": "$$4$$个 "}], [{"aoVal": "D", "content": "$$5$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->函数->一次函数->一次函数基础"], "answer_analysis": ["解方程组$$\\left { \\begin{matrix}y=2x-1 y=kx+k \\end{matrix} \\right.$$， 得$$x=\\frac{-1-k}{k-2}=-1-\\frac{3}{k-2}$$． 因为交点的横坐标$$x$$的值为整数（此时交点的纵坐标$$y$$也一定为整数）， 则$$k-2$$应该是$$3$$的约数， 即$$k-2$$的值可以是$$1$$，$$-1$$，$$3$$，$$-3$$， 所以$$k$$的值可以是$$3$$，$$1$$，$$5$$，$$-1$$，共有$$4$$个． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1587", "queId": "f088b9b741f7482ba86047ecb51fe4cb", "competition_source_list": ["2001年第12届希望杯初一竞赛第10题", "初一上学期单元测试《一元一次方程》第20题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$k$$为整数，则使得方程$$(k-1999)x=2001-2000x$$的解也是整数的$$k$$值有．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$个 "}], [{"aoVal": "B", "content": "$$8$$个 "}], [{"aoVal": "C", "content": "$$12$$个 "}], [{"aoVal": "D", "content": "$$16$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->含参一元一次方程->一元一次方程的含参整数解"], "answer_analysis": ["$$x=\\frac{2001}{k+1}$$为整数，又因为$$2001=1\\times 3\\times 23\\times 29$$，$$k+1$$可取$$\\pm 1$$、$$\\pm 3$$、$$\\pm 23$$、$$\\pm 29$$、$$\\pm (3\\times 23)$$、$$\\pm (3\\times 29)$$、$$\\pm (23\\times 29)$$、$$\\pm 2001$$共$$16$$个值，对应的$$k$$值也有$$16$$个． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "256", "queId": "794952a2416841128d80d9a479aec9b3", "competition_source_list": ["2012年第23届全国希望杯初一竞赛初赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "如图，$$\\triangle ABC$$的面积是$$60$$，$$AD:DC=1:3$$，$$BE:ED=4:1$$，$$EF:FC=4:5$$，则$$\\triangle BEF$$的面积是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->三角形及多边形->与三角形有关的线段->判断三角形的高，中线和角平分线", "课内体系->知识点->几何图形初步->直线、射线、线段->直线、射线、线段的运算->线段中点与等分点"], "answer_analysis": ["因为$$AD:DC=1:3$$，且$${{S}_{\\triangle ABC}}=60$$， 所以$${{S}_{\\triangle BCD}}=\\frac{3}{4}{{S}_{\\triangle ABC}}=45$$． 因为$$BE:ED=4:1$$， 所以$${{S}_{\\triangle BCE}}=\\frac{4}{5}{{S}_{\\triangle BCD}}=\\frac{4}{5}\\times 45=36$$． 因为$$EF:FC=4:5$$， 所以$${{S}_{\\triangle BEF}}=\\frac{4}{9}{{S}_{\\triangle BCE}}=\\frac{4}{9}\\times 36=16$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "566", "queId": "7592cea638e84bb1bb8481c06ad899b0", "competition_source_list": ["2010年第21届全国希望杯初一竞赛初赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$8$$个人用$$35$$天完成了某项工程的$$\\frac{1}{3}$$．此时，又增加$$6$$个人，那么要完成剩余的工程，还需要的天数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$35$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数与实际问题->有理数乘除法与实际问题"], "answer_analysis": ["总工程量的$$\\frac{1}{3}$$为$$8\\times 35=280$$（人$$\\cdot $$天）， 还有$$\\frac{2}{3}$$，即$$2\\times 280=560$$（人$$\\cdot $$天）， 要完成剩余的工程，还需要$$\\frac{560}{8+6}=40$$（天）． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "890", "queId": "a3e7019f6ac94de2bbe662a43b633835", "competition_source_list": ["2009年第20届希望杯初一竞赛第1试第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在数轴上，坐标是整数的点称为``整点''．设数轴的单位长度是$$1$$厘米，在这个数轴上随意画出一条长为$$2008$$厘米的线段$$AB$$，则线段$$AB$$盖住的整点至少有．", "answer_option_list": [[{"aoVal": "A", "content": "$$2006$$个 "}], [{"aoVal": "B", "content": "$$2007$$个 "}], [{"aoVal": "C", "content": "$$2008$$个 "}], [{"aoVal": "D", "content": "$$2009$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->数轴->数轴上整点覆盖问题", "课内体系->能力->运算能力"], "answer_analysis": ["当画的线段的两个端点都在整点上时，线段$$AB$$盖住$$2009$$个整点；当画的线段的两个端点都不在整点上时，线段$$AB$$盖住$$2008$$个整点；所以长为$$2008$$厘米的线段$$AB$$，至少可盖住数轴上$$2008$$个整点． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1373", "queId": "a127093d424e43548f4c680444b49b9d", "competition_source_list": ["2015年第32届全国全国初中数学联赛竞赛A卷第6题7分"], "difficulty": "3", "qtype": "single_choice", "problem": "设$$n$$是小于$$100$$的正整数且使$$5{{n}^{2}}+3n-5$$是$$15$$的倍数，则符合条件的所有正整数$$n$$的和是（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$285$$ "}], [{"aoVal": "B", "content": "$$350$$ "}], [{"aoVal": "C", "content": "$$540$$ "}], [{"aoVal": "D", "content": "$$635$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算", "课内体系->能力->运算能力", "竞赛->知识点->数论->同余->剩余系及其应用", "竞赛->知识点->数论->整除->整除的概念与基本性质"], "answer_analysis": ["∵$$5{{n}^{2}}+3n-5$$是$$15$$的倍数， ∴$$5\\textbar(5{{n}^{2}}+3n-5)$$， ∴$$5\\textbar3n$$， ∴$$5\\textbar n$$． 设$$n=5m$$（$$m$$是正整数）， 则$$5{{n}^{2}}+3n-5=125{{m}^{2}}+15m-5=120{{m}^{2}}+15m+5({{m}^{2}}-1)$$． 又∵$$5{{n}^{2}}+3n-5$$是$$15$$的倍数， ∴$${{m}^{2}}-1$$是$$3$$的倍数， ∴$$m=3k+1$$或$$m=3k+2$$，其中$$k$$是非负整数， ∴$$n=5(3k+1)=15k+5$$或$$n=5(3k+2)=15k+10$$，其中$$k$$是非负整数， ∴符合条件的所有正整数$$n$$的和为 $$(5+20+35+50+65+80+95)+(10+25+40+55+70+85)=635$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "869", "queId": "e40962f4fcf142f19faea625d3b48852", "competition_source_list": ["2016年第27届全国希望杯初一竞赛初赛第2题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "若$$n$$个人完成一项工程需要$$m$$天，则$$(m+n)$$个人完成这项工程需要（~ ）天．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{mn}{m+n}$$ "}], [{"aoVal": "B", "content": "$$\\frac{m-n}{m+n}$$ "}], [{"aoVal": "C", "content": "$$\\frac{m+n}{mn}$$ "}], [{"aoVal": "D", "content": "$$\\frac{mn}{m+2n}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式有关的概念->代数式->列代数式", "课内体系->能力->分析和解决问题能力"], "answer_analysis": ["$$n$$个人完成一项工程需要$$m$$天， 则$$1$$个人完成这项工程需要$$mn$$天， ∴$$(m+n)$$个人完成这项工程需要$$\\frac{mn}{m+n}$$天． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1258", "queId": "8aac50a74e023208014e3f4bf65b18f2", "competition_source_list": ["1995年第6届全国希望杯初一竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a\\textless{}0$$，则下列结论中不成立的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{a}^{2}}={{(-a)}^{2}}$$ "}], [{"aoVal": "B", "content": "$${{a}^{3}}={{(-a)}^{3}}$$ "}], [{"aoVal": "C", "content": "$${{a}^{2}}=\\left\\textbar{} {{a}^{2}} \\right\\textbar$$ "}], [{"aoVal": "D", "content": "$${{a}^{3}}=-\\left\\textbar{} {{a}^{3}} \\right\\textbar$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->数->有理数->绝对值->绝对值的非负性", "课内体系->知识点->数->有理数->绝对值->绝对值的定义", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘法->有理数乘法运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->有理数乘方运算"], "answer_analysis": ["以特殊值$$a=-2$$代入检验，易知$${{(-2)}^{2}}={{(-(-2))}^{2}}$$，$${{(-2)}^{3}}\\ne {{(-(-2))}^{3}}$$，$${{(-2)}^{2}}=\\left\\textbar{} {{(-2)}^{2}} \\right\\textbar$$，$${{(-2)}^{3}}=-\\left\\textbar{} {{(-2)}^{3}} \\right\\textbar$$，所以选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1556", "queId": "e21607ebeae94670a29be9be2185d7ed", "competition_source_list": ["1993年第10届全国初中数学联赛竞赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "多项式$${{x}^{12}}-{{x}^{6}}+1$$除以$${{x}^{2}}-1$$的余式是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$x-1$$ "}], [{"aoVal": "D", "content": "$$x+1$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->整式的乘除运算->综合除法和余数定理 ", "课内体系->知识点->式->整式的乘除->整式的乘除运算->大除法"], "answer_analysis": ["直接利用整式除法比较繁琐，将原式变形得 $${{x}^{12}}-{{x}^{6}}+1={{x}^{6}}({{x}^{6}}-1)+1$$， $$={{x}^{6}}({{x}^{2}}-1)({{x}^{4}}+{{x}^{2}}+1)+1$$． 所以余式为$$1$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "541", "queId": "2d7397681ca74e4eb65e0dc0b71279bc", "competition_source_list": ["2014年第25届全国希望杯初二竞赛初赛第4题4分", "2016~2017学年安徽安庆潜山市潜山第四中学初三上学期期末模拟第2题4分", "2018~2019学年广东深圳福田区深圳实验学校中学部初三上学期开学考试第4题3分", "2015~2016学年江苏苏州相城区初二下学期期中第7题3分", "2018年陕西西安碑林区西安市第六中学初三中考一模第6题3分", "2016~2017学年12月北京海淀区北方交通大学附属中学分校初三上学期月考第6题3分", "2017~2018学年10月山东济南历下区济南燕山学校初三上学期月考第10题3分", "2016~2017学年安徽合肥肥西县肥西县烧脉中学初三上学期期中第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "反比例函数$$y=\\frac{6}{x}$$的图象上有三个点$$({{x}_{1}},{{y}_{1}})$$，$$({{x}_{2}},{{y}_{2}})$$，$$\\left( {{x}_{3}},{{y}_{3}} \\right)$$，其中$${{x}_{1}}\\textless{}{{x}_{2}}\\textless{}0\\textless{}{{x}_{3}}$$，则$${{y}_{1}}$$、$${{y}_{2}}$$、$${{y}_{3}}$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{y}_{1}}\\textless{}{{y}_{2}}\\textless{}{{y}_{3}}$$ "}], [{"aoVal": "B", "content": "$${{y}_{2}}\\textless{}{{y}_{1}}\\textless{}{{y}_{3}}$$ "}], [{"aoVal": "C", "content": "$${{y}_{3}}\\textless{}{{y}_{1}}\\textless{}{{y}_{2}}$$ "}], [{"aoVal": "D", "content": "$${{y}_{3}}\\textless{}{{y}_{2}}\\textless{}{{y}_{1}}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->函数->反比例函数->反比例函数图象与性质->反比例函数增减性"], "answer_analysis": ["∵反比例函数$$y=\\frac{6}{x}$$中，$$k=6\\textgreater0$$， ∴此反比例函数图象的两个分支在一、三象限， ∵$${{x}_{3}}\\textgreater0$$， ∴点$$\\left( {{x}_{3}},{{y}_{3}} \\right)$$在第一象限，$${{y}_{3}}\\textgreater0$$， ∵$${{x}_{1}} ~\\textless{} ~{{x}_{2}} ~\\textless{} ~0$$， ∴点$$\\left( {{x}_{1}},{{y}_{1}} \\right)$$，$$\\left( {{x}_{2}},{{y}_{2}} \\right)$$在第三象限，$$y$$随$$x$$的增大而减小，故$${}{{y}_{2}} ~\\textless{} {} {{y}_{1}}\\textless0$$，于是$${{y}_{2}} ~\\textless{} ~{{y}_{1}} ~\\textless{} ~{{y}_{3}}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "609", "queId": "5eed8813ab244a7d8208aa35e598193a", "competition_source_list": ["2018年重庆中考真题A卷第12题4分", "2018年四川绵阳涪城区绵阳东辰国际学校初三自主招生第3题4分", "2019~2020学年重庆合川区西南大学银翔实验中学初一下学期期末模拟（一）第12题4分", "2018~2019学年广东深圳南山区桃苑学校初二下学期期末第11题3分", "2019年浙江杭州拱墅区杭州市文澜中学初二竞赛第7题3分", "2019~2020学年江苏苏州姑苏区苏州市振华中学校初三下学期单元测试《方程与不等式》第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "若数$$a$$使关于$$x$$的不等式组$$\\begin{cases}\\dfrac{x-1}{2}\\textless{}\\dfrac{1+x}{3} 5x-2\\geqslant x+a \\end{cases}$$有且只有四个整数解，且使关于$$y$$的方程$$\\frac{y+a}{y-1}+\\frac{2a}{1-y}=2$$的解为非负数，则符合条件的所有整数$$a$$的和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-3$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->分式方程->分式方程的解与解分式方程->分式方程与不等式综合", "课内体系->能力->运算能力"], "answer_analysis": ["$$\\begin{cases}\\dfrac{x-1}{2}\\textless{}\\dfrac{1+x}{3} 5x-2\\geqslant x+a \\end{cases}$$， 不等式组整理得：$$\\begin{cases}x\\textless{}5 x\\geqslant \\dfrac{a+2}{4} \\end{cases}$$， 由不等式组有且只有四个整数解，得到$$0\\textless{}\\frac{a+2}{4}\\leqslant 1$$， 解得：$$-2\\textless{}a\\leqslant 2$$，即整数$$a=-1$$，$$0$$，$$1$$，$$2$$， $$\\frac{y+a}{y-1}+\\frac{2a}{1-y}=2$$， 分式方程去分母得：$$y+a-2a=2(y-1)$$， 解得：$$y=2-a$$， 由分式方程的解为非负数以及分式有意义的条件，得到$$a$$为$$-1$$，$$0$$，$$2$$，之和为$$1$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1368", "queId": "aa24b36269f44abc847573cf60190c0c", "competition_source_list": ["2017~2018学年浙江宁波鄞州区宁波市鄞州蓝青学校初二下学期期末第7题3分", "1991年第8届全国初中数学联赛竞赛（第一试）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "设等式$$\\sqrt{a\\left( x-a \\right)}+\\sqrt{a\\left( y-a \\right)}=\\sqrt{x-a}-\\sqrt{a-y}$$在实数范围内成立，其中$$a$$，$$x$$，$$y$$是两两不同的实数，则$$\\frac{3{{x}^{2}}+xy-{{y}^{2}}}{{{x}^{2}}-xy+{{y}^{2}}}$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "~$$\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式化简求值->分式化简求值-条件化简求值", "课内体系->知识点->数->实数->平方根->算术平方根的双重非负性", "课内体系->能力->运算能力"], "answer_analysis": ["据算术根性质，由右端知$$y\\leqslant a\\leqslant x$$， 又由左端知$$a\\geqslant 0$$且$$a\\leqslant 0$$，故$$a=0$$． 由此得$$x=-y$$，代入所求式算值为$$\\frac{1}{3}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1200", "queId": "8aac49074e724b45014e87bf170c50d5", "competition_source_list": ["1996年第7届全国希望杯初一竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$a=-0.01$$时，在$$-{{\\left( -a \\right)}^{2}}$$，$$-\\left\\textbar{} -a \\right\\textbar$$，$$-{{a}^{2}}$$，$$-\\left( -{{a}^{2}} \\right)$$中，其值为正数的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-{{\\left( -a \\right)}^{2}}$$ "}], [{"aoVal": "B", "content": "$$-\\left\\textbar{} -a \\right\\textbar$$ "}], [{"aoVal": "C", "content": "$$-{{a}^{2}}$$ "}], [{"aoVal": "D", "content": "$$-\\left( -{{a}^{2}} \\right)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->正数和负数->正数、负数定义"], "answer_analysis": ["当$$a\\textless{}0$$时，$${{\\left( -a \\right)}^{2}}\\textgreater0$$，$$\\left\\textbar{} -a \\right\\textbar\\textgreater0$$，$${{a}^{2}}\\textgreater0$$， 所以$$-{{\\left( -a \\right)}^{2}}\\textless{}0$$，$$-\\left\\textbar{} -a \\right\\textbar\\textless{}0$$，$$-{{a}^{2}}\\textless{}0$$，因此排除$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$，选$$\\text{D}$$． 事实上，$$a\\textless{}0$$时，$${{a}^{2}}\\textgreater0$$，$$-\\left( -{{a}^{2}} \\right)\\textgreater0$$．当然$$a=-0.01$$时更是如此． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1462", "queId": "c5ee1beca0b2467cb8d890cff13ac4d7", "competition_source_list": ["2017~2018学年浙江金华金东区初二上学期期末第10题3分", "2017~2018学年9月安徽安庆潜山市潜山第四中学初二上学期月考第6题4分", "2017~2018学年浙江宁波镇海区宁波市镇海蛟川书院初二上学期期中第9题4分", "2017~2018学年12月陕西西安碑林区西北工业大学附属��学初二上学期月考第7题3分", "2018~2019学年辽宁大连甘井子区初一下学期期末第10题3分", "2018年湖南长沙初二竞赛长郡教育集团（觉园杯）第1题4分", "2018~2019学年浙江宁波镇海区宁波市镇海蛟川书院初二上学期期中第4题4分", "2017年广西贵港中考真题第6题3分", "2020~2021学年4月浙江杭州上城区杭州市建兰中学初三下学期周测A卷第1题4分", "2018~2019学年湖北武汉武昌区初一下学期期中（十五校联考）第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在平面直角坐标系中，点$$P(m-3,4-2m)$$不可能在．", "answer_option_list": [[{"aoVal": "A", "content": "第一象限 "}], [{"aoVal": "B", "content": "第二象限 "}], [{"aoVal": "C", "content": "第三象限 "}], [{"aoVal": "D", "content": "第四象限 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->函数->平面直角坐标系->坐标系基础->象限内坐标的特征", "课内体系->思想->分类讨论思想"], "answer_analysis": ["①$$m-3\\textgreater0$$，即$$m\\textgreater3$$时， $$-2m\\textless{}-6$$，$$4-2m\\textless{}-2$$， 所以，点$$P(m-3,4-2m)$$在第四象限，不可能在第一象限． ②$$m-3\\textless{}0$$，即$$m\\textless{}3$$时， $$-2m\\textgreater-6$$，$$4-2m\\textgreater-2$$， 点$$P(m-3,4-2m)$$可能在第二或三象限， 综上所述，点$$P$$不可能在第一象限． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "454", "queId": "5e3b1209c18a418d9e833e9f8c64cc41", "competition_source_list": ["1991年第8届全国初中数学联赛竞赛（第一试）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知：$$x=\\frac{1}{2}\\left( {{1991}^{\\frac{1}{n}}}-{{1991}^{-\\frac{1}{n}}} \\right)$$（$$n$$是自然数）．那么$${{\\left( x-\\sqrt{1+{{x}^{2}}} \\right)}^{n}}$$的值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{1991}^{-1}}$$ "}], [{"aoVal": "B", "content": "$$-{{1991}^{-1}}$$ "}], [{"aoVal": "C", "content": "$${{\\left( -1 \\right)}^{n}}1991$$ "}], [{"aoVal": "D", "content": "$${{\\left( -1 \\right)}^{n}}{{1991}^{-1}}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->二次根式->二次根式化简求值", "课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式"], "answer_analysis": ["$$1+{{x}^{2}}=1+\\frac{1}{4}\\left( {{1991}^{\\frac{2}{n}}}-2+{{1991}^{-\\frac{2}{n}}} \\right)=\\frac{1}{4}{{\\left( {{1991}^{\\frac{1}{n}}}+{{1991}^{-\\frac{1}{n}}} \\right)}^{2}}$$， 原式$$={{\\left[ \\frac{1}{2}\\left( {{1991}^{\\frac{1}{n}}}-{{1991}^{-\\frac{1}{n}}} \\right)-\\frac{1}{2}\\left( {{1991}^{\\frac{1}{n}}}+{{1991}^{-\\frac{1}{n}}} \\right) \\right]}^{n}}={{\\left( -1 \\right)}^{n}}{{1991}^{-1}}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "10", "queId": "173009f9de154f9482ecc4e79afc9134", "competition_source_list": ["1999年第10届希望杯初二竞赛第2试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列五个多项式： ①$${{a}^{2}}{{b}^{2}}-{{a}^{2}}-{{b}^{2}}-1$$； ②$${{x}^{3}}-9a{{x}^{2}}+27{{a}^{2}}x-27{{a}^{3}}$$； ③$$x(b+c-d)-y(d-b-c)-2c+2d-2b$$； ④$$3m(m-n)+6n(n-m)$$； ⑤$${{(x-2)}^{2}}+4x$$． 其中在有理数范围内可以进行因式分解的有．", "answer_option_list": [[{"aoVal": "A", "content": "①，②，③ "}], [{"aoVal": "B", "content": "②，③，④ "}], [{"aoVal": "C", "content": "③，④，⑤ "}], [{"aoVal": "D", "content": "①，②，④ "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->公式法->提公因式+平方差", "课内体系->知识点->式->因式分解->提公因式法"], "answer_analysis": ["②式$$={{(x-3a)}^{3}}$$， ③式$$=x(b+c-d)+y(b+c-d)-2(b+c-d)$$ $$=(b+c-d)(x+y-2)$$， ④式$$=(m-n)(3m-6n)=3(m-n)(m-2n)$$． 所以，②、③、④式合乎要求． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1502", "queId": "ef0ec64b8f9b4c11a9cbd02ec364cc1e", "competition_source_list": ["2008年第19届希望杯初二竞赛第2试第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "有面值为$$10$$元、$$20$$元，$$50$$元的人民币（每种至少一张）共$$24$$张，合计$$1000$$元，那么其中面值为$$20$$元的人民币的张数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$或$$4$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$4$$或$$8$$ "}], [{"aoVal": "D", "content": "$$2$$到$$46$$之间的任意偶数 "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->一次方程->方程组"], "answer_analysis": ["设$$10$$元、$$20$$元、$$50$$元分别有$$x$$，$$y$$，$$24-\\left( x+y \\right)$$张， 则$$10x+20y+50\\left( 24-x-y \\right)=1000$$， 即$$40x+30y=200$$，$$4x+3y=20$$， 其中$$x$$，$$y$$都是正整数，由$$x\\geqslant 1$$知$$3y\\leqslant 16$$， 所以$$1\\leqslant y\\leqslant \\frac{16}{3}$$， 所以，$$y$$只能从$$1$$，$$2$$，$$3$$，$$4$$，$$5$$中取， 又$$3y=4\\left( 5-x \\right)$$，其中$$5-x$$是正整数，$$3$$与$$4$$是互质的， 所以，$$y$$中一定有一个因数$$4$$， 因此$$y$$只能取$$4$$， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "328", "queId": "22c9a1fbec934a88b2693cc4e34acee8", "competition_source_list": ["初一上学期单元测试《走进美妙的数学世界》第7题", "2013年第24届全国希望杯初一竞赛初赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a=\\frac{999}{2011}$$，$$b=\\frac{1000}{2012}$$，$$c=\\frac{1001}{2013}$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless{}b\\textless{}c$$ "}], [{"aoVal": "B", "content": "$$\\textasciitilde b\\textless{}c\\textless{}a$$ "}], [{"aoVal": "C", "content": "$$c\\textless{}b\\textless{}a$$ "}], [{"aoVal": "D", "content": "$$a\\textless{}c\\textless{}b$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性"], "answer_analysis": ["解法一： $$a=\\frac{999}{2011}=1-\\frac{1002}{2011}$$， $$b=\\frac{1000}{2012}=1-\\frac{1002}{2012}$$， $$c=\\frac{1001}{2013}=1-\\frac{1002}{2013}$$． ∵$$2011\\textless{}2013\\textless{}2013$$， ∴$$\\frac{1002}{2011}\\textgreater\\frac{1002}{2012}\\textgreater\\frac{1002}{2013}$$， ∴$$1-\\frac{1002}{2011}\\textless{}1-\\frac{1002}{2012}\\textless{}1-\\frac{1002}{2013}$$， ∴$$a\\textless{}b\\textless{}c$$． 解法二： 观察已知三个分数的特征，可设他们分别为$$\\frac{y}{x}$$，$$\\frac{y+1}{x+1}$$，$$\\frac{y+2}{x+2}$$（$$x\\textgreater y\\textgreater0$$）， 则$$\\frac{y}{x}-\\frac{y+1}{x+1}=\\frac{y(x+1)-x(y+1)}{x(x+1)}=\\frac{y-x}{x(x+1)}$$， ∵$$x\\textgreater y\\textgreater0$$， ∴$$y-x\\textless{}0$$， ∴$$\\frac{y-x}{x(x+1)}\\textless{}0$$， ∴$$\\frac{y}{x}\\textless{}\\frac{y+1}{x+1}$$． 同理$$\\frac{y+1}{x+1}\\textless{}\\frac{y+2}{x+2}$$， ∴$$\\frac{y}{x}\\textless{}\\frac{y+1}{x+1}\\textless{}\\frac{y+2}{x+2}$$， ∴$$a\\textless{}b\\textless{}c$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "415", "queId": "1bb2ab3ccc1542c2b8148bebe636bcf3", "competition_source_list": ["1996年第7届希望杯初二竞赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$x$$为整数，且$$\\frac{2}{x+3}+\\frac{2}{3-x}+\\frac{2x+18}{{{x}^{2}}-9}$$为整数，则符合条件的$$x$$的所有值的和为( ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的运算->分式加减混合运算", "课内体系->知识点->式->分式->分式的运算->分式通分"], "answer_analysis": ["原式$$=\\frac{2x-6-2x-6+2x+18}{(x+3)(x-3)}=\\frac{2(x+3)}{(x+3)(x-3)}=\\frac{2}{x-3}$$ $$\\therefore x-3=1$$，$$-1$$，$$-2$$，$$2$$时，原式的值为整数 此时$${{x}_{1}}=4$$，$${{x}_{2}}=2$$，$${{x}_{3}}=1$$，$${{x}_{4}}=5$$ $$\\therefore {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=12$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1164", "queId": "8aac49074e023206014e1e2677235b47", "competition_source_list": ["1993年第4届全国希望杯初一竞赛复赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$n$$为正整数，$$302$$被$$n\\left( n+1 \\right)$$除所得商数$$q$$及余数$$r$$都是正值，则$$r$$的最大值与最小值的和是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$148$$ "}], [{"aoVal": "B", "content": "$$247$$ "}], [{"aoVal": "C", "content": "$$93$$ "}], [{"aoVal": "D", "content": "$$122$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->含乘方的有理数混合运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算"], "answer_analysis": ["$$n\\left( n+1 \\right)$$为偶数． 设$$302$$被$$n\\left( n+1 \\right)$$除商$$q$$余$$r$$，则$$302=n\\left( n+1 \\right)q+r$$知，$$r$$为偶数．显然$$\\text{B}$$、$$\\text{C}$$均应排除． 由除数$$n\\left( n+1 \\right)$$只能取$$6$$，$$12$$，$$20$$，$$30$$，$$42$$，$$56$$，$$72$$，$$90$$，$$110$$，$$132$$，$$156$$，$$182$$，$$210$$，$$240$$，$$272$$这些值，计算得相应的余数中最小的正值为$$2$$，最大正值为$$146$$， 所以$$r$$的正的最小值与最大值的和是$$148$$．选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "49", "queId": "0383131d593a48fb9d0040dd45d7b84b", "competition_source_list": ["2010年第21届全国希望杯初一竞赛初赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a\\textless{}0$$，在$$\\left\\textbar{} a \\right\\textbar$$，$$-a$$，$${{a}^{2009}}$$，$${{a}^{2010}}$$，$$\\left\\textbar{} -a \\right\\textbar$$，$$\\left( \\frac{{{a}^{2}}}{a}+a \\right)$$，$$\\left( \\frac{{{a}^{2}}}{a}-a \\right)$$中负数的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["知识标签->知识点->数->有理数->正数和负数->正数和负数的定义", "知识标签->题型->数->有理数->数轴与有理数有关的概念->题型：区分正负数"], "answer_analysis": ["因为$$a\\textless{}0$$， 所以$$\\left\\textbar{} a \\right\\textbar$$，$$-a$$，$${{a}^{2010}}$$，$$\\left\\textbar{} -a \\right\\textbar$$均为正数， 而$$\\left( \\frac{{{a}^{2}}}{a}-a \\right)=a-a=0$$，$$\\left( \\frac{{{a}^{2}}}{a}+a \\right)=a+a=2a$$， 所以只有$${{a}^{2009}}$$，$$\\left( \\frac{{{a}^{2}}}{a}+a \\right)$$为负数． 故负数的个数是$$2$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "104", "queId": "086083cc85864bc3847ef45743e8a1b6", "competition_source_list": ["2014年第25届全国希望杯初一竞赛初赛第2题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "在下列图形中，恰有三条对称轴的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "平行四边形 "}], [{"aoVal": "B", "content": "圆 "}], [{"aoVal": "C", "content": "等边三角形 "}], [{"aoVal": "D", "content": "正方形 "}]], "knowledge_point_routes": ["课内体系->知识点->几何变换->轴对称->轴对称基础->对称轴条数问题"], "answer_analysis": ["平行四边形不一定是轴对称图形； 圆有无数条对称轴； 等边三角形恰好有三条对称轴； 正方形有四条对称轴． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1121", "queId": "ff8080814d9efd56014daa7edae90ae0", "competition_source_list": ["1993年第4届全国希望杯初一竞赛初赛第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "有理数$$\\frac{1}{2}$$，$$\\frac{11}{5}$$，$$8$$恰是下列三个方程的根：$$\\frac{2x-1}{3}-\\frac{10x+1}{12}=\\frac{2x+1}{4}-1$$，$$3(2y+1)=2(1+y)+3(y+3)$$，$$\\frac{1}{2}\\left[ z-\\frac{1}{2}(z-1) \\right]=\\frac{2}{3}(z-1)$$，则$$\\frac{x}{y}-\\frac{z}{x}$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{171}{40}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{347}{80}$$ "}], [{"aoVal": "C", "content": "$$\\frac{71}{220}$$ "}], [{"aoVal": "D", "content": "$$\\frac{142}{55}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->解一元一次方程->常规一元一次方程解法", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程的相关概念->一元一次方程的解", "课内体系->能力->运算能力"], "answer_analysis": ["以$$\\frac{1}{2}$$，$$\\frac{11}{5}$$，$$8$$分别代入所给三个方程知： $$x=\\frac{1}{2}$$是方程$$\\frac{2x-1}{3}-\\frac{10x+1}{12}=\\frac{2x+1}{4}-1$$的根， $$y=8$$是方程$$3(2y+1)=2(1+y)+3(y+3)$$的根， $$z=\\frac{11}{5}$$是方程$$\\frac{1}{2}\\left[ z-\\frac{1}{2}(z-1) \\right]=\\frac{2}{3}(z-1)$$的根， ∴$$\\frac{x}{y}-\\frac{z}{x}=\\frac{\\frac{1}{2}}{8}-\\frac{\\frac{11}{5}}{\\frac{1}{2}}=\\frac{1}{16}-\\frac{22}{5}=-\\frac{347}{80}$$，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "709", "queId": "5215b6b524964143b94faa8ce715ad32", "competition_source_list": ["2014年第25届全国希望杯初二竞赛复赛第2题4分", "2015~2016学年湖南长沙雨花区湖南广益实验中学初一上学期期中"], "difficulty": "0", "qtype": "single_choice", "problem": "若当$$x=1$$时，代数式$$a{{x}^{3}}+bx+1$$的值是$$5$$，则当$$x=-1$$时，$$a{{x}^{3}}+bx+1$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$-3$$ "}], [{"aoVal": "C", "content": "$$-4$$ "}], [{"aoVal": "D", "content": "$$-5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式的加减运算->整式加减", "课内体系->能力->运算能力"], "answer_analysis": ["当$$x=1$$时，代数式$$a{{x}^{3}}+bx+1$$的值是$$5$$， 所以$$a+b+1=5$$，即$$a+b=4$$． 当当$$x=-1$$时，$$a{{x}^{3}}+bx+1=-(a+b)+1=-4+1=-3$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1056", "queId": "ff8080814d7978b9014d86e31f8425d6", "competition_source_list": ["1991年第2届全国希望杯初一竞赛初赛第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$x=\\frac{1}{2}$$，$$y=-2$$时，代数式$$\\frac{4x-2y}{xy}$$的值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-6$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的基础->分式为特殊值", "课内体系->知识点->式->分式->分式化简求值->分式化简求值-直接代入数值求值"], "answer_analysis": ["$$\\frac{4x-2y}{xy}=\\frac{4\\times \\frac{1}{2}-2\\times (-2)}{\\frac{1}{2}\\times (-2)}=\\frac{2+4}{-1}=-6$$．选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "398", "queId": "2c2cffc0a2684b2d8c969539da8617b2", "competition_source_list": ["2013年竞赛第5题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "对于任意实数$$x$$，$$y$$，$$z$$，定义运算``$$*$$''为：$$x*y=\\frac{3{{x}^{3}}y+3{{x}^{2}}{{y}^{2}}+x{{y}^{3}}+45}{{{\\left( x+1 \\right)}^{3}}+{{\\left( y+1 \\right)}^{3}}-60}$$，且$$x*y*z=\\left( x*y \\right)*z$$，则$$2013*2012*\\cdots *3*2$$的值为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{607}{967}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1821}{967}$$ "}], [{"aoVal": "C", "content": "$$\\frac{5463}{967}$$ "}], [{"aoVal": "D", "content": "$$\\frac{16389}{967}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->综合与实践->规律探究与程序框图->定义新运算", "课内体系->能力->运算能力"], "answer_analysis": ["设$$2013*2012*\\cdots *4=m$$， 则$$\\left( 2013*2012*\\cdots *4 \\right)*3=m*3$$， $$=\\frac{3{{m}^{3}}\\times 3+3{{m}^{2}}\\times 9+m\\times 27+45}{{{m}^{3}}+3{{m}^{2}}+3m+1+64-60}=9$$， 于是$$\\left( 2013*2012*\\cdots *3 \\right)*2=9*2$$， $$=\\frac{3\\times {{9}^{3}}\\times 2+3\\times {{9}^{2}}\\times {{2}^{2}}+9\\times {{2}^{3}}+45}{{{10}^{3}}+{{3}^{3}}-60}$$， $$=\\frac{5463}{967}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "225", "queId": "19227d5a97d74928b00319c994636ada", "competition_source_list": ["2019年浙江杭州拱墅区杭州市文澜中学初二竞赛第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知一次函数$$y=kx+2\\left( k\\textgreater0 \\right)$$有两个不同的点$$({{x}_{1}},{{y}_{1}})$$，$$({{x}_{2}},{{y}_{2}})$$，令$$m=\\left( {{x}_{2}}-{{x}_{1}} \\right)\\left( {{y}_{1}}-{{y}_{2}} \\right)$$，则关于$$x$$的不等式$$mx ~\\textless{} ~-m$$的解是．", "answer_option_list": [[{"aoVal": "A", "content": "$$x ~\\textless{} ~-1$$ "}], [{"aoVal": "B", "content": "$$x\\textgreater-1$$ "}], [{"aoVal": "C", "content": "$$x ~\\textless{} ~1$$ "}], [{"aoVal": "D", "content": "$$x\\textgreater1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->一次函数->一次函数与方程、不等式->一次函数与一元一次不等式"], "answer_analysis": ["∵$$y=kx+2(k\\textgreater0)$$中， $$k\\textgreater0$$． ∴$$y$$随$$x$$增大而增大， ∵$$({{x}_{1}},{{y}_{1}})$$$$({{x}_{2}},{{y}_{2}})$$是$$y=kx+2$$图象上不同两点， ∴$${{x}_{1}}\\textgreater{{x}_{2}}$$时$${{y}_{1}}\\textgreater{{y}_{2}}$$， $${{x}_{1}} ~\\textless{} ~{{x}_{2}}$$时，$${{y}_{1}} ~\\textless{} ~{{y}_{2}}$$． ∴$$({{x}_{2}}-{{x}_{1}})({{y}_{1}}-{{y}_{2}}) ~\\textless{} ~0$$， ∴$$m=({{x}_{2}}-{{x}_{1}})({{y}_{1}}-{{y}_{2}}) ~\\textless{} ~0$$． ∴由$$mx ~\\textless{} ~-m$$， 解得$$x\\textgreater-1$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "28", "queId": "055a5cdc0062482eb035244d84d99b3e", "competition_source_list": ["2019~2020学年3月浙江杭州滨江区杭州二中白马湖学校初一下学期周测C卷第9题", "第7届希望杯初二竞赛第3题", "2017~2018学年5月四川成都天府新区 成都市实验外国语学校（西区）初二下学期月考第10题3分", "上海自主招生近5年（2015-2019）真题题型分类汇编第15题", "初二其它"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$${{x}^{2}}+ax-12$$能分解成两个整数系数的一次因式的乘积，则符合条件的整数$$a$$的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$个 "}], [{"aoVal": "B", "content": "$$4$$个 "}], [{"aoVal": "C", "content": "$$6$$个 "}], [{"aoVal": "D", "content": "$$8$$个 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->因式分解->因式分解的基础->已知因式分解结果求参数"], "answer_analysis": ["设$$x^{2}+ax-12$$能分解成两个整数系数的一次因式的乘积， 即$$x^{2}+ax-12=(x+m)(x+n)$$，$$m$$，$$n$$是整数， $$\\therefore x^{2}+ax-12=x^{2}+(m+n)x+mn$$， $$\\therefore \\begin{cases}mn=-12 \\m+n=a\\end{cases}$$， $$\\because m$$，$$n$$是整数，且$$mn=-12$$， 有$$\\begin{cases}m=12 \\n=-1\\end{cases}$$，$$\\begin{cases}m=-1 \\n=12\\end{cases}$$，$$\\begin{cases}m=6 \\n=-2\\end{cases}$$，$$\\begin{cases}m=-2 \\n=6\\end{cases}$$， $$\\begin{cases}m=4 \\n=-3\\end{cases}$$，$$\\begin{cases}m=-3 \\n=4\\end{cases}$$，$$\\begin{cases}m=3 \\n=-4\\end{cases}$$，$$\\begin{cases}m=-3 \\n=4\\end{cases}$$，$$\\begin{cases}m=2 \\n=-6\\end{cases}$$， $$\\begin{cases}m=-6 \\n=2\\end{cases}$$，$$\\begin{cases}m=1 \\n=-12\\end{cases}$$，$$\\begin{cases}m=-12 \\n=1\\end{cases}$$，共$$12$$种情况． 而$$a=m+n$$，只有$$6$$种结果， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "804", "queId": "a38b3d4287ac473a8ce0dc8499553f26", "competition_source_list": ["1995年第6届希望杯初二竞赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$${{x}_{0}}$$是方程$$\\left\\textbar{} \\frac{1+x}{2} \\right\\textbar-\\left\\textbar{} -x \\right\\textbar=0$$的一个不为$$1$$的根，则．", "answer_option_list": [[{"aoVal": "A", "content": "$${{x}_{0}}\\textgreater2{{x}_{0}}\\textgreater x_{0}^{2}$$ "}], [{"aoVal": "B", "content": "$$x_{0}^{2}\\textgreater{{x}_{0}}\\textgreater2{{x}_{0}}$$ "}], [{"aoVal": "C", "content": "$$x_{0}^{2}\\textgreater2{{x}_{0}}\\textgreater{{x}_{0}}$$ "}], [{"aoVal": "D", "content": "$$2{{x}_{0}}\\textgreater x_{0}^{2}\\textgreater{{x}_{0}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->一次方程->一元一次方程"], "answer_analysis": ["原方程变为$$\\left\\textbar{} \\frac{1+x}{2} \\right\\textbar=\\left\\textbar{} x \\right\\textbar$$，即$$\\frac{1+x}{2}=\\pm x$$，解得$$x=1$$或$$x=-\\frac{1}{3}$$，由题意知$${{x}_{0}}=-\\frac{1}{3}$$，于是$$x_{0}^{2}\\textgreater{{x}_{0}}\\textgreater2{{x}_{0}}$$，故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1456", "queId": "ca67960f8e04435fb969d7963e211afe", "competition_source_list": ["2007年第18届希望杯初一竞赛复赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "将棱长为$$1$$厘米的$$42$$个立方体积木拼在一起，构成一个实心的长方体．如果长方体底面的周长为$$18$$厘米，那么这个长方体的高是．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$厘米 "}], [{"aoVal": "B", "content": "$$3$$厘米 "}], [{"aoVal": "C", "content": "$$6$$厘米 "}], [{"aoVal": "D", "content": "$$7$$厘米 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->几何图形初步->几何图形->立体图形与平面图形->认识图形"], "answer_analysis": ["由构成的长方体的底面矩形周长为$$18$$，得长$$+$$宽$$=9$$厘米， 因为长方体的体积为$$42$$立方厘米， 所以长和宽都应是$$42$$的约数，且底面面积也是$$42$$的约数， 又$$42=1\\times 42=2\\times 3\\times 7$$， 所以长$$=7$$厘米，宽$$=2$$厘米， 此时高$$=42\\div (2\\times 7)=3$$（厘米）. 故选($$\\text{B}$$)． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "255", "queId": "5d7c93abae564f74a75993145063bcaa", "competition_source_list": ["1992年第3届全国希望杯初一竞赛复赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$x-y=2$$，$${{x}^{2}}+{{y}^{2}}=4$$，则$${{x}^{1992}}+{{y}^{1992}}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$${{1992}^{2}}$$ "}], [{"aoVal": "C", "content": "$${{2}^{1992}}$$ "}], [{"aoVal": "D", "content": "$${{4}^{1992}}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的加减->整式有关的概念->整式->整式的定义", "课内体系->知识点->式->整式的乘除->幂的运算->幂的综合运算"], "answer_analysis": ["由$$x-y=2$$,平方得$${{x}^{2}}-2xy+{{y}^{2}}=4$$． 又已知$${{x}^{2}}+{{y}^{2}}=4$$， 两式相减得$$2xy=0$$，∴$$xy=0$$． 所以$$x$$，$$y$$中至少有一个为$$0$$，但$${{x}^{2}}+{{y}^{2}}=4$$． 因此，$$x$$，$$y$$中只能有一个为$$0$$，另一个为$$2$$或$$-2$$． 无论哪种情况，都有$${{x}^{1992}}+{{y}^{1992}}={{0}^{1992}}+{{(\\pm 2)}^{1992}}={{2}^{1992}}$$，选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "403", "queId": "62a2f56439594578aea26609b021b35e", "competition_source_list": ["2010年第21届希望杯初二竞赛第1试第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "某人沿正在向下运动的自动扶梯从楼上走到楼下，用了$$24$$秒；若他站在自动扶梯上不动，从楼上到楼下要用$$56$$秒．若扶梯停止运动，他从楼上走到楼下要用．", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$秒 "}], [{"aoVal": "B", "content": "$$38$$秒 "}], [{"aoVal": "C", "content": "$$42$$秒 "}], [{"aoVal": "D", "content": "$$48$$秒 "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->一次方程->一元一次方程"], "answer_analysis": ["设从楼上到楼下的路程为$$s$$，则自动扶梯每秒向下运动$$\\frac{s}{56}$$的路程． 人沿向下运动的扶梯向下走，经过$$24$$秒，扶梯向下运动了$$24\\times \\frac{s}{56}=\\frac{3s}{7}$$的路程， 人向下走了$$s-\\frac{3}{7}s=\\frac{4}{7}s$$的路程， 所以，人每秒向下走$$\\frac{4}{7}s\\div 24=\\frac{s}{42}$$的路程， 那么，人走$$s$$的路程要用$$42$$秒， 即人沿不动的扶梯从楼上走到楼下要用$$42$$秒． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1078", "queId": "9b73daaa3b3e40d28ffa3eb62f94ee2b", "competition_source_list": ["2019年湖南长沙雨花区湖南广益实验中学初一竞赛（广益杯）第9题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "据统计，$$2017$$``十一''国庆长假期间，长沙市共接待国内外游客约$$319$$万人次，与$$2016$$年相比同比增长$$16.43 \\%$$，数据$$319$$万用科学记数法表示为．", "answer_option_list": [[{"aoVal": "A", "content": "$$3.19\\times {{10}^{6}}$$ "}], [{"aoVal": "B", "content": "$$3.19\\times {{10}^{8}}$$ "}], [{"aoVal": "C", "content": "$$0.319\\times {{10}^{7}}$$ "}], [{"aoVal": "D", "content": "$$319\\times {{10}^{6}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->科学记数法->科学记数法：表示较大的数"], "answer_analysis": ["$$319$$万$$=3190000$$， $$3190000=3.19\\times {{10}^{6}}$$， 其中科学记数法的表示方法为$$\\left\\textbar{} a \\right\\textbar\\times {{10}^{n}}\\left( 1\\leqslant a\\textless{}10 \\right)$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1424", "queId": "c59446af4cf44c27bbd08fe5263c2213", "competition_source_list": ["2009年第20届希望杯初一竞赛第1试第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "用一根长为$$a$$米的细绳围成一个等边三角形，测得它的面积是$$b$$平方米．在这个等边三角形内任取一点$$P$$，则点$$P$$到等边三角形三边的距离的和等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2b}{a}$$米 "}], [{"aoVal": "B", "content": "$$\\frac{4b}{a}$$米 "}], [{"aoVal": "C", "content": "$$\\frac{6b}{a}$$米 "}], [{"aoVal": "D", "content": "$$\\frac{8b}{a}$$米 "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["由面积法知：点$$P$$到等边三角形三边的距离之和等于这个等边三角形的高，而$$b=\\frac{1}{2}\\cdot \\frac{a}{3}\\cdot h$$，所以$$h=\\frac{6b}{a}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1361", "queId": "d2c5afff897a4a6c92349a1337680fba", "competition_source_list": ["1998年第9届希望杯初二竞赛第2试第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "若对于$$\\pm 3$$以外的一切实数$$x$$，等式$$\\frac{m}{x+3}-\\frac{n}{x-3}=\\frac{8x}{{{x}^{2}}-9}$$均成立，则$$mn$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$-8$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$-16$$ "}]], "knowledge_point_routes": ["课内体系->思想->方程思想", "课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->加减消元法解二元一次方程组", "课内体系->知识点->方程与不等式->分式方程->分式方程的解与解分式方程->解可化为一元一次方程的分式方程"], "answer_analysis": ["因为$$\\frac{m}{x+3}-\\frac{n}{x-3}=\\frac{8x}{{{x}^{2}}-9}$$， 所以$$\\frac{m(x-3)-n(x+3)}{{{x}^{2}}-9}=\\frac{8x}{{{x}^{2}}-9}$$． 因为$$x\\ne \\pm 3$$． 所以$$m(x-3)-n(x+3)=8x$$． 即$$(m-n)x-3(m+n)=8x$$． 此式对于一切$$x\\ne \\pm 3$$的值均成立． 所以$$\\begin{cases}m-n=8, m+n=0, \\end{cases}$$ 解得$$\\begin{cases}m=4, n=-4. \\end{cases}$$ 所以$$mn=-16$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "588", "queId": "b0a0f0f391c14c88b5aa2251b105df89", "competition_source_list": ["2019~2020学年10月山西太原杏花岭区太原志达中学初三上学期月考第2题3分", "2019~2020学年10月广东佛山禅城区佛山市惠景中学初三上学期月考第6题3分", "2019~2020学年9月甘肃兰州城关区兰州外国语学校初三上学期月考第8题4分", "2020~2021学年9月四川成都天府新区天府七中初三上学期周测B卷第8题3分", "2020~2021学年10月广东深圳南山区深圳南山外国语学校高新中学初三上学期周测B卷第4题", "2020年湖南长沙岳麓区湖南师范大学附属中学初二竞赛（湖南师范大学附属中学教育集团）（6月攀登杯）第2题4分", "2018~2019学年辽宁沈阳浑南区育才实验学校初二下学期期末第3题", "2020~2021学年9月四川成都天府新区天府七中初三上学期周测C卷第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "若关于$$x$$ 的一元二次方程$$k{{x}^{2}}-x+1=0$$有实数根，则$$k$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$k\\textgreater\\frac{1}{4}$$且$$k\\ne 0$$ "}], [{"aoVal": "B", "content": "$$k\\textless{}\\frac{1}{4}$$且$$k\\ne 0$$ "}], [{"aoVal": "C", "content": "$$k\\leqslant \\frac{1}{4}$$且$$k\\ne 0$$ "}], [{"aoVal": "D", "content": "$$k\\textless{}\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->由一元二次方程根的情况确定参数"], "answer_analysis": ["∵关于$$x$$的一元二次方程$$k{{x}^{2}}-x+1=0$$有实数根， ∴$$k\\ne 0$$且$$\\Delta ={{b}^{2}}-4ac\\geqslant 0$$， ∴$$1-4k\\geqslant 0$$， ∴解得：$$k\\leqslant \\frac{1}{4}$$． 综上，$$k\\leqslant \\frac{1}{4}$$且$$k\\ne 0$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "68", "queId": "4a83933cb8dd4a7c94a9d15a372e6a4b", "competition_source_list": ["2018年浙江宁波余姚市余姚市实验学校初二竞赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知一组数据$$x_1$$，$$x_2$$，$$x_3$$，$$x_4$$，$$x_5$$的平方和为$$200$$，它们的平均数为$$6$$，则这组数据的方差是．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->数据的分析->方差", "课内体系->知识点->统计与概率->数据的分析->平均数/加权平均数"], "answer_analysis": ["∵这五个数的平方和为$$200$$，平均数为$$6$$， ∴$$x_1+x_2+x_3+x_4+x_5=6\\times5=30$$， $$x_1^{2}+x_2^{2}+x_3^{3}+x_4^{4}+x_5^{2}=200$$， ∴这组数据的方差为： $$\\frac15\\times\\left[(x_1-6)^{2}+(x_2-6)^{2}+(x_3-6)^{2}+(x_4-6)^{2}+(x_5-6)^{2}\\right]$$ $$=\\frac15\\times\\left[x_1^{2}+x_2^{2}+x_3^{3}+x_4^{4}+x_5^{2}+5\\times6^{2}-12(x_1+x_2+x_3+x_4+x_5)\\right]$$ $$=\\frac15\\times(200+5\\times36-12\\times30)$$ $$=\\frac15\\times(200+180-360)$$ $$=\\frac15\\times20$$ $$=4$$， 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "15", "queId": "173a1d329f9b4d769de9d9170223d7b7", "competition_source_list": ["2015年第26届全国希望杯初一竞赛复赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "小尼、小浩、小莲、小雪四人中的一人书包里有苹果，老师问：谁的书包里有苹果？四人回答如下： 小尼：苹果不在我这里； 小浩：苹果在小雪那里； 小莲：苹果在小浩那里； 小雪：苹果不在我这里． 若其中只有一人说了假话，则书包里有苹果的是．", "answer_option_list": [[{"aoVal": "A", "content": "小尼 "}], [{"aoVal": "B", "content": "小浩 "}], [{"aoVal": "C", "content": "小莲 "}], [{"aoVal": "D", "content": "小雪 "}]], "knowledge_point_routes": ["知识标签->知识点->命题与证明"], "answer_analysis": ["用假设法： （$$1$$）假设小尼说了假话，那么苹果在小尼那里，小浩和小莲就说的是假话，与题设冲突； （$$2$$）假设小浩说的是假话，其他人说的都是真话，那么苹果在小浩那里； （$$3$$）假设小莲说了假话，则小浩说``苹果在小雪那里''与小雪说``苹果不在我这里''都是真话，互相矛盾； （$$4$$）假设小雪说的是假话，那么小莲说的就是假话，这样小雪、小莲都说假话，与``其中只有一人说的是假话''的题设不符． 因此苹果在小浩那里． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "811", "queId": "a3900fad478e44afb4fe93c5551e0a15", "competition_source_list": ["2015年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "$$1000$$到$$9999$$之间有多少个整数四个数位上各不相同？．", "answer_option_list": [[{"aoVal": "A", "content": "$$3024$$ "}], [{"aoVal": "B", "content": "$$4536$$ "}], [{"aoVal": "C", "content": "$$5040$$ "}], [{"aoVal": "D", "content": "$$6480$$ "}], [{"aoVal": "E", "content": "$$6561$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->数轴->数轴上整点覆盖问题", "美国AMC8->Knowledge Point->Combinatorics->Permutation and Combination->Permutation Problems"], "answer_analysis": ["$$1000$$到$$9999$$之间有多少个整数四个数位上各不相同？ 这个问题可以改为``\\,''有多少四位正整数有四个不同的数，因为数字是$$1000$$到$$9999$$之间的四位整数，对于第一个数有$$9$$中选择，因为它不能是$$0$$，第二个数只有$$9$$种选择，因为它不同于第一个数，第三个数有$$8$$种选择，因为它不同于前两个数，第四个数有$$7$$种选择，因为它不同于前三个数，则$$9\\times 9\\times 8\\times 7=4536$$． 故选$$\\text{B}$$． The question can be rephrased to \\textquotesingle How many four-digit positive integers have four distinct digits?\\textquotesingle, since numbers between $$1000$$ and $$9999$$ are four-digit integers. There are $$9$$ choices for the first number, since it cannot be $$0$$, there are only $$9$$ choices left for the second number since it must differ from the first, $$8$$ choices for the third number, since it must differ from the first two, and $$7$$ choices for the fourth number, since it must differ from all three. This means there are $$9\\times9\\times8\\times7=\\boxed{(\\text{B})4536}$$ integers between $$1000$$ and $$9999$$ with four distinct digits. "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "889", "queId": "89716494f9cc4cfdb2146796b3d85fe6", "competition_source_list": ["2010年第21届希望杯初二竞赛第1试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "若代数式$$\\frac{\\sqrt{2010-x}}{\\textbar x\\textbar-2009}$$有意义，则$$x$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\leqslant 2010$$ "}], [{"aoVal": "B", "content": "$$x\\leqslant 2010$$，且$$x\\ne \\pm 2009$$ "}], [{"aoVal": "C", "content": "$$x\\leqslant 2010$$，且$$x\\ne 2009$$ "}], [{"aoVal": "D", "content": "$$x\\leqslant 2010$$，且$$x\\ne -2009$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->分式->分式的基本运算"], "answer_analysis": ["由题意，得$$\\begin{cases}2010-x\\geqslant 0 \\textbar x\\textbar-2009\\ne 0 \\end{cases}$$， 解得$$x\\leqslant 2010$$，且$$x\\ne \\pm 2009$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "98", "queId": "259363ab5c7d408ea511df3b684e2ece", "competition_source_list": ["1999年第10届希望杯初一竞赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$7-a$$的倒数的相反数是$$-2$$，那么$$a$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$7.5$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6.5$$ "}]], "knowledge_point_routes": ["课内体系->思想->方程思想", "课内体系->知识点->数->有理数->相反数->相反数的性质", "课内体系->知识点->数->有理数->倒数与负倒数"], "answer_analysis": ["$$7-a$$的倒数是$$\\frac{1}{7-a}$$，$$\\frac{1}{7-a}$$的相反数是$$-\\frac{1}{7-a}=\\frac{1}{a-7}$$． 依题意列方程$$\\frac{1}{a-7}=-2$$． 解得$$a=\\frac{13}{2}=6.5$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "274", "queId": "c73fd71416df4e498c6a575f53f2f938", "competition_source_list": ["2004年第15届希望杯初二竞赛第1试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$，$$b\\left( b\\textgreater a \\right)$$是两个任意质数，那么下列四个分数、 ① $$\\frac{a+b}{ab}$$~ ②$$\\frac{b-a}{b+a}$$~ ③$$\\frac{{{b}^{2}}-{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}$$~ ~④$$\\frac{ab}{{{a}^{2}}+{{b}^{2}}}$$中，总是最简分数的有．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$个 "}], [{"aoVal": "B", "content": "$$2$$个 "}], [{"aoVal": "C", "content": "$$3$$个 "}], [{"aoVal": "D", "content": "$$4$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["设$$a=3$$，$$b=5$$， 则$$\\frac{a+b}{ab}=\\frac{3+5}{3\\times 5}=\\frac{8}{15}$$是最简分数． $$\\frac{b-a}{b+a}=\\frac{5-3}{5+3}=\\frac{2}{8}$$不是最简分数． $$\\frac{{{b}^{2}}-{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}==\\frac{{{5}^{2}}-{{3}^{2}}}{{{3}^{2}}+{{5}^{2}}}=\\frac{16}{34}$$不是最简分数． $$\\frac{ab}{{{a}^{2}}+{{b}^{2}}}=\\frac{3\\times 5}{{{3}^{2}}+{{5}^{2}}}=\\frac{15}{34}$$是最简分数． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1126", "queId": "c4716e53801348dfbd34be9f52c6d0db", "competition_source_list": ["2011年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "如下是 Tyler 去年夏天外出捕鱼九次的数量：$$2$$，$$0$$，$$1$$，$$3$$，$$0$$，$$3$$，$$3$$，$$1$$，$$2$$．下列关于平均值，中位数，众数的描述，哪个是正确的？", "answer_option_list": [[{"aoVal": "A", "content": "中位数$$\\textless$$平均值$$\\textless$$众数 "}], [{"aoVal": "B", "content": "平均值$$\\textless$$众数$$\\textless$$中位数 "}], [{"aoVal": "C", "content": "平均值$$\\textless$$中位数$$\\textless$$众数 "}], [{"aoVal": "D", "content": "中位数$$\\textless$$众数$$\\textless$$平均值 "}], [{"aoVal": "E", "content": "众数$$\\textless$$中位数$$\\textless$$平均值 "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Counting, Probability and Statistics->Average Problems->Complex Average Problems", "课内体系->知识点->统计与概率->数据的分析->平均数/加权平均数"], "answer_analysis": ["如下是 Tyler 去年夏天外出捕鱼九次的数量：$$2$$，$$0$$，$$1$$，$$3$$，$$0$$，$$3$$，$$3$$，$$1$$，$$2$$．下列关于均值，中位数，众数的描述哪个正确？ A~ 中位数$$\\textless$$均值$$\\textless$$众数 B~ 均值$$\\textless$$众数$$\\textless$$中位数 C~ 均值数$$\\textless$$中位数$$\\textless$$众数 D~ 中位数$$\\textless$$众数$$\\textless$$均值 E~ 众数$$\\textless$$中位数$$\\textless$$均值 首先，把数字按增加的排列． $$0$$，$$0$$，$$1$$，$$1$$，$$2$$，$$2$$，$$3$$，$$3$$，$$3$$ 故$$\\frac{0+0+1+1+2+2+3+3+3}{9}=\\frac{15}{9}$$，中位数是$$2$$，众数是$$3$$．因为，$$\\frac{15}{9} ~\\textless{} ~2 ~\\textless{} ~3$$，所以答案是mean$$ ~\\textless{} ~$$median$$ ~\\textless{} ~$$mode，即平均数$$ ~\\textless{} ~$$中位数$$ ~\\textless{} ~$$众数． 故选$$\\text{C}$$． First, put the numbers in increasing order. $$0$$, $$0$$, $$1$$, $$1$$, $$2$$, $$2$$, $$3$$, $$3$$, $$3$$ The mean is $$\\frac{0+0+1+1+2+2+3+3+3}{9}= \\frac{15}{9}$$, the median is $$2$$, and the mode is $$3$$. Because, $$\\frac{15}{9}\\textless2\\textless3$$, the answer is mean $$\\textless{}$$ median $$\\textless{}$$ mode. "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1448", "queId": "b3d2e1b745b845e894cfd5b6c95d208f", "competition_source_list": ["2018年全国初中数学联赛竞赛B卷"], "difficulty": "2", "qtype": "single_choice", "problem": "满足$${{(x-1)}^{x+3}}=1$$的整数$$x$$的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["知识标签->题型->方程与不等式->一元二次方程", "知识标签->题型->式->整式的乘除->幂的运算", "知识标签->学习能力->运算能力", "知识标签->学习能力->抽象概括能力", "知识标签->知识点->式->整式的乘除->幂的运算", "知识标签->知识点->方程与不等式->一元二次方程"], "answer_analysis": ["当$$x+3=0$$且$$x-1\\ne 0$$时，$$x=-3$$， 当$$x-1=1$$时，$$x=2$$， 当$$x-1=-1$$且$$x+3$$为偶数时，$$x=0$$不符合题意， 所以，满足条件的整数$$x$$有$$2$$个． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1174", "queId": "8aac49074e023206014e20d832d865d8", "competition_source_list": ["1994年第5届全国希望杯初一竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a\\textless{}0$$，$$b\\textgreater0$$，且$$\\left\\textbar{} a \\right\\textbar\\textless{}\\left\\textbar{} b \\right\\textbar$$，则$$a+b=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left\\textbar{} b \\right\\textbar-\\left\\textbar{} a \\right\\textbar$$ "}], [{"aoVal": "B", "content": "$$-\\left\\textbar{} a \\right\\textbar-\\left\\textbar{} b \\right\\textbar$$ "}], [{"aoVal": "C", "content": "$$\\left\\textbar{} a \\right\\textbar-\\left\\textbar{} b \\right\\textbar$$ "}], [{"aoVal": "D", "content": "$$\\left\\textbar{} a \\right\\textbar+\\left\\textbar{} b \\right\\textbar$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->绝对值的非负性"], "answer_analysis": ["因为$$a\\textless{}0$$，所以$$\\left\\textbar{} a \\right\\textbar=-a$$． 因为$$b\\textgreater0$$，所以$$\\left\\textbar{} b \\right\\textbar=b$$． 所以$$a+b=-\\left\\textbar{} a \\right\\textbar+\\left\\textbar{} b \\right\\textbar$$．选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "250", "queId": "3d43a2bb7db74678a093ac78a63a733a", "competition_source_list": ["2019~2020学年11月浙江杭州拱墅区杭州树兰中学初一上学期周测C卷第4题3分", "2019~2020学年4月浙江温州乐清市乐清市乐成公立寄宿学校初一下学期月考（竞赛班）第7题4分", "2016~2017学年12月江苏镇江句容市句容市华阳学校初一上学期月考第17题3分", "2017~2018学年陕西西安雁塔区西安高新第一中学初中校区东区初级中学初一上学期期末第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某种商品的进价为$$1200$$元，标价为$$1575$$元，后来由于该商品积压，商店准备打折出售，但要保持利润不低于$$5 \\%$$，则至多可打．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$折 "}], [{"aoVal": "B", "content": "$$7$$折 "}], [{"aoVal": "C", "content": "$$8$$折 "}], [{"aoVal": "D", "content": "$$9$$折 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->不等式组应用->不等式组的经济问题", "课内体系->能力->运算能力"], "answer_analysis": ["要保持利润率不低于$$5 \\%$$，设可打$$x$$折． 则$$1575\\times \\frac{x}{10}-1200\\geqslant 1200\\times 5 \\%$$， 解得$$x\\geqslant 8$$， 即要保持利润率不低于$$5 \\%$$，最多可打$$8$$折． 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "780", "queId": "88d776d4e351433594aa13d0c8c74ba6", "competition_source_list": ["2014年第25届全国希望杯初二竞赛复赛第4题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "不等式组$$\\begin{cases}x-4\\leqslant 8-2x x\\textgreater-\\dfrac{2}{3} \\end{cases}$$的最小整数解是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->一元一次不等式组的整数解", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->解一元一次不等式组", "课内体系->能力->运算能力"], "answer_analysis": ["解$$x-4\\leqslant 8-2x$$，得$$x\\leqslant 4$$， 故原不等式的解为$$-\\frac{2}{3}\\textless{}x\\leqslant 4$$， 所以，原不等式组的最小整数解是$$x=0$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1390", "queId": "c9e7933551d74c079448ce3f61c30013", "competition_source_list": ["2010年第15届华杯赛初一竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$x$$，$$y$$满足$$2x+3y=15$$，$$6x+13y=41$$，则$$x+2y$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$\\frac{15}{2}$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->一次方程->方程组"], "answer_analysis": ["$$2x+3y=15$$，$$6x+13y=41$$， 由此可得：$$8x+16y=56$$，两边除以$$8$$得到$$x+2y=7$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "302", "queId": "421f415a5ae349b2bdb1e8d55bf9a802", "competition_source_list": ["1991年第2届希望杯初二竞赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$x$$，$$y$$都是正整数，那么三条边长是$$x$$，$$y$$和$$10$$的三角形有．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$个 "}], [{"aoVal": "B", "content": "$$4$$个 "}], [{"aoVal": "C", "content": "$$5$$个 "}], [{"aoVal": "D", "content": "无数多个 "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->三角形及多边形->与三角形有关的线段->判断能否构成三角形", "课内体系->知识点->三角形->三角形及多边形->与三角形有关的线段->三角形的三边关系"], "answer_analysis": ["因为$$x=y\\textgreater5$$的任何正整数，都可以和$$10$$作为三角形的三条边． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1225", "queId": "bba8cf13fbaa470194995d94e9fe3ade", "competition_source_list": ["1998年第9届希望杯初二竞赛第2试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$$\\sqrt{8+\\sqrt{63}}+\\sqrt{8-\\sqrt{63}}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$3\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$2\\sqrt{3}$$ "}], [{"aoVal": "C", "content": "$$5\\sqrt{2}$$ "}], [{"aoVal": "D", "content": "$$2\\sqrt{5}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算"], "answer_analysis": ["原式$$=\\sqrt{{{(\\sqrt{\\frac{9}{2}}+\\sqrt{\\frac{7}{2}})}^{2}}}+\\sqrt{{{(\\sqrt{\\frac{9}{2}}-\\sqrt{\\frac{7}{2}})}^{2}}}$$$$=2\\sqrt{\\frac{9}{2}}=3\\sqrt{2}$$．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1312", "queId": "a55bcf568aeb4c4d9268840b08447d01", "competition_source_list": ["初一下学期其它", "1997年第8届希望杯初二竞赛第2试第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$${{m}^{2}}+m-1=0$$，那么代数式$${{m}^{3}}+2{{m}^{2}}-1997$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1997$$ "}], [{"aoVal": "B", "content": "$$-1997$$ "}], [{"aoVal": "C", "content": "$$1996$$ "}], [{"aoVal": "D", "content": "$$-1996$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->整式乘除化简求值->整式乘除化简求值-整体代入化简求值", "课内体系->能力->运算能力"], "answer_analysis": ["因为$${{m}^{2}}+m-1=0$$， 所以$${{m}^{2}}+m=1$$， 所以$${{m}^{3}}+2{{m}^{2}}-1997$$$$=m({{m}^{2}}+m)+{{m}^{2}}-1997$$$$=m+{{m}^{2}}-1997$$$$=1-1997$$$$=-1996$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "505", "queId": "555006cff23348068a7890698715dc3a", "competition_source_list": ["2017年全国全国初中数学联赛竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "已知二次函数$$y=a{{x}^{2}}+bx+c$$（$$c\\ne 0$$）的图象与$$x$$轴有唯一交点，则二次函数$$y={{a}^{3}}{{x}^{2}}+{{b}^{3}}x+{{c}^{3}}$$的图象与$$x$$轴的交点个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["课内体系->知识点->函数->二次函数->二次函数与方程、不等式->二次函数与坐标轴交点", "课内体系->能力->运算能力"], "answer_analysis": ["因为二次函数$$y=a{{x}^{2}}+bx+c$$的图象与$$x$$轴有唯一交点， 所以$${{\\Delta }_{1}}={{b}^{1}}-4ac=0$$， 所以$${{b}^{2}}=4ac\\ne 0$$． 故二次函数$$y={{a}^{3}}{{x}^{2}}+{{b}^{3}}x+{{c}^{3}}$$的判别式$${{\\Delta }_{2}}={{({{b}^{3}})}^{2}}-4{{a}^{3}}{{c}^{3}}={{b}^{6}}-\\frac{1}{16}{{(4ac)}^{3}}={{b}^{6}}-\\frac{1}{16}{{({{b}^{2}})}^{3}}=\\frac{15}{16}{{b}^{6}}\\textgreater0$$， 所以，二次函数$$y={{a}^{3}}{{x}^{2}}+{{b}^{3}}x+{{c}^{3}}$$的图象与$$x$$轴有两个交点． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "46", "queId": "12d874d24b904fa881e8200ac3f85b4f", "competition_source_list": ["1990年第1届希望杯初二竞赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个角等于它的余角的$$5$$倍，那么这个角是．", "answer_option_list": [[{"aoVal": "A", "content": "$$45{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$75{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$55{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$65{}^{}\\circ $$ "}]], "knowledge_point_routes": ["竞赛->知识点->几何图形初步->角->角的基础问题"], "answer_analysis": ["因为所求角$$a=5(90{}^{}\\circ -a)$$，解得$$a=75{}^{}\\circ $$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1360", "queId": "b78b34c8529445818297bc53e00f1ef2", "competition_source_list": ["2013年竞赛第2题3分", "2016年安徽芜湖镜湖区芜湖市第一中学初三自主招生第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知关于$$x$$的不等式组$$\\begin{cases}\\dfrac{2x+5}{3}-x\\textgreater-5 \\dfrac{x+3}{2}-t ~\\textless{} ~x \\end{cases}$$，恰有$$5$$个整数解，则$$t$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-6 ~\\textless{} ~t ~\\textless{} ~-\\frac{11}{2}$$ "}], [{"aoVal": "B", "content": "$$-6\\leqslant t ~\\textless{} ~-\\frac{11}{2}$$ "}], [{"aoVal": "C", "content": "$$-6 ~\\textless{} ~t\\leqslant -\\frac{11}{2}$$ "}], [{"aoVal": "D", "content": "$$-6\\leqslant t\\leqslant -\\frac{11}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->不等式->一次不等式组"], "answer_analysis": ["根据题设知不等式组有解，解得，$$3-2t ~\\textless{} ~x ~\\textless{} ~20$$. 由于不等式组恰有$$5$$个整数解，这$$5$$个整数解只能为$$15$$，$$16$$，$$17$$，$$18$$，$$19$$，因此$$14\\leqslant 3-2t ~\\textless{} ~15$$，解得$$-6\\textless t\\leqslant -\\frac{11}{2}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "132", "queId": "0920517f41b94856b59041ecce6682c9", "competition_source_list": ["2018年第29届希望杯初一竞赛初赛第16题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "将一个棱长是整数厘米的长方体的各表面都刷成红色，然后将这个长方体分割成若干个棱长为$$1$$厘米的小正方体，若任何一面都没有涂色的小正方体有$$11$$个，则原来的长方体的体积是~\\uline{~~~~~~~~~~}~立方厘米．", "answer_option_list": [[{"aoVal": "A", "content": "$$117$$ "}], [{"aoVal": "B", "content": "$$99$$ "}], [{"aoVal": "C", "content": "$$96$$ "}], [{"aoVal": "D", "content": "$$84$$ "}], [{"aoVal": "E", "content": "$$48$$ "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->几何图形->立体图形与平面图形->正方体堆积图形的问题", "课内体系->能力->分析和解决问题能力", "课内体系->能力->运算能力"], "answer_analysis": ["设该长方体的棱长分别为$$a$$厘米，$$b$$厘米，$$c$$厘米，$$(a\\geqslant b\\geqslant c)$$， 则由题意得$$(a-2)(b-2)(c-2)=11$$， ∵$$11$$为质数， ∴$$a-2=11$$，$$b-2=1$$，$$c-2=1$$， ∴$$a=13$$，$$b=3$$，$$c=3$$， ∴长方体的体积$$V=abc=117$$立方厘米． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "115", "queId": "17fd018ccac64bb8b1c5a3052e926fc7", "competition_source_list": ["1991年第2届希望杯初二竞赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$${{x}^{2}}+\\textbar x\\textbar+1=0$$的实数根的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->一元二次方程根的判别式", "课内体系->知识点->方程与不等式->一元二次方程->根与系数的关系->一元二次方程根与系数的关系"], "answer_analysis": ["因为不论$$x$$为何实数，$${{x}^{2}}+\\textbar x\\textbar+1$$总是大于零的． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "379", "queId": "50253947a24b46819967d229c250678a", "competition_source_list": ["1994年第11届全国初中数学联赛竞赛第1题6分", "2016~2017学年广东广州天河区华南师范大学附属中学初二上学期期中第3题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$0\\textless{}a\\textless{}1$$，则$$\\sqrt{{{a}^{2}}+\\frac{1}{{{a}^{2}}}-2}\\div \\left( 1+\\frac{1}{a} \\right)\\times \\frac{1}{1+a}$$可化简为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1-a}{1+a}$$ "}], [{"aoVal": "B", "content": "$$\\frac{a-1}{a+1}$$ "}], [{"aoVal": "C", "content": "$$1-{{a}^{2}}$$ "}], [{"aoVal": "D", "content": "$${{a}^{2}}-1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算", "课内体系->能力->运算能力"], "answer_analysis": ["∵$$0\\textless{}a\\textless{}1$$，∴$$a-\\frac{1}{a}\\textless{}0$$， ∴原式$$=\\sqrt{{{\\left( a-\\frac{1}{a} \\right)}^{2}}}\\div \\left( \\frac{a+1}{a} \\right)\\times \\frac{1}{1+a}$$ $$=\\left( \\frac{1}{a}-a \\right)\\times \\frac{a}{a+1}\\times \\frac{1}{1+a}$$ $$=\\frac{\\left( 1+a \\right)\\left( 1-a \\right)}{a}\\times \\frac{a}{a+1}\\times \\frac{1}{1+a}$$ $$=\\frac{1-a}{1+a}$$， 故答案为$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "32", "queId": "20a5539f724b4d5785001a9b6c9cadf6", "competition_source_list": ["2013年第24届全国希望杯初二竞赛初赛第20题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "将不大于$$20$$的正偶数分成两组，使得第一组中数的乘积能被第二组中数的乘积整除，则商的最小值是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数乘法->有理数乘法运算", "课内体系->能力->运算能力", "竞赛->知识点->数论->整除->整除的概念与基本性质", "竞赛->知识点->数论->整除->算术基本定理"], "answer_analysis": ["不大于$$20$$的正偶数的乘积是 $$2\\times 4\\times 6\\times 8\\times 10\\times 12\\times 14\\times 16\\times 18\\times 20$$ $$={{2}^{10}}\\times (1\\times 2\\times 3\\times 4\\times 5\\times 6\\times 7\\times 8\\times 9\\times 10)$$ $$={{2}^{10}}\\times (1\\times 2\\times 3\\times 4\\times 5\\times 6\\times 7\\times 8\\times 9\\times 10)$$ $$={{2}^{18}}\\times (3\\times 5\\times 3\\times 7\\times 9\\times 5)$$ $$={{2}^{18}}\\times {{3}^{4}}\\times {{5}^{2}}\\times 7$$． 设第一、二组中数的乘积分别是$$M$$，$$N$$， 则$$MN={{2}^{18}}\\times {{3}^{4}}\\times {{5}^{2}}\\times 7$$． 题意是要求最小的商，其实就是求最大的$$N$$， 易知$${{N}_{\\max }}={{2}^{9}}\\times {{3}^{2}}\\times 5$$， 故商为$$7$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "85", "queId": "a692c4dfc59c4692b37d84248b540278", "competition_source_list": ["初三上学期其它", "2003年第20届全国初中数学联赛竞赛第4题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "满足等式$$x\\sqrt{y}+\\sqrt{x}y-\\sqrt{2003x}-\\sqrt{2003y}+\\sqrt{2003xy}=2003$$的正整数对$$\\left( x,y \\right)$$的个数是（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->其他方程->无理方程"], "answer_analysis": ["已知等式可化为 $$\\sqrt{y}{{\\left( \\sqrt{x} \\right)}^{2}}+\\left[ {{\\left( \\sqrt{y} \\right)}^{2}}+\\sqrt{2003}\\sqrt{y}-\\sqrt{2003} \\right]\\sqrt{x}$$$$-\\sqrt{2003}\\left( \\sqrt{y}+\\sqrt{2003} \\right)=0$$， 故$$\\left( \\sqrt{y}\\sqrt{x}-\\sqrt{2003} \\right)\\left( \\sqrt{x}+\\sqrt{y}+\\sqrt{2003} \\right)=0$$． 又由题意， $$\\sqrt{x}+\\sqrt{y}+\\sqrt{2003}\\textgreater0$$， 则$$\\sqrt{y}\\sqrt{x}-\\sqrt{2003}=0\\Rightarrow xy=2003$$． 而$$2003$$为质数，且$$x$$、$$y$$为正整数，故$$\\left( x,y \\right)=\\left( 2003,1 \\right)$$，$$\\left( 1,2003 \\right)$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1350", "queId": "c0963efff5d647c7abc4e483e9bd60af", "competition_source_list": ["2010年第21届全国希望杯初一竞赛复赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果在一个正方体的每个面内写一个正整数，然后，在每个顶点处再写一个数，该数等于过这个顶点的三个面内的数的乘积，那么当该正方体各个顶点处的数之和是$$290$$时，各个面内的数之和等于（~ ~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$34$$ "}], [{"aoVal": "B", "content": "$$35$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$37$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->算术基本定理", "课内体系->知识点->几何图形初步->几何图形->立体图形与平面图形->点线面体"], "answer_analysis": ["设正方体的每个面内的数分别为$${{x}_{1}}$$，$${{x}_{2}}$$，$${{x}_{3}}$$，$${{x}_{4}}$$，$${{x}_{5}}$$，$${{x}_{6}}$$， 则顶点上的数分别为$${{x}_{1}}{{x}_{2}}{{x}_{5}}$$，$${{x}_{2}}{{x}_{3}}{{x}_{5}}$$，$${{x}_{3}}{{x}_{4}}{{x}_{5}}$$，$${{x}_{4}}{{x}_{1}}{{x}_{5}}$$，$${{x}_{1}}{{x}_{2}}{{x}_{6}}$$，$${{x}_{2}}{{x}_{3}}{{x}_{6}}$$，$${{x}_{3}}{{x}_{4}}{{x}_{6}}$$，$${{x}_{4}}{{x}_{1}}{{x}_{6}}$$， 则有$$290={{x}_{1}}{{x}_{2}}{{x}_{5}}+{{x}_{2}}{{x}_{3}}{{x}_{5}}+{{x}_{3}}{{x}_{4}}{{x}_{5}}+{{x}_{4}}{{x}_{1}}{{x}_{5}}+{{x}_{1}}{{x}_{2}}{{x}_{6}}+{{x}_{2}}{{x}_{3}}{{x}_{6}}+{{x}_{3}}{{x}_{4}}{{x}_{6}}+{{x}_{4}}{{x}_{1}}{{x}_{6}}$$ $$=({{x}_{5}}+{{x}_{6}})({{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{3}}{{x}_{4}}+{{x}_{4}}{{x}_{1}})$$ $$=({{x}_{1}}+{{x}_{3}})({{x}_{2}}+{{x}_{4}})({{x}_{5}}+{{x}_{6}})$$． 由于$$290=2\\times 5\\times 29$$， 显然在乘积$$({{x}_{1}}+{{x}_{3}})({{x}_{2}}+{{x}_{4}})({{x}_{5}}+{{x}_{6}})$$中，一个因式等于$$2$$，一个因式等于$$5$$，一个因式等于$$29$$， 所以$${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+{{x}_{6}}=2+5+29=36$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1100", "queId": "ff8080814d9efd56014da555a4c6074c", "competition_source_list": ["1992年第3届全国希望杯初一竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "若一个数的立方小于这个数的相反数，那么这个数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "正数 "}], [{"aoVal": "B", "content": "负数 "}], [{"aoVal": "C", "content": "奇数 "}], [{"aoVal": "D", "content": "偶数 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->有理数乘方运算", "课内体系->知识点->数->有理数->相反数->相反数的性质"], "answer_analysis": ["设该数为$$a$$，由题意$$-a$$为$$a$$的相反数，且有$${{a}^{3}}\\textless{}-a$$， ∴$${{a}^{3}}+a\\textless{}0$$，$$a({{a}^{2}}+1)\\textless{}0$$， 因为$${{a}^{2}}+1\\textgreater0$$，所以$$a\\textless{}0$$，即该数一定是负数，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "638", "queId": "63ae037bec934d6f9b1ab825968b0c0b", "competition_source_list": ["1991年第2届希望杯初二竞赛第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "方程$$2{{x}^{5}}+{{x}^{4}}-20{{x}^{3}}-10{{x}^{2}}+2x+1=0$$有一个实数根是（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{5}+\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{5}+\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{3}+\\sqrt{2}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{5}-\\sqrt{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->其他方程->高次方程", "课内体系->方法->降次法"], "answer_analysis": ["原方程可化为 $$\\left( 2{{x}^{5}}-20{{x}^{3}}+2x \\right)+\\left( {{x}^{4}}-10{{x}^{2}}+1 \\right)=0$$． 即$$\\left( 2x+1 \\right)\\left( {{x}^{4}}-10{{x}^{2}}+1 \\right)=0$$． ∴$$2x+1=0$$或$${{x}^{4}}-10{{x}^{2}}+1=0$$， 当$$2x+1=0$$时，解得$$x=-\\frac{1}{2}$$； 当$${{x}^{4}}-10{{x}^{2}}+1=0$$时，$${{x}^{2}}=\\frac{10+\\sqrt{{{10}^{2}}-4}}{2}=5+2\\sqrt{6}$$，或$${{x}^{2}}=\\frac{10-\\sqrt{{{10}^{2}}-4}}{2}=5-2\\sqrt{6}$$， ①当$${{x}^{2}}=5+2\\sqrt{6}$$，得$$x=\\sqrt{3}+\\sqrt{2}$$或$$x=-\\sqrt{3}-\\sqrt{2}$$， ②当$${{x}^{2}}=5-2\\sqrt{6}$$，得$$x=\\sqrt{3}-\\sqrt{2}$$或$$x=-\\sqrt{3}+\\sqrt{2}$$， 综上所述$$x$$可能为$$\\sqrt{3}+\\sqrt{2}$$， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "854", "queId": "729ed27ed13b4ef4ab797c131fb48e99", "competition_source_list": ["1992年第3届希望杯初二竞赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，$$\\angle A=\\theta -\\alpha $$，$$\\angle B=\\theta $$，$$\\angle C=\\theta +\\alpha $$，（$$0{}^{}\\circ \\textless{}\\alpha \\textless{}\\theta \\textless{}90{}^{}\\circ $$） ．若$$\\angle BAC$$与$$\\angle BCA$$的平分线相交于$$P$$点，则$$\\angle APC$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$90{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$105{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$120{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$150{}^{}\\circ $$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->三角形及多边形->与三角形有关的角->三角形内角和的应用", "课内体系->能力->运算能力"], "answer_analysis": ["由$$\\angle A+\\angle B+\\angle C=180{}^{}\\circ $$，即$$\\left( \\theta -\\alpha \\right)+\\theta +\\left( \\theta +\\alpha \\right)=3\\theta =180{}^{}\\circ $$， 所以$$\\theta =\\angle ABC=60{}^{}\\circ $$， 因此$$\\frac{1}{2}\\left( \\angle BAC+\\angle BCA \\right)=\\frac{1}{2}\\left( 180{}^{}\\circ -60{}^{}\\circ \\right)=60{}^{}\\circ $$， 所以$$\\angle APC=180{}^{}\\circ -\\frac{1}{2}\\left( \\angle BAC+\\angle BCA \\right)=180{}^{}\\circ -60{}^{}\\circ =120{}^{}\\circ $$． 故应选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1075", "queId": "85c8af4e8ab54d6783cedf9e0339b5b8", "competition_source_list": ["2003年第14届希望杯初二竞赛第2试第1题", "2020~2021学年江苏苏州高新区苏州市高新区实验初级中学初二下学期期末模拟（一）第7题", "2020~2021学年6月江苏苏州工业园区星港学校初一下学期周测B卷第8题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$y-2x+1$$是$$4xy-4{{x}^{2}}-{{y}^{2}}-k$$的一个因式，则$$k$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->其他方法->待定系数法", "课内体系->能力->运算能力"], "answer_analysis": ["由因式定理可知，当$$y-2x+1=0$$时， $$4xy-4{{x}^{2}}-{{y}^{2}}-k=0$$， 故可令$$\\begin{cases}x=0 y=-1 \\end{cases}$$， 则$$4xy-4{{x}^{2}}-{{y}^{2}}-k=-1-k=0$$， ∴$$k=-1$$． 故答案为：$$k=-1$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1025", "queId": "ff8080814d56502a014d67871c8a2123", "competition_source_list": ["2015年第26届全国希望杯初一竞赛复赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "红光中学初一年级有$$3$$个班，已知一班、二班的平均人数与三班人数之和为$$45$$，二班、三班的平均人数与一班人数之和为$$48$$，一班、三班的平均人数与二班人数之和为$$47$$��则三个班的总人数为（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$68$$ "}], [{"aoVal": "B", "content": "$$70$$ "}], [{"aoVal": "C", "content": "$$72$$ "}], [{"aoVal": "D", "content": "$$74$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->其他方程->三元一次方程组->解三元一次方程组", "课内体系->知识点->方程与不等式->其他方程->三元一次方程组->三元一次方程组的实际应用"], "answer_analysis": ["设一班、二班、三班的人数分别为$$x$$、$$y$$、$$z$$． 由题意，得$$\\left { \\begin{array} ~\\&\\dfrac{x+y}{2}+z=45 \\dfrac{y+z}{2}+x=48 \\dfrac{x+z}{2}+y=47 \\end{array} \\right.$$， 三式相加，得$$2(x+y+z)=140$$， ∴$$x+y+z=70$$． 答：三个班的总人数为$$70$$人． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1015", "queId": "ff8080814d4b1928014d4c0ec4eb052d", "competition_source_list": ["2015年第26届全国希望杯初一竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面有$$4$$个判断： ①互为相反数的两个数的绝对值相等； ②如果$$n$$的绝对值等于$$n$$，则$$n$$一定为正数； ③点$$M$$在数轴上距原点$$2$$个单位长度，且位于原点右侧．若将向左移动$$5$$个单位长度，则此点对应的值为$$-3$$； ④两个数相加，它们的和一定大于其中一个加数． 其中，正确判断的个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->命题与证明->猜想与证明->命题与证明的应用->区分真假命题"], "answer_analysis": ["①正确； ②$$n$$为$$0$$时也满足要求，故错误； ③正确； ④如$$(-1)+(-2)=-3$$，$$-3\\textless{}-2$$且$$-3\\textless{}-1$$，故错误． 综上，正确的个数为两个． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1383", "queId": "9cc4b190243640a1ad3063c857db67a5", "competition_source_list": ["1979年竞赛第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$10+20\\div2+3=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$23$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Number and Operations->Integers->Mixed Basic Operations"], "answer_analysis": ["$$10+20\\div 2+3=10+(20\\div 2)+3=10+10+3=23$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "865", "queId": "b1834863ac8847fcb1043437660fa837", "competition_source_list": ["2012年第17届华杯赛初一竞赛初赛第4题"], "difficulty": "0", "qtype": "single_choice", "problem": "在$$10\\square 10\\square 10\\square 10\\square 10$$的四个``$$\\square $$''中分别填入``$$+$$''、``$$-$$''、``$$\\times $$''、``$$\\div $$''运算符号各一次，所成的算式的值的最小值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-84$$ "}], [{"aoVal": "B", "content": "$$-89$$ "}], [{"aoVal": "C", "content": "$$-94$$ "}], [{"aoVal": "D", "content": "$$-99$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["要使算式的值尽可能小，应当减去一个较大的数，此时应使商尽可能小．如下式所示，所成的算式的值最小：$$10\\div 10\\times 10+10=-89$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "279", "queId": "ab732b18103646509b9e40d79e48039e", "competition_source_list": ["2006年第17届希望杯初一竞赛初赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$$x+y+z=7$$的正整数解有．", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$组 "}], [{"aoVal": "B", "content": "$$12$$组 "}], [{"aoVal": "C", "content": "$$15$$组 "}], [{"aoVal": "D", "content": "$$16$$组 "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->一次方程->特殊方程"], "answer_analysis": ["因为$$x$$，$$y$$ ，$$z$$均为正整数，且$$x+y+z=7$$， 所以$$1\\leqslant x\\leqslant 5$$． 下面分类讨论： 当$$x=1$$时，有$$5$$组解；当$$x=2$$时，有$$4$$组解； 当$$x=3$$时，有$$3$$组解；当$$x=4$$时，有$$2$$组解； 当$$x=5$$时，有$$1$$组解． 共计$$5+4+3+2+1=15$$（组）解． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "320", "queId": "3da01b4f411443aa94c8ba127e057d5c", "competition_source_list": ["2009年第20届希望杯初二竞赛第1试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "复印纸的型号有$$A0$$，$$A1$$，$$A2$$，$$A3$$，$$A4$$等，它们有如下的关系：将上一个型号（例如$$A3$$）的复印纸在长的方向对折后就得到两张下一个型号（得到$$A4$$）的复印纸，且各种型号的复印纸的长与宽的比相等，那么这些型号的复印纸的长与宽的比约为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1.414:1$$ "}], [{"aoVal": "B", "content": "$$2:1$$ "}], [{"aoVal": "C", "content": "$$1:0.618$$ "}], [{"aoVal": "D", "content": "$$1.732:1$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->四边形->特殊平行四边形->矩形->矩形的性质"], "answer_analysis": ["设$$A3$$型号的复印纸长为$$x$$，宽为$$y$$，对折后得到$$A4$$型号的复印纸长为$$y$$，宽为$$\\frac{x}{2}$$， 由题意，得$$\\frac{x}{y}=\\frac{y}{\\dfrac{x}{2}}$$， 即$${{x}^{2}}=2{{y}^{2}}$$， 所以$$x:y=\\sqrt{2}:1\\approx 1.414:1$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "345", "queId": "5dc69d1f521549aca4e9a54fd99cd40f", "competition_source_list": ["2010年第21届全国希望杯初一竞赛复赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有两个两位数的质数，它们的差等于$$6$$，且它们平方的个位数字相同，这样的两位质数的组数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算", "课内体系->能力->运算能力", "课内体系->能力->推理论证能力", "竞赛->知识点->数论->整除->整除的概念与基本性质"], "answer_analysis": ["两位数的质数共有$$21$$个，它们的个位数字只有$$1$$，$$3$$，$$7$$，$$9$$； 差等于$$6$$的有：$$1$$和$$7$$；$$3$$和$$9$$；$$13$$和$$7$$，共三组； 平方数的个位数字相同的只有$$3$$和$$7$$，$$1$$和$$9$$这两组． 因此，符合题设条件的个位数字是$$3$$和$$7$$这一组． 故所求质数是$$23$$和$$17$$；$$43$$和$$37$$，$$53$$和$$47$$，$$73$$和$$67$$，共$$4$$组． ", "平方之后个位数字相同的数有：$$1$$和$$9$$，$$2$$和$$8$$，$$3$$和$$7$$，$$4$$和$$6$$作为两位数质数的个位，不能是偶数，两个两位数相差$$6$$个位一定是$$7$$和$$3$$$$17$$和$$23$$，$$37$$和$$43$$，$$47$$和$$53$$，$$67$$和$$73$$，共$$4$$组．
"], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "506", "queId": "832be9de6c9c402ebd43bc3a29babebf", "competition_source_list": ["2000年第17届全国初中数学联赛竞赛第3题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$，$$b$$是不相等的任意正数，又$$x=\\frac{{{b}^{2}}+1}{a}$$，$$y=\\frac{{{a}^{2}}+1}{b}$$，则$$x$$，$$y$$这两个数一定（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "都不大于$$2$$ "}], [{"aoVal": "B", "content": "都不小于$$2$$ "}], [{"aoVal": "C", "content": "至少有一个大于$$2$$ "}], [{"aoVal": "D", "content": "至少有一个小于$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->分式->分式的基本运算", "课内体系->知识点->式->分式->分式化简求值->分式化简求值-直接代入数值求值"], "answer_analysis": ["令$$a=4$$，$$b=5$$，得到$$x=\\frac{13}{2}$$，$$y=\\frac{17}{5}$$，可排除$$\\text{A}$$，$$\\text{D}$$； 令$$a=1$$，$$b=2$$，得到$$x=5$$，$$y=1$$，可排除$$\\text{B}$$． 故只能选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "772", "queId": "8444265bbc284ec99644a4c7b7dc08b4", "competition_source_list": ["初一上学期单元测试《计算》第12题", "2006年第17届希望杯初二竞赛第1试第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$，$$d$$是互不相等的正整数，且$$abcd=441$$，那么$$a+b+c+d$$的值是（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$32$$ "}], [{"aoVal": "C", "content": "$$34$$ "}], [{"aoVal": "D", "content": "$$36$$ ~ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式的加减运算->整式加减的综合", "课内体系->知识点->式->整式的加减->整式有关的概念->整式->整式的定义", "课内体系->能力->运算能力"], "answer_analysis": ["∵$$441=7\\times 7\\times 3\\times 3$$， 又∵$$a$$，$$b$$，$$c$$，$$d$$互为不相等的正整���， ∴$$a$$，$$b$$，$$c$$，$$d$$为$$21$$，$$7$$，$$3$$，$$1$$， ∴$$a+b+c+d=32$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1235", "queId": "8aac4907507fb88401508043c3d70162", "competition_source_list": ["1997年第8届全国希望杯初一竞赛复赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "有四个关于$$x$$的方程： ①$$x-2=-1$$；②$$\\left( x-2 \\right)+\\left( x-1 \\right)=-1+\\left( x-1 \\right)$$；③$$x=0$$；④$$x-2+\\frac{1}{x-1}=-1+\\frac{1}{x-1}$$． 其中同解的两个方程是．", "answer_option_list": [[{"aoVal": "A", "content": "①与② "}], [{"aoVal": "B", "content": "①与③ "}], [{"aoVal": "C", "content": "①与④ "}], [{"aoVal": "D", "content": "②与④ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->分式方程->分式方程的解与解分式方程->分式方程的解"], "answer_analysis": ["方程①的解$$x=1$$，将$$x=1$$代入方程②，方程②成立， ∴$$x=1$$也是方程②的解． 方程①和②是同解方程，而①与③显然不同解； ①的解代入④，④无意义． ∴$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$都不正确，只有$$\\text{A}$$正确． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1008", "queId": "ff8080814d043e7e014d097728971101", "competition_source_list": ["2010年第21届全国希望杯初一竞赛初赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a\\textless{}0$$，在代数式$$\\left\\textbar{} a \\right\\textbar$$，$$-a$$，$${{a}^{2009}}$$，$${{a}^{2010}}$$，$$\\left\\textbar{} -a \\right\\textbar$$，$$\\left( \\frac{{{a}^{2}}}{a}+a \\right)$$，$$\\left( \\frac{{{a}^{2}}}{a}-a \\right)$$中负数的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->正数和负数->正数、负数定义"], "answer_analysis": ["因为$$a\\textless{}0$$， 所以$$\\left\\textbar{} a \\right\\textbar$$，$$-a$$，$${{a}^{2010}}$$，$$\\left\\textbar{} -a \\right\\textbar$$均为正数， 而$$\\left( \\frac{{{a}^{2}}}{a}-a \\right)=a-a=0$$，$$\\left( \\frac{{{a}^{2}}}{a}+a \\right)=a+a=2a$$， 所以只有$${{a}^{2009}}$$，$$\\left( \\frac{{{a}^{2}}}{a}+a \\right)$$为负数． 故负数的个数是$$2$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "203", "queId": "4f824a78bedb4222a9618de68ecfbd32", "competition_source_list": ["2019~2020学年湖北武汉期末", "1997年第8届全国希望杯初一竞赛复赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "有四个关于$$x$$的方程： ①$$x-2=-1$$；②$$\\left( x-2 \\right)+\\left( x-1 \\right)=-1+\\left( x-1 \\right)$$；③$$x=0$$；④$$x-2+\\frac{1}{x-1}=-1+\\frac{1}{x-1}$$． 其中同解的两个方程是．", "answer_option_list": [[{"aoVal": "A", "content": "①与② "}], [{"aoVal": "B", "content": "①与③ "}], [{"aoVal": "C", "content": "①与④ "}], [{"aoVal": "D", "content": "②与④ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->分式方程->分式方程的解与解分式方程->分式方程的解"], "answer_analysis": ["方程①的解$$x=1$$，将$$x=1$$代入方程②，方程②成立， ∴$$x=1$$也是方程②的解． 方程①和②是同解方程，而①与③显然不同解； ①的解代入④，④无意义． ∴$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$都不正确，只有$$\\text{A}$$正确． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1382", "queId": "f2de14725645445b9261447e92eef433", "competition_source_list": ["2008年第19届希望杯初二竞赛第1试第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a$$，$$b$$，$$c$$，$$d$$，$$e$$只能从$$-3$$，$$-2$$，$$-1$$中取值，又$$x=a-b+c-d+e$$，$$y={{a}^{2}}-{{b}^{2}}+{{c}^{2}}-{{d}^{2}}+{{e}^{2}}$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$x$$的最大值比$$y$$的最大值小 "}], [{"aoVal": "B", "content": "$$x$$的最小值比$$y$$的最小值小 "}], [{"aoVal": "C", "content": "$$x$$的最大值比$$y$$的最小值小 "}], [{"aoVal": "D", "content": "$$x$$的最小值比$$y$$的最大值大 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数与实际问题", "课内体系->能力->运算能力", "课内体系->思想->整体思想", "课内体系->方法->整体法"], "answer_analysis": ["$${{x}_{\\max }}=(-1)\\times 3+3\\times 2=3$$， $${{x}_{\\min }}=(-3)\\times 3+1\\times 2=-7$$， $${{y}_{\\max }}=9\\times 3-1\\times 2=25$$， $${{y}_{\\min }}=1\\times 3-9\\times 2=-15$$． ∴$$x$$的最大值比$$y$$的最大值小， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "909", "queId": "6e89eb3797974bbc999c36930fb7d731", "competition_source_list": ["2013年第30届全国全国初中数学联赛竞赛第1题7分", "2013年全国全国初中数学联赛初一竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$4\\sqrt{3+2\\sqrt{2}}-\\sqrt{41+24\\sqrt{2}}=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{2}-1$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算"], "answer_analysis": ["$$4\\sqrt{3+2\\sqrt{2}}-\\sqrt{41+24\\sqrt{2}}$$ $$=4\\sqrt{{{(\\sqrt{2}+1)}^{2}}}-\\sqrt{{{(4\\sqrt{2}+3)}^{2}}}$$ $$=4(\\sqrt{2}+1)-(4\\sqrt{2}+3)$$ $$=1$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1563", "queId": "f91b180aad8d4c5982602e0699b2035e", "competition_source_list": ["2014年第25届全国希望杯初一竞赛复赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$2013$$表示成两个三位数的正整数的平方的差，这两个三位数中较大的一个是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$671$$ "}], [{"aoVal": "B", "content": "$$337$$ "}], [{"aoVal": "C", "content": "$$183$$ "}], [{"aoVal": "D", "content": "$$107$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->二元一次方程（组）->二元一次方程组应用题->二元一次方程组的数字问题"], "answer_analysis": ["设$$x$$为较大的数，$$y$$为较小的数， 则$$2013={{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$$． ∵$$2013=1\\times 3\\times 11\\times 61$$， ∴满足题意的$$x+y$$是三位数或者四位数， 只能是$$\\left { \\begin{matrix}x+y=671 x-y=3 \\end{matrix} \\right.$$① 或$$\\left { \\begin{matrix}x+y=2013 x-y=1 \\end{matrix} \\right.$$② 或$$\\left { \\begin{matrix}x+y=183 x-y=11 \\end{matrix} \\right.$$③， 解①得$$\\left { \\begin{matrix}x=337 y=334 \\end{matrix} \\right.$$， 解②得$$\\left { \\begin{matrix}x=1007 y=1006 \\end{matrix} \\right.$$，（因为$$x$$，$$y$$都是四位数，故舍去） 解③得$$\\left { \\begin{matrix}x=97 y=86 \\end{matrix} \\right.$$，（因为$$x$$，$$y$$都是两位数，故舍去）． 综上，$$\\left { \\begin{matrix}x=337 y=334 \\end{matrix} \\right.$$， ∴这两个三位数中较大的一个是$$337$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "625", "queId": "b0bce0ae6539425da7a3187b448e9adb", "competition_source_list": ["2003年第14届希望杯初二竞赛第1试第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$x+y=-1$$，则$${{x}^{4}}+5{{x}^{3}}y+{{x}^{2}}y+8{{x}^{2}}{{y}^{2}}+x{{y}^{2}}+5x{{y}^{3}}+{{y}^{4}}$$的值等于（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->整式乘除化简求值->整式乘除化简求值-整体代入化简求值", "课内体系->能力->运算能力"], "answer_analysis": ["$${{x}^{4}}+5{{x}^{3}}y+{{x}^{2}}y+8{{x}^{2}}{{y}^{2}}+x{{y}^{2}}+5x{{y}^{3}}+{{y}^{4}}$$ $$=({{x}^{4}}+2{{x}^{2}}{{y}^{2}}+{{y}^{4}})+(5{{x}^{3}}y+5x{{y}^{3}})+6{{x}^{2}}{{y}^{2}}+{{x}^{2}}y+x{{y}^{2}}$$ $$={{({{x}^{2}}+{{y}^{2}})}^{2}}+5xy({{x}^{2}}+{{y}^{2}})+6{{x}^{2}}{{y}^{2}}+{{x}^{2}}y+x{{y}^{2}}$$ $$=({{x}^{2}}+{{y}^{2}}+2xy)({{x}^{2}}+{{y}^{2}}+3xy)+xy(x+y)$$ $$={{(x+y)}^{2}}({{x}^{2}}+{{y}^{2}}+3xy)+xy(x+y)$$ $$={{x}^{2}}+{{y}^{2}}+3xy-xy$$ $$={{x}^{2}}+{{y}^{2}}+2xy$$ $$={{(x+y)}^{2}}=1$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1016", "queId": "ff8080814d4b1928014d4c1128e90552", "competition_source_list": ["2015年第26届全国希望杯初一竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "小明带$$a$$元钱去超市买文具，买铅笔用去了所带钱数的$$\\frac13$$，买橡皮用去余下钱数的$$\\frac14$$，然后他又用剩下的钱数的$$\\frac12$$买了把尺子．这时小明还剩（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac12a$$元 "}], [{"aoVal": "B", "content": "$$\\frac13a$$元 "}], [{"aoVal": "C", "content": "$$\\frac14a$$元 "}], [{"aoVal": "D", "content": "$$\\frac25a$$元 "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式有关的概念->代数式->列代数式", "课内体系->能力->分析和解决问题能力"], "answer_analysis": ["由题意，小明剩余的钱为$$(1-\\frac{1}{3})\\times (1-\\frac{1}{4})\\times (1-\\frac{1}{2})\\times a=\\frac{1}{4}a$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "730", "queId": "40a63f856f5d4accab0c0a8706ed73f3", "competition_source_list": ["1996年第13届全国初中数学联赛竞赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "设正整数$$a$$，$$m$$，$$n$$满足$$\\sqrt{{{a}^{2}}-4\\sqrt{2}}=\\sqrt{m}-\\sqrt{n}$$，则这样的$$a$$，$$m$$，$$n$$的取值．", "answer_option_list": [[{"aoVal": "A", "content": "有$$1$$组 "}], [{"aoVal": "B", "content": "有$$2$$组 "}], [{"aoVal": "C", "content": "多于$$2$$组 "}], [{"aoVal": "D", "content": "不存在 "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的运算->多重二次根式"], "answer_analysis": ["将原式两边平方得到$${{a}^{2}}-4\\sqrt{2}=m+n-2\\sqrt{mn}$$，由于$$a$$，$$m$$，$$n$$都是正整数，所以$${{a}^{2}}=m+n$$，$$mn=8$$．又$$m\\textgreater n$$，仅当$$m=8$$，$$n=1$$时$$a$$为正整数．所以共有$$1$$组解．所以选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "522", "queId": "63179304286544e894a105f4b97bb48d", "competition_source_list": ["2019~2020学年江苏苏州工业园区苏州星海实验中学初二下学期单元测试《分式》第40题2分", "2005年第16届希望杯初二竞赛初赛第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a\\textless{}b\\textless{}c\\textless{}0$$，则$$\\frac{a}{b+c}$$，$$\\frac{b}{c+a}$$，$$\\frac{c}{a+b}$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{a}{b+c}\\textless{}\\frac{b}{c+a}\\textless{}\\frac{c}{a+b}$$ "}], [{"aoVal": "B", "content": "$$\\frac{a}{b+c}\\textless{}\\frac{c}{a+b}\\textless{}\\frac{b}{c+a}$$ "}], [{"aoVal": "C", "content": "$$\\frac{b}{c+a}\\textless{}\\frac{a}{b+c}\\textless{}\\frac{c}{a+b}$$ "}], [{"aoVal": "D", "content": "$$\\frac{c}{a+b}\\textless{}\\frac{b}{c+a}\\textless{}\\frac{a}{b+c}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->不等式->一次不等式组"], "answer_analysis": ["因为$$a\\textless{}b\\textless{}c\\textless{}0$$， 所以$$0\\textless-c\\textless-b\\textless-a$$， 所以$$0\\textless-b-c\\textless-c-a\\textless-a-b$$， 于是$$\\frac{-c}{-a-b}\\textless\\frac{-b}{-c-a}\\textless\\frac{-a}{-b-c}$$， 所以$$\\frac{c}{a+b}\\textless{}\\frac{b}{c+a}\\textless{}\\frac{a}{b+c}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "803", "queId": "88fabe6565814b01ad8464e9ae5e0538", "competition_source_list": ["2006年竞赛第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个正方形纸片，用剪刀沿一条不过任何顶点的直线将其剪成两部分，拿出其中一部分，再沿一条不过任何顶点的直线将其剪成两部分．又从得到的三部分中拿出其中之一，还是沿一条不过任何顶点的直线将其剪成两部分\\ldots\\ldots 如此下去，最后得到了$$34$$个六十二边形和一些多边形纸片，则至少要剪的刀数是（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2004$$ "}], [{"aoVal": "B", "content": "$$2005$$ "}], [{"aoVal": "C", "content": "$$2006$$ "}], [{"aoVal": "D", "content": "$$2007$$ "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "课内体系->能力->运算能力", "课内体系->知识点->三角形->三角形及多边形->多边形->多边形的内角和定理", "课内体系->知识点->三角形->三角形及多边形->多边形->求多边形的内角和"], "answer_analysis": ["根据题意，用剪刀沿不过顶点的直线剪成两部分时，每剪开一次，使得各部分的内角和增加$$360{}^{}\\circ $$．于是，剪过$$k$$次后，可得$$\\left( k+1 \\right)$$个多边形，这些多边形的内角和为$$\\left( k+1 \\right)\\times 360{}^{}\\circ $$． ∵这$$\\left( k+1 \\right)$$个多边形中有$$34$$个六十二边形，它们的内角和为$$34\\times \\left( 62-2 \\right)\\times 180{}^{}\\circ =34\\times 60\\times 180{}^{}\\circ $$，其余多边形有$$\\left( k+1 \\right)-34=k-33$$（个），而这些多边形的内角和不少于$$\\left( k-33 \\right)\\times 180{}^{}\\circ $$， ∴$$\\left( k+1 \\right)\\times 360{}^{}\\circ \\geqslant 34\\times 60\\times 180{}^{}\\circ +\\left( k-33 \\right)\\times 180{}^{}\\circ $$， 解得$$k\\geqslant 2005$$． 当我们按如下的方式剪$$2005$$刀时，可以得到符合条件的结论．先从正方形上剪下$$1$$个三角形，得到$$1$$个三角形和$$1$$个五边形；再在���边形上剪下$$1$$个三角形，得到$$2$$个三角形和$$1$$个六边形\\ldots\\ldots 如此下去，剪了$$58$$刀后，得到$$58$$个三角形和$$1$$个六十二边形．再取出$$33$$个三角形，在每个三角形上剪一刀，又可得到$$33$$个三角形和$$33$$个四边形，对这$$33$$个四边形，按上述正方形的剪法，再各剪$$58$$刀，便得到$$33$$个六十二边形和$$33\\times 58$$个三角形．于是共剪了$$55+33+33\\times 58=2005$$（刀）． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "898", "queId": "5ca6d9542ba84570815b16642e06a663", "competition_source_list": ["2019~2020学年12月浙江杭州上城区北京师范大学附属杭州中学初三上学期周测A卷第8题3分", "2008年竞赛第2题6分", "2016~2017学年9月浙江杭州拱墅区杭州锦绣·育才中学附属学校初三上学期月考第6题3分", "初三上学期单元测试《概率初步》第23题", "2015~2016学年浙江温州乐清市育英寄宿学校初二上学期期中实验a班第4题4分", "2020~2021学年10月浙江杭州下城区杭州市景成实验学校中学部初三上学期月考第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把一枚六个面编号分别为$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$的质地均匀的正方体骰子先后投掷$$2$$次，若两个正面朝上的编号分别为$$m$$，$$n$$，则二次函数$$y={{x}^{2}}+mx+n$$的图象与$$x$$轴有两个不同交点的概率是 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{12}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{17}{36}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率->概率的计算方法->列举法求概率"], "answer_analysis": ["掷骰子有$$6\\times 6=36$$种情况． 根据题意有：$$4n-{{m}^{2}}\\textless{}0$$， 因此满足的点有：$$n=1$$，$$m=3$$，$$4$$，$$5$$，$$6$$， $$n=2$$，$$m=3$$，$$4$$，$$5$$，$$6$$， $$n=3$$，$$m=4$$，$$5$$，$$6$$， $$n=4$$，$$m=5$$，$$6$$， $$n=5$$，$$m=5$$，$$6$$， $$n=6$$，$$m=5$$，$$6$$， 共有$$17$$种， 故概率为：$$17\\div 36=\\frac{17}{36}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1392", "queId": "dc18f91ec8dd4ac18ad8ced1811c815a", "competition_source_list": ["1998年第15届全国初中数学联赛竞赛第6题"], "difficulty": "3", "qtype": "single_choice", "problem": "满足$${{1998}^{2}}+{{m}^{2}}={{1997}^{2}}+{{n}^{2}}(0\\textless{}m\\textless{}n\\textless{}1998)$$的整数对（$$m$$，$$n$$），共有~\\uline{~~~~~~~~~~}~个．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->同余->完全平方数", "课内体系->知识点->数->有理数->有理数运算巧解->完全平方非负性"], "answer_analysis": ["整理得$$n^{2}-m^{2}=3995=5\\times17\\times47$$，$$(n-m)(n+m)=5\\times17\\times47$$， ∵对$$3995$$的任意整数分拆均可得到$$(m,n)$$，$$0\\textless m\\textless n\\textless1998$$， ∴$$\\begin{cases} n-m=5 n+m=17\\times47 \\end{cases}$$或$$\\begin{cases} n-m=17 n+m=5\\times47 \\end{cases}$$或$$ \\begin{cases} n-m=47 n+m=17\\times5 \\end{cases} $$或$$\\left { \\begin{array}{l} {m+n=5\\times17\\times47} {m-n=1} \\end{array}\\right. $$， ∴满足条件的整数对$$(m,n)$$共$$4$$个． 故答案为$$4$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "517", "queId": "31a03717c36246b6a21d65a17a8e64bd", "competition_source_list": ["2018~2019学年福建泉州南安市初二下学期期中第7题4分", "2018~2019学年山东潍坊高密市初二下学期期末第1题3分", "2019年广东惠州惠城区光正实验学校初二竞赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$y=(m+3){{x}^{{{m}^{2}}-8}}$$是正比例函数，则$$m$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$\\pm 3$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->函数->一次函数->正比例函数->根据正比例函数定义求参数"], "answer_analysis": ["∵$$y=(m+3){{x}^{{{m}^{2}}-8}}$$是正比例函数， ∴$${{m}^{2}}-8=1$$且$$m+3\\ne 0$$， 解得$$m=3$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "846", "queId": "c3cd376094ec45b790c5d7c66f80beab", "competition_source_list": ["初一下学期其它", "2006年第17届希望杯初二竞赛第2试第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "以线段$$a=13$$，$$b=13$$，$$c=10$$，$$d=6$$为边作梯形，其中$$a$$，$$c$$为梯形的两底，这样的梯形（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "能作一个 "}], [{"aoVal": "B", "content": "能作两个 "}], [{"aoVal": "C", "content": "能作无数个 "}], [{"aoVal": "D", "content": "一个也不能作 "}]], "knowledge_point_routes": ["课内体系->知识点->四边形->梯形->梯形的性质"], "answer_analysis": ["由于梯形的两底之差$$a-c=3$$，以及梯形的两腰$$b=13$$，$$d=6$$不能构成三角形；故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "411", "queId": "1b9fc74388bb4cfabf154c72482adb77", "competition_source_list": ["2010年第21届希望杯初二竞赛第1试第6题", "北京初二上学期单元测试《二次根式的运算》第22题"], "difficulty": "3", "qtype": "single_choice", "problem": "设$$p=\\sqrt[3]{7a+1}+\\sqrt[3]{7b+1}+\\sqrt[3]{7c+1}+\\sqrt[3]{7d+1}$$，其中$$a$$、$$b$$、$$c$$、$$d$$是正实数，并且$$a+b+c+d=1$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$p\\textgreater5$$ "}], [{"aoVal": "B", "content": "$$p\\textless{}5$$ "}], [{"aoVal": "C", "content": "$$p\\textless{}4$$ "}], [{"aoVal": "D", "content": "$$p=5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->和与差的立方公式", "课内体系->知识点->式->因式分解->公式法->利用立方和与立方差公式因式分解", "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质", "课内体系->能力->推理论证能力", "课内体系->能力->运算能力"], "answer_analysis": ["∵$$a$$、$$b$$、$$c$$、$$d$$是正实数，且$$a+b+c+d=1$$， ∴$$0\\textless{}a\\textless{}1$$， ∴$$a\\textgreater{{a}^{2}}\\textgreater{{a}^{3}}$$， ∴$$7a+1=a+3a+3a+1\\textgreater{{a}^{3}}+3{{a}^{2}}+3a+1={{(a+1)}^{3}}$$， ∴$$\\sqrt[3]{7a+1}\\textgreater\\sqrt[3]{{{(a+1)}^{3}}}=a+1$$， 同理$$\\sqrt[3]{7b+1}\\textgreater b+1\\sqrt[3]{7c+1}\\textgreater c+1\\sqrt[3]{7d+1}\\textgreater d+1$$， ∴$$p=\\sqrt[3]{7a+1}+\\sqrt[3]{7b+1}+\\sqrt[3]{7c+1}+\\sqrt[3]{7d+1}$$， $$\\textgreater a+1+b+1+c+1+d+1$$， $$=a+b+c+d+4=5$$， ∴$$p\\textgreater5$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "579", "queId": "512d4835e4e547f393e9267ad49d2467", "competition_source_list": ["2007年第18届希望杯初二竞赛第1试第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "若三角形三边的长均能使代数式$${{x}^{2}}-9x+18$$的值为$$0$$，则此三角形的周长是．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$或$$18$$ "}], [{"aoVal": "B", "content": "$$12$$或$$15$$ "}], [{"aoVal": "C", "content": "$$9$$或$$15$$或$$18$$ "}], [{"aoVal": "D", "content": "$$9$$或$$12$$或$$15$$或$$18$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["由$${{x}^{2}}-9x+18=\\left( x-3 \\right)\\left( x-6 \\right)$$知，使代数式$${{x}^{2}}-9x+18$$的值为$$0$$的$$x$$的值是$$3$$或$$6$$，于是三角形三边的长有四种可能情况： ①三边长分别是$$6$$，$$6$$，$$6$$，则三角形的周长为$$18$$； ②三边长分别是$$6$$，$$6$$，$$3$$，则三角形的周长为$$15$$； ③三边长分别是$$3$$，$$3$$，$$3$$，则三角形的周长为$$9$$； ④三边长分别是$$6$$，$$3$$，$$3$$，则不能构成三角形． 综上知，三角形的周长是$$9$$，$$15$$或$$18$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1283", "queId": "8aac50a74e724b3f014e8329d9403c54", "competition_source_list": ["1996年第7届全国希望杯初一竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\alpha $$，$$\\beta $$都是钝角，甲，乙，丙，丁计算$$\\frac{1}{6}(\\alpha +\\beta )$$的结果依次为$$50{}^{}\\circ $$，$$26{}^{}\\circ $$，$$72{}^{}\\circ $$，$$90{}^{}\\circ $$，其中确有正确的结果，那么算得结果正确者是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "丙 "}], [{"aoVal": "D", "content": "丁 "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->角->角的定义和分类->角的分类"], "answer_analysis": ["∵$$90{}^{}\\circ \\textless{}\\alpha \\textless{}180{}^{}\\circ $$，$$90{}^{}\\circ \\textless{}\\beta \\textless{}180{}^{}\\circ $$，∴$$180{}^{}\\circ \\textless{}\\alpha +\\beta \\textless{}360{}^{}\\circ $$， ∴$$30{}^{}\\circ \\textless{}\\frac{1}{6}(\\alpha +\\beta )\\textless{}60{}^{}\\circ $$，可见$$\\frac{1}{6}(\\alpha +\\beta )$$应是$$50{}^{}\\circ $$． 故甲计算正确． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "754", "queId": "68e04fc6f84541f2a01b0919a0fab3c1", "competition_source_list": ["1978年Math League竞赛(8年级[.org])第15题", "1978年竞赛第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "1978年 Math League竞赛(8年级[.org]) Of the numbers $$2$$， $$3$$， $$4$$， and $$5$$， which is (are) the only one(s) which satisfy the inequality $$3x-1\\textless{}11$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$2$$，$$3$$和$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "没有选项 "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Algebra and Sequences->Equations->Inequality", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->一元一次不等式的整数解", "课内体系->知识点->方程与不等式->不等式（组）->不等式组应用->不等式组的其他实际问题"], "answer_analysis": ["The inequality $$3x -1\\textless{} 11$$ is satisfied by only $$2$$ \\& $$3$$， Choice $$\\text{A}$$ is incomplete， choice $$\\text{B}$$ includes $$4$$， \\& choice $$\\text{C}$$ is incorrect． 在数$$2$$，$$3$$，$$4$$，$$5$$中，哪一个是（唯一）满足不等式$$3x-1\\textless11$$的． $$\\text{A}$$．$$2$$ ~ ~ ~ $$\\text{B}$$．$$2$$，$$3$$和$$4$$ ~ ~ ~ $$\\text{C}$$．$$5$$ ~ ~ ~ $$\\text{D}$$．没有选项 满足不等式$$3x-1\\textless{}11$$只有$$2$$和$$3$$．选项$$\\text{A}$$不完整，选项$$\\text{B}$$包含$$4$$，选项$$\\text{C}$$不正确． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "941", "queId": "8542f057c0ba4bddaa314204fe277675", "competition_source_list": ["1992年第3届希望杯初二竞赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$x=6$$，$$y=8$$时，$${{x}^{6}}+{{y}^{6}}+2{{x}^{4}}{{y}^{2}}+2{{x}^{2}}{{y}^{4}}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1200000-254000$$ "}], [{"aoVal": "B", "content": "$$1020000-250400$$ "}], [{"aoVal": "C", "content": "$$1200000-250400$$ "}], [{"aoVal": "D", "content": "$$1020000-254000$$ "}], [{"aoVal": "E", "content": "答案请写在答题纸上 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->因式分解->因式分解：添项、拆项"], "answer_analysis": ["答案请写在答题纸上 "], "answer_value": "E"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "254", "queId": "1db540af0d6b4897b4fbfd2775b43d04", "competition_source_list": ["2002年第13届希望杯初一竞赛第1试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "某品牌的$$\\text{VCD}$$机成本价是每台$$500$$元，$$3$$月份的销售价为每台$$625$$元．经市场预测，该商品销售价在$$4$$月份将降低$$20 \\%$$．而后在$$5$$月份再提高$$8 \\%$$．那么在$$5$$月份销售该品牌的$$\\text{VCD}$$机可获利．", "answer_option_list": [[{"aoVal": "A", "content": "$$25 \\%$$ "}], [{"aoVal": "B", "content": "$$20 \\%$$ "}], [{"aoVal": "C", "content": "$$8 \\%$$ "}], [{"aoVal": "D", "content": "$$12 \\%$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数与实际问题"], "answer_analysis": ["由题意可得，$$5$$月份销售该品牌的$$\\text{VCD}$$机可获利 $$=\\frac{625 \\times(1-20 \\%) \\times(1+8 \\%)-500}{500}$$ $$=\\frac{625 \\times 0.8 \\times 1.08}{500}-1$$ $$=8 \\%$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "561", "queId": "6c66809140124899b294ed587df2aa26", "competition_source_list": ["初一下学期单元测试《质数与合数》第2题", "2005年第16届希望杯初二竞赛初赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个两位数的个位数字和十位数字变换位置后，所得的数比原来的数大$$9$$，这样的两位数中，质数有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$个 "}], [{"aoVal": "B", "content": "$$3$$个 "}], [{"aoVal": "C", "content": "$$5$$个 "}], [{"aoVal": "D", "content": "$$6$$个~ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数基础运算->有理数减法->有理数加减混合运算"], "answer_analysis": ["满足条件的两位数两个数字之差为$$1$$， 这样的两位数有$$12$$、$$23$$、$$34$$、$$45$$、$$56$$、$$67$$、$$78$$、$$89$$， 其中质数有$$23$$，$$67$$、$$89$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "554", "queId": "365775af5d074b88b847035614d22791", "competition_source_list": ["1996年第7届希望杯初二竞赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\frac{1}{a}:\\frac{1}{b}:\\frac{1}{c}=2:3:4$$，则$$a:b:c$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$4:3:2$$ "}], [{"aoVal": "B", "content": "$$6:4:3$$ "}], [{"aoVal": "C", "content": "$$3:4:2$$ "}], [{"aoVal": "D", "content": "$$3:4:6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->分式->倒数通分"], "answer_analysis": ["∵$$\\frac{1}{a}:\\frac{1}{b}:\\frac{1}{c}=2:3:4$$， ∴$$a:b:c=\\frac{1}{2}:\\frac{1}{3}:\\frac{1}{4}=6:4:3$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "813", "queId": "5bde5e2a120d40599cbf8d0881a1cac0", "competition_source_list": ["2009年第20届希望杯初二竞赛第1试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "若分式$$\\frac{\\textbar x\\textbar-2}{3x-2}$$的值是负数，则$$x$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2}{3}\\textless{}x\\textless{}2$$ "}], [{"aoVal": "B", "content": "$$x\\textgreater\\frac{2}{3}$$或$$x\\textless{}-2$$ "}], [{"aoVal": "C", "content": "$$-2\\textless{}x\\textless{}2$$且$$x\\ne \\frac{2}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{3}\\textless{}x\\textless{}2$$或$$x\\textless{}-2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->分式->分式的基本运算"], "answer_analysis": ["因为分式$$\\frac{\\textbar x\\textbar-2}{3x-2}$$的值为负，所以$$\\textbar x\\textbar-2$$和$$3x-2$$异号． 当$$3x-2\\textgreater0$$，即$$x\\textgreater\\frac{2}{3}$$时，应当有 $$\\left\\textbar{} x \\right\\textbar-2\\textless{}0$$，$$\\left\\textbar{} x \\right\\textbar\\textless{}2$$，解得$$-2\\textless{}x\\textless{} 2$$， 所以$$\\frac{2}{3}\\textless{}x\\textless{}2$$ ． 当$$3x-2\\textless{}0$$，即$$x\\textless{}\\frac{2}{3}$$时，应当有 $$\\left\\textbar{} x \\right\\textbar-2\\textgreater0$$，$$\\left\\textbar{} x \\right\\textbar\\textgreater2$$，解得$$x\\textgreater2$$或$$x\\textless{}-2$$， 所以$$x\\textless{}-2$$． 综上可知，$$x$$的取值范围是$$\\frac{2}{3}\\textless{}x\\textless{}2$$或$$x\\textless{}-2$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1050", "queId": "ff8080814d7978b9014d86d752b6258f", "competition_source_list": ["1991年第2届全国希望杯初一竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "以下的运算的结果中，最大的一个数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-13579)+0.2468$$ "}], [{"aoVal": "B", "content": "$$(-13579)+\\frac{1}{2468}$$ "}], [{"aoVal": "C", "content": "$$(-13579)\\times \\frac{1}{2468}$$ "}], [{"aoVal": "D", "content": "$$(-13579)\\div \\frac{1}{2468}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->含乘方的有理数混合运算"], "answer_analysis": ["运算结果对负数来说绝对值越小其值越大． 易见$$(-13579)\\times \\frac{1}{2468}$$的绝对值最小，所以其值最大． 选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1376", "queId": "aa2ed49cdf9f4f0684264b1d19d77189", "competition_source_list": ["1997年第8届希望杯初二竞赛第2试第6题", "初一下学期其它"], "difficulty": "2", "qtype": "single_choice", "problem": "已知：$${{m}^{2}}+m-1=0$$，那么代数式$${{m}^{3}}+2{{m}^{2}}-1997$$的值是（ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1997$$ "}], [{"aoVal": "B", "content": "$$-1997$$ "}], [{"aoVal": "C", "content": "$$1996$$ "}], [{"aoVal": "D", "content": "$$-1996$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->整式乘除化简求值->整式乘除化简求值-整体代入化简求值"], "answer_analysis": ["因为$${{m}^{2}}+m-1=0$$， 所以$${{m}^{2}}+m=1$$， 所以$${{m}^{3}}+2{{m}^{2}}-1997$$ $$=m({{m}^{2}}+m)+{{m}^{2}}-1997$$ $$=m+{{m}^{2}}-1997$$ $$=1-1997$$ $$=-1996$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1042", "queId": "ff8080814d7978b9014d866912cc2431", "competition_source_list": ["1990年第1届全国希望杯初一竞赛初赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "杯子中有大半杯水，第二天较第一天减少了$$10 \\%$$，第三天又较第二天增加了$$10 \\%$$，那么，第三天杯中的水量与第一天杯中的水量相比的结果是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "一样多 "}], [{"aoVal": "B", "content": "多了 "}], [{"aoVal": "C", "content": "少了 "}], [{"aoVal": "D", "content": "多少都可能 "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式有关的概念->代数式->列代数式", "课内体系->能力->分析和解决问题能力"], "answer_analysis": ["设杯中原有水量为$$a$$，依题意可得， 第二天杯中水量为$$a\\times (1-10 \\%)=0.9a$$； 第三天杯中水量为$$(0.9a)\\times (1+10 \\%)=0.9\\times 1.1\\times a$$； 第三天杯中水量与第一天杯中水量之比为$$\\frac{0.9\\times 1.1\\times a}{a}=0.99\\textless1$$． 所以第三天杯中水量比第一天杯中水量少了，选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "22", "queId": "017c542597aa4c02bcb3d316c00c560b", "competition_source_list": ["2015年第32届全国全国初中数学联赛竞赛B卷第4题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知实数$$x$$，$$y$$满足关系式$${{x}^{2}}+xy+{{y}^{2}}=3$$，则$${{(x-y)}^{2}}$$的最大值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["课内体系->方法->换元法", "课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->由一元二次方程根的情况确定参数", "课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->一元二次方程根的判别式"], "answer_analysis": ["设$$x-y=t$$，则$$x=y+t$$， 代入$${{x}^{2}}+xy+{{y}^{2}}=3$$， 得$${{(y+t)}^{2}}+(y+t)y+{{y}^{2}}=3$$， 整理，得$$3{{y}^{2}}+3ty+{{t}^{2}}-3=0$$． 由判别式$$\\Delta ={{(3t)}^{2}}-12({{t}^{2}}-3)\\geqslant 0$$， 得$$-2\\sqrt{3}\\leqslant t\\leqslant 2\\sqrt{3}$$， ∴$${{(x-y)}^{2}}={{t}^{2}}\\leqslant 12$$， 即$${{(x-y)}^{2}}$$的最大值为$$12$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "667", "queId": "63da99c6eb2d4a83b5cbb7a2b4c86083", "competition_source_list": ["2016~2017学年3月湖北武汉武昌区武汉初级中学初二下学期月考第10题3分", "2001年第18届全国初中数学联赛竞赛第1题", "2016~2017学年9月湖北武汉武昌区武汉初级中学初二上学期月考第10题3分", "2019~2020学年12月四川资阳雁江区资阳市雁江区第二中学初三上学期周测D卷第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a$$、$$b$$、$$c$$、为有理数，且等式$$a+b\\sqrt{2}+c\\sqrt{3}=\\sqrt{5+2\\sqrt{6}}$$成立，则$$2a+999b+1001c$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1999$$ "}], [{"aoVal": "B", "content": "$$2000$$ "}], [{"aoVal": "C", "content": "$$2001$$ "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->等式与方程->等式->等式的性质->等式性质1", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算"], "answer_analysis": ["\\textbf{（知识点：根式恒等变形）} 本题需要比较等式两边的各项，利用有理数部分等于理数部分，无理数部分等于无理数部分来求$$a$$、$$b$$、$$c$$的值，由于： $$5+2\\sqrt{6}={{\\left( \\sqrt{3}+\\sqrt{2} \\right)}^{2}}$$． 所以：$$a+b\\sqrt{2}+c\\sqrt{3}=\\sqrt{2}+\\sqrt{3}$$． 则$$a=0$$，$$b=1$$，$$c=1$$，∴$$2a+999b+1001c=2000$$． 所以选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1032", "queId": "8dd73ac27738473d864762fb5885be77", "competition_source_list": ["2014年第31届全国全国初中数学联赛竞赛第4题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$6$$张不同的卡片上分别写有数字$$2$$，$$2$$，$$4$$，$$4$$，$$6$$，$$6$$，从中取出$$3$$张，则这$$3$$张卡片上所写的数字可以作为三角形的三边长的概率是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{4}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率->概率的计算方法->列举法求概率"], "answer_analysis": ["若取出的$$3$$张卡片上的数字互不相同，有$$2\\times 2\\times 2=8$$种取法； 若取出的$$3$$张卡片上的数字有相同的，有$$3\\times 4=12$$种取法． 所以，从$$6$$张不同的卡片中取出$$3$$张，共有$$812=20$$种取法． 要使得三个数字可以构成三角形的三边长， 只可能是：$$(2,4,4)$$，$$(4,4,6)$$，$$(2,6,6)$$，$$(4,6,6)$$， 由于不同的卡片上所写数字有重复， 所以，取出的$$3$$张卡片上所写的数字可以作为三角形的三边长的情况共有$$4\\times 2=8$$种． 因此，所求概率为$$\\frac{8}{20}=\\frac{2}{5}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1036", "queId": "ff8080814d7978b9014d86555ec223bc", "competition_source_list": ["1990年第1届全国希望杯初一竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面说法中不正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "有最小的自然数 "}], [{"aoVal": "B", "content": "没有最小的正有理数 "}], [{"aoVal": "C", "content": "没有最大的负整数 "}], [{"aoVal": "D", "content": "没有最大的非负数 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类"], "answer_analysis": ["$$0$$是最小的自然数，$$\\text{A}$$正确． 可以找到正有理数的无限序列$$1$$，$$\\frac{1}{2}$$，$$\\frac{1}{3}$$，$$\\cdots $$，$$\\frac{1}{n}$$，$$\\cdots $$，没有最小的正有理数，$$\\text{B}$$正确． ``没有最大的负整数''的说法不正确，最大的负整数为$$-1$$． 写出自然数列，$$0$$，$$1$$，$$2$$，$$3$$，$$\\ldots $$，$$n$$，$$\\ldots $$，易知无最大非负数，$$\\text{D}$$正确． 所以不正确的说法应选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "188", "queId": "1063e9c339d64297932f5801e0197437", "competition_source_list": ["2013年第24届全国希望杯初二竞赛复赛第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$3$$个正整数$$a$$，$$b$$，$$c$$，并且$$a\\textgreater b\\textgreater c$$．从中任取$$2$$个，有$$3$$种不同的取法．将每一种取法取出的$$2$$个数分别作和及作差，得到如下$$6$$个数：$$42$$，$$45$$，$$64$$，$$87$$，$$109$$，$$151$$．则$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$12532$$ "}], [{"aoVal": "B", "content": "$$12533$$ "}], [{"aoVal": "C", "content": "$$12534$$ "}], [{"aoVal": "D", "content": "$$12535$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->思想->整体思想", "课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式"], "answer_analysis": ["由题意，可得 $${{(a-b)}^{2}}+{{(a+b)}^{2}}+{{(b-c)}^{2}}+{{(b+c)}^{2}}+{{(a-c)}^{2}}+{{(a+c)}^{2}}$$ $$={{a}^{2}}-2ab+{{b}^{2}}+{{a}^{2}}+2ab+{{b}^{2}}+{{b}^{2}}-2bc+{{c}^{2}}+{{b}^{2}}+2bc+{{c}^{2}}+{{a}^{2}}-2ac+{{c}^{2}}+{{a}^{2}}+2ac+{{c}^{2}}$$ $$=4({{a}^{2}}+{{b}^{2}}+{{c}^{2}})$$， 所以$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$ $$=\\frac{{{42}^{2}}+{{45}^{2}}+{{64}^{2}}+{{87}^{2}}+{{109}^{2}}+{{151}^{2}}}{4}$$ $$=12534$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "558", "queId": "7ec1cea38fea4928af04e0616d5d1bc4", "competition_source_list": ["1994年第5届希望杯初二竞赛第5题"], "difficulty": "0", "qtype": "single_choice", "problem": "有如下命题： ①负数没有立方根 ②一个实数的立方根不是正数就是负数 ③一个正数或负数的立方根和这个数同号，$$0$$的立方根是0 ④如果一个数的立方根是这个数本身，那么这个数必是$$1$$或0 其中错误的是．", "answer_option_list": [[{"aoVal": "A", "content": "①②③ "}], [{"aoVal": "B", "content": "①②④ "}], [{"aoVal": "C", "content": "②③④ "}], [{"aoVal": "D", "content": "①③④ "}]], "knowledge_point_routes": ["课内体系->能力->抽象概括能力", "课内体系->知识点->数->实数->立方根->开立方"], "answer_analysis": ["负数有立方根，$$0$$的立方根是$$0$$，又$$-1$$的立方根也是$$-1$$，所以错误命题是①②④． 应选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "39", "queId": "02fc657a91884990bf00b62b72d4c0f0", "competition_source_list": ["2006年第17届希望杯初二竞赛第1试第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "某校初一，初二年级的学生人数相同，初三年级的学生人数是初二年级学生人数的$$\\frac{4}{5}$$．已知初一年级的男生人数与初二年级的女生人数相同，初三年级男生人数占三个年级男生人数的$$\\frac{1}{4}$$，那么三个年级女生人数占三个年级学生人数的．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{9}{19}$$ "}], [{"aoVal": "B", "content": "$$\\frac{10}{19}$$ "}], [{"aoVal": "C", "content": "$$\\frac{11}{21}$$ "}], [{"aoVal": "D", "content": "$$\\frac{10}{21}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的和差倍分", "竞赛->知识点->方程与不等式->方程应用"], "answer_analysis": ["设初一年级男生有$$m$$人，女生有$$n$$人，则初二年级男生有$$n$$人，女生有$$m$$人；初三年级的学生共有$$\\frac{4}{5}(m+n)$$人；三个年级共有学生$$\\frac{14}{5}(m+n)$$人． 设初三年级男生有$$x$$人，则$$x=\\frac{1}{4}(m+n+x)$$， 解得$$x=\\frac{1}{3}(m+n)$$， 所以，初三年级的女生人数为$$\\frac{4}{5}(m+n)-\\frac{1}{3}(m+n)=\\frac{7}{15}(m+n)$$， 故全校女生人数为$$\\frac{22}{15}(m+n)$$人，占三个年级学生人数的$$\\frac{11}{21}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "736", "queId": "71d7e14d7b404b548f7d9da3a587a498", "competition_source_list": ["2013年山东青岛黄岛区山东省青岛第九中学自主招生", "2012年竞赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果正比例函数$$y=ax\\left( a\\ne 0 \\right)$$与反比例函数$$y=\\frac{b}{x}(b\\ne 0)$$的图象有两个交点，其中一个交点的坐标为$$(-3,-2)$$，那么另一个交点的坐标为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 2,3 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 3,-2 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( -2,3 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( 3,2 \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->反比例函数->反比例函数与一次函数综合"], "answer_analysis": ["由题设知，$$-2=a\\cdot (-3)$$，$$(-3)\\cdot (-2)=b$$， 所以$$a=\\frac{2}{3}$$，$$b=6$$， 解方程组$$\\begin{cases}y=\\dfrac{2}{3}x y=\\dfrac{6}{x} \\end{cases}$$， 得$$\\begin{cases}x=-3 y=-2 \\end{cases}$$， 或$$\\begin{cases}x=3 y=2 \\end{cases}$$， 所以另一个交点的坐标为$$(3,2)$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "218", "queId": "41be413f8c6848f6a0f91ce2ff3b094d", "competition_source_list": ["2020年第22届浙江宁波余姚市余姚市实验学校初三竞赛（实验杯）第11题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知不相等的实数$$a$$，$$b$$满足$$a-b^{2}=b-a^{2}=x$$，则$$x$$．", "answer_option_list": [[{"aoVal": "A", "content": "有最大值，又有最小值 "}], [{"aoVal": "B", "content": "有最大值，没有最小值 "}], [{"aoVal": "C", "content": "有最小值，没有最大值 "}], [{"aoVal": "D", "content": "没有最大值，也没有最小值 "}]], "knowledge_point_routes": ["课内体系->能力->分析和解决问题能力", "课内体系->知识点->式->整式的加减->整式加减化简求值->整式加减化简求值-条件化简求值", "课内体系->知识点->式->整式的乘除->乘法公式->配方求最值"], "answer_analysis": ["∵$$a-b^{2}=b-a^{2}$$， ∴$$a^{2}-b^{2}=b-a$$， $$(a+b)(a-b)=b-a$$ $$(a+b+1)(a-b)=0$$， 又∵$$a\\ne b$$， ∴$$a+b+1=0$$， ∴$$a=-b-1$$， ∴$$x=a-b^{2}$$ $$=-b-1-b^{2}$$ $$=-\\left(b+\\frac{1}{2} \\right)^{2}-\\frac{3}{4}$$， 当$$b=-\\frac{1}{2}$$时，$$x$$取最大值$$-\\frac{3}{4}$$， 但此时$$a=-\\left(-\\frac{1}{2}\\right)-1=-\\frac{1}{2}=b$$， 故最大值取不到，也没有最小值， 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "981", "queId": "cd6a8930b07d4ee3903ca02f6756be9b", "competition_source_list": ["2014年第31届全国全国初中数学联赛竞赛第2题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知非负实数$$x$$，$$y$$，$$z$$满足$$x+y+z=2$$，则$$t=2xy+yz+2zx$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{4}{7}$$ "}], [{"aoVal": "B", "content": "$$\\frac{8}{7}$$ "}], [{"aoVal": "C", "content": "$$\\frac{16}{7}$$ "}], [{"aoVal": "D", "content": "$$\\frac{32}{7}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->因式分解->因式分解：添项、拆项", "课内体系->知识点->方程与不等式->不等式（组）->其它不等式->均值不等式"], "answer_analysis": ["不要搜题 "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1446", "queId": "af4d52b275d042c58468426cf5d27fe5", "competition_source_list": ["2010年全美数学竞赛（AMC）竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "某三角形三边长为连续正整数．最短边占周长$$30 \\%$$．则最长边长度是？．", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}], [{"aoVal": "E", "content": "$$11$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Geometry->Shapes with Straight Sides->Geometric Theorem", "课内体系->知识点->三角形"], "answer_analysis": ["某三角形三边长为连续正整数．最短边占周长$$30 \\%$$．则最长边长度是？ 设$$n$$，$$n+1$$，和$$n+2$$是三角形三边的长度．那么三角形的周长是$$n+(n+1)+(n+2)=3n+3$$利用最小边的长度是周长的$$30 \\%$$这个事实，可以得出这样的结论： $$n=0.3(3n+3)\\Rightarrow 0.9n+0.9\\Rightarrow 0.1n=0.9\\Rightarrow n=9$$．最长边是$$n+2=11$$． 故选$$\\text{E}$$． 由于最短边的长度是整数，等于周界的$$\\frac{3}{10}$$，因此周界必须是$$10$$的倍数．将前面的两个整数添加到每个答案选择中，我们可以看到$$11+10+9=30$$． 故选$$\\text{E}$$． Let $$n$$, $$n+1$$, and $$n+2$$ be the lengths of the sides of the triangle. Then the perimeter of the triangle is $$n+\\left( n+1 \\right)+\\left( n+2 \\right)=3n+3$$. Usin g the fact that the length of the smallest side is $$30 \\%$$ of the perimeter, it follows that: $$n=0.3\\left( 3n +3 \\right)\\Rightarrow n=0.9n+0.9\\Rightarrow 0.1n=0.9\\Rightarrow n=9$$. The longest side is then $$n+2-11$$. Thus, answer choice $$\\left( \\text{E} \\right)11$$ is correct. Since the length of the shortest side is a whole number and is equal to $$\\frac{3}{10}$$ of the perimeter, it follows that the perimeter must be a multiple of $$10$$. Adding the two previous integers to each answer choice, we see that $$11+10+9=30$$. Thus, answer choice $$\\left( \\text{E} \\right)11$$ is correct. "], "answer_value": "E"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "135", "queId": "0f8dd69cfd9a4c8f89b222f12697ce7a", "competition_source_list": ["2019年第2届浙江宁波余姚市余姚市实验学校初二竞赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$n$$为正整数，关于$$x$$的不等式组$$\\begin{cases}\\dfrac{5x+5}{6}-x\\geqslant -\\dfrac{5}{2} \\dfrac{2x+5}{3}\\textless{}nx \\end{cases}$$的整数解的个数不可能为．", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->一元一次不等式组的整数解", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->解一元一次不等式组"], "answer_analysis": ["解不等式组$$\\begin{cases}\\dfrac{5x+5}{6}-x\\geqslant -\\dfrac{5}{2}① \\dfrac{2x+5}{3}\\textless{}nx② \\end{cases}$$， 由①得，$$5x+5-6x\\geqslant -15$$， $$5x-6x\\geqslant -15-5$$， $$-x\\geqslant -20$$， $$x\\leqslant 20$$， 由②得$$2x+5\\textless{}3nx$$， $$2x-3nx\\textless{}-5$$， $$\\left( 2-3n \\right)x\\textless{}-5$$， ∵$$n$$为正整数， ∴$$n\\geqslant 1$$， ∴$$2-3n\\leqslant -1$$， ∴解②，得$$x\\textgreater\\frac{5}{3n-2}$$， ∴原不等式组的解集为$$\\frac{5}{3n-2}\\textless{}x\\leqslant 20$$， ∴当$$n=1$$时，$$5\\textless{}x\\leqslant 20$$，有$$15$$个整数解， 当$$n=2$$时，$$\\frac{5}{4}\\textless{}x\\leqslant 20$$，有$$19$$个整数解， 当$$n=3$$时，$$\\frac{5}{7}\\textless{}x\\leqslant 20$$，有$$20$$个整数解， ∴原不等式组不可能有$$17$$个整数解． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1482", "queId": "eee1d751079941669efdfd8689ebca2c", "competition_source_list": ["2011年第28届全国全国初中数学联赛竞赛第6题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$\\frac{1}{x}+\\frac{1}{y+z}=\\frac{1}{2}$$，$$\\frac{1}{y}+\\frac{1}{z+x}=\\frac{1}{3}$$，$$\\frac{1}{z}+\\frac{1}{x+y}=\\frac{1}{4}$$，则$$\\frac{2}{x}+\\frac{3}{y}+\\frac{4}{z}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式化简求值->分式化简求值-倒数法化简求值"], "answer_analysis": ["法1：由通分然后取倒数得， $$\\frac{xy+xz}{x+y+z}=2$$，$$\\frac{xy+yz}{x+y+z}=3$$，$$\\frac{zx+yz}{x+y+z}=4$$， $$\\therefore \\frac{2}{x}+\\frac{3}{y}+\\frac{4}{z}=\\frac{xy+zx}{x+y+z}\\cdot \\frac{1}{x}+\\frac{xy+yz}{x+y+z}\\cdot \\frac{1}{y}+\\frac{zx+zy}{x+y+z}\\cdot \\frac{1}{z}$$ $$=\\frac{x+y+x+z+y+z}{x+y+z}=2$$． 法2：由已知等式得$$\\frac{xy+zx}{x+y+z}=2$$，$$\\frac{yz+xy}{x+y+z}=3$$，$$\\frac{zx+yz}{x+y+z}=4$$， 所以$$\\frac{xy+yz+zx}{x+y+z}=\\frac{9}{2}$$． 于是，$$\\frac{yz}{x+y+z}=\\frac{5}{2}$$，$$\\frac{zx}{x+y+z}=\\frac{3}{2}$$，$$\\frac{xy}{x+y+z}=\\frac{1}{2}$$． 所以$$\\frac{y}{x}=\\frac{5}{3}$$，$$\\frac{z}{y}=3$$，$$\\frac{y}{x}=\\frac{5}{3}$$， 即$$z=3y=5x$$． 代入$$\\frac{1}{x}+\\frac{1}{y+z}=\\frac{1}{2}$$，得$$\\frac{1}{x}+\\frac{1}{\\dfrac{5}{3}x+5x}=\\frac{1}{2}$$， 解得$$x=\\frac{23}{10}$$． 所以$$\\frac{2}{x}+\\frac{3}{y}+\\frac{4}{z}=\\frac{2}{x}+\\frac{3}{\\dfrac{5}{3}x}+\\dfrac{4}{5x}=\\frac{23}{5x}=2$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1439", "queId": "e56a2ff4982a40c1afd89f15b2641e54", "competition_source_list": ["初一下学期单元测试《不等式与不等式组》一元一次不等式及其应用第28题", "2018~2019学年甘肃兰州城关区兰州市第三十五中学初二下学期期中第12题4分", "2011年第16届华杯赛初一竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "4.设$$a$$，$$b$$是常数，不等式$$\\frac{x}{a}+\\frac{1}{b}\\textgreater0$$的解集为$$x\\textless{}\\frac{1}{5}$$，则关于$$x$$的不等式$$bx-a\\textgreater0$$的解集是．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textgreater\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$x\\textless{}-\\frac{1}{5}$$ "}], [{"aoVal": "C", "content": "$$x\\textgreater-\\frac{1}{5}$$ "}], [{"aoVal": "D", "content": "$$x\\textless{}\\frac{1}{5}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->不等式（组）->含参不等式->由一个已知不等式的解集求另一个不等式的解集"], "answer_analysis": ["不等式$$ \\frac{x}{a}+ \\frac{1}{b}\\textgreater0 $$的解集为$$\\frac{x}{a}\\textgreater- \\frac{1}{b}$$， $$x\\textless{}- \\frac{a}{b}$$， $$x\\textless{} \\frac{1}{5}$$， 所以$$ \\frac{a}{b}=- \\frac{1}{5} $$且$$a\\textless{}0$$，$$b\\textgreater0$$， 所以不等式$$bx-a\\textgreater0$$的解集为$$bx\\textgreater a$$， $$x\\textgreater{} \\frac{a}{b}$$， $$x\\textgreater- \\frac{1}{5}$$， 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "91", "queId": "13480b5ed01849b995575ee532c2c60b", "competition_source_list": ["1991年第2届希望杯初二竞赛第7题"], "difficulty": "0", "qtype": "single_choice", "problem": "两条相交直线所成的各角中（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "必有一个钝角 "}], [{"aoVal": "B", "content": "必有一个锐角 "}], [{"aoVal": "C", "content": "必有一个不是钝角 "}], [{"aoVal": "D", "content": "必有两个锐角 "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->角->角的定义和分类->角的分类"], "answer_analysis": ["相交时两条直线可能互相垂直（垂直是一种特殊的相交）， ∴$$\\text{A}$$、$$\\text{B}$$、$$\\text{D}$$均在垂直的情况下不成立． ", "$$\\text{A}$$．两个$$90{}^\\circ $$，故$$\\text{A}$$错误；
