{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1029", "queId": "6afc3b231a334077b73f51057b53a7d5", "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": "1575", "queId": "ff8080814518d524014519095a860317", "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），$$41\\div7=5$$（周）$$\\cdots\\cdots 6$$ （天），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": "2785", "queId": "acb454229f0b4db98667a42ae6ccd90d", "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7题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知一个三位数的百位、十位和个位分别是$$a，b，c$$，而且$$a\\times b\\times c=a+b+c$$，那么满足上述条件的三位数的和为(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "1032 "}], [{"aoVal": "B", "content": "1132 "}], [{"aoVal": "C", "content": "1232 "}], [{"aoVal": "D", "content": "1332 "}]], "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": ["由$$a\\times b\\times c=a+b+c$$知整数a，b，c都不为零，且$$a\\left( bc-1 \\right)=b+c$$①，从而有$$b+c=a\\left( bc-1 \\right)\\geqslant bc-1$$，即$$2\\geqslant bc-b-c+1=\\left( b-1 \\right)\\left( c-1 \\right)$$，所以我们有$$\\left( b-1 \\right)\\left( c-1 \\right)=0，1，2$$.  ⑴当$$\\left( b-1 \\right)\\left( c-1 \\right)=0$$时，$$b=1$$或$$c=1$$.由$$b=1$$，及①得$$a\\left( c-1 \\right)=c+1$$，则$$a=\\frac{c+1}{c-1}=1+\\frac{2}{c-1}$$，从而$$c=2$$或$$3$$.因此$$\\left( a，b，c \\right)=\\left( 3，1，2 \\right)$$或$$\\left( 2，1，3 \\right)$$.  由$$c=1$$及①得$$a\\left( b-1 \\right)=b+1$$，将上述解法中的b，c互换，即得$$\\left( a，b，c \\right)=\\left( 3，2，1 \\right)$$或$$\\left( 2，3，1 \\right)$$  ⑵当$$\\left( b-1 \\right)\\left( c-1 \\right)=1$$时，$$b=c=2$$带入①得$$3a=4$$，$$a=\\frac{4}{3}$$不为整数，不可.  ⑶当$$\\left( b-1 \\right)\\left( c-1 \\right)=2$$时，$$\\begin{cases}b=2    c=3   \\end{cases}$$或$$\\begin{cases}b=3    c=2   \\end{cases}$$，由此可得$$\\left( a，b，c \\right)=\\left( 1，2，3 \\right)$$或者$$\\left( 1，3，2 \\right)$$  所以和为：$$312+213+321+231+123+132=1332$$ "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1003", "queId": "0e3ea8373e77414398bfb1d37f9c8887", "competition_source_list": ["2017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": "两台同样的拖拉机$$3$$天耕地$$18$$公顷，照这样计算，$$9$$天耕地$$81$$公顷，需要台这样的拖拉机． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["要求需要几台同样的拖拉机，就要先求出每台拖拉机每天耕地多少公顷，已知两台拖拉机$$3$$天耕地$$18$$公顷，用除法求每台拖拉机每天的工作量．然后乘以$$9$$天，求出每台拖拉机$$9$$天的工作量，最后用除法求需要的拖拉机数量．  $$81\\div (18\\div 2\\div 3\\times 9)=81\\div 27=3$$（台）．  答：需要$$3$$台同样的拖拉机．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1489", "queId": "7f2ab9511f084100b0a77d27742464f5", "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14题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "每个空瓶可以装$$2.5$$千克色拉油．王阿姨要把$$96$$千克色拉油全部装在这样的瓶子里，至少需要个这样的瓶子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$38$$ "}], [{"aoVal": "B", "content": "$$39$$ "}], [{"aoVal": "C", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["$$96\\div 2.5=38$$（个）$$\\cdots \\cdots 1$$（千克），  所以至少需要$$38+1=39$$（个）这样的瓶子．  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": "2595", "queId": "c2eca1f9882b4cb388fe3e016f610827", "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": ["知识标签->学习能力->七大能力->逻辑分析"], "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：$$\\text{A}$$选项． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2234", "queId": "3b9c7d2020814abc9241800e7b4c84c2", "competition_source_list": ["2017年IMAS小学中年级竞赛（第二轮）第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "快车与慢车同时从南北两地沿着同一条公路以匀速相向开出．已知快车每小时行驶$$40\\text{km}$$．经过$$2$$小时后，快车已驶过公路中点$$20\\text{km}$$，此时快车与慢车还相距$$6\\text{km}$$．请问慢车每小时行驶多少$$\\text{km}$$？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$23$$ "}], [{"aoVal": "E", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["因快车出发$$2$$小时后，快车驶过公路中点$$20\\text{km}$$，这时快车与慢车相距$$6\\text{km}$$，  由此可判断出$$2$$小时内快车行驶了总路程的一半还多$$20\\text{km}$$，而慢车行驶了总路程的一半还少$$26\\text{km}$$，  即快车$$2$$小时比快车多行驶了$$46\\text{km}$$，  因此两车的速度之差为每小时$$46\\div 2=23\\text{km}$$，  因此慢车每小时行驶$$40-23=17\\text{km}$$． ", "<p>可知$$2$$小时快车总共行驶了$$40\\times 2=80\\text{km}$$，</p>\n<p>由此时快车已驶过公路中点$$20\\text{km}$$知两地相距$$2\\times \\left( 80-20 \\right)=120\\text{km}$$，</p>\n<p>再由快车与慢车此时还相距$$6\\text{km}$$知此时慢车仅行驶了$$120-80-6=34\\text{km}$$，</p>\n<p>所以慢车每小时行驶$$34\\div 2=17\\text{km}$$．</p>"], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "939", "queId": "c644a5bba5e44c889aa4b002df4c92df", "competition_source_list": ["四年级其它", "2013年全国学而思杯五年级竞赛A卷第11题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个五位数，各位数字互不相同，并且满足：从左往右，第一位是$$2$$的倍数，前两位组成的两位数是$$3$$的倍数，前三位组成的三位数是$$5$$的倍数，前四位组成的四位数是$$7$$的倍数，这个五位数是$$11$$的倍数．那么，这个五位数最小是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21067$$ "}], [{"aoVal": "B", "content": "$$21076$$ "}], [{"aoVal": "C", "content": "$$21086$$ "}], [{"aoVal": "D", "content": "$$21087$$ "}]], "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": ["考虑最值确定各位数字：  万位是$$2$$的倍数，故万位最小应为$$2$$；  前两位组成的数是$$3$$的倍数，故前两位最小应为$$21$$；  前三位组成的数是$$5$$的倍数，故前三位最小应为$$210$$；  前四位组成的数是$$7$$的倍数，最小为$$2100$$，但要求各位数字不同，故应为$$2107$$；  这个五位数是$$11$$的倍数，故此数应为$$21076$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2591", "queId": "479285c4292c4202a9912340d0a2b306", "competition_source_list": ["2014年IMAS小学高年级竞赛第二轮检测试题第1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "请问算式$$57.6\\times \\frac{8}{5}+28.8\\times \\frac{184}{5}-14.4\\times 80$$的值为多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$14.4$$ "}], [{"aoVal": "D", "content": "$$38.8$$ "}], [{"aoVal": "E", "content": "$$57.6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数巧算->直接提取公因数/式"], "answer_analysis": ["$$57.6\\times \\frac{8}{5}+28.81\\times \\frac{184}{5}-14.4\\times 80$$  $$=28.8\\times \\frac{16}{5}+28.8\\times \\frac{184}{5}-14.4\\times 80$$  $$=28.8\\times \\left( \\frac{16}{5}+\\frac{184}{5} \\right)-14.4\\times 80$$  $$=28.8\\times \\frac{200}{5}-14.4\\times 80$$  $$=28.8\\times 40-14.4\\times 2\\times 40$$  $$=40\\times (28.8-28.8)$$  $$=0$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "386", "queId": "7c9aafa33eef4dc1a0288f6c100a3fb2", "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": "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"], "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：C。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1865", "queId": "c4c058234a0a481dac81a78151710cec", "competition_source_list": ["2017年湖北武汉中环杯五年级竞赛初赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "小明在黑板上写了一个数，小红把这个数先乘$$2$$，再加上$$4$$，得到的结果为$$8$$．那么，小明在黑板上写的数为~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "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", "课内体系->思想->逆向思想"], "answer_analysis": ["倒推$$(8-4)\\div 2=2$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2430", "queId": "98b7f7cbad2f45658f3944d17c80048e", "competition_source_list": ["2006年全国迎春杯小学中年级竞赛复赛第1题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$347\\times 81+21\\times 925+472\\times 19$$的计算结果等于$$A$$，那么，$$A$$的各位数字之和等于 ( ~ ~ ~ )． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$27$$ "}]], "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": ["原式$$=56500$$，$$5+6+5+0+0=16$$． ", "<p>原式$$=347\\times 81+\\left( 347+125 \\right)\\times 19+21\\times 925$$</p>\n<p>$$=347\\times 81+347\\times 19+125\\times 19+21\\times 925$$</p>\n<p>$$=347\\times \\left( 81+19 \\right)+125\\times 19+21\\times 125+21\\times 800$$</p>\n<p>$$=34700+125\\times \\left( 19+21 \\right)+16800$$</p>\n<p>$$=34700+5000+16800$$</p>\n<p>$$=56500$$，</p>\n<p>$$5+6+5+0+0=16$$.</p>\n<p>因此选$$\\rm C$$．</p>"], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1011", "queId": "2550445e4480431783609754bf114d47", "competition_source_list": ["2014年迎春杯六年级竞赛初赛"], "difficulty": "2", "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->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["解：半径是原来的$$(1+150 \\%)$$，  蛋糕的底面积是原来的$$\\left( 1+150 \\% \\right)^{2}$$，  高不变，那么蛋糕的体积也就是原来的$$\\left( 1+150 \\% \\right)^{2}$$，  $$4\\times \\left( 1+150 \\% \\right)^{2}$$  $$=4\\times 6.25$$  $$=25$$（个）  答：同样高的蛋糕可以让$$25$$个人吃饱。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "88", "queId": "1d5dcf8780394815b1ba633005a29144", "competition_source_list": ["2010年第1届广东深圳启智杯小学中年级竞赛第13题"], "difficulty": "1", "qtype": "single_choice", "problem": ".复旦大学某班$$A$$、$$B$$、$$C$$、$$D$$、$$E$$、$$F$$、$$G$$、$$H$$、$$I$$共$$9$$名同学参加$$2010$$年上海世博会志愿者知识测试．测试合格者进入志愿者选拔范围．测试结果只有一人合格．向他们询问谁合格．他们的回答如下:  $$A$$:``是$$E$$''；$$B$$:``是我''；$$C$$:``是$$B$$''；$$D$$:``不是$$E$$''；  $$E$$:``是$$B$$或$$H$$''；$$F$$:``是$$E$$''；$$G$$:``不是$$B$$''；$$H$$:``不是$$B$$也不是我''；  $$I$$:``$$H$$所说的是事实''.  其中，说实话的只有$$3$$个人，那么请问合格的是？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$A$$ "}], [{"aoVal": "B", "content": "$$B$$ "}], [{"aoVal": "C", "content": "$$C$$ "}], [{"aoVal": "D", "content": "$$D$$ "}], [{"aoVal": "E", "content": "$$E$$ "}], [{"aoVal": "F", "content": "$$F$$ "}], [{"aoVal": "G", "content": "$$G$$ "}], [{"aoVal": "H", "content": "$$H$$ "}]], "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": ["假设$$A$$说实话，则$$F$$、$$G$$、$$H$$、$$I$$也是说实话，这``与说实话的只有$$3$$人''不符，所以$$A$$说了假话，从而$$E$$不合格，$$F$$说了假话，$$D$$说了实话．  假设$$B$$说实话，则$$C$$、$$D$$、$$E$$也是说实话，这``与说实话的只有$$3$$人''不符，所以$$B$$说了假话，从而$$B$$不合格，$$C$$说了假话，$$G$$说了实话．  $$E$$与$$H$$一定是一人说假话，一人说实话，从而$$I$$说假话，继而得出$$H$$说假话，$$E$$说实话．所以$$H$$是合格的． "], "answer_value": "H"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3133", "queId": "00774eb64c794c24af0c7b68f77af6df", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "气象台预报``本市明天降雨概率是$$80 \\%''$$．对此信息，下列说法中~\\uline{~~~~~~~~~~}~正确的． ", "answer_option_list": [[{"aoVal": "A", "content": "本市明天将有$$80 \\%$$的地区降水． "}], [{"aoVal": "B", "content": "本市明天将有$$80 \\%$$的时间降水． "}], [{"aoVal": "C", "content": "明天肯定下雨． "}], [{"aoVal": "D", "content": "明天降水的可能性比较大． "}]], "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->可能性"], "answer_analysis": ["降水概率指的是可能性的大小，并不是降水覆盖的地区或者降水的时间．$$80 \\%$$的概率也不是指肯定下雨，$$100 \\%$$的概率才是肯定下雨．$$80 \\%$$的概率是说明有比较大的可能性下雨． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3140", "queId": "01afb85e7aba416b96a195749107f3b1", "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": "1522", "queId": "c35ac30d4ff74d4c839848f0b5b4832c", "competition_source_list": ["2009年六年级竞赛创新杯", "2009年第7届创新杯六年级竞赛初赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "水果店运来橘子、苹果和梨一共$$320$$千克。橘子和苹果重量之和与梨的重量比是$$11:5$$，橘子的重量是苹果重量的$$\\frac{5}{6}$$，苹果重（  ）千克。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$110$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$130$$ "}]], "knowledge_point_routes": ["课内体系->知识模块->综合与实践", "拓展思维->拓展思维->应用题模块->比例应用题->按比分配"], "answer_analysis": ["苹果、橘子、梨的重量比为$$6:5:5$$，所以苹果的重量为$$320\\times \\frac{6}{16}=120$$（千克）。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "904", "queId": "a5233a8641ad46b5b7f23a204e19fe92", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（四）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列说法正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "最小的自然数与最小的合数的和是$$5$$ "}], [{"aoVal": "B", "content": "在$$45$$的约数中，合数有$$2$$个 "}], [{"aoVal": "C", "content": "在$$20$$以内的质数中，加上$$2$$还是质数的有$$4$$个 "}], [{"aoVal": "D", "content": "对于两个整数$$a$$、$$b$$，若$$a=4b$$，那么$$a$$和$$b$$的最大公约数是$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["$$A$$项，最小的自然数是$$0$$，最小的合数是$$4$$，$$0+4=4$$；  $$B$$项，在$$45$$的约数中，合数有$$9$$、$$15$$、$$45$$共$$3$$个；  $$D$$项，$$a$$和$$b$$的最大公约数是$$b$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "759", "queId": "8c4f26c50b7845c1aa720c05ea7f4d03", "competition_source_list": ["2011年五年级竞赛明心奥数挑战赛"], "difficulty": "1", "qtype": "single_choice", "problem": "有(  )个不同的自然数，它们的平方是2000的因数. ", "answer_option_list": [[{"aoVal": "A", "content": "3 "}], [{"aoVal": "B", "content": "6 "}], [{"aoVal": "C", "content": "10 "}], [{"aoVal": "D", "content": "12 "}], [{"aoVal": "E", "content": "20 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数"], "answer_analysis": ["2000的因数中是完全平方数的有1、4、16、25、100、400六个数，所以共有6个数.选B. "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "441", "queId": "f29bd681670140c294086c01315b0902", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛B卷第8题3分", "2017~2018学年山东聊城莘县六年级下学期期末第21题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "现有$$A$$、$$B$$、$$C$$、$$D$$、$$E$$五个同学，他们分别为来自一中、二中、三中的学生，已知：（$$1$$）每所学校至少有他们中的一名学生；（$$2$$）在二中的晚会上，$$A$$、$$B$$、$$E$$作为被邀请客人演奏了小提琴；（$$3$$）$$B$$过去曾在三中学习，后来转学了，现在同$$D$$在同一个班学习；（$$4$$）$$D$$、$$E$$是同一所学校的三好学生，根据以上叙述可以断定$$A$$所在的学校为． ", "answer_option_list": [[{"aoVal": "A", "content": "一中 "}], [{"aoVal": "B", "content": "二中 "}], [{"aoVal": "C", "content": "三中 "}], [{"aoVal": "D", "content": "不确定 "}]], "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": ["通过分析题意可知：  $$B$$与$$D$$在同一个班学习，  $$D$$和$$E$$是同一所学校的三好学生，  证明$$B$$、$$D$$、$$E$$为同一所学校的人，  在二中的晚会上，  $$A$$、$$B$$、$$E$$作为被邀请演奏了小提琴，  证明他们都不是二中的，  $$B$$过去曾经在三中学习，后来转学了，  证明$$B$$不是三中的，所以$$B$$只能是一中，  $$D$$、$$E$$也只能是一中，  前面已经说过，$$A$$不是二中的，就只能是三中的了，  由$$B$$与$$D$$在同一个班学习，  $$D$$和$$E$$是同一所学校的三好学生可知：  $$B$$．$$D$$．$$E$$为同一所学校的人，  由二中的晚会上，  $$A$$．$$B$$．$$E$$作为被邀请演奏了小提琴可知：  $$A$$．$$B$$．$$E$$都不是二中的，  由$$B$$过去曾经在三中学习，  后来转学了可知$$B$$不是三中的，  所以$$B$$只能是一中，  $$D$$，$$E$$也只能是一中，  又$$A$$不是二中的，  所以$$A$$只能是三中的．  故选：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1434", "queId": "4cd73e9d39244fc89b0af2f09cba0e49", "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": "$$\\frac{1}{2}$$ "}]], "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": "461", "queId": "e51e51948f6c4a65bd0c593f6699827d", "competition_source_list": ["2015年全国迎春杯四年级竞赛初赛第3题"], "difficulty": "3", "qtype": "single_choice", "problem": "五个人站成一列，每人戴一顶不同的帽子，编号为$$1$$、$$2$$、$$3$$、$$4$$、$$5$$的帽子，每人只能看到前面的人的帽子．小王一顶都看不到；小孔只看到$$4$$号帽子；小田没有看到$$3$$号帽子，但看到了$$1$$号帽子；小严看到了有$$3$$顶帽子，但没有看到$$3$$号帽子；小韦看到了$$3$$号帽子和$$2$$号帽子，小田戴号帽子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "以上都不正确 "}]], "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": ["由小王一顶也看不到可知王站在第一个，由小孔只看到了$$4$$号帽子，可以得出小孔站在第二个，且小王戴的是$$4$$号帽子；由小严看到了$$3$$顶帽子却没看到$$3$$号帽子，可知小严站在第四个，且$$3$$号帽子只能是第四或第五个人，由小韦看到$$3$$号帽子，可知，小韦站在第五个，且小严戴$$3$$号帽子；前面出现了$$1、$$2$$、$$3$$、4$$号帽子，所以，小韦戴$$5$$号帽子、小田戴$$2$$号帽子． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1096", "queId": "0c1cbf21bb804766b2ac778e4a899d21", "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10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "做一道减法题时，小明把减数的个位上的$$8$$看作$$3$$，十位上的$$6$$看作$$9$$，算出的差是$$75$$，正确的答案应是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$110$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["把减数$$68$$误看成$$93$$，  则误看成的数比正确的数大了：$$93-68=25$$，  根据：被减数$$-$$减数$$=$$差，  减数变大，则差就变小，所以正确的答案为：$$75+25=100$$．  故选：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1614", "queId": "88da9b68a05f47d8a6935956b55d4c4f", "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": "2", "qtype": "single_choice", "problem": "某公司发年终将一共$$100$$万元，其中一等奖每人$$1.5$$万元，二等奖每人$$1$$万元，三等奖每人$$5000$$元，如果三等奖与一等奖人数之差不少于$$93$$人，但小于$$96$$人，则该公司的总人数为（~ ~ ~ ）人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$137$$ "}], [{"aoVal": "B", "content": "$$146$$ "}], [{"aoVal": "C", "content": "$$147$$ "}], [{"aoVal": "D", "content": "$$148$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设三种奖分别有$$x$$，$$y$$，$$\\text{z}$$人，列代数式得：$$\\begin{cases} 15x+y+0.5z=100    z-x\\geqslant 94    z-x\\leqslant 96 \\end{cases}$$  当$$z-x=94$$，解得：$$\\begin{cases} x=1    y=51    z=95 \\end{cases}$$成立，选$$\\text{C}$$. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1891", "queId": "8aac50a74ff4b16201500e7be0392aae", "competition_source_list": ["2009年全国迎春杯小学中年级四年级竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$60$$名学生，男生、女生各$$30$$名，他们手拉手围成一个圆圈．如果让原本牵着手的男生和女生放开手，可以分成$$18$$个小组．那么，如果原本牵着手的男生和男生放开手时，分成了~\\uline{~~~~~~~~~~}~个小组． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$21$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$42$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["$$60$$名学生围成圈，每个人与相邻的同学牵手，那么有$$60$$对牵着的手，其中男生与女生牵手的有$$18$$对，假设男生与男生牵手的有$$x$$人，那么，参与围圈的男生一共有$$（2x+18）\\div2=x+9$$人，所以$$x+9=30$$，$$x=21$$,那么原来牵手的男生和男生放手，分成了$$21$$个小组． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "714", "queId": "87f11dd0379a47a0a86ca43c1d65b59c", "competition_source_list": ["2020年新希望杯六年级竞赛初赛（团战）第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "算式$$12\\times 43=1000$$是进制的乘法运算． ", "answer_option_list": [[{"aoVal": "A", "content": "五 "}], [{"aoVal": "B", "content": "六 "}], [{"aoVal": "C", "content": "七 "}], [{"aoVal": "D", "content": "八 "}], [{"aoVal": "E", "content": "九 "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["暂无 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2845", "queId": "7be77b4c9c6d4677bcd4eee3e25ef7a1", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一列数，开头四个是$$2$$，$$0$$，$$1$$，$$3$$；从第$$5$$个数开始，每个数是前面四个的和除以$$4$$所得的余数，那么这列数中的第$$2013$$个数是（~ ）． ", "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->应用题模块->周期问题->基本排列的周期问题"], "answer_analysis": ["$$2$$、$$0$$、$$1$$、$$3$$、$$2$$、$$2$$、$$0$$、$$3$$、$$3$$、$$0$$、$$2$$、$$0$$、$$1$$、$$3\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 10$$个数字为一个周期反复出现， 则第$$2013$$个数应该为周期中第三位的$$1$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3434", "queId": "aacc6e352eab46dbb96815ec750bf36e", "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": "1", "qtype": "single_choice", "problem": "将$$180$$写成若干个连续正整数之和（至少两个），共有（~ ）种写法． ", "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->枚举法综合->整数分拆->简单拆分->加法拆数（不完全相同数）"], "answer_analysis": ["$$180=59+60+61=34+35+36+37+38=19+20+\\ldots \\ldots +25+26=16+17+\\ldots \\ldots +23+24=5+6+\\ldots \\ldots +18+19$$共$$5$$种． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "544", "queId": "fe86f52487ae49cca66fa2674f54a682", "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$$2021$$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": ["拓展思维->能力->运算求解"], "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": "3395", "queId": "e0104824ea8d4eca8bc742d4b1eaab57", "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": ["拓展思维->能力->逻辑分析"], "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": "3432", "queId": "c5ba20aeb3bf40f0b5df59c6e0a92744", "competition_source_list": ["2012年全国创新杯五年级竞赛第1题5分", "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": "【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->思想->对应思想"], "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": "446", "queId": "8aac50a74ff4b16201500e8550ac2ad8", "competition_source_list": ["2009年全国迎春杯四年级竞赛初赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一类多位数，从左数第$$3$$位数字开始，每位上的数都等于其左边第$$2$$个数减去左边第$$1$$个数的差．如$$74312$$、$$6422$$．那么这类数中最大的数的前$$2$$位是 ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$96$$ "}], [{"aoVal": "B", "content": "$$85$$ "}], [{"aoVal": "C", "content": "$$95$$ "}], [{"aoVal": "D", "content": "$$99$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["比较两个数的大小首先比较数位,数位相同,然后从首位开始比较相同数位上数字的大小．可构造出满足条件的数位最长的数是$$85321101$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2480", "queId": "1203c8c534504fe38c3c19baaa3b30de", "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": "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->数据处理"], "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": "2104", "queId": "f4e55db9c8c34573ad31f0878c67a685", "competition_source_list": ["2015年北京学而思杯五年级竞赛A卷第16题", "2015年全国学而思杯五年级竞赛A卷第16题", "2015年浙江杭州学而思杯五年级竞赛A卷第10题"], "difficulty": "3", "qtype": "single_choice", "problem": "甲、乙两人从$$A$$地，丙从$$B$$地同时出发，相向而行．当甲、丙相遇时，乙刚好走了全程的$$\\frac{1}{5}$$；此时甲立即掉头向$$\\text{A}$$地行走，乙、丙仍按原方向行走．当甲、乙相遇时，丙距离$$B$$地$$336$$米；三人继续行走，当乙、丙相遇时，甲恰好回到$$A$$地．那么，$$A$$、$$B$$两地之间的距离是米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$480$$ "}], [{"aoVal": "B", "content": "$$560$$ "}], [{"aoVal": "C", "content": "$$720$$ "}], [{"aoVal": "D", "content": "$$840$$ "}], [{"aoVal": "E", "content": "$$1080$$ "}]], "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": ["此类行程题，速度比最重要，注意到甲往返路程相同，而最后乙丙正好相遇．则在甲丙相遇时．乙丙应合走了全程的一半，则在此时刻三人路程比为$$7:2:3$$，即$${{v}_{甲}}:{{v}_{乙}}:{{v}_{丙}}=7:2:3$$，所以甲乙相遇时．丙走了全程的$$\\frac{3}{10}+\\left( \\frac{7}{10}-\\frac{1}{5} \\right) \\times \\frac{3}{7+2}=\\frac{7}{15}$$，$$336\\div \\frac{7}{15}=720$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2270", "queId": "89b8e1e30b984d2f893e9d05d839ead1", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛决赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "一辆长度为$$10$$米的车穿过一个隧道时的速度为$$v$$，用时$$14$$秒；另一辆长度为$$15$$米的车也以速度$$v$$穿过该隧道，用时$$18$$秒．则隧道长度、车的速度$$v$$分别为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7.5$$米，$$4.5$$千米/时 "}], [{"aoVal": "B", "content": "$$7$$米，$$1.25$$千米/时 "}], [{"aoVal": "C", "content": "$$6.5$$米，$$3$$千米/时 "}], [{"aoVal": "D", "content": "$$6$$米，$$2$$千米/时 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->火车问题->火车过桥->完全过桥"], "answer_analysis": ["$$(15-10)\\div (18-14)=\\frac{5}{4}$$米/秒，$$\\frac{5}{4}$$米/秒$$=4.5$$千米/小时，$$\\frac{5}{4}\\times 14-10=7.5$$（米）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2304", "queId": "d78c585cd0ad42e38a836ee3d80f959d", "competition_source_list": ["2011年五年级竞赛创新杯"], "difficulty": "0", "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$$A$$、$$B$$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->比例解行程问题->行程中的比例"], "answer_analysis": ["$$4$$小时甲乙共行了$$AB$$，$$3$$小时甲乙共行了$$\\frac{3}{4}AB$$，$$\\left( 10+80 \\right)\\div \\left( 1-\\frac{3}{4} \\right)=360$$(千米)。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1876", "queId": "b2bfe8da09634ea8b7c08782d2821db5", "competition_source_list": ["2020年长江杯六年级竞赛复赛B卷第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一件商品降价$$110$$元后，又涨价$$110$$元，现价与原价比较：． ", "answer_option_list": [[{"aoVal": "A", "content": "不变 "}], [{"aoVal": "B", "content": "上涨 $$\\frac{1}{100}$$ "}], [{"aoVal": "C", "content": "上涨$$1$$元 "}], [{"aoVal": "D", "content": "下降$$\\frac{1}{100}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["一件商品先降价$$110$$元，又涨价$$110$$元，现价与原价比较，是不变的．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3259", "queId": "3642e2df25894b12bc10eae9d67b2c4f", "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->能力->实践应用"], "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": "2796", "queId": "b5eb93cbb912476083d3f250086aec3f", "competition_source_list": ["2013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": "请问下列哪一项时间最接近于一天的时间？ ", "answer_option_list": [[{"aoVal": "A", "content": "半天 "}], [{"aoVal": "B", "content": "$$2$$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->思想->对应思想"], "answer_analysis": ["解法$$1$$：我们首先把所有时间都化成以分钟为单位．一天$$=1440$$分钟，半天$$=720$$分钟，$$2$$天$$=2880$$分钟，$$23$$小时$$=1380$$分钟，$$26$$小时$$=1560$$分钟．而：  $$1440-720=720$$  $$2880-1440=1440$$  $$1440-1380=60$$  $$1560-1440=120$$  $$1440-1410=30$$  所以$$1410$$分钟最接近于一天的时间．故选$$\\text{E}$$．  解法$$2$$：一天$$=24$$小时$$=1440$$分钟，而半天与一天相差半天，即相差$$720$$分钟；$$2$$天与一天相差$$1$$天，即相差$$1440$$分钟；$$23$$小时与一天相差$$1$$小时，即相差$$60$$分钟；$$26$$小时与一天相差$$2$$小时，即相差$$120$$分钟；$$1410$$分钟与一天相差$$30$$分钟．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": "2199", "queId": "359f896aff6e45a29f896f54d4f5206a", "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7题"], "difficulty": "0", "qtype": "single_choice", "problem": "复仇者们都是说英语的，灭霸说：``leishen'' flies at $$800$$ km/hour and have $$4400$$ km to travel． It will take him  hours to complete his trip． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$4.5$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$5.5$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->实践应用"], "answer_analysis": ["雷神每小时飞行$$800$$千米，它还有$$4400$$千米要飞，它还需要花多久？  $$4400\\div 800=5.5$$（小时）． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "379", "queId": "857a6b70d91a403483b52e48f540bb72", "competition_source_list": ["其它小学奥数优秀生培养教程8级", "2010年全国华杯赛竞赛复赛第1题", "2019年北京海淀区北京一零一中学六年级下学期小升初模拟（九）第15题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$10$$个盒子中放乒乓球，每个盒子中的球的个数不能少于$$11$$，不能是$$13$$，也不能是$$5$$的倍数，且彼此不同，那么至少需要~\\uline{~~~~~~~~~~}~个乒乓球． ", "answer_option_list": [[{"aoVal": "A", "content": "$$156$$ "}], [{"aoVal": "B", "content": "$$200$$ "}], [{"aoVal": "C", "content": "$$173$$ "}], [{"aoVal": "D", "content": "$$110$$ "}], [{"aoVal": "E", "content": "$$214$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["想要乒乓球总数尽量少，则每个盒子里的乒乓球数应尽量少．  考虑极端情况，不少于$$11$$则最少为$$11$$，不能是$$13$$也不能是$$5$$的倍数，  即不能为$$13$$，$$15$$，$$20\\cdot \\cdot \\cdot $$，故至少需要$$11+12+14+16+17+18+19+21+22+23=173$$（个）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2056", "queId": "d446037c711c4a46b5fe477a352b1660", "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": "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": "2312", "queId": "b3ed1ddcdd5540cc85da575e1ce5817a", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（四）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$A$$、$$B$$两个损坏的时钟，每分钟$$A$$时钟秒针能走$$70$$秒，$$B$$时钟秒针能走$$50$$秒．现在两个时钟都停留在$$3$$点整，一段时间以后，$$A$$、$$B$$两个时钟对应的指针恰好指向相同的位置，这段时间最短是（~ ）分钟．~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$720$$ "}], [{"aoVal": "C", "content": "$$2160$$ "}], [{"aoVal": "D", "content": "$$3600$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["$$A$$比$$B$$每分钟多跑$$20$$秒，$$A$$比$$B$$多跑一圈需要多跑$$12$$小时，$$12\\times 60\\times 60\\div 20=2160$$分钟． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1336", "queId": "28789c3b14ce4bf88baebef09b72edfd", "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": "一块均匀生长的草地按照$$1:2:3$$的面积比分成三块．一群牛先用$$12$$天时间吃完了第一块草地的草，接着又用$$48$$天吃完了第二块草地的草．此时，这群牛需要（~ ）天能够吃完第三块草地的草． ", "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": "1110", "queId": "1012565037d24b1681b2ef6eae823875", "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．  Grandma said: \"there are $53$ Sundays in $2006$.\"~So the first day of $2007$ is~\\uline{~~~~~~~~~~}~. ", "answer_option_list": [[{"aoVal": "A", "content": "Monday "}], [{"aoVal": "B", "content": "Tuesday "}], [{"aoVal": "C", "content": "Saturday "}], [{"aoVal": "D", "content": "Sunday "}]], "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", "Overseas Competition->知识点->应用题模块->周期问题"], "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．  There are 365 days in a year, $$365=7\\times 52+1$$. Because there are $$53$$ Sundays in $2006$, so the first day of $2006$ is Sunday and the last day of $2006$ is Sunday. So the first day of $2007$ is Monday. "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1479", "queId": "75f40a739bc34d27bae8c8ad270d677c", "competition_source_list": ["2011年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "为了节约用水，市政府规定：家庭月用水不超过$$60$$立方米（含$$60$$立方米）按照每立方米$$1.5$$元计算，超过$$60$$立方米但不超过$$80$$立方米（含$$80$$立方米）的部分按照每立方米$$2.5$$元计算，超过$$80$$立方米的部分按照每立方米$$5.5$$元计算．小丽家$$5$$月份用水量为$$87$$立方米，那么小丽家本月要缴水费元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$67.5$$ "}], [{"aoVal": "B", "content": "$$130.5$$ "}], [{"aoVal": "C", "content": "$$178.5$$ "}], [{"aoVal": "D", "content": "$$215$$ "}]], "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": ["$$60\\times 1.5+\\left( 80-60 \\right)\\times 2.5+\\left( 87-80 \\right)\\times 5.5$$  $$=90+50+38.5$$  $$=178.5$$（元）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3445", "queId": "dc9a67dd8d4f4df884eb8c3b7641f676", "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7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小红有$$3$$种不同颜色的上衣，$$4$$种不同颜色的半身裙，问她共有种不同的穿法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["分步用乘法，$$3\\times 4=12$$，共有$$12$$种穿法． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2291", "queId": "9757c7f1167344dabb10923b74b99212", "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2题2分", "2019年四川成都锦江区嘉祥外国语学校小升初第4题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两人在$$100$$米长的跑道上赛跑．如果他们同时从起点出发，都以均匀的速度跑向终点．当甲跑完$$80$$米时，乙在甲身后$$10$$米；当甲到达终点时，乙距离终点还有米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11.875$$ "}], [{"aoVal": "D", "content": "$$12.5$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["$$80:(80-10)=8:7$$，  $$100-100\\div 8\\times 7=12.5$$（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": "2963", "queId": "d701eee73c304ca981627128bca9f7fb", "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": "3418", "queId": "c9ac9584872e4207898479889506b2d3", "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，黄色的面有$$2$$个，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": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->七大能力->逻辑分析", "拓展思维->思想->分类讨论思想"], "answer_analysis": ["两个黄色面有相邻和相对两种位置关系 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "620", "queId": "c28f29ea6d464bdbbbdc2e5b00857d94", "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": "$$127$$ "}], [{"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": "3429", "queId": "aef6d8010e2149ba972ba8b3d243939f", "competition_source_list": ["2017年全国亚太杯四年级竞赛初赛第28题"], "difficulty": "3", "qtype": "single_choice", "problem": "将 $$1，$$2$$，$$3$$，$$ \\ldots\\ldots 依次写下去组成一个数$$12345678910111213$$ ···．如果写到某个自然数时，所组成的数恰好第一次能被 $$225$$ 整除，那么这个组成的数的各位数字之和是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1053$$ "}], [{"aoVal": "B", "content": "$$1054$$ "}], [{"aoVal": "C", "content": "$$1055$$ "}], [{"aoVal": "D", "content": "$$1056$$ "}], [{"aoVal": "E", "content": "$$1057$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->计数模块", "拓展思维->能力->数据处理"], "answer_analysis": ["$$225=9\\times 25$$  这个数能被 $$225$$ 整除，所以这个数能被 $$9$$ 和 $$25$$ 整除  这个数能被 $$9$$ 整除，则这个数的末两位能被 $$25$$ 整除，然后开始枚举，  ①写到 $$25$$ 时，利用乱切法，$$\\left( 1+25 \\right)\\times 25\\div 2=13\\times 25$$不能被 $$9$$ 整除，不符合．  ②写到 $$50$$ 时，利用乱切法，$$\\left( 1+50 \\right)\\times 50\\div 2=51\\times 25$$不能被 $$9$$ 整除，不符合．  ③写到 $$75$$ 时，利用乱切法，$$\\left( 1+75 \\right)\\times 75\\div 2=38\\times 25$$ 不能被 $$9$$ 整除，不符合．  ④写到 $$100$$ 时，利用乱切法，$$\\left( 1+100 \\right)\\times 100\\div 2=101\\times 50$$ 不能被 $$9$$ 整除，不符合．  ⑤写到 $$125$$ 时，利用乱切法，$$\\left( 1+125 \\right)\\times 125\\div 2=63\\times 125$$能被 $$9$$ 整除，符合．  所以写到 $$125$$ 时，所组成的数能被 $$225$$ 整除．  $$1\\sim 99$$ ：变成$$0\\sim 99$$ ，则数码和为$$\\left( 9+9 \\right)\\times 50=900$$ ．  $$100\\sim 125$$ ：  $$1\\times 26+1\\times 10+2\\times 6+\\left( 1+2+\\cdots \\cdots +9 \\right)\\times 2+1+2+3+4+5$$  $$=26+10+12+5\\times 9\\times 2+3\\times 5$$  $$=26+10+12+90+15$$  $$=153$$  所以$$900+153=1053$$ ．  这个组成的数的各位数字之和是 $$1053$$ ． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2116", "queId": "f9b577bccce746c78c5cae31ab7a2127", "competition_source_list": ["2016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": "0", "qtype": "single_choice", "problem": "有甲、乙两个仓库，甲仓库存粮$$30$$吨，如果从甲仓库中取出$$\\frac{1}{10}$$放入乙仓库，则两仓存粮相等，两仓一共存粮吨． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$27$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$54$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["甲仓存粮$$30$$吨，如果从甲仓中取出$$\\frac{1}{10}$$后，则甲仓还剩下全部的$$1-\\frac{1}{10}$$，  即还剩$$30\\times \\left( 1-\\frac{1}{10} \\right)$$吨，此时两仓存粮数相等，  所以两个仓库一共存粮$$30\\times \\left( 1-\\frac{1}{10} \\right)\\times 2=54$$吨．  故选：$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "669", "queId": "6bd5917dada14bf9ad6be0265d5cc7d9", "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": "1", "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": ["知识标签->拓展思维->数论模块->整除->整除特征->整除初识"], "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": "143", "queId": "17294fa6ba19451babc5b53eb5032659", "competition_source_list": ["2008年二年级竞赛学而思杯"], "difficulty": "2", "qtype": "single_choice", "problem": "一些麋鹿的骨盆骨与所有猪的骨盆骨具有许多相同的特征.虽然不是所有的麋鹿都有这些特征，但是一些动物学家声称，所有具有这些特征的动物都是麋鹿. 如果以上陈述和动物学家的声明都是真的，则以下（ ）也一定为真. ", "answer_option_list": [[{"aoVal": "A", "content": "麋鹿与猪的相似之处要多于它与其他动物的相似之处. "}], [{"aoVal": "B", "content": "一些麋鹿与猪在其他方面的不同之处要少得多. "}], [{"aoVal": "C", "content": "所有动物，如果它们的骨盆骨具有相同的特征，那么它们的其他骨骼部位一般也会具有相同或相似的特征. "}], [{"aoVal": "D", "content": "所有的猪都是麋鹿. "}], [{"aoVal": "E", "content": "所有的麋鹿都是猪. "}]], "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": ["所有猪的骨盆骨都具有那些特征，而所有具有这些特征的动物都是麋鹿，故所有的猪都是麋鹿，选D. "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2033", "queId": "e6165ea248634e7a8bd0c0f235829df7", "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": "【举一反三1】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->能力->逻辑分析"], "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": "2212", "queId": "6c4651e5bb144e398cde849279391cfb", "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": "$$75$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->转化与化归的思想"], "answer_analysis": ["剩下的路程为：$$300-120=180$$（千米），计划总时间为：$$300÷50=6$$（小时），前$$120$$千米已用去$$120÷40=3（$$小时），所以剩下路程的速度为： $$（300-120）÷（6-3）=60$$（千米/时）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1729", "queId": "b1aec7bae406482eaac7625009683830", "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": "日 "}], [{"aoVal": "B", "content": "六 "}], [{"aoVal": "C", "content": "五 "}], [{"aoVal": "D", "content": "四 "}]], "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": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1952", "queId": "ca0c901b226e440494929065eb386b0d", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "等腰三角形中大角是小角的$$2$$倍，则这个三角形是什么三角形（按角分类）？ ", "answer_option_list": [[{"aoVal": "A", "content": "等腰锐角三角形 "}], [{"aoVal": "B", "content": "等腰直角三角形 "}], [{"aoVal": "C", "content": "等腰钝角三角形 "}], [{"aoVal": "D", "content": "等腰直角三角形或者等腰锐角三角形 "}]], "knowledge_point_routes": ["知识标签->数学思想->数形结合思想"], "answer_analysis": ["这个等腰三角形的三个内角之比为$$225$$或者$$255$$，前一种情况下，三角形最大的内角为$$180\\div \\left( 2+2+5 \\right)\\times 5=100$$°；后一种情况下，三角形最大的内角为$$180\\div \\left( 2+5+5 \\right)\\times 5=75$$．所以这个等腰三角形是钝角三角形或者锐角三角形． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3298", "queId": "68762ca39e6442efaf977619ed4e729a", "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": "【S】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": "$$98$$ "}], [{"aoVal": "C", "content": "$$480$$ "}], [{"aoVal": "D", "content": "$$720$$ "}]], "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", "课内体系->思想->对应思想"], "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": "2611", "queId": "8335b63ae51846ce9aa7d127e2d29705", "competition_source_list": ["2013年第14届上海中环杯小学中年级三年级竞赛初赛第3题"], "difficulty": "0", "qtype": "single_choice", "problem": "定义新运算：$$a*b=5a-4b$$，则$$3*2*1=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$31$$ "}], [{"aoVal": "D", "content": "$$35$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计算模块->定义新运算->直接运算型->普通型"], "answer_analysis": ["$$3*2=5\\times 3-4\\times 2=7$$，$$7*1=5\\times 7-4\\times 1=31$$．  知识点：直接计算型的定义新运算 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2660", "queId": "3b428ffccc8c413992fd7a26b5fb5856", "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13题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$0.\\underbrace{00\\cdots 0}_{10个0}3\\times 0.\\underbrace{00\\cdots 0}_{10个0}25$$的积为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.\\underbrace{00\\cdots 0}_{10个0}75$$ "}], [{"aoVal": "B", "content": "$$0.\\underbrace{00\\cdots 0}_{20个0}75$$ "}], [{"aoVal": "C", "content": "$$0.\\underbrace{00\\cdots 0}_{21个0}75$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$0.\\underbrace{00\\cdots 0}_{10个0}3$$有$$11$$位小数，$$0.\\underbrace{00\\cdots 0}_{10个0}25$$有$$12$$位小数，  故两数相乘有$$11+12=23$$位小数，  而$$3\\times 25=75$$占了$$2$$位，故前面有$$23-2=21$$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": "1318", "queId": "30f8077df077407b899c42edc09ce79e", "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2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "买$$2$$瓶汽水和$$1$$瓶矿泉水一共要花$$7$$元，买$$4$$瓶汽水和$$3$$瓶矿泉水一共要花$$16$$元，那么买$$10$$瓶汽水和$$10$$瓶矿泉水一共要花元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$45$$ "}], [{"aoVal": "B", "content": "$$46$$ "}], [{"aoVal": "C", "content": "$$47$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["$$2$$汽$$+1$$水$$=7$$①，  $$4$$汽$$+3$$水$$=16$$②，  ②$$-$$①$$\\times 2$$得：水$$=2$$，所以汽$$=2.5$$，  故：$$10$$汽$$+10$$水$$=10\\times 2.5+10\\times 2=45$$（元）．  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": "813", "queId": "b5facd2a399942818cbb2c8ddc8ce603", "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": ["知识标签->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": ["枚举得$$42$$的因数有$$1$$、$$2$$、$$3$$、$$6$$、$$7$$、$$14$$、$$21$$、$$42$$，共$$8$$个． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1849", "queId": "bb8d4964e3774fcfbd9361427c7c0b20", "competition_source_list": ["小学高年级六年级其它新题型大集结第48题", "2015~2016学年浙江台州椒江区三甲中心校六年级上学期统测第16题2分", "2014~2015学年浙江台州椒江区三甲中心校六年级上学期统测第16题1分", "2007年第12届全国华杯赛竞赛初赛第2题10分"], "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": ["课内体系->思想->对应思想", "拓展思维->拓展思维->应用题模块->工程问题"], "answer_analysis": ["$1\\div\\left( \\dfrac{1}{30}+\\dfrac{1}{45}\\right)=1\\div\\dfrac{1}{18}=18$（分钟） "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "587", "queId": "d05f975ec95d460b990cd44017d32b37", "competition_source_list": ["2013年第11届创新杯四年级竞赛初赛第1题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2000$$年后为三个连续自然数乘积的第一个年份是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2013$$ "}], [{"aoVal": "B", "content": "$$2048$$ "}], [{"aoVal": "C", "content": "$$2146$$ "}], [{"aoVal": "D", "content": "$$2184$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["根据题意，只要把给出的这四个选项中的数分解质因数，进而把质因数重新组合，即可找出符合题意的数．  $$\\text{A}$$选项：$$2013=3\\times 671$$，故$$\\text{A}$$错误；  $$\\text{B}$$选项：$$2048=2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2\\times 2$$，故$$\\text{B}$$错误；  $$\\text{C}$$选项：$$2146=2\\times 1073$$，故$$\\text{C}$$错误；  $$\\text{D}$$选项：$$2184=2\\times 2\\times 2\\times 3\\times 7\\times 13=(2\\times 2\\times 3)\\times 13\\times (2\\times 7)=12\\times 13\\times 14$$，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": "2248", "queId": "ff8080814518d52401451908239902f4", "competition_source_list": ["2014年全国迎春杯五年级竞赛初赛第13题"], "difficulty": "3", "qtype": "single_choice", "problem": "甲、乙两人比赛折返跑，同时从$$A$$出发，到达$$B$$点后，立即返回，先回到$$A$$点的人获胜．甲先到达$$B$$点，在距离$$B$$点$$24$$米的地方遇到乙．相遇后，甲的速度减为原来的一半，乙的速度保持不变．在距离终点$$48$$米的地方，乙追上甲．那么，当乙到达终点时，甲距离终点还有（~~~~ ）米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["注意到第一次遇到时乙走了一个全程少$$24$$米，而两次加起来，乙一共走了两个全程少$$48$$米，则第二次遇到时乙也是走了一个全程少$$24$$米，而乙的速度不变，所以两人这两次相遇追及时间是相同的，而甲两次的路程分别为全程多$$24$$米和全程少$$72$$米，两次甲的路程差了$$96$$米，速度比$$2:1$$，则路程比$$2:1$$，说明甲第二次走了$$96$$米，乙走了$$96+24\\times 2=144$$米，甲乙速度比$$2:3$$，所以乙走完剩下的$$48$$米时甲应该只走了$$48\\div3\\times 2=32$$米，剩$$16$$米． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "149", "queId": "34f472d517c34effb1f0e746000caedb", "competition_source_list": ["2020年广东广州羊排赛三年级竞赛第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "动物园每天早上$$8:30$$开放，中午$$12$$点关闭，下午$$14:30$$开放，下午$$16:30$$关闭，那么动物园每天开放的时长是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$小时 "}], [{"aoVal": "B", "content": "$$6$$小时$$30$$分 "}], [{"aoVal": "C", "content": "$$6$$小时 "}], [{"aoVal": "D", "content": "$$5$$小时$$30$$分 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意分析可知，上午开放的时长是：$$12:00-8:30=3:30$$；  下午开放的时长是：$$16:30-14:30=2:00$$，  所以每天开放的时长是：$$3:30+2:00=5:30$$，即$$5$$小时$$30$$分钟，  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": "2395", "queId": "08d61224c5de4574926fcd3427d6554f", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$\\frac{\\dfrac{7}{18}\\times 4\\dfrac{1}{2}+\\dfrac{1}{6}}{13\\dfrac{1}{2}-3\\dfrac{3}{4}\\div \\dfrac{5}{16}}\\times 18=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$2\\frac{5}{9}$$ "}], [{"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$$=\\frac{\\dfrac{7}{18}\\times \\dfrac{9}{2}+\\dfrac{1}{6}}{13\\dfrac{1}{2}-\\dfrac{15}{4}\\times \\dfrac{16}{5}}\\times 18$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\frac{\\dfrac{7}{4}+\\dfrac{1}{6}}{13\\dfrac{1}{2}-12}\\times 18$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\dfrac{\\dfrac{7}{4}+\\dfrac{1}{6}}{1\\dfrac{1}{2}}\\times 18$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\dfrac{\\left( \\dfrac{7}{4}+\\dfrac{1}{6} \\right)\\times 2}{3}\\times 18$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=21+2$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=23$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2267", "queId": "6a1e2d92c9c4465ab8a576b80da62f92", "competition_source_list": ["2010年第8届创新杯六年级竞赛初赛第9题4分", "2010年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "某人所在单位每天早上有车$$7$$点准时到他家接他上班．有一天，他提前一个小时步行出门，朝单位方向走去，途中遇到接他上班的车，他坐上车到达单位比平时早$$10$$分钟，此人早上步行了分钟. ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$55$$ "}]], "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": ["本题关键是某人走了多少时间的路．某人走的路汽车走$$5$$分钟（提前$$10$$分钟，但汽车少走了某人所走路程的一个来回），汽车走$$5$$分钟说明某人与汽车提前$$5$$分钟相遇，也就是某人只走了$$60-5=55$$分钟即与汽车相遇．即某人走$$55$$分钟的路程汽车走$$5$$分钟．据此解答．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2351", "queId": "8aac50a7519fa10a01519fd6a0e20033", "competition_source_list": ["2016年全国华杯赛小学高年级竞赛初赛第2题"], "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": ["设甲速$${{v}_{1}}$$乙速$${{v}_{2}}$$，由第一次运动过程可得：$$\\frac{{{v}_{1}}}{{{v}_{2}}}=\\frac{140}{300-140}=\\frac{7}{8}$$，由第二次运动过程可得$$\\frac{{{v}_{1}}}{{{v}_{2}}+1}=\\frac{300-180}{180}=\\frac{2}{3}$$．两次过程中甲速度不变，化为连比为：$${{V}_{1}}:{{V}_{2}}:({{V}_{2}}+1)=14:16:21$$，由此可知这$$1$$米相当于$$21-16=5$$份，每一份为$$\\frac{1}{5}$$．故乙原来为$$\\frac{16}{5}$$米/秒． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1999", "queId": "d810d69820ef4fecbecc39205ec205a8", "competition_source_list": ["2008年第6届创新杯六年级竞赛初赛B卷第4题5分"], "difficulty": "2", "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，要赢得$$96 \\%$$的比赛，他至少还要赢． ", "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->能力->运算求解"], "answer_analysis": ["已经进行$$20$$场比赛，赢了$$95 \\%$$，  赢了$$20\\times 95 \\%=19$$（场）．  想要达到$$96 \\%$$的胜率，  并且以后每场都获胜，设再赢$$x$$场，  $$\\left( x+19 \\right)\\div \\left( 20+x \\right)=96 \\%$$．  解得$$x=5$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "208", "queId": "50f64b1a42874d9686106aaa82eac408", "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": ["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": ["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": "1330", "queId": "2ca801c0e6324939aaf3df5a1b8f70ae", "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$$18$$厘米，已知长是宽的$$2$$倍，那么长方形的长是（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$厘米 "}], [{"aoVal": "B", "content": "$$12$$厘米 "}], [{"aoVal": "C", "content": "$$6$$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：$$18\\div 2\\div (1+2)\\times 2$$  $$=9\\div 3\\times 2$$  $$=6$$ （厘米）  答：长方形的长是$$6$$厘米。  故选：C。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "833", "queId": "9f50c50e36a246719b723d0929c5aadf", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "张华、赵敏和钱勇三人隔不同的天数去图书馆一次：张华$$4$$天去一次；赵敏$$5$$天去一次；钱勇$$6$$天去一次．上次他们三人是星期天相遇的，再过（~ ）天他们才能再次相遇． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "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": ["$$4$$、$$5$$、$$6$$的最小公倍数是$$60$$，即再过$$60$$天再次相遇． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "538", "queId": "de3cae5278324d92a37e818bc846d4b9", "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->思想->逐步调整思想"], "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": "3340", "queId": "ff808081477bd84c014790a0440b346c", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "气象台预报``本市明天降雨概率是$$80 \\%''$$．对此信息，下列说法中正确的是（ ~ ~ ~)． ", "answer_option_list": [[{"aoVal": "A", "content": "本市明天将有$$80 \\%$$的地区降水． "}], [{"aoVal": "B", "content": "本市明天将有$$80 \\%$$的时间降水． "}], [{"aoVal": "C", "content": "明天肯定下雨． "}], [{"aoVal": "D", "content": "明天降水的可能性比较大． "}]], "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": ["降水概率指的是可能性的大小，并不是降水覆盖的地区或者降水的时间．$$80 \\%$$的概率也不是指肯定下雨，$$100 \\%$$的概率才是肯定下雨．$$80 \\%$$的概率是说明有比较大的可能性下雨． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2161", "queId": "389739d7bee54932bea5d17fa15a4c4c", "competition_source_list": ["2016年新希望杯小学高年级六年级竞赛训练题（六）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "小明在路边捡到一台怪钟，这个钟比正常的钟慢$$\\frac{1}{6}$$，现在的时间是中午$$12$$点整，怪钟也恰好是$$12$$点整，再过（~ ）分钟这台钟的时针和分针再次重合． ", "answer_option_list": [[{"aoVal": "A", "content": "$$65\\frac{3}{11}$$ "}], [{"aoVal": "B", "content": "$$65\\frac{6}{11}$$ "}], [{"aoVal": "C", "content": "$$78\\frac{6}{11}$$ "}], [{"aoVal": "D", "content": "$$78\\frac{9}{11}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["可以判断重合时时针在$$1$$点到$$2$$点之间，假设怪钟走过了$$x$$分钟，$$\\frac{x}{60}\\times 5=x-60$$，解得$$x=\\frac{720}{11}$$，怪钟走过$$\\frac{720}{11}$$分钟实际经过的时间是$$\\frac{720}{11}\\times \\frac{6}{5}=\\frac{864}{11}=78\\frac{6}{11}$$（分）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2088", "queId": "d02b98aeb50542b6af5e1646248b75e4", "competition_source_list": ["2017年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": ["拓展思维->能力->实践应用"], "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": "975", "queId": "002c7d6b12cd4c30afb33b6789ace1ad", "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3题"], "difficulty": "1", "qtype": "single_choice", "problem": "某年某月有30天，4个星期六，5个星期五，那么这年这个月10号是星期(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "四 "}], [{"aoVal": "B", "content": "五 "}], [{"aoVal": "C", "content": "六 "}], [{"aoVal": "D", "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": ["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.(如下为某月日历) 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号应为星期五，3号、10号、17号、24号，就为4个星期六，即10号为星期六，选C ", "<p>该月有$$30$$天，一个星期有$$7$$天，$$30\\div7=4$$（个）$$\\cdots\\cdots2$$（天），有$$4$$个星期六，$$5$$个星期五，所以该月$$1$$号，不能是星期五，不然就是$$5$$个星期六，产生矛盾．那么该月$$2$$号就是星期五，$$3$$号是星期六，所以$$10$$号也是星期六，选$$\\text{C}$$．</p>"], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1159", "queId": "58cb8c59b70d423295e4002326ef551c", "competition_source_list": ["2017年北京小升初", "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7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$3$$台同样的车床$$6$$小时可加工零件$$1440$$个。如果增加$$2$$台同样的车床，且每台车床每小时多加工$$12$$个零件，加工$$3680$$个零件需要（  ）小时。 ", "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->归一归总问题->归总问题"], "answer_analysis": ["原来$$1$$台车床$$1$$小时加工零件$$1440\\div 3\\div 6=80$$（个），现有车床$$3+2=5$$（台），且$$1$$台车床$$1$$小时可加工$$80+12=92$$（个）零件，所以加工$$3680$$个零件需要$$3680\\div 5\\div 92=8$$（小时）。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3182", "queId": "1558244f82fe453a9507746d8600674e", "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": "整数$$N=1357911131517\\ldots \\ldots 201320152017$$是由$$1 \\tilde{  } 2017$$中所有的奇数由小到大的顺序依次写出得到，那么这个整数中出现了（~ ）次``$$17$$''（相邻两位组成的两位数是$$17$$）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$21$$ "}], [{"aoVal": "C", "content": "$$31$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["第一类，奇数末位$$17$$形式结尾，形如$$\\square \\square 17$$，首位是$$0$$，$$1$$，第二位是$$0 \\tilde{  } 9$$有$$20$$种，首位是$$2$$，第三位是$$0$$，只有$$2017$$，共$$21$$种；第二类，奇数末位以$$1$$结尾，且首位是$$7$$，那么有$$7173$$，$$701703$$，$$711713$$，$$721723$$，$$\\ldots \\ldots $$，共$$32$$种． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1517", "queId": "99e5f1a49ae441a2a07ba566eac2b441", "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$$1$$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": "$$20$$ "}], [{"aoVal": "E", "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$$10\\div2=5$$（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": "829", "queId": "bf4a58bbcd0e4f71b1fade3d8a953315", "competition_source_list": ["2017年华杯赛六年级竞赛初赛", "2017年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "小明把$$6$$个数分别写在三张卡片的正面和反面，每个面上写一个数，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$$28\\text{、}40\\text{、}49$$ ，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": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->整除"], "answer_analysis": ["解：因为$$28$$、$$40$$、$$49$$奇偶性不一样，根据卡片正反面上两个数字和相等，  若数字和为偶数，则$$28$$的反面为偶数且至少为$$22$$，与反面上的数都只能被$$1$$和它自己整除相矛盾，  所以数字和为奇数，所以$$49$$的背面是$$2$$，和为$$49+2=51$$，  从而反面上的平均数是$$\\left( 51\\times 3-28-40-49 \\right)\\div 3=36\\div 3=12$$。  故选：B。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "83", "queId": "a206df9548104d23a5e6da7f47fa2302", "competition_source_list": ["2017年河南郑州东风杯竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "蛋君在桌上摆了一些不同形状的积木，如下所示：■◇◇●●●■◇◇●●●■◇◇●●●$$\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot $$，照这样的规律摆，蛋君摆第$$40$$个时，需要选择的图形是． ", "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": "1788", "queId": "9b3eb73a74444b96bbf2d21624af4cbd", "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5题3分"], "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$$1$$个．如果每天吃$$4$$个，不到$$11$$天就吃完了，最后一天也是只吃了$$1$$个．那么，这些糖果原来有个 ", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$37$$ "}], [{"aoVal": "D", "content": "$$41$$ "}]], "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": ["暂无 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2701", "queId": "90c6eb38fcc14e9798f827e5126900ed", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "计算$$0.1\\dot{6}+0.\\dot{3}+0.5=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.\\dot{9}\\dot{6}$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$1.\\dot{3}$$ "}], [{"aoVal": "D", "content": "$$1.5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["$$0.1\\dot{6}+0.\\dot{3}+0.5$$  $$=\\frac{16-1}{90}+\\frac{1}{3}+\\frac{1}{2}$$  $$=\\frac{16-1}{90}+\\frac{30}{90}+\\frac{45}{90}=1$$ "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3058", "queId": "f385e68acbf34b678c1d2dff8e0ae2be", "competition_source_list": ["2016年全国华杯赛小学中年级四年级竞赛初赛B卷"], "difficulty": "1", "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，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，如抽出的牌是$$3$$，$$8$$，$$8$$，$$9$$，那么算式为（$$9$$-$$8$$）$$\\times $$$$8$$$$\\times $$$$3$$或（$$9$$-$$8$$÷$$8$$）$$\\times $$$$3$$等．在下面四个选项中，唯一无法凑出$$24$$点的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$，$$2$$， $$3$$，$$3$$ "}], [{"aoVal": "B", "content": "$$1$$，$$5$$，$$5$$，$$5$$ "}], [{"aoVal": "C", "content": "$$2$$，$$2$$，$$2$$，$$2$$ "}], [{"aoVal": "D", "content": "$$3$$，$$3$$，$$3$$，$$3$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["选项$$A$$：$$\\left( {1+3} \\right)\\times2\\times3=24$$  选项$$B$$：$$5\\times\\left[ {5-\\left( {1÷5} \\right) } \\right] =24$$  选项$$D$$：$$3\\times3\\times3-3=24$$．  也可使用排除法，没有$$1$$时，$$2\\times2\\times2\\times2=16$$，乘法达不到$$24$$，加减除更达不到$$24$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1026", "queId": "06628bcaff494432af59bf3f0cb23552", "competition_source_list": ["其它改编自2015年全国希望杯六年级竞赛初赛第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个分数，若分母减$$1$$，化简后得$$\\frac{1}{3}$$；若分子加$$4$$，化简后得$$\\frac{1}{2}$$，则这个分数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{16}$$ "}], [{"aoVal": "B", "content": "$$\\frac{7}{22}$$ "}], [{"aoVal": "C", "content": "$$\\frac{9}{28}$$ "}], [{"aoVal": "D", "content": "$$\\frac{11}{34}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["$$\\left { \\begin{matrix}\\frac{b}{a-1}=\\frac{1}{3}    \\frac{b+4}{a}=\\frac{1}{2}   \\end{matrix} \\right.$$  得：$$\\left { \\begin{matrix}a=22    b=7   \\end{matrix} \\right.$$  这个分数是$$\\frac{7}{22}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2150", "queId": "2a97d1435e3844c1810ea28254a9e40f", "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": ["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->平均速度->公式法"], "answer_analysis": ["假设距离为$$20$$千米，则去用$$5$$小时，回来$$1$$小时，共六小时，共$$40$$千米，所以平均速度是：$$40\\div 6=\\frac{20}{3}$$时． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "981", "queId": "04d9cf1fb40545d4b03c39fac47781ae", "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->思想->对应思想"], "answer_analysis": ["根据题意分析可知，从当哥哥像弟弟现在这么大时，弟弟的年龄恰好是哥哥的年龄的一半这句话可知，可以设哥哥今年$$x$$岁，则弟弟今年$$\\left( 35-x \\right)$$岁，根据题意得：  $$\\begin{eqnarray} x-\\left( 35-x \\right)\\&=\\&35-x-\\frac{1}{2}\\left( 35-x \\right)   x-35+x\\&=\\&35-x-17.5+\\frac{1}{2}x   x+x+x-\\frac{1}{2}x\\&=\\&35-17.5+35   2.5x\\&=\\&52.5   x\\&=\\&21\\end{eqnarray}$$  哥哥为$$21$$岁，则弟弟今年$$35-21=14$$（岁）．  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": "1188", "queId": "d9c992e6675d40c0adee49155f80543b", "competition_source_list": ["2008年陈省身杯小学高年级六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "中心小学六年级学生参加：``信息学奥林匹克竞赛''，获奖的学生占六年级学生总数的$$\\frac{1}{40}$$，其中获一等奖的占获奖学生人数的$$\\frac{11}{17}$$．已知六年级学生总人数是$$680$$人，求获一等奖的学生人数．下面列式中正确的有（~~ ）个？  （$$1$$）$$680\\times \\frac{1}{40}\\times \\frac{11}{17}$$（$$2$$）$$680\\times \\left( \\frac{1}{40}\\times \\frac{11}{17} \\right)$$（$$3$$）$$680\\times \\frac{1}{40}$$（$$4$$）$$680\\times \\frac{11}{17}$$ ", "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->分百应用题->量率对应已知单位1"], "answer_analysis": ["根据题意得，获奖的学生占六年级学生总数的$$\\frac{1}{40}$$，其中获一等奖的占获奖学生人数的$$\\frac{11}{17}$$．已知六年级学生总人数是$$680$$人，求获一等奖的学生人数．可列式 $$680\\times \\frac{1}{40}\\times \\frac{11}{17}$$即$$680\\times \\left( \\frac{1}{40}\\times \\frac{11}{17} \\right)$$，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": "831", "queId": "b6244f35668641239a90e212123a91e3", "competition_source_list": ["2005年第3届创新杯五年级竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$1$$，$$2$$，$$3$$，$$5$$，$$7$$，$$8$$中不是一个整数的平方的个位数，共有． ", "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": ["拓展思维->思想->对应思想"], "answer_analysis": ["因为整数的平方的个位数不能为$$2$$，$$3$$，$$7$$，$$8$$，所以在$$1$$，$$2$$，$$3$$，$$5$$，$$7$$，$$8$$中，不可能是一个整数的平方的个位数的数共有$$4$$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": "3034", "queId": "bca2e7e7261e42e4ab6ffa40ead44f01", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有如下四个结论：  ①~~ 两个方程是同解方程，则有相同的解；  ②~~ 有相同解的两个方程组是同解方程组；  ③~~ 一个一元一次方程，如果有解，就一定有整数解；  ④~~ 一个方程，如果有整数解，则必定有无穷多个非整数解．  错误结论是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "①，②和③ "}], [{"aoVal": "B", "content": "①，②和④ "}], [{"aoVal": "C", "content": "①，③和④ "}], [{"aoVal": "D", "content": "②，③和④ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["①正确；②不正确，因为两个方程虽然有相同的解，也可能还有不同的解，不一定是同解方程；③不正确，例如$$4x=3+2x$$有解，但没有整数解；④不正确，如$$2x=4x-4$$仅有$$1$$个解． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1700", "queId": "8cfafa945b9f4f1d8a784282fc087063", "competition_source_list": ["2013年第11届全国创新杯五年级竞赛第8题5分", "2016年创新杯五年级竞赛训练题（四）第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "四位歌手轮唱一首含有四个相等乐段的歌曲，每人把这首歌曲唱三遍就结束，第一位歌手开始唱第二乐段时第二位歌手开始唱，第一位歌手开始长第三乐段时第三位歌手开始唱，第一位歌手开始唱第四个乐段时第四人同时开始唱．则第四个人同时唱的时间占总的歌唱时间的（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{15}$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{15}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{5}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["所有歌手都将整个歌曲唱了三遍，故易知总时间段有$$15$$段，其中$$9$$段是重合（同时在唱）． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3444", "queId": "f7fe949062504c55a7832281eeed199c", "competition_source_list": ["2017年河南郑州豫才杯五年级竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "《红楼梦》第二十回中，写贾环和丫鬟莺儿掷骰子的情节．骰子，古代汉族民间娱乐用来投掷的博具．骰子是很容易制作和取得的乱数产生器．如若当两枚骰子的点数之和大于$$7$$时，莺儿得$$1$$分，否则贾环得$$1$$分，这个游戏公平吗？（~ ） ", "answer_option_list": [[{"aoVal": "A", "content": "公平~~~~~~ "}], [{"aoVal": "B", "content": "贾环获胜可能性大 "}], [{"aoVal": "C", "content": "莺儿获胜的可能性大 "}], [{"aoVal": "D", "content": "无法判断 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["点数之和大于$$7$$的可能情况有$$15$$种，而点数之和不大于$$7$$的可能情况有$$21$$种，所以贾环获胜的可能性大． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "210", "queId": "5e9a64a9d40e444781e5d75506989551", "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~\\uline{~~~~~~~~~~}~． ", "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->整体思想"], "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": "2643", "queId": "43da80bfddbc4685a5e645e3cbc0caee", "competition_source_list": ["2016年全国AMC六年级竞赛8第19题"], "difficulty": "2", "qtype": "single_choice", "problem": "\\textbf{2016年 AMC 8 第$$19$$题}  The sum of $$25$$ consecutive even integers is $$10000$$. What is the largest of these $$25$$ consecutive integers? ", "answer_option_list": [[{"aoVal": "A", "content": "$$360$$ "}], [{"aoVal": "B", "content": "$$388$$ "}], [{"aoVal": "C", "content": "$$412$$ "}], [{"aoVal": "D", "content": "$$416$$ "}], [{"aoVal": "E", "content": "$$424$$ "}]], "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->等差数列"], "answer_analysis": ["翻译：$$25$$个连续的偶数和是$$10000$$，那么这$$25$$个连续偶数中最大的一个是？  等差数列的中间项是$$10000\\div 25=400$$，比$$400$$大的还有$$(25-1)\\div 2=12$$（项），所以最大的一项是$$400+12\\times 2=424$$． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2147", "queId": "a69cf88fda124133975b20b1c9c5b10e", "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20题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "自动扶梯由下向上行驶，男孩沿着扶梯从顶向下走到底，他走了$$150$$级．女孩沿着扶梯从下向上走到顶，走了$$75$$级．如果男孩行走的速度是女孩的$$3$$倍．那么，可以看到的自动扶梯的级数是~\\uline{~~~~~~~~~~}~级． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$150$$ "}], [{"aoVal": "D", "content": "$$90$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["假设女孩速度是$$1$$级$$/$$秒，则男孩速度为$$3$$级$$/$$秒．  则男孩逆扶梯行走时间：$$150\\div3=50$$秒，  女孩顺扶梯行走时间为：$$75$$秒．  则扶梯长度$$=50\\times $$（$$3-$$扶梯速度）①，  扶梯长度$$=75\\times$$（$$1+$$扶梯速度）②，  ①$$\\times2+$$②$$\\times3$$得：$$5$$扶梯长度$$=600$$，可得扶梯长度$$=120$$级． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2139", "queId": "61c38ae5ff084241be19eb11019cc55d", "competition_source_list": ["2015年第4届广东广州羊排赛六年级竞赛第7题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "小青从家里走去学校，速度为$$30$$米/分；放学回家走同一段路，速度为$$60$$米/分．那么小青走这段路时的平均速度是米/分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$48$$ "}]], "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": ["设这段路长度为$$60$$，则平均速度为$$(60\\times 2)\\div (60\\div 30+60\\div 60)=40$$（米/分）．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3359", "queId": "badaf632626d4e79a03923a7bcb8e8fc", "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->拓展思维->计数模块->加乘原理->组数问题->一般组数问题"], "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": "3172", "queId": "10bf0bae5ec84e2dab65b9b3f2562fe5", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "暗箱中有五张分别写$$1$$、$$2$$、$$3$$、$$4$$、$$5$$数字的卡片，从中随机摸出三张，由这三张上的数字组成三位数中，能被$$3$$整除的可能性是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{10}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{5}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["五张卡片取三张共有$$\\text{C}_{5}^{3}=10$$种取法，又因为$$1+2+3+4+5=15$$是$$3$$的倍数，故取走$$2$$张和为$$3$$的倍数的卡片都满足条件．枚举可得$$1+2$$、$$1+5$$、$$2+4$$、$$4+5$$共有$$4$$种情况，可能性为$$\\frac{4}{10}=\\frac{2}{5}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3032", "queId": "d336be10b8d04d5282e6a2e418046e66", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a$$，$$b$$都表示数，规定：若$$a\\textgreater b$$，则$$\\left\\textbar{} a-b \\right\\textbar=a-b$$；若$$a=b$$，则$$\\left\\textbar{} a-b \\right\\textbar=0$$；若$$a\\textless{}b$$，则$$\\left\\textbar{} a-b \\right\\textbar=b-a$$．  现在请你从$$1\\backsim 100$$这$$100$$个自然数中任选$$50$$个数，从小到大记为$${{a}_{1}}$$，$${{a}_{2}}$$，$$\\cdots $$，$${{a}_{50}}$$；再将剩下的$$50$$个数从大到小记为$${{b}_{1}}$$，$${{b}_{2}}$$，$$\\cdots $$，$${{b}_{50}}$$，计算：$$\\left\\textbar{} {{a}_{1}}-{{b}_{1}} \\right\\textbar+\\left\\textbar{} {{a}_{2}}-{{b}_{2}} \\right\\textbar+\\cdots +\\left\\textbar{} {{a}_{50}}-{{b}_{50}} \\right\\textbar=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$900$$ "}], [{"aoVal": "B", "content": "$$1600$$ "}], [{"aoVal": "C", "content": "$$2500$$ "}], [{"aoVal": "D", "content": "$$2700$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "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->分类讨论型定义新运算"], "answer_analysis": ["构造$${{a}_{1}}$$，$${{a}_{2}}\\cdots \\cdots $$，$${{a}_{50}}$$分别是$$51$$，$$52$$，$$\\cdots \\cdots 100$$，$${{b}_{1}}$$ ，$${{b}_{2}}$$，$$\\cdots \\cdots {{b}_{50}}$$分别是$$50$$，$$49\\cdots \\cdots 1$$，则原式$$=51-50+52-49+53-48+\\cdots \\cdots +100-50=1+3+5+\\cdots \\cdots 100=2500$$．  答案：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1319", "queId": "2404caee6c0a43cbaf0449edc5765126", "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4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "本学期前五次测验的平均分为$$92$$，第六次的测验成绩比六次测验成绩的平均分高$$5$$分，请问第六次的测验成绩是分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$95$$ "}], [{"aoVal": "B", "content": "$$97$$ "}], [{"aoVal": "C", "content": "$$98$$ "}], [{"aoVal": "D", "content": "$$99$$ "}]], "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": ["因为由题干可知，本学期前五次测验的平均分为$$92$$，  则前五次测验的总分为：$$92\\times 5=460$$（分），  第六次的测验成绩比六次测验成绩的平均分高$$5$$分，  设第六次的测验成绩为$$x$$分，  则可列方程为：$$x-\\left( 460+x \\right)\\div 6=5$$，  解得$$x=98$$，  所以第六次的测验成绩是$$98$$分．  故选$$\\text{C}$$． ", "<p>移多补少：现在的平均数是$92+5\\div5=93$，</p>\n<p>所以第六次测验成绩是$93+5=98$．</p>\n<p>故选$$\\text{C}$$．</p>"], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3267", "queId": "32258151ec184f978aa541929de57f38", "competition_source_list": ["2016年创新杯小学高年级五年级竞赛训练题（二）第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$、$$2$$、$$3$$、$$4$$、$$5$$中任取$$3$$个组成一个三位数，其中不能被$$3$$整除的三位数有（~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$42$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["不能取$$\\left( 1,2,3 \\right)$$、$$\\left( 1,3,5 \\right)$$、$$\\left( 23,4 \\right)$$、$$\\left( 3,4,5 \\right)$$，$$A_{5}^{3}-4\\times A_{3}^{3}=60-24=36$$（个）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "287", "queId": "f19febc7eeaa4576aef83f87b48de7e2", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛复赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "若一个整数中某个数字等于其他所有数字之和，则称这样的整数为``$$S$$数''（如$$871$$，$$8=7+1$$）．那么最大的四位``$$S$$数''与最小的四位``$$S$$数''的差是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8899$$ "}], [{"aoVal": "B", "content": "$$7789$$ "}], [{"aoVal": "C", "content": "$$6689$$ "}], [{"aoVal": "D", "content": "$$5589$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意可得最大的``$$S$$数''为$$9900$$，最小的``$$S$$数''为$$1001$$，则两者之间的差为$$8899$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1604", "queId": "49c9569050314b37b1dd09763f393313", "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复赛第7题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "一块肥皂使用一次，它的体积减少10\\%，当肥皂使用n次后，它的体积小于原来的一半，那么n的最小值是(  ). ", "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->分百应用题->转化单位1"], "answer_analysis": ["由于$${{\\left( 1-10 \\% \\right)}^{6}}\\approx53.1 \\%$$，$${{\\left( 1-10 \\% \\right)}^{7}}\\approx47.8 \\%$$，所以$$n$$的最小值为7. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "591", "queId": "ab49bc9989f4475aa8e6ef037237443b", "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18题8分"], "difficulty": "3", "qtype": "single_choice", "problem": "有$$2012$$名学生排成一行，从左向右依次编成$$1$$，$$2$$，\\ldots，$$2012$$号．第一次从左向右``$$1$$，$$2$$''报数，凡报到$$2$$的学生留下；从第二次起，每次都是让留下的学生从左向右``$$1$$，$$2$$，$$3$$''报数，凡报到$$3$$的学生留下，直到只留下$$1$$名学生．请问这名最后留下的学生编号是？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1279$$ "}], [{"aoVal": "B", "content": "$$1458$$ "}], [{"aoVal": "C", "content": "$$2000$$ "}]], "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": ["第一次报数后留下：$$2$$、$$4$$、$$6$$、\\ldots\\ldots（$$2$$的倍数）；  第二次报数后留下：$$6$$、$$12$$、$$18$$、\\ldots\\ldots（$$6=2\\times 3$$的倍数）；  第三次报数后留下：$$18$$、$$36$$、$$54$$、\\ldots\\ldots（$$18=2\\times {{3}^{2}}$$的倍数）；  第四次报数后留下：$$54$$、$$108$$、$$162$$、\\ldots\\ldots（$$54=2\\times {{3}^{3}}$$的倍数）；  第五次报数后留下：$$162$$、$$324$$、$$486$$、\\ldots\\ldots（$$162=2\\times {{3}^{4}}$$的倍数）；  第六次报数后留下：$$486$$、$$972$$、$$1458$$、\\ldots\\ldots（$$486=2\\times {{3}^{5}}$$的倍数）；  第七次报数后留下：$$1458$$（$$1458=2\\times {{3}^{6}}$$的倍数）；  即最后留下的学生编号是$$1458$$．  或者可以按照规律，最后留下的学生的编号是$$2\\times {{3}^{n}}$$，$$n$$要尽可能大；  取$$n=6$$，可得最后留下的学生编号是$$2\\times {{3}^{6}}=1458$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1120", "queId": "33cbeebb7d0b4643a928575cb325fc19", "competition_source_list": ["2006年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "某班在一次数学考试中，平均成绩为78分，男生的平均成绩为75.5分，女生的平均成绩为81分，则班上男、女生人数之比为(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "5:4 "}], [{"aoVal": "B", "content": "6:5 "}], [{"aoVal": "C", "content": "7:6 "}], [{"aoVal": "D", "content": "7:5 "}]], "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": ["设男生人数为$$a$$人，女生人数$$b$$人. $$75.5a+81b=78 \\times \\left( a+b \\right)$$， $$25a=30b$$， $$\\frac{a}{b}=\\frac{30}{25}=\\frac{6}{5}$$ "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1186", "queId": "2facccd9ebd44ad28089da85a8b30404", "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3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "大李和小李比赛吃包子，大李比小李多吃$$20$$个，大李比小李的$$4$$倍多$$2$$个，大李和小李一共吃了个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$26$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "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->非整数倍差倍剩余"], "answer_analysis": ["根据题意大李比小李的$$4$$倍多$$2$$个，  则大李比小李多$$3$$倍加$$2$$个，  即小李吃的包子数$$\\times3+2=20$$（个）．  小李吃了$$(20-2)\\div3=6$$（个），大李吃了$$6+20=26$$（个），  两个人一共吃了$$26+6=32$$（个）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2858", "queId": "80916f33508f404f8330941848e1fdbd", "competition_source_list": ["2016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": "1", "qtype": "single_choice", "problem": "古希腊著名的毕达哥拉斯学派把$$1$$、$$3$$、$$6$$、$$10$$．这样的数称为``三角形数''，而把$$1$$、$$4$$、$$9$$、$$16$$$$\\cdots \\cdots $$这样的数称为``正方形数''，从上图中可以发现，任何一个大于$$1$$的``正方形数''都可以看作两个相邻的``三角形数''之和．下列等式中，符合这一个规律的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$13=3+10$$ "}], [{"aoVal": "B", "content": "$$25=9+16$$ "}], [{"aoVal": "C", "content": "$$36=15+21$$ "}], [{"aoVal": "D", "content": "$$49=24+25$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->数形结合找规律"], "answer_analysis": ["任何一个大于$$1$$的``正方形数''都可以看作两个相邻``三角形数''之和，由于``正方形数''为两个``三角形数''之和，正方形数可以用代数式表示为：$${{\\left( n+1 \\right)}^{2}}$$，两个三角形数分别表示为$$\\frac{1}{2}n(n+1)$$和$$\\frac{1}{2}(n+1)(n+2)$$，所以由正方形数可以推得$$n$$的值，然后求得三角形数的值，  $$\\text{A}$$项，$$13$$不是``正方形数''，故$$\\text{A}$$项错误；  $$\\text{B}$$项，$$9$$与$$16$$不是``三角形数''，故$$\\text{B}$$项错误；  $$\\text{C}$$项，$$36$$为``正方形数''，$$15$$、$$21$$为两个相邻``三角形数''，故$$\\text{C}$$项正确；  $$\\text{D}$$项，$$24$$与$$25$$不是``三角形数''，故$$\\text{D}$$项错误．  