\n$$\\text{B}$$．两个$$90{}^\\circ $$，故$$\\text{B}$$错误；
\n$$\\text{C}$$．一钝角、一锐角或两个直角，故$$\\text{C}$$正确；
\n$$\\text{D}$$．和小于$$180{}^\\circ $$，故$$\\text{D}$$错误．
\n故选$$\\text{C}$$．
"], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "585", "queId": "55b88f51d95d4b63958859c6ad6aaf45", "competition_source_list": ["2017年第1届重庆全国初中数学联赛初一竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a\\textless{}b\\textless{}0$$，$${{a}^{2}}+{{b}^{2}}=4ab$$，则$$\\frac{a+b}{a-b}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{6}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式"], "answer_analysis": ["$$\\because {{\\left( a+b \\right)}^{2}}=6ab$$，$${{\\left( a-b \\right)}^{2}}=2ab$$,且$$a\\textless{}b\\textless{}0$$ $$\\therefore a+b=-\\sqrt{6ab}$$，$$a-b=-\\sqrt{2ab}\\therefore \\frac{a+b}{a-b}=\\sqrt{3}$$ "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "123", "queId": "cbc065c0f3534caa9d2236faf758a6a9", "competition_source_list": ["初一上学期其它", "1992年第9届全国初中数学联赛竞赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$${{x}^{2}}-13x+1=0$$，则$${{x}^{4}}+{{x}^{-4}}$$的个位数字是（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-含分式", "课内体系->能力->运算能力"], "answer_analysis": ["由$${{x}^{2}}-13x+1$$可得到$$x+\\dfrac{1}{x}=13$$，所以$${{x}^{2}}+\\dfrac{1}{{{x}^{2}}}={{(x+\\dfrac{1}{x})}^{2}}-2={{13}^{2}}-2=167$$． 同样的$${{x}^{4}}+\\dfrac{1}{{{x}^{4}}}={{({{x}^{2}}+\\dfrac{1}{{{x}^{2}}})}^{2}}-2$$，可以根据个位数字可以直接判断结果的个位数字为$$7$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1209", "queId": "7d6777976eaa4386b5e9c287af53e279", "competition_source_list": ["2012年第23届全国希望杯初一竞赛复赛第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$77$$可以表示成$$n$$（$$n\\geqslant 2$$）个连续自然数的和，则$$n$$的值的个数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["$$77=\\frac{11\\times 7\\times 2}{2}$$． $$77=\\frac{(11\\times 7)\\times 2}{2}=\\frac{(38+39)\\times 2}{2}=38+39$$； $$77=\\frac{(11\\times 2)\\times 7}{2}=\\frac{(8+14)\\times 7}{2}=8+9+10+11+12+13+14$$； $$77=\\frac{(2\\times 7)\\times 11}{2}=\\frac{(2+12)\\times 11}{2}=2+3+4+5+6+7+8+9+10+11+12$$． 综上，$$n$$的值的个数是$$3$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1362", "queId": "c5245674e0914c77b047fe393f566639", "competition_source_list": ["2020~2021学年6月江苏苏州工业园区星港学校初一下学期周测B卷第8题2分", "2003年第14届希望杯初二竞赛第2试第1题", "2020~2021学年江苏苏州高新区苏州市高新区实验初级中学初二下学期期末模拟（一）第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$y-2x+1$$是$$4xy-4{{x}^{2}}-{{y}^{2}}-k$$的一个因式，则$$k$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->因式分解->其他方法->待定系数法"], "answer_analysis": ["由因式定理可知，当$$y-2x+1=0$$时， $$4xy-4{{x}^{2}}-{{y}^{2}}-k=0$$， 因此可令$$\\begin{cases}x=0 y=-1 \\end{cases}$$， 则$$4xy-4{{x}^{2}}-{{y}^{2}}-k=-1-k=0$$， ∴$$k=-1$$． 所以答案为：$$k=-1$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "805", "queId": "4e936c86cb2d4cd1b106f83756f03747", "competition_source_list": ["2017~2018学年江苏无锡滨湖区无锡外国语学校初一下学期期中第10题3分", "2007年竞赛第1题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "方程组$$\\begin{cases}\\left\\textbar{} x \\right\\textbar+y=12 x+\\left\\textbar{} y \\right\\textbar=6 \\end{cases}$$的解的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->运算能力", "知识标签->数学思想->分类讨论思想", "知识标签->题型->方程与不等式->二元一次方程（组）->解二元一次方程组->题型：解含绝对值的方程组", "知识标签->知识点->方程与不等式->二元一次方程（组）->二元一次方程组->二元一次方程组的解", "知识标签->知识点->方程与不等式->二元一次方程（组）->二元一次方程组->加减消元法解二元一次方程组"], "answer_analysis": ["①当$$x\\textgreater0$$，$$y\\textgreater0$$时，原方程组为$$\\begin{cases}x+y=12 x+y=6 \\end{cases}$$显然无解； ②当$$x ~\\textless{} ~0$$，$$y\\textgreater0$$时，原方程组为$$\\begin{cases}-x+y=12 x+y=6 \\end{cases}$$，解得$$\\begin{cases}x=-3 y=9 \\end{cases}$$； ③当$$x\\textgreater0$$，$$y ~\\textless{} ~0$$时，原方程组为$$\\begin{cases}x+y=12 x-y=6 \\end{cases}$$，解得$$\\begin{cases}x=9 y=3 \\end{cases}$$，舍去； ④当$$x ~\\textless{} ~0$$，$$y ~\\textless{} ~0$$时，原方程组为$$\\begin{cases}-x+y=12 x-y=6 \\end{cases}$$显然无解； 综上，只有$$1$$组解．故选$$\\text{A}$$． 若$$x\\geqslant 0$$，则$$\\begin{cases}x+y=12 x+\\left\\textbar{} y \\right\\textbar=6 \\end{cases}$$，于是$$\\left\\textbar{} y \\right\\textbar-y=-6$$，显然不可能． 若$$x ~\\textless{} ~0$$，则$$\\begin{cases}-x+y=12 x+\\left\\textbar{} y \\right\\textbar=6 \\end{cases}$$， 于是$$\\left\\textbar{} y \\right\\textbar+y=18$$，解得$$y=9$$，进而求得$$x=-3$$． 所以，原方程组的解为$$\\begin{cases}x=-3 y=9 \\end{cases}$$，只有$$1$$个解． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1040", "queId": "ff8080814d7978b9014d865fecf823ff", "competition_source_list": ["1990年第1届全国希望杯初一竞赛初赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$代表有理数，那么，$$a$$和$$-a$$的大小关系是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$大于$$-a$$ "}], [{"aoVal": "B", "content": "$$a$$小于$$-a$$ "}], [{"aoVal": "C", "content": "$$a$$大于$$-a$$或$$a$$小于$$-a$$ "}], [{"aoVal": "D", "content": "$$a$$不一定大于$$-a$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->相反数->相反数的性质"], "answer_analysis": ["令$$a=0$$，马上可以排除$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$，应选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "312", "queId": "2715996e9a84456a84f6ded26d8342df", "competition_source_list": ["2019年全美数学竞赛（AMC）竞赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "海滩上有$$50$$人戴墨镜，有$$35$$人戴帽子．而有些人既戴墨镜又戴帽子．如果从戴帽子的人里随机选一个，选中戴墨镜的概率是$$\\frac{2}{5}$$．相反地，如果从戴墨镜的人里随机选一个，这个人戴帽子的概率是多大？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{14}{85}$$ "}], [{"aoVal": "B", "content": "$$\\frac{7}{25}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{7}$$ "}], [{"aoVal": "E", "content": "$$\\frac{7}{10}$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Counting, Probability and Statistics->Classical Probability", "课内体系->知识点->统计与概率"], "answer_analysis": ["海滩上有$$50$$人戴墨镜，有$$35$$人戴帽子．而有些人既戴墨镜又戴帽子．如果从戴帽子的人里随机选一个，选中戴墨镜的概率是$$\\frac{2}{5}$$．相反地，如果从戴墨镜的人里随机选一个，这个人戴帽子的概率是多大？ This is a problem of conditional probability, a good application of Bayes formula. $$P(G\\textbar C)=\\frac{2}{5}$$， $$\\textbar G\\cap C\\textbar=\\textbar C\\textbar\\cdot P(G\\textbar C)=14$$， $$P(C\\textbar G)=\\frac{\\textbar G\\cap C\\textbar}{\\textbar G\\textbar}=\\frac{7}{25}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1411", "queId": "ce90042078714a05a72e8655abd5fa4f", "competition_source_list": ["1993年第4届希望杯初二竞赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果实数$$x$$，$$y$$满足等式$$2x+{{x}^{2}}+{{x}^{2}}{{y}^{2}}+2=-2xy$$，那么$$x+y$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->因式分解->因式分解：公式法"], "answer_analysis": ["等式$$2x+{{x}^{2}}+{{x}^{2}}{{y}^{2}}+2=-2xy$$即$${{(x+1)}^{2}}+{{(xy+1)}^{2}}=0$$， 所以$$x+1=0$$，$$xy+1=0$$， 解得$$x=-1$$，$$y=1$$，则$$x+y=0$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "835", "queId": "b165a8f3f957411fb438212f43247c7d", "competition_source_list": ["2010年第21届全国希望杯初一竞赛初赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若以$$x$$为未知数的方程$$x-2a+4=0$$的根是负数，则（~ ~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( a-1 \\right)\\left( a-2 \\right)\\textless{}0$$ "}], [{"aoVal": "B", "content": "$$\\left( a-1 \\right)\\left( a-2 \\right)\\textgreater0$$ "}], [{"aoVal": "C", "content": "$$\\left( a-3 \\right)\\left( a-4 \\right)\\textless{}0$$ "}], [{"aoVal": "D", "content": "$$\\left( a-3 \\right)\\left( a-4 \\right)\\textgreater0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->一元一次方程的相关概念->一元一次方程的解", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程的相关概念->由一元一次方程的解求参数的值"], "answer_analysis": ["由$$x-2a+4=0$$，得$$x=2a-4=2(a-2)$$， 因为方程$$x-2a+4=0$$的根是负数， 所以$$a\\textless{}2$$， 所以$$a-3\\textless{}0$$，$$a-4\\textless{}0$$， 所以$$\\left( a-3 \\right)\\left( a-4 \\right)\\textgreater0$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "639", "queId": "b0c6eaa4265342ce80ff0e7f37078e1c", "competition_source_list": ["1999年竞赛（全国初中数学竞赛）第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知三个两两互���的正整数$$x$$，$$y$$，$$z$$满足方程组$$\\begin{cases}{{x}^{3}}+{{y}^{3}}+3xyz={{z}^{3}} {{x}^{2}}+7{{y}^{2}}={{z}^{2}} \\end{cases}$$，则$${{x}^{2}}+{{y}^{2}}+{{x}^{2}}$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$26$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->其他方程->多元二次方程（组）", "课内体系->能力->运算能力"], "answer_analysis": ["由$${{x}^{3}}+{{y}^{3}}+3xyz={{z}^{3}}$$， 得$${{x}^{3}}+{{y}^{3}}+{{\\left( -z \\right)}^{3}}-3xy\\left( -z \\right)=0$$， 所以$$\\left( x+y-z \\right)\\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy+yz+zx \\right)=0$$， 因为，$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy+yz+zx=\\frac{1}{2}\\left[ {{\\left( x-y \\right)}^{2}}+{{\\left( y+z \\right)}^{2}}+{{\\left( z+x \\right)}^{2}} \\right]\\textgreater0$$， 所以$$x+y-z=0$$，即$$z=x+y$$， 所以$$7{{x}^{2}}={{z}^{2}}-{{x}^{2}}=\\left( z-x \\right)\\left( z+x \\right)=y\\left( 2z+y \\right)$$， ∴$$7y=2x+y$$， 所以$$x=3y$$，$$z=4y$$， 由于$$1=\\left( x,y \\right)=\\left( 3y,y \\right)=y$$， 所以，$$x=3$$，$$y=1$$，$$z=4$$，于是$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=26$$． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1031", "queId": "8a37c6c0712341dca6171b287a9452e4", "competition_source_list": ["2012年第29届全国全国初中数学联赛竞赛第7题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知互不相等的实数$$a$$，$$b$$，$$c$$满足$$a+\\frac{1}{b}=b+\\frac{1}{c}=c+\\frac{1}{a}=t$$，则$$t=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$+1$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$\\pm 2$$ "}], [{"aoVal": "D", "content": "$$\\pm 1$$ "}]], "knowledge_point_routes": ["知识标签->题型->方程与不等式->其它方程->一次不定方程", "知识标签->知识点->方程与不等式->不定方程"], "answer_analysis": ["由$$a+\\frac{1}{b}=b+\\frac{1}{c}=c+\\frac{1}{a}=t$$， 得$$b=\\frac{1}{t-a}$$，$$c=\\frac{at-1}{a}$$，$$\\frac{1}{t-a}+\\frac{at-1}{a}=t$$， 整理得$$a{{t}^{3}}-{{t}^{2}}-{{a}^{2}}{{t}^{2}}+{{a}^{2}}-at+1=0$$， 即$$({{t}^{2}}-1)({{a}^{2}}-at+1)=0$$． 同理得：$$({{t}^{2}}-1)({{b}^{2}}-bt+1)=0$$，$$({{t}^{2}}-1)({{c}^{2}}-ct+1)=0$$． 若$${{t}^{2}}-1\\ne 0$$，则$$a$$，$$b$$，$$c$$为二次方程$${{x}^{2}}-xt+1=0$$的解， 这与$$a$$，$$b$$，$$c$$互不相等矛盾，不满足题意， 故$${{t}^{2}}-1=0$$， 即$$t=\\pm 1$$． $$\\frac{1}{b}=t-a$$，∴$$b=\\frac{1}{t-a}$$， $$c=t-\\frac{1}{a}$$，∴$$\\frac{1}{c}=\\frac{1}{t-\\dfrac{1}{a}}$$， ∴$$t=b+\\frac{1}{c}=\\frac{1}{t-a}+\\frac{1}{t-\\dfrac{1}{a}}$$， ∴$$t\\left( t-a \\right)\\left( t-\\dfrac{1}{a} \\right)=t-\\dfrac{1}{a}+t-a$$， 整理得$$\\left( t-a-\\frac{1}{a} \\right)\\left( {{t}^{2}}-1 \\right)=0$$， $$a$$，$$ b$$，$$ c$$互不相等，∴$$t\\ne a+\\frac{1}{a}$$， ∴$${{t}^{2}}=1$$， ∴$$t=\\pm 1$$． 故答案为：$$\\pm 1$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "668", "queId": "3bc27a2368204f6484b4559293d5063d", "competition_source_list": ["2018年全国初中数学联赛初一竞赛"], "difficulty": "3", "qtype": "single_choice", "problem": "已知实数$$a$$，$$b$$，$$c$$满足$$a+b+c=0$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$$，则$$\\frac{{{a}^{5}}+{{b}^{5}}+{{c}^{5}}}{abc}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{2}$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式有关的概念->多项式", "竞赛->知识点->数与式->整式->对称多项式 "], "answer_analysis": ["不要搜题 "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "499", "queId": "678f7cf6db0048f4841d7959bf4603fc", "competition_source_list": ["1997年第8届希望杯初二竞赛第1试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$P$$是线段$$AB$$上的一点，$$AB=1$$，以$$AP$$和$$BP$$为边分别作两个正方形，当这两个正方形的面积的差的绝对值为$$\\frac{1}{2}$$时，$$AP$$的长是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{4}$$或$$\\frac{3}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{3}$$或$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{5}$$或$$\\frac{4}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{7}$$或$$\\frac{5}{7}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->四边形->特殊平行四边形"], "answer_analysis": ["两正方形的面积差$$=A{{P}^{2}}-{{\\left( 1-AP \\right)}^{2}}=2AP-1$$， 由题意$$\\left\\textbar{} 2AP-1 \\right\\textbar=\\frac{1}{2}$$，则有 $$\\begin{cases}2AP-1=\\frac{1}{2},\\Rightarrow AP=\\frac{3}{4} 2AP-1=-\\frac{1}{2},\\Rightarrow AP=\\frac{1}{4} \\end{cases}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "286", "queId": "8afd85b991584674bfdca5910c365bfc", "competition_source_list": ["2018年第29届希望杯初一竞赛初赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$$2017\\times20182018-2018\\times20172017$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2017$$ "}], [{"aoVal": "D", "content": "$$2018$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->含乘方的有理数混合运算", "课内体系->能力->运算能力"], "answer_analysis": ["$$2017\\times20182018-2018\\times20172017$$ $$=2017\\times2018\\times10001-2018\\times2017\\times10001$$ $$=0$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "633", "queId": "714e62427a884441ba1068f24020faae", "competition_source_list": ["2015年第32届全国全国初中数学联赛竞赛A卷第1题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "(★)设实数$$a$$，$$b$$，$$c$$满足：$$a+b+c=3$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=4$$，则$$\\frac{{{a}^{2}}+{{b}^{2}}}{2-c}+\\frac{{{b}^{2}}+{{c}^{2}}}{2-a}+\\frac{{{c}^{2}}+{{a}^{2}}}{2-b}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->运算能力", "知识标签->题型->式->整式的乘除->乘法公式->题型：利用平方差公式计算", "知识标签->题型->式->分式->分式的运算->题型：分式加减、乘除混合运算", "知识标签->知识点->式->分式->分式的运算->分式的混合运算", "知识标签->知识点->式->整式的乘除->乘法公式->平方差公式"], "answer_analysis": ["∵$$a+b+c=3$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=4$$， ∴$$\\frac{{{a}^{2}}+{{b}^{2}}}{2-c}+\\frac{{{b}^{2}}+{{c}^{2}}}{2-a}+\\frac{{{c}^{2}}+{{a}^{2}}}{2-b}$$ $$=\\frac{4-{{c}^{2}}}{2-c}+\\frac{4-{{a}^{2}}}{2-a}+\\frac{4-{{b}^{2}}}{2-b}$$ $$=2+c+2+a+2+b$$ $$=a+b+c+6$$ $$=3+6$$ $$=9$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "975", "queId": "bfb174450c784ee2b39149f7386e6adb", "competition_source_list": ["2002年第13届希望杯初二竞赛第1试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2001$$年$$7$$月$$13$$日，北京市获得了第$$29$$届奥运会的主办权，这一天是星期五，那么第$$29$$届奥运会在北京市举办的那一年的$$7$$月$$13$$日是．", "answer_option_list": [[{"aoVal": "A", "content": "星期四 "}], [{"aoVal": "B", "content": "星期五 "}], [{"aoVal": "C", "content": "星期六 "}], [{"aoVal": "D", "content": "星期日 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->综合除法和余数定理 "], "answer_analysis": ["因为$$2001$$年$$7$$月$$13$$日是星期五．又$$365\\div 7=52\\cdots 1$$， $$2004$$年和$$2008$$年均为闰年，比其他几年的$$2$$月份多$$1$$天． 又$$2008$$年距$$2001$$年共$$7$$年时间． 所以，$$2008$$年的$$7$$月$$13$$日应是$$(5+7+2)\\div 7=2\\cdots 0=1\\cdots 7$$， 即$$2008$$年的$$7$$月$$13$$日是星期日． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1205", "queId": "8aac49074e724b45014e87c836655131", "competition_source_list": ["1996年第7届全国希望杯初一竞赛复赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "设关于$$x$$的方程$$a\\left( x-a \\right)+b\\left( x+b \\right)=0$$有无穷多个解，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a+b=0$$ "}], [{"aoVal": "B", "content": "$$a-b=0$$ "}], [{"aoVal": "C", "content": "$$ab=0$$ "}], [{"aoVal": "D", "content": "$$\\frac{a}{b}=0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->含参一元一次方程->解的情况", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程的相关概念->一元一次方程的解"], "answer_analysis": ["整理原方程得$$\\left( a+b \\right)x={{a}^{2}}-{{b}^{2}}$$． 要使该方程有无穷多解，只当$$a+b=0$$且$${{a}^{2}}-{{b}^{2}}=0$$， 当$$a+b=0$$时$$a=-b$$，$${{a}^{2}}-{{b}^{2}}=0$$． 所以当$$a+b=0$$时，原方程有无穷多个解，选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "299", "queId": "4219a6c31c754585a01202ae92d7ee93", "competition_source_list": ["2019年广东惠州惠城区光正实验学校初二竞赛第1题3分", "2018~2019学年山东日照莒县初二上学期期末第6题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "若式子$$\\frac{\\sqrt{m+1}}{\\textbar m-3\\textbar}$$有意义，则实数$$m$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$m\\geqslant -1$$ "}], [{"aoVal": "B", "content": "$$m\\textgreater-1$$ "}], [{"aoVal": "C", "content": "$$m\\textgreater-1$$且$$m\\ne 3$$ "}], [{"aoVal": "D", "content": "$$m\\geqslant -1$$且$$m\\ne 3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的基础->分式有意义的条件", "课内体系->知识点->式->二次根式->二次根式的基础->二次根式有意义的条件", "课内体系->能力->运算能力"], "answer_analysis": ["要使式子$$\\frac{\\sqrt{m+1}}{\\textbar m-3\\textbar}$$有意义， ∴$$m+1\\geqslant 0$$且$$m-3\\ne 0$$， ∴$$m\\geqslant -1$$且$$m\\ne 3$$． 选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "388", "queId": "2c18aaa59fee4452a29dc4200d2aea95", "competition_source_list": ["2010年竞赛第4题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "在一列数$${{x}_{1}}$$，$${{x}_{2}}$$，$${{x}_{3}}$$，$$\\cdots $$中，已知$${{x}_{1}}=1$$，且当$$k\\geqslant 2$$时， $${{x}_{k}}={{x}_{k-1}}+1-4\\left( \\left[ \\frac{k-1}{4} \\right]-\\left[ \\frac{k-2}{4} \\right] \\right)$$ （$$[a]$$表示不超过实数$$a$$的最大整数，例如$$[2.6]=2$$，$$[0.2]=0$$），则$${{x}_{2010}}$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->含参方程"], "answer_analysis": ["由$${{x}_{1}}=1$$和$${{x}_{k}}={{x}_{k-1}}+1-4\\left( \\left[ \\frac{k-1}{4} \\right]-\\left[ \\frac{k-2}{4} \\right] \\right)$$可得 $${{x}_{1}}=1$$，$${{x}_{2}}=2$$，$${{x}_{3}}=3$$，$${{x}_{4}}=4$$， $${{x}_{5}}=1$$，$${{x}_{6}}=2$$，$${{x}_{7}}=3$$，$${{x}_{8}}=4$$， $$\\cdots $$ 因为$$2010=4\\times 502+2$$，所以$${{x}_{2010}}=2$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "463", "queId": "9de67eee035e4bc8a0b7e42b76914859", "competition_source_list": ["2016年第27届全国希望杯初一竞赛复赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "马小虎计算一个数乘$$8$$，再减$$63$$，由于粗心，把乘号看成除号，减号看成了加号，但得数是正确的，这道题的正确得数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$56$$ "}], [{"aoVal": "C", "content": "$$63$$ "}], [{"aoVal": "D", "content": "$$65$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->能力->分析和解决问题能力", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的数字问题"], "answer_analysis": ["设这个数为$$x$$， 由题意，得$$8x-63=\\frac{x}{8}+63$$， 解得$$x=16$$， ∴这道题的正确得数是$$8\\times 16-63=65$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1443", "queId": "a1d7fa802d084da19f61d9047ecfada8", "competition_source_list": ["2003年第14届希望杯初二竞赛第2试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "不等式$$0\\leqslant ax+5\\leqslant 4$$的整数解是$$1$$，$$2$$，$$3$$，$$4$$，则$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\leqslant -\\frac{5}{4}$$ "}], [{"aoVal": "B", "content": "$$a\\textless{}-1$$ "}], [{"aoVal": "C", "content": "$$-\\frac{5}{4}\\leqslant a ~\\textless{} ~-1$$ "}], [{"aoVal": "D", "content": "$$a\\geqslant -\\frac{5}{4}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->含参不等式->由不等式（组）的整数解情况求参数范围"], "answer_analysis": ["由$$0\\leqslant ax+5\\leqslant 4$$得$$-5\\leqslant ax\\leqslant -1$$． 若$$a\\textgreater0$$，则$$-\\frac{5}{a}\\leqslant x\\leqslant -\\frac{1}{a}$$，$$x$$不可能取到正整数解； 若$$a=0$$，不等式无解． 所以必有$$a\\textless{}0$$，此时$$-\\frac{1}{a}\\leqslant x\\leqslant -\\frac{5}{a}$$． 又因原不等式的整数解是$$1$$、$$2$$、$$3$$、$$4$$， 所以$$0\\textless{}-\\frac{1}{a}\\leqslant 1$$，且$$4\\leqslant -\\frac{5}{a}\\textless{}5$$，解得$$-\\frac{5}{4}\\leqslant a\\textless{}-1$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "495", "queId": "62fcd602ce57483ea29c6e8055b166c9", "competition_source_list": ["2013年第24届全国希望杯初二竞赛初赛第3题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "下列命题中，正确的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "若$$a\\textgreater0$$，则$$a\\textgreater\\frac{1}{a}$$ "}], [{"aoVal": "B", "content": "若$$a\\textgreater{{a}^{2}}$$，则$$a\\textgreater1$$． "}], [{"aoVal": "C", "content": "若$$0\\textless{}a\\textless{}1$$，则$$a\\textgreater{{a}^{2}}$$ "}], [{"aoVal": "D", "content": "若$$\\left\\textbar{} a \\right\\textbar=a$$，则$$a\\textgreater0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["取特殊值． 当$$a=\\frac{1}{2}$$时，$$a\\textless{}\\frac{1}{a}$$，可知$$\\text{A}$$错误； 当$$a=\\frac{1}{2}\\textless{}1$$时，$$\\frac{1}{2}\\textgreater{{\\left( \\frac{1}{2} \\right)}^{2}}=\\frac{1}{4}$$，可知$$\\text{B}$$错误； 当$$a=0$$时，$$\\left\\textbar{} 0 \\right\\textbar=0$$，可知$$\\text{D}$$错误． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "102", "queId": "2a334921c6a647e09164b3520a9c01c8", "competition_source_list": ["2016年第33届全国全国初中数学联赛竞赛第6题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "设实数$$x$$，$$y$$，$$z$$，满足$$x+y+z=1$$，则$$M=xy+2yz+3xz$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["课内体系->方法->配方法", "课内体系->知识点->式->整式的乘除->整式乘除化简求值->整式乘除化简求值-整体代入化简求值", "课内体系->知识点->式->整式的加减->整式有关的概念->整式->整式的定义", "课内体系->能力->运算能力"], "answer_analysis": ["方法一：$$M=xy+(2y+3x)z$$ $$=xy+(2y+3x)(1-x-y)$$ $$=-3{{x}^{2}}-4xy-2{{y}^{2}}+3x+2y$$ $$=-2\\left[ {{y}^{2}}+2\\left( x-\\frac{1}{2} \\right)y+{{\\left( x-\\frac{1}{2} \\right)}^{2}} \\right]-3{{x}^{2}}+3x+2{{\\left( x-\\frac{1}{2} \\right)}^{2}}$$ $$=-2{{\\left( y+x-\\frac{1}{2} \\right)}^{2}}-{{x}^{2}}+x+\\frac{1}{2}$$ $$=-2{{\\left( y+x-\\frac{1}{2} \\right)}^{2}}-{{\\left( x-\\frac{1}{2} \\right)}^{2}}+\\frac{3}{4}$$ $$\\leqslant \\frac{3}{4}$$． 当且仅当$$x=\\frac{1}{2}$$，$$y=0$$时，不等式取等号， 所以$${{M}_{\\max }}=\\frac{3}{4}$$． 方法二：令$$y=1-x-z$$，代入$$M$$，则： $$M=x\\left( 1-x-z \\right)+2\\left( 1-x-z \\right)z+3xz=x-{{x}^{2}}-xz+2z-2xz-2{{z}^{2}}+3xz$$ $$=-{{x}^{2}}+x-2{{z}^{2}}+2z=-{{\\left( x-\\frac{1}{2} \\right)}^{2}}-2{{\\left( z-\\frac{1}{2} \\right)}^{2}}+\\frac{3}{4}$$，$${{M}_{\\max }}=\\frac{3}{4}$$． 所以选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "159", "queId": "142426ab9a8d4844bd19b24ce55c2bed", "competition_source_list": ["2016年第27届全国希望杯初一竞赛复赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$3m+5$$与$$m-1$$互为相反数，则$$\\frac{\\left\\textbar{} m \\right\\textbar}{2016}$$的倒数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2016}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{1}{2016}$$ "}], [{"aoVal": "C", "content": "$$2016$$ "}], [{"aoVal": "D", "content": "$$-2016$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->相反数->相反数的性质"], "answer_analysis": ["∵$$3m+5$$与$$m-1$$互为相反数， ∴$$3m+5+m-1=0$$， 解得$$m=-1$$， ∴$$\\frac{\\left\\textbar{} m \\right\\textbar}{2016}=\\frac{1}{2016}$$， ∴$$\\frac{\\left\\textbar{} m \\right\\textbar}{2016}$$的倒数是$$2016$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1088", "queId": "ff8080814d9539f1014d9b56fb7809e4", "competition_source_list": ["1992年第3届全国希望杯初一竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "有理数$$-\\frac{1}{a}$$一定不是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "正整数 "}], [{"aoVal": "B", "content": "负整数 "}], [{"aoVal": "C", "content": "负分数 "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类"], "answer_analysis": ["因$$-\\frac{1}{a}$$为有理数，所以$$a\\ne 0$$． 若$$a=-1$$，$$-\\frac{1}{a}=1$$，排除$$\\text{A}$$； 若$$a=1$$，$$-\\frac{1}{a}=-1$$，排除$$\\text{B}$$； 若$$a=2$$，$$-\\frac{1}{a}=-\\frac{1}{2}$$，排除$$\\text{C}$$．故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1122", "queId": "ff8080814d9efd56014daa8228bf0af5", "competition_source_list": ["1993年第4届全国希望杯初一竞赛初赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "在自然数：$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$\\cdots $$中，前$$15$$个质数之和的负倒数等于（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{1}{328}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{1}{329}$$ "}], [{"aoVal": "C", "content": "$$-\\frac{1}{337}$$ "}], [{"aoVal": "D", "content": "$$-\\frac{1}{340}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->倒数与负倒数"], "answer_analysis": ["前$$15$$个质数为$$2$$，$$3$$，$$5$$，$$7$$，$$11$$，$$13$$，$$17$$，$$19$$，$$23$$，$$29$$，$$31$$，$$37$$，$$41$$，$$43$$，$$47$$． 它们的和为$$328$$，$$328$$的负倒数为$$-\\frac{1}{328}$$，选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "763", "queId": "56fab50b9ddb4558b1919921ee118dd3", "competition_source_list": ["2009年第20届希望杯初二竞赛第2试第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$x$$的整数部分记为$$\\left[ x \\right]$$,$$x$$的小数部分记为$$\\left { x \\right }$$，易知$$x=\\left[ x \\right]+\\left { x \\right }\\left( 0 ~\\textless{} ~\\left { x \\right } ~\\textless{} ~1 \\right)$$.若$$x=\\sqrt{3-\\sqrt{5}}-\\sqrt{3+\\sqrt{5}}$$，那么$$\\left[ x \\right]$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->方法->代入法", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算", "课内体系->思想->整体思想"], "answer_analysis": ["\\textbf{（知识点：多重二次根式）} 因为$$\\sqrt{3\\pm \\sqrt{5}}=\\sqrt{\\frac{6\\pm 2\\sqrt{5}}{2}}=\\frac{\\sqrt{{{\\left( \\sqrt{5}\\pm 1 \\right)}^{2}}}}{\\sqrt{2}}=\\frac{\\sqrt{5}\\pm 1}{\\sqrt{2}}$$， 所以$$x=\\frac{\\sqrt{5}-1}{\\sqrt{2}}-\\frac{\\sqrt{5}+1}{\\sqrt{2}}=\\frac{-2}{\\sqrt{2}}=-\\sqrt{2}\\approx -1.41$$， 所以$$\\left[ x \\right]=-2$$． 所以选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1180", "queId": "8aac49074e023206014e20ea73f3662d", "competition_source_list": ["1994年第5届全国希望杯初一竞赛复赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "当$$-1\\textless{}a\\textless{}0$$时，则有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{a}\\textgreater a$$ "}], [{"aoVal": "B", "content": "$$\\left\\textbar{} {{a}^{3}} \\right\\textbar\\textgreater{{a}^{3}}$$ "}], [{"aoVal": "C", "content": "$$-a\\textgreater{{a}^{2}}$$ "}], [{"aoVal": "D", "content": "$${{a}^{3}}\\textless{}-{{a}^{2}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性"], "answer_analysis": ["采用特殊值法： 取$$a=-\\frac{1}{2}$$，则$$\\frac{1}{a}=-2$$，$${{a}^{2}}={(-\\frac{1}{2})^{2}}=\\frac{1}{4}$$，$${{a}^{3}}={(-\\frac{1}{2})^{3}}=-\\frac{1}{8}$$． 由$$-2\\textless{}-\\frac{1}{2}$$，排除$$\\text{A}$$； 由$$\\left\\textbar{} -\\frac{1}{8} \\right\\textbar=-(-\\frac{1}{8})$$，排除$$\\text{B}$$； 由$$-\\frac{1}{8}\\textgreater-\\frac{1}{4}$$，排除$$\\text{D}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1097", "queId": "ff8080814d9539f1014d9b6467eb0a71", "competition_source_list": ["1992年第3届全国希望杯初一竞赛初赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "张梅写出了五个有理数，前三个有理数的平均值为$$15$$，后两个有理数的平均值是$$10$$，那么张梅写出的五个有理数的平均值是（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$8\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$12\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算", "课内体系->知识点->统计与概率->数据的分析->平均数/加权平均数"], "answer_analysis": ["前三个数之和$$=15\\times 3$$，后两个数之和$$=10\\times 2$$． 所以五个有理数的平均数为$$\\frac{15\\times 3+10\\times 2}{5}=13$$．选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "138", "queId": "0943b32c6eae4081843cb846475fcfb8", "competition_source_list": ["2017年第1届重庆全国初中数学联赛初一竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a\\textless{}b\\textless{}0$$，$${{a}^{2}}+{{b}^{2}}=4ab$$，则$$\\frac{a+b}{a-b}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{6}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式"], "answer_analysis": ["$$\\because {{\\left( a+b \\right)}^{2}}=6ab$$，$${{\\left( a-b \\right)}^{2}}=2ab$$,且$$a\\textless{}b\\textless{}0$$ $$\\therefore a+b=-\\sqrt{6ab}$$，$$a-b=-\\sqrt{2ab}\\therefore \\frac{a+b}{a-b}=\\sqrt{3}$$ "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "549", "queId": "a2b77e2cfb764c92b77b89c5065ecdba", "competition_source_list": ["2019~2020学年重庆合川区西南大学银翔实验中学初一下学期期末模拟（一）第12题4分", "2018~2019学年广东深圳南山区桃苑学校初二下学期期末第11题3分", "2019年浙江杭州拱墅区杭州市文澜中学初二竞赛第7题3分", "2018年重庆中考真题A卷第12题4分", "2019~2020学年江苏苏州姑苏区苏州市振华中学校初三下学期单元测试《方程与不等式》第10题3分", "2018年四川绵阳涪城区绵阳东辰国际学校初三自主招生第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若数$$a$$使关于$$x$$的不等式组$$\\begin{cases}\\dfrac{x-1}{2}\\textless{}\\dfrac{1+x}{3} 5x-2\\geqslant x+a \\end{cases}$$有且只有四个整数解，且使关于$$y$$的方程$$\\frac{y+a}{4}+\\frac{2a}{3}=2$$的解为非负数，则符合条件的所有整数$$a$$的和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-3$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->分式方程->分式方程的解与解分式方程->由分式方程的解确定参数"], "answer_analysis": ["$$\\begin{cases}\\dfrac{x-1}{2}\\textless{}\\dfrac{1+x}{3} 5x-2\\geqslant x+a \\end{cases}$$， 不等式组整理得：$$\\begin{cases}x\\textless{}5 x\\geqslant \\dfrac{a+2}{4} \\end{cases}$$， 由不等式组有且只有四个整数解，得到$$0\\textless{}\\frac{a+2}{4}\\leqslant 1$$， 解得：$$-2\\textless{}a\\leqslant 2$$，即整数$$a=-1$$，$$0$$，$$1$$，$$2$$， $$\\frac{y+a}{y-1}+\\frac{2a}{1-y}=2$$， 分式方程去分母得：$$y+a-2a=2(y-1)$$， 解得：$$y=2-a$$， 由分式方程的解为非负数以及分式有意义的条件，得到$$a$$为$$-1$$，$$0$$，$$2$$，之和为$$1$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "44", "queId": "09ec479e6a4e4ad1997850d617081bad", "competition_source_list": ["1997年第8届希望杯初二竞赛第2试第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "有男女两个运动队，男队有队员$$m$$人，女队有队员$$n$$人（$$m\\textgreater10$$，$$n\\textgreater10$$），先从男队中调$$10$$人到女队帮助训练，训练后又从女队中调$$10$$人（这$$10$$人中可以有原来男队中的队员）去男队参加总结．这时，男队中有$$a$$个女队员，女队中有$$b$$个男队员，那么$$a$$，$$b$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b$$ "}], [{"aoVal": "B", "content": "$$a\\textless{}b$$ "}], [{"aoVal": "C", "content": "$$a=b$$ "}], [{"aoVal": "D", "content": "当$$m\\geqslant n$$时，$$a\\geqslant b$$，当$$m\\textless{}n$$时，$$a\\textless{}b$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->操作与游戏"], "answer_analysis": ["设训练后回男队的$$10$$人中有$$c$$个女队员，则男队中女队员的人数$$a=c$$．此时，女队中应有男队员的人数为$$b=10-(10-c)=c$$（人）． 所以$$a=b$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1534", "queId": "ef6a1786d2c2461d90d25e54af9f71e9", "competition_source_list": ["2017~2018学年北京朝阳区朝阳外国语学校初二上学期单元测试《数学基础知识竞赛》第17题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$${{(2a-3b)}^{2}}-(2a+3b)(2a-3b)$$的结果是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$18{{b}^{2}}-12ab$$ "}], [{"aoVal": "B", "content": "$$18{{b}^{2}}+12ab$$ "}], [{"aoVal": "C", "content": "$$9{{b}^{2}}-6ab$$ "}], [{"aoVal": "D", "content": "$$9{{b}^{2}}+6ab$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->整式的乘除运算->整式乘除的综合", "课内体系->能力->运算能力"], "answer_analysis": ["$$18{{b}^{2}}-12ab$$． $${{(2a-3b)}^{2}}-(2a+3b)(2a-3b)$$， $$=4{{a}^{2}}-12ab+9{{b}^{2}}-(4{{a}^{2}}-9{{b}^{2}})$$， $$=4{{a}^{2}}-12ab+9{{b}^{2}}-4{{a}^{2}}+9{{b}^{2}}$$， $$=18{{b}^{2}}-12ab$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "74", "queId": "074e77d8f25d4d75b1914287179a1d81", "competition_source_list": ["2016~2017学年10月河南郑州二七区郑州实验外国语中学初一上学期月考第11题3分", "2008年竞赛第7题6分", "2016~2017学年福建泉州南安市南安市实验中学初一下学期期中第16题4分", "2018~2019学年浙江杭州西湖区杭州市之江实验中学初一下学期期末第18题4分"], "difficulty": "4", "qtype": "single_choice", "problem": "小王沿街匀速行走，发现每隔$$6$$分钟从背后驶过一辆$$18$$路公交车，每隔$$3$$分钟从迎面驶来一辆$$18$$路公交车．假设每辆$$18$$路公交车行驶速度相同，而且$$18$$路公交车总站每隔固定时间发一辆车，那么发车间隔的时间是~\\uline{~~~~~~~~~~}~分钟．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->二元一次方程（组）->二元一次方程组应用题->二元一次方程组的行程问题"], "answer_analysis": ["设$$18$$路公交车的速度是$$x$$米/分，小王行走的速度是$$y$$米/分，同向行驶的相邻两车的间距为$$s$$米． 每隔$$6$$分钟从背后开过一辆$$18$$路公交车，则 $$6x-6y=s$$．① 每隔$$3$$分钟从迎面驶来一辆$$18$$路公交车，则 $$3x+3y=s$$．② 由①②得，$$s=4x$$，所以$$\\frac{s}{x}=4$$. 即$$18$$路公交车总站发车间隔的时间是$$4$$分钟． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "253", "queId": "1daf4ccc49694f779a695f26ad45a85b", "competition_source_list": ["2019~2020学年重庆九龙坡区重庆市育才中学初三上学期期中第11题4分", "2020~2021学年广东深圳福田区深圳市实验学校初二下学期期中第10题3分", "2019~2020学年5月河南郑州中牟县郑州枫杨外国语学校(东校区)初二下学期月考第7题3分", "2020年湖南长沙岳麓区湖南师范大学附属中学初二竞赛（湖南师范大学附属中学教育集团）（6月攀登杯）第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果关于$$x$$的不等式组$$\\begin{cases}m-5x\\geqslant 2 x-\\dfrac{11}{2} ~\\textless{} ~3\\left( x+\\dfrac{1}{2} \\right) \\end{cases}$$有且仅有四个整数解，且关于$$y$$的分式方程$$\\frac{2-my}{2-y}-\\frac{8}{y-2}=1$$有非负数解，则符合条件的所有整数$$m$$的和是．", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$22$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->分式方程->分式方程的解与解分式方程->分式方程的特殊解问题", "课内体系->知识点->方程与不等式->分式方程->分式方程的解与解分式方程->分式方程与不等式综合"], "answer_analysis": ["解不等式组$$\\begin{cases}m-5x\\geqslant 2① x-\\dfrac{11}{2} ~\\textless{} ~3\\left( x+\\dfrac{1}{2} \\right)② \\end{cases}$$， 解不等式①得：$$-5x\\geqslant 2-m$$ $$ x\\leqslant \\frac{m-2}{5}$$， 解不等式②得：$$2x-11 ~\\textless{} ~6x+3$$ $$ 2x-6x ~\\textless{} ~14$$ $$ -4x ~\\textless{} ~14$$ $$ x\\textgreater-\\frac{7}{2}$$， ∴原不等式组的解集为$$-\\frac{7}{2} ~\\textless{} ~x\\leqslant \\frac{m-2}{5}$$， ∵原不等式组有且仅有四个整数解，它们分别为$$-3$$、$$-2$$、$$-1$$，$$0$$， ∴$$0\\leqslant \\frac{m-2}{5} ~\\textless{} ~1$$， ∴$$0\\leqslant m-2 ~\\textless{} ~5$$， ∴$$2\\leqslant m ~\\textless{} ~7$$， 解分式方程$$\\frac{2-my}{2-y}-\\frac{8}{y-2}=1$$， 去分母得$$my-2-8=y-2$$， $$\\left( m-1 \\right)y=8$$ $$y=\\frac{8}{m-1}$$， ∵分式方程有非负数解， ∴$$\\frac{8}{m-1}\\geqslant 0$$且$$\\frac{8}{m-1}\\ne 2$$， ∴$$m\\textgreater1$$且$$m\\ne 5$$， ∵$$m$$为整数， ∴$$m$$可以取$$2$$、$$3$$、$$4$$、$$6$$， 它们的和为$$2+3+4+6=15$$， ∴符合条件的所有整数$$m$$的和为$$15$$， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1515", "queId": "f3cc0ff8e43b4c94ba07125fb88e6b5d", "competition_source_list": ["2005年第16届希望杯初二竞赛复赛第10题4分", "初一下学期其它第17题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知整数$$x$$，$$y$$，$$z$$满足$$x\\leqslant y\\textless{} z$$，且$$\\begin{cases}\\left\\textbar{} x+y \\right\\textbar+\\left\\textbar{} y+z \\right\\textbar+\\left\\textbar{} z+x \\right\\textbar=4 ① \\left\\textbar{} x-y \\right\\textbar+\\left\\textbar{} y-z \\right\\textbar+\\left\\textbar{} z-x \\right\\textbar=2 ②\\end{cases}$$，那么$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}$$的值等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$2$$或$$14$$ "}], [{"aoVal": "D", "content": "$$14$$或$$17$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->解含绝对值的方程组", "课内体系->能力->运算能力"], "answer_analysis": ["由②可知$$\\left\\textbar{} x-y \\right\\textbar$$、$$\\left\\textbar{} y-z \\right\\textbar$$、$$\\left\\textbar{} z-x \\right\\textbar$$中必须有一个为$$0$$，只有$$x-y=0$$，$$x=y$$，代入②中得$$\\left\\textbar{} z-x \\right\\textbar=1$$ $$z=x+1$$将$$x=y$$，$$z=x+1$$代入①中，得$$\\left\\textbar{} x \\right\\textbar+\\left\\textbar{} 2x+1 \\right\\textbar=2$$，$$\\left\\textbar{} x \\right\\textbar=0$$，$$1$$，$$2$$．经检验$$\\left\\textbar{} x \\right\\textbar=1$$，依据题意得． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1306", "queId": "f2ab48ac822741978be4464ad537d297", "competition_source_list": ["2001年第18届全国初中数学联赛竞赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x$$、$$y$$是正整数，并且$$xy+x+y=23$$，$${{x}^{2}}y+x{{y}^{2}}=120$$，则$${{x}^{2}}+{{y}^{2}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$43$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$34$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["知识标签->题型->方程与不等式->一元二次方程->根与系数的关系->题型：韦达定理应用", "知识标签->知识点->方程与不等式->一元二次方程->一元二次方程根的判别式", "知识标签->知识点->方程与不等式->一元二次方程->一元二次方程的根与系数的关系"], "answer_analysis": ["由于$$xy+\\left( x+y \\right)=23$$，$$xy\\cdot \\left( x+y \\right)=120$$， 则$$x$$，$$y$$与$$\\left( x+y \\right)$$为方程$${{t}^{2}}-23t+120=0$$的两个根，得到$${{t}_{1}}=8$$，$${{t}_{2}}=15$$， 即$$xy=8$$，$$x+y=15$$ ①，或者$$xy=15$$，$$x+y=8$$ ②， ①的时候$$x$$，$$y$$为方程$${{u}^{2}}-15u+8=0$$的根，$${{\\Delta }_{1}}={{15}^{2}}-32=193$$，不是完全平方数，$$xy$$不可能为题目中要求的正整数，舍， ②的时候$$x$$，$$y$$为方程$${{u}^{2}}-8u+15=0$$的根，$${{u}_{1}}=3$$，$${{u}_{2}}=5$$， 故$${{x}^{2}}+{{y}^{2}}={{\\left( x+y \\right)}^{2}}-2xy=34$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1266", "queId": "d71ee2fd3f344e8aaefce7c10d7363c3", "competition_source_list": ["2017~2018学年广东汕头金平区初二上学期期末第7题3分", "2014~2015学年重庆开县初二上学期期末第7题3分", "2017~2018学年北京朝阳区朝阳外国语学校初二上学期单元测试《数学基础知识竞赛》第18题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "下列各式中，计算结果是$${{x}^{2}}+7x-18$$的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$(x-2)(x+9)$$ "}], [{"aoVal": "B", "content": "$$(x+2)(x+9)$$ "}], [{"aoVal": "C", "content": "$$(x-3)(x+6)$$ "}], [{"aoVal": "D", "content": "$$(x-1)(x+18)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->整式的乘除运算->多项式乘多项式", "课内体系->能力->运算能力"], "answer_analysis": ["$$\\text{B}$$、$$(x+2)(x+9)={{x}^{2}}+11x+18$$， $$\\text{C}$$、$$(x-3)(x+6)={{x}^{2}}+3x-18$$， $$\\text{D}$$、$$(x-1)(x+18)={{x}^{2}}+17x-18$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "712", "queId": "763bb39db02c4e47b1b29c9bf74ea4e0", "competition_source_list": ["2013年第30届全国全国初中数学联赛竞赛第2题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "满足等式$${{\\left( 2-m \\right)}^{{{m}^{2}}-m-2}}=1$$的所有实数$$m$$的和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["课内体系->思想->分类讨论思想", "课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->幂的运算->零指数幂", "课内体系->知识点->式->整式的乘除->幂的运算->幂的综合运算"], "answer_analysis": ["分三种情况进行讨论： ①若$$2-m=1$$，即$$m=1$$时，满足已知等式； ②若$$2-m=-1$$，即$$m=3$$时， $${{\\left( 2-m \\right)}^{{{m}^{2}}-m-2}}={{(-1)}^{4}}=1$$满足已知等式； ③若$$2-m\\ne \\pm 1$$，即$$m\\ne 1$$且$$m\\ne 3$$时， 由已知，得$$\\begin{cases}2-m\\ne 0 {{m}^{2}}-m-2=0 \\end{cases}$$，解得$$m=-1$$， 所以满足等式$${{\\left( 2-m \\right)}^{{{m}^{2}}-m-2}}=1$$的所有实数$$m$$的和$$1+3+(-1)=3$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "708", "queId": "5f8619b1ddd14a099cd09493185758d8", "competition_source_list": ["2001年第12届希望杯初一竞赛第2试第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "两个正整数的和是$$60$$，它们的最小公倍数是$$273$$，则它们的乘积是．", "answer_option_list": [[{"aoVal": "A", "content": "$$273$$ "}], [{"aoVal": "B", "content": "$$819$$ "}], [{"aoVal": "C", "content": "$$1911$$ "}], [{"aoVal": "D", "content": "$$3549$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->因数与倍数"], "answer_analysis": ["两个正整数为$$a$$与$$b$$， 则$$a+b=60={{2}^{2}}\\times 3\\times 5$$， $$[a,b]=273=3\\times 7\\times 13$$， 显然$$a$$，$$b$$的最大公约数是$$1$$或$$3$$， 如果$$(a,b)=1$$，则$$[a,b]=a\\times b$$， $$a$$，$$b$$只能取$$(21,13)$$，$$(7,39)$$，$$(1,273)$$，$$(3,91)$$，其和均不为$$60$$， 因此$$(a,b)=3$$， 于是$$a=3\\times 7$$，$$b=3\\times 13$$， 所以$$a\\times b=\\left( 3\\times 7 \\right)\\times \\left( 3\\times 13 \\right)=819$$， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "546", "queId": "70f1e67b332340ae8c4cb0f06a23d0da", "competition_source_list": ["2018~2019学年9月湖南长沙雨花区湖南广益实验中学初一上学期月考第3题3分", "2019年湖南长沙雨花区湖南广益实验中学初一竞赛（广益杯）第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "无论$$x$$为何有理数，$${{x}^{2}}+2$$的值总是．", "answer_option_list": [[{"aoVal": "A", "content": "不大于$$2$$ "}], [{"aoVal": "B", "content": "小于$$2$$ "}], [{"aoVal": "C", "content": "不小于$$2$$ "}], [{"aoVal": "D", "content": "大于$$2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性"], "answer_analysis": ["∵$${{x}^{2}}\\geqslant 0$$， ∴$${{x}^{2}}+2\\geqslant 2$$， 即$${{x}^{2}}+2$$的值总是不小于$$2$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1027", "queId": "cd83f936829647f4b3ac78d69e8eaaad", "competition_source_list": ["2000年希望杯初二竞赛二试"], "difficulty": "3", "qtype": "single_choice", "problem": "一三角形的三边长分别是$$a$$，$$b$$，$$c$$（$$a$$，$$b$$，$$c$$都是质数）且$$a+b+c=16$$，则此三角形是．", "answer_option_list": [[{"aoVal": "A", "content": "直角三角形 "}], [{"aoVal": "B", "content": "等腰三角形 "}], [{"aoVal": "C", "content": "等边三角形 "}], [{"aoVal": "D", "content": "直角三角形或等腰三角形 "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "课内体系->知识点->三角形->三角形及多边形->与三角形有关的线段->三角形的三边关系", "课内体系->知识点->三角形->三角形及多边形->与三角形有关的线段->与三边关系有关的证明"], "answer_analysis": ["因为$$a$$，$$b$$，$$c$$均为质数且$$a+b+c=16$$，所以$$a$$，$$b$$，$$c$$中有一数为$$2$$，设$$a=2$$，则$$b+c=14$$， 所以$$\\textbar b-c\\textbar\\textless{}2$$．从而有$$\\textbar b-c\\textbar=0$$或$$\\textbar b-c\\textbar=1$$．当$$\\textbar b-c\\textbar=1$$时，$$b$$，$$c$$均不是整数，不合题意．因 此，只有$$\\textbar b-c\\textbar=0$$，即$$a=2$$，$$b=c=7$$，所以三角形是等腰三角形． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1271", "queId": "8aac50a74e442d83014e4cd195651d91", "competition_source_list": ["1995年第6届全国希望杯初一竞赛复赛第5题", "初一上学期单元测试《有理数》数轴的概念及应用第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "数轴上坐标是整数的点称为整点，某数轴的单位长度是$$1$$厘米，若在这个数轴上随意画出一条长为$$1995$$厘米的线段$$AB$$，则线段$$AB$$盖住的整点有个．", "answer_option_list": [[{"aoVal": "A", "content": "$$1994$$或$$1995$$ "}], [{"aoVal": "B", "content": "$$1994$$或$$1996$$ "}], [{"aoVal": "C", "content": "$$1995$$或$$1996$$ "}], [{"aoVal": "D", "content": "$$1995$$或$$1997$$ "}]], "knowledge_point_routes": ["课内体系->能力->空间想象能力", "课内体系->能力->运算能力", "课内体系->知识点->数->有理数->数轴->数轴上的规律探究"], "answer_analysis": ["若所画的长为$$1995$$厘米的线段的两个端点$$A$$与$$B$$均为整点时， 此时线段$$AB$$盖住的整点个数是$$1995+1=1996$$个． 若$$A$$点不是整点，则$$B$$点也不是整点，此时线段$$AB$$盖住的整点个数为$$1995$$个， 所以长为$$1995$$厘米的线段盖住的整点是$$1995$$个， 所以长为$$1995$$厘米的线段盖住的整点是$$1995$$或$$1996$$个．选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "27", "queId": "174e4c3698d943d1a472875c6be0ef96", "competition_source_list": ["2014年第25届全国希望杯初一竞赛初赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知长方体的长、宽、高都是整数厘米，将长、宽、高都增加$$1$$厘米后，长方体的表面积可能增加（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$平方厘米 "}], [{"aoVal": "B", "content": "$$103$$平方厘米 "}], [{"aoVal": "C", "content": "$$214$$平方厘米 "}], [{"aoVal": "D", "content": "$$400$$平方厘米 "}]], "knowledge_point_routes": ["课内体系->思想->方程思想", "课内体系->知识点->式->整式的乘除->整式乘除化简求值->整式乘除化简求值-整体代入化简求值", "课内体系->能力->运算能力", "课内体系->能力->推理论证能力"], "answer_analysis": ["设长方体的长、宽、高分别为$$a$$厘米，$$b$$厘米，$$c$$厘米， 则长方体的表面积为$$2(ab+bc+ac)$$平方厘米， 将长方体的长、宽、高都增加$$1$$厘米后， 长方体的表面积增加的值如下： $$2[(a+1)(b+1)+(b+1)(c+1)+(a+1)(c+1)]-2(ab+bc+ac)$$ $$=2(a+b+1+b+c+1+a+c+1)$$ $$=4(a+b+c)+6$$． ∵$$a$$，$$b$$，$$c$$都是正整数， ∴$$4(a+b+c)+6\\geqslant 4(1+1+1)+6=18$$，且能写成$$4m+6$$的形式，其中$$m$$为正整数， 选项中所给数字，只有$$214$$可以表示为$$4\\times 52+6$$的形式，故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1290", "queId": "c06b759f36ff4ce68a689849408b3f3c", "competition_source_list": ["2018~2019学年甘肃兰州城关区兰州市第三十五中学初二下学期期中第12题4分", "初一下学期单元测试《不等式与不等式组》一元一次不等式及其应用第28题", "2011年第16届华杯赛初一竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a$$，$$b$$是常数，不等式$$\\frac{x}{a}+\\frac{1}{b}\\textgreater0$$的解集为$$x\\textless{}\\frac{1}{5}$$，则关于$$x$$的不等式$$bx-a\\textgreater0$$的解集是．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textgreater\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$x\\textless{}-\\frac{1}{5}$$ "}], [{"aoVal": "C", "content": "$$x\\textgreater-\\frac{1}{5}$$ "}], [{"aoVal": "D", "content": "$$x\\textless{}\\frac{1}{5}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->解一元一次不等式"], "answer_analysis": ["方法一：原不等式变形得：$$\\frac{x}{a}\\textgreater-\\frac{1}{b}$$， ∵$$x\\textless{}\\frac{1}{5}$$， ∴$$a\\textless{}0$$． 解不等式得：$$x\\textless{}-\\frac{a}{b}$$，$$-\\frac{a}{b}=\\frac{1}{5}$$，即$$b=-5a$$． ∴$$-5ax-a\\textgreater0$$， ∴$$x\\textgreater-\\frac{1}{5}$$． 方法二：因为不等式等$$\\frac{x}{a}+\\frac{1}{b}\\textgreater0$$的解集为$$x ~\\textless{} ~\\frac{1}{5}$$，因此必有$$a ~\\textless{} ~0$$，所以$$\\frac{-a}{b}=\\frac{1}{5}$$，并且$$b\\textgreater0$$，所以由$$bx-a\\textgreater0$$得到$$x\\textgreater-\\frac{1}{5}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "892", "queId": "6106d6148a044ef6904235e3436b1964", "competition_source_list": ["2017年湖南长沙天心区湘郡培粹实验中学初二竞赛（觉园杯）第5题4分", "2019年湖南长沙天心区湘郡培粹实验中学初二竞赛初赛(觉园杯)第5题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$均不为$$0$$，若$$\\frac{x-y}{a}=\\frac{y-z}{b}=\\frac{z-x}{c}=abc ~\\textless{} ~0$$，则$$P\\left( ab,bc \\right)$$不可能在．", "answer_option_list": [[{"aoVal": "A", "content": "第一象限 "}], [{"aoVal": "B", "content": "第二象限 "}], [{"aoVal": "C", "content": "第三象限 "}], [{"aoVal": "D", "content": "第四象限 "}]], "knowledge_point_routes": ["课内体系->思想->整体思想", "课内体系->知识点->函数->平面直角坐标系->坐标系基础->象限内坐标的特征", "课内体系->能力->运算能力", "课内体系->方法->整体法"], "answer_analysis": ["∵$$abc \\textless{} 0$$， ∴$$a$$，$$b$$，$$c$$中三个都是负数或两正数，一个是负数，当三个都是负数时：若``$$\\frac{x-y}{a}=abc$$，则$$x-y=a^{2}bc\\textgreater0$$，即$$x \\textgreater y$$，同理可得：$$y \\textgreater z$$，$$z \\textgreater x$$这三个式子不能同时成立，即$$a$$，$$b$$，$$c$$不能同时是负数．则$$P\\left( ab,bc \\right)$$不可能在第一象限． 所以选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1519", "queId": "caf2eaa2c8734d0d95924c3c73511e01", "competition_source_list": ["1985年全美数学竞赛（AMC）竞赛第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\frac{1}{7}$$的小数展开后 ， 小数点右边第$$1985$$位是多少．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数的概念及分类->有理数的定义", "美国AMC8->Knowledge Point->Number and Operations->Decimals->Recurring Decimal"], "answer_analysis": ["$$\\frac{1}{7}= 0.142857$$$$142857142857$$$$\\cdots $$，所以有$$6$$位重复的小数 ，所以第$$1985$$位是$$5$$. 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1529", "queId": "f3f5bcea5080424ba92bbacae82b37dd", "competition_source_list": ["1992年第9届全国初中数学联赛竞赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "在一个由$$8\\times 8$$个方格组成的边长为$$8$$的正方形棋盘内放一个半径为$$4$$的圆，若把圆周经过的所有小方格的圆内部分的面积之和记为$${{S}_{1}}$$，把圆周经过的所有小方格的圆外部分的面积之和记为$${{S}_{2}}$$，则$$\\frac{{{S}_{1}}}{{{S}_{2}}}$$的整数部分是（~ ）.", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->四边形->特殊平行四边形->正方形->正方形的性质"], "answer_analysis": ["据正方形的对称性，只需考虑它的$$\\frac{1}{4}$$部分即可，记圆周经过的所有小方格的圆内部分的$$4$$面积之和为$$S_{1}^{\\^{\\prime }}$$，圆周经过的所有小方格的圆外部分的面积之和为$$S_{2}^{\\^{\\prime }}$$， 则$$S_{1}^{\\^{\\prime }}=4\\pi -8$$，$$S_{2}^{\\^{\\prime }}=15-4\\pi $$， $$\\therefore \\frac{S_{1}^{\\^{\\prime }}}{S_{2}^{\\^{\\prime }}}=\\frac{4{{S}_{1}}}{4{{S}_{2}}}=\\frac{4\\pi -8}{15-4\\pi }\\approx \\frac{4.56}{2.44}$$ 故$$\\frac{{{S}_{1}}}{{{S}_{2}}}$$的整数部分是$$1$$ . 故选$$\\text{B}$$ . "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "476", "queId": "5e4da096387b4be98ec040c357756efa", "competition_source_list": ["1999年第10届希望杯初二竞赛第2试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$，$$c$$为正数，且$$a\\ne b$$，若$$x=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$$，$$y=\\frac{1}{\\sqrt{ab}}+\\frac{1}{\\sqrt{bc}}+\\frac{1}{\\sqrt{ca}}$$，则$$x$$与$$y$$的大小关系是（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textgreater y$$ "}], [{"aoVal": "B", "content": "$$x \\textless{} y$$ "}], [{"aoVal": "C", "content": "$$x-y$$ "}], [{"aoVal": "D", "content": "随$$a$$，$$b$$，$$c$$的取值而变化 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->分式->分式的运算->分式加减、乘除、乘方混合运算"], "answer_analysis": ["由题意有$$2x-2y=\\frac{2}{a}+\\frac{2}{b}+\\frac{2}{c}-\\frac{2}{\\sqrt{ab}}-\\frac{2}{\\sqrt{bc}}-\\frac{2}{\\sqrt{ac}}$$ $$=(\\frac{1}{a}-\\frac{2}{\\sqrt{ab}}+\\frac{1}{b})+(\\frac{1}{b}-\\frac{2}{\\sqrt{bc}}+\\frac{1}{c})+(\\frac{1}{c}-\\frac{2}{\\sqrt{ac}}-+\\frac{1}{a})$$ $$={{(\\frac{1}{\\sqrt{a}}-\\frac{1}{\\sqrt{b}})}^{2}}+{{(\\frac{1}{\\sqrt{b}}-\\frac{1}{\\sqrt{c}})}^{2}}+{{(\\frac{1}{\\sqrt{c}}-\\frac{1}{\\sqrt{a}})}^{2}}$$ 又∵$$a\\ne b$$， ∴$${{(\\frac{1}{\\sqrt{a}}-\\frac{1}{\\sqrt{b}})}^{2}}\\textgreater0$$，$${{(\\frac{1}{\\sqrt{b}}-\\frac{1}{\\sqrt{c}})}^{2}}\\geqslant 0$$，$${{(\\frac{1}{\\sqrt{c}}-\\frac{1}{\\sqrt{a}})}^{2}}\\geqslant 0$$． ∴$$x-y\\textgreater0$$，$$x\\textgreater y$$．选$$\\text A$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "351", "queId": "2303af7982004d7ebf4badfd0f676756", "competition_source_list": ["2013年第24届全国希望杯初二竞赛复赛第7题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁$$4$$名跑步运动员的速度依次是$${{v}_{1}}$$，$${{v}_{2}}$$，$${{v}_{3}}$$，$${{v}_{4}}$$，且$${{v}_{1}}\\textgreater{{v}_{2}}\\textgreater{{v}_{3}}\\textgreater{{v}_{4}}\\textgreater0$$，他们沿直跑道进行追逐赛的规则如下： ①$$4$$人在同一起跑线上，同时同向出发； ②经过一段时间后，甲、乙、丙同时反向，谁先遇到丁，谁就是冠军．则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "冠军是甲 "}], [{"aoVal": "B", "content": "冠军是乙 "}], [{"aoVal": "C", "content": "冠军是丙 "}], [{"aoVal": "D", "content": "甲、乙、丙同时遇到丁 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的行程问题->一元一次方程的行程问题-变速问题"], "answer_analysis": ["出发时间$$t$$后，甲、乙、丙与丁的距离为$${{s}_{i}}=({{v}_{i}}-{{v}_{4}})t$$，$$i=1$$，$$2$$，$$3$$． 甲、乙、丙反向后，遇到丁的时间为$${{t}_{i}}^{\\prime }=\\frac{({{v}_{i}}-{{v}_{4}})t}{{{v}_{i}}+{{v}_{4}}}=\\left( 1-\\frac{2{{v}_{4}}}{{{v}_{i}}+{{v}_{4}}} \\right)t$$． 已知$${{v}_{1}}\\textgreater{{v}_{2}}\\textgreater{{v}_{3}}\\textgreater{{v}_{4}}\\textgreater0$$， 故有$${{t}_{1}}\\textgreater{{t}_{2}}\\textgreater{{t}_{3}}$$， 即丙先遇到丁，丙是冠军． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1249", "queId": "8aac490751148307015127b9a84a3662", "competition_source_list": ["2015~2016学年浙江温州乐清市乐清市育英寄宿学校初二上学期期中实验a班第4题4分", "2019~2020学年12月浙江杭州上城区北京师范大学附属杭州中学初三上学期周测A卷第8题3分", "初三上学期单元测试《概率初步》第23题", "2008年竞赛第2题6分", "2020~2021学年10月浙江杭州下城区杭州市景成实验学校中学部初三上学期月考第8题3分", "2016~2017学年9月浙江杭州拱墅区杭州锦绣·育才中学附属学校初三上学期月考第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把一枚六个面编号分别为$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$的质地均匀的正方体骰子先后投掷$$2$$次，若两个正面朝上的编号分别为$$m$$，$$n$$，则二次函数$$y={{x}^{2}}+mx+n$$的图象与$$x$$轴有两个不同交点的概率是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{12}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{17}{36}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率->概率的计算方法->列举法求概率"], "answer_analysis": ["掷骰子有$$6\\times 6=36$$种情况． 根据题意有：$$4n-{{m}^{2}}\\textless{}0$$， 因此满足的点有：$$n=1$$，$$m=3$$，$$4$$，$$5$$，$$6$$， $$n=2$$，$$m=3$$，$$4$$，$$5$$，$$6$$， $$n=3$$，$$m=4$$，$$5$$，$$6$$， $$n=4$$，$$m=5$$，$$6$$， $$n=5$$，$$m=5$$，$$6$$， $$n=6$$，$$m=5$$，$$6$$， 共有$$17$$种， 故概率为：$$17\\div 36=\\frac{17}{36}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "540", "queId": "251e42ca14194a4cbcf9af072c53fdbf", "competition_source_list": ["2005年第16届希望杯初二竞赛复赛第10题4分", "初一下学期其它第17题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知整数$$x$$，$$y$$，$$z$$满足$$x\\leqslant y\\leqslant z$$，且$$\\begin{cases}\\left\\textbar{} x+y \\right\\textbar+\\left\\textbar{} y+z \\right\\textbar+\\left\\textbar{} z+x \\right\\textbar=4 ① \\left\\textbar{} x-y \\right\\textbar+\\left\\textbar{} y-z \\right\\textbar+\\left\\textbar{} z-x \\right\\textbar=2 ②\\end{cases}$$，那么$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}$$的值等于（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$2$$或$$14$$ "}], [{"aoVal": "D", "content": "$$14$$或$$17$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->解含绝对值的方程组", "课内体系->能力->运算能力"], "answer_analysis": ["由②可知$$\\left\\textbar{} x-y \\right\\textbar$$、$$\\left\\textbar{} y-z \\right\\textbar$$、$$\\left\\textbar{} z-x \\right\\textbar$$中必须有一个为$$0$$，只有$$x-y=0$$，$$x=y$$，代入②中得$$\\left\\textbar{} z-x \\right\\textbar=1$$ $$z=x+1$$将$$x=y$$，$$z=x+1$$代入①中，得$$\\left\\textbar{} x \\right\\textbar+\\left\\textbar{} 2x+1 \\right\\textbar=2$$，$$\\left\\textbar{} x \\right\\textbar=0$$，$$1$$，$$2$$．经检验$$\\left\\textbar{} x \\right\\textbar=1$$，依据题意得． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1241", "queId": "8aac4907507fb88401508a7fcce12217", "competition_source_list": ["2018~2019学年甘肃兰州城关区兰州市第三十五中学初二下学期期中第12题4分", "2011年第16届华杯赛初一竞赛初赛第3题", "初一下学期单元测试《不等式与不等式组》一元一次不等式及其应用第28题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a$$，$$b$$是常数，不等式$$\\frac{x}{a}+\\frac{1}{b}\\textgreater0$$的解集为$$x\\textless{}\\frac{1}{5}$$，则关于$$x$$的不等式$$bx-a\\textgreater0$$的解集是．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textgreater\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$x\\textless{}-\\frac{1}{5}$$ "}], [{"aoVal": "C", "content": "$$x\\textgreater-\\frac{1}{5}$$ "}], [{"aoVal": "D", "content": "$$x\\textless{}\\frac{1}{5}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->不等式（组）->含参不等式->由一个已知不等式的解集求另一个不等式的解集"], "answer_analysis": ["不等式$$ \\frac{x}{a}+ \\frac{1}{b}\\textgreater0 $$的解集为$$\\frac{x}{a}\\textgreater- \\frac{1}{b}$$， $$x\\textless{}- \\frac{a}{b}$$， $$x\\textless{} \\frac{1}{5}$$， 所以$$ \\frac{a}{b}=- \\frac{1}{5} $$且$$a\\textless{}0$$，$$b\\textgreater0$$， 所以不等式$$bx-a\\textgreater0$$的解集为$$bx\\textgreater a$$， $$x\\textgreater{} \\frac{a}{b}$$， $$x\\textgreater- \\frac{1}{5}$$， 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1216", "queId": "eddaf324d7854339aa4fb3872c104066", "competition_source_list": ["2013年第24届全国希望杯初二竞赛初赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "点$$(-7,m)$$和点$$(-8,n)$$都在直线$$y=-2x-6$$上，则$$m$$和$$n$$的大小关系是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$m\\textgreater n$$ "}], [{"aoVal": "B", "content": "$$m\\textless{}n$$ "}], [{"aoVal": "C", "content": "$$m=n$$ "}], [{"aoVal": "D", "content": "不能确定的 "}]], "knowledge_point_routes": ["课内体系->知识点->函数->一次函数->一次函数基础", "课内体系->能力->运算能力"], "answer_analysis": ["因为直线$$y=-2x-6$$的斜率$$k=-2\\textless{}0$$， 所以$$y$$随$$x$$的增大而减小． 由于$$-7\\textgreater-8$$， 所以$$m\\textless{}n$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "555", "queId": "3f1fc556b197469c8bb8f49f765b297d", "competition_source_list": ["2008年第19届希望杯初二竞赛第1试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a$$，$$b$$和$$\\sqrt{a}+\\sqrt{b}$$都是有理数，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{a}$$，$$\\sqrt{b}$$都是有理数 "}], [{"aoVal": "B", "content": "$$\\sqrt{a}$$，$$\\sqrt{b}$$都是无理数 "}], [{"aoVal": "C", "content": "$$\\sqrt{a}$$，$$\\sqrt{b}$$都是有理数或都是无理数 "}], [{"aoVal": "D", "content": "$$\\sqrt{a}$$，$$\\sqrt{b}$$中有理数和无理数各有一个 "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->无理数有关的计算->无理数的定义"], "answer_analysis": ["$$a$$，$$b$$都是有理数， ∴$$a-b$$是有理数． 又$$\\sqrt{a}+\\sqrt{b}$$是有理数， ∴$$\\sqrt{a}-\\sqrt{b}=\\frac{a-b}{\\sqrt{a}+\\sqrt{b}}$$是有理数． 于是$$\\sqrt{a}=\\frac{1}{2}\\left[ (\\sqrt{a}+\\sqrt{b})+(\\sqrt{a}-\\sqrt{6}) \\right]$$是有理数 ，$$\\sqrt{b}=\\frac{1}{2}\\left[ (\\sqrt{a}+\\sqrt{b})-(\\sqrt{a}-\\sqrt{b}) \\right]$$是有理数． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1505", "queId": "e5f901de3e994e77bd61f09db8f46f7e", "competition_source_list": ["2008年第19届希望杯初二竞赛第1试第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$，$$c$$都是正整数，并且$$abc=2008$$，则$$a+b+c$$的最小值是~\\uline{~~~~~~~~~~}~． Given that $$a$$, $$b$$, $$c$$ are all positive intergers and $$abc=2008$$, then the minimum value of $$a+b+c$$ is~\\uline{~~~~~~~~~~}~.", "answer_option_list": [[{"aoVal": "A", "content": "$$243$$ "}], [{"aoVal": "B", "content": "$$252$$ "}], [{"aoVal": "C", "content": "$$257$$ "}], [{"aoVal": "D", "content": "$$506$$ "}], [{"aoVal": "E", "content": "$$2010$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->数的特征->因数和倍数", "竞赛->知识点->数��->整除->分解素因数"], "answer_analysis": ["$$2008$$分解成$$2\\times 4\\times 251$$时， $$a+b+c=2+4+251=257$$． 这时$$a+b+c$$的值最小． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "932", "queId": "616a49db47e74327999ad11b87123606", "competition_source_list": ["2012年第23届全国希望杯初一竞赛初赛第2题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "北京景山公园中的景山的相对高度（即从北京的地平面到山顶的垂直距离）是$$45.7$$米，海拔高度是$$94.2$$米，而北京香山公园中的香炉峰（俗称``鬼见愁''）的海拔高度是$$557$$米，则香炉峰的相对高度是（~ ）米．", "answer_option_list": [[{"aoVal": "A", "content": "$$508.5$$ "}], [{"aoVal": "B", "content": "$$511.3$$ "}], [{"aoVal": "C", "content": "$$462.8$$ "}], [{"aoVal": "D", "content": "$$605.5$$ "}]], "knowledge_point_routes": ["课内体系->能力->分析和解决问题能力", "课内体系->能力->运算能力", "课内体系->能力->推理论证能力", "课内体系->知识点->数->有理数->有理数与实际问题->有理数加减法与实际问题"], "answer_analysis": ["由景山的资料知道，北京地平面的海拔高度是$$94.2-45.7=48.5$$（米）， 而$$557-48.5=508.5$$（米），这就是香炉峰的相对高度． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1223", "queId": "8e827a62d976431d8617e2fe8f00d5d0", "competition_source_list": ["2019年湖南长沙雨花区湖南广益实验中学初一竞赛（广益杯）第5题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "在数轴上$$-1$$与$$1$$之间的有理数有．", "answer_option_list": [[{"aoVal": "A", "content": "无数个 "}], [{"aoVal": "B", "content": "$$3$$个 "}], [{"aoVal": "C", "content": "$$2$$个 "}], [{"aoVal": "D", "content": "$$1$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用数轴"], "answer_analysis": ["有无数个有理数． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "35", "queId": "5d000c894b7c426f99da537562e8abc2", "competition_source_list": ["2012年第23届全国希望杯初二竞赛初赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "$${{x}_{1}}$$，$${{x}_{2}}$$，$$\\cdots $$，$${{x}_{100}}$$是自然数，且$${{x}_{1}}\\textless{}{{x}_{2}}\\textless{}\\cdots \\textless{}{{x}_{100}}$$，若$${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{100}}=7001$$，那么，$${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{50}}$$的最大值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2225$$ "}], [{"aoVal": "B", "content": "$$2226$$ "}], [{"aoVal": "C", "content": "$$2227$$ "}], [{"aoVal": "D", "content": "$$2228$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->能力->推理论证能力", "课内体系->知识点->综合与实践->规律探究与程序框图->规律探究->算式找规律"], "answer_analysis": ["因为$${{x}_{1}}$$，$${{x}_{2}}$$，$$\\cdots $$，$${{x}_{100}}$$是自然数，且$${{x}_{1}}\\textless{}{{x}_{2}}\\textless{}\\cdots \\textless{}{{x}_{100}}$$， 所以$$7001={{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{100}}\\geqslant {{x}_{1}}+({{x}_{1}}+1)+({{x}_{1}}+2)+\\cdots +({{x}_{1}}+99)$$， 即$$7001\\geqslant 100{{x}_{1}}+4950$$， 解得$${{x}_{1}}\\leqslant 20$$． 又因为$$20+21+22+\\cdots +119=6950=7001-51$$， $$21+22+23+\\cdots +120=7050\\textgreater7001$$， 所以，当$${{x}_{1}}=20$$，$${{x}_{2}}=21$$，$$\\cdots $$，$${{x}_{49}}=68$$，$${{x}_{50}}=70$$，$${{x}_{51}}=71$$，$$\\cdots $$，$${{x}_{100}}=120$$时， $${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{50}}$$取得最大值$$2226$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "778", "queId": "a36fe09b03c449eb80bcc69012020188", "competition_source_list": ["2011年第22届全国希望杯初一竞赛初赛第10题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两人沿同一路线骑车（匀速）从$$A$$到$$B$$，甲需要$$30$$分钟，乙需要$$40$$分钟，如果乙比甲早出发$$6$$分钟，则甲追上乙以后，乙再经过（~ ）分钟到达$$B$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的和差倍分"], "answer_analysis": ["甲每分钟走全程的$$\\frac{1}{30}$$，乙每分钟走全程的$$\\frac{1}{40}$$， 设甲$$t$$分钟追上乙， 则$$\\frac{t}{30}=\\frac{(6+t)}{40}$$，解得$$t=18$$（分钟）． $$40-6-18=16$$��分钟）． 此后，乙还要走$$16$$分钟才能到达$$B$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "14", "queId": "04f3d76d27ee4c90afb164d96c79ad26", "competition_source_list": ["1996年第13届全国初中数学联赛竞赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$$20$$个点将某圆周$$20$$等分，那么顶点只能在这$$20$$个点中选取的正多边形的个数有（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$个 "}], [{"aoVal": "B", "content": "$$8$$个 "}], [{"aoVal": "C", "content": "$$12$$个 "}], [{"aoVal": "D", "content": "$$24$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->圆->圆与多边形->正多边形与圆->圆内接正多边形相关计算"], "answer_analysis": ["$$\\text{C}$$：$$12$$个；$$\\frac{20}{4}+\\frac{20}{5}+\\frac{20}{10}+\\frac{20}{20}=12$$． ", "设正$$k$$边形满足条件，则除去$$k$$个顶点外的$$20-k$$个点均匀的分布在正$$k$$边形各边所对的劣弧上，则有：
\n于是$$\\frac{20-k}{k}=\\frac{20}{k}-1$$ 是整数，故$$k$$能整除$$20$$，同时$$k\\geqslant 3$$3．
\n所以$$k=4$$或$$5$$或$$10$$或$$20$$，
\n故所求正多边形的个数为$$\\frac{20}{4}+\\frac{20}{5}+\\frac{20}{10}+\\frac{20}{20}=12$$．
\n故选$$C$$．
"], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1206", "queId": "8aac49074e724b45014e87c9ad1d5141", "competition_source_list": ["1996年第7届全国希望杯初一竞赛复赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{6}+\\frac{1}{8}+\\frac{1}{10}+\\frac{1}{12}$$中删去两个加数后使余下的四个加数之和恰等于$$1$$，那么删去的两个加数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{4}$$，$$\\frac{1}{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{4}$$，$$\\frac{1}{12}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{6}$$，$$\\frac{1}{10}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{10}$$，$$\\frac{1}{8}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数加法->有理数加法运算", "课内体系->能力->运算能力"], "answer_analysis": ["易知$$\\frac{1}{4}+\\frac{1}{6}+\\frac{1}{12}=\\frac{1}{2}$$，所以$$\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{6}+\\frac{1}{12}=1$$， 因此删去的两个加数为$$\\frac{1}{8}$$和$$\\frac{1}{10}$$，选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1102", "queId": "ff8080814d9efd56014da5587fc20762", "competition_source_list": ["1992年第3届全国希望杯初一竞赛复赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$1992$$个自然数：$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$1991$$，$$1992$$的每一个数前面任意添上``$$+$$''号或``$$-$$''号，则其代数和一定是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "奇数 "}], [{"aoVal": "B", "content": "偶数 "}], [{"aoVal": "C", "content": "负整数 "}], [{"aoVal": "D", "content": "非负整数 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数基础运算->有理数减法->有理数加减混合运算"], "answer_analysis": ["由于两个整数$$a$$，$$b$$前面任意添加``$$+$$''号或``$$-$$''号，其代数和的奇偶性不变． 这个性质对$$n$$个整数也是正确的． 因此，$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$1991$$，$$1992$$的每一个数前面任意添上``$$+$$''号或``$$-$$''号，其代数和的奇偶性与$$(-1)+2-3+4-5+6-7+8-\\cdots -1991+1992=996$$的奇偶性相同，是偶数，所以选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1292", "queId": "c4f21ef452454b67afd21dd62e58a2ed", "competition_source_list": ["1999年第10届希望杯初一竞赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "$$\\frac{1}{1999}$$的相反数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1999$$ "}], [{"aoVal": "B", "content": "$$-1999$$ "}], [{"aoVal": "C", "content": "$$-\\frac{1}{1999}$$ "}], [{"aoVal": "D", "content": "$$\\left\\textbar{} -\\frac{1}{1999} \\right\\textbar$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->相反数->相反数的性质"], "answer_analysis": ["根据相反数的定义，$$\\frac{1}{1999}$$的相反数是$$-\\frac{1}{1999}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "337", "queId": "b0250beeba4b463eb32d72040b129a3a", "competition_source_list": ["1998年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "对于$$x=7$$，以下哪一项是最小的？．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac 6x$$ "}], [{"aoVal": "B", "content": "$$\\frac 6{x+1}$$ "}], [{"aoVal": "C", "content": "$$\\frac 6{x-1}$$ "}], [{"aoVal": "D", "content": "$$\\frac x6$$ "}], [{"aoVal": "E", "content": "$$\\frac {x+1}6$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Number and Operations->Fractions->Fraction Operations", "课内体系->知识点->数->有理数->分数"], "answer_analysis": ["设$$x=7$$，下面哪个最小？ 最小的分数形式是$$b$$，其中$$b$$大于$$a$$． 在这个问题中，我们需要所有给定值中的最大可能值以分母表示．这个值是$$x+1$$或8 分子越小，就是$$6$$． 带有$$\\frac {6}{x+1}$$的答案选项是$$\\rm B$$． 为每个答案选项代入$$x$$会给出 ($$\\rm A$$)$$\\frac 67$$ ($$\\rm B$$)$$\\frac 68$$ ($$\\rm C$$)$$\\frac 66$$ ($$\\rm D$$)$$\\frac 76$$ ($$\\rm E$$)$$\\frac 86$$ 从这里，我们可以看到最小的是答案选项$$\\rm B$$． The smallest fraction would be in the form $$\\frac{a}{b}$$ where $$b$$ is larger than $$a$$. In this problem, we would need the largest possible value out of all the given values to be in the denominator. This value is $$x+1$$ or $$8$$. The smaller would go on the numerator, which is $$6$$. The answer choice with $$\\frac{6}{x+1}$$ is $$\\text{B}$$. Plugging $$x$$ in for every answer choice would give ($$\\rm A$$)$$\\frac 67$$ ($$\\rm B$$)$$\\frac 68$$ ($$\\rm C$$)$$\\frac 66$$ ($$\\rm D$$)$$\\frac 76$$ ($$\\rm E$$)$$\\frac 86$$ From here, we can see that the smallest is answer choice $$\\text{B}$$. "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1477", "queId": "e5b2e4bbeffa4f329086fd36df93a8c8", "competition_source_list": ["1996年第7届希望杯初二竞赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "化简分式$$\\left( x-y+\\frac{4xy}{x-y} \\right)\\left( x+y-\\frac{4xy}{x+y} \\right)$$的结果是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{y}^{2}}-{{x}^{2}}$$ "}], [{"aoVal": "B", "content": "$${{x}^{2}}-{{y}^{2}}$$ "}], [{"aoVal": "C", "content": "$${{x}^{2}}-4{{y}^{2}}$$ "}], [{"aoVal": "D", "content": "$$4{{x}^{2}}-{{y}^{2}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->分式->分式的基本运算", "课内体系->知识点->式->分式->分式的基础->分式的性质", "课内体系->知识点->式->分式->分式的运算->分式约分"], "answer_analysis": ["原式$$=\\left( x-y+\\frac{4xy}{x-y} \\right)\\left( x+y-\\frac{4xy}{x+y} \\right)$$ $$=\\frac{{{x}^{2}}-2xy+{{y}^{2}}+4xy}{x-y}\\cdot \\frac{{{x}^{2}}+2xy+{{y}^{2}}-4xy}{x+y}$$ $$=\\frac{{{(x+y)}^{2}}}{x-y}\\cdot \\frac{{{(x-y)}^{2}}}{x+y}$$ $$=(x+y)(x-y)$$ $$={{x}^{2}}-{{y}^{2}}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "688", "queId": "566e2d1d43254ddb82412eace033e3b0", "competition_source_list": ["初一下学期其它", "1994年第11届全国初中数学联赛竞赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$x=\\frac{1+\\sqrt{1994}}{2}$$时，多项式$${{(4{{x}^{3}}-1997x-1994)}^{2001}}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$${{2}^{2001}}$$ "}], [{"aoVal": "D", "content": "$$-{{2}^{2001}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式化简求值->二次根式的化简求值——利用完全平方"], "answer_analysis": ["\\textbf{（知识点：二次根式化简求值）} ∵$$x=\\frac{1+\\sqrt{1994}}{2}$$， ∴$${{(2x-1)}^{2}}=1994$$，即$$4{{x}^{2}}-4x-1993=0$$， ∴$${{(4{{x}^{3}}-1997x-1994)}^{2001}}$$ $$={{[(4{{x}^{2}}-4x-1993)x+(4{{x}^{2}}-4x-1993)-1]}^{2001}}$$ $$={{(-1)}^{2001}}$$ $$=-1$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "371", "queId": "1aedc073565b4968aed8ffdb0fb73e90", "competition_source_list": ["2002年第13届希望杯初一竞赛第1试第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "$$\\frac{1}{a}$$是有理数，则它的相反数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$ "}], [{"aoVal": "B", "content": "$$-a$$ "}], [{"aoVal": "C", "content": "$$-\\frac{1}{a}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{a}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["由相反数定义可知，$$\\frac{1}{a}$$的相反数是$$-\\frac{1}{a}$$， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "31", "queId": "0281f223455c40b49bb24a7ffd53191a", "competition_source_list": ["2016年全国华杯赛初一竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x+y+z=5$$，$$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=5$$，$$xyz=1$$，则$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->运算能力", "知识标签->题型->式->整式的乘除->乘法公式->题型：利用完全平方公式计算", "知识标签->知识点->式->分式->分式的运算->分式的加减", "知识标签->知识点->式->整式的乘除->乘法公式->完全平方公式"], "answer_analysis": ["∵$$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=5$$，$$xyz=1$$， ∴$$\\frac{yz+xz+xy}{xyz}=5$$，即$$xy+yz+xz=5$$． ∵$$x+y+z=5$$， ∴$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{\\left( x+y+z \\right)}^{2}}-2\\left( xy+yz+xz \\right)=15$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1214", "queId": "d264bb5b0aa94776b8817c95f8b89575", "competition_source_list": ["2014年第25届全国希望杯初一竞赛初赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$A$$和$$B$$都是$$6$$次多项式，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$A-B$$一定是多项式 "}], [{"aoVal": "B", "content": "$$A-B$$是次数不低于$$6$$的整式 "}], [{"aoVal": "C", "content": "$$A+B$$一定是单项式 "}], [{"aoVal": "D", "content": "$$A+B$$是次数不高于$$6$$的整式 "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式有关的概念->整式->整式的定义", "课内体系->知识点->式->整式的加减->整式有关的概念->多项式->多项式的项"], "answer_analysis": ["若$$A=B$$，则$$A-B=0$$，为单项式，不是多项式，次数为$$0$$； 若$$A=2{{a}^{6}}+{{a}^{5}}$$，$$B=3{{a}^{6}}$$，则$$A+B=5{{a}^{6}}+{{a}^{5}}$$，为多项式． 由上可知，$$\\text{D}$$选项正确． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "569", "queId": "3ac721d239474c32b0950caab657b93d", "competition_source_list": ["2018~2019学年5月天津南开区天津市南开中学初一下学期月考第12题3分", "2000年竞赛（全国初中数学竞赛）第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$，$$b$$，$$c$$的平均数为$$M$$，$$a$$与$$b$$的平均数为$$N$$，$$N$$与$$c$$的平均数为$$P$$，若$$a\\textgreater b\\textgreater c$$，则$$M$$与$$P$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$M=P$$ "}], [{"aoVal": "B", "content": "$$M\\textgreater P$$ "}], [{"aoVal": "C", "content": "$$M \\textless{} P$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-作差法", "课内体系->能力->运算能力"], "answer_analysis": ["由题意得：$$a+b+c=3M$$， $$a+b=2N$$，$$N+c=2P$$， ∴$$M=\\frac{a+b+c}{3}$$， $$P=\\frac{N+c}{2}$$， $$N=\\frac{a+b}{2}$$， ∴将$$N$$代入$$P$$可得： $$P=\\frac{a+b+2c}{4}$$， $$M-P=\\frac{a+b-2c}{12}$$， 又∵$$a\\textgreater b\\textgreater c$$， ∴$$a++c\\textgreater3c$$， ∴$$M-P\\textgreater0$$， ∴$$M\\textgreater P$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1566", "queId": "e6ce5474719a4b06a5493fe5cdcc1679", "competition_source_list": ["2010年第21届全国希望杯初一竞赛初赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\angle AOB$$和$$\\angle BOC$$互补，且$$\\angle AOB$$比$$\\angle BOC$$大$$18{}^{}\\circ $$，则$$\\angle AOB$$的度数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$54{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$81{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$99{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$162{}^{}\\circ $$ "}]], "knowledge_point_routes": ["知识标签->题型->几何图形初步->相交线与平行线->相交线->题型：计算对顶角、邻补角度数", "知识标签->知识点->几何图形初步->相交线与平行线->相交线->邻补角", "知识标签->知识点->几何图形初步->角->角的和差"], "answer_analysis": ["∵$$\\angle AOB$$和$$\\angle BOC$$互为邻补角， ∴$$\\angle AOB+\\angle BOC=180{}^{}\\circ $$． ∵$$\\angle AOB$$比$$\\angle BOC$$大$$18{}^{}\\circ $$， ∴$$\\angle AOB-\\angle BOC=18{}^{}\\circ $$． 两式相加，化简得$$\\angle AOB=99{}^{}\\circ $$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1115", "queId": "ff8080814d9efd56014daa76494d0aa4", "competition_source_list": ["1993年第4���全国希望杯初一竞赛初赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$是有理数，则在下列说法中正确的一个是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-a$$是负数 "}], [{"aoVal": "B", "content": "$${{a}^{2}}$$是正数 "}], [{"aoVal": "C", "content": "$$-\\left\\textbar{} {{a}^{2}} \\right\\textbar$$是负数 "}], [{"aoVal": "D", "content": "$${{(a-1993)}^{2}}+0.001$$是正数 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->有理数乘方运算"], "answer_analysis": ["当$$a=0$$，显然$$\\text{A}$$，$$\\text{B}$$，$$\\text{C}$$，均不正确，应排除，所以选$$\\text{D}$$． 事实上，对任意有理数$$a$$，都有$${{(a-1993)}^{2}}\\geqslant 0$$， 所以$${{(a-1993)}^{2}}+0.001\\textgreater0$$是正数． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "152", "queId": "215f0857bfa742ad823e88d300229584", "competition_source_list": ["2013年第24届全国希望杯初一竞赛复赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$$\\left\\textbar{} x+1 \\right\\textbar+\\left\\textbar{} 2x-1 \\right\\textbar=1$$的整数解的个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["知识标签->知识点->方程与不等式->一元一次方程->解一元一次方程", "知识标签->学习能力->运算能力", "知识标签->题型->方程与不等式->一元一次方程->解一元一次方程->题型：含绝对值的一元一次方程"], "answer_analysis": ["由$$\\left\\textbar{} x+1 \\right\\textbar$$依次取$$0$$，$$\\pm1$$， 可得方程组$$\\begin{cases}x+1=0 2x-1=\\pm 1 \\end{cases}$$或$$\\begin{cases}x+1=\\pm1 2x-1=0 \\end{cases}$$， 解之，可得两方程均无解． 所以，原方程整数解的个数为$$0$$ "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "821", "queId": "69648d5657094f4d80a7a53d2e2d1833", "competition_source_list": ["2018~2019学年9月福建福州鼓楼区福州励志中学初三上学期月考第10题4分", "2017年浙江宁波鄞州区鄞州中学自主招生三位一体第8题5分", "2019~2020学年11月浙江杭州江干区杭州市采荷中学初三上学期月考（杭州采荷中学教育集团）第10题3分", "2006年竞赛第3题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\text{Rt}\\triangle ABC$$在三个顶点$$A$$、$$B$$、$$C$$均在抛物线$$y={{x}^{2}}$$上，并且斜边$$AB$$平行于$$x$$轴．若斜边上的高为$$h$$，则( )．", "answer_option_list": [[{"aoVal": "A", "content": "$$h \\textless{} 1$$ "}], [{"aoVal": "B", "content": "$$h=1$$ "}], [{"aoVal": "C", "content": "$$1 \\textless{} h \\textless{} 2$$ "}], [{"aoVal": "D", "content": "$$h\\textgreater2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->二次函数->二次函数与几何综合->二次函数与直角三角形结合", "课内体系->能力->运算能力", "课内体系->能力->推理论证能力"], "answer_analysis": ["设点$$A$$的坐标为$$\\left( a,{{a}^{2}} \\right)$$，点$$C$$的坐标为$$\\left( c,{{c}^{2}} \\right)$$，$$\\left( \\left\\textbar{} c \\right\\textbar{} \\textless{} \\left\\textbar{} a \\right\\textbar{} \\right)$$，则点$$B$$的坐标为$$\\left( -a,{{a}^{2}} \\right)$$，由勾股定理，得$$A{{C}^{2}}={{\\left( c-a \\right)}^{2}}+{{\\left( {{c}^{2}}-{{a}^{2}} \\right)}^{2}}$$，$$B{{C}^{2}}={{\\left( c+a \\right)}^{2}}+{{\\left( {{c}^{2}}+{{a}^{2}} \\right)}^{2}}$$，$$A{{C}^{2}}+B{{C}^{2}}=A{{B}^{2}}$$，所以$${{\\left( {{a}^{2}}-{{c}^{2}} \\right)}^{2}}={{a}^{2}}-{{c}^{2}}$$．由于$${{a}^{2}}\\textgreater{{c}^{2}}$$，所以$${{a}^{2}}-{{c}^{2}}=1$$，根据图象知斜边$$AB$$上高$${{h}^{2}}={{a}^{2}}-{{c}^{2}}=1$$，所以$$h=1$$，故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1332", "queId": "8aac50a751148307015114e61a5503ff", "competition_source_list": ["2015年第26届全国希望杯初二竞赛复赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$n$$个球放在$$100$$个箱子中（箱子里可以不放球）．若无论怎样放都有$$4$$个箱子里的球的个数相同，则$$n$$的最大值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1616$$ "}], [{"aoVal": "B", "content": "$$1689$$ "}], [{"aoVal": "C", "content": "$$2689$$ "}], [{"aoVal": "D", "content": "$$2616$$ "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "课内体系->知识点->几何图形初步->命题与证明"], "answer_analysis": ["假设可以使得每$$4$$个箱子里的球个数完全相同，那么此���最少有$$3\\times (0+1+2+\\cdots +32)+33=1617$$个球，于是所求的$$n$$的最大值为$$1616$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "377", "queId": "2bf44f63fb964f26a3cb2f8bc5ae68c4", "competition_source_list": ["1992年第3届希望杯初二竞赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "正整数$$a$$被$$7$$除，得到余数$$4$$，则$${{a}^{3}}+5$$被$$7$$除，得到的余数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->综合除法和余数定理 "], "answer_analysis": ["设$$a=7k+4$$（$$k$$为正整数），则 $${{a}^{3}}+5={{\\left( 7k+4 \\right)}^{3}}+5$$ $$={{\\left( 7k \\right)}^{3}}+3\\times {{\\left( 7k \\right)}^{2}}\\times 4+3\\times \\left( 7k \\right)\\times {{4}^{2}}+{{4}^{3}}+5$$ $$=7\\left( {{7}^{2}}{{k}^{3}}+3\\times 7{{k}^{2}}\\times 4+3k\\times {{4}^{2}}+9 \\right)+6$$， 因此，$${{a}^{3}}+5$$被$$7$$除余$$6$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "997", "queId": "8596dd213b9d4a9c9286fce6d168223a", "competition_source_list": ["2010年第15届华杯赛初一竞赛初赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "满足$$\\left\\textbar{} \\left\\textbar{} x-1 \\right\\textbar-\\left\\textbar{} x \\right\\textbar-\\left\\textbar{} x-1 \\right\\textbar+\\left\\textbar{} x \\right\\textbar=1$$的$$x$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$\\pm \\frac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$\\pm \\frac{3}{4}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->解一元一次方程->含绝对值的一元一次方程", "课内体系->知识点->数->有理数->绝对值->绝对值综合", "课内体系->知识点->数->有理数->绝对值->绝对值的非负性", "竞赛->知识点->数与式->绝对值->给定范围绝对值化简"], "answer_analysis": ["当$$x\\leqslant 0$$，原方程为：$$\\left\\textbar{} (-x+1)-(-x) \\right\\textbar-(-x+1)+(-x)=1$$， 即$$1+x-1-x=1$$，无解； 当$$0 ~\\textless{} ~x\\leqslant 1$$，原方程为：$$\\left\\textbar{} (1-x)-x \\right\\textbar-(1-x)+x=1$$， 即$$\\left\\textbar{} 1-2x \\right\\textbar-1+2x=1$$，这种情况下当$$0 ~\\textless{} ~x\\leqslant \\frac{1}{2}$$ 时，$$1-2x-1+2x=1$$无解； 当$$\\frac{1}{2} ~\\textless{} ~x\\leqslant 1$$，$$2x-1-1+2x=1$$，$$x=\\frac{3}{4}$$； 当$$x ~\\textless{} ~1$$，原方程为：$$\\left\\textbar{} (x-1)-x \\right\\textbar-(x-1)+x=1$$， 即$$1-(x-1)+x=1$$，无解． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1033", "queId": "7cc87d2132b24abb8167950d217e67ff", "competition_source_list": ["2008年第19届希望杯初一竞赛第2试第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$$\\left\\textbar{} xy \\right\\textbar-2\\left\\textbar{} x \\right\\textbar+\\left\\textbar{} y \\right\\textbar=4$$的整数解有．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$组 "}], [{"aoVal": "B", "content": "$$4$$组 "}], [{"aoVal": "C", "content": "$$6$$组 "}], [{"aoVal": "D", "content": "$$8$$组 "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->二次方程->二元二次方程组"], "answer_analysis": ["原方程变形为$$\\left( \\left\\textbar{} x \\right\\textbar+1 \\right)\\left( \\left\\textbar{} y \\right\\textbar-2 \\right)=2$$， 因为$$\\left\\textbar{} x \\right\\textbar+1\\geqslant 1$$， 所以$$\\begin{cases}\\left\\textbar{} x \\right\\textbar+1=1 \\left\\textbar{} y \\right\\textbar-2=2 \\end{cases}$$①，或$$\\begin{cases}\\left\\textbar{} x \\right\\textbar+1=2 \\left\\textbar{} y \\right\\textbar-2=1 \\end{cases}$$②， 解①得$$\\begin{cases}x=0 y=4 \\end{cases}$$，或$$\\begin{cases}x=0 y=-4 \\end{cases}$$， 解②得$$\\begin{cases}x=1 y=3 \\end{cases}$$，或$$\\begin{cases}x=1 y=-3 \\end{cases}$$，或$$\\begin{cases}x=-1 y=-3 \\end{cases}$$，或$$\\begin{cases}x=-1 y=3 \\end{cases}$$． 综上所述，原方程组的整数解有$$6$$组． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "963", "queId": "e44c04b574b34353a951be650fad88b7", "competition_source_list": ["2018年第29届希望杯初一竞赛初赛第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$是三个大于$$3$$的质数，则下列判断中一定正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a+b+c$$是偶数 "}], [{"aoVal": "B", "content": "$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$是偶数 "}], [{"aoVal": "C", "content": "$$a+b+c$$是$$3$$的倍数 "}], [{"aoVal": "D", "content": "$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$是$$3$$的倍数 "}], [{"aoVal": "E", "content": "$$a+b+c$$是质数 "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->整除的概念与基本性质", "竞赛->知识点->数论->整除->素数与合数", "课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类", "课内体系->知识点->数->有理数->有理数基础运算->有理数加法->有理数加法运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数除法运算", "课内体系->能力->运算能力", "课内体系->思想->整体思想"], "answer_analysis": ["$$\\text{A}$$选项：因为$$a$$，$$b$$，$$c$$是三个大于$$3$$的质数， 所以$$a$$，$$b$$，$$c$$一定均为奇数， 所以$$a+b+c$$是奇数．错误． $$\\text{B}$$选项：因为$$a$$，$$b$$，$$c$$均为奇数， 所以$$a^{2}$$、$$b^{2}$$，$$c^{2}$$也为奇数， 所以$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$是奇数．错误． $$\\text{C}$$选项：若$$a=5$$、$$b=7$$、$$c=11$$， 则$$a+b+c=23$$，不是$$3$$的倍数．错误． $$\\text{D}$$选项：因为除了$$3$$，所有质数的平方除以$$3$$余数均为$$1$$， 所以$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$是$$3$$的倍数，正确． 故选$$\\text{D}$$. "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1085", "queId": "787103a11ae14229884883745a5e798c", "competition_source_list": ["1997年第14届全国初中数学联赛竞赛"], "difficulty": "3", "qtype": "single_choice", "problem": "给定平面上$$n$$个点，已知$$1$$，$$2$$，$$4$$，$$8$$，$$16$$，$$32$$都是其中两点之间的距离，那么点数$$n$$的最小可能值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["竞赛->知识点->几何杂题->几何最值"], "answer_analysis": ["观察出这些距离都是$$2$$的$$n$$次幂，而$$1+2\\textless{}4$$，$$1+2+4\\textless{}8$$，$$\\cdots $$，$$1+2+4+\\cdots +16\\textless{}32$$，即这些线段中的任几条都不能构成一个多边形，这六条线段也不能构成六边形，所以至少要$$7$$个点才能满足条件． 选择$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "96", "queId": "07f7f530a2f8444a8b4a70b5510a6b38", "competition_source_list": ["初一单元测试《因式分解的应用》第24题", "2005年竞赛第4题6分", "初三上学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$A=48\\times \\left( \\frac{1}{{{3}^{2}}-4}+\\frac{1}{{{4}^{2}}-4}+\\cdots +\\frac{1}{{{100}^{2}}-4} \\right)$$，则与$$A$$最接近的正整数是（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数运算巧解->裂项", "课内体系->能力->运算能力"], "answer_analysis": ["∵$$\\frac{1}{{{a}^{2}}-4}=\\frac{1}{(a+2)(a-2)}=\\frac{1}{4}\\left( \\frac{1}{a+2}-\\frac{1}{a-2} \\right)$$， ∴$$A=48\\times \\left( \\frac{1}{{{3}^{2}}-4}+\\frac{1}{{{4}^{2}}-4}+\\cdots +\\frac{1}{{{100}^{2}}-4} \\right)$$ $$=48\\times \\frac{1}{4}\\left( \\frac{1}{3-2}-\\frac{1}{3+2}+\\frac{1}{4-2}-\\frac{1}{4+2}+\\frac{1}{5-2}-\\frac{1}{5+2}+\\cdots +\\frac{1}{100-2}-\\frac{1}{100+2} \\right)$$ $$=48\\times \\frac{1}{4}\\left( 1-\\frac{1}{5}+\\frac{1}{2}-\\frac{1}{6}+\\frac{1}{3}-\\frac{1}{7}+\\cdots +\\frac{1}{98}-\\frac{1}{102} \\right)$$ $$=12\\times \\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{99}-\\frac{1}{100}-\\frac{1}{101}-\\frac{1}{102} \\right)$$ $$=12\\times \\left( \\frac{25}{12}-\\frac{1}{99}-\\frac{1}{100}-\\frac{1}{101}-\\frac{1}{102} \\right)$$． ∵$$\\frac{1}{99}+\\frac{1}{100}+\\frac{1}{101}+\\frac{1}{102}\\textless{}\\frac{4}{99}\\textless{}\\frac{4}{96}=\\frac{1}{24}$$， ∴$$\\frac{25}{12}\\textgreater\\frac{25}{12}-\\frac{1}{99}-\\frac{1}{100}-\\frac{1}{101}-\\frac{1}{102}\\textgreater\\frac{25}{12}-\\frac{1}{24}=\\frac{49}{24}$$， ∴$$12\\times \\frac{25}{12}\\textgreater12\\times \\left( \\frac{25}{12}-\\frac{1}{99}-\\frac{1}{100}-\\frac{1}{101}-\\frac{1}{102} \\right)\\textgreater12\\times \\frac{49}{24}$$． 即$$25\\textgreater A\\textgreater24.5$$， ∴与$$A$$最接近的正整数是$$25$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "448", "queId": "20134a8cdb6e49b2902ce48cc4850d1e", "competition_source_list": ["1991年第2届希望杯初二竞赛第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "质数$$a$$和$$b$$是$$x$$的整系数方程$${{x}^{2}}-21x+t=0$$的两个根，则$$\\frac{b}{a}+\\frac{a}{b}$$等于（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2213$$ "}], [{"aoVal": "B", "content": "$$\\frac{58}{21}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2402}{49}$$ "}], [{"aoVal": "D", "content": "$$\\frac{365}{38}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->思想->方程思想", "课内体系->知识点->方程与不等式->一元二次方程->根与系数的关系->一元二次方程根与系数的关系"], "answer_analysis": ["由$$a+b=21$$，$$a$$，$$b$$是质数可知$$a$$，$$b$$必为$$2$$与$$19$$两数， 而$$\\frac{b}{a}+\\frac{a}{b}=\\frac{{{a}^{2}}+{{b}^{2}}}{ab}=\\frac{{{2}^{2}}+{{19}^{2}}}{2\\times 19}=\\frac{365}{38}$$．故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "51", "queId": "12e5aa883b0d47d88bfbcbfbbfcd9ad6", "competition_source_list": ["2018年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{2018年全美数学竞赛（AMC）竞赛} Tyler is tiling the floor of his $$12$$ foot by $$16$$ foot living room. He plans to place one-foot by one-foot square tiles to form a border along the edges of the room and to fill in the rest of the floor with two-foot by two-foot square tiles. How many tiles will he use?．", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$87$$ "}], [{"aoVal": "C", "content": "$$91$$ "}], [{"aoVal": "D", "content": "$$96$$ "}], [{"aoVal": "E", "content": "$$120$$ "}]], "knowledge_point_routes": ["课内体系->知识点->四边形->特殊平行四边形->矩形->矩形的周长与面积", "美国AMC8->Knowledge Point->Word Problem->Word Problems Combining with Geometry->Tiles Word Problems"], "answer_analysis": ["Tyler要给自己$$12$$英尺$$\\times 16$$英尺的房间铺地板．为此他打算用边长一英尺的方砖沿墙边铺，并用边长两英尺的地砖铺剩下的地面．他需要多少块地砖？ 他将在边上周围放置$$(12\\cdot 2)+(14\\cdot 2)=52$$块瓷砖．在房间的内部，我们有$$10\\cdot 14=140$$平方英尺．每个瓷砖占$$4$$个平方英尺，所以他会在房间的内部使用$$\\frac {140}{4}=35$$块瓷砖．因此，答案是$$52+35=87$$． 故选$$\\text{B}$$． He will place $$(12\\cdot 2)+(14\\cdot 2)=52$$ tiles around the border. For the inner part of the room, we have $$10\\cdot 14=140$$ square feet. Each tile takes up $$4$$ square feet, so he will use $$\\frac{140}{4}=35$$ tiles for the inner part of the room. Thus, the answer is $$52+35=87$$. "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1010", "queId": "c8e99a304048454a8b5c8eb8b760b868", "competition_source_list": ["2013年竞赛第28题"], "difficulty": "1", "qtype": "single_choice", "problem": "2013年$$AMC8$$竞赛第$$28$$题 On a number line， is the same distance from $$1.75$$ as it is from $$7.25$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$2.75$$ "}], [{"aoVal": "B", "content": "$$3.25$$ "}], [{"aoVal": "C", "content": "$$3.75$$ "}], [{"aoVal": "D", "content": "$$4.5$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Counting, Probability and Statistics->Average Problems->Average Formulas", "课内体系->知识点->数->有理数->有理数与实际问题->有理数乘除法与实际问题"], "answer_analysis": ["The average of $$1.75$$ and $$7.25$$ is equidistant from them， The average is $$(1.75 + 7.25)\\div2=4.5$$． 在数轴上，到$$1.75$$的距离和它到$$7.25$$的距离相同． $$\\text{A}$$．$$2.75$$~ ~ ~ ~ ~ $$\\text{B}$$．$$3.25$$~ ~ ~ ~ ~ $$\\text{C}$$．$$3.75$$~ ~ ~ ~ ~ $$\\text{D}$$．$$4.5$$ $$1.75$$和$$7.25$$的平均值与它们的距离相等，平均值是$$(1.75+7.25)\\div2=4.5$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1591", "queId": "f09fb4012bd444368aa953dd6fc3e3f0", "competition_source_list": ["2013年第24届全国希望杯初一竞赛复赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$是整数，则下列代数式中，值不可能是整数的为（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2a-1}{9}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3a-2}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{{{a}^{2}}-6a-10}{6}$$ "}], [{"aoVal": "D", "content": "$$\\frac{{{a}^{2}}-2}{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类", "竞赛->知识点->数论->整除->整除的概念与基本性质"], "answer_analysis": ["当$$a=5$$时，$$\\frac{2a-1}{9}=1$$，是整数； 当$$a=0$$时，$$\\frac{3a-2}{2}=-1$$，是整数； 当$$a=4$$时，$$\\frac{{{a}^{2}}-6a-10}{6}=-5$$，是整数． 而对任意整数$$a$$，$${{a}^{2}}$$是被$$3$$整除或被$$3$$除余$$1$$的整数， 所以$${{a}^{2}}-2$$是被$$3$$除余$$1$$或余$$2$$的整数， 因此对任意整数$$a$$，$$\\frac{{{a}^{2}}-2}{3}$$都不是整数． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1341", "queId": "8aac50a7511483070151197094871188", "competition_source_list": ["2015年第26届全国希望杯初三竞赛初赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x$$，$$y$$，$$z$$都是正整数，代数式$${{x}^{2}}y-{{y}^{2}}z+{{z}^{2}}x-{{x}^{2}}z+{{y}^{2}}x+{{z}^{2}}y-2xyz$$的值是质数，则$${{z}^{x+y}}$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的加减->整式的加减运算->整式加减的综合", "课内体系->知识点->式->整式的乘除->幂的运算->幂的综合运算", "课内体系->知识点->式->整式的乘除->整式乘除化简求值"], "answer_analysis": ["∵$${{x}^{2}}y-{{y}^{2}}z+{{z}^{2}}x-{{x}^{2}}z+{{y}^{2}}x+{{z}^{2}}y-2xyz=(y+x)(z-x)(z-y)$$， 且代数式的值为质数，$$x$$，$$y$$，$$z$$都是正整数， ∴$$x+y\\geqslant 2$$， ∴$$\\left { \\begin{matrix}z-x=1 z-y=1 \\end{matrix} \\right.$$， 即$$x=y$$， 且$$x+y$$为质数， 又$$x+y=2x$$， ∴$$x=y=1$$， ∴$$z=x+1=2$$， ∴$${{z}^{x+y}}={{2}^{2}}=4$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "692", "queId": "4d8b95fa73c045c583ee13c945060448", "competition_source_list": ["1991年第2届希望杯初二竞赛第4题"], "difficulty": "0", "qtype": "single_choice", "problem": "$$a=\\frac{\\left\\textbar{} -5 \\right\\textbar}{\\sqrt{{{\\left( -5 \\right)}^{2}}}}$$，$$b=\\frac{\\sqrt{{{\\left( 1-3 \\right)}^{2}}}}{1-3}$$，$$c=\\frac{\\left\\textbar{} 4-\\dfrac{3}{4} \\right\\textbar}{3+\\dfrac{1}{4}}$$，则$$a$$，$$b$$，$$c$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b\\textgreater c$$ "}], [{"aoVal": "B", "content": "$$a=b=c$$ "}], [{"aoVal": "C", "content": "$$a=c\\textgreater b$$ "}], [{"aoVal": "D", "content": "$$a=b\\textgreater c$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->实数与根式运算问题"], "answer_analysis": ["因为$$a=1$$，$$b=-1$$，$$c=1$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "150", "queId": "1cd76f8884a449519b1362fab3506651", "competition_source_list": ["1998年第9届希望杯初二竞赛第1试第5题"], "difficulty": "0", "qtype": "single_choice", "problem": "要使分式$$\\frac{1}{\\frac{1-\\left\\textbar{} x \\right\\textbar}{\\left\\textbar{} x \\right\\textbar}}$$有意义，则$$x$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\ne 0$$ "}], [{"aoVal": "B", "content": "$$x\\ne 1$$且$$x\\ne 0$$ "}], [{"aoVal": "C", "content": "$$x\\ne 0$$或$$x\\ne \\pm 1$$ "}], [{"aoVal": "D", "content": "$$x\\ne 0$$且$$x\\ne \\pm 1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->分式->分式的基本运算"], "answer_analysis": ["要使分式$$\\frac{1}{\\frac{1-\\left\\textbar{} x \\right\\textbar}{\\left\\textbar{} x \\right\\textbar}}$$有意义， 必须使$$\\left\\textbar{} x \\right\\textbar\\ne 0$$， 且$$\\frac{1-\\left\\textbar{} x \\right\\textbar}{\\left\\textbar{} x \\right\\textbar}\\ne 0$$， 即$$x\\ne 0$$且$$1-\\left\\textbar{} x \\right\\textbar\\ne 0$$， 所以$$x$$的取值范围是$$x\\ne 0$$且$$x\\ne \\pm 1$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "528", "queId": "556c0998702840b9b3d9894e9da6dd59", "competition_source_list": ["1991年第2届希望杯初二竞赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$$\\textbar{{x}^{2}}-1\\textbar=\\frac{1}{10}\\left( x+\\frac{9}{10} \\right)$$的实根的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元二次方程->根与系数的关系->一元二次方程根与系数的关系"], "answer_analysis": ["当$$\\textbar x\\textbar\\textgreater1$$时，有方程$${{x}^{2}}-1=\\frac{1}{10}\\left( x+\\frac{9}{10} \\right)$$，由根与系数关系可知，方程有一正根一负根，且正根符合要求；当$$\\textbar x\\textbar\\textless{}1$$时，有方程$$1-{{x}^{2}}=\\frac{1}{10}\\left( x+\\frac{9}{10} \\right)$$，同理可知也是一正根一负根，正根符合要求，所以共有$$2$$个根． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1227", "queId": "8aac49074ff4b1620150268a7165710c", "competition_source_list": ["初二上学期单元测试《分式》分式的运算第14题", "1996年第7届希望杯初二竞赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x$$、$$y$$、$$a$$、$$b$$都是正数，且$$a\\textless{}b$$，$$\\frac{x}{y}=\\frac{a}{b}$$，如果$$x+y=c$$，则$$x$$与$$y$$中较大的一个的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{ab}{a+b}$$ "}], [{"aoVal": "B", "content": "$$\\frac{ab}{b+c}$$ "}], [{"aoVal": "C", "content": "$$\\frac{ac}{a+b}$$ "}], [{"aoVal": "D", "content": "$$\\frac{bc}{a+b}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小", "课内体系->知识点->数->实数->无理数有关的计算->无理数大小的比较"], "answer_analysis": ["∵$$x$$、$$y$$、$$a$$、$$b$$都是正数，且$$a\\textless{}b$$， 由$$\\frac{x}{y}=\\frac{a}{b}\\textless{}1$$，得$$x\\textless{}y$$． ∴$$x$$，$$y$$中较大的数是$$y$$． 又$$x+y=c$$，$$x=\\frac{a}{b}y$$， ∴$$\\frac{a}{b}y+y=c$$， 得$$\\frac{a+b}{c}y=c$$， ∴$$y=\\frac{bc}{a+b}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "455", "queId": "abc47ab2859e42e4bffe60ff257e568a", "competition_source_list": ["1998年第9届希望杯初一竞赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a=\\left( -\\frac{1}{6} \\right)+\\left( -\\frac{1}{5} \\right)-\\left( -\\frac{1}{4} \\right)$$．则$$a$$的相反数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{17}{60}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{7}{60}$$ "}], [{"aoVal": "C", "content": "$$\\frac{17}{60}$$ "}], [{"aoVal": "D", "content": "$$\\frac{7}{60}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["$$a=\\left( -\\frac{1}{6} \\right)+\\left( -\\frac{1}{5} \\right)-\\left( -\\frac{1}{4} \\right)$$ $$\\textasciitilde\\textasciitilde=\\frac{(-10)+(-12)-(-15)}{60}$$ $$\\textasciitilde\\textasciitilde=\\frac{-10-12+15}{60}=-\\frac{7}{60}$$． 所以$$a$$的相反数是$$\\frac{7}{60}$$，选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "726", "queId": "44f4b41516ca4c54b9ed6310ddd2ce4f", "competition_source_list": ["2001年第18届全国初中数学联赛竞赛第6题", "2001年第18届全国全国初中数学联赛初一竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "某商场对顾客实行优惠，规定：①如一次购物不超过$$200$$元，则不予折扣；②如一次购物超过$$200$$元但不超过$$500$$元的，按标价给予九折优惠；③如一次购物超过$$500$$元的，其中$$500$$元按第②条给予优惠，超过$$500$$元的部分则给予八折优惠．某人两次去购物，分别付款$$168$$元和$$423$$元；如果他只去一次购物同样的商品，则应付款是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$522.8$$元 "}], [{"aoVal": "B", "content": "$$510.4$$元 "}], [{"aoVal": "C", "content": "$$560.4$$元 "}], [{"aoVal": "D", "content": "$$472.8$$元 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数与实际问题", "课内体系->能力->分析和解决问题能力", "课内体系->能力->运算能力"], "answer_analysis": ["显然，$$168$$小于$$200\\times 0.9=180$$ ，没有经过打折；$$423$$小于$$500\\times 0.9=450$$ ，且大于$$200$$，所以这是经过$$9$$折后的价格；合在一起是$$168+423\\div 0.9=638\\textgreater500$$ ，按照③，可是应付款为：$$500\\times 0.9+138\\times 0.8=560.4$$ 元． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "567", "queId": "5eb318f7a3174e4a8ea377e8087d2bf0", "competition_source_list": ["2014年第31届全国全国初中数学联赛竞赛第2题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知非负实数$$x$$，$$y$$，$$z$$满足$$x+y+z=1$$，则$$t=2xy+yz+2zx$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{4}{7}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{9}{16}$$ "}], [{"aoVal": "D", "content": "$$\\frac{12}{25}$$ "}], [{"aoVal": "E", "content": "答案请写在答题纸上 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->因式分解->因式分解：添项、拆项"], "answer_analysis": ["答案请写在答题纸��� "], "answer_value": "E"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1395", "queId": "a167c87efdce4a5c9733275241ab6c73", "competition_source_list": ["1988年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "在集合 $ {3,6,9,10 }$ 中加入第五个数字$n$，使得这五个数字的平均值等于中位数$$.$$ 则 $n$ 有种可能的取值．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "more than $$4$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Counting, Probability and Statistics->Average Problems->Complex Average Problems", "课内体系->知识点->统计与概率->数据的分析->平均数/加权平均数"], "answer_analysis": ["数列$$ {3,6,9,10 }$$中增加第五个数$$n$$，使得五个数的平均数等于其中位数．$$n$$可能值的个数是？ 添加$$n$$后可能的中间值为$$6$$、$$n$$或$$9$$．现在我们举例分析． 例$$1$$：中位数是$$6$$， 在本例中，$$n\\textless{}6$$，$$\\frac{3+n+6+9+10}{5}=6\\Rightarrow n=2$$，因此本例贡献了$$1$$． 例$$2$$：中位数是$$n$$， 我们有$$6\\textless{}n\\textless{}9$$，$$\\frac{3+6+n+9+10}{5}=n\\Rightarrow n=7$$，所以本例也贡献了$$1$$． 例$$3$$：中位数是$$9$$， 我们有$$9\\textless{}n$$，$$\\frac{3+6+9+n+10}{5}=9\\Rightarrow 17$$，所以本例也贡献$$1$$． 总共有$$3$$个可能的$$n$$值． 故选$$\\text{C}$$． The possible medians after $$n$$ is added are $$6$$, $$n$$, or $$9$$. Now we use casework Case$$1$$: The median is $$6$$, In this case, $$n\\textless{}6$$ and $$\\frac{3+n+6+9+10}{5}=6\\Rightarrow n=2$$, so this case contributes $$1$$. Case$$2$$: The median is $$n$$, We have $$6\\textless{}n\\textless{}9$$ and $$\\frac{3+6+n+9+10}{5}=n\\Rightarrow n=7$$, so this case also contributes$$1$$. Case$$3$$: The median is $$9$$, We have $$9\\textless{}n$$ and $$\\frac{3+6+9+n+10}{5}=9\\Rightarrow 17$$, so this case adds $$1$$. "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "129", "queId": "0f7bf79d1ed848329d07a1366336bbcc", "competition_source_list": ["1997年第8届希望杯初二竞赛第2试第5题", "2020~2021学年安徽合肥包河区初一下学期期末第10题3分", "2016~2017学年河南洛阳新安县初一下学期期末第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a\\textgreater0\\textgreater b\\textgreater c$$，$$a+b+c=1$$，$$M=\\frac{b+c}{a}$$，$$N=\\frac{a+c}{b}$$，$$P=\\frac{a+b}{c}$$，则$$M$$，$$N$$，$$P$$之间的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$M\\textgreater N\\textgreater P$$ "}], [{"aoVal": "B", "content": "$$N\\textgreater P\\textgreater M$$ "}], [{"aoVal": "C", "content": "$$P\\textgreater M\\textgreater N$$ "}], [{"aoVal": "D", "content": "$$M\\textgreater P\\textgreater N$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性"], "answer_analysis": ["$$M+1=\\frac{b+c}{a}+1=\\frac{b+c+a}{a}=\\frac{1}{a}$$， 同理$$N+1=\\frac{1}{b}$$，$$P+1=\\frac{1}{c}$$， 因为$$a\\textgreater0\\textgreater b\\textgreater c$$， 所以$$\\frac{1}{a}\\textgreater\\frac{1}{c}\\textgreater\\frac{1}{b}$$． 所以$$M+1\\textgreater P+1\\textgreater N+1$$． 所以$$M\\textgreater P\\textgreater N$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "631", "queId": "ac27d293406741a5a304ebe7c935f07e", "competition_source_list": ["2001年第12届希望杯初一竞赛第10题", "初一上学期单元测试《一元一次方程》第20题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$k$$为整数，则使得方程$$(k-1999)x=2001-2000x$$的解也是整数的$$k$$值有．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$个 "}], [{"aoVal": "B", "content": "$$8$$个 "}], [{"aoVal": "C", "content": "$$12$$个 "}], [{"aoVal": "D", "content": "$$16$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->含参一元一次方程->一元一次方程的含参整数解"], "answer_analysis": ["$$x=\\frac{2001}{k+1}$$为整数，又$$2001=1\\times 3\\times 23\\times 29$$， $$k+1$$可取$$\\pm 1$$、$$\\pm 3$$、$$\\pm 23$$、$$\\pm 29$$、$$\\pm (3\\times 23)$$、$$\\pm (3\\times 29)$$、$$\\pm (23\\times 29)$$、$$\\pm 2001$$共$$16$$个值，对应的$$k$$值也有$$16$$个． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1219", "queId": "d6fc20bcbe714b448db1f2b609f1fd5a", "competition_source_list": ["初一上学期其它", "1992年第9届全国初中数学联赛竞赛���3题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$${{x}^{2}}-13x+1=0$$，则$${{x}^{4}}-{{x}^{-4}}$$的个位数字是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式"], "answer_analysis": ["由$${{x}^{2}}-13x+1$$可得到$$x+\\dfrac{1}{x}=13$$，所以$${{x}^{2}}+\\dfrac{1}{{{x}^{2}}}={{(x+\\dfrac{1}{x})}^{2}}-2={{13}^{2}}-2=167$$． 同样的$${{x}^{4}}+\\dfrac{1}{{{x}^{4}}}={{({{x}^{2}}+\\dfrac{1}{{{x}^{2}}})}^{2}}-2$$，可以根据个位数字可以直接判断结果的个位数字为$$7$$． "], "answer_value": "D"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "641", "queId": "32e85daae9eb4ed8947ff9e541b3ae39", "competition_source_list": ["2001年第12届希望杯初二竞赛第2试第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知多项式$$a{{x}^{3}}+b{{x}^{2}}+cx+d$$除以$$x-1$$时，所得的余数是$$1$$，除以$$x-2$$时所得的余数是$$3$$，那么多项式$$a{{x}^{3}}+b{{x}^{2}}+cx+d$$除以$$\\left( x-1 \\right)\\left( x-2 \\right)$$时，所得的余式是( )", "answer_option_list": [[{"aoVal": "A", "content": "$$2x-1$$ "}], [{"aoVal": "B", "content": "$$2x+1$$ "}], [{"aoVal": "C", "content": "$$x+1$$ "}], [{"aoVal": "D", "content": "$$x-1$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->整式的乘除运算->综合除法和余数定理 ", "课内体系->知识点->式->整式的乘除->整式的乘除运算->多项式除以多项式", "课内体系->知识点->式->整式的乘除->整式的乘除运算->整式乘除的综合"], "answer_analysis": ["设$$y=a{{x}^{3}}+b{{x}^{2}}+cx+d$$，除以$$\\left( x-1 \\right)\\left( x-2 \\right)$$时所得的余式为$$mx+n$$，商式为$$q\\left( x \\right)$$， 当$$y=1$$时，$$\\left( x-1 \\right)\\cdot q\\left( x \\right)+m+n=1$$， 当$$y=2$$时，$$\\left( x-2 \\right)\\cdot q\\left( x \\right)+2m+n=3$$， 所以$$m=2,n=-1$$ 所以多项式$$a{{x}^{3}}+b{{x}^{2}}+cx+d$$除以$$\\left( x-1 \\right)\\left( x-2 \\right)$$时所得的余式为$$2x-1$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "227", "queId": "21e74ef77ad348b58ed4a87a0797d24b", "competition_source_list": ["2017年浙江宁波鄞州区宁波兴宁中学自主招生第4题5分", "2020年湖南长沙天心区湘郡培粹实验中学初一竞赛初赛（9月）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "有铅笔、练习本、圆珠笔三种学习用品．若购铅笔$$3$$支，练习本$$7$$本，圆珠笔$$1$$支共需$$3.15$$元；若购铅笔$$4$$支，练习本$$10$$本，圆珠笔$$1$$支共需$$4.2$$元．现购铅笔，练习本，圆珠笔各$$1$$个，共需．", "answer_option_list": [[{"aoVal": "A", "content": "$$1.2$$元 "}], [{"aoVal": "B", "content": "$$1.05$$元 "}], [{"aoVal": "C", "content": "$$0.95$$元 "}], [{"aoVal": "D", "content": "$$0.9$$元 "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->其他方程->三元一次方程组->三元一次方程组的实际应用"], "answer_analysis": ["设一支铅笔、一本练习本和一支圆珠笔的单价分别为$$x$$、$$y$$和$$z$$元， 根据题意得：$$\\begin{cases}3x+7y+z=3.15① 4x+10y+z=4.2 ②\\end{cases}$$， $$3\\times$$①$$-$$$$2\\times$$②可得：$$x+y+z=1.05$$． 故选：$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "128", "queId": "0b96a1cb764841b4bd0c9f443c378481", "competition_source_list": ["2001年第12届希望杯初一竞赛第2试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "用一根长为$$a$$米的线围成一个等边三角形，测知这个等边三角形的面积为$$b$$平方米．现在这个等边三角形内任取一点$$P$$，则点$$P$$到等边三角形三边距离之和为（ ）米．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2b}{a}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4b}{a}$$ "}], [{"aoVal": "C", "content": "$$\\frac{6b}{a}$$ "}], [{"aoVal": "D", "content": "$$\\frac{8b}{a}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->等腰三角形->等边三角形->等边三角形的性质", "课内体系->知识点->三角形->三角形及多边形->与三角形有关的线段->三角形的周长与面积问题", "课内体系->知识点->三角形->三角形及多边形->多边形->求多边形周长与面积", "课内体系->能力->推理论证能力"], "answer_analysis": ["等边三角形周长为$$a$$，则边长为$$\\frac{a}{3}$$， 设$$P$$到等边三角���的三边分别为$$x$$、$$y$$、$$z$$， 则等边三角形的面积为$$b=\\frac{1}{2}\\times \\frac{a}{3}\\times \\left( x+y+z \\right)$$ 解得$$x+y+z=\\frac{6b}{a}$$， 故选C "], "answer_value": "C"} +{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "563", "queId": "3ac0af4ca727410c99b2a5444b83cf25", "competition_source_list": ["2015年第32届全国全国初中数学联赛竞赛B卷第1题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "设实数$$a$$，$$b$$，$$c$$满足：$$a+b+c=6$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=4$$，则$$\\frac{{{a}^{2}}+{{b}^{2}}}{2-c}+\\frac{{{b}^{2}}+{{c}^{2}}}{2-a}+\\frac{{{c}^{2}}+{{a}^{2}}}{2-b}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->分式->分式的基本运算"], "answer_analysis": ["∵$$a+b+c=6$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=4$$， ∴$$\\frac{{{a}^{2}}+{{b}^{2}}}{2-c}+\\frac{{{b}^{2}}+{{c}^{2}}}{2-a}+\\frac{{{c}^{2}}+{{a}^{2}}}{2-b}$$ $$=\\frac{4-{{c}^{2}}}{2-c}+\\frac{4-{{a}^{2}}}{2-a}+\\frac{4-{{b}^{2}}}{2-b}$$ $$=2+c+2+a+2+b$$ $$=a+b+c+6$$ $$=6+6$$ $$=12$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "540", "queId": "3fa3ca53534544b4965d9303e491d08d", "competition_source_list": ["2009年黑龙江全国高中数学联赛竞赛初赛第9题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "某电影院第一排共有$$9$$个座位，现有$$3$$名观众就座，若他们每两人都不能相邻且要求每人左右至多只有两个空位，那么不同的坐法种数共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$42$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$54$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["将座位从左至右编为$$1,2,\\cdots ,9$$，考虑从左至右第一个有人坐的座位号： ①如果是$$1$$，则另两个人的座位号只能是$$4$$和$$7$$，只有$$1$$种方法； ②如果是$$2$$，则下一个人可以坐$$4$$或$$5$$．若是$$4$$，则第三个人只能坐$$7$$号；若是$$5$$，则第三个人可以坐$$7$$或$$8$$．共有$$3$$种方法； ③如果是$$3$$，则下一个人可以坐$$5$$或$$6$$．若是$$5$$，则第三个人可以坐$$7$$或$$8$$；若是$$6$$，则第三个人可以坐$$8$$或$$9$$．共有$$2+2=4$$种方法． 所以，不同的坐法种数为$$\\left( 1+3+4 \\right)\\text{A}_{3}^{3}=48$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "686", "queId": "52c23f2c492748e4ab76aef707ff1d70", "competition_source_list": ["2021~2022学年安徽阜阳太和县安徽省太和中学高一下学期月考（竞赛考试）第8~8题"], "difficulty": "3", "qtype": "single_choice", "problem": "在锐角$$\\vartriangle ABC$$中，$$a，b，c$$分别为角$$A，B，C$$的对边，已知$${{b}^{2}}+{{c}^{2}}={{a}^{2}}+bc，b=2$$，则$$\\vartriangle ABC$$的面积\\emph{S}的取值范围是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ \\frac{\\sqrt{3}}{2}，2\\sqrt{3} \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( \\frac{\\sqrt{3}}{2}，2\\sqrt{2} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{\\sqrt{3}}{2}，2\\sqrt{3} \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{\\sqrt{3}}{2}，2\\sqrt{2} \\right)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->解三角形"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据条件求出$$A=\\frac{\\text{ } ! !\\pi ! !\\text{ }}{3}$$，利用三角形面积公式得到$${{S}_{\\vartriangle ABC}}=\\frac{1}{2}bc\\sin A=\\frac{\\sqrt{3}}{2}c$$，采用极端值方法求出$$c$$的最值，进而得到$$c$$的范围，求出面积的取值范围.\\\\ 【详解】\\\\ $$\\cos A=\\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=\\frac{1}{2}$$，因为$$\\vartriangle ABC$$为锐角三角形，故$$A=\\frac{\\text{ } ! !\\pi ! !\\text{ }}{3}$$，\\\\ $${{S}_{\\vartriangle ABC}}=\\frac{1}{2}bc\\sin A=\\frac{\\sqrt{3}}{2}c$$，当\\emph{BC}⊥\\emph{AB}时，$$c=b\\cos A=1$$，当\\emph{CB}⊥\\emph{AC}时，$$c=\\frac b{\\cos A}=4$$，故$$c\\in \\left( 1,4 \\right)$$，所以$${{S}_{\\vartriangle ABC}}=\\frac{\\sqrt{3}}{2}c\\in \\left( \\frac{\\sqrt{3}}{2},2\\sqrt{3} \\right)$$.\\\\ 故选：C "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "713", "queId": "9a6acc98891c444ab037d9f2efeb4aab", "competition_source_list": ["2011年山东全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$z={{(\\sqrt{3}-3\\text i)}^{n}}$$, 若$$z$$为实数，则最小的正整数$$n$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["$$z={{(\\sqrt{3}-3\\text i)}^{n}}={{(-2\\sqrt{3})}^{n}}{{(-\\frac{1}{2}+\\frac{\\sqrt{3}}{2}\\text i)}^{n}}$$，$$n=3$$是使$$z$$为实数的最小的正整数． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "915", "queId": "dfe339efc17f4690a9ef18752acbcb8b", "competition_source_list": ["2015年吉林全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$f(x)=x\\left\\textbar{} x \\right\\textbar$$，若对任意的$$x\\geqslant 1$$有$$f(x+m)+mf(x)\\textless{}0$$恒成立，则实数$$m$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\infty ,-1)$$ "}], [{"aoVal": "B", "content": "$$(-\\infty ,-1]$$ "}], [{"aoVal": "C", "content": "$$(-\\infty ,-2)$$ "}], [{"aoVal": "D", "content": "$$(-\\infty ,-2]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["显然$$m\\textless{}0$$，所以$$f\\left( x+m \\right)\\textless{}-mf\\left( x \\right)=f\\left( \\sqrt{-m}x \\right)$$． 因为$$f\\left( x \\right)$$是单调增的奇函数，所以$$x+m\\textless{}\\sqrt{-m}x$$，即$$\\left( \\sqrt{-m}-1 \\right)x\\textgreater m$$． 所以必须$$\\sqrt{-m}-1\\geqslant 0$$，$$m\\leqslant -1$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1129", "queId": "e18b3207e5a740c59fa460158c0a3b9e", "competition_source_list": ["2015年黑龙江全国高中数学联赛竞赛初赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a\\textgreater0,b\\textgreater0$$，则下列不等式中不恒成立的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2}{\\frac{1}{a}+\\frac{1}{b}}\\geqslant \\sqrt{ab}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{a}+\\frac{1}{b}\\geqslant \\frac{4}{a+b}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{\\textbar a-b\\textbar}\\geqslant \\sqrt{a}-\\sqrt{b}$$ "}], [{"aoVal": "D", "content": "$${{a}^{2}}+{{b}^{2}}+1\\textgreater a+b$$ "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->等式与不等式->不等式->不等式的性质->针对不等式变形判断正误", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件", "课内体系->知识点->等式与不等式->不等式->基本不等式->利用基本不等式证明其它不等式", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的实际应用"], "answer_analysis": ["$$\\frac{1}{a}+\\frac{1}{b}\\geqslant 2\\sqrt{\\frac{1}{ab}}$$，则$$\\sqrt{ab}\\left( \\frac{1}{a}+\\frac{1}{b} \\right)\\geqslant 2$$，即$$\\frac{2}{\\frac{1}{a}+\\frac{1}{b}}\\leqslant \\sqrt{ab}$$，当且仅当$$a=b$$时取等号，故不等式不恒成立. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "158", "queId": "4ad0aed456b848718fcb3553ee471743", "competition_source_list": ["2021年贵州全国高中数学联赛竞赛初赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$G$$和$$O$$分别是$$\\triangle ABC$$的重心和外心，且$$AB=\\sqrt{3}$$，$$AC=3$$，则$$\\overrightarrow{AG}\\cdot \\overrightarrow{AO}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->平面向量", "竞赛->知识点->复数与平面向量->平面向量的概念与运算"], "answer_analysis": ["因为$$O$$是$$\\triangle ABC$$的外心，故$$O$$在$$AB$$边上的投影为$$AB$$的中点， 故$$\\overrightarrow{AB}\\cdot \\overrightarrow{AO}=\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar\\left\\textbar{} \\overrightarrow{AO} \\right\\textbar\\cdot \\cos \\angle BAO=\\frac{1}{2}{{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}^{2}}$$， 同理可得，$$\\overrightarrow{AC}\\cdot \\overrightarrow{AO}=\\frac{1}{2}{{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}^{2}}$$， 所以$$\\overrightarrow{AG}\\cdot \\overrightarrow{AO}=\\frac{1}{3}(\\overrightarrow{AB}+\\overrightarrow{AC})\\cdot \\overrightarrow{AO}=\\frac{1}{6}\\left( {{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}^{2}}+{{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}^{2}} \\right)=\\frac{1}{6}(3+9)=2$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "236", "queId": "98dae8ac5e28461083e119e1eb95faab", "competition_source_list": ["2012年吉林全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "(★★)若一系列函数的解析式相同，值域相同，但其定义域不同，则称这些函数为``同族函数''，那么函数解析式为$$y=-{{x}^{2}}$$，值域为$$\\left { 0, -1, -9 \\right }$$的``同族函数''共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$个 "}], [{"aoVal": "B", "content": "$$8$$个 "}], [{"aoVal": "C", "content": "$$9$$个 "}], [{"aoVal": "D", "content": "$$10$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->函数的定义", "竞赛->知识点->函数->函数的概念"], "answer_analysis": ["$$1\\times 3\\times 3=9$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "504", "queId": "3239768bc25f4cb6bd420e810cb9be59", "competition_source_list": ["2013年辽宁全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "原点的直线$$l$$交双曲线$$xy=-2\\sqrt{2}$$于$$P,Q$$两点，其中$$P$$点在第二象限，将下半平面沿$$x$$轴折起使之与上半平面成直二面角，线段$$PQ$$的最短长度是．", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$3\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$4\\sqrt{2}$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->双曲线"], "answer_analysis": ["设点$$P$$的坐标为$$\\left( x,y \\right)$$，由对称性$$Q$$的坐标为$$\\left( -x,-y \\right)$$，分别过$$P,Q$$作$$x$$轴的垂线，垂足分别为$${{P}_{1}}\\left( x,0 \\right){ }{{Q}_{1}}\\left( -x,0 \\right)$$，折成直二面角后， $${{\\left\\textbar{} PQ \\right\\textbar}^{2}}={{\\left\\textbar{} P{{P}_{1}} \\right\\textbar}^{2}}+{{\\left\\textbar{} {{P}_{1}}{{Q}_{1}} \\right\\textbar}^{2}}+{{\\left\\textbar{} {{Q}_{1}}Q \\right\\textbar}^{2}}$$ $$={{y}^{2}}+{{\\left( 2x \\right)}^{2}}+{{\\left( -y \\right)}^{2}}=4{{x}^{2}}+2{{\\left( \\frac{-2\\sqrt{2}}{x} \\right)}^{2}}$$ $$=4{{x}^{2}}+\\frac{16}{{{x}^{2}}}{\\geqslant }2\\sqrt{64}=16$$， 即$$\\left\\textbar{} PQ \\right\\textbar{\\geqslant }4$$；当$$4{{x}^{2}}=\\frac{16}{{{x}^{2}}}$$即$$x=\\sqrt{2}$$时，$$\\left\\textbar{} PQ \\right\\textbar=4$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "301", "queId": "cbfebe1748534f2bb8fcb14432cd10e0", "competition_source_list": ["竞赛第19题"], "difficulty": "2", "qtype": "single_choice", "problem": "正六棱柱的12个顶点的任意2个项点所在直线中，异面直线的对数为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "1125 "}], [{"aoVal": "B", "content": "1278 "}], [{"aoVal": "C", "content": "1350 "}], [{"aoVal": "D", "content": "1542 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 先考虑任取4个点，求出不共面的情形的个数，从而可求异面直线的对数.\\\\ 【详解】\\\\ 12个顶点构成的4点组有${\\mathrm{C}}_{12}^{4}=495$个，其中共面的4点组可以按在上下底面的个数分类讨论．\\\\ \\textbf{情形一}~~~全部在上底面或全部在下底面．共面4点组数为$2{\\mathrm{C}}_{6}^{4}=30$．\\\\ \\textbf{情形二}~~~两个底面各2个，共面4点组数按在上底面的2个点间的位置关系分类（此时上下底面的点分别连线，则连线平行），为$6\\times 3+6\\times 2+3\\times 3=39$．\\\\ 因此共面的4点组共有69个，进而不共面的4点组有426个，每个不共面的4点组贡献3对异面直线，且不同的4点组贡献的异面直线不同，\\\\ 因此所求异面直线的对数为$426\\times 3=1278$．\\\\ 故选：B. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "386", "queId": "82fd9ed9d65e4e0aa950afc9bf57470b", "competition_source_list": ["2008年湖南全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a=\\sin (\\sin 2008{}^{}\\circ )$$，$$b=\\sin (\\cos 2008{}^{}\\circ )$$，$$c=\\cos (\\sin 2008{}^{}\\circ )$$，$$d=\\cos (\\cos 2008{}^{}\\circ )$$，则$$a$$、$$b$$、$$c$$、$$d$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless{}b\\textless{}c\\textless{}d$$ "}], [{"aoVal": "B", "content": "$$b\\textless{}a\\textless{}d\\textless{}c$$ "}], [{"aoVal": "C", "content": "$$c\\textless{}d\\textless{}b\\textless{}a$$ "}], [{"aoVal": "D", "content": "$$d\\textless{}c\\textless{}a\\textless{}b$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与��质"], "answer_analysis": ["因为$$2008{}^{}\\circ =5\\times 360{}^{}\\circ +180{}^{}\\circ +28{}^{}\\circ $$，所以， $$a=\\sin (-\\sin 28{}^{}\\circ )=-\\sin (\\sin 28{}^{}\\circ )\\textless{}0$$； $$b=\\sin (-\\cos 28{}^{}\\circ )=-\\sin (\\cos 28{}^{}\\circ )\\textless{}0$$； $$c=\\cos (-\\sin 28{}^{}\\circ )=\\cos (\\sin 28{}^{}\\circ )\\textgreater0$$； $$d=\\cos (-\\cos 28{}^{}\\circ )=\\cos (\\cos 28{}^{}\\circ )\\textgreater0$$． 又$$\\sin 28{}^{}\\circ \\textless{}\\cos 28{}^{}\\circ $$，故$$b\\textless{}a\\textless{}d\\textless{}c.$$故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "38", "queId": "061c167dfb5e4bcdb4a8dc742d2921a9", "competition_source_list": ["2008年天津全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "设不定方程$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xyz+10=0$$的正整数解$$\\left( x, y, z \\right)$$中满足$$x, y, z$$均大于$$2008$$的不同解的数目为$$k$$，则$$k$$满足．", "answer_option_list": [[{"aoVal": "A", "content": "$$k=0$$ "}], [{"aoVal": "B", "content": "$$1\\leqslant k\\leqslant 2008$$ "}], [{"aoVal": "C", "content": "$$k\\textgreater2008$$，但$$k$$是有限的数 "}], [{"aoVal": "D", "content": "$$k$$是无穷大 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->韦达定理"], "answer_analysis": ["由题意知$$\\left( {{x}_{0}}, {{y}_{0}}, {{z}_{0}} \\right)=\\left( 3, 4, 5 \\right)$$是原不定方程的一个特解．对于原不定方程的任意一个正整数解$$\\left( {{x}_{1}}, y, z \\right)$$，假设$${{x}_{1}}\\leqslant y\\leqslant z$$，且$${{x}_{1}}z$$．设关于$$x$$的二次方程$${{x}^{2}}-yz\\cdot x+{{y}^{2}}+{{z}^{2}}+10=0$$的两个根为$${{x}_{1}}, {{x}_{2}}$$，由韦达定理，$${{x}_{1}}+{{x}_{2}}=yz, {{x}_{1}}{{x}_{2}}={{y}^{2}}+{{z}^{2}}+10{{z}^{2}}$$，因此$${{x}_{2}}$$是正整数，且大于$$z$$，于是$$\\left( {{x}_{2}}, y, z \\right)$$也是原不定方程的一个解，由于原不定方程是轮换对称的，所以$$\\left( y, z, {{x}_{2}} \\right)$$也是它的解，并且它是由小到大排列的． 如此反复利用上面的结论，可以由一个特解得到无穷多个解，因此满足$$x, y, z$$均大于$$2008$$的解有无穷多个．故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1200", "queId": "fe3d8dfb91ad4ce5b5d0c479666a1074", "competition_source_list": ["2018年湖北全国高中数学联赛竞赛初赛第1题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "若对任意的$$\\theta \\in \\left[ 0,\\frac{\\pi }{2} \\right]$$，不等式$$4+2\\sin \\theta \\cos \\theta -a\\sin \\theta a\\cos \\theta \\leqslant 0$$恒成立，则实数$$a$$的最小值为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["知识标签->知识点->三角函数->三角函数的图象与性质->正弦型函数->正弦型函数的图象与性质", "知识标签->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正弦", "知识标签->题型->三角函数->三角函数的图象与性质->正弦型函数->正弦型三角函数图像与性质问题", "知识标签->题型->三角函数->三角恒等变换->倍角、半角公式问题", "知识标签->素养->数学运算"], "answer_analysis": ["设$$x=\\sin \\theta +\\cos \\theta =\\sqrt{2}\\sin \\left( \\theta +\\frac{\\pi }{4} \\right)$$，则$$2\\sin \\theta \\cdot \\cos \\theta ={{x}^{2}}-1$$， 当$$\\theta \\in \\left[ 0,\\frac{\\pi }{2} \\right]$$时，得$$1\\leqslant x\\leqslant \\sqrt{2}$$， 故原不等式变为：$${{x}^{2}}-ax+3\\leqslant 0\\Rightarrow a\\geqslant x+\\frac{3}{x}$$， 此时，函数$$f\\left( x \\right)=x+\\frac{3}{x}$$单调递减， 从而，实数$$a$$的最小值为$$f\\left( 1 \\right)=4$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1205", "queId": "fe536d4330f64edea89419a3d2c06eb1", "competition_source_list": ["竞赛第12题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知一圆锥曲面顶点\\emph{S}，其母线与轴所成的角为$30{^{\\circ}}$，在轴线上取一点\\emph{C}，使得$SC=5$，过点\\emph{C}作一个与轴线夹角为$45{^{\\circ}}$的截面，则截得的曲线方程可表示为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$x^{2}+2y^{2}=25$ "}], [{"aoVal": "B", "content": "$x^{2}+3y^{2}=50$ "}], [{"aoVal": "C", "content": "$2x^{2}+5y^{2}=50$ "}], [{"aoVal": "D", "content": "$2x^{2}+6y^{2}=75$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据平面截圆锥所得曲线形状的判断法则可得截得的��线为椭圆且离心率为$e=\\frac{\\cos 45{^{\\circ}}}{\\cos 30{^{\\circ}}}$，求出其长轴后可得曲线方程.\\\\ 【详解】\\\\ 根据平面截圆锥所得曲线形状的判断法则，截得的曲线为椭圆，离心率$e=\\frac{\\cos 45{^{\\circ}}}{\\cos 30{^{\\circ}}}=\\frac{\\sqrt{6}}{3}$，\\\\ 且根据正弦定理，可得椭圆的长轴长$2a=SC\\cdot \\sin 30{^{\\circ}}\\left(\\frac{1}{\\sin 75{^{\\circ}}}+\\frac{1}{\\sin 15{^{\\circ}}}\\right)=5\\sqrt{6}$，\\\\ 因此椭圆的半焦距$c=5$，进而所求曲线方程为$\\frac{x^{2}}{\\frac{75}{2}}+\\frac{y^{2}}{\\frac{25}{2}}=1\\Rightarrow 2x^{2}+6y^{2}=75$．\\\\ 故选：D. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "746", "queId": "9a9a036237e146479dbf32341380e5b9", "competition_source_list": ["2008年山西全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$${{a}_{n}}=\\frac{n(n+1)}{2}$$，从数列$$ {{{a}_{n}} }$$中，去掉所有能被$$3$$整除的数后，剩下的项自小到大排成数列$$ {{{b}_{n}} }$$，则$${{b}_{200}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$200808$$ "}], [{"aoVal": "B", "content": "$$179101$$ "}], [{"aoVal": "C", "content": "$$153201$$ "}], [{"aoVal": "D", "content": "$$116808$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["当且仅当$$n=3k-2(k=1,2,\\cdots )$$时能被$$3$$整除， 故$${{b}_{k}}=\\frac{(3k-2)(3k-1)}{2}=1+\\frac{9k(k-1)}{2}$$， 所以 $${{b}_{200}}=1+\\frac{9\\times 200\\times 199}{2}=179101$$， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "10", "queId": "29da40057dbe4ce99e3400bf9d5755a3", "competition_source_list": ["2010年辽宁全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "某天下午要排物理、化学、生物和两节自习共$$5$$节课，如果第一节不排生物，最后一节不排物理，那么不同的排法共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$种 "}], [{"aoVal": "B", "content": "$$39$$种 "}], [{"aoVal": "C", "content": "$$60$$种 "}], [{"aoVal": "D", "content": "$$78$$种 "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["不加限制的排法共有$$\\frac{\\text{A}_{5}^{5}}{2}=60$$种；生物排在第一节，物理排在最后一节各有$$\\frac{\\text{A}_{4}^{4}}{2}=12$$种，其中生物排在第一节且物理排在最后一节有$$\\frac{\\text{A}_{3}^{3}}{2}=3$$种．因此符合要求的排法共有$$60-12\\times 2+3=39$$种． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "136", "queId": "100ccf0d9b0a48ff8d3bd3dc17413ee5", "competition_source_list": ["2008年山东全国高中数学联赛竞赛初赛第10题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "对于实数$$x$$，$$[x]$$表示不超过$$x$$的最大整数．已知正数数列$$ {{{a}_{n}} }$$满足：$${{a}_{1}}=1$$， $${{S}_{n}}=\\frac{1}{2}\\left( {{a}_{n}}+\\frac{1}{{{a}_{n}}} \\right)$$，其中$${{S}_{n}}$$为数列$$ {{{a}_{n}} }$$的前$$n$$项和，则$$\\left[ \\frac{1}{{{S}_{1}}}+\\frac{1}{{{S}_{2}}}+\\ldots +\\frac{1}{{{S}_{100}}} \\right]=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["由$${{S}_{n}}=\\frac{1}{2}\\left( {{a}_{n}}+\\frac{1}{{{a}_{n}}} \\right)=\\frac{1}{2}\\left( {{S}_{n}}-{{S}_{n-1}}+\\frac{1}{{{S}_{n}}-{{S}_{n-1}}} \\right)$$， 得$${{S}_{n}}+{{S}_{n-1}}=\\frac{1}{{{S}_{n}}-{{S}_{n-1}}}$$， 即$$S_{n}^{2}=S_{n-1}^{2}+1$$． 因为$${{S}_{1}}={{a}_{1}}=1$$，所以$$S_{n}^{2}=n{{S}_{n}}=\\sqrt{n}$$． 因为$$\\sqrt{n}+\\sqrt{n-1}\\textless{}2\\sqrt{n}\\textless{}\\sqrt{n+1}+\\sqrt{n}$$， 所以$$\\frac{1}{\\sqrt{n+1}+\\sqrt{n}}=\\sqrt{n+1}-\\sqrt{n}\\textless{}\\frac{1}{2\\sqrt{n}}$$ $$\\textless{}\\frac{1}{\\sqrt{n}+\\sqrt{n-1}}=\\sqrt{n}-\\sqrt{n-1}$$， 令$$S=\\frac{1}{{{S}_{1}}}+\\frac{1}{{{S}_{2}}}+\\ldots +\\frac{1}{{{S}_{100}}}$$， 则$$\\frac{S}{2}\\textgreater\\sqrt{101}-1\\textgreater9$$， 解得$$S\\textgreater18$$． 又因为$${{S}_{1}}={{a}_{1}}=1$$，所以$$\\frac{S}{2}-\\frac{1}{2{{S}_{1}}}=\\frac{1}{2{{S}_{2}}}+\\frac{1}{2{{S}_{3}}}+\\ldots +\\frac{1}{2{{S}_{100}}}\\textless{}\\sqrt{100}-1=9$$， 即$$S\\textless{}2\\left( 9+\\frac{1}{2} \\right)=19$$，从而得$$[S]=18$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "211", "queId": "1131e97a4a29405f93e10f6b35b53296", "competition_source_list": ["2008年江西全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$n$$为正整数，且$$3n+1$$与$$5n-1$$皆为完全平方数，对于以下两个命题：（甲）$$7n+13$$必为合数；（乙）$$8\\left( 17{{n}^{2}}+3n \\right)$$必为两个平方数的和． 你的判断是．", "answer_option_list": [[{"aoVal": "A", "content": "甲对乙错 "}], [{"aoVal": "B", "content": "甲错乙对 "}], [{"aoVal": "C", "content": "甲、乙都对 "}], [{"aoVal": "D", "content": "甲、乙都不一定对 "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->整除->质数（算数基本定理）"], "answer_analysis": ["设$$3n+1={{a}^{2}}$$，$$5n-1={{b}^{2}}$$，$$a$$，$$b$$为正整数， 则$$7n+13=9\\left( 3n+1 \\right)-4\\left( 5n-1 \\right)={{\\left( 3a \\right)}^{2}}-{{\\left( 2b \\right)}^{2}}$$ $$=\\left( 3a-2b \\right)\\left( 3a+2b \\right)$$① 由此知，$$3a-2b$$为正整数，且$$3a-2b\\ne 1$$． 因为若$$3a-2b=1$$， 则$$27n+9={{\\left( 3a \\right)}^{2}}={{\\left( 2b+1 \\right)}^{2}}=4{{b}^{2}}+4b+1$$， 即$$27n=4\\left( {{b}^{2}}+b-2 \\right)$$，则$$4\\left\\textbar{} n \\right.$$，记$$n=4k$$， 得$$5n-1=20k-1$$不为平方数，矛盾！ 所以$$3a-2b\\geqslant 2$$，故由①式得，$$7n+13$$为合数； 又因为$$8\\left( 17{{n}^{2}}+3n \\right)=\\left[ \\left( 3n+1 \\right)+\\left( 5n-1 \\right) \\right]\\left[ 4\\left( 3n+1 \\right)+\\left( 5n-1 \\right) \\right]$$ $$=\\left( {{a}^{2}}+{{b}^{2}} \\right)\\left[ {{\\left( 2a \\right)}^{2}}+{{b}^{2}} \\right]={{\\left( 2{{a}^{2}}+{{b}^{2}} \\right)}^{2}}+{{\\left( ab \\right)}^{2}}$$， 故选$$\\text{C}$$．（例如$$65$$即可为上述$$n$$的一个取值） "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1009", "queId": "e05ac10cdd7040ff96a9cb32b7400ce0", "competition_source_list": ["2008年黑龙江全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "设集合$$P= {(x,y)\\textbar y=k }$$，$$Q= {(x,y)\\textbar y={{a}^{x}}+1$$，$$a\\textgreater0$$且$$a\\ne 1 }$$，已知$$P\\cap Q$$只有一个子集，那么$$k$$的取值范围．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\infty ,1)$$ "}], [{"aoVal": "B", "content": "$$\\left( -\\infty ,+1 \\right]$$ "}], [{"aoVal": "C", "content": "$$(1,+\\infty )$$ "}], [{"aoVal": "D", "content": "$$(+\\infty ,+\\infty )$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算", "竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["$$P\\cap Q$$只有一个子集即$$P\\cap Q$$为空集， 数形结合可知$$k\\leqslant 1$$时，直线$$y=k$$与曲线$$y={{a}^{x}}+1$$无交点．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "680", "queId": "faf98c239e6b4daa997a39d37e8a386c", "competition_source_list": ["第二十届全国希望杯高一竞赛复赛邀请赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "设函数$$f(x)$$，$$g(x)$$的定义域为$$D$$，又$$h(x)=f(x)+g(x)$$．若$$f(x),g(x)$$的最大值分别是$$M$$，$$N$$，最小值分别是$$m$$，$$n$$，则下面的结论中正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$h(x)$$的最大值是$$M+N$$ "}], [{"aoVal": "B", "content": "$$h(x)$$的最小值是$$m+n$$ "}], [{"aoVal": "C", "content": "$$h(x)$$的值域为$$ {x\\textbar m+n\\leqslant x\\leqslant M+N }$$ "}], [{"aoVal": "D", "content": "$$h(x)$$的值域为$$ {x\\textbar m+n\\leqslant x\\leqslant M+N }$$的一个子集 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["取$$f(x)=\\sin x,g(x)=\\cos x$$，$$D=\\mathbf{R}$$，即可知$$\\text A\\text B\\text C$$都不正确． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "895", "queId": "9b590c238eaf4e9fbb3aff474ed4f210", "competition_source_list": ["1999年全国全国高中数学联赛竞赛一试第1题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "给定公比为$$q\\left( q\\ne 1 \\right)$$的等比数列$$\\left { {{a}_{n}} \\right }$$，设$${{b}_{1}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}$$，$${{b}_{2}}={{a}_{4}}+{{a}_{5}}+{{a}_{6}}$$，\\ldots，$${{b}_{n}}={{a}_{3n}}_{-2}+{{a}_{3n}}_{-1}+{{a}_{3n}}$$，\\ldots，则数列$$\\left { {{b}_{n}} \\right }$$（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "是等差数列 "}], [{"aoVal": "B", "content": "是公比为$$q$$的等比数列 "}], [{"aoVal": "C", "content": "是公比为$${{q}^{3}}$$的等比数列 "}], [{"aoVal": "D", "content": "既非等差数列也非等比数列 "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["由题设，$${{a}_{n}}={{a}_{1}}{{q}^{n-1}}$$， $$\\frac{{{b}_{n+1}}}{{{b}_{n}}}=\\frac{{{a}_{3n+1}}+{{a}_{3n+2}}+{{a}_{3n+3}}}{{{a}_{3n-2}}+{{a}_{3n-1}}+{{a}_{3n}}}=\\frac{{{a}_{1}}{{q}^{3n}}+{{a}_{1}}{{q}^{3n+1}}+{{a}_{1}}{{q}^{3n+2}}}{{{a}_{1}}{{q}^{3n-3}}+{{a}_{1}}{{q}^{3n-2}}+{{a}_{1}}{{q}^{3n-1}}}$$， $$=\\frac{{{a}_{1}}{{q}^{3n}}\\left( 1+q+{{q}^{2}} \\right)}{{{a}_{1}}{{q}^{3n-3}}\\left( 1+q+{{q}^{2}} \\right)}={{q}^{3}}$$， 因此，$$\\left { {{b}_{n}} \\right }$$是公比为$${{q}^{3}}$$的等比数列． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "404", "queId": "79d6503bfc7a4500831da0ddd27da6b2", "competition_source_list": ["2008年陕西全国高中数学联赛竞赛初赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f(x)={{x}^{3}}-{{\\log }_{2}}(\\sqrt{{{x}^{2}}+1}-x)$$，则对于任意实数$$a$$、$$b(a+b\\ne 0)$$，$$\\frac{f(a)+f(b)}{{{a}^{3}}+{{b}^{3}}}$$的值．", "answer_option_list": [[{"aoVal": "A", "content": "恒大于零 "}], [{"aoVal": "B", "content": "恒等于零 "}], [{"aoVal": "C", "content": "恒小于零 "}], [{"aoVal": "D", "content": "符号不确定 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["因为$$f(-x)+f(x)$$ $$=-{{x}^{3}}-{{\\log }_{2}}(\\sqrt{{{x}^{2}}+1}+x)+{{x}^{3}}-{{\\log }_{2}}(\\sqrt{{{x}^{2}}+1}-x)$$ $$=-{{\\log }_{2}}[(\\sqrt{{{x}^{2}}+1}+x)(\\sqrt{{{x}^{2}}+1}-x)]=0$$， 所以$$f(-x)=-f(x)$$，即$$f(x)$$为奇函数． 又因为$$f(x)={{x}^{3}}-{{\\log }_{2}}(\\sqrt{{{x}^{2}}+1}-x)={{x}^{3}}+{{\\log }_{2}}(\\sqrt{{{x}^{2}}+1}+x)$$， 在$$(0,+\\infty )$$上单调递增，所以，$$f(x)$$在$$\\mathbf{R}$$上是增函数． 注意到$$f(a)-f(-b)$$与$$a-(-b)$$同号， 所以$$\\frac{f(a)+f(b)}{a+b}\\textgreater0$$． 又因为$${{a}^{3}}+{{b}^{3}}$$与$$a+b$$同号， 故$$\\frac{f(a)+f(b)}{{{a}^{3}}+{{b}^{3}}}\\textgreater0$$．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "902", "queId": "7cc1ca7283584029901d95edb7093879", "competition_source_list": ["2019~2020学年3月湖北武汉洪山区武汉市洪山高级中学（武汉外国语学校光谷分校）高二下学期月考第10题", "2008年全国高中数学联赛竞赛一试第3题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙两人进行乒乓球比赛，约定每局胜者得$$1$$分，负者得$$0$$分，比赛进行到有一人比对方多$$2$$分或打满$$6$$局时停止．设甲在每局中获胜的概率为$$\\frac{2}{3}$$，乙在每局中获胜的概率为$$\\frac{1}{3}$$，且各局胜负相互独立，则比赛停止时已打局数$$\\xi $$的期望$$E\\xi $$为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{241}{81}$$ "}], [{"aoVal": "B", "content": "$$\\frac{266}{81}$$ "}], [{"aoVal": "C", "content": "$$\\frac{274}{81}$$ "}], [{"aoVal": "D", "content": "$$\\frac{670}{243}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->概率初步"], "answer_analysis": ["解法一： 依题意知，$$\\xi $$的所有可能值为$$2$$，$$4$$，$$6$$， 设每两局比赛为一轮，则该轮结束时比赛停止的概率为$${{\\left( \\frac{2}{3} \\right)}^{2}}+{{\\left( \\frac{1}{3} \\right)}^{2}}=\\frac{5}{9}$$， 若该轮结束时比赛还将继续，则甲、乙在该轮中必是各得一分，此时，该轮比赛结果对下轮比赛是否停止没有影响，从而有 $$P(\\xi =2)=\\frac{5}{9}$$， $$P(\\xi =4)=\\left( \\frac{4}{9} \\right)\\left( \\frac{5}{9} \\right)=\\frac{20}{81}$$， $$P(\\xi =6)={{\\left( \\frac{4}{9} \\right)}^{2}}=\\frac{16}{81}$$， 故$$E\\xi =2\\times \\frac{5}{9}+4\\times \\frac{20}{81}+6\\times \\frac{16}{81}=\\frac{266}{81}$$； 解法二： 依题意知，$$\\xi $$的所有可能值为$$2$$，$$4$$，$$6$$， 令$${{A}_{k}}$$表示甲在第$$k$$局比赛中获胜，则$${{\\bar{A}}_{k}}$$表示乙在第$$k$$局比赛中获胜， 由独立性与互不相容性得 $$P(\\xi =2)=P({{A}_{1}}{{A}_{2}})+P({{\\bar{A}}_{1}}{{\\bar{A}}_{2}})=\\frac{5}{9}$$， $$P(\\xi =4)=P({{A}_{1}}{{\\bar{A}}_{2}}{{A}_{3}}{{A}_{4}})+P({{A}_{1}}{{\\bar{A}}_{2}}{{\\bar{A}}_{3}}{{\\bar{A}}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{A}_{3}}{{A}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{\\bar{A}}_{3}}{{\\bar{A}}_{4}})$$ $$=2\\left[ {{\\left( \\frac{2}{3} \\right)}^{3}}\\left( \\frac{1}{3} \\right)+{{\\left( \\frac{1}{3} \\right)}^{3}}\\left( \\frac{2}{3} \\right) \\right]=\\frac{20}{81}$$， $$P(\\xi =6)=P({{A}_{1}}{{\\bar{A}}_{2}}{{A}_{3}}{{\\bar{A}}_{4}})+P({{A}_{1}}{{\\bar{A}}_{2}}{{\\bar{A}}_{3}}{{A}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{A}_{3}}{{\\bar{A}}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{\\bar{A}}_{3}}{{A}_{4}})$$ $$=4{{\\left( \\frac{2}{3} \\right)}^{2}}{{\\left( \\frac{1}{3} \\right)}^{2}}=\\frac{16}{81}$$， 故$$E\\xi =2\\times \\frac{5}{9}+4\\times \\frac{20}{81}+6\\times \\frac{16}{81}=\\frac{266}{81}$$�� 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "830", "queId": "d1d8411b4cc244d2898f061be9b42b32", "competition_source_list": ["2013年黑龙江全国高中数学联赛竞赛初赛第9题5分", "高二上学期单元测试《三角恒等变形》自招第17题"], "difficulty": "1", "qtype": "single_choice", "problem": "化简$$\\frac{\\sin 4\\alpha }{4{{\\sin }^{2}}\\left( \\frac{ \\pi }{4}+\\alpha \\right)\\tan \\left( \\frac{ \\pi }{4}-\\alpha \\right)}$$=．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\cos 2\\alpha $$ "}], [{"aoVal": "B", "content": "$$\\sin 2\\alpha $$ "}], [{"aoVal": "C", "content": "$$\\cos \\alpha $$ "}], [{"aoVal": "D", "content": "$$\\sin \\alpha $$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["$$\\frac{\\sin 4\\alpha }{4{{\\sin }^{2}}\\left( \\frac{ \\pi }{4}+\\alpha \\right)\\tan \\left( \\frac{ \\pi }{4}-\\alpha \\right)}=\\frac{\\sin 4\\alpha }{4\\sin \\left( \\frac{ \\pi }{4}+\\alpha \\right)\\sin \\left( \\frac{ \\pi }{4}-\\alpha \\right)}=\\frac{\\sin 4\\alpha }{2\\sin \\left( \\frac{ \\pi }{2}+2\\alpha \\right)}=\\sin 2\\alpha $$ "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "503", "queId": "2df3aed39a514933860f65944addaed2", "competition_source_list": ["2017年四川全国高中数学联赛竞赛初赛第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设集合$$M=\\left { 1,2,3,4,5,6,7,8,9,10 \\right }$$，$$A=\\left { \\left( x,y,z \\right)\\left\\textbar{} x,y,z\\in M,9\\left\\textbar{} \\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}} \\right) \\right. \\right. \\right }$$，则集合$$A$$中元素的个数是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$143$$ "}], [{"aoVal": "B", "content": "$$224$$ "}], [{"aoVal": "C", "content": "$$243$$ "}], [{"aoVal": "D", "content": "$$268$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->同余->同余的概念与性质"], "answer_analysis": ["因为$${{1}^{3}}\\equiv {{4}^{3}}\\equiv {{7}^{3}}\\equiv {{10}^{3}}\\equiv 1\\left( \\bmod 9 \\right)$$，$${{2}^{3}}\\equiv {{5}^{3}}\\equiv {{8}^{3}}\\equiv -1\\left( \\bmod 9 \\right)$$，$${{3}^{3}}\\equiv {{6}^{3}}\\equiv {{9}^{3}}\\equiv 0\\left( \\bmod 9 \\right)$$， 所以若$$9\\left\\textbar{} {{x}^{3}}+{{y}^{3}}+{{z}^{3}} \\right.$$，则$$x,y,z$$都是$$3$$的倍数或者$$3$$类数一样一个． 因此，$$A$$中元素的个数为$${{3}^{3}}+4\\times 3\\times 3\\times 3!=243$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "776", "queId": "bf508dcf9365410b9d41e10c7592b396", "competition_source_list": ["2022年江苏徐州贾汪区江苏省贾汪中学高一竞赛（下学期春季）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "定义在$R$上的函数$f\\left( x \\right)$满足：$f\\left( x+3 \\right)+f\\left( x \\right)=0$，且函数$f\\left( x-\\frac{3}{2} \\right)$为奇函数$.$给出以下3个命题：\\\\ $①$函数$f\\left( x \\right)$的周期是6；~~~~~~$②$函数$f\\left( x \\right)$的图象关于点$\\left( -\\frac{3}{2},0 \\right)$对称；\\\\ $③$函数$f\\left( x \\right)$的图象关于$y$轴对称$.$其中，真命题的个数是$($ $)$", "answer_option_list": [[{"aoVal": "A", "content": "3 "}], [{"aoVal": "B", "content": "2 "}], [{"aoVal": "C", "content": "1 "}], [{"aoVal": "D", "content": "0 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【详解】\\\\ 由$f\\left( x+3 \\right)=-f\\left( x \\right)$，知$f\\left( x+6 \\right)=f\\left( x \\right)$.所以，①正确；\\\\ 将$f\\left( x-\\frac{3}{2} \\right)$的图像向左平移$\\frac{3}{2}$ 个单位，即为$f\\left( x \\right)$的图像，而$f\\left( x-\\frac{3}{2} \\right)$的图像关于原点对称，所以，②正确；\\\\ 由②知，$f\\left( -x \\right)=-f\\left( -3+x \\right)=f\\left( x \\right)$ ，则$f\\left( x \\right)$为偶函数，所以，③正确. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "536", "queId": "4cf62f294aa949f4b0f6bb32d59bef45", "competition_source_list": ["2010年山东全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若集合$$M=\\left { x\\left\\textbar{} \\frac{\\left\\textbar{} 3-x \\right\\textbar}{\\left\\textbar{} 5-x \\right\\textbar}\\leqslant \\frac{1}{2} \\right. \\right }$$和集合$$N= {x\\textbar{{x}^{2}}-2x+c\\leqslant 0 }$$满足$$M\\cap N=M$$，则实数$$c$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$c\\leqslant -\\frac{44}{9}$$ "}], [{"aoVal": "B", "content": "$$c\\leqslant -\\frac{55}{9}$$ "}], [{"aoVal": "C", "content": "$$c\\leqslant -\\frac{66}{9}$$ "}], [{"aoVal": "D", "content": "$$c\\leqslant -\\frac{77}{9}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["因$$x\\ne 5$$，有$$\\frac{\\left\\textbar{} 3-x \\right\\textbar}{\\left\\textbar{} 5-x \\right\\textbar}\\leqslant \\frac{1}{2}\\Leftrightarrow 2\\left\\textbar{} 3-x \\right\\textbar\\leqslant \\left\\textbar{} 5-x \\right\\textbar$$． 当$$x\\leqslant 3$$时，$$2(3-x)\\leqslant (5-x)$$，解得$$x\\geqslant 1$$； 当$$3\\textless{}x\\textless{}5$$时，$$2(x-3)\\leqslant (5-x)$$，解得$$x\\leqslant \\frac{11}{3}$$； 当$$x\\textgreater5$$时，$$2(x-3)\\leqslant x-5$$，无解． 所以$$M=\\left { x\\textbar1\\leqslant x\\leqslant \\frac{11}{3} \\right }$$． 又因$${{x}^{2}}-2x+c=0,x=1\\pm \\sqrt{1-c}$$， 所以$$N=\\left { x\\textbar1-\\sqrt{1-c}\\leqslant x\\leqslant 1+\\sqrt{1-c} \\right }$$． 显然，$$1-\\sqrt{1-c}\\leqslant 1$$， 所以只要$$1+\\sqrt{1-c}\\geqslant \\frac{11}{3}$$，即$$c\\leqslant -\\frac{55}{9}$$，即有$$M\\cap N=M$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "997", "queId": "a54b8f2fd0dd49b0a2e8595ca247ffe1", "competition_source_list": ["2022~2023学年陕西西安高一上学期期末（高新唐南中学）第7题", "2021~2022学年重庆高一上学期期末第6题", "2022~2023学年山东滨州高一上学期期末（北镇中学）第6题", "2022~2023学年甘肃兰州高一上学期期末（第六十中学）第6题", "2021~2022学年福建莆田城厢区莆田第一中学高一下学期开学考试（学科素养能力竞赛）第5~5题", "2022~2023学年甘肃兰州高一上学期期末（第六十中学）第6题", "2022~2023学年山东滨州高一上学期期末（北镇中学）第6题", "2022~2023学年甘肃兰州高一上学期期末（第六十中学）第6题", "2022~2023学年陕西西安高一上学期期末（高新唐南中学）第7题", "2022~2023学年陕西西安高一上学期期末（高新唐南中学）第7题", "2022~2023学年陕西西安高一上学期期末（高新唐南中学）第7题", "2022~2023学年广东广州荔湾区广州市协和中学高二上学期期末第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a={{2}^{0.3}}$$，$$b={{3}^{0.4}}$$，$$c={{\\log }_{0.2}}0.3$$，则（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b\\textgreater c$$ "}], [{"aoVal": "B", "content": "$$b\\textgreater c\\textgreater a$$ "}], [{"aoVal": "C", "content": "$$c\\textgreater b\\textgreater a$$ "}], [{"aoVal": "D", "content": "$$b\\textgreater a\\textgreater c$$ "}]], "knowledge_point_routes": ["课内体系->知识点->基本初等函数"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 比较大小，可先与常见的常数$$0,1$$进行比较，然后根据函数的单调性进行比较大小\\\\ 【详解】\\\\ $$c={{\\log }_{0.2}}0.3\\textless{} {{\\log }_{0.2}}0.2=1$$\\\\ $$a={{2}^{0.3}}\\textgreater1$$\\\\ $$b={{3}^{0.4}}\\textgreater1$$\\\\ 则有：$$a\\textgreater c,b\\textgreater c$$\\\\ $$a={{2}^{0.3}}\\textless{} {{3}^{0.3}}\\textless{} {{3}^{0.4}}$$\\\\ 故有：$$b\\textgreater a\\textgreater c$$\\\\ 故选：D "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "897", "queId": "cd800a13c3f64f9eb98c7d445d7a8af2", "competition_source_list": ["2008年黑龙江全国高中数学联赛竞赛初赛第8题5分", "2017~2018学年江西赣州寻乌县寻乌县第二中学高三上学期期中理科第11题5分", "2005年高考真题福建卷理科第10题5分", "2021年陕西宝鸡渭滨区高三二模理科第6题5分", "2013~2014学年12月辽宁沈阳沈河区沈阳市第二中学高二上学期月考理科第10题5分", "2016~2017学年12月北京海淀区中央民族大学附属中学高二上学期月考理科第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$${{F}_{1}}$$、$${{F}_{2}}$$是双曲线$$\\frac{{{x}^{2}}}{{{a}^{2}}}-\\frac{{{y}^{2}}}{{{b}^{2}}}=1(a\\textgreater0$$，$$b\\textgreater0)$$的两焦点，以线段$${{F}_{1}}{{F}_{2}}$$为边作正三角形$$M{{F}_{1}}{{F}_{2}}$$，若边$$M{{F}_{1}}$$的中点在双曲线上，则双曲线的离心率是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4+2\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{3}-1$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{3}+1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{3}+1$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学抽象", "课内体系->素养->数学运算", "课内体系->知识点->圆锥曲线->双曲线->双曲线的几何性质->双曲线的离心率->求双曲线的离心率"], "answer_analysis": ["$$\\triangle M{{F}_{1}}{{F}_{2}}$$是正三角形，且边$$M{{F}_{1}}$$的中点在双曲线上， 设边$$M{{F}_{1}}$$的中点为$$P$$，则有$$\\angle {{F}_{1}}P{{F}_{2}}=90{}^{}\\circ $$， 从而$$\\left\\textbar{} P{{F}_{2}} \\right\\textbar=\\sqrt{3}c$$，$$\\left\\textbar{} P{{F}_{1}} \\right\\textbar=c$$． 根据双曲线的定义可知$$2a=\\left\\textbar{} P{{F}_{2}} \\right\\textbar-\\left\\textbar{} P{{F}_{1}} \\right\\textbar=\\left( \\sqrt{3}-1 \\right)c$$， 解得$$e=\\frac{c}{a}=\\frac{2}{\\sqrt{3}-1}=\\sqrt{3}+1$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "875", "queId": "8582733c3a9e43858046d08252c5bfe4", "competition_source_list": ["1985年全国高中数学联赛竞赛一试第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$Z,W,\\lambda $$为复数$$\\lambda \\ne 1$$，关于$$Z$$的方程$$\\overline{Z}-\\lambda Z=W$$有下面四个结论： （1）．$$Z=\\frac{\\overline{ \\lambda }W=\\overline{W}}{1-\\textbar\\lambda{{\\textbar}^{2}}}$$是这个方程的解； （2）这个方程只有一个解； （3）这个方程有两个解； （4）这个方程有无穷多解．则．", "answer_option_list": [[{"aoVal": "A", "content": "只有（1）和（2）是正确的 "}], [{"aoVal": "B", "content": "只有（1）和（3）是正确的 "}], [{"aoVal": "C", "content": "只有（1）和（4）是正确的 "}], [{"aoVal": "D", "content": "以上$$A$$、$$B$$、$$C$$都不正确 "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["由题中给出的四个结论，可知本题需根据解方程的情况作出选择，于是考虑在方程 $$\\overline{Z}-\\lambda Z=W$$ 的两端同取共轭，得 $$Z-\\overline{\\lambda Z}=\\overline{W}$$， 以$$\\lambda $$乘两端，得 $$\\lambda Z-{{\\left\\textbar{} \\lambda \\right\\textbar}^{2}}\\overline{Z}=\\lambda \\overline{W}$$ 与原方程两端分别相加，得 $$\\overline{Z}(1-{{\\left\\textbar{} \\lambda \\right\\textbar}^{2}})=\\lambda \\overline{W}+W$$ 两端在取共轭，得 $$\\overline{Z}(1-{{\\left\\textbar{} \\lambda \\right\\textbar}^{2}})=\\overline{\\lambda }W+\\overline{W}$$ ∵$$ {{\\left\\textbar{} \\lambda \\right\\textbar}^{2}}\\ne 1$$，∴$$Z=\\frac{\\overline{\\lambda }W+\\overline{W}}{1-{{\\left\\textbar{} \\lambda \\right\\textbar}^{2}}}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "433", "queId": "4c34b2c005be4421bf113da45f7c432e", "competition_source_list": ["2012年山东全国高中数学联赛竞赛初赛第8题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "称分子和分母的最大公约数为$$1$$的分数为既约分数，所有分母为$$100$$的正的既约真分数之和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$35$$ "}], [{"aoVal": "D", "content": "$$45$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->整除->最大公约数和最小公倍数"], "answer_analysis": ["由$$100={{2}^{2}}{{5}^{2}}$$，令 $$A=\\left { n\\left\\textbar{} 2 \\right\\textbar n, n\\leqslant 100, n\\in {{\\mathbf{N}}^{*}} \\right }$$， $$B=\\left { n\\left\\textbar{} 5 \\right\\textbar n, n\\leqslant 100, n\\in {{\\mathbf{N}}^{*}} \\right }$$， 则$$\\left\\textbar{} A\\cup B \\right\\textbar=\\left\\textbar{} A \\right\\textbar+\\left\\textbar{} B \\right\\textbar-\\left\\textbar{} A\\cap B \\right\\textbar=50+20-10=60$$． 因此符合条件的真分数共有$$40$$个． 又若正数$$i$$和$$100$$互质，则$$100-i$$和$$100$$也互质，因此若$$\\frac{i}{100}$$是一个正的既约真分数，则$$\\frac{100-i}{100}$$也为一正的既约真分数，则$$\\frac{100-i}{100}+\\frac{i}{100}=1$$．因此所有真分数之和为$$20$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "475", "queId": "a2afc80629d94e45bef8fdceade30c34", "competition_source_list": ["2017年四川全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f\\left( x \\right)=a\\ln x+{{x}^{2}}$$在$$x=1$$处有极值，则实数$$a$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->导数"], "answer_analysis": ["$${f}'\\left( x \\right)=\\frac{a}{x}+2x$$，由题设$${f}'\\left( 1 \\right)=0$$，解得$$a=-2$$ "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "98", "queId": "0b70d044123f4da7bf58cfa5f8c0bf51", "competition_source_list": ["1986年全国高中数学联赛竞赛一试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "边长为$$a$$、$$b$$、$$c$$的三角形，其面积等于$$\\frac{1}{4}$$，而外接圆半径为$$1$$，若$$s=\\sqrt{a}+\\sqrt{b}+\\sqrt{c}$$，$$t=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$$，则$$s$$与$$t$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$s\\textgreater t$$ "}], [{"aoVal": "B", "content": "$$s=t$$ "}], [{"aoVal": "C", "content": "$$s\\textless{}t$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角形中的问题->解三角形", "竞赛->知识点->不等式->几个重要的不等式->均值"], "answer_analysis": ["∵$$c=2R\\sin C=2\\sin C$$，又$$\\frac{1}{2}ab\\sin C=\\frac{1}{4}$$， ∴$$abc=1$$，于是 $$t=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$$ $$=\\frac{1}{2}\\left( \\frac{1}{a}+\\frac{1}{b} \\right)+\\frac{1}{2}\\left( \\frac{1}{b}+\\frac{1}{c} \\right)+\\frac{1}{2}\\left( \\frac{1}{c}+\\frac{1}{a} \\right)$$ $$\\mathsf{\\geqslant }\\sqrt{\\frac{\\mathsf{1}}{ab}}+\\sqrt{\\frac{1}{bc}}+\\sqrt{\\frac{1}{ca}}$$ $$=\\frac{\\sqrt{c}+\\sqrt{a}+\\sqrt{b}}{\\sqrt{abc}}$$ $$=\\sqrt{a}+\\sqrt{b}+\\sqrt{c}$$ $$=s$$． 且其中等号不可能取到，否则$$a=b=c=R=1$$，这显然是不可能的． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "447", "queId": "5550e1cfc3654d8c859fea8d447e43ab", "competition_source_list": ["2015年河南全国高中数学联赛高二竞赛初赛第5题8分", "2016~2017学年江苏无锡高三上学期期中第13题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知正数$$a,b$$满足：$$a+3b=7$$，则$$\\frac{1}{1+a}+\\frac{4}{2+b}$$的最小值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{13+2\\sqrt{3}}{14}$$ "}], [{"aoVal": "B", "content": "$$\\frac{13+3\\sqrt{3}}{14}$$ "}], [{"aoVal": "C", "content": "$$\\frac{13+4\\sqrt{3}}{14}$$ "}], [{"aoVal": "D", "content": "$$\\frac{13+5\\sqrt{3}}{14}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->柯西不等式", "竞赛->知识点->不等式->几个重要的不等式->柯西"], "answer_analysis": ["由柯西不等式，$$\\left[ 1+a+3\\left( 2+b \\right) \\right]\\left( \\frac{1}{1+a}+\\frac{4}{2+b} \\right)\\geqslant {{\\left( 1+2\\sqrt{3} \\right)}^{2}}$$， 所以$$\\frac{1}{1+a}+\\frac{4}{2+b}\\geqslant \\frac{13+4\\sqrt{3}}{14}$$． $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\frac{1}{1+a}+\\frac{4}{2+b}$$ $$=\\frac{1}{14}\\left[ \\left( a+1 \\right)+3\\left( 2+b \\right) \\right]\\cdot \\left( \\frac{1}{1+a}+\\frac{4}{2+b} \\right)$$ $$=\\frac{1}{14}\\left[ 13+\\frac{3\\left( 2+b \\right)}{a+1}+\\frac{4\\left( a+1 \\right)}{2+b} \\right]$$ $$\\geqslant \\frac{1}{14}\\left[ 13+2\\sqrt{\\frac{3\\left( 2+b \\right)}{a+1}\\cdot \\frac{4\\left( a+1 \\right)}{2+b}} \\right]$$ $$=\\frac{13+4\\sqrt{3}}{14}$$， 当且仅当$$\\frac{3\\left( 2+b \\right)}{a+1}=\\frac{4\\left( a+1 \\right)}{2+b}$$， 即$$a=\\frac{\\sqrt{3}}{2}b+\\sqrt{3-1}$$时等号成立， 故应填$$\\frac{13+4\\sqrt{3}}{14}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1100", "queId": "ca8c43dea02442aaa64eedf289d800c6", "competition_source_list": ["2011年全国高中数学联赛竞赛复赛一试第8题8分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知$${{a}_{n}}=\\text{C}_{200}^{n}{{\\left( \\sqrt[3]{6} \\right)}^{200-n}}{{\\left( \\frac{1}{\\sqrt{2}} \\right)}^{n}}\\left( n=1,2,\\cdots ,95 \\right)$$，则数列$$\\left { {{a}_{n}} \\right }$$中整数项有个.", "answer_option_list": [[{"aoVal": "A", "content": "$10$ "}], [{"aoVal": "B", "content": "$15$ "}], [{"aoVal": "C", "content": "$20$ "}], [{"aoVal": "D", "content": "$25$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->二项式定理及其应用"], "answer_analysis": ["$${{a}_{n}}=\\text{C}_{200}^{n}\\cdot {{3}^{\\frac{200-n}{3}}}\\cdot {{2}^{\\frac{400-5 \\pi }{6}}}$$， 要使$${{a}_{n}}(1\\leqslant n\\leqslant 95)$$为整数，必有$$\\frac{200-n}{3}$$，$$\\frac{400-5n}{6}$$均为整数，从而$$6\\textbar n+4$$． 当$$n=2,8,14,20,26,32,38,44,50,56,62,68,74,80$$时，$$\\frac{200-n}{3}$$和$$\\frac{400-5n}{6}$$均为非负整数，所以$${{a}_{n}}$$为整数，共有$$14$$个． 当$$n=86$$时，$${{a}_{n}}=\\text{C}_{200}^{86}\\cdot {{3}^{38}}\\cdot 2_{{}}^{-5}$$，在$$\\text{C}_{200}^{86}=\\frac{200!}{86!\\cdot 114!}$$中， $$200$$中因数$$2$$的个数为$$\\left[ \\frac{200}{2} \\right]+\\left[ \\frac{200}{{{2}^{2}}} \\right]+\\left[ \\frac{200}{{{2}^{3}}} \\right]+\\left[ \\frac{200}{{{2}^{4}}} \\right]+\\left[ \\frac{200}{{{2}^{5}}} \\right]+\\left[ \\frac{200}{{{2}^{6}}} \\right]+\\left[ \\frac{200}{{{2}^{7}}} \\right]=197$$， 同理可计算得$$86$$中因数$$2$$的个数为$$82$$，$$114$$中因数$$2$$的个数为$$110$$， 所以$$\\text{C}_{200}^{86}$$中因数$$2$$的个数为$$197-82-110=5$$，故$${{a}_{86}}$$是整数． 当$$n=92$$时，$${{a}_{92}}=\\text{C}_{200}^{92}\\cdot {{3}^{36}}\\cdot {{2}^{-10}} $$在$$\\text{C}_{200}^{92}=\\frac{200!}{92!108!}$$中，同样可求得$$92$$中因数$$2$$的个数为$$88$$，$$108$$中���数$$2$$的个数为$$105$$，故$$\\text{C}_{200}^{86}$$中因数$$2$$的个数为$$197-88-105=4$$，故不是整数． 因此，整数项的个数为$$14+1=15$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "373", "queId": "506027439ec34c6a8270c63ce79c26ac", "competition_source_list": ["2021年福建全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$f\\left( x \\right)$$和$$g\\left( x \\right)$$)是两个二次项系数均为$$1$$的二次函数．若$$g\\left( 6 \\right)=35$$﹐$$\\frac{f\\left( -1 \\right)}{g\\left( -1 \\right)}=\\frac{f\\left( 1 \\right)}{g\\left( 1 \\right)}=\\frac{21}{20}$$，则$$f\\left( 6 \\right)=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$31$$ "}], [{"aoVal": "B", "content": "$$32$$ "}], [{"aoVal": "C", "content": "$$33$$ "}], [{"aoVal": "D", "content": "$$34$$ "}], [{"aoVal": "E", "content": "$$35$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质", "竞赛->知识点->函数->二次函数"], "answer_analysis": ["设$$f\\left( x \\right)={{x}^{2}}+ax+b$$，$$g\\left( x \\right)={{x}^{2}}+cx+d$$， 则由题设条件可得 $$20\\left( 1-a+b \\right)=21\\left( 1-c+d \\right)\\cdots \\cdots $$①﹐$$20\\left( 1+a+b \\right)=21\\left( 1+c+d \\right)\\cdots \\cdots $$②． ①、②两式左右两边分别相加，得$$40+40b=42+42d$$，$$20b=1+21d$$﹔ ①、②两式左右两边分别相减，得$$-40a=-42c$$，$$20a=21c$$． 另由$$g\\left( 6 \\right)=35$$，得$$36+6c+d=35$$． 所以，$$36+6\\times \\frac{20}{21}a+\\frac{20b-1}{21}=35$$，$$6a+b=-1$$， 所以，$$f\\left( 6 \\right)=36+6a+b=35$$． 设$$h\\left( x \\right)=21g\\left( x \\right)-20f\\left( x \\right)$$， 则由条件知$$h\\left( x \\right)$$是二次项系数为$$1$$的二次函数． 又$$h\\left( -1 \\right)=21g\\left( -1 \\right)-20f\\left( -1 \\right)=0$$，$$h\\left( 1 \\right)=21g\\left( 1 \\right)-20f\\left( 1 \\right)=0$$， 所以，$$h(x)=(x+1)(x-1)={{x}^{2}}-1$$． 因此，$$h\\left( 6 \\right)=21g\\left( 6 \\right)-20f\\left( 6 \\right)={{6}^{2}}-1=35$$． 所以，$$21\\times 35-20f\\left( 6 \\right)=35$$，$$f\\left( 6 \\right)=35$$． "], "answer_value": "E"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "810", "queId": "6a08a426b0614c16875d740e354edb3c", "competition_source_list": ["全国高中数学联赛竞赛模拟一试（十三）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$，$$c$$，$$d$$为正数，且$$a+2b=c+2d=1$$．则$$\\frac{1}{a}+\\frac{1}{bcd}$$的最小值为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->多元函数极值（一试）", "竞赛->知识点->不等式->几个重要的不等式->柯西", "课内体系->知识点->等式与不等式->不等式->柯西不等式", "课内体系->方法->均值不等式法"], "answer_analysis": ["由均值不等式及题设条件知 $$2cd\\leqslant {{\\left( \\frac{c+2d}{2} \\right)}^{2}}=\\frac{1}{4}\\Rightarrow cd\\leqslant \\frac{1}{8}$$①． 当且仅当$$c=\\frac{1}{2}$$，$$d=\\frac{1}{4}$$时，以上两式的等号成立． 此时，由式①得$$\\frac{1}{a}+\\frac{1}{bcd}\\geqslant \\frac{1}{a}+\\frac{8}{b}$$． 又由条件及柯西不等式得 $$\\frac{1}{a}+\\frac{8}{b}=\\left( \\frac{1}{a}+\\frac{8}{b} \\right)\\left( a+2b \\right)\\geqslant {{\\left( 1+4 \\right)}^{2}}=25$$， 当且仅当$$b=2a$$时，上式等号成立． 结合条件，知此时$$a=\\frac{1}{5}$$，$$b=\\frac{2}{5}$$． 于是，$$\\frac{1}{a}+\\frac{1}{bcd}\\geqslant 25$$． 当 $$a=\\frac{1}{5}$$，$$b=\\frac{2}{5}$$，$$c=\\frac{1}{2}$$，$$d=\\frac{1}{4}$$时， $$\\frac{1}{a}+\\frac{1}{bcd}$$取得最小值$$25$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "214", "queId": "113a508bb2cf4164b4e51d1f236fb695", "competition_source_list": ["2008年全国高中数学联赛竞赛一试第10题9分", "2010年河南全国高中数学联赛竞赛初赛第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设数列$$ {{{a}_{n}} }$$的前$$n$$项和$${{S}_{n}}$$满足：$${{S}_{n}}+{{a}_{n}}=\\frac{n-1}{n(n+1)}$$，$$n=1,2,\\cdots $$，则通项$${{a}_{n}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{{{2}^{n}}}+\\frac{1}{n(n+1)}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{{{2}^{n}}}-\\frac{1}{n(n+1)}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{{{2}^{n}}}+\\frac{1}{n(n-1)}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{{{2}^{n}}}-\\frac{1}{n(n-1)}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列���通项与求和", "课内体系->知识点->数列"], "answer_analysis": ["$${{a}_{n+1}}={{S}_{n+1}}-{{S}_{n}}=\\frac{n}{(n+1)(n+2)}-{{a}_{n+1}}-\\frac{n-1}{n(n+1)}+{{a}_{n}}$$， 即$$2{{a}_{n+1}}=\\frac{n+2-2}{(n+1)(n+2)}-\\frac{1}{n+1}+\\frac{1}{n(n+1)}+{{a}_{\\begin{smallmatrix} n \\end{smallmatrix}}}$$ $$=\\frac{-2}{(n+1)(n+2)}+{{a}_{n}}+\\frac{1}{n(n+1)}$$， 由此得$$2\\left( {{a}_{n+1}}+\\frac{1}{(n+1)(n+2)} \\right)={{a}_{n}}+\\frac{1}{n(n+1)}$$， 令$${{b}_{n}}={{a}_{n}}+\\frac{1}{n(n+1)}$$，$${{b}_{1}}={{a}_{1}}+\\frac{1}{2}=\\frac{1}{2}({{a}_{1}}=0)$$， 有$${{b}_{n+1}}=\\frac{1}{2}{{b}_{n}}$$，故$${{b}_{n}}=\\frac{1}{{{2}^{n}}}$$，所以$${{a}_{n}}=\\frac{1}{{{2}^{n}}}-\\frac{1}{n(n+1)}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "150", "queId": "14401eaf8eab4ee28ce7ff62fa2c6767", "competition_source_list": ["2019年上海全国高中数学联赛高三竞赛初赛第4题7分", "2019年上海全国高中数学联赛竞赛初赛第4题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "设等差数列$$\\left { {{a}_{n}} \\right }$$的公差为$$d\\left( d\\ne 0 \\right)$$，前$$n$$项和为$${{S}_{n}}$$．若数列$$\\left { \\sqrt{8{{S}_{n}}+2n} \\right }$$也是公差为$$d$$的等差数列，则数列$$\\left { {{a}_{n}} \\right }$$的通项$${{a}_{n}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$4n-\\frac{3}{4}$$ "}], [{"aoVal": "B", "content": "$$4n-\\frac{5}{4}$$ "}], [{"aoVal": "C", "content": "$$4n-\\frac{7}{4}$$ "}], [{"aoVal": "D", "content": "$$4n-\\frac{9}{4}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列", "课内体系->知识点->数列->等差数列->等差数列的概念与通项公式->等差数列求通项问题"], "answer_analysis": ["由$$\\sqrt{8{{S}_{n+1}}+2\\left( n+1 \\right)}-\\sqrt{8{{S}_{n}}+2n}=d\\left( n\\geqslant 1 \\right)$$ $$\\Rightarrow 8{{S}_{n+1}}-8{{S}_{n}}+2=d+2d\\sqrt{8{{S}_{n}}+2n}$$ $$\\Rightarrow 8{{a}_{n+1}}+2=d+2d\\sqrt{8{{S}_{n}}+2n}$$， 类似地， $$8{{a}_{n+2}}+2=d+2d\\sqrt{8{{S}_{n+1}}+2\\left( n+1 \\right)}$$， 将以上两式相减得 $$8\\left( {{a}_{n+2}}-{{a}_{n+1}} \\right)=2d\\left( \\sqrt{8{{S}_{n+1}}+2\\left( n+1 \\right)}-\\sqrt{8{{S}_{n}}+2n} \\right)$$ $$\\Rightarrow 8d=2{{d}^{2}}$$， 而$$d\\ne 0$$，故$$d=4$$， 又$$\\sqrt{8{{S}_{2}}+4}-\\sqrt{8{{S}_{1}}+2}=\\sqrt{16{{a}_{1}}+36}-\\sqrt{8{{a}_{1}}+2}=4$$， 由此解得$${{a}_{1}}=\\frac{7}{4}$$， 从而，$${{a}_{n}}=\\frac{7}{4}+4\\left( n-1 \\right)=4n-\\frac{9}{4}$$． 故答案为：$$4n-\\frac{9}{4}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "675", "queId": "4574c1ca60094c808583788fbb75fce1", "competition_source_list": ["2010年AMC10竞赛B第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2010-AMC10B-11$$ A shopper plans to purchase an item that has a listed price greater than ＄$$100$$ and can use any one of the three coupons. Coupon $$A$$ gives $$15$$\\% off the listed price, Coupon $$B$$ gives ＄$$30$$ off the listed price, and Coupon $$C$$ gives $$25$$\\% off the amount by which the listed price exceeds ＄$$100$$. Let $$x$$ and $$y$$ be the smallest and largest prices, respectively, for which Coupon $$A$$ saves at least as many dollars as Coupon $$B$$ or Coupon $$C$$. What is $$y- x$$? 购物者计划购买标价大于 ＄100 的商品，并且可以使用三种优惠券中的任何一种。 优惠券$$A$$ 优惠$$15$$ 元，优惠券$$B$$ 优惠$$30$$ 元，优惠券$$C$$ 优惠25\\%，超过$$100$$ 元的优惠券。 设 X 和 Y 分别是最小和最大的价格，对于该价格，优惠券 A 至少节省与优惠券 B 或优惠券 C 一样多的美元。y-x 是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$75$$ "}], [{"aoVal": "D", "content": "$$80$$ "}], [{"aoVal": "E", "content": "$$100$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的应用->函数的实际应用->一次函数模型", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"], "answer_analysis": ["Let the listed price be $$(100+p)$$, where $$p\\textgreater0$$. Coupon $$A$$ saves us: $$0.15(100+p)=(0.15 p+15)$$. Coupon $$B$$ saves us: $$30$$. Coupon $$C$$ saves us: $$0.25p$$. Now, the condition is that $$A$$ has to be greater than or equal to either $$B$$ or $$C$$ which give us the following inequalities: $$A≥B⇒0.15 p+15≥30⇒p≥100$$, $$A≥C⇒0.15 p+15≥0.25p⇒p≤150$$. We see here that the greatest possible value for $$p$$ is $$150$$,~ thus $$y= 100+150=250$$ and the smallest value for $$p$$ is $$100$$ so $$x= 100+100=200$$. The difference between $$y$$ and $$x$$ is $$y-x=250-200=(\\rm A) 50$$. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "92", "queId": "210e9a45d67446cab2d715742c6d7231", "competition_source_list": ["2017年天津全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设复数$$z$$满足$$\\left\\textbar{} z-\\left\\textbar{} z+1 \\right\\textbar{} \\right\\textbar=\\left\\textbar{} z+\\left\\textbar{} z-1 \\right\\textbar{} \\right\\textbar$$，则下列判断错误的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$z$$可能为纯虚数 "}], [{"aoVal": "B", "content": "$$z$$可能为实数 "}], [{"aoVal": "C", "content": "$$z$$的实部小于$$2$$ "}], [{"aoVal": "D", "content": "$$z$$的辐角可能为$$\\frac{ \\pi }{4}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的应用", "竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["令$$z=b\\text{i}$$，不难算出$$\\left\\textbar{} z-\\left\\textbar{} z+1 \\right\\textbar{} \\right\\textbar=\\sqrt{{{b}^{2}}+{{b}^{2}}+1}=\\left\\textbar{} z+\\left\\textbar{} z-1 \\right\\textbar{} \\right\\textbar$$，$$A$$对； 令$$z=0$$，显然$$\\left\\textbar{} z-\\left\\textbar{} z+1 \\right\\textbar{} \\right\\textbar=\\left\\textbar{} z+\\left\\textbar{} z-1 \\right\\textbar{} \\right\\textbar=1$$，$$B$$对； 设$$z=a+b\\text{i}$$，由题意，$${{\\left( a-\\sqrt{{{\\left( a+1 \\right)}^{2}}+{{b}^{2}}} \\right)}^{2}}={{\\left( a+\\sqrt{{{\\left( a-1 \\right)}^{2}}+{{b}^{2}}} \\right)}^{2}}$$， 若$$a\\geqslant 2$$，则$${{\\left( a-\\sqrt{{{\\left( a+1 \\right)}^{2}}+{{b}^{2}}} \\right)}^{2}}\\textless{}{{\\left( a+1 \\right)}^{2}}+{{b}^{2}}$$， 但$${{\\left( a+\\sqrt{{{\\left( a-1 \\right)}^{2}}+{{b}^{2}}} \\right)}^{2}}\\geqslant {{\\left( a+\\sqrt{1+{{b}^{2}}} \\right)}^{2}}={{a}^{2}}+1+{{b}^{2}}+2a\\sqrt{1+{{b}^{2}}}\\geqslant {{a}^{2}}+1+{{b}^{2}}+2a={{\\left( a+1 \\right)}^{2}}+{{b}^{2}}$$， 矛盾，所以$$a\\textless{}2$$，$$C$$对； "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "773", "queId": "84d0181f56f041fba26956f813a926fe", "competition_source_list": ["2008年甘肃全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若实数$$x$$、$$y$$满足$${{(x-3)}^{2}}+4{{(y-1)}^{2}}=4$$，则$$\\frac{x+y-3}{x-y+1}$$的最大值和最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$，$$0$$ "}], [{"aoVal": "B", "content": "$$0$$，$$-1$$ "}], [{"aoVal": "C", "content": "$$1$$，$$-1$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$，$$-\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->换元技巧->三角换元"], "answer_analysis": ["设$$\\begin{cases}x=3+2\\cos \\theta y=1+\\sin \\theta \\end{cases}$$， 则$$t=\\frac{x+y-3}{x-y+1}=\\frac{2\\cos \\theta +\\sin \\theta +1}{2\\cos \\theta -\\sin \\theta +3}$$， 化为$$(2t-2)\\cos \\theta -(t+1)\\sin \\theta +3t-1=0$$， 由万能公式得$$(t+1){{\\tan }^{2}}\\frac{\\theta }{2}-2(t+1)\\tan \\frac{\\theta }{2}+5t-3=0$$． 当$$t\\ne 1$$时，$$\\frac{1}{4}\\Delta ={{(t+1)}^{2}}-(t+1)(5t-3)\\geqslant 0$$， 所以$$-1\\textless{}t\\leqslant 1$$； 当$$t=-1$$时，解得$$\\cos \\theta =-1$$有意义．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "129", "queId": "0c40a6bafec84cf3a486847b2b22e3ef", "competition_source_list": ["2012年AMC10竞赛A第16题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2012-AMC10A-16$$ Three runners start running simultaneously from the same point on a $$500-$$meter circular track. They each run clockwise around the course maintaining constant speeds of $$4.4$$, $$4.8$$, and $$5.0$$ meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run? 三名跑者在一条 500 米的环形跑道上从同一点同时开始跑步。 他们每个人都绕着赛道顺时针运行，保持每秒 4.4、$$4.8$$ 和 5.0 米的恒定速度。 一旦他们再次聚集在圆形路线上的某个地方，跑步者就会停下来。 跑者跑多少秒？", "answer_option_list": [[{"aoVal": "A", "content": "$$1000$$ "}], [{"aoVal": "B", "content": "$$1250$$ "}], [{"aoVal": "C", "content": "$$2500$$ "}], [{"aoVal": "D", "content": "$$5000$$ "}], [{"aoVal": "E", "content": "$$10000$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->等式->方程组的解集", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"], "answer_analysis": ["First consider the first two runners. The faster runner will lap the slower runner exactly once, or run $$500$$ meters farther. Let $$x$$ be the time these runners run in seconds. Because $$4.8x-4.4x=500\\Rightarrow x=1250$$ is a multiple of $$500$$, it turns out they just meet back at the start line. Now we must find a time that is a multiple of $$1250$$ and results in the $$5.0\\rm m/s$$ runner to end up on the start line. Every $$1250$$ seconds, that fastest runner goes $$5.0(1250)=6250$$ meters. In $$2(1250)=2500$$ seconds, he goes $$5.0(2500)=12500$$ meters. Therefore the runners run $$\\boxed{\\rm(C)\\textasciitilde2500} $$seconds. Working backwards from the answers starting with the smallest answer, if they had run $$1000$$ seconds, they would have run $$4400$$, $$4800$$, $$5000$$ meters, respectively. The first two runners have a difference of $$400$$ meters, which is not a multiple of $$500$$ (one lap), so they are not in the same place. If they had run $$1250$$ seconds, the runners would have run $$5500$$, $$6000$$, $$6250$$ meters, respectively. The last two runners have a difference of $$250$$ meters, which is not a multiple of $$500$$. If they had run $$2500$$ seconds, the runners would have run $$11000$$, $$12000$$, $$12500$$ meters, respectively. The distance separating each pair of runners is a multiple of $$500$$, so the answer is $$\\boxed{\\rm(C)\\textasciitilde2500}$$ seconds. Let $$t$$ be the time run in seconds, then the difference in meters run between the three runners is $$0.2t$$, $$0.4t$$, $$0.6t$$. For them to be at the same location all of them need to be multiples of $$500$$. It is now easy to see that $$0.2t=500$$, $$0.4t=1000$$, $$0.6t=1500$$, so $$t=\\boxed{\\rm(C)\\textasciitilde2500}$$. After $$t$$ seconds, respectively the runners would\\textquotesingle ve ran $$4.4t$$, $$4.8t$$ and $$5t$$ meters. Their current positions on the track are these values $$(\\rm mod\\textasciitilde500)$$. We\\textquotesingle re trying to find the value of $$t$$ such that $$4.4t\\equiv 4.8t\\equiv 5t\\textasciitilde{} \\rm (mod \\textasciitilde500)$$ Subtracting $$4.4t$$ on all sides, we get $$0\\equiv 0.4t\\equiv 0.6t\\textasciitilde{} \\rm (mod \\textasciitilde500)$$ Now, we must find a value for $$t$$ such that both $$0.6t$$ and $$0.4t$$ are simultaneously multiples of $$500$$. Plugging in $$500$$ for $$0.4t$$ we get $$t=1250$$, but this does not work for $$0.6t$$ ( $$750$$ isn\\textquotesingle t a multiple of $$500$$). Plugging in $$0.4t=1000$$, we get $$t=2500$$, and this does work for $$0.6t$$. Therefore, $$t=2500$$ and the answer is $$\\rm(C)\\textasciitilde2500$$. $$\\cdot$$ Note: Modular Arithmetic works only for integral values, so my usage of decimals is technically incorrect but the intuition leads to the right answer. Similar to the solution above, but is much quicker and does not involve trial and error. This uses decimal mod arithmetic, which can be justified by intuition $$\\cdots$$~~After $$t$$ seconds, respectively the runners would\\textquotesingle ve ran $$4.4t$$, $$4.8t$$, and $$5t$$ meters. These three values are congruent $$\\rm(mod\\textasciitilde500)$$, so $$4.4t\\equiv 4.8t\\equiv 5t\\textasciitilde{} \\rm (mod \\textasciitilde500)$$. Subtract $$4.4t$$ from all three sides to get $$0$$, $$0.4t$$, and $$0.6t$$ are congruent. Now all we need to find is a value of t for which $$0.4t$$ and $$0.6t$$ are congruent mod $$500$$. Subtract $$0.4t$$ from both sides to get $$0.2t$$ and $$0$$ are congruent mod $$500$$, or that $$0.2t=\\frac t5$$ is a multiple of $$500$$. Let $$t=500k$$, so we want $$100k$$ to be a multiple of $$500$$, or $$k$$ to be a multiple of $$5$$. Therefore, thes mallest value of $$t$$ is when $$k=5$$, and when $$t=500k=500(5)=2500\\rm(C)$$. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "358", "queId": "674011ecdd154aefbc33befad21694ad", "competition_source_list": ["2009年吉林全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$I$$是$$\\triangle ABC$$的内心，$$AC=2,BC=3,AB=4$$，若$$\\overrightarrow{AI}=x+y\\overrightarrow{AC}$$，则$$x+y$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{4}{9}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{9}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算", "竞赛->知识点->平面几何->内心相关问题（二试）"], "answer_analysis": ["在$$\\triangle ABC$$中，$$I$$为内心，连$$AI$$并延长交$$BC$$于$$D$$点， 则$$D$$分$$BC$$的比$$\\lambda =\\frac{AB}{AC}=\\frac{4}{2}=2$$． 故$$\\overrightarrow{AD}=\\frac{1}{3}\\overrightarrow{AB}+\\frac{2}{3}\\overrightarrow{AC}$$．又$$BC=3$$，故$$BD=2, DC=1$$． 又在$$\\triangle ABD$$中，$$I$$分$$AD$$的比$${\\lambda }'=\\frac{AB}{BD}=\\frac{4}{2}=2$$， 即$$\\overrightarrow{AI}=\\frac{2}{3}\\overrightarrow{AD}=\\frac{2}{9}\\overrightarrow{AB}+\\frac{4}{9}\\overrightarrow{AC}$$，所以$$x+y=\\frac{2}{3}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "420", "queId": "47920f405e6b4c16b9e9f846f25754d3", "competition_source_list": ["2002年全国全国高中数学联赛竞赛一���第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知两个实数集合$$A=\\left { {{a}_{1}},{{a}_{2}},\\cdots ,{{a}_{100}} \\right }$$与$$B=\\left { {{b}_{1}},{{b}_{2}},\\cdots ,{{b}_{50}} \\right }$$，若从$$A$$到$$B$$的映射$$f$$使得$$B$$中的每一个元素都有原象，且$$f\\left( {{a}_{1}} \\right)\\leqslant f\\left( {{a}_{2}} \\right)\\leqslant \\cdots \\leqslant f\\left( {{a}_{100}} \\right)$$，则这样的映射共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\text{C}_{100}^{50}$$ "}], [{"aoVal": "B", "content": "$$\\text{C}_{90}^{50}$$ "}], [{"aoVal": "C", "content": "$$\\text{C}_{100}^{49}$$ "}], [{"aoVal": "D", "content": "$$\\text{C}_{99}^{49}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->映射", "课内体系->知识点->计数原理->排列与组合->组合->隔板法", "课内体系->素养->逻辑推理"], "answer_analysis": ["不妨设$${{b}_{1}}\\textless{}{{b}_{2}}\\textless{}\\cdots \\textless{}{{b}_{50}}$$，将$$A$$中元素$${{a}_{1}}$$，$${{a}_{2}}$$，\\ldots，$${{a}_{100}}$$按顺序分为非空的$$50$$组，定义映射$$f:A\\to B$$，使得第$$i$$组的元素在$$f$$之下的象都是$${{b}_{i}}\\left( i=1,2, \\cdots ,50 \\right)$$，易知这样的$$f$$满足题设要求，每个这样的分组都一一对应满足条件的映射，于是满足题设要求的映射$$f$$的个数与$$A$$按足码顺序分为$$50$$组的分法数相等，而$$A$$的分法数为$$\\text{C}_{99}^{49}$$，则这样的映射共有$$\\text{C}_{99}^{49}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "145", "queId": "1cee0f83b227494a951c7573cafea3ff", "competition_source_list": ["2023年江苏连云港灌南县灌南县中学高三竞赛（下学期3月解题能力竞赛）第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "已知集合$A=\\left {x\\left\\textbar{} x\\textless{} 2\\right.\\right }$，$B=\\left {x\\left\\textbar{} x^{2}-2x-3\\leq 0\\right.\\right }$，则$A\\cup B=$（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\left[-1,2\\right)$ "}], [{"aoVal": "B", "content": "$\\left(2,3\\right]$ "}], [{"aoVal": "C", "content": "$\\left(-1,3\\right]$ "}], [{"aoVal": "D", "content": "$\\left(-\\mathrm{\\infty },3\\right]$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 先求出集合$B$，再依据并集的定义求并集.\\\\ 【详解】\\\\ $B=\\left {x\\left\\textbar{} x^{2}-2x-3\\leq 0\\right.\\right }=\\left {x\\left\\textbar{} -1\\leq x\\leq 3\\right.\\right }$，又$A=\\left {x\\left\\textbar{} x\\textless{} 2\\right.\\right }$，\\\\ 所以$A\\cup B=\\left(-\\mathrm{\\infty }\\mathrm{，}3\\right]$\\\\ 故选：D "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "819", "queId": "8d4177ff1279401a96c785c9b3c4b54d", "competition_source_list": ["2009年河北全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果函数$$f(x)={{a}^{x}}({{a}^{x}}-3{{a}^{2}}-1)(a\\textgreater0$$且$$a\\ne 1)$$在区间$$[0,+\\infty ]$$上是增函数，那么实数$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 0,\\frac{2}{3} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left[ \\frac{\\sqrt{3}}{3},1 \\right)$$ "}], [{"aoVal": "C", "content": "$$(0,\\sqrt{3}]$$ "}], [{"aoVal": "D", "content": "$$\\left[ \\frac{3}{2},+\\infty \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["利用复合函数的单调性求解，令$${{a}^{x}}=t$$，则 $$y=g(t)={{t}^{2}}-(3{{a}^{2}}+1)t$$． ①当$$a\\textgreater1$$时，$$t$$是关于$$x$$的增函数，欲使$$f(x)$$在$$x\\in [0,+\\infty )$$上是增函数，需$$g(t)$$在$$t\\in [1,+\\infty )$$上是增函数，故$$\\frac{3{{a}^{2}}+1}{2}\\leqslant 1$$，即$${{a}^{2}}\\leqslant \\frac{1}{3}$$，矛盾． ②当$$0\\textless{}a\\textless{}1$$时，$$t$$是关于$$x$$的减函数，欲使$$f(x)$$在$$x\\in [0,+\\infty )$$上是增函数，需$$g(t)$$在$$t\\in (0,1]$$上是减函数，故$$\\frac{3{{a}^{2}}+1}{2}\\geqslant 1$$，即$${{a}^{2}}\\geqslant \\frac{1}{3}$$，所以$$\\frac{\\sqrt{3}}{3}\\leqslant a\\textless{}1$$． 综上可知选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1074", "queId": "d7d0bcc25c024651a1708865633015f5", "competition_source_list": ["竞赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知集合$A=\\left {a_{1},a_{2},\\cdots ,a_{n}\\right }$，任取$1\\leq i\\textless{} j\\textless{} k\\leq n,a_{i}+a_{j}\\in A,a_{j}+a_{k}\\in A,a_{i}+a_{k}\\in A$中至少有一个成立，则\\emph{n}的最大值为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "3 "}], [{"aoVal": "B", "content": "5 "}], [{"aoVal": "C", "content": "7 "}], [{"aoVal": "D", "content": "9 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 可证明集合\\emph{A}的正数至多有3个，负数至多有3个，故可判断\\emph{n}的最大值.\\\\ 【详解】\\\\ 不妨设$a_{1}\\textgreater{} a_{2}\\textgreater{} \\cdots \\textgreater{} a_{n}$，若集合\\emph{A}中的正数个数不小于4，取$(i,j,k)=(1,2,3)$，\\\\ 可得$a_{2}+a_{3}=a_{1}$，取$(i,j,k)=(1,2,4)$，可得$a_{2}+a_{4}=a_{1}$，因此$a_{3}=a_{4}$，矛盾．\\\\ 因此集合\\emph{A}的正数至多有3个，同理，集合\\emph{A}中的负数至多有3个．\\\\ 又考虑$A= {3,2,1,0,-1,-2,-3 }$，\\\\ 符合题意，因此\\emph{n}的最大值为7．\\\\ 故选：C. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1198", "queId": "f0621a4b7a49472fa0d5ac0e1ca8673f", "competition_source_list": ["2005年全国高中数学联赛竞赛一试第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "空间四点$$A$$、$$B$$、$$C$$、$$D$$满足$$\\textbar\\overrightarrow{AB}\\textbar=3$$，$$\\textbar\\overrightarrow{BC}\\textbar=7$$，$$\\textbar\\overrightarrow{CD}\\textbar=11$$，$$\\textbar\\overrightarrow{DA}\\textbar=9$$，则$$\\overrightarrow{AC}\\cdot \\overrightarrow{BD}$$=~\\uline{~~~~~~~~~~}~", "answer_option_list": [[{"aoVal": "A", "content": "只有一个 "}], [{"aoVal": "B", "content": "有二个 "}], [{"aoVal": "C", "content": "有四个 "}], [{"aoVal": "D", "content": "有无穷多个 "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间向量"], "answer_analysis": ["注意到$${{3}^{2}}+{{11}^{2}}=1130={{7}^{2}}+{{9}^{2}}$$，由于$$\\overrightarrow{AB}+\\overrightarrow{BC}+\\overrightarrow{CD}+\\overrightarrow{DA}=\\vec{0}$$， 则$$D{{A}^{2}}={{\\overrightarrow{DA}}^{2}}={{(\\overrightarrow{AB}+\\overrightarrow{BC}+\\overrightarrow{CD})}^{2}}=A{{B}^{2}}+B{{C}^{2}}+C{{D}^{2}}+2(\\overrightarrow{AB}\\cdot \\overrightarrow{BC}+\\overrightarrow{BC}\\cdot \\overrightarrow{CD}+\\overrightarrow{CD}\\cdot \\overrightarrow{AB})$$ $$=A{{B}^{2}}-B{{C}^{2}}+C{{D}^{2}}+2({{\\overline{BC}}^{2}}+\\overrightarrow{AB}\\cdot \\overrightarrow{BC}+\\overrightarrow{BC}\\cdot \\overrightarrow{CD}+\\overrightarrow{CD}\\cdot \\overrightarrow{AB})$$ $$=A{{B}^{2}}-B{{C}^{2}}+C{{D}^{2}}+2(\\overrightarrow{AB}+\\overrightarrow{BC})\\cdot (\\overrightarrow{BC}+\\overrightarrow{CD})$$， 即$$2\\overrightarrow{AC}\\cdot \\overrightarrow{BD}=A{{D}^{2}}+B{{C}^{2}}-A{{B}^{2}}-C{{D}^{2}}=0$$，∴$$\\overrightarrow{AC}\\cdot \\overrightarrow{BD}$$只有一个值得$$0$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "47", "queId": "93f64c91ac624999b0562988739ffa14", "competition_source_list": ["2016年天津全国高中数学联赛竞赛初赛第6题6分", "2016年高考真题天津卷"], "difficulty": "1", "qtype": "single_choice", "problem": "设函数$$f\\left( x \\right)$$的定义域是$$\\left( -\\infty ,+\\infty \\right)$$，对于下列（$$1$$）$$\\sim $$（$$4$$）的四个命题： （$$1$$）若$$f\\left( x \\right)$$是奇函数，则$$f\\left( f\\left( x \\right) \\right)$$也是奇函数； （$$2$$）若$$f\\left( x \\right)$$是周期函数，则$$f\\left( f\\left( x \\right) \\right)$$也是周期函数； （$$3$$）若$$f\\left( x \\right)$$是单调递减函数，则$$f\\left( f\\left( x \\right) \\right)$$是单调递增函数； 正确的命题共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$个 "}], [{"aoVal": "B", "content": "$$2$$个 "}], [{"aoVal": "C", "content": "$$3$$个 "}], [{"aoVal": "D", "content": "$$0$$个 "}]], "knowledge_point_routes": ["知识标签->知识点->函数->函数的应用->简单的函数方程", "知识标签->知识点->函数->函数及其表示->函数的概念->复合函数", "知识标签->知识点->函数->函数的性质->单调性", "知识标签->知识点->函数->函数的性质->奇偶性", "知识标签->知识点->函数->函数的性质->周期性", "知识标签->素养->数学运算", "知识标签->素养->数据分析", "知识标签->素养->逻辑推理", "知识标签->题型->函数->函数的性质->周期性->函数周期性判断", "知识标签->题型->函数->函数的性质->单调性->用定义法证明函数的单调性", "知识标签->题型->函数->函数的性质->奇偶性->利用定义判断函数奇偶性"], "answer_analysis": ["若$$f\\left( x \\right)$$是奇函数，则$$f\\left( f\\left( -x \\right) \\right)=f\\left( -f\\left( x \\right) \\right)=-f\\left( f\\left( x \\right) \\right)$$，（$$1$$）对； 若$$f\\left( x+T \\right)=f\\left( x \\right)$$，则$$f\\left( f\\left( x+T \\right) \\right)=f\\left( f\\left( x \\right) \\right)$$，（$$2$$）对； 对任意$${{x}_{1}}\\textless{}{{x}_{2}}$$，有$$f\\left( {{x}_{1}} \\right)\\textgreater f\\left( {{x}_{2}} \\right)$$，则$$f\\left( f\\left( {{x}_{1}} \\right) \\right)\\textless{}f\\left( f\\left( {{x}_{2}} \\right) \\right)$$，（$$3$$）对； 取$$f\\left( x \\right)=\\begin{cases}1,x=0 0,x=1 \\left\\textbar{} x \\right\\textbar+1,其它 \\end{cases}$$，则$$f\\left( f\\left( x \\right) \\right)=\\begin{cases}0,x=0 1,x=1 \\left\\textbar{} x \\right\\textbar+2,其它 \\end{cases}$$，（$$4$$）错． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "455", "queId": "3a5da8ec9c1d4521a0f276ded97a6195", "competition_source_list": ["2018年山西全国高中数学联赛竞赛初赛第6题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$$\\cos \\frac{2\\pi }{7}\\cos \\frac{4\\pi }{7}\\cos \\frac{6\\pi }{7}$$的值为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正弦", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->半角公式", "课内体系->素养->数学运算"], "answer_analysis": ["记$$S=\\cos \\frac{2\\pi }{7}\\cos \\frac{4\\pi }{7}\\cos \\frac{6\\pi }{7}$$，则$$S=-\\cos \\frac{\\pi }{7}\\cos \\frac{2\\pi }{7}\\cos \\frac{4\\pi }{7}$$， $$-S\\cdot 8\\sin \\frac{\\pi }{7}=8\\sin \\frac{\\pi }{7}\\cdot \\cos \\frac{\\pi }{7}\\cos \\frac{2\\pi }{7}\\cos \\frac{4\\pi }{7}=4\\sin \\frac{2\\pi }{7}\\cdot \\cos \\frac{2\\pi }{7}\\cos \\frac{4\\pi }{7}=2\\sin \\frac{4\\pi }{7}\\cos \\frac{4\\pi }{7}$$ $$=\\sin \\frac{8\\pi }{7}=-\\sin \\frac{\\pi }{7}$$，所以，$$S=\\frac{1}{8}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "512", "queId": "5a4c2c653f9e4d30b2a30aacc7937a0c", "competition_source_list": ["2004年高考真题天津卷文科第9题5分", "2008年贵州全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "函数$$y={{3}^{x+1}}(-1\\leqslant x\\textless{}0)$$的反函数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$y=1+{{\\log }_{3}}x(x\\textgreater0)$$ "}], [{"aoVal": "B", "content": "$$y=-1+{{\\log }_{3}}x(x\\textgreater0)$$ "}], [{"aoVal": "C", "content": "$$y=1+{{\\log }_{3}}x(1\\leqslant x\\textless{}3)$$ "}], [{"aoVal": "D", "content": "$$y=-1+{{\\log }_{3}}x(1\\leqslant x\\textless{}3)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["由$$y={{3}^{x+1}}$$解得，$$x=-1+{{\\log }_{3}}y$$， 因为$$-1\\leqslant x\\textless{}0$$，得$$1\\textless{}y\\textless{}3$$， 所以函数$$y={{3}^{x+1}}(-1\\leqslant x\\textless{}0)$$的反函数是$$y=-1+{{\\log }_{3}}x(1\\leqslant x\\textless{}3)$$， 故选$$\\text{D}$$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "131", "queId": "8705857d51684aea8292bc2bf34fa058", "competition_source_list": ["2008年贵州全国高中数学联赛竞赛初赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "把函数$$y=\\sin \\left( 2x+\\frac{ \\pi }{6} \\right)-1$$的图象按向量$$\\overrightarrow{a}=\\left( \\frac{ \\pi }{6},1 \\right)$$平移，再把所得图象上各点的横坐标缩短为原来的$$\\frac{1}{2}$$，则所得图象的函数解析式是．", "answer_option_list": [[{"aoVal": "A", "content": "$$y=\\sin \\left( 4x+\\frac{2 \\pi }{3} \\right)-2$$ "}], [{"aoVal": "B", "content": "$$y=\\sin \\left( 4x-\\frac{ \\pi }{6} \\right)$$ "}], [{"aoVal": "C", "content": "$$y=\\sin \\left( 2x+\\frac{ \\pi }{6} \\right)$$ "}], [{"aoVal": "D", "content": "$$y=\\cos \\left( 4x+\\frac{2 \\pi }{3} \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["由题意，把函数$$y=\\sin \\left( 2x+\\frac{ \\pi }{6} \\right)-1$$的图象按向量$$\\overrightarrow{a}=\\left( \\frac{ \\pi }{6},1 \\right)$$平移， 可得$$y=\\sin \\left[ 2\\left( x-\\frac{ \\pi }{6} \\right)+\\frac{ \\pi }{6} \\right]=\\sin \\left( 2x-\\frac{ \\pi }{6} \\right)$$， 再把所得图象上各点的横坐标缩短为原来的$$\\frac{1}{2}$$， 可得$$y=\\sin \\left( 4x-\\frac{ \\pi }{6} \\right)$$，故选$$\\text{B}$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "625", "queId": "e84f00d563e840d99296186bef846387", "competition_source_list": ["2008年辽宁全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "从正方体的$$8$$个顶点的任意两个所确定的所有直线中取出两条，则这两条��线是异面直线的概率是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{29}{189}$$ "}], [{"aoVal": "B", "content": "$$\\frac{29}{63}$$ "}], [{"aoVal": "C", "content": "$$\\frac{34}{63}$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{7}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的角与距离", "竞赛->知识点->排列组合与概率->概率初步"], "answer_analysis": ["从$$8$$个顶点中任取两点可确定$$\\text{C}_{8}^{2}=28$$（条）直线， 从$$8$$个顶点中任取$$4$$个不共面的点共有$$\\text{C}_{8}^{4}-12$$组， 而其中每一组不共面的$$4$$点可出现$$3$$对异面直线， 所以所求的概率为$$\\frac{3(\\text{C}_{8}^{4}-12)}{\\text{C}_{28}^{2}}=\\frac{29}{63}$$. 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "624", "queId": "5f8ed4e0cbe34685a843d40f72a5f12b", "competition_source_list": ["2011年甘肃全国高中数学联赛竞赛初赛第8题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$[x]$$表示不超过实数$$x$$的最大整数，则在平面上，由满足$${{[x]}^{2}}+{{[y]}^{2}}=50$$的点所形成的图形的面积是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}], [{"aoVal": "E", "content": "$$12$$ "}]], "knowledge_point_routes": ["课内体系->知识点->几何证明选讲", "竞赛->知识点->组合->容斥原理与极端原理（二试）", "竞赛->知识点->数论模块->取整函数->与[x]有关的方程和不等式"], "answer_analysis": ["在平面上，由满足$${{[x]}^{2}}+{{[y]}^{2}}=50$$的点所形成的图形关于$$x$$轴,y轴对称，可先考察第一象限的情况，即当$$x\\textgreater0, y\\textgreater0$$时，由$${{[x]}^{2}}+{{[y]}^{2}}=50$$得 $$\\begin{cases}{{[x]}^{2}}=49 {{[y]}^{2}}=1 \\end{cases}\\begin{cases}{{[x]}^{2}}=25 {{[y]}^{2}}=25 \\end{cases}\\begin{cases}{{[x]}^{2}}=1 {{\\left[ y \\right]}^{2}}=49 \\end{cases}$$ 所以相应有 $$\\begin{cases}[x]=7 [y]=1 \\end{cases}\\begin{cases}[x]=5 [y]=5 \\end{cases}\\begin{cases}[x]=1 [y]=7 \\end{cases}$$ 因此 $$\\begin{cases}7\\leqslant x\\textless{}8 1\\leqslant y\\textless{}2 \\end{cases}\\begin{cases}5\\leqslant x\\textless{}6 5\\leqslant y\\textless{}6 \\end{cases}\\begin{cases}1\\leqslant x\\textless{}2 7\\leqslant y\\textless{}8 \\end{cases}$$ 所以在第一象限形成的图形的面积由容斥原理计算为$$3$$． 故在平面上，由满足$${{[x]}^{2}}+{{[y]}^{2}}=50$$的点所形成的图形的面积是$$12$$． "], "answer_value": "E"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "258", "queId": "2b4e09c82a4e426f804fcd4a4e3d5fb3", "competition_source_list": ["1995年全国高中数学联赛竞赛一试第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "设复平面上单位圆内接正$$20$$边形的$$20$$个顶点所对应的复数依次为$${{Z}_{1}}{{Z}_{2}}\\ldots {{Z}_{20}}$$，则复数$$Z_{1}^{1995}$$，$$Z_{2}^{1995}\\ldots Z_{20}^{1995}$$所对应的不同点的个数是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->模、辐角与单位根", "竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["设$${{Z}_{1}}=\\cos \\theta +\\text{i}\\sin \\theta $$，则： $${{Z}_{k}}=(\\cos \\theta +\\text{i}\\sin \\theta )(\\cos \\frac{2(k-1)\\mathsf{\\pi }}{20}+\\text{isin }\\frac{2(k-1)\\mathsf{\\pi }}{20})$$，$$1\\mathsf{\\leqslant }k\\mathsf{\\leqslant }20$$由$$1995=20\\times 99+15$$，得 $$\\mathop{Z}_{k}^{1995}=(\\cos 1995\\theta +\\text{isin 1995}\\theta ){{(\\cos \\frac{3\\mathsf{\\pi }}{2}+\\text{isin }\\frac{3\\mathsf{\\pi }}{2})}^{k-1}}$$ $$=(\\cos 1995\\theta +\\text{isin 1995}\\theta ){{(-i)}^{k-1}}$$，$$k=1$$，$$2$$，$$\\ldots $$，$$20$$， 共有四个不同的值． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "254", "queId": "26d9a99e444440708ee3c903243fc30d", "competition_source_list": ["1999年全国全国高中数学联赛竞赛一试第4题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "给定下列两个关于异面直线的命题： 命题Ⅰ：若平面$$\\alpha $$上的直线$$a$$与平面$$\\beta $$上的直线$$b$$为异面直线，直线$$c$$是$$\\alpha $$与$$\\beta $$的交线，那么，$$c$$至多与$$a,b$$中的一条相交； 命题Ⅱ：不存在这样的无穷多条直线，它们中的任意两条都是异面直线． 那么（ ）.", "answer_option_list": [[{"aoVal": "A", "content": "命题Ⅰ正确，命题Ⅱ不正确 "}], [{"aoVal": "B", "content": "命题Ⅱ正确，命题Ⅰ不正确 "}], [{"aoVal": "C", "content": "两个命题都正确 "}], [{"aoVal": "D", "content": "两个命题都不正确 "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的平行和垂直"], "answer_analysis": ["易知命题Ⅰ不正确； 又可以取无穷多个平行平面， 在每个平面上取一条直线， 且使这些直线两两不同向， 则这些直线中的任意两条都是异面直线， 从而命题Ⅱ也不正确． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "173", "queId": "109afa59a7fd4f1ca82dca1e823e9bc4", "competition_source_list": ["2012年陕西全国高中数学联赛竞赛初赛第10题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "从公路旁的材料工地沿笔直公路向同一方向运送电线杆到$$500\\text{m}$$以外的公路边埋栽，在$$500\\text{m}$$处栽一根，然后每间隔$$50\\text{m}$$在公路边栽一根．已知运输车辆一次最多只能运$$3$$根，要完成运栽$$20$$根电线杆的任务，并返回材料工地，则运输车总的行程最小为~ ．", "answer_option_list": [[{"aoVal": "A", "content": "$$13000$$ "}], [{"aoVal": "B", "content": "$$14000$$ "}], [{"aoVal": "C", "content": "$$15000$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和", "课内体系->知识点->数列"], "answer_analysis": ["先假设要完成$$21$$根电线杆的运栽，每次$$3$$根，设第$$k$$次往返的路程为$${{a}_{k}}\\left( k=1, 2,\\cdots , 7 \\right)$$，则$${{a}_{1}}=2\\times 600=1200$$，且 $${{a}_{k+1}}={{a}_{k}}+2\\times 150\\left( k=1, 2,\\cdots , 6 \\right)$$． 所以$$\\left { {{a}_{n}} \\right }$$是首项为$$1200$$，公差为$$300$$的等差数列． 故$${{S}_{7}}=7\\times 1200+\\frac{7\\times 6}{2}\\times 300=14700$$（m）． 但实际只运栽$$20$$根，那么必有一次运$$2$$根，其余$$6$$次均运$$3$$根． 若将运$$2$$根的安排在第$$k$$次（$$k=1, 2,\\cdots , 7$$），则$${{a}_{1}}, {{a}_{2}},\\cdots , {{a}_{k-1}}$$均不变，$${{a}_{k}}, {{a}_{k+1}},\\cdots , {{a}_{7}}$$各减少$$100\\text{m}$$，所以 $${{S}_{7}}\\left( k \\right)={{S}_{7}}-100\\left( 8-k \\right)=13800+100k$$. 显然，当$$k=1$$，即第$$1$$次运$$2$$根，其余$$6$$次各运$$3$$根时，总的行程最小，最小值为$$14000\\text{m}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "342", "queId": "1f62468d2e59472cad7d21bc8a0c0c36", "competition_source_list": ["2005年全国高中数学联赛竞赛一试第6题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "记集合$$T= {0,1,2,3,4,5,6 }$$，$$M=\\left { \\frac{{{a}_{1}}}{7}+\\frac{{{a}_{2}}}{{{7}^{2}}}+\\frac{{{a}_{3}}}{{{7}^{3}}}+\\frac{{{a}_{4}}}{{{7}^{4}}}\\textbar{{a}_{i}}\\in T,i=1,2,3,4 \\right }$$，将$$M$$中的元素按从大到小的顺序排列，则第$$2005$$个数是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{7}+\\frac{5}{{{7}^{2}}}+\\frac{6}{{{7}^{3}}}+\\frac{3}{{{7}^{4}}}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{7}+\\frac{5}{{{7}^{2}}}+\\frac{6}{{{7}^{3}}}+\\frac{2}{{{7}^{4}}}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{7}+\\frac{1}{{{7}^{2}}}+\\frac{0}{{{7}^{3}}}+\\frac{4}{{{7}^{4}}}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{7}+\\frac{1}{{{7}^{2}}}+\\frac{0}{{{7}^{3}}}+\\frac{3}{{{7}^{4}}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->整除->质数（算数基本定理）", "竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["用$${{[{{a}_{1}}{{a}_{2}}\\ldots {{a}_{k}}]}_{p}}$$表示$$k$$位$$p$$进制数，将集合$$M$$中的每个数乘以$${{7}^{4}}$$，得 $${{M}^{\\prime }}= {{{a}_{1}}\\cdot {{7}^{3}}+{{a}_{2}}\\cdot {{7}^{2}}+{{a}_{3}}\\cdot 7+{{a}_{4}}\\textbar{{a}_{i}}\\in T,i=1,2,3,4 }= {{{[{{a}_{1}}{{a}_{2}}{{a}_{3}}{{a}_{4}}]}_{7}}\\textbar{{a}_{i}}\\in T,i=1,2,3,4 }$$． $${{M}^{\\prime }}$$中的最大数为$${{[6666]}_{7}}={{[2400]}_{10}}$$． 在十进制数中，从$$2400$$起从大到小顺序排列的第$$2005$$个数是$$2400-2004=396$$． 而$${{[396]}_{10}}={{[1104]}_{7}}$$将此数除以$${{7}^{4}}$$，便得$$M$$中的数$$\\frac{1}{7}+\\frac{1}{{{7}^{2}}}+\\frac{0}{{{7}^{3}}}+\\frac{4}{{{7}^{4}}}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "975", "queId": "edf277585f0b410c86cb3d78db0c16d1", "competition_source_list": ["2009年山东全国高中数学联赛竞赛初赛第9题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "以四面体的顶点和各棱中点为顶点的空间四边形有 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$141$$个 "}], [{"aoVal": "B", "content": "$$144$$个 "}], [{"aoVal": "C", "content": "$$423$$个 "}], [{"aoVal": "D", "content": "$$432$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->两个基本计数原理", "竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["四面体有$$4$$个顶点和$$6$$条棱，每条棱有$$1$$个中点，其有$$10$$个点，从其中任取$$4$$个不共面的点有以下几种情况： ①这$$4$$个点都是顶点的取法只有$$1$$种； ②这$$4$$个点中有$$3$$个顶点的取法共有$$4\\times 3$$=12（种）； ③这$$4$$个点中有$$2$$顶点的取法共有$$6\\times (\\text{C}_{5}^{2}-2)=48$$（种）； ④这$$4$$个点中有$$1$$个顶点的取法共有$$4\\times (\\text{C}_{6}^{3}-3)=68$$（种）； ⑤这$$4$$个点中没有顶点的取法共有$$\\text{C}_{6}^{4}-3=12$$（种）． 综上所述，从这$$10$$个点中取$$4$$个不共面的点的取法共有 $$1+12+48+68+12=141$$（种）． 每不共面的$$4$$个点可以构成$$3$$个不同的空间四边形，则以这$$10$$个点为顶点的空间四边形共有$$141\\times 3=423$$（个）． ", "从这$$10$$个点中任取$$4$$个，共有$$\\text{C}_{10}^{4}=210$$（种）取法，其中取出的$$4$$个点共面的情况共有以下$$3$$种：
\n①从每个面上的$$6$$个点中任取$$4$$个点都共面，这样的取法共有$$4\\text{C}_{6}^{4}=60$$（种）；
\n②每个棱上的$$3$$个点与相对棱的中点共面，这样的取法共有$$6$$种；
\n③去掉$$1$$组相对棱后，其余四棱的中点共面，这样的取法共有$$3$$种．
\n综上所述，所取$$4$$点共面的取法共$$60+6+3=69$$（种）．所以，所取$$4$$点不共面的取法共有$$210-69=141$$（种），以下同方法$$1$$．故选$$\\text{C}$$．
\n"], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "329", "queId": "27c49ff13b074fdd8b8bc29df3c47760", "competition_source_list": ["2012年天津全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "设椭圆与$$x$$轴交于$$A$$、$$B$$两点，已知对于椭圆上不同于$$A$$、$$B$$的任意一点$$P$$，直线$$AP$$与$$BP$$的斜率之积均为$$-\\frac{1}{2}$$，则椭圆的离心率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{\\sqrt{3}}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{2}}{\\sqrt{3}}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{\\sqrt{2}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->椭圆"], "answer_analysis": ["设椭圆方程为$$\\frac{{{x}^{2}}}{{{a}^{2}}}+\\frac{{{y}^{2}}}{{{b}^{2}}}=1$$，则$$-\\frac{{{b}^{2}}}{{{a}^{2}}}=-\\frac{1}{2}$$，于是$$e=\\sqrt{\\frac{{{c}^{2}}}{{{a}^{2}}}}=\\sqrt{\\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}}}=\\sqrt{\\frac{1}{2}}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1011", "queId": "b2fcc21b93074c5ab13d330047150335", "competition_source_list": ["2003年AMC12竞赛B第11题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2003-AMC12B-11$$ Cassandra sets her watch to the correct time at noon. At the actual time of $$1:00$$ PM, she notices that her watch reads $$12:57$$ and $$36$$ seconds. Assuming that her watch loses time at a constant rate, what will be the actual time when her watch first reads $$10:00$$ PM? 卡桑德拉将手表调到中午的正确时间。 在 $$1:00$$ PM 的实际时间，她注意到她的手表显示 $$12:57$$ : $$36$$ 。 假设她的手表以恒定的速率走时，当她的手表第一次显示 $$10:00$$PM 时的实际时间是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$10:22$$ PM and $$24$$ seconds "}], [{"aoVal": "B", "content": "$$10:24$$ PM "}], [{"aoVal": "C", "content": "$$10:25$$ PM "}], [{"aoVal": "D", "content": "$$10:27$$ PM "}], [{"aoVal": "E", "content": "$$10:30$$ PM "}]], "knowledge_point_routes": ["课内体系->知识点->函数的应用->函数的实际应用->一次函数模型", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Simple Logical Reasoning"], "answer_analysis": ["$\\dfrac{57.6}{60}=\\dfrac{600}{X}$ $X=625$ "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "66", "queId": "20dbfc38ad8d48bfb497fc722a6dc1cd", "competition_source_list": ["2020~2021学年12月贵州贵阳观山湖区贵阳市第一中学高一上学期月考第12题3分", "2017年天津全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$x\\in \\left[ 0,1 \\right]$$，且$${{\\log }_{2}}{{\\log }_{2}}\\left( 2x+2 \\right)+{{2}^{2x+2}}$$为整数，则满足此条件的实数$$x$$有．", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$个 "}], [{"aoVal": "B", "content": "$$13$$个 "}], [{"aoVal": "C", "content": "$$14$$个 "}], [{"aoVal": "D", "content": "$$15$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数", "竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["因为$$f\\left( x \\right)={{\\log }_{2}}{{\\log }_{2}}\\left( 2x+2 \\right)+{{2}^{2x+2}}$$在$$\\left[ 0,1 \\right]$$上单调增，且$$f\\left( 0 \\right)=4$$，$$f\\left( 1 \\right)=17$$． 所以，$$f\\left( x \\right)$$可以取到$$4$$至$$17$$之间的所有整数，共$$14$$个． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "60", "queId": "0a69a63af48b4bbfb0da6ab776e66741", "competition_source_list": ["2008年湖南全国高中数学联赛竞赛初赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设函数$$f(x)={{x}^{3}}+3{{x}^{2}}+6x+14$$，且$$f(a)=1$$，$$f(b)=19$$，则$$a+b=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$-2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["由$$f(x)={{x}^{3}}+3{{x}^{2}}+6x+14={{\\left( x+1 \\right)}^{3}}+3\\left( x+1 \\right)+10$$， 令$$g(y)={{y}^{3}}+3y$$，则$$g(y)$$为奇函数且单调递增． 而$$f(a)={{\\left( a+1 \\right)}^{3}}+3\\left( a+1 \\right)+10=1$$， $$f(b)={{\\left( b+1 \\right)}^{3}}+3\\left( b+1 \\right)+10=19$$， 所以$$g(a+1)=-9$$，$$g(b+1)=9$$，$$g(-b-1)=-9$$， 从而$$g(a+1)=g(-b-1)$$， 即$$a+1=-b-1$$，故$$a+b=-2$$．选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "878", "queId": "9fc8c52661bf4a52b4097c9b5db78ee5", "competition_source_list": ["2017年AMC12竞赛B第16题", "2017年AMC10竞赛B第20题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2017~ AMC10B$$ P20 The number $$21!=51,090,942,171,709,440,000$$, has over $$60,000$$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd? 数字21！=51,090,942,171,709,440,000，有超过$60,000$ 个的正整数因数。随机选择其中一个，它是奇数的概率是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{21}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{19}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{18}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "E", "content": "$$\\frac{11}{21}$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities"], "answer_analysis": ["Solution 1 We note that the only thing that affects the parity of the factor are the powers of $$2$$. There are $$10+5+2+1=18$$ factors of $$2$$ in the number. Thus, there are $$18$$ cases in which a factor of $$21!$$ would be even (have a factor of $$2$$ in its prime factorization), and $$1$$ case in which a factor of $$21!$$ would be odd. Therefore, the answer is $$\\frac 1{19}$$. Solution 2 Consider how to construct any divisor $$D$$ of $$21!$$. First by Legendre\\textquotesingle s theorem for the divisors of a factorial , we have that there are a total of $$18$$ factors of $$2$$ in the number. $$D$$ can take up either $$0$$,$$1$$,$$2$$,$$3$$,$$\\cdots$$, or all $$18$$ factors of $$2$$, for a total of $$19$$ possible cases. In order for $$D$$ to be odd, however, it must have $$0$$ factors of $$2$$, meaning that there is a probability of $$1$$ case$$/19$$ cases$$=\\frac 1{19}$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1066", "queId": "ceac6bffc7524a0cb7f95d9303326db7", "competition_source_list": ["1998年全国高中数学联赛竞赛一试第13题20分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知复数$$z=1-\\sin \\theta +\\text{i}\\cos \\theta \\left( \\frac{ \\pi }{2} ~\\textless{} ~\\theta ~~\\textless{} ~ \\pi \\right)$$，求$$z$$的共轭复数$$\\overline{z}$$的辐角主值．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{ \\pi }{4}-\\frac{\\theta }{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{ \\pi }{4}+\\frac{\\theta }{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3 \\pi }{4}-\\frac{\\theta }{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3 \\pi }{4}+\\frac{\\theta }{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->复数->复数的三角形式", "竞赛->知识点->复数与平面向量->模、辐角与单位根"], "answer_analysis": ["$$z=1+\\cos \\left( \\frac{ \\pi }{2}+\\theta \\right)+\\text{i}\\sin \\left( \\frac{ \\pi }{2}+\\theta \\right)=2{{\\cos }^{2}}\\frac{\\frac{ \\pi }{2}+\\theta }{2}+2\\text{i}\\sin \\frac{\\frac{ \\pi }{2}+\\theta }{2}\\cos \\frac{\\frac{ \\pi }{2}+\\theta }{2}$$ $$=2\\cos \\frac{\\frac{ \\pi }{2}+\\theta }{2}\\left( \\cos \\frac{\\frac{ \\pi }{2}+\\theta }{2}+\\text{i}\\sin \\frac{\\frac{ \\pi }{2}+\\theta }{2} \\right)$$． 当$$\\frac{ \\pi }{2} ~\\textless{} ~\\theta ~~\\textless{} ~ \\pi $$时，$$\\overline{z}=-2\\cos \\frac{\\frac{ \\pi }{2}+\\theta }{2}\\left( -\\cos \\frac{\\frac{ \\pi }{2}+\\theta }{2}+\\text{i}\\sin \\frac{\\frac{ \\pi }{2}+\\theta }{2} \\right)$$ $$=-2\\cos \\left( \\frac{ \\pi }{4}+\\frac{\\theta }{2} \\right)\\left( \\cos \\left( \\frac{3 \\pi }{4}-\\frac{\\theta }{2} \\right)+\\text{i}\\sin \\left( \\frac{3 \\pi }{4}-\\frac{\\theta }{2} \\right) \\right)$$． ∴辐角主值为$$\\frac{3 \\pi }{4}-\\frac{\\theta }{2}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "886", "queId": "96bb1534f35f4ee4b1787ad3b8aa25d3", "competition_source_list": ["高二上学期单元测试《代数变形（3）》自招第11题", "1987年全国高中数学联赛竞赛一试第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "对任意给定的自然数$$n$$，若$${{n}^{6}}+3a$$为正整数的立方，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "这样的$$a$$有无穷多个 "}], [{"aoVal": "B", "content": "这样的$$a$$存在，但只有有限个 "}], [{"aoVal": "C", "content": "这样的$$a$$不存在 "}], [{"aoVal": "D", "content": "以上$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$的结论都不正确 "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->不定方程->因式分解与恒等变形"], "answer_analysis": ["对任意自然数$$k$$，取 $$a=3{{n}^{4}}k+9{{n}^{2}}{{k}^{2}}+9{{k}^{2}}$$， 则$${{n}^{6}}+3a={{({{n}^{2}}+3k)}^{3}}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "574", "queId": "ba02b8c75db2419195dbc3e0d52a97bc", "competition_source_list": ["1983年全国高中数学联赛竞赛一试第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a$$、$$b$$、$$c$$、$$d$$、$$m$$、$$n$$都是正实数．$$P=\\sqrt{ab}+\\sqrt{cd}，Q=\\sqrt{ma+nc}\\cdot \\sqrt{\\frac{b}{m}+\\frac{d}{n}}$$，那么．", "answer_option_list": [[{"aoVal": "A", "content": "$$P\\mathsf{\\geqslant }Q$$ "}], [{"aoVal": "B", "content": "$$P\\mathsf{\\leqslant }Q$$ "}], [{"aoVal": "C", "content": "$$P ~\\textless{} ~Q$$ "}], [{"aoVal": "D", "content": "$$P，Q$$间的大小关系确定，而与$$m，n$$的大小有关 "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的概念", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式与恒成立问题", "课内体系->知识点->等式与不等式->不等式->基本不等式->利用基本不等式求最值"], "answer_analysis": ["$$P=\\sqrt{ab}+\\sqrt{cd}$$，$$Q=\\sqrt{am+nc}\\cdot \\sqrt{\\frac{b}{m}+\\frac{d}{n}}$$． $${{P}^{2}}=ab+cd+2\\sqrt{abcd}$$． $${{Q}^{2}}=(am+nc)(\\frac{b}{m}+\\frac{d}{n})=ab+cd+\\frac{nbc}{m}+\\frac{mad}{n}$$． $$\\therefore \\frac{nbc}{m}+\\frac{mad}{n}\\mathsf{\\geqslant 2}\\sqrt{\\frac{nbc}{m}\\cdot \\frac{mad}{n}}=2\\sqrt{abcd}$$． $$\\because {{Q}^{2}}\\mathsf{\\geqslant }{{P}^{2}}$$， 由于$$P\\mathsf{，}Q$$均为正数． $$\\therefore Q\\mathsf{\\geqslant P}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "925", "queId": "9b9c064151ef4b9687cb0ea061d675af", "competition_source_list": ["2019年全国高中数学联赛竞赛初赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "设整数$$n\\textgreater4$$，$${{(x+2\\sqrt{y}-1)}^{n}}$$的展开式中$${{x}^{n-4}}$$与$$xy$$两项的系数相等，则$$n$$的值为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$31$$ "}], [{"aoVal": "B", "content": "$$41$$ "}], [{"aoVal": "C", "content": "$$51$$ "}], [{"aoVal": "D", "content": "$$61$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->二项式定理", "竞赛->知识点->排列组合与概率->二项式定理及其应用"], "answer_analysis": ["根据二项式定理，通项$${{T}_{r}}=\\text{C}_{n}^{r}{{x}^{n-r}}{{(2\\sqrt{y}-1)}^{r}}$$， 于是根据题意得 $${{T}_{4}}={{T}_{n-1}}\\Rightarrow \\text{C}_{n}^{4}=\\text{C}_{n}^{n-1}\\text{C}_{n-1}^{2}\\cdot 4\\cdot {{(-1)}^{n-3}}$$， 解得$$n=51$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "51", "queId": "048cc396a0c74868ab6065d454fe33fc", "competition_source_list": ["2012年湖南全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "若实数$$x$$满足：对任意正数$$a$$，均有$${{x}^{2}}\\textless{}1+a$$，则$$x$$的最小值是", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "不存在 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->二次函数"], "answer_analysis": ["因为$$a\\textgreater0$$，所以$$a+1\\textgreater1$$，$${{x}^{2}}\\textless{}1+a$$等价于$${{x}^{2}}\\leqslant 1$$，$$-1\\leqslant x\\leqslant 1$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "964", "queId": "e4adc205019e485bbd30705678c30ef7", "competition_source_list": ["2002年AMC10竞赛B第21题", "2002年AMC12竞赛B第17题"], "difficulty": "3", "qtype": "single_choice", "problem": "$$2002-AMC12B-17$$ Andy's lawn has twice as much area as Beth's lawn and three times as much area as Carlos' lawn. Carlos' lawn mower cuts half as fast as Beth's mower and one third as fast as Andy's mower. If they all start to mow their lawns at the same time, who will finish first? 安迪的草坪面积是贝丝草坪的两倍，是卡洛斯草坪的三倍。 卡洛斯的割草机的割草速度是贝丝的割草机的一半，是安迪的割草机的三分之一。 如果他们同时开始修剪草坪，谁会先完成？", "answer_option_list": [[{"aoVal": "A", "content": " Andy "}], [{"aoVal": "B", "content": " Beth "}], [{"aoVal": "C", "content": " Carlos "}], [{"aoVal": "D", "content": "Andy and Carlos tie for first "}], [{"aoVal": "E", "content": "All three tie "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Application->Proportion Word Problems", "课内体系->知识点->函数的应用->函数的实际应用->分式函数模型"], "answer_analysis": ["We say Andy\\textquotesingle s lawn has an area of $$x$$. Beth\\textquotesingle s lawn thus has an area of $$\\dfrac{x}{2}$$, and Carlos\\textquotesingle s lawn has an area of $$\\dfrac{x}{3}$$. We say Andy\\textquotesingle s lawn mower cuts at a speed of $$y$$. Carlos\\textquotesingle s cuts at a speed of $$\\dfrac{y}{3}$$, and Beth\\textquotesingle s cuts at a speed $$\\dfrac{2y}{3}$$. Each person\\textquotesingle s lawn is cut at a speed of $$\\dfrac{\\rm area}{\\rm rate}$$, so Andy\\textquotesingle s is cut in $$\\dfrac{x}{y}$$ time, as is Carlos\\textquotesingle s. Beth\\textquotesingle s is cut in $$\\dfrac{3}{4}\\times\\dfrac{x}{y}$$, so the first one to finish is Beth. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "929", "queId": "92abc587a5e14747bd2d865aa2e8bf22", "competition_source_list": ["1996年全国高中数学联赛竞赛一试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "存在整数$$n$$，使$$\\sqrt{p+n}+\\sqrt{n}$$是整数的质数$$p$$（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "不存在 "}], [{"aoVal": "B", "content": "只有一个 "}], [{"aoVal": "C", "content": "多于一个，但为有限个 "}], [{"aoVal": "D", "content": "有无穷多个 "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->不定方程->因式分解与恒等变形"], "answer_analysis": ["设$$p$$为任意奇质数，且$$p=2k+1$$，于是$$n={{k}^{2}}$$，便有： $$\\sqrt{p+n}+\\sqrt{n}=\\sqrt{2k+1+{{k}^{2}}}+\\sqrt{{{k}^{2}}}=2k+1$$， 所以，每个奇质数都有题设的性质． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "904", "queId": "8a3bb6c59daa4c879e3c708784a96baa", "competition_source_list": ["2015年吉林全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$f(x)=x\\left\\textbar{} x \\right\\textbar$$，若对任意的$$x\\geqslant 1$$有$$f(x+m)+mf(x)\\textless{}0$$恒成立，则实数$$m$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\infty ,-1)$$ "}], [{"aoVal": "B", "content": "$$(-\\infty ,-1]$$ "}], [{"aoVal": "C", "content": "$$(-\\infty ,-2)$$ "}], [{"aoVal": "D", "content": "$$(-\\infty ,-2]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["显然$$m\\textless{}0$$，所以$$f\\left( x+m \\right)\\textless{}-mf\\left( x \\right)=f\\left( \\sqrt{-m}x \\right)$$． 因为$$f\\left( x \\right)$$是单调增的奇函数，所以$$x+m\\textless{}\\sqrt{-m}x$$，即$$\\left( \\sqrt{-m}-1 \\right)x\\textgreater m$$． 所以必须$$\\sqrt{-m}-1\\geqslant 0$$，$$m\\leqslant -1$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "961", "queId": "b2857b62b8b14721bee93beae2ac2a53", "competition_source_list": ["2011年山东全国高中数学联赛竞赛初赛第7题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$6$$名同学咨询成绩．老师说：甲不是$$6$$人中成绩最好的，乙不是$$6$$人中成绩最差的，而且$$6$$人的成绩各不相同．那么他们$$6$$人的成绩不同的可能排序共有 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$种 "}], [{"aoVal": "B", "content": "$$216$$ 种 "}], [{"aoVal": "C", "content": "$$384$$ 种 "}], [{"aoVal": "D", "content": "$$504$$种 "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["解法一 以$$A$$记甲成绩排名第一的所有可能的排序之集， 以$$B$$记乙成绩排名为最后的所有可能的排序之集，则$$\\left\\textbar{} A \\right\\textbar=\\left\\textbar{} B \\right\\textbar=5!$$，$$\\left\\textbar{} A\\cap B \\right\\textbar=4!$$． 甲排名第一或乙排名最后的所有可能的排序数为 $$\\left\\textbar{} A\\cup B \\right\\textbar=\\left\\textbar{} A \\right\\textbar+\\left\\textbar{} B \\right\\textbar-\\left\\textbar{} A\\cap B \\right\\textbar=216$$． 按照老师所述，这６位同学成绩可能的排序数为$$6!-216=504$$． 解法二 以乙的成绩不在最后为前提，考虑甲的成绩不在第一的所有可能排序． （$$1$$）甲的成绩排在最后的所有可能的排序数为$$A_{5}^{5}=120$$； （$$2$$）甲的成绩不在最后，又不在第一的所有可能排序数为$$C_{4}^{1}\\cdot C_{4}^{1}\\cdot A_{4}^{4}=384$$． 所以甲不在首，乙不在尾的所有可能排序数为$$120+384=504$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1105", "queId": "eeeab4aa05af44109e7a5ac0ca514591", "competition_source_list": ["2008年四川全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设有一体积为$$54$$的正四面体，若以它的四个面的中心为顶点作一个四面体，则所作四面体的体积为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["由对称性可知，所作的四面体也是正四面体， 设原四面体的棱长、体积分别为$${{a}_{1}}$$、$${{V}_{1}}$$， 所作四面体的棱长、体积分别为$${{a}_{2}}$$、$${{V}_{2}}$$， 则由几何性质得$$\\frac{{{a}_{2}}}{{{a}_{1}}}=\\frac{1}{3}$$，故$$\\frac{{{V}_{2}}}{{{V}_{1}}}={{\\left( \\frac{1}{3} \\right)}^{3}}$$． 所以，所作的四面体的体积$${{V}_{2}}=\\frac{1}{27}{{V}_{1}}=2$$．故选$$\\text{B}$$ "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "853", "queId": "8d7ad978eb6e4b87ba295d59e2a15696", "competition_source_list": ["2015年黑龙江全国高中数学联赛竞赛初赛第11题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "四棱锥$$P-ABCD$$中，底面$$ABCD$$是正方形，边长为$$a,PD=a,PA=PC=\\sqrt{2}a$$，在这个四棱锥中放入一个球，则球的最大半径为．", "answer_option_list": [[{"aoVal": "A", "content": "$$(\\sqrt{2}-1)a$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{2}a$$ "}], [{"aoVal": "C", "content": "$$(1-\\frac{\\sqrt{2}}{2})a$$ "}], [{"aoVal": "D", "content": "$$a$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["采用体积法. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1059", "queId": "c5878b1e839d4300b843800839434436", "competition_source_list": ["2017年陕西全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$n$$是正整数，以下各组数$$a$$，$$b$$中，使$$\\frac{b}{a}$$为既约分数的是（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a=n+1$$，$$b=2n-1$$ "}], [{"aoVal": "B", "content": "$$a=2n-1$$，$$b=5n+2$$ "}], [{"aoVal": "C", "content": "$$a=n+1$$，$$b=3n+1$$ "}], [{"aoVal": "D", "content": "$$a=3n+1$$，$$b=5n+2$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->函数的应用->函数与方程"], "answer_analysis": ["$$(3n+1,5n+2)=(3n+1,2n+1)=(n,2n+1)=1$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "437", "queId": "8b7e0ffdfefa4b93bd5fe47d730154e0", "competition_source_list": ["2010年山东全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$0\\textless{}a\\textless{}b$$，在$$a,b$$之间插入一个正数$$k$$，使$$a,k,b$$成等比数列；在$$a,b$$之间插入两个正数$$m,n$$，使$$a,m,n,b$$成等差数列，则$${{(k+1)}^{2}}$$与$$(m+1)(n+1)$$的大小关系为．", "answer_option_list": [[{"aoVal": "A", "content": "$${{(k+1)}^{2}}\\textless{}(m+1)(n+1)$$ "}], [{"aoVal": "B", "content": "$${{(k+1)}^{2}}=(m+1)(n+1)$$ "}], [{"aoVal": "C", "content": "$${{(k+1)}^{2}}\\textgreater(m+1)(n+1)$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["$$a,k,b$$成等比数列，$${{k}^{2}}=ab$$， 所以$${{(k+1)}^{2}}={{k}^{2}}+2k+1=ab+2\\sqrt{ab}+1$$ $$\\textless{}ab+a+b+1=(a+1)(b+1).$$ $$a,m,n,b$$成等差数列，所以$$a+b=m+n$$，且$$b-a\\textgreater n-m$$， 所以由$$(m+1)+(n+1)=(a+1)+(b+1)$$， 知$$(m+1)(n+1)\\textgreater(a+1)(b+1)$$． 所以$${{(k+1)}^{2}}\\textless{}(a+1)(b+1)\\textgreater(m+1)(n+1)$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "490", "queId": "439f1f4bba3e4d04b6467a6001afe50f", "competition_source_list": ["2016年AMC10竞赛B第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "有五颗球分别标有 {$$1,2,3,4,5$$}, 从中一次性取出两个球，并且将球上的数字相乘，乘积是偶数的概率是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$0.2$$ "}], [{"aoVal": "B", "content": "$$0.3$$ "}], [{"aoVal": "C", "content": "$$0.5$$ "}], [{"aoVal": "D", "content": "$$0.7$$ "}], [{"aoVal": "E", "content": "$$0.8$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities", "课内体系->知识点->统计与概率->概率"], "answer_analysis": ["The product will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both numbers are odd is $$\\dfrac{\\left( \\begin{array}{l} {3} {2} \\end{array}\\right)}{\\left( \\begin{array}{l} {5} {2} \\end{array}\\right)}=\\dfrac{3}{10}$$, so the answer is $$1-0.3$$ which is $$\\left(\\text{D}\\right) 0.7$$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1087", "queId": "d7f595fecc56462fa01544eabd8e1dcc", "competition_source_list": ["2009年湖南全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知数列$$ {{{a}_{n}} }$$满足$${{a}_{1}}=0, {{a}_{n+1}}={{a}_{n}}+1+2\\sqrt{1+{{a}_{n}}}(n=1, 2,\\cdots )$$，则$${{a}_{2009}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$4036080$$ "}], [{"aoVal": "B", "content": "$$4036078$$ "}], [{"aoVal": "C", "content": "$$4036082$$ "}], [{"aoVal": "D", "content": "$$4036099$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["由已知得 $${{a}_{n+1}}+1={{a}_{n}}+1+2\\sqrt{1+{{a}_{n}}}+1={{(\\sqrt{{{a}_{n}}+1}+1)}^{2}}$$． 又因为$${{a}_{n+1}}\\textgreater0$$，所以 $$\\sqrt{{{a}_{n+1}}+1}=\\sqrt{{{a}_{n}}+1}+1$$ 故数列$$ {\\sqrt{{{a}_{n}}+1} }$$是首项为$$1$$，公差为$$1$$的等差数列，$$\\sqrt{{{a}_{n}}+1}=n$$， 即$${{a}_{n}}={{n}^{2}}-1=(n-1)(n+1)$$． 所以$${{a}_{2009}}=2008\\times 2010=4 036 080.$$ 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "616", "queId": "b592502baa994c2c9c4187f549d71692", "competition_source_list": ["2010年黑龙江全国高中数学联赛竞赛初赛第10题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "将一骰子抛掷两次，所得向上点数分别为$$m$$和$$n$$，则函数$$y=\\frac{2}{3}m{{x}^{3}}-nx+1$$在$$\\left[ 1,\\left. +\\infty \\right) \\right.$$上为增函数的概率是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{6}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->概率初步", "竞赛->知识点->导数模块->导数"], "answer_analysis": ["$${f}'(x)=2m{{x}^{2}}-n\\geqslant 0$$ 在$$\\left[ 1, +\\infty \\right]$$恒成立，即$$2{{x}^{2}}\\geqslant \\frac{n}{m}$$恒成立， 即$$2m\\geqslant n$$，但符合$$2m\\textless{}n$$条件的有($$1,3$$),($$1,4$$),($$1,5$$),($$1,6$$),($$2,5$$),($$2,6$$)共$$6$$种，故$$1-\\frac{6}{36}=\\frac{5}{6}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "826", "queId": "a8a1d71bd7e94694bf0eaac7a4c8d2b6", "competition_source_list": ["2015~2016学年2月湖南长沙开福区长沙市第一中学高三下学期月考理科第8题5分", "2017~2018学年12月北京海淀区北京市育英中学高二上学期月考理科第5题", "2014年黑龙江全国高中数学联赛竞赛初赛第3题5分", "2009年高考真题山东卷理科第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设双曲线$$\\frac{{{x}^{2}}}{{{a}^{2}}}-\\frac{{{y}^{2}}}{{{b}^{2}}}=1$$的一条渐近线与抛物线$$y={{x}^{2}}+1$$只有一个公共点，则双曲线的离心率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{4}$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{5}}{2}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{5}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->圆锥曲线->双曲线->双曲线的几何性质->双曲线的渐近线", "课内体系->知识点->圆锥曲线->双曲线->双曲线的几何性质->双曲线的离心率->求双曲线的离心率", "课内体系->知识点->圆锥曲线->双曲线->双曲线的定义、标准方程->双曲线的标准方程", "课内体系->知识点->圆锥曲线->抛物线->直线和抛物线的位置关系", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理"], "answer_analysis": ["根据双曲线和抛物线的对称性，可设双曲线的渐近线$$y=\\frac{b}{a}x$$与抛物线$$y={{x}^{2}}+1$$只有一个公共点. 将$$y=\\frac{b}{a}x\\left( a\\textgreater0 \\right)$$代入$$y={{x}^{2}}+1$$，得$$a{{x}^{2}}-bx+a=0$$，$$a\\textgreater0$$.令$${{\\left( -b \\right)}^{2}}-4{{a}^{2}}=0$$，得$${{b}^{2}}=4{{a}^{2}}$$，$$a\\textgreater0$$， 所以$${{c}^{2}}={{a}^{2}}+{{b}^{2}}={{a}^{2}}+4{{a}^{2}}=5{{a}^{2}}$$，$$a\\textgreater0$$，$$b\\textgreater0$$，所以$$\\frac{c}{a}=\\sqrt{5}$$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "594", "queId": "90b5234a263a481587d8391d35b05ad3", "competition_source_list": ["2016~2017学年广东深圳福田区深圳市红岭中学高二下学期期中理科第7题5分", "2019~2020学年12月福建厦门思明区福建省厦门第二中学高三上学期月考理科第4题5分", "2006年全国高中数学联赛竞赛一试第5题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "设$$f(x)={{x}^{3}}+{{\\log }_{2}}(x+\\sqrt{{{x}^{2}}+1})$$，则对任意实数$$a$$，$$b$$，$$a+b\\geqslant 0$$是$$f(a)+f(b)\\geqslant 0$$的（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "充分必要条件 "}], [{"aoVal": "B", "content": "充分而不必要条件 "}], [{"aoVal": "C", "content": "必要而不充分条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的性质->单调性", "课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与函数结合", "课内体系->素养->逻辑推理"], "answer_analysis": ["$$f(x)$$为单调递增函数，且$$f(-x)=-{{x}^{3}}+{{\\log }_{2}}\\left( -x+\\sqrt{{{x}^{2}}+1} \\right)=-{{x}^{3}}+{{\\log }_{2}}\\left( \\frac{1}{x+\\sqrt{{{x}^{2}}+1}} \\right)=-f(x)$$， $$f(x)$$为奇函数，则对任意实数$$a$$，$$b$$，由$$a+b\\geqslant 0$$，得$$a\\geqslant -b$$， ∴$$f(a)\\geqslant f(-b)=-f(b)$$，∴$$f(a)+f(b)\\geqslant 0$$； 由$$f(a)+f(b)\\geqslant 0$$得$$f(a)\\geqslant -f(b)=f(-b)$$， 所以$$a\\geqslant -\\frac{1}{b}$$，∴$$a+b\\geqslant 0$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "775", "queId": "72c38e5166b5464cbe02cac3229886a4", "competition_source_list": ["2015年福建全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知实数$$x$$，$$y$$，$$z$$满足$${{x}^{2}}+2{{y}^{2}}+3{{z}^{2}}=24$$，则$$x+2y+3z$$的最小值为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$-12$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$-16$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->柯西", "课内体系->知识点->等式与不等式->不等式->柯西不等式"], "answer_analysis": ["由柯西不等式，$${{\\left( x+2y+3z \\right)}^{2}}\\leqslant \\left( 1+2+3 \\right)\\left( {{x}^{2}}+2{{y}^{2}}+3{{z}^{2}} \\right)=144$$， 所以$$-12\\leqslant x+2y+3z\\leqslant 12$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1044", "queId": "aa5354bede64475796cb31c6c57db669", "competition_source_list": ["2018~2019学年浙江杭州拱墅区杭州第十四中学康桥校区高一上学期期中第13题3分", "2012年浙江全国高中数学联赛竞赛初赛第10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$f\\left( x \\right)={{x}^{2}}+bx+c$$，若方程$$f\\left( x \\right)=x$$无实根，则方程$$f\\left( f\\left( x \\right) \\right)=x$$．", "answer_option_list": [[{"aoVal": "A", "content": "有四个相异实根 "}], [{"aoVal": "B", "content": "有两个相异实根 "}], [{"aoVal": "C", "content": "有一个实根 "}], [{"aoVal": "D", "content": "无实数根 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->二次函数"], "answer_analysis": ["$$f\\left( x \\right)=x$$无实根，则二次函数图象$$f\\left( x \\right)={{x}^{2}}+bx+c$$在直线$$y=x$$上方，即$$f\\left( x \\right)\\textgreater x$$，所以$$f\\left( f\\left( x \\right) \\right)\\textgreater f\\left( x \\right)\\textgreater x$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "712", "queId": "4eaa93cb239043559f6f78dc3ffd4311", "competition_source_list": ["2016年湖南全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设函数$$f\\left( x \\right)=2x-\\cos x$$，$$\\left { {{a}_{n}} \\right }$$是公差为$$\\frac{ \\pi }{8}$$的等差数列，$$f\\left( {{a}_{1}} \\right)+f\\left( {{a}_{2}} \\right)+\\cdots +f\\left( {{a}_{5}} \\right)=5 \\pi $$，则$${{\\left[ f\\left( {{a}_{3}} \\right) \\right]}^{2}}-{{a}_{1}}{{a}_{5}}=$$（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{16} \\pi $$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{8} \\pi $$ "}], [{"aoVal": "D", "content": "$$\\frac{13}{16}{{ \\pi }^{2}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角函数的图象与性质->正弦型函数的图象与性质", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦函数的图象和性质", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的正弦", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的余弦", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->和差角公式化简求值综合运用", "课内体系->知识点->数列->等差数列->等差数列的概念与通项公式->等差数列求通项问题", "课内体系->知识点->数列->等差数列->等差数列的性质及应用", "课内体系->素养->数学运算"], "answer_analysis": ["$$f\\left( {{a}_{1}} \\right)+f\\left( {{a}_{2}} \\right)+\\cdots +f\\left( {{a}_{5}} \\right)=\\sum\\limits_{i=1}^{5}{\\left( 2{{a}_{i}}-\\cos {{a}_{i}} \\right)}$$ $$=10{{a}_{3}}-\\left[ \\cos \\left( {{a}_{3}}-\\frac{ \\pi }{4} \\right)+\\cos \\left( {{a}_{3}}+\\frac{ \\pi }{4} \\right) \\right]-\\left[ \\cos \\left( {{a}_{3}}-\\frac{ \\pi }{8} \\right)+\\cos \\left( {{a}_{3}}+\\frac{ \\pi }{8} \\right) \\right]-\\cos {{a}_{3}}$$ $$=10{{a}_{3}}-2\\cos {{a}_{3}}\\cos \\frac{ \\pi }{4}-2\\cos {{a}_{3}}\\cos \\frac{ \\pi }{8}-\\cos {{a}_{3}}$$ 令$$g\\left( {{a}_{3}} \\right)=10{{a}_{3}}-\\left( \\sqrt{2}+2\\cos \\frac{ \\pi }{8}+1 \\right)\\cos {{a}_{3}}$$，则$${g}'\\left( {{a}_{3}} \\right)=10+\\left( \\sqrt{2}+2\\cos \\frac{ \\pi }{8}+1 \\right)\\sin {{a}_{3}}\\textgreater0$$，$$g\\left( {{a}_{3}} \\right)$$在$$\\mathbf{R}$$上单调增，又$$g\\left( \\frac{ \\pi }{2} \\right)=5 \\pi $$，所以$${{a}_{3}}=\\frac{ \\pi }{2}$$． 从而，$${{a}_{1}}=\\frac{ \\pi }{4}{{a}_{5}}=\\frac{3}{4} \\pi $$，$${{\\left[ f\\left( {{a}_{3}} \\right) \\right]}^{2}}-{{a}_{1}}{{a}_{5}}={{\\left( ~\\pi -\\cos \\frac{ \\pi }{2} \\right)}^{2}}-\\frac{ \\pi }{4}\\cdot \\frac{3}{4} \\pi =\\frac{13}{16}{{ \\pi }^{2}}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "203", "queId": "0dd33cb549a94f8e88f935e1b5a66384", "competition_source_list": ["2020~2021学年河北石家庄新华区石家庄市第二中学高二上学期期中（竞赛班）第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$${{a}_{0}}{{x}^{2020}}+{{a}_{1}}{{x}^{2019}}\\left( 1-x \\right)+{{a}_{2}}{{x}^{2018}}{{\\left( 1-x \\right)}^{2}}+\\cdots +{{a}_{2020}}{{\\left( 1-x \\right)}^{2020}}=1$$，则$${{a}_{0}}+{{a}_{1}}+\\cdots +{{a}_{2020}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$${{2}^{2020}}$$ "}], [{"aoVal": "D", "content": "$${{2}^{2021}}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->知识点->计数原理->二项式定理->二项式系数的性质", "课内体系->知识点->计数原理->二项式定理->二项式定理的展开式"], "answer_analysis": ["因为$${{\\left[ x+\\left( 1-x \\right) \\right]}^{2020}}$$ $$={{a}_{0}}{{x}^{2020}}+{{a}_{1}}{{x}^{2019}}\\left( 1-x \\right)+{{a}_{2}}{{x}^{2018}}{{\\left( 1-x \\right)}^{2}}+\\cdots +{{a}_{2020}}{{\\left( 1-x \\right)}^{2020}}$$ $$=1$$． 且由二项式定理得，当$$0\\leqslant k\\leqslant 2020$$， 且$$k\\in \\mathbf{Z}$$时，$${{a}_{k}}=\\text{C}_{2020}^{k}$$， 所以$${{a}_{0}}+{{a}_{1}}+\\cdots +{{a}_{2020}}$$ $$=\\text{C}_{2020}^{0}+\\text{C}_{2020}^{1}+\\text{C}_{2020}^{2}+\\cdots +\\text{C}_{2020}^{2020}$$ $$={{2}^{2020}}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "310", "queId": "8b265def31ab4473a1e2a1680482ea22", "competition_source_list": ["2008年黑龙江全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "给定数列$$ {{{x}_{n}} }$$，$${{x}_{1}}=1$$，且$${{x}_{n+1}}=\\frac{\\sqrt{3}{{x}_{n}}+1}{\\sqrt{3}-{{x}_{n}}}$$，则$$\\sum\\limits_{n=1}^{2008}{{{x}_{n}}=}$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$2+\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$-2+\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->（模）周期数列"], "answer_analysis": ["由于$${{x}_{n+1}}=\\frac{{{x}_{n}}+\\frac{\\sqrt{3}}{3}}{1-\\frac{\\sqrt{3}}{3}{{x}_{n}}}$$，令$${{x}_{n}}=\\tan {{a}_{n}}$$， 因此$${{x}_{n+1}}=\\tan \\left( {{a}_{n}}+\\frac{ \\pi }{6} \\right)$$，$${{x}_{n+6}}={{x}_{n}}$$． 易算得$${{x}_{1}}=1$$，$${{x}_{2}}=2+\\sqrt{3}$$，$${{x}_{3}}=-2-\\sqrt{3}$$， $${{x}_{4}}=-1$$，$${{x}_{5}}=-2+\\sqrt{3}$$，$${{x}_{6}}=2-\\sqrt{3}$$，$${{x}_{7}}=1$$，\\ldots， 所以$$\\sum\\limits_{n=1}^{2008}{{{x}_{n}}={{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=0}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "640", "queId": "a344f4062fc747fb96c3e635c0d4880b", "competition_source_list": ["2020~2021学年北京高二单元测试", "竞赛", "2020年北京海淀区北京大学自主招生（强基计划）第7题5分", "2020~2021学年北京高三单元测试"], "difficulty": "2", "qtype": "single_choice", "problem": "方程$$19x+93y=4xy$$的整数解个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "前三个答案都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->解方程（组）"], "answer_analysis": ["当$$x$$、$$y$$中有一个为$$0$$时，$$\\left( x,y \\right)$$只能为$$\\left( 0,0 \\right)$$． 当$$x$$、$$y$$均不为$$0$$时，由原式可以得到$$\\begin{cases}19\\frac{x}{y}+93=4x 19+93\\frac{y}{x}=4y \\end{cases}$$，两式左右两端均为整数，设$$k=\\frac{x}{y}$$，则$$19k\\in \\mathbf{Z}$$，$$\\frac{93}{k}\\in \\mathbf{Z}\\Rightarrow k=\\pm \\frac{1}{19}$$，$$\\pm \\frac{3}{19}$$，$$\\pm 1$$，$$\\pm \\frac{31}{19}$$，$$\\pm 3$$，$$\\pm \\frac{93}{19}$$，$$\\pm 31$$，$$\\pm 93$$，共$$8$$组． 由于$$19\\times 93\\equiv 3\\left( \\bmod 4 \\right)$$，$$93\\equiv 1\\left( \\bmod 4 \\right)$$，$$19\\equiv 3\\left( \\bmod 4 \\right)$$，故$$19\\frac{x}{y}\\equiv 3\\left( \\bmod 4 \\right)$$，$$93\\frac{y}{x}\\equiv 1\\left( \\bmod 4 \\right)$$，在$$k$$的$$8$$组取值中，每一组恰有一个取值能同时满足上述两同余方程， 但由于$$x$$、$$y$$均不为$$0$$，须舍去$$k=-\\frac{93}{19}$$，故此时$$\\left( x,y \\right)$$共有$$7$$组． 综上所述，原方程整数解的组数为$$8$$，故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "596", "queId": "51f2223572054bbb8cb5a4a57211764f", "competition_source_list": ["2009年山东全国高中数学联赛竞赛初赛第8题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "在复平面上，复数$${{z}_{1}}$$对应的点在连结$$1$$和$$\\text{i}$$两点的线段上运动，复数$${{z}_{2}}$$对应的点在以原点为圆心．半径等于$$1$$的圆上运动．则复数$${{z}_{1}}+{{z}_{2}}$$对应的点所在区域的面积为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$4+ \\pi $$ "}], [{"aoVal": "B", "content": "$$2\\sqrt{2}+\\pi $$ "}], [{"aoVal": "C", "content": "$$\\pi $$ "}], [{"aoVal": "D", "content": "$$\\frac{3\\pi }{2}+1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的应用"], "answer_analysis": ["由已知可得$${{z}_{1}}=t+(1-t)i(0\\leqslant t\\leqslant 1)$$．$${{z}_{2}}=\\cos \\theta +i\\sin \\theta $$，得 $${{z}_{1}}+{{z}_{2}}=t+\\cos \\theta +(1-t+\\sin \\theta )i$$． 设$${{z}_{1}}+{{z}_{2}}=x+yi$$，则$$\\begin{cases}x=t+\\cos \\theta y=1-t+\\sin \\theta \\end{cases}$$．消去$$\\theta $$，得 $${{(x-t)}^{2}}+{{[y-(1-t)]}^{2}}=1$$． 即$${{z}_{1}}+{{z}_{2}}$$对应的点以$$(t,1-t)$$为圆心，半径等于$$1$$的圆上，而因为 （$$0\\leqslant t\\leqslant 1$$）．故$${{z}_{1}}+{{z}_{2}}$$对应点所在区域为图中阴影部分．其面积为 $$2\\cdot \\frac{\\pi }{2}+2\\cdot \\sqrt{2}=2\\sqrt{2}+\\pi $$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1036", "queId": "aeb394b959bc49b6a0b0b29005a90ad4", "competition_source_list": ["2021年吉林全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "关于$$x$$的方程$${{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-a\\cos \\left( 1-x \\right)=-1$$只有一个实数解，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$a=-1$$ "}], [{"aoVal": "B", "content": "$$a=1$$ "}], [{"aoVal": "C", "content": "$$a=2$$ "}], [{"aoVal": "D", "content": "$$a$$的值不唯一 "}]], "knowledge_point_routes": ["课内体系->知识点->函数的应用->函数的零点->函数零点的概念", "竞赛->知识点->函数->函数方程"], "answer_analysis": ["函数$$f\\left( x \\right)={{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-a\\cos \\left( 1-x \\right)+1$$的图象关于直线$$x=1$$对称， 又方程只有一个实数解， ∴$$f\\left( 1 \\right)=0$$，得$$a=2$$， 当$$a=1$$时，$$f\\left( x \\right)={{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-2\\cos \\left( 1-x \\right)+1\\geqslant 1-2+1=0$$， 当且仅当$$x=1$$时取等号， 即方程$${{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-a\\cos \\left( 1-x \\right)=-1$$只有一个实数解，符合题设． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1114", "queId": "ea6ebb272dc9469794d3ee1c10b4c1b2", "competition_source_list": ["2015年黑龙江全国高中数学联赛竞赛初赛第12题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "设等差数列$$ {{{a}_{n}} }$$满足：$$\\frac{{{\\sin }^{2}}{{a}_{3}}-{{\\cos }^{2}}{{a}_{3}}+{{\\cos }^{2}}{{a}_{3}}{{\\cos }^{2}}{{a}_{6}}-{{\\sin }^{2}}{{a}_{3}}{{\\sin }^{2}}{{a}_{6}}}{\\sin ({{a}_{4}}+{{a}_{5}})}=1$$，公差$$d\\in (-1,0)$$，若当且仅当$$n=9$$时数列$$ {{{a}_{n}} }$$的前$$n$$项和$${{S}_{n}}$$取得最大值，则首项$${{a}_{1}}$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(\\frac{7 \\pi }{6},\\frac{4 \\pi }{3})$$ "}], [{"aoVal": "B", "content": "$$(\\frac{4 \\pi }{3},\\frac{3 \\pi }{2})$$ "}], [{"aoVal": "C", "content": "$$[\\frac{7 \\pi }{6},\\frac{4 \\pi }{3}]$$ "}], [{"aoVal": "D", "content": "$$[\\frac{4 \\pi }{3},\\frac{3 \\pi }{2}]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的概念", "竞赛->知识点->数列与数学归纳法->数列的通项与求和", "竞赛->知识点->数列与数学归纳法->数列的综合应用", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["$${{\\sin }^{2}}{{a}_{3}}-{{\\cos }^{2}}{{a}_{3}}+{{\\cos }^{2}}{{a}_{3}}{{\\cos }^{2}}{{a}_{6}}-{{\\sin }^{2}}{{a}_{3}}{{\\sin }^{2}}{{a}_{6}}={{\\sin }^{2}}{{a}_{3}}-{{\\sin }^{2}}{{a}_{6}}$$， 而$${{\\sin }^{2}}{{a}_{3}}-{{\\sin }^{2}}{{a}_{6}}=\\frac{1-\\cos 2{{a}_{3}}}{2}-\\frac{1-\\cos 2{{a}_{6}}}{2}=\\sin \\left( {{a}_{6}}+{{a}_{3}} \\right)\\sin \\left( {{a}_{3}}-{{a}_{6}} \\right)$$， 所以$$\\sin \\left( {{a}_{3}}-{{a}_{6}} \\right)=1$$，$${{a}_{3}}-{{a}_{6}}=-3d=2k \\pi +\\frac{ \\pi }{2}$$，所以$$d=-\\frac{2}{3}k \\pi -\\frac{ \\pi }{6}$$，$$k\\in \\mathbf{Z}$$． 由$$-1\\textless{}d\\textless{}0$$，可得$$-\\frac{1}{4}\\textless{}k\\textless{}-\\frac{1}{4}+\\frac{3}{2 \\pi }$$，所以$$k=0$$，$$d=-\\frac{ \\pi }{6}$$． 由$${{a}_{9}}\\textgreater0$$，$${{a}_{10}}\\textless{}0$$，可得$$\\begin{cases}{{a}_{1}}+8d\\textgreater0 {{a}_{1}}+9d\\textless{}0 \\end{cases}$$，解得$$\\frac{4}{3} \\pi \\textless{}{{a}_{1}}\\textless{}\\frac{3}{2} \\pi $$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "890", "queId": "859bd0a3e90d4056a703510642b208ce", "competition_source_list": ["第二十届全国希望杯高一竞赛复赛邀请赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$x$$是某个三角形的最小内角，则$$y=\\frac{\\cos x}{\\cos \\frac{x}{2}-\\sin \\frac{x}{2}}$$的值域是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\sqrt{2},\\sqrt{2}]$$ "}], [{"aoVal": "B", "content": "$$\\left( -\\sqrt{2},\\frac{\\sqrt{3}+1}{2} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left( 1,\\frac{\\sqrt{3}+1}{2} \\right]$$ "}], [{"aoVal": "D", "content": "$$(0,\\sqrt{2}]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["$$0\\textless{}x\\leqslant \\frac{ \\pi }{3}$$，$$y=\\frac{{{\\cos }^{2}}\\frac{x}{2}-{{\\sin }^{2}}\\frac{x}{2}}{\\cos \\frac{x}{2}-\\sin \\frac{x}{2}}=\\cos \\frac{x}{2}+\\sin \\frac{x}{2}=\\sqrt{2}\\sin \\left( \\frac{x}{2}+\\frac{ \\pi }{4} \\right)$$，即可得$$y$$的值域． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "458", "queId": "59e9f02806b34c859178a6223ea073d5", "competition_source_list": ["2008年天津全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知二次函数$$f\\left( x \\right)={{x}^{2}}-3x+2$$，则方程$$f\\left( f\\left( x \\right) \\right)=0$$不同实数根的数目为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->方法->图象法", "课内体系->方法->换元法", "课内体系->知识点->函数的应用->函数的零点->函数零点的概念", "课内体系->知识点->函数的应用->函数的零点->零点的个数问题->零点、交点、根的等价转化", "课内体系->知识点->函数的应用->函数的实际应用->二次函数模型", "课内体系->素养->数学抽象", "课内体系->素养->逻辑推理", "课内体系->思想->数形结合思想"], "answer_analysis": ["因为$$f\\left( f\\left( x \\right) \\right)={{\\left( {{x}^{2}}-3x+2 \\right)}^{2}}-3\\left( {{x}^{2}}-3x+2 \\right)+2={{x}^{4}}-6{{x}^{3}}+10{{x}^{2}}-3x$$， 所以有$$x\\left( x-3 \\right)\\left( {{x}^{2}}-3x+1 \\right)=0,{{x}_{1}}=0,{{x}_{2}}=3,{{x}_{34}}=\\frac{3\\pm \\sqrt{5}}{2}$$， 因此原方程有$$4$$个不同实根．故选$$\\text{D}$$． 注 也可以讨论$$f\\left( x \\right)=0$$根的分布情况． 因为当$$x\\leqslant \\frac{3}{2}$$时，函数$$f\\left( x \\right)$$单调递减，当$$x\\textgreater\\frac{3}{2}$$时，函数$$f\\left( x \\right)$$单调递增，且$$f\\left( x \\right)=0$$的两个根为$$1,2$$，所以当$$x\\leqslant \\frac{3}{2}$$时，函数$$f\\left( x \\right)\\in \\left[ -\\frac{1}{4},+\\infty \\right)\\supset \\left[ 1,2 \\right]$$，因此$$f\\left( f\\left( x \\right) \\right)=0$$有两个不同实根；当$$x\\textgreater\\frac{3}{2}$$时，函数$$f\\left( x \\right)\\in \\left( -\\frac{1}{4},+\\infty \\right)\\supset \\left[ 1,2 \\right]$$，因此$$f\\left( f\\left( x \\right) \\right)=0$$也有两个不同实根．综上所述，原方程有$$4$$个不同实根． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "808", "queId": "e420994965964bd9880b490bdd3ecc5a", "competition_source_list": ["2012年吉林全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$${{A}_{1}}, {{A}_{2}},\\cdots , {{A}_{n}}$$为集合$$S=\\left { 1, 2,\\cdots , n \\right }$$的$$n$$个不同子集$$\\left( n\\geqslant 4 \\right)$$，为了表示这些子集，作$$n$$行$$n$$列的数阵，规定第$$i$$行与第$$j$$列的数为$${{a}_{ij}}=\\begin{cases}0, i\\notin {{A}_{j}}, ,1, i\\in {{A}_{j}},\\end{cases}$$则下列说法错误的是", "answer_option_list": [[{"aoVal": "A", "content": "数阵中第一列的数全是$$0$$当且仅当$${{A}_{1}}=\\varnothing $$ "}], [{"aoVal": "B", "content": "数阵中第$$n$$列的数全是$$1$$当且仅当$${{A}_{n}}=S$$ "}], [{"aoVal": "C", "content": "数阵中第$$j$$行的数字和表明集合$${{A}_{j}}$$含有几个元素 "}], [{"aoVal": "D", "content": "数阵中所有的$${{n}^{2}}$$个数字之和不超过$${{n}^{2}}-n+1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["当$${{A}_{1}}, {{A}_{2}},\\cdots , {{A}_{n}}$$中一个为$$S$$本身，其余$$n-1$$个子集为$$S$$的互不相同的$$n-1$$元子集时，数阵中所有的$${{n}^{2}}$$个数字之和最大，为$${{n}^{2}}-n+1$$，因此，$$D$$是正确的．数阵中第$$j$$行的数字和表明元素$$j$$属于几个子集，因此，$$C$$是错误的． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1071", "queId": "b8280458692c418893cfebe0e2614b5f", "competition_source_list": ["2022~2023学年湖南永州宁远县高一上学期月考（明德湘南中学基础知识竞赛）第8题", "2022~2023学年10月广东佛山禅城区佛山市第一中学高一上学期月考第7题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "若两个正实数$$x$$，$$y$$满足$$2x+y=1$$且存在这样的$$x$$，$$y$$使不等式$$\\dfrac{1}{x}+\\dfrac{2}{y}\\textless{} {{m}^{2}}+2m$$有解，则实数$$m$$的取值范围是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$$ {m\\textbar m\\textless{} -4$$或$$m\\textgreater2 }$$ "}], [{"aoVal": "B", "content": "$$\\left { m\\left\\textbar{} -2 \\right.\\textless{} m\\textless{} 4 \\right }$$ "}], [{"aoVal": "C", "content": "$$ {m\\textbar m\\textless{} -2$$或$$m\\textgreater4 }$$ "}], [{"aoVal": "D", "content": "$$\\left { m\\left\\textbar{} -4 \\right.\\textless{} m\\textless{} 2 \\right }$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->解不等式", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件"], "answer_analysis": ["由题意，$$\\dfrac{1}{x}+\\dfrac{2}{y}=\\left(\\frac{1}{x}+\\dfrac{2}{y}\\right)(2x+y)$$ $$=4+\\dfrac{4x}{y}+\\dfrac{y}{x}\\geqslant 4+2\\sqrt{\\dfrac{4x}{y}\\times \\dfrac{y}{x}}=8$$， 当且仅当$$\\dfrac{4x}{y}=\\dfrac{y}{x}$$，即$$y=\\dfrac{1}{2},x=\\dfrac{1}{4}$$时等号成立． 故若存在这样的$$x$$，$$y$$使不等式$$\\dfrac{1}{x}+\\dfrac{2}{y}\\textless{} {{m}^{2}}+2m$$有解． 即$${{m}^{2}}+2m\\textgreater8\\Leftrightarrow (m-2)(m+4)\\textgreater0\\Leftrightarrow m\\textgreater2$$或$$m\\textless{} -4$$． 故选：$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "963", "queId": "e4ac866820054340bf0afeb881c2de79", "competition_source_list": ["第二十届全国希望杯高二竞赛初赛邀请赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "设直线$$l$$：$$x+y=-2$$，抛物线$$C$$：$${{y}^{2}}=2x$$，当点$$P\\in l$$，点$$Q\\in C$$时，线段$$PQ$$的最小长度等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3\\sqrt{2}}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5\\sqrt{2}}{8}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆锥曲线"], "answer_analysis": ["不难算出与$$x+y=-2$$平行且与$${{y}^{2}}=2x$$相切的直线方程为$$x+y=-\\frac{1}{2}$$，$$PQ$$的最短距离就是$$x+y=-2$$与$$x+y=-\\frac{1}{2}$$之间的距离，易算得为$$\\frac{\\left\\textbar{} -\\frac{1}{2}-(-2) \\right\\textbar}{\\sqrt{2}}=\\frac{3\\sqrt{2}}{4}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1206", "queId": "fe5d355b7e8f4dae9ccf9f01aeed85ae", "competition_source_list": ["2019年全国高中数学联赛竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "平面直角坐标系中，$$\\overrightarrow{e}$$是单位向量，向量$$\\overrightarrow{a}$$满足$$\\overrightarrow{a}\\cdot \\overrightarrow{e}=2$$且$$\\textbar\\overrightarrow{a}{{\\textbar}^{2}}\\leqslant 5\\textbar\\overrightarrow{a}+t\\overrightarrow{e}\\textbar$$对任意实数$$t$$成立，则$$\\textbar\\overrightarrow{a}\\textbar$$的取值范围是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$[1,4]$$ "}], [{"aoVal": "B", "content": "$$[5,8]$$ "}], [{"aoVal": "C", "content": "$$[5,20]$$ "}], [{"aoVal": "D", "content": "$$[\\sqrt{5},2\\sqrt{5}]$$ "}]], "knowledge_point_routes": ["课内体系->知识点->平面向量->平面向量基本定理及其坐标表示", "竞赛->知识点->复数与平面向量->平面向量的概念与运算"], "answer_analysis": ["设$$\\overrightarrow{e}=\\left( 1,0 \\right)$$，$$\\overrightarrow{a}=\\left( 2,x \\right)$$，则 $$\\forall t\\in \\mathbf{R}$$，$$4+{{x}^{2}}\\leqslant 5\\sqrt{{{\\left( 2+t \\right)}^{2}}+{{x}^{2}}}$$， 即$$4+{{x}^{2}}\\leqslant 5\\textbar x\\textbar\\Leftrightarrow 1\\textless{}\\textbar x\\textbar\\textless{}4$$， 于是$$\\textbar\\overrightarrow{a}\\textbar=\\sqrt{4+{{x}^{2}}}$$的取值范围是$$\\left[ \\sqrt{5},2\\sqrt{5} \\right]$$． 故答案为：$$\\left[ \\sqrt{5},2\\sqrt{5} \\right]$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "367", "queId": "79b544af048843ecac49ce42d515d764", "competition_source_list": ["2014年浙江全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知等比数列$$\\left { {{a}_{n}} \\right }:{{a}_{1}}=5$$，$${{a}_{4}}=625$$，则$$\\sum\\limits_{k=1}^{2014}{\\frac{1}{{{\\log }_{5}}{{a}_{k}}{{\\log }_{5}}{{a}_{k+1}}}=}$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2014}{2015}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2013}{2014}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2012}{4028}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2013}{4030}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["设等比数列的公比为$$q$$，则 $$625=5{{q}^{3}}\\Rightarrow q=5\\Rightarrow \\sum\\limits_{k=1}^{2014}{\\frac{1}{{{\\log }_{5}}{{a}_{k}}{{\\log }_{5}}{{a}_{k+1}}}=\\sum\\limits_{k=1}^{2014}{\\frac{1}{k\\left( k+1 \\right)}}=\\frac{2014}{2015}}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "212", "queId": "cbdb3f303635477087b9736b6dd06f40", "competition_source_list": ["2014年天津全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "等比数列$$\\left { {{a}_{n}} \\right }$$的前$$n$$项和为$${{S}_{n}}$$，并且对任意正整数$$n$$成立$${{S}_{n+2}}=4{{S}_{n}}+3$$，则$${{a}_{2}}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "2 "}], [{"aoVal": "B", "content": "6 "}], [{"aoVal": "C", "content": "$$2$$或6 "}], [{"aoVal": "D", "content": "$$2$$或$$-6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["设公比为$$q$$．由于$$q{{S}_{n}}=q\\left( {{a}_{1}}+{{a}_{2}}+\\ldots +{{a}_{n}} \\right)={{a}_{2}}+{{a}_{3}}+\\ldots +{{a}_{n+1}}$$，所以$${{S}_{n+1}}=q{{S}_{n}}+{{a}_{1}}$$．进而$${{S}_{n+2}}=q\\left( q{{S}_{n}}+{{a}_{1}} \\right)+{{a}_{1}}={{q}^{2}}{{S}_{n}}+{{a}_{1}}\\left( q+1 \\right)$$．与已知条件比较可知$${{q}^{2}}=4$$，$${{a}_{1}}\\left( q+1 \\right)=3$$．所以$$q=2$$，$${{a}_{1}}=1$$，或$$q=-2$$，$${{a}_{1}}=-3$$．相应地，$${{a}_{2}}=2$$或$$6$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1192", "queId": "e2a73912929e4d788bbb89ca7c65fbb4", "competition_source_list": ["1998年全国高中数学联赛竞赛一试第2题6分", "2009年黑龙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "若非空集合$$A=\\left { x\\textbar2a+1\\leqslant x\\leqslant 3a-5 \\right }$$，$$B=\\left { x\\textbar3\\leqslant x\\leqslant 22 \\right }$$，则能使$$A\\subseteq A\\cap B$$成立的所有$$a$$的集合是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left { a\\textbar{} 1\\leqslant a\\leqslant 9 \\right }$$ "}], [{"aoVal": "B", "content": "$$\\left { a\\textbar{} 6\\leqslant a\\leqslant 9 \\right }$$ "}], [{"aoVal": "C", "content": "$$\\left { a\\textbar a\\leqslant 9 \\right }$$ "}], [{"aoVal": "D", "content": "$$\\varnothing $$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["$$A\\subseteq B$$，$$A\\ne \\varnothing $$．$$\\Rightarrow $$$$3\\leqslant 2a+1\\leqslant 3a-5\\leqslant 22$$，$$\\Rightarrow $$$$6\\leqslant a\\leqslant 9$$． 故选$$B$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "299", "queId": "1ede31281ee145bdb8926bbda549b285", "competition_source_list": ["2014年浙江全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$\\angle A$$、$$\\angle B$$、$$C$$为$$\\triangle ABC$$的三个内角，命题$$P:\\angle A=\\angle B$$；命题$$Q:\\sin \\angle A=\\sin \\angle B$$，则$$\\neg P$$是$$\\neg Q$$的．", "answer_option_list": [[{"aoVal": "A", "content": "充分非必要条件 "}], [{"aoVal": "B", "content": "必要非充分条件 "}], [{"aoVal": "C", "content": "充分必要条件 "}], [{"aoVal": "D", "content": "既非充分又非必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->常用逻辑用语"], "answer_analysis": ["在$$\\triangle ABC$$中$$\\angle A\\ne \\angle B\\Leftrightarrow \\sin \\angle A\\ne \\sin \\angle B$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "16", "queId": "01b34e5961ed4506baf18d12832334b5", "competition_source_list": ["2017年AMC10竞赛A第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "2017 AMC10A P13 Define a sequence recursively by $$F_0 =0$$, $$F_1=1$$, and $$F_n=$$the remainder when $$F_{n-1}+ F_{n-2}$$ is divided by $$3$$, for all $$n≥ 2$$. Thus the sequence starts $$0$$, $$1$$, $$1$$, $$2$$, $$0$$, $$2$$, $$\\cdots$$ What is $$F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}$$? 用$F_0=0$$，$$F1=1$$，以及Fn=F_{n-1}+ F_{n-2}$除以$3$$时的余数来递归地定义一个数列(n≥2)。$$F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}$$?=?", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}], [{"aoVal": "E", "content": "$$10$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Reasoning->Recurrence and Recursion"], "answer_analysis": ["A patten starts to emerge as the function is continued. The repeating patten is $$0$$, $$1$$, $$1$$, $$2$$, $$0$$, $$2$$, $$2$$, $$1$$ $$\\ldots$$. The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is $$(\\rm D) 9$$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "761", "queId": "918a0493e73141ddaf89a4c1ff225d38", "competition_source_list": ["2012年黑龙江全国高中数学联赛竞赛初赛第6题5分", "2016~2017学年6月陕西西安莲湖区西安市第一中学高二下学期月考理科第9题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$10$$个相同的小球装入$$3$$个编号为$$1$$，$$2$$，$$3$$的盒子（每次要把$$10$$个球装完），要求每个盒子里球的个数不少于盒子的编号数，这样的装法种数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->知识点->计数原理->排列与组合->组合->隔板法"], "answer_analysis": ["根据题意， 先在编号为$$2$$、$$3$$的三个盒子中分别放入$$1$$、$$2$$个小球，编号为$$1$$的盒子里不���， 再将剩下的$$7$$个小球放入$$3$$个盒子里，每个盒子里至少一个， 分析可得，$$7$$个小球排好，有$$6$$个空位，在$$6$$个空位中任选$$2$$个， 插入挡板，共$$\\text{C}_{6}^{2}=15$$种装法， 即可得符合题目要求的装法共有$$15$$种， 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "981", "queId": "d28894d0a8ad4482903fc63fd9dac71e", "competition_source_list": ["2018年陕西全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知四面体$$ABCD$$内接于球$$O$$，且$$AD$$是球$$O$$的直径．若$$\\triangle ABC$$和$$\\triangle BCD$$都是边长为$$1$$的等边三角形，则四面体$$ABCD$$的体积是( )", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{2}}{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{2}}{12}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{3}}{6}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{3}}{12}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量"], "answer_analysis": ["解：四面体以等腰直角$$\\triangle ABD$$为底，高为$$\\frac{\\sqrt{2}}{2}$$，故体积为$$\\frac{1}{3}\\times \\frac{\\sqrt{2}}{2}\\times \\frac{1}{2}=\\frac{\\sqrt{2}}{12}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "793", "queId": "7bdfb56f83684b2b88623835af87eae1", "competition_source_list": ["2015年黑龙江全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设变量$$x,y$$满足约束条件$$\\begin{cases}2x-y+2\\geqslant 0 8x-y-4\\leqslant 0 x\\geqslant 0,y\\geqslant 0 \\end{cases}$$，目标函数$$z=abx+y(a\\textgreater0,b\\textgreater0)$$的最大值是$$8$$，则$$a+b$$的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->线性规划", "竞赛->知识点->不等式->几个重要的不等式->均值"], "answer_analysis": ["$$ab+4=8$$，$$ab=4$$，所以$$a+b\\geqslant 2\\sqrt{ab}=4$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "845", "queId": "f6b9b9784d824da7b3a44e65a701f3fa", "competition_source_list": ["2011年浙江全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知$$\\theta \\in [\\frac{5 \\pi }{4},\\frac{3 \\pi }{2}]$$，则$$\\sqrt{1-\\sin 2\\theta }-\\sqrt{1+\\sin 2\\theta }$$可化简为．", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\sin \\theta $$ "}], [{"aoVal": "B", "content": "$$-2\\sin \\theta $$ "}], [{"aoVal": "C", "content": "$$-2\\cos \\theta $$ "}], [{"aoVal": "D", "content": "$$2\\cos \\theta $$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["因为$$\\theta \\in [\\frac{5 \\pi }{4},\\frac{3 \\pi }{2}]$$，所以$$\\sqrt{1-\\sin 2\\theta }-\\sqrt{1+\\sin 2\\theta }$$ $$=\\left\\textbar{} \\cos \\theta -\\sin \\theta \\right\\textbar-\\left\\textbar{} \\cos \\theta +\\sin \\theta \\right\\textbar=2\\cos \\theta $$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "731", "queId": "4ee2f148dce44d2faf880d2a8e6da5c1", "competition_source_list": ["2018年吉林全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f(x)=\\frac{{{({{2}^{x}}+1)}^{2}}}{{{2}^{x}}x}+1$$在区间$$[-2018,0)\\cup (0,2018]$$上的最大值为$$M$$，最小值为$$N$$．则$$M+N=$$ ．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["因为 $$y=\\frac{{{({{2}^{x}}+1)}^{2}}}{{{2}^{x}}\\cdot x}=\\left( {{2}^{x}}+\\frac{1}{{{2}^{x}}}+2 \\right)\\cdot \\frac{1}{x}$$ 为$$[-\\infty ,0)\\cup (0,+\\infty ]$$上的奇函 数，故 $$f(x)=\\frac{{{({{2}^{x}}+1)}^{2}}}{{{2}^{x}}\\cdot x}+1$$ 的图象关于点$$(0,1)$$对称，所以$$M+N=2$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "715", "queId": "b1560421fcb048699931465c4d69eff3", "competition_source_list": ["2012年辽宁全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "设双曲线$$\\frac{{{x}^{2}}}{{{a}^{2}}}-\\frac{{{y}^{2}}}{{{b}^{2}}}=1\\left( a\\textgreater0, b\\textgreater0 \\right)$$的右焦点为$$F$$，过点$$F$$作���$$x$$轴垂直的直线$$l$$交两条渐近线于$$A, B$$两点，$$P$$是$$l$$与双曲线的一个交点．设$$O$$为坐标原点，若有实数$$m, n$$使得$$\\overrightarrow{OP}=m\\overrightarrow{OA}+n\\overrightarrow{OB}$$，且$$mn=\\frac{2}{9}$$，则该双曲线的离心率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3\\sqrt{2}}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{9}{8}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3\\sqrt{5}}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3\\sqrt{2}}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算", "竞赛->知识点->解析几何->双曲线"], "answer_analysis": ["将$$A\\left( c,\\frac{bc}{a} \\right), B\\left( c, -\\frac{bc}{a} \\right)$$代入$$\\overrightarrow{OP}=m\\overrightarrow{OA}+n\\overrightarrow{OB}$$， 得$$P\\left( \\left( m+n \\right)c,\\left( m-n \\right)\\frac{bc}{a} \\right)$$，代入双曲线方程，得 $$\\frac{{{\\left( m+n \\right)}^{2}}{{c}^{2}}}{{{a}^{2}}}-\\frac{{{\\left( m-n \\right)}^{2}}{{b}^{2}}{{c}^{2}}}{{{a}^{2}}{{b}^{2}}}=1$$， 化简得$$4mn{{c}^{2}}={{a}^{2}}$$， $$e=\\frac{c}{a}=\\frac{1}{\\sqrt{4mn}}=\\sqrt{\\frac{9}{8}}=\\frac{3\\sqrt{2}}{4}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "372", "queId": "b04c901e1ad746ab93d6ff9fe99bba5d", "competition_source_list": ["1991年全国高中数学联赛竞赛一试第4题"], "difficulty": "0", "qtype": "single_choice", "problem": "设函数$$y=f\\left( x \\right)$$对一切实数$$x$$都满足$$f\\left( 3+x \\right)=f\\left( 3-x \\right)$$，且方程$$f\\left( x \\right)=0$$恰有$$6$$个不同的实根，则这$$6$$个实根的和为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["若$$3+\\alpha $$是$$f(x)=0$$的一个根，则由已知 $$f\\left( 3-\\alpha \\right)=f\\left( 3+\\alpha \\right)=0$$ 即$$3-\\alpha $$也是一个根．因此可设方程$$f\\left( x \\right)=0$$的六个根为 $$3\\mathsf{\\pm }{{\\alpha }_{1}}$$，$$3\\mathsf{\\pm }{{\\alpha }_{2}}$$，$$3\\mathsf{\\pm }{{\\alpha }_{3}}$$． 于是它们的行等于$$18$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "474", "queId": "557fb421bef045228d15712fb322df4b", "competition_source_list": ["2013年四川全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "当平面上的点$$\\left( x,y \\right)$$的坐标$$x,y$$都为有理数时，该点称为有理点，设$$r$$是给定的正实数，则圆$${{\\left( x-1 \\right)}^{2}}+{{\\left( y-\\sqrt{2} \\right)}^{2}}={{r}^{2}}$$上的有理点．", "answer_option_list": [[{"aoVal": "A", "content": "最多有一个 "}], [{"aoVal": "B", "content": "最多有两个 "}], [{"aoVal": "C", "content": "最多有四个 "}], [{"aoVal": "D", "content": "可以有无穷多个 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->圆与方程", "竞赛->知识点->逻辑->逻辑推理"], "answer_analysis": ["设$$\\left( a,b \\right),\\left( c,d \\right)$$为这个圆上的两个有理点，则 $${{\\left( a-1 \\right)}^{2}}+{{\\left( b-\\sqrt{2} \\right)}^{2}}={{\\left( c-1 \\right)}^{2}}+{{\\left( d-\\sqrt{2} \\right)}^{2}}$$， 整理得$$2\\sqrt{2}\\left( b-d \\right)={{a}^{2}}+{{b}^{2}}-{{c}^{2}}-{{d}^{2}}+2\\left( c-a \\right)$$. 显然等式右边是有理数，因此当且仅当$$b=d$$时，等式的左边是有理数，此时$$a\\ne c,a+c=2$$.由此得这两个有理点关于直线$$x=1$$对称，而圆上与$$\\left( a,b \\right)$$关于$$x=1$$对称的点最多只有一个. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "741", "queId": "c3cfa74083b44602acb07dd0565f07d7", "competition_source_list": ["高一上学期单元测试《二次与对勾函数》竞赛第7题"], "difficulty": "0", "qtype": "single_choice", "problem": "若全集为$$X$$，$$A$$是$$X$$的一个真子集，定义$$A$$的特征函数：$${{f}_{A}}(x)=\\begin{cases}1，x\\in A; 0，x\\in {{C}_{X}}A. \\end{cases}$$ 则对$$X$$的两个真子集$$A，B$$，下列命题中不准确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$A\\subseteq B\\Leftrightarrow {{f}_{A}}(x)\\leqslant {{f}_{B}}(x)，x\\in X$$ "}], [{"aoVal": "B", "content": "$${{f}_{{{C}_{X}}A}}(x)=1-{{f}_{A}}(x)，x\\in X$$ "}], [{"aoVal": "C", "content": "$${{f}_{A\\cap B}}(x)={{f}_{A}}(x){{f}_{B}}(x)，x\\in X$$ "}], [{"aoVal": "D", "content": "$${{f}_{A\\cup B}}(x)={{f}_{A}}(x)+{{f}_{B}}(x)，x\\in X$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->二次函数", "竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["若$$x\\in A\\cap B$$，则$${{f}_{A\\cup B}}(x)={{f}_{A}}(x)={{f}_{B}}(x)=1$$，故$\\text{D}$不正确． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "266", "queId": "3d85381944e341ae85c01b3ae749a66e", "competition_source_list": ["2012年吉林全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "若一系列函数的解析式相同，值域相同，但其定义域不同，则称这些函数为``同族函数''，那么函数解析式为$$y=-{{x}^{2}}$$，值域为$$\\left { 0, -1, -9 \\right }$$的``同族函数''共有", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$个 "}], [{"aoVal": "B", "content": "$$8$$个 "}], [{"aoVal": "C", "content": "$$9$$个 "}], [{"aoVal": "D", "content": "$$10$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的概念"], "answer_analysis": ["$$1\\times 3\\times 3=9$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1178", "queId": "eb650d9a34764b56987885118c6a1889", "competition_source_list": ["2009年浙江全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "若等差数列$$ {{{a}_{n}} }$$满足$$3{{a}_{8}}=5{{a}_{13}}$$，且$${{a}_{1}}\\textgreater0$$，则前$$n$$项之和$${{S}_{n}}$$的最大值是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{S}_{10}}$$ "}], [{"aoVal": "B", "content": "$${{S}_{11}}$$ "}], [{"aoVal": "C", "content": "$${{S}_{20}}$$ "}], [{"aoVal": "D", "content": "$${{S}_{21}}$$ "}]], "knowledge_point_routes": ["知识标签->素养->数学运算", "知识标签->题型->数列->等差数列->等差数列的性质问题->求等差数列前n项和的最值 ", "知识标签->知识点->数列->等差数列->等差数列的前n项和", "知识标签->知识点->数列->等差数列->等差数列的概念通项公式", "知识标签->知识点->数列->等差数列->等差数列的性质及应用"], "answer_analysis": ["由$$3{{a}_{8}}=5{{a}_{13}}$$可得$${{a}_{20}}\\textgreater0$$，$${{a}_{21}} ~\\textless{} ~0$$，∴$${{S}_{20}}$$最大． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1162", "queId": "c26080e1ceec485e801f614906f0e3e3", "competition_source_list": ["2014年AMC12竞赛A第21题"], "difficulty": "3", "qtype": "single_choice", "problem": "2014AMC12A, 21 For every real number $$x$$, let $$\\left\\lfloor {x} \\right\\rfloor$$ denote the greatest integer not exceeding $$x$$, and let $$f(x)= \\left\\lfloor {x} \\right\\rfloor(2014^{x-\\left\\lfloor {x} \\right\\rfloor}- 1)$$. The set of all numbers $$x$$ such that $$1\\leqslant x\\textless{} 2014$$ and $$f(x)\\leqslant1$$ is a union of disjoint intervals. What is the sum of the lengths of those intervals? 对于任意实数$$x$$，令$$\\left\\lfloor {x} \\right\\rfloor$$ 表示不超过$$x$$ 的最大整数，令$$f(x)= \\left\\lfloor {x} \\right\\rfloor(2014^{x-\\left\\lfloor {x} \\right\\rfloor}- 1)$$. 满足 $$1\\leqslant x\\textless{} 2014$$ 和 $$f(x)\\leqslant1$$ 的所有 $$x$$ 的集合是一些不相交区间的并集。 这些区间的长度之和是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\frac{ \\log 2015}{ \\log 2014}$$ "}], [{"aoVal": "C", "content": "$$\\frac{ \\log 2014}{ \\log 2013}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2014}{2013}$$ "}], [{"aoVal": "E", "content": "$$2014^{\\frac{1}{2014}}$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Number Theory->Gaussian Function", "课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算"], "answer_analysis": ["记$\\left { x\\right }=x-\\left\\lfloor x\\right\\rfloor\\textgreater0$. 则$f(x)\\leq 1\\Leftrightarrow 2014^{\\left {x\\right }}-1\\leq \\frac{1}{\\left\\lfloor x\\right\\rfloor}$ $\\Leftrightarrow \\left {x\\right }\\leq\\log_{2014}{\\frac{\\left\\lfloor x\\right\\rfloor+1}{\\left\\lfloor x\\right\\rfloor}}$ 于是单个满足$f(x)\\leq 1$的区间长度为$\\log_{2014}{\\frac{\\left\\lfloor x\\right\\rfloor+1}{\\left\\lfloor x\\right\\rfloor}}$ 区间总长度$=\\sum_{x=1}^{2013}\\log_{2014}{\\frac{\\left\\lfloor x\\right\\rfloor+1}{\\left\\lfloor x\\right\\rfloor}}=\\log_{2014}{\\frac{2}{1}\\times\\frac{3}{2}\\times\\cdots\\times\\frac{2014}{2013}}=\\log_{2014}{2014}=1$, 选$$\\text{A}$$. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "479", "queId": "d0e2d6c360c94509804d7d14711b5f86", "competition_source_list": ["2012年天津全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "数列$$\\left { {{a}_{n}} \\right }$$的前$$n$$项和$${{S}_{n}}={{n}^{2}}-2n$$，则$${{a}_{3}}+{{a}_{17}}$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$35$$ "}], [{"aoVal": "C", "content": "$$34$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["$${{a}_{n}}=\\begin{cases}{{S}_{1}},n=1 {{S}_{n}}-{{S}_{n-1}},n\\geqslant 2 \\end{cases}$$当$$n\\textgreater1$$时，$${{a}_{n}}={{S}_{\\text{n}}}-{{S}_{\\text{n-1}}}=2\\text{n}-3$$，因此$${{a}_{3}}+{{a}_{17}}=34$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "451", "queId": "318779969b3044fda4483e5c2cabd7a1", "competition_source_list": ["2019年吉林全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "若复数$$z$$满足$$\\left\\textbar{} z \\right\\textbar{} ~\\textless{} ~1$$且$$\\left\\textbar{} \\overline{z}+\\frac{1}{z} \\right\\textbar=\\frac{5}{2}$$，则$$\\left\\textbar{} z \\right\\textbar=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{4}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["注意到， $$\\left\\textbar{} \\overline{z}+\\frac{1}{z} \\right\\textbar=\\left\\textbar{} \\frac{\\overline{z}z+1}{z} \\right\\textbar=\\frac{{{\\left\\textbar{} z \\right\\textbar}^{2}}+1}{\\left\\textbar{} z \\right\\textbar}=\\frac{5}{2}$$， 解得$$\\left\\textbar{} z \\right\\textbar=2$$ （舍去）或$$\\left\\textbar{} z \\right\\textbar=\\frac{1}{2}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "151", "queId": "0ca6c46dd5f248feb1aec0a7f97b56e1", "competition_source_list": ["2022年江苏徐州贾汪区江苏省贾汪中学高一竞赛（下学期春季）第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$\\omega \\textgreater0$，函数$f\\left( x \\right)=\\text{sin}\\left( \\omega x+\\frac{\\text{ }\\pi\\text{ }}{\\text{6}} \\right)$在$\\left( \\frac{\\text{ }\\pi\\text{ }}{\\text{2}},\\text{ }\\pi\\text{ } \\right)$上单调递减，则$\\omega $的取值范围是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\left[ \\frac{2}{3},\\frac{4}{3} \\right]$ "}], [{"aoVal": "B", "content": "$\\left[ \\frac{2}{3},\\frac{3}{4} \\right]$ "}], [{"aoVal": "C", "content": "$\\left( 0,\\frac{2}{3} \\right]$ "}], [{"aoVal": "D", "content": "$\\left( 0,\\frac{3}{2} \\right]$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 由题意可得$t=\\omega x+\\frac{\\text{ }\\pi\\text{ }}{\\text{6}}\\in (\\frac{\\text{ }\\pi\\text{ }\\omega }{2}+\\frac{\\text{ }\\pi\\text{ }}{\\text{6}},\\text{ }\\pi\\text{ }\\omega +\\frac{\\text{ }\\pi\\text{ }}{\\text{6}})$，再根据$y=\\sin t$的单调区间，列出不等式组求解即可.\\\\ 【详解】\\\\ 解：因为$\\omega \\textgreater0$，$x\\in \\left( \\frac{\\text{ }\\pi\\text{ }}{\\text{2}}\\text{, }\\pi\\text{ } \\right)$，\\\\ 所以$t=\\omega x+\\frac{\\text{ }\\pi\\text{ }}{\\text{6}}\\in (\\frac{\\text{ }\\pi\\text{ }\\omega }{2}+\\frac{\\text{ }\\pi\\text{ }}{\\text{6}},\\text{ }\\pi\\text{ }\\omega +\\frac{\\text{ }\\pi\\text{ }}{\\text{6}})$，\\\\ 又因为$y=\\sin t$在$(\\frac{\\text{ }\\pi\\text{ }}{2}+2k\\text{ }\\pi\\text{ },\\frac{3\\text{ }\\pi\\text{ }}{2}+2k\\text{ }\\pi\\text{ }),k\\in \\text{Z}$上单调递减，\\\\ 所以$\\text{ } {\\text{ }\\begin{array}{*{35}{l}} \\frac{\\text{ }\\pi\\text{ }\\omega }{2}+\\frac{\\text{ }\\pi\\text{ }}{\\text{6}}\\ge \\frac{\\text{ }\\pi\\text{ }}{2}+2k\\text{ }\\pi\\text{ } \\text{ }\\pi\\text{ }\\omega +\\frac{\\text{ }\\pi\\text{ }}{\\text{6}}\\le \\frac{3\\text{ }\\pi\\text{ }}{2}+2k\\text{ }\\pi\\text{ } \\end{array}\\text{ }$，$k\\in \\text{Z}$,\\\\ 解得：$\\frac{2}{3}+2k\\text{ }\\pi\\text{ }\\le \\omega \\le \\frac{4}{3}+2k\\text{ }\\pi\\text{ },k\\in \\text{Z}$.\\\\ 当$k=0$时，$\\frac{2}{3}\\le \\omega \\le \\frac{4}{3}$.\\\\ 故选：A "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "300", "queId": "34ce6bf66e244404aec9e0fc3d49b151", "competition_source_list": ["2017年AMC10竞赛B第14题"], "difficulty": "2", "qtype": "single_choice", "problem": "2017 AMC10B P14 An integer $$N$$ is selected at random in the range $$1\\leqslant N\\leqslant2020$$. What is the probability that the remainder when $$N^{16}$$ is divided by $$5$$ is $$1$$? 在$1\\leqslant N\\leqslant2020$的范围内随机选择一个整数$N$。当$N^{16}$除以$5$时，余数是$1$的概率是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac 15$$ "}], [{"aoVal": "B", "content": "$$\\frac 25$$ "}], [{"aoVal": "C", "content": "$$\\frac 35$$ "}], [{"aoVal": "D", "content": "$$\\frac 45$$ "}], [{"aoVal": "E", "content": "$$1$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Geometric Models of Probabilities"], "answer_analysis": ["By Fermat\\textquotesingle s Little Theorem, $$N^{16}=\\left( N^{4}\\right)^{4}\\equiv1$$(mod $$5$$), when $$N$$ is relatively prime to $$5$$. Hence, this happens with probability $$\\frac 45$$. Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits $$0-9$$. The pattern for $$0$$ is $$0$$, no matter what power, so $$0$$ doesn\\textquotesingle t work. Likewise, the pattern for $$5$$ is always $$5$$. Doing the same for the rest of the digits, we find that the units digits of $$1^{16}$$,$$2^{16}$$,$$3^{16}$$,$$4^{16}$$,$$6^{16}$$,$$7^{16}$$,$$8^{16}$$ and $$9^{16}$$ all have the remainder of $$1$$ when divided by $$5$$, so $$\\frac 45$$. We can use modular arithmetic for each residue of $$n$$ (mod $$5$$). If $$n\\equiv0$$ (mod $$5$$), then $$n^{16}\\equiv0^{16}\\equiv0$$(mod $$5$$), If $$n\\equiv1$$ (mod $$5$$), then $$n^{16}\\equiv1^{16}\\equiv0$$(mod $$5$$), If $$n\\equiv2$$ (mod $$5$$), then $$n^{16}\\equiv\\left( n^{2}\\right)^{8}\\equiv\\left( 2^{2}\\right)^{8}\\equiv4^{8}\\equiv\\left( -1\\right)^{8}\\equiv1$$ (mod $$5$$). If $$n\\equiv3$$ (mod $$5$$), then $$n^{16}\\equiv\\left( n^{2}\\right)^{8}\\equiv\\left( 3^{2}\\right)^{8}\\equiv9^{8}\\equiv\\left( -1\\right)^{8}\\equiv1$$ (mod $$5$$). If $$n\\equiv4$$ (mod $$5$$), then $$n^{16}\\equiv4^{16}\\equiv\\left( -1\\right)^{16}=1$$ (mod $$5$$). In $$4$$ out of the $$5$$ cases, the result was $$1$$(mod $$5$$), and since each case occurs equally as $$2020\\equiv0$$ (mod $$5$$), the answer is $$\\frac 45$$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "796", "queId": "cd1e7fd4f5ff4202ba88b1fad192677a", "competition_source_list": ["2008年黑龙江全国高中数学联赛竞赛初赛第12题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "实数$$x$$、$$y$$满足$$1+{{\\cos }^{2}}(2x+3y-1)=\\frac{{{x}^{2}}+{{y}^{2}}+2(x+1)(1-y)}{x-y+1}$$，则$$xy$$的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{25}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->均值"], "answer_analysis": ["因为$$1\\leqslant 1+{{\\cos }^{2}}(2x+3y-1)\\leqslant 2$$， 且$$\\frac{{{x}^{2}}+{{y}^{2}}+2(x+1)(1-y)}{x-y+1}=\\frac{1}{x-y+1}+x-y+1\\geqslant 2$$， 所以$$\\begin{cases}x-y+1=1 2x+3y-1=k \\pi ~\\end{cases}$$， 即$$x=y=\\frac{k \\pi +1}{5}$$， 其中$$k\\in Z$$．所以$$xy$$的最小值为$${{\\left( \\frac{1}{5} \\right)}^{2}}=\\frac{1}{25}$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "936", "queId": "973b41e541894f069ceac6902d75f76b", "competition_source_list": ["2009年第二十届全国希望杯高二竞赛复赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "椭圆$$\\frac{{{x}^{2}}}{{{a}^{2}}}+\\frac{{{y}^{2}}}{{{b}^{2}}}=1(a\\textgreater b\\textgreater0)$$中长度为整数的焦点弦（过焦点的弦）称为``好弦''．则当$$a=6,b=3$$时，该椭圆中所有好弦的长度之和是．", "answer_option_list": [[{"aoVal": "A", "content": "$$132$$ "}], [{"aoVal": "B", "content": "$$138$$ "}], [{"aoVal": "C", "content": "$$252$$ "}], [{"aoVal": "D", "content": "$$258$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->椭圆", "竞赛->知识点->解析几何->直线与圆锥曲线"], "answer_analysis": ["过两个焦点的弦为长轴，长度是$$12$$．仅过左焦点的好弦中，长度最长为$$11$$，最短为$$3$$（垂直于长轴的焦点弦），且长度为$$45\\cdots 11$$的弦均有$$2$$条，故只过左焦点的好弦长度之和为$$3+2\\times (4+5+\\cdots +11)=123$$，因此所有好弦的长度之和是$$123\\times 2+12=258$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "79", "queId": "4153e3ba172a4147b7fe1228392d6e5e", "competition_source_list": ["2008年天津全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "若方程$${{a}^{x}}+2x-4=0\\left( a\\textgreater0, a\\ne 1 \\right)$$的所有根为$${{u}_{1}},{{u}_{2}},\\cdots ,{{u}_{k}}$$，其中$$k$$为正整数，方程$${{\\log }_{a}}2x+x-2=0\\left( a\\textgreater0, a\\ne 1 \\right)$$的所有根为$${{v}_{1}}， {{v}_{2}}，\\cdots ，{{v}_{l}}$$，其中$$l$$为正整数，则 $$\\frac{{{u}_{1}}+{{u}_{2}}+\\cdots +{{u}_{k}}+{{v}_{1}}+{{v}_{2}}+\\cdots +{{v}_{l}}}{k+l}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "2 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["方程$${{a}^{x}}+2x-4=0$$等价于$$\\frac{{{a}^{x}}}{2}=2-x$$， 其根即为$$y=\\frac{{{a}^{x}}}{2}$$与$$y=2-x$$的交点的横坐标． $${{\\log }_{a}}^{2x}+x-2=0$$等价于$${{\\log }_{a}}^{2x}=2-x$$， 其根即为$$y={{\\log }_{a}}^{2x}$$与$$y=2-x$$的交点的横坐标． 因为$$y=\\frac{{{a}^{x}}}{2}$$与$$y={{\\log }_{a}}^{2x}$$互为反函数， 所以此它们的图像关于$$y=x$$对称， 因此所有根的算术平均就是$$y=x$$与$$y=2-x$$交点的横坐标$$1$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "571", "queId": "cc7a84dfc90b45b686f8e1f797092c05", "competition_source_list": ["2010年河南全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$5$$名同学分配到$$A$$，$$B$$，$$C$$三个宿舍中，每个宿舍至少安排$$1$$名，其中甲同学不能分配到$$A$$宿舍，则不同的分配方案种数是(~ ~ )", "answer_option_list": [[{"aoVal": "A", "content": "$$76$$种 "}], [{"aoVal": "B", "content": "$$100$$种 "}], [{"aoVal": "C", "content": "$$132$$种 "}], [{"aoVal": "D", "content": "$$150$$种 "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->计数原理->排列与组合->组合->分组分配法", "课内体系->知识点->计数原理->排列与组合->排列组合综合"], "answer_analysis": ["分两步：第一步，分配甲，有${\\rm{A}}_{\\rm{2}}^{\\rm{1}}$种方法，设 甲被分配到$$B$$宿舍； 第二步，分配其余$$4$$名同学，分两类：第一类，其余$$4$$名 同学都不分配到$$B$$宿舍，有${\\rm{C}}_{\\rm{4}}^{\\rm{3}}{\\rm{A}}_{\\rm{2}}^{\\rm{2}} + {\\rm{C}}_{\\rm{4}}^{\\rm{1}}{\\rm{C}}_{\\rm{2}}^{\\rm{2}}$种方法； 第二类，其余$$4$$名同学有人分配到$$B$$宿舍，有${\\rm{C}}_{\\rm{4}}^{\\rm{2}}{\\rm{A}}_{\\rm{3}}^{\\rm{3}}$种 方法．故分配其余$$4$$名同学有${\\rm{C}}_{\\rm{3}}^{\\rm{2}}{\\rm{A}}_{\\rm{2}}^{\\rm{2}} + {\\rm{C}}_{\\rm{4}}^{\\rm{2}}{\\rm{C}}_{\\rm{2}}^{\\rm{2}} + {\\rm{C}}_{\\rm{4}}^{\\rm{2}}{\\rm{C}}_{\\rm{3}}^{\\rm{3}}$种方法 所以不同的分配方案种数是$${N = {\\rm{A}}_{\\rm{2}}^{\\rm{1}}{\\rm{(C}}_{\\rm{4}}^{\\rm{3}}{\\rm{A}}_{\\rm{2}}^{\\rm{2}} + {\\rm{C}}_{\\rm{4}}^{\\rm{2}}{\\rm{C}}_{\\rm{2}}^{\\rm{2}} + {\\rm{C}}_{\\rm{4}}^{\\rm{2}}{\\rm{A}}_{\\rm{3}}^{\\rm{3}}{\\rm{)}} = 100}$$．故选$${\\rm{B}}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "988", "queId": "a53fae6840f64f51af60e072f4e958d2", "competition_source_list": ["2009年浙江全国高中数学联赛竞赛初赛第3题5分", "2017~2018学年6月河北石家庄辛集市河北辛集中学高一下学期月考第14题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "若等差数列$$ {{{a}_{n}} }$$满足$$3{{a}_{8}}=5{{a}_{13}}$$，且$${{a}_{1}}\\textgreater0$$，则前$$n$$项之和$${{S}_{n}}$$的最大值是（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{S}_{10}}$$ "}], [{"aoVal": "B", "content": "$${{S}_{11}}$$ "}], [{"aoVal": "C", "content": "$${{S}_{20}}$$ "}], [{"aoVal": "D", "content": "$${{S}_{21}}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->知识点->数列->等差数列->等差数列的前n项和->求等差数列前n项和的最值 "], "answer_analysis": ["由$$3{{a}_{8}}=5{{a}_{13}}$$可得$${{a}_{20}}\\textgreater0$$，$${{a}_{21}} ~\\textless{} ~0$$，∴$${{S}_{20}}$$最大． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "65", "queId": "6fadea55c09a488fba3169c6152e74d8", "competition_source_list": ["2023年江苏连云港灌南县灌南县中学高三竞赛（下学期3月解题能力竞赛）第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知数列$\\left {a_{n}\\right }$的前$n$项和为$S_{n}$，且满足$a_{1}=1$，$a_{n}a_{n+1}=2S_{n}$，设$b_{n}=\\frac{a_{n}}{3^{n}}$，若存在正整数$p,q\\left(p\\textless{} q\\right)$，使得$b_{1}$，$b_{p}$，$b_{q}$成等差数列，则（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$p=1$ "}], [{"aoVal": "B", "content": "$p=2$ "}], [{"aoVal": "C", "content": "$p=3$ "}], [{"aoVal": "D", "content": "$p=4$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据数列的递推公式得出$b_{n}=\\frac{a_{n}}{3^{n}}=\\frac{n}{3^{n}}$，然后根据等差数列的性质进项求解即可得出结果.\\\\ 【详解】\\\\ 数列$\\left {a_{n}\\right }$满足$a_{1}=1$，$a_{n}a_{n+1}=2S_{n}$，\\\\ 当$n=1$时，$a_{1}a_{2}=2S_{1}=2a_{1}$��解得：$a_{2}=2$；\\\\ 当$n\\geq 2$时，$2a_{n}=2(S_{n}-S_{n-1})=a_{n}(a_{n+1}-a_{n-1})$，\\\\ 因为$a_{n}\\neq 0$，所以$a_{n+1}-a_{n-1}=2$，所以数列$\\left {a_{n}\\right }$是首项为1，公差为1的等差数列，\\\\ 所以$a_{n}=1+(n-1)=n$，$b_{n}=\\frac{a_{n}}{3^{n}}=\\frac{n}{3^{n}}$，\\\\ 若存在正整数$p,q\\left(p\\textless{} q\\right)$，使得$b_{1}$，$b_{p}$，$b_{q}$成等差数列，\\\\ 则$2b_{p}=b_{1}+b_{q}$，所以$\\frac{2p}{3^{p}}=\\frac{1}{3}+\\frac{q}{3^{q}}$~~~~~~~~①\\\\ 因为数列$ {b_{n} }$是单调递减数列，\\\\ 当$p=1$时，由$\\frac{2}{3}=\\frac{1}{3}+\\frac{q}{3^{q}}$，解得：$q=1$，舍去；\\\\ 当$2\\leq p\\textless{} q$时，则$\\frac{1}{3}\\geq \\frac{p-1}{3^{p-1}}$，$\\frac{p-1}{3^{p-1}}-\\frac{2p}{3^{p}}=\\frac{p-3}{3^{p}}$；\\\\ 当$3\\leq p$时，$\\frac{1}{3}\\geq \\frac{p-1}{3^{p-1}}\\geq \\frac{2p}{3^{p}}$，$\\frac{q}{3^{q}}\\textgreater{} 0$，所以$\\frac{2p}{3^{p}}\\textless{} \\frac{1}{3}+\\frac{q}{3^{q}}$，①式不成立，\\\\ 所以$p=2$，则有$\\frac{4}{9}=\\frac{1}{3}+\\frac{q}{3^{q}}$，解得：$q=3$，\\\\ 故选：$\\mathrm{B}$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "882", "queId": "d69aeba2e9c14513b17f8a0b0eca5ffa", "competition_source_list": ["2017~2018学年山东青岛崂山区青岛第二中学高一上学期期末第7题5分", "2008年吉林全国高中数学联赛竞赛初赛第1题6分", "2017~2018学年12月山西太原迎泽区太原市第五中学高一上学期月考第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "为了得到函数$$y=\\sin \\left( 2x-\\frac{\\pi }{6} \\right)$$的图象，可以将函数$$y=\\cos 2x$$的图象（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "向右平移$$\\frac{\\pi }{6}$$个单位 "}], [{"aoVal": "B", "content": "向右平移$$\\frac{\\pi }{3}$$个单位 "}], [{"aoVal": "C", "content": "向左平移$$\\frac{\\pi }{6}$$个单位 "}], [{"aoVal": "D", "content": "向左平移$$\\frac{\\pi }{3}$$个单位 "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->诱导公式", "课内体系->知识点->三角函数->三角函数的图象与性质->正余弦型、正切型函数图象变换", "课内体系->素养->逻辑推理"], "answer_analysis": ["$$y=\\sin \\left( 2x-\\frac{\\pi }{6} \\right)=\\cos \\left( \\frac{2\\pi }{3}-2x \\right)=\\cos \\left( 2x-\\frac{2\\pi }{3} \\right)=\\cos \\left( 2\\left( x-\\frac{\\pi }{3} \\right) \\right)$$． 故选择$$B$$选项． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "817", "queId": "5cd0a7a8862d494789620386a2fa9c0e", "competition_source_list": ["2011年浙江全国高中数学联赛竞赛初赛第2题5分", "2018年广东深圳龙岗区深圳科学高中高三四模文科第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果复数$$\\left( a+2\\text{i} \\right)\\left( 1+\\text{i}\\right)$$的模为$$4$$，则实数$$a$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$2\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$\\pm 2$$ "}], [{"aoVal": "D", "content": "$$\\pm 2\\sqrt{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["由题意得$$\\sqrt{2}\\cdot \\sqrt{{{a}^{2}}+4}=4\\Rightarrow a=\\pm 2$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "620", "queId": "99f71f635cb44c7db444ed9f4763531c", "competition_source_list": ["全国全国高中数学联赛竞赛一试"], "difficulty": "1", "qtype": "single_choice", "problem": "已知原点在椭圆$${{k}^{2}}{{x}^{2}}+{{y}^{2}}-4kx+2ky+{{k}^{2}}-1=0$$的内部，那么参数$$k$$的取值范围是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\textbar k\\textbar\\textgreater1$$ "}], [{"aoVal": "B", "content": "$$\\textbar k\\textbar\\ne 1$$ "}], [{"aoVal": "C", "content": "$$-1\\textless{}k\\textless{}1$$ "}], [{"aoVal": "D", "content": "$$0\\textless{}\\textbar k\\textbar\\textless{}1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->圆锥曲线->椭圆->椭圆的定义、标准方程->椭圆的定义", "课内体系->素养->数学抽象"], "answer_analysis": ["椭圆外部的点可以离原点很远， 它的坐标$$x$$、$$y$$的绝对值可以很大， 使得方程左边大于$$0$$， 所以内部点的坐标使方程左边小于$$0$$， 用原点的坐标代入方程左边得$${{k}^{2}}-1\\textless{}0$$ 又$$k\\ne 0$$，（否则方程不表示椭圆）， 所以$$0\\textless{}\\textbar k\\textbar\\textless{}1$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "165", "queId": "262993ef53d94b4a86231fc1193b19f3", "competition_source_list": ["2020~2021学年河南郑州中原区郑州市第十九中学高二上学期期中理科第11题5分", "2020~2021学年河南郑州郑东新区郑州市第四十七中学高二上学期期中理科第11题5分", "2020~2021学年河南郑州高新区郑州中学高二上学期期中理科第11题5分", "2020~2021学年河南郑州二七区郑州市第二中学高二上学期期中理科第11题5分", "2009年辽宁全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\triangle ABC$$的三边$$a,b,c$$成等比数列，$$a,b,c$$所对的角依次为$$A,B,C$$．则$$\\sin B+\\cos B$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 1,1+\\frac{\\sqrt{3}}{2} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left[ \\frac{1}{2},1+\\frac{\\sqrt{3}}{2} \\right]$$ "}], [{"aoVal": "C", "content": "$$(1,\\sqrt{2}]$$ "}], [{"aoVal": "D", "content": "$$\\left[ \\frac{1}{2},\\sqrt{2} \\right]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列", "竞赛->知识点->三角函数->三角形中的问题->解三角形", "竞赛->知识点->三角函数->三角函数的图像与性质", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["$$ac={{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\\cos B\\geqslant 2ac-2ac\\cos B\\Rightarrow \\cos B\\geqslant \\frac{1}{2}\\Rightarrow 0\\textless{}B\\leqslant \\frac{ \\pi }{3}$$， 于是$$\\sin B+\\cos B=\\sqrt{2}\\sin \\left( B+\\frac{ \\pi }{4} \\right)\\in (1,\\sqrt{2}]$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "193", "queId": "2656611fc3934281ba60097e41ef38e6", "competition_source_list": ["2012年黑龙江全国高中数学联赛竞赛初赛第11题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$A, B$$为抛物线$$C:{{y}^{2}}=4x$$上的不同两点，$$F$$为抛物线$$C$$的焦点，若$$\\overrightarrow{FA}=-4\\overrightarrow{FB}$$，则直线$$AB$$的斜率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\pm \\frac{2}{3}$$ "}], [{"aoVal": "B", "content": "$$\\pm \\frac{3}{2}$$ "}], [{"aoVal": "C", "content": "$$\\pm \\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$\\pm \\frac{4}{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆锥曲线"], "answer_analysis": ["设$$A\\left( {{x}_{1}}, {{y}_{1}} \\right), B\\left( {{x}_{2}}, {{y}_{2}} \\right)$$，由$$\\overrightarrow{FA}=-4\\overrightarrow{FB}$$，得 $${{y}_{1}}=-4{{y}_{2}}$$．~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ① 设$$AB$$直线为$$y=k\\left( x-1 \\right)$$，与抛物线联立得$$k{{y}^{2}}-4y-4k=0$$．则 $${{y}_{1}}+{{y}_{2}}=4/k$$，~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ② $${{y}_{1}}\\cdot {{y}_{2}}=-4$$.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ③ 由①②③联立得出$$k$$值． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "241", "queId": "7df13990c75b4af1a8b513bdb6f91bd4", "competition_source_list": ["2011年吉林全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$${{a}_{1}}, {{a}_{2}},\\cdots , {{a}_{2011}}$$是一列互不相等的正整数，如果任意改变这$$2011$$个数的顺序，把它们记为$${{b}_{1}}, {{b}_{2}},\\cdots , {{b}_{2011}}$$，则数$$M=({{a}_{1}}-{{b}_{1}})({{a}_{2}}-{{b}_{2}})({{a}_{3}}-{{b}_{3}})\\cdots ({{a}_{2011}}-{{b}_{2011}})$$的值．", "answer_option_list": [[{"aoVal": "A", "content": "必为$$0$$ "}], [{"aoVal": "B", "content": "必为$$1$$ "}], [{"aoVal": "C", "content": "是奇数 "}], [{"aoVal": "D", "content": "是偶数 "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的综合应用"], "answer_analysis": ["假设$$M$$是奇数，则$${{a}_{i}}-{{b}_{i}}(i=1, 2,\\cdots , 2011)$$必定都是奇数，从而这$$2011$$个奇数的和也是奇数；另一方面，其和 $$({{a}_{1}}-{{b}_{1}})+({{a}_{2}}-{{b}_{2}})+({{a}_{3}}-{{b}_{3}})+\\cdots +({{a}_{2011}}-{{b}_{2011}})=({{a}_{1}}+{{a}_{2}}+{{a}_{3}}+\\cdots +{{a}_{2011}})-({{b}_{1}}+{{b}_{2}}+{{b}_{3}}+\\cdots +{{b}_{2011}})=0$$为偶数，矛盾，所以假设不成立，即$$M$$是偶数． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "838", "queId": "d1e00f3ea2b746dfa05540d934e0283b", "competition_source_list": ["1986年全国高中数学联赛竞赛一试第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "平面上有一个点集$$M$$和七个不同的圆$${{C}_{1}}$$，$${{C}_{2}}$$，\\ldots$${{C}_{7}}$$，其中圆$${{C}_{7}}$$恰好经过$$M$$中的$$7$$个点，圆$${{C}_{6}}$$恰好经过$$M$$中的$$6$$个点，\\ldots，圆$${{C}_{1}}$$恰好经过$$M$$中的$$1$$个点，那么$$M$$中的点数最少为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->图论（二试）"], "answer_analysis": ["要使$$M$$中的点的个数最少，各圆要尽可能多地共同经过$$M$$中的点．因为不相同的两圆最多只有两个公共点，而$${{C}_{7}}$$经过$$M$$中的$$7$$个点，所以$${{C}_{6}}$$至少要经过$$M$$中的另外四个点，而$${{C}_{5}}$$除了和$${{C}_{7}}$$、$${{C}_{6}}$$各有两个公共点外，还要经过$$M$$中的一个点，则$$M$$中至少要有$$12$$个点． 另一方面，如果有$$12$$个点，就能满足要求： 先作两两相交的三个圆，记为$${{C}_{7}}$$、$${{C}_{6}}$$、$${{C}_{5}}$$．两作圆$${{C}_{4}}$$与两圆$${{C}_{6}}$$、$${{C}_{7}}$$都相交于另外的点（但与$${{C}_{5}}$$不相交），又过$${{C}_{5}}$$与$${{C}_{7}}$$的一个交点作圆$${{C}_{3}}$$，使与$${{C}_{5}}$$及$${{C}_{7}}$$各交一个新的点，将所有的交点都算作$$M$$中的点，此时$$M$$中的点为$$12$$个．$${{C}_{7}}$$与$${{C}_{6}}$$、$${{C}_{5}}$$、$${{C}_{4}}$$各有两个交点，又与$${{C}_{3}}$$有一个新交点，共$$7$$个点；$${{C}_{6}}$$与$${{C}_{7}}$$、$${{C}_{5}}$$、$${{C}_{4}}$$各有两个交点，共$$6$$个点；$${{C}_{5}}$$与$${{C}_{7}}$$、$${{C}_{6}}$$各有两个交点，且与$${{C}_{3}}$$有一个新交点，共$$5$$个点；$${{C}_{4}}$$与$${{C}_{7}}$$、$${{C}_{6}}$$各有两个交点，共$$4$$个点；$${{C}_{3}}$$是过$${{C}_{7}}$$与$${{C}_{5}}$$的一个交点，且与其各交得一个新点，共有$$3$$个点．最后在$$M$$的$$12$$个点中，选取两个点、一个点分别作满足要求的圆$${{C}_{2}}$$、$${{C}_{1}}$$是容易的事． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "959", "queId": "8655b2d1b92340d2912a61f594977a64", "competition_source_list": ["2018年上海杨浦区高三二模第16题5分", "2017年辽宁全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知长方体的表面积为$$\\frac{45}{2}$$，棱长的总和为$$24$$，则长方体的体对角线与棱所成角的最大值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\arccos \\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\arccos \\frac{\\sqrt{2}}{3}$$ "}], [{"aoVal": "C", "content": "$$\\arccos \\frac{\\sqrt{3}}{9}$$ "}], [{"aoVal": "D", "content": "$$\\arccos \\frac{\\sqrt{6}}{9}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->立体几何初步->基本立体图形->空间几何体的概念->棱柱、棱锥、棱台的结构特征", "课内体系->知识点->立体几何初步->基本图形位置关系->空间中的基本事实与定理->点、直线、平面之间的位置关系", "课内体系->知识点->立体几何初步->基本图形位置关系->探索性问题->几何法求空间角"], "answer_analysis": ["设三条棱$$a\\leqslant b\\leqslant c$$， ∴$$ab+ac+bc=\\frac{45}{4}$$，$$a+b+c=6$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=\\frac{27}{2}$$， $${{a}^{2}}+{{b}^{2}}+{{c}^{2}}\\geqslant {{a}^{2}}+2bc={{a}^{2}}+2\\left[ \\frac{45}{4}-a\\left( 6-a \\right) \\right]$$， 整理得$${{a}^{2}}-4a+3\\leqslant 0$$， ∴$$1\\leqslant a\\leqslant 3$$， ∴最短棱长为$$1$$，体对角线长为$$\\frac{3\\sqrt{6}}{2}$$，$$\\cos \\theta =\\frac{2}{3\\sqrt{6}}=\\frac{\\sqrt{6}}{9}$$， 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "6", "queId": "009b64a94ce64a9a83f0476619b785c0", "competition_source_list": ["2014年湖南全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "设$$M=\\left { a\\left\\textbar{} a={{x}^{2}}-{{y}^{2}}, x, y\\in Z \\right. \\right }$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$9\\in M$$，$$10\\in M$$ "}], [{"aoVal": "B", "content": "$$9\\notin M$$，$$10\\in M$$ "}], [{"aoVal": "C", "content": "$$9\\in M$$，$$10\\notin M$$ "}], [{"aoVal": "D", "content": "$$9\\notin M$$，$$10\\notin M$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["因为$$9={{3}^{2}}-{{0}^{2}}$$，所以$$9\\in M$$；假设$$10\\in M$$，则存在整数$$m$$、$$n$$使得 $$10={{m}^{2}}-{{n}^{2}}=\\left( m+n \\right)\\left( m-n \\right)$$， 则因为将$$10$$分解为两个整数因子之积，必定一个因子为奇数，另一个因子为偶数，而$$\\left( m+n \\right)$$与$$\\left( m-n \\right)$$同奇同偶，矛盾，所以$$10\\notin M$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "812", "queId": "ad20b593d2784f21976829fbea08396b", "competition_source_list": ["1990��全国高中数学联赛竞赛一试第2题", "2006年上海复旦大学自主招生千分考第24题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$f\\left( x \\right)$$是定义在实数集上的周期为$$2$$的周期函数，且是偶函数．已知当$$x\\in \\left[ 2,3 \\right]$$时，$$f\\left( x \\right)=x$$，则当$$x\\in \\left[ -2,0 \\right]$$时，$$f\\left( x \\right)$$的解析式为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$x+4$$ "}], [{"aoVal": "B", "content": "$$2-x$$ "}], [{"aoVal": "C", "content": "$$3-\\left\\textbar{} x+1 \\right\\textbar$$ "}], [{"aoVal": "D", "content": "$$2+\\left\\textbar{} x+1 \\right\\textbar$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["当$$x\\in \\left[ -2,-1 \\right]$$时，$$f\\left( x \\right)=f\\left( x+4 \\right)=x+4$$， 当$$x\\in \\left[ -1,0 \\right]$$时，$$f\\left( x \\right)=f\\left( -x \\right)=f\\left( 2-x \\right)=2-x$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "848", "queId": "6ee483faf2eb40f0be777a295f5d6b17", "competition_source_list": ["2019~2020学年10月广东深圳南山区深圳市南山区育才中学高三上学期月考理科第10题5分", "2017~2018学年广东深圳龙岗区深圳科学高中高二上学期期中第10题5分", "2013年黑龙江全国高中数学联赛竞赛初赛第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知各项均为正数的等比数列$$\\left { {{a}_{n}} \\right }$$满足$${{a}_{7}}={{a}_{6}}+2{{a}_{5}}$$，若存在两项$${{a}_{m}}$$，$${{a}_{n}}$$使得$$\\sqrt{{{a}_{m}}{{a}_{n}}}=4{{a}_{1}}$$，则$$\\frac{1}{m}+\\frac{4}{n}$$的最小值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{9}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{25}{6}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->数列->等比数列->等比数列的概念与通项公式", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的实际应用", "课内体系->知识点->等式与不等式->不等式->基本不等式->利用基本不等式求最值", "课内体系->知识点->基本初等函数->实数指数幂运算", "课内体系->知识点->基本初等函数->指数函数->指数函数的图象及性质"], "answer_analysis": ["由各项均为正数的等比数列$$\\left { {{a}_{n}} \\right }$$满足$${{a}_{7}}={{a}_{6}}+2{{a}_{5}}$$， 可得$${{a}_{1}}{{q}^{6}}={{a}_{1}}{{q}^{5}}+2{{a}_{1}}{{q}^{4}}$$， 所以$${{q}^{2}}-q-2=0$$，解得$$q=2$$或$$q=-1$$（舍去）． 因为$$\\sqrt{{{a}_{m}}{{a}_{n}}}=4{{a}_{1}}$$，所以$${{q}^{m+n-2}}=16$$， 所以$${{2}^{m+n-2}}={{2}^{4}}$$，所以$$m+n=6$$． 所以$$\\frac{1}{m}+\\frac{4}{n}=\\frac{1}{6}\\left( m+n \\right)\\left( \\frac{1}{m}+\\frac{4}{n} \\right)$$ $$=\\frac{1}{6}\\left( 5+\\frac{n}{m}+\\frac{4m}{n} \\right)$$ $$\\geqslant \\frac{1}{6}\\left( 5+2\\sqrt{\\frac{n}{m}\\cdot \\frac{4m}{n}} \\right)$$ $$=\\frac{3}{2}$$． 当且仅当$$\\frac{n}{m}=\\frac{4m}{n}$$，即$$n=2m$$时，等号成立， 又$$m+n=6$$，所以$$m=2$$，$$n=4$$，符合题意． 故最小值为$$\\frac{3}{2}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "391", "queId": "20049da57d90488985f0e8cb0e1f7be4", "competition_source_list": ["2022~2023学年湖南永州宁远县高一上学期月考（明德湘南中学基础知识竞赛）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列函数中，既是奇函数又是增函数的为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$y=-\\frac{1}{x}$ "}], [{"aoVal": "B", "content": "$y=-{{x}^{3}}$ "}], [{"aoVal": "C", "content": "$y=x+1$ "}], [{"aoVal": "D", "content": "$y=x\\left\\textbar{} x \\right\\textbar$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的性质->奇偶性->函数奇偶性的应用"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据基本函数的奇偶性、单调性逐项判断即可.\\\\ 【详解】\\\\ 对于A：$y=-\\frac{1}{x}$为反比例函数，为奇函数，在区间$\\left( -\\infty ,0 \\right)$以及$\\left( 0,+\\infty \\right)$上都是增函数，但在定义域内不是增函数，故A错误；\\\\ 对于B：$y=-{{x}^{3}}$为幂函数，既是奇函数又是减函数，不符合题意，故B错误；\\\\ 对于C：$y=x+1$为一次函数，不是奇函数，不符合题意，故C错误；\\\\ 对于D： $y=x\\left\\textbar{} x \\right\\textbar=\\left { \\begin{array}{*{35}{l}} {{x}^{2}}, \\& x\\ge 0 -{{x}^{2}}, \\& x\\textless{} 0 \\end{array} \\right.$为奇函数；$x\\ge 0$时， $y={{x}^{2}}$为增函数，$x\\textless{} 0$时，$y=-{{x}^{2}}$ 为增函数，且${{x}^{2}}\\ge -{{x}^{2}}$该函数在\\emph{R}上为增函数，故D正确；\\\\ 故选： D "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "841", "queId": "8544895800b342b3aa351396e0e8b1aa", "competition_source_list": ["2009年高二竞赛广州市第1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知复数$$z$$满足$$\\frac{\\text{i}}{z-1}=3$$，则复数$$z$$的实部与虚部之和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$3+\\text{i}$$ "}], [{"aoVal": "B", "content": "$$1+\\frac{1}{3}\\text{i}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->复数->复数的概念及几何意义->复数的基本概念", "课内体系->知识点->复数->复数的运算->复数的四则运算综合", "课内体系->知识点->复数->复数的运算->复数的乘法和除法", "课内体系->素养->数学运算"], "answer_analysis": ["$$z=1+\\frac{\\text{i}}{3}$$，故实部与虚部之和为$$1+\\frac{1}{3}=\\frac{4}{3}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "85", "queId": "86f3a127f4a448669348cd5ad9cea3b0", "competition_source_list": ["2017~2018学年广东广州白云区艺术中学高二下学期期中文科第6题5分", "2010年黑龙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知复数$${{z}_{1}}=m+2\\text{i}$$，$${{z}_{2}}=3-4\\text{i}$$．若$$\\frac{{{z}_{1}}}{{{z}_{2}}}$$为实数，则实数$$m$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{8}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "C", "content": "$$-\\frac{8}{3}$$ "}], [{"aoVal": "D", "content": "$$-\\frac{3}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->复数->复数的运算->复数的乘法和除法", "课内体系->知识点->复数->复数的概念及几何意义->复数的基本概念->实部与虚部", "课内体系->素养->数学运算"], "answer_analysis": ["$$\\frac{{{z}_{1}}}{{{z}_{2}}}=\\frac{m+2\\text{i}}{3-4\\text{i}}=\\frac{(m+2\\text{i})(3+4\\text{i})}{(3-4\\text{i})(3+4\\text{i})}=\\frac{3m-8+(6+4m)\\text{i}}{25}$$， ∵$$\\frac{{{z}_{1}}}{{{z}_{2}}}$$为实数， ∴$$6+4m=0$$即$$m=-\\frac{3}{2}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "360", "queId": "1ba5ddc32454496780dd7d197239992d", "competition_source_list": ["2012年四川全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "设集合$$S=\\left { x\\textbar{{x}^{2}}-5x-6\\textless{}0 \\right }, T=\\left { x\\textbar\\left\\textbar{} x+2 \\right\\textbar\\leqslant 3 \\right }$$，则$$S\\cap T=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left { x\\textbar-5\\leqslant x\\textless{}-1 \\right }$$ "}], [{"aoVal": "B", "content": "$$\\left { x\\textbar-5\\leqslant x\\textless{}5 \\right }$$ "}], [{"aoVal": "C", "content": "$$\\left { x\\textbar-1\\textless{}x\\leqslant 1 \\right }$$ "}], [{"aoVal": "D", "content": "$$\\left { x\\textbar1\\leqslant x\\textless{}5 \\right }$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["$$S=\\left { x\\textbar-1\\textless{}x\\textless{}6 \\right }, T=\\left { x\\textbar-5\\leqslant x\\leqslant 1 \\right }$$，故$$S\\cap T=\\left { x\\textbar-1\\textless{}x\\leqslant 1 \\right }$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "99", "queId": "0f5b19518eec4cddbccc7f4b210e90cf", "competition_source_list": ["2020年江苏全国高中数学联赛竞赛初赛第9题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知正整数$$m+n$$均为质数，且$$7m+n$$和$$mn+11$$也都是质数，则$$m^{n}+n^{m}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$37$$ "}], [{"aoVal": "C", "content": "$$57$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质", "竞赛->知识点->数论模块->整除->质数（算数基本定理）"], "answer_analysis": ["无 "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "7", "queId": "00a18e29b03b437b8a05631dbfeba28d", "competition_source_list": ["2011年吉林全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在正五棱柱$$ABCDE-{{A}_{1}}{{B}_{1}}{{C}_{1}}{{D}_{1}}{{E}_{1}}$$的侧棱$$C{{C}_{1}}$$上有一点$$P$$，若截面$$PA{{E}_{1}}$$与侧面$$AE{{E}_{1}}{{A}_{1}}$$互相垂直，则这样的$$P$$点．", "answer_option_list": [[{"aoVal": "A", "content": "有且仅有一个 "}], [{"aoVal": "B", "content": "有时有两个，有时有一个 "}], [{"aoVal": "C", "content": "恰有两个 "}], [{"aoVal": "D", "content": "有时不存在 "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的平行和垂直"], "answer_analysis": ["在面$$ABCDE$$内过$$A$$作$$AE$$的垂线交$$DC$$延长线于$$F$$，在面$${{A}_{1}}{{B}_{1}}{{C}_{1}}{{D}_{1}}{{E}_{1}}$$内过$${{E}_{1}}$$作$${{E}_{1}}{{A}_{1}}$$的垂线交$${{B}_{1}}{{C}_{1}}$$的延长线与点$${{F}_{1}}$$，则面$$AF{{F}_{1}}{{E}_{1}}$$与$$C{{C}_{1}}$$的交点为$$P$$，由正五边形的图形特征知只有一个交点，即$$P$$恰有一个． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "573", "queId": "6cceff81d9d448eeb46ddd69f4a62957", "competition_source_list": ["2004年全国高中数学联赛竞赛一试第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "不等式$$\\sqrt{{{\\log }_{2}}x-1}+\\frac{1}{2}{{\\log }_{\\frac{1}{2}}}{{x}^{3}}+2\\textgreater0$$的解集为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$[2,3)$$ "}], [{"aoVal": "B", "content": "$$(2,3]$$ "}], [{"aoVal": "C", "content": "$$[2,4)$$ "}], [{"aoVal": "D", "content": "$$(2,4]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["原不等式等价于$$\\begin{cases}\\sqrt{{{\\log }_{2}}x-1}-\\dfrac{3}{2}{{\\log }_{2}}x+\\dfrac{3}{2}+\\dfrac{1}{2}\\textgreater0 {{\\log }_{2}}x-1\\geqslant 0 \\end{cases}$$， 设$$\\sqrt{{{\\log }_{2}}x-1}=t$$，则有$$\\begin{cases}t-\\dfrac{3}{2}{{t}^{2}}+\\dfrac{1}{2}\\textgreater0 t\\geqslant 0 \\end{cases}$$，解得$$0\\leqslant t\\textless{}1$$， 即$$0\\leqslant {{\\log }_{2}}x-1\\textless{}1$$，$$2\\leqslant x\\textless{}4$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "362", "queId": "59777b331bc04ef7921dd20aab46477a", "competition_source_list": ["1998年全国高中数学联赛竞赛一试第1题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$a\\textgreater1$$，$$b\\textgreater{} 1$$，且$$\\text{lg}\\left( a+b \\right)=\\text{lg}a+\\text{lg}b$$，则$$\\lg \\left( a-1 \\right)+\\lg \\left( b-1 \\right)$$的值（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "等于$$\\text{lg2}$$ "}], [{"aoVal": "B", "content": "等于$$1$$ "}], [{"aoVal": "C", "content": "等于$$0$$ "}], [{"aoVal": "D", "content": "不是与$$a$$，$$b$$无关的常数 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["$$a+b=ab$$，$$\\left( a-1 \\right)\\left( b-1 \\right)=1$$，由$$a-1\\textgreater0$$，$$b-1\\textgreater0$$，故$$\\text{lg}\\left( a-1 \\right)\\left( b-1 \\right)=0$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "239", "queId": "343bd5af4a554a61a6be66e2288d73f2", "competition_source_list": ["2014年浙江全国高中数学联赛竞赛初赛第10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f\\left( x \\right)=\\cos \\left( a\\sin x \\right)-\\sin \\left( b\\cos x \\right)$$无零点，则$${{a}^{2}}+{{b}^{2}}$$的取值范围为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ 0,\\frac{ \\pi }{4} \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left[ 0,\\frac{{{ \\pi }^{2}}}{2} \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left[ 0,\\frac{{{ \\pi }^{2}}}{4} \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left[ 0,\\frac{ \\pi }{2} \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["由已知得$2k \\pi +b\\cos x=\\frac{ \\pi }{2}-a\\sin x$，$\\left( 2k+1 \\right) \\pi -b\\cos x=\\frac{ \\pi }{2}-a\\sin x\\left( k\\in \\mathbf{Z} \\right)$， 无解，所以方程$\\sqrt{{{a}^{2}}+{{b}^{2}}}\\sin \\left( x+\\varphi~ \\right)=\\frac{ \\pi }{2}+2k \\pi $，或 $\\sqrt{{{a}^{2}}+{{b}^{2}}}$$\\sqrt{{{a}^{2}}+{{b}^{2}}}\\sin \\left( x-\\varphi~ \\right)=\\frac{ \\pi }{2}-\\left( 2k+1 \\right) \\pi $ 其中$\\sin \\varphi =\\frac{b}{\\sqrt{{{a}^{2}}+{{b}^{2}}}}$，$\\cos \\varphi =\\frac{a}{\\sqrt{{{a}^{2}}+{{b}^{2}}}}$，无解． 所以$\\sqrt{{{a}^{2}}+{{b}^{2}}}\\textless\\left\\textbar{} \\frac{ \\pi }{2}+2k \\pi \\right\\textbar$或$\\left( \\left\\textbar{} \\frac{ \\pi }{2}-\\left( 2k+1 \\right) \\pi \\right\\textbar{} \\right)\\Rightarrow {{a}^{2}}+{{b}^{2}}$取值范围为$\\left[ 0 \\frac{{{ \\pi }^{2}}}{4} \\right)$． $$.$$ "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "894", "queId": "6f4e8d82d69a4ed9a34d2835df17b541", "competition_source_list": ["2022~2023学年12月安徽滁州定远县��徽省定远县民族中学高一上学期月考第5题", "2022~2023学年湖南邵阳隆回县湖南省隆回县第二中学高一上学期月考（竞赛）第3题", "2021~2022学年江苏南京鼓楼区金陵中学高一上学期期中第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "若正实数\\emph{a}，\\emph{b}满足lg \\emph{a}＋lg \\emph{b}＝1，则$\\frac{2}{a}+\\frac{5}{b}$的最小值为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\sqrt{2}$ "}], [{"aoVal": "B", "content": "2$\\sqrt{2}$ "}], [{"aoVal": "C", "content": "$\\frac{\\sqrt{10}}{2}$ "}], [{"aoVal": "D", "content": "2 "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->基本不等式"], "answer_analysis": ["\\hfill\\break 应用对数运算得到$ab=10$，由目标式结合基本不等式有$\\frac{2}{a}+\\frac{5}{b}\\ge 2\\sqrt{\\frac{2}{a}\\cdot \\frac{5}{b}}$即可求其最小值.\\\\ 【详解】\\\\ ∵$\\lg a+\\lg b=1$，即$\\lg ab=1$，\\\\ ∴$ab=10$，而$a\\textgreater0,b\\textgreater0$，\\\\ ∴$\\frac{2}{a}+\\frac{5}{b}\\ge 2\\sqrt{\\frac{2}{a}\\cdot \\frac{5}{b}}=2$当且仅当$a=2,b=5$时等号成立.\\\\ ∴$\\frac{2}{a}+\\frac{5}{b}$的最小值为2.\\\\ 故选：D\\\\ 【点睛】\\\\ 易错点睛：利用基本不等式求最值时，须满足的三个条件：\\\\ （1）``一正二定三相等''\\,``一正''就是各项必须为正数；\\\\ （2）``二定''就是要求和的最小值，必须把构成和的二项之积转化成定值；要求积的最大值，则必须把构成积的因式的和转化成定值；\\\\ （3）``三相等''是利用基本不等式求最值时，必须验证等号成立的条件，若不能取等号则这个定值就不是所求的最值，这也是最容易发生错误的地方 "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "691", "queId": "c39d2cd0c11a4186b6536af0a50da4ea", "competition_source_list": ["2008年安徽全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$\\frac{1}{2008}$$写成十进制循环小数的形式 $$\\frac{1}{2008}=0.000\\underbrace{498\\ldots 625}_{{}}\\underbrace{498\\ldots 625}_{{}}\\ldots $$其循环节的长度为．", "answer_option_list": [[{"aoVal": "A", "content": "30 "}], [{"aoVal": "B", "content": "40 "}], [{"aoVal": "C", "content": "50 "}], [{"aoVal": "D", "content": "60 "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->同余->几个重要的定理（欧拉定理、费尔马小定理、威尔逊定理）"], "answer_analysis": ["$$\\frac{1}{2008}=\\frac{1}{8\\cdot 251}=\\frac{1}{1000}\\cdot \\frac{125}{251}$$，则$$\\frac{125}{251}=0.\\underbrace{498\\ldots 625}_{{}}\\underbrace{498\\ldots 625}_{{}}\\ldots $$． 设循环节长度为$$k$$，则 $$\\frac{125}{251}\\cdot {{10}^{k}}=\\underbrace{498\\ldots 625}_{{}}+0.\\underbrace{498\\ldots 625}_{{}}\\underbrace{498\\ldots 625}_{{}}\\ldots =\\underbrace{498\\cdots 625}_{{}}+\\frac{125}{251}=n+\\frac{125}{251}$$，$$n\\in {{\\mathbf{N}}^{+}}$$． 于是，$$251n=125\\left( {{10}^{k}}-1 \\right)$$，则$$251\\left\\textbar{} {{10}^{k}}-1 \\right.$$． 由费马小定理，$$251\\left\\textbar{} {{10}^{250}}-1 \\right.$$，则$$k\\left\\textbar{} 250 \\right.$$，只能选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "820", "queId": "6ea26e05bea64408a26781c9f9a5ec14", "competition_source_list": ["2021~2022学年安徽阜阳太和县安徽省太和中学高一下学期月考（竞赛考试）第3~3题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知向量$$\\vec{a}=(1，m)，\\vec{b}=(3，-2)，$$且$$(\\vec{a}-\\vec{b})\\bot \\vec{b}$$，则\\emph{m}＝（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "－5 "}], [{"aoVal": "B", "content": "－3 "}], [{"aoVal": "C", "content": "3 "}], [{"aoVal": "D", "content": "5 "}]], "knowledge_point_routes": ["课内体系->知识点->平面向量->平面向量基本定理及其坐标表示"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据向量的线性坐标运算求得向量$$\\vec{a}-\\vec{b}$$，再由向量垂直的坐标表示建立方程，求解即可.\\\\ 【详解】\\\\ 解：由题意得$$\\vec{a}-\\vec{b}=(-2，m\\text{+}2)$$.\\\\ 又$$(\\vec{a}-\\vec{b})\\bot \\vec{b}$$，∴$$(\\vec{a}-\\vec{b})\\cdot \\vec{b}=-6-2(m\\text{+}2)=0$$ ，解得\\emph{m}＝－5.\\\\ 故选：A. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "884", "queId": "bb1d540199df462fb431ddf72fd8e1ab", "competition_source_list": ["2015年辽宁全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f\\left( x \\right)={{x}^{2}}-53x+196+\\left\\textbar{} {{x}^{2}}-53x+196 \\right\\textbar$$，则$$f\\left( 1 \\right)+f\\left( 2 \\right)+\\cdots +f\\left( 50 \\right)=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$660$$ "}], [{"aoVal": "B", "content": "$$664$$ "}], [{"aoVal": "C", "content": "$$668$$ "}], [{"aoVal": "D", "content": "$$672$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->二次函数", "竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["$${{x}^{2}}-53x+196=\\left( x-4 \\right)\\left( x-49 \\right)$$， 当$$x\\textless{}4$$或$$x\\textgreater49$$时，$${{x}^{2}}-53x+196\\textgreater0$$，$$f\\left( x \\right)=2\\left( {{x}^{2}}-53x+196 \\right)$$； 当$$4\\leqslant x\\leqslant 49$$时，$${{x}^{2}}-53x+196\\leqslant 0$$，$$f\\left( x \\right)=0$$． $$f\\left( 1 \\right)+f\\left( 2 \\right)+f\\left( 3 \\right)+f\\left( 50 \\right)=330$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "260", "queId": "9449e3230a1140d9b8f7fd7349689f57", "competition_source_list": ["2008年安徽全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "在平面中，到两条相交直线的距离之和为$$1$$的点的轨迹为．", "answer_option_list": [[{"aoVal": "A", "content": "椭圆 "}], [{"aoVal": "B", "content": "双曲线的一部分 "}], [{"aoVal": "C", "content": "抛物线的一部分 "}], [{"aoVal": "D", "content": "矩形 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->曲线与方程"], "answer_analysis": ["以两条定直线所成的角的两条角平分线所在的直线分别为$$x,y$$轴建立平面直角坐标系，设两条直线方程为$$y=kx$$和$$y=-kx$$，$$k\\textgreater0$$． 点$$\\left( x,y \\right)$$到$$y=\\pm kx$$的距离和为$$1$$，则$$\\frac{\\left\\textbar{} y-kx \\right\\textbar+\\left\\textbar{} y+kx \\right\\textbar}{\\sqrt{{{k}^{2}}+1}}=1$$． 轨迹由$$4$$条线段组成，构成一个矩形． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "82", "queId": "45f8ad4ca7334162b98048b60b01086d", "competition_source_list": ["2015年天津全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$$\\left\\textbar{} y \\right\\textbar=1+\\sqrt{2x-{{x}^{2}}}$$表示的曲线是．", "answer_option_list": [[{"aoVal": "A", "content": "一个圆 "}], [{"aoVal": "B", "content": "两个半圆 "}], [{"aoVal": "C", "content": "一个椭圆 "}], [{"aoVal": "D", "content": "以上结论都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->曲线与方程"], "answer_analysis": ["$$y\\textgreater0$$时，$${{\\left( y-1 \\right)}^{2}}+{{\\left( x-1 \\right)}^{2}}=1$$，$$y\\geqslant 1$$，表示一个半圆； $$y\\textless{}0$$时，$${{\\left( y+1 \\right)}^{2}}+{{\\left( x-1 \\right)}^{2}}=1$$，$$y\\leqslant -1$$，表示一个半圆． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "370", "queId": "fa714f1db370492cb4703e3830c0f5fe", "competition_source_list": ["2019~2020学年江西新余高二上学期期末理科第6题5分", "2009年高考真题湖北卷理科第5题5分", "2018~2019学年云南昆明高二下学期期中理科昆明三中、滇池中学第9题5分", "2013年黑龙江全国高中数学联赛竞赛初赛第5题5分", "2016~2017学年12月广东广州天河区天河中学高三上学期月考理科第6题5分", "2017~2018学年8月四川成都龙泉驿区成都龙泉一中高三上学期月考文科第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "将甲、乙、丙、丁四名学生分到三个不同的班，每个班至少分到一名学生，且甲、乙两名学生不能分到一个班，则不同分法的种数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->排列与组合->排列", "课内体系->知识点->计数原理->排列与组合->组合->分组分配法", "课内体系->素养->逻辑推理"], "answer_analysis": ["如丙、丁分到同一个班级，则方法数就是三个元素的一个全排列，即$$\\text{A}_{3}^{3}$$；若丙分到甲或乙所在的班级，则丁只能独自一个班级，方法数是$$2\\text{A}_{3}^{3}$$；同理，若丁分到甲或乙所在的班级，则丙独自一个班级，方法数是$$2\\text{A}_{3}^{3}$$．根据分类加法计数原理，得总的方法数是$$5\\text{A}_{3}^{3}=30$$． ", "总的方法数是$$\\text{C}_{4}^{2}\\text{A}_{3}^{3}=36$$，甲、乙被分到同一个班级的方法数是$$\\text{A}_{3}^{3}=6$$，故甲、乙不分到同一个班级的方法数是$$36-6=30$$．
"], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "638", "queId": "3c69cb1d68e1493787702094161b38b5", "competition_source_list": ["1993年全国全国高中数学联赛竞赛一试第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "集合$$A$$，$$B$$的并集$$A\\cup B= {{{a}_{1}},{{a}_{2}},{{a}_{3}} }$$，当$$A\\ne B$$时，$$(A,B)$$与$$(B,A)$$视为不同的对，则这样的$$(A,B)$$对的个数有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$26$$ "}], [{"aoVal": "D", "content": "$$27$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["已知$$A\\cup B= {{{a}_{1}},{{a}_{2}},{{a}_{3}} }$$，则作为其子的$$A$$，$$B$$最多只有$$3$$元素． （$$1$$）若$$A= {{{a}_{1}},{{a}_{2}},{{a}_{3}} }$$则满足题意得$$B$$可以使空集，或者是单元素的集合，或是二元素的集合，或是三元素的集合，这样的$$B$$有$$\\text{C}_{3}^{0}+\\text{C}_{3}^{1}+\\text{C}_{3}^{2}+\\text{C}_{3}^{3}={{2}^{3}}$$个，这时$$(A,B)$$有$$\\text{C}_{3}^{3}\\cdot {{2}^{3}}$$对． （$$2$$）若$$A$$为二元素的集合，则有$$\\text{C}_{3}^{2}$$种，其对应的$$B$$的$${{2}^{3}}$$个（$$\\text{C}_{2}^{0}+\\text{C}_{2}^{1}+\\text{C}_{2}^{2}={{2}^{2}}$$），这时$$(A,B)$$有$$C_{3}^{2}\\cdot {{2}^{2}}$$时． （$$3$$）若$$A$$为单元素的集合，则有$$\\text{C}_{3}^{1}$$种，其对应的$$B$$有$$2$$个，这时$$(A,B)\\text{C}_{3}^{1}\\cdot 2$$对． （$$4$$）若$$A$$是空集，则有$$\\text{C}_{3}^{0}$$种，其对应的$$B$$有一个，这时$$(A,B)$$有$$\\text{C}_{3}^{0}\\cdot 1$$对． ∴ 这样的$$(A,B)$$共有 $$\\text{C}_{3}^{3}\\cdot {{2}^{3}}+\\text{C}_{3}^{2}\\cdot {{2}^{2}}+\\text{C}_{3}^{1}\\cdot 2+\\text{C}_{3}^{0}\\cdot {{2}^{0}}={{3}^{3}}=27$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "346", "queId": "1f7f8f47e86b468e95efdc9691ffff5d", "competition_source_list": ["2022~2023学年广东深圳南山区深圳实验学校高二上学期段考（三）第8题", "2010年四川全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$${{A}_{1}}, {{A}_{2}}$$为椭圆$$\\frac{{{x}^{2}}}{{{a}^{2}}}+\\frac{{{y}^{2}}}{{{b}^{2}}}=1(a\\textgreater b\\textgreater0)$$的左、右顶点，若在椭圆上存在异于$${{A}_{1}}, {{A}_{2}}$$的点$$P$$，使得$${PO}\\bot {P{{A}_{2}}}$$，其中$$O$$为坐标原点，则椭圆的离心率$$e$$的取值范围是", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 0,\\frac{1}{2} \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 0,\\frac{\\sqrt{2}}{2} \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{2}, 1 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{\\sqrt{2}}{2}, 1 \\right)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->圆锥曲线->椭圆->椭圆的简单几何性质"], "answer_analysis": ["由题设知$$\\angle OP{{A}_{2}}=90{}^{}\\circ $$，设$$P(x, y)(x\\textgreater0)$$， 以$$O{{A}_{2}}$$为直径的圆方程为$${{\\left( x-\\frac{a}{2} \\right)}^{2}}+{{y}^{2}}=\\frac{{{a}^{2}}}{4}$$， 与椭圆方程联立得$$\\left( 1-\\frac{{{b}^{2}}}{{{a}^{2}}} \\right){{x}^{2}}-ax+{{b}^{2}}=0$$． 由题设知，要求此方程在$$(0, a)$$上有实根． 由此得$$0\\textless{}\\frac{a}{2\\left( 1-\\frac{{{b}^{2}}}{{{a}^{2}}} \\right)}\\textless{}a$$化简得$${{e}^{2}}\\textgreater\\frac{1}{2}$$， 所以$$e$$的取值范围为$$\\left( \\frac{\\sqrt{2}}{2}, 1 \\right)$$．故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "87", "queId": "0b412861836747eebba1833491fed159", "competition_source_list": ["2012年河北全国高中数学联赛高二竞赛初赛第7题8分", "2015~2016学年天津和平区天津市耀华中学高二上学期期中文科第18题4分", "2017~2018学年天津和平区天津市耀华中学高二上学期期中理科第14题4分", "2017~2018学年天津和平区天津市耀华中学高二上学期期中文科第14题4分", "2017~2018学年天津和平区天津市耀华中学高一下学期期中第14题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在三棱锥$A-BCD$中，侧棱$AB$、$AC$、$AD$两两垂直，$\\vartriangle ABC$、$\\vartriangle ACD$、$\\vartriangle ADB$的面积分别为$\\frac{\\sqrt{2}}{2}$、$\\frac{\\sqrt{3}}{2}$、$\\frac{\\sqrt{6}}{2}$，则三棱锥$A-BCD$的外接球的体积为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$\\sqrt{2}\\pi $ "}], [{"aoVal": "B", "content": "$\\sqrt{3}\\pi $ "}], [{"aoVal": "C", "content": "$\\sqrt{6}\\pi $ "}], [{"aoVal": "D", "content": "$3\\pi $ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["在三棱锥$$A-BCD$$中，侧棱$$AB$$、$$AC$$、$$AD$$两两垂直， 补成长方体，两���的外接球是同一个，长方体的体对角线即为球的直径， 设长方体的三度分别为$$a$$、$$b$$、$$c$$， 则有$$\\frac{1}{2}ab=\\frac{\\sqrt{6}}{2}$$，$$\\frac{1}{2}bc=\\frac{\\sqrt{3}}{2}$$，$$\\frac{1}{2}bc=\\frac{\\sqrt{2}}{2}$$， 解得：$$a=\\sqrt{3}$$，$$b=\\sqrt{2}$$，$$c=1$$， 所以球的直径$$d=\\sqrt{3+2+1}=\\sqrt{6}$$， 球的半径$$r=\\frac{d}{2}=\\frac{\\sqrt{6}}{2}$$， ∴三棱锥$$A-BCD$$的外接球的体积为： $$V=\\frac{4}{3} \\pi \\cdot {{\\left( \\frac{\\sqrt{6}}{2} \\right)}^{3}}=\\sqrt{6} \\pi $$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "677", "queId": "4579f339405b46b8a8a2784b7aaeea26", "competition_source_list": ["2016~2017学年9月天津静海区天津市静海区第一中学高三上学期月考理科第7题5分", "2010年浙江全国高中数学联赛竞赛初赛第8题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$f(x)=-\\textbar x\\textbar$$，$$a=f\\left( {{\\log }_{e}}\\frac{1}{ \\pi } \\right)$$，$$b=f\\left( {{\\log }_{ \\pi }}\\frac{1}{e} \\right)$$，$$c=f\\left( {{\\log }_{\\frac{1}{e}}}\\frac{1}{{{ \\pi }^{2}}} \\right)$$，则下述关系式正确的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b\\textgreater c$$ "}], [{"aoVal": "B", "content": "$$b\\textgreater c\\textgreater a$$ "}], [{"aoVal": "C", "content": "$$c\\textgreater a\\textgreater b$$ "}], [{"aoVal": "D", "content": "$$b\\textgreater a\\textgreater c$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->基本初等函数->指对幂比较大小", "课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算", "课内体系->知识点->函数的概念与性质->函数的性质->单调性->函数单调性的应用->用单调性比较大小"], "answer_analysis": ["$$a=f\\left( {{\\log }_{e}}\\frac{1}{ \\pi } \\right)=-{{\\log }_{e}} \\pi \\in (-2,-1)$$，$$b=f({{\\log }_{ \\pi }}\\frac{1}{e})=-{{\\log }_{ \\pi }}e\\in (-1,0)$$，$$c=f\\left( {{\\log }_{\\frac{1}{e}}}\\frac{1}{{{ \\pi }^{2}}} \\right)=-2{{\\log }_{2}} \\pi \\in (-2,-4)$$，由此能求出结果． 【解答】解：∵$$f(x)=-\\textbar x\\textbar$$， ∴$$a=f(lo{{g}_{e}}\\frac{1}{ \\pi })=-\\left\\textbar{} {{\\log }_{e}}\\frac{1}{ \\pi } \\right\\textbar=-{{\\log }_{e}} \\pi \\in (-2,-1)$$， $$b=f\\left( {{\\log }_{ \\pi }}\\frac{1}{e} \\right)=-\\left\\textbar{} {{\\log }_{ \\pi }}\\frac{1}{e} \\right\\textbar=-{{\\log }_{ \\pi }}e\\in (-1,0)$$， $$c=f\\left( {{\\log }_{\\frac{1}{e}}}\\frac{1}{{{ \\pi }^{2}}} \\right)=-\\left\\textbar{} {{\\log }_{\\frac{1}{e}}}\\frac{1}{{{ \\pi }^{2}}} \\right\\textbar=-{{\\log }_{e}}{{ \\pi }^{2}}=-2{{\\log }_{e}} \\pi \\in (-2,-4)$$， ∴$$b\\textgreater a\\textgreater c$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "979", "queId": "bbbf9ec6f4b8494087414cabae68567f", "competition_source_list": ["2018年辽宁全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设三位数$$n=\\overline{abc}$$，若以$$a$$、$$b$$、$$c$$为三条边的长可构成一个等腰（含等边）三角形，则这样的三位数有个．", "answer_option_list": [[{"aoVal": "A", "content": "$$45$$ "}], [{"aoVal": "B", "content": "$$81$$ "}], [{"aoVal": "C", "content": "$$165$$ "}], [{"aoVal": "D", "content": "$$216$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["依题意，等边三角形共有$$9$$个，等腰非等边三角形共有$$3\\times \\left( 8+8+8+8+8+6+4+2 \\right)=156$$个， 因此，所求为$$9+156=165$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "789", "queId": "fb368c09207a470e9955b7ebabd754aa", "competition_source_list": ["2017~2018学年江西南昌西湖区南昌市第八中学高一上学期期末第10题5分", "2018~2019学年广东深圳高一上学期期末高中联考联盟第8题5分", "2018~2019学年浙江杭州江干区杭州第四中学下沙校区高二下学期期中第12题4分", "2018~2019学年江苏盐城高一上学期期末第5题5分", "2017~2018学年北京海淀区中央民族大学附属中学高一下学期期中第7题5分", "2008年黑龙江全国高中数学联赛竞赛初赛第4题5分", "2015年北京海淀区高三三模第4题"], "difficulty": "0", "qtype": "single_choice", "problem": "若$$\\sin \\left( \\frac{ \\pi }{4}-x \\right)=\\frac{3}{5}$$，则$$\\sin 2x$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{19}{25}$$ "}], [{"aoVal": "B", "content": "$$\\frac{16}{25}$$ "}], [{"aoVal": "C", "content": "$$\\frac{14}{25}$$ "}], [{"aoVal": "D", "content": "$$\\frac{7}{25}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式->正余弦和差积相互转化求值", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正弦", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的正弦->利用正弦和差角公式凑角求值"], "answer_analysis": ["$$\\sin \\left( \\frac{ \\pi }{4}-x \\right)=\\frac{\\sqrt{2}}{2}(\\cos x-\\sin x)=\\frac{3}{5}$$，所以$$\\cos x-\\sin x=\\frac{3\\sqrt{2}}{5}$$， 所以$${{(\\text{cos }x-\\text{sin }x)}^{2}}=1-\\sin 2x=\\frac{18}{25}$$，所以$$\\sin 2x=\\frac{7}{25}$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "438", "queId": "4c3857c6de28417584d81009f89a6e6c", "competition_source_list": ["1995年全国高中数学联赛竞赛一试第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "$${{\\log }_{\\sin 1}}\\cos 1$$，$${{\\log }_{\\sin 1}}\\text{tg1}$$，$${{\\log }_{\\cos 1}}\\sin 1$$，$$\\text{lo}{{\\text{g}}_{\\cos 1}}\\text{tg1}$$的大小关系是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{\\log }_{\\sin 1}}\\cos 1\\textless{}{{\\log }_{\\cos 1}}\\sin 1\\textless{}{{\\log }_{\\sin 1}}\\text{tg1lo}{{\\text{g}}_{\\cos 1}}\\text{tg1}$$ "}], [{"aoVal": "B", "content": "$${{\\log }_{\\cos 1}}\\sin 1\\textless{}{{\\log }_{\\cos 1}}\\text{tg1lo}{{\\text{g}}_{\\sin 1}}\\cos 1\\textless{}{{\\log }_{\\sin 1}}\\text{tg1}$$ "}], [{"aoVal": "C", "content": "$${{\\log }_{\\sin 1}}\\text{tg}1\\textless{}{{\\log }_{\\cos 1}}\\text{tg1lo}{{\\text{g}}_{\\cos 1}}\\sin 1\\textless{}{{\\log }_{\\sin 1}}\\cos 1$$ "}], [{"aoVal": "D", "content": "$${{\\log }_{\\cos 1}}\\text{tg}1\\textless{}{{\\log }_{\\sin 1}}\\text{tg1lo}{{\\text{g}}_{\\sin 1}}\\cos 1\\textless{}{{\\log }_{\\cos 1}}\\sin 1$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->三角函数线", "课内体系->知识点->基本初等函数->指对幂比较大小", "课内体系->知识点->基本初等函数->对数的概念及其运算", "课内体系->知识点->基本初等函数->实数指数幂运算"], "answer_analysis": ["由$$\\frac{ \\pi }{4}\\textless{}1$$知$$\\cos 1\\textless{}\\sin 1\\textless{}1\\textless{}\\text{tg}1$$，从而$${{\\log }_{\\sin 1}}\\text{tg1}\\textless{}0$$且$${{\\log }_{\\cos 1}}\\text{tg1}\\textless{}0$$，而$${{\\log }_{\\sin 1}}\\text{cos 1}\\textgreater0$$且$${{\\log }_{\\cos 1}}\\text{sin 1}\\textgreater0$$，于是可排除$$\\text{A}$$和$$\\text{B}$$，只剩$$\\text{C}$$和$$\\text{D}$$又由$${{\\log }_{\\cos 1}}\\text{sin 1}\\textless{}{{\\log }_{\\cos 1}}\\text{cos 1=}{{\\log }_{\\sin 1}}\\text{sin 1}{{\\log }_{\\sin 1}}\\text{cos 1}$$即知应． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "718", "queId": "c3b7c968732f4ec89387e7ca65e45991", "competition_source_list": ["2016年AMC10竞赛A第20题"], "difficulty": "1", "qtype": "single_choice", "problem": "若某个特殊的正整数$$N$$使得$$(a+b+c+d+1)^{}N$$ 的展开式中，合并同类项后发现同时含有 $$a$$, $$b$$, $$c$$, and $$d$$四个变量正整数乘幂的项恰好有$$1001$$项，求$$N$$的值？ For some particular value of $$N$$, when $$(a+b+c+d+1)^{}N$$ is expanded and like terms are combined, the resulting expression contains exactly $$1001$$ terms that include all four variables $$a$$, $$b$$, $$c$$, and $$d$$, each to some positive power. What is $$N$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$17$$ "}], [{"aoVal": "E", "content": "$$19$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["All the desired terms are in the form $$a^{}xb^{}yc^{}zd^{}w1^{}t$$, where $$x+y+z+w+t=N$$ (the $$1^{}t$$ part is necessary to make stars and bars work better.) Since $$x$$, $$y$$, $$z$$, and~ $$w$$ must be at least $$1$$ ($$t$$ can be $$0$$), let $$x^{}\\prime =x-1$$, $$y^{}\\prime =y-1$$, $$z^{}\\prime =z-1$$, and $$w^{}\\prime =w-1$$, so $$x^{}\\prime +y^{}\\prime+z^{}\\prime+w^{}\\prime+t=N-4$$. Now, we use stars and bars to see that there are $$\\left(\\frac {(N-4)+4}4\\right)$$ or $$\\left(\\frac N4\\right)$$ solutions to this equation. We notice that $$1001=7\\cdot 11\\cdot 13$$, which leads us to guess that $$N$$ is around these numbers. This suspicion proves to be correct, as we see that $$\\left(\\frac {14}4\\right)=1001$$, giving us our answer of $$N=14$$. An alternative is to instead make the transformation $$6^{}\\prime =t+1$$, so $$x+y+z+w+t^{}\\prime =N+1$$, and all variables are positive integers. The solution to this, by Stars and Bars is $$\\left(\\frac {(N+1)-1}4\\right)=\\left(\\frac N4\\right)$$ and we can proceed as above. ", "By Hockey Stick Identity, the number of terms that have all $$a$$, $$b$$, $$c$$, $$d$$ raised to a positive power is $$\\left(\\frac {N-1}3\\right)+\\left(\\frac {N-2}3\\right)+\\cdots +\\left(\\frac 43\\right)+\\left(\\frac 33\\right)=\\left(\\frac N4\\right)$$. We now want to find some $$N$$ such that $$\\left(\\frac N4\\right)=1001$$. As mentioned above, after noticing that $$1001=7\\cdot 11\\cdot 13$$, and some trial and error, we find that $$\\left(\\frac {14}4\\right)=1001$$, giving us our answer of $$N=14$$.
"], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "644", "queId": "f1a38976bc27442d900f640e01acd328", "competition_source_list": ["2010年辽宁全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "集合$$A=\\left { 1,3,5,7 \\right },B=\\left { 2,4,6,8,20 \\right }$$，若$$C= {s\\left\\textbar{} s=a+b, a\\in A, b\\in B \\right. }$$，则集合$$C$$的元素个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算", "竞赛->知识点->集合->集合的划分与覆盖"], "answer_analysis": ["$$s$$是奇数，且最小值是$$3$$，最大值是$$27$$，中间缺少$$17$$、$$19$$，故集合$$C$$共有$$11$$个元素． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1006", "queId": "f2b477fcab63427dbf32a8d14f21577c", "competition_source_list": ["2018年天津全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设复数$$z$$满足$$\\left\\textbar{} z \\right\\textbar=1$$，则$$\\left\\textbar{} \\left( z+1 \\right)+\\text{i}\\left( 7-z \\right) \\right\\textbar$$的值不可能为．", "answer_option_list": [[{"aoVal": "A", "content": "$$4\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$4\\sqrt{3}$$ "}], [{"aoVal": "C", "content": "$$5\\sqrt{2}$$ "}], [{"aoVal": "D", "content": "$$5\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的应用"], "answer_analysis": ["$$\\left( z+1 \\right)+\\text{i}\\left( 7-z \\right)=\\left( 1-\\text{i} \\right)z+\\left( 1+7\\text{i} \\right)=\\left( 1-\\text{i} \\right)\\left( z-3+4\\text{i} \\right)$$， 即$$\\left\\textbar{} \\left( z+1 \\right)+\\text{i}\\left( 7-z \\right) \\right\\textbar=\\left\\textbar{} 1-\\text{i} \\right\\textbar\\cdot \\left\\textbar{} z-3+4\\text{i} \\right\\textbar=\\sqrt{2}\\cdot \\left\\textbar{} z-\\left( 3-4\\text{i} \\right) \\right\\textbar$$， 即可以表示点$$z$$到$$3-4\\text{i}$$的距离的$$\\sqrt{2}$$倍， ∵$$z$$在单位圆上， ∴$$\\left\\textbar{} \\left( z+1 \\right)+\\text{i}\\left( 7-z \\right) \\right\\textbar$$的取值范围是$$\\left[ 4\\sqrt{2},6\\sqrt{2} \\right]$$． 故选：$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "375", "queId": "e33d3d6c1bc747438ee2f13f5866016e", "competition_source_list": ["2008年全国高中数学联赛竞赛一试第6题6分", "2016年辽宁全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\triangle ABC$$的内角$$A$$，$$B$$，$$C$$所对的边$$a$$，$$b$$，$$c$$成等比数列，则$$\\frac{\\sin A\\cot C+\\cos A}{\\sin B\\cot C+\\cos B}$$的取值范围是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 0,+\\infty \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 0,\\frac{\\sqrt{5}+1}{2} \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{\\sqrt{5}-1}{2},\\frac{\\sqrt{5}+1}{2} \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{\\sqrt{5}-1}{2},+\\infty \\right)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数列->等比数列->等比数列的性质及应用", "课内体系->知识点->解三角形->解三角形的实际应用", "课内体系->素养->数据分析", "课内体系->素养->数学运算"], "answer_analysis": ["$$\\frac{\\sin A\\cot C+\\cos A}{\\sin B\\cot C+\\cos B}=\\frac{\\sin A\\cos C+\\sin C\\cos A}{\\sin B\\cos C+\\sin C\\cos B}=\\frac{\\sin \\left( A+C \\right)}{\\sin \\left( B+C \\right)}=\\frac{\\sin B}{\\sin A}=\\frac{b}{a}$$， 由题设有$${{b}^{2}}=ac$$，则$$a\\left\\textbar{} a-b \\right\\textbar\\textless{}{{b}^{2}}\\textless{}a\\left( a+b \\right)$$，解得$$\\frac{\\sqrt{5}-1}{2}\\textless{}\\frac{b}{a}\\textless{}\\frac{\\sqrt{5}+1}{2}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "647", "queId": "4dfb54a90c4344088f2dd0a35ca02f16", "competition_source_list": ["2011年天津全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若直线$$y=x-3$$与曲线$$y={{\\text{e}}^{x+a}}$$相切，则实数$$a$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-4$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->导数", "竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["设切点的横坐标为$${{x}_{0}}$$．在$$x={{x}_{0}}$$处，曲线$$y={{\\text{e}}^{x+a}}$$的斜率为$${{\\text{e}}^{{{x}_{0}}+a}}$$．而直线$$y=x-3$$的斜率为$$1$$．因此$${{\\text{e}}^{{{x}_{0}}+a}}=1$$，得$${{x}_{0}}=-a$$．因此，切点的纵坐标$${{\\text{e}}^{{{x}_{0}}+a}}=1={{x}_{0}}-3$$，即$$1=-a-3$$，所以$$a=-4$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "401", "queId": "2cb82a786b98483fb49bfc5e5d59c1da", "competition_source_list": ["2008年浙江全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$P$$为三角形$$ABC$$内部任一点（不包括边界），且满足$$(-)(+-2)=0$$，则$$\\triangle ABC$$一定为 ．", "answer_option_list": [[{"aoVal": "A", "content": "直角三角形 "}], [{"aoVal": "B", "content": "等边三角形 "}], [{"aoVal": "C", "content": "等腰直角三角形 "}], [{"aoVal": "D", "content": "等腰三角形 "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角形中的问题->解三角形"], "answer_analysis": ["因为$$\\overrightarrow{PB}-\\overrightarrow{PA}=\\overrightarrow{AB}$$，$$\\overrightarrow{PB}+\\overrightarrow{PA}-2\\overrightarrow{PC}=\\overrightarrow{CB}+\\overrightarrow{CA}$$， 所以已知条件可改写为$$\\overrightarrow{AB}\\cdot (\\overrightarrow{CB}+\\overrightarrow{CA})=0$$． 容易得到此三角形为等腰三角形．故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "531", "queId": "2e5763e4951c45078d8657b122e829ad", "competition_source_list": ["2002年AMC12竞赛B第22题"], "difficulty": "3", "qtype": "single_choice", "problem": "2002AMC12B, 22 For all integers $$n$$ greater than $$1$$, define $$a_{n}=\\frac{1}{\\log _{n}2002}$$. Let $$b=a_{2}+a_{3}+a_{4}+a_{5}$$ and $$c=a_{10}+a_{11}+a_{12}+a_{13}+a_{14}$$. Then $$b-c$$ equals． 对于大于$$1$$的所有整数$$n$$，定义$$a_{n}=\\frac{1}{\\log _{n}2002}$$。令 $$b=a_{2}+a_{3}+a_{4}+a_{5}$$，$$c=a_{10}+a_{11}+a_{12}+a_{13} +a_{14}$$。则 $$b-c$$ 等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2002}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{1001}$$ "}], [{"aoVal": "E", "content": "$$\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算", "美国AMC10/12->Knowledge Point->Algebra->Sequence->Sum of the First n Terms for Geometric Sequences"], "answer_analysis": ["$\\frac{1}{\\log_a{b}}=\\log_b{a}$ "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "533", "queId": "3b396ace61fc43c7a6dea62851a6c84f", "competition_source_list": ["2010年河北全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "将正偶数集合$$ {2,4,6,8,\\cdots }$$从小到大按第$$n$$组有$$2n-1$$个数进行分组：$$ {2 }$$，$$ {4,6,8 }$$， $$ {10,12,14,16,18 }$$，\\ldots，问$$2010$$位于第组中．", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的划分与覆盖"], "answer_analysis": ["2010是数列$${{a}_{n}}=2n$$的第$$1005$$项， 设$$2010$$位于第$$n$$组， 则$$\\begin{cases}1+3+5+\\cdots +(2n-1)\\geqslant 1005 1+3+5+\\cdots +(2n-3)\\textless{}1005 \\end{cases}$$， 即$$\\begin{cases}{{n}^{2}}\\geqslant 1005 {{(n-1)}^{2}}\\textless{}1005 \\end{cases}$$， 所以$$n=32$$，故$$2010$$位于第$$32$$组． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "403", "queId": "202fcf396fd640d29f7772e1e973708a", "competition_source_list": ["2008年山东全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\vec{a}=(\\cos \\alpha ,\\sin \\alpha )$$，$$\\vec{b}=(\\cos \\beta ,\\sin \\beta )$$，$$\\textbar\\vec{a}-\\vec{b}\\textbar=\\frac{2\\sqrt{5}}{5}$$，若$$0\\textless{}\\alpha \\textless{}\\frac{ \\pi }{2}$$，$$-\\frac{ \\pi }{2}\\textless{}\\beta \\textless{}0$$，且$$\\sin \\beta =-\\frac{5}{13}$$，则$$\\sin \\alpha =$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{31}{65}$$ "}], [{"aoVal": "B", "content": "$$\\frac{32}{65}$$ "}], [{"aoVal": "C", "content": "$$\\frac{33}{65}$$ "}], [{"aoVal": "D", "content": "$$\\frac{34}{65}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["由题意知$$\\textbar\\vec{a}-\\vec{b}{{\\textbar}^{2}}={{(\\cos \\alpha -\\cos \\beta )}^{2}}+{{(\\sin \\alpha -\\sin \\beta )}^{2}}$$ $$=2-2(\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta )=2-2\\cos (\\alpha -\\beta )$$， 则$$2-2\\cos (\\alpha -\\beta )=\\frac{4}{5}$$，$$\\cos (\\alpha -\\beta )=\\frac{3}{5}$$． 由$$\\alpha =(\\alpha -\\beta )+\\beta $$，且$$0\\textless{}\\alpha \\textless{}\\frac{ \\pi }{2}$$，$$-\\frac{ \\pi }{2}\\textless{}\\beta \\textless{}0$$及$$\\sin \\beta =-\\frac{5}{13}$$，$$\\cos (\\alpha -\\beta )=\\frac{3}{5}$$， 知$$\\cos \\beta =\\frac{12}{13}$$，$$\\sin (\\alpha -\\beta )=\\frac{4}{5}$$， 从而得$$\\sin \\alpha =\\sin [\\beta +(\\alpha -\\beta )]=\\sin \\beta \\cos (\\alpha -\\beta )+\\cos \\beta \\sin (\\alpha -\\beta )$$ $$=-\\frac{5}{13}\\times \\frac{3}{5}+\\frac{12}{13}\\times \\frac{4}{5}=\\frac{33}{65}$$， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "719", "queId": "8cbcb50d149247c4b7da906e65a857ac", "competition_source_list": ["2009年河北全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "方程$$\\frac{{{x}^{2}}}{\\sin ({{19}^{n}}){}^{}\\circ }+\\frac{{{y}^{2}}}{\\cos ({{19}^{n}}){}^{}\\circ }=1(n\\in N*)$$所表示的曲线为．", "answer_option_list": [[{"aoVal": "A", "content": "焦点在$$x$$轴上的双曲线 "}], [{"aoVal": "B", "content": "双曲线，其焦点所在的轴与$$n$$有关 "}], [{"aoVal": "C", "content": "焦点在$$y$$轴上的椭圆 "}], [{"aoVal": "D", "content": "椭圆，其焦点所在的轴与$$n$$有关 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->椭圆"], "answer_analysis": ["当$$n=1$$时，$$\\sin 19{}^{}\\circ \\textless{}\\cos 19{}^{}\\circ $$； $$n=2$$时， $$\\sin ({{19}^{2}}){}^{}\\circ =\\sin 361{}^{}\\circ =\\sin 1{}^{}\\circ \\textless{}\\cos 1{}^{}\\circ $$； $$n=2k(k\\in N*)$$时， $$\\sin ({{19}^{2k}}){}^{}\\circ =\\sin ({{361}^{k}}){}^{}\\circ =\\sin 1{}^{}\\circ \\textless{}\\cos 1{}^{}\\circ $$； $$n=2k+1(k\\in N*)$$时， $$\\sin ({{19}^{2k+1}}){}^{}\\circ =\\sin [19\\times {{(361)}^{k}}]{}^{}\\circ =\\sin 19{}^{}\\circ \\textless{}\\cos 19{}^{}\\circ $$． 所以方程表示焦点在$$y$$轴上的椭圆．选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "590", "queId": "7f35761862144d3aba07adbd7dad983b", "competition_source_list": ["2011年浙江全国高中数学联赛竞赛初赛第9题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知函数$$f\\left( x \\right)=\\sin \\left( 2x-\\frac{\\pi }{6} \\right)-m$$在$$\\left[ 0,\\frac{\\pi }{2} \\right]$$上有有两个零点，则$$m$$的取值范围是（ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( \\frac{1}{2},1 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left[ \\frac{1}{2},1 \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left[ \\frac{1}{2},1 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{1}{2},1 \\right]$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角函数的图象与性质->正弦型函数的图象与性质", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦函数的图象和性质", "课内体系->知识点->函数的应用->函数的零点->零点的个数问题->已知零点情况求参数的取值范围", "课内体系->知识点->函数的应用->函数的零点->函数零点的概念", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的定义域", "课内体系->素养->直观想象", "课内体系->素养->数学运算", "课内体系->思想->数形结合思想"], "answer_analysis": ["$$\\because x\\in \\left[ 0,\\frac{\\pi }{2} \\right]$$，$$\\therefore 2x-\\frac{\\pi }{6}\\in \\left[ -\\frac{\\pi }{6},\\frac{5\\pi }{6} \\right]$$，$$\\therefore \\sin \\left( 2x-\\frac{\\pi }{6} \\right)\\in \\left[ -\\frac{1}{2},1 \\right]$$，令$$t=2x-\\frac{\\pi }{6}$$，$$y=m$$， 在同一直角坐标系中作出$$y=\\sin t$$（$$t\\in \\left[ -\\frac{\\pi }{6},\\frac{5\\pi }{6} \\right]$$）与$$y=m$$的图象，由图象可知， 当$$\\frac{1}{2}\\leqslant m \\textless{} 1$$时，$$y=\\sin t$$（$$t\\in \\left[ -\\frac{\\pi }{6},\\frac{5\\pi }{6} \\right]$$）的图象与直线$$y=m$$有两个交点， 即函数$$f(x)=\\sin \\left( 2x-\\frac{\\pi }{6}-m \\right)$$在$$\\left[ 0,\\frac{\\pi }{2} \\right]$$上有两个零点． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "96", "queId": "5d25c1df230f4d069082eb57ac72e231", "competition_source_list": ["2009年湖南全国高中数学联赛竞赛初赛第11题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "以双曲线$$\\frac{{{x}^{2}}}{4}-\\frac{{{y}^{2}}}{{{b}^{2}}}=1(b\\textgreater0)$$的离心率为半径，右焦点为圆心的圆与双曲线的渐近线相切，则双曲线的离心率为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2\\sqrt{3}}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4\\sqrt{3}}{3}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5\\sqrt{3}}{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->双曲线"], "answer_analysis": ["易知$$a=2, c=\\sqrt{4+{{b}^{2}}}, e=\\frac{\\sqrt{4+{{b}^{2}}}}{2}$$． 渐近线方程为$$\\frac{x}{2}\\pm \\frac{y}{b}=0$$，即$$bx\\pm 2y=0$$． 右焦点$$(\\sqrt{4+{{b}^{2}}}, 0)$$到渐近线的距离$$d=b$$，从而得方程$$b=\\frac{\\sqrt{4+{{b}^{2}}}}{2}$$． 解方程得，$$b=\\frac{2}{3}\\sqrt{3}, e=\\frac{\\sqrt{4+\\frac{4}{3}}}{2}=\\frac{2\\sqrt{3}}{3}$$．故填$$\\frac{2\\sqrt{3}}{3}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "423", "queId": "3e9830b6bc3844e0bf1d26c974d35d3b", "competition_source_list": ["2016年吉林全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$x\\in \\left( -\\frac{3}{4} \\pi ，\\frac{ \\pi }{4} \\right)$$，且$$\\cos \\left( \\frac{ \\pi }{4}-x \\right)=-\\frac{3}{5}$$，则$$\\cos 2x$$的值是（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{7}{25}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{24}{25}$$ "}], [{"aoVal": "C", "content": "$$\\frac{24}{25}$$ "}], [{"aoVal": "D", "content": "$$\\frac{7}{25}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式的化简和求值", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->半角公式", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正弦", "课内体系->素养->数学运算"], "answer_analysis": ["$$\\cos 2x=\\cos \\left( 2x-\\frac{\\mathrm{ }\\pi }{2}+\\frac{\\mathrm{ }\\pi }{2} \\right)$$ $$=-\\sin \\left( 2x-\\frac{\\mathrm{ }\\pi }{2} \\right)$$ $$=-2\\sin \\left( x-\\frac{\\mathrm{ }\\pi }{4} \\right)\\cos \\left( x-\\frac{\\mathrm{ }\\pi }{4} \\right)$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "336", "queId": "a25f1331184f4cc88c82ea1d9b9aa9b1", "competition_source_list": ["2017~2018学年3月天津河西区天津市新华中学高一下学期月考第8题", "2009年湖南全国高中数学联赛竞赛初赛第6题5分", "2018~2019学年陕西西安碑林区西安建筑科技大学附属中学高一下学期期中（必修4）第10题5分", "2016~2017学年湖北黄冈黄州区湖北省黄冈中学高一上学期期末第7题5分", "2018~2019学年山东济南历城区济南外国语学校高二上学期开学考试第9题5分", "2019~2020学年山东济南历城区济南历城一中高二上学期开学考试第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知非零向量$$\\overrightarrow{AB}$$与$$\\overrightarrow{AC}$$满足$$\\left( \\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB}\\textbar}+\\frac{\\overrightarrow{AC}}{\\textbar\\overrightarrow{AC}\\textbar} \\right)\\cdot \\overrightarrow{BC}=0$$且$$\\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB}\\textbar}\\cdot \\frac{\\overrightarrow{AC}}{\\textbar\\overrightarrow{AC}\\textbar}=\\frac{1}{2}$$，则$$\\triangle ABC$$为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "三边均不相等的三角形 "}], [{"aoVal": "B", "content": "直角三角形 "}], [{"aoVal": "C", "content": "等腰非等边三角形 "}], [{"aoVal": "D", "content": "等边三角形 "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算", "竞赛->知识点->复数与平面向量->平面向量的应用"], "answer_analysis": ["设$$\\frac{\\overrightarrow{AB}}{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}=\\overrightarrow{AE}$$，$$\\frac{\\overrightarrow{AC}}{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}=\\overrightarrow{AF}$$， 则原式$$=\\left( \\overrightarrow{AE}+\\overrightarrow{AF} \\right)\\cdot \\overrightarrow{BC}=0$$， 即$$\\overrightarrow{AD}\\cdot \\overrightarrow{BC}=0$$， ∴$$AD\\bot BC$$， ∴四边形$$AEDF$$为菱形． ∵$$\\overrightarrow{AE}\\cdot \\overrightarrow{AF}=\\left\\textbar{} \\overrightarrow{AE} \\right\\textbar\\cdot \\left\\textbar{} \\overrightarrow{AF} \\right\\textbar\\cos \\angle BAC=\\frac{1}{2}$$， ∴$$\\cos \\angle BAC=\\frac{1}{2}$$， ∴$$\\angle BAC=60{}^{}\\circ $$， ∴$$\\angle BAD=\\angle DAC=30{}^{}\\circ $$， ∴$$\\triangle ABH$$ ≌ $$\\triangle AHC$$， ∴$$AB=AC$$， ∴$$\\triangle ABC$$为等边三角形． 故选$$\\text{D}$$． 设$$\\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB\\textbar}}$$为$$\\overrightarrow{AB}$$上的单位向量，$$\\frac{\\overrightarrow{AC}}{\\left\\textbar{} AC \\right\\textbar}$$为$$\\overrightarrow{AC}$$上的单位向量． 则$$\\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB}\\textbar}+\\frac{\\overrightarrow{AC}}{\\textbar\\overrightarrow{AC}\\textbar}$$的方向为$$\\angle BAC$$的角平分线$$\\overrightarrow{AO}$$的方向． 而$$\\overrightarrow{AO}\\cdot \\overrightarrow{BC}=0$$，∴$$\\overrightarrow{AO}\\bot \\overrightarrow{BC}$$． ∴$$\\triangle ABC$$为等腰三角形，$$AB=AC$$． $$\\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB}\\textbar}\\cdot \\frac{\\overrightarrow{AC}}{\\textbar\\overrightarrow{AC}\\textbar}=\\left\\textbar{} \\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB}\\textbar} \\right\\textbar\\left\\textbar{} \\frac{\\overrightarrow{AC}}{\\textbar\\overrightarrow{AC}\\textbar} \\right\\textbar\\cos \\angle BAC=\\frac{1}{2}$$． ∵$$\\left\\textbar{} \\frac{\\overrightarrow{AB}}{\\textbar\\overrightarrow{AB}\\textbar} \\right\\textbar=\\left\\textbar{} \\frac{\\overrightarrow{AC}}{\\textbar\\overrightarrow{AC}\\textbar} \\right\\textbar=1$$，∴$$\\angle BAC=60{}^{}\\circ $$． 综上所述，$$\\triangle ABC$$为等边三角形． ", "由$$\\left( \\frac{\\overrightarrow{AB}}{|\\overrightarrow{AB}|}+\\frac{\\overrightarrow{AC}}{|\\overrightarrow{AC}|} \\right)\\cdot \\overrightarrow{BC}=0$$，得$$\\angle BAC$$的平分线垂直于$$BC$$，
\n∴$$AB=AC$$，
\n∵$$\\frac{\\overrightarrow{AB}}{|\\overrightarrow{AB}|}\\cdot \\frac{\\overrightarrow{AC}}{|\\overrightarrow{AC}|}=\\cos \\left\\langle \\overrightarrow{AB},\\overrightarrow{AC} \\right\\rangle =\\frac{1}{2}$$，$$\\left\\langle \\overrightarrow{AB},\\overrightarrow{AC} \\right\\rangle =\\in (0{}^\\circ ,180{}^\\circ )$$，
\n∴$$\\angle BAC=60{}^\\circ $$，$$\\triangle ABC$$为等边三角形．
\n故选$$\\text{D}$$．
\n"], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "178", "queId": "10bfd60622624e25980f8affd8c80d29", "competition_source_list": ["2015年辽宁全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "长方体$$ABCD-{{A}_{1}}{{B}_{1}}{{C}_{1}}{{D}_{1}}$$中，$$AB=A{{A}_{1}}=1$$，$$AD=2$$，则异面直线$${{A}_{1}}D$$与$${{B}_{1}}{{D}_{1}}$$间的距离为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间向量"], "answer_analysis": ["用向量法. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "179", "queId": "fee1377fbd2a40289fd7d110d8308bc5", "competition_source_list": ["2010年浙江全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$\\left { {{a}_{n}} \\right }$$，$$\\left { {{b}_{n}} \\right }$$分别为等差数列与等比数列，且$${{a}_{1}}={{b}_{1}}=4, {{a}_{4}}={{b}_{4}}=1$$，则以下结论正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{a}_{2}}\\textgreater{{b}_{2}}$$ "}], [{"aoVal": "B", "content": "$${{a}_{3}}\\textless{}{{b}_{3}}$$ "}], [{"aoVal": "C", "content": "$${{a}_{5}}\\textgreater{{b}_{5}}$$ "}], [{"aoVal": "D", "content": "$${{a}_{6}}\\textgreater{{b}_{6}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["设等差数列的公差为$$d$$，等比数列公比为$$q$$，由$${{a}_{1}}={{b}_{1}}=4,{{a}_{4}}={{b}_{4}}=1,$$ 得$$d=-1, q=\\frac{\\sqrt[3]{2}}{2},$$得$${{a}_{2}}=3, {{b}_{2}}=2\\sqrt[3]{2}$$；$${{a}_{3}}=2, {{b}_{3}}=\\sqrt[3]{4}$$；$${{a}_{5}}=0,{{b}_{5}}=\\frac{\\sqrt[3]{2}}{2}$$；$${{a}_{6}}=-1,,$$ $$ {{b}_{6}}=\\frac{\\sqrt[3]{4}}{4}$$ "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "858", "queId": "cd57f74aaa9445be96625852e0049805", "competition_source_list": ["2010年高考真题江苏卷理科第10题5分", "2018~2019学年3月河南郑州金水区郑州市第七中学高一下学期月考第16题5分", "2018~2019学年11月河北石家庄辛集市河北辛集中学高一上学期月考第22题5分", "2010年湖南全国高中数学联赛竞赛初赛第7题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "设定义在区间$$\\left( 0,\\frac{ \\pi }{2} \\right)$$上的函数$$y=6\\cos x$$的图象与$$y=5\\tan x$$的图象的交点为$$P$$，过点$$P$$作$$x$$轴的垂线，垂足为$${{P}_{1}}$$，直线$$P{{P}_{1}}$$与函数$$y=\\sin x$$的图象交于点$${{P}_{2}}$$，则线段$${{P}_{1}}{{P}_{2}}$$的长为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{4}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{6}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质", "课内体系->知识点->三角函数"], "answer_analysis": ["由$$\\begin{cases}y=6\\cos x y=5\\tan x \\end{cases}$$消$$y$$得$$6\\cos x=5\\tan x$$．整理得$$6{{\\cos }^{2}}x=5\\sin x$$， $$6{{\\sin }^{2}}x+5\\sin x-6=0$$．$$\\left( 3\\sin x-2 \\right)\\left( 2\\sin x+3 \\right)=0$$，所以$$\\sin x=\\frac{2}{3}$$或$$\\sin x=-\\frac{3}{2}$$（舍去）． 故点$${{P}_{2}}$$的纵坐标为$${{y}_{2}}=\\frac{2}{3}$$．所以$$\\left\\textbar{} {{P}_{1}}{{P}_{2}} \\right\\textbar=\\frac{2}{3}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1149", "queId": "bd9f0e823a5342e292e9d384105acd68", "competition_source_list": ["2015年天津全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "设$$A$$、$$B$$和$$C$$是三个集合，则$$B$$和$$C$$都是$$A$$的子集是$$\\left( A\\cap B \\right)\\cup \\left( A\\cap C \\right)=B\\cup C$$成立的．", "answer_option_list": [[{"aoVal": "A", "content": "充分条件但不是必要条件 "}], [{"aoVal": "B", "content": "必要条件但不是充分条件 "}], [{"aoVal": "C", "content": "充分必要条件 "}], [{"aoVal": "D", "content": "既非充分也非必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->常用逻辑用语", "竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["如果$$B$$和$$C$$都是$$A$$的子集，则$$A\\cap B=B$$，$$A\\cap C=C$$，从而$$\\left( A\\cap B \\right)\\cup \\left( A\\cap C \\right)=B\\cup C$$成立；反过来，如果$$\\left( A\\cap B \\right)\\cup \\left( A\\cap C \\right)=B\\cup C$$成立，则由$$A\\cap B$$和$$A\\cap C$$都是$$A$$的子集知$$B\\cup C$$是$$A$$的子集，即$$B$$和$$C$$都是$$A$$的子集．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1027", "queId": "982df62f8cef42089ca583ece1e69d57", "competition_source_list": ["1985年全国高中数学联赛竞赛一试第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$PQ$$为经过抛物线$${{y}^{2}}=2px$$焦点的任意一条弦，$$MN$$为$$PQ$$大准线$$l$$上的射影，$$PQ$$绕$$l$$转一周所得的旋转面面积为$${{S}_{1}}$$，以$$MN$$为直径的球面面积为$${{S}_{2}}$$，则下面的结论中，正确的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{S}_{1}}\\textgreater{{S}_{2}}$$ "}], [{"aoVal": "B", "content": "$${{S}_{1}}\\textless{}{{S}_{2}}$$ "}], [{"aoVal": "C", "content": "$${{S}_{1}}\\mathsf{\\geqslant }{{S}_{2}}$$ "}], [{"aoVal": "D", "content": "有时$${{S}_{1}}\\textgreater{{S}_{2}}$$，有时$${{S}_{1}}={{S}_{2}}$$，有时$${{S}_{1}}\\textless{}{{S}_{2}}$$． "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->抛物线"], "answer_analysis": ["在$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$四个答案中，$$\\text{C}$$包含着$$\\text{A}$$，即若$$\\text{A}$$正确，则$$\\text{C}$$必定正确，因此不必考虑$$\\text{A}$$． 当过焦点的弦$$PQ$$的倾角（与$$x$$轴的夹角）很小时，$$PQ$$绕$$l$$旋转一周所得旋转体（圆台）的母线很长，而相应的高却很短，亦即球的直径很小，因此$$\\text{B}$$不成立． 我们知道，圆台的侧面积由母线$$PQ$$和两底半径$$PMQN$$的长确定．根据抛物线的定义， $$\\left\\textbar{} PM \\right\\textbar=\\left\\textbar{} PF \\right\\textbar\\mathsf{}\\left\\textbar{} QN \\right\\textbar=\\left\\textbar{} QF \\right\\textbar$$ （其中$$F$$为焦点） 若取$$PQ$$的倾角$$\\alpha $$为参数，设 $$\\left\\textbar{} PF \\right\\textbar={{\\rho }_{1}}\\mathsf{} \\left\\textbar{} QF \\right\\textbar={{\\rho }_{2}}$$ 则$$\\left\\textbar{} PM \\right\\textbar={{\\rho }_{1}}\\mathsf{} \\left\\textbar{} QN \\right\\textbar={{\\rho }_{2}}\\mathsf{} \\left\\textbar{} PQ \\right\\textbar={{\\rho }_{1}}+{{\\rho }_{2}}$$ 于是 $${{S}_{1}}= \\pi ({{\\rho }_{1}}+{{\\rho }_{2}}{{)}^{2}}$$， $${{S}_{2}}= \\pi \\centerdot {{\\left\\textbar{} MN \\right\\textbar}^{2}}= \\pi {{\\sin }^{2}}\\alpha {{({{\\rho }_{1}}+{{\\rho }_{2}})}^{2}}$$ 显然$${{S}_{1}}\\mathsf{\\geqslant }{{S}_{2}}$$，且当$$a=\\frac{ \\pi }{2}$$时取到等号． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "582", "queId": "5639193a518e4cecb5563e13013ed1e6", "competition_source_list": ["2009年浙江全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$${{x}^{3}}-12x+a=0$$有三个不同的实数根，则实数$$a$$的取值范围为．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-16,16)$$ "}], [{"aoVal": "B", "content": "$$\\left[ -16, 16 \\right]$$ "}], [{"aoVal": "C", "content": "$$(-\\infty ,-8)$$ "}], [{"aoVal": "D", "content": "$$(8,+\\infty )$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->导数"], "answer_analysis": ["令$$f(x)={{x}^{3}}-12x+a$$，则$$f\\prime (x)=3{{x}^{2}}-12$$， $$f\\prime (x)=3{{x}^{2}}-12=0\\Rightarrow x=\\pm 2$$． 要使$$f(x)=0$$有三个不同的实数根，则必须有$$f(2)f(-2)\\textless{}0$$，即$$(a-16)(a+16)\\textless{}0$$，也即有$$-16\\textless a\\textless16$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "526", "queId": "75c31f1c03a7454fa68f26c5dada9d64", "competition_source_list": ["2018年四川全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$S_n$$、$$T_n$$分别为等差数列$$ {a_n }$$、$$ {b_n }$$的前$$n$$项和，且对任意的正整数$$n$$，均有$$\\frac{S_{n}}{T_{n}}= \\frac{2n+6}{n+1}$$．若$$\\frac{a_{m}}{b_{m}}$$ 为素数，则正整数$$m$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["由条件设$$S_n=kn(2n+6)$$，$$T_n=kn(n+1)$$． 当$$m=1$$时，$$\\frac{a_{1}}{b_{1}}= \\frac{S_{1}}{T_{1}}= \\frac{8}{2}=4$$ ，不满足题中条件，舍去． 当$$m\\geqslant 2$$时， $$\\frac{a_{m}}{b_{m}}= \\frac{S_{m}-S_{m-1}}{T_{m}-T_{m-1}}= \\frac{4m+4}{2m}=2+ \\frac{2}{m}$$． 于是，仅在$$m=2$$时， $$\\frac{a_{m}}{b_{m}}=3$$ 为素数． 因此，所求正整数$$m=2$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "849", "queId": "735af1b5f05c4b1f8c38e793bd23f7cf", "competition_source_list": ["2007年全国全国高中数学联赛竞赛一试第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设函数$$f(x)=3\\sin x+2\\cos x+1$$．若实数$$a$$、$$b$$、$$c$$使得$$af(x)+bf(x-c)=1$$对任意实数$$x$$恒成立，则$$\\frac{b\\cos c}{a}$$的值等于（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "−1 "}], [{"aoVal": "D", "content": "1 "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角恒等变换->和差角公式->辅助角公式", "课内体系->知识点->三角函数->三角恒等变换->倍角、和差角公式综合", "课内体系->素养->数学运算"], "answer_analysis": ["令$$c= \\pi $$，则对任意的$$x\\in \\mathbf{R}$$，都有$$f(x)+f(x-c)=2$$，于是取$$a=b=\\frac{1}{2}$$，$$c= \\pi $$，则对任意的$$x\\in \\mathbf{R}$$，$$af(x)+bf(x-c)=1$$，由此得$$\\frac{b\\cos c}{a}=-1$$． 一般地，由题设可得$$f(x)=\\sqrt{13}\\sin (x+\\phi )+1$$，$$f(x-c)=\\sqrt{13}\\sin (x+\\phi -c)+1$$，其中$$0 ~\\textless{} ~\\phi ~~\\textless{} ~\\frac{\\pi }{2}$$且$$\\tan \\phi =\\frac{2}{3}$$，于是$$af(x)+bf(x-c)=1$$可化为 $$\\sqrt{13}a\\sin (x+\\phi )+\\sqrt{13}b\\sin (x+\\phi -c)+a+b=1$$，即 $$\\sqrt{13}a\\sin (x+\\phi )+\\sqrt{13}b\\sin (x+\\phi )\\cos c-\\sqrt{13}b\\sin c\\cos (x+\\phi )+(a+b-1)=0$$，所以 $$\\sqrt{13}(a+b\\cos c)\\sin (x+\\phi )-\\sqrt{13}b\\sin c\\cos (x+\\phi )+(a+b-1)=0$$． 由已知条件，上式对任意$$x$$∈$$R$$恒成立，故必有$$\\begin{cases} a+b\\cos c=0 （1） b\\sin c=0 （2） a+b-1=0 （3） \\end{cases}$$， 若$$b$$=0，则由（$$1$$）知$$a$$=0，显然不满足（$$3$$）式，故$$b\\ne 0$$．所以，由（$$2$$）知$$\\sin c$$=0，故$$c=2k \\pi + \\pi $$或$$c=2k \\pi $$（$$k\\in \\mathbf{Z}$$）．当$$c=2k \\pi $$时，$$\\cos c=1$$，则（$$1$$）、（$$3$$）两式矛盾．故$$c=2k \\pi + \\pi $$（$$k\\in \\mathbf{Z}$$），$$\\cos c$$ =−1．由（$$1$$）、（$$3$$）知$$a=b=\\frac{1}{2}$$，所以$$\\frac{b\\cos c}{a}=-1$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "456", "queId": "70d050d99db249dfabc570918cb2c529", "competition_source_list": ["2005年全国高中数学联赛竞赛一试第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "空间四点$$A$$、$$B$$、$$C$$、$$D$$满足$$\\textbar\\overrightarrow{AB}\\textbar=3$$，$$\\textbar\\overrightarrow{BC}\\textbar=7$$，$$\\textbar\\overrghtarrow{CD}\\textbar=11$$，$$\\textbar\\overrightarrow{DA}\\textbar=9$$，则$$\\overrightarrow{AC}\\cdot \\overrightarrow{BD}$$=~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "只有一个 "}], [{"aoVal": "B", "content": "有二个 "}], [{"aoVal": "C", "content": "有四个 "}], [{"aoVal": "D", "content": "有无穷多个 "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间向量"], "answer_analysis": ["注意到$${{3}^{2}}+{{11}^{2}}=1130={{7}^{2}}+{{9}^{2}}$$，由于$$\\overrightarrow{AB}+\\overrightarrow{BC}+\\overrightarrow{CD}+\\overrightarrow{DA}=\\vec{0}$$， 则$$D{{A}^{2}}={{\\overrightarrow{DA}}^{2}}={{(\\overrightarrow{AB}+\\overrightarrow{BC}+\\overrightarrow{CD})}^{2}}=A{{B}^{2}}+B{{C}^{2}}+C{{D}^{2}}+2(\\overrightarrow{AB}\\cdot \\overrightarrow{BC}+\\overrightarrow{BC}\\cdot \\overrightarrow{CD}+\\overrightarrow{CD}\\cdot \\overrightarrow{AB})$$ $$=A{{B}^{2}}-B{{C}^{2}}+C{{D}^{2}}+2({{\\overline{BC}}^{2}}+\\overrightarrow{AB}\\cdot \\overrightarrow{BC}+\\overrightarrow{BC}\\cdot \\overrightarrow{CD}+\\overrightarrow{CD}\\cdot \\overrightarrow{AB})$$ $$=A{{B}^{2}}-B{{C}^{2}}+C{{D}^{2}}+2(\\overrightarrow{AB}+\\overrightarrow{BC})\\cdot (\\overrightarrow{BC}+\\overrightarrow{CD})$$， 即$$2\\overrightarrow{AC}\\cdot \\overrightarrow{BD}=A{{D}^{2}}+B{{C}^{2}}-A{{B}^{2}}-C{{D}^{2}}=0$$，∴$$\\overrightarrow{AC}\\cdot \\overrightarrow{BD}$$只有一个值得$$0$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "584", "queId": "b0d881d495034013a0d96ed2206151eb", "competition_source_list": ["2017年福建全国高中数学联赛竞赛初赛第9题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)=\\sqrt{2x-7}+\\sqrt{12-x}+\\sqrt{44-x}$$的最大值为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->柯西", "课内体系->知识点->等式与不等式->不等式->柯西不等式"], "answer_analysis": ["由柯西不等式，$${{\\left( f\\left( x \\right) \\right)}^{2}}\\leqslant \\left( 3+2+6 \\right)\\left( \\frac{2x-7}{3}+\\frac{12-x}{2}+\\frac{44-x}{6} \\right)={{11}^{2}}$$，所以$$f\\left( x \\right)\\leqslant 11$$，当且仅当$$x=8$$时等号成立． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "117", "queId": "6b1ffd3ce8e24f09a6154f1eb190b293", "competition_source_list": ["2009年AMC10竞赛A第19题"], "difficulty": "3", "qtype": "single_choice", "problem": "圆$$A$$的半径为$$100$$.圆$$B$$的半径为整数$$r\\textless100$$，当它绕着圆$$A$$的圆周滚动一次时，仍然与圆$$A$$保持内切．这两个圆在$$B$$的行程的起点和终点有相同的切点.$$r$$可能有多少个值？", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$50$$ "}], [{"aoVal": "E", "content": "$$90$$ "}]], "knowledge_point_routes": ["课内体系->知识点->圆锥曲线", "美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Circle"], "answer_analysis": ["The circumference of circle $$A$$ is $$200\\pi$$ , and the circumference of circle $$B$$ with radius r is $$2r\\pi$$. Since circle $$B$$ makes a complete revolution and ends up on the same point, the circumference of $$A$$ must be a multiple of the circumference of $$B$$, therefore the quotient must be an integer. Thus, $$\\dfrac{200 \\pi}{2 \\pi \\cdot r}= \\dfrac{100}{r}$$. Therefore $$r$$ must then be a factor of $$100$$, excluding $$100$$ (because then circle $$B$$ would be the same size as circle $$A$$). $$100 =2^{2}\\cdot5^{2}$$. Therefore $$100$$ has $$(2+1)\\cdot (2+1)$$ factors*. But you need to subtract $$1$$ from $$9$$, in order to exclude $$100$$. Therefore the answer is $$8$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1023", "queId": "fff26bac954b41ec9c27cfc4969f79c5", "competition_source_list": ["2008年山东全国高中数学联赛竞赛初赛第17题12分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$x\\textgreater0,y\\textgreater0,z\\textgreater0$$，且$$xyz=1$$，则$$\\frac{1}{1+x}+\\frac{1}{1+y}+\\frac{1}{1+z}$$可以取的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "前三个答案都不对 "}]], "knowledge_point_routes": ["竞��->知识点->不等式->换元技巧->代数换元", "竞赛->知识点->不等式->不等式的证明", "课内体系->方法->换元法"], "answer_analysis": ["任取$$a\\textgreater0$$，令$$b=ax，c=by$$，由$$xyz=1$$ 得$$x=\\frac{b}{a}，y=\\frac{c}{b}，z=\\frac{a}{c}$$， 从而有$$\\frac{1}{1+x}+\\frac{1}{1+y}+\\frac{1}{1+z}=\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{a+c}$$ $$\\textgreater\\frac{a}{a+b+c}+\\frac{b}{a+b+c}+\\frac{c}{a+b+c}=1$$， 又 $$\\frac{a}{a+b}+\\frac{b}{b+c}+\\frac{c}{a+c} ~\\textless{} ~\\frac{a+c}{a+b+c}+\\frac{a+b}{a+b+c}+\\frac{b+c}{a+b+c}=2$$， 所以 $$1 ~\\textless{} ~\\frac{1}{1+x}+\\frac{1}{1+y}+\\frac{1}{1+z} ~\\textless{} ~2$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "755", "queId": "5382c0080e644477a62ee14e9b058d6e", "competition_source_list": ["2011年天津全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$x\\in \\left( 0,\\frac{ \\pi }{2} \\right)$$时总有$$\\sin x\\textgreater kx$$成立，则实数$$k$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( -\\infty ,\\frac{\\pi }{2} \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( -\\infty ,\\left. \\frac{ \\pi }{2} \\right] \\right.$$ "}], [{"aoVal": "C", "content": "$$\\left( -\\infty ,\\frac{2}{\\pi } \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( -\\infty ,\\left. \\frac{2}{ \\pi } \\right] \\right.$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["作出$$y=\\sin x$$和$$y=kx$$的图象，易知选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "180", "queId": "6b45aa62a917456fa82f93f3f02458a2", "competition_source_list": ["2007年全国全国高中数学联赛竞赛一试第6题6分", "2019~2020学年上海浦东新区上海市建平中学高一上学期期中第16题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$A$$与$$B$$是集合{$$1$$，$$2$$，$$3$$，\\ldots，$$100$$}的两个子集，满足：$$A$$与$$B$$的元素个数相同，且$$A\\cap B$$为空集．若$$n\\in A$$时总有$$2n+2\\in B$$，则集合$$A\\cup B$$的元素个数最多为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$62$$ "}], [{"aoVal": "B", "content": "$$66$$ "}], [{"aoVal": "C", "content": "$$68$$ "}], [{"aoVal": "D", "content": "$$74$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["先证$$\\textbar A\\cup B\\textbar\\leqslant 66$$，只须证$$\\left\\textbar{} A \\right\\textbar\\leqslant 33$$，为此只须证若$$A$$是{$$1$$，$$2$$，\\ldots，$$49$$}的任一个$$34$$元子集，则必存在$$n\\in A$$，使得$$2n+2\\in B$$，证明如下： 将{$$1$$，$$2$$，\\ldots，$$49$$}分成如下$$33$$个集合：{$$1$$，$$4$$}，{$$3$$，$$8$$}，{$$5$$，$$12$$}，\\ldots，{$$23$$，$$48$$}共$$12$$个；{$$2$$，$$6$$}，{$$10$$，$$22$$}，{$$14$$，$$30$$}，{$$18$$，$$38$$}共$$4$$个；{$$25$$}，{$$27$$}，{$$29$$}，\\ldots，{$$49$$}共$$13$$个；{$$26$$}，{$$34$$}，{$$42$$}，{$$46$$}共$$4$$个．由于$$A$$是{$$1$$，$$2$$，\\ldots，$$49$$}的$$34$$元子集，从而由抽屉原理可知上述$$33$$个集合中至少有一个$$2$$元集合中的数均属于$$A$$，即存在$$n\\in A$$，使得$$2n+2\\in B$$， 如取$$A$$={$$1$$，$$3$$，$$5$$，\\ldots，$$23$$，$$2$$，$$10$$，$$14$$，$$18$$，$$25$$，$$27$$，$$29$$，\\ldots，$$49$$，$$26$$，$$34$$，$$42$$，$$46$$}，$$B$$=$$ {2n+2\\textbar n\\in A }$$，则$$AB$$满足题设且$$\\textbar A\\cup B\\textbar\\leqslant 66$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "369", "queId": "4bd140ebd0e34989be770f055867ef28", "competition_source_list": ["2011年辽宁全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "一个盒子里有$$3$$个黑球和$$4$$个白球，现从盒子里随机每次取出一个球，取出后不再放回，每个球被取出的可能性相等．直到某种颜色的球全部被取出．最后取出的是黑球的概率是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{7}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{7}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->概率初步"], "answer_analysis": ["$$7$$个球全被取出的方式共有$$\\text{C}_{7}^{3}=\\text{C}_{7}^{4}=35$$（种），$$3$$个黑球全部被取出的方式有$$\\text{C}_{6}^{3}=20$$（种），故所求概率为$$\\frac{20}{35}=\\frac{4}{7}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "704", "queId": "4e98822b117e4b719119baf3c52fd11a", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$m, n$$为实数，且直线$$mx+ny=4$$和圆$${{x}^{2}}+{{y}^{2}}=4$$没有公共点，则关于$$x$$的方程$${{x}^{2}}+2mx+{{n}^{2}}=0$$有实根的概率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{16}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆综合"], "answer_analysis": ["直线$$mx+ny=4$$和圆$${{x}^{2}}+{{y}^{2}}=4$$相离，则$$\\frac{\\left\\textbar{} 4 \\right\\textbar}{\\sqrt{{{m}^{2}}+{{n}^{2}}}}\\textgreater2$$，得$${{m}^{2}}+{{n}^{2}}\\textless{}4$$． 方程$${{x}^{2}}+2mx+{{n}^{2}}=0$$有实根时，判别式$$\\Delta =4{{m}^{2}}-4{{n}^{2}}\\geqslant 0\\Rightarrow \\left\\textbar{} m \\right\\textbar\\geqslant \\left\\textbar{} n \\right\\textbar$$． 所以有实根的概率为$$\\frac{1}{2}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "664", "queId": "ff80808145933b5101459341d737002c", "competition_source_list": ["2016~2017学年10月北京西城区北京市第十四中学高三上学期月考理科第1题5分", "2014年北京海淀区高三一模理科第1题", "2016年山西太原高三三模文科第1题5分", "2014年黑龙江全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知集合$$A=\\left { 1,2,\\frac{1}{2} \\right }$$，集合$$B=\\left { y\\left\\textbar{} y={{x}^{2}},x\\in A \\right. \\right }$$，则$$A\\cap B=$$（~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left { \\frac{1}{2}\\right }$$ "}], [{"aoVal": "B", "content": "$$\\left { 2 \\right }$$ "}], [{"aoVal": "C", "content": "$$\\left { 1 \\right }$$ "}], [{"aoVal": "D", "content": "$$\\varnothing $$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->集合->集合的基本运算->交集", "课内体系->知识点->集合->集合的基本运算->交、并、补集混合运算"], "answer_analysis": ["由题可知：$$B=\\left { 1,4,\\frac{1}{4} \\right }$$，所以 $$A\\cap B=\\left {1 \\right }$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "480", "queId": "f5f499d4824f43d78fd3851f4f80c44d", "competition_source_list": ["2016年北京海淀区中国人民大学附属中学高三零模理科第6题5分", "2014年吉林全国高中数学联赛竞赛初赛第4题6分", "2016~2017学年2月湖南长沙岳麓区湖南师范大学附属中学高三上学期月考理科第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "某地举行一次民歌大奖赛，六个省各有一对歌手参加决赛，现要选出$$4$$名优胜者，则选出的$$4$$名选手中有且只有两个人是同一省份的歌手的概率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{16}{33}$$ "}], [{"aoVal": "B", "content": "$$\\frac{33}{128}$$ "}], [{"aoVal": "C", "content": "$$\\frac{32}{33}$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{11}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->知识点->统计与概率->概率->事件与概率->古典概型->古典概型的概率计算（涉及计数原理）"], "answer_analysis": ["由题意知本题是一个古典概型， 试验发生的总事件是从$$12$$名选手中选出$$4$$个优胜者，共有$$\\text{C}_{12}^{4}$$种结果， 而满足条件的是选出的$$4$$名选手中恰有且只有两个人 是同一省份的歌手表示从$$6$$个省中选一个省， 它的两名选手都获奖，同时从余下的$$10$$名选手中选一个， 再从剩下的$$4$$个省中选一个，共有$$\\text{C}_{6}^{1}\\text{C}_{10}^{1}\\text{C}_{4}^{1}$$种选法， ∴$$P=\\frac{\\text{C}_{6}^{1}\\text{C}_{10}^{1}\\text{C}_{4}^{1}}{\\text{C}_{12}^{4}}=\\frac{16}{33}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "322", "queId": "4b9bce326bff47af9dd7ef863ca82c7e", "competition_source_list": ["2015年吉林全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "(★★)已知$$f(x)=x\\left\\textbar{} x \\right\\textbar$$，若对任意的$$x\\geqslant 1$$有$$f(x+m)+mf(x)\\textless{}0$$恒成立，则实数$$m$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\infty ,-1)$$ "}], [{"aoVal": "B", "content": "$$(-\\infty ,-1]$$ "}], [{"aoVal": "C", "content": "$$(-\\infty ,-2)$$ "}], [{"aoVal": "D", "content": "$$(-\\infty ,-2]$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的性质->单调性->函数单调性的应用->利用函数单调性解不等式", "竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["显然$$m\\textless{}0$$，所以$$f\\left( x+m \\right)\\textless{}-mf\\left( x \\right)=f\\left( \\sqrt{-m}x \\right)$$． 因为$$f\\left( x \\right)$$是单调增的奇函数，所以$$x+m\\textless{}\\sqrt{-m}x$$，即$$\\left( \\sqrt{-m}-1 \\right)x\\textgreater m$$． 所以必须$$\\sqrt{-m}-1\\geqslant 0$$，$$m\\leqslant -1$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1147", "queId": "cf9c7765077f4560a4d082c8f67b09c7", "competition_source_list": ["2011年AMC10竞赛A第20题"], "difficulty": "3", "qtype": "single_choice", "problem": "在半径为$r$的圆周上独立随机地选择两点．从每一个点沿顺时针方向画出一条长为$r$的弦．这两条弦相交的概率是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\dfrac{1}{6}$$ "}], [{"aoVal": "B", "content": "$$\\dfrac{1}{5}$$ "}], [{"aoVal": "C", "content": "$$\\dfrac{1}{4}$$ "}], [{"aoVal": "D", "content": "$$\\dfrac{1}{3}$$ "}], [{"aoVal": "E", "content": "$$\\dfrac{1}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率->事件与概率->几何概型->与角度有关的几何概率的计算", "课内体系->知识点->直线和圆的方程->圆与方程->直线与圆的位置关系->圆的弦长的相关问题", "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Geometric Models of Probabilities"], "answer_analysis": ["Fix a point $$A$$ from which we draw a clockwise chord. In order for the clockwise chord from another point $$B$$ to intersect that of point $$A$$, $$A$$ and $$B$$ must be no more than $$r$$ units apart.By drawing the circle, we quickly see that $$B$$ can be on $$\\dfrac{120}{360}=\\dfrac{1}{3}(\\rm D)$$ of the perimeter of the crcle. (lmagine a regular hexagon inscribed in the circle). "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1208", "queId": "f9eb5503b2b14a3ba6705d886ad70dc5", "competition_source_list": ["2022年浙江宁波竞赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$a,b\\in R$，则``$a\\textgreater b$''是``$\\left\\textbar{} a \\right\\textbar\\textgreater b$''的（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "充分不必要条件 "}], [{"aoVal": "B", "content": "必要不充分条件 "}], [{"aoVal": "C", "content": "充要条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 判断条件间的推出关系，根据充分必要性的定义判断即可.\\\\ 【详解】\\\\ 当$a\\textgreater b$：\\\\ 若$a,b$异号，即$a\\textgreater0\\textgreater b$，显然$\\left\\textbar{} a \\right\\textbar\\textgreater b$成立；\\\\ 若$a\\textgreater b\\ge 0$或$0\\ge a\\textgreater b$，均有$\\left\\textbar{} a \\right\\textbar\\textgreater b$成立；\\\\ 所以充分性成立；\\\\ 当$\\left\\textbar{} a \\right\\textbar\\textgreater b$：若$a=-2$，$b=1$，显然$a\\textgreater b$不成立，故必要性不成立.\\\\ 所以``$a\\textgreater b$''是``$\\left\\textbar{} a \\right\\textbar\\textgreater b$''的充分不必要条件.\\\\ 故选：A "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "311", "queId": "1ae5ee1c26ba46e3a1c6d108a7110f9d", "competition_source_list": ["2015年吉林全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f(x)=4{{x}^{3}}-3x$$在$$(a,a+2)$$上存在最大值，则实数$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\frac{5}{2},-1)$$ "}], [{"aoVal": "B", "content": "$$(-\\frac{5}{2},-1]$$ "}], [{"aoVal": "C", "content": "$$(-\\frac{5}{2},-\\frac{1}{2})$$ "}], [{"aoVal": "D", "content": "$$(-\\frac{5}{2},-\\frac{1}{2}]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->导数"], "answer_analysis": ["$${f}'\\left( x \\right)=12{{x}^{2}}-3=3\\left( 2x+1 \\right)\\left( 2x-1 \\right)$$，$$f\\left( x \\right)$$在$$\\left( -\\infty ,-\\frac{1}{2} \\right)$$单调增，在$$\\left( -\\frac{1}{2},\\frac{1}{2} \\right)$$单调减，在$$\\left( \\frac{1}{2},+\\infty \\right)$$单调增，所以$$f\\left( x \\right)$$在$$\\left( a,a+2 \\right)$$上的最大值只能是$$f\\left( -\\frac{1}{2} \\right)=1$$． 因此，$$a\\textless{}-\\frac{1}{2}\\textless{}a+2$$，另外$$f\\left( 1 \\right)=1$$，所以$$a+2\\leqslant 1$$，综合有$$-\\frac{5}{2}\\textless{}a\\leqslant -1$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "736", "queId": "53494af0bb9045a792783054347cd6bd", "competition_source_list": ["2016年AMC10竞赛第23题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某運算◇對所有的非零實數$$a$$、$$b$$、$$c$$滿足：$$a$$◇$$(b$$ ◇$$c)=(a$$◇$$b)\\cdot c$$(此處 $$\\cdot$$ 是一般的乘號)，且 $$a$$◇$$a=1$$ ．若方程式$$2016$$◇$$(6$$◇$$x)=100$$ 的解爲$$\\frac pq$$,其中$$p$$、$$q$$爲互質的正整數，則 $$p+q$$之値爲何？ A binary operation ◇ has the properties that $$a$$◇$$(b$$ ◇$$c)=(a$$◇$$b)\\cdot c$$ and that $$a$$◇$$a=1$$ for all nonzero real numbers $$a$$, $$b$$, and $$c$$. (Here represents multiplication). The solution to the equation $$2016$$◇$$(6$$◇$$x)=100$$ can be written as $$\\frac pq$$, where $$p$$ and $$q$$ are relatively prime positive integers. What is $$p+q$$?+q?", "answer_option_list": [[{"aoVal": "A", "content": "$$109$$ "}], [{"aoVal": "B", "content": "$$201$$ "}], [{"aoVal": "C", "content": "$$301$$ "}], [{"aoVal": "D", "content": "$$3049$$ "}], [{"aoVal": "E", "content": "33,601 "}]], "knowledge_point_routes": [], "answer_analysis": ["We see that $$a$$◇$$a=1$$, and think of division. Testing, we see that the first condition $$a$$◇$$(b$$◇$$c)=(a$$◇$$b)\\cdot c$$ is satisfied, because $$\\frac {a}{\\dfrac bc}=\\frac ab\\cdot c$$. Therefore, division can be the operation ◇. Solving the equation, $$\\frac {2016}{\\dfrac 6x}=\\frac {2016}6\\cdot x=336x=100\\Rightarrow x=\\frac {100}{336}=\\frac {25}{84}$$, so the answer is $$25+84=109$$. We can manipulate the given identities to arrive at a conclusion about the binary operator ◇. Substituting $$b=c$$ into the first identity yields $$(a$$◇$$b)\\cdot b=a$$◇$$(b$$◇$$b)=a$$◇$$1=a$$◇$$(a$$◇$$a)=(a$$◇$$a)\\cdot a=a$$. Hence, $$(a$$◇$$b)\\cdot b=a$$, or, dividing both sides of the equation by $$b$$$, $$(a$$◇$$b)=\\frac ab$$. Hence, the given equation becomes $$\\frac {2016}{\\dfrac 6x}=100$$. Solving yields $$x=\\frac {100}{336}=\\frac {25}{84}$$, so the answer is $$25+84=109$$. One way to eliminate the ◇ in this equation is to make $$a=b$$ so that $$a$$◇$$(b$$◇$$c)=c$$. In this case, we can make $$b=2016$$. $$2016$$◇$$(6$$◇$$x)=100\\Rightarrow (2016$$◇$$6)\\cdot x=100$$ By multiplying both sides by $$x$$, we get: $$(2016$$◇$$6)\\cdot 6=\\frac {600}x\\Rightarrow 2016$$◇$$(6$$◇$$6)=\\frac {600}x$$ Because $$6$$◇$$6=2016$$◇$$2016=1$$: $$2016$$◇$$(2016$$◇$$2016)=\\frac {600}x\\Rightarrow (2016$$◇$$2016)\\cdot 2016=\\frac {600}x\\Rightarrow 2016=\\frac {600}x$$ Therefore, $$x=\\frac {600}{2016}=\\frac {25}{84}$$, so the answer is $$25+84=109$$. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "999", "queId": "ee08a7610f984e638ba96461883ab75c", "competition_source_list": ["2014年吉林全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "若五项的数列$$\\left { {{a}_{n}} \\right }$$：$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$${{a}_{4}}$$，$${{a}_{5}}$$满足$$0\\leqslant {{a}_{1}}\\textless{}{{a}_{2}}\\textless{}{{a}_{3}}\\textless{}{{a}_{4}}\\textless{}{{a}_{5}}$$，且对任意的$$i$$，$$j$$（$$1\\leqslant i\\leqslant j\\leqslant 5$$），均有$${{a}_{j}}-{{a}_{i}}$$在该数列中． ①$${{a}_{1}}=0$$； ②$${{a}_{5}}=4{{a}_{2}}$$； ③$$\\left { {{a}_{n}} \\right }$$为等差数列； ④集合$$A=\\left { {{a}_{i}}+{{a}_{j}}\\left\\textbar{} 1\\leqslant i\\leqslant j\\leqslant 5 \\right. \\right }$$含$$9$$个元素． 则上述论断正确的有个．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["论断正确的有①②③④． 因为$${{a}_{1}}-{{a}_{1}}=0\\in \\left { {{a}_{n}} \\right }$$，所以$${{a}_{1}}=0$$； 因为 $$a={{a}_{1}}\\textless{}{{a}_{3}}-{{a}_{2}}\\textless{}{{a}_{4}}-{{a}_{2}}\\textless{}{{a}_{5}}-{{a}_{2}}\\textless{}{{a}_{5}}$$， 且 $${{a}_{3}}-{{a}_{2}}$$，$${{a}_{4}}-{{a}_{2}}$$，$${{a}_{5}}-{{a}_{2}}\\in \\left { {{a}_{n}} \\right }$$， 所以 $${{a}_{3}}-{{a}_{2}}={{a}_{2}}$$，$${{a}_{4}}-{{a}_{2}}={{a}_{3}}$$，$${{a}_{5}}-{{a}_{2}}={{a}_{4}}$$； 于是 $${{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}={{a}_{4}}-{{a}_{3}}={{a}_{5}}-{{a}_{4}}$$． 所以$$\\left { {{a}_{n}} \\right }$$为等差数列，且$$\\left { {{a}_{n}} \\right }$$：$$0$$,d,2d,3d,4d，因此$${{a}_{5}}=4{{a}_{2}}$$；集合$$A=\\left { {{a}_{i}}+{{a}_{j}}\\left\\textbar{} 1\\leqslant i\\leqslant j\\leqslant 5 \\right. \\right }$$含$$9$$个元素． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "985", "queId": "bbcbfc617c83423d867ebca871031ccc", "competition_source_list": ["2009年新疆全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$f(x)$$的定义在实数上的函数，$$f(3)=-\\sqrt{3}$$，且$$f(x+2)\\left[ 1-f(x) \\right]=1+f(x)$$，则$$f(2009)$$=．", "answer_option_list": [[{"aoVal": "A", "content": "$$2+\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$2-\\sqrt{3}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$-\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->函数综合"], "answer_analysis": ["当$$x=1$$时， $$f(3)\\left[ 1-f(x) \\right]=1+f(1)$$ 所以$$-\\sqrt{3}\\left[ 1-f(1) \\right]=1+f(1)$$ 所以$$f(1)=\\frac{1+\\sqrt{3}}{\\sqrt{3}-1}=2+\\sqrt{3}$$ 又因为$$f(x+2)=\\frac{1+f(x)}{1-f(x)}$$ 所以$$f(x+4)=\\frac{1+f(x+2)}{1-f(x+2)}$$ $$=\\frac{1+\\frac{1+f(x)}{1-f(x)}}{1-\\frac{1+f(x)}{1-f(x)}} $$ $$=\\frac{2}{-2f(x)}$$ $$=-\\frac{1}{f(x) }$$ 所以$$f(x+8)=-\\frac{1}{f(x+4)}=-\\frac{1}{-\\frac{1}{f(x)}}=f(x)$$ 所以$$f(2009)=f(251\\times 8+1)=f(1)=2+\\sqrt{3}$$ 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "665", "queId": "455c5891e04a4ec991833679f6853267", "competition_source_list": ["2021年吉林全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知数列$$\\left { {{a}_{n}} \\right }$$的通项公式为$${{a}_{n}}=\\frac{2n-17}{2n-19}\\left( n=1,2,\\cdots \\right)$$，则$$\\left { {{a}_{n}} \\right }$$中最大的项是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{a}_{1}}$$ "}], [{"aoVal": "B", "content": "$${{a}_{9}}$$ "}], [{"aoVal": "C", "content": "$${{a}_{10}}$$ "}], [{"aoVal": "D", "content": "$${{a}_{12}}$$ "}]], "knowledge_point_routes": ["课内体系->思想->函数思想", "课内体系->素养->逻辑推理", "课内体系->素养->数学抽象", "课内体系->方法->分离常数法", "课内体系->知识点->数列->数列的概念->数列的函数特性->数列中最大项与最小项的求解问题", "课内体系->知识点->数列->数列的概念->数列的函数特性->数列单调性问题"], "answer_analysis": ["通项公式$${{a}_{n}}=1+\\frac{1}{n-9.5}$$， 当$$n\\leqslant 9$$时，有$${{a}_{n}}\\textless{}1$$， 当$$n\\geqslant 10$$时，有$${{a}_{n}}\\textgreater1$$，且$$\\left { {{a}_{n}} \\right }$$递减，从而$$\\left { {{a}_{n}} \\right }$$中最大的项是$${{a}_{10}}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1028", "queId": "8f447fef9546406fb2b6a038df3ff75f", "competition_source_list": ["2010年AMC12竞赛B第16题"], "difficulty": "2", "qtype": "single_choice", "problem": "2010AMC12B, 16 Positive integers $$a$$, $$b$$, and $$c$$ are randomly and independently selected with replacement from the set $$ {1,2,3,\\cdots ,2010 }$$. What is the probability that $$abc+ab+a$$ is divisible by $$3$$? 从集合$$ {1,2,3,\\cdots ,2010 }$$中独立随机有放回地选取正整数$$a$$, $$b$$, 和$$c$$。$$abc+ab+a$$被$$3$$整除的概率是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{29}{81}$$ "}], [{"aoVal": "C", "content": "$$\\frac{31}{81}$$ "}], [{"aoVal": "D", "content": "$$\\frac{11}{27}$$ "}], [{"aoVal": "E", "content": "$$\\frac{13}{27}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率->事件与概率->事件的独立性->相互独立事件的概率乘法公式", "美国AMC10/12->Knowledge Point->Number Theory->Aliquot Theory->Divisibility Rules of Certain Numbers (2/3/5/9/11)", "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities"], "answer_analysis": ["首先, 注意到$abc+ab+a=a(bc+b+1)$, 因此当$3\\textbar a$时，有$3\\textbar abc+ab+a$. 然后, 当$3\\nmid a$时, $3\\textbar abc+ab+a\\Leftrightarrow 3\\textbar bc+b+1\\Leftrightarrow b(c+1)\\equiv 2\\pmod 3$. 在模$$3$$的意义下, $(b,c)$共有$$3\\times 3=9$$种可能的取值, 其中只有$(2, 0)$和$(1, 1)$满足$b(c+1)\\equiv 2\\pmod 3$. 因此, 总的概率为$\\frac{1}{3}+\\frac{2}{3}\\times\\frac{2}{9}=\\frac{13}{27}$. "], "answer_value": "E"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "544", "queId": "55fbd1762bb14d72b90edb32df788fcc", "competition_source_list": ["全国高中数学联赛竞赛模拟一试（十五）第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)={{x}^{4}}-{{x}^{2}}-\\frac{1}{{{x}^{2}}}+\\frac{1}{{{x}^{4}}}$$的值域为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ 0,+\\infty \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left[ 1,+\\infty \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left[ 2,+\\infty \\right)$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["课内体��->知识点->函数的概念与性质->函数的概念及其表示->函数的值域", "竞赛->知识点->函数->基本初等函数", "竞赛->知识点->函数->二次函数"], "answer_analysis": ["注意到， $$f\\left( x \\right)={{\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}} \\right)}^{2}}-\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}} \\right)-2$$ $$={{\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}}-\\frac{1}{2} \\right)}^{2}}-\\frac{9}{4}$$． 由于$${{x}^{2}}+\\frac{1}{{{x}^{2}}}\\geqslant 2$$， 从而，所求函数的值域为$$\\left[ 0,+\\infty \\right)$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "900", "queId": "bfd08f555c1e485a9aedf91421aadbd1", "competition_source_list": ["全国高中数学联赛竞赛模拟一试（十五）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)={{x}^{4}}-{{x}^{2}}-\\frac{1}{{{x}^{2}}}+\\frac{1}{{{x}^{4}}}$$的值域为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ 2,+\\infty \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left[ 0,+\\infty \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left[ -2,+\\infty \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left[ 1,+\\infty \\right)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域", "竞赛->知识点->函数->二次函数", "竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["注意到， $$f\\left( x \\right)={{\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}} \\right)}^{2}}-\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}} \\right)-2$$ $$={{\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}}-\\frac{1}{2} \\right)}^{2}}-\\frac{9}{4}$$． 由于$${{x}^{2}}+\\frac{1}{{{x}^{2}}}\\geqslant 2$$， 从而，所求函数的值域为$$\\left[ 0,+\\infty \\right)$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "188", "queId": "1d548a44a8234f548ab99df07aa9da0e", "competition_source_list": ["2017~2018学年浙江温州瓯海区浙江省瓯海中学高三上学期期中理科第9题4分", "2007年全国全国高中数学联赛竞赛一试第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设实数$$a$$使得不等式$$\\textbar2x-a\\textbar+\\textbar3x-2a\\textbar\\geqslant {{a}^{2}}$$对任意实数$$x$$恒成立，则满足条件的$$a$$所组成的集合是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ -\\frac{1}{3}，\\frac{1}{3} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left[ -\\frac{1}{2}，\\frac{1}{2} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left[ -\\frac{1}{4}，\\frac{1}{3} \\right]$$ "}], [{"aoVal": "D", "content": "$$[-3，3]$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->等式与不等式->不等式->解不等式->含绝对值的不等式", "课内体系->知识点->等式与不等式->不等式->解不等式->一元二次不等式"], "answer_analysis": ["令$$x=\\frac{2}{3}a$$，则有$$\\textbar a\\textbar\\leqslant \\frac{1}{3}$$，排除$$\\text{B}$$、$$\\text{D}$$，由对称性排除$$\\text{C}$$，从而只有$$\\text{A}$$正确． 一般地，对$$k\\in \\mathbf{R}$$，令$$x=\\frac{1}{2}ka$$，则原不等式为$$\\textbar a\\textbar\\cdot \\textbar k-1\\textbar+\\frac{3}{2}\\textbar a\\textbar\\cdot \\left\\textbar{} k-\\frac{4}{3} \\right\\textbar\\geqslant \\textbar a{{\\textbar}^{2}}$$，由此易知原不等式等价于$$\\textbar a\\textbar\\leqslant \\textbar k-1\\textbar+\\frac{3}{2}\\left\\textbar{} k-\\frac{4}{3} \\right\\textbar$$，对任意的$$k\\in \\mathbf{R}$$成立．由于 $$\\textbar k-1\\textbar+\\frac{3}{2}\\left\\textbar{} k-\\frac{4}{3} \\right\\textbar=\\begin{cases} \\frac{5}{2}k-3， k\\geqslant \\frac{4}{3} 1-\\frac{1}{2}k， 1\\leqslant k ~\\textless{} ~\\frac{4}{3} 3-\\frac{5}{2}k， k ~\\textless{} ~1 \\end{cases}$$， 所以$$_{k\\in \\mathbf{R}}^{\\min }\\left { \\textbar k-1\\textbar+\\frac{3}{2}\\left\\textbar{} k-\\frac{4}{3} \\right\\textbar{} \\right }=\\frac{1}{3}$$，从而上述不等式等价于$$\\textbar a\\textbar\\leqslant \\frac{1}{3}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "593", "queId": "99dbb27c2db64580b29e119e2d2e24fa", "competition_source_list": ["2009年浙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知椭圆$$\\frac{{{x}^{2}}}{25}+\\frac{{{y}^{2}}}{9}=1$$上一点$$P$$到点$$(4,0)$$距离等于$$4$$，则$$P$$点到直线$$x=-\\frac{25}{4}$$的距离为．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$\\frac{15}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{4}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->椭圆"], "answer_analysis": ["因为$$a=5,b=3$$，则$$c=4$$．于是$$P$$到另一个焦点$$(-4,0)$$的距离等于$$2\\times 5-4=6$$．由于直线$$x=-\\frac{25}{4}$$为椭圆的左准线方程，则$$P$$到直线$$x=-\\frac{25}{4}$$的距离为$$d=\\frac{6}{e}=7.5$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "333", "queId": "2c1c66a3578d4e4985b3afa0e4415587", "competition_source_list": ["1992年全国高中数学联赛竞赛一试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$f\\left( x \\right)$$是定义在实数集$$R$$上的函数，且满足下列关系：$$f\\left( 10+x \\right)=f\\left( 10-x \\right)$$．$$f\\left( 20-x \\right)=-f\\left( 20+x \\right)$$．则$$f\\left( x \\right)$$是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "偶函数，又是周期函数 "}], [{"aoVal": "B", "content": "偶函数，但不是周期函数 "}], [{"aoVal": "C", "content": "奇函数，又是周期函数 "}], [{"aoVal": "D", "content": "奇函数，但不是周期函数 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["由所给第一式得 $$f\\left[ 10+\\left( 10-x \\right) \\right]=f\\left[ 10-\\left( 10-x \\right) \\right]$$， ∴~~~~ $$f\\left( x \\right)=f\\left( 20-x \\right)$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ① 又由所给第二式得 $$f\\left( x \\right)=-f\\left( 20-x \\right)$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ② ∴~~~~ $$\\begin{matrix}f\\left( 40+x \\right)=f\\left[ 20+\\left( 20+x \\right) \\right] =-f\\left( 20+x \\right)=f\\left( x \\right) \\end{matrix}$$ 可见$$f(x)$$是周期函数． 由①，②得 $$f\\left( -x \\right)=f\\left( 20+x \\right)=-f\\left( x \\right)$$， ∴$$f\\left( x \\right)$$是奇函数．因此答案是（$$\\text{C}$$）． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "982", "queId": "97a6d6c9af8340fc847212b408e7c105", "competition_source_list": ["2010年河南全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$f(x)=\\frac{x-1}{x+1}$$，记$${{f}_{1}}(x)=f(x)$$，若$${{f}_{n+1}}(x)=f({{f}_{n}}(x))$$，则$${{f}_{2010}}(x)=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$x$$ "}], [{"aoVal": "B", "content": "$$-\\frac{1}{x}$$ "}], [{"aoVal": "C", "content": "$$\\frac{x-1}{x+1}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1+x}{1-x}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["本题从题目上可预知$${{f}_{n}}(x)$$具有周期性，为了寻找周期，采用赋值法容易计算．记$${{a}_{1}}={{f}_{1}}(2), {{a}_{n}}={{f}_{n}}(2)$$，则$${{a}_{1}}=\\frac{1}{3},{{a}_{2}}=\\frac{\\frac{1}{3}-1}{\\frac{1}{3}+1}=-\\frac{2}{4}=-\\frac{1}{2},{{a}_{3}}=\\frac{-\\frac{1}{2}-1}{-\\frac{1}{2}+1}=-3,$$ $${{a}_{4}}=\\frac{-3-1}{-3+1}=2, {{a}_{5}}=\\frac{2-1}{2+1}=\\frac{1}{3},\\cdots $$． 故可知周期$$T=4$$，所以$${{a}_{2010}}={{f}_{2010}}(2)={{f}_{2}}(2)={{a}_{2}}=-\\frac{1}{2}$$，将$$x=2$$代入可得选项B. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "309", "queId": "46f55a4b882c492d890533d4f11aabd0", "competition_source_list": ["2003年AMC10竞赛B第22题"], "difficulty": "3", "qtype": "single_choice", "problem": "一个钟在半点时响一次，在整点时，几点钟就响几次．例如，在下午$$1$$点有一次报时，在正午$$12$$点和午夜$$12$$点各有$$12$$次报时．从$$2003$$年$$2$$月$$26$$日上午$$11$$点$$15$$分开始，第$$2003$$次报时将在哪一天发生？", "answer_option_list": [[{"aoVal": "A", "content": "March $$8$$ "}], [{"aoVal": "B", "content": "March $$9$$ "}], [{"aoVal": "C", "content": "March $$10$$ "}], [{"aoVal": "D", "content": "March $$20$$ "}], [{"aoVal": "E", "content": "March $$21$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->等式->方程组的解集", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Recurrence and Recursion"], "answer_analysis": ["First, find how many chimes will have already happened before midnight (the beginning of the day) of February $$27$$, $$2003$$. $$13$$ half-hours have passed, and the number of chimes according to the hour is $$1+2+3+\\cdots +12$$. The total number of chimes is $$13+78=91$$. Every day, there will be $$24$$ half-hours and $$2(1+2+3+\\cdots 12)$$ chimes according to the arrow, resulting in $$24+156=180$$ total chimes. On February $$26$$, the number of chimes that still need to occur is $$2003-91=1912$$. $$1912\\div 180=10\\rm R112$$. Rounding up, it is $$11$$ days past February $$26$$, which is March $$9$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "784", "queId": "e41381c012024cd1b4044f09291f86a5", "competition_source_list": ["1997年全国高中数学联赛竞赛一试第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "在平面直角坐标系中，若方程$$m\\left( {{x}^{2}}+{{y}^{2}}+2y+1 \\right)={{\\left( x-2y+3 \\right)}^{2}}$$表示的曲线为椭圆，则$$m$$的取值范围为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$(0,1)$$ "}], [{"aoVal": "B", "content": "$$(1,+\\infty )$$ "}], [{"aoVal": "C", "content": "$$(0,5)$$ "}], [{"aoVal": "D", "content": "$$(5,+\\infty )$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->圆锥曲线->椭圆->椭圆的定义、标准方程->椭圆的标准方程", "课内体系->知识点->直线和圆的方程->直线与方程->平面中的距离->点到直线的距离公式"], "answer_analysis": ["看成是轨迹上点到$$(0,-1)$$的距离与到直线$$x-2y+3=0$$的距离的比： $$\\frac{\\sqrt{{{x}^{2}}+{{\\left( y+1 \\right)}^{2}}}}{\\frac{\\left\\textbar{} x-2y+3 \\right\\textbar}{\\sqrt{{{1}^{2}}+{{\\left( -2 \\right)}^{2}}}}}=\\sqrt{\\frac{5}{m}}\\textless{}1\\Rightarrow m\\textgreater5$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "486", "queId": "365fad487d6c4a0fbcdbad340907824b", "competition_source_list": ["2008年AMC10竞赛B第18题", "2008年AMC12竞赛B第10题"], "difficulty": "3", "qtype": "single_choice", "problem": "$$2008-AMC10B-18$$ Bricklayer Brenda would take $$9$$ hours to build a chimney alone, and bricklayer Brandon would take $$10$$ hours to build it alone. When they work together they talk a lot, and their combined output is decreased by $$10$$ bricks per hour. Working together, they build the chimney in $$5$$ hours. How many bricks are in the chimney? 瓦工布伦达单独建造一个烟囱需要花费 9小时，而瓦工布兰登单独建造它需要花费 10小时。 当他们一起工作时，他们会说很多话，他们的总效率每小时减少 10 块砖。 他们一起工作，在 小时内建造了烟囱。 烟囱里有多少块砖？", "answer_option_list": [[{"aoVal": "A", "content": "$$500$$ "}], [{"aoVal": "B", "content": "$$900$$ "}], [{"aoVal": "C", "content": "$$950$$ "}], [{"aoVal": "D", "content": "$$1000$$ "}], [{"aoVal": "E", "content": "$$1900$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->等式->方程组的解集", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"], "answer_analysis": ["Let $$x$$ be the number of bricks in the chimney, Using $$d=vt$$, we get $$x=\\left( \\frac{x}{9}+\\frac{x}{10}-10 \\right)\\cdot \\left( 5 \\right)$$. Solving for $$x$$, we get $$900\\Rightarrow \\text{B}$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "695", "queId": "45b53d99a0d648099dd19ca782044c9b", "competition_source_list": ["1998年全国高中数学联赛竞赛一试第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设命题$$P$$：关于$$x$$的不等式$${{a}_{1}}{{x}^{2}}+{{b}_{1}}{{x}}+{{c}_{1}}\\textgreater0$$与$${{a}_{2}}{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}\\textgreater0$$的解集相同；命题$$Q$$：$$\\frac{{{a}_{1}}}{{{a}_{2}}}=\\frac{{{b}_{1}}}{{{b}_{2}}}=\\frac{{{c}_{1}}}{{{c}_{2}}}$$． 则命题$$Q$$．", "answer_option_list": [[{"aoVal": "A", "content": "是命题$$P$$的充分必要条件 "}], [{"aoVal": "B", "content": "是命题$$P$$的充分条件但不是必要条件 "}], [{"aoVal": "C", "content": "是命题$$P$$的必要条件但不是充分条件 "}], [{"aoVal": "D", "content": "既不是是命题$$P$$的充分条件也不是命题$$P$$的必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["若两个不等式的解集都是$$\\mathbf{R}$$，否定$$\\text{A}$$、$$\\text{C}$$，若比值为$$-1$$，否定$$\\text{A}$$、$$\\text{B}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1056", "queId": "ee633f00f517474a979807407e6f5502", "competition_source_list": ["2006年上海复旦大学自主招生千分考第16题", "1981年全国高中数学联赛竞赛一试第2题"], "difficulty": "3", "qtype": "single_choice", "problem": "条件甲：$$\\sqrt{1+\\sin \\theta }=a$$．条件乙：$$\\sin \\frac{\\theta }{2}+\\cos \\frac{\\theta }{2}=a$$．则下列（~ ）是正确的．", "answer_option_list": [[{"aoVal": "A", "content": "甲是乙的充分必要条件 "}], [{"aoVal": "B", "content": "甲是乙的必要条件 "}], [{"aoVal": "C", "content": "甲是乙的充分条件 "}], [{"aoVal": "D", "content": "甲不是乙的必要条件，也不是充分条件 "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["因为$$\\sqrt{1+\\sin \\theta }=\\sqrt{{{\\sin }^{2}}\\frac{\\theta }{2}+2\\sin \\frac{\\theta }{2}\\cos \\frac{\\theta }{2}+{{\\cos }^{2}}\\frac{\\theta }{2}}=\\sqrt{{{\\left( \\sin \\frac{\\theta }{2}+\\cos \\frac{\\theta }{2} \\right)}^{2}}}=\\left\\textbar{} \\sin \\frac{\\theta }{2}+\\cos \\frac{\\theta }{2} \\right\\textbar$$ 所以条件甲等价于：$$\\left\\textbar{} \\sin \\frac{\\theta }{2}+\\cos \\frac{\\theta }{2} \\right\\textbar=a$$． 在条件甲中，易知$$a\\geqslant 0$$，而在条件乙中，$$a$$可能取负值． 所以由条件甲推不出乙，由乙也推不出甲， 即甲不是乙的必要条件，也不是充分条件． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "424", "queId": "2cf3496e464740a280b4fdb8c1853a0d", "competition_source_list": ["2014年辽宁全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\triangle ABC$$的三个内角为$$A$$、$$B$$、$$C$$，若$$\\frac{\\sin A+\\sqrt{3}\\cos A}{\\cos A-\\sqrt{3}\\sin A}=\\tan \\frac{7 \\pi }{12}$$，则$$\\sin 2B+2\\cos C$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["因为 $$\\frac{\\sin A+\\sqrt{3}\\cos A}{\\cos A-\\sqrt{3}\\sin A}=\\frac{\\tan A+\\tan \\frac{ \\pi }{3}}{1-\\tan A\\tan \\frac{ \\pi }{3}}=\\tan \\left( A+\\frac{ \\pi }{3} \\right)$$， 故$$A+\\frac{ \\pi }{3}=\\frac{7 \\pi }{12}$$，$$A=\\frac{ \\pi }{4}$$． 由于$$2B+2C=\\frac{3 \\pi }{2}$$，故 $$\\sin 2B+2\\cos C=\\sin \\left( \\frac{3 \\pi }{2}-2C \\right)+2\\cos C$$ $$=-\\cos 2C+2\\cos C$$ $$=1-2{{\\cos }^{2}}C+2\\cos C$$ $$=-2{{\\left( \\cos C-\\frac{1}{2} \\right)}^{2}}+\\frac{3}{2}\\leqslant \\frac{3}{2}$$． 当$$C=\\frac{ \\pi }{3}$$时取等号．故$$\\sin 2B+2\\cos C$$的最大值为$$\\frac{3}{2}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1173", "queId": "eb5553dd1e034f6f853a8fc3fd04f9fe", "competition_source_list": ["全国高中数学联赛竞赛模拟一试（五）第3题"], "difficulty": "0", "qtype": "single_choice", "problem": "从$$0$$，$$1$$，$$\\cdots $$，$$9$$这十个号码中任意抽取三个，其中至少有两个号码是连续整数的概率为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$ \\frac{7}{15}$$ "}], [{"aoVal": "B", "content": "$$ \\frac{8}{15}$$ "}], [{"aoVal": "C", "content": "$$ \\frac{2}{5}$$ "}], [{"aoVal": "D", "content": "$$ \\frac{3}{5}$$ "}], [{"aoVal": "E", "content": "$$ \\frac{1}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->两个基本计数原理", "竞赛->知识点->排列组合与概率->概率初步", "竞赛->知识点->排列组合与概率->排列与组合", "课内体系->知识点->计数原理"], "answer_analysis": ["设抽出的号码$$i$$，$$j$$，$$k$$（$$0\\leqslant i\\textless{}j\\textless{}k\\leqslant 9$$）中任何两个均不相邻， 则$$i$$，$$j-1$$，$$k-2$$互不相邻，且只能从$$ {0,1,\\cdots ,7 }$$中取值， 故所求的概率$$ p=1- \\frac{\\text{C}_{8}^{3}}{\\text{C}_{15}^{3}}= \\frac{8}{15}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "655", "queId": "a7ef7dfc33c348c580baa65ba85126f1", "competition_source_list": ["2008年山东全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知函数$$y=\\frac{x-a}{x-a-1}$$的反函数的图象关于点$$\\left( -1,3 \\right)$$成中心对称图形，则实数$$a$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$-2$$ "}], [{"aoVal": "D", "content": "$$-3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["方法一：在原函数图象上取点$$\\left( 1,\\frac{1}{a}-1 \\right)$$， 则点$$\\left( \\frac{1}{a}-1,1 \\right)$$在其反函数图象上， 它关于点$$(-13)$$的对称点为$$\\left( -1-\\frac{1}{a},5 \\right)$$，从而点$$\\left( 5,-1-\\frac{1}{a} \\right)$$在原函数图象上， 所以有$$-1-\\frac{1}{a}=\\frac{5-a}{5-a-1}$$， 解得$$a=2$$．故选$$\\text{A}$$． 方法二：因函数$$y=-\\frac{x-a}{x-a-1}$$与其反函数的图象关于直线$$y=x$$对称， 故它们的对称中心必也关于$$y=x$$对称． 可知点$$\\left( -1,3 \\right)$$关于直线$$y=x$$的对称点为$$(3-1)$$． 又因为函数$$y=-\\frac{x-a}{x-a-1}$$，即$$y=-1+\\frac{1}{x-(a+1)}$$， 图象关于点$$(a+1,-1)$$成中心对称图���，故$$a+1=3$$，得$$a=2$$．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "195", "queId": "d076cf8331204114be69026b3d210b67", "competition_source_list": ["2010年黑龙江全国高中数学联赛竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "若把函数$$y=\\sqrt{3}\\cos x-\\sin x$$的图象向右平移$$m$$（$$m\\textgreater0$$）个单位长度后，所得到的图象关于$$y$$轴对称，则$$m$$的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{ \\pi }{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3} \\pi $$ "}], [{"aoVal": "C", "content": "$$\\frac{ \\pi }{6}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{6} \\pi $$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["$$y=\\sqrt{3}\\cos x-\\sin x=2\\cos \\left( x+\\frac{ \\pi }{6} \\right),$$对称轴方程$$x=k \\pi -\\frac{ \\pi }{6}, k\\in Z$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "5", "queId": "0959c4ee5706421a878d77b9fe75c8ab", "competition_source_list": ["2002年全国高中数学联赛竞赛一试第2题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "若实数$$x$$，$$y$$满足$${{\\left( x+5 \\right)}^{2}}+{{\\left( y+12 \\right)}^{2}}={{14}^{2}}$$，则$${{x}^{2}}+{{y}^{2}}$$的最小值为（~ ~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->基本不等式->利用基本不等式求最值", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的概念", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件", "课内体系->素养->数学运算"], "answer_analysis": ["令$$x+5=14\\cos \\theta $$,$$y-12=14\\sin \\theta $$ ,则$${{x}^{2}}+{{y}^{2}}=196+28\\left( 5\\cos \\theta -12\\sin \\theta \\right)+169=365+364\\sin \\left( \\theta +\\varphi \\right)\\geqslant 1$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "104", "queId": "fa328434c6d745de9d1f367639a116ad", "competition_source_list": ["2017年山西全国高中数学联赛竞赛初赛第8题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$M=\\left { 1,2,\\cdots ,2017 \\right }$$是前$$2017$$个正整数构成的集合，若从$$M$$中去掉一个元素后，$$M$$中剩下的元素之和恰为一个平方数，则去掉的元素是 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$65$$ "}], [{"aoVal": "C", "content": "$$1008$$ "}], [{"aoVal": "D", "content": "$$1677$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和", "竞赛->知识点->集合->集合的划分与覆盖"], "answer_analysis": ["$$S=1+2+\\cdots +2017=\\frac{2018\\cdot 2017}{2}=1009\\cdot 2017=2035153\\in \\left( {{1400}^{2}},{{1500}^{2}} \\right)$$， 而$${{1450}^{2}}={{\\left( 1400+50 \\right)}^{2}}=1960000+140000+2500=2102500\\textgreater S$$， 又$${{1425}^{2}}={{\\left( 1400+25 \\right)}^{2}}=1960000+70000+625=2030625\\textless{}S$$， $${{1426}^{2}}={{\\left( 1425+1 \\right)}^{2}}=2030625+2850+1=2033476$$， 所以$$S={{1426}^{2}}+1677$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "537", "queId": "75d1bef60e4c4b9495c0dd39cb48151d", "competition_source_list": ["2016年AMC10竞赛第16题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某個三角形的頂點座標爲 $$A(0,2)$$、 $$B(-3,2)$$、$$C(-3,0)$$ ，將此三角形以$$x$$軸爲對稱軸作對稱得到三角形$$\\triangle A^{}\\prime B^{}\\prime C^{}\\prime $$ ．再將她以原點爲中心，逆時針方向旋轉90°．得到$$\\triangle A^{\\prime \\prime }B^{\\prime \\prime }C^{\\prime \\prime }$$．試問下列哪一個敘述能將變換回原來的？ $$\\text{A}$$选项．繞原點逆時針旋轉$$90^{}\\circ $$ $$\\text{B}$$选项．繞原點順時針旋轉$$90^{}\\circ $$ $$\\text{C}$$选项．關於$$x$$軸的對稱 $$\\text{D}$$选项．關於直線$$y=x$$對稱 $$\\rm E$$选项．關於$$y$$軸的對稱 A triangle with vertices $$A(0,2)$$, $$B(-3,2)$$, and $$C(-3,0)$$ is reflected about the $$x$$-axis, then the image $$\\triangle A^{}\\prime B^{}\\prime C^{}\\prime $$ is rotated counterclockwise about the origin by $$90^{}\\circ $$ to produce $$\\triangle A^{\\prime \\prime }B^{\\prime \\prime }C^{\\prime \\prime }$$. Which of the following transformations will return $$\\triangle A^{\\prime \\prime }B^{\\prime \\prime }C^{\\prime \\prime }$$ to $$\\triangle ABC$$?", "answer_option_list": [[{"aoVal": "A", "content": "counterclockwise retation about the origin by $$90^{}\\circ $$ "}], [{"aoVal": "B", "content": "clockwise rotation about the origin by $$90^{}\\circ $$ "}], [{"aoVal": "C", "content": "reflection about the x---axis "}], [{"aoVal": "D", "content": "reflection about the liney=x "}], [{"aoVal": "E", "content": "reflection about the y---axis. "}]], "knowledge_point_routes": [], "answer_analysis": ["Consider a point $$(x,y)$$. Reflecting it about the $$x$$-axis will map it to $$(x,-y)$$, and rotating it counterclockwise about the origin by $$90^{}\\circ $$will map it to $$(y,x)$$. The operation that undoes this is a reflection about the $$y=x$$, so the answer is $$\\rm D$$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "831", "queId": "ff8080814a85cc99014a89235e00081a", "competition_source_list": ["2010年黑龙江全国高中数学联赛竞赛初赛第12题5分", "2020~2021学年广东深圳福田区深圳市高级中学高中部高二下学期期中第8题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知集合$$M= {1,2,3 }$$，$$N= {1,2,3,4 }$$，定义函数$$f:M\\to N$$．若点 $$A(1,f(1))$$，$$B(2,f(2))$$，$$C(3,f(3))$$，$$\\Delta ABC$$的外接圆圆心为$$D$$，且$$\\overrightarrow{DA}+\\overrightarrow{DC}=\\lambda \\overrightarrow{DB}(\\lambda\\in R)$$ ，则满足条件的函数$$f(x)$$有（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$个 "}], [{"aoVal": "B", "content": "$$10$$个 "}], [{"aoVal": "C", "content": "$$12$$个 "}], [{"aoVal": "D", "content": "$$16$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->两个基本计数原理->分类加法计数原理", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->函数的定义", "课内体系->素养->数学运算"], "answer_analysis": ["由$$\\overrightarrow{DA}+\\overrightarrow{DC}=\\overrightarrow{\\gamma DB}$$（$$\\gamma \\in \\mathbf{R}$$），分析可得$$\\triangle ABC$$是等腰三角形， 且$$BA=BC$$，必有$$f(1)=f(3)$$，$$f(1)\\ne f(2)$$． 点$$A(1,f(1))$$、当$$f(1)=1=f(3)$$时$$f(2)=2$$、$$3$$、$$4$$，三种情况． $$f(1)=f(3)=2$$；$$f(2)=1$$、$$3$$、$$4$$，有三种． $$f(1)=f(3)=3$$；$$f(2)=2$$、$$1$$、$$4$$，有三种． $$f(1)=f(3)=4$$；$$f(2)=2$$、$$3$$、$$1$$，有三种． 因而满足条件的函数$$f(x)$$有$$12$$种． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "889", "queId": "d6a08daab1b84a229b1ee7fb117ec969", "competition_source_list": ["1988年全国高中数学联赛竞赛一试第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "在坐标平面上，纵横坐标都是整数的点叫做整点．我们用Ⅰ表示所有直线的集合，$$M$$表示恰好通过一个整点的直线的集合，$$N$$表示不通过任何整点的直线的集合，$$P$$表示通过无穷多个整点的直线的集合，那么表达式：． （1）$$M{\\cup }N{\\cup }P=$$Ⅰ；（3）$$N\\ne \\varnothing $$； （2）$$M\\ne \\varnothing $$；~~~~~~~~~~ （4）$$P\\ne \\varnothing $$ 中正确的个数是", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算", "竞赛->知识点->解析几何->直线与方程"], "answer_analysis": ["直线$$y=\\frac{1}{2}$$不通过任何整点，所以表达式（2）正确； 直线$$y=\\sqrt{2}x$$无穷多个，恰好通过一个整点$$\\left( 0,0 \\right)$$，所以表达式（$$3$$）正确， 直线$$y=x$$通过多个整点，所以表达式（$$4$$）正确； 我们来证明，若 $$ax+by+c=0$$ 通过两个整点$$\\left( {{x}_{1}},{{y}_{1}} \\right)$$、$$\\left( {{x}_{2}},{{y}_{2}} \\right)$$， 则通过无穷多个整点 $$\\left( {{x}_{2}}+k\\left( {{x}_{2}}-{{x}_{1}} \\right),{{y}_{1}}+k\\left( {{y}_{2}}-{{y}_{1}} \\right) \\right)$$，$$k\\in \\bf Z$$． 事实上 $$a\\left[ {{x}_{1}}+k\\left( {{x}_{2}}-{{x}_{1}} \\right) \\right]+b\\left[ {{y}_{1}}+k\\left( {{y}_{2}}-{{y}_{1}} \\right) \\right]+c$$ $$=a{{x}_{1}}+b{{y}_{1}}+c+k\\left( a{{x}_{2}}+b{{y}_{2}}+c \\right)-k\\left( a{{x}_{1}}+b{{y}_{1}}+c \\right)$$ $$=0$$， 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "896", "queId": "812fcc3d76cd4826a2b5e6966a64818d", "competition_source_list": ["1989年全国高中数学联赛竞赛一试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)=\\text{arctan}x+\\frac{1}{2}\\arcsin x$$的值域是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(- \\pi , \\pi )$$ "}], [{"aoVal": "B", "content": "$$\\left[ -\\frac{3 \\pi }{4},\\frac{3 \\pi }{4} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left( -\\frac{3 \\pi }{4},\\frac{3 \\pi }{4} \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left[ -\\frac{ \\pi }{2},\\frac{ \\pi }{2} \\right]$$ "}]], "knowledge_point_routes": ["知识标签->题型->函数->函数及其表示->函数的值域->用单调性观察法求值域", "知识标签->素养->数学运算", "知识标签->知识点->三角函数->三角函数的图象与性质->反三角函数"], "answer_analysis": ["$$f(x)$$的定义域是$$[-1,1]$$，此时 $$-\\frac{ \\pi }{4}{\\leqslant }\\text{arctg}x{\\leqslant }\\frac{ \\pi }{4}$$，$$-\\frac{ \\pi }{4}{\\leqslant }\\frac{1}{2}\\arcsin x{\\leqslant }\\frac{ \\pi }{4}$$． 而且$$\\text{arctg}x$$和$$\\arcsin x$$是单调增加的，从而$$f(x)$$的值域是$$\\left[ -\\frac{ \\pi }{2},-\\frac{ \\pi }{2} \\right]$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "989", "queId": "a9cb8a6c7f39491788e6c7c186200683", "competition_source_list": ["2018年AMC10竞赛A第14题"], "difficulty": "2", "qtype": "single_choice", "problem": "不超过 $$\\frac{4^{100}+3^{100}}{4^{96}+3^{96}}$$ 的最大整数是? (改编自 2018 AMC10 A 第$$14$$题)", "answer_option_list": [[{"aoVal": "A", "content": "$$255$$ "}], [{"aoVal": "B", "content": "$$256$$ "}], [{"aoVal": "C", "content": "$$270$$ "}], [{"aoVal": "D", "content": "$$264$$ "}], [{"aoVal": "E", "content": "$$265$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Calculation->Exponentiation", "课内体系->知识点->集合->集合的基本关系->子集个数的计算"], "answer_analysis": ["We write $$\\frac{4^{100}+3^{100}}{4^{96}+3^{96}}=\\frac{4^{96}}{4^{96}+3^{96}}\\cdot \\frac{4^{100}}{4^{96}}+\\frac{3^{96}}{4^{96}+3^{96}}\\cdot \\frac{3^{100}}{3^{96}}=\\frac{4^{96}}{4^{96}+3^{96}}\\cdot 256+\\frac{3^{96}}{4^{96}+3^{96}}\\cdot 81$$. Hence we see that our number is less than $256$ but more than $81$. So the answer is $$\\boxed{\\rm (A)\\textasciitilde255}$$. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "507", "queId": "880d6e5276024ccb95db40d76fc19bba", "competition_source_list": ["2018年吉林全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$x$$、$$y$$、$$z\\textgreater0$$，满足$$x+y=xy$$，$$x+y+z=xyz$$．则$$z$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(0,\\sqrt{3}]$$ "}], [{"aoVal": "B", "content": "$$(1,\\sqrt{3}]$$ "}], [{"aoVal": "C", "content": "$$\\left( 0,\\frac{4}{3} \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left( 1,\\frac{4}{3} \\right]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->换元技巧->代数换元"], "answer_analysis": ["由已知$$y=\\frac{x}{x-1}$$得，$$z=\\frac{x+y}{xy-1}=\\frac{xy}{xy-1}={{\\left( 1-\\frac{1}{xy} \\right)}^{-1}}$$ $$={{\\left( 1-\\frac{1}{x}+\\frac{1}{{{x}^{2}}} \\right)}^{-1}}$$． 因为$$x\\textgreater1$$， 即$$0\\textless{}\\frac{1}{x}\\textless{}1$$， 所以，$$\\frac{3}{4}\\leqslant 1-\\frac{1}{x}+\\frac{1}{{{x}^{2}}}\\textless{}1$$． 故$$1\\textless{}z\\leqslant \\frac{4}{3}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "754", "queId": "69ad4e8d2c0648f28cd79083e185f5f1", "competition_source_list": ["2000年AMC10竞赛第17题"], "difficulty": "3", "qtype": "single_choice", "problem": "$$2000-AMC10-17$$ Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly? 鲍里斯有一台令人难以置信的硬币更换机。 当他投入$$25$$美分时，它返回五个$$5$$分币； 当他投入$$1$$个五分硬币时，它返回$$5$$个人$$1$$分硬币； 当他投入一分钱时，它会返回五个$$25$$美分硬币。 鲍里斯从一分钱开始。 鲍里斯在反复使用机器后可能会获得以下哪些金额？", "answer_option_list": [[{"aoVal": "A", "content": "＄$$3.63$$ "}], [{"aoVal": "B", "content": "＄$$5.13$$ "}], [{"aoVal": "C", "content": "＄$$6.30$$ "}], [{"aoVal": "D", "content": "＄$$7.45$$ "}], [{"aoVal": "E", "content": "＄$$9.07$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的应用->函数与方程", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"], "answer_analysis": ["Consider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn\\textquotesingle t change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn\\textquotesingle t change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by~~cents. This implies that the only possible values, in cents, he can have are the ones one more than a multiple of . Of the choices given, the only one is D "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "828", "queId": "77a73cbc5992493e868ab3113dba5f1d", "competition_source_list": ["2019年江西全国高中数学联赛竞赛初赛改编第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "数列$ {a_n }$满足: $a_0=\\sqrt{3}$, $a_{n+1}=[a_n]+\\dfrac{1}{ {a_n }}$(其中$[a_n]$和${a_n}$分别表示实数$x$的整数部分与小数部分), 则$a_{2020}=$．", "answer_option_list": [[{"aoVal": "A", "content": "$3029+\\dfrac{\\sqrt{3}-1}{2}$ "}], [{"aoVal": "B", "content": "$3032+\\dfrac{\\sqrt{3}-1}{2}$ "}], [{"aoVal": "C", "content": "$3035+\\dfrac{\\sqrt{3}-1}{2}$ "}], [{"aoVal": "D", "content": "前三个答案都不对 "}]], "knowledge_point_routes": ["知识标签->知识点->数列->数列的概念->数列的表示方法->通项公式"], "answer_analysis": ["$$1^{\\circ}$$迭代几项尝试一下， $$a_1=\\sqrt{3}=1+(\\sqrt{3}-1)$$， $$\\Rightarrow{a_2}=[a_1]+\\frac{1}{\\left { {{a}_{n}} \\right }}=1+\\frac{1}{\\sqrt{3}-1}=2+\\frac{\\sqrt{3}-1}{2}$$， $$\\Rightarrow{{a}_{3}}=[{{a}_{2}}]+\\frac{1}{\\left { {{a}_{2}} \\right }}=2+\\frac{2}{\\sqrt{3}-1}=4+(\\sqrt{3}-1)$$， $$\\Rightarrow{{a}_{4}}=[{{a}_{3}}]+\\frac{1}{\\left { {{a}_{3}} \\right }}=4+\\frac{1}{\\sqrt{3}-1}=5+\\frac{\\sqrt{3}-1}{2}$$， $$\\Rightarrow{{a}_{5}}=[{{a}_{4}}]+\\frac{1}{\\left { {{a}_{4}} \\right }}=5+\\frac{2}{\\sqrt{3}-1}=7+(\\sqrt{3}-1)$$， $$\\Rightarrow{{a}_{6}}=[{{a}_{5}}]+\\frac{1}{\\left { {{a}_{5}} \\right }}=7+\\frac{1}{\\sqrt{3}-1}=8+\\frac{\\sqrt{3}-1}{2}$$． $$2^{\\circ}$$猜想$$a_{2n}=3n-1+\\frac{\\sqrt{3}-1}{2}$$， 用数学归纳法证明， ①$$n=1$$时显然； ②若$$n=k$$时成立，即$$a_{2k}=3k-1+\\frac{\\sqrt{3}-1}{2}$$， 则$$a_{2k+1}=\\left[a_{2k}\\right]+\\frac{1}{ {a_{2k} }}=3k-1+\\frac{2}{\\sqrt{3}-1}=3k+1+\\left(\\sqrt{3}-1\\right)$$， $$\\Rightarrow a_{2k+2}=\\left[a_{2k+1}\\right]+\\frac{1}{ {a_{2k+1} }}=3k+1+\\frac{1}{\\sqrt{3}-1}=3k+2+\\frac{\\sqrt{3}-1}{2}$$，得证． 故，$${{a}_{2020}}=3\\times 1010-1+\\frac{\\sqrt{3}-1}{2}=3029+\\frac{\\sqrt{3}-1}{2}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "893", "queId": "cd77cc8ca1cb49ff9d1ec5c7ddfc37a9", "competition_source_list": ["2017年天津南开区高三二模文科第1题5分", "2015年黑龙江全国高中数学联赛竞赛初赛第1题5分", "2014年重庆高三零模文科第4题5分", "2017年天津南开区高三二模理科第1题5分", "2014年重庆高三零模理科第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设集合$$A= {-1,0,2 }$$，集合$$B= {-x\\textbar x\\in A,2-x\\notin A }$$，则$$B=$$（ ~ ~）", "answer_option_list": [[{"aoVal": "A", "content": "$$ {1 }$$ "}], [{"aoVal": "B", "content": "$$ {-2 }$$ "}], [{"aoVal": "C", "content": "$$ {-1,-2 }$$ "}], [{"aoVal": "D", "content": "$$ {-1,0 }$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->思想->分类讨论思想", "课内体系->知识点->集合->集合的概念与表示方法->集合的含义、元素与集合->元素与集合之间的关系->元素与集合的关系判断"], "answer_analysis": ["∵集合$$A= {-1,0,2 }$$，集合$$B= {-x\\textbar x\\in A,2-x\\notin A }$$， ∴$$-1\\in A$$，且$$2-\\left( -1 \\right)=3\\notin A$$，故$$1\\in B$$； $$0\\in A$$，但$$2-0=2\\in A$$，不满足题意； $$2\\in A$$，但$$2-2=0\\in A$$，不满足题意； 故$$B= {1 }$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "670", "queId": "6d824925a05e48bd8b0a1f8c98a2a937", "competition_source_list": ["2017~2018学年江西南昌西湖区南昌市第八中学高一上学期期末第10题5分", "2017~2018学年北京海淀区中央民族大学附属中学高一下学期期中第7题5分", "2015年北京海淀区高三三模第4题", "2018~2019学年浙江杭州江干区杭州第四中学下沙校区高二下学期期中第12题4分", "2018~2019学年广东深圳高一上学期期末高中联考联盟第8题5分", "2008年黑龙江全国高中数学联赛竞赛初赛第4题5分", "2018~2019学年江苏盐城高一上学期期末第5题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "若$$\\sin \\left( \\frac{ \\pi }{4}-x \\right)=\\frac{3}{5}$$，则$$\\sin 2x$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{19}{25}$$ "}], [{"aoVal": "B", "content": "$$\\frac{16}{25}$$ "}], [{"aoVal": "C", "content": "$$\\frac{14}{25}$$ "}], [{"aoVal": "D", "content": "$$\\frac{7}{25}$$ "}]], "knowledge_point_routes": ["课��体系->素养->数学运算", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正弦", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的正弦->利用正弦和差角公式凑角求值", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式->正余弦和差积相互转化求值"], "answer_analysis": ["$$\\sin \\left( \\frac{ \\pi }{4}-x \\right)=\\frac{\\sqrt{2}}{2}(\\cos x-\\sin x)=\\frac{3}{5}$$，所以$$\\cos x-\\sin x=\\frac{3\\sqrt{2}}{5}$$， 所以$${{(\\cos x-\\sin x)}^{2}}=1-\\sin 2x=\\frac{18}{25}$$，所以$$\\sin 2x=\\frac{7}{25}$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "589", "queId": "686029d5c6bc4281ac28fc2486c119da", "competition_source_list": ["2015年四川全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知二面角$$\\alpha -l-\\beta $$的大小为$$30{}^{}\\circ $$，则由平面$$\\alpha $$上的圆在平面$$\\beta $$上的正射影得到的椭圆的离心率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{3}}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{3}}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的角与距离", "竞赛->知识点->解析几何->椭圆"], "answer_analysis": ["设圆的半径为$$1$$，则椭圆的长轴长为$$2$$，短轴长为$$2\\cos 30{}^{}\\circ =\\sqrt{3}$$，则$$c=\\frac{1}{2}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "293", "queId": "16ac2bcfa03d4514b73cb1f5594fcf36", "competition_source_list": ["2016年陕西全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$\\vec{a},\\vec{b},\\vec{c}$$是同一平面内的三个单位向量，且$$\\vec{a}\\bot \\vec{b}$$，则$$(\\vec{c}-\\vec{a})\\cdot (\\vec{c}-\\vec{b})$$的最大值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1+\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$1-\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{2}-1$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积的运算律", "课内体系->知识点->平面向量->平面向量的运算->数量积->向量的数量积的定义", "课内体系->知识点->平面向量->平面向量的运算->数量积->利用向量数量积求模长", "课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积运算（非坐标）", "课内体系->知识点->平面向量->平面向量的运算->数量积->利用向量数量积求夹角", "课内体系->知识点->平面向量->平面向量的基本概念->向量的概念->向量的夹角的判断", "课内体系->知识点->平面向量->平面向量的基本概念->向量的模", "课内体系->素养->数学运算"], "answer_analysis": ["$$(\\vec{c}-\\vec{a})\\cdot (\\vec{c}-\\vec{b})=1-\\overrightarrow{c}\\left( \\overrightarrow{a}+\\overrightarrow{b} \\right)=1-\\left\\textbar{} \\overrightarrow{c} \\right\\textbar\\left\\textbar{} \\overrightarrow{a}+\\overrightarrow{b} \\right\\textbar\\text{cos }\\theta \\leqslant 1+\\sqrt{2}$$，当$$\\overrightarrow{c}$$与$$\\overrightarrow{a}+\\overrightarrow{b}$$的夹角为$$180{}^{}\\circ $$时取等号． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "294", "queId": "626c476492ae4b79991ce8633ab2cad4", "competition_source_list": ["2019~2020学年4月湖南长沙天心区长郡中学高一下学期月考第12题3分", "1998年全国高中数学联赛竞赛一试第3题6分", "2014~2015学年10月辽宁沈阳和平区东北育才中学高二上学期月考理科第7题5分", "2018~2019学年10月山东济宁市中区济宁市第一中学高二上学期月考第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "各项均为实数的等比数列$$\\left { {{a}_{n}} \\right }$$前$$n$$项之和记为$${{S}_{n}}$$，若$${{S}_{10}}=10$$，$${{S}_{30}}=70$$，则$${{S}_{40}}$$等于（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$150$$ "}], [{"aoVal": "B", "content": "$$-200$$ "}], [{"aoVal": "C", "content": "$$150$$或$$-200$$ "}], [{"aoVal": "D", "content": "$$-50$$或$$400$$~ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["首先$$q\\ne 1$$，于是$$\\frac{{{a}_{1}}}{q-1}\\left( {{q}^{10}}-1 \\right)=10$$，$$\\frac{{{a}_{1}}}{q-1}\\left( {{q}^{30}}-1 \\right)=70$$， ∴$$\\textasciitilde{{q}^{20}}+{{q}^{10}}+1=7\\Rightarrow {{q}^{10}}=2$$（$$-3$$舍）． ∴$${{S}_{40}}=10\\left( {{q}^{40}}1 \\right)=150$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1188", "queId": "eb94ba5fffff4e51ae3d04f5ff36a4d3", "competition_source_list": ["2006年上海复旦大学自主招生千分考第6题", "2010年浙江全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在正三棱柱$$ABC-{{A}_{1}}{{B}_{1}}{{C}_{1}}$$中，若$$AB=\\sqrt{2}B{{B}_{1}}$$，则$$A{{B}_{1}}$$与$${{C}_{1}}B$$所成的角的大小是．", "answer_option_list": [[{"aoVal": "A", "content": "$$60{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$75{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$90{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$105{}^{}\\circ $$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的角与距离"], "answer_analysis": ["取$${{B}_{1}}{{A}_{1}}$$的中点$$M$$，易知$${{C}_{1}}M\\bot $$平面$$A{{A}_{1}}{{B}_{1}}B$$，所以$${{C}_{1}}M\\bot A{{B}_{1}}$$， 在矩形$$AB{{B}_{1}}{{A}_{1}}$$中，$$\\tan \\angle {{B}_{1}}BM=\\tan \\angle {{B}_{1}}AB=\\frac{\\sqrt{2}}{2}$$， 于是$$\\angle {{B}_{1}}BM=\\angle {{B}_{1}}AB$$，从而$$MB\\bot A{{B}_{1}}$$， 因此，$$A{{B}_{1}}\\bot $$平面$$BM{{C}_{1}}$$，所以$$A{{B}_{1}}\\bot B{{C}_{1}}$$，即$$A{{B}_{1}}$$与$${{C}_{1}}B$$所成的角的大小为$$90{}^{}\\circ $$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "950", "queId": "e4a0014aaee64726b5ef689dd1b7fa1c", "competition_source_list": ["2010年四川全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "设正三棱锥$$S-ABC$$的底面边长为$$3$$，侧棱长为$$2$$，则侧棱$$SA$$与底面$$ABC$$所成的角的大小是．", "answer_option_list": [[{"aoVal": "A", "content": "$$30{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$45{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$60{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$\\arctan \\sqrt{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的角与距离"], "answer_analysis": ["设顶点$$S$$在底面$$\\triangle ABC$$的射影是$$H$$，则$$H$$为$$\\triangle ABC$$的外心．从而$$AH=\\frac{2}{3}\\times 3\\times \\frac{\\sqrt{3}}{2}=\\sqrt{3}$$，于是可得$$\\angle SAH={{30}^{\\circ }}$$．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "199", "queId": "0dbe980901f846d0ab2b16f22243f0af", "competition_source_list": ["1994年全国高中数学联赛竞赛一试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$0\\textless{}b\\textless{}1$$，$$0\\textless{}a\\textless{}\\frac{ \\pi }{4}$$，则下列三数：$$x={{(\\sin a)}^{{{\\log }_{b}}\\sin a}}$$，$$y={{(\\cos a)}^{{{\\log }_{b}}\\cos a}}$$，$$z={{(\\sin a)}^{{{\\log }_{b}}\\cos a}}$$大小关系是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textless{}z\\textless{}y$$ "}], [{"aoVal": "B", "content": "$$y\\textless{}z\\textless{}x$$ "}], [{"aoVal": "C", "content": "$$z\\textless{}x\\textless{}y$$ "}], [{"aoVal": "D", "content": "$$x\\textless{}y\\textless{}z$$ "}]], "knowledge_point_routes": ["课内体系->知识点->基本初等函数->对数的概念及其运算", "课内体系->知识点->基本初等函数->指对幂比较大小", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->三角函数线", "课内体系->素养->数学运算"], "answer_analysis": ["∵$$0\\textless{}b\\textless{}1$$， ∴$$f(x)={{\\log }_{b}}x$$是减函数，又 ∵$$0\\textless{}a\\textless{}\\frac{ \\pi }{4}$$， ∴$$0\\textless{}\\sin a\\textless{}\\cos a\\textless{}1$$， ∴$${{\\log }_{b}}\\sin a\\textgreater{{\\log }_{b}}\\cos a\\textgreater0$$， ∴$${{(\\sin a)}^{{{\\log }_{b}}\\sin a}}\\textless{}{{(\\sin a)}^{{{\\log }_{b}}\\cos a}}$$，即$$x\\textgreater z$$； 又$${{(\\sin a)}^{{{\\log }_{b}}\\sin a}}\\textless{}{{(\\cos a)}^{{{\\log }_{b}}\\cos a}}$$，即$$z\\textless{}y$$，故有$$x\\textless{}z\\textless{}y$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1035", "queId": "e084cb5e007a498297ffc8dc860afe39", "competition_source_list": ["2008年四川全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在公差为$$4$$的正项等差数列$$ {{{a}_{n}} }$$中，$${{a}_{3}}$$与$$2$$的算术平均值等于$${{S}_{3}}$$与$$2$$的几何平均值，其中$${{S}_{3}}$$表示数列前$$3$$项的和，则$${{a}_{10}}$$的值．", "answer_option_list": [[{"aoVal": "A", "content": "$$38$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$44$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->思想->方程思想", "课内体系->知识点->数列->等差数列->等差数列的概念与通项公式->等差数列求通项问题", "课内体系->知识点->数列->等差数列->等差数列的前n项和->等差数列求和问题", "课内体系->知识点->数列->等差数列->等差数列的性质及应用", "课内体系->知识点->数列->数列的概念->数列的前n项和", "课内体系->知识点->数列->数列的概念->数列的定义", "课内体系->知识点->数列->数列的概念->数列的表示方法->通项公式->常见数列的通项公式", "课内体系->方法->方程组法"], "answer_analysis": ["因为等差数列$$ {{{a}_{n}} }$$的公差$$d=4$$， 则$${{a}_{3}}={{a}_{1}}+8$$，$${{S}_{3}}=3{{a}_{1}}+12$$， 从而$$\\frac{({{a}_{1}}+8)+2}{2}=\\sqrt{2(3{{a}_{1}}+12)}$$， 解得$${{a}_{1}}=2$$．所以$${{a}_{10}}=2+9\\times 4=38$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "442", "queId": "24b84d997f8742eda421b32b67a4e6e5", "competition_source_list": ["2012年吉林全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设方程$${{\\log }_{4}}x-{{\\left( \\frac{1}{4} \\right)}^{x}}=0$$与$${{\\log }_{\\frac{1}{4}}}x-{{\\left( \\frac{1}{4} \\right)}^{x}}=0$$的根分别为$${{x}_{1}}, {{x}_{2}}$$，则", "answer_option_list": [[{"aoVal": "A", "content": "$$0\\textless{}{{x}_{1}}{{x}_{2}}\\textless{}1$$ "}], [{"aoVal": "B", "content": "$${{x}_{1}}{{x}_{2}}=1$$ "}], [{"aoVal": "C", "content": "$$1\\textless{}{{x}_{1}}{{x}_{2}}\\textless{}2$$ "}], [{"aoVal": "D", "content": "$${{x}_{1}}{{x}_{2}}\\geqslant 2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["显然$${{x}_{2}}=\\frac{1}{2}$$．设$$f\\left( x \\right)={{\\log }_{4}}x-{{\\left( \\frac{1}{4} \\right)}^{x}}$$，则$$f\\left( 1 \\right)\\cdot f\\left( 2 \\right)\\textless{}0$$，所以$$1\\textless{}{{x}_{2}}\\textless{}2$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "415", "queId": "2cdee0b2f87744e9905b50e2a21dc9ca", "competition_source_list": ["2017年湖南全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设集合$$X=\\left { 1,2,3,\\cdots ,2017 \\right }$$，集合 $$S=\\left { \\left( x,y,z \\right)\\left\\textbar{} x,y,z\\in X,且三条件x\\textless{}y\\textless{}z,y\\textless{}z\\textless{}x,z\\textless{}x\\textless{}y 恰有一个成立\\right. \\right }$$ 若$$\\left( x,y,z \\right)\\in S$$且$$\\left( z,w,x \\right)\\in S$$，则下列选项正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( y,z,w \\right)\\in S$$且$$\\left( x,y,w \\right)\\notin S$$ "}], [{"aoVal": "B", "content": "$$\\left( y,z,w \\right)\\in S$$且$$\\left( x,y,w \\right)\\in S$$ "}], [{"aoVal": "C", "content": "$$\\left( y,z,w \\right)\\notin S$$且$$\\left( x,y,w \\right)\\in S$$ "}], [{"aoVal": "D", "content": "$$\\left( y,z,w \\right)\\notin S$$且$$\\left( x,y,w \\right)\\notin S$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的划分与覆盖", "竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["特殊值排除法， 取$$x=2$$，$$y=3$$，$$z=4$$，$$w=1$$，显然满足$$(x,y,z)$$和$$(z,w,x)$$都在$$S$$中， 此时$$(y ,z,w)=(3,4,1)\\in S$$， $$(x,y,w)=(2,3,1)\\in S$$，故$$\\text A$$、$$\\text C$$、$$\\text D$$均错误； 只有$$\\text B$$成立，故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1076", "queId": "b82fd682192444ccac52211ea4f5a164", "competition_source_list": ["竞赛第20题"], "difficulty": "2", "qtype": "single_choice", "problem": "考虑三维空间中任意给定的空间四边形$ABCD$，\\emph{A}，\\emph{B}，\\emph{C}，\\emph{D}是四个顶点，四条线段$AB,BC,CD,DA$依次首尾相接，将点\\emph{A}的内角定义为射线$AD$和射线$AB$的夹角，其补角为角\\emph{A}的外角，其他顶点定义类似，考虑这种空间四边形的外角和\\emph{x}，则有（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$x=2\\pi$ "}], [{"aoVal": "B", "content": "$x\\geq 2\\pi$ "}], [{"aoVal": "C", "content": "$x\\leq 2\\pi$ "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 首先外角和与内角和之和为$4\\pi$，再根据各角的不等关系可证内角和不大于$2\\pi$，外角和不小于$2\\pi$．\\\\ 【详解】\\\\ 对于空间四边形，外角和与内角和之和为$4\\pi$，因此考虑内角和与$2\\pi$的大小关系．\\\\ 取\\emph{A}，\\emph{B}，\\emph{C}，\\emph{D}为正四面体的四个顶点，则内角和为$\\frac{\\pi }{3}\\times 4=\\frac{4\\pi }{3}\\textless{} 2\\pi$，\\\\ 因此选项AC都错误，\\\\ 接下来证明内角和不大于$2\\pi$．\\\\ 事实上，有$\\left {\\begin{array}{l} \\angle ABD+\\angle CBD\\geq \\angle ABC, \\angle ADB+\\angle CDB\\geq \\angle CDA, \\end{array}\\right.$\\\\ 从而$2x=(\\angle ABD+\\angle BDA+\\angle DAB)+(\\angle DBC+\\angle BDC+\\angle BCD)$\\\\ $\\geq \\angle DAB+\\angle ABC+\\angle BCD+\\angle CDA$，\\\\ 因此内角和不大于$2\\pi$，外角和不小于$2\\pi$．\\\\ 故选：B. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "706", "queId": "9145a7720ff84c54ad057b4e3dd6790c", "competition_source_list": ["2009年黑龙江全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知平面区域$$\\Omega = {(x,y)\\textbar x+y\\leqslant 6,x\\geqslant 0,y\\geqslant 0 }$$，$$A= {(x,y)\\textbar x\\leqslant 4,y\\geqslant 0,x-2y\\geqslant 0 }$$，若向区域$$\\Omega $$上随机投一点$$P$$，则点$$P$$落入区域$$A$$的概率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{9}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{9}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->概率初步", "竞赛->知识点->集合->集合的划分与覆盖", "竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["不难算出平面区域$$\\Omega $$的面积为$$\\frac{1}{2}\\cdot 6\\cdot 6=18$$，$$A$$的面积为$$\\frac{1}{2}\\cdot 4\\cdot 2=4$$， 所以所求概率为$$\\frac{4}{18}=\\frac{2}{9}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "282", "queId": "2b9f734382eb4cd0bf828aeccf2cfff0", "competition_source_list": ["2016年天津全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)=\\left\\textbar{} \\sin 2x+\\cos 2x \\right\\textbar$$的最小正周期是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2 \\pi $$ "}], [{"aoVal": "B", "content": "$$ \\pi $$ "}], [{"aoVal": "C", "content": "$$\\frac{ \\pi }{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{ \\pi }{4}$$ "}]], "knowledge_point_routes": ["课内体系->素养->直观想象", "课内体系->素养->数学运算", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦型函数的图象与性质->求正弦型函数的周期", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦函数的图象和性质", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->辅助角公式", "课内体系->知识点->函数的应用->函数与方程->函数的图象变换问题->翻折变换问题", "课内体系->思想->函数思想"], "answer_analysis": ["$$f\\left( x \\right)=\\left\\textbar{} \\sin 2x+\\cos 2x \\right\\textbar$$可以改写为 $$f\\left( x \\right)=\\left\\textbar{} \\sqrt{2}\\sin \\left( 2x+\\frac{\\pi }{4} \\right) \\right\\textbar$$， 由函数图象之间的关系可见$$f\\left( x \\right)$$与$$g\\left( x \\right)=\\left\\textbar{} \\sin 2x \\right\\textbar$$有相同的最小正周期． 因为$$\\left\\textbar{} \\sin x \\right\\textbar$$的最小正周期为$$\\pi $$，所以$$g\\left( x \\right)$$的最小正周期为$$\\frac{\\pi }{2}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "797", "queId": "bac0b1e060794d7e931069c115726bc7", "competition_source_list": ["2022~2023学年3月江苏常州武进区江苏省前黄高级中学高一下学期月考第2题", "2022~2023学年3月江苏常州武进区江苏省前黄高级中学高一下学期月考第2题", "2022~2023学年湖南邵阳隆回县湖南省隆回县第二中学高一上学期月考（竞赛）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知角\\emph{α}终边上一点\\emph{M}的坐标为$(1,\\sqrt{3})$，则$\\sin 2\\alpha =$（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$-\\frac{1}{2}$ "}], [{"aoVal": "B", "content": "$\\frac{1}{2}$ "}], [{"aoVal": "C", "content": "$-\\frac{\\sqrt{3}}{2}$ "}], [{"aoVal": "D", "content": "$\\frac{\\sqrt{3}}{2}$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据题意，结合$\\alpha $所在象限，得到$\\sin \\alpha $和$\\cos \\alpha $的值，再根据公式，求得答案.\\\\ 【详解】\\\\ 由角$\\alpha $终边上一点\\emph{M}的坐标为$\\left( 1,\\sqrt{3} \\right)$，\\\\ 得$\\sin \\alpha =\\frac{\\sqrt{3}}{2}$，$\\cos \\alpha =\\frac{1}{2}$，\\\\ 故$\\sin 2\\alpha =2\\sin \\alpha \\cos \\alpha =\\frac{\\sqrt{3}}{2}$，\\\\ 故选D.\\\\ 【点睛】\\\\ 本题考查已知角的终边求对应的三角函数值，二倍角公式，属于简单题. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "172", "queId": "262f0636a3dc45fa9554ebb269caf0cf", "competition_source_list": ["2008年河南全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\triangle ABC$$的三个内角$$A$$、$$B$$、$$C$$依次成等差数列，$$\\sin A$$、$$\\sin B$$、$$\\sin C$$依次成等比数列，则$$\\triangle ABC$$．", "answer_option_list": [[{"aoVal": "A", "content": "是直角三角形，但不是等腰三角形 "}], [{"aoVal": "B", "content": "不是等腰三角形，也不是直角三角形 "}], [{"aoVal": "C", "content": "是等腰直角三角形 "}], [{"aoVal": "D", "content": "是等边三角形 "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["因为$$B=60{}^{}\\circ $$，$${{\\sin }^{2}}B=\\sin A\\sin C$$， 所以$${{\\sin }^{2}}60{}^{}\\circ =\\sin A\\sin (120{}^{}\\circ -A)$$， 可知$$\\frac{3}{4}=\\sin A\\left( \\frac{\\sqrt{3}}{2}\\cos A+\\frac{1}{2}\\sin A \\right)$$ $$=\\frac{\\sqrt{3}}{2}\\sin A\\cos A+\\frac{1}{2}{{\\sin }^{2}}A$$ $$=\\frac{\\sqrt{3}}{4}\\sin 2A+\\frac{1-\\cos 2A}{4}$$ 因此$$\\sin (2A-30{}^{}\\circ )=1$$，而$$0{}^{}\\circ \\textless{}A\\textless{}120{}^{}\\circ $$， 易知$$2A-30{}^{}\\circ =90{}^{}\\circ $$，即$$A=60{}^{}\\circ $$， 因此$$\\triangle ABC$$是等边三角形．故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "226", "queId": "22172070d1f04eb3bd337f3de4b53cba", "competition_source_list": ["2020年贵州全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\text{i}$$是虚数单位，则$$\\sum\\limits_{k=1}^{2020}{\\left( k\\cdot {{\\text{i}}^{k+1}} \\right)}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1010-1010\\text{i}$$ "}], [{"aoVal": "B", "content": "$$1010+1010\\text{i}$$ "}], [{"aoVal": "C", "content": "$$-1010+1010\\text{i}$$ "}], [{"aoVal": "D", "content": "$$1010-1010\\text{i}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和", "课内体系->知识点->复数->复数的运算->复数的四则运算综合"], "answer_analysis": ["设$$S=\\text{i}+2{{\\text{i}}^{2}}+3{{\\text{i}}^{3}}+\\cdots +2019{{\\text{i}}^{2019}}+2020{{\\text{i}}^{2020}}$$， 则$$\\text{i}S={{\\text{i}}^{2}}+2{{\\text{i}}^{3}}+3{{\\text{i}}^{4}}+\\cdots +2019{{\\text{i}}^{2020}}+2020{{\\text{i}}^{2021}}$$， 两式相减，得： $$S-\\text{i}S=\\text{i}+{{\\text{i}}^{2}}+{{\\text{i}}^{3}}+\\cdots +{{\\text{i}}^{2020}}-{{2020}^{2021}}$$ $$=\\dfrac{\\text{i}\\left( 1-{{\\text{i}}^{2020}} \\right)}{1-\\text{i}}-2020{{\\text{i}}^{2021}}$$ $$=-2021\\text{i}$$， 故$$S=-\\dfrac{2020\\text{i}}{1-\\text{i}}=1010-1010\\text{i}$$， 即$$\\sum\\limits_{k=1}^{2020}{\\left( k\\cdot {{\\text{i}}^{k+1}} \\right)}=1010+1010\\text{i}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "252", "queId": "38e3e419980d48049c12cc46a3de88f9", "competition_source_list": ["1994年全国高中数学联赛竞赛一试第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "给出下列两个命题： （$$1$$）设$$a$$、$$b$$、$$c$$都是复数，如果$${{a}^{2}}+{{b}^{2}}\\textgreater{{c}^{2}}$$，则$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}\\textgreater0$$． （$$2$$）设$$a$$、$$b$$、$$c$$都是复数，如果$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}\\textgreater0$$，则$${{a}^{2}}+{{b}^{2}}\\textgreater{{c}^{2}}$$． 那么下述说法正确的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "命题（$$1$$）正确，命题（$$2$$）也正确 "}], [{"aoVal": "B", "content": "命题（$$1$$）正确，命题（$$2$$）错误 "}], [{"aoVal": "C", "content": "命题（$$1$$）错误，命题（$$2$$）也错误 "}], [{"aoVal": "D", "content": "命题（$$1$$）错误，命题（$$2$$）正确 "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["命题（$$1$$）是正确的．$${{a}^{2}}+{{b}^{2}}\\textgreater{{c}^{2}}$$表明$${{a}^{2}}+{{b}^{2}}$$与$${{c}^{2}}$$都是实数，因此，根据移项法则有$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}\\textgreater0$$． 命题（$$2$$）是错误的．$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}\\textgreater0$$仅表明$${{a}^{2}}+{{b}^{2}}-{{c}^{2}}$$是实数，并不能保证$${{a}^{2}}+{{b}^{2}}$$与$${{c}^{2}}$$都是实数，故$${{a}^{2}}+{{b}^{2}}\\textgreater{{c}^{2}}$$不一定成立． 例如，取$$a=2+\\text{i}$$，$$b=\\text{i}$$，$$c=\\sqrt{2}+\\sqrt{\\text{2i}}$$，则有 $${{a}^{2}}+{{b}^{2}}-{{c}^{2}}=(3+4\\text{i})+(-1)-4\\text{i}=2\\textgreater0$$，但并没有$${{a}^{2}}+{{b}^{2}}=2+4\\text{i}\\textgreater4\\text{i}={{c}^{2}}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1067", "queId": "a631421546d845c39b6e28842ff98619", "competition_source_list": ["2021年贵州全国高中数学联赛竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知集合$$M=\\left { \\left( x,y \\right)\\textbar\\textbar x\\textbar+\\textbar y\\textbar\\leqslant 1 \\right }$$，集合$$N=\\left { \\left( x,y \\right)\\textbar{{x}^{2}}+{{y}^{2}}\\leqslant \\textbar x\\textbar+\\textbar y\\textbar{} \\right }$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$M=N$$ "}], [{"aoVal": "B", "content": "$$M\\subseteq N$$ "}], [{"aoVal": "C", "content": "$$N\\subseteq M$$ "}], [{"aoVal": "D", "content": "前三个都不正确 "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["对于任意$$\\left( x,y \\right)\\in M$$，均有$$\\textbar x\\textbar+\\textbar y\\textbar\\leqslant 1$$，故$$\\begin{cases}\\textbar x\\textbar\\leqslant 1 \\textbar y\\textbar\\leqslant 1 \\end{cases}$$，故有$$\\begin{cases}{{x}^{2}}\\leqslant \\textbar x\\textbar{} {{y}^{2}}\\leqslant \\textbar y\\textbar{} \\end{cases}\\Rightarrow {{x}^{2}}+{{y}^{2}}\\leqslant \\textbar x\\textbar+\\textbar y\\textbar$$， 故$$\\left( x,y \\right)\\in N$$，所以$$M\\subseteq N$$； 又因为$$\\left( 1,1 \\right)\\in N$$，$$\\left( 1,1 \\right)\\notin M$$， 故$$M\\subsetneqq N$$，选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "20", "queId": "0e236c8568224a9d92310b5202a16bcf", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "若点$$P$$在椭圆$$\\frac{{{x}^{2}}}{16}+\\frac{{{y}^{2}}}{9}=1$$上，它到直线$$\\frac{x}{4}+\\frac{y}{3}=1$$的距离为$$\\frac{6}{5}$$，则点$$P$$的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆锥曲线"], "answer_analysis": ["由题知，可转换为该椭圆与直线平行并且到直线$$\\frac{x}{4}+\\frac{y}{3}=1$$的距离为$$\\frac{6}{5}$$的直线与椭圆的交点个数，与直线平行并且到直线$$\\frac{x}{4}+\\frac{y}{3}=1$$的距离为$$\\frac{6}{5}$$的直线方程为$$3x+4y-6=0$$或$$3x+4y-18=0$$，由$$\\begin{cases}\\frac{{{x}^{2}}}{16}+\\frac{{{y}^{2}}}{9}=1 3x+4y-18=0 \\end{cases}$$，消去$$y$$可得$$9{{x}^{2}}+{{\\left( 3x-18 \\right)}^{2}}=144$$，化简可得$${{x}^{2}}-6x+10=0$$，由判别式可知该方程无解，所以直线$$3x+4y-6=0$$与椭圆必有$$2$$个交点，故选$$\\text{B}$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "103", "queId": "25b5944565bd427aaa1ecd915dfcb63f", "competition_source_list": ["2010年四川全国高中数学联赛竞赛初赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "记$$F(x,y)={{(x-y)}^{2}}+{{\\left( \\frac{x}{3}+\\frac{3}{y} \\right)}^{2}}(y\\ne 0)$$，则$$F(x, y)$$的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{12}{8}$$ "}], [{"aoVal": "B", "content": "$$\\frac{16}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{18}{5}$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->双曲线"], "answer_analysis": ["设动点$$P\\left( x, -\\frac{x}{3} \\right)$$与$$Q\\left( y,\\frac{3}{y} \\right)$$，则$$F(x, y)={{\\left\\textbar{} PQ \\right\\textbar}^{2}}$$，点$$P$$的轨迹为直线$$y=-\\frac{x}{3}$$，点$$Q$$的轨迹为双曲线$$y=\\frac{3}{x}$$，双曲线上的任一点$$\\left( {{x}_{0}},\\frac{3}{{{x}_{0}}} \\right)$$到直线$$x+3y=0$$的距离: $$d=\\frac{\\left\\textbar{} {{x}_{0}}+3\\cdot \\frac{3}{{{x}_{0}}} \\right\\textbar}{\\sqrt{10}}\\geqslant \\frac{6}{\\sqrt{10}}$$， 当$${{x}_{0}}=\\pm 3$$时等号成立．