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": "2075", "queId": "d90bdd52501a4edfae2d415ce934fa66", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（一）第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "车间里，甲和乙每小时都能装配$$1$$辆汽车．甲装配$$3$$辆汽车要用$$1$$小时维护，乙装配$$4$$辆汽车要用$$1.5$$小时维护．甲、乙同时工作，装配$$200$$辆汽车最少要用小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$134$$ "}], [{"aoVal": "B", "content": "$$135$$ "}], [{"aoVal": "C", "content": "$$136$$ "}], [{"aoVal": "D", "content": "$$137$$ "}]], "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": ["$$3+1=4$$与$$4+1.5=5.5$$的最小公倍数是$$44$$，前$$44\\times 3=132$$个小时共装配$$\\frac{132}{4}\\times 3+\\frac{132}{5.5}\\times 4=195$$辆，再过$$4$$小时，前$$2.5$$小时装$$2.5\\times 2=5$$辆，后$$1.5$$小时装完．共需$$132+4=136$$小时． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "443", "queId": "8ecf325916944fb29448db9c437cf826", "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8题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲，乙，丙，丁四个人比赛乒乓球，每两个人都要赛一场，结果甲胜丁，并且甲，乙，.丙胜的场数相同，那么丁胜的场数是(  )场. ", "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->条件型逻辑推理->神推理"], "answer_analysis": ["依题意可知甲、乙、丙、丁四人的比赛一共要赛6场.因为，甲胜丁，所以甲至少胜一场.①当甲胜一场时，由于乙、丙与甲胜的场数相同，所以乙、丙也各胜一场，则甲、乙、丙三人共胜3场，则还剩下三场.又因为乒乓球比赛不存在着平局的可能，所以丁胜3场，与``甲胜丁''矛盾.② 当甲胜2场时，则乙、丙也各胜2场，此时，甲、乙、丙三人已胜了6场，故丁只能胜0场.因此，由以上可知丁胜的场数是0场，故选A. "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2251", "queId": "ff8080814670afec01467438443b0557", "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->拓展思维->行程模块->比例解行程问题->行程中的比例"], "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": "1805", "queId": "85b00a85d149422f9dd9b152002dc124", "competition_source_list": ["2018年湖北武汉新希望杯六年级竞赛训练题（一）第5题", "2017年新希望杯六年级竞赛训练题（一）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个等腰三角形的周长是$$80$$厘米，其中两条边长度的比是$$1:2$$，则底长厘米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$16$$或$$20$$ "}], [{"aoVal": "D", "content": "$$\\frac{80}{3}$$ "}]], "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": ["三边之比为$$1:2:2$$，$$80\\div (1+2+2)=16$$（厘米）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "658", "queId": "34ba917c29b64c239a7873af4d8900ae", "competition_source_list": ["五年级其它五升六讲义", "2019年陕西西安长安区交通大学附属中学航天学校小升初入学真卷6第13题7分", "2016年第3届广东深圳鹏程杯六年级竞赛集训材料第十章从算术到代数第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个两位数，如果把数码$$1$$加写在它的前面，那么可以得到一个三位数，如果把$$1$$写在它的后面，那么也可以得到一个三位数，而且这两个三位数相差$$414$$，求原来的两位数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$57$$ "}], [{"aoVal": "C", "content": "$$58$$ "}], [{"aoVal": "D", "content": "$$59$$ "}], [{"aoVal": "E", "content": "$$60$$ "}]], "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->位值原理的整体换元"], "answer_analysis": ["设原来的两位数为$$x$$则依题意可得$$x+100+414=10x+1$$解得$$x=57$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "157", "queId": "1f60f5b1402a4cdeb69ef0410e39b75f", "competition_source_list": ["2012年全国迎春杯四年级竞赛初赛第3题"], "difficulty": "0", "qtype": "single_choice", "problem": "在下图的每个方框中填入一个数字，使得乘法竖式成立．那么，这个算式的乘积是 . ", "answer_option_list": [[{"aoVal": "A", "content": "$$837$$ "}], [{"aoVal": "B", "content": "$$827$$ "}], [{"aoVal": "C", "content": "$$297$$ "}], [{"aoVal": "D", "content": "$$397$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["如图 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1104", "queId": "c27d6e25f2f94cf9bb5b3cb73d7905cc", "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": "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": ["设甲、乙两根绳子的长度都为$$9$$米，则：  甲剩下：  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde9\\times \\left( 1-\\frac{1}{3} \\right)-\\frac{1}{3}$$  $$=6-\\frac{1}{3}$$  $$=5\\frac{2}{3}$$（米），  乙剩下：  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left( 9-\\frac{1}{3} \\right)\\times \\left( 1-\\frac{1}{3} \\right)$$  $$=\\frac{26}{3}\\times \\frac{2}{3}$$  $$=5\\frac{7}{9}$$（米），  $$5\\frac{2}{3}$$米$$ ~\\textless{} ~5\\frac{7}{9}$$米．  答：乙绳剩下部分长．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2416", "queId": "0c36da6268164557a5db93721ef2dc44", "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$$10$$个数的整数部分为． ", "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->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": ["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": "2160", "queId": "2f6d2f7baa904eb092f39282fbf2ffef", "competition_source_list": ["2014年第13届上海小机灵杯小学高年级五年级竞赛初赛第12题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "从甲地到乙地的路只有上坡与下坡，全程$$21$$千米．如果上坡的速度是$$4$$千米/时，下坡的速度是$$6$$千米/时，从甲地到乙地需$$4.25$$小时，那么从乙地到甲地需要~\\uline{~~~~~~~~~~}~小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3.5$$ "}], [{"aoVal": "B", "content": "$$4.5$$ "}], [{"aoVal": "C", "content": "$$5.5$$ "}]], "knowledge_point_routes": ["课内体系->思想->转化与化归的思想", "拓展思维->七大能力->实践应用"], "answer_analysis": ["假设从甲到乙全为上坡：$$21\\div 4=5.25\\rm h$$，$$5.25-4.25=1\\rm h$$，  每$$1\\rm km$$下坡算成上坡，多算$$\\dfrac{1}{4}-\\dfrac{1}{6}=\\dfrac{1}{12}\\rm h$$，$$1\\div \\dfrac{1}{12}=12\\left( \\rm km \\right)$$．  乙到甲上坡$$12\\rm km$$，下坡$$9\\rm km$$，用时$$12\\div 4+9\\div 6=4.5\\left( \\rm h \\right)$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1915", "queId": "9ca2bcf9d26e4af8996e5059c7813cb2", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "小笨以$$60$$元的价格卖了两块猪肉．其中一块盈利$$20  \\% $$，另一块亏损$$20  \\% $$，则小笨最后~\\uline{~~~~~~~~~~}~（填``盈利''或``亏损''）了~\\uline{~~~~~~~~~~}~元． ", "answer_option_list": [[{"aoVal": "A", "content": "盈利；5 "}], [{"aoVal": "B", "content": "亏损；5 "}], [{"aoVal": "C", "content": "盈利；12 "}], [{"aoVal": "D", "content": "亏损；12 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["盈利$$20  \\% $$，则原价是$$50$$元；亏损$$20  \\% $$，则原价是$$75$$元，所以原价总共是$$125$$元，而卖出 价总共是$$120$$元，所以小笨最后亏损了$$5$$元钱．  方程解法：盈利部分，$$1.2x=60$$，得$$x=50$$，亏损部分，$$0.8x=60$$，$$x=75$$，盈利$$10$$元和亏损$$15$$元，一共亏损$$5$$元。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1706", "queId": "a3cec83556944d6b89285598e636f285", "competition_source_list": ["2018年IMAS小学高年级竞赛（第一轮）第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "餐厅有面粉$$240\\text{kg}$$，原计划使用$$8$$天．改变食谱后，每天比原计划少用$$6\\text{kg}$$．请问这批面粉实际使用了多少天？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["按照原计划，每天使用$$240\\div 8=30\\text{kg}$$．  改变食谱后，每天使用$$30-6=24\\text{kg}$$．  得知这批面粉实际使用了$$240\\div 24=10$$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": "3044", "queId": "eeabab1324354d62a181518d07ace50f", "competition_source_list": ["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，$$a\\oplus b=2\\times b-1$$；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": ["知识标签->拓展思维->计算模块->定义新运算->直接运算型"], "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": "2021", "queId": "ea724b1bdd8e4ed98f9f9920ce2d1f60", "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25题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "用$$300$$粒种子做发芽试验，结果有$$5 \\%$$的种子没有发芽，发芽的种子有多少粒？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$600$$ "}], [{"aoVal": "C", "content": "$$285$$ "}], [{"aoVal": "D", "content": "$$275$$ "}]], "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": ["因为有$$5 \\%$$的种子没有发芽，那么有$$1-5 \\%=95 \\%$$的种子发芽了，  所以发芽的种子有$$300\\times95 \\%=285$$（粒）．  故答案为：$$285$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "839", "queId": "8d3e25ee49c1433cae2e4108cedb1ae4", "competition_source_list": ["2021年新希望杯六年级竞赛初赛第22题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2021$$最多可以表示成~\\uline{~~~~~~~~~~}~个连续自然数的和． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$42$$ "}], [{"aoVal": "C", "content": "$$58$$ "}], [{"aoVal": "D", "content": "$$47$$ "}], [{"aoVal": "E", "content": "$$39$$ "}]], "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": ["本题考查学生的拆分乘积的能力和计算能力  $$2021=43\\times47$$，所以将$$43$$看作连续自然数的中间数，所以连续的数为$$20$$，$$21$$，$$22$$，$$\\cdots~ $$，$$43$$，$$\\cdots $$，$$66$$，此和为$$2021$$.  因此，$$2021$$最多可表示$$47$$个自然数的和． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "574", "queId": "0a9c74bbf7ea4becb2051ff36e9d9a3f", "competition_source_list": ["2011年五年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "已知十位数$$\\overline{1a2a3a4a5a}$$能被$$11$$整除，则$$a=$$（  ）。 ", "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->数论模块->整除->整除特征->基础复合数字的整除特征"], "answer_analysis": ["根据能被$$11$$整除的数的特征可得．奇数位上的数字之和：$$a+a+a+a+a=5a$$，偶数位上的数字之和：$$5+4+3+2+1=15$$，因为$$\\textbar5a-15\\textbar=11n$$，$$n$$为整数；当$$n=0$$时，$$a=3$$，当$$n\\text{\\textgreater}0$$时，$$a$$没有小于$$10$$的整数解，所以$$a=3$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "628", "queId": "d51dee6df0fd4c489a10ddb8d9a49c36", "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": "一个物体的$$2016$$度逆时针旋转，与它的多少度顺时针旋转是一样的？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$72$$ "}], [{"aoVal": "C", "content": "$$144$$ "}], [{"aoVal": "D", "content": "$$252$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["逆时针旋转的度数$$+$$顺时针旋转的度数和是$$360$$的倍数，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": "1877", "queId": "f29b19c6ba8c4de59fb6a384bea8fbba", "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": "【能1-2】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": "1889", "queId": "8eefb2e622a7465f8b0eae79c236c1a9", "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->对应思想"], "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": "3474", "queId": "eb9fd917c17b4790b6e782974015acc8", "competition_source_list": ["2011年全国创新杯五年级竞赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "将各位数字的和为$$4$$的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": "$$26$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["枚举法：各位数字和为$$4$$的自然数依次为$$4$$，$$13$$，$$22$$，$$31$$，$$40$$，$$103$$，$$112$$，$$121$$，$$130$$，$$202$$，$$ 211$$，$$220$$，$$301$$，$$310$$，$$400$$，$$1003$$，$$1012$$，$$1021$$，$$1030$$，$$1102$$，$$1111$$，$$1210$$，$$1300$$，$$2002$$，$$2011$$，即$$2011$$是第$$26$$个． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2166", "queId": "26826d7b0f7149f3b0701a7521c92f59", "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，$$1.5$$分钟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：$$180\\div 1.5=120$$（米/分）$$=2$$（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": "2502", "queId": "1e6a0ba0fe68451b80c7450e1edc7eef", "competition_source_list": ["2015年走美杯六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "用$$2015$$减去它的$$\\frac{1}{2}$$，再减去余下的$$\\frac{1}{3}$$，再减去余下的$$\\frac{1}{4}$$，$$\\cdots $$，以此类推，一直减去余下的$$\\frac{1}{31}$$，那么最后的得数为（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$31$$ "}], [{"aoVal": "B", "content": "$$65$$ "}], [{"aoVal": "C", "content": "$$62$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数巧算->连锁约分"], "answer_analysis": ["解：$$2015\\times \\left( 1-\\frac{1}{2} \\right)\\times \\left( 1-\\frac{1}{3} \\right)\\times \\left( 1-\\frac{1}{4} \\right)\\times \\cdots \\times \\left( 1-\\frac{1}{31} \\right)$$  $$=2015\\times \\frac{1}{2}\\times \\frac{2}{3}\\times \\frac{3}{4}\\times \\cdots \\times \\frac{30}{31}$$  $$=2015\\times \\frac{1}{31}$$  $$=65$$  故答案为：$$65$$。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1258", "queId": "1ad07204d5344856bad683311c7e256d", "competition_source_list": ["2013年走美杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "两数之和为$$17$$，两数之差为$$7$$，较小的数为． ", "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->和差倍问题->和差问题->二量和差问题->明和明差"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde(17-7)\\div 2$$  $$=10\\div 2$$  $$=5$$  因此较大数是$$12$$，较小数是$$5$$．选$$\\text{B}$$．  $$\\text{A}$$、$$\\text{C}$$为干扰项，排除． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2448", "queId": "10bfbcb000654e17a6b37f7de1dfd9b4", "competition_source_list": ["2013年IMAS小学高年级竞赛第一轮检测试题第8题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "将分数$$\\frac{22}{7}$$化成小数后，请问应取到小数点后第几位才能使得结果与$$3.14159$$的差最小（不需要做四舍五入且用大数减小数）？ ", "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->分数->分数基础->分、百、小相互转化"], "answer_analysis": ["$$\\frac{22}{7}=3.\\overset{\\centerdot }{\\mathop{1}} ,4285\\overset{\\centerdot }{\\mathop{7}} ,$$，逐一取小数点后的数，得到  $$3.14159-3.1=0.04159$$  $$3.14159-3.14=0.00159$$  $$3.142-3.14159=0.00041$$  $$3.1428-3.14159=0.00121$$  $$3.14258-3.14159=0.00126$$  $$\\cdots$$  由上面可知，取到小数后第$$3$$位时结果与$$3.14159$$的差最小．  若取小数点后的数位超过$$3$$，则结果与$$3.14159$$的差不断增大．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "383", "queId": "e8f7e862da384970adf4899770cdf856", "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": "甲乙丙站成一排，甲比乙矮，丙比乙高．他们三人分别穿红黄蓝的衣服．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．甲乙丙的衣服颜色分别是． ", "answer_option_list": [[{"aoVal": "A", "content": "蓝红黄 "}], [{"aoVal": "B", "content": "蓝黄红 "}], [{"aoVal": "C", "content": "红黄蓝 "}], [{"aoVal": "D", "content": "黄蓝红 "}], [{"aoVal": "E", "content": "红蓝黄 "}]], "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": ["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．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1183", "queId": "2220f6c1b32e4970abe9009cd71e4cf3", "competition_source_list": ["2013年第25届广东广州五羊杯六年级竞赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "张迪的书架分为上、下两层，共有书$$90$$本，若把上层的书取$$7$$本放至下层，则下层的书的数量比上层的两倍还多$$9$$本．则原书架下层的书有本． ", "answer_option_list": [[{"aoVal": "A", "content": "$$27$$ "}], [{"aoVal": "B", "content": "$$34$$ "}], [{"aoVal": "C", "content": "$$56$$ "}], [{"aoVal": "D", "content": "$$63$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["解：设原来上层有书$$x$$本，则原来下层有书$$(90-x)$$本．  $$2(x-7)+9=90-x+7$$，解得$$x=34$$，  $$90-34=56$$（本）  所以原来下层有书$$56$$本．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1629", "queId": "76c060df203040f5a9d83816c785a8ce", "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，丙在$$A$$地植树棵． ", "answer_option_list": [[{"aoVal": "A", "content": "$$150$$ "}], [{"aoVal": "B", "content": "$$300$$ "}], [{"aoVal": "C", "content": "$$450$$ "}], [{"aoVal": "D", "content": "$$600$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "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，甲在$$A$$地植树$$28\\times 25=700$$棵，则丙在$$A$$地植树$$1000-700=300$$棵． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1509", "queId": "63ea2a9ae1ab45e7b61d6cc6f39fcb53", "competition_source_list": ["2017年河南郑州K6联赛竞赛模拟第六套第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一根电线截去$$\\frac{3}{4}$$后，还剩下$$\\frac{3}{4}$$米，截去的和剩下的相比（~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "截去的长 "}], [{"aoVal": "B", "content": "剩下的长 "}], [{"aoVal": "C", "content": "一样长 "}], [{"aoVal": "D", "content": "无法比较 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["截去$$\\frac{3}{4}$$，则剩下$$1-\\frac{3}{4}=\\frac{1}{4}$$，所以截去的多． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1869", "queId": "978425f2c7104d0083803cebcac8192d", "competition_source_list": ["2004年希望杯二年级竞赛复赛", "2004年希望杯三年级竞赛复赛", "2004年希望杯一年级竞赛复赛", "2004年希望杯四年级竞赛复赛", "2004年希望杯五年级竞赛复赛", "2004年第2届希望杯五年级竞赛复赛第3题", "2004年希望杯六年级竞赛复赛"], "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，哪几个数在重复循环呢？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->归纳总结->归纳推理"], "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": "797", "queId": "5737a3d44980494cbe10e0c74343784c", "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28题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个奇质数和一个偶质数的和是$$9$$．这个奇质数是？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"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->质数与合数判定->质数与合数的认识"], "answer_analysis": ["$$9=7+2$$ "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "619", "queId": "d5189d2b853d4dbea527995b9f85c247", "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": "$${{\\left( 11101 \\right)}_{2}}$$ "}], [{"aoVal": "B", "content": "$${{\\left( 10011 \\right)}_{2}}$$ "}], [{"aoVal": "C", "content": "$${{\\left( 10101 \\right)}_{2}}$$ "}], [{"aoVal": "D", "content": "$${{\\left( 11001 \\right)}_{2}}$$ "}]], "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": ["$$25=16+8+1$$，则$$25$$转换成二进制数是$${{\\left( 11001 \\right)}_{2}}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "415", "queId": "9730d6bde0a64c3088febd2524d716be", "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": "\\textbf{(2019 World Mathematical Olympiad,~Primary 4, \\#10)}  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（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$）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->拓展思维->组合模块->智巧趣题->数学趣题->空瓶换水问题->不可借空瓶"], "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": "3357", "queId": "b64bf98299e1433cb7a92270890d632a", "competition_source_list": ["2012年全国希望杯五年级竞赛复赛第7题"], "difficulty": "0", "qtype": "single_choice", "problem": "在平面上有$$7$$个点，其中任意三个点都不在同一条直线上，如果连接这$$7$$个点中的每两个点，最多可以得到~\\uline{~~~~~~~~~~}~条线段；以这些线段为边，最多能构成~\\uline{~~~~~~~~~~}~个三角形． ", "answer_option_list": [[{"aoVal": "A", "content": "$21,35$ "}], [{"aoVal": "B", "content": "$$21,50$$ "}], [{"aoVal": "C", "content": "$$35,50$$ "}], [{"aoVal": "D", "content": "$24,35$ "}]], "knowledge_point_routes": ["拓展思维->能力->抽象概括"], "answer_analysis": ["每两个点确定一条线段，共有$$\\text{C}_{7}^{2}=7\\times 6\\div 2=21$$条线段 ；  每三个点确定一个三角形，共有$$\\text{C}_{7}^{3}=7\\times 6\\times 5\\div (3\\times 2\\times 1)=35$$个三角形． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1623", "queId": "4e5c3c9fa30b4458ae98421d0f66726d", "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$$10$$千克油，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->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": ["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": "905", "queId": "8aac49075139269a015186811970221c", "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": "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->拓展思维->数论模块->整除->整除特征->整除特征综合"], "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": "1759", "queId": "96886c53a6dd4f9e91e994c2071de2e2", "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"], "difficulty": "2", "qtype": "single_choice", "problem": "有一种特殊的计算器，当输入一个$$10\\sim 49$$的自然数后，计算器会先将这个数乘以$$2$$，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->两量还原问题"], "answer_analysis": ["解：A：$$44-2=42$$，颠倒后是$$24$$，$$24\\div 2=12$$；$$12$$是$$10\\sim 49$$的自然数，符合要求；  B：$$43-2=41$$，颠倒后是$$14$$，$$14\\div 2=7$$，$$7$$不是$$10\\sim 49$$的自然数，不符合要求；  C：$$42-2=40$$，颠倒后是$$4$$，$$4\\div 2=2$$，$$2$$不是$$10\\sim 49$$的自然数，不符合要求；  D：$$41-2=39$$，颠倒后是$$93$$，$$93\\div 2=46.5$$，$$46.5$$不是$$10\\sim 49$$的自然数，不符合要求。  故选：A。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3318", "queId": "ff808081451d64f901451d90b3b40037", "competition_source_list": ["2014年全国迎春杯三年级竞赛复赛第14题"], "difficulty": "2", "qtype": "single_choice", "problem": "把$$1\\sim 5$$这$$5$$个自然数从左到右排成一排，要求从第三个数起，每个数都是前两个数的和或差，那么一共有（~~~~~~~ ）种放法． ", "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->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["$$5$$只能在首位或者末位，$$52314$$，$$54132$$，$$41325$$，$$23145$$，所以选B． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2082", "queId": "fda64268520a4ee6943cf4aca2a2b5a8", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（四）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一圆形花坛，绕它走一圈是$$120$$米，如果在花坛周围每隔$$6$$米栽一株丁香花，再在每  相邻的两株丁香之间栽$$2$$株月季花，共栽月季花朵． ", "answer_option_list": [[{"aoVal": "A", "content": "$$38$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$44$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["这是一个环形植树问题，棵树$$=$$段数，故丁香花的数量为$$120\\div 6=20$$（株），月季花的数量为$$20\\times 2=40$$（株）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3367", "queId": "661dac6c726a4e94bceb5aed73b861a4", "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$$1$$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": "892", "queId": "edc13111a2654349bf43ea0ac4ee347b", "competition_source_list": ["2017年河南郑州K6联赛竞赛模拟第八套第8题4分", "2017年河南郑州K6联赛竞赛模拟第九套第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "小亚和小巧玩猜数游戏，每人每次出$$1$$至$$4$$中的一个数字．如果两人出的数字相加，和是奇数就算小亚赢，和是偶数就算小巧赢．那么，小亚赢的可能性（ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "比小巧大 "}], [{"aoVal": "B", "content": "比小巧小 "}], [{"aoVal": "C", "content": "和小巧一样大 "}], [{"aoVal": "D", "content": "无法确定 "}]], "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": ["$$1$$至$$5$$的$$5$$个数中相加是偶数的是偶$$+$$偶$$=$$偶，奇$$+$$奇$$=$$偶，是奇数的可能性是奇$$+$$偶$$=$$奇，偶数的可能性比奇数的可能性大． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1324", "queId": "2c984e439c16479baa255a2ab784ec71", "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": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题"], "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": "526", "queId": "eb3f4b3ec6744b38802feb6e62d8825f", "competition_source_list": ["2017年新希望杯六年级竞赛训练题（二）第6题", "2018年湖北武汉新希望杯六年级竞赛训练题（二）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "一副扑克牌共有$$54$$张（包括大王、小王），至少从中取张牌，才能保证其中必有$$2$$种花色． ", "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->抽屉原理->简单抽屉原理"], "answer_analysis": ["取完一种花色加大王和小王，再取一张一定满足，$$13+2+1=16$$（张）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "601", "queId": "418f2a7222264bf9bf72bcafdee9af01", "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": "$$200$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "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": "2335", "queId": "0520635d9ee24e3fa1a6e74af13445b5", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "海南省$$2015$$年最高寿的一个老人年龄是$$116$$岁，如果用字母$$S$$代表海南人的年龄，则（~ ）正确． ", "answer_option_list": [[{"aoVal": "A", "content": "$$S$$可以``赋值''$$3$$ "}], [{"aoVal": "B", "content": "$$S$$代表$$116$$这个特定的数 "}], [{"aoVal": "C", "content": "$$S$$可以取任何的整数 "}], [{"aoVal": "D", "content": "$$S$$仅代表一个特定的数 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->等量代换->字母表示数"], "answer_analysis": ["按题意，$$S$$代表人的年龄，不是特定的数，不能任意取值；$$S$$代表海南人的年龄，不可以取任何整数，也不是仅代表特定的数，故（$$B$$），（$$C$$）和（$$D$$）都不正确． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1875", "queId": "86b031f774f346268e765a6e15d20e50", "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10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "科学家研究表明，$$10000{{\\text{m}}^{2}}$$的森林在生长季节每周可吸收$$6.3$$吨二氧化碳．城北森林的面积是$$50000{{\\text{m}}^{2}}$$，请计算一下，预计$$2020$$年$$2$$月这片森林一共可以吸收二氧化碳． ", "answer_option_list": [[{"aoVal": "A", "content": "$$130.5$$吨 "}], [{"aoVal": "B", "content": "$$126$$吨 "}], [{"aoVal": "C", "content": "$$135$$吨 "}], [{"aoVal": "D", "content": "$$882$$吨 "}]], "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": ["$$10000{{\\text{m}}^{2}}$$的森林每天可吸收二氧化碳：$$6.3\\div 7=0.9$$（吨），  $$10000{{\\text{m}}^{2}}$$的森林$$29$$天可吸收二氧化碳：$$0.9\\times 29=26.1$$（吨），  则$$50000{{\\text{m}}^{2}}$$，$$2020$$年$$2$$月可吸收二氧化碳：$$26.1\\times 5=130.5$$（吨）．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2672", "queId": "6cc9f51fe33e45a7b8b2b33b8e28d5d2", "competition_source_list": ["2008年第6届创新杯五年级竞赛初赛B卷第2题5分", "2008年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "小鲸鱼说:``妈妈，我长到您现在这么大年龄时，您就31岁啦!''大鲸鱼说:``我像你这么大年龄时，你只有1岁''，则大鲸鱼现在(  )岁. ", "answer_option_list": [[{"aoVal": "A", "content": "18 "}], [{"aoVal": "B", "content": "20 "}], [{"aoVal": "C", "content": "21 "}], [{"aoVal": "D", "content": "22 "}]], "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": ["设大鲸鱼现在$$x$$岁，小鲸鱼现在$$y$$岁，则 $$\\begin{cases} x+\\left( x-y \\right)=31    y-\\left( x-y \\right)=1   \\end{cases}$$ 即$$\\begin{cases} 2x-y=31    2y-x=1   \\end{cases}$$ 由两式化简可以解得$$x=21$$，所以大鲸鱼现在$$21$$岁. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2705", "queId": "a3289a82dc834f21b459d954e6147034", "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3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "将从$$1$$开始的自然数分组如下：$$\\left( 1 \\right)$$，$$\\left( 2,3,4 \\right)$$，$$\\left( 5,6,7,8,9 \\right)$$，$$\\left( 10,11,12,13,14,15,16\\right)\\cdots $$按此规律第$$15$$组第七个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$203$$ "}], [{"aoVal": "B", "content": "$$204$$ "}], [{"aoVal": "C", "content": "$$205$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["每组的个数呈现等差数列，且公差为$$2$$，所以第$$14$$组的最后一个数为：$$1+3+5+7+\\cdots \\cdots +27=196$$，  故第$$15$$组的第$$7$$个数是：$$196+7=203$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1534", "queId": "d13aa9c33dac41b792548d93c25eb5ca", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第14题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$3$$根吸管连在一块，好像一根魔术棒，每根吸管长$$20\\operatorname{cm}$$，两根之间的接头重叠部分是$$2\\operatorname{cm}$$，请问这根``魔术棒''的长度为多少$$\\operatorname{cm}$$？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$54$$ "}], [{"aoVal": "B", "content": "$$55$$ "}], [{"aoVal": "C", "content": "$$56$$ "}], [{"aoVal": "D", "content": "$$58$$ "}], [{"aoVal": "E", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["若没有接头，$$3$$根吸管共长$$20\\times 3=60(\\text{cm})$$，一个接头重叠部分需$$2\\text{cm}$$，两个接头重叠部分需$$4\\text{cm}$$，则``魔术棒''长$$60-4=56(\\text{cm})$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2264", "queId": "ff80808148c43ff50148c9712ff205f3", "competition_source_list": ["2014年全国华杯赛小学中年级竞赛初赛第5题"], "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->能力->运算求解"], "answer_analysis": ["到达$$\\text{A}$$地实际是$$10$$点$$16$$分，到达$$\\text{C}$$地是$$13$$点$$04$$分，从$$\\text{A}$$地到$$\\text{B}$$地共耗时$$168$$分钟；即实际比原计划少用时间$$12$$分钟．那么在远足过程中本来是应该要走$$180$$分钟，现在节约了$$12$$分钟，则每走一分钟就节约时间$$12\\div 180=\\frac{1}{15}$$分钟；如果按照正常情况，晚出发$$6$$分钟，应该晚$$6$$分钟到达$$\\text{B}$$点，现在准时到达$$\\text{B}$$点，那么从$$\\text{A}$$到$$\\text{B}$$的过程中节约了$$6$$分钟，则所用的时间应该为$$6$$除以$$\\frac{1}{15}$$，则$$\\text{A}$$到$$\\text{B}$$耗时$$90$$分钟，则是$$11$$点$$40$$分钟到达$$\\text{B}$$点． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2634", "queId": "b09e2fec68744a548593839e289ac190", "competition_source_list": ["2021年鹏程杯六年级竞赛初赛第20题"], "difficulty": "3", "qtype": "single_choice", "problem": "我们用$$(x,y)$$表示$$x$$，$$y$$这两数中较大的数减去较小的数所得的结果，比如：  $$(2,3)=(3,2)=3-2=1$$，$$(5,5)=5-5=0$$，$$\\cdots $$  现将$$1\\sim 2020$$这$$2020$$个自然数分成$$A$$，$$B$$两组（每组$$1010$$个数），并把$$A$$组的数从小到大排列得到$$a_1\\textless a_2\\textless\\cdots \\textless a_{1010}$$．而将$$B$$组的数从大到小排列得到$$b_1\\textgreater b_2\\textgreater\\cdots \\textgreater b_{1010}$$．则$$(a_1,b_1)+(a_2,b_2)+(a_3,b_3)+\\cdots +(a_{1009},b_{1009})+(a_{1010},b_{1010})$$的值为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1000000$$ "}], [{"aoVal": "B", "content": "$$1010025$$ "}], [{"aoVal": "C", "content": "$$1020100$$ "}], [{"aoVal": "D", "content": "$$1030225$$ "}], [{"aoVal": "E", "content": "$$4080400$$ "}]], "knowledge_point_routes": ["拓展思维->能力->符号代换->代数运算"], "answer_analysis": ["首先我们来证明，$$1\\sim 2020$$中的后$$1010$$个数一定不会放到一组$$(x,y)$$中来计算．设$$1011\\sim 2020$$这$$1010$$个数，有$$m$$个出现在$$A$$组，有$$n$$个出现在$$B$$组，则$$m+n=1010$$．  由于$$A$$组中的数是从小到大排列的，$$B$$组中的数是从大到小排列的，所以$$1011\\sim 2020$$中的数如果出现在$$A$$组，其中最小的数应为$$a_{1010-m+1}=a_{1011-m}$$；这些数出现在$$B$$组，其中最小的数为$$b_n=b_{1010-m}$$．由此可以看出，不管$$m$$和$$n$$怎么取值，$$1011\\sim 2020$$中的数都不会放到一组$$(x,y)$$中来计算．同理$$1\\sim 1010$$中的数也不会放到一组$$(x,y)$$中来计算．那么每一组$$(x,y)$$中，必有一个来自于$$1\\sim 1010$$，另一个来自于$$1011\\sim 2020$$，得到的结果必然是后一半的数减前一半的数，我们要计算的算式可以改为：$$(1011+1012+\\cdots+2019+2020)-(1+2+3+\\cdots+1009+1010)$$$$=1010\\times1010=1020100$$．  故选 $$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3065", "queId": "e5fcc7aa09e1464ea2ab9da72ef108b2", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛决赛第16题"], "difficulty": "3", "qtype": "single_choice", "problem": "在等式$$\\frac{1}{18}+\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=1$$中，符号$$x$$，$$y$$，$$z$$分别代表三个不同的自然数，那么，这三个数的和为~\\uline{~~~~~~~~~~}~． ", "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->能力->逻辑分析"], "answer_analysis": ["首先估算，若没有$$\\frac{1}{2}$$，则三个分数最大为：$$\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{18}\\textless{}1$$，则必有$$\\frac{1}{2}$$，依次类推，当有$$\\frac {1}{3}$$时，满足题目条件，原式$$=\\frac {1}{18}+\\frac {1}{2}+\\frac {1}{3}+\\frac {1}{z}=1$$，解得$$z=9$$，则$$x+y+z=2+3+9=14$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2757", "queId": "57114ca142c448689182875daaf5ef59", "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->数学思想->逐步调整思想"], "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": "1065", "queId": "0b3af1887bac4d1186898d3f6bf058f0", "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8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一根粗细均匀的木头长$$48$$分米，要锯成$$4$$分米长的木棍，每锯一次要$$3$$分钟，锯完一次要休息$$2$$分钟，全部锯完要用分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$53$$ "}], [{"aoVal": "B", "content": "$$55$$ "}], [{"aoVal": "C", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["先用总长度除以每段的长度，求出一共有多少段，然后减去$$1$$，就是需要锯的次数；  用每锯一次需要的时间乘上锯的次数就是锯木头需要的时间，每锯一次需要休息$$2$$分钟，那么锯$$11$$次需要休息个$$2$$分钟，再由此求出休息的时间，然后加上锯木头的时间即可．  $$48$$米需要据$$48\\div 4=12$$（段），而$$12$$段需要据：$$12-1=11$$（次），  锯木头需要的时间为：$$3\\times 11=33$$（分钟），  休息的时间：$$2\\times (11-1)=20$$（分钟），  总共需要：$$33+20=53$$（分钟）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "273", "queId": "568f86d41087481fae3eade65ae2fbf5", "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．  It takes $$2$$ people $$2$$ minutes to eat $$2$$ buns at the same time.  $$10$$ people will take  minutes to eat $$10$$ buns at the same time. ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "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．  The time taken for two people to eat two buns at the same time is $$2$$ minutes.  The time taken is the same for one person to eat one bun.  As such, the time taken for $$10$$ people to eat $$10$$ buns is still $$2$$ minutes. "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3456", "queId": "bd94e12a91f74e709746e60a769a1ee8", "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7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{(2019 World Mathematical Olympiad, Primary 1, Question \\#7)}  A square has $$4$$ angles, cut $$1$$ angle, and there are still （$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$）angles left.  一个正方形有$$4$$个角，剪去$$1$$个角，还剩（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$）个角． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "以上都对 "}]], "knowledge_point_routes": ["拓展思维->能力->图形认知", "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->角计数"], "answer_analysis": ["一个正方形剪去一个角有$$3$$种情况，沿对角线剪开，则剩下$$3$$个角．从对角开始剪到一边，剩$$4$$个角．从一边剪到另一边，剩$$5$$个角．  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": "621", "queId": "192be516aefa4ab88054cf927a54dbf8", "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，组中的数都是质数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$，$$21$$，$$17$$ "}], [{"aoVal": "B", "content": "$$91$$，$$71$$，$$51$$ "}], [{"aoVal": "C", "content": "$$43$$，$$53$$，$$73$$ "}], [{"aoVal": "D", "content": "$$17$$，$$37$$，$$85$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$\\text{A}$$中$$21$$不是质数，故$$\\text{A}$$错误；  $$\\text{B}$$中$$51$$，$$91$$不是质数，故$$\\text{B}$$错误．  $$\\text{D}$$中$$85$$不是质数，故$$\\text{D}$$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": "1660", "queId": "76f23b9379b54e07ac5a0ec10e42c310", "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": "五年级$$2$$班的同学们一起去划船，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->知识点->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": "2583", "queId": "42f829b4dd6841a389d0969f5dbae72e", "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$$2.857142$$的两个数字上方加$$2$$个循环点，得到以下循环小数，其中最小的是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2.\\dot{8}5714\\dot{2}$$ "}], [{"aoVal": "B", "content": "$$2.85\\dot{7}14\\dot{2}$$ "}], [{"aoVal": "C", "content": "$$2.857\\dot{1}4\\dot{2}$$ "}], [{"aoVal": "D", "content": "$$2.8571\\dot{4}\\dot{2}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "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$$2.857\\dot{1}4\\dot{2}$$. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1449", "queId": "ac1ed6a33b8f4c1db2b7dd8411313375", "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25题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$8$$和(~ ~ ~)的平均数是$$10$$? ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "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": ["$$10\\times2-8=12$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "302", "queId": "84403b8955d1457ca019ecbc39ab3ef3", "competition_source_list": ["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，如果只允许在天平的一边放砝码，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->思想->枚举思想"], "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": "2192", "queId": "1bb311ecc44f460b99b2c2d0e684acd1", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙三人每分钟分别行$$60$$米、$$50$$米和$$40$$米，甲从$$B$$地、乙和丙从$$A$$地同时出发相向而行，途中甲遇到乙后$$15$$分又遇到丙．$$A$$，$$B$$两地的距离为米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1500$$ "}], [{"aoVal": "B", "content": "$$16500$$ "}], [{"aoVal": "C", "content": "$$15000$$ "}], [{"aoVal": "D", "content": "$$1650$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["甲遇到乙后$$15$$分钟，甲遇到了丙，所以遇到乙的时候，甲和丙之间的距离为：$$（60+40）\\times15=1500$$（米），而乙丙之间拉开这么大的距离一共要$$1500\\div（50-40）=150$$（分），即从出发到甲与乙相遇一共经过了$$150$$分钟，所以$$A$$、$$B$$之间的距离为：$$（60+50）\\times150=16500$$（米）．  