故$$F(x, y)$$的最小值为$$\\frac{18}{5}$$．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "162", "queId": "106cf35a477843498a80e17651c54a13", "competition_source_list": ["2010年浙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$p:({{x}^{2}}+x+1)\\sqrt{x+3}\\geqslant 0, q:x\\geqslant -2$$，则$$p$$是$$q$$的．", "answer_option_list": [[{"aoVal": "A", "content": "充分而不必要条件 "}], [{"aoVal": "B", "content": "必要而不充分条件 "}], [{"aoVal": "C", "content": "充要条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["$$p$$成立$$\\Leftrightarrow x\\geqslant -3$$，所以$$p $$成立，推不出$$q$$一定成立． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "130", "queId": "21594be7a7564473bb576e1f553a5ccc", "competition_source_list": ["2009年浙江全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知平面上单位向量$$\\overrightarrow{a}=\\left( \\frac{5}{13},\\frac{12}{13} \\right),\\overrightarrow{b}=\\left( \\frac{4}{5},\\frac{3}{5} \\right)$$，则下列关系式正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\overrightarrow{a}\\bot \\overrightarrow{b}$$ "}], [{"aoVal": "B", "content": "$$(\\overrightarrow{a}+\\overrightarrow{b})\\bot (\\overrightarrow{a}-\\overrightarrow{b})$$ "}], [{"aoVal": "C", "content": "$$(\\overrightarrow{a}+\\overrightarrow{b})//(\\overrightarrow{a}-\\overrightarrow{b})$$ "}], [{"aoVal": "D", "content": "$$\\overrightarrow{a}\\bot (\\overrightarrow{a}+\\overrightarrow{b})$$ "}]], "knowledge_point_routes": ["课内体系->方法->图象法", "课内体系->知识点->平面向量->向量应用->平面几何中的向量方法", "课内体系->知识点->平面向量->平面向量的运算->数量积->数量积的坐标表达式", "课内体系->知识点->平面向量->平面向量的运算->数量积->向量的数量积的定义", "课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积运算（非坐标）", "课内体系->知识点->平面向量->平面向量的运算->数量积->利用数量积解决向量垂直问题（非坐标运算）", "课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积的运算律", "课内体系->知识点->平面向量->平面向量的运算->数量积->线性运算和数量积综合问题", "课内体系->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量的加法运算及运算规则", "课内体系->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量的减法运算及运算规则", "课内体系->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量线性运算综合（非坐标）", "课内体系->知识点->平面向量->平面向量基本定理及其坐标表示->平面向量基本定理及其意义", "课内体系->知识点->平面向量->平面向量基本定理及其坐标表示->平面向量的正交分解及坐标表示", "课内体系->知识点->平面向量->平面向量基本定理及其坐标表示->平面向量加、减、数乘的坐标运算", "课内体系->知识点->平面向量->平面向量基本定理及其坐标表示->坐标表示平面向量的垂直", "课内体系->知识点->平面向量->平面向量的基本概念->单位向量", "课内体系->知识点->平面向量->平面向量的基本概念->向量的概念", "课内体系->素养->直观想象", "课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->思想->数形结合思想"], "answer_analysis": ["因为$$\\overrightarrow{a},\\overrightarrow{b}$$都是非零单位向量，则以$$\\overrightarrow{a},\\overrightarrow{b}$$为边，$$\\overrightarrow{a}-\\overrightarrow{b},\\overrightarrow{a}+\\overrightarrow{b}$$为对角线构成一个菱形，所以$$(\\overrightarrow{a}+\\overrightarrow{b})\\bot (\\overrightarrow{a}-\\overrightarrow{b})$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "520", "queId": "d0fe4a56a1d04af7bd17f208adf3afcb", "competition_source_list": ["2009年四川全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设数列$$ {{{a}_{n}} }$$满足：$${{a}_{1}}=2,{{a}_{n+1}}=1-\\frac{1}{{{a}_{n}}}$$，记数列$$ {{{a}_{n}} }$$的前$$n$$项之积为$${{P}_{n}}$$，则$${{P}_{2009}}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->（模）周期数列"], "answer_analysis": ["因为 $${{a}_{n+2}}=1-\\frac{1}{{{a}_{n+1}}}=1-\\frac{1}{1-\\frac{1}{{{a}_{n}}}}=\\frac{-1}{{{a}_{n}}-1}$$， 于是 $${{a}_{n+3}}=1-\\frac{1}{{{a}_{n+2}}}=1-\\frac{1}{\\frac{-1}{{{a}_{n}}-1}}={{a}_{n}}$$， 故$$ {{{a}_{n}} }$$是以$$3$$为周期的周期数列． 又$${{a}_{1}}=2,{{a}_{2}}=\\frac{1}{2},{{a}_{3}}=-1$$，从而$${{P}_{3}}=-1$$，所以， $${{P}_{2009}}={{(-1)}^{669}}{{P}_{2}}=-1$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "965", "queId": "8aac49074f72f20f014f8174050d36b5", "competition_source_list": ["2010~2011学年北京海淀区高三上学期期末理科第7题", "2018~2019学年北京西城区北京市第十三中学高二上学期期中第10题3分", "2019~2020学年11月重庆沙坪坝区重庆市南开中学高二上学期周测A卷第6题5分", "2019~2020学年10月北京海淀区北京市第五十七中学高二上学期月考第10题", "2016年吉林全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知椭圆$$E:\\frac{{{x}^{2}}}{m}+\\frac{{{y}^{2}}}{4}=1$$，对于任意实数$$k$$，下列直线被椭圆$$E$$截得的弦长与$$l:y=kx+1$$被椭圆$$E$$截得的弦长不可能相等的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$kx+y+k=0$$ "}], [{"aoVal": "B", "content": "$$kx-y-1=0$$ "}], [{"aoVal": "C", "content": "$$kx+y-k=0$$ "}], [{"aoVal": "D", "content": "$$kx+y-2=0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->圆锥曲线->直线与圆锥曲线问题->弦长求解问题", "课内体系->知识点->圆锥曲线->椭圆->直线和椭圆的位置关系", "课内体系->素养->数学运算"], "answer_analysis": ["由数形结合可知，当$$l$$过点$$(-1,0)$$时，直线$$l$$和选项A中的直线关于$$x$$对称，被椭圆$$E$$所截得的弦长相同，故不能选A． 当$$l$$过点$$(1,0)$$时，直$$l$$和选项C中的直线关于$$x$$轴对称，被椭圆$$E$$所截得的弦长相同，故不能选C． 当$$k=0$$时，直线$$l$$和选项B中的直线关于$$x$$轴对称，被椭圆$$E$$所截得的弦长相同，故不能选B． 直线$$l$$斜率为$$k$$，在$$y$$轴上的截距为$$1$$；选项D中的直线$$kx+y-2=0$$ 斜率为$$-k$$，在$$y$$轴上的截距为$$2$$，这两直线不关于$$x$$轴、$$y$$轴、原点对称，故被椭圆$$E$$所截得的弦长不可能相等．故选D． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "662", "queId": "d5f6a51360534d7faaa0a61715943d2c", "competition_source_list": ["2005年高考真题福建卷理科第12题5分", "2009年黑龙江全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$f(x)$$是定义在$$\\text{R}$$上的以$$3$$为周期的奇函数，且$$f(2)=0$$，则方程$$f(x)=0$$在区间$$\\left( 0,6 \\right)$$内解的个数的最小值是（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的性质->周期性->函数周期性与奇偶性综合问题", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->抽象函数", "课内体系->知识点->函数的应用->函数的零点->零点的个数问题->求零点个数问题（不含参）", "课内体系->素养->数学抽象", "课内体系->素养->数学运算"], "answer_analysis": ["$$f(x)$$是定义在$$\\text{R}$$周期为$$3$$的奇函数，∴$$f(0)=0=f(3)$$，$$f(5)=f(2)=-f(-2)=-f(1)=0$$，∴$$f(1)=0=f(4)$$，∴选D. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "69", "queId": "079344bc716a4dcfa2c4c285d30c220f", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "6个人围成一圈，若$$2$$种站法经过旋转可以重合则算同$$1$$种，则不同的站法有．", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$种 "}], [{"aoVal": "B", "content": "$$60$$种 "}], [{"aoVal": "C", "content": "$$120$$种 "}], [{"aoVal": "D", "content": "$$720$$种 "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->两个基本计数原理", "竞赛->知识点->排列组合与概率->排列与组合", "课内体系->知识点->计数原理->两个基本计数原理->分步乘法计数原理"], "answer_analysis": ["总共$$10$$个球，先考虑选$$5$$个位置放篮球，共有$$\\text{C}_{10}^{5}=252$$种，再从剩下的$$5$$个位置种选$$3$$个位置放排球，总共有$$10$$种方法，剩下的$$2$$个位置放橄榄球，故有$$252\\times 10=2520$$种. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "834", "queId": "77bd9818a7b74b3781318bdaf0e37387", "competition_source_list": ["2008年贵州全国高中数学联赛竞赛初赛第10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知双曲线$$\\frac{{{x}^{2}}}{6}-\\frac{{{y}^{2}}}{3}=1$$的焦点$${{F}_{1}}$$、$${{F}_{2}}$$，点$$M$$在双曲线上，且$$M{{F}_{1}}\\bot x$$轴，则$${{F}_{1}}$$到直线$${{F}_{2}}M$$的距离为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{6}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3\\sqrt{6}}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5\\sqrt{6}}{6}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->双曲线"], "answer_analysis": ["由已知可得$$M\\left( 3,\\frac{\\sqrt{6}}{2} \\right)$$，则$$M{{F}_{1}}=\\frac{\\sqrt{6}}{2}$$， 故$$M{{F}_{2}}=2\\sqrt{6}+\\frac{\\sqrt{6}}{2}=\\frac{5\\sqrt{6}}{2}$$， 故$${{F}_{1}}$$到直线$${{F}_{2}}M$$的距离为： $$\\frac{\\left\\textbar{} {{F}_{1}}{{F}_{2}} \\right\\textbar\\cdot \\left\\textbar{} M{{F}_{1}}_{2} \\right\\textbar}{\\left\\textbar{} M{{F}_{2}} \\right\\textbar}=\\frac{6\\times \\frac{\\sqrt{6}}{2}}{\\frac{5\\sqrt{6}}{2}}=\\frac{6}{5}$$，故选$$\\text{A}$$. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "411", "queId": "be4956589d424dc5b715233b87fc9499", "competition_source_list": ["2009年河北全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "将数列$$ {2n-1 }(n\\in N*)$$依原顺序按第$$n$$组有$${{2}^{n}}$$项的要求分组，则$$2009$$在第组．", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的综合应用", "竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["前$$n$$组所含项数之和为$$2+{{2}^{2}}+\\cdots +{{2}^{n}}={{2}^{n+1}}-2$$， $$2009$$是数列的第$$1005$$项． $$510={{2}^{9}}-2\\textless{}1005\\textless{}{{2}^{10}}-2=1022$$． 因此，$$2009$$在第$$9$$组，选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "922", "queId": "8e1032fee70a43c3b7335b02cc1b0898", "competition_source_list": ["2008年河北全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$y=f(x+2)$$的图象过点$$(-1,3)$$，则函数$$f(x)$$的图象关于$$y$$轴对称的图形一定过点．", "answer_option_list": [[{"aoVal": "A", "content": "$$(1,-3)$$ "}], [{"aoVal": "B", "content": "$$(-1,3)$$ "}], [{"aoVal": "C", "content": "$$(-3,-3)$$ "}], [{"aoVal": "D", "content": "$$(-3,3)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->函数综合"], "answer_analysis": ["函数$$y=f(x+2)$$的图象过点$$(-1,3)$$，则$$f(-1+2)=3$$，即$$f(1)=3$$，所以数$$f(x)$$的图象关于$$y$$轴对称的图形一定过点$$f(-1)=3$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "171", "queId": "1d2b0485674e496ebfdfe4358d643b77", "competition_source_list": ["2009年AMC10竞赛B第23题"], "difficulty": "3", "qtype": "single_choice", "problem": "瑞秋和罗伯特在一个圆形跑道上跑步．瑞秋逆时针跑，每隔$90$秒完成一圈，罗伯特顺时针跑，每隔$80$秒完成一圈．两人在同一时间从同一条线上出发．在他们开始跑步后的$10$分钟和$11$分钟之间的某个随机时间，一个站在赛道内的摄影师拍了一张照片，照片显示了以起跑线为中心的四分之一的赛道．瑞秋和罗伯特都在照片上的概率是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{16}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{16}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "E", "content": "$$\\frac{5}{16}$$ "}]], "knowledge_point_routes": ["课内体系->学科符号->数与式->新定义符号->取最值函数->max{f(x),g(x)}", "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Geometric Models of Probabilities"], "answer_analysis": ["After $$10$$ minutes ($$600$$ seconds), Rachel will have completed $$6$$ laps and be $$30$$ seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in $$22.5$$ seconds, she will be in the picture between $$18.75$$ seconds and $$41.25$$ seconds of the tenth minute. After $$10$$ minutes Robert will have completed $$7$$ laps and will be $$40$$ seconds past the starting line. Because Robert runs one-fourth of a lap in $$20$$ seconds, he will be in the picture between $$30$$ and $$50$$ seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between $$30$$ and $$41.25$$ seconds of the tenth minute. So the probability that both runners are in the picture is $$\\frac{41 .25-30}{60}= \\boxed{\\frac{3}{16}}$$. The answer is $$\\rm(C)$$. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "371", "queId": "1bc4df3971b146dc892cc35a3d5426de", "competition_source_list": ["2018~2019学年10月四川成都金牛区成都市实验外国语学校高二上学期周测C卷第9题5分", "2017年河南洛阳高三二模文科第9题5分", "2016年广东揭阳高三一模理科第12题5分", "2014年黑龙江全国高中数学联赛竞赛初赛第10题5分", "2016年山东日照高三二模理科第9题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知直线$$x+y-k=0\\left( k\\textgreater0 \\right)$$与圆$${{x}^{2}}+{{y}^{2}}=4$$交于不同的两点$$A$$，$$B$$，$$O$$是坐标原点，且有$$\\left\\textbar{} \\overrightarrow{OA}+\\overrightarrow{OB}\\left\\textbar{} \\geqslant \\frac{\\sqrt{3}}{3} \\right\\textbar\\overrightarrow{AB} \\right\\textbar$$，那么$$k$$的取值范围是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( \\sqrt{3},+\\infty \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left[ \\sqrt{2},+\\infty \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left[ \\sqrt{2},2\\sqrt{2} \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left[ \\sqrt{3},2\\sqrt{2} \\right)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->直线和圆的方程->圆与方程->直线与圆的位置关系->圆的切线的相关问题", "课内体系->素养->数学运算"], "answer_analysis": ["设$$AB$$中点为$$D$$，则$$OD\\bot AB$$， ∵$$\\left\\textbar{} \\overrightarrow{OA}+\\overrightarrow{OB}\\left\\textbar{} \\geqslant \\frac{\\sqrt{3}}{3} \\right\\textbar\\overrightarrow{AB} \\right\\textbar$$， ∴$$\\left\\textbar{} 2\\overrightarrow{OD}\\left\\textbar{} \\geqslant \\frac{\\sqrt{3}}{3} \\right\\textbar\\overrightarrow{AB} \\right\\textbar$$， ∴$$\\left\\textbar{} \\overrightarrow{AB}\\left\\textbar{} \\leqslant 2\\sqrt{3} \\right\\textbar\\overrightarrow{OD} \\right\\textbar$$， ∵$$\\left\\textbar{} \\overrightarrow{OD}{{\\textbar}^{2}}+\\frac{1}{4} \\right\\textbar\\overrightarrow{AB}{{\\textbar}^{2}}=4$$， ∴$$\\textbar\\overrightarrow{OD}{{\\textbar}^{2}}\\geqslant 1$$， ∵直线$$x+y-k=0\\left( k\\textgreater0 \\right)$$与圆$${{x}^{2}}+{{y}^{2}}=4$$交于不同的两点$$A,B$$， ∴$$\\textbar\\overrightarrow{OD}{{\\textbar}^{2}}\\textless{}4$$ ∴$$4\\textgreater\\textbar\\overrightarrow{OD}{{\\textbar}^{2}}\\geqslant 1$$ ∴$$4\\textgreater{{\\left( \\frac{\\left\\textbar{} -k \\right\\textbar}{\\sqrt{2}} \\right)}^{2}}\\geqslant 1$$ ∵$$k\\textgreater0$$， ∴$$\\sqrt{2}\\leqslant k\\textless{}2\\sqrt{2}$$. 故选：$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "919", "queId": "fb96b44cd915485d822b88451afe11cb", "competition_source_list": ["2018年黑龙江全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "$${{\\left( a+2b-3c \\right)}^{4}}$$的展开式中$$ab{{c}^{2}}$$的系数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$208$$ "}], [{"aoVal": "B", "content": "$$216$$ "}], [{"aoVal": "C", "content": "$$217$$ "}], [{"aoVal": "D", "content": "$$218$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->计数原理->二项式定理->求项的系数或二项式系数", "课内体系->知识点->计数原理->二项式定理->求二项式展开式的特定项"], "answer_analysis": ["依题意，知$$ab{{c}^{2}}$$的系数为$$\\text{C}_{4}^{2}{{(-3)}^{2}}\\text{C}_{2}^{1}\\times 2\\times 1=216$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "218", "queId": "9432909b0d48427eaaee5bc460b25255", "competition_source_list": ["1989年全国高中数学联赛竞赛一试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "(★★)对任意的函数$$y=f\\left( x \\right)$$，在同一个直角坐标系中，函数$$y=f\\left( x-1 \\right)$$与函数$$y=f\\left( -x+1 \\right)$$的图象恒．", "answer_option_list": [[{"aoVal": "A", "content": "关于$$x$$轴对称； "}], [{"aoVal": "B", "content": "关于直线$$x=1$$对称 "}], [{"aoVal": "C", "content": "关于直线$$x=-1$$对称 "}], [{"aoVal": "D", "content": "关于$$y$$轴对称 "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的性质->对称性->两个函数的互对称问题", "竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["$$f(x)$$和$$f(-x)$$的图象关于直线$$x=0$$对称， $$f(x-1)$$与$$f(-x+1)$$的图象关于直线$$x=1$$对称． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1034", "queId": "c54152a426ae440687e703a262ce3f9c", "competition_source_list": ["2018年天津全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "以下四个数中，最大的为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\ln \\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{\\text{e}}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\ln \\pi }{\\pi }$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{10}\\ln 10}{20}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->导数"], "answer_analysis": ["考虑函数$$f\\left( x \\right)=\\frac{\\ln x}{x}$$． 则四个选项分别为$$f\\left( 2 \\right)$$、$$f\\left( \\text{e} \\right)$$、$$f\\left( \\pi \\right)$$、$$f\\left( \\sqrt{10} \\right)$$， 又$${f}'\\left( x \\right)=\\frac{1-\\ln x}{{{x}^{2}}}$$，则$$f\\left( x \\right)$$在区间$$\\left( 0,\\text{e} \\right)$$上单调递增，在区间$$\\left( \\text{e},+\\infty \\right)$$上单调递减． 从而，$$f\\left( x \\right)$$在区间$$\\left( 0,+\\infty \\right)$$上的最大值为$$f\\left( \\text{e} \\right)$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "89", "queId": "087bc89ba3294db9a5015dfb5baddd03", "competition_source_list": ["2022年湖南湘西吉首市竞赛（第一届中小学生教师解题大赛）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知复数$z$满足$2\\le \\left\\textbar{} z-\\left( 1-\\text{i} \\right) \\right\\textbar\\le 3$，则复数$z$在复平面内对应的点$Z$所在区域的面积为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\text{ } ! !\\pi ! !\\text{ }$ "}], [{"aoVal": "B", "content": "$3\\text{ } ! !\\pi ! !\\text{ }$ "}], [{"aoVal": "C", "content": "$5\\text{ } ! !\\pi ! !\\text{ }$ "}], [{"aoVal": "D", "content": "$6\\text{ } ! !\\pi ! !\\text{ }$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 令$z=a+b\\text{i}$且$a,b\\in \\text{R}$，可得$1\\le {{(a-1)}^{2}}+{{(b+1)}^{2}}\\le 4$，然后根据复数模的几何意义结合条件即得.\\\\ 【详解】\\\\ 令$z=a+b\\text{i}$且$a,b\\in \\text{R}$，则$2\\le \\left\\textbar{} (a-1)+(b+1)\\text{i} \\right\\textbar\\le 3$，\\\\ 所以$4\\le {{(a-1)}^{2}}+{{(b+1)}^{2}}\\le 9$，\\\\ 所以复数$z$在复平面内对应的点$Z$所在区域是圆${{(a-1)}^{2}}+{{(b+1)}^{2}}=4$和圆${{(a-1)}^{2}}+{{(b+1)}^{2}}=9$围成的圆环，\\\\ 所以点$Z$所在区域的面积为$9\\text{ } ! !\\pi ! !\\text{ }-4\\text{ } ! !\\pi ! !\\text{ }=5\\text{ } ! !\\pi ! !\\text{ }$.\\\\ 故选：C. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "427", "queId": "479ba9a7bad34506a145194f37d17d97", "competition_source_list": ["2016年高考真题天津卷", "2016年天津全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$0\\textless{}a\\textless{}b$$，已知$$a$$，$$s$$，$$t$$，$$b$$依次成等差数列，$$a$$，$$u$$，$$v$$，$$b$$依次成等比数列，记$$x=st\\left( s+t \\right)$$，$$y=uv\\left( u+v \\right)$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textgreater y$$ "}], [{"aoVal": "B", "content": "$$x=y$$ "}], [{"aoVal": "C", "content": "$$x\\textless{}y$$ "}], [{"aoVal": "D", "content": "既有$$x\\textgreater y$$的情形，也有$$x\\textless{}y$$的情形 "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->排序", "竞赛->知识点->不等式->几个重要的不等式->均值", "竞赛->知识点->不等式->不等式的综合应用（数列与不等式）", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件"], "answer_analysis": ["由题设有$$x=\\frac{2a+b}{3}\\cdot \\frac{a+2b}{3}\\left( a+b \\right)$$，$$y=ab\\left( \\sqrt[3]{{{a}^{2}}b}+\\sqrt[3]{a{{b}^{2}}} \\right)$$， 由均值不等式，$$\\frac{2a+b}{3}\\geqslant \\sqrt[3]{{{a}^{2}}b}$$，$$\\frac{a+2b}{3}\\geqslant \\sqrt[3]{a{{b}^{2}}}$$（等号都不成立）， 由排序不等式，$$a+b=\\sqrt[3]{{{a}^{3}}}+\\sqrt[3]{{{b}^{3}}}\\geqslant \\sqrt[3]{{{a}^{2}}b}+\\sqrt[3]{a{{b}^{2}}}$$（等号不成立）， 所以，$$x\\textgreater y$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1209", "queId": "fe86cb491f514cd9a3f2dedb9a82d301", "competition_source_list": ["2014年四川全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "半径为$$6$$的球，则该球内接正三棱锥的体积的最大值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$32\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$54\\sqrt{3}$$ "}], [{"aoVal": "C", "content": "$$64\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$72\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["设棱锥的高与一条侧棱的夹角为$$\\theta $$，侧棱长$$l=12\\cos \\theta $$，高$$h=12{{\\cos }^{2}}\\theta $$，底面正三角形的外接圆半径$$R=12\\sin \\theta \\cos \\theta $$，底面边长为$$a=12\\sqrt{2}\\sin \\theta \\cos \\theta $$，从而 $$V=432\\sqrt{3}{{\\cos }^{4}}\\theta {{\\sin }^{2}}\\theta $$ $$=216\\sqrt{3}\\cdot {{\\cos }^{2}}\\theta \\cdot {{\\cos }^{2}}\\theta \\cdot 2{{\\sin }^{2}}\\theta $$ $$\\leqslant 216\\sqrt{3}\\cdot {{\\left( \\frac{{{\\cos }^{2}}\\theta +{{\\cos }^{2}}\\theta +2{{\\sin }^{2}}\\theta }{3} \\right)}^{3}}$$ $$=64\\sqrt{3}$$， 当$${{\\tan }^{2}}\\theta =\\frac{1}{2}$$时等号成立． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "779", "queId": "581227768c344530932c1599aa34ab0c", "competition_source_list": ["2009年河北全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设函数$$f(x)=\\ln x$$，$$g(x)=ax+\\frac{b}{x}$$，它们的图象在$$x$$轴上的公共点处有公切线，则当$$x\\textgreater1$$时，$$f(x)$$与$$g(x)$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$f(x)\\textgreater g(x)$$ "}], [{"aoVal": "B", "content": "$$f(x)\\textless{}g(x)$$ "}], [{"aoVal": "C", "content": "$$f(x)=g(x)$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->导数模块->导数"], "answer_analysis": ["$$f(x)$$与$$g(x)$$的图象在$$x$$轴上有唯一公共点$$(1,0)$$， 所以$$g(1)=0$$，即$$a+b=0$$． 因为$${{f}^{\\prime }}(x)=\\frac{1}{x}$$，$${{g}^{\\prime }}(x)=a-\\frac{b}{{{x}^{2}}}$$， 由题意$${{f}^{\\prime }}(1)={{g}^{\\prime }}(1)=1$$，即$$a-b=1$$． 所以$$a=\\frac{1}{2}$$，$$b=-\\frac{1}{2}$$． 令$$F(x)=f(x)-g(x)=\\ln x-\\left( \\frac{1}{2}x-\\frac{1}{2x} \\right)$$， 则$${{F}^{\\prime }}(x)=\\frac{1}{x}-\\frac{1}{2}-\\frac{1}{2{{x}^{2}}}=-\\frac{1}{2}{{\\left( \\frac{1}{x}-1 \\right)}^{2}}\\leqslant 0$$． 所以$$F(x)$$在$$(1,+\\infty )$$内单调递减． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "937", "queId": "81b81f3da59f4f0ca68c7c41e877a5a6", "competition_source_list": ["2002年全国高中数学联赛竞赛一试第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)=\\frac{x}{1-{{2}^{x}}}-\\frac{x}{2}$$（~ ~ ）", "answer_option_list": [[{"aoVal": "A", "content": "是偶函数但不是奇函数 "}], [{"aoVal": "B", "content": "是奇函数但不是偶函数 "}], [{"aoVal": "C", "content": "既是奇函数又是偶函数 "}], [{"aoVal": "D", "content": "既不是奇函数又不是偶函数 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["$$f\\left( x \\right)$$定义域为$$(-\\infty ,0)\\cup (0,+\\infty )$$；$$f\\left( x \\right)-f\\left( -x \\right)=\\frac{x}{1-{{2}^{x}}}-\\frac{x}{2}-\\frac{-x}{1-{{2}^{-x}}}+\\frac{-x}{2}=\\frac{x-x{{2}^{x}}}{1-{{2}^{x}}}-x=0$$．即$$f\\left( x \\right)$$是偶函数. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "852", "queId": "6ee6a08223cc4da786ff41f3090d1c03", "competition_source_list": ["2019年吉林全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在正方形$$ABCD$$中，$$M$$为边$$BC$$的中点．若$$\\overrightarrow{AC}=\\lambda \\overrightarrow{AM}+\\mu \\overrightarrow{BD}$$，则$$\\lambda +2\\mu =$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{4}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{15}{8}$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算", "课内体系->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量的加法运算及运算规则"], "answer_analysis": ["注意到， $$\\overrightarrow{AC}=\\lambda \\overrightarrow{AM}+\\mu \\overrightarrow{BD}$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\lambda \\left( \\overrightarrow{AB}+\\overrightarrow{BM} \\right)+\\mu \\left( \\overrightarrow{BA}+\\overrightarrow{AD} \\right)$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\lambda \\left( \\overrightarrow{AB}+\\dfrac{1}{2}\\overrightarrow{AD} \\right)+\\mu \\left( -\\overrightarrow{AB}+\\overrightarrow{AD} \\right)$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\left( \\lambda -\\mu \\right)\\overrightarrow{AB}+\\left( \\dfrac{1}{2}\\lambda +\\mu \\right)\\overrightarrow{AD}$$． 又$$\\overrightarrow{AC}=\\overrightarrow{AB}+\\overrightarrow{AD}$$， 则$$\\begin{cases}\\lambda -\\mu =1 \\dfrac{1}{2}\\lambda +\\mu =1 \\end{cases}\\Rightarrow \\begin{cases}\\lambda =\\dfrac{4}{3} \\mu =\\dfrac{1}{3} \\end{cases}$$， 故$$\\lambda +2\\mu =2$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "814", "queId": "61301464eee94f0d8ebaed46945786db", "competition_source_list": ["高二下学期单元测试《整除与同余》自招", "第二十届全国希望杯高一竞赛复赛邀请赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "关于$$x$$的整系数一元二次方程$$a{{x}^{2}}+bx+c=0(a\\ne 0)$$中，若$$a+b$$是偶数，$$c$$是奇数，则．", "answer_option_list": [[{"aoVal": "A", "content": "方程没有整数根 "}], [{"aoVal": "B", "content": "方程有两个相等的整数根 "}], [{"aoVal": "C", "content": "方程有两个不相等的整数根 "}], [{"aoVal": "D", "content": "不能判定方程整数根的情况 "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->解方程（组）"], "answer_analysis": ["若方程有整数根，则根据题意 $$a{{x}^{2}}+bx+c=\\begin{cases}c(\\text{mod}2)\\left. 2 \\right\\textbar x a+b+c(\\text{mod}2)\\left. 2 \\right\\textbar x \\end{cases}\\equiv 1(\\text{mod}2)$$ 矛盾，因此题中方程没有整数根． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "385", "queId": "3e55864c3db744fe838230dc96a86068", "competition_source_list": ["2019~2020学年4月山西太原迎泽区太原市第五中学高二下学期周测C卷理科第6题6分", "2012年黑龙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设集合$$A=\\left { 1, 2, 3, 4, 5, 6 \\right }$$，$$B=\\left { 4, 5, 6, 7 \\right }$$，则满足$$S\\subseteq A$$且$$S\\cap B\\ne \\varnothing $$的集合$$S$$的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$57$$ "}], [{"aoVal": "B", "content": "$$56$$ "}], [{"aoVal": "C", "content": "$$49$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["$$S\\subseteq A$$，且$$S\\cap B\\ne \\varnothing $$，说明$$S$$是$$A$$的子集，且$$S$$与$$B$$有公共元素； ∴$$A$$的构成情况为：①含一个元素：从$$4$$，$$5$$，$$6$$中选一个元素，个数为$$\\text{C}_{3}^{1}=3$$； ②含两个元素:从$$4$$，$$5$$，$$6$$选两个元素，或从$$1$$，$$2$$，$$3$$选一个，从$$4$$，$$56$$选一个，个数为：$$\\text{C}_{3}^{2}+\\text{C}_{3}^{1}\\text{C}_{3}^{1}=12$$；③含三个元素：从$$4$$，$$5$$，$$6$$选三个，或从$$4$$，$$5$$，$$6$$选两个，从$$1$$，$$2$$，$$3$$选一个或从$$4$$，$$5$$，$$6$$选一个，从$$1$$，$$2$$，$$3$$选两个，个数为：$$\\text{C}_{3}^{3}+\\text{C}_{3}^{2}\\text{C}_{3}^{1}+\\text{C}_{3}^{1}\\text{C}_{3}^{2}=19$$； ④含四个元素：从$$4$$，$$5$$，$$6$$选三个从$$1$$，$$2$$，$$3$$选一个，或从$$4$$，$$5$$，$$6$$选两个，或从$$4$$，$$5$$，$$6$$选一个，从$$1$$，$$2$$，$$3$$选三个个数为：$$\\text{C}_{3}^{3}\\text{C}_{3}^{1}+\\text{C}_{3}^{2}\\text{C}_{3}^{2}+\\text{C}_{3}^{1}\\text{C}_{3}^{3}=15$$； ⑤含五个元素：从$$4$$，$$5$$，$$6$$选三个，从$$1$$，$$2$$，$$3$$选两个，或从$$4$$，$$5$$，$$6$$选两个，从$$1$$，$$2$$，$$3$$选三个，个数为：$$\\text{C}_{3}^{3}\\text{C}_{3}^{2}+\\text{C}_{3}^{2}\\text{C}_{3}^{3}=6$$含\"6\"个元素：从$$4$$，$$5$$，$$6$$选三个，从$$1$$，$$2$$，$$3$$选三个，个数为$$\\text{C}_{3}^{3}\\text{C}_{3}^{3}=1$$； ∴集合$$S$$的个数为：$$2+12+19+15+6+1=56$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "891", "queId": "78324fcaf23d42feb7034bbb85022a54", "competition_source_list": ["2020~2021学年浙江杭州萧山区杭州第二中学钱江学校高一下学期期中第5题5分", "2012年浙江全国高中数学联赛竞赛初赛第3题5分", "2016年上海闸北区高三二模理科第11题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\overrightarrow{a}$$与$$\\overrightarrow{b}$$均为单位向量，其夹角为$$\\theta $$，则命题$$P:\\textbar\\overrightarrow{a}-\\overrightarrow{b}\\textbar\\textgreater1$$是命题$$Q:\\theta \\in \\left[ \\frac{ \\pi }{2},\\frac{5 \\pi }{6} \\right)$$的（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "充分非必要条件 "}], [{"aoVal": "B", "content": "必要非充分条件 "}], [{"aoVal": "C", "content": "充分且必要条件 "}], [{"aoVal": "D", "content": "非充分且非必要条件 "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积的运算律", "课内体系->知识点->平面向量->平面向量的基本概念->向量的概念->向量的夹角的判断", "课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与向量结合"], "answer_analysis": ["若$$\\left\\textbar{} \\overrightarrow{a}-\\overrightarrow{b} \\right\\textbar\\textgreater1$$，则$${{\\overrightarrow{a}}^{2}}-2\\overrightarrow{a}\\cdot \\overrightarrow{b}+{{\\overrightarrow{b}}^{2}}=2-2\\overrightarrow{a}\\cdot \\overrightarrow{b}\\textgreater1$$，即$$\\overrightarrow{a}\\cdot \\overrightarrow{b}\\textless{}\\frac{1}{2}$$， 则$$\\cos \\theta =\\frac{\\overrightarrow{a}\\cdot \\overrightarrow{b}}{\\left\\textbar{} \\overrightarrow{a} \\right\\textbar\\cdot \\left\\textbar{} \\overrightarrow{b} \\right\\textbar}=\\overrightarrow{a}\\cdot \\overrightarrow{b}\\textless{}\\frac{1}{2}$$，∴$$\\theta \\in \\left( \\frac{ \\pi }{3}, \\pi ~\\right]$$，即$$P:\\theta \\in \\left( \\frac{ \\pi }{3}, \\pi ~\\right]$$， ∵$$Q:\\theta \\in \\left[ \\frac{ \\pi }{2},\\frac{5 \\pi }{6} \\right)$$，∴$$P$$是$$Q$$的必要不充分条件， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "960", "queId": "92eb030c0c0248c280733d98e16a5358", "competition_source_list": ["2014年AMC12竞赛A第11题", "2014年AMC10竞赛A第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2014-AMC10A-15$$ David drives from his home to the airport to catch a flight. He drives $$35$$ miles in the first hour, but realizes that he will be $$1$$ hour late if he continues at this speed. He increases his speed by $$15$$ miles per hour for the rest of the way to the airport and arrives $$30$$ minutes early. How many miles is the airport from his home? 大卫从他家开车去机场赶飞机。 他在第一个小时内行驶了 35 英里，但他意识到如果他继续保持这个速度，他将迟到 1小时。 在前往机场的剩余路程中，他将速度提高了每小时 15英里，并提前了 30分钟。 机场离他家多少英里？", "answer_option_list": [[{"aoVal": "A", "content": "$$140$$ "}], [{"aoVal": "B", "content": "$$175$$ "}], [{"aoVal": "C", "content": "$$210$$ "}], [{"aoVal": "D", "content": "$$245$$ "}], [{"aoVal": "E", "content": "$$280$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->等式->方程组的解集", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"], "answer_analysis": ["Note that he drives at $$50$$ miles per hour after the first hour and continues doing so until he arrives. Let $$d$$ be the distance still needed to travel after the frst $$1$$ hour.~ We have that $$\\dfrac{d}{50}+1.5= \\dfrac{d}{35}$$, ~where the $$1.5$$ comes from $$1$$ hour late decreased to $$0.5$$ hours early. Simplifying gives $$7d+525=10 d$$, or $$d=175$$. Now, we must add an extra $$35$$ miles traveled in the frst hour, giving a total or $$(\\rm C) 210$$ miles. Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices. Quickly checking, we know that neither choice $$(\\rm A)$$ or choice $$(\\rm B)$$ work, but $$(\\rm C)$$ does. We can verify as follows. After l hour at $$35 \\rm mph$$, David has $$175$$ miles left. This then takes him $$3.5$$ hours at $$50 \\rm mph$$. But $$210/35=6$$ hours. Since $$1+3.5=4.5$$ hours is $$1.5$$ hours less than $$6$$ our anwer is $$(\\rm C)210$$. Let us call the total number of miles after David changes speed $$x$$. We realize that if David is driving $$50$$ miles an hour, and he arrives $$30$$ minutes early, $$x≡25 (\\rm mod 50)$$ The only solution that fits the descripion is $$210$$. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "273", "queId": "5479f4bb7ae94dd9a312faa6b41898a7", "competition_source_list": ["2017年陕西全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "在空间直角坐标系中，$$\\triangle ABC$$的三个顶点坐标分别为$$A\\left( 3,4,1 \\right)$$，$$B\\left( 0,4,5 \\right)$$，$$C\\left( 5,2,0 \\right)$$，则$$\\tan \\frac{A}{2}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{5}}{5}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{6}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{6}}{6}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间向量", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["$$\\cos A=\\frac{\\overrightarrow{AB}\\cdot \\overrightarrow{AC}}{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}=\\frac{\\left( -3,0,4 \\right)\\cdot \\left( 2,-2,-1 \\right)}{\\sqrt{{{\\left( -3 \\right)}^{2}}+{{4}^{2}}}\\cdot \\sqrt{{{2}^{2}}+{{\\left( -2 \\right)}^{2}}+{{\\left( -1 \\right)}^{2}}}}=-\\frac{2}{3}$$， 又$$\\cos A=\\frac{1-{{\\tan }^{2}}\\frac{A}{2}}{1+{{\\tan }^{2}}\\frac{A}{2}}$$，解得$$\\tan \\frac{A}{2}=\\sqrt{5}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1194", "queId": "fe2ba6e532c54158a24ad5c61acf05cb", "competition_source_list": ["1999年全国全国高中数学联赛竞赛一试第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "在某次乒乓球单打比赛中，原计划每两名选手恰比赛一场，但有$$3$$名选手各比赛了$$2$$场之后就退出了，这样，全部比赛只进行了$$50$$场．那么，在上述$$3$$名选手之间比赛的场数是（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["解法一： 设这三名选手之间的比赛场数是$$r$$，共$$n$$名选手参赛． 由题意，可得$$\\text{C}_{n-3}^{2}+6-r=50$$， 即$$\\frac{\\left( n-3 \\right)\\left( n-4 \\right)}{2}=44+r$$． 由于$$0\\leqslant r\\leqslant 3$$， 经检验可知，仅当$$r=1$$时，$$n=13$$为正整数． 解法二： $$3$$名选手之间比赛的可能场数为$$0$$、$$1$$、$$2$$、$$3$$，设总人数为$$N$$人． 那么除这$$3$$人外的$$N-3$$人中比赛场数为$$C_{N-3}^{2}=\\frac{(N-3)(N-4)}{2}$$． ①当这$$3$$人之间比赛$$0$$场时，他们每人与另外$$N-3$$人（以下称为``局外人''）要比赛两场，这些比赛没有重合，共计$$6$$场，则有方程：$$\\frac{(N-3)(N-4)}{2}+6=50$$，$$N$$无整数解，故舍去． ②当这$$3$$人之间比赛$$1$$场时，他们有两人与``局内人''分别比赛一场，另一人两场都是和局内人比赛的，所以共计$$5$$场，则有方程：$$\\frac{(N-3)(N-4)}{2}+5=50$$，$$N=13$$，是整数解，满足条件． ③当这$$3$$人之间比赛$$2$$场时，他们有$$1$$人与另两人分别比赛一场，另两人都有一场与局内人的比赛，所以共计$$4$$场，则有方程：$$\\frac{(N-3)(N-4)}{2}+4=50$$，$$N$$无整数解，故舍去． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "262", "queId": "124d4a1f19f54e4fa8654685f6722570", "competition_source_list": ["2019~2020学年上海浦东新区华东师范大学第二附属中学高一上学期单元测试第5题", "2003年全国全国高中数学联赛竞赛一试第13题20分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$\\frac{3}{2}\\leqslant x\\leqslant 5$$，则下式$$2\\sqrt{x+1}+\\sqrt{2x-3}+\\sqrt{15-3x}$$最大不超过．", "answer_option_list": [[{"aoVal": "A", "content": "$${}2\\sqrt{17}$$ "}], [{"aoVal": "B", "content": "$${}2\\sqrt{19}$$ "}], [{"aoVal": "C", "content": "$${}2\\sqrt{21}$$ "}], [{"aoVal": "D", "content": "$${}2\\sqrt{23}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->柯西", "课内体系->知识点->等式与不等式->不等式->柯西不等式"], "answer_analysis": ["由$${{\\left( a+b+c+d \\right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+2\\left( ab+bc+cd+da+ac+bd \\right)$$可得 $$a+b+c+d\\leqslant 2\\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}}$$当且仅当$$a=b=c=d$$时取等号， 由柯西不等式 $${{\\left( 2\\sqrt{x+1}+\\sqrt{2x-3}+\\sqrt{15-3x} \\right)}^{2}}=\\left( \\sqrt{x+1}+\\sqrt{x+1}+\\sqrt{2x-3}+\\sqrt{15-3x} \\right)$$ $$\\leqslant 4\\left( x+1+x+1+2x-3+15-3x \\right)$$ $$=4(x+14)\\leqslant 4\\times 19$$ ∵$$\\sqrt{x+1}$$，$$\\sqrt{2x-3}$$，$$\\sqrt{15-3x}$$不能同时相等， ∴$$2\\sqrt{x+1}+\\sqrt{2x-3}+\\sqrt{15-3x}\\textless{}2\\sqrt{19}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "862", "queId": "77f524f6ccfe4a319049484dc828a2a8", "competition_source_list": ["2008年福建全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "在直角坐标平面$$xOy$$中，点$$A\\left( 5,0 \\right)$$．对于某个正实数$$k$$，存在函数$$f(x)=a{{x}^{2}}$$，$$a\\textgreater0$$，使得$$\\angle QOA=2\\angle POA$$，这里$$P(1,f(1))$$、$$Q(k,f(k))$$，则$$k$$的取值范围为．", "answer_option_list": [[{"aoVal": "A", "content": "$$(2,+\\infty )$$ "}], [{"aoVal": "B", "content": "$$(3,+\\infty )$$ "}], [{"aoVal": "C", "content": "$$[4,+\\infty )$$ "}], [{"aoVal": "D", "content": "$$[8,+\\infty )$$ "}]], "knowledge_point_routes": ["课内体系->知识点->直线和圆的方程->直线与方程->倾斜角和斜率的概念->斜率计算", "课内体系->知识点->三角函数->三角函数的图象与性质->正切函数的图象和性质", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正切", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->半角公式", "课内体系->素养->数学运算"], "answer_analysis": ["由于$$0\\textless{}\\angle QOA$$，$$\\angle POA\\textless{}\\frac{ \\pi }{2}$$， 因此$$\\angle QOA=2\\angle POA$$的充要条件是$$\\tan \\angle QOA=\\tan 2\\angle POA$$． 由$$\\tan \\angle QOA=\\tan 2\\angle POA=\\frac{2\\tan \\angle POA}{1-{{\\tan }^{2}}\\angle POA}$$， 结合$$\\tan \\angle POA=a$$，$$\\tan \\angle QOA=ak$$， 得$$ak=\\frac{2a}{1-{{a}^{2}}}$$，即$$a=\\sqrt{1-\\frac{2}{k}}$$，因此$$k\\textgreater2$$．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "223", "queId": "d51d9cef28454ea695b0ccc877bd266e", "competition_source_list": ["2008年贵州全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f(x)=\\sqrt{3}\\sin 2x+\\cos 2x$$的最小正周期是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{ \\pi }{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{ \\pi }{2}$$ "}], [{"aoVal": "C", "content": "$$ \\pi $$ "}], [{"aoVal": "D", "content": "$$2 \\pi $$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["$$f\\left( x \\right)={{\\cos }^{2}}2x+\\sqrt{3}\\sin 2x\\cos 2x$$ $$=\\frac{1+\\cos 4x}{2}+\\frac{\\sqrt{3}}{2}\\sin 4x$$ $$=\\sin \\left( 4x+\\frac{ \\pi }{6} \\right)+\\frac{1}{2}$$， 所以最小正周期$$T=\\frac{2 \\pi }{4}=\\frac{ \\pi }{2}$$，故选$$\\text{B}$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "101", "queId": "2ee9c14234fb4d89982890d59a6c659a", "competition_source_list": ["1988年全国高中数学联赛竞赛一试第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "已知原点在椭圆$${{k}^{2}}{{x}^{2}}+{{y}^{2}}-4kx+2ky+{{k}^{2}}-1=0$$的内部，那么参数$$k$$的取值范围是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\textbar k\\textbar\\textgreater1$$； "}], [{"aoVal": "B", "content": "$$\\textbar k\\textbar\\ne 1$$； "}], [{"aoVal": "C", "content": "$$-1\\textless{}k\\textless{}1$$； "}], [{"aoVal": "D", "content": "$$0\\textbar k\\textbar1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->椭圆"], "answer_analysis": ["椭圆外部的点可以离原点很远，它的坐标$$x\\mathsf{}y$$的绝对值可以很大，使得方程左边大于$$0$$，所以内部点的坐标使方程左边小于$$0$$，用原点的坐标代入方程左边得 $${{K}^{2}}-1\\textless{}0$$ 又$$k\\ne 0$$，（否则方程不表示椭圆），所以答$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "233", "queId": "156e6b80522441668e6f3d1873ca532f", "competition_source_list": ["2003年AMC10竞赛B第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$[x]$$ 表示不超过$$x$$的最大整数$$.$$ 例如, $$[3]=3$$, $$\\left[\\frac 92\\right]=4$$. 求$$[\\sqrt 1]+[\\sqrt 2]+[\\sqrt 3]+\\cdots +[\\sqrt {16}]$$的值.", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$ "}], [{"aoVal": "B", "content": "$$38$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$42$$ "}], [{"aoVal": "E", "content": "$$136$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Reasoning->Information Migration (new definition)", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->取整函数（高斯函数）"], "answer_analysis": ["The first three values in the sum are equal to $$1$$, the next five equal to $$2$$, the next seven equal to $$3$$, and the last one equal to $$4$$, For example, since $$2^{2}=4$$ any square root of a number less than $$4$$ must be less than $$2$$. Sum them all together to get $$3\\times 1+5\\times 2+7\\times 3+1\\times 4=3+10+21+4=38$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "738", "queId": "ace0c84fa43f4097a90536384ac7d9a0", "competition_source_list": ["2008年黑龙江全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$P$$为三角形$$ABC$$内部任一点（不包括边界），且满足$$(\\overrightarrow{PB}-\\overrightarrow{PA})\\cdot (\\overrightarrow{PB}+\\overrightarrow{PA}-2\\overrightarrow{PC})=0$$,则$$\\triangle ABC$$一定为．", "answer_option_list": [[{"aoVal": "A", "content": "直角三角形 "}], [{"aoVal": "B", "content": "等边三角形 "}], [{"aoVal": "C", "content": "等腰三角形 "}], [{"aoVal": "D", "content": "等腰直角三角形 "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算"], "answer_analysis": ["因为$$\\begin{cases}\\overrightarrow{PB}-\\overrightarrow{PA}=\\overrightarrow{AB} \\overrightarrow{PB}+\\overrightarrow{PA}-2\\overrightarrow{PC}=\\overrightarrow{CB}+\\overrightarrow{CA} \\end{cases}$$， 所以已知条件可改写为$$\\overrightarrow{AB}\\cdot (\\overrightarrow{CB}+\\overrightarrow{CA})=0$$． 容易得到此三角形为等腰三角形．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "157", "queId": "104d9364e5d84d70ab6c19cb8a9f091b", "competition_source_list": ["竞赛第9题"], "difficulty": "3", "qtype": "single_choice", "problem": "已知\\emph{x}，\\emph{y}，\\emph{z}是非负实数，且$x+y+z=2$，则$x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}+xyz$的最大值为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "1 "}], [{"aoVal": "B", "content": "2 "}], [{"aoVal": "C", "content": "$\\frac{5}{4}$ "}], [{"aoVal": "D", "content": "以上答案都不对 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 利用基本不等式可求最大值.\\\\ 【详解】\\\\ $2x^{2}y^{2}\\leq xy\\left(x^{2}+y^{2}\\right)$，$2y^{2}z^{2}\\leq yz\\left(y^{2}+z^{2}\\right)$，$2z^{2}x^{2}\\leq zx\\left(z^{2}+x^{2}\\right)$，\\\\ 所以$2x^{2}y^{2}+2y^{2}z^{2}+2z^{2}x^{2}+2xyz\\leq xy\\left(x^{2}+y^{2}\\right)+yz\\left(y^{2}+z^{2}\\right)+zx\\left(z^{2}+x^{2}\\right)+2xyz$，\\\\ $\\leq xy\\left(x^{2}+y^{2}\\right)+yz\\left(y^{2}+z^{2}\\right)+zx\\left(z^{2}+x^{2}\\right)+\\left(x+y+z\\right)xyz$\\\\ $=\\left(xy+yz+zx\\right)\\left(x^{2}+y^{2}+z^{2}\\right)$\\\\ $=\\frac{1}{2}\\left(2xy+2yz+2zx\\right)\\left(x^{2}+y^{2}+z^{2}\\right)$\\\\ $\\leq \\frac{1}{2}\\frac{\\left[\\left(2xy+2yz+2zx\\right)+\\left(x^{2}+y^{2}+z^{2}\\right)\\right]^{2}}{2}=2$，\\\\ 因此所求代数式的最大值为1．\\\\ 故选：A. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1081", "queId": "b843f51510874d5394e94aee56f39950", "competition_source_list": ["1986年全国高中数学联赛竞赛一试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果四面体的每一个面都不是等腰三角形，那么其长度不等的棱的条数最少为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->逻辑推理"], "answer_analysis": ["因四面体每个面都不是等腰三角形，故至少有三条棱的长度互不相等． 另一方面，每两条对棱都相等的四面体可以是仅长度互不相等且每个面都不是等腰三角形的四面体． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "567", "queId": "9091b114620845d7a4bdaf634971fc71", "competition_source_list": ["2003年AMC12竞赛B第4题", "2003年AMC10竞赛B"], "difficulty": "0", "qtype": "single_choice", "problem": "$$2003-AMC10B-8$$ Moe uses a mower to cut his rectangular $$90$$-foot by $$150$$-foot lawn. The swath he cuts is $$28$$ inches wide, but he overlaps each cut by $$4$$ inches to make sure that no grass is missed. he walks at the rate of $$5000$$ feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn? Moe 使用割草机修剪 90 英尺 x 150 英尺的长方形草坪。 他切割的条带宽度为 28 英寸，但他将每个切割条重叠 4 英寸以确保不会遗漏任何草。 他一边推着割草机，一边以每小时 5000 英尺的速度行走。 以下哪一项最接近 Moe 修剪草坪所需的小时数？", "answer_option_list": [[{"aoVal": "A", "content": "$$0.75$$ "}], [{"aoVal": "B", "content": "$$0.8$$ "}], [{"aoVal": "C", "content": "$$1.35$$ "}], [{"aoVal": "D", "content": "$$1.5$$ "}], [{"aoVal": "E", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的应用->函数的实际应用->一次函数模型", "美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Quadrilateral"], "answer_analysis": ["Since the swath Moe actually mows is $$24$$ inches, or $$2$$ feet wide, he mows $$10000$$ square feet in one hour. His lawn has an area of $$13500$$, so it will take Moe $$1.35$$ hours to finish mowing the lawn. Thus the answer is $$\\boxed{(\\text{C})1.35}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "361", "queId": "1f9e530d52c647e1804700101bd86a53", "competition_source_list": ["2010年全国全国高中数学联赛竞赛复赛一试第2题8分", "高二上学期单元测试《导数及其应用》自招第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知函数$$y=(a{{\\cos }^{2}}x-3)\\sin x$$的最小值为$$-3$$，则实数$$a$$的取值范围是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{3}{2}\\leqslant a\\leqslant 12$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{2}\\leqslant a\\leqslant 12$$ "}], [{"aoVal": "C", "content": "$$-12\\leqslant a\\leqslant \\frac{3}{2}$$ "}], [{"aoVal": "D", "content": "$$-12\\leqslant a\\leqslant- \\frac{3}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域->求函数的值域->用换元法求值域", "课内体系->知识点->等式与不等式->不等式->解不等式->一元二次不等式", "课内体系->素养->数学运算"], "answer_analysis": ["令$$\\sin x=t$$，则原函数化为$$g(t)=(-a{{t}^{2}}+a-3)t$$，即$$g(t)=-a{{t}^{3}}+(a-3)t$$， 由$$-a{{t}^{3}}+(a-3)t\\geqslant -3$$，$$-at({{t}^{2}}-1)-3(t-1)\\geqslant 0$$，$$(t-1)(-at(t+1)-3)\\geqslant 0$$及$$t-1\\leqslant 0$$，知$$-at(t+1)-3\\leqslant 0$$即$$a({{t}^{2}}+t)\\geqslant -3$$，（$$1$$） 当$$t=0$$，$$1$$时（$$1$$）总成立； 对$$0\\textless{}t\\leqslant 1$$，$$0\\textless{}{{t}^{2}}+t\\leqslant 2$$，对$$-1\\textless{}t\\textless{}0$$，$$-\\frac{1}{4}\\leqslant {{t}^{2}}+t\\textless{}0$$， 从而可知$$-\\frac{3}{2}\\leqslant a\\leqslant 12$$． 问题即函数$$f\\left( x \\right)=-a{{x}^{3}}+(a-3)x$$在区间$$[-1,1]$$上的最小值为$$-3$$．函数$$f\\left( x \\right)$$的导函数为$${{f}^{\\prime }}\\left( x \\right)=-3a{{x}^{2}}+(a-3)$$． 注意到$$f(1)=-3$$，于是$${{f}^{\\prime }}\\left( 1 \\right)\\leqslant 0$$，可得$$a\\geqslant -\\frac{3}{2}$$． 情形一：$$-\\frac{3}{2}\\leqslant a\\leqslant 3$$， 此时恒有$${{f}^{\\prime }}\\left( x \\right)\\leqslant 0$$，函数$$f(x)$$在区间$$[-1,1]$$上单调递减，符合题意． 情形二：$$a\\textgreater3$$． 此时函数$$f(x)$$在区间$$[-1,1]$$上有极小值点$$x=-\\sqrt{\\frac{a-3}{3a}}$$， 于是根据题意，有$$f\\left( -\\sqrt{\\frac{a-3}{3\\alpha }} \\right)\\geqslant -3$$， 即$$-\\frac{2(a-3)}{3}\\cdot \\sqrt{\\frac{a-3}{3a}}\\geqslant -3$$， 整理得$$(a-12)(4{{a}^{2}}+12a+9)\\leqslant 0$$， 解得$$a\\leqslant 12$$． 综上所述，$$a$$的取值范围是$$\\left[ -\\frac{3}{2},12 \\right]$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "502", "queId": "3f49d6d2bb5d4441a2db8beef617040e", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "体育课下课后，老师要求体育委员把$$5$$个相同的篮球、$$3$$个相同的排球、$$2$$个相同的橄榄球排成一排放好，则不同的放法有．", "answer_option_list": [[{"aoVal": "A", "content": "$$420$$种 "}], [{"aoVal": "B", "content": "$$1260$$种 "}], [{"aoVal": "C", "content": "$$5040$$种 "}], [{"aoVal": "D", "content": "$$2520$$种 "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->两个基本计数原理", "竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["总共$$10$$个球，先考虑选$$5$$个位置放篮球，共有$$\\text{C}_{10}^{5}=252$$种，再从剩下的$$5$$个位置种选$$3$$个位置放排球，总共有$$10$$种方法，剩下的$$2$$个位置放橄榄球，故有$$252\\times 10=2520$$种. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "497", "queId": "87fbbd8459af449c83bc4909774def30", "competition_source_list": ["1984年全国高中数学联赛竞赛一试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$F\\left( \\frac{1-x}{1+x} \\right)=x$$，则下列等式中正确的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$F\\left( -2-x \\right)=-2-F\\left( x \\right)$$ "}], [{"aoVal": "B", "content": "$$F\\left( -x \\right)=F\\left( \\frac{1+x}{1-x} \\right)$$ "}], [{"aoVal": "C", "content": "$$F\\left( {{x}^{-1}} \\right)=F\\left( x \\right)$$ "}], [{"aoVal": "D", "content": "$$F\\left( F\\left( x \\right) \\right)=-x$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的概念", "竞赛->知识点->不等式->换元技巧->代数换元"], "answer_analysis": ["令 $$u=\\frac{1-x}{1+x}$$，解得 $$x=\\frac{1-u}{1+u}$$． 所以 $$F\\left( u \\right)=\\frac{1-u}{1+u}$$，即 $$F\\left( x \\right)=\\frac{1-x}{1+x}$$ 然后逐一验证 $$F\\left( -2-x \\right)=\\frac{1-\\left( -2-x \\right)}{1+\\left( -2-x \\right)}=-\\frac{3+x}{1+x}$$ $$=-2-\\frac{1-x}{1+x}=-2-F\\left( x \\right)$$ 选择$$A$$． 又本题也可用特殊值验证排除法． 由 $$F\\left( \\frac{1-x}{1+x} \\right)=x$$中． 令 $$x=1$$，得 $$F\\left( 0 \\right)=1$$ 令 $$x=0$$，得 $$F\\left( 1 \\right)=0$$． 而在$$\\text{B}$$的等式中，令$$x=0$$，有$$F\\left( 0 \\right)=F\\left( 1 \\right)$$，所以应排除$$\\text{B}$$，在$$\\text{C}$$中的等式中，$$x=0$$无意义，排除$$\\text{C}$$．在等式$$\\text{D}$$中，左边$$F\\left( F\\left( 1 \\right) \\right)=F\\left( 0 \\right)=1\\ne -1$$��所以也排除$$\\text{D}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "330", "queId": "ec76c8006220487f9e172d707ff9fc55", "competition_source_list": ["2012年黑龙江全国高中数学联赛竞赛初赛第9题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知不等式$$\\frac{x-5}{x+1}\\textless{}0$$的解集为$$M$$，若$${{x}_{0}}\\in M$$，则$${{\\log }_{2}}\\left( {{x}_{0}}+1 \\right)\\textless{}1$$的概率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{5}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->概率初步", "竞赛->知识点->函数->基本初等函数", "竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["$$M=\\left { x\\textbar-1\\textless{}x\\textless{}5 \\right }, {{\\log }_{2}}\\left( x+1 \\right)\\textless{}1$$的解集为$$\\left { x\\textbar-1\\textless{}x\\textless{}1 \\right }$$，所以概率为$$\\frac{1}{3}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "768", "queId": "9aa94c040db64133ab20cda1b167951b", "competition_source_list": ["2021年新疆全国高中数学联赛竞赛初赛第3题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "将正整数中所有数码不超过$$5$$的数从小到大排成一列，则第$$2021$$个数是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$10000$$ "}], [{"aoVal": "B", "content": "$$13205$$ "}], [{"aoVal": "C", "content": "$$13321$$ "}], [{"aoVal": "D", "content": "$$14102$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->排列与组合", "竞赛->知识点->排列组合与概率->两个基本计数原理"], "answer_analysis": ["方法一：所有数码不超过$$5$$的一位正整数有$$5$$个，两位正整数有$$5\\times 6=30$$个，三位正整数有$$5\\times {{6}^{2}}=180$$个，四位正整数有$$5\\times {{6}^{3}}=1080$$个，共有$$1295$$个； 万位数为$$1$$，千位为$$0$$，共$$216$$个； 万位数为$$1$$，千位为$$1$$，共$$216$$个； 万位数为$$1$$，千位为$$2$$，共$$216$$个；共$$1943$$个， 万位数为$$1$$，千位为$$3$$，百位是$$0$$，$$1$$各$$36$$个，共$$72$$个， 一共$$1943+72=2015$$个，还差$$6$$个，百位是$$2$$，个位取$$0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$， 所以第$$2021$$个数是$$13205$$． 方法二：数码不超过$$5$$的数可以与一个六进制数建立一一对应关系，$$2021=1\\times {{6}^{4}}+3\\times {{6}^{3}}+2\\times {{6}^{2}}+0\\times 6+5$$，利用除$$6$$取余法可得， 即$${{\\left( 2021 \\right)}_{10}}={{\\left( 13205 \\right)}_{6}}$$， 所以答案是：$$13205$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "363", "queId": "280e72fb79ac499e969926e0d700a530", "competition_source_list": ["第二十届全国希望杯高一竞赛复赛邀请赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f(x)=a{{x}^{3}}+b{{x}^{2}}+cx-1(a\\textless{}0)$$，且$$f(5)=3$$，则使$$f(x)=0$$成立的$$x$$的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "不确定的 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["注意到$$x\\to -\\infty $$时，$$f(x)\\textgreater0$$；$$f(0)=-1\\textless{}0$$；$$f(5)=3\\textgreater0$$；$$x\\to +\\infty $$时，$$f(x)\\textless{}0$$，于是$$f(x)=0$$有$$3$$个实根． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "200", "queId": "21dbc6782f2d4d93876e00b4039a4ce8", "competition_source_list": ["2005年AMC12竞赛A第21题"], "difficulty": "3", "qtype": "single_choice", "problem": "2005AMC12A, 21 How many ordered triples of integers $$(a, b, c)$$, with $$a\\geqslant2$$, $$b\\geqslant1$$, and $$c\\geqslant0$$, satisfy both $$\\log_ab=c^{2005}$$ and $$a+b+c= 2005$$? 有多少个整数 $$(a, b, c)$$ 的有序三元组，其中 $$a\\geqslant2$$、$$b\\geqslant1$$ 和 $$c\\geqslant0$$ 满足 $$\\log_ab =c^{2005}$$ 和 $$a+b+c= 2005$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}], [{"aoVal": "E", "content": "$$4$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Calculation->Solve Quadratic Equations", "课内体系->知识点->基本初等函数->对数的概念及其运算->指对化简求值"], "answer_analysis": ["由$\\log_a{b}=c^{2005}$可知$b=a^{c\\^{2005}}$. 又由$a+b+c=2005$知$a\\textless2005$, 故$c=0$或$c=1$. $c=0$时, $b=a^{0}=1$, $a=2005-b-c=2004$. $c=1$时, $b=a^{1}=a$, 再由$a+b+c=2005$知$(a,b,c)=(1002, 1002, 1)$. 共两种, 选$$\\text{C}$$. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "398", "queId": "4bfdbd424d7140aa82b43d3b084e4c3e", "competition_source_list": ["2015年黑龙江全国高中数学联赛竞赛初赛第15题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$x$$，$$y\\in \\left[ -\\frac{\\pi }{4},\\frac{\\pi }{4} \\right]$$，且$${{x}^{3}}+\\sin x-2a=0$$，$$4{{y}^{3}}+\\frac{1}{2}\\sin 2y+a=0$$，则$$\\cos (x+2y)=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->函数的概念与性质->函数的性质->奇偶性"], "answer_analysis": ["解析：第二个式子可以化成$${{(2y)}^{3}}+\\sin 2y+2a=0$$，由于$${{x}^{3}}+\\sin x$$是奇函数，且$$2y\\in \\left[ -\\frac{\\pi }{2},\\frac{\\pi }{2} \\right]$$， 所以$$x=-2y$$，$$\\cos (x+2y)=\\cos 0=1$$． 故答案为：$$1$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "348", "queId": "39a2fe5b074647c3838f0d4df64d6a53", "competition_source_list": ["2009年浙江全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$${{\\left( x-\\frac{1}{{{x}^{6}}} \\right)}^{2009}}$$的二项展开式中常数项为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\text{C}_{2009}^{286}$$ "}], [{"aoVal": "B", "content": "$$\\text{C}_{2009}^{287}$$ "}], [{"aoVal": "C", "content": "$$-\\text{C}_{2009}^{286}$$ "}], [{"aoVal": "D", "content": "$$-\\text{C}_{2009}^{287}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->二项式定理->求二项式展开式的特定项", "课内体系->知识点->计数原理->二项式定理->二项式定理的展开式", "课内体系->知识点->计数原理->二项式定理->求项的系数或二项式系数", "课内体系->素养->数学运算"], "answer_analysis": ["由于 $${{\\left( x-\\frac{1}{{{x}^{6}}} \\right)}^{2009}}=\\sum\\limits_{k=0}^{2009}{\\text{C}_{2009}^{k}{{x}^{2009-k}}{{(-{{x}^{-6}})}^{k}}}$$， 则若要出现常数项，须满足$$2009-k-6k=0\\Rightarrow k=287$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1061", "queId": "af170110e88e4191b7b3eb644f3ad89e", "competition_source_list": ["2014年天津全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "直线$$l$$在平面$$ \\pi $$上，直线$$m$$平行于平面$$ \\pi $$，并与直线$$l$$异面．动点$$P$$在平面$$ \\pi $$上，且到$$l$$和$$m$$的距离相等．则$$P$$点的轨迹是．", "answer_option_list": [[{"aoVal": "A", "content": "直线 "}], [{"aoVal": "B", "content": "椭圆 "}], [{"aoVal": "C", "content": "抛物线 "}], [{"aoVal": "D", "content": "双曲线 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->轨迹方程（二试）"], "answer_analysis": ["设$$m$$在平面$$ \\pi $$上的投影为$${m}'$$，$${m}'$$交$$l$$于$$O$$点．在平面$$ \\pi $$上，以$$O$$为原点，$$l$$为$$y$$轴建立直角坐标系，则可设$${m}'$$的方程为$$y=k \\pi $$．又设$$P$$点的坐标为$$\\left( x, y \\right)$$，则$$P$$到$$l$$的距离为$$\\left\\textbar{} x \\right\\textbar$$；它到$${m}'$$的距离为$$\\frac{\\left\\textbar{} y-kx \\right\\textbar}{\\sqrt{1+{{k}^{2}}}}$$，从而$$P$$到$$m$$的距离平方等于 $$\\frac{{{\\left( y-kx \\right)}^{2}}}{1+{{k}^{2}}}+{{a}^{2}}$$， 其中$$a$$为直线$$m$$到平面$$ \\pi $$的距离．因此，$$P$$点的轨迹方程是 $$\\frac{{{\\left( y-kx \\right)}^{2}}}{1+{{k}^{2}}}+{{a}^{2}}={{x}^{2}}$$． 可见轨迹是双曲线． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "220", "queId": "115223573a894ddf85a7eafa12cfba67", "competition_source_list": ["1999年全国全国高中数学联赛竞赛一试第3题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$${{\\left( {{\\log }_{2}}3 \\right)}^{x}}-{{\\left( {{\\log }_{5}}3 \\right)}^{x}}\\geqslant {{\\left( {{\\log }_{2}}3 \\right)}^{-y}}-{{\\left( {{\\log }_{5}}3 \\right)}^{-y}}$$，则（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$x-y\\geqslant 0$$ "}], [{"aoVal": "B", "content": "$$x+y\\geqslant 0$$ "}], [{"aoVal": "C", "content": "$$x-y\\leqslant 0$$ "}], [{"aoVal": "D", "content": "$$x+y\\leqslant 0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性��->函数的性质->单调性->函数单调性的应用->利用函数单调性解不等式", "课内体系->素养->数学运算"], "answer_analysis": ["记$$f\\left( t \\right)={{\\left( {{\\log }_{2}}3 \\right)}^{t}}-{{\\left( {{\\log }_{5}}3 \\right)}^{t}}$$， 则$$f\\left( t \\right)$$在$$\\mathbf{R}$$上是严格增函数． 原不等式即$$f\\left( x \\right)\\geqslant f\\left( -y \\right)$$． 故$$x\\geqslant -y$$， 即$$x+y\\geqslant 0$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "232", "queId": "26a5ca2c4a18445d9914ab54df84e07e", "competition_source_list": ["2011年吉林全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，已知$$6\\overrightarrow{AC}\\cdot \\overrightarrow{AB}=2\\overrightarrow{AB}\\cdot \\overrightarrow{BC}=3\\overrightarrow{BC}\\cdot \\overrightarrow{CA}$$，则$$\\angle A=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$30{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$45{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$60{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$135{}^{}\\circ $$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算", "竞赛->知识点->三角函数->三角形中的问题->解三角形"], "answer_analysis": ["设$$\\triangle ABC$$的三边分别为$$a,b,c,$$ 由已知$$6\\overrightarrow{AC}\\cdot \\overrightarrow{AB}=2\\overrightarrow{AB}\\cdot \\overrightarrow{BC}=3\\overrightarrow{BC}\\cdot \\overrightarrow{CA}$$， 可得$$6bc\\cos A=2ac\\cos \\left( ~\\pi -B \\right)=3ab\\cos \\left( ~\\pi -C \\right)$$ 即$$6bc\\cos A=-2ac\\cos B=-3ab\\cos C$$， 再利用余弦定理可得$$6bc\\cdot \\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=-2ac\\cdot \\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}=-3ab\\cdot \\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$$， 化简可得$${{a}^{2}}=5{{b}^{2}},{{c}^{2}}=2{{b}^{2}},$$ 所以$$\\cos A=\\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}=-\\frac{\\sqrt{2}}{2}$$， 故$$A=135{}^{}\\circ $$，故选$$\\text{D}$$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "628", "queId": "3c3c2ceff3864c45a85b5ace1df39c1e", "competition_source_list": ["1993年全国高中数学联赛竞赛一试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "若直线$$x=\\frac{ \\pi }{4}$$被曲线$$C$$：$$\\left( x-\\arcsin \\alpha \\right)\\left( x-\\arccos \\alpha \\right)+\\left( y-\\arcsin \\alpha \\right)\\left( y+\\arccos \\alpha \\right)=0$$所截得的弦长为$$d$$，当$$a$$变化时$$d$$的最小值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{ \\pi }{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{ \\pi }{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{ \\pi }{2}$$ "}], [{"aoVal": "D", "content": "$$ \\pi $$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->圆与方程", "竞赛->知识点->解析几何->直线与方程"], "answer_analysis": ["由题设知，曲线$$C$$是以$${{P}_{1}}\\arcsin a,\\arcsin a$$，$${{P}_{2}}\\arccos a-\\arccos a$$两点为直径端点的圆．其圆心的横坐标为 $${{x}_{0}}=\\frac{\\arcsin a+\\arccos a}{2}=\\frac{\\mathsf{\\pi }}{4}$$， 故直线$$x=\\frac{\\pi }{4}$$过圆心，$$d$$就是该圆的直径．而 $${{d}^{2}}=\\left[ {{(\\arcsin a)}^{2}}+(\\arccos a) \\right]\\mathsf{\\geqslant }{{(\\arcsin a+\\arccos a)}^{2}}={{\\frac{ \\pi }{4}}^{2}}$$． ∴$$d\\mathsf{\\geqslant }\\frac{ \\pi }{2}$$，又当$$\\arcsin a=\\arccos a$$时相应的$$d\\mathsf{\\geqslant }\\frac{ \\pi }{2}$$， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1038", "queId": "b33bd35ffcb346bbb248a4d8f2a0ef87", "competition_source_list": ["2018年河北唐山高三二模理科第13题5分", "2017~2018学年陕西西安碑林区西北工业大学附属中学高二上学期期中理科第14题3分", "2018~2019学年山东济南高三上学期期末理科第13题5分", "2020年山东日照高三一模第14题5分", "2020~2021学年11月广东广州海珠区广州市第五中学高三上学期月考第14题5分", "2008年贵州全国高中数学联赛竞赛初赛第13题4分", "2017年湖北武汉高三三模理科第13题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$${{\\left( {{x}^{2}}-\\frac{1}{x} \\right)}^{6}}$$的展开式中，常数项为 ．（用数字作答）", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$45$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->二项式定理->二项式定理的展开式", "课内体系->知识点->计数原理->二项式定理->求项的系数或二项式系数", "课内体系->素养->数学运算"], "answer_analysis": ["$${{T}_{r+1}}=\\text{C}_{6}^{r}{{({{x}^{2}})}^{6-r}}{{\\left( -\\frac{1}{x} \\right)}^{2}}$$ $$={{(-1)}^{r}}\\text{C}_{6}^{r}{{x}^{12-2r}}\\cdot {{x}^{-r}}$$ $$={{(-1)}^{r}}\\text{C}_{6}^{r}{{x}^{12-3r}}$$， ∵求常数项， ∴令$$12-3r=0$$，则$$r=4$$， ∴$${{T}_{5}}={{(-1)}^{4}}\\cdot \\text{C}_{6}^{4}{{x}^{0}}=15$$． 故答案为：$$15$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "377", "queId": "1bd8bb55f36f4573a32f6adce857bcee", "competition_source_list": ["2014年AMC10竞赛A第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2014-AMC10A-11$$ A customer who intends to purchase an appliance has three coupons, only one of which may be used: Coupon $$1$$: $$10$$\\% off the listed price if the listed price is at least ＄$$50$$. Coupon $$2$$: ＄$$20$$ off the listed priceif the listed price is at least＄$$100$$. Coupon $$3$$: $$18$$\\% off the amount by which the listed price exceeds ＄$$100$$. For which of the following listed prices will coupon $$1$$ offer a greater price reduction than either coupon $$2$$ or coupon $$3$$? 打算购买电器的客户有三张优惠券，只能使用其中一张： 优惠券 $$1$$：如果标价至少为 ＄$$50$$，则比标价低 10\\%。 优惠券$$2$$：如果标价至少为$$100$$，则标价可减$20。 优惠券 $$3$$：标价超过$$100$$的部分可享受 $$18$$\\% 的折扣。 对于以下哪个列出的价格，优惠券 $$1$$ 比优惠券 $$2$$ 或优惠券 $$3$$ 的降价幅度更大？", "answer_option_list": [[{"aoVal": "A", "content": "＄$$179.95$$ "}], [{"aoVal": "B", "content": "＄$$199.95$$ "}], [{"aoVal": "C", "content": "＄$$219.95$$ "}], [{"aoVal": "D", "content": "＄$$239.95$$ "}], [{"aoVal": "E", "content": "＄$$259.95$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的应用->函数的实际应用->一次函数模型", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"], "answer_analysis": ["Let the listed price be $$x$$. Since all the answer choices are above ＄$$100$$, we can assume $$x\\textgreater{} 100$$. Thus the discounts after the coupons are used will be as follows: Coupon $$1$$: $$x\\times10$$\\%$$=$$$$0.1 x$$. Coupon $$2$$: $$20$$. Coupon $$3$$: $$18$$\\%$$\\times (x-100)=0.18 x-18$$. For coupon $$1$$ to give a greater price reduction than the other coupons, we must have $$0.1 x\\textgreater20 \\Rightarrow x\\textgreater200$$ and $$0.1x\\textgreater0.18 x-18 \\Rightarrow0 .08 x\\textless18 \\Rightarrow x\\textless225$$. From the first inequality, the listed price must be greater than ＄$$200$$, so answer choices $$(\\rm A)$$ and $$(\\rm B)$$ are eliminated. From the second inequality, the listed price must be less than ＄$$225$$, so answer choices $$(\\rm D)$$ and $$(\\rm E)$$ are eliminated. The only answer choice that remains is $$(\\rm C)$$＄$$219.95$$. For coupon $$1$$ to be the most effective, we want $$10$$\\% of the price to be greater than $$20$$. This clearly ~occurs if the value is over $$200$$. For coupon $$1$$ to be more effective than coupon $$3$$, we want to minimize the value over $$200$$, so $$(\\rm C)$$＄$$219.95$$ is the smallest number over $$200$$. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "591", "queId": "83c96124ba954cfea44c793e087b34d6", "competition_source_list": ["2018年安徽全国高中数学联赛竞赛初赛第3题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)=\\left\\textbar{} \\sin 2x+\\sin 3x+\\sin 4x \\right\\textbar$$的最小正周期为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\pi$$ "}], [{"aoVal": "B", "content": "$$2\\pi$$ "}], [{"aoVal": "C", "content": "$$3\\pi$$ "}], [{"aoVal": "D", "content": "$$4\\pi$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["由和差化积公式得： $$\\sin 2x+\\sin 4x=2\\cos x\\sin 3x$$， ∴原式$$=\\left\\textbar{} 2\\cos x\\sin 3x+\\sin 3x \\right\\textbar$$$$=\\left\\textbar{} 2\\cos +1 \\right\\textbar\\cdot \\left\\textbar{} \\sin 3x \\right\\textbar$$， 又$$\\left\\textbar{} 2\\cos x+1 \\right\\textbar$$最小正周期为$$2\\pi $$， $$\\left\\textbar{} \\sin 3x \\right\\textbar$$最小正周期为$$\\frac{\\pi }{3}$$， 故$$f\\left( x \\right)$$最小正周期为$$2\\pi $$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "956", "queId": "d6efe0c0de544db4a6c104088bff2fdc", "competition_source_list": ["1997年全国高中数学联赛竞赛一试第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$f(x)={{x}^{2}}- \\pi x$$，$$\\alpha =\\arcsin \\frac{1}{3}$$，$$\\beta =\\arctan \\frac{5}{4}$$，$$\\gamma =\\arccos \\left( -\\frac{1}{3} \\right)$$，$$\\delta =\\text{arc}\\cot \\left( -\\frac{5}{4} \\right)$$，则（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$f\\left( \\alpha \\right)\\textgreater f\\left( \\beta \\right)\\textgreater f\\left( \\delta \\right)\\textgreater f\\left( \\gamma \\right)$$ "}], [{"aoVal": "B", "content": "$$f\\left( \\alpha \\right)\\textgreater f\\left( \\delta \\right)\\textgreater f\\left( \\beta \\right)\\textgreater f\\left( \\gamma \\right)$$ "}], [{"aoVal": "C", "content": "$$f\\left( \\delta \\right)\\textgreater f\\left( \\alpha \\right)\\textgreater f\\left( \\beta \\right)\\textgreater f\\left( \\gamma \\right)$$ "}], [{"aoVal": "D", "content": "$$f\\left( \\delta \\right)\\textgreater f\\left( \\alpha \\right)\\textgreater f\\left( \\gamma \\right)\\textgreater f\\left( \\beta \\right)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->反三角函数", "课内体系->知识点->函数的概念与性质->二次函数->二次函数的图象及性质->二次函数相关的恒成立问题", "课内体系->知识点->函数的概念与性质->二次函数->二次函数的图象及性质->含参二次函数的图象及性质", "课内体系->素养->直观想象"], "answer_analysis": ["$$f(x)$$的对称轴为$$x=\\frac{ \\pi }{2}$$， 易得$$0\\textless{}\\alpha \\textless{}\\frac{ \\pi }{6}\\textless{}\\frac{ \\pi }{4}\\textless{}\\beta \\textless{}\\frac{ \\pi }{3}\\textless{}\\frac{ \\pi }{2}\\textless{}\\gamma \\textless{}\\frac{2 \\pi }{3}\\textless{}\\frac{3 \\pi }{4}\\textless{}\\delta \\textless{}\\frac{5 \\pi }{6}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "617", "queId": "406c03eca2c0491b8b2336041f07b08b", "competition_source_list": ["2012年天津全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "若$$x\\textgreater1$$，则$${{x}^{\\ln \\ln x}}-{{\\left( \\ln x \\right)}^{\\ln x}}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "正数 "}], [{"aoVal": "B", "content": "零 "}], [{"aoVal": "C", "content": "负数 "}], [{"aoVal": "D", "content": "以上皆有可能 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["令$$m=\\ln x$$，则$${{x}^{\\ln \\ln x}}-{{\\left( \\ln x \\right)}^{\\ln x}}={{\\left( {{\\text{e}}^{m}} \\right)}^{\\ln m}}-{{m}^{m}}=0$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1167", "queId": "d90494d0f3df480c80afdcdecbbd5c5c", "competition_source_list": ["第二十届全国希望杯高一竞赛复赛邀请赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "定义集合$$M,N$$的一种运算*：$$M*N= {x\\textbar x={{x}_{1}}{{x}_{2}},{{x}_{1}}\\in M,{{x}_{2}}\\in N }$$，若$$M= {1,2,3 }$$，$$N= {0,1,2 }$$，则$$M*N$$中的所有元素的和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["$$M*N= {0,1,2,3,4,6 }$$，故所求的和为$$16$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "395", "queId": "201430f86d7c4054a28c8ce1252e8349", "competition_source_list": ["2019年吉林全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "设集合$$A=\\left { 2,0,1,3 \\right }$$，$$B=\\left { x\\left\\textbar{} -x\\in A,2-{{x}^{2}}\\notin A \\right. \\right }$$，则集合$$B$$中所有元素的和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-4$$ "}], [{"aoVal": "B", "content": "$$-5$$ "}], [{"aoVal": "C", "content": "$$-6$$ "}], [{"aoVal": "D", "content": "$$-7$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["易知，$$B=\\left { -2,-3 \\right }$$， 则集合$$B$$中所有元素的和为$$-5$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "757", "queId": "72af90b6511c403eb78595c440dfdb70", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知三个不同的平面$$\\alpha ,\\beta ,\\gamma $$和两条不重合的直线$$m, n$$，有下列$$4$$个命题： ①$$m//\\alpha $$，$$\\alpha \\cap \\beta =n$$，则$$m//n$$ ②$$m\\bot \\alpha , m//n, n\\subset \\beta $$，则$$\\alpha \\bot \\beta $$ ③$$\\alpha \\bot \\beta ,\\gamma \\bot \\beta $$，则$$\\alpha //\\gamma $$ ④$$\\alpha \\cap \\beta =m, m\\bot \\gamma $$，则$$\\alpha \\bot \\gamma $$ 其中正确命题的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的平行和垂直"], "answer_analysis": ["①~~ 错误，$$\\beta $$不一定是经过直线$$m$$的平面； ②~~ 正确； ③错误，例如教室的墙角，不妨设$$\\alpha $$为东墙面，$$\\gamma $$为北墙面，$$\\beta $$为底面，满足$$\\alpha \\bot \\beta ,\\gamma \\bot \\beta $$，但$$\\alpha $$与$$\\gamma $$相交； ④正确，故选$$\\text{A}$$. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "585", "queId": "ac42c008ca624034b51473b6399a16e8", "competition_source_list": ["2015年吉林全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$f(x)=\\begin{cases}{{\\left( \\frac{1}{4} \\right)}^{x}}, , x\\in ( ,-\\infty , 1 ) {{\\log }_{\\frac{1}{2}}}x, x\\in \\left[ ,1 , +\\infty \\right) \\end{cases}$$，则$$f\\left[ f( ,-1) \\right]=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "D", "content": "$$-\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的概念", "竞赛->知识点->函数->函数方程", "竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["$$f\\left[ f\\left( -1 \\right) \\right]=f\\left[ {{\\left( \\frac{1}{4} \\right)}^{-1}} \\right]=f\\left( 4 \\right)={{\\log }_{\\frac{1}{2}}}4=-2$$ "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "935", "queId": "a95bfbd396184fbb900671303482aa6a", "competition_source_list": ["2015年广东全国高中数学联赛竞赛初赛第3题8分"], "difficulty": "0", "qtype": "single_choice", "problem": "若函数$$y={{\\log }_{a}}\\left( {{x}^{2}}-ax+1 \\right)$$有最小值，则$$a$$的取值范围是~\\uline{~~~~~~~~~~}~", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless{}1$$ "}], [{"aoVal": "B", "content": "$$1\\textless{}a\\textless{}2$$ "}], [{"aoVal": "C", "content": "$$a\\textgreater{}2$$ "}], [{"aoVal": "D", "content": "以上选项都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数", "课内体系->知识点->函数的概念与性质->二次函数"], "answer_analysis": ["因为$${{x}^{2}}-ax+1$$无最大值，所以$$a\\textgreater1$$．函数有最小值，则$${{x}^{2}}-ax+1$$有最小值，$$\\Delta \\textless{}0$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "846", "queId": "ad4bb8e9063648e88b56d8ff53f9b963", "competition_source_list": ["2004年AMC10竞赛A第17题", "2004年AMC12竞赛A第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2004-AMC10A-18$$ Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run $$100$$ meters. They next meet after Sally has run $$150$$ meters past their first meeting point. Each girl runs at a constant speed. What is the length of the track in meters? Brenda 和 Sally 在圆形轨道上以相反的方向奔跑，分别从一条直径的两个端点开始。 他们在 Brenda 跑了 100 米后第一次见面。 在 Sally 跑过他们的第一个会面点 150 米后，他们下一次见面。 每个女孩都以恒定的速度奔跑。 轨道的长度是多少米？", "answer_option_list": [[{"aoVal": "A", "content": "$$250$$ "}], [{"aoVal": "B", "content": "$$300$$ "}], [{"aoVal": "C", "content": "$$350$$ "}], [{"aoVal": "D", "content": "$$400$$ "}], [{"aoVal": "E", "content": "$$500$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems", "课内体系->知识点->等式与不等式->等式->方程组的解集"], "answer_analysis": ["Call the length of the race track $$x$$. When they meet at the first meeting point, Brenda has run $$100$$ meters, while Sally has run $$\\dfrac{x}{2}-100$$ meters. By the second meeting point, Sally has run $$150$$ meters, while Brenda has run $$x-150$$ meters. Since they run at a constant speed, we can set up a proportion: $$\\dfrac{100}{x-150}=\\dfrac{\\dfrac{x}{2}-100}{150}$$. Cross-multiplying, we get that $$x=350\\Rightarrow\\boxed {(\\text{C})350}$$. Sidenote by carlos $$8$$: Since they run at constant speeds, Brenda must\\textquotesingle ve ran $$200$$ meters to get to the second meeting point, therefore we can make an equation $$200=x-150$$, solving for $$x$$, gives us our answer $$350$$. The total distance the girls run between the start and the first meeting is one half of the track length. The total distance they run between the two meetings is the track length. As the girls run at constant speeds, the interval between the meetings is twice as long as the interval between the start and the first meeting. Thus between the meetings Brenda will run $$2\\times100=200$$ meters. Therefore the length of the track is $$150+200=350$$ meters $$\\Rightarrow \\boxed {(\\text{C})350}$$. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "632", "queId": "d5de8fbb9b584de880ac7a312811f805", "competition_source_list": ["2018年辽宁全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知点$$P$$、$$Q$$在$$\\triangle ABC$$内，且$$\\overrightarrow{PA}+2\\overrightarrow{PB}+3\\overrightarrow{PC}=2\\overrightarrow{QA}+3\\overrightarrow{QB}+5\\overrightarrow{QC}=\\overrightarrow{0}$$，则$$\\frac{\\left\\textbar{} \\overrightarrow{PQ} \\right\\textbar}{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{30}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{31}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{32}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{33}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量线性运算综合（非坐标）", "竞赛->知识点->复数与平面向量"], "answer_analysis": ["由题设知 $${{S}_{\\triangle BCP}}:{{S}_{\\triangle CAP}}:{{S}_{\\triangle ABP}}=1:2:3$$， $${{S}_{\\triangle BCQ}}:{{S}_{\\triangle CAQ}}:{{S}_{\\triangle ABQ}}=2:3:5$$， 则$${{S}_{\\triangle ABP}}={{S}_{\\triangle ABQ}}=\\frac{1}{2}{{S}_{\\triangle ABC}}$$，于是，$$PQ\\text{//}AB$$， 且$${{S}_{\\triangle BCP}}=\\frac{1}{6}{{S}_{\\triangle ABC}}$$，$${{S}_{\\triangle BCQ}}=\\frac{1}{5}{{S}_{\\triangle ABC}}$$， 故$$\\frac{\\textbar\\overrightarrow{PQ}\\textbar}{\\textbar\\overrightarrow{AB}\\textbar}=\\frac{1}{5}-\\frac{1}{6}=\\frac{1}{30}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "599", "queId": "37c04d78f5534ae68f5f0c82d76faed2", "competition_source_list": ["2022~2023学年湖南永州宁远县高一上学期月考（明德湘南中学基础知识竞赛）第3题"], "difficulty": "0", "qtype": "single_choice", "problem": "函数$f(x)=\\frac{\\sqrt{2x-1}}{{{x}^{2}}-1}$的定义城为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$[\\frac{1}{2},+\\infty )$ "}], [{"aoVal": "B", "content": "(1+∞) "}], [{"aoVal": "C", "content": "$\\text{ } ! !\\textasciitilde ! !\\text{ (-1,}\\frac{\\text{1}}{\\text{2}}\\text{)}\\cup \\text{(1,+}\\infty \\text{) } ! !\\textasciitilde ! !\\text{ }$ "}], [{"aoVal": "D", "content": "$[\\frac{\\text{1}}{\\text{2}}\\text{,1)U(1,+}\\infty \\text{)}$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 根据偶次方根下非负，且分母不为零，列式即可得解.\\\\ 【详解】\\\\ 由$f(x)=\\frac{\\sqrt{2x-1}}{{{x}^{2}}-1}$，\\\\ 可得：$\\left { \\begin{align} \\& 2x-1\\ge 0 \\& {{x}^{2}}-1\\ne 0 \\end{align} \\right.$，解得：$x\\ge \\frac{1}{2}$且$x\\ne 1$，\\\\ 故选：D. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1060", "queId": "af0c2a8f0a2b4bbcab75a296abda128a", "competition_source_list": ["2008年安徽全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若函数$$y=f(x)$$的图象绕原点顺时针旋转$$90{}^{}\\circ $$后，与函数$$y=g(x)$$的图象重合，则有．", "answer_option_list": [[{"aoVal": "A", "content": "$$g(x)={{f}^{-1}}(-x)$$ "}], [{"aoVal": "B", "content": "$$g(x)={{f}^{-1}}(x)$$ "}], [{"aoVal": "C", "content": "$$g(x)=-{{f}^{-1}}(-x)$$ "}], [{"aoVal": "D", "content": "$$g(x)=-{{f}^{-1}}(x)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->平面几何->几何变换（二试）->平移和旋转（二试）"], "answer_analysis": ["若原函数为$$f\\left( a,b \\right)$$，则其反函数为$$f\\left( b,a \\right)$$， 所以原函数顺时针旋转$$90{}^{}\\circ $$后为$$f\\left( b,-a \\right)$$， 所以$$g\\left( x \\right)$$为$$-{{f}^{-1}}(x)$$，故选$$\\text{D}$$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "966", "queId": "e4b27b8c72c542d88f57f64de48039ff", "competition_source_list": ["2008年黑龙江全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "将正方形的每条边$$8$$等分，再取分点为顶点（不包括正方形的顶点），可以得到不同的三角形个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1372$$ "}], [{"aoVal": "B", "content": "$$2024$$ "}], [{"aoVal": "C", "content": "$$3136$$ "}], [{"aoVal": "D", "content": "$$4495$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->图论（二试）", "竞赛->知识点->排列组合与概率->两个基本计数原理"], "answer_analysis": ["首先注意到三角形的三个顶点不在正方形的同一边上，任选正方形的三边，使三个顶点分别在其上，有$$4$$种方法；再在选出的三条边上各选一点，有$${{7}^{3}}$$种方法．这类三角形共有$$4\\times {{7}^{3}}=1372$$（个）．另外，若三角形有两个顶点在正方形的一条边上，第三个顶点在另一条边上，则先取一边使其上有三角形的两个顶点，有$$4$$种方法，再在这条边上任取两点有$$21$$种方法，然后在其余的$$21$$个分点中任取一点作为第三个顶点，这类三角形共有$$4\\times 21\\times 21=1764$$（个）．综上可知，不同三角形的个数为$$1 372+1 764=3 136$$．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "765", "queId": "ff80808148ac4b390148b09c965d07d4", "competition_source_list": ["2011年浙江全国高中数学联赛竞赛初赛第8题5分", "2018~2019学年浙江宁波北仑区浙江省北仑中学高一下学期期中A卷第10题4分", "高考真题"], "difficulty": "2", "qtype": "single_choice", "problem": "在平面区域$$\\left { (x,y)\\textbar\\left\\textbar{} x \\right\\textbar\\leqslant 1,\\left\\textbar y \\right\\textbar\\leqslant 1 \\right }$$上恒有$$ax-2by\\leqslant 2$$，则动点$$P(a,b)$$所形成平面区域的面积为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->线性规划->二元一次不等式组的概念", "课内体系->知识点->等式与不等式->不等式->线性规划->含参线性规划相关问题->参数与可行域面积", "课内体系->素养->直观想象"], "answer_analysis": ["平面区域$$\\left { (x,y)\\textbar\\left\\textbar{} x \\right\\textbar\\leqslant 1,\\left\\textbar{} y \\right\\textbar\\leqslant1 \\right }$$的四个边界点$$(-1,-1)$$，$$(-1,1)$$，$$(1,-1)$$，$$(1,1)$$满足$$ax-2by\\leqslant 2$$，即有$$a+2b\\leqslant 2,a-2b\\leqslant 2,-a-2b\\leqslant 2,-a+2b\\leqslant 2$$，由此计算动点$$P(a,b)$$所形成平面区域的面积为$$4$$．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1015", "queId": "ee2416b12e254d25bb5f795dfb55e323", "competition_source_list": ["2013年天津全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "等差数列$$\\left { {{a}_{n}} \\right }$$前$$n$$项的和为$${{S}_{n}}$$，已知$$\\frac{{{S}_{25}}}{{{a}_{23}}}=5$$，$$\\frac{{{S}_{45}}}{{{a}_{33}}}=25$$，则$$\\frac{{{S}_{65}}}{{{a}_{43}}}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$125$$ "}], [{"aoVal": "B", "content": "$$85$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$35$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列", "竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["由于$${{S}_{25}}=25\\cdot {{a}_{13}}$$，所以由已知条件可得$${{a}_{13}}:{{a}_{23}}=1:5$$，从而$${{a}_{23}}:{{a}_{33}}=5:9$$，$${{a}_{33}}:{{a}_{43}}=9:13$$．现在，$${{S}_{65}}=65{{a}_{33}}$$，所以$${{S}_{65}}:{{a}_{43}}=45:1$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "334", "queId": "1b4e2d425f544560b0f37ff9b0b4ea91", "competition_source_list": ["高一上学期单元测试《等差、等比数列、递推数列1》竞赛第2题", "2018~2019学年内蒙古呼和浩特回民区呼和浩特市回民中学高二上学期期中（体、艺班）第16题5分", "2017~2018学年浙江杭州上城区杭州第四中学高一下学期期中第14题4分", "2007年高考真题江西卷文科第13题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知等差数列$$\\left { {{a}_{n}} \\right }$$的前$$n$$项和为$${{S}_{n}}$$，若$${{S}_{12}}=21$$，则$${{a}_{2}}+{{a}_{5}}+{{a}_{8}}+{{a}_{11}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$\\dfrac{7}{2}$ "}], [{"aoVal": "D", "content": "$\\dfrac{7}{4}$ "}]], "knowledge_point_routes": ["知识标签->知识点->数列->等差数列->等差数列的概念通项公式", "知识标签->知识点->数列->等差数列->等差数列的性质及应用", "知识标签->知识点->数列->等差数列->等差数列的前n项和", "知识标签->素养->数学运算", "知识标签->题型->数列->等差数列->等差数列的性质问题->等差数列前n项和的性质", "知识标签->题型->数列->等差数列->等差数列的性质问题->求等差数列的通项公式", "知识标签->题型->数列->等差数列->等差数列的性质问题->等差数列的前n项和计算 "], "answer_analysis": ["$${{a}_{1}}+{{a}_{4}}+{{a}_{7}}+{{a}_{10}}$$，$${{a}_{2}}+{{a}_{5}}+{{a}_{8}}+{{a}_{11}}$$，$${{a}_{3}}+{{a}_{6}}+{{a}_{9}}+{{a}_{12}}$$构成等差数列， ∴原式$$=21\\div 3=7$$． 因为$${{a}_{1}}+{{a}_{12}}={{a}_{2}}+{{a}_{11}}={{a}_{3}}+{{a}_{10}}={{a}_{4}}+{{a}_{9}}={{a}_{5}}+{{a}_{8}}={{a}_{6}}+{{a}_{7}}$$， 且他们加起来为$${{S}_{12}}=21$$，故$${{a}_{2}}+{{a}_{5}}+{{a}_{8}}+{{a}_{11}}=\\frac{2}{6}\\cdot {{S}_{12}}=7$$． $${{S}_{12}}=6\\left( {{a}_{1}}+{{a}_{12}} \\right)=21$$，$${{a}_{1}}+{{a}_{12}}=\\frac{7}{2}$$，$${{a}_{2}}+{{a}_{5}}+{{a}_{8}}+{{a}_{11}}=2\\left( {{a}_{1}}+{{a}_{12}} \\right)=7$$． 故答案为：$$7$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1196", "queId": "e73e14064e2449d0a9edf81bd95f6527", "competition_source_list": ["2016年浙江全国高中数学联赛竞赛初赛第8题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "记$$[x]$$为不超过$$x$$的最大整数，若集合$$S=\\left { (x，y)\\left\\textbar{} \\textbar[x+y]\\textbar+\\textbar[x-y]\\textbar\\leqslant 1 \\right. \\right }$$，则集合$$S$$所表示的平面区域的面积为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{2}$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$\\frac{9}{2}$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->直观想象", "课内体系->特色题型->新定义", "课内体系->思想->分类讨论思想", "课内体系->知识点->等式与不等式->不等式->线性规划->简单线性规划->画出二元一次不等式（组）所表示的平面区域", "课内体系->知识点->等式与不等式->不等式->解不等式->含绝对值的不等式"], "answer_analysis": ["当$$\\begin{cases}\\textbar[x+y]\\textbar=1 \\textbar[x-y]\\textbar=0 \\end{cases}$$时，$$\\begin{cases}1\\leqslant x+y ~\\textless{} ~2 0\\leqslant x-y ~\\textless{} ~1 \\end{cases}$$或$$\\begin{cases}-1\\leqslant x+y ~\\textless{} ~0 0\\leqslant x-y ~\\textless{} ~1 \\end{cases}$$ 同理$$\\begin{cases}\\textbar[x+y]\\textbar=0 \\textbar[x-y]\\textbar=1 \\end{cases}$$或$$\\begin{cases}\\textbar[x+y]\\textbar=0 \\textbar[x-y]\\textbar=0 \\end{cases}$$都可以类似表示，画图即可求出面积. 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "833", "queId": "8d591f5225744be4beef84200ac31ecc", "competition_source_list": ["2008年天津全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "抛物线$$y=a{{x}^{2}}+bx+1$$的参数$$a$$、$$b$$满足$$8{{a}^{2}}+4ab={{b}^{3}}$$，则当$$a$$、$$ b$$变动时，抛物线的顶点一定在上．", "answer_option_list": [[{"aoVal": "A", "content": "抛物线 "}], [{"aoVal": "B", "content": "双曲线 "}], [{"aoVal": "C", "content": "圆或椭圆 "}], [{"aoVal": "D", "content": "直线 "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->圆锥曲线->双曲线->双曲线的定义、标准方程->双曲线的标准方程", "课内体系->知识点->函数的概念与性质->二次函数->二次函数的图象及性质->含参二次函数的图象及性质", "课内体系->思想->方程思想", "课内体系->方法->换元法"], "answer_analysis": ["抛物线$$y=a{{x}^{2}}+bx+1$$顶点的坐标为$$\\left( -\\frac{b}{2a},\\frac{4a-{{b}^{2}}}{4a} \\right)$$，设$$x=-\\frac{b}{2a},y=\\frac{4a-{{b}^{2}}}{4a}$$，则有$$\\frac{b}{a}=-2x, y=1-\\frac{{{b}^{2}}}{4a}=1+\\frac{bx}{2}$$．因为$$a\\ne 0$$，所以$$a, b$$满足的条件等价于$$8+4\\frac{b}{a}=b{{\\left( \\frac{b}{a} \\right)}^{2}}$$，于是有$$8+4\\left( -2x \\right)=b{{\\left( -2x \\right)}^{2}}=4x\\left( 2y-2 \\right)$$，即$$xy=1$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "234", "queId": "1dce3b4a1ff345939258f8bfa3ed6865", "competition_source_list": ["2016年湖南全国高中数学联赛竞赛初赛第5题5分", "2010年上海复旦大学自主招生千分考第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "给定平面向量$$\\left( 1,1 \\right)$$，那么，平面向量$$\\left( \\frac{1-\\sqrt{3}}{2},\\frac{1+\\sqrt{3}}{2} \\right)$$是将向量$$\\left( 1,1 \\right)$$经过（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "顺时针旋转$$60{}^{}\\circ $$所得 "}], [{"aoVal": "B", "content": "顺时针旋转$$120{}^{}\\circ $$所得 "}], [{"aoVal": "C", "content": "逆时针旋转$$60{}^{}\\circ $$所得 "}], [{"aoVal": "D", "content": "逆时针旋转$$120{}^{}\\circ $$所得 "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的应用"], "answer_analysis": ["两向量对应的角度分别是$$45{}^{}\\circ $$和$$105{}^{}\\circ $$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "974", "queId": "8aac49075139269a01514c260c2033eb", "competition_source_list": ["1989年全国高中数学联赛竞赛一试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$A$$，$$B$$是锐角$$\\triangle ABC$$的两内角，则复数$$z=\\left( \\cos B-\\sin A \\right)+\\text{i}\\left( \\sin B-\\cos A \\right)$$在复平面上所对应的点位于（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "第一象限 "}], [{"aoVal": "B", "content": "第二象限 "}], [{"aoVal": "C", "content": "第三象限 "}], [{"aoVal": "D", "content": "第四象限 "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["因为$$A+B\\textgreater\\frac{ \\pi }{2}$$，进而$$0\\textless{}\\frac{ \\pi }{2}-A\\textless{}B\\textless{}\\frac{ \\pi }{2}$$， 所以$$\\cos B\\textless{}\\cos \\left( \\frac{ \\pi }{2}-A \\right)=\\sin A$$， $$\\sin B\\textgreater\\sin \\left( \\frac{ \\pi }{2}-A \\right)=\\cos A$$， $$\\cos B-\\sin A\\textless{}0$$，$$\\sin B-\\cos A\\textgreater0$$． 说明$$z$$位于第二象限． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "58", "queId": "130e35ae2a4d4849ba80348b2032ce03", "competition_source_list": ["2021年吉林全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "关于$$x$$的方程$${{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-a\\cos \\left( 1-x \\right)=0$$只有一个实数解，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$a=-1$$ "}], [{"aoVal": "B", "content": "$$a=1$$ "}], [{"aoVal": "C", "content": "$$a=2$$ "}], [{"aoVal": "D", "content": "$$a$$的值不唯一 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数方程"], "answer_analysis": ["函数$$f\\left( x \\right)={{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-a\\cos \\left( 1-x \\right)$$的图象关于直线$$x=1$$对称， 又方程只有一个实数解， ∴$$f\\left( 1 \\right)=0$$，得$$a=1$$， 当$$a=1$$时，$$f\\left( x \\right)={{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-\\cos \\left( 1-x \\right)\\geqslant 1-1=0$$， 当且仅当$$x=1$$时取等号， 即方程$${{2}^{\\left\\textbar{} 2x-2 \\right\\textbar}}-a\\cos \\left( 1-x \\right)=0$$只有一个实数解，符合题设． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "482", "queId": "295ba4065ea7489d93110ef6ba3fa444", "competition_source_list": ["1990年全国高中数学联赛竞赛一试第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$\\alpha \\in \\left( \\frac{ \\pi }{4}，\\frac{ \\pi }{2} \\right)$$，则$${{\\left( \\cos \\alpha \\right)}^{\\cos \\alpha }}$$，$${{\\left( \\sin \\alpha \\right)}^{\\cos \\alpha }}$$，$${{\\left( \\cos \\alpha \\right)}^{\\sin \\alpha }}$$的大小顺序是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{\\left( \\cos \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\sin \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\cos \\alpha \\right)}^{\\sin \\alpha }}$$； "}], [{"aoVal": "B", "content": "$${{\\left( \\cos \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\cos \\alpha \\right)}^{\\sin \\alpha }} ~\\textless{} ~{{\\left( \\sin \\alpha \\right)}^{\\cos \\alpha }}$$； "}], [{"aoVal": "C", "content": "$${{\\left( \\sin \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\cos \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\cos \\alpha \\right)}^{\\sin \\alpha }}$$； "}], [{"aoVal": "D", "content": "$${{\\left( \\cos \\alpha \\right)}^{\\sin \\alpha }} ~\\textless{} ~{{\\left( \\cos \\alpha \\right)}^{\\cos \\alpha }} ~\\textless{} ~{{\\left( \\sin \\alpha \\right)}^{\\cos \\alpha }}$$． "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["∵$$\\alpha \\in \\left( \\frac{ \\pi }{4}，\\frac{ \\pi }{2} \\right)$$， ∴ $$0 ~\\textless{} ~\\cos \\alpha ~~\\textless{} ~\\frac{\\sqrt{2}}{2} ~\\textless{} ~\\sin \\alpha ~~\\textless{} ~1$$． 故 $$\\cos \\alpha {{}^{\\sin \\alpha }} ~\\textless{} ~\\cos \\alpha {{}^{\\cos \\alpha }}$$． ∴ $$\\cos \\alpha {{}^{\\cos \\alpha }} ~\\textless{} ~{{(\\sin \\alpha )}^{\\cos \\alpha }}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "811", "queId": "85061f1372e74785b875b48ad33b92b4", "competition_source_list": ["1989年第7届美国数学邀请赛（AIME）竞赛第10题"], "difficulty": "4", "qtype": "single_choice", "problem": "设$$a$$，$$b$$，$$c$$是三角形的三边，$$\\alpha $$，$$\\beta $$，$$\\gamma $$分别是相对于这三边的角，若$${{a}^{2}}+{{b}^{2}}=2021{{c}^{2}}$$，则$$\\frac{\\cot \\gamma }{\\cot \\alpha +\\cot \\beta }=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$\\dfrac{1}{2021}$ "}], [{"aoVal": "B", "content": "$\\dfrac{1}{1010}$ "}], [{"aoVal": "C", "content": "$1010$ "}], [{"aoVal": "D", "content": "$2021$ "}]], "knowledge_point_routes": ["知识标签->题型->解三角形->正余弦定理的简单应用->利用正、余弦定理求解边角", "知识标签->知识点->解三角形->正弦定理", "知识标签->知识点->解三角形->余弦定理", "知识标签->素养->数学运算"], "answer_analysis": ["由余弦定理$${{a}^{2}}+{{b}^{2}}-2ab\\cos \\gamma ={{c}^{2}}$$，代入条件式得 $$2ab\\cos \\gamma =1988{{c}^{2}}$$，或$$\\left( \\frac{a}{b} \\right)\\left( \\frac{b}{c} \\right)\\cos \\gamma =994$$． 由下正弦定理$$\\frac{a}{c}=\\frac{\\sin \\alpha }{\\sin \\gamma }$$，$$\\frac{b}{c}=\\frac{\\sin \\beta }{\\sin \\gamma }$$， 于是$$\\frac{\\sin \\alpha \\sin \\beta \\cos \\gamma }{{{\\sin }^{2}}\\gamma }=994$$． 因为$$\\alpha +\\beta +\\gamma = \\pi $$，所以$$\\alpha +\\beta = \\pi -\\gamma $$．从而$$\\sin \\left( \\alpha +\\beta \\right)=\\sin \\gamma $$． $$\\frac{\\cot \\gamma }{\\cot \\alpha +\\cot \\beta }=\\frac{\\sin \\alpha \\sin \\beta \\cos \\gamma }{\\sin \\left( \\alpha +\\beta \\right)\\sin \\gamma }=994$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "560", "queId": "3b6debd1e0c54f5da2e637755f3c1b03", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "体育课下课后，老师要求体育委员把$$3$$个相同的篮球、$$2$$个相同的排球、$$1$$个相同的橄榄球排成一排放好，则不同的放法有．", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$种 "}], [{"aoVal": "B", "content": "$$60$$种 "}], [{"aoVal": "C", "content": "$$120$$种 "}], [{"aoVal": "D", "content": "$$720$$种 "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->两个基本计数原理", "竞赛->知识点->排列组合与概率->排列与组合", "课内体系->知识点->计数原理->两个基本计数原理->分步乘法计数原理"], "answer_analysis": ["总共$$10$$个球，先考虑选$$5$$个位置放篮球，共有$$\\text{C}_{10}^{5}=252$$种，再从剩下的$$5$$个位置种选$$3$$个位置放排球，总共有$$10$$种方法，剩下的$$2$$个位置放橄榄球，故有$$252\\times 10=2520$$种. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1033", "queId": "aa382179f818427487cdbf5240ac6f45", "competition_source_list": ["2009年吉林全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "若函数$$f(x)=\\begin{cases}{{\\left( \\frac{1}{2} \\right)}^{x}},x\\geqslant 4 f(x+1),x\\textless{}4 \\end{cases}$$，则$$f({{\\log }_{2}}3)=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{23}{8}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{11}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{19}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{24}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["因为$$1\\textless{}{{\\log }_{2}}3\\textless{}2$$，所以 $$f\\left( {{\\log }_{2}}3 \\right)=f\\left( {{\\log }_{2}}3+1 \\right)=f\\left( {{\\log }_{2}}3+2 \\right)=f\\left( {{\\log }_{2}}3+3 \\right)={{\\left( \\frac{1}{2} \\right)}^{{{\\log }_{2}}3+3}}=\\frac{1}{3}\\cdot \\frac{1}{8}=\\frac{1}{24}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "409", "queId": "79d8d8595bb9441f96028c4126c2f6c1", "competition_source_list": ["1982年全国高中数学联赛竞赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$${{x}_{1}}$$，$${{x}_{2}}$$是方程$${{x}^{2}}-(k-2)x+({{k}^{2}}+3k+5)=0$$（$$k$$为实数）的两个实数根，$$x_{1}^{2}+x_{2}^{2}$$的最大值是(~ ~ )．", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$5\\frac{5}{9}$$ "}], [{"aoVal": "D", "content": "不存在 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->二次函数"], "answer_analysis": ["由韦达定理得 $$x_{1}^{2}+x_{2}^{2}={{({{x}_{1}}+{{x}_{2}})}^{2}}-2{{x}_{1}}{{x}_{2}}$$ $$={{(k-2)}^{2}}-2({{k}^{2}}+3k+5)$$ $$=-{{(k+5)}^{2}}+19$$． 又因方程有两个实根，所以$$k$$值要受下面条件的约束： $$\\Delta ={{(k-2)}^{2}}-4({{k}^{2}}+3k+5)\\mathsf{\\geqslant }0$$， 即 $$3{{k}^{2}}+16k+16\\mathsf{\\leqslant }0$$， 解得 $$-4\\mathsf{\\leqslant }k\\mathsf{\\leqslant }-\\frac{4}{3}$$ 因此，求$$x_{1}^{2}+x_{2}^{2}$$的最大值问题，就是当$$-4\\mathsf{\\leqslant }k\\mathsf{\\leqslant }-\\frac{4}{3}$$的条件下，求函数 $$y=-{{(k+5)}^{2}}+19$$ 的最大值． 由于对称轴$${{a}_{1}}$$在$$[-4\\mathsf{，}-\\frac{4}{3}]$$的外面，所以在$$[-4\\mathsf{，}-\\frac{4}{3}]$$上函数单调递减，在$$k=-4$$取得最大值$$18$$． $$\\therefore $$ $$x_{1}^{2}+x_{2}^{2}$$的最大值是$$18$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "44", "queId": "06516305335a425b904737dd8da7fb5c", "competition_source_list": ["2016年浙江全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知正三棱锥$$S-ABC$$，底面是边长为$$1$$的正三角形，侧棱长为$$2$$，若过直线$$AB$$的截面，将正三棱锥的体积分成两个相等的部分，则截面与底面所成二面角的平面角的余弦值为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{15}}{10}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4\\sqrt{15}}{15}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{15}}{15}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2\\sqrt{15}}{15}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->立体几何初步->基本立体图形->空间几何体的体积、表面积->组合体求体积、表面积问题", "课内体系->知识点->立体几何初步->基本立体图形->空间几何体的概念->棱柱、棱锥、棱台的结构特征", "课内体系->知识点->立体几何初步->基本立体图形->平面图形、空间几何体的直观图认识", "课内体系->知识点->立体几何初步->基本图形位置关系->空间中的基本事实与定理->点、直线、平面之间的位置关系", "课内体系->知识点->立体几何初步->基本图形位置关系->空间中的基本事实与定理->已知线面位置关系判定结论的问题", "课内体系->知识点->立体几何初步->基本图形位置关系->探索性问题->几何法求空间角", "课内体系->素养->逻辑推理", "课内体系->素养->直观想象"], "answer_analysis": ["设截面与棱$$SC$$交于$$D$$点，由已知条件可知，点$$D$$为棱$$SC$$的中点． 取$$AB$$的中点$$E$$，连结$$EC$$、$$DE$$、$$SE$$， 则$$\\angle DEC$$为截面与底面所成二面角的平面角，设为$$\\theta $$． 在$$\\triangle SEC$$中，$$SE=\\frac{\\sqrt{15}}{2}$$，$$EC=\\frac{\\sqrt{3}}{2}$$，$$SC=2$$，所以中线$$DE=\\frac{\\sqrt{5}}{2}$$． 在$$\\triangle DEC$$应用余弦定理得$$\\cos \\theta =\\frac{2\\sqrt{15}}{15}$$．正确答案为$$D$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "847", "queId": "e43f29d6fd524f478ab76bf392636f23", "competition_source_list": ["2022~2023学年湖南永州宁远县高一上学期月考（明德湘南中学基础知识竞赛）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "定义在\\emph{R}上的偶函数\\emph{f}(\\emph{x})满足:对任意的${{x}_{1}},{{x}_{2}}\\in [0,+\\infty ),{{x}_{1}}\\ne {{x}_{2}}$，有$({{x}_{2}}-{{x}_{1}})[f({{x}_{2}})-f({{x}_{1}})]\\textless{} 0$，则（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "\\emph{f}(3)\\textless{}\\emph{f}(-2)\\textless{}\\emph{f}(1) "}], [{"aoVal": "B", "content": "\\emph{f}(1)\\textless{}\\emph{f}(-2)\\textless{}\\emph{f}(3) "}], [{"aoVal": "C", "content": "\\emph{f}(3)\\textless{}\\emph{f}(1)\\textless{}\\emph{f}(-2) "}], [{"aoVal": "D", "content": "\\emph{f}(-2)\\textless{}\\emph{f}(1)\\textless{}\\emph{f}(3) "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的性质->单调性"], "answer_analysis": ["\\hfill\\break 已知不等关系说明函数是减函数，再由偶函数得$f(-2)=f(2)$，然后由单调性可得大小关系．\\\\ 【详解】\\\\ ∵对任意的${{x}_{1}},{{x}_{2}}\\in [0,+\\infty ),{{x}_{1}}\\ne {{x}_{2}}$，有$({{x}_{2}}-{{x}_{1}})[f({{x}_{2}})-f({{x}_{1}})]\\textless{} 0$，\\\\ ∴$f(x)$在$[0,+\\infty )$上是减函数．\\\\ ∴$f(3) \\textless{} f(2) \\textless{} f(1)$，\\\\ ∵$f(x)$是偶函数，∴$f(-2)=f(2)$\\\\ ∴$f(3)\\textless{} f(-2)\\textless{} f(1)$．\\\\ 故选：A． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "710", "queId": "ff808081475cea960147620b8b2e0969", "competition_source_list": ["高考真题", "2009年第二十届全国希望杯高二竞赛复赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知两个等差数列$$\\left { {{a}_{n}} \\right }$$与$$\\left { {{b}_{n}} \\right }$$的前$$n$$项和分别为$${{S}_{n}}$$与$${{T}_{n}}$$，并且$$\\frac{{{S}_{n}}}{{{T}_{n}}}=\\frac{2n+4}{3n+7}$$，则$$\\frac{{{a}_{5}}}{{{b}_{7}}}$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{7}{11}$$ "}], [{"aoVal": "B", "content": "$$\\frac{7}{13}$$ "}], [{"aoVal": "C", "content": "$$\\frac{11}{23}$$ "}], [{"aoVal": "D", "content": "$$\\frac{9}{23}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->数列->等差数列->等差数列的前n项和->等差数列前n项和的性质"], "answer_analysis": ["因为等差数列$$n$$项和公式形如$$A{{n}^{2}}+Bn$$，可设$${{S}_{n}}=kn(2n+4)$$；$${{T}_{n}}=kn(3n+7)$$，则$${{a}_{5}}=22k$$，$${{b}_{7}}=46k$$，所以$$\\frac{{{a}_{5}}}{{{b}_{7}}}=\\frac{11}{23}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "575", "queId": "68425d90fbe04e4db6584fc6ae173714", "competition_source_list": ["2018年吉林全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "三棱锥$$P-ABC$$的底面$$\\triangle ABC$$是边长为$$3$$的正三角形，$$PA=3$$，$$PB=4$$，$$PC=5$$，则三棱锥$$P-ABC$$的体积为( )", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{10}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{11}$$ "}], [{"aoVal": "D", "content": "$$2\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量"], "answer_analysis": ["解：$${{V}_{P-ABC}}={{V}_{A-PBC}}=\\frac{1}{3}\\times \\left( \\frac{1}{2}\\times 3\\times 4 \\right)\\times \\sqrt{{{3}^{2}}-{{\\left( \\frac{5}{2} \\right)}^{2}}}=\\sqrt{11}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "133", "queId": "186b4b111ab44e3cb5e88560aff159f6", "competition_source_list": ["1992年全国高中数学联赛竞赛一试第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "设四面体四个面的面积分别为$${{S}_{1}}$$，$${{S}_{2}}$$，$${{S}_{3}}$$，$${{S}_{4}}$$，它们的最大值为$$S$$，记$$\\lambda =\\frac{\\sum\\limits_{i=1}^{4}{{{S}_{i}}}}{S}$$，则$$\\lambda $$一定满足（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\textless{}\\lambda \\mathsf{\\leqslant }4$$ "}], [{"aoVal": "B", "content": "$$3\\textless{}\\lambda \\textless{}4$$ "}], [{"aoVal": "C", "content": "$$2.5\\textless{}\\lambda \\mathsf{\\leqslant }4.5$$ "}], [{"aoVal": "D", "content": "$$3.5\\textless{}\\lambda \\textless{}5.5$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["因为 $${{S}_{i}}\\mathsf{\\leqslant }S$$ $$\\left( i=1,2,3,4 \\right)$$ 所以 $$\\sum\\limits_{i=1}^{4}{{{S}_{i}}}\\mathsf{\\leqslant }4S$$． 进而 $$\\lambda =\\frac{\\sum\\limits_{i=1}^{4}{{{S}_{i}}}}{S}\\mathsf{\\leqslant }4$$． 特别地，当四面体为正四面体时，式中的等号成立，从而否定（$$\\text{B}$$）． 考察侧面与底面所成的二面角的平面角均等于$$45{}^{}\\circ $$的任一个正三棱锥，以$${{S}_{4}}$$表示该正三棱锥的底面面积，则 $$S={{S}_{4}}=\\left( {{S}_{1}}+{{S}_{2}}+{{S}_{3}} \\right)\\cos {{45}^{\\circ }}=\\frac{\\sqrt{2}}{2}\\left( {{S}_{1}}+{{S}_{2}}+{{S}_{3}} \\right)$$， 于是 $$\\left( {{S}_{1}}+{{S}_{2}}+{{S}_{3}} \\right)=\\sqrt{2}S$$， $${{S}_{1}}+{{S}_{2}}+{{S}_{3}}+{{S}_{4}}=\\left( 1+\\sqrt{2} \\right)S$$． 此时 $$\\lambda =\\frac{\\sum\\limits_{i=1}^{4}{{{S}_{i}}}}{S}=1+\\sqrt{2}\\textless{}2.5$$， 从而否定（$$\\text{C}$$）与（$$\\text{D}$$）．因此答案是（$$\\text{A}$$）． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1116", "queId": "e15971a4412d4e52a21d6818e333422b", "competition_source_list": ["2014年江西全国高中数学联赛竞赛初赛第6题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "等差数列$$ {{{a}_{n}} }$$、$$ {{{b}_{n}} }$$的前$$n$$项和分别为$${{S}_{n}}$$、$${{T}_{n}}$$，若对任意的正整数$$n$$都有$$\\frac{{{S}_{n}}}{{{T}_{n}}}=\\frac{5n-3}{2n+1}$$，则$$\\frac{{{a}_{20}}}{{{b}_{7}}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{20}{7}$$ "}], [{"aoVal": "B", "content": "$$\\frac{40}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{64}{9}$$ "}], [{"aoVal": "D", "content": "$$\\frac{72}{7}$$ "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->数列->等差数列->等差数列的性质及应用", "课内体系->知识点->数列->等差数列->等差数列的前n项和->等差数列前n项和的性质"], "answer_analysis": ["因为$$ {{{a}_{n}} }$$，$$ {{{b}_{n}} }$$为等差数列，故可设 $${{S}_{n}}=kn(5n-3)$$，$${{T}_{n}}=kn(2n+1)$$， 当$$n\\geqslant 2$$时， $${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}=k(10n-8)$$，$${{b}_{n}}={{T}_{n}}-{{T}_{n-1}}=k(4n-1)$$， 所以$$\\frac{{{a}_{20}}}{{{b}_{7}}}=\\frac{k(200-8)}{k(28-1)}=\\frac{64}{9}$$． 故答案为��$$\\frac{64}{9}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "306", "queId": "2323a87460994780a961d448fed97a0d", "competition_source_list": ["2016~2017学年北京东城区北京市第一七一中学高一下学期期中第8题", "1997年全国高中数学联赛竞赛一试第1题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知数列$$\\left { {{x}_{n}} \\right }$$满足$${{x}_{n+1}}={{x}_{n}}-{{x}_{n-1}}(n\\geqslant 2)$$，$${{x}_{1}}=a$$，$${{x}_{2}}=b$$，记$${{S}_{n}}={{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{n}}$$，则下列结论正确的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{x}_{100}}=-a$$，$${{S}_{100}}=2b-a$$ "}], [{"aoVal": "B", "content": "$${{x}_{100}}=-b$$，$${{S}_{100}}=2b-a$$ "}], [{"aoVal": "C", "content": "$${{x}_{100}}=-b$$，$${{S}_{100}}=b-a$$ "}], [{"aoVal": "D", "content": "$${{x}_{100}}=-a$$，$${{S}_{100}}=b-a$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["已知中$${{x}_{n+1}}={{x}_{n}}-{{x}_{n-1}}$$（$$n\\geqslant 2$$），$${{x}_{1}}=a$$，$${{x}_{2}}=b$$． ∴$${{x}_{3}}={{x}_{2}}-{{x}_{1}}=b-a$$， $${{x}_{4}}={{x}_{3}}-{{x}_{2}}=-a$$， $${{x}_{5}}={{x}_{4}}-{{x}_{3}}=-b$$， $${{x}_{6}}={{x}_{5}}-{{x}_{4}}=a-b$$， $${{x}_{7}}={{x}_{6}}-{{x}_{5}}=a$$， $${{x}_{8}}={{x}_{7}}-{{x}_{6}}=b$$， $$\\vdots $$ 由此可知：$${{x}_{n}}$$每隔六项开始重复， 即$${{x}_{i}}={{x}_{i}}+6(i\\geqslant 1)$$． ∴$${{S}_{100}}=16({{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+{{x}_{6}})+{{x}_{97}}+{{x}_{98}}+{{x}_{99}}+{{x}_{100}}$$ $$=16(a+b+b-a-a-b+a-b)+a+b+b-a-a$$ $$=2b-a$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "709", "queId": "64d6bc9ba27c43929cc2cf435ab0c9de", "competition_source_list": ["2011年AMC10竞赛A第10题", "2011年AMC12竞赛A第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "2011 AMC10A 10 A majority of the $$30$$ students in Ms. Demeanor\\textquotesingle s class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $$1$$. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $$$17.71$$. What was the cost of a pencil in cents? 德米诺尔女士班级的 30 个学生中的大多数在学校书店购买铅笔。 这些学生中的每个人都购买了相同数量的铅笔，而且这个数量大于 1。 一支铅笔的成本（以美分计）大于每个学生购买的铅笔数量，所有铅笔的总成本为 $17.71。 一支铅笔的成本是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$23$$ "}], [{"aoVal": "E", "content": "$$77$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->等式->方程组的解集", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"], "answer_analysis": ["Let $$x\\textgreater15$$ be the number of students at the bookstore， $$p\\textgreater1$$ be the number of pencils each student bought, and $$c\\textgreater p$$ be the price of one pencil. We see that $$xcp=1771=7\\cdot 11\\cdot 23$$. Then we must have $$x=23$$，$$c=11$$，$$p=7$$ by the original conditions and the answer is $$c=11$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "402", "queId": "39fb59e9e5554b14b94e540fdf497067", "competition_source_list": ["2008年AMC10竞赛B第18题", "2008年AMC12竞赛B第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2008-AMC12B-10$$ Bricklayer Brenda would take $$9$$ hours to build a chimney alone, and bricklayer Brandon would take $$10$$ hours to build it alone. When they work together they talk a lot, and their combined output is decreased by $$10$$ bricks per hour. Working together, they build the chimney in $$5$$ hours. How many bricks are in the chimney? 瓦工布伦达单独建造一个烟囱需要花费 $9$$ 小时，而瓦工布兰登单独建造它需要花费 $10$$ 小时。 当他们一起工作时，他们会说很多话，他们的总产量每小时减少 10 美元。 他们一起工作，在 $5$$ 小时内建造了烟囱。 