甲遇到乙后$$15$$分钟，甲遇到了丙，设甲、乙相遇用时$$t$$分钟，则甲、丙相遇用时（$$t$$+$$15$$）分钟，再根据路程和相等，可列方程$$$$（$$60$$+$$50$$）$$\\times t$$$$=$$$$$$（$$60$$+$$40$$）$$\\times $$（$$t$$+$$15$$）$$$$ 解出：$$t=150$$，所以$$A$$、$$B$$之间的距离为：$$$$（$$60$$+$$50$$）$$\\times150$$$$=$$$$16500$$（米）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "749", "queId": "c7f6368a9ee34549a963510616efe0a1", "competition_source_list": ["2014年中环杯五年级竞赛决赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$420\\times 814\\times 1616$$除以$$13$$的余数为（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}]], "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": ["解：$$420\\times 814\\times 1616\\equiv 4\\times 8\\times 4\\equiv 128\\equiv 11$$（$$\\text{mod }13$$）  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": "1202", "queId": "3462d4be1b1a46f6ad4a3968655e2c99", "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$$1$$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": "$$20$$ "}]], "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$$10\\div2=5$$（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": "1306", "queId": "30d44af9ef92410bb6eed1f838aeea76", "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. A 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->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": ["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": "1020", "queId": "0635e5290f5b4985a8d71fb0e5f562f5", "competition_source_list": ["2012年第10届创新杯四年级竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "杰克和吉莉每人各有一只水壶，其中都装有$$1$$升水．第一天，杰克把他壶中的$$1$$毫升水倒入吉莉的壶中，第二天吉莉把她的壶中的$$3$$毫升水倒入杰克的壶中，第三天杰克把他壶中的$$5$$毫升水倒入吉莉的壶中，这样继续做下去，其中每个人倒出的水比前一天从对方得到的水多$$2$$毫升．那么第$$101$$天结束后，杰克壶中有毫升水？（$$1$$升$$=1000$$毫升） ", "answer_option_list": [[{"aoVal": "A", "content": "$$799$$ "}], [{"aoVal": "B", "content": "$$899$$ "}], [{"aoVal": "C", "content": "$$900$$ "}], [{"aoVal": "D", "content": "$$1000$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde1000-1+3-5+7-9+11-13+\\cdots \\cdots +199-201$$  $$=1000+\\left( 3-1 \\right)+\\left( 7-5 \\right)+\\left( 11-9 \\right)+\\cdots \\cdots +\\left( 199-197 \\right)-201$$  $$=1000+\\underbrace{2+2+\\cdots +2}_{50个2}-201$$  $$=1000+100-201$$  $$=899$$（毫升）．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1746", "queId": "9f81403edcca4a81839eea20e07b56bb", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "当天，游乐场的游客有很多，过山车项目排起了长队，小明数了数共有$$60$$人，其中男生人数有$$x$$人，且男生人数是女生人数的$$\\frac{1}{3}$$，则下列关系式正确的是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$x+\\frac{1}{3}x=60$$ ~~~~~~~~~~~~~~~~~ "}], [{"aoVal": "B", "content": "$$x+3x=60$$~~~~~~~~~~ "}], [{"aoVal": "C", "content": "$$x:60=1:\\left( 1+\\frac{1}{3} \\right)$$~~~~~~~~~ "}], [{"aoVal": "D", "content": "$$\\frac{x}{60}=\\frac{3}{1+3}$$ "}]], "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": ["用字母表示数；男生人数有$$x$$人，女生人数是$$3x$$人，则关系式为$$x+3x=60$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "367", "queId": "f6b7cc96b8b5473f9b3ac0dc8df36998", "competition_source_list": ["2016年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "从自然数$$1$$，$$2$$，$$3$$，\\ldots，$$2015$$，$$2016$$中，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": ["课内体系->知识点->数学广角->鸽巢问题->利用抽屉原理解决实际问题", "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": ["$$1$$到$$2016$$中，数字和最大$$28$$。数字和为$$28$$的只有$$1999$$这$$1$$个数，  最坏情况：取数字和$$1$$到$$27$$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。$$n=110$$，选B。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "726", "queId": "6ca792581ed444cdb17911e4a23aec00", "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5题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "有一个两位数，如果在两个数字中间加一个$$0$$，所得的三位数比原数大$$6$$倍．那么，这个两位数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$19$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["设这个两位数是$$\\overline{ab}$$，则$$\\overline{a0b}=7\\times \\overline{ab}$$，拆开得$$b=5a$$，$$a=1$$，$$b=5$$．这个两位数是$$15$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1192", "queId": "26c7b2e595f7485196c6688e0a1cb6bf", "competition_source_list": ["2012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": "假期里小丁在一家餐馆里做兼职洗碗．每洗好一个碗可得到$$3$$元，每打破一个碗不仅不能得到洗这个碗的工资，还要另赔$$9$$元．一星期之内，他总共洗了$$500$$个碗，共领到了$$1368$$元．请问他一共打破了多少个碗？ ", "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->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": ["洗$$500$$个碗，应该领到$$3\\times 500=1500$$（元），但是小丁只领到了$$1368$$元，少领了$$1500-1368=132$$（元），那么原因是小丁打破了一些碗，洗一个碗，应得$$3$$元，但打破一个，需要赔偿$$9$$元，也就是说打破一个碗实际减少$$12$$元工资，故小丁共打破$$132\\div 12=11$$个碗． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "116", "queId": "7e0505d15d5045868ed4fb4148285d94", "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）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$$(9-8)\\times 8\\times 3$$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": "$$1$$，$$5$$，$$5$$，$$5$$ "}], [{"aoVal": "D", "content": "$$3$$，$$3$$，$$7$$，$$7$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["选项$$\\rm B$$：$$(7+1-4)\\times 6=24$$；  选项$$\\rm C$$：$$(5-1\\div 5)\\times 5=24$$；  选项$$\\rm D$$：$$(3+3\\div 7)\\times 7=24$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "999", "queId": "12bbfd988e1d4af9baea48e1883904a6", "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7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "三年级分成$$8$$个小队去种树，一共种了$$129$$棵，各小队种的棵数都不相同，其中种树最多的小队种了$$20$$棵，种树最少的小队最少种了棵． ", "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->加减法应用->加减法应用顺口溜"], "answer_analysis": ["要使某一队种的树量少，则其它队伍种树越多越好，已知种树最多的小队种了$$20$$棵，且各小队种的棵树都不相同，则令其它几个小队的种树数量为：$$19$$，$$18$$，$$17$$，$$16$$，$$15$$，$$14$$．已知$$8$$个小队一共种$$129$$棵树，故最少的小队种树：  $$129-(20+19+18+17+16+15+14)$$  $$=129-(20+14)\\times7\\div 2$$  $$=129-34\\times7\\div 2$$  $$=129-238\\div 2$$  $$=129-119$$  $$=10$$（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": "755", "queId": "64274ab6846a4c9b8230da5cf039b2fb", "competition_source_list": ["2008年第6届创新杯六年级竞赛复赛第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一房间中有红、黄，蓝三种灯，当房间所有灯都关闭时，拉一次开关，红灯亮；第二次拉开关，红、黄灯都亮；第三次拉开关，红、黄、蓝灯都亮；第四次拉开关，三灯全关闭．现在从$$1\\sim 100$$编号的同学走过该房间，并将开关各拉若干次，他们拉开关的方式为：编号为奇数者，他拉的次数都是他的编号数；编号为偶数者，其编号可以写成$${{2}^{r}}\\cdot p$$（其中$$p$$为正奇数，$$r$$为正整数），就拉$$p$$次．$$100$$人都走过房间后，房间灯的情况为． ", "answer_option_list": [[{"aoVal": "A", "content": "只有红灯亮 "}], [{"aoVal": "B", "content": "只有红、黄灯亮 "}], [{"aoVal": "C", "content": "三灯都亮 "}], [{"aoVal": "D", "content": "三灯都不亮 "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["本题主要考查数的运算．  奇数和为$$1+3+5+\\cdots +99=2500$$，  $$2$$的$$r$$次方$$\\times P=2$$、$$4$$、$$6$$、$$8$$、$$10\\cdots $$，  $$r$$分别为$$1\\sim 6$$，$$6$$种．  其中$$r=1$$时，有$$25$$种，$$p$$值的和为$$1\\sim 50$$内奇数的和．  $$r=2$$时，有$$13$$种，$$p$$值的和为$$1\\sim 25$$内奇数的和．  $$r=3$$时，有$$6$$种，$$p$$值的和为$$1\\sim 12$$内奇数的和．  $$r=4$$时，有$$3$$种，$$p$$值为$$1$$、$$3$$、$$5$$．  $$r=5$$时，有$$2$$种，$$p$$值为$$1$$、$$3$$．  $$r=6$$时，$$p=1$$．  所有$$p$$值的和为  $$1\\times 6+3\\times 5+5\\times 4+\\left[ \\left( 11+7 \\right)\\times \\frac{3}{2} \\right]\\times 3+\\left[ \\left( 25+13 \\right)\\times \\frac{7}{2} \\right]\\times 2+\\left( 27+49 \\right)\\times \\frac{12}{2}$$  $$=6+15+20+81+266+456$$  $$=844$$．  总数$$2500+884=3344$$，  $$3344\\div 4=838$$．  即灯全灭．  故本题正确答案为都不亮．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2732", "queId": "a7e47398c7e34189944bb0ce69615fac", "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": "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": ["拓展思维->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": "2979", "queId": "c4f31e911c89417683d72538156333ea", "competition_source_list": ["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", "2006年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "练1：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->计算模块->方程基础->不定方程->不定方程组"], "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$$\\text{C}$$.  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": "688", "queId": "be4c73af794941aaacf074457a159c33", "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7题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "下面的数中一个``零''也不读出来的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$235001$$ "}], [{"aoVal": "B", "content": "$$60015$$ "}], [{"aoVal": "C", "content": "$$107000$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$235001$$读作二十三万五千零一，故$$\\text{A}$$错误；  $$60015$$读作六万零一十五，故$$\\text{B}$$错误；  $$107000$$读作十万七千，故$$\\text{C}$$正确．  所以一个``零''也读不出来的是$$\\text{C}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1140", "queId": "748aec6981444d389b6b48d61a0ef070", "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": "\\textbf{(2019 Youth Mathematics Olympics, Primary 4, Question \\#3)~}  There are $$4$$ distinct natural numbers whose average is $$10$$. The largest of them is at least（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$）．  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（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->应用题模块->加减法应用->最值问题", "拓展思维->思想->对应思想"], "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": "1262", "queId": "1f096d5638a647f285f9406cca3ca704", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第15题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "某一年的二月份，星期六的天数都比其余星期中的日子之天数还多，请问这个月最后一天是星期几？ ", "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 "}], [{"aoVal": "E", "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": ["若为平年，则二月份有$$28$$天．  因$$28$$为$$7$$的倍数，故在二月份无论是星期几，都各有$$4$$天，  即不会有一个日子比其余星期中的日子之天数还多，  所以该年为闰年，且由$$29=4\\times7+1$$知共有$$5$$个星期六，  故最后一天是星期六，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": "3136", "queId": "1740e6cb4a844e2ba4f43969a197ba6f", "competition_source_list": ["2018年第23届华杯赛小学中年级竞赛初赛第5题", "2018年华杯赛小学中年级竞赛初赛第5题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "数字和等于$$218$$的最小自然数是个$$n$$位数，则$$n=$$（~ ）． ", "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->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（无要求）"], "answer_analysis": ["数字和一定，且数最小，则要保证数位最少，因此要尽可能多的$$9$$  $$218\\div 9=24\\cdots \\cdots 2$$,故该数为$$2\\underbrace{999999\\cdots \\cdots 999}_{24}$$  则$$n=25$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2195", "queId": "506b8ddc3f0e4f2e897dabbc389564ff", "competition_source_list": ["2006年第4届创新杯六年级竞赛初赛B卷第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两人进行$$100$$米赛跑，甲冲过终点线时，乙正好在甲后面$$20$$米处，第二次比赛时甲的起跑线比原起跑线推后$$20$$米，且两次比赛中各自速度不变，问第二次比赛结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "两人同时到达 "}], [{"aoVal": "B", "content": "甲到终点线时，乙正好在甲后面$$2$$米 "}], [{"aoVal": "C", "content": "甲到终点线时，乙正好在甲后面$$4$$米 "}], [{"aoVal": "D", "content": "乙到终点线时，甲正好在乙后面$$2$$米 "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["由题意得每跑$$1$$米，  甲胜的距离为：$$\\frac{20}{100}=0.2$$米，  $$20\\times0.2=4$$（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": "840", "queId": "a89ec20e93c64ced9e85c16f7c9a126d", "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": "在一个两位数的两个数字中间加一个$$0$$，所得三位数比原数大$$6$$倍．这个两位数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$19$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["设这个两位数为$$\\overline{ab}$$，根据题意有$$\\overline{aob}=\\left( 6+1 \\right)\\times \\overline{ab}$$，位值原理拆开化简得：$$5a=b$$，又因为$$a$$，$$b$$均为$$0 \\tilde{  } 9$$的数字，所以得$$a=1$$，$$b=5$$，则这个两位数是$$15$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2263", "queId": "9193e5ccdc164bd0a017cfc2230f710c", "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7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "一艘客船在长为$$176$$千米的河流中顺流而下再逆流而上共用时$$19$$小时，逆流用时比顺流用时多$$3$$小时．有一只船速为每小时$$12$$千米的小船在这条河中顺流而下$$10$$小时走了千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$140$$ "}], [{"aoVal": "C", "content": "$$150$$ "}], [{"aoVal": "D", "content": "$$160$$ "}]], "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": ["先分别求出顺流与逆流用时：  顺流：$$(19-3)\\div 2$$  $$=16\\div 2$$  $$=8$$（小时）  逆流：$$8+3=11$$（小时）  再求出水流速度：  顺流速度：$$176\\div 8=22$$（千米每小时）  逆流速度：$$176\\div 11=16$$（千米每小时）  则水流速度：$$(22-16)\\div 2$$  $$=6\\div 2$$  $$=3$$（千米每小时）  则$$(12+3)\\times 10$$  $$=15\\times 10$$  $$=150$$（千米）  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "407", "queId": "ff8080814db3e529014e0055847a59ac", "competition_source_list": ["2015年北京华杯赛小学高年级竞赛初赛C卷第3题", "2015年全国华杯赛小学高年级竞赛初赛第3题"], "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->能力->推理推导->言语逻辑推理"], "answer_analysis": ["因为恰有$$2$$位同学捐了款，据丙所说知甲、乙、丁就至少$$2$$人捐款，所以丙没捐款；  再据甲所说知丙、丁之中至少有$$1$$人捐了款，现在丙没捐款，所以丁一定捐款了；  再据乙所说知丁、甲之中至多有$$1$$人捐了款，现在丁捐款了，所以甲一定没捐款；  恰有$$2$$位同学捐了款，即恰有$$2$$位同学没捐款，现在甲、丙都没捐款，所以乙、丁都捐款了． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1351", "queId": "7546b98d6f7c406bbf511663a0d0806b", "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10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "将非$$0$$自然数依照下列形式不断写下去：$$123456\\cdots \\cdots $$，第$$2019$$个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["一位数有数字：$$9\\times1=9$$（个），  两位数有数字：$$90\\times2=180$$（个），  三位数有数字：$$900\\times3=1800$$（个）  则四位数有数字：$$2019-9-180-1800=30$$（个）．  $$30\\div4=7$$（组）$$\\cdots \\cdots2$$（个），  则四位数有$$1000$$，$$1001$$，$$1002$$，$$1003$$，$$1004$$，$$1005$$，$$1006$$，  所以还剩下的$$2$$个数字分别为$$1$$，$$0$$．  故第$$2019$$个数字为$$0$$．  所以答案选择$$\\text{A} $$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3147", "queId": "03e69bea907a4b7d9f1ccd012da73fcb", "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10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "二（$$1$$）班有$$50$$名同学参加语文、数学期末测试，每个学生至少有一门是优．语文得优的有$$38$$人，数学得优的有$$42$$人，语文、数学都得优的有人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$80$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["把语文得优秀的人数与数学得优秀的人数相加，即：$$38+42=80$$（人），$$80$$人比参加语文、数学期末测试的$$50$$人多$$80-50=30$$（人），多出的$$30$$人就是语文、数学都得优秀的人数．  $$38+42-50=30$$（人）．  答：语文、数学都得优秀的有$$30$$人． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "953", "queId": "d43e370b3cc14219bb4d7fb43cb198b0", "competition_source_list": ["2018年华杯赛六年级竞赛模拟卷"], "difficulty": "2", "qtype": "single_choice", "problem": "是否存在两个数，使得它们的和为$$222$$，差为111. ", "answer_option_list": [[{"aoVal": "A", "content": "有 "}], [{"aoVal": "B", "content": "没有 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的应用"], "answer_analysis": ["偶数$$+$$偶数$$=$$偶数，偶数$$-$$偶数$$=$$偶数  奇数$$+$$奇数$$=$$偶数，奇数$$-$$奇数$$=$$奇数  观察可得，如果两个数的和为偶数，则差不可能为奇数。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3218", "queId": "1b3df3b2835d4a40a6d56169fb6a8082", "competition_source_list": ["2013年第11届创新杯三年级竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "四位数$$2013$$的各位数字和为$$6$$，且各位数字均不相同．在具有这些性质的四位整数中，按由小到大顺序排列，$$2013$$是第个． ", "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->对应思想"], "answer_analysis": ["根据题意，只能由$$0$$，$$1$$，$$2$$，$$3$$四个数字来组成四位数．  以``1''开头的四位数有$$3\\times2=6个$$，  因此以``$$2$$''开头的最小数应是$$2013$$，  因此$$2013$$是第$$7$$个．  答︰$$2013$$是第$$7$$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": "2668", "queId": "560b1370fff14c089e79efd41871d196", "competition_source_list": ["2018年第23届华杯赛小学中年级竞赛初赛第1题", "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"], "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． ", "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． ", "<p>因为$$A$$，$$B$$都小于$$1$$，</p>\n<p>所以无法确定$$A\\times B$$与$$0.9$$的大小关系．</p>\n<p>①当$$A\\times B&lt;{}0.9$$，则$$A\\times B+0.1&lt;{}1$$，</p>\n<p>②当$$A\\times B=0.9$$，则$$A\\times B+0.1=1$$，</p>\n<p>③当$$A\\times B&gt;0.9$$，则$$A\\times B+0.1&gt;1$$．</p>\n<p>故选$$\\text{D}$$．</p>"], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "54", "queId": "61ccc9455158466985f7c7338bee9707", "competition_source_list": ["2015年第27届广东广州五羊杯小学高年级竞赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$10$$种不同的小球各$$100$$个放入同一个袋子里．从袋子中取出若干个小球，要想在取出的小球中必须有$$3$$种同样的球并有$$10$$个以上（含$$10$$个）的话，那么最少要从袋子中取出个小球． ", "answer_option_list": [[{"aoVal": "A", "content": "$$274$$ "}], [{"aoVal": "B", "content": "$$273$$ "}], [{"aoVal": "C", "content": "$$272$$ "}], [{"aoVal": "D", "content": "$$271$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["考察抽屉原理中的最不利原则，  $$2\\times100+8\\times9+1=273$$（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": "2686", "queId": "9e70db2e23344e1e927080d4a0446113", "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": "895", "queId": "78c60b715b7e4b998f572e3119405cb7", "competition_source_list": ["2017年湖北武汉中环杯五年级竞赛初赛第17题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果一个数的所有因数中恰好有$$3$$个因数为合数．这样的数就称为``思维数''．比如$$16$$就是一个``思维数''，因为它的所有因数中，正好有$$4$$，$$6$$，$$8$$为合数．在所有小于$$100$$的正整数中，``思维数''一共有~\\uline{~~~~~~~~~~}~个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["知识标签->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": ["分析易知思维数质因数分解后形式只有$${{a}^{4}}$$，$$a{{b}^{2}}$$两种形式．  $${{a}^{4}}$$，此时$$a$$只能是$$2，3$$  $$a{{b}^{2}}$$，当$$b$$是$$2$$，$$a$$可以是$$3$$，$$5$$，$$7$$，$$11$$，$$13$$，$$17$$，$$19$$，$$23$$共$$8$$个；  当$$b$$是$$3$$，$$a$$可以是$$2$$，$$5$$，$$7$$，$$11$$共$$4$$个；  当$$b$$是$$5$$，$$a$$可以是$$2$$，$$3$$共$$2$$个；  当$$b$$是$$7$$，$$a$$可以是$$1$$共$$1$$个；  综上共有$$17$$个． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2918", "queId": "edc00644f9a84377a9c83ac2c2eb593e", "competition_source_list": ["2019年第24届YMO一年级竞赛决赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在一个减法算式中，如果被减数减少$$10$$，差减少$$20$$，要使等式成立，减数应该． ", "answer_option_list": [[{"aoVal": "A", "content": "减少$$10$$ "}], [{"aoVal": "B", "content": "减少$$30$$ "}], [{"aoVal": "C", "content": "增大$$30$$ "}], [{"aoVal": "D", "content": "增大$$10$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["设这个算式原来是：$$50-30=20$$，  被减数减少$$10$$，差减少$$20$$，  $$40-40=0$$，  减数由原来的$$30$$变为$$40$$，增加了$$10$$，  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3352", "queId": "9acdf8aae0ce49dc8493a9ea341d73de", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "把$$1$$到$$5$$这$$5$$个自然数从左到右排成一排，要求从第三个数起，每个数都是前两个数的和或差，那么一共有（  ）种排法。 ", "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->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"], "answer_analysis": ["解：$$5$$只能在首位或者末位，$$52314$$，$$54132$$，$$41325$$，$$23145$$。  故选:B。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3389", "queId": "bb6045c68e0e4453a072bec18fd77afc", "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2题"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$10-1000$$之间，个位数是$$3$$或$$8$$的数的个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$200$$ "}], [{"aoVal": "B", "content": "$$198$$ "}], [{"aoVal": "C", "content": "$$196$$ "}], [{"aoVal": "D", "content": "$$194$$ "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["每十个数之间，就包含个位数是$$3$$或者$$8$$的数各一个，$$(1000-10)\\div 10\\times 2=198$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "252", "queId": "3fee9be63fc84ad5b36ff84684f0c1c5", "competition_source_list": ["2009年第14届全国华杯赛竞赛复赛第6题", "2021年新希望杯五年级竞赛模拟（考前培训100题）第17题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知三个合数$$A$$，$$B$$，$$C$$两两互质，且$$A\\times B\\times C=11011\\times 28$$，那么$$A+B+C$$的最大值为~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1601$$ "}], [{"aoVal": "B", "content": "$$1626$$ "}], [{"aoVal": "C", "content": "$$1673$$ "}], [{"aoVal": "D", "content": "$$14027$$ "}], [{"aoVal": "E", "content": "$$22031$$ "}], [{"aoVal": "F", "content": "$$77081$$ "}], [{"aoVal": "G", "content": "我不会 "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["因为$$11011\\times 28=11\\times 11\\times 13\\times 7\\times 2\\times 2\\times 7=2^{2}\\times7^{2}\\times11^{2}\\times13$$，要使$$A+B+C$$最大，且$$A$$、$$B$$、$$C$$为两两互质的合数，则$$A=11\\times 11\\times 13=1573$$，$$B=7\\times 7=49$$，$$C=2\\times 2=4$$，那么$$A+B+C=1626$$．  故答案为：$$1626$$．  乘积一定时，差小和小，差大和大.  $$11011\\times 28=2^{2}\\times 7^{2}\\times 11^{2}\\times 13$$，三个合数之间两两互质，所以$$A=2^{2}$$，$$B=7^{2}$$，$$C=11^{2}\\times 13$$时，$$(A+B+C)$$最大为$$1626$$. "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2864", "queId": "615fb52001cd493f86439cc9b2ae52c1", "competition_source_list": ["2017年河南郑州联合杯竞赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "根据规律，选出你认为括号中正确的数：（~ ~ ~ ），$$1$$，$$1$$，$$1$$，$$\\frac{9}{11}$$，$$\\frac{11}{13}$$，$$\\frac{13}{17}$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$~~~~~ "}], [{"aoVal": "B", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$~~~~~~ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "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": ["找规律；将中间三个$$1$$化成分数是$$\\frac{3}{3}$$，$$\\frac{5}{5}$$，$$\\frac{7}{7}$$，分子是连续的奇数，分母是从$$2$$开始不断递增的质数，则第一个数是$$\\frac{1}{2}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1445", "queId": "950cfa18113a4bd09d1728c0f002634a", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "等腰三角形的两个角之比为$$2:5$$，则这个三角形是什么三角形（按角分类）？ ", "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": "钝角三角形或者锐角三角形 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["这个等腰三角形的三个内角之比为$$2:2:5$$或者$$2:5:5$$，前一种情况下，三角形最大的内角为$$180\\div \\left( 2+2+5 \\right)\\times 5={100}^{}\\circ$$；后一种情况下，三角形最大的内角为$$180\\div \\left( 2+5+5 \\right)\\times 5={75}^{}\\circ$$．所以这个等腰三角形是钝角三角形或者锐角三角形． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "502", "queId": "cf5869d6f5b0400bbb10124a13e1f2fb", "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几个不同的数，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->拓展思维->组合模块->抽屉原理->构造型抽屉原理", "课内体系->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．  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": "1399", "queId": "294ebb92ee324723bea69fadb174b682", "competition_source_list": ["2018~2019学年四川成都双流县四川师范大学附属圣菲小学六年级上学期开学考试第1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模拟（五）第5题2分", "2017年四川成都六年级竞赛“全能明星”选拔赛第5题2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第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "母亲今年$$37$$岁，女儿今年$$7$$岁，年后母亲的年龄是女儿的$$4$$倍． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["年龄问题，$$(37-7)\\div (4-1)=10$$（岁）．  年龄问题的特点是两人的年龄差始终不变，即几年后两人增加的年龄是相同的，根据这个特点解决题目中的问题即可．设$$x$$年后母亲的年龄是女儿的$$4$$倍，可得方程$$(7+x)\\times 4=37-x$$，解得$$x=3$$．即$$3$$年后母亲的年龄是女儿的$$4$$倍． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "753", "queId": "689f32a580644b6384eed5e5d903bd5e", "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": "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$$6n$$的因数个数为（~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$个 "}], [{"aoVal": "B", "content": "$$25$$个 "}], [{"aoVal": "C", "content": "$$30$$个 "}], [{"aoVal": "D", "content": "$$40$$个 "}]], "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": ["依题意不妨假设$$n=a{{b}^{4}}\\left( a\\ne b \\right)$$，则$$2n=2a{{b}^{4}}\\left( a\\ne b\\ne 2 \\right)$$，$$3n=3a{{b}^{4}}={{a}^{2}}{{b}^{4}}\\left( a=3 \\right)$$，所以$$6n=6a{{b}^{4}}=2{{a}^{2}}{{b}^{4}}$$．因此$$6n$$的因数个数为：$$\\left( 1+1 \\right)\\times \\left( 2+1 \\right)\\times \\left( 4+1 \\right)=30$$（个）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2244", "queId": "a32837abf3604a978305d9d1a5e6cafa", "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": "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->能力->实践应用"], "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": "2424", "queId": "14128b52d7504b249f12a1c386eb5cad", "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，$$7$$天没有回家，回来后一次撕下这$$7$$张日历，发现这$$7$$张日期数相加得$$119$$，那么，王老师回家这天是（  ）号。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->等差数列实际应用"], "answer_analysis": ["因为$$119\\div 7=17$$，所以第四天也就是中间一天是$$17$$号，故这$$7$$天的日期为：$$14\\text{、}15\\text{、}16\\text{、}17\\text{、}18\\text{、}19\\text{、}20$$，因此，回家是$$20+1=21$$（号）。故选D。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "740", "queId": "63d5e84c2804411b87d6eacfd853c789", "competition_source_list": ["2020年新希望杯五年级竞赛第29题", "2020年希望杯五年级竞赛模拟第29题"], "difficulty": "2", "qtype": "single_choice", "problem": "【$$2020$$五年级卷第$$29$$题】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$${{2}^{77232917}}-1$$．以下各数中，能被$$15$$整除的是． ", "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->逻辑分析"], "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": "3400", "queId": "8e80d41ee6964d689fe0cbd6927ecd10", "competition_source_list": ["2007年第5届创新杯五年级竞赛第3题5分", "2007年五年级竞赛创新杯", "2007年第5届创新杯五年级竞赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$3$$名医生、$$3$$名护士被分配到$$3$$所小学为学生体检，每所学校$$1$$名医生、$$1$$名护士。 共有~\\uline{~~~~~~~~}~种不同的分配方法。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$45$$ "}]], "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"], "answer_analysis": ["设三所学校为$$A$$校，$$B$$校和$$C$$校，先考虑$$3$$名医生分配方法：第一名医生有$$3$$种分配方法；第二名有$$2$$种分配方法，第三名有$$1$$种分配方法，所以有$$3\\times 2\\times 1=6$$（种）分配方法。同理，$$3$$名护士也有$$6$$种分配方法。因此，共有$$6\\times6=36$$（种）不同的分配方法。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1861", "queId": "b715710b0e8e40ae862d425346c0c429", "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->七大能力->逻辑分析"], "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": "2124", "queId": "fa223e09af2f42d28806f2ba53f413bf", "competition_source_list": ["2016年河南郑州K6联赛六年级竞赛第18题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一种录音机，现在售机$$90$$元，比原来降低$$10  \\% $$，降低了（~ ）元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$81$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["原价为$$90\\div \\left( 1-10  \\% ~\\right)=100$$(元)，降低了$$100-90=10$$(元)． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1496", "queId": "c34dcfe9fd3946b490a24e6b11ff5ef5", "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->数学思想->整体思想"], "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": "2538", "queId": "2c03993b08a94a73af3428cd810057ea", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛"], "difficulty": "3", "qtype": "single_choice", "problem": "设$$a\\ne 2$$，在下列结论中，（~ ）是错误结论． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2x+3=ax-3$$和$$\\left( 2x+3 \\right)\\left( {{m}^{2}}+1 \\right)=\\left( ax-3 \\right)\\left( {{m}^{2}}+1 \\right)$$是同解方程 "}], [{"aoVal": "B", "content": "$$2x+3=ax-3$$和$$2\\left( 2x+3 \\right)+{{m}^{2}}-1=2\\left( ax-3 \\right)+{{m}^{2}}+1$$是同解方程 "}], [{"aoVal": "C", "content": "$$2x+3=ax-3$$和$$2\\left( 2x+3 \\right)\\left( {{m}^{2}}-1 \\right)=2\\left( ax-3 \\right)\\left( {{m}^{2}}-1 \\right)$$是同解方程 "}], [{"aoVal": "D", "content": "$$2x+3=ax-3$$和$$\\left( a-2 \\right)x=6$$是同解方程 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["对于任何$$m$$的值，$${{m}^{2}}+1$$都不等于$$0$$，由原理二，（$$\\text{A}$$）正确；由原理一和二，$$\\text{B}$$）正确；由原理一，（$$\\text{D}$$）正确；当$$m=1$$时，$${{m}^{2}}-1=0$$，所以（$$\\text{C}$$）是错误结论． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1721", "queId": "6e59122f9bb04470bde3ef89d8c16f3f", "competition_source_list": ["2017年IMAS小学高年级竞赛（第二轮）第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "某班级的数学期末考试的结果为满分$$100$$分有$$4$$位学生﹑得$$90$$分至$$99$$分有$$6$$位学生、得$$80$$分至$$89$$分有$$18$$位学生、得$$70$$分至$$79$$分有$$12$$位学生、得$$69$$分以下有$$10$$位学生．已知全班平均分数为$$81.4$$分．请问该班学业生数学期末考试的总得分为多少分？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4050$$ "}], [{"aoVal": "B", "content": "$$3750$$ "}], [{"aoVal": "C", "content": "$$4070$$ "}], [{"aoVal": "D", "content": "$$3820$$ "}], [{"aoVal": "E", "content": "不能确定 "}]], "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": ["由题意可知这个班级总共有$$4+6+18+12+10=50$$位学生，所以该班学生数学期末考试的总得分为$$81.4\\times 50=4070$$分，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": "803", "queId": "8467cc4189c04d0b9531acd36eb22fc6", "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": "$$19$$ "}], [{"aoVal": "E", "content": "$$20$$ "}]], "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": ["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": "486", "queId": "af5e722f7d4e47488f5525b35024bc2d", "competition_source_list": ["2013年全国华杯赛小学高年级竞赛初赛C卷第9题"], "difficulty": "4", "qtype": "single_choice", "problem": "黑板上有$$11$$个$$1$$，$$22$$个$$2$$，$$33$$个$$3$$，$$44$$个$$4$$．做以下操作：每次擦掉$$3$$个不同的数字，并且把没擦掉的第四种数字多写$$2$$个．例如：某次操作擦掉$$1$$个$$1$$，$$1$$个$$2$$，$$1$$个$$3$$，那就再写上$$2$$个$$4$$．经过若干次操作后，黑板上只剩下$$3$$个数字，而且无法继续进行操作，那么最后剩下的三个数字的乘积是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["这类题需要抓住操作的本质：无论如何操作，任意两种数的个数的差只有不变，加$$3$$和减$$3$$三种情况，  （比如说$$1$$和$$2$$的个数，要么同时减少一个；要么$$1$$的个数加$$2$$个，$$2$$的个数少一个；要么$$1$$的个数少$$1$$个，$$2$$的个数加$$2$$个，分别对应：个数差不变，加$$3$$，减$$3$$）  而事实上，更进一步刻画这个特征：任意两个数的个数差，同余$$3$$（即：除以$$3$$的余数保持不变）  那么继续，看$$1$$的个数和$$2$$的个数，用$$2$$的个数减去$$1$$的个数，他们的差是$$22-11=11$$，$$11$$除以$$3$$的余数是$$2$$，那就是说当数字$$2$$的个数比$$1$$的个数多时，他们的差一定是除以$$3$$余$$2$$的；但如果数字$$1$$的个数比$$2$$多的话，那么这个差除以$$3$$的余数一定是$$1$$，（可以试着操作几次，让$$1$$的个数多于$$2$$的个数）．这样都算下来比较麻烦，观察一些特殊的，不用分类讨论的，可以发现$$1$$的个数和$$4$$的个数，它们的差是$$33$$，正好是$$3$$的倍数，也就是说，操作到最后，剩下的$$1$$的个数和$$4$$的个数要么一样多，要么个数差$$3$$．再根据题意，最后只剩下$$3$$个数，如果$$1$$的个数为$$3$$，那么$$2$$的个数必然为$$2$$，不合题意．如果$$3$$的个数为$$3$$，而$$1$$和$$2$$不可能都为$$0$$个，所以操作到最后时$$1$$的个数和$$4$$的个数都只能为$$0$$，而$$2$$的个数一定是$$2$$，那么$$3$$的个数就是$$1$$，所以剩下的三个数是$$2$$个$$2$$和$$1$$个$$3$$，他们乘积是$$12$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "360", "queId": "7347549f3cd14ee1a98e8cf249a73fdc", "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": "$$92$$ "}], [{"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": "3284", "queId": "75e34f59649b4a3b89adcc4db5dc3fd8", "competition_source_list": ["2009年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "恰好有两位数字相同的三位数共有（ ）个。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$270$$ "}], [{"aoVal": "B", "content": "$$243$$ "}], [{"aoVal": "C", "content": "$$240$$ "}], [{"aoVal": "D", "content": "$$267$$ "}]], "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": ["若三位数中恰有两位数字相同，那么相同的两个数位有以下三种情况：  ①百位数与十位数相同：因为百位数不能为$$0$$，所以百位数与十位数有$$1$$、$$2$$、$$3$$、$$\\cdots$$、$$9$$，有$$9$$种可能；个位数可在$$0\\sim 9$$中去掉一个已经作为百位数外的$$9$$个数中选取，也有$$9$$种可能，因此这样的三位数有$$9\\times 9=81$$（个）。  ②百位数与个位数相同：同理有$$81$$个三位数。  ③十位数与个位数相同：分两种情况。  （I）个位与百位都为$$0$$，百位有$$9$$种；  （II）当个位与百位不为$$0$$时，个位与百位有$$9$$种，百位有$$8$$种，共$$9\\times 8=72$$（种）；  所以共$$72+9=81$$（种）。  因此，恰有两位数字相同的三位数共有$$81\\times 3=243$$（个）。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1211", "queId": "8b0b19aed88b4524a8ae974cc7eac3c0", "competition_source_list": ["2016年北京学而思杯小学高年级五年级竞赛第11题7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14题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "一项工作，甲单独完成需要$$24$$天，乙单独完成需要$$36$$天，丙单独完成需要$$48$$天．现在甲乙丙三人轮流单独工作，甲、乙工作的天数比为$$1:2$$，乙、丙工作的天数比为$$3:5$$，那么，完成这项工作一共用了多少天？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$天 "}], [{"aoVal": "B", "content": "$$38$$天 "}], [{"aoVal": "C", "content": "$$19$$天 "}], [{"aoVal": "D", "content": "$$48$$天 "}]], "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->方程法解其他问题"], "answer_analysis": ["甲乙丙的工作天数之比是$$3:6:10$$，  设甲工作$$3x$$天，乙工作$$6x$$天，丙工作$$10x$$天，  那么$$\\frac{1}{24}\\times 3x+\\frac{1}{36}\\times 6x+\\frac{1}{48}\\times 10x=1$$，  解得：$$x=2$$，  故完成这项工作一共用了$$\\left( 3+6+10 \\right)\\times 2=38$$（天）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "808", "queId": "576a393b432d44298e3432e1fb107b9c", "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6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$6$$个互不相同的正整数的和是$$2002$$，则这$$6$$个数的最大公因数最大是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$182$$ "}], [{"aoVal": "B", "content": "$$154$$ "}], [{"aoVal": "C", "content": "$$143$$ "}], [{"aoVal": "D", "content": "$$91$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["设$$6$$个互不相同的正整数的最大公因数为$$a$$，则$$a$$为正整数，  $$6$$个互不相同的正整数为$$a{{x}_{1}}$$，$$a{{x}_{2}}$$，$$a{{x}_{3}}$$，$$a{{x}_{4}}$$，$$a{{x}_{5}}$$，$$a{{x}_{6}}$$，  则$${{x}_{1}}$$，$${{x}_{2}}$$，$${{x}_{3}}$$，$${{x}_{4}}$$，$${{x}_{5}}$$，$${{x}_{6}}$$为$$6$$个互不相同的正整数，  ∴$$a{{x}_{1}}+a{{x}_{2}}+a{{x}_{3}}+a{{x}_{4}}+a{{x}_{5}}+a{{x}_{6}}=2002$$，  ∴$$a({{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+{{x}_{6}})=2002$$，  ∵$${{x}_{1}}$$到$${{x}_{6}}$$为$$6$$个互不相同的正整数，  则$${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{6}}$$最小为$$1+2+3+4+5+6=21$$，  $$2002\\div 21=95\\cdots \\cdots 7$$，  故$${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{6}}=21$$时不满足题意，  当$${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{6}}=22$$时，  $$2002\\div 22=91$$，  ∴$$a$$最大为$$91$$．  故选：$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "76", "queId": "61e68f7fa6a44ef3b509010befd38e18", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛A卷第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "淘气家的挂钟时针长$$7\\text{cm}$$，一昼夜这根时针扫过的面积为平方厘米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$153.86$$ "}], [{"aoVal": "B", "content": "$$87.92$$ "}], [{"aoVal": "C", "content": "$$307.72$$ "}], [{"aoVal": "D", "content": "$$76.43$$ "}]], "knowledge_point_routes": ["拓展思维->思想->数形结合思想"], "answer_analysis": ["首先要明确的是：经过一昼夜，时针围钟面转了两周，而转一周所走过的面积是以时针长度$$7$$厘米为半径的圆的面积．利用圆的面积$$S=\\pi {{r}^{2}}$$即可求解．  $$3.14\\times {{7}^{2}}\\times 2=3.14\\times 49\\times 2=307.72$$（平方厘米）．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2446", "queId": "10a375051f5745f1a53a9ef7618a1b00", "competition_source_list": ["2011年全国创新杯五年级竞赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "兰佛德数字是这样一种数字，它的数字中每一个数码都出现两次，并且数码$$1$$被一个其他数码分开，数码$$2$$被两个其他数码分开，以此类推．