烟囱里有多少块砖？", "answer_option_list": [[{"aoVal": "A", "content": "$$500$$ "}], [{"aoVal": "B", "content": "$$900$$ "}], [{"aoVal": "C", "content": "$$950$$ "}], [{"aoVal": "D", "content": "$$1000$$ "}], [{"aoVal": "E", "content": "$$1900$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems", "课内体系->知识点->等式与不等式->等式->方程组的解集"], "answer_analysis": ["Let $$x$$ be the number of bricks in the chimney, Using $$d=vt$$, we get $$x=\\left( \\frac{x}{9}+\\frac{x}{10}-10 \\right)\\cdot \\left( 5 \\right)$$. Solving for $$x$$, we get $$900\\Rightarrow \\text{B}$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "23", "queId": "023647ff6461474685bf00297ba804b3", "competition_source_list": ["2015年四川全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，内角$$ABC$$的对边长分别为$$abc$$，若$$\\frac{a}{\\cos A}=\\frac{b}{2\\cos B}=\\frac{c}{3\\cos C}$$，则$$\\angle A$$的大小为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{ \\pi }{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{ \\pi }{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{ \\pi }{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5 \\pi }{12}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->三角函数->三角形中的问题->解三角形"], "answer_analysis": ["结合正弦定理，有$$\\tan A=\\frac{1}{2}\\tan B=\\frac{1}{3}\\tan C$$， 熟知$$\\tan A+\\tan B+\\tan C=\\tan A\\tan B\\tan C$$，代入解得$$\\tan A=1$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "883", "queId": "8da9bbda8c884bb98ee07d346323fd62", "competition_source_list": ["2018年吉林全国高中数学联赛竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "已知 $$f(x)=\\frac{sinx}{2+cosx}$$ ，则对任意 $$x\\in R$$ ，下列说法错误的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$f(x)\\geqslant\\frac13sinx$$ "}], [{"aoVal": "B", "content": "$$\\textbar f(x)\\textbar\\leqslant\\textbar x\\textbar$$ "}], [{"aoVal": "C", "content": "$$\\textbar f(x)\\textbar\\leqslant \\frac{\\sqrt3}3$$ "}], [{"aoVal": "D", "content": "$$f(\\pi+x)+f(\\pi-x)=0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角函数的图象与性质"], "answer_analysis": ["$$A$$ "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "280", "queId": "7e12a4a050c64281bd96ac7c94b9d87d", "competition_source_list": ["2018年辽宁全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知点$$P$$、$$Q$$在$$\\triangle ABC$$内，且$$\\overrightarrow{PA}+2\\overrightarrow{PB}+3\\overrightarrow{PC}=2\\overrightarrow{QA}+\\overrightarrow{QB}+5\\overrightarrow{QC}=\\overrightarrow{0}$$，则$$\\frac{\\left\\textbar{} \\overrightarrow{PQ} \\right\\textbar}{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}$$等于( )", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{30}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{31}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{32}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{33}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量"], "answer_analysis": ["由题设知 $${{S}_{\\triangle BCP}}:{{S}_{\\triangle CAP}}:{{S}_{\\triangle ABP}}=1:2:3$$， $${{S}_{\\triangle BCQ}}:{{S}_{\\triangle CAQ}}:{{S}_{\\triangle ABQ}}=2:3:5$$， 则$${{S}_{\\triangle ABP}}={{S}_{\\triangle ABQ}}=\\frac{1}{2}{{S}_{\\triangle ABC}}$$，于是，$$PQ\\text{//}AB$$， 且$${{S}_{\\triangle BCP}}=\\frac{1}{6}{{S}_{\\triangle ABC}}$$，$${{S}_{\\triangle BCQ}}=\\frac{1}{5}{{S}_{\\triangle ABC}}$$， 故$$\\frac{\\textbar\\overrightarrow{PQ}\\textbar}{\\textbar\\overrightarrow{AB}\\textbar}=\\frac{1}{5}-\\frac{1}{6}=\\frac{1}{30}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "432", "queId": "50b82489ad3b4d9d8109a6ce91918859", "competition_source_list": ["2015年四川全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知正三棱锥$$P-ABC$$的底面$$ABC$$是正三角形，该三棱锥的外接球的球心$$O$$满足$$\\overrightarrow{OA}+\\overrightarrow{OB}+\\overrightarrow{OC}=\\overrightarrow{0}$$，则二面角$$A-PB-C$$的余弦值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{2}}{8}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{3}}{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的角与距离"], "answer_analysis": ["由题意，$$O$$应该在$$P-ABC$$的底面$$ABC$$上，设底面边长为$$2$$，则外接球半径等于$$\\frac{2}{3}\\sqrt{3}$$，于是侧棱长$$\\frac{2}{3}\\sqrt{6}$$．作$$CH\\bot PB$$于$$H$$，由对称性知$$AH\\bot PB$$，所以$$\\angle AHC$$就是二面角$$A-PB-C$$的平面角，不难算出$$CH=\\frac{\\sqrt{10}}{2}$$． 于是由余弦定理，可得$$\\cos \\angle AHC=\\frac{1}{5}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "448", "queId": "c7973693f0214599aeb3f0d0fe3a6669", "competition_source_list": ["2021年第31届浙江绍兴上虞区希望杯高一竞赛复赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "已知集合$$A=\\left { \\frac{1}{9},\\frac{1}{3},1,3,9,27 \\right }$$，$$B= {y\\left\\textbar{} y={{\\log }_{3}}x,x\\in A \\right. }$$，则$$A\\cap B$$=．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$或$$3$$ "}], [{"aoVal": "B", "content": "$$\\left { 1,3 \\right }$$ "}], [{"aoVal": "C", "content": "$$\\varnothing $$ "}], [{"aoVal": "D", "content": "$$\\left { 0,1,2,3 \\right }$$ "}]], "knowledge_point_routes": ["课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算", "课内体系->知识点->集合->集合的基本运算->交集", "课内体系->素养->数学运算"], "answer_analysis": ["由题意知$$B=\\left { -2,-1,0,1,2,3 \\right }$$所以$$ A \\cap B= \\left { 1,3\\right } $$，故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "973", "queId": "b2affc6c875f4a4cb2c6e43870d571b7", "competition_source_list": ["2002年AMC10竞赛A第17题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2002-AMC10A-17$$ Sarah pours $$4$$ ounces of coffee into a cup that can hold $$8$$ ounces. Then she pours $$4$$ ounces of cream into a second cup that can also hold $$8$$ ounces. She then pours half of the contents of the first cup into the second cup, completely mixes the contents of the second cup, then pours half of the contents of the second cup back into the first cup. What fraction of the contents in the first cup is cream? 莎拉将 4盎司的咖啡倒入一个可容纳 8盎司的杯子中。 然后她将 4盎司的奶油倒入第二个杯子中，该杯子也可以装 8 盎司。 然后她将第一个杯子的一半内容倒入第二个杯子，将第二个杯子的内容完全混合，然后将第二个杯子的一半内容倒回第一个杯子。 第一杯中的混合物有多少是奶油？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac 14$$ "}], [{"aoVal": "B", "content": "$$\\frac 13$$ "}], [{"aoVal": "C", "content": "$$\\frac 38$$ "}], [{"aoVal": "D", "content": "$$\\frac 25$$ "}], [{"aoVal": "E", "content": "$$\\frac 12$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems", "课内体系->知识点->集合->容斥原理"], "answer_analysis": ["We will simulate the process in steps. In the beginning, we have: ▪ $$4$$ ounces of coffee in cup $$1$$ ▪ $$4$$ ounces of cream in cup $$2$$ In the first step we pour $$\\frac 42=2$$ ounces of coffee from cup $$1$$ to cup $$2$$; getting: ▪ $$2$$ ounces of coffee in cup $$1$$ ▪ $$2$$ ounces of coffee and $$4$$ ounces of cream in cup $$2$$; In the second step we pour $$\\frac 22=1$$ ounce of coffee and $$\\frac 42=2$$ ounces of cream from cup $$2$$ to cup $$1$$, getting: ▪ $$2+1=3$$ ounces of coffee and $$0+2=2$$ ounces of cream in cup $$1$$ ▪ the rest in cup $$2$$ Hence at the end we have $$3+2=5$$ ounces of liquid in cup $$1$$, and out of these $$2$$ ounces is cream. Thus the answer is $$\\frac 25$$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "658", "queId": "6d6e46aafbfa4344bc7eeede1cc3b6a6", "competition_source_list": ["2010年河北全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "函数$$f(x)={{x}^{3}}-3{{x}^{2}}+3x+1$$的图象的对称中心为．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-1,2)$$ "}], [{"aoVal": "B", "content": "$$(1,2)$$ "}], [{"aoVal": "C", "content": "$$(-1,-2)$$ "}], [{"aoVal": "D", "content": "$$(1,-2)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["因为$$f(x)={{(x-1)}^{3}}+2$$，所以函数图象的对称中心为$$(1,2)$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "905", "queId": "814ae37bff0b4d1d905de7214affc8fa", "competition_source_list": ["2022~2023学年湖南邵阳隆回县湖南省隆回县第二中学高一上学期月考（竞赛）第5题", "2022~2023学年四川泸州泸县泸县第一中学高一上学期期末第5题", "2022~2023学年甘肃兰州高一上学期期末（兰州市第六十三中学 兰化三中 线上 考试）第10题"], "difficulty": "0", "qtype": "single_choice", "problem": "酒驾是严重危害交通安全的违法行为．为了保障交通安全，根据国家有关规定：$100\\text{ml}$血液中酒精含量低于$20\\text{mg}$的驾驶员可以驾驶汽车，酒精含量达到$20\\text{mg}$一一$79\\text{mg}$的驾驶员即为酒后驾车，$80\\text{mg}$及以上认定为醉酒驾车．假设某驾驶员喝了一定量的酒后，其血液上升到了$1\\text{mg}/\\text{ml}$.如果停止喝酒以后，他血液中的酒精含量会以每小时$30 ! \\% ! $的速度减少，那么他至少经过几个小时才能驾驶汽车？（参考数据：$\\lg 0.2\\approx -0.7$，$\\lg 0.3\\approx -0.5,\\lg 0.7\\approx -0.15,\\lg 0.8\\approx -0.1$）", "answer_option_list": [[{"aoVal": "A", "content": "1 "}], [{"aoVal": "B", "content": "3 "}], [{"aoVal": "C", "content": "5 "}], [{"aoVal": "D", "content": "7 "}]], "knowledge_point_routes": ["课内体系->知识点->函数的应用->函数的实际应用"], "answer_analysis": ["由题意得：$100\\text{ml}$血液中酒精含量低于$20\\text{mg}$的驾驶员可以驾驶汽车故${{\\left( 1-30 ! \\% ! \\right)}^{x}}\\textless{} 0.2$，即${{0.7}^{x}}\\textless{} 0.2$两边取对数即可得$\\lg {{0.7}^{x}}\\textless{} \\lg 0.2$，即$x\\textgreater\\frac{\\lg 0.2}{\\lg 0.7}\\approx 4.67$那么他至少经过$$5$$个小时才能驾驶汽车故选：$$\\text{C}$$ "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "751", "queId": "960b28d85c504577ae00c4cc8c16994b", "competition_source_list": ["2016年湖南全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在某次乒乓球单打比赛中，原计划每两名选手各比赛一场，但有$$3$$名选手各比赛了两场之后就退出了，这样全部比赛只进行了$$50$$场，那么上述$$3$$名选手之间比赛的场数是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->排列与组合->组合", "课内体系->知识点->计数原理->两个基本计数原理->分类加法计数原理", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理"], "answer_analysis": ["设选手人数为$$n$$，$$3$$名退出的选手之间的比赛场数为$$x$$，$$0\\leqslant x\\leqslant 3$$，由题意 $$\\mathrm{C}_{n}^{2}+\\left( 2\\times 3-x \\right)=50$$ 则$$x=\\mathrm{C}_{n}^{2}-44$$． 注意到$$\\mathrm{C}_{10}^{2}=45\\mathrm{C}_{11}^{2}=55$$，所以$$x=1$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1179", "queId": "d942b2ae4c9645edbd842d454807887a", "competition_source_list": ["2019年全国高中数学联赛竞赛初赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a$$，$$b$$，$$c$$均大于$$1$$，满足$$\\begin{cases}\\textasciitilde\\lg a+\\log _{b}c=3 \\lg b+\\log _{a}c=4 \\end{cases}$$， 则$$\\lg a\\cdot \\lg c$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$\\frac{8}{3}$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$\\frac{16}{3}$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["记分别记$$\\lg a$$，$$\\lg b$$，$$\\lg c$$为$$x$$，$$y$$，$$z$$，则$$x$$，$$y$$，$$z\\textgreater0$$且$$\\begin{cases}x+\\frac{z}{y}=3 y+\\frac{z}{x}=4 \\end{cases}\\Rightarrow \\begin{cases}xy+z=3y xy+z=4x \\end{cases}$$，设$$x=3t$$，$$y=4t$$，则$$z=12t-12{{t}^{2}}$$， 其中$$t\\in (0,1)$$， 因此$$\\lg a\\cdot \\lg c=xz=3t(12t-12{{t}^{2}})=18\\cdot (t\\cdot t\\cdot (2-2t))\\leqslant 18\\cdot {{\\left( \\frac{2}{3} \\right)}^{3}}=\\frac{16}{3}$$， 等号当$$t=\\frac{2}{3}$$时取得，因此所求最大值为$$\\frac{16}{3}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "64", "queId": "0eb48b13fce34060b09cdc0870550840", "competition_source_list": ["高二上学期单元测试《不等式专题》自招", "高二上学期单元测试《不等式专题》自招第3题", "高二上学期单元测试《不等式专题（1）》自招第1题", "2009年黑龙江全国高中数学联赛竞赛初赛第10题5分", "2020~2021学年浙江杭州西湖区杭州学军中学高一上学期单元测试《基本不等式》第43题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a\\textgreater b\\textgreater0$$，那么$${{a}^{2}}+\\frac{1}{b(a-b)}$$的最小值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->思想->转化化归思想", "课内体系->知识点->等式与不等式->不等式->基本不等式->利用基本不等式求最值", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的实际应用"], "answer_analysis": ["由$$a\\textgreater b\\textgreater0$$知$$a-b\\textgreater0$$，故有$$b(a-b)\\leqslant \\dfrac{{{a}^{2}}}{4}$$． ∴$${{a}^{2}}+\\dfrac{1}{b(a-b)}\\geqslant {{a}^{2}}+\\dfrac{4}{{{a}^{2}}}\\geqslant 4$$． 当且仅当$$\\begin{cases}b=a-b {{a}^{2}}=\\dfrac{4}{{{a}^{2}}} \\end{cases}$$，即$$\\begin{cases}a=\\sqrt{2} b=\\dfrac{\\sqrt{2}}{2} \\end{cases}$$时等号成立． ∴$${{a}^{2}}+\\dfrac{1}{b(a-b)}$$的最小值为$$4$$，故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "390", "queId": "8ffd56644e874087938630f5caa311a1", "competition_source_list": ["2010年河南全国高中数学联赛竞赛初赛第6题5分", "2018~2019学年上海静安区上海市市北中学高三上学期期中第16题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$\\triangle ABC$$的内角$$A, B, C$$所对的边$$a, b, c$$成等比数列，则$$\\frac{\\sin A+\\cos A\\tan C}{\\sin B+\\cos B\\tan C}$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left(0, +\\infty \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 0,\\frac{\\sqrt{5}+1}{2} \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{\\sqrt{5}-1}{2},\\frac{\\sqrt{5}+1}{2} \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{\\sqrt{5}-1}{2}, +\\infty \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法", "竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->三角函数->三角形中的问题->解三角形"], "answer_analysis": ["设$$a, b, c$$的公比为$$q$$，则$$b=aq, c=a{{q}^{2}}$$，而 $$\\frac{\\sin A+\\cos A\\tan C}{\\sin B+\\cos B\\tan C}$$ $$=\\frac{\\sin A\\cos C+\\cos A\\sin C}{\\sin B\\cos C+\\cos B\\sin C}$$ $$=\\frac{\\sin (A+C)}{\\sin (B+C)}=\\frac{\\sin ( \\pi -B)}{\\sin ( \\pi -A)}$$ $$=\\frac{\\sin B}{\\sin A}=\\frac{b}{a}=q.$$ 因此，只需求$$q$$的取值范围． 因$$a, b, c$$成等比数列，最大边只能是$$a$$或$$c$$，因此$$a, b, c$$要构成三角形的三边，必须且只需$$a+bc$$且$$b+ca$$．即有不等式组$$\\begin{cases}a+aq\\textgreater a{{q}^{2}} aq+a{{q}^{2}}\\textgreater a \\end{cases}$$，即$$\\begin{cases}{{q}^{2}}-q-1\\textless{}0 {{q}^{2}}+q-1\\textgreater0 \\end{cases}$$ 解得$$\\begin{cases}\\frac{1-\\sqrt{5}}{2}\\textless{}q\\textless{}\\frac{\\sqrt{5}+1}{2} q\\textgreater\\frac{\\sqrt{5}-1}{2} \\end{cases}$$或$$\\begin{cases}\\frac{1-\\sqrt{5}}{2}\\textless{}q\\textless{}\\frac{\\sqrt{5}+1}{2} q\\textless{}-\\frac{\\sqrt{5}+1}{2} \\end{cases}$$ 从而$$\\frac{\\sqrt{5}-1}{2}\\textless{}q\\textless{}\\frac{\\sqrt{5}+1}{2},$$ 因此所求的取值范围是$$\\left( \\frac{\\sqrt{5}-1}{2},\\frac{\\sqrt{5}+1}{2} \\right)$$．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "228", "queId": "269c238cd8a24819bfb9abf79d7833ce", "competition_source_list": ["2008年河北全国高中数学联赛竞赛初赛第3题6分", "2019~2020学年重庆沙坪坝区重庆市第七中学高一上学期单元测试《集合与函数》第29题"], "difficulty": "1", "qtype": "single_choice", "problem": "若函数$$y={{\\log }_{a}}({{x}^{2}}-ax+1)$$有最小值，则$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0 \\textless{} a \\textless{} 1$$ "}], [{"aoVal": "B", "content": "$$0 \\textless{} a \\textless{} 2,a\\ne 1$$ "}], [{"aoVal": "C", "content": "$$1 \\textless{} a \\textless{} 2$$ "}], [{"aoVal": "D", "content": "$$a\\geqslant 2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->复合函数", "课内体系->知识点->基本初等函数->对数函数->对数函数的图象及性质", "课内体系->知识点->基本初等函数->指数函数->利用指数函数性质求最值", "课内体系->素养->数学运算"], "answer_analysis": ["当$$a\\textgreater1$$时，$$y$$有最小值，则说明$${{x}^{2}}-ax+1$$有最小值，故$${{x}^{2}}-ax+1=0$$中$$\\Delta ~~\\textless{} ~0$$， 即$${{a}^{2}}-4 ~\\textless{} ~0$$，所以$$2\\textgreater a\\textgreater1$$． 当$$0 ~\\textless{} ~a ~\\textless{} ~1$$时，$$y$$有最小值， 则说明$${{x}^{2}}-ax+1$$有最大值，与二次函数性质相互矛盾，舍去． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "298", "queId": "23145819aeb544eaafaa13815566cc88", "competition_source_list": ["竞赛"], "difficulty": "0", "qtype": "single_choice", "problem": "已知非零实数$x,y$满足$(3x+y)^{5}+x^{5}+4x+y=0$，则$\\dfrac{x}{y}=$$\\underline{ }$.", "answer_option_list": [[{"aoVal": "A", "content": "$-\\dfrac{1}{3}$ "}], [{"aoVal": "B", "content": "$-\\dfrac{1}{4}$ "}], [{"aoVal": "C", "content": "$\\dfrac{1}{3}$ "}], [{"aoVal": "D", "content": "$\\dfrac{1}{4}$ "}]], "knowledge_point_routes": ["知识标签->知识点->函数->函数的性质"], "answer_analysis": ["条件等式变形为$(3x+y)^{5}+(3x+y)=-(x^{5}+x)$，设$f(t)=t^{5}+t$，显然$f(t)$是单调奇函数，又$f(3x+y)=-f(x)=f(-x)$，故$3x+y=-x,\\dfrac{x}{y}=-\\dfrac{1}{4}$ "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "742", "queId": "5364a69521e64a13b87768da929acc0c", "competition_source_list": ["2008年AMC12竞赛A第17题"], "difficulty": "2", "qtype": "single_choice", "problem": "2008AMC12A, 17 Let $$a_{1}$$, $$a_{2}$$, $$\\cdots$$ be a sequence determined by the rule $$a_{n}=\\frac{a_{n-1}}{2}$$ if $$a_{n-1}$$ is even and $$a_{n}=3a_{n-1}+1$$ if $$a_{n-1}$$ is odd. For how many positive integers $$a_{1}\\leqslant 2008$$ is it true that $$a_{1}$$ is less than each of $$a_{2}$$, $$a_{3}$$, and $$a_{4}$$? 令数列$$a_{1}$$, $$a_{2}$$, $$\\cdots$$满足: $$a_{n}=\\frac{a_{n-1}}{ 2}$$若$$a_{n-1}$$是偶数; $$a_{n}=3a_{n-1}+1$$若$$a_{n-1}$$是奇数。 有多少个正整数$$a_{1}\\leqslant 2008$$满足$$a_{1}$$小于$$a_{2}$$, $$a_{3}$$和$$a_{4}$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$250$$ "}], [{"aoVal": "B", "content": "$$251$$ "}], [{"aoVal": "C", "content": "$$501$$ "}], [{"aoVal": "D", "content": "$$502$$ "}], [{"aoVal": "E", "content": "$$1004$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数列->数列的概念->数列的定义", "美国AMC10/12->Knowledge Point->Geometry->Basic Analytic Geometry->Curves and Equations"], "answer_analysis": ["首先, 若$a_1$是偶数, 则$a_2=\\frac{a_1}{2}\\textless a_1$, 不满足条件. 若$a_1$是奇数, 则$a_2=3a_1+1$, $a_3=\\frac{3}{2}a_1+\\frac{1}{2}$必然都大于$a_1$, 问题出在$a_4$上$$.$$ 我们不知道$a_3$的奇偶, 也就无法计算$a_4$, 更无法讨论它和$a_1$的大小关系$$.$$ 因此, 接下来, 我们先对$a_3$的奇偶性进行讨论. 若$a_3$为偶数, 这等价于$3a_1+1$是$$4$$的倍数, 也就是说$a_1\\equiv 1\\pmod 4$. 此时$a_4=\\frac{a_3}{2}=\\frac{3}{4}a_1+\\frac{1}{4}\\leq a_1$, 不满足条件. 若$a_3$为奇数, 也就是说$a_1\\equiv 3\\pmod 4$时, $a_4=3a_3+1\\textgreater a_1$, 满足条件. 综上可知有且只有模$$4$$余$$3$$的$a_1$满足条件, 这些数在总共$$2008$$个数中占了$\\frac{1}{4}$, 有$\\frac{1}{4}\\times 2008=502$个, 选$$\\text{D}$$. 拓展: 数学上有一个著名的问题, 叫$3n+1$猜想, 讲的是: 在本题中所给的数列中, 无论$a_1$取什么正整数, $1$最终都会出现在数列的某一项$$.$$ 这一猜想至今未得到证明. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1058", "queId": "c58111881d5b44d3b9214267f7927f61", "competition_source_list": ["2015年天津全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$\\triangle ABC$$的周长为$$12$$，内切圆的半径为$$1$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\triangle ABC$$必为直角三角形 "}], [{"aoVal": "B", "content": "$$\\triangle ABC$$必为锐角三角形 "}], [{"aoVal": "C", "content": "$$\\triangle ABC$$必为直角三角形或锐角三角形 "}], [{"aoVal": "D", "content": "以上结论都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角形中的问题->解三角形"], "answer_analysis": ["设三边为$$abc$$，则$$\\tan \\frac{A}{2}=\\frac{1}{p-a}=\\frac{1}{6-a}$$（$$p$$是半周长）． 当$$a\\leqslant 5$$时，$$0\\textless{}\\tan \\frac{A}{2}\\leqslant 1$$，则$$A\\leqslant 90{}^{}\\circ $$； 当$$a\\textgreater5$$时，$$\\tan \\frac{A}{2}\\textgreater1$$，则$$A\\textgreater90{}^{}\\circ $$． 所以若有一条边的长度大于$$5$$，则$$\\triangle ABC$$是钝角三角形，否则$$\\triangle ABC$$是直角或锐角三角形． 设$$a=5+x$$，$$b=c=\\frac{7-x}{2}$$，内切圆半径为$$r$$，由面积法，得 $$\\frac{1}{2}\\cdot \\left( 5+x \\right)\\cdot \\sqrt{{{\\left( \\frac{7-x}{2} \\right)}^{2}}-{{\\left( \\frac{5+x}{2} \\right)}^{2}}}=\\frac{1}{2}\\cdot r\\cdot 12$$ 化简可得$$r=\\frac{\\left( 5+x \\right)\\sqrt{1-x}}{2\\sqrt{6}}$$ 当$$r=1$$时，化简得方程$${{x}^{3}}+9{{x}^{2}}+15x-1=0$$，此方程在$$\\left( 0,1 \\right)$$之间有解$${{x}_{0}}$$，即存在三边长为$$5+{{x}_{0}}\\frac{7-{{x}_{0}}}{2}\\frac{7-{{x}_{0}}}{2}$$，内切圆半径为$$1$$的$$\\triangle ABC$$，是一个钝角三角形． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1080", "queId": "d352c562097d44a4a2282f77f570d8b3", "competition_source_list": ["2009年辽宁全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$${{a}_{n}}={{2}^{n}}$$，$${{b}_{n}}=n$$，（$$n=1$$，$$2$$，$$3$$\\ldots），$${{A}_{n}}$$、$${{B}_{n}}$$分别为数列$$\\left { {{a}_{n}} \\right }$$、$$\\left { {{b}_{n}} \\right }$$的前$$n$$项和，记$${{c}_{n}}={{a}_{n}}{{B}_{n}}+{{b}_{n}}{{A}_{n}}-{{a}_{n}}{{b}_{n}}$$，则数列$$\\left { {{c}_{n}} \\right }$$的前$$10$$项和为（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{2}^{10}}+53$$ "}], [{"aoVal": "B", "content": "$${{2}^{11}}+53$$ "}], [{"aoVal": "C", "content": "$$110\\times \\left( {{2}^{9}}-1 \\right)$$ "}], [{"aoVal": "D", "content": "$$110\\times \\left( {{2}^{10}}-1 \\right)$$~ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["$${{a}_{n}}={{2}^{n}}$$，$${{b}_{n}}=n$$，$${{A}_{n}}={{2}^{n+1}}-2$$，$${{B}_{n}}=\\frac{n\\left( n+1 \\right)}{2}$$，于是$${{c}_{n}}=\\frac{n\\left( n+1 \\right)}{2}{{2}^{n}}+n\\cdot {{2}^{n}}-2n$$， 不妨设$${{P}_{n}}=\\sum\\limits_{i=1}^{n}{i\\cdot {{2}^{i}}}$$，$${{Q}_{n}}=\\sum\\limits_{i=1}^{n}{\\frac{i\\left( i+1 \\right)}{2}\\cdot {{2}^{i}}}$$，于是$$\\sum\\limits_{i=1}^{n}{{{c}_{i}}={{P}_{n}}+{{Q}_{n}}-n\\left( n+1 \\right)}$$， $${{Q}_{n}}=1\\times 2+3\\times {{2}^{2}}+\\cdots +\\frac{n\\left( n+1 \\right)}{2}\\cdot {{2}^{n}}$$① $$2{{Q}_{n}}=1\\times {{2}^{2}}+3\\times {{2}^{3}}+\\cdots +\\frac{n\\left( n-1 \\right)}{2}\\cdot {{2}^{n}}+\\frac{n\\left( n+1 \\right)}{2}\\cdot {{2}^{n+1}}$$② ①$$-$$②得$$-{{Q}_{n}}=1\\times {{2}^{1}}+2\\times {{2}^{2}}+\\cdots +n\\cdot {{2}^{n}}-\\frac{n\\left( n+1 \\right)}{2}\\cdot {{2}^{n+1}}$$ 即$${{Q}_{n}}=\\frac{n\\left( n+1 \\right)}{2}\\cdot {{2}^{n+1}}-{{P}_{n}}$$，即$${{Q}_{n}}+{{P}_{n}}=\\frac{n\\left( n+1 \\right)}{2}\\cdot {{2}^{n+1}}=n\\left( n+1 \\right)\\cdot {{2}^{n}}$$ ∴$$\\sum\\limits_{i=1}^{n}{{{c}_{i}}={{P}_{n}}+{{Q}_{n}}-n\\left( n+1 \\right)=n\\left( n+1 \\right)\\left( {{2}^{n}}-1 \\right)}$$，将$$n=10$$代入即可． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "461", "queId": "2d51981d3ff94efcbc68289ccc809784", "competition_source_list": ["2008年江苏全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知关于$$x$$的方程$${{x}^{2}}-2ax+{{a}^{2}}-4a=0$$至少有一个模为$$3$$的复数根，则实数$$a$$的所有取值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$，$$9$$ "}], [{"aoVal": "B", "content": "$$-1$$，$$ 9 $$，$$2-\\sqrt{13}$$ "}], [{"aoVal": "C", "content": "$$1 $$，$$9$$， $$2+\\sqrt{13}$$ "}], [{"aoVal": "D", "content": "$$1$$，$$ 9$$，$$ 2-\\sqrt{13}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["将方程写为$${{(x-a)}^{2}}=4a$$．当$$a\\geqslant 0$$时，此时方程有实根， 因为该实根模为$$3$$，故方程有一根为$$3$$或$$-3$$．代入， 由$${{(a\\pm 3)}^{2}}=4a$$，得$$a=1$$或$$9$$； 当$$a\\textless{}0$$时，得$$x=a\\pm 2\\sqrt{\\textbar a\\textbar}\\text{i}$$， 故$$\\textbar x{{\\textbar}^{2}}={{a}^{2}}-4a=9$$，得$$a=2-\\sqrt{13}$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "229", "queId": "198134db724a40f297dbb9cbec789838", "competition_source_list": ["2019年贵州全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$x$$、$$y$$、$$z$$均为素数，且$$x\\leqslant~ y\\leqslant z$$．则不定方程$$x^{2}+y^{2}+z^{2}=2019$$的正整数解有组．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->同余->平方剩余", "竞赛->知识点->排列组合与概率->概率初步"], "answer_analysis": ["注意到，素数$$x\\leqslant y\\leqslant z$$． 由$$3x^{2}\\leqslant 2019\\Rightarrow 2\\leqslant x\\leqslant 25$$， 故$$x$$、$$y$$的所有可能取值为：$$2$$、$$3$$、$$5$$、$$7$$、$$11$$、$$13$$、$$17$$、$$19$$、$$23$$． 由$$3z^{2}\\geqslant 2019$$，$$z^{2}\\textless{}2019\\Rightarrow 26\\leqslant z\\textless{}45$$， 故$$z$$的所有可能取值为：$$29$$、$$31$$、$$37$$、$$41$$、$$43$$． 当$$z=29$$时， $$x^{2}+y^{2}=2019-29^{2}=1178$$，无解； 当$$z=31$$时， $$x^{2}+y^{2}=2019-31^{2}=1058=23^{2}+23^{2}$$； 当$$z=37$$时， $$x^{2}+y^{2}=2019-37^{2}=650=11^{2}+23^{2}=17^{2}+19^{2}$$； 当$$z=41$$时， $$x^{2}+y^{2}=2019-41^{2}=338=7^{2}+17^{2}=13^{2}+13^{2}$$； 当$$z=43$$时， $$x^{2}+y^{2}=2019-43^{2}=170$$，无解． 综上，原不定方程共有五个解： $$(x,y,z)=(23,23,31)$$，$$(11,23,37)$$，$$(17,19,37)$$，$$(7,17,41)$$，$$(13,13,41)$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "81", "queId": "0b24d1e9f3b74f13b1642a3898e75d32", "competition_source_list": ["2021年AMC10竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "2021 AMC 10A Problem 25 How many ways are there to place $$3$$ indistinguishable red chips, $$3$$ indistinguishable blue chips, and $$3$$indistinguishablegreenchipsin thesquaresofa $$3\\times3$$ grid so thatnotwochips of the same color are directly adjacent to each other, either vertically or horizontally? 将 $$3$$ 枚不可区分的红色筹码, $$3$$ 枚不可区分的蓝色筹码和 $$3$$ 枚不可区分的绿色筹码分别放在 $$3\\times3$$ 方格表的各个小方格中，使得无论是垂直方向还是水平方向，都没有两个相同颜色的筹码相邻，共问有多少种放法?", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$30$$ "}], [{"aoVal": "E", "content": "$$36$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Number Theory"], "answer_analysis": ["N\\A "], "answer_value": "E"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1143", "queId": "e1b02aaf30a84f6383c5524113c292e8", "competition_source_list": ["2021~2022学年福建莆田城厢区莆田第一中学高一下学期开学考试（学科素养能力竞赛）第7~7题"], "difficulty": "1", "qtype": "single_choice", "problem": "某地新能源汽车工厂2017年生产新能源汽车的年产量为260万辆，根据前期市场调研，为满足市场需求，以后每一年的产量都比上一年产量提高25\\%，那么该工厂到哪一年的产量才能首次超过800万辆（参考数据：$$\\lg 1.25\\approx 0.097,\\lg 1.3\\approx 0.11,\\lg 4\\approx 0.60$$）（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "2021年 "}], [{"aoVal": "B", "content": "2022年 "}], [{"aoVal": "C", "content": "2023年 "}], [{"aoVal": "D", "content": "2024年 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 由题意确定函数模型，再根据指数与对数运算解不等式.\\\\ 【详解】\\\\ 设产量首次超过800万辆的年份为$$x$$，则$$260\\times {{\\left( 1+25\\% \\right)}^{x-2017}}\\textgreater800$$，\\\\ 即$$\\lg 260+\\left( x-2017 \\right)\\lg 1.25\\textgreater\\lg 800$$，\\\\ $$x-2017\\textgreater\\frac{\\lg \\frac{800}{260}}{\\lg 1.25}=\\frac{\\lg 4-\\lg 1.3}{\\lg 1.25}\\approx 5.05$$，\\\\ 即$$x\\textgreater2017+5.05=2022.05$$，\\\\ 所以$$x=2023$$，\\\\ 故选：C. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1168", "queId": "e69faf08d2384ec7bad8dcb01b43f543", "competition_source_list": ["2008年江西全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "设$${{a}^{2}}+{{b}^{2}}=1$$，$$\\left( b\\ne 0 \\right)$$，若直线$$ax+by=2$$和椭圆$$\\frac{{{x}^{2}}}{6}+\\frac{{{y}^{2}}}{2}=1$$有公共点，则$$\\frac{a}{b}$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ -\\frac{1}{2},\\frac{1}{2} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left[ -1,1 \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left( -\\infty ,-1 \\right]\\cup \\left[ 1,+\\infty \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left[ -2,2 \\right]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆锥曲线"], "answer_analysis": ["将$$y=\\frac{2-ax}{b}$$代入椭圆方程并整理得$$\\left( 3{{a}^{2}}+{{b}^{2}} \\right){{x}^{2}}-12ax+12-6{{b}^{2}}=0$$， 因直线和椭圆有公共点，则判别式$${{\\left( 12a \\right)}^{2}}-4\\left( 3{{a}^{2}}+{{b}^{2}} \\right)\\left( 12-6{{b}^{2}} \\right)\\geqslant 0$$， 利用$${{a}^{2}}+{{b}^{2}}=1$$，化简得$${{a}^{2}}\\geqslant {{b}^{2}}$$， 所以$$\\left\\textbar{} \\frac{a}{b} \\right\\textbar\\geqslant 1$$．即$$\\frac{a}{b}\\in \\left( -\\infty ,-1 \\right]\\cup \\left[ 1,+\\infty \\right)$$．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "955", "queId": "8aac49074e724b45014e8c4d77ae61f4", "competition_source_list": ["2015年黑龙江全国高中数学联赛竞赛初赛第9题5分", "2017~2018学年北京东城区北京市第五中学高二下学期期中理科第6题6分", "2015年黑龙江全国高中数学联赛竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两人玩猜数字游戏，先由甲心中想一个数字，记为$$a$$，再由乙猜甲刚才想到的数字，把乙猜的数字记为$$b$$，且$$a$$、$$b\\in {1,2,\\ldots ,6 }$$． 若$$\\left\\textbar{} a-b \\right\\textbar\\leqslant 1$$，则称甲、乙``心有灵犀''． 现任意找两个人玩这个游戏，得出他们``心有灵犀''的概率为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{9}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{7}{18}$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{9}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->统计与概率->概率->事件与概率->古典概型->古典概型的概率计算（不涉及计数原理）"], "answer_analysis": ["若``心有灵犀''，则当$$a=1$$时，$$b=1$$或$$b=2$$； 当$$a=2$$时，$$b=1$$或$$b=2$$或$$b=3$$；当$$a=3$$时，$$b=2$$或$$b=3$$或$$b=4$$； 当$$a=4$$时，$$b=3$$或$$b=4$$或$$b=5$$；当$$a=5$$时，$$b=4$$或$$b=5$$或$$b=6$$； 当$$a=6$$时，$$b=5$$或$$b=6$$，即有$$16$$种满足题意的结果，∴$$P=\\frac{16}{36}=\\frac{4}{9}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "414", "queId": "62e656e0845847a6b1f1a52932bc488c", "competition_source_list": ["2012年江西全国高中数学联赛竞赛初赛第4题10分"], "difficulty": "0", "qtype": "single_choice", "problem": "若实数$$a, b, c$$满足：$$a+b+c={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$，则$$a+b+c$$的最大值是~\\uline{~~~~~~~~~~}~.", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->柯西", "课内体系->知识点->等式与不等式->不等式->柯西不等式"], "answer_analysis": ["提示：据柯西不等式， $$3\\left( a+b+c \\right)=\\left( {{1}^{2}}+{{1}^{2}}+{{1}^{2}} \\right)\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\right)\\geqslant {{\\left( a+b+c \\right)}^{2}}$$，所以$$a+b+c\\leqslant 3$$，当$$a=b=c=1$$时取得等号． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "508", "queId": "55c4c04a6e2b43dcad60a523b11bcf45", "competition_source_list": ["全国全国高中数学联赛竞赛一试"], "difficulty": "2", "qtype": "single_choice", "problem": "已知三个平面$$\\alpha $$、$$\\beta $$、$$\\lambda $$，每两个平面之间的夹角都是$$\\theta $$，且$$\\alpha \\cap \\beta =a$$，$$\\beta \\cap \\gamma =b$$，$$\\gamma \\cap \\alpha =c$$．若有： 命题甲：$$\\theta \\textgreater\\frac{ \\pi }{3}$$． 命题乙：$$a$$，$$b$$，$$c$$相交于一点． 则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "甲是乙的充分条件但不必要 "}], [{"aoVal": "B", "content": "甲是乙的必要条件但不充分 "}], [{"aoVal": "C", "content": "甲是乙的充分必要条件 "}], [{"aoVal": "D", "content": "$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$都不对 "}]], "knowledge_point_routes": ["课内体系->知识点->立体几何初步->基本图形位置关系->空间中的基本事实与定理", "课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与立体几何结合", "课内体系->素养->逻辑推理"], "answer_analysis": ["设$$a$$，$$b$$，$$c$$交于一点， 由所成三面角内部一点引三条射线分别垂直于$$\\alpha $$、$$\\beta $$、$$\\lambda $$， 其中每两条射线所成的角都是$$ \\pi -\\varphi $$， $$\\varphi $$为三面角中两两相等的二面角的平面角， 总和$$3( \\pi -\\varphi )\\textless{}2 \\pi $$， ∴$$\\varphi \\textgreater\\frac{1}{3} \\pi $$． 当$$\\varphi \\textgreater\\frac{2}{3} \\pi $$时，$$\\varphi \\textgreater\\frac{1}{3} \\pi $$． 若$$a$$，$$b$$，$$c$$不交于一点，则互相平行， 这时$$\\varphi \\textgreater\\frac{1}{3} \\pi $$． 故选$$\\text{A}$$． （注：认定两平面间夹角范围是$$0\\textless{}\\theta \\mathsf{\\leqslant }\\frac{1}{2} \\pi $$） "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1120", "queId": "e5fc7af1ccd34835be767374b29bbf52", "competition_source_list": ["2016年天津全国高中数学联赛竞赛初赛第3题6分", "2016年高考真题天津卷"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$0\\textless{}a\\textless{}b$$，已知，$$s$$，$$t$$，$$b$$依次成等差数列，$$a$$，$$u$$，$$v$$，$$b$$依次成等比数列，记$$x=st\\left( s+t \\right)$$，$$y=uv\\left( u+v \\right)$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textgreater y$$ "}], [{"aoVal": "B", "content": "$$x=y$$ "}], [{"aoVal": "C", "content": "$$x\\textless{}y$$ "}], [{"aoVal": "D", "content": "既有$$x\\textgreater y$$的情形，也有$$x\\textless{}y$$的情形 "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->均值", "竞赛->知识点->不等式->几个重要的不等式->排序", "竞赛->知识点->不等式->不等式的综合应用（数列与不等式）"], "answer_analysis": ["由题设有$$x=\\frac{2a+b}{3}\\cdot \\frac{a+2b}{3}\\left( a+b \\right)$$，$$y=ab\\left( \\sqrt[3]{{{a}^{2}}b}+\\sqrt[3]{a{{b}^{2}}} \\right)$$， 由均值不等式，$$\\frac{2a+b}{3}\\geqslant \\sqrt[3]{{{a}^{2}}b}$$，$$\\frac{a+2b}{3}\\geqslant \\sqrt[3]{a{{b}^{2}}}$$（等号都不成立）， 由排序不等式，$$a+b=\\sqrt[3]{{{a}^{3}}}+\\sqrt[3]{{{b}^{3}}}\\geqslant \\sqrt[3]{{{a}^{2}}b}+\\sqrt[3]{a{{b}^{2}}}$$（等号不成立）， 所以，$$x\\textgreater y$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "219", "queId": "6b589a42a187467db467f6f0c5a169ba", "competition_source_list": ["2001年全国高中数学联赛竞赛一试第3题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "在四个函数$$y=\\sin \\left\\textbar{} x \\right\\textbar$$、$$y=\\cos \\left\\textbar{} x \\right\\textbar$$、$$y=\\left\\textbar{} \\cot x \\right\\textbar$$、$$y=\\lg \\left\\textbar{} \\sin x \\right\\textbar$$中，以$$ \\pi $$为周期、在$$\\left( 0,\\frac{ \\pi }{2} \\right)$$上单调递增的偶函数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$y=\\sin \\left\\textbar{} x \\right\\textbar$$ "}], [{"aoVal": "B", "content": "$$y=\\cos \\left\\textbar{} x \\right\\textbar$$ "}], [{"aoVal": "C", "content": "$$y=\\left\\textbar{} \\text{cot}x \\right\\textbar$$ "}], [{"aoVal": "D", "content": "$$y=\\lg \\left\\textbar{} \\sin x \\right\\textbar$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["可考虑用排除法．$$y=\\sin \\left\\textbar{} x \\right\\textbar$$不是周期函数（可通过作图判断），故$$\\text{A}$$错误； $$y=\\cos \\left\\textbar{} x \\right\\textbar$$的最小正周期为$$2 \\pi $$，且在$$\\left( 0,\\frac{ \\pi }{2} \\right)$$上是减函数，故$$\\text{B}$$错误； $$y=\\left\\textbar{} \\text{cot}x \\right\\textbar$$在$$\\left( 0,\\frac{ \\pi }{2} \\right)$$上是减函数，故$$\\text{C}$$错误． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "580", "queId": "ff5e9487665f4073bc8ded14e34e899e", "competition_source_list": ["2016年陕西全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$1，$$2$$，\\cdots ，20$$这$$20$$个数中，任取$$3$$个不同的数，则这$$3$$个数构成等差数列的概率为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{10}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{19}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{38}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->数据分析", "课内体系->素养->逻辑推理", "课内体系->知识点->数列->等差数列->等差数列的概念与通项公式->等差数列求通项问题", "课内体系->知识点->数列->等差数列->等差数列的性质及应用", "课内体系->知识点->统计与概率->概率->事件与概率->古典概型", "课内体系->知识点->计数原理->排列与组合->组合"], "answer_analysis": ["设等差数列的三个数为$${{a}_{1}}$$，$${{a}_{1}}+d$$，$${{a}_{1}}+2d$$，其中$${{a}_{1}}$$，$$d$$都是正整数． 由$${{a}_{1}}+2d\\leqslant 20$$，得$${{a}_{1}}\\leqslant 20-2d$$． 等差数列的个数为$$\\sum\\limits_{d=1}^{9}{\\left( 20-2d \\right)}=180-9\\times 10=90$$． 故所求概率为$$\\frac{90}{\\mathrm{C}_{20}^{3}}=\\frac{3}{38}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "343", "queId": "39957d4412fb4d5d9ef560e96288722c", "competition_source_list": ["2008年山东全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "以双曲线$$\\frac{{{x}^{2}}}{4}-\\frac{{{y}^{2}}}{m}=1$$的离心率为半径，以右焦点为圆心的圆与双曲线的渐近线相切，则$$m=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{6}{5}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->直线和圆的方程->圆与方程->直线与圆的位置关系->圆的切线的相关问题", "课内体系->知识点->直线和圆的方程->直线与方程->平面中的距离->点到直线的距离公式", "课内体系->知识点->圆锥曲线->双曲线->双曲线的几何性质->双曲线的渐近线", "课内体系->知识点->圆锥曲线->双曲线->双曲线的几何性质->双曲线的离心率->求双曲线的离心率", "课内体系->素养->数学运算"], "answer_analysis": ["易知$$a=2$$，$$b=\\sqrt{m}$$，$$c=4+m$$，$$e=\\frac{\\sqrt{4+m}}{2}$$． 渐近线方程：$$\\frac{x}{2}\\pm \\frac{y}{\\sqrt{m}}=0$$，即$$\\sqrt{m}x\\pm 2y=0$$． 右焦点$$\\left( \\sqrt{4+m},0 \\right)$$到渐近线的距离$$d=b=\\sqrt{m}$$． 从而得方程$$\\sqrt{m}=\\frac{\\sqrt{4+m}}{2}$$，解得$$m=\\frac{4}{3}$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "84", "queId": "0f2bad531ae348c3b83876d20f438e30", "competition_source_list": ["2020年贵州全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "点$$A$$，$$B$$，$$C$$均位于单位圆上，且 $$\\textbar{} \\overrightarrow{AB}\\textbar= \\sqrt{3}$$，则 $$\\overrightarrow{AB}\\cdot \\overrightarrow{AC}$$ 的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{3}+ \\frac{3}{2}$$ "}], [{"aoVal": "B", "content": "$$2 \\sqrt{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3 \\sqrt{3}}{2}$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算"], "answer_analysis": ["由已知，可设$$A\\left( -1,0 \\right)$$，$$B\\left( \\frac{1}{2},\\frac{\\sqrt{3}}{2} \\right)$$，$$C\\left( \\cos \\theta , \\sin \\theta \\right)\\left( 0\\leqslant \\theta \\leqslant 2\\pi \\right)$$，则$$\\overrightarrow{AB}\\cdot \\overrightarrow{AC}=\\left( \\frac{3}{2},\\frac{\\sqrt{3}}{2} \\right)\\cdot \\left( \\cos \\theta +1,\\sin \\theta \\right)=\\frac{3}{2}\\cos \\theta +\\frac{\\sqrt{3}}{2}\\sin \\theta +\\frac{3}{2}$$ $$=\\sqrt{3}\\sin \\left( \\theta +\\frac{\\pi }{3} \\right)+\\frac{3}{2}\\leqslant \\sqrt{3}+\\frac{3}{2}$$， 所以，$$\\overrightarrow{AB}\\cdot \\overrightarrow{AC}$$的最大值为$$\\sqrt{3}+\\frac{3}{2}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "307", "queId": "39565f31da02438faf7ce8b3fa8f40d5", "competition_source_list": ["2021~2022学年广东肇庆德庆县广东省德庆县香山中学高一下学期期中第4题", "2021~2022学年安徽阜阳太和县安徽省太和中学高一下学期月考（竞赛考试）第1~1题", "2021~2022学年广东肇庆德庆县广东省德庆县香山中学高一下学期期中第4题"], "difficulty": "0", "qtype": "single_choice", "problem": "在$$\\vartriangle ABC$$中，已知$$B=60{}^{}\\circ ,AC=\\sqrt{3},AB=1$$，则$$BC=$$（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "1 "}], [{"aoVal": "B", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "2 "}], [{"aoVal": "D", "content": "4 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 直接利用余弦定理即可求得.\\\\ 【详解】\\\\ 在$$\\vartriangle ABC$$中，已知$$B=60{}^{}\\circ ,AC=\\sqrt{3},AB=1$$，即为$$B=60{}^{}\\circ ,b=\\sqrt{3},c=1$$，\\\\ 由余弦定理$${{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\\cos B$$得：$$3={{a}^{2}}+1-2a\\times 1\\times \\frac{1}{2}$$,解得：$$a=2$$（边长大于0，所以$$a=-1$$舍去）\\\\ 即$$BC=2$$.\\\\ 故选：C "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "700", "queId": "df461f4915464e4da121d8bb2d724b15", "competition_source_list": ["2012年辽宁全国高中数学联赛竞赛初赛第1题6分", "2012~2013学年北京东城区高二下学期期末理科第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "用数字$$0$$，$$1$$，$$2$$，$$3$$组成无重复数字的四位数，这样的四位数的个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->计数原理->排列与组合->排列->特殊元素优先法"], "answer_analysis": ["最高位只能填$$1$$、$$2$$、$$3$$，有$$3$$种方法，其余的位置没有限制， 任意排有$$A_{3}^{3}=6$$种方法， 这样的四位数的个数为$$3\\times 6=18$$种． 故选B． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "34", "queId": "0333cc23d84641a095cda26a2ac238b3", "competition_source_list": ["2021年第31届浙江绍兴上虞区希望杯高一竞赛复赛第6题"], "difficulty": "0", "qtype": "single_choice", "problem": "在边长为$$a$$的立方体的表面及内部可以找到$$9$$个点，使得其中任意两个点的距离至少为$$1$$，则$$a$$的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2\\sqrt{2}}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2\\sqrt{3}}{3}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{3}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->立体几何初步->基本立体图形->空间几何体的概念->棱柱、棱锥、棱台的结构特征"], "answer_analysis": ["由于是$$9$$个点，故$$4$$个为上平面的顶点，下平面$$4$$个顶点，正方体中心为$$1$$个点， 故只需���$$\\begin{cases} a\\geqslant 1 \\sqrt{3}a\\geqslant 2 \\end{cases}$$，解得$$a\\geqslant \\frac{2\\sqrt{3}}{3}$$，故当$$a=\\frac{2\\sqrt{3}}{3}$$时， 一定存在$$9$$个点使得两两距离至少为$$1$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1172", "queId": "d92a0175518d4971b1ddf158816ef5f6", "competition_source_list": ["2009年河北全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$A$$、$$B$$是球的一条直径的两个端点，$$ \\pi $$是过点$$A$$且垂直于$$AB$$的一个平面．$$C$$、$$D$$是球面上不同于$$A$$、$$B$$的两个点，延长$$BC$$、$$BD$$，分别交平面$$ \\pi $$于$$E$$、$$F$$两点．若$$A$$、$$E$$、$$F$$三点构成等腰三角形，则$$AB$$与$$CD$$的位置关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$AB$$与$$CD$$垂直 "}], [{"aoVal": "B", "content": "$$AB$$与$$CD$$平行 "}], [{"aoVal": "C", "content": "$$AB$$与$$CD$$垂直且异面 "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的平行和垂直"], "answer_analysis": ["若$$AE=EF\\ne AF$$，则$$AB$$与$$CD$$不垂直，排除选项$$\\text{A}$$和$$\\text{C}$$； 若$$AB$$与$$CD$$平行，则$$A$$、$$E$$、$$F$$共线，构不成三角形，排除$$\\text{B}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1091", "queId": "f7f387aea7db4c5f972211af4a1efbb5", "competition_source_list": ["2004年AMC10竞赛A第6题", "2004年AMC12竞赛A第4题"], "difficulty": "0", "qtype": "single_choice", "problem": "$$2004-AMC10A-6$$ Bertha has $$6$$ daughters and no sons. Some of her daughters have $$6$$ daughters, and the rest have none. Bertha has a total of $$30$$ daughters and granddaughters, and no great-granddaughters. How many of Bertha\\textquotesingle s daughters and grand-daughters have no children? 伯莎有 $6$$ 个女儿，没有儿子。 她的一些女儿有 $6$$ 个女儿，其余的没有。 伯莎总共有 30个女儿和孙女，没有曾孙女。 伯莎的女儿和孙女有多少没有孩子？", "answer_option_list": [[{"aoVal": "A", "content": "$$22$$ "}], [{"aoVal": "B", "content": "$$23$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$25$$ "}], [{"aoVal": "E", "content": "$$26$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->排列与组合->排除法", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Simple Logical Reasoning"], "answer_analysis": ["Since Bertha has $$6$$ daughters, she has $$30-6=24$$ granddaughters, of which none have daughters. Of Bertha\\textquotesingle s daughters, $$\\dfrac{24}{6}=4$$ have daughters, so $$6-4=2$$ do not have daughters. Therefore, of Bertha\\textquotesingle s daughters and granddaughters, $$24+2=26$$ do not have daughters. $$\\boxed{ (\\text{E}) 26}$$. ▪ ▪ Alcumus** Bertha has $$30-6=24$$ granddaughters, none of whom have any daughters. The granddaughters are the children of $$\\dfrac{24}{6}=4$$ of Bertha\\textquotesingle s daughters, so the number of women having no daughters is $$30-4=\\boxed {26}$$. Draw a tree diagram and see that the answer can be found in the sum of $$6+6$$ granddaughters, $$5+5$$ daughters, and $$4$$ more daughters. Adding them together gives the answer of $$\\boxed{ (\\text{E}) 26}$$. "], "answer_value": "E"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "70", "queId": "133df186e82d481e8df9f5c79a12032d", "competition_source_list": ["2008年吉林全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "若存在钝角$$\\alpha $$，使得$$\\sin \\alpha -\\sqrt{3}\\cos \\alpha ={{\\log }_{2}}({{x}^{2}}-x+2)$$成立，则实数$$x$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$ {x\\textbar-1\\leqslant x\\textless{}0$$或$$1\\textless{}x\\leqslant 2 }$$ "}], [{"aoVal": "B", "content": "$$ {x\\textbar-1\\textless{}x\\textless{}0$$或$$1\\textless{}x\\textless{}2 }$$ "}], [{"aoVal": "C", "content": "$$ {x\\textbar0\\leqslant x\\leqslant 1 }$$ "}], [{"aoVal": "D", "content": "$$ {x\\textbar-1\\textless{}x\\textless{}2 }$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数综合"], "answer_analysis": ["设$$\\frac{ \\pi }{2}\\textless{}\\alpha \\textless{} \\pi $$，则$$\\sin \\alpha -\\sqrt{3}\\cos \\alpha =2\\sin \\left( \\alpha -\\frac{ \\pi }{3} \\right)\\in \\left( 1,2 \\right]$$， 可知$$1\\textless{}{{\\log }_{2}}({{x}^{2}}-x+2)\\leqslant 2$$， 即$$2\\textless{}{{x}^{2}}-x+2\\leqslant 4$$， 解得$$-1\\leqslant x\\textless{}0$$或$$1\\textless{}x\\leqslant 2$$．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "72", "queId": "989ded5c88b543609d717432a35144fc", "competition_source_list": ["2012年浙江全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知集合$$P=\\left { x\\textbar1\\leqslant x\\leqslant 2 \\right }$$，$$M=\\left { x\\textbar2-a\\leqslant x\\leqslant 1+a \\right }$$，若$$P\\cap M=P$$，则实数$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( -\\infty , 1 \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left[ 1, +\\infty \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left[ -1, 1 \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left[ -1, +\\infty \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["由$$P\\cap M=P\\Rightarrow P\\subseteq M\\Rightarrow 2-a\\leqslant 1$$，$$1+a\\geqslant 2\\Rightarrow a\\geqslant 1$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "22", "queId": "022bab6814ea4683ae204b00f0bd72bc", "competition_source_list": ["第二十届全国希望杯高一竞赛复赛邀请赛第2题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "方程$${{a}^{-x}}={{\\log }_{a}}x(a\\textgreater0,a\\ne 1)$$的实数根的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["画图可知，$$y={{a}^{-x}}$$与$$y={{\\log }_{a}}x$$的图象只有$$1$$个交点． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "726", "queId": "accbbf7046ed476f8165c18c6f01c83a", "competition_source_list": ["2018年天津全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "在等差数列$$\\left { {{a}_{n}} \\right }$$中，$${{a}_{2}}+{{a}_{11}}+{{a}_{14}}=-6$$，则前$$17$$项的和$$\\sum\\limits_{i=1}^{17}{{{a}_{i}}}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$-34$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$34$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["设公差为$$d$$，则 $${{a}_{2}}+{{a}_{11}}+{{a}_{14}}=3{{a}_{1}}+24d=3{{a}_{9}}$$， 结合已知条件得$${{a}_{9}}=-2$$， 从而，前$$17$$项的和为$$17{{a}_{9}}=-34$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "733", "queId": "c3c4a2d2a8ec41b89eb059192d050ebc", "competition_source_list": ["1994年全国高中数学联赛竞赛一试第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "在平面直角坐标系中，方程$$\\frac{\\left\\textbar{} x+y \\right\\textbar}{2a}+\\frac{\\left\\textbar{} x-y \\right\\textbar}{2b}=1$$（$$a$$、$$b$$为不相等的两个正数）所代表的曲线是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "三角形 "}], [{"aoVal": "B", "content": "正方形 "}], [{"aoVal": "C", "content": "非正方形的长方形 "}], [{"aoVal": "D", "content": "非正方形的菱形 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与方程"], "answer_analysis": ["将直角坐标系$$XOY$$绕原点逆时针旋转$$45{}^{}\\circ $$，得到新坐标系$${{X}^{\\prime }}O{{Y}^{\\prime }}$$，点$$P$$在坐标系$${{X}^{\\prime }}O{{Y}^{\\prime }}$$中的坐标为（$${{x}^{\\prime }}$$，$${{y}^{\\prime }}$$）在坐标系$$XOY$$中的坐标为（$$x$$，$$y$$），则 $$\\begin{cases}{{x}^{\\prime }}=\\frac{1}{\\sqrt{2}}(x+y) {{y}^{\\prime }}=\\frac{1}{\\sqrt{2}}(-x+y) \\end{cases}$$ 题中方程$$\\frac{\\left\\textbar{} x+y \\right\\textbar}{2a}+\\frac{\\left\\textbar{} x-y \\right\\textbar}{2b}=1$$① 化成$$\\frac{\\left\\textbar{} {{x}^{\\prime }} \\right\\textbar}{a}+\\frac{\\left\\textbar{} {{y}^{\\prime }} \\right\\textbar}{b}=\\sqrt{2}$$② 显然，②代表的曲线关于$${{X}^{\\prime }}$$轴、$${{Y}^{\\prime }}$$轴对称，在$${{X}^{\\prime }}O{{Y}^{\\prime }}$$的第Ⅰ象限内，②成为$$\\frac{{{x}^{\\prime }}}{a}+\\frac{{{y}^{\\prime }}}{b}=\\sqrt{2}$$，即为线段$$AB$$，其中$$A$$（$$\\sqrt{2}a$$，$$0$$），$$B$$（$$0$$，$$\\sqrt{2}b$$）． 据对称性，在第Ⅱ象限内方程②是线段$$BC$$，其中$$C$$（$$-\\sqrt{2}a$$，$$0$$）； 在第Ⅲ象限内方程②是线段$$CD$$，其中$$D$$（$$0$$，$$\\sqrt{2}b$$）； 在第Ⅳ象限内方程②是线段$$AD$$． 由对称性知，$$AB=BC=CD=AD$$，又由于$$a\\ne b$$，故$$AC\\ne BD$$，所以$$ABCD$$是非正方形的菱形． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "783", "queId": "c8820d660ffe443b8e7f26f97b423f98", "competition_source_list": ["2002年全国全国高中数学联赛竞赛一试第8题9分"], "difficulty": "2", "qtype": "single_choice", "problem": "将二项式$${{\\left( \\sqrt{x}+\\frac{1}{2\\sqrt[4]{x}} \\right)}^{n}}$$的展开式按$$x$$的降幂排列，若前三项系数成等差数列，则该展开式中$$x$$的指数是整数的项共有个．", "answer_option_list": [[{"aoVal": "A", "content": "$2$ "}], [{"aoVal": "B", "content": "$3$ "}], [{"aoVal": "C", "content": "$4$ "}], [{"aoVal": "D", "content": "以上答案均不对 "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->二项式定理", "竞赛->知识点->排列组合与概率->二项式定理及其应用"], "answer_analysis": ["不难求出前三项的系数分别是$$1,\\frac{1}{2}n,\\frac{1}{8}n(n-1)$$， ∵$$2\\cdot \\frac{1}{2}n=1+\\frac{1}{8}n(n-1)$$， ∴当$$n=8$$时，$${{T}_{r+1}}=C_{n}^{r}{{\\left( \\frac{1}{2} \\right)}^{r}}{{x}^{\\frac{16-3r}{4}}}$$ （$$r=0$$，$$1$$，$$2$$，$$\\cdots $$，$$8$$） ∴$$r=0,4,8$$，即有$$3$$个． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "821", "queId": "ff8080814a39795c014a3bff668e00ac", "competition_source_list": ["2014年黑龙江全国高中数学联赛竞赛初赛第11题5分", "2013~2014学年北京朝阳区高三上学期期末理科第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "已知数列$$\\left { {{a}_{n}} \\right }$$满足$${{a}_{n}}=n\\cdot {{k}^{n}}(n\\in {{N}^{*}},0\\textless{}k\\textless{}1)$$下面说法正确的是（~ ）． ①$$k=\\frac{1}{2}$$时，数列$$\\left { {{a}_{n}} \\right }$$为递减数列； ②$$\\frac{1}{2}\\textless{}k\\textless{}1$$时，数列$$\\left { {{a}_{n}} \\right }$$不一定有最大项； ③$$0\\textless{}k\\textless{}\\frac{1}{2}$$时，数列$$\\left { {{a}_{n}} \\right }$$为递减数列； ④当$$\\frac{k}{1-k}$$为正整数时，数列$$\\left { {{a}_{n}} \\right }$$必有两项相等的最大项.", "answer_option_list": [[{"aoVal": "A", "content": "①② "}], [{"aoVal": "B", "content": "②④ "}], [{"aoVal": "C", "content": "③④ "}], [{"aoVal": "D", "content": "②③ "}]], "knowledge_point_routes": ["课内体系->知识点->数列->数列的概念->数列的函数特性->数列中最大项与最小项的求解问题", "课内体系->知识点->数列->数列的概念->数列的函数特性->数列单调性问题", "课内体系->知识点->函数的概念与性质->函数的性质->单调性", "课内体系->思想->函数思想", "课内体系->素养->数学运算", "课内体系->特色题型->解答压轴"], "answer_analysis": ["①当$$k=\\frac{1}{2}$$时，$${{a}_{n}}=n\\cdot {{\\left( \\frac{1}{2} \\right)}^{n}}(n\\in {{N}^{*}})$$，$$\\because {{a}_{1}}=\\frac{1}{2},{{a}_{2}}=2\\times \\frac{1}{4}=\\frac{1}{2}$$，$$\\therefore {{a}_{1}}={{a}_{2}}$$， 即数列$$\\left { {{a}_{n}} \\right }$$不是递减数列，$$\\therefore $$①错误； ②当$$\\frac{1}{2}\\textless{}k\\textless{}1$$时，$$\\frac{{{a}_{n+1}}}{{{a}_{n}}}=\\frac{(n+1)\\cdot {{k}^{n+1}}}{n\\cdot {{k}^{n}}}=\\frac{(n+1)k}{n}$$， $$\\therefore k\\textless\\frac{k(n+1)}{n}\\textless2k$$，因此数列$$\\left { {{a}_{n}} \\right }$$可有最大项，因此错误； ③当$$0\\textless{}k\\textless{}\\frac{1}{2}$$时，$$\\frac{{{a}_{n+1}}}{{{a}_{n}}}=\\frac{(n+1)\\cdot {{k}^{n+1}}}{n\\cdot {{k}^{n}}}=\\frac{(n+1)k}{n}\\textless{}\\frac{n+1}{2n}\\leqslant 1$$， $$\\therefore {{a}_{n+1}}\\textless{{a}_{n}}$$，故数列$$\\left { {{a}_{n}} \\right }$$为递减数列； ④当$$\\frac{k}{1-k}$$为正整数时，$$\\frac{{{a}_{n+1}}}{{{a}_{n}}}=\\frac{(n+1)\\cdot {{k}^{n+1}}}{n\\cdot {{k}^{n}}}=\\frac{(n+1)k}{n}$$， $$\\therefore $$当$$\\frac{k}{1-k}$$为正整数时，$$1\\textgreater k\\geqslant \\begin{matrix} 1 2 \\end{matrix}$$， 即当$$k=\\frac{1}{2}$$时，$${{a}_{1}}={{a}_{2}}\\textgreater{{a}_{3}}\\textgreater{{a}_{4}}\\textgreater\\cdots $$ 当$$1\\textgreater k\\textgreater\\frac{1}{2}$$时，令$$\\frac{k}{1-k}=m\\in {{\\mathbf{N}}^{*}}$$，解得$$k=\\frac{m}{1+m}$$，则$$\\frac{{{a}_{n+1}}}{{{a}_{n}}}=\\frac{(n+1)m}{n(1+m)}$$， 综上数列$$\\left { {{a}_{n}} \\right }$$必有两项相等的最大项． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1042", "queId": "b34505e0b05c4e98bee80ae6a45f4dd0", "competition_source_list": ["2016年AMC10竞赛第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "從$$1$$到$$2016$$的整數中隨意地選出三個不同的整數，若這三個數的乘積爲奇數的機率爲$$p$$，則下列何者正確？ Three distinct integers are selected at random between $$1$$ and $$2016$$, inclusive. Which of the following is a correct statement about the probability $$p$$ that the product of the three integers is odd?", "answer_option_list": [[{"aoVal": "A", "content": "$$ p\\textless{} \\frac{1}{8} $$ "}], [{"aoVal": "B", "content": "$$ p= \\frac{1}{8} $$ "}], [{"aoVal": "C", "content": "$$ \\frac{1}{8} \\frac{1}{3} $$ "}], [{"aoVal": "D", "content": "$$p=\\frac 13$$ "}], [{"aoVal": "E", "content": "$$p\\textgreater\\frac 13$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["For the product to be odd, all three factors have to be odd. The probability of this is $$\\frac {1008}{2016}\\cdot \\frac {1007}{2015}\\cdot \\frac {1006}{2014}$$. $$\\frac {1008}{2016}=\\frac 12$$, but $$\\frac {1007}{2015}$$ and $$\\frac {1006}{2014}$$ are slightly less than $$\\frac 12$$. Thus, the whole product is slightly less than $$\\frac 12\\cdot \\frac 12\\cdot \\frac 12=\\frac 18$$, so $$p\\textless\\frac 18$$. For the product to be odd, all three factors have to be odd. There are a total of $$\\left(\\frac {2016}3\\right)$$ ways to choose $$3$$ numbers at random, and there are $$\\left(\\frac {1008}3\\right)$$ to choose $$3$$ odd numbers. Therefore, the probability of choosing $$3$$ odd numbers is $$\\frac {(\\frac {1008}3)}{(\\frac {2016}3)}$$. Simplifying this, we obtain $$\\frac {1008\\cdot 1007\\cdot 1006}{2016\\cdot 2015\\cdot 2014}$$, which is slightly less than $$\\frac 18$$, so our answer is $$p\\textless\\frac 18$$. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "711", "queId": "914b4490c5924315bad3cff69e001046", "competition_source_list": ["2003年全国全国高中数学联赛竞赛一试第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "删去正整数数列$$1$$，$$2$$，$$3$$，$$\\cdots $$中的所有完全平方数，得到一个新数列，这个新数列的第$$2003$$项是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2046$$ "}], [{"aoVal": "B", "content": "$$2047$$ "}], [{"aoVal": "C", "content": "$$2048$$ "}], [{"aoVal": "D", "content": "$$2049$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["注意到$${{45}^{2}}=2025$$，$${{46}^{2}}=2116$$，∴$$2026={{a}_{2026-45}}={{a}_{1981}}$$，$$2115={{a}_{2115-45}}={{a}_{2070}}$$， 而且在从第$$1981$$项到第$$2070$$项之间的$$90$$项中没有完全平方数， 又$$1981+22=2003$$，∴$${{a}_{2003}}={{a}_{1981}}+22=2026+22=2048$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "440", "queId": "e7f47efc0d794af4b8a7edf0ea86af35", "competition_source_list": ["2013年AMC10竞赛B第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "2013年$$AMC10$$竞赛$$B$$第$$15$$题 A wire is cut into two pieces, one of length $$a$$ and the other of length $$b$$. The piece of length $$a$$ is bent to form an equilateral triangle, and the piece of length $$b$$ is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is $$\\frac ab$$? 一根电线被切成两块，一块长度为 $$a$$，另一块长度为 $$b$$。 长度$$a$$ 弯曲形成等边三角形，长度$$b$$ 弯曲形成正六边形。 三角形和六边形的面积相等。 求 $$\\frac ab$$？", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{6}}{2}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$2$$ "}], [{"aoVal": "E", "content": "$$\\frac{3\\sqrt{2}}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->解三角形->三角形面积公式", "美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Regular Polygon"], "answer_analysis": ["Using the area formulas for an equilateral triangle $$\\left(\\frac{s^{2}\\sqrt{3}}{4}\\right)$$ and regular hexagon $$\\left(\\frac{3s^{2}\\sqrt{3}}{2}\\right)$$ , with side length $$s$$ and plugging $$\\frac a3$$ and $$\\frac b6$$ into each equation, we find that $$\\frac{a^{2}\\sqrt{3}}{36}= \\frac{b^{2}\\sqrt{3}}{24}$$. Simplifying this, we get $$\\frac ab=\\boxed{\\rm(B)\\textasciitilde\\frac{\\sqrt{6}}{2}}$$. The regular hexagon can be broken into $$6$$ small equilateral triangles, each of which is similar to the big equilateral triangle. The big triangle\\textquotesingle s area is $$6$$ times the area of one of the little triangles. Therefore each side of the big triangle is $$\\sqrt{6}$$ times the side of the small triangle. The desired ratio is $$\\frac{3 \\sqrt{6}}{6}= \\frac{\\sqrt{6}}{2}\\Rightarrow (B)$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1093", "queId": "bce6ab0ab1fa42acb98e89ebfd521d75", "competition_source_list": ["2008年辽宁全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知抛物线$${{x}^{2}}=2py(p\\textgreater0)$$，过点$$M\\left( 0,-\\frac{p}{2} \\right)$$向抛物线引两条切线，$$A$$、$$B$$为切点，则线段$$AB$$的长度是（ ．", "answer_option_list": [[{"aoVal": "A", "content": "$$3p$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{2}p$$ "}], [{"aoVal": "C", "content": "$$2p$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{2}p$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆锥曲线"], "answer_analysis": ["切线方程为$$y=kx-\\frac{p}{2}$$，代入抛物线方程得$$\\frac{{{x}^{2}}}{2p}=kx-\\frac{p}{2}$$， 即$${{x}^{2}}-2pkx+{{p}^{2}}=0$$有一个实根， 故$$4{{p}^{2}}{{k}^{2}}-4{{p}^{2}}=0$$， 解得$$k=\\pm 1$$，$$x=\\pm p$$， 所以$$AB$$的长度为$$2p$$．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "305", "queId": "3dd8ba49c33d4f91bf19358e28dd4293", "competition_source_list": ["2012年黑龙江全国高中数学联赛竞赛初赛第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "数列$$\\left { {{a}_{n}} \\right }$$满足$${{a}_{1}}=1, {{a}_{2}}=\\frac{1}{2}$$，并且$${{a}_{n}}\\left( {{a}_{n-1}}+{{a}_{n+1}} \\right)=2{{a}_{n+1}}\\cdot {{a}_{n-1}}\\left( n\\geqslant 2 \\right)$$，则数列$$\\left { {{a}_{n}} \\right }$$的第$$2012$$项为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2010}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2011}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2012}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2013}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["$${{a}_{n}}\\left( {{a}_{n-1}}+{{a}_{n+1}} \\right)=2{{a}_{n+1}}{{a}_{n-1}}\\Rightarrow \\frac{1}{{{a}_{n+1}}}+\\frac{1}{{{a}_{n-1}}}=2\\frac{1}{{{a}_{n}}}\\left( n\\geqslant 2 \\right)$$． 所以$$\\left { \\frac{1}{{{a}_{n}}} \\right }$$是等差数列，且$$\\frac{1}{{{a}_{1}}}=1, d=1$$，则数列的通项公式$${{a}_{n}}=\\frac{1}{n}$$，故第$$2012$$项为$$\\frac{1}{2012}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1123", "queId": "f3c5fe5f1b014af88eeab461ce50add1", "competition_source_list": ["2008年AMC12竞赛B第23题"], "difficulty": "3", "qtype": "single_choice", "problem": "2008AMC12B, 23 The sum of the base-$$10$$ logarithms of the divisors of $$10^{n}$$ is $$792$$. What is $$n$$? $$10^{n}$$的所有因数的以$$10$$为底的对数之和是$$792$$. 则$$n$$的值为? ．", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$14$$ "}], [{"aoVal": "E", "content": "$$15$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Calculation->Exponentiation", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Simple Logical Reasoning", "课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算"], "answer_analysis": ["$$792=\\sum_{i=0}^{}n\\sum_{j=0}^{}n\\left(i\\log{2}+j\\log{5}\\right)=\\sum_{i=0}^{}n\\left((n+1)i\\log{2}+\\frac{n(n+1)}{2}\\log{5}\\right)=\\frac{n(n+1)^{2}}{2}(\\log{2}+\\log{5})=\\frac{n(n+1)^{2}}{2}$$, 其中$\\log$指$\\log_{10}$. 于是有$n(n+1)^{2}=1584=2^{4}\\times 3^{2}\\times 11$. 可以看出$11\\textbar n$, 然后如果再有$2^{2}\\textbar n$或$3^{2}\\textbar n$, $(n+1)^{2}$就太小了, 于是只有$n=11$, 选$$\\text{A}$$. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "643", "queId": "7f866dd4bfb44a7c92640956d486e1d1", "competition_source_list": ["2010年山东全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f(x)={{x}^{3}}+(a+1){{x}^{2}}+(a+1)x+a$$在其定义域内既有极大值又有极小值，则实数$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1\\textless{}a\\textless{}2$$ "}], [{"aoVal": "B", "content": "$$a\\textgreater2$$ "}], [{"aoVal": "C", "content": "$$a\\textless{}-1$$ "}], [{"aoVal": "D", "content": "$$a\\textgreater2$$或$$a\\textless{}-1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->导数"], "answer_analysis": ["由$$f(x)={{x}^{3}}+(a+1){{x}^{2}}+(a+1)x+a,$$ 得$$f'(x)=3{{x}^{2}}+2(a+1)x+(a+1)$$． 已知$$f(x)$$在其定义域内既有极大值又有极小值， 所以$$f\\prime (x)=0$$一定有两个不相等的实数根． 从而$$\\Delta =4{{(a+1)}^{2}}-12(a+1)\\textgreater0$$， 解得$$a\\textgreater2$$或$$a\\textless{}-1$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "39", "queId": "0623eb90032940f7afc14c843aefcb02", "competition_source_list": ["1991年全国高中数学联赛竞赛一试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$是正整数，$$a\\textless{}100$$．并且$${{a}^{3}}+23$$能被$$24$$整除，那么这样的$$a$$的个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->同余->同余的概念与性质"], "answer_analysis": ["考虑使得$$24\\textbar{{a}^{3}}-1$$成立的$$a$$． $${{a}^{3}}-1=\\left( a-1 \\right)\\left( {{a}^{2}}+a+1 \\right)=\\left[ \\left( a-1 \\right)a\\left( a+1 \\right)+1 \\right]$$． 因$$a\\left( a+1 \\right)+1$$是奇数， 若要$$24\\textbar\\left( {{a}^{3}}-1 \\right)$$，必有$${{2}^{3}}\\left( a-1 \\right)$$． 若$$a-1$$不能被$$3$$整除，则$$3\\textbar a\\left( a+1 \\right)$$， 从而$$a\\left( a+1 \\right)+1$$不能被$$3$$整除． 因此，若要$$24\\textbar\\left( {{a}^{3}}-1 \\right)$$，必有$$3\\left( a-1 \\right)$$．这样就有 $$24\\textbar\\left( a-1 \\right)$$，即 $$a=24k+1$$， $$\\left( k\\in Z \\right)$$． 由$$24k+1\\textless{}100$$，$$k$$可能取的一切值为$$0,1,2,3,4$$，也就是这样的$$a$$有$$1,25,49,73,97$$五个． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "61", "queId": "070bed4109b14d0882452921d79e5d2d", "competition_source_list": ["2009年黑龙江全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "下面的命题中，正确的个数为． （1） 将函数$$y=4\\sin 2x$$图象向左平移$$\\frac{\\pi }{3}$$个单位得到函数$$y=4\\sin \\left( 2x+\\frac{\\pi }{3} \\right)$$的图象； （2） 函数$$y=4\\cos (2x+\\varphi )$$图象关于点$$\\left( \\frac{\\pi }{6},0 \\right)$$对称的充要条件是$$\\varphi =k\\pi +\\frac{\\pi }{6}(k\\in Z)$$； （3） 函数$$y=\\frac{4\\tan x}{1-{{\\tan }^{2}}x}$$的周期为$$\\frac{\\pi }{2}$$； （4） 化简$$\\sqrt{1+\\sin 2}-\\sqrt{1-\\sin 2}$$等于$$2\\sin 1$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角不等式", "竞赛->知识点->三角函数->三角函数的图像与性质", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["$$y=4\\sin 2x$$向左平移$$\\frac{\\text{ }\\pi }{3}$$个单位，得到$$y=4\\sin 2\\left( x+\\frac{\\text{ }\\pi }{3} \\right)=4\\sin \\left( 2x+\\frac{2}{3}\\text{ }\\pi ~\\right)$$，⑴错； 由$$4\\cos \\left( 2\\cdot \\frac{\\text{ }\\pi }{6}+\\varphi \\right)=0$$，得$$\\frac{\\text{ }\\pi }{3}+\\varphi =k\\text{ }\\pi +\\frac{\\text{ }\\pi }{2}$$，即$$\\varphi =k\\text{ }\\pi +\\frac{\\text{ }\\pi }{6}$$，$$k\\in \\mathbf{Z}$$，⑵对； $$y=\\frac{4\\tan x}{1-{{\\tan }^{2}}x}=2\\tan 2x$$，但$$x\\ne k\\text{ }\\pi \\pm \\frac{\\text{ }\\pi }{4},k\\text{ }\\pi +\\frac{\\text{ }\\pi }{2}$$，$$y$$不是周期函数，⑶错； $$\\sqrt{1+\\sin 2}-\\sqrt{1-\\sin 2}=\\sqrt{{{\\left( \\sin 1+\\cos 1 \\right)}^{2}}}-\\sqrt{{{\\left( \\sin 1-\\cos 1 \\right)}^{2}}}=2\\cos 1$$，⑷错． 正确的命题个数只有$$1$$个． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "11", "queId": "0985c6a316a84320a4c75a746f28b2d9", "competition_source_list": ["2018年四川全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "对任意正整数$$n$$，定义$$Z(n)$$为使得$$1+2+\\cdots +m$$是$$n$$的倍数的最小正整数$$m$$．关于下列三个命题： ①若$$p$$为奇素数，则$$Z(p)=p-1$$； ②对任意正整数$$a$$，均有$$Z(2^{}a)\\textgreater2^{}a$$； ③对任意正整数$$a$$，均有$$Z(3^{}a)=3^{}a-1$$， 其中，真命题的序号为．", "answer_option_list": [[{"aoVal": "A", "content": "①② "}], [{"aoVal": "B", "content": "①③ "}], [{"aoVal": "C", "content": "②③ "}], [{"aoVal": "D", "content": "①②③ "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->整除->质数（算数基本定理）"], "answer_analysis": ["注意到，$$1+2+ \\cdots +m= \\frac{m(m+1)}{2}$$， 在命题①中，由$$p\\left\\textbar{} \\frac{m(m+1)}{2} \\right.$$，知$$2p\\textbar m(m+1)$$， 又$$p$$为奇素数，则$$p\\textbar m$$或$$p\\textbar(m+1)$$， 故$$m$$的最小值为$$p-1$$， 因此，命题①正确． 在命题②中，由$$2^{a}\\left\\textbar{} \\frac{m(m+1)}{2} \\right.$$，知$$2^{a+1}\\textbar m(m+1)$$， 注意到，$$(m,m+1)=1$$， 则$$2^{a+1}\\textbar m$$或$$2^{a+1}\\textbar(m+1)$$， 故$$Z(2^{}a)=2^{a+1}-1\\textgreater2^{}a$$， 因此，命题②正确． 在命题③中，由$$3^{a}\\left\\textbar\\frac{m(m+1)}{2} \\right.$$，知$$2\\times 3^{a}\\textbar m(m+1)$$， 注意到，$$2\\textbar m(m+1)$$，$$(2,3)=1$$， 则$$2\\times3^{}a\\textbar m(m+1)\\Leftrightarrow 3^{}a\\textbar m(m+1)$$， 又$$(m,m+1)=1$$， 于是，$$3^{}a\\textbar m$$或$$3^{}a\\textbar(m+1)$$， 故$$Z(3^{}a)=3^{}a-1$$， 因此，命题③正确． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "887", "queId": "d20873b7e67a4813a1456880d3a3055f", "competition_source_list": ["2017~2018学年4月北京西城区北京市第八中学高三下学期月考理科第8题5分", "2009年竞赛珠海市第12题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "对于直角坐标平面内的任意两点$$A({{x}_{1}} ,{{y}_{1}})$$、$$B({{x}_{2}},{{y}_{2}})$$，定义，它们之间的一种``距离''：$$\\textbar\\textbar AB\\textbar\\textbar=\\textbar{{x}_{1}}-{{x}_{2}}\\textbar+\\textbar{{y}_{1}}-{{y}_{2}}\\textbar$$．给出下列三个命题： ①若点$$C$$在线段$$AB$$上，则$$\\textbar\\textbar AC\\textbar\\textbar+\\textbar\\textbar CB\\textbar\\textbar=\\textbar\\textbar AB\\textbar\\textbar$$； ②在$$\\triangle ABC$$中，若$$\\angle C=90{}^{}\\circ $$，则$$\\textbar\\textbar AC\\textbar{{\\textbar}^{2}}+\\textbar\\textbar CB\\textbar{{\\textbar}^{2}}=\\textbar\\textbar AB\\textbar{{\\textbar}^{2}}$$； ③在$$\\triangle ABC$$中，$$\\textbar\\textbar AC\\textbar\\textbar+\\textbar\\textbar CB\\textbar\\textbar\\textgreater\\textbar\\textbar AB\\textbar\\textbar$$． 其中真命题的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角形中的问题->解三角形"], "answer_analysis": ["注意到$$C$$点坐标的大小介于$$A,B$$点之间，易知①正确； 取点$$C$$为原点，$$A,B$$点分别在$$x,y$$轴上， 则$$\\textbar\\textbar AB\\textbar{{\\textbar}^{2}}=\\textbar AB{{\\textbar}^{2}}+2\\textbar AC\\textbar\\textbar BC\\textbar\\textgreater\\textbar AC{{\\textbar}^{2}}+\\textbar BC{{\\textbar}^{2}}=\\textbar\\textbar AC\\textbar{{\\textbar}^{2}}+\\textbar\\textbar CB\\textbar{{\\textbar}^{2}}$$，②不正确； 当$$AC$$、$$CB$$分别与$$x$$、$$y$$轴平行时，$$\\textbar\\textbar AC\\textbar\\textbar+\\textbar\\textbar CB\\textbar\\textbar=\\textbar\\textbar AB\\textbar\\textbar$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "857", "queId": "7370631685e84c8c9fc7dd916de9bcc3", "competition_source_list": ["2012年浙江全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$y=\\frac{\\sqrt{3}}{2}\\sin \\left( x+\\frac{ \\pi }{2} \\right)+\\cos \\left( \\frac{ \\pi }{6}-x \\right)$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{13}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{13}}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{13}}{2}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{13}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["$$y=\\frac{\\sqrt{3}}{2}\\sin \\left( x+\\frac{ \\pi }{2} \\right)+\\cos \\left( \\frac{ \\pi }{6}-x \\right)=\\sqrt{3}\\cos x+\\frac{1}{2}\\sin x=\\frac{\\sqrt{13}}{2}\\sin \\left( x+\\varphi \\right)$$，其中$$\\sin \\varphi =\\frac{2\\sqrt{39}}{13}$$，$$\\cos \\varphi =\\frac{\\sqrt{13}}{13}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "124", "queId": "2a6ee04e44814636a32300a25839dd56", "competition_source_list": ["2009年吉林全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "称横坐标为整数的点为``次整点''，过曲线$$y=\\sqrt{9-{{x}^{2}}}$$上任意两个次整点作直线，则倾斜角大于$$30{}^{}\\circ $$的直线条数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆综合"], "answer_analysis": ["次整点有$$\\left( \\pm 3,0 \\right)$$，$$\\left( \\pm 2,\\sqrt{5} \\right)$$，$$\\left( \\pm 1,2\\sqrt{2} \\right)$$，$$\\left( 0,3 \\right)$$，斜率为负的直线都满足要求，共有$$5+3+1=9$$条，斜率为正的只有$$5$$条满足要求，总共$$9+5=14$$条． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "938", "queId": "8631003a12d349029ecc49e71d98b16d", "competition_source_list": ["2008年江苏全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知平面上点的集合$$M= {(x,y)\\textbar y=\\sqrt{2x-{{x}^{2}}} }$$，$$N= {(x,y)\\textbar y=k(x+1) }$$．当$$M\\cap N\\ne \\varnothing $$时，$$k$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ -\\frac{\\sqrt{3}}{3},\\frac{\\sqrt{3}}{3} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left[ 0,\\frac{\\sqrt{3}}{3} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left[ -\\frac{\\sqrt{3}}{3},0 \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left[ \\frac{\\sqrt{3}}{3},+\\infty \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["集合$$M$$的图形为以$$\\left( 1,0 \\right)$$为圆心、$$1$$为半径的圆的上半圆， 集合$$N$$的图形为过$$\\left( -1,0 \\right)$$的直线． 若直线与圆有公共点，则易得其倾斜角在$$\\left[ 0,\\frac{ \\pi }{6} \\right]$$内， 即$$k\\in \\left[ 0,\\frac{\\sqrt{3}}{3} \\right]$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "913", "queId": "7402f4a9833d431ca5609d2a3eefca92", "competition_source_list": ["2021~2022学年福建莆田城厢区莆田第一中学高一下学期开学考试（学科素养能力竞赛）第1~1题"], "difficulty": "0", "qtype": "single_choice", "problem": "已知集合$$M=\\left { -1,0,1 \\right }$$，$$N=\\left { y\\left\\textbar{} y={{x}^{2}} \\right. \\right }$$，则$$M\\bigcap N=$$（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left { 0 \\right }$$ "}], [{"aoVal": "B", "content": "$$\\left { -1,1 \\right }$$ "}], [{"aoVal": "C", "content": "$$\\left { 0,1 \\right }$$ "}], [{"aoVal": "D", "content": "$$\\left { -1,0,1 \\right }$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 求出集合$$N$$，利用交集的定义可得结果.\\\\ 【详解】\\\\ $$N=\\left { y\\left\\textbar{} y={{x}^{2}} \\right. \\right }=\\left { y\\left\\textbar{} y\\ge 0 \\right. \\right }$$，故$$M\\bigcap N=\\left { 0,1 \\right }$$.\\\\ 故选：C. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "572", "queId": "3b861111940842e2bd3c880a2bf665b3", "competition_source_list": ["2018~2019学年浙江宁波北仑区浙江省北仑中学高一下学期期中A卷第1题4分", "2014年浙江全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知圆$${{\\left( x+2 \\right)}^{2}}+{{\\left( y-1 \\right)}^{2}}=1$$与圆$${{x}^{2}}+{{\\left( y+1 \\right)}^{2}}=1$$关于直线$$l$$对称，则$$l$$的方程为．", "answer_option_list": [[{"aoVal": "A", "content": "$$x+y+1=0$$ "}], [{"aoVal": "B", "content": "$$x-y+1=0$$ "}], [{"aoVal": "C", "content": "$$x-y-1=0$$ "}], [{"aoVal": "D", "content": "$$x+y-1=0$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆综合"], "answer_analysis": ["直线$$l$$即为两圆心$$\\left( -2, 1 \\right)$$、$$\\left( 0, -1 \\right)$$的垂直平分线，即为 $$x-y+1=0$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "71", "queId": "0ad2e2d4b7d84e3e89cfeccc7a8dec55", "competition_source_list": ["2016年吉林全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个有限项的数列满足：任何$$3$$个连续项之和都是负数，且任何$$4$$个连续项之和都是正数，则此数列项数的最大值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->思想->转化化归思想", "课内体系->思想->函数思想", "课内体系->知识点->数列->数列的概念->数列的表示方法->通项公式", "课内体系->知识点->数列->数列的概念->数列的函数特性->数列中最大项与最小项的求解问题", "课内体系->知识点->数列->等差数列->等差数列的性质及应用"], "answer_analysis": ["如果有$$6$$项，则 $$0 ~\\textless{} ~\\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}} \\right)+\\left( {{a}_{2}}+{{a}_{3}}+{{a}_{4}}+{{a}_{5}} \\right)+\\left( {{a}_{3}}+{{a}_{4}}+{{a}_{5}}+{{a}_{6}} \\right)$$ $$=\\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}} \\right)+\\left( {{a}_{2}}+{{a}_{3}}+{{a}_{4}} \\right)+\\left( {{a}_{3}}+{{a}_{4}}+{{a}_{5}} \\right)+\\left( {{a}_{4}}+{{a}_{5}}+{{a}_{6}} \\right)$$ $$ ~\\textless{} ~0$$． 矛盾，所以最多$$5$$项． 数列$$2$$，$$2$$，$$-5$$，$$2$$，$$2$$满足要求，所以数列项数最大值为$$5$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "518", "queId": "67ff83dbf1ed4a7e874311cfd914287b", "competition_source_list": ["2016~2017学年广东深圳福田区深圳市红岭中学高二下学期期中理科第8题5分", "2014年天津和平区高三二模文科第6题5分", "2008年全国高中数学联赛竞赛一试第2题6分", "2014年天津和平区高三二模理科第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$A=[-2,4)$$，$$B= {x\\textbar{{x}^{2}}-ax-4\\leqslant 0 }$$，若$$B\\subseteq A$$，则实数$$a$$的取值范围为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ -1,2 \\right)$$ "}], [{"aoVal": "B", "content": "$$[-1,2]$$ "}], [{"aoVal": "C", "content": "$$[0,3]$$ "}], [{"aoVal": "D", "content": "$$[0,3)$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->数学抽象", "课内体系->知识点->等式与不等式->不等式->解不等式->一元二次不等式", "课内体系->知识点->集合->集合的基本关系->子集", "课内体系->知识点->集合->集合的基本关系->集合关系中的含参问题"], "answer_analysis": ["∵$$B\\subseteq A$$， ∴①$$B=\\varnothing $$，$$\\Delta \\textless{}0$$，无解， ②$$B\\ne \\varnothing $$，$${{x}^{2}}-ax-4=0$$有两根$${{x}_{1}}$$，$${{x}_{2}}\\in [-2,4)$$， 设$$f(x)=x^{2}-ax-4$$， ∴$$\\begin{cases}f(-2)\\geqslant 0 f(4)\\textgreater0 -2\\textless\\dfrac{a}{2}\\textless{}4 \\end{cases}\\Rightarrow 0\\leqslant a\\textless{}3$$． 故选$$\\text{D}$$． ", "因$${{x}^{2}}-2ax-4=0$$有两个实根$${{x}_{1}}=\\frac{a}{2}-\\sqrt{4+\\frac{{{a}^{2}}}{4}}$$，$${{x}_{2}}=\\frac{a}{2}+\\sqrt{4+\\frac{{{a}^{2}}}{4}}$$，故$$B\\subseteq A$$等价于$${{x}_{1}}\\mathsf{\\geqslant }-2$$且$${{x}_{2}}<{}4$$，即$$\\frac{a}{2}-\\sqrt{4+\\frac{{{a}^{2}}}{4}}\\mathsf{\\geqslant }-2$$且$$\\frac{a}{2}+\\sqrt{4+\\frac{{{a}^{2}}}{4}}<{}4$$，解之得$$0\\mathsf{\\leqslant }a<{}3$$．
\n故选$$\\text{D}$$．
\n"], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1187", "queId": "e27452cbbb3741b7b9b59e9e55dec680", "competition_source_list": ["2005年全国高中数学联赛竞赛一试第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "空间四点$$A$$、$$B$$、$$C$$、$$D$$满足$$\\textbar\\overrightarrow{AB}\\textbar=3$$，$$\\textbar\\overrightarrow{BC}\\textbar=7$$，$$\\textbar\\overrightarrow{CD}\\textbar=11$$，$$\\textbar\\overrightarrow{DA}\\textbar=9$$，则$$\\overrightarrow{AC}\\cdot \\overrightarrow{BD}$$的取值（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "只有一个 "}], [{"aoVal": "B", "content": "有二个 "}], [{"aoVal": "C", "content": "有四个 "}], [{"aoVal": "D", "content": "有无穷多个 "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间向量"], "answer_analysis": ["注意到$${{3}^{2}}+{{11}^{2}}=1130={{7}^{2}}+{{9}^{2}}$$，由于$$\\overrightarrow{AB}+\\overrightarrow{BC}+\\overrightarrow{CD}+\\overrightarrow{DA}=\\vec{0}$$， 则$$D{{A}^{2}}={{\\overrightarrow{DA}}^{2}}={{(\\overrightarrow{AB}+\\overrightarrow{BC}+\\overrightarrow{CD})}^{2}}=A{{B}^{2}}+B{{C}^{2}}+C{{D}^{2}}+2(\\overrightarrow{AB}\\cdot \\overrightarrow{BC}+\\overrightarrow{BC}\\cdot \\overrightarrow{CD}+\\overrightarrow{CD}\\cdot \\overrightarrow{AB})$$ $$=A{{B}^{2}}-B{{C}^{2}}+C{{D}^{2}}+2({{\\overline{BC}}^{2}}+\\overrightarrow{AB}\\cdot \\overrightarrow{BC}+\\overrightarrow{BC}\\cdot \\overrightarrow{CD}+\\overrightarrow{CD}\\cdot \\overrightarrow{AB})$$ $$=A{{B}^{2}}-B{{C}^{2}}+C{{D}^{2}}+2(\\overrightarrow{AB}+\\overrightarrow{BC})\\cdot (\\overrightarrow{BC}+\\overrightarrow{CD})$$， 即$$2\\overrightarrow{AC}\\cdot \\overrightarrow{BD}=A{{D}^{2}}+B{{C}^{2}}-A{{B}^{2}}-C{{D}^{2}}=0$$，∴$$\\overrightarrow{AC}\\cdot \\overrightarrow{BD}$$只有一个值得$$0$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "687", "queId": "7fc540273d734edc9ed8098991f6d17b", "competition_source_list": ["2016~2017学年12月北京顺义区顺义区第一中学高一上学期月考第7题5分", "1984年全国高中数学联赛竞赛一试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "方程$$\\sin x=\\lg x$$的根的个数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质", "竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["方程$$\\sin x=\\lg x$$的实根个数即函数$$y=\\sin x$$与函数$$y=\\lg x$$的交点的个数， 由函数$$y=\\sin x$$与函数$$y=\\lg x$$的图像可得图像交点为$$3$$个， 即方程$$\\sin x=\\lg x$$的实根的个数是$$3$$．~~~~~~~ "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "286", "queId": "8b152272e2f745bd85753799c562bbf4", "competition_source_list": ["2010年山东全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知整数集合$$M= {m\\textbar{{x}^{2}}+mx-36=0$$有整数解$$ }$$，集合$$A$$满足条件：①$$\\varnothing \\subset A\\subseteq M$$；②若$$a\\in A$$，则$$-a\\in A$$．那么所有这样的集合$$A$$的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$31$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["设$$\\alpha ,\\beta $$为方程$${{x}^{2}}+mx-36=0$$的两根，则$$\\alpha \\cdot \\beta =-36$$．因此， $$\\left\\textbar{} \\alpha \\right\\textbar=1,\\left\\textbar{} \\beta \\right\\textbar=36$$，有$$m=\\pm 35$$； $$\\left\\textbar{} \\alpha \\right\\textbar=2,\\left\\textbar{} \\beta \\right\\textbar=18$$，有$$m=\\pm 16$$； $$\\left\\textbar{} \\alpha \\right\\textbar=3,\\left\\textbar{} \\beta \\right\\textbar=12$$，有$$m=\\pm 9$$； $$\\left\\textbar{} \\alpha \\right\\textbar=4,\\left\\textbar{} \\beta \\right\\textbar=9$$，有$$m=\\pm 5$$； $$\\left\\textbar{} \\alpha \\right\\textbar=6,\\left\\textbar{} \\beta \\right\\textbar=6$$，有$$m=0$$． $$M=\\left { 0 \\right }\\cup \\left { -5, 5 \\right }\\cup \\left { -9, 9 \\right }\\cup \\left { -16, 16 \\right }\\cup \\left { -35, 35 \\right }$$． 由条件①知$$A\\ne \\varnothing $$．由条件②知$$A$$是由一些成对的相反数所成之集，所以$$M$$的$$5$$对相反数共能组成$${{2}^{5}}-1$$个不同的非空集合$$A$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "102", "queId": "139e95baef9040ab8ccde7ee12c6ad8b", "competition_source_list": ["2011年山东全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f(x)={{\\log }_{0.3}}({{x}^{2}}+x-2)$$的单调递增区间是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\infty ,-2)$$ "}], [{"aoVal": "B", "content": "$$(-\\infty ,1)$$ "}], [{"aoVal": "C", "content": "$$(-2,1)$$ "}], [{"aoVal": "D", "content": "$$(1, +\\infty )$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数", "竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["由对数函数的性质知，$${{x}^{2}}+x-2\\textgreater0$$，则$$x\\textgreater1$$或$$x\\textless{}-2$$．当$$x\\textless{}-2$$时，$$f(x)$$为增函数；当$$x\\textgreater1$$时，$$f(x)$$为减函数． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "703", "queId": "6dc5f0a3abe64ff98a6f627895087309", "competition_source_list": ["2022~2023学年湖南永州宁远县高一上学期月考（明德湘南中学基础知识竞赛）第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$y=f(x)$是定义在$\\text{R}$上的奇函数，当$x\\textgreater0$时，$f(x)=x-2$，那么不等式$f(x)\\textless{} \\frac{1}{2}$的解集是（~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$ {x\\textbar0\\textless{} x\\textless{} \\frac{5}{2} , }$ "}], [{"aoVal": "B", "content": "$ {x\\textbar-\\frac{3}{2}\\textless{} x\\le 0 , }$ "}], [{"aoVal": "C", "content": "$ {x\\textbar-\\frac{3}{2}\\textless{} x\\textless{} 0或0\\le x\\textless{} \\frac{5}{2} , }$ "}], [{"aoVal": "D", "content": "$ {x\\textbar x\\textless{} -\\frac{3}{2}或0\\le x\\textless{} \\frac{5}{2} , }$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据$y=f(x)$是定义在上$\\text{R}$的奇函数，求得函数解析式，然后再由分段函数的定义域，分$x\\textless{} 0$，$x\\textgreater0,x=0$三种情况讨论求解不等式即可.\\\\ 【详解】\\\\ 因为$y=f(x)$是定义在$\\text{R}$上的奇函数，所以$f(0)=0$，\\\\ 当$x\\textless{} 0$时，则$-x\\textgreater0$，所以$f(-x)=-x-2=-f\\left( x \\right)$，所以$f(x)=x+2$，\\\\ 综上所述，$f(x)=\\text{ } ! ! { ! !\\text{ }\\begin{array}{*{35}{l}} x+2,x\\textless{} 0 0,x=0 x-2,x\\textgreater0 \\end{array}\\text{ }$，\\\\ 不等式$f(x)\\textless{} \\frac{1}{2}$等价于$\\text{ } ! ! { ! !\\text{ }\\begin{array}{*{35}{l}} x\\textless{} 0 x+2\\textless{} \\frac{1}{2} \\end{array}\\text{ }$或$\\text{ } ! ! { ! !\\text{ }\\begin{array}{*{35}{l}} x=0 0\\textless{} \\frac{1}{2} \\end{array}\\text{ }$或$\\text{ } ! ! { ! !\\text{ }\\begin{array}{*{35}{l}} x\\textgreater0 x-2\\textless{} \\frac{1}{2} \\end{array}\\text{ }$，\\\\ 解得$x\\textless{} -\\frac{3}{2}$或$x=0$或$0\\textless{} x\\textless{} \\frac{5}{2}$，\\\\ 综上：不等式$f(x)\\textless{} \\frac{1}{2}$的解集是$\\text{ } ! ! { ! !\\text{ }x\\textbar\\text{ } ! !\\textbar ! !\\text{ }x\\textless{} -\\frac{3}{2}\\text{ }$或$\\text{ }0\\le x\\textless{} \\frac{5}{2}\\text{ } ! ! } ! !\\text{ }$.\\\\ 故选：D. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1201", "queId": "fe3e7d4cfe3d418d94d27bca3755e7ce", "competition_source_list": ["1989年全国高中数学联赛竞赛一试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "对任意的函数$$y=f\\left( x \\right)$$，在同一个直角坐标系中，函数$$y=f\\left( x-1 \\right)$$与函数$$y=f\\left( -x+1 \\right)$$的图象恒．", "answer_option_list": [[{"aoVal": "A", "content": "关于$$x$$轴对称； "}], [{"aoVal": "B", "content": "关于直线$$x=1$$对称 "}], [{"aoVal": "C", "content": "关于直线$$x=-1$$对称 "}], [{"aoVal": "D", "content": "关于$$y$$轴对称 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["$$f(x)$$和$$f(-x)$$的图象关于直线$$x=0$$对称， $$f(x-1)$$与$$f(-x+1)$$的图象关于直线$$x=1$$对称． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1000", "queId": "b2de3e4f81c4412ab8ec703432c6a13d", "competition_source_list": ["竞赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "若兩個鋭角的度數比是$$5$$ ：$$4$$，其中一個角的餘角是另一個角的餘角之$$2$$倍，則此兩角的度數和為何？ The ratio of the neasures of two acute angles is $$5:4$$, and the complement of one of these two angles is twice as large as the complement of the other. What is the sun of the degree measures of the two angles?", "answer_option_list": [[{"aoVal": "A", "content": "$$75$$ "}], [{"aoVal": "B", "content": "$$90$$ "}], [{"aoVal": "C", "content": "$$135$$ "}], [{"aoVal": "D", "content": "$$150$$ "}], [{"aoVal": "E", "content": "$$270$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["We can set up a system of equations where $$x$$ and $$y$$ are the two acute angles. WLOG, assume that $$x\\textless y$$ in order for the complement of $$x$$ to be greater than the complement of $$y$$. Therefore, $$5x=4y$$ and $$\\begin{array}{l} {90-x=2\\left( 90-y\\right)} {90-x=2\\left( 90-1.25x\\right)} {1.5x=90} {x=60} \\end{array}$$ Solving for $$y$$ in the first equation and substituting into the second equation yields Substituting this $$x$$ value back into the first equation yields $$y=75$$, leaving $$x+y$$ equal to $$\\left(\\text{C}\\right) 135$$. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "692", "queId": "60218af5240541e68240f1655a09248a", "competition_source_list": ["1984年全国高中数学联赛竞赛一试第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$a\\textgreater0,a\\ne 1,F\\left( x \\right)$$是一奇数，则$$G\\left( x \\right)=F\\left( x \\right)\\cdot \\left( \\frac{1}{{{a}^{x}}-1}+\\frac{1}{2} \\right)$$是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "奇函数 "}], [{"aoVal": "B", "content": "偶函数 "}], [{"aoVal": "C", "content": "不是奇函数也不是偶函数 "}], [{"aoVal": "D", "content": "奇偶性与$$a$$的具体数值有关 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数综合"], "answer_analysis": ["∵$$F\\left( x \\right)$$是奇函数，$$F\\left( -x \\right)=-F\\left( x \\right)$$ 令 $$g\\left( x \\right)=\\frac{1}{{{a}^{x}}-1}+\\frac{1}{2}=\\frac{{{a}^{x}}+1}{2\\left( {{a}^{x}}-1 \\right)}$$， 则 $$g\\left( -x \\right)=\\frac{{{a}^{-x}}+1}{2\\left( {{a}^{-x}}-1 \\right)}=\\frac{\\frac{1}{{{a}^{x}}}+1}{2\\left( \\frac{1}{{{a}^{x}}}+1 \\right)}$$， $$=\\frac{{{a}^{x}}+1}{2(1-{{a}^{x}})}=-g\\left( x \\right)$$， 即$$g\\left( x \\right)$$也是奇函数． 当$$a\\textgreater0$$，$$a\\ne 1$$时，$$G\\left( x \\right)=F\\left( x \\right)g\\left( x \\right)$$在$$R$$上有意义． 而 $$G\\left( -x \\right)=F\\left( -x \\right)g\\left( -x \\right)=\\left[ -F\\left( -x \\right) \\right]\\left[ -g\\left( -x \\right) \\right]$$ $$=F\\left( x \\right)g\\left( x \\right)$$， ∴$$G\\left( x \\right)$$是偶函数． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "967", "queId": "9bfb9658a7de421fb3230f6a7785c7f3", "competition_source_list": ["2008年河北全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\cos x+\\cos y=1$$，则$$\\sin x-\\sin y$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$[-1,1]$$ "}], [{"aoVal": "B", "content": "$$[-2,2]$$ "}], [{"aoVal": "C", "content": "$$[0,\\sqrt{3}]$$ "}], [{"aoVal": "D", "content": "$$[-\\sqrt{3},\\sqrt{3}]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->换元技巧->代数换元", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["设$$\\sin x-\\sin y=t$$，易得 $$\\cos x\\cos y-\\sin x\\sin y=\\frac{{{t}^{2}}-1}{2}$$， 即$$\\cos (x+y)=\\frac{{{t}^{2}}-1}{2}$$． 由于$$-1\\leqslant \\cos (x+y)\\leqslant 1$$， 所以$$-1\\leqslant \\frac{{{t}^{2}}-1}{2}\\leqslant 1$$， 解得$$-\\sqrt{3}\\leqslant t\\leqslant \\sqrt{3}$$．故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "277", "queId": "300f6a1040344d4bbdbc4471462990fa", "competition_source_list": ["2019年吉林全国高中数学联赛竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$x$$、$$y$$满足$$\\textbar y\\textbar{} \\leqslant2-x$$ ~，且 $$x\\geqslant -1$$ ，则 $$2x+y$$~ 的最小值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-7$$ "}], [{"aoVal": "B", "content": "$$-5$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->线性规划"], "answer_analysis": ["略 "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "888", "queId": "e45c7fe317c7408ba87cae4978c8c730", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$f(x)=2{{x}^{2}}+3px+2q$$和$$\\varphi (x)=x+\\frac{4}{x}$$是定义在集合$$M=\\left { x\\left\\textbar{} 1\\leqslant x\\leqslant \\frac{9}{4} \\right. \\right }$$上的函数，对任意的$$x\\in M$$，存在常数$${{x}_{0}}\\in M$$，使得$$f(x)\\geqslant f({{x}_{0}}),\\varphi (x)\\geqslant \\varphi ({{x}_{0}})$$，且$$f({{x}_{0}})=\\varphi ({{x}_{0}})$$，则函数$$f(x)$$在$$M$$上的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$\\frac{33}{8}$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->函数的概念与性质->函数的性质->单调性", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件", "课内体系->知识点->等式与不等式->不等式->解不等式->不等式中的恒成立与能成立问题", "课内体系->方法->定义法"], "answer_analysis": ["易知$$x=2$$时，$$\\varphi \\left( x \\right)$$取得最小值$$4$$，所以$${{x}_{0}}=2$$． 在$$\\left[ 1,\\frac{9}{4} \\right]$$上，$$f\\left( x \\right)$$在$$x=2$$取得最小值，所以$$-\\frac{3}{4}p=2$$，解得$$p=-\\frac{8}{3}$$． 又$$f\\left( 2 \\right)=\\varphi \\left( 2 \\right)=4$$，$$8+6p+2q=4$$，解得$$q=6$$． $$f\\left( x \\right)=2{{x}^{2}}-8x+12=2{{\\left( x-2 \\right)}^{2}}+4$$，在$$M$$上最大值为$$f\\left( 1 \\right)=6$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "339", "queId": "1b594e6302f5494c915b18c219aba25e", "competition_source_list": ["1992年全国高中数学联赛竞赛一试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "在$${\\triangle }ABC$$中，角$$A$$，$$B$$，$$C$$的对边分别记为$$a$$，$$b$$，$$c\\left( b\\ne 1 \\right)$$，且$$\\frac{C}{A}$$，$$\\frac{\\sin B}{\\sin A}$$都是$${{\\log }_{\\sqrt{b}}}x={{\\log }_{{b}}}\\left( 4x-4 \\right)$$的根，则$${\\triangle }ABC$$．", "answer_option_list": [[{"aoVal": "A", "content": "是等腰三角形，但不是直角三角形 "}], [{"aoVal": "B", "content": "是直角三角形，但不是等腰三角形 "}], [{"aoVal": "C", "content": "是等腰直角三角形 "}], [{"aoVal": "D", "content": "不是等腰三角表，也不是直角三角形 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数", "竞赛->知识点->三角函数->三角形中的问题->解三角形"], "answer_analysis": ["由已知方程得 $${{\\log }_{b}}{{x}^{2}}={{\\log }_{b}}\\left( 4x-4 \\right)$$， 即 $${{x}^{2}}-4x+4=0$$， 求得的根 $${{x}_{1}}={{x}_{2}}=2$$． 故 $$C=2A$$ 以及 $$\\sin B=2\\sin A$$， 因 $$A+B+C=180{}^{}\\circ $$，所以$$3A+B={{180}^{\\circ }}$$， 因此 $$\\sin B=\\sin 3A$$， 所以 $$2\\sin A-4{{\\sin }^{3}}A=2\\sin A$$， 即 $$\\sin A\\left( 1-4{{\\sin }^{2}}A \\right)=0$$，但 $$\\sin A{\\ne }0$$， 所以 $${{\\sin }^{2}}A=\\frac{1}{4}$$， 而 $$\\sin A\\textgreater0$$， 所以 $$\\sin A=\\frac{1}{2}$$， 从而 $$A=30{}^{}\\circ $$，$$C=60{}^{}\\circ $$，$$B=90{}^{}\\circ $$． 因此答案是（$$\\text{B}$$）． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "840", "queId": "65ef759db2d94b70ae84d1cd5fe88a2f", "competition_source_list": ["第二十届全国希望杯高二竞赛初赛邀请赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$x$$表示三角形一个内角的大小，并且$$\\sin x+\\cos x={{\\sin }^{3}}x+{{\\cos }^{3}}x$$，则该三��形是（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "直角三角形或钝角三角形 "}], [{"aoVal": "B", "content": "直角三角形或锐角三角形 "}], [{"aoVal": "C", "content": "钝角三角形 "}], [{"aoVal": "D", "content": "直角三角形 "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->三角函数->三角恒等变换->利用三角恒等变换判断三角形形状"], "answer_analysis": ["$$\\sin x+\\cos x={{\\sin }^{3}}x+{{\\cos }^{3}}x\\Leftrightarrow \\sin x(1-{{\\sin }^{2}}x)=\\cos x({{\\cos }^{2}}x-1)$$$$\\Leftrightarrow \\sin x{{\\cos }^{2}}x=-\\cos x{{\\sin }^{2}}x\\Leftrightarrow \\sin x\\cos x(\\cos x+\\sin x)=0$$． ∵$$x$$表示三角形一个内角，则$$x\\in (0, \\pi )$$． ∴$$x$$为直角或$$\\cos x+\\sin x=0$$． 当$$\\cos x+\\sin x=0$$时，则$$\\cos x=-\\sin x ~\\textless{} ~0$$，（$$x\\in (0, \\pi )$$） ， ∴$$x$$为钝角，此时$$x=\\frac{3}{4} \\pi $$． 故三角形为直角三角形或钝角三角形． 故选$$\\rm A$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "816", "queId": "ad2457af62fd4a1c80e60618f29b77e2", "competition_source_list": ["1983年全国高中数学联赛竞赛一试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$x=\\frac{1}{{{\\log }_{\\frac{1}{2}}}\\frac{1}{3}}+\\frac{1}{{{\\log }_{\\frac{1}{5}}}\\frac{1}{3}}$$的值是属于区间(~ ~ )．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-2,-1)$$ "}], [{"aoVal": "B", "content": "$$(1,2)$$ "}], [{"aoVal": "C", "content": "$$(-3,-2)$$ "}], [{"aoVal": "D", "content": "$$(2,3)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["$$x=\\frac{1}{{{\\log }_{\\frac{1}{2}}}\\frac{1}{3}}+\\frac{1}{{{\\log }_{\\frac{1}{5}}}\\frac{1}{3}}=\\frac{1}{{{\\log }_{{{2}^{-1}}}}{{3}^{-1}}}+\\frac{1}{{{\\log }_{{{5}^{-1}}}}{{3}^{-1}}}$$ $$=\\frac{1}{{{\\log }_{2}}3}+\\frac{1}{{{\\log }_{5}}3}={{\\log }_{3}}2+{{\\log }_{3}}5={{\\log }_{3}}10$$， ∵$${{3}^{2}} ~\\textless{} ~10 ~\\textless{} ~{{3}^{3}}$$， ∴$$x\\in \\left( 2,3 \\right)$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "944", "queId": "f26e644958fe4b309aa036aaff2333bc", "competition_source_list": ["2019~2020学年4月山西太原迎泽区太原市第五中学校高二下学期周测C卷理科第6题6分", "2012年黑龙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设集合$$A=\\left { 1, 2, 3, 4, 5, 6 \\right }$$，$$B=\\left { 4, 5, 6, 7, 8$$ \\right }$$，则满足$$S\\subseteq A$$且$$S\\cap B\\ne \\varnothing $$的集合$$S$$的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$57$$ "}], [{"aoVal": "B", "content": "$$56$$ "}], [{"aoVal": "C", "content": "$$49$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->两个基本计数原理->分步乘法计数原理", "竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["$$S\\subseteq A$$，且$$S\\cap B\\ne \\varnothing $$，说明$$S$$是$$A$$的子集，且$$S$$与$$B$$有公共元素； ∴$$A$$的构成情况为：①含一个元素：从$$4$$，$$5$$，$$6$$中选一个元素，个数为$$\\text{C}_{3}^{1}=3$$； ②含两个元素:从$$4$$，$$5$$，$$6$$选两个元素，或从$$1$$，$$2$$，$$3$$选一个，从$$4$$，$$56$$选一个，个数为：$$\\text{C}_{3}^{2}+\\text{C}_{3}^{1}\\text{C}_{3}^{1}=12$$；③含三个元素：从$$4$$，$$5$$，$$6$$选三个，或从$$4$$，$$5$$，$$6$$选两个，从$$1$$，$$2$$，$$3$$选一个或从$$4$$，$$5$$，$$6$$选一个，从$$1$$，$$2$$，$$3$$选两个，个数为：$$\\text{C}_{3}^{3}+\\text{C}_{3}^{2}\\text{C}_{3}^{1}+\\text{C}_{3}^{1}\\text{C}_{3}^{2}=19$$； ④含四个元素：从$$4$$，$$5$$，$$6$$选三个从$$1$$，$$2$$，$$3$$选一个，或从$$4$$，$$5$$，$$6$$选两个，或从$$4$$，$$5$$，$$6$$选一个，从$$1$$，$$2$$，$$3$$选三个个数为：$$\\text{C}_{3}^{3}\\text{C}_{3}^{1}+\\text{C}_{3}^{2}\\text{C}_{3}^{2}+\\text{C}_{3}^{1}\\text{C}_{3}^{3}=15$$； ⑤含五个元素：从$$4$$，$$5$$，$$6$$选三个，从$$1$$，$$2$$，$$3$$选两个，或从$$4$$，$$5$$，$$6$$选两个，从$$1$$，$$2$$，$$3$$选三个，个数为：$$\\text{C}_{3}^{3}\\text{C}_{3}^{2}+\\text{C}_{3}^{2}\\text{C}_{3}^{3}=6$$含\"6\"个元素：从$$4$$，$$5$$，$$6$$选三个，从$$1$$，$$2$$，$$3$$选三个，个数为$$\\text{C}_{3}^{3}\\text{C}_{3}^{3}=1$$； ∴集合$$S$$的个数为：$$2+12+19+15+6+1=56$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "454", "queId": "87cdde50dae44467b1ac359b5d5cbebb", "competition_source_list": ["2017年四川全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$${{F}_{1}},{{F}_{2}}$$为椭圆$$\\frac{{{x}^{2}}}{{{a}^{2}}}+\\frac{{{y}^{2}}}{{{b}^{2}}}=1\\left( a\\textgreater b\\textgreater0 \\right)$$的左右焦点，该椭圆上存在两点$$A,B$$，使得$$\\overrightarrow{{{F}_{1}}A}=3\\overrightarrow{{{F}_{2}}B}$$，则该椭圆的离心率的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 0,\\frac{1}{2} \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 0,\\frac{1}{3} \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{2},1 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{1}{3},1 \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->椭圆"], "answer_analysis": ["因为$$\\left\\textbar{} \\overrightarrow{{{F}_{1}}A} \\right\\textbar=a+e{{x}_{A}}\\leqslant a+e\\cdot a=a+c$$，$$\\left\\textbar{} \\overrightarrow{{{F}_{2}}B} \\right\\textbar=a-e{{x}_{B}}\\geqslant a-e\\cdot a=a-c$$， 所以$$a+c\\geqslant 3\\left( a-c \\right)$$，取等号时$$AB$$重合，不成立，因此$$a+c\\textgreater3\\left( a-c \\right)$$， 化简得$$\\frac{c}{a}\\textgreater\\frac{1}{2}$$ "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "118", "queId": "61db9ac847ab48b2909b6c924d6d8d62", "competition_source_list": ["2003年全国全国高中数学联赛竞赛一试第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$x\\in \\left[ -\\frac{5 \\pi }{12},-\\frac{ \\pi }{3} \\right]$$，则$$y=\\tan \\left( x+\\frac{2 \\pi }{3} \\right)-\\tan \\left( x+\\frac{ \\pi }{6} \\right)+\\cos \\left( x+\\frac{ \\pi }{6} \\right)$$的最大值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{12\\sqrt{2}}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{11\\sqrt{2}}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{11\\sqrt{3}}{6}$$ "}], [{"aoVal": "D", "content": "$$\\frac{12\\sqrt{3}}{5}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["$$y=\\tan \\left( x+\\frac{2 \\pi }{3} \\right)+\\cot \\left( x+\\frac{2 \\pi }{3} \\right)+\\cos \\left( x+\\frac{ \\pi }{6} \\right)=\\frac{1}{\\cos \\left( x+\\frac{2 \\pi }{3} \\right)\\sin \\left( x+\\frac{2 \\pi }{3} \\right)}+\\cos \\left( x+\\frac{ \\pi }{6} \\right)$$ $$=\\frac{2}{\\sin \\left( 2x+\\frac{4 \\pi }{3} \\right)}+\\cos \\left( x+\\frac{ \\pi }{6} \\right)$$ 因为$$x\\in \\left[ -\\frac{5 \\pi }{12},-\\frac{ \\pi }{3} \\right]$$，∴$$2x+\\frac{4 \\pi }{3}\\in \\left[ \\frac{ \\pi }{2},\\frac{2 \\pi }{3} \\right]$$，$$x+\\frac{ \\pi }{6}\\in \\left[ -\\frac{ \\pi }{4},-\\frac{ \\pi }{6} \\right]$$， 因此$$\\frac{2}{\\sin \\left( 2x+\\frac{4 \\pi }{3} \\right)}$$与$$\\cos \\left( x+\\frac{ \\pi }{6} \\right)$$在$$\\left[ -\\frac{5 \\pi }{12},-\\frac{ \\pi }{3} \\right]$$上同为增函数， 故当$$x=-\\frac{ \\pi }{3}$$时，$$y$$取最大值$$\\frac{11\\sqrt{3}}{6}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1160", "queId": "e674d0387e4644209924b497bf33c13a", "competition_source_list": ["2014年辽宁全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$${\\triangle }ABC$$的三边$$a$$，$$b$$，$$c$$成等比数列，$$a$$，$$b$$，$$c$$的对角依次为$$\\angle A$$，$$\\angle B$$，$$\\angle C$$．则$$\\sin B+\\cos B$$的范围是( )．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ \\frac{1}{2} , 1+\\frac{\\sqrt{3}}{2} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left( 1 , \\sqrt{2} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left( 1 , 1+\\frac{\\sqrt{3}}{2} \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left[ \\frac{1}{2} , \\sqrt{2} \\right]$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角恒等变换->倍角、和差角公式综合", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->辅助角公式", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦函数的图象和性质", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦型函数的图象与性质", "课内体系->知识点->数列->等比数列->等比数列的性质及应用", "课内体系->知识点->解三角形->正弦定理", "课内体系->素养->数学运算"], "answer_analysis": ["因为$$a$$，$$b$$，$$c$$成等比数列，所以$${{b}^{2}}=ac$$． 则$${{b}^{2}}=ac={{a}^{2}}+{{c}^{2}}-2ac\\cos B{\\geqslant }2ac-2ac\\cos B$$， $$\\cos B{\\geqslant }\\frac{1}{2}$$． 于是，$$0 \\textless{} B{\\leqslant }60{}^{}\\circ $$． 由$$\\frac{1}{2}{\\leqslant }\\cos B \\textless{} 1$$及$$0 \\textless{} \\sin B{\\leqslant }\\frac{\\sqrt{3}}{2}$$，得 $$\\frac{1}{2} \\textless{} \\cos B+\\sin B \\textless{} \\frac{\\sqrt{3}}{2}+1$$． ① 另一方面$$\\sin B+\\cos B=\\sqrt{2}\\sin (B+45{}^{}\\circ )$$． 而$$45{}^{}\\circ \\textless{} \\angle B+45{}^{}\\circ {\\leqslant }105{}^{}\\circ $$． 故$$1 \\textless{} \\sin B+\\cos B{\\leqslant }\\sqrt{2}$$． ② 综合①、②，有$$1 \\textless{} \\sin B+\\cos B{\\leqslant }\\sqrt{2}$$． 故选$$\\text{B}$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "506", "queId": "5a4069179ee1462d84bd72d368987db2", "competition_source_list": ["2021年吉林全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "方程组$$\\begin{cases}\\left( x+y-1 \\right)\\sqrt{x-1}=0 {{x}^{2}}+{{y}^{2}}+2x-4=0 \\end{cases}$$的解的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->解方程（组）"], "answer_analysis": ["由题设得：$$\\begin{cases}x-1=0 {{x}^{2}}+{{y}^{2}}+2x-4=0 \\end{cases}$$或$$\\begin{cases}x-1\\textgreater0 x+y-1=0 {{x}^{2}}+{{y}^{2}}+2x-4=0 \\end{cases}$$， 有三组解$$\\left( 1,1 \\right)$$，$$\\left( 1,-1 \\right)$$，$$\\left( \\frac{\\sqrt{6}}{2},1-\\frac{\\sqrt{6}}{2} \\right)$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1193", "queId": "f4f1a09d4d5e4322a8a2538145053644", "competition_source_list": ["2012年浙江全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$\\text{i}$$为虚数单位，则复数$$\\frac{1+2\\text{i}}{\\text{i}-2}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\text{i}$$ "}], [{"aoVal": "B", "content": "$$-\\text{i}$$ "}], [{"aoVal": "C", "content": "$$-\\frac{4}{5}-\\frac{3}{5}\\text{i}$$ "}], [{"aoVal": "D", "content": "$$-\\frac{4}{5}+\\frac{3}{5}\\text{i}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->复数->复数的运算->复数的四则运算综合", "课内体系->知识点->复数->复数的运算->复数的乘法和除法", "课内体系->知识点->复数->复数的概念及几何意义->复数的基本概念", "课内体系->知识点->复数->复数的概念及几何意义->复数的代数表示法"], "answer_analysis": ["$$\\frac{1+2\\text{i}}{\\text{i}-2}=\\frac{\\left( 1+2\\text{i} \\right)\\left( \\text{i}+2 \\right)}{\\left( \\text{i}-2 \\right)\\left( \\text{i}+2 \\right)}=-\\text{i}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "441", "queId": "755b9978ecc341efa7bbc97bfcfd8601", "competition_source_list": ["2016年天津河北区高三一模理科第13题5分", "2017~2018学年浙江杭州西湖区杭州学军中学高二上学期期中理科第14题4分", "2016年天津河北区高三二模文科第14题5分", "2015年甘肃全国高中数学联赛竞赛初赛第9题7分", "2016年天津河北区高三二模理科第14题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$x$$，$$y$$是正实数，且$$x+y=1$$，则$$\\frac{{{x}^{2}}}{x+2}+\\frac{{{y}^{2}}}{y+1}$$的最小值是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{15}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{3}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的概念", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件"], "answer_analysis": ["设$$x+2=s$$，$$y+1=t$$，则$$s+t=x+y+3=4$$， 所以$$\\frac{{{x}^{2}}}{x+2}+\\frac{{{y}^{2}}}{y+1}$$ $$=\\frac{{{(s-2)}^{2}}}{s}+\\frac{{{(t-1)}^{2}}}{t}$$ $$=\\left( s-4+\\frac{4}{s} \\right)+\\left( t-2+\\frac{1}{t} \\right)$$ $$=(s+t)+\\left( \\frac{4}{s}+\\frac{1}{t} \\right)-6$$ $$=\\left( \\frac{4}{s}+\\frac{1}{t} \\right)-2$$． 因为$$\\frac{4}{s}+\\frac{1}{t}=\\frac{1}{4}\\left( \\frac{4}{s}+\\frac{1}{t} \\right)(s+t)$$ $$=\\frac{1}{4}\\left( \\frac{4t}{s}+\\frac{s}{t}+5 \\right)\\geqslant \\frac{9}{4}$$， 当且仅当$\\left {\\begin{align}\\&\\frac{4t}{s}=\\frac{s}{t} \\&s+t=4\\end{align}\\right.$即$\\left {\\begin{align}\\&s=\\frac83 \\&t=\\frac43\\end{align}\\right.$也即$\\left {\\begin{align}\\&x=\\frac23 \\&y=\\frac13\\end{align}\\right.$时取等， 故当$x=\\frac23,y=\\frac13$时，$\\frac{{{x}^{2}}}{x+2}+\\frac{{{y}^{2}}}{y+1}$取得最小值$\\frac{1}{4}$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "263", "queId": "3468020679fc4b2296f617c01db9f801", "competition_source_list": ["2020~2021学年河北石家庄新华区石家庄市第二中学高二上学期期中（竞赛班）第11题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$\\triangle ABC$$的顶点$$B\\left( -3,0 \\right)$$和$$C(3,0)$$，顶点$$A$$在椭圆 $$\\frac{{{x}^{2}}}{16}+\\frac{{{y}^{2}}}{7}=1$$上，则$$\\frac{\\sin B+\\sin C}{\\sin A}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->圆锥曲线->椭圆->椭圆的定义、标准方程->椭圆的定义", "课内体系->知识点->解三角形->正弦定理->利用正弦定理求解边角->边角互化（利用正弦定理）", "课内体系->素养->数学抽象", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理"], "answer_analysis": ["由题设知：椭圆的长半轴长$$a=4$$，半焦距$$c=3$$， $$B$$、$$C$$为椭圆的左右焦点， ∵顶点$$A$$在椭圆上， ∴$$\\left\\textbar{} AB \\right\\textbar+\\left\\textbar{} AC \\right\\textbar=2a=8$$， $$\\textbar BC\\textbar=2c=6$$， 由正弦定理可得： $$\\frac{\\sin B+\\sin C}{\\sin A}=\\frac{\\left\\textbar{} AC \\right\\textbar+\\left\\textbar{} AB \\right\\textbar}{\\left\\textbar{} BC \\right\\textbar}=\\frac{8}{6}=\\frac{4}{3}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "554", "queId": "c3373a7adf1845309aa00c2a53305309", "competition_source_list": ["2015年天津全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "用$$[x]$$表示不大于$$x$$的最大整数，方程$${{x}^{2}}-[x]-2=0$$共有（ ~ ~）个不同的实根．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->取整函数->与[x]有关的方程和不等式"], "answer_analysis": ["$$x-1\\textless{}{{x}^{2}}-2=[x]\\leqslant x$$，解得$$\\frac{1+\\sqrt{5}}{2}\\textless{}x\\leqslant 2$$或者$$-1\\leqslant x\\textless{}\\frac{1-\\sqrt{5}}{2}$$，所以$$[x]=2$$、$$1$$或$$-1$$．代入原方程，解得$$x=2$$、$$\\sqrt{3}$$或$$-1$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "901", "queId": "a477013e530544f191f731b4c2dd5825", "competition_source_list": ["2014年浙江全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知集合$$P=\\left { 1,\\left\\textbar{} a \\right\\textbar{} \\right }$$，$$Q=\\left { 2, {{b}^{2}} \\right }$$为全集$$U=\\left { 1, 2, 3, {{a}^{2}}+{{b}^{2}}+a+b \\right }$$的子集，且$${{\\complement }_{U}}\\left { P\\cup Q \\right }=\\left { 6 \\right }$$，则下面结论正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a=3$$，$$b=1$$ "}], [{"aoVal": "B", "content": "$$a=3$$，$$b=-1$$ "}], [{"aoVal": "C", "content": "$$a=-3$$，$$b=1$$ "}], [{"aoVal": "D", "content": "$$a=-3$$，$$b=-1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["由$${{\\complement }_{U}}\\left { P\\cup Q \\right }=\\left { 6 \\right }\\Rightarrow {{a}^{2}}+{{b}^{2}}+a+b=6$$，显然只有答案$$D$$符合． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1068", "queId": "d7c1d0b750bc4419b42c6835754ee65e", "competition_source_list": ["2015年福建全国高中数学联赛竞赛初赛第8题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知函数$$f(x)={{\\text{e}}^{x}}(x-a{{\\text{e}}^{x}})$$恰有两个极值点$${{x}_{1}}$$， $${{x}_{2}}({{x}_{1}}\\textless{}{{x}_{2}})$$，则$$a$$的取值范围是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 0,\\frac{1}{2} \\right)\\cup\\left( \\frac{1}{2},1 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 0,1\\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{2},1 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( 0,\\frac{1}{2} \\right)$$ "}]], "knowledge_point_routes": ["知识标签->素养->数学运算", "知识标签->知识点->函数->函数的应用->函数的零点->函数零点的概念", "知识标签->知识点->导数->导数的应用->导数与极值", "知识标签->知识点->导数->导数的应用->导数与单调性", "知识标签->题型->导数->导数的应用->利用导数研究函数的极值问题->已知极值情况求参数的取值范围"], "answer_analysis": ["$${{f}^{\\prime }}(x)={{\\text{e}}^{x}}(x-a{{\\text{e}}^{x}})+{{\\text{e}}^{x}}(1-a{{\\text{e}}^{x}})={{\\text{e}}^{x}}[-2a{{\\text{e}}^{x}}+x+1]=0$$有两个相异的实根， 即$$2a=\\frac{x+1}{{{\\text{e}}^{x}}}=g(x)$$有两个解， $${{g}^{\\prime }}(x)=\\frac{-x}{{{\\text{e}}^{x}}}$$， $$g(x)$$在$$(-\\infty ,0)$$单调递增；在$$(0,+\\infty )$$单调递减， $$x\\textless{}0$$，$$g(x)\\textless{}1$$，$$x\\textgreater0$$时，$$0\\textless{}g(x)\\textless{}1$$， 因此$$2a\\in (0,1)$$，即$$a\\in \\left( 0,\\frac{1}{2} \\right)$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "530", "queId": "713b01badad04b808f26b11e1f654bdb", "competition_source_list": ["2013年浙江全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知等比数列$${a}_{n}{:}{{a}_{1}}=3$$，且第一项至第八项的几何平均数为$$9$$，则第三项是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3\\sqrt[9]{81}$$ "}], [{"aoVal": "B", "content": "$$3\\sqrt[7]{81}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt[3]{9}$$ "}], [{"aoVal": "D", "content": "$$3\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["计算得$$q={{3}^{\\frac{2}{7}}},{{a}_{3}}=3\\sqrt[7]{81}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "185", "queId": "affa3932fb284e6a887b98391a7ee667", "competition_source_list": ["2010年AMC10竞赛A第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "马文在闰年$$2008$$年的$$5$$月$$27$$日星期二过生日．他下一个星期六的生日是在哪一年？", "answer_option_list": [[{"aoVal": "A", "content": "$$2011$$ "}], [{"aoVal": "B", "content": "$$2012$$ "}], [{"aoVal": "C", "content": "$$2013$$ "}], [{"aoVal": "D", "content": "$$2015$$ "}], [{"aoVal": "E", "content": "$$2017$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Reasoning->Recurrence and Recursion", "课内体系->知识点->推理与证明->逻辑推理题"], "answer_analysis": ["$$\\boxed{\\rm(E)}2017$$. There are $$365$$ days in a non-leap year. There are $$7$$ days in a week. Since $$365=52\\cdot7+1$$(or $$365$$ is congruent to $$1\\textasciitilde\\rm mod\\textasciitilde7$$), the same date (after February) moves \"forward\" one day in the subsequent year, if that year is not a leap year. For example: $$5/27/08$$ Tue $$5/27/09$$ Wed However, a leap year has $$366$$ days, and $$366=52\\cdot7+2$$. So the same date (after February) moves \"forward\" two days in the subsequent year, if that year is a leap year. For example: $$5/27/11$$ Fri $$5/27/12$$ Sun You can keep count forward to find that the first time this date falls on a Saturday is in $$2017$$: $$5/27/13$$ Mon $$5/27/14$$ Tue $$5/27/15$$ Wed $$5/27/16$$ Fri $$5/27/17$$ Sat "], "answer_value": "E"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "685", "queId": "d16ae53f987b4d74956b21869d003b8b", "competition_source_list": ["2021~2022学年福建莆田城厢区莆田第一中学高一下学期开学考试（学科素养能力竞赛）第3~3题"], "difficulty": "0", "qtype": "single_choice", "problem": "方程$${{\\log }_{3}}x=x-4$$的一个实根所在的区间是", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 2,3 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 3,4 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( 5,6 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( 6,7 \\right)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的应用->函数的零点->求零点所在区间的问题"], "answer_analysis": ["\\hfill\\break 设$$f(x)={{\\log }_{3}}x-x+4$$，证明$$f(5)\\cdot f(6)\\textless{} 0$$即得解.