下面四个数是兰佛德数字的一个是（~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12142334$$ "}], [{"aoVal": "B", "content": "$$41312432$$ "}], [{"aoVal": "C", "content": "$$14132342$$ "}], [{"aoVal": "D", "content": "$$32432141$$ "}]], "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": ["$$1\\times 1$$  $$2\\times \\times 2$$  $$3\\times \\times \\times 3$$  $$4\\times \\times \\times \\times 4$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "533", "queId": "e2b009b7f7414a9ea27b0698df6a5f3e", "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{~~~~~~~~~~}~ ", "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->数学思想->逐步调整思想"], "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，如果第三句是对的，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": "500", "queId": "bd3fe530c5ef4ba9b673127eb18b1190", "competition_source_list": ["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": "熊大和熊二打算在$$1$$小时后去市场买菜，现在是下午$$5$$时，所以熊大和熊二在几点钟去市场买菜？ ", "answer_option_list": [[{"aoVal": "A", "content": "下午$$4$$时 "}], [{"aoVal": "B", "content": "下午$$5$$时 "}], [{"aoVal": "C", "content": "下午$$6$$时 "}], [{"aoVal": "D", "content": "我不知道 "}]], "knowledge_point_routes": ["知识标签->数学思想->整体思想"], "answer_analysis": ["现在是下午$$5$$时，一小时后是下午$$6$$时． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2960", "queId": "c039e12e41eb4cfa9e2eb9c82f5a6fd8", "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": "$$0.000002005\\times 0.000004$$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->思想->整体思想"], "answer_analysis": ["因为$$0.\\underbrace{000002005}_{9位}\\times 0.\\underbrace{000004}_{6位}=0.\\underbrace{00000000000802}_{14位}$$，  故小数部分有$$12$$个$$0$$，  故答案是：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2710", "queId": "68ae348d9e02414bb08ba610a930fdd5", "competition_source_list": ["2015年第27届广东广州五羊杯小学高年级竞赛第4题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "下面几个分数中不能化成有限小数的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{52}{65}$$ "}], [{"aoVal": "B", "content": "$$\\frac{13}{25}$$ "}], [{"aoVal": "C", "content": "$$\\frac{14}{35}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{12}$$ "}]], "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": ["观察约分后分数的分母中，是否含有除了$$2$$、$$5$$以外的因数即可．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3344", "queId": "57bece96827145cfa92b1fe45f8ece45", "competition_source_list": ["2017年全国华杯赛竞赛初赛模拟试卷2第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "欧洲杯欧洲杯小组赛中，$$a，b，c，d$$四支足球队分别在一个小组，已知最终比赛结果中，$$a$$的排名高于$$b$$，$$c$$的排名高于$$d$$，无排名相同的结果，则符合这种情况的小组赛的排名有（ ）种． ", "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": ["用$$A\\textgreater B$$表示$$A$$的排名高于$$B$$．那么根据题目条件，$$C$$相对于A和B的位置有$$3$$种可能：  （$$1$$）$$C\\textgreater A\\textgreater B$$，此时，$$D$$的位置有$$3$$种可能：$$D\\textgreater A，A\\textgreater D\\textgreater B，B\\textgreater D$$；  （$$2$$）$$A\\textgreater C\\textgreater B$$，此时，$$D$$的位置有$$2$$种可能：$$D\\textgreater B，B\\textgreater D$$； （$$3$$）$$A\\textgreater B\\textgreater C$$，此时，$$D$$的位置有$$1$$种可能．  总的可能情况有$$3+2+1=6$$（种）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2392", "queId": "1c91ee87d4474c0693c31787f6b9bb77", "competition_source_list": ["2009年第7届创新杯六年级竞赛初赛第4题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知下面四个算式中，得数最大的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2008+\\frac{2008}{2009}$$ "}], [{"aoVal": "B", "content": "$$2008-\\frac{2008}{2009}$$ "}], [{"aoVal": "C", "content": "$$2008\\div \\frac{2008}{2009}$$ "}], [{"aoVal": "D", "content": "$$2008\\times \\frac{2008}{2009}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由题意，只需比较$$\\text{A}$$与$$\\text{C}$$的大小，  $$2008\\div \\frac{2008}{2009}=2008\\times \\frac{2009}{2008}=2009$$，  ∵$$\\frac{2008}{2009} ~\\textless{} ~1$$，  ∴$$2008\\frac{2008}{2009} ~\\textless{} ~2008\\div \\frac{2008}{2009}$$，  ∴得数最大的是$$2008\\div \\frac{2008}{2009}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2487", "queId": "2b55051ea3e9428c814e1d3b202bac4f", "competition_source_list": ["2014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": "当$$10000000$$减去$$10101$$，请问所得到的差中总共有多少个数码$$9$$？ ", "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->拓展思维->计算模块->整数->多位数计算->多位数的加减运算"], "answer_analysis": ["$$10000000-10101=9989899$$，所以差中总共有$$5$$个数码$$9$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1577", "queId": "ff8080814518d524014519096b8c031c", "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$$2$$，第二个数减$$2$$，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": "$$8$$，$$12$$，$$5$$，$$20$$ "}], [{"aoVal": "D", "content": "$$9$$，$$11$$，$$12$$，$$13$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["神蒙法：将答案一一代入检验，最后答案为C．  设相同的结果为$$2x$$，根据题意有：$$2x-2+2x+2+x+4x=45$$，$$\\Rightarrow x=5$$  易知原来的$$4$$ 个数依次是$$8$$，$$12$$，$$5$$，$$20$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "378", "queId": "9b3032d94ce7499b9078d93d5b71f9d8", "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": ["拓展思维->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": "2101", "queId": "f03994e1c31d450f8e6852c041896173", "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3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "大李和小李比赛吃包子，大李比小李多吃$$20$$个，大李比小李的$$4$$倍多$$2$$个，大李吃了个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$26$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "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->非整数倍差倍剩余"], "answer_analysis": ["根据题意大李比小李的$$4$$倍多$$2$$个，  则大李比小李多$$3$$倍加$$2$$个，  即小李吃的包子数$$\\times3+2=20$$（个）．  小李吃了$$(20-2)\\div3=6$$（个），大李吃了$$6+20=26$$（个），  两个人一共吃了$$26+6=32$$（个）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3399", "queId": "a4fd9eeb26ae4ddab65ffe96c52cebc2", "competition_source_list": ["2014年全国走美杯六年级竞赛初赛A卷第12题"], "difficulty": "2", "qtype": "single_choice", "problem": "用$$6$$颗颜色不同的彩色珠子串成一个手链，有~\\uline{~~~~~~~~~~}~种不同的串法（旋转、翻转后相同算一种）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$720$$ "}], [{"aoVal": "D", "content": "$$360$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["先选定一颗珠子，其他珠子在其后边开始全排列有5！种．手链可以翻转，再除以$$2$$为$$5！\\div 2=60$$，故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3160", "queId": "087eff41146447c498b2a1a88fe23725", "competition_source_list": ["2009年第7届创新杯四年级竞赛初赛第5题4分", "2009年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$2009\\square \\square \\square 02009\\approx 2010$$亿（四舍五入）。那么其中的三位数$$\\square \\square \\square $$有(  )种填写方法。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$1000$$ "}], [{"aoVal": "B", "content": "$$999$$ "}], [{"aoVal": "C", "content": "$$500$$ "}], [{"aoVal": "D", "content": "$$499$$ "}]], "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": ["三位数$$\\square \\square \\square $$中可填入$$500\\text{，}501\\text{，}502\\text{，}\\cdots \\text{，}999$$，共有$$999-500+1=500$$（种）填写方法。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3059", "queId": "f3947de40dac471e9626ec85cb9ce960", "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": ["知识标签->学习能力->七大能力->逻辑分析"], "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$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2439", "queId": "21918d7225804a6f943e7b46e0828241", "competition_source_list": ["2016年全国AMC六年级竞赛8第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "\\textbf{(2016 American Mathematics Competition 8, Question \\#15)}  What is the largest power of $$2$$ that is a divisor of $${{13}^{4}}-{{11}^{4}}$$?  $${{13}^{4}}-{{11}^{4}}$$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": ["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", "Overseas Competition->知识点->数论模块->分解质因数"], "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": "3276", "queId": "3294fe45f020456f8c0609a300865fb8", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁做竞猜游戏，在一个箱子中装有大小一样的二十个球，编号分别为$$1$$、$$2$$、$$3$$、$$\\cdot \\cdot \\cdot $$、$$20$$，从中任意取出一个球，猜测球的号码的情况．甲猜：奇数；乙猜：偶数；丙猜：$$3$$的倍数；丁猜：含有数字$$1$$．赢的可能性最大的是． ", "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->计数模块->统计与概率->概率->基本概率->可能性"], "answer_analysis": ["甲：$$\\frac{10}{20}$$；乙：$$\\frac{10}{20}$$；丙：$$\\frac{6}{20}$$；丁：$$\\frac{11}{20}$$；丁最大． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1584", "queId": "841a0168695041998b70e2c5aa142a9e", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "等腰三角形的两个角之比为$$2:5$$，则这个三角形是什么三角形（按角分类）？ ", "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": "不确定 "}]], "knowledge_point_routes": ["知识标签->数学思想->数形结合思想"], "answer_analysis": ["这个等腰三角形的三个内角之比为$$225$$或者$$255$$，前一种情况下，三角形最大的内角为$$180\\div \\left( 2+2+5 \\right)\\times 5=100$$°；后一种情况下，三角形最大的内角为$$180\\div \\left( 2+5+5 \\right)\\times 5=75$$．所以这个等腰三角形是钝角三角形或者锐角三角形． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1136", "queId": "41adf57db93c4574872c58302e70912b", "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2题1分"], "difficulty": "2", "qtype": "single_choice", "problem": "从前有一位老人，临终时，他把$$17$$匹马留给$$3$$个儿子，他说：``老大出力最多，得总数的$$\\dfrac{1}{2}$$；老二得总数的$$\\dfrac{1}{3}$$；老三最小，就拿总数的$$\\dfrac{1}{9}$$．''那么老大、老二、老三分别分到匹马． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$、$$6$$、$$3$$ "}], [{"aoVal": "B", "content": "$$9$$、$$6$$、$$2$$ "}], [{"aoVal": "C", "content": "$$9$$、$$5$$、$$3$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["老大、老二、老三分得马的匹数之比为$$\\dfrac{1}{2}:\\dfrac{1}{3}:\\dfrac{1}{9}=9:6:2$$，  故老大、老二、老三分别分到$$9$$、$$6$$、$$2$$匹马；故选$$\\text{B}$$．  此题，还有一个趣解：先向一个人借$$1$$匹马，这样总共有$$17+1=18$$匹马．  老大分得$$18\\times \\dfrac{1}{2}=9$$匹马，老二分得$$18\\times \\dfrac{1}{3}=6$$匹马，老三分得$$18\\times \\dfrac{1}{9}=2$$匹马；  这样，$$18$$匹马，还恰好多$$18-9-6-2=1$$匹马，正好还掉． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "616", "queId": "a6b7ba55b8524163955f0840b0eb9387", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛A卷第3题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "某校五年级同学分组展开课外活动，每组$$5$$人多$$4$$人；每组$$6$$人多$$5$$人；每组$$7$$人多$$6$$人．全年级至少有人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$280$$ "}], [{"aoVal": "B", "content": "$$271$$ "}], [{"aoVal": "C", "content": "$$209$$ "}], [{"aoVal": "D", "content": "$$265$$ "}]], "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": ["假如再多加$$1$$人，则按照每组$$5$$人，每组$$6$$人，每组$$7$$人都能刚好分，  而$$5$$、$$6$$、$$7$$的最小公倍数是$$210$$，  所以原来人数为：$$210-1=209$$人，  故全年级至少有$$209$$人． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2015", "queId": "b42afc49f6c34445a2782f77d2157f44", "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->学习能力->七大能力->逻辑分析"], "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": "3457", "queId": "c69a4c467bb14956a25779074e16a47b", "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6题"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$1\\sim 100$$这$$100$$个数中，所有不能被$$7$$和$$11$$整数的自然数的和是（~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3820$$ "}], [{"aoVal": "B", "content": "$$3897$$ "}], [{"aoVal": "C", "content": "$$4315$$ "}], [{"aoVal": "D", "content": "$$4555$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["总和为$$1+2+\\cdot \\cdot \\cdot +100=5050$$，  $$5050-7\\times (1+2+\\cdot \\cdot \\cdot +14)-11\\times (1+2+\\cdot \\cdot \\cdot +9)+77$$  $$=5050-7\\times 105-11\\times 45+77=3897$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "910", "queId": "e0541e80cea54fcf83b2e89fd346adb8", "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": "$$2005$$年$$7$$月$$25$$日，黑龙江发生$$5$$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黑龙江地震的~\\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->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": ["$$8$$级和$$5$$级差了$$3$$级，那么30*30*30=27000 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "292", "queId": "ff8080814518d52401451b56a194069d", "competition_source_list": ["2014年全国迎春杯三年级竞赛复赛第15题"], "difficulty": "3", "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->思想->整体思想"], "answer_analysis": ["无论甲怎么走，乙只要让最后火柴棒剩两根，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． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3077", "queId": "bd77d1db033b434e8731d5b9c23b4c9e", "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年陕西西安雁塔区西安铁一中小升初（二十六）第20题5分"], "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": ["知识标签->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": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2442", "queId": "2f44fb504b5f448d9224dae3ac042a1b", "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$$\\textasciitilde3+7\\times 4$$的正确结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$41$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["先算乘除，再算加减. "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1984", "queId": "bcdbf507e0d64f948549687b3be5c1a9", "competition_source_list": ["2014年第2届广东广州羊排赛六年级竞赛第6题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "一群小猫按照黑猫、白猫、花猫、花猫、灰猫、黑猫、白猫、花猫、花猫、灰猫、黑猫、白猫$$\\cdots \\cdots $$的规律列队站好，那么第$$2014$$只猫是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "黑猫 "}], [{"aoVal": "B", "content": "白猫 "}], [{"aoVal": "C", "content": "花猫 "}], [{"aoVal": "D", "content": "灰猫 "}]], "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": ["周期为$$5$$，$$2014\\div 5=402\\cdots \\cdots 4$$，对应的是花猫． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3225", "queId": "cc166083d1de48c4933777a5770d1644", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "实验二校一个歌舞表演队里，能表演独唱的有$$10$$人，能表演跳舞的有$$18$$人，两种都能表演的有$$7$$人．这个表演队共有多少人能登台表演歌舞？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$21$$ "}], [{"aoVal": "C", "content": "$$22$$ "}], [{"aoVal": "D", "content": "$$23$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据包含排除法，这个表演队能登台表演歌舞的人数为：$$10+18-7=21$$（人）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2620", "queId": "29486909a8ef49dcb821203bc0a060ac", "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": "$$502$$ "}], [{"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->方程思想"], "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": "171", "queId": "de9853d73887413da13f47f55c338059", "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->思想->整体思想"], "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": "120", "queId": "4fe649f460fe42a8b73236b36cf15f66", "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": "【热身2】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->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": "2217", "queId": "9986c6ec991c435fb5ba754b49b15659", "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": "2608", "queId": "f5ea61990c864dd6b61d8c5ab54ce3e3", "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1题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面的$$\\triangle$$，$$\\bigcirc$$，$$\\square$$各代表一个数，如果$$\\triangle +\\triangle +\\triangle =36$$，$$\\square \\times \\triangle =240$$，$$\\bigcirc \\div \\square =6$$，那么，$$\\bigcirc$$的值是（ ） ", "answer_option_list": [[{"aoVal": "A", "content": "120 "}], [{"aoVal": "B", "content": "100 "}], [{"aoVal": "C", "content": "130 "}], [{"aoVal": "D", "content": "124 "}]], "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": ["由$$\\triangle +\\triangle +\\triangle =36$$可得$$\\triangle =12$$，$$\\square \\times \\triangle =240$$可得$$\\square =20$$，$$\\bigcirc \\div \\square =6$$，$$\\bigcirc =120$$ "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2083", "queId": "dda0482516814d8e8f4c8af6dfae2924", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "张师傅$$2013$$年的工资为$$3000$$元/月，以后每年增加$$20  \\% $$，那么$$2016$$年他的工资比$$2013$$年是否增加了$$60  \\% $$？为什么？（~ ~） ", "answer_option_list": [[{"aoVal": "A", "content": "增加了$$60  \\% $$ "}], [{"aoVal": "B", "content": "不是增加$$60  \\% $$ "}], [{"aoVal": "C", "content": "不确定 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["$$2016$$年他的工资比$$2013$$年不是增加了$$60  \\% $$．因为$$2014$$年工资为$$3000\\times \\left( 1+20  \\% ~\\right)=3000\\times 1.2$$，$$2015$$年工资为$$3000\\times \\left( 1+20  \\% ~\\right)\\times \\left( 1+20  \\% ~\\right)=3000\\times 1.44$$， $$2016$$年工资为$$3000\\times 1.44\\times \\left( 1+20  \\% ~\\right)=3000\\times 1.728=3000\\times \\left( 1+72.8  \\% ~\\right)$$， 由此可见，张师傅的工资增加了$$72.8  \\% $$，而不是$$60  \\% $$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2326", "queId": "f4db406d70b9400c8b7e6a95e4d18e77", "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9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲船逆水航行$$360$$千米需$$18$$小时，返回原地需$$10$$小时，乙船逆水航行同样一段距离需$$15$$小时，返回原地需． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$小时 "}], [{"aoVal": "B", "content": "$$9$$小时 "}], [{"aoVal": "C", "content": "$$10$$小时 "}], [{"aoVal": "D", "content": "$$11$$小时 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$({{V}_{{甲}}}-{{V}_{{水}}})\\times 18=360$$，$$({{V}_{{甲}}}+{{V}_{{水}}})\\times 10=360$$，$${{V}_{{水}}}=8$$千米/小时；$$({{V}_{{乙}}}-{{V}_{{水}}})\\times 15=360$$，$${{V}_{{乙}}}=32$$千米/小时，$$360\\div ({{V}_{{乙}}}+{{V}_{{水}}})=9$$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": "2375", "queId": "0abe165615c04387b2993fe8930f70b3", "competition_source_list": ["2011年六年级竞赛创新杯"], "difficulty": "0", "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->拓展思维->计算模块->定义新运算->反解未知数型"], "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": "3071", "queId": "b8ea79d2377b47e085e98c430b98fe1b", "competition_source_list": ["2016年创新杯小学高年级六年级竞赛训练题（一）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "在不等式$$\\frac{5}{22}\\textless{}\\frac{23}{\\square }\\textless{}\\frac{4}{17}$$的方框中填入一个自然数，使得不等号成立，这样的自然数一共有（~ ~ ~）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->逻辑分析"], "answer_analysis": ["$$\\frac{23\\times 17}{4}\\textless{}\\square \\textless{}\\frac{22\\times 23}{5}$$，有$$97.75\\textless{}\\square \\textless{}101.2$$，那么一共有$$4$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1181", "queId": "118e725814cc4f9dbcc939f62365c6e8", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "某楼盘按照分期付款的形式售房，张明家购买了一套现价位$$12$$万的新房，购房时需首付（第一年）款$$3$$万元，从第二年起，以后每年应付房款$$5000$$元，与上一年剩余欠款的利息之和，已知剩余欠款的年利率为$$0.4  \\% $$，第年张明家需要交房款$$5200$$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->能力->抽象概括"], "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": "3409", "queId": "c97ead11b7b745fb9df08aec9c2d58a9", "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，每题的评分是$$0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$或$$7$$．有一群学生每人得分的乘积都是$$36$$，而且任意两天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->能力->数据处理"], "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": "685", "queId": "35a5b8434d1646559bdb2bebb1e3f9af", "competition_source_list": ["2016年创新杯小学高年级五年级竞赛训练题（三）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$198$$块水果糖和$$132$$块巧克力都可以分别平均分给四年级（$$1$$）班的每位同学而无剩余，四年级（$$1$$）班最多有（~ ）名学生． ", "answer_option_list": [[{"aoVal": "A", "content": "$$22$$ "}], [{"aoVal": "B", "content": "$$33$$ "}], [{"aoVal": "C", "content": "$$66$$ "}], [{"aoVal": "D", "content": "$$99$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$198+132=330$$（块），∵$$198$$块水果糖和$$132$$块巧克力都可以分别平均分给四年级（$$1$$）班的每位同学而无剩余，∴ 学生人数可以被$$330$$整除，经验算最多有$$66$$人． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1260", "queId": "426d642e637e423eaf59fbe4e54b5fbd", "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2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$A$$、$$B$$两个仓库有同样多的货物，甲单独搬完一个仓库需要$$10$$小时，乙单独搬完一个仓库需要$$12$$小时，丙单独搬完一个仓库需要$$15$$小时，现在甲搬$$A$$仓库，乙搬$$B$$仓库，丙一会帮甲，一会帮乙，最后两个仓库同时搬完，三个人同时开始，协同搬完两个仓库，总共用时间（~ ）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->能力->实践应用"], "answer_analysis": ["假设两个仓库的工作总量为单位``$$1$$''，则三人合作完成总做总量为单位``$$2$$''，则所需要的时间为$$2\\div \\left( \\frac{1}{10}+\\frac{1}{12}+\\frac{1}{15} \\right)= 8$$小时． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "912", "queId": "97ef3c5a7c144ac0adb1e002d8fd61a6", "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6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，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->思想->逐步调整思想"], "answer_analysis": ["要最大，考虑首位为$$9$$，前五位不能为$$0$$，否则三位数的百位会为$$0$$；因为$$7\\times 11\\times 13=1001$$，考虑$$1001-11=990$$，$$1001-13=988$$．则只能$$988$$作为开始；$$881$ $889$$中最大有$$884$$为$$13$$的倍数；$$841$ $849$$中最大有$$847$$为$$11$$的倍数；$$470$ $479$$中最大有$$473$$为$$11$$的倍数；$$730$ $739$$中最大有$$737$$为$$11$$的倍数；综上，这个六位数为$$9884737$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1749", "queId": "d1d6ea569d6f4c67b2a74300f909448a", "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4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小丽在计算除法时，把除数$$530$$末尾的$$0$$漏写了，得到的商是$$60$$．正确的商应该是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$600$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["除数的末尾的$$0$$漏掉，则正确的结果应该为$$6$$．  故选：$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1327", "queId": "476cb31342ad49da868f6b97be0f9fde", "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9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果四个人的平均年龄是$$23$$岁，四个人中没有小于$$18$$岁的，那么四个人中年龄最大的可能是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$32$$岁 "}], [{"aoVal": "B", "content": "$$35$$岁 "}], [{"aoVal": "C", "content": "$$38$$岁 "}], [{"aoVal": "D", "content": "$$40$$岁 "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["四个人的平均年龄是$$23$$岁，他们的年龄和是$$23\\times4=92$$岁，要求最大的年龄是多少岁，必须使另外三人年龄最小，又没有小于$$18$$岁的，那么其他三人的年龄最小都是$$18$$岁，然后再用四人的年龄和减去最小$$3$$人的年龄和即可．  $$23\\times 4=92$$（岁），  $$92-18\\times 3=92-54=38$$（岁），  答：四个人中年龄最大的可能是$$38$$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": "2875", "queId": "618b909bc47449a398072333a320a9a6", "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：$$\\text{A}$$选项． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2814", "queId": "fb19bedf994e4a108ddf95816925ea47", "competition_source_list": ["2015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": "请问算式$$100\\times 100-2015$$的值的各位数码之和是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$27$$ "}], [{"aoVal": "B", "content": "$$29$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$34$$ "}], [{"aoVal": "E", "content": "$$39$$ "}]], "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": ["$$100\\times 100-2015=7985$$，各位数码和为：$$7+8+9+5=29$$，故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1043", "queId": "072fde29bf1347aa8c05ee1e46a6a3ea", "competition_source_list": ["2019年陕西西安雁塔区西安交通大学附属中学 小升初入学真卷10第4题3分", "2016年世界少年奥林匹克数学竞赛六年级竞赛初赛A卷第7题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "四袋水果共$$46$$个，如果第一袋增加$$1$$个，第二袋减少$$2$$个，第三袋增加$$1$$倍，第四袋减少一半，那么四袋水果的个数就相等了，设谁为未知数x ", "answer_option_list": [[{"aoVal": "A", "content": "原先每袋的水果数量 "}], [{"aoVal": "B", "content": "现在每袋的水果数量 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["设后来每袋水果有$$x$$个，那么第一袋原先的数量为$$(x-1)$$个，  第二袋原先的数量为$$(x+2)$$个，第三袋原先的数量为$$0.5x$$个，第四袋原先的数量为$$2x$$个，  根据题意得：$$x-1+x+2+0.5x+2x=46$$，  解得：$$x=10$$，  则第四袋水果原有$$2\\times 10=20$$（个）．  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": "1954", "queId": "c1047a68e2384007ab5695fbf2ba58d6", "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31题"], "difficulty": "2", "qtype": "single_choice", "problem": "拉里可能在说谎．别人问他年龄时，他把自己的真实年龄加$$10$$，除以$$2$$，减$$10$$，再乘$$2$$，得到$$30$$，由此他告诉别人他今年$$30$$岁．请问拉里的真实年龄是？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["还原法，$$(30\\div2+10)\\times2-10=40$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3031", "queId": "f7bafeaa82144e4eaa5cdf477f86d564", "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1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$\\left( 2021+2020+2019+2018+2017 \\right)\\div 5=$$． ", "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->拓展思维->计算模块->整数->整数提取公因数->整数构造提取->整数倍数关系"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left( 2021+2020+2019+2018+2017 \\right)\\div 5$$  $$=2019\\times 5\\div 5$$  $$=2019$$，  故选：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1251", "queId": "500a824b5cc14ae9998282e43305af00", "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5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两筐苹果共重$$46$$千克，从甲筐中取出$$8$$千克苹果放入乙筐，这样乙筐比甲筐重$$2$$千克，甲筐苹果原来重千克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$29$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$31$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["因为由题干可知，甲、乙两筐苹果共重$$46$$千克，从甲筐中取出$$8$$千克苹果放入乙筐，这样乙筐比甲筐重$$2$$千克，设甲筐中原来有苹果$$x$$千克，则可列方程为：  $$46-x+8=x-8+2$$，解得$$x=30$$，所以甲筐苹果原来重$$30$$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": "25", "queId": "02cde06cff2346c797eca46df1d7457e", "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", "课内体系->能力->逻辑分析"], "answer_analysis": ["每局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$$11：8$$，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": "2551", "queId": "6bcd9ba048e645c088ec72fe0154a807", "competition_source_list": ["2006年走美杯四年级竞赛", "2006年走美杯六年级竞赛", "2006年走美杯五年级竞赛"], "difficulty": "0", "qtype": "single_choice", "problem": "$$a\\vee b$$表示$$a$$、$$b$$两个数中取较大的一个，$$a\\wedge b$$表示$$a$$、$$b$$两个数中取较小的一个，则$$\\left( 666\\vee 888 \\right)\\wedge \\left( 777\\vee 555 \\right)$$等于（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$555$$ "}], [{"aoVal": "B", "content": "$$666$$ "}], [{"aoVal": "C", "content": "$$777$$ "}], [{"aoVal": "D", "content": "$$888$$ "}]], "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": ["$$666\\vee 888$$两个数中取较大的是$$888$$，$$777\\vee 555$$两个数中取较大的是$$777$$，$$888\\wedge 777$$两个数中取较小的是$$777$$。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "472", "queId": "c0fa9b19cc6b46878fdd9e6f42540e60", "competition_source_list": ["2019年美国数学大联盟杯三年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "从下午$$12:30$$到下午$$4:00$$经过了多少分钟 ", "answer_option_list": [[{"aoVal": "A", "content": "$$180$$ "}], [{"aoVal": "B", "content": "$$210$$ "}], [{"aoVal": "C", "content": "$$240$$ "}], [{"aoVal": "D", "content": "$$480$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["从下午$$12:30$$到下午$$4:00$$经过了$$3$$个小时$$30$$分钟，也就是$$3\\times60+30=210$$（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": "3469", "queId": "efce77eb760e4c91b0dee8ad94b2408b", "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6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "用数字$$0$$、$$3$$、$$4$$、$$5$$、$$6$$最多可以组成个无重复数字的三位数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$64$$ "}], [{"aoVal": "D", "content": "$$100$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["∵首先百位数不能选$$0$$有$$4$$种选择，  当百位数确定后十位数有$$4$$种选择，  当十位数确定后个位数有$$3$$种选择，  ∴根据分步计数原理所组成的三位数为$$4\\times4\\times3=68$$种．  故选：$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2662", "queId": "6393e4f40ff2466a88bc1c123c24475d", "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": ["拓展思维->能力->运算求解"], "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": "2466", "queId": "f0fa01cc4a5d4ea3a17cbded820938b3", "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"], "difficulty": "3", "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": ["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"], "answer_analysis": ["先比较$$a$$和$$b$$， $$a=\\frac{2005\\times 2006}{2007\\times 2008}=\\frac{2005}{2007}\\times \\frac{2006}{2008} $$，$$b=\\frac{2006\\times 2007}{2008\\times 2009}=\\frac{2006}{2008}\\times \\frac{2007}{2009}$$ 比较$$\\frac{2005}{2007}$$和$$\\frac{2007}{2009}$$，$$1-\\frac{2005}{2007}=\\frac{2}{2007}$$，$$1-\\frac{2007}{2009}=\\frac{2}{2009}$$，由于$$\\frac{2}{2007}\\textgreater\\frac{2}{2009}$$，所以 $$\\frac{2005}{2007} \\textless{} \\frac{2007}{2009}$$，则$$\\frac{2005}{2007}\\times \\frac{2006}{2008} \\textless{} \\frac{2006}{2008}\\times\\frac{2007}{2009}$$，$$a \\textless{} b$$。 同样的道理：$$b \\textless{} c$$。则$$a \\textless{} b \\textless{} c$$，选D。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "258", "queId": "ff5f1cc6cf2d42f6b0ecb1bf885dd74d", "competition_source_list": ["全国二年级下学期其它尖子生题库北师大版第四单元尖子生竞赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "小桶一次能装$$5$$千克的油，大桶一次能装$$7$$千克的油．你能用这$$2$$个桶往空桶里倒入$$9$$千克的油吗？ ", "answer_option_list": [[{"aoVal": "A", "content": "可以 "}], [{"aoVal": "B", "content": "不可以 "}]], "knowledge_point_routes": ["知识标签->拓展思维->组合模块->操作与策略->操作问题->倒水问题"], "answer_analysis": ["先准备好$$1$$个空桶，把大桶装满油，倒入小桶，这时大桶里还剩$$2$$千克的油，倒进空桶里，再把大桶里装满油，倒进桶里，这样空桶里就有$$9$$千克油． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "691", "queId": "24a7e5b9bbaf4489909e4de0c9a543d4", "competition_source_list": ["2009年第14届全国华杯赛竞赛初赛第4题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$19$$、$$197$$、$$2009$$这三个数中，质数的个数是(  )． ", "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->逐步调整思想"], "answer_analysis": ["$$19$$是常见的质数，$$197$$容易检验知也是质数，本题主要是考查$$2009$$这个数是否是质数$$.$$实际上，$$2009=7\\times 7\\times 41$$ ，是个合数，所以在$$19$$，$$197$$，$$2009$$这三个数中有$$2$$个质数．正确答案为$$\\rm C$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "425", "queId": "8e6b1a666d0a4049bfa2accc6cbdee3e", "competition_source_list": ["2012年全国迎春杯四年级竞赛初赛第7题"], "difficulty": "3", "qtype": "single_choice", "problem": "红、黄、蓝$$3$$种颜色的球分别有$$11$$、$$12$$、$$17$$个，每次操作可以将$$2$$个不同颜色的球换成$$2$$个第三种颜色的球，则在操作过程中，红色球至多有~\\uline{~~~~~~~~~~}~个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$37$$ "}], [{"aoVal": "C", "content": "$$38$$ "}], [{"aoVal": "D", "content": "$$39$$ "}], [{"aoVal": "E", "content": "$$40$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->抽屉原理->构造型抽屉原理"], "answer_analysis": ["（1）红球最多，那么就要黄、蓝球最少，而黄、蓝球最少则取决于这两种球数之差最小．  （2）一开始，黄蓝球数差$$5$$，每次操作，黄球和蓝球的差要么不变，要么改变$$3$$，所以变化后，这两种球数之差最小为$$1$$．  （3）当黄、蓝球数差$$1$$时，我们就一直把它们换成红色，最后只剩下一个黄球，剩下的$$39$$个全是红球，此时红球会最多．  （4）下面构造说明红球最多可以有$$39$$个．（括号内三个数依次表示红、黄、蓝$$3$$种颜色的球的个数）  （$$11$$，$$12$$，$$17）→（10$$，$$14$$，$$16）→（9$$，$$16$$，$$15）→（11$$，$$15$$，$$14）→（13$$，$$14$$，$$13）→（15$$，$$13$$，$$12）→（17$$，$$12$$，$$11）→（19$$，$$11$$，$$10$$）→$$\\cdots$$ →（$$39$$，$$1$$，$$0$$）． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "58", "queId": "0f51060d4e2444fdbb3deee235593e5f", "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、豹子和大象中的两只动物去参加．  （1）如果派狮子去，那么也要派老虎去.  （2）如果不派豹子去，那么也不能派老虎去.  （3）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": ["假设狮子去，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；  假设包子不去，则狮子不去，那么只有大象去，矛盾，所以豹子一定去；  豹子去，则大象不去，狮子也不去，所以豹子和老虎去，故选$$\\text{C}$$ "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2285", "queId": "a92e340a3db741e081e4ce66000f308f", "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": "罗伯开车 $$4$$ 小时走了 $$120$$ 千米．请问他开车的平均速度是多少米/分钟？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$105$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$500$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["略 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2414", "queId": "1cc7f3ae861e4288af1a90306fd4225e", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$\\bigcirc$，$$★$$，$\\triangle$代表三个数，  $\\bigcirc+\\bigcirc=12$，  $$★+★+★=15$$，  $$△+△+△+△=24$$  $\\bigcirc+\\bigstar+\\triangle=$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "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->代数逻辑推理"], "answer_analysis": ["根据题意分析可知，  $\\bigcirc+\\bigcirc=12$，即$2\\bigcirc=12$，$\\bigcirc=12\\div2=6$，  $$★+★+★=15$$，$$3★=15$$，$\\bigstar=15\\div3=5$，  $$△+△+△+△=24$$，$$4△=24$$，$\\triangle=24\\div4=6$，  由此可知，$\\bigcirc+\\bigstar+\\triangle=6+5+6=17$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "592", "queId": "0fa18bc0830c4601bd134accd1ba384f", "competition_source_list": ["2016年创新杯五年级竞赛训练题（二）第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$39$$个偶数的平均数，如果保留一位小数是$$23.4$$，如果保留两位小数，得数最小的是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$23.32$$ "}], [{"aoVal": "B", "content": "$$23.35$$ "}], [{"aoVal": "C", "content": "$$23.38$$ "}], [{"aoVal": "D", "content": "$$23.41$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["设这$$39$$个偶数分别是$${{a}_{1}}$$，$${{a}_{2}}$$，$$\\cdots $$，$${{a}_{39}}$$，则有$$23.35\\leqslant \\square \\textless{}23.45$$．所以有$$910.65\\leqslant {{a}_{1}}+{{a}_{2}}+\\cdots +{{a}_{39}}\\textless{}914.55$$．从而$$39$$个偶数的和最小是$$912$$．因为$$912\\div 39\\approx 23.38$$，即为所求的两位小数． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3051", "queId": "eec767520a8c4497b4a0b0a085906421", "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": "一个长方体的长、宽、高恰好是3个连续的自然数，并且它的体积的数值等于它的所有棱长之和的数值的2倍，那么这个长方体的表面积是(  ) ", "answer_option_list": [[{"aoVal": "A", "content": "74 "}], [{"aoVal": "B", "content": "148 "}], [{"aoVal": "C", "content": "150 "}], [{"aoVal": "D", "content": "154 "}]], "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": ["设长方体的三条棱长分别为a-1，a，a+1，则它的体积为$${{a}^{3}}-a$$， 它的所有棱长之和为$$\\left[ \\left( a-1 \\right)+a+\\left( a+1 \\right) \\right]\\times 4=12a$$ 于是有$${{a}^{3}}-a=12a\\times 2$$，即$${{a}^{3}}=25a$$，$${{a}^{2}}=25$$，$$a=5$$， 即这个长方体的棱长分别为4，5，6 所以，它的表面积为$$\\left(4\\times 5+4\\times 6+5\\times 6\\right)\\times 2=148$$ "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2034", "queId": "c669d8cd6ca54beca5453ee3f06c4005", "competition_source_list": ["竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "有两根同样长的绳子，从第一根中先用去$$\\frac{1}{3}$$，再用去$$\\frac{1}{3}$$米；从第二根中先用去$$\\frac{1}{3}$$米，再用去余下长度的$$\\frac{1}{3}$$，仍都有剩余，第一根所剩部分与第二根所剩部分相比较(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "第一根长 "}], [{"aoVal": "B", "content": "第二根长 "}], [{"aoVal": "C", "content": "两根一样长 "}], [{"aoVal": "D", "content": "无法确定 "}]], "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": ["设两根绳都长$$a$$米，第一根绳剩下$$a\\times \\left( 1-\\frac{1}{3} \\right)-\\frac{1}{3}=\\frac{2}{3}a-\\frac{1}{3}$$（米），  第二根绳剩下$$\\left( a-\\frac{1}{3} \\right)\\times (1-\\frac{1}{3})=\\frac{2}{3}a-\\frac{2}{9}$$（米），比较知第二根绳剩下的比第一根绳剩下的长. "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "789", "queId": "a80805b69aff4422af72598bc8f1e5d3", "competition_source_list": ["2005年第3届创新杯五年级竞赛初赛第4题", "2005年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "如果a，b，c是三个任意的整数，那么$$\\frac{a+b}{2}$$，$$\\frac{b+c}{2}$$，$$\\frac{c+a}{2}$$(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "都不是整数 "}], [{"aoVal": "B", "content": "至少有两个整数 "}], [{"aoVal": "C", "content": "至少有一个整数 "}], [{"aoVal": "D", "content": "都是整数 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的加减规律"], "answer_analysis": ["当a，b，c这三个数中没有奇数时，显然$$\\frac{a+b}{2}$$，$$\\frac{b+c}{2}$$，$$\\frac{c+a}{2}$$这三个数均为整数; 当a，b，c这三个数中只有一个奇数时，不妨设a为奇数，那么b，c为偶数，则$$\\frac{a+b}{2}$$，$$\\frac{b+c}{2}$$，$$\\frac{c+a}{2}$$三个数中只有$$\\frac{b+c}{2}$$为整数; 当a，b，c这三个数中只有两个奇数时，不妨设a，b为奇数，那么c为偶数，则$$\\frac{a+b}{2}$$，$$\\frac{b+c}{2}$$，$$\\frac{c+a}{2}$$三个数中只有$$\\frac{a+b}{2}$$为整数; 当a，b，c这三个数均为奇数时，显然$$\\frac{a+b}{2}$$，$$\\frac{b+c}{2}$$，$$\\frac{c+a}{2}$$均为整数; 故无论a，b，c为任何整数时，$$\\frac{a+b}{2}$$，$$\\frac{b+c}{2}$$，$$\\frac{c+a}{2}$$三个数中至少有一个为整数，选C. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2873", "queId": "6a4fa91b249746588f85fdcecb6f907d", "competition_source_list": ["2014年中环杯三年级竞赛初赛"], "difficulty": "0", "qtype": "single_choice", "problem": "对于两个数字$$a$$、$$b$$，定义新运算：$$a*b=a\\times b+a+b$$，则$$9*11=$$（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$99$$ "}], [{"aoVal": "B", "content": "$$109$$ "}], [{"aoVal": "C", "content": "$$119$$ "}], [{"aoVal": "D", "content": "$$129$$ "}]], "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": ["根据定义，得$$9*11=9\\times 11+9+11=119$$。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2665", "queId": "8833de08840b40a0b1c04d2a9c2147ff", "competition_source_list": ["2018年IMAS小学高年级竞赛（第一轮）第17题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个三位数除以$$37$$后所得到的商是$$a$$、余数是$$b$$，其中$$a$$、$$b$$都是非负整数．请问$$a+b$$的最大值是多少？． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$62$$ "}], [{"aoVal": "C", "content": "$$64$$ "}], [{"aoVal": "D", "content": "$$66$$ "}], [{"aoVal": "E", "content": "$$68$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由题意可知，$$b$$的最大值为$$36$$，而$$999=37\\times27$$、$$998=37\\times26+36$$，因此$$a+b$$的最大值是$$26+36=62$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "640", "queId": "15f83c895b714c5faeff433c2f78772e", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛A卷第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小红去买醋，她带的钱如果买每瓶$$7$$元的醋，还剩$$3$$元，如果买每瓶$$6$$元的醋，也还剩$$3$$元，小红带了． ", "answer_option_list": [[{"aoVal": "A", "content": "$$45$$元 "}], [{"aoVal": "B", "content": "$$42$$元 "}], [{"aoVal": "C", "content": "$$39$$元 "}], [{"aoVal": "D", "content": "$$36$$元 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["买每瓶$$7$$元的和每瓶$$6$$元的都剩$$3$$元，总价减去$$3$$元后是$$6$$和$$7$$的公倍数，$$[6,7]=6\\times 7=42$$，$$42+3=45$$（元）．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3134", "queId": "0df5499567fb4e4cbfb78a78e1876bd3", "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决赛", "2011年中环杯四年级竞赛初赛"], "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": "$$160$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$112$$ "}]], "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，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": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2982", "queId": "ce0f95fc17244be4ac0ea198c5b55527", "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": "$$224$$ "}], [{"aoVal": "E", "content": "$$230$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "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": "1963", "queId": "e5662913057e480898ab6f2d21c5db23", "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->思想->逆向思想"], "answer_analysis": ["$$\\left( 2014\\div 22 \\right)\\times 2+2=2012$$. "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1581", "queId": "ff8080814518d5240145201abacc0a6c", "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9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第7题3分", "2014年全国迎春杯三年级竞赛复赛第11题"], "difficulty": "1", "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->假设法解鸡兔同笼->基本型->变型题"], "answer_analysis": ["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$$2800-200=2600$$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$$2600$$个钱币全部是是一元的，  那么可得红蓝钱币一共有$$(6800-2600\\times 1)\\div (4-1)=1400$$（个），  则红色钱币有$$1400\\div2=700$$（个）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3154", "queId": "07bb975f50404a87a576b2e35794192e", "competition_source_list": ["2020年少年袋鼠竞赛（Junior  Kangaroo）小学中年级竞赛第22题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "（KMC-2020-Question 22）  Several teams came to the summer Kangaroo camp. Each team has $$5$$ or $$6$$ members. There are $$43$$ people in total. How many teams are at this camp？  有几支队伍来到了袋鼠夏令营．每支队伍会有$$5$$或$$6$$名队员．如果他们一共有$$43$$人．请问袋鼠夏令营里将会有多少支队伍？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"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->整体思想", "Overseas Competition->知识点->应用题模块"], "answer_analysis": ["暂无 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2613", "queId": "94caac7e0b4142bc9baa0bf35bdc358d", "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": "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->拓展思维->计算模块->方程基础->不定方程->不定方程组"], "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": "3247", "queId": "2d1fd03cf71d424d877d93b8bc5e8938", "competition_source_list": ["其它", "2015年湖北武汉世奥赛六年级竞赛模拟训练题（三）第10题"], "difficulty": "3", "qtype": "single_choice", "problem": "一个骰子六个面上的数字分别为$$0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，现在来掷这个骰子，把每次掷出的点数依次求和，当总点数超过$$12$$时就停止不再掷了，这种掷法最有可能出现的总点数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计数模块->统计与概率->概率->典型问题->掷骰子"], "answer_analysis": ["掷的总点数在$$8$$至$$12$$之间时，再掷一次，总点数才有可能超过$$12$$（至多是$$17$$）．当总点数是$$8$$时，再掷一次，总点数是$$13$$的可能性比总点数超过$$13$$的可能性大．当总点数在$$9$$至$$12$$之间时，再掷一次，总点数是$$13$$的可能性不比总点数是$$14$$，$$15$$，$$16$$，$$17$$的可能性小．  例如，总点数是$$11$$时，再掷一次，出现$$0\\sim 5$$的可能性相同，所以总点数是$$11\\sim 16$$的可能性相同，即总数是$$13$$的可能性不比总数点数分别是$$14$$，$$15$$，$$16$$的可能性小，综上所述，总点数是$$13$$的可能性最大． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2702", "queId": "7f520fd163df44da9686bca4aa19cf17", "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9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一条大鱼，鱼尾的质量是$$1$$千克，鱼头的质量等于鱼尾的质量与鱼身的质量的一半的和，而鱼身的质量又等于鱼头和鱼尾质量的和．那么这条鱼重千克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["由题意可以设鱼身的质量为$$x$$千克，鱼头的质量为$$y$$千克，  那么可以列方程$$\\frac{1}{2}+\\frac{1}{2}x=y$$，$$x=1+y$$，  解得$$x=3$$，$$y=2$$，所以这条鱼重$$1+2+3=6$$（千克）．  故选：$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "950", "queId": "d8b2b130bcc84a2a989d8bcc958c64fa", "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": "$$\\overline{**45}$$，$$\\overline{19*8}$$，$$\\overline{23*1}$$，$$\\overline{3*49}$$，$$\\overline{5*63}$$ are five four-digit numbers, where $$*$$ represents unrecognizable numbers, if one of them is a perfect square number, then this number may be ．  $$\\overline{**45}$$，$$\\overline{19*8}$$，$$\\overline{23*1}$$，$$\\overline{3*49}$$，$$\\overline{5*63}$$是五个四位数，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": "$$\\overline{23*1}$$ "}], [{"aoVal": "D", "content": "$$\\overline{3*49}$$ "}], [{"aoVal": "E", "content": "$$\\overline{5*63}$$ "}]], "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", "海外竞赛体系->知识点->数论模块->完全平方数"], "answer_analysis": ["完全平方数的个位只能为$$0$$、$$1$$、$$4$$、$$5$$、$$6$$、$$9$$，因此排除$$\\text{B}$$选项和$$E$$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": "481", "queId": "af47ec6783c84e16a032213e4a3f2c09", "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->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": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1853", "queId": "f717dcbe71e94b8ebe54ecee564b8e02", "competition_source_list": ["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", "2004年第2届创新杯六年级竞赛初赛第5题", "2004年六年级竞赛创新杯"], "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": ["设甲、乙两根绳子的长度都为$$9$$米，则：  甲剩下：  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde9\\times \\left( 1-\\frac{1}{3} \\right)-\\frac{1}{3}$$  $$=6-\\frac{1}{3}$$  $$=5\\frac{2}{3}$$（米），  乙剩下：  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left( 9-\\frac{1}{3} \\right)\\times \\left( 1-\\frac{1}{3} \\right)$$  $$=\\frac{26}{3}\\times \\frac{2}{3}$$  $$=5\\frac{7}{9}$$（米），  $$5\\frac{2}{3}$$米$$ ~\\textless{} ~5\\frac{7}{9}$$米．  答：乙绳剩下部分长．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1468", "queId": "5a94ec874a5d4e6aadaff80c8d0b812c", "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": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde(18-4)\\div (8-6)$$，  $$=14\\div 2$$，  $$=7$$（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": "3243", "queId": "24a549d324c64d8287edbdce41f47fbb", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "二年级从$$5$$个学生中选择$$2$$人合作主持节目，共有种不同的选择方法． ", "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->能力->实践应用"], "answer_analysis": ["由于每个主持人都有可能和另外$$1$$个主持人去主持节目，  一共有$$5\\times (5-1)=20$$（种），  去掉重复情况共有$$20\\div 2=10$$（种），  选择$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1942", "queId": "98888fe6a32c4340a0bc7030f83eae16", "competition_source_list": ["2014年全国迎春杯竞赛", "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，~\\uline{~~~~~~~~~~}~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->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": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2881", "queId": "6aac8801c54c446f9c6c516c2a6c3df2", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛B卷第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "分子、分母是相邻的自然数，这个分数一定是． ", "answer_option_list": [[{"aoVal": "A", "content": "真分数 "}], [{"aoVal": "B", "content": "假分数 "}], [{"aoVal": "C", "content": "带分数 "}], [{"aoVal": "D", "content": "最简分数 "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["做这种题之前一定要先明确这几个概念：  真分数是指分子小于分母的分数．真分数的分数值小于一．  分子大于或者等于分母的分数叫假分数，假分数大于$$1$$或等于$$1$$．  分数值大于$$1$$或等于$$1$$的分数，带分数是假分数的一种形式．  非零自然数与真分数相加（负整数时与真分数相减）所成的分数（或真分数与假分数相加减化简后的数），一般读作几又几分之几，假分数的倒数一定不大于一．  分子、分母是相邻的自然数，  这个分数有可能是真分数比如$$\\frac{1}{2}$$，  也可能是假分数$$\\frac{3}{2}$$；  又可能是带分数$$1\\frac{1}{2}$$；  但是无论是什么分数，都是最简分数．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2756", "queId": "ac9602a2d1de4e1cb2fbe208810135b3", "competition_source_list": ["2008年第6届创新杯六年级竞赛初赛B卷第7题5分"], "difficulty": "2", "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，李平又喝了第二天所剩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->运算求解"], "answer_analysis": ["设妈妈买的饮料一共有$$x$$瓶，  则第一天喝了$$\\left( \\frac{1}{2}x+0.5 \\right)$$瓶，  那么剩下$$\\left( x-\\frac{1}{2}x-0.5 \\right)$$瓶，  则第二天喝了$$\\frac{1}{2}\\left( x-\\frac{1}{2}x-0.5 \\right)+0.5$$（瓶），  那么剩下$$\\left( x-\\frac{1}{2}x-0.5 \\right)-\\left[ \\frac{1}{2}\\left( x-\\frac{1}{2}x-0.5 \\right)+0.5 \\right]$$（瓶），  所以第三天喝了$$\\frac{1}{2}\\left { \\left( x-\\frac{1}{2}x-0.5 \\right)-\\left[ \\frac{1}{2}\\left( x-\\frac{1}{2}x-0.5 \\right)+0.5 \\right] \\right }+0.5$$（瓶），  $$\\left( \\frac{1}{2}x+0.5 \\right)+\\left[ \\frac{1}{2}\\left( x-\\frac{1}{2}x-0.5 \\right)+0.5 \\right]+\\frac{1}{2}\\left { \\left( x-\\frac{1}{2}x-0.5 \\right)-\\left[ \\frac{1}{2}\\left( x-\\frac{1}{2}x-0.5 \\right)+0.5 \\right] \\right }+0.5=x$$，  解得$$x=7$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2241", "queId": "51fdcdba1c134190a36e5c17f9ebfa36", "competition_source_list": ["2016年第16届世奥赛六年级竞赛决赛第4题", "2016年全国世奥赛竞赛A卷第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "中央一台的《焦点访谈》是一档时事性、政治性较强的电视节目，约在晚间$$7$$点$$38$$分，时针与分针重合时播出．用时针与分针的重合来比喻时事、政治的焦点．则《焦点访谈》播出的准确时间是$$19$$时（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$38\\frac{5}{11}$$ "}], [{"aoVal": "B", "content": "$$38\\frac{3}{11}$$ "}], [{"aoVal": "C", "content": "$$38\\frac{2}{11}$$ "}], [{"aoVal": "D", "content": "$$38\\frac{1}{11}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["我们知道分针每分钟走$$6{}^{}\\circ $$，时针每分钟走$$0.5{}^{}\\circ $$．$$7$$点整时，分针和时针之间的夹角是$$210{}^{}\\circ $$，时针与分针重合也就是分针追上时针，需要分钟$$210{}^{}\\circ \\div \\left( 6{}^{}\\circ -0.5{}^{}\\circ \\right)=38\\frac{2}{11}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2722", "queId": "9eaac274872b4049876f8d38afba491d", "competition_source_list": ["2011年全国世奥赛三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$100 - 99 - 98 + 97 +96 -95 -94+93~ \\cdots~ +4 - 3 - 2 + 1$$的结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$150$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["$$00000$$ "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "464", "queId": "d770a2ad1c674692b76f5ecb9f2c9f72", "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．那么绳子有段． ", "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->操作与策略->归纳递推"], "answer_analysis": ["略 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2599", "queId": "24add561c6484027850c7f6b08291b4f", "competition_source_list": ["2015年第11届全国新希望杯五年级竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "解方程：$$2(x+1)=4\\times (8-x)$$，则$$x$$等于多少？ ", "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->计算模块->方程基础->一元一次方程->整数系数方程"], "answer_analysis": ["$$2(x+1)=4\\times (8-x)$$  $$2x+2=32-4x$$  $$6x=30$$  $$x=$$$$5$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "545", "queId": "f55cdc1cf66548cd8274ff04cdc015ec", "competition_source_list": ["2013年全国华杯赛小学中年级竞赛初赛A卷第3题", "2016年全国华杯赛小学中年级竞赛在线模拟第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．  故选：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1075", "queId": "4603510f2a5a45428ef32bfc5d80d64e", "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1题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明妈妈花$$8$$元买了一条鱼，以$$9$$元价格卖掉，然后觉得不合算，又花$$10$$元买回来，以$$11$$元卖给另一个人，那么小明妈妈赚了元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解->程序性计算"], "answer_analysis": ["$$\\left( 9-8 \\right)+\\left( 11-10 \\right)=2$$（元）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1949", "queId": "a606c85411c44b058f3e75309b69316d", "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$$25  \\% $$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": "$$40$$ "}], [{"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->认识单位1"], "answer_analysis": ["$$20\\div \\left( 125  \\% -75  \\% \\right)=40$$（米）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1645", "queId": "b151fbc64ded47a6bf334658745107ea", "competition_source_list": ["2020年长江杯六年级竞赛复赛B卷第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "新星小学开展广播操比赛，三年级学生站成一个实心方阵（正方形队列）时，还多$$10$$人，如果站成一个每边多$$1$$人的实心方阵，则还缺少$$15$$人．该校三年级有人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$154$$ "}], [{"aoVal": "B", "content": "$$144$$ "}], [{"aoVal": "C", "content": "$$169$$ "}], [{"aoVal": "D", "content": "$$174$$ "}]], "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": ["此题解答的关键是扩大的方阵共有多少人，要求扩大的方阵共有多少人，就要求出扩大的方阵每边上的人数．  扩大的方阵每边上有：  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left( 10+15+1 \\right)\\div 2$$  $$=16\\div 2$$  $$=13$$（人），  原来人数：  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde13\\times 13-15$$  $$=169-15$$  $$=154$$（人）．  答：原来有$$154$$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": "1199", "queId": "2fd580bdac4548778615f645a3e9096f", "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": ["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": "2750", "queId": "4e0b315c53d544ea8e0f83c3c91b158b", "competition_source_list": ["2017年全国希望杯小学高年级六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a=2015\\times 2017$$，$$b=2014\\times 2018$$，$$c=2016\\times 2016$$，将$$a$$，$$b$$，$$c$$从大到小排列． ", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b\\textgreater c$$ "}], [{"aoVal": "B", "content": "$$c\\textgreater b\\textgreater a$$ "}], [{"aoVal": "C", "content": "$$b\\textgreater a\\textgreater c$$ "}], [{"aoVal": "D", "content": "$$c\\textgreater a\\textgreater b$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$a=2015\\times 2017=\\left( 2016-1 \\right)\\times \\left( 2016+1 \\right)$$  $$=2016\\times 2016-1=c-1$$  $$b=2014\\times 2018=\\left( 2016-2 \\right)\\times \\left( 2016+2 \\right)$$  $$=2016\\times 2016-4=c-4$$  所以$$c\\textgreater a\\textgreater b$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "144", "queId": "30727229e0954a758b91d4138eab2216", "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": "现在是星期六上午$$6$$时，小丽同学准备在$$3$$小时后去奶奶家打扫卫生，请问小丽几点钟去奶奶家打扫卫生？ ", "answer_option_list": [[{"aoVal": "A", "content": "上午$$6$$时 "}], [{"aoVal": "B", "content": "上午$$8$$时 "}], [{"aoVal": "C", "content": "上午$$9$$时 "}], [{"aoVal": "D", "content": "下午$$9$$时 "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["3小时后是上午$$9$$时． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2843", "queId": "ff80808149608ac60149658a27050f97", "competition_source_list": ["2011年全国华杯赛竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "若连续的四个自然数都为合数，那么这四个数之和的最小值为（ ~ ~~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$101$$ "}], [{"aoVal": "C", "content": "$$102$$ "}], [{"aoVal": "D", "content": "$$103$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["任何四个连续自然数之和一定被$$4$$除余$$2$$，所以只有$$102$$满足条件．``都为合数''这个条件可以被无视了． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1917", "queId": "b79fcf68610d4f94ac00f268b142d21b", "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", "2014年迎春杯四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "为了减少城市交通拥堵的情况，某城市拟定从$${2014}$$年$${1}$$月$${1}$$日起开始试行新的限行规则，规定尾号为$$1$$、$$6$$的车辆周一、周二限行，尾号$$2$$、$$7$$的车辆周二、周三限行，尾号$$3$$、$$8$$的车辆周三、周四限行，尾号$$4$$、$$9$$的车辆周四、周五限行，尾号$$5$$、$$0$$的车辆周五、周一限行，周六、周日不限行。由于$$1$$月$$31$$日是春节，因此，$$1$$月$$30$$日和$$1$$月$$31$$日两天不限行。已知$$2014$$年$$1$$月$$1$$日是周三并且限行，那么$$2014$$年$$1$$月份~\\uline{~~~~}~组尾号可出行的天数最少。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$、$$6$$ "}], [{"aoVal": "B", "content": "$$2$$、$$7$$ "}], [{"aoVal": "C", "content": "$$4$$、$$9$$ "}], [{"aoVal": "D", "content": "$$5$$、$$0$$ "}]], "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": ["$$1$$月份共$$31$$天，$$31\\div 7=4$$(周)$$\\cdots \\cdots 3$$(天)。由于$$1$$月$$1$$日是周三，所以$$1$$月份周三、周四、周五各$$5$$天，周一、周二各$$4$$天。其中$$1$$月$$30$$日周四、$$1$$月$$31$$日周五。所以只看周三即可。周三$$2$$、$$7$$以及$$3$$、$$8$$限行。所以此题B组尾号可出行的天数最少。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3473", "queId": "f95ddc9ae9b943c5840d91db12064a1a", "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}  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->逐步调整思想", "Overseas Competition->知识点->计数模块->加乘原理"], "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": "2772", "queId": "bf07860697e141ffb411cda60e97851f", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛", "六年级其它"], "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": ["Overseas Competition->知识点->计算模块->定义新运算", "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": "3282", "queId": "3b5e0023c8954441a6c750e4aeb62fd6", "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决赛", "2011年中环杯四年级竞赛初赛"], "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": "$$180$$ "}], [{"aoVal": "B", "content": "$$150$$ "}], [{"aoVal": "C", "content": "$$160$$ "}]], "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": "3339", "queId": "c84ccb9c10de4cf88c62108cf0ba9ad4", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "二年级从$$5$$个学生中选择$$2$$人合作主持节目，共有种不同的选择方法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["由于每个主持人都有可能和另外$$1$$个主持人去主持节目，  一共有$$5\\times (5-1)=20$$（种），  去掉重复情况共有$$20\\div 2=10$$（种），  选择$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "842", "queId": "c40b1fcf98f649aeacd8dd525affc757", "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 "}]], "knowledge_point_routes": ["拓展思维->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": "913", "queId": "c9a9ba6106854825b2c85b382851eecd", "competition_source_list": ["2012年全国华杯赛小学高年级竞赛复赛A卷第12题"], "difficulty": "3", "qtype": "single_choice", "problem": "已知$$98$$个互不相同的质数$$p_{1}^{2}$$，$$p_{2}^{2}$$，$$p_{3}^{2}$$，$$\\cdots$$，$$p_{98}^{2}$$，记$$N=p_{1}^{2}+p_{2}^{2}+\\cdots +p_{98}^{2}$$，问：$$N$$被$$3$$除的余数是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "1或2 "}]], "knowledge_point_routes": ["知识标签->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": ["（1）如果这些质数中不含质数$$3$$，那么这些数平方被$$3$$除的余数就是$$1$$，所以$$N$$被$$3$$除的余数就是$$98$$被$$3$$除的余数，是$$2$$；  （2）如果有$$3$$，那么这些数的平方除以$$3$$余数是$$1$$，所以$$N$$被$$3$$除的余数就是剩下$$97$$个数除以$$3$$的余数，是$$1$$．  根据余数的可乘性：  若一个数除以$$3$$余$$0$$，则其平方除以$$3$$也余$$0$$；  若一个数除以$$3$$余$$1$$，则其平方除以$$3$$也余$$1$$；  若一个数除以$$3$$余$$2$$，则其平方除以$$3$$余$$1$$；  即一个平方数除以$$3$$的余数只可能是$$0$$或$$1$$．  若则$$98$$个质数中不含质数$$3$$，则每个质数都不是$$3$$的倍数，那么每个质数的平方除以$$3$$均余$$1$$，根据余数的可加性，$$98$$除以$$3$$余$$2$$，则$$N$$除以$$3$$余$$2$$；若这$$98$$个质数中含有$$3$$，则$$97$$除以$$3$$余$$1$$，$$N$$除以$$3$$也余$$1$$．  综上，$$N$$被$$3$$除的余数可能是$$1$$或$$2$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1772", "queId": "b67918189cef4ec9b0af28d9af1b196a", "competition_source_list": ["2018年四川成都五年级上学期单元测试《第六单元综合能力擂台及竞赛加分》尖子生题库第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "买$$40$$元$$1$$千克的茶叶和$$60$$元$$1$$千克的茶叶共$$20$$千克，一共用去$$900$$元，$$40$$元$$1$$千克和$$60$$元$$1$$千克的茶叶分别买了． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$千克和$$5$$千克 "}], [{"aoVal": "B", "content": "$$5$$千克和$$15$$千克 "}], [{"aoVal": "C", "content": "$$12$$千克和$$8$$千克 "}], [{"aoVal": "D", "content": "$$8$$千克和$$12$$千克 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数学广角->鸡兔同笼->假设法解鸡兔同笼"], "answer_analysis": ["假设$$20$$千克的茶叶全是$$40$$元$$1$$千克，则应花$$40\\times 20=800$$元，  而实际花了$$900$$元，多用了$$100$$元，则这$$100$$元是由$$60$$元$$1$$千克的茶叶的差价造成的，  则$$60$$元$$1$$千克的茶叶有$$100\\div (60-40)=5$$千克，  $$40$$元$$1$$千克的茶叶有：$$20-5=15$$千克．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2152", "queId": "0cfc0fcf1d804290abb42fd4738e1fda", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在环形跑道上，两人在一处背靠背站好，然后开始跑，每隔$$4$$分钟相遇一次；如果两人从同处同向同时跑，每隔$$20$$分钟相遇一次，已知环形跑道的长度是$$1600$$米，那么两人的速度分别是~\\uline{~~~~~~~~~~}~米/分、~\\uline{~~~~~~~~~~}~米/分． ", "answer_option_list": [[{"aoVal": "A", "content": "240；160 "}], [{"aoVal": "B", "content": "250；160 "}], [{"aoVal": "C", "content": "240；150 "}], [{"aoVal": "D", "content": "250；150 "}]], "knowledge_point_routes": ["知识标签->拓展思维->行程模块->环形跑道->相遇与追及结合"], "answer_analysis": ["两人反向沿环形跑道跑步时，每隔$$4$$分钟相遇一次，即两人$$4$$分钟共跑完一圈；当两人同向跑步时，每$$20$$分钟相遇一次，即其中的一人比另一人多跑一圈需要$$20$$分钟．两人速度和为：$$1600 \\div 4 = 400$$(米/分)，两人速度差为：$$1600 \\div 20 = 80$$(米/分)，所以两人速度分别为：$$400 + 80 \\div 2 = 240$$(米/分)，$$400 - 240 = 160$$(米/分) "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2865", "queId": "732f5f02a6c74a818f648f5a90c9be14", "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1题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\triangle$$, $$\\bigcirc$$, and $$\\square$$ each represents a number. If $$\\triangle +\\triangle +\\triangle =60$$, $$\\square \\times \\triangle =260$$, and $$\\bigcirc \\div \\square =6$$, what is the value of $\\bigcirc$?  $$\\triangle$$, $$\\bigcirc$$, 和 $$\\square$$ 各代表一个数$$.$$ 若 $$\\triangle +\\triangle +\\triangle =60$$, $$\\square \\times \\triangle =260$$, 且 $$\\bigcirc \\div \\square =6$$,~ $\\bigcirc$的值为? ", "answer_option_list": [[{"aoVal": "A", "content": "$$72$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$78$$ "}], [{"aoVal": "D", "content": "$$108$$ "}], [{"aoVal": "E", "content": "$$56$$ "}]], "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", "Overseas Competition->知识点->计算模块->方程基础->等量代换"], "answer_analysis": ["$$\\triangle +\\triangle +\\triangle =60$$, $$\\triangle =20$$.  $$\\square \\times \\triangle =260$$, $$\\square =13$$, $$\\bigcirc \\div \\square =6$$, $$\\bigcirc =78$$ "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2030", "queId": "c65a16672cf04ff980b6d52504619000", "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2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "同学们排队做操，小军的前面有$$5$$人，后面有$$6$$人，这一排一共有人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$人 "}], [{"aoVal": "B", "content": "$$11$$人 "}], [{"aoVal": "C", "content": "$$10$$人 "}]], "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"], "answer_analysis": ["小军前面$$5$$人，后面有$$6$$人，小军自己$$1$$人，  一共有：$$5+6+1=12$$（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": "2509", "queId": "1a64be4943e7482f8c9efd67d4e28508", "competition_source_list": ["2016年全国美国数学大联盟杯五年级竞赛初赛第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "【复习7】  $$2\\times 4\\times 8 \\times 16 \\times32 \\times64=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2^{10}$$ "}], [{"aoVal": "B", "content": "$$2^{15}$$ "}], [{"aoVal": "C", "content": "$$2^{21}$$ "}], [{"aoVal": "D", "content": "$$2^{120}$$ "}]], "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": ["$$2\\times 4\\times 8 \\times 16 \\times32 \\times64=2^{(1+2+3+4+6)}=2^{21}$$ "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2761", "queId": "6d8ae6a3d1744c0f9c40bd06eabbb606", "competition_source_list": ["2010年第8届创新杯六年级竞赛初赛第10题4分"], "difficulty": "1", "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->能力->运算求解"], "answer_analysis": ["$$1$$号混合液中，水:油:醋$$=1:2:3=3:6:9$$ ，$$2$$号混合液中，水:油:醋$$=3:4:5$$ 由上可知，$$1$$号、$$2$$号按上面比例混合可得到水:油:醋$$=6:10:14=3:5:7$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "881", "queId": "e47d4add013c48bf8f581a451c0fbd84", "competition_source_list": ["2005年第3届创新杯六年级竞赛复赛第1题", "2005年六年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "四个不同的质数之和为$$31$$，其中最小的质数是（  ）。 ", "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->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": ["这四个数的和为$$31$$是奇数，那么这四个数不可能都为奇数。即这四个数中含有偶数，又偶数中质数只有$$2$$，且$$2$$为最小的质数，故这四个数中最小的质数为$$2$$，故选A。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3285", "queId": "c338460443f7495a9f663f7bd5beb308", "competition_source_list": ["2013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$$36$$位客人用餐，一张圆桌可坐$$5$$人，一张方桌可坐$$4$$人．请问下列哪一项安排桌子的方式可以使每位客人都有座位，且没有多余的座位？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$张圆桌、$$2$$张方桌 "}], [{"aoVal": "B", "content": "$$2$$张圆桌、$$4$$张方桌 "}], [{"aoVal": "C", "content": "$$3$$张圆桌、$$5$$张方桌 "}], [{"aoVal": "D", "content": "$$4$$张圆桌、$$4$$张方桌 "}], [{"aoVal": "E", "content": "$$5$$张圆桌、$$3$$张方桌 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["若有$$1$$张圆桌、$$2$$张方桌，共可坐$$1\\times 5+2\\times 4=13$$人，少于客人人数，故不合；  若有$$2$$张圆桌、$$4$$张方桌，共可坐$$2\\times 5+4\\times 4=26$$人，少于客人人数，故不合；  若有$$3$$张圆桌、$$5$$张方桌，共可坐$$3\\times 5+6\\times 4=39$$人，多余客人人数，故不合；  若有$$4$$张圆桌、$$4$$张方桌，共可坐$$4\\times 5+4\\times 4=36$$人，恰等于客人人数，故满足；  若有$$5$$张圆桌、$$3$$张方桌，共可坐$$6\\times 5+2\\times 4=38$$人，多余客人人数，故不合；  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3478", "queId": "e2d8fc22a8b24d84ae508be98f0968b3", "competition_source_list": ["2019年第36届迎春杯小学中年级竞赛决赛A卷第8题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "老师每天给学生的作业评星．小明本周周一至周五各交了一次作业，共得到了$$15$$颗星，他最多一次作业得过$$5$$颗星，最少一次作业得过$$2$$颗星、那么，他这$$5$$次作业得到的星数顺次排成的五位数有~\\uline{~~~~~~~~~~}~种不同可能． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$50$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数学广角->排列组合->组合数", "拓展思维->拓展思维->计数模块->排列组合"], "answer_analysis": ["$$5$$次作业可能的数量为$$5+4+2+2+2$$或者$$5+3+3+2+2$$，如果是$$5+4+2+2+2$$，有$$5\\times 4=20$$（种）；如果是$$5+3+3+2+2$$，有$$5\\times (3+2+1)=30$$（种），共$$50$$种．  $$15=5+4+2+2+2=5+3+3+2+2$$；第一种情况有$$A^{2}_5=20$$种可能；第二种情况有$$C_5^{1}\\times C_4^{2}=30$$种可能，第二种情况还可以用除序做，有$$\\frac{A_5^{5}}{A_2^{2}\\times A_2^{2}}=30$$种.  总共有$$20+30=50$$种不同可能. "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1174", "queId": "46749fc2f0064276ab399b34f0fec258", "competition_source_list": ["2017年上海中环杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "某老师被学生问到自己的年龄时不愿意公开，但更不愿意说谎．于是她就对学生说：``我$$6$$年后年龄的$$9$$倍，减去我$$6$$年前年龄的$$9$$倍，等于我现在年龄的$$4$$倍少$$8$$岁．''想一想，如果设该老师今年$$x$$岁，那么$$6$$年后年龄的$$9$$倍应该表示为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9x+6$$ "}], [{"aoVal": "B", "content": "$$x+6\\times9$$ "}], [{"aoVal": "C", "content": "$$9\\left( x+6\\right)$$ "}]], "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": ["今年老师$$x$$岁，则$$6$$年后老师的年龄是$$(x+6)$$岁，$$9$$倍就是$$9(x+6)$$岁．故选 C． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3420", "queId": "9cc461b0bc1d4009af57ae26779635ac", "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6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果从$$4$$本不同的语文书、$$5$$本不同的数学书、$$6$$本不同的外语书中选取$$2$$本不同学科的书阅读，那么共有种不同的选择． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$74$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["在语文和数学中选择$$2$$本，则有：$$4\\times5=20$$（种），  在语文和外语中选择，则有：$$4\\times6=24$$（种），  在数学和外语中选择，则有：$$5\\times6=30$$（种）．  共计：$$20+24+30=74$$（种），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": "3312", "queId": "52578b093910497fb6f0647447bf30f1", "competition_source_list": ["2005年第4届小机灵杯四年级竞赛初赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "在所有三位数中，各数位上的和是$$9$$的数有~\\uline{~~~~~~~~~~}~个．（每个数字上没有数字$$0$$） ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$45$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想", "Overseas Competition->知识点->计数模块->枚举法综合->整数分拆->简单拆分->加法拆数（指定数）"], "answer_analysis": ["枚举，$$7+6+5+4+3+2+1=28$$（个）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2813", "queId": "84952f69e437485fbbcfe4fc57917b64", "competition_source_list": ["2016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": "小亮看一本故事书的正文，他从第二天开始每天比前一天多看一页，他已经连续看了$$9$$天，这本书的正文还剩$$48$$页没看．已知他第四天看了$$39$$页，请问这本数的正文总共有多少页？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$351$$ "}], [{"aoVal": "B", "content": "$$399$$ "}], [{"aoVal": "C", "content": "$$360$$ "}], [{"aoVal": "D", "content": "$$408$$ "}], [{"aoVal": "E", "content": "$$432$$ "}]], "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": ["由等差数列的知识：末项$$=$$首项$$+$$（项数$$-1$$）$$\\times $$公差  可知，第一天看了$$36$$页，第九天看了：$$36+(9-1)\\times 1=44$$（页）  $$9$$天看的总和$$=$$（首项$$+$$末项）$$\\times $$项数$$\\div 2$$  $$=(36+44)\\times 9\\div 2$$  $$=360$$（页）．  书共有：$$360+48=408$$（页）． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "182", "queId": "311c5713fa9541beba065dd6cb47512c", "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": ["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：$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1767", "queId": "660ff9ee4dde49e5b26091e2368e4241", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有五位小孩围成一圈在玩报数游戏，$$A$$报``$$1$$''，$$B$$报``$$2$$''，$$C$$报``$$3$$''，$$D$$报``$$4$$''，$$E$$报``$$5$$''，然后又回到$$A$$报``$$6$$''，依次报数，每个人报的数比前一个人多$$1$$．请问$$99$$是位小孩报的． ", "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": ["由题意可知$$1$$、$$2$$、$$3$$、$$4$$、$$5$$这五个数为一个周期依序循环重复出现．再由$$99=5\\times19+4$$可以判断出$$99$$是$$D$$小孩报的．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2572", "queId": "8ffb1e6308f14f02a1e8d3ccbb0971c4", "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：$$\\frac{3}{{{1}^{2}}+1}-\\frac{5}{{{2}^{2}}+2}+\\frac{7}{{{3}^{2}}+3}-\\frac{9}{{{4}^{2}}+4}+\\cdots +\\frac{4039}{{{2019}^{2}}+2019}-\\frac{4041}{{{2020}^{2}}+2020}$$的值是  ． ", "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": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\frac{3}{{{1}^{2}}+1}-\\frac{5}{{{2}^{2}}+2}+\\frac{7}{{{3}^{2}}+3}-\\frac{9}{{{4}^{2}}+4}+\\cdots +\\frac{4039}{{{2019}^{2}}+2019}-\\frac{4041}{{{2020}^{2}}+2021}$$  $$=\\frac{3}{1\\times 2}-\\frac{5}{2\\times 3}+\\frac{7}{3\\times 4}-\\frac{9}{4\\times 5}+\\cdots +\\frac{4039}{2019\\times 2020}-\\frac{4041}{2020\\times 2021}$$  $$=\\left( 1+\\frac{1}{2} \\right)-\\left( \\frac{1}{2}+\\frac{1}{3} \\right)+\\left( \\frac{1}{3}+\\frac{1}{4} \\right)-\\left( \\frac{1}{4}+\\frac{1}{5} \\right)+\\cdots +\\left( \\frac{1}{2019}+\\frac{1}{2020} \\right)-\\left( \\frac{1}{2020}+\\frac{1}{2021} \\right)$$  $$=1+\\frac{1}{2}-\\frac{1}{2}-\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{4}-\\frac{1}{5}+\\cdots +\\frac{1}{2019}+\\frac{1}{2020}-\\frac{1}{2020}-\\frac{1}{2021}$$  $$=1-\\frac{1}{2021}$$  $$=\\frac{2020}{2021}$$ "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "170", "queId": "3541c7971ef049b1a89289059a31a9b1", "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．