\\\\ 【详解】\\\\ 因为$${{\\log }_{3}}x=x-4$$，所以$${{\\log }_{3}}x-x+4=0$$.\\\\ 设$$f(x)={{\\log }_{3}}x-x+4$$，\\\\ 所以$$f(5)={{\\log }_{3}}5-5+4={{\\log }_{3}}5-1\\textgreater0$$，\\\\ $$f(6)={{\\log }_{3}}6-6+4={{\\log }_{3}}6-2={{\\log }_{3}}\\frac{6}{9}\\text{=}{{\\log }_{3}}\\frac{2}{3}\\textless{} 0$$，\\\\ 所以$$f(5)\\cdot f(6)\\textless{} 0$$.\\\\ 故选：C\\\\ 【点睛】\\\\ 本题主要考查零点问题，考查零点区间的确定，意在考查学生对这些知识的理解掌握水平. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "525", "queId": "d100b705324649a59cc9ba345805c33b", "competition_source_list": ["2015年辽宁全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\triangle ABC$$的三边$$abc$$成等比数列，边$$abc$$所对的角依次为$$ABC$$，且$$\\sin A\\sin B+\\sin B\\sin C+\\cos 2B=1$$，则角$$B$$为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{ \\pi }{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{ \\pi }{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{ \\pi }{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2 \\pi }{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角形中的问题->解三角形", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["$$\\sin A\\sin B+\\sin B\\sin C=1-\\cos 2B=2{{\\sin }^{2}}B\\Rightarrow a+c=2b$$，又$$ac={{b}^{2}}$$， 则$${{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\\cos B={{\\left( a+c \\right)}^{2}}-2ac\\left( 1+\\cos B \\right)=4{{b}^{2}}-2{{b}^{2}}\\left( 1+\\cos B \\right)$$， 解得$$B=\\frac{ \\pi }{3}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "328", "queId": "7e3c9b52da2642238cbe6931526e3bd1", "competition_source_list": ["2008年福建全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知实数$$a$$使得只有一个实数$$x$$满足不等式$$\\left\\textbar{} {{x}^{2}}+2ax+3a \\right\\textbar\\leqslant 2$$，则满足条件的所有的实数$$a$$的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "无穷多 "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["令$$f(x)={{x}^{2}}+2ax+3a$$，由$$f\\left( -\\frac{3}{2} \\right)=\\frac{9}{4}$$知， 函数$$f(x)$$的图象经过点$$\\left( -\\frac{3}{2},\\frac{9}{4} \\right)$$． 欲使得不等式$$\\left\\textbar{} {{x}^{2}}+2ax+3a \\right\\textbar\\leqslant 2$$只有一个解， 则抛物线$$f(x)={{x}^{2}}+2ax+3a$$的图像必须与直线$$y=2$$相切， 即$${{x}^{2}}+2ax+3a-2=0$$的差别式$$\\Delta =4{{a}^{2}}-4(3a-2)=0$$． 解得$$a=1$$，$$2$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "264", "queId": "227742c468904e73a9a6ef24f08a4fd4", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "在平面直角坐标系中，已知$$\\triangle ABC$$三个顶点的坐标分别为$$A(2, 1), B(-1, -1), C(1, 3)$$，点$$P$$在直线$$BC$$上运动，动点$$Q$$满足，则点$$Q$$的轨迹方程为．", "answer_option_list": [[{"aoVal": "A", "content": "$$2x-y-3=0$$ "}], [{"aoVal": "B", "content": "$$x-2y-3=0$$ "}], [{"aoVal": "C", "content": "$$x+2y-4=0$$ "}], [{"aoVal": "D", "content": "$$2x+y-3=0$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与方程"], "answer_analysis": ["$${{k}_{BC}}=\\frac{3+1}{1+1}=2$$，可得$$BC:y+1=2\\left( x+1 \\right)$$，即$$2x-y+1=0$$， 设$$P\\left( m,2m+1 \\right)$$为直线$$BC$$上一点，$$Q\\left( x,y \\right)$$， 由$$\\overrightarrow{PQ}=\\overrightarrow{PA}+\\overrightarrow{PB}+\\overrightarrow{PC}$$，可得$$\\left( x-m,y-2m-1 \\right)$$ $$=\\left( 2-m,1-2m-2 \\right)+\\left( -1-m,-1-2m-1 \\right)+\\left( 1-m,3-2m-2 \\right)$$ $$=\\left( 2-3m,-6m \\right)$$， 所以有$$\\left( x-m,y-2m-1 \\right)=\\left( 2-3m,-6m \\right)$$， 即$$\\begin{cases}x-m=2-3m y-2m-1=-6m \\end{cases}$$，即$$\\begin{cases}2m=2-x 4m=1-y \\end{cases}$$， 所以$$1-y=2\\left( 2-x \\right)$$，即$$2x-y-3=0$$，故选$$\\text{A}$$. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "813", "queId": "65a32d7f06d64f6899cc11f7d9d72ea8", "competition_source_list": ["1995年全国高中数学联赛竞赛一试第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果甲的身高数或体重数至少有一项比乙大，则称甲不亚于乙．在$$100$$个小伙子中，如果某人不亚于其他$$99$$人，就称他为棒小伙子．那么，$$100$$个小伙子中的棒小伙子最多可能有（~~~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$个 "}], [{"aoVal": "B", "content": "$$2$$个 "}], [{"aoVal": "C", "content": "$$50$$个 "}], [{"aoVal": "D", "content": "$$100$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->逻辑推理"], "answer_analysis": ["取$$100$$个小伙子为这样一种特殊情况，他们的身高与体重互不相等，并且最高者同时也就是最轻者，次高者同时也就是次轻者，$$\\ldots $$，第$$k$$高者同时也就是第$$k$$轻者（$$k=1$$，$$2$$，$$\\ldots $$，$$100$$），显然这$$100$$个小伙子都是棒小伙子． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "866", "queId": "9fb9936fbdcf4382901dd1ca767fb4d4", "competition_source_list": ["2008年黑龙江全国高中数学联赛竞赛初赛第11题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$${{x}_{1}}$$是方程$$x+\\lg x=3$$的根，$${{x}_{2}}$$是方程$$x+{{10}^{x}}=3$$的根，那么$${{x}_{1}}+{{x}_{2}}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["由于$$\\lg x=3-x$$，$${{10}^{x}}=3-x$$，令$$y=3-x$$， 则$${{y}_{1}}=\\lg x$$，$${{y}_{2}}={{10}^{x}}$$互为反函数． 所以$${{y}_{1}}=\\lg x$$与$${{y}_{2}}={{10}^{x}}$$和直线$$y=3-x$$的交点$$A$$、$$B$$关于直线$$y=x$$对称， 由$$y=x$$与$$y=3-x$$联立解得线段$$AB$$的中点$$C\\left( \\frac{3}{2},\\frac{3}{2} \\right)$$， 因此$${{x}_{1}}+{{x}_{2}}=3$$．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "565", "queId": "7f1327ad43e2433388afaec06c19f518", "competition_source_list": ["2016年广东全国高中数学联赛高三竞赛初赛第7题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知锐角$$\\alpha $$，$$\\beta $$满足条件：$$\\frac{{{\\sin }^{4}}\\alpha }{{{\\cos }^{2}}\\beta }+\\frac{{{\\cos }^{4}}\\alpha }{{{\\sin }^{2}}\\beta }=1$$，则$$\\alpha +\\beta =$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{ \\pi }{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{ \\pi }{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{ 2\\pi }{3}$$ "}], [{"aoVal": "D", "content": "$${ \\pi }$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的实际应用", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件", "课内体系->知识点->等式与不等式->不等式->不等式的性质", "课内体系->知识点->等式与不等式->不等式->柯西不等式", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式的化简和求值", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->和差角公式化简求值综合运用", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的余弦", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理"], "answer_analysis": ["∵$$\\frac{{{\\sin }^{4}}\\alpha }{{{\\cos }^{2}}\\beta }+\\frac{{{\\cos }^{4}}\\alpha }{{{\\sin }^{2}}\\beta }=1$$， 即$$\\frac{{{\\sin }^{4}}\\alpha }{{{\\cos }^{2}}\\beta }+\\frac{{{\\cos }^{4}}\\alpha }{{{\\sin }^{2}}\\beta }-1={{\\left( \\frac{{{\\sin }^{2}}\\alpha }{\\cos \\beta }-\\cos \\beta \\right)}^{2}}+{{\\left( \\frac{{{\\cos }^{2}}\\alpha }{\\sin \\beta }-\\sin \\beta \\right)}^{2}}=0$$， ∴$$\\frac{{{\\sin }^{2}}\\alpha }{\\cos \\beta }-\\cos \\beta =0$$，$$\\frac{{{\\cos }^{2}}\\alpha }{\\sin \\beta }-\\sin \\beta =0$$， ∴$${{\\sin }^{2}}\\alpha ={{\\cos }^{2}}\\beta $$，$${{\\cos }^{2}}\\alpha ={{\\sin }^{2}}\\beta $$， 根据$$\\alpha $$、$$\\beta $$为锐角，可得$$\\alpha +\\beta =90{}^{}\\circ $$． 故答案为：$$90{}^{}\\circ $$． 由柯西不等式，$$\\left( {{\\cos }^{2}}\\beta +{{\\sin }^{2}}\\beta \\right)\\left( \\frac{{{\\sin }^{4}}\\alpha }{{{\\cos }^{2}}\\beta }+\\frac{{{\\cos }^{4}}\\alpha }{{{\\sin }^{2}}\\beta } \\right)\\geqslant {{\\left( {{\\sin }^{2}}\\alpha +{{\\cos }^{2}}\\alpha \\right)}^{2}}=1$$， 因此，$$\\frac{{{\\cos }^{4}}\\beta }{{{\\sin }^{4}}\\alpha }=\\frac{{{\\sin }^{4}}\\beta }{{{\\cos }^{4}}\\alpha }$$，即$$\\cos \\alpha \\cos \\beta =\\sin \\alpha \\sin \\beta $$，$$\\cos \\left( \\alpha +\\beta \\right)=0$$，故$$\\alpha +\\beta =\\frac{ \\pi }{2}$$． 故答案为：$$\\frac{ \\pi }{2}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "444", "queId": "55508b008823452b96a614f82e303501", "competition_source_list": ["2012年湖南全国高中数学联赛竞赛初赛第3题6分", "2011年北京清华大学自主招生夏令营第8题"], "difficulty": "3", "qtype": "single_choice", "problem": "数列$$\\left { {{a}_{n}} \\right }$$共有$$11$$项，$${{a}_{1}}=0$$，$${{a}_{11}}=4$$，$$\\left\\textbar{} {{a}_{k+1}}-{{a}_{k}} \\right\\textbar=1$$，$$k=1$$，$$2$$，$$\\cdots$$，$$10$$，满足这样的条件的不同数列的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$140$$ "}], [{"aoVal": "D", "content": "$$160$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["设$${{b}_{k+1}}={{a}_{k+1}}-{{a}_{k}}$$，则$${{b}_{k+1}}=\\pm 1$$，且$${{b}_{2}}+{{b}_{3}}+\\cdots+{{b}_{11}}=4$$，只需$${{b}_{2}}$$，$${{b}_{3}}$$，$$\\cdots$$，$${{b}_{11}}$$中有$$7$$个$$1$$和$$3$$个$$-1$$即可，所求的方法数为$$\\text{C}_{10}^{3}=120$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1210", "queId": "f55b90ec301048b587968b04a7b713d0", "competition_source_list": ["2011年高考真题湖北卷理科第1题", "2012年黑龙江全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\text{i}$$为虚数单位，则$${{\\left( \\frac{1+\\text{i}}{1-\\text{i}}\\right)}^{2011}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\text{i}$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$\\text{i}$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["知识标签->题型->复数->复数的运算->复数中的周期问题", "知识标签->素养->数学运算", "知识标签->知识点->复数->复数的运算->复数的乘法和除法"], "answer_analysis": ["因为$$\\frac{1+\\text{i}}{1-\\text{i}}=\\frac{{{\\left( 1+\\text{i} \\right)}^{2}}}{1-{{\\text{i}}^{2}}}=\\text{i}$$，所以$${{\\left(\\frac{1+\\text{i}}{1-\\text{i}} \\right)}^{2011}}={{\\text{i}}^{2011}}={{\\text{i}}^{4\\times502+3}}={{\\text{i}}^{3}}=-\\text{i}$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1001", "queId": "ee0dbdf123e04a30b7145fc376b73b75", "competition_source_list": ["全国高中数学联赛竞赛模拟一试（十八）第3题"], "difficulty": "3", "qtype": "single_choice", "problem": "已知非零实数$$x$$，$$y$$，$$z$$满足$$\\frac{x-y}{x+y}\\cdot\\frac{y-z}{y+z}\\cdot\\frac{z-x}{z+x}=\\frac{14}{15}$$．则$$\\frac{y}{x+y}+\\frac{z}{y+z}+\\frac{x}{z+x}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{53}{30}$$ "}], [{"aoVal": "B", "content": "$$\\frac{59}{30}$$ "}], [{"aoVal": "C", "content": "$$\\frac{61}{30}$$ "}], [{"aoVal": "D", "content": "前三个答案都不对 "}]], "knowledge_point_routes": ["课内体系->方法->换元法", "竞赛->知识点->不等式->换元技巧->代数换元"], "answer_analysis": ["令$$\\frac{y}{x+y}=m$$，$$\\frac{z}{y+z}=n$$，$$\\frac{x}{z+x}=l$$． 则$$\\frac{14}{15}=\\frac{x-y}{x+y}\\cdot\\frac{y-z}{y+z}\\cdot\\frac{z-x}{z+x}=(1-2m)(1-2n)(1-2l )$$． 即$$1-2(m+n+l)+4(mn+nl+lm)-8mnl=\\frac{14}{15}$$． 又$$\\left( {\\frac{1}{m}-1} \\right)\\left( {\\frac{1}{n}-1} \\right)\\left( {\\frac{1}{l}-1} \\right)=\\frac{x}{y}\\cdot\\frac{y}{z}\\cdot\\frac{z}{x}=1$$ $$\\Rightarrow 1-\\left( m+n+l \\right)+\\left( mn+nl+lm \\right)-mnl=mnl$$， 故$$1-2(m+n+l)-4+4(m+n+l)=\\frac{14}{15}$$ $$\\Rightarrow m+n+l=\\frac{59}{30}$$ $$\\Rightarrow \\frac{y}{x+y}+\\frac{z}{y+z}+\\frac{x}{z+x}=\\frac{59}{30}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "55", "queId": "cbb6442bfd65471282c9d8c197bc55ad", "competition_source_list": ["2008年浙江全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$f\\left( x \\right)={{x}^{2}}+\\left( {{a}^{2}}+{{b}^{2}}-1 \\right)x+{{a}^{2}}+2ab-{{b}^{2}}$$是偶函数，则函数图象与$$y$$轴交点的纵坐标的最大值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$2\\sqrt{2}$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->二次函数", "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["由已知条件可知$${{a}^{2}}+{{b}^{2}}-1=0$$， 函数图象与$$y$$轴交点的纵坐标为$${{a}^{2}}+2ab-{{b}^{2}}$$． 令$$a=\\cos \\theta $$，$$ b=\\sin \\theta $$， 则$${{a}^{2}}+2ab-{{b}^{2}}={{\\cos }^{2}}\\theta +2\\sin \\theta \\cos \\theta -{{\\sin }^{2}}\\theta $$ $$=\\cos 2\\theta +\\sin 2\\theta \\leqslant \\sqrt{2}$$， 故选$$\\text{A}$$. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "721", "queId": "4ebfcfcaf0af4c45b537fcfbeee9f9f8", "competition_source_list": ["2008年河南全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "抛物线$${{y}^{2}}=4x$$的焦点为$$F$$，点$$A$$、$$B$$在抛物线上，且$$\\angle AFB=\\frac{2}{3} \\pi $$，弦$$AB$$中点$$M$$在准线$$l$$上的射影为$${M}'$$，则$$\\frac{\\textbar M{M}'\\textbar}{AB}$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{4\\sqrt{3}}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{3}}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2\\sqrt{3}}{3}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆锥曲线"], "answer_analysis": ["设$$AB$$在准线$$l$$上的射影分别为$${A}'$$、$${B}'$$， 则$$\\textbar M{M}'\\textbar=\\frac{\\textbar A{A}'\\textbar+\\textbar B{B}'\\textbar}{2}$$． 又���为$$\\textbar A{A}'\\textbar=\\textbar AF\\textbar$$，$$\\textbar B{B}'\\textbar=\\textbar BF\\textbar$$， 所以$$\\textbar M{M}'\\textbar=\\frac{\\textbar AF\\textbar+\\textbar BF\\textbar}{2}$$． 因为$$\\angle AFB=\\frac{2}{3} \\pi $$， 所以$$\\textbar AB\\textbar=\\textbar FA{{\\textbar}^{2}}+\\textbar FB{{\\textbar}^{2}}-2\\textbar FA\\textbar\\cdot \\textbar FB\\textbar\\cos \\angle AFB$$ $$=\\textbar FA{{\\textbar}^{2}}+\\textbar FB{{\\textbar}^{2}}+\\textbar FA\\textbar\\cdot \\textbar FB\\textbar$$ 因此$$\\textbar AB{{\\textbar}^{2}}+\\textbar M{M}'{{\\textbar}^{2}}=\\textbar AB{{\\textbar}^{2}}+{{\\left( \\frac{\\textbar AF\\textbar+\\textbar BF\\textbar}{2} \\right)}^{2}}$$ $$\\geqslant \\textbar AB{{\\textbar}^{2}}+\\textbar FA\\textbar\\cdot \\textbar FB\\textbar$$ $$={{(\\textbar FA\\textbar+\\textbar FB\\textbar)}^{2}}=4\\textbar M{M}'{{\\textbar}^{2}}$$， 则$$\\textbar AB{{\\textbar}^{2}}\\geqslant 3\\textbar M{M}'{{\\textbar}^{2}}$$，即$$\\frac{\\textbar M{M}'\\textbar}{\\textbar AB\\textbar}\\leqslant \\frac{\\sqrt{3}}{3}$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "434", "queId": "70c04c8bb15c4fec95429b88c65f2fcf", "competition_source_list": ["1985年全国高中数学联赛竞赛一试第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "假如有两个命题；甲：$$a$$是大于零的实数；乙：$$a\\textgreater b$$且$${{a}^{-1}}\\textgreater{{b}^{-1}}$$，那么（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "甲是乙的充分而不必要条件 "}], [{"aoVal": "B", "content": "甲是乙的必要而不充分条件 "}], [{"aoVal": "C", "content": "甲是乙的充分必要条件 "}], [{"aoVal": "D", "content": "甲既不是乙的充分条件也不是乙的必要条件． "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->常用逻辑用语"], "answer_analysis": ["由于``甲''成立时，``乙''不一定成立，（例如$$a=3$$，$$b=2$$时甲成立，但乙不成立），因此可知``甲''不是``乙''的充分条件， 接着看``乙''在什么情况下成立，很明显，当且仅当$$a\\textgreater0$$且$$b\\textless{}0$$时``乙''才能成立，由此可知，``甲''是``乙''成立的必要条件，综上所述，``甲''是``乙''的必要而不充分条件． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "613", "queId": "4063a6ebee1c470192768eb8938554b9", "competition_source_list": ["2008年陕西全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$y=\\tan x-\\frac{2}{\\left\\textbar{} \\cos x \\right\\textbar}$$的最大值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1-2\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$1+2\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$-\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["要使$$y$$最大，应有$$\\tan x\\textgreater0$$． 不妨设$$0\\textless{}x\\textless{}\\frac{ \\pi }{2}$$，则$$y=\\tan x-\\frac{2}{\\cos x}=\\frac{\\sin x-2}{\\cos x}$$， 即$$\\sin x-y\\cos x=2$$．所以$$\\sin (x-\\theta )=\\frac{2}{\\sqrt{1+{{y}^{2}}}}$$，其中$$\\tan \\theta =y$$． 由$$\\sin (x-\\theta )\\leqslant 1$$，得$$\\frac{2}{\\sqrt{1+{{y}^{2}}}}\\leqslant 1$$， 解得$$y\\leqslant -\\sqrt{3}$$或$$y\\geqslant 3$$． 易知$$y\\textless{}0$$，所以$$y\\leqslant -\\sqrt{3}$$． 将$$y=-\\sqrt{3}$$代入$$\\sin x-y\\cos x=2$$，得$$x=\\frac{ \\pi }{6}$$． 故$${{y}_{\\max }}=-\\sqrt{3}$$．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "344", "queId": "27e1d67904194ca992aa99ea253ed822", "competition_source_list": ["2016年四川全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f(x)={{x}^{2}}-2tx+t$$，当$$x\\in [-1,1]$$时，记$$f(x)$$的最小值为$$m$$，则$$m$$的最大值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["课内体系->素养->直观想象", "课内体系->素养->数学运算", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->分段函数", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->函数求值问题", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域", "课内体系->知识点->函数的概念与性质->函数的性质->单调性->函数单调性的应用->利用单调性求函数最值", "课内体系->知识点->函数的概念��性质->二次函数->二次函数的图象及性质", "课内体系->知识点->函数的概念与性质->二次函数->二次函数的概念"], "answer_analysis": ["$$f\\left( x \\right)={{\\left( x-t \\right)}^{2}}+t-{{t}^{2}}$$，$$m=\\begin{cases}t-{{t}^{2}},-1\\leqslant t\\leqslant 1 1-t,t\\textgreater1 1+3t,t\\textless{}-1 \\end{cases}$$ 因此，$${{m}_{\\max }}=t-{{t}^{2}}=-{{\\left( t-\\frac{1}{2} \\right)}^{2}}+\\frac{1}{4}\\leqslant \\frac{1}{4}$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "487", "queId": "298df3791f1c45c495e128dafae49608", "competition_source_list": ["2008年全国高中数学联赛竞赛一试第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "若三个棱长均为整数（单位：$$\\text{cm}$$）的正方体的表面积之和为$$564\\text{c}{{\\text{m}}^{2}}$$，则这三个正方体的体积之和为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$764\\text{c}{{\\text{m}}^{3}}$$或$$586\\text{c}{{\\text{m}}^{3}}$$ "}], [{"aoVal": "B", "content": "$$764\\text{c}{{\\text{m}}^{3}}$$ "}], [{"aoVal": "C", "content": "$$586\\text{c}{{\\text{m}}^{3}}$$或$$564\\text{c}{{\\text{m}}^{3}}$$ "}], [{"aoVal": "D", "content": "$$586\\text{c}{{\\text{m}}^{3}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["设这三个正方体的棱长分别为$$a,b,c$$，则有$$6({{a}^{2}}+{{b}^{2}}+{{c}^{2}})=564$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=94$$， 不妨设$$1\\leqslant a\\leqslant b\\leqslant c\\textless{}10$$，从而$$3{{c}^{2}}\\geqslant {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=94$$，$${{c}^{2}}\\textgreater31$$， 故$$6\\leqslant c\\leqslant 10$$．$$c$$只能取$$9$$，$$8$$，$$7$$，$$6$$， 若$$c=9$$，则$${{a}^{2}}+{{b}^{2}}=94-{{9}^{2}}=13$$，易知$$a=2$$，$$b=3$$，得一组解$$(a,b,c)=(2,3,9)$$， 若$$c=8$$，则$${{a}^{2}}+{{b}^{2}}=94-64=30$$，$$b\\leqslant 5$$，但$$2{{b}^{2}}\\geqslant 30$$，$$b\\geqslant 4$$，从而$$b=4$$或$$5$$， 若$$b=5$$，则$${{a}^{2}}=5$$无解，若$$b=4$$，则$${{a}^{2}}=14$$无解，此时无解， 若$$c=7$$，则$${{a}^{2}}+{{b}^{2}}=94-49=45$$，有唯一解$$a=3$$，$$b=6$$， 若$$c=6$$，则$${{a}^{2}}+{{b}^{2}}=94-36=58$$，此时$$2{{b}^{2}}\\geqslant {{a}^{2}}+{{b}^{2}}=58$$，$${{b}^{2}}\\geqslant 29$$，故$$b\\geqslant 6$$，但$$b\\leqslant c=6$$，故$$b=6$$，此时$${{a}^{2}}=58-36=22$$无解， 综上，共有两组解$$\\begin{cases}a=2 b=3 c=9 \\end{cases}$$或$$\\begin{cases}a=3 b=6 c=7 \\end{cases}$$， 体积为$${{V}_{1}}={{2}^{3}}+{{3}^{3}}+{{9}^{3}}=764\\text{c}{{\\text{m}}^{3}}$$或$${{V}_{2}}={{3}^{3}}+{{6}^{3}}+{{7}^{3}}=586\\text{c}{{\\text{m}}^{3}}$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1163", "queId": "eb1c4eae085d48ec80b05c56df1c23e8", "competition_source_list": ["2016年浙江全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "曲线$$\\left( x+2y+a \\right)\\left( {{x}^{2}}-{{y}^{2}} \\right)=0$$为平面上交于一点的三条直线的充要条件是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a=0$$ "}], [{"aoVal": "B", "content": "$$a=1$$ "}], [{"aoVal": "C", "content": "$$a=-1$$ "}], [{"aoVal": "D", "content": "$$a\\in \\mathbf{R}$$~ "}]], "knowledge_point_routes": ["课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与函数结合", "课内体系->知识点->函数的应用->函数与方程", "课内体系->知识点->直线和圆的方程->直线与方程->直线方程的五种形式->直线过定点问题", "课内体系->知识点->直线和圆的方程->直线与方程->直线方程的五种形式->直线的一般式方程", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理"], "answer_analysis": ["三条直线过原点. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "553", "queId": "3fbc7005b22e42488656fc5f15d0aa0e", "competition_source_list": ["2011年福建全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$y=$$$f(x)$$满足： （$$1$$）$$f(1)=1$$（表示$x=1$时对应的$y$的值，下同）；（$$2$$）当$$0\\textless{}x\\textless{}1$$时，$$f(x)\\textgreater0$$；（$$3$$）对任意的实数$$x, y$$均有$$f(x+y)-f(x-y)=2f(1-x)f(y),$$则$$f\\left( \\frac{1}{3} \\right)=$$ ．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$\\dfrac{1}{2}$ "}], [{"aoVal": "C", "content": "$\\dfrac{\\sqrt{2}}{2}$ "}], [{"aoVal": "D", "content": "$\\dfrac{\\sqrt{3}}{3}$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质", "竞赛->知识点->函数->函��的图像与性质"], "answer_analysis": ["在条件等式（$$3$$）中，令$$x=\\frac{2}{3}, y=\\frac{1}{3},$$得 $$f(1)-f\\left( \\frac{1}{3} \\right)=2f\\left( \\frac{1}{3} \\right)f\\left( \\frac{1}{3} \\right),$$ 结合$$f(1)=1, f\\left( \\frac{1}{3} \\right)\\textgreater0$$，解得$$f\\left( \\frac{1}{3} \\right)=\\frac{1}{2}$$ "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "25", "queId": "5cfe23a809bf4bb3aaae6781dbb81399", "competition_source_list": ["高考真题", "2008年贵州全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "在等差数列$$\\left { {{a}_{n}} \\right }$$中，$${{a}_{1}}+2{{a}_{8}}+{{a}_{15}}=96$$，则$$2{{a}_{9}}-{{a}_{10}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->数列->等差数列->等差数列的性质及应用", "课内体系->知识点->数列->等差数列->等差数列的概念与通项公式->等差数列求通项问题"], "answer_analysis": ["$${{a}_{1}}+2{{a}_{8}}+{{a}_{15}}=4{{a}_{8}}=96$$，$${{a}_{8}}=24$$． $$2{{a}_{9}}-{{a}_{10}}={{a}_{8}}=24$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "321", "queId": "1b232580c3fe4620953684c5f0d544e8", "competition_source_list": ["2015年黑龙江全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f(x)={{\\sin }^{5}}x+1$$，根据函数的性质，积分的性质和积分的几何意义，探求$$\\int_{-\\frac{ \\pi }{2}}^{\\frac{ \\pi }{2}}{f(x)\\text{d}x}$$的值，结果是", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{6}+\\frac{ \\pi }{2}$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$ \\pi $$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->积分", "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["因为$$\\sin 5x$$是奇函数，关于原点对称，且积分限关于原点对称，积分就是面积， 所以$$x\\textless{}0$$和$$x\\textgreater0$$时积分的值一正一负，且面积相等， 所以正负抵消，即$$\\int_{-\\frac{ \\pi }{2}}^{\\frac{ \\pi }{2}}{\\sin^{5}x\\text{d}x}=0$$， 所以$$\\int_{-\\frac{ \\pi }{2}}^{\\frac{ \\pi }{2}}{f(x)\\text{d}x}=\\int_{-\\frac{ \\pi }{2}}^{\\frac{ \\pi }{2}}{\\sin^{5}x\\text{d}x}+\\int_{-\\frac{ \\pi }{2}}^{\\frac{ \\pi }{2}}{\\text{1d}x}=x\\textbar_{-\\frac{ \\pi }{2}}^{\\frac{ \\pi }{2}}=\\frac{ \\pi }{2}-\\left( -\\frac{ \\pi }{2} \\right)= \\pi $$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1020", "queId": "aa1b7904861642a19f17c713ea239584", "competition_source_list": ["2014年湖南全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "设条件$$p$$：实数$$m$$，$$n$$满足$$\\begin{cases}2\\textless{}m+n\\textless{}4 0\\textless{}mn\\textless{}3 \\end{cases}$$条件$$q$$：实数$$m$$、$$n$$满足$$\\begin{cases}0\\textless{}m\\textless{}1 2\\textless{}n\\textless{}3. \\end{cases}$$则．", "answer_option_list": [[{"aoVal": "A", "content": "$$p$$是$$q$$的充分不必要条件 "}], [{"aoVal": "B", "content": "$$p$$是$$q$$的充要不充分条件 "}], [{"aoVal": "C", "content": "$$p$$是$$q$$的充要条件 "}], [{"aoVal": "D", "content": "$$p$$既不是$$q$$的充分条件又不是$$q$$的必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->常用逻辑用语"], "answer_analysis": ["由$$q:\\begin{cases}0\\textless{}m\\textless{}1 2\\textless{}n\\textless{}3 \\end{cases}\\Rightarrow p:\\begin{cases}2\\textless{}m+n\\textless{}4 0\\textless{}mn\\textless{}3 \\end{cases}$$ 则$$p$$是$$q$$的必要条件；注意到当$$\\begin{cases}m=1 n=2 \\end{cases}$$时，满足$$p:\\begin{cases}2\\textless{}m+n\\textless{}4 0\\textless{}mn\\textless{}3 \\end{cases}$$而不满足$$q:\\begin{cases}0\\textless{}m\\textless{}1 2\\textless{}n\\textless{}3 \\end{cases}$$则$$p$$不是$$q$$的充分条件．所以$$p$$是$$q$$的必要不充分条件． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "324", "queId": "7999e94481fe4306944726914fa76e09", "competition_source_list": ["2015年四川全国高中数学联赛竞赛初赛第1题5分", "2017年北京西城区高三三模理科第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "设$$n$$为正整数，二项式$${{\\left( {{x}^{2}}+\\frac{1}{{{x}^{3}}} \\right)}^{n}}$$的展开式��含有$${{x}^{7}}$$的项，则$$n$$的最小值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->计数原理->二项式定理->二项式定理的展开式", "课内体系->知识点->计数原理->二项式定理->求项的系数或二项式系数"], "answer_analysis": ["$${{T}_{r+1}}=C_{n}^{r}{{({{x}^{2}})}^{n-r}}{{\\left( \\frac{1}{{{x}^{3}}} \\right)}^{r}}=C_{n}^{r}{{x}^{2}}^{n-5r}$$．令$$2n-5r=7$$，所以$$r=1$$时，$$n$$取得最小值$$6$$．选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1065", "queId": "b39fb4710ee041699a9e0b547d6be154", "competition_source_list": ["2006年AMC10竞赛B第21题"], "difficulty": "3", "qtype": "single_choice", "problem": "对于一对特殊的骰子，在每个骰子上掷出$1$、$2$、$3$、$4$、$5$和$6$的概率的比例是$1:2:3:4:5:6$.两颗骰子共掷出$7$的概率是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\dfrac{4}{63}$$ "}], [{"aoVal": "B", "content": "$$\\dfrac{1}{8}$$ "}], [{"aoVal": "C", "content": "$$\\dfrac{8}{63}$$ "}], [{"aoVal": "D", "content": "$$\\dfrac{1}{6}$$ "}], [{"aoVal": "E", "content": "$$\\dfrac{2}{7}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率->事件与概率->互斥事件->互斥事件的概率加法公式", "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities"], "answer_analysis": ["Let $$x$$ be the \\uline{probability} of rolling a $$1$$. The probabilities of rolling a $$2$$, $$3$$, $$4$$, $$5$$, and $$6$$ are $$2x$$, $$3x$$, $$4x$$, $$5x$$, and $$6x$$, respectively. The sum of the probabilties of rolling each number must equal $$1$$, so $$x+2x+3x+4x+5x+6x=1$$, $$21x=1$$, $$x= \\dfrac{1}{21}$$, So the probabilities of rolling a $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, and $$6$$ are respectively $$\\dfrac{1}{21}$$,~ $$\\dfrac{2}{21}$$, $$\\dfrac{3}{21}$$, $$\\dfrac{4}{21}$$, $$\\dfrac{5}{21}$$, and, $$\\dfrac{6}{21}$$. The possible combinations of two rolls that total $$7$$ are: $$(1,6)$$; $$(2,5)$$; $$(3,4)$$; $$(4,3)$$; $$(5,2)$$; $$(6,1)$$. The probability of rolling a total of $$7$$ on the two dice is equal to the sum of the probabilities of rolling each combination. $$P= \\dfrac{1}{21}\\cdot \\dfrac{6}{21}+ \\dfrac{2}{21}\\cdot \\dfrac{5}{21}+ \\dfrac{3}{21}\\cdot \\dfrac{4}{21}+ \\dfrac{4}{21}\\cdot \\dfrac{3}{21}+ \\dfrac{5}{21}\\cdot \\dfrac{2}{21}+ \\dfrac{6}{21}\\cdot \\dfrac{1}{21}= \\dfrac{8}{63}\\Rightarrow C$$ "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1037", "queId": "fc1daea602ee472289c84eb0aebfe066", "competition_source_list": ["2018年天津全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若集合$$A=\\left { 1,2,\\cdots ,10 \\right }$$，$$B=\\left { 1,2,3,4 \\right }$$，$$C$$为$$A$$的子集，且$$C\\cap B\\ne \\varnothing $$， 则这样的子集$$C$$有个．", "answer_option_list": [[{"aoVal": "A", "content": "$$256$$ "}], [{"aoVal": "B", "content": "$$959$$ "}], [{"aoVal": "C", "content": "$$960$$ "}], [{"aoVal": "D", "content": "$$961$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["由满足$$C\\cap B=\\varnothing $$的子集$$C$$有$${{2}^{6}}$$个， 知满足$$C\\cap B\\ne \\varnothing $$的子集$$C$$有$${{2}^{10}}-{{2}^{6}}=960$$个． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1109", "queId": "fcc6c7e7384c4a0aa0515c504ed54772", "competition_source_list": ["2016年陕西全国高中数学联赛竞赛初赛第8题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设非负实数$$a$$，$$b$$，$$c$$满足$$ab+bc+ca=a+b+c\\textgreater0$$，则$$\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca}$$的最小值为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$2\\sqrt{2}$$ "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->等式与不等式->不等式->不等式的性质", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的概念", "课内体系->知识点->等式与不等式->不等式->基本不等式->利用基本不等式求最值", "课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式成立的条件"], "answer_analysis": ["不妨设$$a\\geqslant b\\geqslant c$$ $$\\left( a+b+c \\right)\\left( \\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca} \\right)=\\left( a+b \\right)\\sqrt{ab}+\\left( b+c \\right)\\sqrt{bc}+\\left( c+a \\right)\\sqrt{ca}+\\left( c\\sqrt{ab}+a\\sqrt{bc}+b\\sqrt{ca} \\right)$$ $$\\geqslant \\left( a+b \\right)\\sqrt{ab}+\\left( b+c \\right)\\sqrt{bc}+\\left( c+a \\right)\\sqrt{ca}$$ $$\\geqslant 2\\sqrt{ab}\\sqrt{ab}+2\\sqrt{bc}\\sqrt{bc}+2\\sqrt{ca}\\sqrt{ca}$$ $$=2\\left( ab+bc+ca \\right)$$ 所以，$$\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca}\\geqslant 2$$，当且仅当$$c=0$$，$$a=b$$时等号成立． 由$$c=0$$，$$a=b$$，解得$$a=b=2$$，$$c=0$$． 所以，$$\\sqrt{ab}+\\sqrt{bc}+\\sqrt{ca}$$的最小值为$$2$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "892", "queId": "f237c03cbafa4bc8ad45d7cc23a8c5c3", "competition_source_list": ["2008年河北全国高中数学联赛竞赛初赛第4题6分", "高二上学期单元测试《代数变形1》自招第7题", "2019~2020学年天津滨海新区塘沽第一中学高一上学期期中第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a\\textgreater b$$，$$ab=1$$，则$$\\frac{{{a}^{2}}+{{b}^{2}}}{a-b}$$的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->换元技巧->代数换元"], "answer_analysis": ["记$$a-b=t$$，则$$t\\textgreater0$$， $$\\frac{{{a}^{2}}+{{b}^{2}}}{a-b}=\\frac{{{t}^{2}}+2}{t}=t+\\frac{2}{t}\\geqslant 2\\sqrt{2}$$， 当且仅当$$a=\\frac{\\sqrt{6}+\\sqrt{2}}{2}$$，$$b=\\frac{\\sqrt{6}-\\sqrt{2}}{2}$$时取等号．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1150", "queId": "c6ac19a0b8184b3c9fa61aaf40a45a0f", "competition_source_list": ["2009年湖南全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$f(x)$$为$$R\\to R$$函数，且对任意实数，有 $$f({{x}^{2}}+x)+2f({{x}^{2}}-3x+2)=9{{x}^{2}}-15x$$ 则$$f(50)$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$72$$ "}], [{"aoVal": "B", "content": "$$73$$ "}], [{"aoVal": "C", "content": "$$144$$ "}], [{"aoVal": "D", "content": "$$146$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["用$$(1-x)$$代替条件等式中的$$x$$，得 $$f({{x}^{2}}-3x+2)+2f({{x}^{2}}+x)=9{{x}^{2}}-3x-6$$． 由上式及原式消去$$f({{x}^{2}}-3x+2)$$，得 $$f({{x}^{2}}+x)=3{{x}^{2}}+3x-4=3({{x}^{4}}+x)-4$$． 故$$f(50)=150-4=146$$，故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "949", "queId": "b6fa23581e664c898a9a01106fb200bf", "competition_source_list": ["2011年AMC10竞赛A第18题"], "difficulty": "2", "qtype": "single_choice", "problem": "2011年$$AMC10$$竞赛$$A$$第$$18$$题 Circles $$A$$, $$B$$, and $$C$$ each have radius $$1$$. Circles $$A$$ and $$B$$ share one point of tangency. Circle $$C$$ has a point of tangency with the midpoint of $$\\overline{AB}$$. What is the area inside Circle $$C$$ but outside circle $$A$$ and circle $$B$$? 圆 $$A$$、$$B$$ 和 $$C$$ 的半径都是 $$1$$。 圆 $$A$$ 和 $$B$$ 有一个公共点。 圆 $$C$$ 与 $$\\overline{AB}$$ 的中点有一个切点。 $$C$$ 圆内并且在$$A$$ 圆外和$$B$$ 圆外的面积是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$3- \\dfrac{\\pi}{2}$$ "}], [{"aoVal": "B", "content": "$$\\dfrac{\\pi}{2}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$\\dfrac{3 \\pi}{4}$$ "}], [{"aoVal": "E", "content": "$$1+ \\dfrac{\\pi}{2}$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Circle", "课内体系->知识点->三角函数->三角函数的概念->任意角与弧度制->弧长公式与扇形面积"], "answer_analysis": ["Not specific: Draw a rectangle with vertices at the centers of $$A$$ and $$B$$ and the intersection of $$A$$, $$C$$ and $$B$$, $$C$$. Then, we can compute the shaded area as the area of half of $$C$$ plus the area of the rectangle minus the area of the two sectors created by $$A$$ and $$B$$. This is $$\\dfrac{\\pi(1)^{2}}{2}+(2)(1)-2 \\cdot \\dfrac{\\pi(1)^{2}}{4}= (\\rm C)2$$. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "205", "queId": "1d798df11da4409b89b2a1844dc8aa9b", "competition_source_list": ["2018~2019学年北京东城区北京市第一六六中学高二上学期期中第12题5分", "2006年全国高中数学联赛竞赛一试第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "在直三棱柱中，$$\\angle BAC=\\frac{ \\pi }{2}$$，$$AB=AC=A{{A}_{1}}=1$$．已知$$G$$与$$E$$分别为$${{A}_{1}}{{B}_{1}}$$和$$C{{C}_{1}}$$的中点，$$D$$与$$F$$分别为线段$$AC$$和$$AB$$上的动点（不包括端点）．若$$GD\\bot EF$$，则线段$$DF$$的长度的取值范围为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ \\frac{1}{\\sqrt{5}}, 1 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left[ \\frac{1}{5}, 2 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left[ 1, \\sqrt{2} \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left[ \\frac{1}{\\sqrt{5}}, \\sqrt{2} \\right)$$ "}]], "knowledge_point_routes": ["课内体系->素养->直观想象", "课内体系->素养->数学运算", "课内体系->知识点->空间向量与立体几何->空间向量的应用->向量法求空间距离", "课内体系->思想->数形结合思想"], "answer_analysis": ["建立直角坐标系，以$$A$$为坐标原点，$$AB$$为$$x$$轴，$$AC$$为$$y$$轴，$$A{{A}_{{}}}$$为$$z$$轴，则$$F({{t}_{1}},0,0)$$（$$0\\textless{}{{t}_{1}}\\textless{}1$$），$$E\\left( 0,1,\\frac{1}{2} \\right)$$，$$G\\left( \\frac{1}{2},0,1 \\right)$$，$$D\\left( 0,{{t}_{2}},0 \\right)$$（$$0\\textless{}{{t}_{2}}\\textless{}1$$）．所以$$\\overrightarrow{EF}=\\left( {{t}_{1}},-1,-\\frac{1}{2} \\right)$$，$$\\overrightarrow{GD}=\\left( -\\frac{1}{2},{{t}_{2}},-1 \\right)$$．因为$$GD\\bot EF$$，所以$${{t}_{1}}+2{{t}_{2}}=1$$，由此推出$$0\\textless{}{{t}_{2}}\\textless{}\\frac{1}{2}$$．又$$\\overrightarrow{DF}=({{t}_{1}},-{{t}_{2}},0)$$，$$\\left\\textbar{} \\overrightarrow{DF} \\right\\textbar=\\sqrt{{{t}_{1}}^{2}+{{t}_{2}}^{2}}=\\sqrt{5{{t}_{2}}^{2}-4{{t}_{2}}+1}=\\sqrt{5{{({{t}_{2}}-\\frac{2}{5})}^{2}}+\\frac{1}{5}}$$，从而有$$\\sqrt{\\frac{1}{5}}\\leqslant \\left\\textbar{} \\overrightarrow{DF} \\right\\textbar\\textless{}1$$． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "970", "queId": "8e8c7ae0e5354f499a0cb01a1f73714a", "competition_source_list": ["1997年全国高中数学联赛竞赛一试第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设等差数列的首项及公差均为非负整数，项数不少于$$3$$，且各项的和为$$9{{7}^{2}}$$，则这样的数列共有（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$个 "}], [{"aoVal": "B", "content": "$$3$$个 "}], [{"aoVal": "C", "content": "$$4$$个 "}], [{"aoVal": "D", "content": "$$5$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->不定方程->因式分解与恒等变形"], "answer_analysis": ["设首项为$$a$$，公差为$$d$$，项数为$$n$$，则$$na+\\frac{1}{2}n\\left( n-1 \\right)d={{97}^{2}}$$，$$n\\left[ 2a+\\left( n-1 \\right)d \\right]=2\\times 9{{7}^{2}}$$，即$$n$$为$$2\\times 9{{7}^{2}}$$的大于$$3$$的约数． ∴①$$n=9{{7}^{2}}$$，$$2a+\\left( 9{{7}^{2}}-1 \\right)d=2$$，$$d=0$$，$$a=1$$；$$d\\geqslant 1$$时$$a\\textless{}0$$．有$$1$$解； ②$$n=97$$，$$2a+96d=194$$，$$d=0$$，$$a=97$$；$$d=1$$，$$a=49$$；$$d=2$$，$$a=1$$．有$$3$$解； ③$$n=2\\times 97$$，$$n=2\\times 9{{7}^{2}}$$，无解．$$n=1$$，$$2$$时$$n\\textless{}3$$． 所以符合要求的数列共有$$1+3=4$$个． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "941", "queId": "a04ead5e1b6d48d18e84a11a215482de", "competition_source_list": ["第二十届全国希望杯高二竞赛初赛邀请赛第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果实数$$xy$$满足关系$$\\left( \\sqrt{{{x}^{2}}+1}-x \\right)\\left( \\sqrt{{{y}^{2}}+1}-y \\right)\\leqslant 1$$，那么．", "answer_option_list": [[{"aoVal": "A", "content": "$$x+y\\geqslant 0$$ "}], [{"aoVal": "B", "content": "$$x+y\\leqslant 0$$ "}], [{"aoVal": "C", "content": "$$x-y\\geqslant 0$$ "}], [{"aoVal": "D", "content": "$$x-y\\leqslant 0$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法", "竞赛->知识点->不等式->换元技巧->三角换元"], "answer_analysis": ["由$$xy$$的对称性，显然大小不能确定，即$$C$$、$$\\text{D}$$错误； 设$$x=\\tan \\alpha ,y=\\tan \\beta ,\\alpha ,\\beta \\in \\left( -\\frac{ \\pi }{2},\\frac{ \\pi }{2} \\right)$$，于是原不等式可化简为$$\\frac{1-\\sin \\alpha }{\\cos \\alpha }\\cdot \\frac{1-\\sin \\beta }{\\cos \\beta }\\leqslant 1$$，整理可得$$\\sin \\alpha +\\sin \\beta \\geqslant 1-\\cos (\\alpha +\\beta )\\geqslant 0$$，即$$2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha -\\beta }{2}\\geqslant 0$$，又$$\\cos \\frac{\\alpha -\\beta }{2}\\textgreater0$$，故$$\\sin \\frac{\\alpha +\\beta }{2}\\geqslant 0$$．另外，易知$$\\cos \\frac{\\alpha +\\beta }{2}\\textgreater0$$，从而$$\\sin (\\alpha +\\beta )=2\\sin \\frac{\\alpha +\\beta }{2}\\cos \\frac{\\alpha +\\beta }{2}\\geqslant 0$$，于是$$x+y=\\tan \\alpha +\\tan \\beta =\\frac{\\sin (\\alpha +\\beta )}{\\cos \\alpha \\cos \\beta }\\geqslant 0$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "756", "queId": "57e5828096e34f1cb9eca5cd0c2ebd29", "competition_source_list": ["2015年天津全国高中数学联赛竞赛初赛第12题9分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a,b,c,d$$都是实数，$$a+2b+3c+4d=\\sqrt{10}$$，则$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+{{\\left( a+b+c+d \\right)}^{2}}$$的最小值是~\\uline{~~~~~~~~~~}~", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->柯西", "课内体系->知识点->等式与不等式->不等式->柯西不等式"], "answer_analysis": ["待定系数，由柯西不等式， $${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+{{\\left( a+b+c+d \\right)}^{2}}\\geqslant \\frac{{{\\left[ \\left( {{x}_{1}}+{{x}_{5}} \\right)a+\\left( {{x}_{2}}+{{x}_{5}} \\right)b+\\left( {{x}_{3}}+{{x}_{5}} \\right)c+\\left( {{x}_{4}}+{{x}_{5}} \\right)d \\right]}^{2}}}{{{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}+{{x}_{4}}^{2}+{{x}_{5}}^{2}}$$令$${{x}_{1}}+{{x}_{5}}=1$$，$${{x}_{2}}+{{x}_{5}}=2$$，$${{x}_{3}}+{{x}_{5}}=3$$，$${{x}_{4}}+{{x}_{5}}=4$$， 则$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+{{\\left( a+b+c+d \\right)}^{2}}\\geqslant \\frac{10}{{{\\left( 1-{{x}_{5}} \\right)}^{2}}+{{\\left( 2-{{x}_{5}} \\right)}^{2}}+{{\\left( 3-{{x}_{5}} \\right)}^{2}}+{{\\left( 4-{{x}_{5}} \\right)}^{2}}+{{x}_{5}}^{2}}$$ $$=\\frac{10}{5{{x}_{5}}^{2}-20{{x}_{5}}+30}=\\frac{10}{5{{\\left( {{x}_{5}}-2 \\right)}^{2}}+10}$$ 取$${{x}_{5}}=2$$，得$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{d}^{2}}+{{\\left( a+b+c+d \\right)}^{2}}\\geqslant 1$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "194", "queId": "4f8f11d3bdce49469a5bb05693f68334", "competition_source_list": ["2018年黑龙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知集合$$A=\\left { \\left( x,y \\right)\\left\\textbar{} x+{{a}^{2}}y+6=0 \\right. \\right }$$，集合$$B=\\left { \\left( x,y \\right)\\left\\textbar{} \\left( a-2 \\right)x+3ay+2a \\right.=0 \\right }$$，若$$A\\cap B=\\varnothing $$，则$$a$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$或$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "$$0$$或$$-1$$ "}]], "knowledge_point_routes": ["课内体系->思想->转化化归思想", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->知识点->直线和圆的方程->直线与方程->直线的位置关系->判定两条直线的位置关系"], "answer_analysis": ["由题意知$$x+{{a}^{2}}y+6=0$$与$$\\left( a-2 \\right)x+3ay+2a=0$$均为直线方程． 若$$A\\cap B=\\varnothing $$，则这两条直线斜率相同且两条直线不重合． 由斜率相同得$${{a}^{2}}=\\frac{3a}{a-2}$$． 从而，$${{a}^{3}}-2{{a}^{2}}-3a=0$$．① （1）当$$a=0$$时，经验证成立． （2）当$$a\\ne 0$$时，由式①得， $${{a}^{2}}-2a-3=0\\Rightarrow a=-1$$，$$3$$． 红验证$$a=-1$$满足条件； 当$$a=3$$时，两直线重合，舍去． 综上，$$a=0$$或$$-1$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "994", "queId": "a5478d2d5525431e89be84d450aa599f", "competition_source_list": ["2016年四川全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$${{F}_{1}}$$，$${{F}_{2}}$$是椭圆和双曲线的公共焦点，$$P$$是它们的一个公共点，且$$\\angle {{F}_{1}}P{{F}_{2}}=60{}^{}\\circ $$，则该椭圆和双曲线的离心率之积的最小值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{3}}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{3}}{2}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->圆锥曲线->椭圆->椭圆的简单几何性质->椭圆的离心率->求椭圆的离心率", "课内体系->知识点->圆锥曲线->双曲线->双曲线的几何性质->双曲线的离心率->求双曲线的离心率", "课内体系->知识点->解三角形->余弦定理", "课内体系->知识点->解三角形->正余弦定理的综合应用->正余弦定理综合求解边角", "课内体系->素养->直观想象", "课内体系->素养->数学运算"], "answer_analysis": ["设$$\\left\\textbar{} P{{F}_{1}} \\right\\textbar=x$$，$$\\left\\textbar{} P{{F}_{2}} \\right\\textbar=y$$，$$\\left\\textbar{} {{F}_{1}}{{F}_{2}} \\right\\textbar=2c$$，则$${{e}_{1}}{{e}_{2}}=\\frac{2c}{x+y}\\cdot \\frac{2c}{x-y}$$（不妨设$$x\\textgreater y$$）． 在$$\\triangle P{{F}_{1}}{{F}_{2}}$$��，由余弦定理，$$4{{c}^{2}}={{x}^{2}}+{{y}^{2}}-xy$$． 于是，$${{e}_{1}}{{e}_{2}}=\\frac{4{{c}^{2}}}{{{x}^{2}}-{{y}^{2}}}=\\frac{{{x}^{2}}+{{y}^{2}}-xy}{{{x}^{2}}-{{y}^{2}}}$$，去分母，移项得$$\\left( 1-{{e}_{1}}{{e}_{2}} \\right){{x}^{2}}-xy+\\left( 1+{{e}_{1}}{{e}_{2}} \\right){{y}^{2}}=0$$， 判别式$$\\Delta =1-4\\left( 1-{{e}_{1}}{{e}_{2}} \\right)\\left( 1+{{e}_{1}}{{e}_{2}} \\right)\\geqslant 0$$，解得$${{e}_{1}}{{e}_{2}}\\geqslant \\frac{\\sqrt{3}}{2}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "649", "queId": "6d62485f03d24facb31e5e894e16c219", "competition_source_list": ["2009年AMC10竞赛B第14题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2009-AMC10B-14$$ On Monday, Millie puts a quart of seeds, $$25 \\%$$ of which are millet, into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only $$25 \\%$$ of the millet in the feeder, but they eat all of the other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet? 星期一，米莉将一夸脱种子（其中 25\\% 是小米）放入喂鸟器中。 在连续的每一天，她添加另一夸脱相同的种子混合物，但没有去除任何剩余的种子。 这些鸟每天只吃喂食器中 25\\% 的小米，但它们吃掉所有其他种子。 哪一天，米莉刚放完种子，鸟儿会发现喂食器里一半以上的种子是小米？", "answer_option_list": [[{"aoVal": "A", "content": "Tuesday "}], [{"aoVal": "B", "content": "Wednesday "}], [{"aoVal": "C", "content": "Thursday "}], [{"aoVal": "D", "content": "Friday "}], [{"aoVal": "E", "content": "Saturday "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Sequence->Sum of the First n Terms for Geometric Sequences", "课内体系->知识点->数列->数列的实际应用"], "answer_analysis": ["On Monday, day $$1$$, the birds find $$\\frac14$$ quart of millet in the feeder. On Tuesday they find $$\\frac{1}{4}+ \\frac{3}{4}\\cdot \\frac{1}{4}$$ quarts of millet. On Wednesday, day $$3$$, they find $$\\frac{1}{4}+ \\frac{3}{4}\\cdot \\frac{1}{4}+\\left(\\frac{3}{4}\\right)^{2}\\cdot \\frac{1}{4}$$ quarts of millet. The number of quarts of millet they find on day $$n$$ is $$\\frac{1}{4}+ \\frac{3}{4}\\cdot \\frac{1}{4}+\\left(\\frac{3}{4}\\right)^{2}\\cdot \\frac{1}{4}+ \\cdots +\\left(\\dfrac{3}{4}\\right)^{n-1}\\cdot \\frac{1}{4}= \\dfrac{\\left(\\dfrac{1}{4}\\right)\\left(1-\\left(\\dfrac{3}{4}\\right)^{n}\\right)}{1- \\dfrac{3}{4}}=1-\\left(\\frac{3}{4}\\right)^{n}$$. The birds always find $$\\frac34$$ quart of other seeds, so more than half the seeds are millet if $$1-\\left(\\frac{3}{4}\\right)^{n}\\textgreater\\frac34$$ , that is, when $$\\left(\\frac{3}{4}\\right)^{n}\\textless{} \\frac{1}{4}$$. Because $$\\left(\\frac{3}{4}\\right)^{4}= \\frac{81}{256}\\textgreater{} \\frac{1}{4}$$ and $$\\left(\\frac{3}{4}\\right)^{5}= \\frac{243}{1024}\\textless{} \\frac{1}{4}$$, this will first occur on day $$5$$ which is $$\\boxed{\\rm Friday}$$. The answer is $$\\rm(D)$$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "350", "queId": "7075cd60d41a469299331e2639419e74", "competition_source_list": ["2013年辽宁全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知集合$$A= {x{}\\textbar{{x}^{2}}-3x-10{\\leqslant }0 }{,}B= {x\\textbar m+1{\\leqslant }x{\\leqslant }2m-1 }$$．当$$A{\\cap }B=\\varnothing $$时，实数$$m$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\textless{}m\\textless{}4$$ "}], [{"aoVal": "B", "content": "$$m\\textless{}2$$或$$m\\textgreater4$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}\\textless{}m\\textless{}4$$ "}], [{"aoVal": "D", "content": "$$m\\textless{}-\\frac{1}{2}$$或$$m\\textgreater4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["$$A=\\left { x\\textbar-2{\\leqslant }x{\\leqslant }5 \\right }$$，$$A{\\cap }B=\\varnothing $$，当$$B=\\varnothing $$时，$$2m-1\\textless{}m+1$$，即$$m\\textless{}2$$;当$$B\\ne \\varnothing $$时，$$m{\\geqslant }2$$，$$m+1\\textgreater5$$，即$$m\\textgreater4$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1103", "queId": "ca9ca81d400c437596e50c8677448a2a", "competition_source_list": ["2008年贵州全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "给定两个向量$$\\overrightarrow{a}=\\left( 1,2 \\right)$$，$$\\overrightarrow{b}=\\left( x,1 \\right)$$，若$$(\\overrightarrow{a}+2\\overrightarrow{b})//(2\\overrightarrow{a}-2\\overrightarrow{b})$$，则$$x$$的值等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算"], "answer_analysis": ["因为$$\\overrightarrow{a}=\\left( 1,2 \\right)$$，$$\\overrightarrow{b}=\\left( x,1 \\right)$$， 所以有$$(\\overrightarrow{a}+2\\overrightarrow{b})=\\left( 1+2x,4 \\right)$$， $$(2\\overrightarrow{a}-2\\overrightarrow{b})=\\left( 2-2x,2 \\right)$$， 因为$$(\\overrightarrow{a}+2\\overrightarrow{b})//(2\\overrightarrow{a}-2\\overrightarrow{b})$$， 所以$$\\left( 1+2x \\right)\\times 2=4\\times \\left( 2-2x \\right)$$， 解得$$x=\\frac{1}{2}$$，故选$$\\text{C}$$. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "143", "queId": "d9b1ad008b2647c3b23659a2dcad10c1", "competition_source_list": ["2009年第二十届全国希望杯高二竞赛复赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知函数$$f(x)$$对一切实数$$a,b$$都满足$$f(a+b)=f(a)+f(b)$$，则不恒为零的函数$$f(x)$$是．", "answer_option_list": [[{"aoVal": "A", "content": "奇函数 "}], [{"aoVal": "B", "content": "偶函数 "}], [{"aoVal": "C", "content": "奇函数也是偶函数 "}], [{"aoVal": "D", "content": "非奇非偶函数 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["令$$a=b=0$$，可得$$f(0)=0$$，令$$a=-b$$，可得$$f(a)=-f(-a)$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "304", "queId": "704f40b106c14fa2beb5a00d64d87994", "competition_source_list": ["2022~2023学年湖南邵阳隆回县湖南省隆回县第二中学高一上学期月考（竞赛）第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$f\\left( x \\right)=\\sqrt{x-2}$，若$f\\left( 2{{a}^{2}}-5a+4 \\right)\\textless{} f\\left( {{a}^{2}}+a+4 \\right)$，则实数$a$的取值范围是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\left( -\\infty ,\\frac{1}{2} \\right)\\cup \\left( 2,+\\infty \\right)$ "}], [{"aoVal": "B", "content": "$\\left[ 2,6 \\right)$ "}], [{"aoVal": "C", "content": "$\\left( 0,\\frac{1}{2} \\right]\\cup \\left[ 2,6 \\right)$ "}], [{"aoVal": "D", "content": "$\\left( 0,6 \\right]$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的性质->单调性"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据$f\\left( x \\right)$的单调性，以及定义域，结合一元二次不等式的求解，直接计算即可.\\\\ 【详解】\\\\ 对$f\\left( x \\right)=\\sqrt{x-2}$，且定义域为$\\left[ 2,+\\infty \\right)$，由复合函数单调性可知其在定义域单调递增，\\\\ 故$f\\left( 2{{a}^{2}}-5a+4 \\right)\\textless{} f\\left( {{a}^{2}}+a+4 \\right)$，等价于$2\\le 2{{a}^{2}}-5a+4\\textless{} {{a}^{2}}+a+4$，\\\\ 由$2\\le 2{{a}^{2}}-5a+4$，即$2{{a}^{2}}-5a+2\\ge 0$，$\\left( 2a-1 \\right)\\left( a-2 \\right)\\ge 0$，解得$a\\in \\left( -\\infty ,\\frac{1}{2} \\right]\\cup \\left[ 2,+\\infty \\right)$；\\\\ 由$2{{a}^{2}}-5a+4\\textless{} {{a}^{2}}+a+4$，即${{a}^{2}}-6a\\textless{} 0$，解得$a\\in \\left( 0,6 \\right)$；\\\\ 故实数$a$的取值范围为$\\left( 0,\\frac{1}{2} \\right]\\cup \\left[ 2,6 \\right)$.\\\\ 故选：C. "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1026", "queId": "d2d050a7a846460fac8d79ff6496df35", "competition_source_list": ["2005年AMC12竞赛B第23题"], "difficulty": "3", "qtype": "single_choice", "problem": "2005AMC12B, 23 Let $$S$$ be the set of ordered triples $$(x, y,z)$$ of real numbers for which $$\\log_{10}(x+y)=z$$ and $$\\log_{10}(x^{2}+y^{2})=z+1$$. There are real numbers $$a$$ and $$b$$ such that for all ordered triples $$(x, y,z)$$ in $$S$$ we have $$x^{3}+y^{3}=a\\cdot10^{3z}+b\\cdot10^{2z}$$. What is the value of $$a+b$$? 令 $$S$$ 为有序实数三元组 $$(x, y,z)$$ 的集合，其中 $$\\log_{10}(x+y)=z$$ 且 $$\\log_{ 10}(x^{2}+y^{2})=z+1$$。 存在实数 $$a$$ 和 $$b$$ ，使得对于 $$S$$ 中的所有有序三元组 $$(x, y,z)$$ ， 有$$x^{3}+y^{} {3}=a\\cdot10^{3z}+b\\cdot10^{2z}$$。 $$a+b$$ 的值是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{15}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{29}{2}$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$\\frac{39}{2}$$ "}], [{"aoVal": "E", "content": "$$24$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Calculation->Exponentiation", "课内体系->知识点->基本初等函数->实数指数幂运算"], "answer_analysis": ["这题主要考代数变形. 由题意, $x+y=10^{}z$, $x^{2}+y^{2}=10^{z+1}$. 由此可得$xy=\\frac{(x+y)^{2}-(x^{2}+y^{2})}{2}=\\frac{1}{2}10^{2z}-5\\times 10^{}z$. 代入$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$, 得 $x^{3}+y^{3}=10^{}z(10\\times 10^{}z-\\frac{1}{2}10^{2z}+5\\times 10^{}z)=-\\frac{1}{2}10^{3z}+15\\times 10^{2z}$. 故$a+b=15-\\frac{1}{2}=\\frac{29}{2}$, 选$$\\text{B}$$. "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "410", "queId": "3a0b481eb63c41a2a5a530aff1ff7fd3", "competition_source_list": ["2000年AMC12竞赛第13题"], "difficulty": "3", "qtype": "single_choice", "problem": "$$2000-AMC12-13$$ One morning each member of Angela's family drank an $$8$$-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family? 一天早上，$$Angela$$ 的每个家庭成员都喝了 8 美元一盎司的咖啡加牛奶混合物。 咖啡和牛奶的量因杯子而异，但从未为零。 安吉拉喝了牛奶总量的四分之一，咖啡总量的六分之一。 家里有多少人？", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->等式->方程组的解集", "美国AMC10/12->Knowledge Point->Algebra->Application->Linear Function Word Problems"], "answer_analysis": ["Let $$c$$ be the total amount of coffee, $$m$$ of milk, and $$p$$ the number of people in the family. Then each person drinks the same total amount of coffee and milk ($$8$$ ounces), so $$\\left(\\frac{c}{6}+\\frac{m}{4} \\right)p=c+m$$. Regrouping, we get $$2c(6-p)=3m(p-4)$$. Since both $$c$$, $$m$$ are positive, it follows that $$6-p$$ and $$p-4$$ are also positive, which is only possible when $$p=5$$ ($$\\text{C}$$). One could notice that (since there are only two components to the mixture) Angela must have more than her \"fair share\" of milk and less then her \"fair share\" of coffee in order to ensure that everyone has $$8$$ ounces. The \"fair share\" is $$\\frac{1}{p}$$. So, $$\\frac{1}{6}\\textless\\frac{1}{p}\\textless\\frac{1}{4}$$ Which requires that $$p$$ be $$p=5$$ ($$\\text{C}$$), since $$p$$ is a whole number. Again, let $$c$$, $$m$$, sand $$p$$ be the total amount of coffee, total amount of milk, and number of people in the family, respectively. Then the total amount that is drunk is $$8p$$, and also $$c+m$$. Thus, $$c+m=8p$$, so $$m=8p-c$$ and $$c=8p-m$$. We also know that the amount Angela drank, which is is $$\\frac{c}{6}+\\frac{m}{4}$$, equal to $$8$$ ounces, thus $$\\frac{c}{6}+\\frac{m}{4}=8$$. Rearranging gives $$24p-c=96$$. Now notice that $$c\\textgreater0$$ (by the problem statement). In addition, $$m\\textgreater0$$ so $$c=8p-m\\textless8p$$. Therefore, $$0\\textless c\\textless8p$$ ,and so We know that so $$24p\\textgreater24p-c\\textgreater16p$$. We know that $$24p-c=96$$, so $$24p\\textgreater96\\textgreater16p$$. From the leftmost inequality, we get $$p\\textgreater4$$, and from the rightmost inequality, we get $$p\\textless6$$. The only possible value of $$p$$ is $$p=5$$ ($$\\text{C}$$). Let $$c$$, $$m$$, and $$p$$ be the total amount of coffee, total amount of milk, and number of people in the family, respectively. $$c$$ and $$m$$ obviously can\\textquotesingle t be $$0$$. We know $$\\frac{c}{6}+\\frac{m}{4}=8$$ or $$3c+2m=96$$ and $$c+m=8p$$ or $$2c+2m=16p$$. Then, $$(2c+2m)+c=16p+c=96$$. Because $$16p$$ and $$96$$ are both divisible by $$16$$, $$c$$ must also be divisible by $$16$$. Let $$c=16k$$. Now, $$3(16k)+2m=2m=48k+2m=96k$$ can\\textquotesingle t be $$0$$, otherwise $$c$$ is $$0$$, and $$k$$ can\\textquotesingle t be $$2$$, otherwise $$m$$ is $$0$$. Therefore $$k$$ must be $$1$$, $$c=16$$ and $$m=24$$. $$c+m=16+24=40=8p$$. Therefore, $$p=5$$ ($$\\text{C}$$). Let $$m$$,$$c$$ be the total amounts of milk and coffee, respectively. In order to know the number of people, we first need to find the total amount of mixture $$t=m+c$$. We are given that $$\\frac{m}{4}+\\frac{c}{6}=8$$. Multiplying the equation by $$4$$ to get $$m+\\frac{2}{3}x=(m+c)-\\frac{1}{3}c=t-\\frac{1}{3}c=32$$. Since $$\\frac{1}{3}m\\textgreater0$$, we have $$t\\textgreater32$$.~ Now multiplying the equation by $$6$$ to get $$\\frac{3}{2}m+c=(m+c)+\\frac{1}{3}m=t+\\frac{1}{3}m=48$$. Since, $$\\frac{1}{3}m\\textgreater0$$, we have $$t\\textless48$$. Thus, $$32\\textless t\\textless48$$. Since $$t$$ is a multiple of $$8$$, the only possible value for in that range is $$40$$. Therefore, there are $$\\frac{40}{8}=5$$ people in Angela\\textquotesingle s family. ($$\\text{C}$$). "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "806", "queId": "5caaa585f0bd47c0b394a7f6ae9abdb5", "competition_source_list": ["2012年浙江全国高中数学联赛竞赛初赛第8题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$\\overrightarrow{i},\\overrightarrow{j}$$分别表示平面直角坐标系$$x, y$$轴上的单位向量，且$$\\left\\textbar{} \\overrightarrow{a}-\\overrightarrow{i} \\right\\textbar+\\left\\textbar{} \\overrightarrow{a}-\\overrightarrow{2j} \\right\\textbar=\\sqrt{5}$$，则$$\\left\\textbar{} \\overrightarrow{a}+\\overrightarrow{2i} \\right\\textbar$$取值范围为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ 2\\sqrt{3}, 3 \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left[ \\frac{6\\sqrt{5}}{5}, 2\\sqrt{2} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left[ \\sqrt{5}, 4 \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left[ \\frac{6\\sqrt{5}}{5}, 3 \\right]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算"], "answer_analysis": ["满足$$\\left\\textbar{} \\overrightarrow{a}-\\overrightarrow{i} \\right\\textbar+\\left\\textbar{} \\overrightarrow{a}-\\overrightarrow{2j} \\right\\textbar=\\sqrt{5}$$的向量$$\\overrightarrow{a}$$的终点在线段$$2x+y-2=0\\left( 0\\leqslant x\\leqslant 1 \\right)$$上，所以$$\\left\\textbar{} \\overrightarrow{a}+\\overrightarrow{2i} \\right\\textbar$$的最大值就是点$$\\left( -2, 0 \\right)$$与$$\\left( 1, 0 \\right)$$距离$$3$$，最小值就是点$$\\left( -2, 0 \\right)$$到线段的距离$$\\frac{6\\sqrt{5}}{5}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1157", "queId": "cb4492840e77422c93c97a9c1e30e8eb", "competition_source_list": ["1989年全国高中数学联赛竞赛一试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "以长方体$$8$$个顶点中的任意$$3$$个为顶点的所有三角形中，锐角三角形的个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["以长方体的$$8$$个顶点中的任意$$3$$个为顶点的三角形共有$$\\text{C}_{8}^{3}$$个，且这些三角形只能是直角三角形或锐角三角形．而以长方体的每个顶点为直角顶点的三角形都是$$6$$个，所以在$$\\text{C}_{8}^{3}$$个三角形中共有$$6\\times 8=48$$（个）直角三角形．从而所求锐角三角形的个数为$$\\text{C}_{8}^{3}-48=8$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "651", "queId": "6459ffe253464bd98df07477e2459e14", "competition_source_list": ["2009年高二竞赛广州市第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$\\frac{\\sin \\left( \\alpha +2\\beta \\right)}{\\sin \\alpha }=3$$，且$$\\beta \\ne \\frac{1}{2}k \\pi, \\alpha +\\beta \\ne n \\pi +\\frac{ \\pi }{2}\\left( n,k\\in Z \\right)$$，则$$\\frac{\\tan \\left( \\alpha +\\beta \\right)}{\\tan \\beta }$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$-2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式的化简和求值", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->三角函数的定义->计算任意角的三角函数值", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式", "课内体系->知识点->三角函数->三角函数的概念->任意角与弧度制->弧度制", "课内体系->知识点->三角函数->三角函数的概念->任意角与弧度制->角的定义和分类->任意角的表示", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->和差角公式化简求值综合运用", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的正弦", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的余弦", "课内体系->方法->构造法", "课内体系->素养->数学运算", "课内体系->思想->转化化归思想"], "answer_analysis": ["$$3=\\frac{\\sin (\\alpha +\\beta +\\beta )}{\\sin (\\alpha +\\beta -\\beta )}=\\frac{\\sin (\\alpha +\\beta )\\cos \\beta +\\cos (\\alpha +\\beta )\\sin \\beta }{\\sin (\\alpha +\\beta )\\cos \\beta -\\cos (\\alpha +\\beta )\\sin \\beta }$$， 分子分母同时除以$$\\cos \\beta \\cos (\\alpha +\\beta )$$， 有$$\\frac{\\tan (\\alpha +\\beta )+\\tan \\beta }{\\tan (\\alpha +\\beta )-\\tan \\beta }=3$$，解得$$\\tan (\\alpha +\\beta )=2\\tan \\beta $$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "207", "queId": "ab64e5de181344b3abae198f4969e2e3", "competition_source_list": ["2004年全国高中数学联赛竞赛一试第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设锐角$$\\theta $$使关于$$x$$的方程$${{x}^{2}}+4x\\cos \\theta +\\cot \\theta =0$$有重根，则$$\\theta $$的弧度数为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{ \\pi }{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{ \\pi }{12}$$或$$\\frac{5 \\pi }{12}$$ "}], [{"aoVal": "C", "content": "$$\\frac{ \\pi }{6}$$或$$\\frac{5 \\pi }{12}$$ "}], [{"aoVal": "D", "content": "$$\\frac{ \\pi }{12}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->解方程（组）"], "answer_analysis": ["由方程有重根，故$$\\frac{1}{4}\\Delta =\\text{4co}{{\\text{s}}^{2}}\\theta -\\cot \\theta =0$$， ∵$$0\\textless{}\\theta \\textless{}\\frac{ \\pi }{2}\\Rightarrow 2\\sin 2\\theta =1$$，$$\\Rightarrow \\theta =\\frac{ \\pi }{12}$$或$$\\frac{5 \\pi }{12}$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "161", "queId": "6b38c80bbb7b4730a0a30d43f1bef4c2", "competition_source_list": ["2017年天津全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "正四棱锥的底面边长为$$2017$$，侧棱长为$$2000$$，则侧棱与底面所成的角与下面哪个角的差的绝对值最小．", "answer_option_list": [[{"aoVal": "A", "content": "$$30{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$40{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$50{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$60{}^{}\\circ $$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["正四棱锥的高为$$\\sqrt{{{2000}^{2}}-\\frac{{{2017}^{2}}}{2}}$$， 所以，侧棱与底面所成的角的正弦值为$$\\frac{\\sqrt{{{2000}^{2}}-\\frac{{{2017}^{2}}}{2}}}{2000}\\approx \\frac{1400}{2000}=0.7$$，比$$\\frac{\\sqrt{2}}{2}$$稍微小点，也就是角度比$$45{}^{}\\circ $$小点，因此是$$40{}^{}\\circ $$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "349", "queId": "39a36593b52a42e7bf47e077d7ad6fa3", "competition_source_list": ["2009年山东全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "若集合$$M= {x-{{\\log }_{\\frac{1}{2}}}(x-1)\\textgreater-1 }$$，$$N= {x\\textbar1\\textless{}{{2}^{x}}\\textless{}4 }$$，则$$M\\cap N$$= ．", "answer_option_list": [[{"aoVal": "A", "content": "$$ {x\\textbar1\\textless{}x\\textless{}3 }$$ "}], [{"aoVal": "B", "content": "$$ {x\\textbar1\\textless{}x\\textless{}2 }$$ "}], [{"aoVal": "C", "content": "$$ {x\\textbar0\\textless{}x\\textless{}3 }$$ "}], [{"aoVal": "D", "content": "$$ {x\\textbar0\\textless{}x\\textless{}2 }$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["$$x\\in M\\Leftrightarrow {{\\log }_{\\frac{1}{2}}}(x-1)\\textgreater-1\\Leftrightarrow \\begin{cases}x-1\\textgreater0 x-1\\textless{}{{\\left( \\frac{1}{2} \\right)}^{-1}}=2 \\end{cases}\\Leftrightarrow 1\\textless{}x\\textless{}3$$；$$x\\in N\\Leftrightarrow 1\\textless{}{{2}^{x}}\\textless{}4\\Leftrightarrow 0\\textless{}x\\textless{}2$$，所以$$M\\cap N= {x\\textbar1\\textless{}x\\textless{}2 }$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "577", "queId": "4d373854e7494284a03978aae5250a58", "competition_source_list": ["2008年甘肃全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$$10\\sin \\left( x+\\frac{ \\pi }{6} \\right)=x$$根的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的应用->函数的零点->函数零点的概念", "课内体系->知识点->函数的应用->函数的零点->零点的个数问题->利用函数图象研究方程根的分布问题（图象与零点综合）", "课内体系->知识点->函数的应用->函数的零点->零点的个数问题->零点、交点、根的等价转化", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦函数的图象和性质", "课内体系->素养->数学抽象", "课内体系->素养->直观想象"], "answer_analysis": ["由正弦函数的性质同时绘出$$y=10\\sin \\left( x+\\frac{ \\pi }{6} \\right)$$和$$y=x$$的图象， 注意到$$3 \\pi -\\frac{ \\pi }{6}\\textless{}10\\textless{}3 \\pi +\\frac{ \\pi }{2}$$， 及$$\\left\\textbar{} -\\frac{ \\pi }{6}-3 \\pi ~\\right\\textbar\\textless{}10\\textless{}\\left\\textbar{} -\\frac{ \\pi }{6}-3 \\pi -\\frac{ \\pi }{2} \\right\\textbar$$， 可知两条曲线有$$7$$个交点．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "446", "queId": "832c2af0944a480e882ae1fd9627c4c7", "competition_source_list": ["2021年吉林全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f(x)=\\begin{cases}{{x}^{2}},x\\textless{}0 -{{x}^{2}},x\\geqslant 0 \\end{cases}$$，若对$$\\forall x\\in ({{t}^{2}}-4,{{t}^{2}})$$，不等式$$f(x+t)\\textless{}4f(x)$$恒成立，则实数$$t$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( \\frac{1-\\sqrt{17}}{2},\\frac{1+\\sqrt{17}}{2} \\right)$$ "}], [{"aoVal": "B", "content": "$$(0,1)$$ "}], [{"aoVal": "C", "content": "$$\\left[ \\frac{1-\\sqrt{17}}{2},\\frac{1+\\sqrt{17}}{2} \\right]$$ "}], [{"aoVal": "D", "content": "$$[0,1]$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->函数的概念与性质->函数的性质->单调性->函数单调性的应用->利用函数单调性解不等式", "课内体系->方法->单调性法"], "answer_analysis": ["函数$$f(x)$$在$$\\mathbf{R}$$上递减，且$$4f(x)=f(2x)$$， 所以$$f(x+t)\\textless{}4f(x)\\Leftrightarrow f(x+t)\\textless{}f(2x)\\Leftrightarrow x+t\\textgreater2x\\Leftrightarrow t\\textgreater x$$， 所以$$t\\geqslant {{t}^{2}}$$，解得$$0\\leqslant t\\leqslant 1$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "154", "queId": "189a1fc41d8d48759875807e1687b678", "competition_source_list": ["2019~2020学年3月湖北武汉洪山区武汉市洪山高级中学（武汉外国语学校光谷分校）高二下学期月考第10题", "2008年全国高中数学联赛竞赛一试第3题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙两人进行乒乓球比赛，约定每局胜者得$$1$$分，负者得$$0$$分，比赛进行到有一人比对方多$$2$$分或打满$$6$$局时停止．设甲在每局中获胜的概率为$$\\frac{2}{3}$$，乙在每局中获胜的概率为$$\\frac{1}{3}$$，且各局胜负相互独立，则比赛停止时已$$4$$局的概率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2}{9}$$ "}], [{"aoVal": "B", "content": "$$\\frac{20}{81}$$ "}], [{"aoVal": "C", "content": "$$\\frac{7}{27}$$ "}], [{"aoVal": "D", "content": "$$\\frac{16}{81}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->概率初步", "课内体系->知识点->统计与概率->概率->事件与概率->事件的独立性->相互独立事件的概率乘法公式", "课内体系->知识点->统计与概率->概率->事件与概率->互斥事件->互斥事件的概率加法公式"], "answer_analysis": ["解法一：设比赛结束时已打局数为随机变量$$\\xi $$ 依题意知，$$\\xi $$的所有可能值为$$2$$，$$4$$，$$6$$， 设每两局比赛为一轮，则该轮结束时比赛停止的概率为$${{\\left( \\frac{2}{3} \\right)}^{2}}+{{\\left( \\frac{1}{3} \\right)}^{2}}=\\frac{5}{9}$$， 若该轮结束时比赛还将继续，则甲、乙在该轮中必是各得一分，此时，该轮比赛结果对下轮比赛是否停止没有影响，从而有 $$P(\\xi =2)=\\frac{5}{9}$$， $$P(\\xi =4)=\\left( \\frac{4}{9} \\right)\\left( \\frac{5}{9} \\right)=\\frac{20}{81}$$， $$P(\\xi =6)={{\\left( \\frac{4}{9} \\right)}^{2}}=\\frac{16}{81}$$， 故选$$\\text{B}$$ 解法二： 依题意知，$$\\xi $$的所有可能值为$$2$$，$$4$$，$$6$$， 令$${{A}_{k}}$$表示甲在第$$k$$局比赛中获胜，则$${{\\bar{A}}_{k}}$$表示乙在第$$k$$局比赛中获胜， 由独立性与互不相容性得 $$P(\\xi =2)=P({{A}_{1}}{{A}_{2}})+P({{\\bar{A}}_{1}}{{\\bar{A}}_{2}})=\\frac{5}{9}$$， $$P(\\xi =4)=P({{A}_{1}}{{\\bar{A}}_{2}}{{A}_{3}}{{A}_{4}})+P({{A}_{1}}{{\\bar{A}}_{2}}{{\\bar{A}}_{3}}{{\\bar{A}}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{A}_{3}}{{A}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{\\bar{A}}_{3}}{{\\bar{A}}_{4}})$$ $$=2\\left[ {{\\left( \\frac{2}{3} \\right)}^{3}}\\left( \\frac{1}{3} \\right)+{{\\left( \\frac{1}{3} \\right)}^{3}}\\left( \\frac{2}{3} \\right) \\right]=\\frac{20}{81}$$， $$P(\\xi =6)=P({{A}_{1}}{{\\bar{A}}_{2}}{{A}_{3}}{{\\bar{A}}_{4}})+P({{A}_{1}}{{\\bar{A}}_{2}}{{\\bar{A}}_{3}}{{A}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{A}_{3}}{{\\bar{A}}_{4}})+P({{\\bar{A}}_{1}}{{A}_{2}}{{\\bar{A}}_{3}}{{A}_{4}})$$ $$=4{{\\left( \\frac{2}{3} \\right)}^{2}}{{\\left( \\frac{1}{3} \\right)}^{2}}=\\frac{16}{81}$$， 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "920", "queId": "f6f9630c58a643abb419b4ad403c1cfe", "competition_source_list": ["2018~2019学年四川成都青羊区成都市树德中学（光华校区）高一下学期期中理科第6题5分", "2008年全国高中数学联赛竞赛一试第1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "函数$$f(x)=\\frac{5-4x+{{x}^{2}}}{2-x}$$在$$(-\\infty ,2)$$上的最小值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的概念", "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->不等式->几个重要的不等式->均值"], "answer_analysis": ["当$$x\\textless{}2$$时，$$2-x\\textgreater0$$，因此$$f(x)=\\frac{1+(4-4x+{{x}^{2}})}{2-x}=\\frac{1}{2-x}+(2-x)\\geqslant 2\\cdot \\sqrt{\\frac{1}{2-x}\\cdot (2-x)}=2$$， 当且仅当$$\\frac{1}{2-x}=2-x$$时上式取等号， 而此方程有解$$x=1\\in (-\\infty ,2)$$，因此$$f(x)$$在$$(-\\infty ,2)$$上的最小值为$$2$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "928", "queId": "a4b6a461670648f6b12bd017a643a777", "competition_source_list": ["竞赛第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "随机投掷三颗骰子，下列说法中正确的是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "有两颗骰子之和为7的概率是$\\frac{7}{12}$ "}], [{"aoVal": "B", "content": "有两颗骰子之和为8的概率是$\\frac{19}{54}$ "}], [{"aoVal": "C", "content": "所有骰子中最小值为2的概率是$\\frac{65}{216}$ "}], [{"aoVal": "D", "content": "所有骰子中最小值为3的概率是$\\frac{41}{216}$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 利用古典概率的概率公式可得各选项中的概率.\\\\ 【详解】\\\\ 有两颗骰子之和为7的可能为$1+6,2+5,3+4$，对应的基本事件数为$(4\\times 3!+2\\times 3)\\times 3=90$，\\\\ 因此所求概率为$\\frac{90}{6^{3}}=\\frac{5}{12}$．\\\\ 有两颗骰子之和为8的可能为$2+6,3+5,4+4$，对应的基本事件数为$(4\\times 3!+2\\times 3)\\times 2+(5\\times 3+1)=76$，\\\\ 因此所求概率为$\\frac{76}{6^{3}}=\\frac{19}{54}$．\\\\ 所有骰子中最小值为2，则按2出现次数分类，对应的基本事件数为$(6\\times 3!+4\\times 3)+4\\times 3+1=61$，\\\\ 因此所求概率为$\\frac{61}{6^{3}}=\\frac{61}{216}$．\\\\ 所有骰子中最小值为3，则按3出现次数分类，对应的基本事件数为$(3\\times 3!+3\\times 3)+3\\times 3+1=37$，\\\\ 因此所求概率为$\\frac{37}{6^{3}}=\\frac{37}{216}$．\\\\ 故选：B "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "682", "queId": "88da430137e44e7983d19c9fa0547fbf", "competition_source_list": ["2002年AMC12竞赛A第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2002-AMC12A-10$$ Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream? 莎拉将 4 盎司咖啡放入一个 8 盎司的杯子中，将 4 盎司奶油放入第二个相同大小的杯子中。 然后，她将第一杯咖啡的一半倒入第二杯，充分搅拌后，将第二杯中的一半液体倒回第一杯。 第一杯里的液体有多少是奶油？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{5}$$ "}], [{"aoVal": "E", "content": "$$\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Application->Proportion Word Problems", "课内体系->知识点->集合->容斥原理"], "answer_analysis": ["她将第一杯咖啡的一半倒入第二杯，充分搅拌后，第二杯中有$$2$$盎司咖啡，$$4$$盎司奶油。将第二杯中的一半液体倒回第一杯，第一杯中的奶油有$$2$$盎司。 第一杯里的液体有$$5$$盎司，有2/5是奶油 "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "4", "queId": "0072b1acdd2d4679b762bd6ded9b7bf4", "competition_source_list": ["1989年全国高中数学联赛竞赛一试第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个正数，若其小数部分、整数部分和其自身成等比数列，则该数为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{-1+\\sqrt{5}}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{-1+\\sqrt{3}}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1+\\sqrt{5}}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1+\\sqrt{3}}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["设该数为$$x$$，则其整数部分为$$[x]$$，小数部分为$$x-[x]$$，由已知得 $$x(x-[x])={{[x]}^{2}}$$， 其中$$[x]\\textgreater0$$，$$0\\textless{}x-[x]\\textless{}1$$，解得 $$x=\\frac{1+\\sqrt{5}}{2}[x]$$． 由$$0\\textless{}x-[x]\\textless{}1$$知， $$0\\textless{}\\frac{\\sqrt{5}-1}{2}[x]\\textless{}1$$，$$0\\textless{}[x]\\textless{}\\frac{1+\\sqrt{5}}{2}\\textless{}2$$， 即$$[x]=1$$，$$x=\\frac{1+\\sqrt{5}}{2}$$． 故答案为：$$\\frac{1+\\sqrt{5}}{2}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "12", "queId": "0179735dd1914d0b91a7bc2e6ae0a77c", "competition_source_list": ["全国全国高中数学联赛竞赛一试"], "difficulty": "2", "qtype": "single_choice", "problem": "已知三个平面$$\\alpha $$、$$\\beta $$、$$\\lambda $$，每两个平面之间的夹角都是$$\\theta $$，且$$\\alpha \\cap \\beta =a$$，$$\\beta \\cap \\gamma =b$$，$$\\gamma \\cap \\alpha =c$$．若有： 命题甲：$$\\theta \\textgreater\\frac{ \\pi }{3}$$． 命题乙：$$a$$，$$b$$，$$c$$相交于一点． 则．", "answer_option_list": [[{"aoVal": "A", "content": "甲是乙的充分非必要条件 "}], [{"aoVal": "B", "content": "甲是乙的必要非充分条件 "}], [{"aoVal": "C", "content": "甲是乙的充分必要条件 "}], [{"aoVal": "D", "content": "甲是乙的非充分非必要条件 "}]], "knowledge_point_routes": ["课内体系->知识点->立体几何初步->基本图形位置关系->空间中的基本事实与定理", "课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与立体几何结合", "课内体系->素养->逻辑推理"], "answer_analysis": ["设$$a$$，$$b$$，$$c$$交于一点， 由所成三面角内部一点引三条射线分别垂直于$$\\alpha $$、$$\\beta $$、$$\\lambda $$， 其中每两条射线所成的角都是$$ \\pi -\\varphi $$， $$\\varphi $$为三面角中两两相等的二面角的平面角， 总和$$3( \\pi -\\varphi )\\textless{}2 \\pi $$， ∴$$\\varphi \\textgreater\\frac{1}{3} \\pi $$． 当$$\\varphi \\textgreater\\frac{2}{3} \\pi $$时，$$\\varphi \\textgreater\\frac{1}{3} \\pi $$． 若$$a$$，$$b$$，$$c$$不交于一点，则互相平行， 这时$$\\varphi \\textgreater\\frac{1}{3} \\pi $$． 故选$$\\text{A}$$． （注：认定两平面间夹角范围是$$0\\textless{}\\theta {\\leqslant }\\frac{1}{2} \\pi $$） "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "353", "queId": "87897ab21073454596d556d1857e5eb0", "competition_source_list": ["2019年吉林全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "设集合$$A=\\left { 2,0,1,3 \\right }$$，$$B=\\left { x\\left\\textbar{} -x\\in A,2-{{x}^{2}}\\notin A \\right. \\right }$$，则集合$$B$$中所有元素的积为．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$-5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$-7$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算", "课内体系->知识点->集合->集合的概念与表示方法->集合的含义、元素与集合->集合的概念"], "answer_analysis": ["易知，$$B=\\left { -2,-3 \\right }$$， 则集合$$B$$中所有元素的积为$$6$$． 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1084", "queId": "fc88375282354841b50f2beb347a866b", "competition_source_list": ["1999年全国全国高中数学联赛竞赛一试第6题6分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知点$$A\\left( 1,2 \\right)$$，过点$$\\left( 5,-2 \\right)$$的直线与抛物线$${{y}^{2}}=4x$$交于另外两点$$B,C$$，那么，$$\\triangle ABC$$是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "锐角三角形 "}], [{"aoVal": "B", "content": "钝角三角形 "}], [{"aoVal": "C", "content": "直角三角形 "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->抛物线", "竞赛->知识点->解析几何->直线与圆锥曲线"], "answer_analysis": ["设$$B\\left( {{t}^{2}},2t \\right)$$，$$C\\left( {{s}^{2}},2s \\right)$$，$$s\\ne t$$，$$s\\ne 1$$，$$t\\ne 1$$， 则直线$$BC$$的方程为$$y-2s=\\frac{2s-2t}{{{s}^{2}}-{{t}^{2}}}\\left( x-{{s}^{2}} \\right)$$， 化简得$$2x-\\left( s+t \\right)y+2st=0$$． 由于直线$$BC$$过点$$\\left( 5,-2 \\right)$$， 故$$2\\times 5-\\left( s+t \\right)\\left( -2 \\right)+2st=0$$， 即$$\\left( s+1 \\right)\\left( t+1 \\right)=-4$$． 因此，$${{k}_{AB}}{{k}_{AC}}=\\frac{4}{\\left( t+1 \\right)\\left( s+1 \\right)}=-1$$， 所以，$$\\angle BAC=90{}^{}\\circ $$， 从而$$\\triangle ABC$$是直角三角形． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "791", "queId": "a87aae7f87854c088b4cd7c8033b6d30", "competition_source_list": ["2008年浙江全国高中数学联赛竞赛初赛第3题6分", "2019~2020学年辽宁高一上学期期中（辽南协作体）第11题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$f(x)$$在$$[0,1]$$上有定义，要使函数$$f(x-a)+f(x+a)$$有定义，则$$a$$的取值范围为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( -\\infty ,-\\frac{1}{2} \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left[ -\\frac{1}{2},\\frac{1}{2} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{2},+\\infty \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( -\\infty ,-\\frac{1}{2} \\right]\\cup \\left[ \\frac{1}{2},+\\infty \\right)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->不等式的性质", "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->函数的概念"], "answer_analysis": ["函数$$f(x-a)+f(x+a)$$的定义域为$$[a,1+a]\\cap [-a,1-a]$$． 当$$a\\geqslant 0$$时，应有$$a\\leqslant 1-a$$，即$$a\\leqslant \\frac{1}{2}$$； 当$$a\\leqslant 0$$时，应有$$-a\\leqslant 1+a$$，即$$a\\geqslant -\\frac{1}{2}$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "313", "queId": "1af001605a7a41b7bf5f56dee30ec628", "competition_source_list": ["2009年AMC10竞赛A第20题"], "difficulty": "3", "qtype": "single_choice", "problem": "$$2009-AMC10A-20$$ Andrea and Lauren are $$20$$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of~ $$1$$ kilometer per minute. After $$5$$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea? 安德里亚和劳伦相距 20 公里。 他们骑自行车相向而行，$$Andrea$$ 的速度是 Lauren 的三倍，而且他们之间的距离以每分钟 1 公里的速度减少。 5 分钟后，安德里亚因为轮胎漏气而停止骑自行车并等待劳伦。 在他们开始骑自行车几分钟后，劳伦与安德里亚相遇？", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$55$$ "}], [{"aoVal": "D", "content": "$$65$$ "}], [{"aoVal": "E", "content": "$$80$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->等式->方程组的解集", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"], "answer_analysis": ["Let their speeds in kilometers per hour be $$v_A$$ and $$v_L$$. We know that $$vA=3vL$$ and that $$v_A+v_L=60$$. (The second equation follows from the fact that $$1\\rm km/min=60\\rm km/h)$$. This solves to $$v_A=45$$ and $$v_L=15$$. As the distance decreases at a rate of $$1$$ kilometer per minute, after $$5$$ minutes the distance between them will be $$20-5=15$$ kilometers. From this point on, only~ Lauren will be riding her bike. As there are $$15$$ kilometers remaining and $$vL=15$$, she will need exactly an hour to get to Andrea. Therefore the total time in minutes is $$5+60= 65$$. Because the speed of Andrea is $$3$$ times as fast as Lauren and the distance between them is decreasing at a rate of $$1$$ kilometer per minute, Andrea\\textquotesingle s speed is $$\\dfrac{3}{4}\\rm km/min$$,and Lauen\\textquotesingle s $$\\dfrac{1}{4}\\rm km/min$$. Therefore, after $$5$$ minutes, Andrea will have biked $$\\dfrac{3}{4}\\cdot 5= \\dfrac{15}{4}\\rm km$$. In all, Lauren will have to bike $$20- \\dfrac{15}{4}= \\dfrac{80}{4}- \\dfrac{15}{4}= \\dfrac{65}{4}\\rm km$$. Because her speed is$$\\dfrac{1}{4}\\rm km/\\min$$, the time elapsed will be$$\\dfrac{65}{\\dfrac{4}{\\dfrac{1}{4}}}=\\left( \\rm D\\right)65$$. "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1010", "queId": "a0ed9c5484a74687b0b18aa595cddce7", "competition_source_list": ["2008年四川全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知椭圆$$\\frac{{{x}^{2}}}{4}+\\frac{{{y}^{2}}}{3}=1$$的左顶点为$${{A}_{1}}$$，右焦点为$${{F}_{2}}$$，点$$P$$为该椭圆上一动点，则当$$\\overrightarrow{P{{A}_{1}}}\\cdot \\overrightarrow{P{{F}_{2}}}$$取最小值时，$$\\textbar\\overrightarrow{P{{A}_{1}}}+\\overrightarrow{P{{F}_{2}}}\\textbar$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$2\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{13}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->椭圆"], "answer_analysis": ["由条件知$$a=2$$，$$b=\\sqrt{3}$$，则$$c=1$$， 故$${{A}_{1}}(-2,0)$$���$${{F}_{2}}\\left( 1,0 \\right)$$．设$$P(xy)$$， 则$$\\overrightarrow{P{{A}_{1}}}=(-2-x,-y)$$，$$\\overrightarrow{P{{F}_{2}}}=(1-x,-y)$$． 于是$$\\overrightarrow{P{{A}_{1}}}\\cdot \\overrightarrow{P{{F}_{2}}}=(x+2)(x-1)+{{y}^{2}}$$ $$={{x}^{2}}+x-2+3\\left( 1-\\frac{{{x}^{2}}}{4} \\right)$$ $$=\\frac{1}{4}{{x}^{2}}+x+1=\\frac{1}{4}{{(x+2)}^{2}}\\geqslant 0$$， 当$$x=-2$$时等号成立． 即当$$P$$与$${{A}_{1}}$$重合时，$$\\overrightarrow{P{{A}_{1}}}\\cdot \\overrightarrow{P{{F}_{2}}}$$取最小值， 此时$$\\textbar\\overrightarrow{P{{A}_{1}}}+\\overrightarrow{P{{F}_{2}}}\\textbar=\\textbar\\overrightarrow{{{A}_{1}}{{F}_{2}}}\\textbar=3$$． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "714", "queId": "c84ebb904af84e89b11282c668bbd9c7", "competition_source_list": ["2014年吉林全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列函数既是奇函数，又在区间$$\\left[ -1, 1 \\right]$$上单调递减的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$f\\left( x \\right)=\\sin x$$ "}], [{"aoVal": "B", "content": "$$f\\left( x \\right)=-\\left\\textbar{} x+1 \\right\\textbar$$ "}], [{"aoVal": "C", "content": "$$f\\left( x \\right)=\\ln \\frac{2-x}{2+x}$$ "}], [{"aoVal": "D", "content": "$$f\\left( x \\right)=\\frac{1}{2}\\left( {{a}^{x}}+{{a}^{-x}} \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["设$$f\\left( x \\right)=\\ln \\frac{2-x}{2+x}$$， 则$$f\\left( -x \\right)=\\ln \\frac{2+x}{2-x}=-\\ln \\frac{2-x}{2+x}=-f\\left( x \\right)$$， 因此，$$f\\left( x \\right)=\\ln \\frac{2-x}{2+x}$$是奇函数． 又$$t=\\frac{2-x}{2+x}=-1+\\frac{4}{2+x}$$为区间$$\\left[ -1, 1 \\right]$$上的单调递减函数，$$y=\\ln t$$为区间$$\\left( 0, +\\infty \\right)$$上的单调递增函数，而$$f\\left( x \\right)=\\ln \\frac{2-x}{2+x}$$为$$y=\\ln t$$与$$t=\\frac{2-x}{2+x}$$的复合函数，因此函数$$f\\left( x \\right)=\\ln \\frac{2-x}{2+x}$$在区间$$\\left[ -1, 1 \\right]$$上单调递减． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "408", "queId": "87ad8aac924a4e20a90e13249d40fbfb", "competition_source_list": ["2008年浙江全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "圆锥的轴截面$$SAB$$是边长为$$2$$的正三角形，$$O$$为底面的中心，$$M$$为$$SO$$的中点，动点$$P$$在圆锥的底面内（包括圆周），若$$AM\\bot MP$$，则点$$P$$形成的轨迹长度为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{7}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{7}}{2}$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->立体几何初步->基本图形位置关系->探索性问题->截面的探索问题", "课内体系->知识点->立体几何初步->基本立体图形->空间几何体的概念->圆柱、圆锥、圆台和球的结构特征", "课内体系->素养->数学运算", "课内体系->素养->直观想象"], "answer_analysis": ["略． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "727", "queId": "a83ff6b7e6c847eb92c546c4f233b081", "competition_source_list": ["2008年福建全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知一个正三棱柱的底面边长为$$1$$，且两个侧面的异面对角线互相垂直．则它的侧棱长为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{2}}{2}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{3}}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["设$$ABC-{{A}_{1}}{{B}_{1}}{{C}_{1}}$$是正三棱柱，侧棱的长为$$a$$， 侧面对角线$$A{{B}_{1}}$$与$$B{{C}_{1}}$$互相垂直．则$$\\overrightarrow{{{B}_{1}}A}\\cdot \\overrightarrow{B{{C}_{1}}}=0$$， 即$$(\\overrightarrow{{{B}_{1}}B}+\\overrightarrow{BA})\\cdot (\\overrightarrow{B{{B}_{1}}}+\\overrightarrow{{{B}_{1}}{{C}_{1}}})=0$$， 所以$$\\overrightarrow{{{B}_{1}}B}\\cdot \\overrightarrow{B{{B}_{1}}}+\\overrightarrow{{{B}_{1}}B}\\cdot \\overrightarrow{{{B}_{1}}{{C}_{1}}}+\\overrightarrow{BA}\\cdot \\overrightarrow{B{{B}_{1}}}+\\overrightarrow{BA}\\cdot \\overrightarrow{{{B}_{1}}{{C}_{1}}}=0$$， 即$$-{{a}^{2}}+0+0+\\cos 60{}^{}\\circ =0$$． 可得$$a=\\frac{\\sqrt{2}}{2}$$．故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1183", "queId": "f4a59369699743a6a9a637f198890f8b", "competition_source_list": ["2008年山东全国高中数学联赛竞赛初赛第9题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "向量$$\\overrightarrow{a}=(1,-1)$$，$$\\overrightarrow{b}=(-1,2)$$，则$$\\left( 2a+b \\right)\\cdot a=$$( )", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量的加法运算及运算规则", "课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积运算（非坐标）", "课内体系->知识点->平面向量->平面向量的运算->数量积->数量积的坐标表达式", "课内体系->知识点->平面向量->平面向量的运算->数量积->线性运算和数量积综合问题", "课内体系->知识点->平面向量->平面向量的运算->数量积->向量的数量积的定义"], "answer_analysis": ["$$a=\\left( 1,-1 \\right)$$，$$b=\\left( -1,2 \\right)$$，$$\\therefore \\left( 2a+b \\right)\\cdot a=\\left( 1,0 \\right)\\cdot \\left( 1,-1 \\right)=1$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "659", "queId": "7b0b30cce2a642f2b98d5d3e5c40daee", "competition_source_list": ["2020年北京海淀区北京大学自主招生（强基计划）第2题5分", "2020~2021学年北京高三单元测试", "2020~2021学年北京高二单元测试", "竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在$${{\\left( 2019\\times 2020 \\right)}^{2021}}$$的全体正因数中选出若干个，使得其中任意两个的乘积都不是平方数，则最多可选因数个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "前三个答案都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->整除->带余除法（辗转相除法）"], "answer_analysis": ["将各正因数的幂指数记为数组$$\\left( {{\\alpha }_{1i}},{{\\alpha }_{2i}},{{\\alpha }_{3i}},{{\\alpha }_{4i}},{{\\alpha }_{5i}} \\right)$$ 即任取正因数$${{a}_{i}}$$，有$${{a}_{i}}={{2}^{{{\\alpha }_{1i}}}}\\times {{3}^{{{\\alpha }_{2i}}}}\\times {{5}^{{{\\alpha }_{3i}}}}\\times {{101}^{{{\\alpha }_{4i}}}}\\times {{673}^{{{\\alpha }_{5i}}}}$$（质数分解唯一定理）， 只需考虑幂指数的奇偶性，故设$${{b}_{i}}=\\left( {{\\beta }_{1i}},{{\\beta }_{2i}},{{\\beta }_{3i}},{{\\beta }_{4i}},{{\\beta }_{5i}} \\right)$$，其中$$\\beta \\in \\left { 0,1 \\right }$$且$${{\\beta }_{ji}}\\equiv {{\\alpha }_{ji}}\\left( \\bmod 2 \\right)$$， 若两正因数$${{a}_{i}}$$，$${{a}_{j}}$$为完全平方数， $$\\because {{a}_{i}}\\times {{a}_{j}}={{2}^{{{\\alpha }_{1i}}+\\alpha 1j}}\\times {{3}^{{{\\alpha }_{2i}}\\times {{\\alpha }_{2j}}}}\\times {{5}^{{{\\alpha }_{3i}}+{{\\alpha }_{3j}}}}\\times {{101}^{{{\\alpha }_{4i}}+{{\\alpha }_{4j}}}}\\times {{673}^{{{\\alpha }_{3i}}+{{d}_{5j}}}}$$， $$\\therefore $$$$2\\left\\textbar{} {{\\alpha }_{ki}}+{{\\alpha }_{kj}} \\right.$$，$$k=1$$，$$2$$，$$3$$，$$4$$，$$5$$ $$\\Leftrightarrow 2\\left\\textbar{} {{\\beta }_{ki}}+{{\\beta }_{kj}} \\right.$$，$$k=1$$，$$2$$，$$3$$，$$4$$，$$5$$， $$\\because \\beta \\in \\left { 0,1 \\right }$$， $$\\therefore $$必有$${{\\beta }_{ki}}={{\\beta }_{kj}}$$， $$\\therefore $$当且仅当$${{\\beta }_{ki}}\\pm {{\\beta }_{kj}}\\left( k=1,2,3,4,5 \\right)$$，即任意两正因数，其各质因数的幂指数奇偶性均不同时，才有$${{a}_{i}}\\times {{a}_{j}}$$不是完全平方数， $$\\because {{b}_{i}}$$为$$5$$组数组，$${{\\beta }_{i}}\\in \\left { 0,1 \\right }$$， $$\\therefore $$共有$${{2}^{5}}$$种，选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "859", "queId": "f222d4dd12ad43079b00215e5bff23f9", "competition_source_list": ["2023年江苏连云港灌南县灌南县中学高三竞赛（下学期3月解题能力竞赛）第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知椭圆$C$：$\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1\\left(a\\textgreater{} b\\textgreater{} 0\\right)$的左、右焦点为$F_{1}$，$F_{2}$，点$A\\left(-2,2\\right)$为椭圆$C$内一点，点$Q\\left(a,b\\right)$在双曲线$E$：$\\frac{x^{2}}{4}-\\frac{y^{2}}{4}=1$上，若椭圆上存在一点$P$，使得$\\left\\textbar{} PA\\right\\textbar{} +\\left\\textbar{} PF_{2}\\right\\textbar{} =8$，则$a$的取值范围是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\left(\\sqrt{5}+1,5\\right]$ "}], [{"aoVal": "B", "content": "$\\left[3,5\\right]$ "}], [{"aoVal": "C", "content": "$\\left(\\sqrt{5}+1,2\\sqrt{5}\\right]$ "}], [{"aoVal": "D", "content": "$\\left[\\sqrt{3},\\sqrt{5}\\right]$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 先求出椭圆左焦点$F_{1}$坐标为$(-2,0)$，由题得$\\textbar{} \\left\\textbar{} PA\\right\\textbar{} -\\left\\textbar{} PF_{1}\\right\\textbar{} \\textbar{} =\\textbar{} 8-2a\\textbar{} \\leq \\textbar{} AF_{1}\\textbar{} =2$，解不等式得到$3\\leq a\\leq 5$，再解不等式$\\frac{4}{a^{2}}+\\frac{4}{4-a^{2}}\\textless{} 1$即得解.\\\\ 【详解】\\\\ 点$Q\\left(a,b\\right)$在双曲线$E$：$\\frac{x^{2}}{4}-\\frac{y^{2}}{4}=1$上，所以$a^{2}-b^{2}=4$.\\\\ 所以椭圆左焦点$F_{1}$坐标为$(-2,0)$.\\\\ 因为$\\left\\textbar{} PA\\right\\textbar{} +\\left\\textbar{} PF_{2}\\right\\textbar{} =8$，所以$\\left\\textbar{} PA\\right\\textbar{} +2a-\\left\\textbar{} PF_{1}\\right\\textbar{} =8,\\therefore \\textbar{} \\left\\textbar{} PA\\right\\textbar{} -\\left\\textbar{} PF_{1}\\right\\textbar{} \\textbar{} =\\textbar{} 8-2a\\textbar{} \\leq \\textbar{} AF_{1}\\textbar{} =2$,\\\\ 所以$3\\leq a\\leq 5$.\\\\ 因为$a^{2}-b^{2}=4$，所以$b^{2}=a^{2}-4$.\\\\ 点$A\\left(-2,2\\right)$为椭圆$C$内一点，所以$\\frac{4}{a^{2}}+\\frac{4}{b^{2}}\\textless{} 1,\\therefore \\frac{4}{a^{2}}+\\frac{4}{a^{2}-4}\\textless{} 1$，\\\\ 所以$a^{4}-12a^{2}+16\\textgreater{} 0,\\therefore a\\textgreater{} \\sqrt{5}+1$或$a\\textless{} \\sqrt{5}-1$.\\\\ 综上：$\\sqrt{5}+1\\textless{} a\\leq 5$.\\\\ 故选：A "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "468", "queId": "6c45861e406149e28eb1d32091a60c67", "competition_source_list": ["2010年河南全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$${{F}_{1}}, {{F}_{2}}$$是双曲线$${{x}^{2}}-{{y}^{2}}=4$$的两个焦点，$$P$$是双曲线上任意一点，从$${{F}_{1}}$$引$$\\angle {{F}_{1}}P{{F}_{2}}$$平分线的垂线，垂足为$$M$$，则点$$M$$的轨迹方程是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{x}^{2}}+{{y}^{2}}=1$$ "}], [{"aoVal": "B", "content": "$${{x}^{2}}+{{y}^{2}}=2$$ "}], [{"aoVal": "C", "content": "$${{x}^{2}}+{{y}^{2}}=4$$ "}], [{"aoVal": "D", "content": "$${{x}^{2}}+{{y}^{2}}=8$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->曲线与方程", "竞赛->知识点->解析几何->双曲线"], "answer_analysis": ["延长$${{F}_{1}}M$$交$$P{{F}_{2}}$$（或$$P{{F}_{2}}$$的延长线）于点$$N$$，则$$\\triangle P{{F}_{1}}N$$是等腰三角形，$$M$$是底边$${{F}_{1}}N$$的中点，连结$$OM$$，$$OM=\\frac{1}{2}{{F}_{2}}N=\\frac{1}{2}\\left\\textbar{} P{{F}_{1}}-P{{F}_{2}} \\right\\textbar=2$$，故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "351", "queId": "4bc05d1eaef143fe9cffb2fd4ac24e09", "competition_source_list": ["1986年全国高中数学联赛竞赛一试第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "设$$Z$$为复数，$$M= {Z\\textbar{{\\left( Z-1 \\right)}^{2}}=\\textbar Z-1{{\\textbar}^{2}} }$$，那么（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$M=$${纯虚数} "}], [{"aoVal": "B", "content": "$$M=$${实数} "}], [{"aoVal": "C", "content": "{实数}$$\\subset M\\subset $${复数} "}], [{"aoVal": "D", "content": "$$M=$${复数} "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["∵$${{(Z-1)}^{2}}=\\textbar Z-1{{\\textbar}^{2}}$$， 即$${{(Z-1)}^{2}}=(Z-1)(\\overline{Z}\\mathsf{-1})$$， ∴$$(Z-1)(\\overline{Z}\\mathsf{-1})=\\mathsf{0}$$， 因此$$Z=1$$或$$Z=\\overline{Z}$$， 即$$Z$$为实数． 故选$$\\text{B}$$． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "795", "queId": "ad0d365e1ad5478dba4211174f1f0f25", "competition_source_list": ["2012年黑龙江全国高中数学联赛竞赛初赛第1题5分", "2011年高考真题湖北卷理科第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\text{i}$$为虚数单位，则$${{\\left( \\frac{1+\\text{i}}{1-\\text{i}}\\right)}^{2011}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\text{i}$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$\\text{i}$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["知识标签->知识点->复数->复数的运算->复数的乘法和除法", "知识标签->素养->数学运算", "知识标签->题型->复数->复数的运算->复数中的周期问题"], "answer_analysis": ["因为$$\\frac{1+\\text{i}}{1-\\text{i}}=\\frac{{{\\left( 1+\\text{i} \\right)}^{2}}}{1-{{\\text{i}}^{2}}}=\\text{i}$$，所以$${{\\left(\\frac{1+\\text{i}}{1-\\text{i}} \\right)}^{2011}}={{\\text{i}}^{2011}}={{\\text{i}}^{4\\times502+3}}={{\\text{i}}^{3}}=-\\text{i}$$， 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "137", "queId": "4625184e0f894baab4c6ed7494cf98bb", "competition_source_list": ["2008年福建全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$$\\left\\textbar{} {{x}^{2}}-3x+2 \\right\\textbar+\\left\\textbar{} {{x}^{2}}+2x-3 \\right\\textbar=11$$实数解的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->解方程（组）"], "answer_analysis": ["原方程可以转化为$$\\left\\textbar{} x-1 \\right\\textbar(\\left\\textbar{} x-2 \\right\\textbar+\\left\\textbar{} x+3 \\right\\textbar)=11$$． 当$$x\\leqslant -3$$时，方程转化为$$2{{x}^{2}}-x-12=0$$， 它的两个根都大于$$-3$$，此时无解； 当$$-3\\textless{}x\\leqslant 1$$时，方程转化为$$-5x+5=11$$， 所以$$x=-\\frac{6}{5}$$，满足题意； 当$$1x\\leqslant 2$$时，方程转化为$$5x-5=11$$， 所以$$x=\\frac{16}{5}\\textgreater2$$，此时无解； 当$$x\\textgreater2$$时，方程转化为$$2{{x}^{2}}-x-12=0$$， 其中一个根$$x=\\frac{1+\\sqrt{97}}{4}$$满足题意． 综上所述，满足方程的实数解有$$2$$个．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1092", "queId": "af756b2b80a84da1b652758ae71115c0", "competition_source_list": ["2008年山西全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "不等边$$\\triangle ABC$$中，角$$B=60{}^{}\\circ $$，则直线$${{l}_{1}}$$：$$x+\\frac{\\sin A+\\sin C}{\\sqrt{3}}y+1=0$$与直线$${{l}_{2}}$$：$$x\\cdot \\tan (60{}^{}\\circ -C)+y\\cdot (\\sin A-\\sin C)-\\tan \\frac{C-A}{2}=0$$的位置关系是．", "answer_option_list": [[{"aoVal": "A", "content": "垂直 "}], [{"aoVal": "B", "content": "平行 "}], [{"aoVal": "C", "content": "重合 "}], [{"aoVal": "D", "content": "其他情况 "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->解析几何->直线与方程"], "answer_analysis": ["由于$$A+C=120{}^{}\\circ $$，则 $$\\tan (60{}^{}\\circ -C)=\\tan \\left( \\frac{A+C}{2}-C \\right)=\\tan \\frac{A-C}{2}$$， $$\\sin A-\\sin C=2\\sin \\frac{A-C}{2}\\cos \\frac{A+C}{2}=\\sin \\frac{A-C}{2}$$， $$\\sin A+\\sin C=2\\sin \\frac{A+C}{2}\\cos \\frac{A-C}{2}=\\sqrt{3}\\cos \\frac{A-C}{2}$$， 故直线$${{l}_{1}}$$、$${{l}_{2}}$$的方程皆可化为$$x+y\\cos \\frac{A-C}{2}+1=0$$， 因此两直线重合．故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1097", "queId": "e116cbcbc794479b84dbf0267b8a886c", "competition_source_list": ["2005年全国高中数学联赛竞赛一试第5题6分", "2019~2020学年上海浦东新区华东师范大学第二附属中学高一下学期单元测试《任意角的三角比》（实验班）第7题", "2019~2020学年上海浦东新区华东师范大学第二附属中学高二上学期单元测试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$$\\frac{{{x}^{2}}}{\\sin \\sqrt{2}-\\sin \\sqrt{3}}+\\frac{{{y}^{2}}}{\\cos \\sqrt{2}-\\cos \\sqrt{3}}=1$$表示的曲线是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "焦点在$$x$$轴上的椭圆 "}], [{"aoVal": "B", "content": "焦点在$$x$$轴上的双曲线 "}], [{"aoVal": "C", "content": "焦点在$$y$$轴上的椭圆 "}], [{"aoVal": "D", "content": "焦点在$$y$$轴上的双曲线 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->轨迹方程（二试）"], "answer_analysis": ["∵$$\\sqrt{2}+\\sqrt{3}\\textgreater{} \\pi $$，$$\\therefore 0\\textless{}\\frac{ \\pi }{2}-\\sqrt{2}\\textless{}\\sqrt{3}-\\frac{ \\pi }{2}\\textless{}\\frac{ \\pi }{2}$$，$$\\therefore \\cos \\left( \\frac{ \\pi }{2}-\\sqrt{2} \\right)\\textgreater\\cos \\left( \\sqrt{3}-\\frac{ \\pi }{2} \\right)$$ 即$$\\sin \\sqrt{2}\\textgreater\\sin \\sqrt{3}$$． 又$$0\\textless{}\\sqrt{2}\\textless{}\\frac{ \\pi }{2}$$，$$\\frac{ \\pi }{2}\\textless{}\\sqrt{3}\\textless{} \\pi $$， $$\\therefore \\cos \\sqrt{2}\\textgreater0$$，$$\\cos \\sqrt{3}\\textless{}0$$，$$\\therefore \\cos \\sqrt{2}-\\cos \\sqrt{3}\\textgreater0$$， 方程表示的曲线是椭圆． ∵$$(\\sin \\sqrt{2}-\\sin \\sqrt{3})-(\\cos \\sqrt{2}-\\cos \\sqrt{3})$$ $$=2\\sqrt{2}\\sin \\frac{\\sqrt{2}-\\sqrt{3}}{2}\\sin \\left( \\frac{\\sqrt{2}+\\sqrt{3}}{2}+\\frac{ \\pi }{4} \\right)$$\\ldots\\ldots$$(*)$$ $$-\\frac{ \\pi }{2}\\textless{}\\frac{\\sqrt{2}-\\sqrt{3}}{2}\\textless{}0$$，$$\\therefore \\sin \\frac{\\sqrt{2}-\\sqrt{3}}{2}\\textless{}0$$，$$\\frac{ \\pi }{2}\\textless{}\\frac{\\sqrt{2}+\\sqrt{3}}{2}\\textless{}\\frac{3 \\pi }{4}$$，$$\\therefore \\frac{3 \\pi }{4}\\textless{}\\frac{\\sqrt{2}+\\sqrt{3}}{2}+\\frac{ \\pi }{4}\\textless{} \\pi $$． $$\\therefore \\sin \\left( \\frac{\\sqrt{2}+\\sqrt{3}}{2}+\\frac{ \\pi }{4} \\right)\\textgreater0$$，$$\\therefore (*)$$式$$\\textless{}0$$． 即$$\\sin \\sqrt{2}-\\sin \\sqrt{3}\\textless{}\\cos \\sqrt{2}-\\cos \\sqrt{3}$$． ∴曲线表示焦点在$$y$$轴上的椭圆， 故选$$\\text{C}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "926", "queId": "8a7f12d35aa34652b81219aeb24b58b4", "competition_source_list": ["2011年山东全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$pa, b, c, d$$成等比数列，$$q:ad=bc$$， 则$$p$$是$$q$$的．", "answer_option_list": [[{"aoVal": "A", "content": "充分不必要条件 "}], [{"aoVal": "B", "content": "必要不充分条件 "}], [{"aoVal": "C", "content": "充分且必要条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->常用逻辑用语", "竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["充分性显然成立，必要性不成立．例：$$a=1, b=2, c=5, d=10$$. "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "611", "queId": "44bd142564114a478aad912c768207b7", "competition_source_list": ["2008年吉林全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "$$2008$$年中国北京奥运会吉祥物由$$5$$个``中国福娃''组成，分别叫贝贝、晶晶、欢欢、迎迎、妮妮．现有两套不同大小的福娃（共$$10$$个福娃），从两套福娃中任意选出$$5$$个福娃，恰好缺一个组成完整``奥运会吉祥物''的选法有．", "answer_option_list": [[{"aoVal": "A", "content": "$$160$$种 "}], [{"aoVal": "B", "content": "$$320$$种 "}], [{"aoVal": "C", "content": "$$32$$种 "}], [{"aoVal": "D", "content": "$$120$$种 "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["$$\\text{C}_{5}^{1}\\cdot \\text{C}_{4}^{1}\\cdot 2\\cdot 2\\cdot 2=160$$．故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "94", "queId": "25aa649767e6429da76ed262b9f0d1e3", "competition_source_list": ["2014~2015学年浙江杭州高一下学期期末理科第24题3分", "2011年辽宁全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)=\\sqrt{x-3}+\\sqrt{12-3x}$$的值域为( )．", "answer_option_list": [[{"aoVal": "A", "content": "$$[1,\\sqrt{2}]$$ "}], [{"aoVal": "B", "content": "$$[1,\\sqrt{3}]$$ "}], [{"aoVal": "C", "content": "$$\\left[ 1,\\frac{3}{2} \\right]$$ "}], [{"aoVal": "D", "content": "$$[1,2]$$ "}]], "knowledge_point_routes": ["课内体系->方法->换元法", "课内体系->素养->逻辑推理", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域->求函数的值域->用换元法求值域"], "answer_analysis": ["对于$$f(x)$$，有$$3\\leqslant x\\leqslant 4$$，则$$0\\leqslant x-3\\leqslant 1$$， 令$$x-3={{\\sin }^{2}}\\theta $$，$$0\\leqslant \\theta \\leqslant \\frac{ \\pi }{2}$$， 则$$f\\left( x \\right)=\\sqrt{x-3}+\\sqrt{3\\left( 4-x \\right)}=\\sin \\theta +\\sqrt{3\\left( 1-{{\\sin }^{2}}\\theta \\right)}=\\sin \\theta +\\sqrt{3}\\cos \\theta =2\\sin \\left( \\theta +\\frac{ \\pi }{3} \\right)$$， ∵$$\\frac{ \\pi }{3}\\leqslant \\theta +\\frac{ \\pi }{3}\\leqslant \\frac{5 \\pi }{6}$$， ∴$$\\frac{1}{2}\\leqslant \\sin \\left( \\theta +\\frac{ \\pi }{3} \\right)\\leqslant 1$$，$$1\\leqslant 2\\sin \\left( \\theta +\\frac{ \\pi }{3} \\right)\\leqslant 2$$． 函数$$f\\left( x \\right)=\\sqrt{x-3}+\\sqrt{12-3x}$$的值域为$$[1,2]$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "265", "queId": "2b5a836ad27e4147bb30c95b900be98d", "competition_source_list": ["2019年福建全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知复数$$z$$、$${{z}_{1}}$$、$${{z}_{2}}({{z}_{1}}\\ne {{z}_{2}})$$满足：$$z_{1}^{2}=z_{2}^{2}=-2-2\\sqrt{3}\\text{i}$$，$$\\textbar z-{{z}_{1}}\\textbar=\\textbar z-{{z}_{2}}\\textbar=4$$．则$$\\textbar z\\textbar=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$2\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$2\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->复数->复数的概念及几何意义", "竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["先求复数$$-2-2\\sqrt{3}\\text{i}$$的平方根． 设$${{(x+y\\text{i)}}^{2}}=-2-2\\sqrt{3}\\text{i(}x,y\\in \\mathbf{R})$$． 则$$({{x}^{2}}-{{y}^{2}})+2xy\\text{i=}-2-2\\sqrt{3}\\text{i}$$ $$\\Rightarrow \\begin{cases}{{x}^{2}}-{{y}^{2}}=-2 2xy=-2\\sqrt{3} \\end{cases}$$， $$\\Rightarrow \\begin{cases}{{x}_{1}}=1 {{y}_{1}}=-\\sqrt{3} \\end{cases}$$，$$\\begin{cases}{{x}_{2}}=-1 {{y}_{2}}=\\sqrt{3} \\end{cases}$$． 由$$z_{1}^{2}=z_{2}^{2}=-2-2\\sqrt{3}\\text{i}$$，$${{z}_{1}}\\ne {{z}_{2}}$$，知$${{z}_{1}}$$、$${{z}_{2}}$$为复数$$-2-2\\sqrt{3}\\text{i}$$的两个平方根． 由对称性，不妨设$${{z}_{1}}=1-\\sqrt{3}\\text{i}$$，$${{z}_{2}}=-1+\\sqrt{3}\\text{i}$$． 于是，$$\\textbar{{z}_{1}}-{{z}_{2}}\\textbar=\\textbar z-{{z}_{1}}\\textbar=\\textbar z-{{z}_{2}}\\textbar=4$$． 从而，复数$$z$$、$${{z}_{1}}$$、$${{z}_{2}}$$对应的点$$Z$$、$${{Z}_{1}}$$、$${{Z}_{2}}$$构成边长为$$4$$的正三角形． 又复数$${{z}_{1}}$$、$${{z}_{2}}$$对应的点$${{Z}_{1}}$$、$${{Z}_{2}}$$关于原点$$0$$对称，从而，$$OZ$$为$$\\triangle Z{{Z}_{1}}{{Z}_{2}}$$的高． 故$$\\textbar z\\textbar=\\textbar OZ\\textbar=2\\sqrt{3}$$． "], "answer_value": "C"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "586", "queId": "377ba1396fe444c5bed5c2c042afa220", "competition_source_list": ["2015年湖南全国高中数学联赛竞赛初赛第5题5分", "2005年全国高中数学联赛竞赛一试第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "使关于$$x$$的不等式$$\\sqrt{x-3}+\\sqrt{6-x}\\geqslant k$$有解的实数$$k$$的最大值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{6}-\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{6}+\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{6}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["令$$y=\\sqrt{x-3}+\\sqrt{6-x}$$，$$3\\leqslant x\\leqslant 6$$，则 $${{y}^{2}}=(x-3)+(6-x)+2\\sqrt{(x-3)(6-x)}\\leqslant 2[(x-3)+(6-x)]=6$$． ∴$$0\\textless{}y\\leqslant \\sqrt{6}$$，实数$$k$$的最大值为$$\\sqrt{6}$$． 故选$$\\text{D}$$． "], "answer_value": "D"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "422", "queId": "2cedfdd20fee40ee923fab48908157f2", "competition_source_list": ["2010年山东全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$ {{{a}_{n}} }$$是一等差数列，$${{S}_{n}}$$是其前$$n$$项之和，则$$-{{a}_{m}}\\textless{}{{a}_{1}}\\textless{}-{{a}_{m+1}}$$是$${{S}_{m}}\\textgreater0,{{S}_{m+1}}\\textless{}0$$的．", "answer_option_list": [[{"aoVal": "A", "content": "充分必要条件 "}], [{"aoVal": "B", "content": "充分而不必要条件 "}], [{"aoVal": "C", "content": "必要而不充分条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列", "竞赛->知识点->逻辑->常用逻辑用语"], "answer_analysis": ["事实上， $$\\left. \\begin{matrix}{{S}_{m}}=\\frac{m}{2}({{a}_{1}}+{{a}_{m}})\\textgreater0\\Leftrightarrow {{a}_{1}}+{{a}_{m}}\\textgreater0 {{S}_{m+1}}=\\frac{m+1}{2}({{a}_{1}}+{{a}_{m+1}})\\textless{}0\\Leftrightarrow {{a}_{1}}+{{a}_{m+1}}\\textless{}0 \\end{matrix} \\right }$$ $$\\Leftrightarrow {{a}_{1}}+{{a}_{m+1}}\\textless{}0\\textless{}{{a}_{1}}+{{a}_{m}}\\Leftrightarrow -{{a}_{m}}\\textless{}{{a}_{1}}\\textless{}-{{a}_{m+1}}.$$ 所以是充分必要条件． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "684", "queId": "a80cce6d36af44babe964792d448882f", "competition_source_list": ["2012年黑龙江全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "``$$\\lg x$$，$$\\lg y$$，$$\\lg z$$成等差数列''是``$${{y}^{2}}=xz$$''成立的（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "充分不必要条件 "}], [{"aoVal": "B", "content": "必要不充分条件 "}], [{"aoVal": "C", "content": "充要条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与数列结合", "课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算"], "answer_analysis": ["因为$$\\lg x$$，$$\\lg y$$，$$\\lg z$$成等差数列， 所以$$2\\lg y=\\lg x+\\lg z$$，所以$${{y}^{2}}=xz$$； 若$${{y}^{2}}=xz$$，当$$x\\textless{}0$$，$$z\\textless{}0$$时， $$\\lg x$$，$$\\lg z$$无意义． 所以``$$\\lg x$$，$$\\lg y$$，$$\\lg z$$成等差数列''是``$${{y}^{2}}=xz$$''成立的充分不必要条件． 故选$$\\text{A}$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "699", "queId": "575361495cd1406cb6928ed952d5c84a", "competition_source_list": ["2010年辽宁全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "``函数$$f(x)$$在$$\\left[ a,b \\right]$$上单调''是``函数$$f(x)$$在$$\\left[ a,b \\right]$$上有最大值和最小值''的．", "answer_option_list": [[{"aoVal": "A", "content": "充分条件而非必要条件 "}], [{"aoVal": "B", "content": "必要条件而非充分条件 "}], [{"aoVal": "C", "content": "充分必要条件 "}], [{"aoVal": "D", "content": "即非充分条件亦非必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["若$$f(x)$$在$$\\left[ a,b \\right]$$上单调，则$$f(a)$$和$$f(b)$$一个是最大值，另一个是最小值，故是充分条件；如，$$\\sin x$$在$$\\left[ 0, \\pi ~\\right]$$上有最大值$$1$$和最小值$$0$$，但不单调，故不是必要条件． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1043", "queId": "ee4dbf91acc042e2be88137392ff9c06", "competition_source_list": ["2016年河北全国高中数学联赛高三竞赛初赛第5题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "求$$y=\\sqrt{{{x}^{2}}-{{x}^{4}}}+\\sqrt{2{{x}^{2}}-{{x}^{4}}}$$的最大值为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$2$ "}], [{"aoVal": "B", "content": "$\\sqrt{2}$ "}], [{"aoVal": "C", "content": "$2\\sqrt{2}$ "}], [{"aoVal": "D", "content": "$4$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->柯西不等式", "竞赛->知识点->不等式->几个重要的不等式->柯西"], "answer_analysis": ["由柯西不等式，$${{\\left( \\sqrt{{{x}^{2}}\\left( 1-{{x}^{2}} \\right)}+\\sqrt{\\left( 2-{{x}^{2}} \\right){{x}^{2}}} \\right)}^{2}}\\leqslant \\left( {{x}^{2}}+2-{{x}^{2}} \\right)\\left( 1-{{x}^{2}}+{{x}^{2}} \\right)=2$$，当$$x=\\pm \\frac{\\sqrt{6}}{3}$$时取等号． "], "answer_value": "B"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "267", "queId": "26ff59619e5941d4b713608c85d37478", "competition_source_list": ["2010年河北全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "对于任意的整数$$n\\left( n{\\geqslant }2 \\right)$$，满足$${{a}^{n}}=a+1$$，$${{b}^{2n}}=b+3a$$的正数$$a$$与$$b$$的大小关系是( )", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b\\textgreater1$$ "}], [{"aoVal": "B", "content": "$$b\\textgreater a\\textgreater1$$ "}], [{"aoVal": "C", "content": "$$a\\textgreater1$$，$$0 \\textless{} b \\textless{} 1$$ "}], [{"aoVal": "D", "content": "$$0 \\textless{} a \\textless{} 1$$，$$b\\textgreater1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->证明不等式的基本方法->比较法", "课内体系->知识点->等式与不等式->不等式->不等式的性质", "课内体系->知识点->基本初等函数->指数函数->指数函数的图象及性质", "课内体系->素养->逻辑推理", "课内体系->素养->数学运算"], "answer_analysis": ["首先，$$a\\textgreater1$$，$$b\\textgreater1$$．否则， 若$$0 \\textless{} a \\textless{} 1$$，则$${{a}^{n}}=a+1\\textgreater1$$，从而$$a\\textgreater1$$，矛盾； 若$$0 \\textless{} b \\textless{} 1$$，$${{b}^{2n}}=b+3a\\textgreater3a\\textgreater3$$，从而$$b\\textgreater1$$，矛盾． 故$$a$$、$$b$$均大于1． 一方面，$${{a}^{2n}}-{{b}^{2n}}={{\\left( a+1 \\right)}^{2}}-\\left( b+3a \\right)={{a}^{2}}-a-b+1$$． 另一方面，$${{a}^{2n}}-{{b}^{2n}}=\\left( a-b \\right)\\left( {{a}^{2n-1}}+{{a}^{2n-2}}b+\\cdots +{{b}^{2n-1}} \\right)$$． 故$${{a}^{2}}-a-b+1$$ $$=\\left( a-b \\right)\\left( {{a}^{2n-1}}+{{a}^{2n-2}}b+\\cdots +{{b}^{2n-1}} \\right)$$， 即$$\\frac{{{a}^{2}}-a-b+1}{a-b}$$ $$={{a}^{2n-1}}+{{a}^{2n-2}}b+\\cdots +{{b}^{2n-1}}\\textgreater1$$． $$\\therefore \\frac{{{a}^{2}}-2a+1}{a-b}\\textgreater0$$，即$$\\frac{{{\\left( a-1 \\right)}^{2}}}{a-b}\\textgreater0$$．故$$a\\textgreater b$$． 综上所述，有$$a\\textgreater b\\textgreater1$$． "], "answer_value": "A"} +{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "249", "queId": "98e2e56bb1504ec495dc75e4d4d87ef4", "competition_source_list": ["2021~2022学年安徽阜阳太和县安徽省太和中学高一下学期月考（竞赛考试）第2~2题"], "difficulty": "0", "qtype": "single_choice", "problem": "函数$$f(x)=\\frac{1}{\\sqrt{2-x}}+{{(x+2)}^{0}}$$的定义域为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\infty ,2)\\cup (2,+\\infty )$$ "}], [{"aoVal": "B", "content": "$$(-\\infty ,-2)\\cup (-2,2)$$ "}], [{"aoVal": "C", "content": "$$(-\\infty ,-2)$$ "}], [{"aoVal": "D", "content": "$$(-\\infty ,2)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的定义域"], "answer_analysis": ["\\hfill\\break 根据函数解析式，只需解析式有意义，即$$\\left { \\begin{array}{*{35}{l}} 2-x\\textgreater0 x+2\\ne 0 \\end{array} \\right.$$，解不等式即可求解.\\\\ 【详解】\\\\ 由$$f(x)=\\frac{1}{\\sqrt{2-x}}+{{(x+2)}^{0}}$$，\\\\ 则$$\\left { \\begin{array}{*{35}{l}} 2-x\\textgreater0 x+2\\ne 0 \\end{array} \\right.$$，解得$$x\\textless{} 2$$且$$x\\ne -2$$，\\\\ 所以函数的定义域为$$\\left( -\\infty ,-2 \\right)\\cup \\left( -2,2 \\right)$$.\\\\ 故选：B "], "answer_value": "B"}