它们所有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->能力->构造模型->模型思想"], "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": "3122", "queId": "f9b4fb17b4f844d496a9010598406669", "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$$\\square =$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "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": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3324", "queId": "e3d313592d48441c95733e28ee7c75b1", "competition_source_list": ["2008年第6届创新杯四年级竞赛复赛第5题4分", "2008年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "四年级（$$1$$）班有$$46$$人，喜欢打乒乓球的有$$32$$人，喜欢打羽毛球的有$$26$$人，既喜欢打乒乓球又喜欢打羽毛球的至少有（  ）人。 ", "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->计数模块->容斥原理->二量容斥"], "answer_analysis": ["把喜欢打乒乓球的$$32$$人和喜欢打羽毛球的$$26$$人加起来，$$32+26=58$$（人），比全班$$46$$人多$$12$$人，所以既喜欢打乒乓球又喜欢打羽毛球的至少有$$12$$人。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1395", "queId": "31c321dc911b4d6d9d75d9f88e607b09", "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5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一些学生做广播操，正好排成相等的四列，小芳所在的那一列，无论是从前面数还是从后面数小芳都是在第$$6$$个，这个班共有学生个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$44$$ "}], [{"aoVal": "C", "content": "$$45$$ "}]], "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->矩形队伍求总数"], "answer_analysis": ["从题干可以知道，一些学生做广播操，正好排成相等的四列，也就是每一列人数相等；  小芳所在的那一列，无论是从前面数还是从后面数小芳都是在第$$6$$个，  $$5+5+1=11$$（个），那么每一列$$11$$人；  $$11\\times 4=44$$（人）．  答：这个班共有学生$$44$$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": "1421", "queId": "8805dcb73d604a39b72cbbfc2123f39c", "competition_source_list": ["2010年第11届上海中环杯小学中年级三年级竞赛初赛第3题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$7$$个数的平均数是$$11$$，前$$4$$个数的平均数是$$8$$，后$$4$$个数的平均数是$$13$$，第$$4$$个数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "又搞事！ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["$$7$$个数的平均数是$$11$$，所以$$7$$个数的总和是$$77$$．前$$4$$个数平均数是$$8$$，所以前$$4$$个数总和是$$32$$，同样的，后$$4$$个数总和是$$52$$．所以第$$4$$个数是：$$32+52-77=7$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2051", "queId": "e65b443b6f0a4b0e8f647c3dc86e0506", "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": "放学后，小明从学校步行回家．当小明走了全程的一半时下雨了:当雨停了时，剩下路程是他雨中步行路程的$$\\frac{5}{7}$$．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->逻辑分析"], "answer_analysis": ["设雨中步行路程为$$7$$份，剩下路程为$$5$$份，即路程的一半是$$12$$份，所以全程$$24$$份，雨中步行路程是全程的$$\\frac{7}{24}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1672", "queId": "b16d50a2c7784e99b61566c418f405fa", "competition_source_list": ["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$$2018$$年$$11$$月$$1$$日是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "四 "}], [{"aoVal": "B", "content": "三 "}], [{"aoVal": "C", "content": "二 "}], [{"aoVal": "D", "content": "一 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "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": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2655", "queId": "32921ead7e5249a7a8dc4851f5b6a53a", "competition_source_list": ["2017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$$2017$$除以$$9$$的余数为多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "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": ["$$2017=9\\times 224+1$$．  故选$$\\text{B}$$． ", "<p>一个正整数除以$$9$$的余数等于这个正整数的各个数码之和除以$$9$$的余数，即$$2017$$除以$$8$$的余数等于$$2+0+1+7=10$$除以$$9$$的余数，因此所求之余数为$$1$$．</p>\n<p>故选$$\\text{B}$$．</p>"], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "702", "queId": "fa8d8e8ccc094244a9d5d17ae8f407a7", "competition_source_list": ["2016年第12届全国新希望杯五年级竞赛决赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "平面上有$$2015$$个红色和蓝色的点，用线段连接所有的这些染色的点，则两个端点颜色不同的线段总数是（~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "奇数 "}], [{"aoVal": "B", "content": "偶数 "}], [{"aoVal": "C", "content": "无法确定 "}], [{"aoVal": "D", "content": "不填 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["由于$$2015$$是奇数，所以蓝红点的奇偶性也不相同，不同的点为一红一篮，由于其中一种为偶数个，则最后线段也为偶数个． ~ "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2760", "queId": "49dd67eecaf14050a3c4bd199ac408b6", "competition_source_list": ["2006年第4届创新杯六年级竞赛复赛第4题", "1993年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "把足够大的一张厚度为$$0.1$$毫米的纸连续对折，要使对折后的整叠纸总厚度超过$$1$$厘米，至少要对折(  ) ", "answer_option_list": [[{"aoVal": "A", "content": "7次 "}], [{"aoVal": "B", "content": "8次 "}], [{"aoVal": "C", "content": "9次 "}], [{"aoVal": "D", "content": "10次 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->等比数列->分数等比数列"], "answer_analysis": ["每折一次厚度为原来的$$2$$倍，折$$6$$次后，厚度为$$0.1\\times {{2}^{6}}=6.4$$ 毫米，小于$$1$$厘米。折$$7$$次后，厚度为$$0.1\\times {{2}^{7}}=12.8$$毫米，大于$$1$$厘米。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1859", "queId": "8aac49074f5e941e014f644613660eca", "competition_source_list": ["2010年全国华杯赛竞赛初赛第2题", "2016年全国华杯赛小学高年级竞赛在线模拟第2题", "2017年全国小升初八中入学备考课程"], "difficulty": "2", "qtype": "single_choice", "problem": "两条纸带，较长的一条为$$23\\text{cm}$$，较短的一条为$$15\\text{cm}$$．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 （ ~ ~ ）$$\\text{cm}$$． ", "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": ["设剪下的长度为$$x$$厘米则可以列出不等式：$$23-x \\geqslant 2（15-x）$$，整理得$$x \\geqslant 7$$，所以剪下的长度至少是$$7$$厘米． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "812", "queId": "578bd4de6a5342e28dbe5710c68208a8", "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->运算求解"], "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": "636", "queId": "9443b71578d6421490e87969dd1326e1", "competition_source_list": ["2011年全国学而思杯五年级竞赛第5题", "2011年全国学而思杯四年级竞赛第5题", "2011年北京学而思综合能力诊断六年级竞赛第3题"], "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黑龙江地震的~\\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->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": ["$$8$$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": "586", "queId": "0873ba1e52b14c9484bc14feaef6ac21", "competition_source_list": ["2015年全国美国数学大联盟杯五年级竞赛初赛第40题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$，$$B$$两个袋子里分别有若干个球，$$A$$袋球和$$B$$袋球的数量比是$$4:1$$．从$$A$$袋中取出若干个球放入$$B$$袋后，$$A$$袋球和$$B$$袋球的数量之比变为$$8:3$$；再从$$B$$袋中取出若干个球放入$$A$$袋后，$$A$$袋球和$$B$$袋球的数量之比变为$$9:2$$．两个袋子里最少一共有几只球? ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$44$$ "}], [{"aoVal": "C", "content": "$$55$$ "}], [{"aoVal": "D", "content": "$$110$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["统一份数，最终发现，$$5$$份和$$11$$份的最小公倍数是55． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "670", "queId": "99236df45f5148c6ae44a1dc4d8830cd", "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": "$${{\\left( 11101 \\right)}_{2}}$$ "}], [{"aoVal": "B", "content": "$${{\\left( 1011 \\right)}_{2}}$$ "}], [{"aoVal": "C", "content": "$${{\\left( 10101 \\right)}_{2}}$$ "}], [{"aoVal": "D", "content": "$${{\\left( 11001 \\right)}_{2}}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$25=16+8+1$$，则$$25$$转换成二进制数是$${{\\left( 11001 \\right)}_{2}}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "855", "queId": "663035bef1a84af19bf93fa85fa54ffe", "competition_source_list": ["2021年第8届鹏程杯四年级竞赛初赛第12题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "一串数：$$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$\\cdots $$即从第三个数起，每个数恰好等于它前面相邻两个数之和．则第$$2021$$个数被$$6$$除的余数是． ", "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->归纳总结->归纳推理"], "answer_analysis": ["利用余数的周期性和可加性解题．  这串数除以$$6$$的余数依次是$$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$2$$，$$1$$，$$3$$，$$4$$，$$1$$，$$5$$，$$0$$，$$5$$，$$5$$，$$4$$，$$3$$，$$1$$，$$4$$，$$5$$，$$3$$，$$2$$，$$5$$，$$1$$，$$0$$，$$1$$，$$1\\cdots \\cdots $$，  余数的周期是$$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$2$$，$$1$$，$$3$$，$$4$$，$$1$$，$$5$$，$$0$$，$$5$$，$$5$$，$$4$$，$$3$$，$$1$$，$$4$$，$$5$$，$$3$$，$$2$$，$$5$$，$$1$$，$$0$$，  共$$24$$个，  $$2021\\div24=84\\cdots \\cdots5$$，  第五个是$$5$$，所以选$$\\text E$$，  故选$$\\text E$$． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3199", "queId": "82a821a2f8ac48a8820a365804e4a03a", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（一）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "由数字$$0-9$$组成一个三位数，三位数有数字重复的组合有种． ", "answer_option_list": [[{"aoVal": "A", "content": "$$220$$ "}], [{"aoVal": "B", "content": "$$255$$ "}], [{"aoVal": "C", "content": "$$252$$ "}], [{"aoVal": "D", "content": "$$225$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$9+9\\times 9\\times 3=252$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "19", "queId": "8f4758092efc461fb9e662e7905cf484", "competition_source_list": ["2017年北京学而思杯小学中年级四年级竞赛年度教学质量测评第18题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "一群小朋友手拉手围成一圈做游戏．现在从其中的一个小朋友开始，按顺时针的顺序报数$$1$$，$$2$$，$$3$$，$$4$$，$$5\\cdots \\cdots $$报$$8$$的小朋友和报$$17$$的小朋友是同一个人．已知小朋友的人数不少于$$5$$人，不多于$$20$$人．请问：共有（~ ~ ~ ）个小朋友在手拉手做游戏． ", "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": ["略 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1044", "queId": "0a92617ea4d74d339cd5673e27106bcc", "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": "一本书的正文共$$193$$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": ["知识标签->学习能力->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": "3062", "queId": "d3c26dc709f747f29868b4cac4f367aa", "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8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知一组等差数列的第$$17$$项为$$73$$，第$$27$$项为$$113$$，这组等差数列的第$$7$$项是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$23$$ "}], [{"aoVal": "B", "content": "$$33$$ "}], [{"aoVal": "C", "content": "$$43$$ "}], [{"aoVal": "D", "content": "$$53$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["根据中项定理，第$$7$$项$$+$$第$$27$$项$$=2$$倍的第$$17$$项．  故第$$7$$项为：$$73\\times 2-113=33$$．  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": "1602", "queId": "68f5a80273d445a9bda86d7061d55a96", "competition_source_list": ["2017年新希望杯六年级竞赛训练题（三）第4题", "2018年湖北武汉新希望杯六年级竞赛训练题（三）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "贝贝看一本$$320$$页的故事书，第一天看了全书的$$\\frac{1}{4}$$，第二天看了余下的$$\\frac{1}{5}$$，两天一共看了页． ", "answer_option_list": [[{"aoVal": "A", "content": "114 "}], [{"aoVal": "B", "content": "120 "}], [{"aoVal": "C", "content": "128 "}], [{"aoVal": "D", "content": "132 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$320\\times \\frac{1}{4}+320\\times \\left( 1-\\frac{1}{4} \\right)\\times \\frac{1}{5}=80+48=128$$（页）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3055", "queId": "b893b7bd132344ed818219956edda44e", "competition_source_list": ["2016年全国迎春杯五年级竞赛初赛A卷第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "算式$$\\left( 19\\times 19-12\\times 12 \\right)\\div \\left( \\frac{19}{12}-\\frac{12}{19} \\right)$$的计算结果是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$144$$ "}], [{"aoVal": "B", "content": "$$228$$ "}], [{"aoVal": "C", "content": "$$361$$ "}], [{"aoVal": "D", "content": "$$456$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["原式$$=\\left( {{19}^{2}}-{{12}^{2}} \\right)\\div \\left( \\frac{{{19}^{2}}-{{12}^{2}}}{12\\times 19} \\right)=\\left( {{19}^{2}}-{{12}^{2}} \\right)\\times \\frac{12\\times 19}{\\left( {{19}^{2}}-{{12}^{2}} \\right)}=228$$．  原式$$=\\left( 19\\times 19-12\\times 12 \\right)\\times\\left( \\frac{12\\times 19}{19\\times 19-12\\times 12} \\right)=228$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1912", "queId": "e0779ce0819f46738d5cb7467cf3c7f8", "competition_source_list": ["2011年六年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "某人年初买了一支股票，该股票当年下跌了$$20 \\%$$，第二年应上涨(  )\\%才能回到初始值。 ", "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->认识单位1"], "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": "3235", "queId": "2caaa6695c754a3dacf11642e4596b53", "competition_source_list": ["2017年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": "1", "qtype": "single_choice", "problem": "五年级三班有$$50$$名学生，参加语文竞赛的有$$28$$人，参加数学竞赛的有$$22$$人，参加英语竞赛的有$$20$$人．如果每人最多参加两科竞赛，那么该班未参加竞赛人数最多可能有（~ ）人． ", "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->对应思想"], "answer_analysis": ["总人数一定，要使未参加竞赛的人数最多，那么参加两科的人数就要最多，只参加一科的人数最少，假设全是参加两科的人，一共$$\\left( 28+22+20 \\right)\\div 2=35$$（人），此时未参加竞赛的人最多$$5035=15$$人． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3105", "queId": "cb9742165b1944e9900d976213d6ebe9", "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": ["知识标签->拓展思维->计算模块->定义新运算->直接运算型->普通型"], "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": "29", "queId": "0620ef4eb34a40218d9b8c36ff6d3692", "competition_source_list": ["2018年华杯赛四年级竞赛模拟卷"], "difficulty": "2", "qtype": "single_choice", "problem": "用$$6$$个$$2$$，$$3$$个$$7$$组成的$$9$$位数中能被$$8$$整除的有(  )个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数的认识->数的特征->因数与倍数->3、5的倍数特征", "拓展思维->拓展思维->组合模块->数字谜->竖式数字谜"], "answer_analysis": ["略 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2585", "queId": "3a0da813d2b94bb58e76a784b6eb748a", "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$$14$$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$$13$$名选手的号码之总和恰好是$$100$$．请问小明的号码是什么？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$14$$ "}], [{"aoVal": "E", "content": "$$17$$ "}]], "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$$14$$号选手号码成等差数列，那么$$1$$号到$$14$$号的总和是$$(1+14)\\times 14\\div 2=105$$，则小明号码：$$105-100=5$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "929", "queId": "b859c5960c0a413490efab5ebfa0267e", "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4题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$2000$$、$$2001$$、$$2002\\cdots \\cdots 2017$$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->数论模块->完全平方数->平方数的简单应用"], "answer_analysis": ["$${{a}^{2}}-{{b}^{2}}=\\left( a+b \\right)\\left( ~a-b \\right)$$，考虑奇偶性则该数应为一个奇数或者$$4$$的倍数，而$$2000-2017$$之间，$$2002$$、$$2006$$、$$2010$$、$$2014$$不满足，则共有$$18-4=14$$个． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "946", "queId": "e61760829176455f8e004c6c101a2103", "competition_source_list": ["2011年五年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "设a是一个满足下列条件的最大的正整数:使得a除64的余数是4，用a除155的余数是5，用a除187的余数是7.则$$a=$$(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "10 "}], [{"aoVal": "B", "content": "15 "}], [{"aoVal": "C", "content": "30 "}], [{"aoVal": "D", "content": "60 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->余数问题->余数的性质->余数的可加性"], "answer_analysis": ["$$a\\left\\textbar{} 60 \\right.$$，$$a\\left\\textbar{} 150 \\right.$$，$$a\\left\\textbar{} 180 \\right.$$，$$\\left( 60，150，180 \\right)=30$$ "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1562", "queId": "56c6f988f77847fdb793d00000b0cc9e", "competition_source_list": ["2016年全国希望杯六年级竞赛复赛第12题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "两根粗细相同，材料相同的蜡烛，长度比是$$21:16$$，它们同时开始燃烧，$$18$$分钟后，长蜡烛与短蜡烛的长度比是$$15:11$$，则较长的那根蜡烛还能燃烧~\\uline{~~~~~~~~~~}~分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$140$$ "}], [{"aoVal": "E", "content": "$$150$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["差不变问题，化连比：$$21-16=5$$，$$15-11=4$$，$$21:16=\\left( 21\\times 4 \\right):\\left( 16\\times 4 \\right)=84:64$$；  $$15:11=\\left( 15\\times 5 \\right):\\left( 11\\times 5 \\right)=75:55$$．$$84-75=9$$，$$18\\div 9=2$$（分钟）  $$75\\times 2=150$$（分钟）. "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "882", "queId": "c47600c5801b4cc1a10c022dd8c74379", "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": "1530", "queId": "99f8c636e3794ae6baa46883b64a95cb", "competition_source_list": ["2013年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "水池$$A$$和$$B$$同为长$$3$$米，宽$$2$$米，深$$1.2$$米的长方体。$$1$$号阀门用来向$$A$$池注水，$$18$$分钟可将无水的$$A$$池注满；$$2$$号阀门用来从$$A$$池向$$B$$池放水，$$24$$分钟可将$$A$$池中满池水放入$$B$$池。若同时打开$$1$$号和$$2$$号阀门，那么当$$A$$池水深$$0.4$$米时，$$B$$池有(  )立方米的水。 ", "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": ["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": ["解：设水池$$A$$和$$B$$的容积为``$$1$$''，  同时打开$$1$$号和$$2$$号阀门，则$$A$$池每分钟进水效率为：$$\\frac{\\text{1}}{\\text{18}}-\\frac{1}{24}=\\frac{1}{72}$$，  $$A$$池水深$$0.4$$米，则$$A$$池进水：$$0.4\\div 1.2=\\frac{1}{3}$$，  需要时间：$$\\frac{1}{3}\\div \\frac{1}{72}=24$$（分钟），  $$B$$池进水：$$24\\times \\frac{1}{24}=1$$，  所以$$B$$池有水：$$3\\times 2\\times 1.2=7.2$$（立方米）。  故选：D。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1255", "queId": "5933fdf789a641d1ab6fda96605b6c3a", "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2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "相传有个人不讲究说话艺术常引起误会．一天他摆宴席请客，他看到还有几个人没来，就自言自语：``怎么该来的还不来呢?''客人听了，心想难道我们是不该来的，于是有一半客人走了，他一看十分着急，又说：``不该走的倒走了！''剩下的人一听，是我们该走啊！又有剩下的三分之二的人离开了，他着急地一拍大腿，连说：``我说的不是他们．''于是最后剩下的四个人也都告辞走了，聪明的你能知道刚开始来了位客人吗？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->能力->符号代换->代数运算"], "answer_analysis": ["可以设原来有$$x$$人，第一批走了$$\\frac{1}{2}x$$，第二批走了$$\\frac{2}{3}\\left( x-\\frac{1}{2}x \\right)$$，剩下四人，以人数为等量关系可列方程求解．  解：设原来有$$x$$人，  $$\\begin{eqnarray}\\frac{1}{2}x+\\frac{2}{3}\\left( x-\\frac{1}{2}x \\right)+4\\&=\\&x   x\\&=\\&24\\end{eqnarray}$$  ∴开始来了$$24$$个客人．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "772", "queId": "526951072eaf4c10892800b229d1320f", "competition_source_list": ["2013年IMAS小学高年级竞赛第二轮检测试题第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "请问下面哪一项叙述一定是正确的？ ", "answer_option_list": [[{"aoVal": "A", "content": "几个奇数之和一定是奇数 "}], [{"aoVal": "B", "content": "一个偶数加上一个奇数结果为偶数 "}], [{"aoVal": "C", "content": "两个质数之和一定是偶数 "}], [{"aoVal": "D", "content": "两个奇数之和一定是合数 "}], [{"aoVal": "E", "content": "连续三个正整数中一定有一个数能被$$3$$整除 "}]], "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": ["$$\\text{A}$$：$$1+3=4$$×  $$\\text{B}$$：$$2+3=5$$×  $$\\text{C}$$：$$2+3=5$$×  $$\\text{D}$$：$$1+1=2$$×． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3253", "queId": "99656be02ec441e5966da5eb0c0a30bf", "competition_source_list": ["其它改编自2014年全国希望杯六年级竞赛初赛第20题"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$1$$，$$2$$，$$3\\cdots$$，$$50$$中，任取$$10$$个连续的数，则其中恰有$$3$$个质数的概率是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{15}{41}$$ "}], [{"aoVal": "B", "content": "$$\\frac{18}{41}$$ "}], [{"aoVal": "C", "content": "$$\\frac{20}{41}$$ "}], [{"aoVal": "D", "content": "$$\\frac{22}{41}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["在$$1$$至$$50$$的数中共有$$15$$个质数，任意连续的$$10$$个数排列的共有$$41$$种情况（以$$1---9$$为开始，$$41---50$$结束），那么列举其中的$$3$$个质数的组合分别为：$$6---15$$，$$7---16$$，$$8---17$$，$$9---18$$，$$12---21$$，$$13---22$$，$$14---23$$，$$15---24$$，$$16---25$$，$$17---26$$，$$22---31$$，$$23---32$$，$$28---37$$，$$29---38$$，$$34---43$$，$$35---44$$，$$36---45$$，$$37---46$$，$$38---47$$，$$39---48$$，$$40---49$$，$$41---50$$，共$$22$$组三个质数的连续$$10$$个数．所以概率为$$\\frac{22}{41}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1736", "queId": "7788460692c14c3dabdf69d7d77891a0", "competition_source_list": ["2020年第25届YMO五年级竞赛决赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "在一次数学考试中，已知每个同学的分数都是整数，最高分为$$90$$分，老师随便挑选出$$4$$人，他们的得分各不相同，且平均分为$$87$$分，那么这$$4$$位同学中分数最低的那一位至少得~\\uline{~~~~~~~~~~}~分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$81$$ "}], [{"aoVal": "B", "content": "$$83$$ "}], [{"aoVal": "C", "content": "$$79$$ "}], [{"aoVal": "D", "content": "$$80$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->平均数问题->综合题->平均数最值"], "answer_analysis": ["根据题意分析可知，  设这四个人为$$A$$、$$B$$、$$C$$、$$D$$，$$4$$个人的平均分为$$87$$分，  则四个人的总分为：$$87\\times 4=348$$（分），  分数各不相同且为整数，  所以假设$$A$$：$$90$$分，$$B$$：$$89$$分，$$C$$：$$88$$分，  则最低分为$$348-90-89-88=81$$（分）．  故答案为：$$81$$分． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3074", "queId": "fd08aa7be7b9439990a08b5c3d97d969", "competition_source_list": ["2017年河南郑州K6联赛竞赛模拟第六套第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列哪一个算式最适合估计$$4.9\\div \\frac{1}{4}-6\\frac{11}{12}$$的值？（ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5\\div 4-6$$ "}], [{"aoVal": "B", "content": "$$5\\times 4-7$$ "}], [{"aoVal": "C", "content": "$$5\\div 4-7$$ "}], [{"aoVal": "D", "content": "$$5\\times 4-6$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["把$$4.9$$看作$$5$$，$$6\\frac{11}{12}$$看作$$7$$，所以$$4.9\\div \\frac{1}{4}-6\\frac{11}{12}\\approx 5\\times 4-7$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1180", "queId": "1db766ae2e184f2388a09f81b1414c31", "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": "甲乙二人同时开始加工一批零件，每人加工零件总数的一半，甲完成任务的$$1/3$$时，乙加工了$$50$$个零件，甲完成$$3/5$$时乙完成了一半．这批零件共有（~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$360$$ "}], [{"aoVal": "B", "content": "$$420$$ "}], [{"aoVal": "C", "content": "$$600$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["甲完成$$2/3$$时乙完成了一半，则有甲完成任务的$$1/3$$时，乙应该完成任务的$$\\left( 1/2 \\right)/2=1/4$$，即是$$45$$个．所以，零件共有：$$45/\\left( 1/4 \\right)*2=360$$个． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1411", "queId": "439a1a8a6cee4810a88356f157e7309e", "competition_source_list": ["六年级其它", "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3题3分", "2017年四川成都六年级竞赛“全能明星”选拔赛第9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9题2分", "2018~2019学年四川成都双流区四川师范大学附属圣菲小学六年级上学期开学考试第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "往浓度为$$10 \\%$$，质量为$$400$$克的糖水中加入克水，就可以得到浓度为$$8 \\%$$的糖水． ", "answer_option_list": [[{"aoVal": "A", "content": "$$90$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$110$$ "}], [{"aoVal": "D", "content": "$$120$$ "}]], "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->已知溶液溶剂求溶质"], "answer_analysis": ["浓度问题，$$400\\times 10 \\%\\div 8 \\%-400=100\\text{g}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2601", "queId": "24b697c3607745efb8669ad8505467c0", "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21题2分", "小学高年级五年级下学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$，$$d$$是非$$0$$自然数，已知$$a\\textgreater b\\textgreater c\\textgreater d$$，那么$$\\frac{1}{a}$$，$$\\frac{1}{b}$$，$$\\frac{1}{c}$$，$$\\frac{1}{d}$$中最小的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{a}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{b}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{c}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{d}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$a$$、$$b$$、$$c$$、$$d$$都是非零自然数，$$a\\textgreater b\\textgreater c\\textgreater d$$，  利用：分子相同，分母越大，分数越小，  可知：$$\\frac{1}{a}\\textless\\frac{1}{b}\\textless\\frac{1}{c}\\textless\\frac{1}{d}$$，  所以$$\\frac{1}{a}$$最小．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3273", "queId": "9e3606b711374767bdd56e877e79f134", "competition_source_list": ["2017年陕西西安小升初某工大附中", "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", "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$$4$$个小球，魔术师从中任取几个小球，把每一个小球都变成$$9$$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$$9$$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": "$$2020$$ "}], [{"aoVal": "B", "content": "$$2019$$ "}], [{"aoVal": "C", "content": "$$2018$$ "}], [{"aoVal": "D", "content": "$$2017$$ "}]], "knowledge_point_routes": ["知识标签->课内知识点->随机现象->随机现象的可能性"], "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": "3173", "queId": "d9b9ab6cb226474fa61255c56dc12d94", "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~\\uline{~~~~~~~~~~}~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->能力->数感认知->整数数字加工"], "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": "1843", "queId": "a9655933c8d84eac8103da922e62ead8", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（三）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "市政府大楼和某大学之间是一条$$204$$米长的道路，现要在这条路的两侧种树，且每一侧每两棵树的间隔为$$12$$米．若每棵树需要$$50$$元人工费，则至少需要元预算． ", "answer_option_list": [[{"aoVal": "A", "content": "$$800$$ "}], [{"aoVal": "B", "content": "$$1600$$ "}], [{"aoVal": "C", "content": "$$1700$$ "}], [{"aoVal": "D", "content": "$$1800$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["间隔刚好为$$12$$米时，预算最少．  $$2\\times \\left( \\frac{204}{12}-1 \\right)\\times 50=1600$$（元）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1755", "queId": "d67392e4c0df4919946686cbd6807534", "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7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "今年哥哥$$15$$岁，弟弟$$7$$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": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["根据题意分析可知，哥哥$$15$$岁，弟弟$$7$$岁，哥哥比弟弟大$$15-7=8$$（岁），  当哥哥的年龄是弟弟的$$3$$倍时，弟弟的年龄为：$$8\\div 2=4$$（岁），  故：$$7-4=3$$（年）前哥哥的年龄是弟弟的$$3$$倍，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": "3203", "queId": "272bbfb5ae32408eb1b78a33daef898a", "competition_source_list": ["2017年河南郑州K6联赛竞赛模拟第7题", "2017年河南郑州模拟考试k6考试第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$5$$个轻重、大小完全相同的小球，编号分别为$$1$$、$$2$$、$$3$$、$$m$$和$$n$$，把它们放入不透明的盒子中，任意摸出两个球，如果是$$m$$和$$n$$，则有奖．那么，摸一次，获奖的可能性是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{20}$$~~~ "}], [{"aoVal": "B", "content": "$$\\frac{1}{10}$$~~~ "}], [{"aoVal": "C", "content": "$$\\frac{1}{5}$$~~~~~~ "}], [{"aoVal": "D", "content": "$$\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->计数模块->统计与概率->概率->典型问题->摸小球"], "answer_analysis": ["一共有$$5$$个球，每次摸两个，则这些球中任意球都可能和其它的球组合，每一个球和其它球的组合方式有$$4$$种，所以$$4\\times 5=20$$种，排除重复的，还剩$$20\\div 2=10$$个球，而同时摸到$$m$$和$$n$$的只有一种，所以可能性是$$\\frac{1}{10}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "230", "queId": "4cd8998c7d7842898667433919f663f7", "competition_source_list": ["2005年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "设$${{a}_{1}}，{{a}_{2}}，{{a}_{3}}，{{a}_{4}}，{{a}_{5}}$$是任意五个奇数，且$${{a}_{1}} \\textless{} {{a}_{2}} \\textless{} {{a}_{3}} \\textless{} {{a}_{4}} \\textless{} {{a}_{5}}$$，$${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+{{a}_{5}}=85$$，符合这些条件的五个奇数显然有很多，如$${{a}_{1}}=1，{{a}_{2}}=3，{{a}_{3}}=7，{{a}_{4}}=11，{{a}_{5}}=63$$，或$${{a}_{1}}=5，{{a}_{2}}=7，{{a}_{3}}=13，{{a}_{4}}=21，{{a}_{5}}=39$$，等等，在这些答案中，记$${{a}_{5}}$$的最大值和最小值分别为M和m，则（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "$$M=67，m=23$$ "}], [{"aoVal": "B", "content": "$$M=67，m=19$$ "}], [{"aoVal": "C", "content": "$$M=69，m=21$$ "}], [{"aoVal": "D", "content": "$$M=69，m=17$$ "}]], "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5个奇数的和不变，那么当其中最大的奇数越大越好时，其余4个奇数越小越好，即当$${{a}_{1}}，{{a}_{2}}，{{a}_{3}}，{{a}_{4}}$$取最小值时，$${{a}_{5}}$$有最大值，那么$${{a}_{5}}$$有最大值为$$85-\\left( 1+3+5+7 \\right)=69$$.同样当其中最大奇数越小越好时，这5个奇数越接近越好，当这5个奇数最接近时每两个数之间的差为2，这时这5个数的和为5的倍数，又85刚好能被5整除，且$$85\\div 5=17$$为奇数正好成立，那么$${{a}_{5}}$$的最小值为$$17+2+2=21$$，故$$M=69，m=21$$选C. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2135", "queId": "0aa029a0c68143b5b8e6b986064326c0", "competition_source_list": ["2020年长江杯六年级竞赛复赛A卷第2题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙三个人进行$$100$$米赛跑，假设他们三人都是匀速前进，当甲到终点时，乙离终点还有$$10$$米，丙离终点还有$$19$$米．当乙到达终点时，丙离终点还有米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$\\frac{8}{9}$$ "}], [{"aoVal": "D", "content": "$$9.5$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["甲、乙、丙三个人进行$$100$$米赛跑，注意到他们三人都是匀速前进，当甲到 终点时，乙距离终点还有$$10$$米，丙距离终点还有$$19$$米，这样可以知道甲、乙、丙三人的路程，他们在相同时间内，甲跑了$$100$$米，乙跑了$$90$$米，丙跑了$$81$$米，因此他们的路程比为$$100:90:81$$，当乙到达终点时，乙的路程为$$10$$米，由于乙和丙的路程比为，乙$$:$$丙$$=90:81=10:9$$，所以乙的路程为$$10$$米时，丙的路程为$$9$$米，由于丙离终点还有$$19$$米，所以当乙到达终点时，最后的丙离终点：$19-9=10$（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": "1770", "queId": "8d85b338295e4c739d2cb8ea2b34e65c", "competition_source_list": ["2008年第6届创新杯六年级竞赛初赛A卷第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$、$$B$$、$$C$$，$$D$$四个数，每次去掉一个数，将其余$$3$$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->能力->运算求解"], "answer_analysis": ["不妨设$$A\\geqslant B\\geqslant C\\geqslant D$$，  则$$B+C+D=23\\times 3=69$$，  $$A+C+D=26\\times 3=78$$，  $$A+B+D=30\\times 3=90$$，  $$A+B+C=33\\times 3=99$$四个等式相加得$$3\\times \\left( A+B+C+D \\right)=336$$，  从而$$A+B+C+D=112$$，  所以$$A$$、$$B$$、$$C$$、$$D$$的平均数为$$112\\div 4=28$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2208", "queId": "3a55ef65b91b4326a11eedd6b5c4bc3e", "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": "甲乙两人比赛折返跑，同时从$$A$$点出发，到达$$B$$点后，立即返回，先回到$$A$$点的人获胜。甲先到达$$B$$点，在距离$$B$$点$$24$$米的地方遇到乙。相遇后，甲的速度减为原来的一半，乙的速度保持不变。在距离终点$$48$$米的地方，乙追上甲。那么，当乙到达终点时，甲距离终点还有（  ）米。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "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": ["解：设$$A$$、$$B$$之间的距离是$$x$$米，则第一次相遇时，甲跑的路程是$$x+24$$米，乙跑的路程是$$x-24$$米，  所以第一次相遇时甲乙的速度之比是：$$(x+24):\\left( x-24 \\right)$$，  第二次相遇时甲乙的速度之比是：$$\\left( x-24-48 \\right):(x+24-48)=\\left( x-72 \\right):\\left( x-24 \\right)$$；  所以$$(x+24):\\left( x-24 \\right)=2\\left( x-72 \\right):\\left( x-24 \\right)$$，  因此$$x+24=2\\left( x-72 \\right)$$，  解得$$x=168$$，  即两地之间的距离是$$168$$米，  所以第二次相遇时甲乙的速度之比是：  $$\\left( 168-72 \\right):\\left( 168-24 \\right)$$  $$=96:144$$  $$=2:3$$  所以乙到达终点时，甲跑的路程是：  $$(168+24)\\times \\frac{2}{3}$$  $$=192\\times \\frac{2}{3}$$  $$=128$$（米），  因此当乙到达终点时，甲距离终点：  $$168-24-128=16$$（米）  答：当乙到达终点时，甲距离终点$$16$$米。  故选：D。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1781", "queId": "b1f50503a93b48c8807d5fa282c6ffbf", "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10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某游乐场在开门前有$$400$$人排队等待，开门后每分钟来的人数是固定的，$$1$$个入口每分钟可以进入$$10$$个游客。如果开放了$$4$$个入口，$$20$$分钟后就没有人排队。现在开放$$6$$个入口，那么开门（  ）分钟后没有人排队。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "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": ["$$4$$个入口$$20$$分钟可以进游客$$4\\times 10\\times 20=800$$（人），即$$20$$分钟来了游客$$800-400=400$$（人），所以每分钟来游客$$20$$人。 设开放$$6$$个入口，开门$$x$$分钟后没有游客排队，则$$400+20x=6\\times 10\\times x$$ 解得$$x=10$$，故开放$$6$$个入口，开门$$10$$分钟后就没有游客排队。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "185", "queId": "5e43171900dc4deca88db614a618cd97", "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": "497", "queId": "f38ec5acc4b24acca819d109a8b73c3f", "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": "$$92$$ "}], [{"aoVal": "D", "content": "$$146$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "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": "3402", "queId": "c4c9c1e81f544d68ba13e8a477284c5d", "competition_source_list": ["2014年第12届全国创新杯小学高年级六年级竞赛第1题", "小学高年级六年级其它2014年数学思维能力等级测试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "将六个分数$$\\frac{8}{35}$$、$$\\frac{3}{8}$$、$$\\frac{1}{45}$$、$$\\frac{11}{120}$$、$$\\frac{4}{9}$$、$$\\frac{5}{21}$$分成三组，使每组的两个分数的和相等，那么与$$\\frac{1}{45}$$分在同一组的那个分数是（~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{8}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{21}$$ "}], [{"aoVal": "D", "content": "$$\\frac{11}{120}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["$$\\frac{8}{35}+\\frac{5}{21}=\\frac{24+25}{105}=\\frac{7}{15}$$，$$\\frac{1}{45}+\\frac{4}{9}=\\frac{1+20}{45}=\\frac{7}{15}$$，$$\\frac{3}{8}+\\frac{11}{120}=\\frac{45+11}{120}=\\frac{7}{15}$$，故与$$\\frac{1}{45}$$分在同一组的是$$\\frac{4}{9}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2985", "queId": "c4ff99cefdc64b29acc9003de377a48a", "competition_source_list": ["2003年第1届创新杯六年级竞赛复赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$1+2-3\\times 4\\div 5+6-7\\times 8\\div 9$$的计算结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{17}{45}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{37}{45}$$ "}], [{"aoVal": "D", "content": "$$\\frac{8}{9}$$ "}]], "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": ["因为$$3\\times 4\\div 5$$和$$7\\times 8\\div 9$$的计算结果都除不尽，  因此可把这两项的结果写成分数形式，运用加法交换与结合律简算即可．  $$1+2-3\\times 4\\div 5+6-7\\times 8\\div 9$$  $$=3-\\frac{12}{5}+6-\\frac{56}{9}$$  $$=(3+6)-\\left( \\frac{12}{5}+\\frac{56}{9} \\right)$$  $$=9-\\frac{388}{45}$$  $$=\\frac{17}{45}$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1858", "queId": "8e782042b2e24ca7916bd77576ee8a43", "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1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个最简分数，如果分子减$$a$$，分母不变，则等于$$\\frac{1}{8}$$，如果分母减$$a$$，分子不变，则等于$$\\frac{1}{2}$$，那么这个分数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{16}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{8}$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{16}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["设原分数为$$\\frac{y}{x}$$  $$\\left { \\begin{array}{*{35}{l}} \\frac{y-a}{x}-\\frac{1}{8}\\Rightarrow 8y-8a    x-a\\frac{y}{x-a}=\\frac{1}{2}\\Rightarrow 2y=x-a\\Rightarrow 16y=8x-8a   \\end{array} \\right.$$  $$16y-x=8y-8a$$  $$9x=24y$$  $$\\frac{y}{x}=\\frac{9}{24}=\\frac{3}{8}$$  选$$\\rm B$$． ", "<p>设：第一次约分后分子为$$x$$，分母为$$8x$$，</p>\n<p>则原来的分数是$$\\frac{x+a}{8x}$$，</p>\n<p>$$\\begin{eqnarray} \\frac{x+a}{8x-a}&amp;=&amp;\\frac{1}{2}\\\\2x+2a&amp;=&amp;8x-a\\\\3a&amp;=&amp;6x\\\\a&amp;=&amp;2x \\end{eqnarray}$$</p>\n<p>原来的分数是$$\\frac{x+2x}{8x}=\\frac{3x}{8x}=\\frac{3}{8}$$，</p>\n<p>故选 B．</p>"], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1455", "queId": "9e4775ddc3824e5e98fba97ebe2f094a", "competition_source_list": ["2014年广东广州羊排赛六年级竞赛第7题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "水结成冰，体积会增大$$\\frac{1}{11}$$；那么冰化成水，体积会减少． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{10}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{11}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{12}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{11}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["设原来的水体积为``$$1$$''，变成冰后体积为$$1\\times (1+\\frac{1}{11})=\\frac{12}{11}$$，冰变成水后，体积减少$$(\\frac{12}{11}-1)\\div \\frac{12}{11}=\\frac{1}{12}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3368", "queId": "bb0652fa867a46a5a3c06a6222bfc4cc", "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9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个自然数，如果它顺着数和倒过来数都是一样的，则称这个数为``对称数''．例如，$$2$$，$$33$$，$$101$$，$$1331$$是对称数，但$$220$$不是对称数．由数字$$1$$、$$2$$、$$3$$、$$4$$组成的不超过三位数的对称数有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$21$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "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": ["一位数有：$$1$$，$$2$$，$$3$$，$$4$$共计$$4$$个；  两位数：$$11$$，$$22$$，$$33$$，$$44$$共计$$4$$个；  三位数：$$111$$，$$121$$，$$131$$，$$141$$，$$212$$，$$222$$，$$232$$，$$242$$，$$313$$，$$323$$，$$333$$，$$343$$，$$414$$，$$424$$，$$434$$，$$444$$共计$$16$$个．  合计：$$4+4+16=24$$（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": "1927", "queId": "b7b446e0f50c4e25a43d9228f86ca82f", "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8题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一个杯子装满了浓度为$$15 \\%$$的盐水，有大、中、小铁球各一个，它们的体积比为$$10:5:3$$，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{~~~~~~~~~~}~$$ \\%$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["大球体积：$$10 \\%\\times \\frac{10}{3}=\\frac{1}{3}$$  总共溢出$$\\frac{1}{3}$$的水  浓度：$$15 \\%\\times (1-\\frac{1}{3})=10 \\%$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "168", "queId": "ff0fa961e16c42d4b1ea2717d4d33d2d", "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，嘉泽同学依次编号为$$1$$、$$2$$、$$3$$、$$\\cdots$$、$$n$$，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": "3373", "queId": "6ac867bcee0b4a80ad4cca90ee84d04f", "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8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$7$$张卡片上分别写着数字$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$，$$8$$，从中抽出两张，组成的所有的两位数中，奇数的个数有． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$个 "}], [{"aoVal": "B", "content": "$$18$$个 "}], [{"aoVal": "C", "content": "$$24$$个 "}], [{"aoVal": "D", "content": "$$42$$个 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["两位数的个数可以是$$3$$，$$5$$，$$7$$有$$3$$个情况．  十位有$$7-1=6$$（种）情况．可以组成$$6\\times 3=18$$个两位奇数．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1940", "queId": "988112519008440f958f8fce87b90dcb", "competition_source_list": ["2015年世界少年奥林匹克数学竞赛六年级竞赛复赛A卷第12题12分"], "difficulty": "1", "qtype": "single_choice", "problem": "张飞、关羽、刘备三人栽树，已知张飞、关羽合栽$$15$$天栽完，关羽、刘备合栽$$12$$天栽完，刘备，张飞合栽$$8$$天栽完．若按张飞、关羽、刘备的顺序轮流各栽一天，栽完这批树苗共需~\\uline{~~~~~~~~~~}~天？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$21\\dfrac{9}{13}$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$21\\dfrac{13}{9}$$ "}], [{"aoVal": "D", "content": "$$21\\dfrac{9}{11}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$21\\dfrac{9}{13}$ "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3162", "queId": "ab4b1cc9738844cc99089e4af79580f4", "competition_source_list": ["2017年第11届北京学而思杯小学高年级五年级竞赛第13题8分", "2017年第11届北京学而思综合能力诊断五年级竞赛4月第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个各位数字互不相同的五位数，如果万位，千位，百位的数字依次增大；百位，十位，个位的数字依次减小，我们称这样的数为``宝塔数''．那么符合条件的``宝塔数''有~\\uline{~~~~~~~~~~}~个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$375$$ "}], [{"aoVal": "B", "content": "$$756$$ "}], [{"aoVal": "C", "content": "$$1134$$ "}], [{"aoVal": "D", "content": "$$2600$$ "}]], "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": ["设这个五位数为$$\\overline{abcde}$$，则有$$a\\textless{}b\\textless{}c$$，$$c\\textgreater d\\textgreater e$$．分为两类：  情况$$1$$：含有数字$$0$$，那么数字$$0$$一定在个位，共有$$C_{9}^{4}\\times C_{3}^{2}=378$$种情况．  情况$$2$$：不含数字$$0$$，那么共有$$C_{9}^{5}\\times C_{4}^{2}=756$$种情况．  因此共有$$378+756=1134$$个宝塔数． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1377", "queId": "2d29072e38ed4d5d9a660f5bb9ef55b3", "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": "\\textbf{(2019 World Mathematical Olympiad,~Primary 4, \\#10)}  A brewery has a promotion where $$3$$ empty bottles can be exchanged for $$1$$ bottle of beer. Dad bought $$13$$ bottles of beer, so he can drink a total of ($$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$$) bottles of beer.  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（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$）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", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "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": "1360", "queId": "87be108813d547b78626fc2b8000c9e4", "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": "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->拓展思维->应用题模块->比例应用题->基础份数思想"], "answer_analysis": ["前一半时乙的工作量是甲的$$2$$倍，所以后一半甲的工作量应是乙的$$2$$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$$100\\times 3\\times 5=1500$$（个）字，故总工作量为$$1500\\times 6\\times 2=18000$$（个）字。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1528", "queId": "7f580e1ec6954eb591bdce162aad42e5", "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": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["倒推．$$44$$ 对应的是$$44-2=42$$，颠倒后是$$24$$，则原数为$$12$$，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": "3187", "queId": "11b588b6b13e4d3e8fb1db70efd2652a", "competition_source_list": ["2017年四川成都六年级竞赛“全能明星”选拔赛第10题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "从成都到上海开行了动车组，全程共停靠$$11$$个站，则共需种不同的车票． ", "answer_option_list": [[{"aoVal": "A", "content": "$$55$$ "}], [{"aoVal": "B", "content": "$$105$$ "}], [{"aoVal": "C", "content": "$$110$$ "}], [{"aoVal": "D", "content": "$$122$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["两站之间的往返车票各一种，即$$2$$种，  $$n$$个车站每两站之间有$$2$$种，  则$$n$$个车站的车票种类数：$$n(n-1)$$，  则$$11$$个车站的车票种类数：$$11\\times (11-1)=110$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1768", "queId": "80dd864c9c29458d96eaf241c154b78a", "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4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$30$$个小朋友排一队，从前面数小红排在第$$15$$个，小军排在小红后面第$$4$$个，那么小军从后往前数排第个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["课内体系->七大能力->实践应用", "拓展思维->七大能力->实践应用"], "answer_analysis": ["从前往后数，小军排第$$15+4=19$$个，则从后往前数，小军排：$$30-19+1=12$$个．  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": "3423", "queId": "bc4cc9b5e332484c9395b8747ad94caf", "competition_source_list": ["2010年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "一个骰子六个面上分别写着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->拓展思维->计数模块->统计与概率->概率->基本概率"], "answer_analysis": ["数字和是3的倍数有:$$\\left( 1,2 \\right),\\left( 1,5 \\right),\\left( 2,4 \\right),\\left( 3,6 \\right),\\left( 3,3 \\right),\\left( 4,5 \\right),\\left( 6,6 \\right)$$,共有7种可能,投掷$$\\left( 2,1 \\right)$$和$$\\left( 1,2 \\right)$$算一种,投掷两次共有$$6+5+4+3+2+1=21$$种可能,则两数之和为3的倍数的可能性是$$\\frac{7}{21}=\\frac{1}{3}$$. "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2435", "queId": "1060cfda7cd441e09e2fcde0247f030f", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛"], "difficulty": "0", "qtype": "single_choice", "problem": "一枚标准的人民币$$1$$元硬币约重$$6$$克，则$$15$$亿枚这样的硬币约重（~ ）吨． ", "answer_option_list": [[{"aoVal": "A", "content": "$$90$$ "}], [{"aoVal": "B", "content": "$$900$$ "}], [{"aoVal": "C", "content": "$$9000$$ "}], [{"aoVal": "D", "content": "$$90000$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$15$$亿$$=1500000000$$，$$6\\times 1500000000=9000000000$$克$$=9000000$$千克$$=9000$$吨． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1081", "queId": "25bd3481cf084c6ca40f7925cb92571d", "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7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "周末，小明乘$$45$$路公交车回家，当车到游乐园站时，他发现车上人数的$$\\frac{1}{6}$$下车后，这时又上来了车上人数的$$\\frac{1}{6}$$，那么现在车上的人数． ", "answer_option_list": [[{"aoVal": "A", "content": "增加了 "}], [{"aoVal": "B", "content": "减少了 "}], [{"aoVal": "C", "content": "同样多 "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->思想->赋值思想"], "answer_analysis": ["设公交车的原有人数看做单位``$$1$$''，  $$1\\times \\left( 1-\\frac{1}{6} \\right)=\\frac{5}{6}$$，  $$\\frac{5}{6}\\times \\left( 1+\\frac{1}{6} \\right)=\\frac{35}{36}$$，  $$1\\textgreater\\frac{35}{36}$$，故车上人减少了．  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2581", "queId": "2430c0d014304e31b10c37cf3ff0b3dc", "competition_source_list": ["2015年美国数学大联盟杯六年级竞赛初赛第34题5分(每题5分)"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$0.75$$转换为分数，这些分数的分子和分母都是小于$$100$$的正整数．这样的分数共有多少个？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$25$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意可得，$$0.75$$转化成为分数为$$\\frac{3}{4}$$，要求分数的分子和分母都是小于$$100$$的正整数，因此这样的分数有$$\\frac{3}{4}$$、$$\\frac{6}{8}$$、$$\\frac{9}{12}$$、$$\\frac{12}{16}$$、$$\\frac{15}{20}$$、$$\\frac{18}{24}$$、$$\\frac{21}{28}$$、$$\\frac{24}{32}$$、$$\\frac{27}{36}$$、$$\\frac{30}{40}$$、$$\\frac{33}{44}$$、$$\\frac{36}{48}$$、$$\\frac{39}{52}$$、$$\\frac{42}{56}$$、$$\\frac{45}{60}$$、$$\\frac{48}{64}$$、$$\\frac{51}{68}$$、$$\\frac{54}{72}$$、$$\\frac{57}{76}$$、$$\\frac{60}{80}$$、$$\\frac{63}{84}$$、$$\\frac{66}{88}$$、$$\\frac{69}{92}$$、$$\\frac{72}{96}$$，这样的分数总共有$$24$$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": "2910", "queId": "8dff5446c7eb419f901d7d20ed77879f", "competition_source_list": ["2006年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "下面的△，○，□各代表一个数，如果，$$\\bigcirc \\div \\triangle =10$$，$$\\square \\times \\triangle =240$$，$$\\bigcirc \\div \\square =6$$，那么，○的值是(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "100 "}], [{"aoVal": "C", "content": "130 "}], [{"aoVal": "D", "content": "140 "}]], "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": ["由题中可得$$\\triangle=12$$，$$\\square =20$$，$$\\bigcirc =120$$选A "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2361", "queId": "043c9a23518541b3904defe150d75831", "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．  $$13+1+2+3$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$14+2$$ "}], [{"aoVal": "B", "content": "$$1+2$$ "}], [{"aoVal": "C", "content": "$$2+3$$ "}], [{"aoVal": "D", "content": "$$13+3$$ "}]], "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": "1682", "queId": "ccfdfd7219f6477bae39f3185eedbb83", "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": "一项工程，甲做了它的$$\\frac{1}{3}$$后由乙做，乙做了剩下的$$\\frac{1}{4}$$后，全由丙完成，则甲、乙、丙所做的工程比是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3:6:2$$ "}], [{"aoVal": "B", "content": "$$1:3:6$$ "}], [{"aoVal": "C", "content": "$$2:1:3$$ "}], [{"aoVal": "D", "content": "$$3:4:3$$ "}]], "knowledge_point_routes": ["拓展思维->思想->赋值思想"], "answer_analysis": ["乙做了$$\\left( 1-\\frac{1}{3} \\right)\\times \\frac{1}{4}=\\frac{1}{6}$$，则丙做了$$1-\\frac{1}{3}-\\frac{1}{6}=\\frac{1}{2}$$，所以甲、乙、丙工程比为$$\\frac{1}{3}:\\frac{1}{6}:\\frac{1}{2}=2:1:3$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1280", "queId": "629cafc1167045129ca509f78e96d666", "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10题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$6$$年前，母亲的年龄是儿子的$$5$$倍．$$6$$年后母子年龄总和是$$78$$岁．问：母亲今年岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$66$$ "}], [{"aoVal": "B", "content": "$$54$$ "}], [{"aoVal": "C", "content": "$$51$$ "}], [{"aoVal": "D", "content": "$$45$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["$$6$$年前，母子的年龄和为$$78-\\left( 6+6 \\right)\\times 2=54$$（岁），儿子那时$$54\\div \\left( 5+1 \\right)=9$$（岁），母亲那时$$9\\times 5=45$$（岁），母亲今年$$45+6=51$$（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": "231", "queId": "328aa883e9f34930ba265eb0533b9883", "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$$\\div $$ "}], [{"aoVal": "E", "content": "$$\\bigcirc $$填入$$\\times $$，$$\\square $$填入$$+$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->巧填算符->等式填符号->必填符号型"], "answer_analysis": ["等式左边等于$$13$$，当算式右边依次填入×、－时，2×8-3=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": "297", "queId": "56e6ce4d075c4a0fa647c005c88f0ebb", "competition_source_list": ["小学高年级六年级其它小高著名杯赛拉分题第9题", "2009年第14届全国华杯赛竞赛初赛第3题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "按照中国篮球职业联赛组委会的规定，各队队员的号码可以选择的范围是$$0$$～$$55$$号，但选择两位数的号码时，每位数字均不能超过$$5$$．那么，可供每支球队选择的号码共（~ ~ ~）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$34$$ "}], [{"aoVal": "B", "content": "$$35$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$56$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["根据题意，可供选择的号码可以分为一位数和两位数两大类，其中一位数可以为$$0$ $9$$，有$$10$$种选择；两位数的十位可以为$$1$ $5$$，个位可以为$$0$ $5$$，根据乘法原理，两位数号码有$$5\\times 6=30$$种选择．所以可供选择的号码共有$$10+30=40$$种． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1124", "queId": "4638f3fe8afc4c70971496945af546cd", "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B卷第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "期末考试五年级（I）班数学平均分为90分，全班总分为4950分，则五（I）班有(  )个同学. ", "answer_option_list": [[{"aoVal": "A", "content": "50 "}], [{"aoVal": "B", "content": "54 "}], [{"aoVal": "C", "content": "55 "}], [{"aoVal": "D", "content": "56 "}]], "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": ["总分$$\\div $$平均分为该班的人数，即$$4950\\div 90=55$$(人)，因此，该班有$$55$$个学生. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "817", "queId": "acdcc5d885874c7387ce3f0809854a07", "competition_source_list": ["2017年迎春杯五年级竞赛", "2017年迎春杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "村长让喜羊羊买东西，他把若干个羊币分成$$6$$份，分别包在$$6$$个小纸包里，村长告诉喜羊羊，从$$1$$羊币到$$63$$羊币，不管是多少，都能从这$$6$$包中挑出一包或几包来支付并且不用找钱；那么这$$6$$包中最大的那一包装了个羊币 . ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$63$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "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": ["$${{2}^{6}}=64$$，可以转化成二进制来解决，那么这$$6$$包所包装的羊币分别为$$1,2,4,8,16,32,$$则可以满足$$1$$个羊币到$$63$$个羊币都可取出，最大的那一包装了$$32$$个羊币。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "361", "queId": "7347a700e981457db14aa97003209b86", "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3题"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$1$$，$$2$$，$$4$$，$$5$$，$$7$$克的砝码各一个，丢失了其中一个砝码，结果天平无法称出$$10$$克的重量（砝码必须放在天平的一边），丢失的砝码重（  ）克。 ", "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->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["由于$$10=1+2+7\\cdots\\cdots$$①，或$$10=1+4+5\\cdots\\cdots$$②。  所以失去$$2$$克或$$7$$克砝码时，可用②称出$$10$$克的重量；失去$$4$$克或$$5$$克砝码时，可用①称出$$10$$克的重量，因此排除$$B$$、$$C$$、$$D$$。  故选：$$A$$。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2749", "queId": "ff808081453c067901453fffcb460303", "competition_source_list": ["2014年全国迎春杯三年级竞赛复赛第4题"], "difficulty": "1", "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->计算模块->方程基础->等量代换->文字转化为算式代换"], "answer_analysis": ["等量代换，设小猪为$$a$$，鸡为$$b$$，鸭为$$c$$，鱼为$$d$$，根据条件可得：$$a=6b$$，$$3b=4c$$，$$2c=6d$$，则：$$2a=12b=16c=48d$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2847", "queId": "8d28ff15a04e45d5bf990657c4bb3958", "competition_source_list": ["2013年IMAS小学高年级竞赛第一轮检测试题第3题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "小王从家到学校只有两种方式可供选择：（$$a$$）步行$$3$$分钟到离家较近的公车站，然后乘$$15$$分钟的公车到学校；（$$b$$）步行$$5$$分钟到离家较近的地铁站，然后乘$$6$$分钟的地铁到离学校较近的地铁站，再步行$$5$$分钟到学校．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": "$$15$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$18$$ "}], [{"aoVal": "E", "content": "$$19$$ "}]], "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": ["从家到学校，若小王选择第一种方式需要$$3+15=18$$分钟，选择第二种方式则需要$$5+6+5=16$$分钟，所以小王从家到学校至少需要$$16$$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": "3035", "queId": "c1379ac15b8844aab203f35018b2d229", "competition_source_list": ["2014年迎春杯六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "算式$$2013\\times \\frac{2015}{2014}+2014\\times \\frac{2016}{2015}+\\frac{4029}{2014\\times 2015}$$的计算结果是（  ） ", "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->比较与估算->比较大小综合->分数基准数法"], "answer_analysis": ["解：$$\\frac{2015}{2014}\\textgreater1$$，那么$$2013\\times \\frac{2015}{2014}\\textgreater2013$$；  $$\\frac{2016}{2015}\\textgreater1$$，那么$$2014\\times \\frac{2016}{2015}\\textasciitilde\\textgreater2014$$，  $$2013+2014=4027$$，  则$$2013\\times \\frac{2015}{2014}+2014\\times \\frac{2016}{2015}+\\frac{4029}{2014\\times 2015}\\textgreater4027$$  选项中只有$$4029\\textgreater4027$$，所以只能是B这个选项正确。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1290", "queId": "1b7c36ad9bc74323a3ca5839c5897e05", "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1题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙三种商品单价的比是$$6:5:4$$，已知甲商品比丙商品的单价多$$12$$元，则三种商品的单价之和为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$75$$元 "}], [{"aoVal": "B", "content": "$$90$$元 "}], [{"aoVal": "C", "content": "$$95$$元 "}], [{"aoVal": "D", "content": "$$100$$元 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["设甲商品的单价为$$6x$$元，则乙商品的单价为$$5x$$元，丙商品的单价为$$4x$$元，根据甲商品比丙商品的单价多$$12$$元，即可得出关于$$x$$的一元一次方程，解之即可得出$$x$$值，再将三种商品的单价相加即可得出结论．  设甲商品的单价为$$6x$$元，则乙商品的单价为$$5x$$元，丙商品的单价为$$4x$$元，  依题意，得：$$6x-4x=12$$，  解得：$$x=6$$，  ∴$$6x+5x+4x=90$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3012", "queId": "e52223c5cecf47878268c0edda664106", "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": "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->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": "2500", "queId": "22a8db40bfe94bb4ba5bd28d44b30e38", "competition_source_list": ["2016年创新杯小学高年级五年级竞赛训练题（三）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "下面是一连串有规律的数列：$$9$$、$$22$$、$$39$$、$$60$$、$$85$$、$$114\\cdots \\cdots $$数列第$$30$$个数是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2010$$ "}], [{"aoVal": "B", "content": "$$2011$$ "}], [{"aoVal": "C", "content": "$$2015$$ "}], [{"aoVal": "D", "content": "$$2016$$ "}]], "knowledge_point_routes": ["拓展思维->能力->归纳总结->归纳推理"], "answer_analysis": ["由题意可得，$$9$$与$$22$$之间相差$$13$$；$$22$$与$$39$$相差$$17$$；$$\\cdots $$依次可得每两个数之间的差值都比前面相邻两数之间的差值多$$4$$，因此可得，第$$30$$个数为$$2010$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1628", "queId": "6da91c9c9e1c46c9aed12c7454f05b86", "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3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "快车和慢车同时从甲、乙两地相对开出，快车每小时行$$33$$千米，相遇行了全程的$$\\frac{4}{7}$$，已知慢车行完全程需要$$8$$小时，则甲、乙两地相距千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$198$$ "}], [{"aoVal": "B", "content": "$$200$$ "}], [{"aoVal": "C", "content": "$$186$$ "}]], "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": ["两车同时开出到相遇，快车行了全程的$$\\frac{4}{7}$$，  则慢车行了全程的$$\\frac{3}{7}$$，故速度比为$$\\frac{4}{7}:\\frac{3}{7}=4:3$$，  所以快车行完全程所需时间：慢车行完全程所需时间$$=3:4$$，  则快车行完全程时间：$$8\\div 4\\times 3=6$$（人），  故全程：$$33\\times 6=198\\text{km}$$，故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1856", "queId": "86610319242548e29fe8b46dd34bf133", "competition_source_list": ["2017年第13届湖北武汉新希望杯五年级竞赛决赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "数学期末考试中，甲、乙、丙的平均分是$$91$$分，若甲的分数是$$90$$分，则乙和丙的平均分是（~ ）分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$91.5$$ "}], [{"aoVal": "B", "content": "$$92$$~~~ "}], [{"aoVal": "C", "content": "$$92.5$$ "}], [{"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->综合题->综合题普通型"], "answer_analysis": ["乙丙和为：$$91\\times 3-90=183$$，平均数：$$183\\div 2=91.5$$ "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "667", "queId": "428fb0cff4df45c28f10a54e83150c96", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "用$$\\text{a}$$，$$\\text{b}$$，$$c$$能组成$$6$$个无重复数字的三位数，如$$\\overline{abc}$$，$$\\overline{acb}$$等，且这$$6$$个数的和是$$4662$$， 问：这$$6$$个数都是$$3$$的倍教吗？ ", "answer_option_list": [[{"aoVal": "A", "content": "全部都是 "}], [{"aoVal": "B", "content": "不全是 "}], [{"aoVal": "C", "content": "都不是 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["由数字$$\\text{a}$$，$$\\text{b}$$，$$c$$组成的$$6$$个不同的三位数的和是  $$\\overline{abc}+\\overline{acb}+\\overline{bac}+\\overline{bca}+\\overline{cab}+\\overline{cba}$$  $$=\\left( a+b+c \\right)\\times 100\\times 2+\\left( a+b+c \\right)\\times 10\\times 2+\\left( a+b+c \\right)\\times 1\\times 2$$  $$=\\left( a+b+c \\right)\\times 222=4662$$， 即$$a+b+c=4662\\div 222=21$$，所以这$$6$$个数都是$$3$$的倍数． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2054", "queId": "c6bb42de45164192b53ff9c44cfc23cc", "competition_source_list": ["2015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": "2", "qtype": "single_choice", "problem": "八年后我的年龄是两年前我的年龄的两倍．请问两年后我是几岁？ ", "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": ["可以用解方程的方法，设我今年$$x$$岁，那么$$x+8=2\\left( x-2 \\right)$$，解得$$x=12$$，所以两年后我是$$12+2=14$$（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": "1270", "queId": "27b7cf8b0ef34b54803de967237b47be", "competition_source_list": ["2017年新希望杯六年级竞赛训练题（二）第4题", "2018年湖北武汉新希望杯六年级竞赛训练题（二）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙三人共同加工一批零件．甲比乙多加工零件$$100$$个，丙加工零件是乙加工零件的$$\\frac{3}{4}$$，甲加工零件是乙、丙两人加工零件总数的$$\\frac{2}{3}$$．甲、乙、丙一共加工了（~~~ ）个零件． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1750$$ "}], [{"aoVal": "B", "content": "$$1800$$ "}], [{"aoVal": "C", "content": "$$1850$$ "}], [{"aoVal": "D", "content": "$$1900$$ "}]], "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": ["$$\\left( \\frac{3}{4}+1 \\right)\\times \\frac{2}{3}=\\frac{7}{6}$$，乙：$$100\\div \\left( \\frac{7}{6}-1 \\right)=600$$（个），甲：$$600+100=700$$（个），丙：$$600\\times \\frac{3}{4}=450$$（个），$$600+700+450=1750$$（个）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3240", "queId": "3e8bb559c437425f8dce190dd02159f8", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛B卷第9题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "随机投掷一枚均匀的硬币两次，两次都正面朝上的概率是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{3}$$ "}]], "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": ["因为由题干可知，现随机投掷一枚均匀的硬币两次，  则可能的情况为：``正正''、``正负''、``负正''、``负负''，  所以两次都正面朝上的概率是$$\\frac{1}{4}$$，  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1799", "queId": "ed991b6b1d3e41f3b437cad33c4802b8", "competition_source_list": ["2011年全国创新杯五年级竞赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "为了节约用水，市政府规定：家庭用水在$$60$$立方米以内（含$$60$$立方米）的按照每立方米$$1.5$$元计算，超过$$60$$立方米到$$80$$立方米（含$$80$$立方米）的按照$$2.5$$元计算，超过$$80$$立方米按照每立方米$$5.5$$元计算，小丽家$$5$$月份用水为$$87$$立方米，那么小丽家本月要收缴水费（~ ）元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$67.5$$ "}], [{"aoVal": "B", "content": "$$130.5$$ "}], [{"aoVal": "C", "content": "$$178.5$$ "}], [{"aoVal": "D", "content": "$$215$$ "}]], "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": ["$$60\\times1.5+(80-60)\\times2.5+(87-80)\\times 5.5=90+50+38.5=178.5$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "649", "queId": "1a55210a09154f2e95b195b04b7650d1", "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", "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"], "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->能力->运算求解"], "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": "1897", "queId": "8aac50a7508d5d410150d535456d774e", "competition_source_list": ["2015年北京华杯赛小学高年级竞赛初赛A卷第4题"], "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~\\uline{~~~~~~~~~~}~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->思想->方程思想"], "answer_analysis": ["设共有$$x$$张票，赛前一小时的余票降价$$y$$元．由题意得$$\\frac{1}{4}\\times \\left( x\\times 50 \\right)=\\frac{1}{3}\\left[ x\\times \\left( 50-y \\right) \\right]$$；$$y=\\frac{25}{2}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2269", "queId": "a41bfe9a3492438baf829ae4560a3570", "competition_source_list": ["2016年IMAS小学高年级竞赛第二轮检测试题第5题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "小杰与小乔沿着公园的环形人行道跑步，小杰差$$400\\text{m}$$就跑完$$2$$圈，小乔还差$$500\\text{m}$$就跑完$$3$$圈，两人路程合起来共跑了$$4$$圈还多$$100\\text{m}$$，请问这条环形人行道一圈为多少$$\\text{m}$$？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1000$$ "}], [{"aoVal": "B", "content": "$$900$$ "}], [{"aoVal": "C", "content": "$$800$$ "}], [{"aoVal": "D", "content": "$$750$$ "}], [{"aoVal": "E", "content": "$$700$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->方程解行程问题->一元一次方程解行程问题"], "answer_analysis": ["解：设这条环形人行道一圈$$x\\text{m}$$，  $$2x-400+3x-500=4x+100$$  $$5x-4x=100+900$$  $$x=1000$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "908", "queId": "c06e48d9c95546c884fdaf2a9cb80d95", "competition_source_list": ["2022年第九届鹏程杯四年级竞赛初赛第24题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$，$$3$$，$$9$$，$$27$$，$$81$$这五个数中，每次或者取一个、或者取几个相加，得到一个新数，把这些新数从小到大排列起来是$$1$$，$$3$$，$$4$$，$$9$$，$$10$$，$$12$$，······，那么第$$27$$个数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1177$$ "}], [{"aoVal": "B", "content": "$$114$$ "}], [{"aoVal": "C", "content": "$$112$$ "}], [{"aoVal": "D", "content": "$$111$$ "}], [{"aoVal": "E", "content": "$$109$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["无 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "176", "queId": "1ff795fcc97d4cf9ad73e4087e0712f3", "competition_source_list": ["2017年IMAS小学中年级竞赛（第一轮）第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在算式$$\\overline{1A}+\\overline{B1}=100$$中，请问字母$$B$$所代表的数码是什么？ ", "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": ["$$100$$的个位数码为$$0$$﹐故$$\\text{A}$$代表的数码一定是$$9$$，因为$$100-19=81$$﹐所以$$B=8$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2503", "queId": "1e6ccadc52fa496ca2d9af7d817a7d8a", "competition_source_list": ["2013年华杯赛四年级竞赛初赛"], "difficulty": "2", "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->拓展思维->计算模块->整数->多位数计算->多位数的运算和数字和相关->求计算结果的数字和"], "answer_analysis": ["解：根据题意可得：  $$45\\times 40=1800$$  $$1800$$的数字和是：$$1+8+0+0=9$$  所以，$$45$$与$$40$$的积的数字和是$$9$$。  故选：A。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2372", "queId": "0a98ce343edd48ba9db1840119cacc14", "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1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "小数乘法：$$0.025\\times 0.04$$的结果（不计末尾的$$0$$）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": ["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": ["$$0.025\\times 0.04=0.001$$ "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1179", "queId": "2f99c72aaad0452190027be7b1d605e8", "competition_source_list": ["2018年IMAS小学中年级竞赛（第二轮）第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "某年的一月份共有$$4$$个星期六与$$5$$个星期日，请问这年的$$1$$月$$17$$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->能力->实践应用"], "answer_analysis": ["一月份有$$31$$天，一共四个星期多出三天，而共有$$4$$个星期六与$$5$$个星期日，故该月第一天为星期日，因此$$1$$月$$17$$日是星期二．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1069", "queId": "0f393ab356fa4912a7bda683ef9788cf", "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->知识点->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": "2840", "queId": "91a7b2ade3544687af081af1677b670d", "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": "已知$$\\frac{41}{60}=\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}$$，那么满足算式的（$$a$$，$$b$$，$$c$$）的解有（~ ~ ~）组．（顺序不同则为不同的解） ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["这道题我们可以利用放缩法求得范围再去求解，但是太过繁琐，我们换个思路，我们将式子重新编写为：$$\\frac{41}{60}=\\frac{x}{60}+\\frac{y}{60}+\\frac{z}{60}$$；此时要使的每个分数变为分子为$$1$$的最简真分数，那么必有$$x$$，$$y$$，$$z$$为$$60$$的因数，且$$x+y+z=41$$；  $$60$$的因数有$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$10$$，$$12$$，$$15$$，$$20$$，$$30$$，$$60$$；不妨令$$x\\geqslant y\\geqslant z$$；  $$x=30$$时，此时有$$y+z=11=10+1=6+5$$有$$2$$种拆分方式  $$x=20$$时，此时有$$y+z=21=20+1=6+15$$有$$2$$种拆分方式；  当$$x\\leqslant 15$$时，此时有$$y+z\\geqslant 31$$，根据平均数原则，那么必有一个数超过$$15$$，和假设矛盾，不成立:  综上, $$\\frac{41}{60}=\\frac{1}{2}+\\frac{1}{6}+\\frac{1}{60}=\\frac{1}{2}+\\frac{1}{10}+\\frac{1}{12}=\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{60}=\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{10}$$;  考虑有顺序，那么一共有$$3\\times A_{3}^{2}+3=21$$组解． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "535", "queId": "e2bdbe7969374cfead638228b4d1151c", "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": "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->数据处理"], "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）．  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": "3226", "queId": "54ec17f34a9a438c8818fda94ae8f5ab", "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": "有几个正三位数满足个位数是十位数和百位数的积？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$~~ "}], [{"aoVal": "B", "content": "$$23$$~ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["枚举法有序枚举：$$100$$、$$111$$、$$122$$、$$133$$、$$144$$、$$155$$、$$166$$、$$177$$、$$188$$、$$199$$、$$200$$、$$212$$、$$224$$、$$236$$、$$248$$、$$300$$、$$313$$、$$326$$、$$339$$、$$400$$、$$414$$、$$428$$、$$500$$、$$515$$、$$600$$、$$616$$、$$700$$、$$717$$、$$800$$、$$818$$、$$900$$、$$919$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3477", "queId": "f9aaaf008d35459081eea9b38a35c33d", "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": ["拓展思维->能力->逻辑分析"], "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": "2767", "queId": "b13d16f5fc9f47ad90e1a1fba814fa38", "competition_source_list": ["2006年第4届创新杯五年级竞赛初赛A卷第1题", "2006年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "数列$$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$21$$，$$34$$，$$55$$，$$89$$，$$\\cdots$$的前$$2006$$个数中，偶数有（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$667$$个 "}], [{"aoVal": "B", "content": "$$668$$个 "}], [{"aoVal": "C", "content": "$$669$$个 "}], [{"aoVal": "D", "content": "$$670$$个 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["数列中每三个数一循环，一个循环中仅最后一个为偶数，前$$2006$$个数中有$$668$$个循环余$$2$$个数，即有$$668$$个偶数，故选B。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "242", "queId": "32b812470e81403eb00839de93fe276f", "competition_source_list": ["2016年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁四支足球队进行比赛，小明说：``甲第一，丁第四。''小伟说：``丁第二，丙第三。''小凡说：``丙第二，乙第一。''每个人的预测都只对了一半，那么，实际的第一名至第四名的球队依次是（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "甲乙丁丙 "}], [{"aoVal": "B", "content": "甲丁乙丙 "}], [{"aoVal": "C", "content": "乙甲丙丁 "}], [{"aoVal": "D", "content": "丙甲乙丁 "}]], "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"], "answer_analysis": ["解：根据分析，假设小明说的第一句是对的，即甲是第一，则小凡说的乙是第一是错的，  则丙是第二是对的，就可以推测出小伟说的丙第三是错的，则小伟说的丁第二是对的，  与丙第二矛盾，故假设不成立，故小明说的甲第一是错的，丁第四是对的；  由此可以推测出乙是第一，丙是第三，则甲是第二。  故排名是：乙甲丙丁。  故选：C。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "643", "queId": "6250d9f2f564421fb132e0daabbff724", "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7题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一个箱子，它的底部是正方形，长，宽，高都是整数，它的体积为144，则这个箱子的尺寸可以有(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "4种 "}], [{"aoVal": "B", "content": "5种 "}], [{"aoVal": "C", "content": "6种 "}], [{"aoVal": "D", "content": "8种 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数"], "answer_analysis": ["因为长方体的体积=长×宽×高，底部为正方形，即长、宽相等.$$144=1\\times 1\\times 144=2\\times 2\\times 36=3\\times 3\\times 16=4\\times 4\\times 9=6\\times 6\\times 4=12\\times 12\\times 1$$，所以共6种 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3005", "queId": "c0a8ffa31bd8416f9c7d82ee4a3764dc", "competition_source_list": ["2020年第1届广东深圳超常思维竞赛六年级竞赛初赛第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "给定六个数：$$1$$、$$3$$、$$9$$、$$27$$、$$81$$、$$243$$，从中每次取出若干个数相加，可以得出一个新数，这样共得到$$63$$个新数．把它们从小到大排列起来是$$1$$、$$3$$、$$4$$、$$9$$、$$10$$、$$\\cdots $$．那么，第$$39$$个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$244$$ "}], [{"aoVal": "B", "content": "$$246$$ "}], [{"aoVal": "C", "content": "$$252$$ "}], [{"aoVal": "D", "content": "$$256$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->逻辑分析"], "answer_analysis": ["$$1$$，$$3$$，$$9$$，$$27$$，$$81$$，$$243$$这$$6$$个数是$$3$$的次方数，  即为$${{3}^{0}}$$，$${{3}^{1}}$$，$${{3}^{2}}$$，$${{3}^{3}}$$，$${{3}^{4}}$$，$${{3}^{5}}$$，  那么选择任意个数的数求和的个数为$$\\text{C}_{6}^{1}+\\text{C}_{6}^{2}+\\text{C}_{6}^{3}+\\text{C}_{6}^{4}$$$$+\\text{C}_{6}^{5}+\\text{C}_{6}^{6}={{2}^{6}}-1=63$$个．  所以从小到大的排列可以按照二进制的个数排列．  则从小到大第$$39$$个数转化成二进制是第$${{\\left( 39 \\right)}_{10}}={{\\left( 100111 \\right)}_{2}}$$个，  则该数为：$$1\\times {{3}^{5}}+1\\times {{3}^{2}}+1\\times {{3}^{1}}+1\\times {{3}^{0}}=256$$．  故选$$\\text{E}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1030", "queId": "042280e3f9dd4e9fa280a778b8ddf44c", "competition_source_list": ["2020年长江杯六年级竞赛复赛B卷第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两个长方形，它们的周长相等，甲的长与宽之比为$$5:2$$，乙的长与宽之比为$$9:5$$．甲与乙的面积之比． ", "answer_option_list": [[{"aoVal": "A", "content": "$$45:10$$ "}], [{"aoVal": "B", "content": "$$10:45$$ "}], [{"aoVal": "C", "content": "$$8:9$$ "}], [{"aoVal": "D", "content": "$$9:10$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->几何模块->直线型->图形认知->正方形和长方形->长方形周长"], "answer_analysis": ["假设两个周长相等的长方形周长都是$$a$$，  甲的长是$$a\\div 2\\times \\frac{5}{5+2}=\\frac{5}{14}a$$，  甲的宽是$$a\\div 2\\times \\frac{2}{5+2}=\\frac{1}{7}a$$，  $${{S}_{{甲}}}=\\frac{5}{14}a\\times \\frac{1}{7}a=\\frac{5}{98}{{a}^{2}}$$，  乙的长是$$a\\div 2\\times \\frac{9}{5+9}=\\frac{9}{28}a$$，  乙的宽是$$a\\div 2\\times \\frac{5}{5+9}=\\frac{5}{28}a$$，  $${{S}_{{乙}}}=\\frac{9}{28}a\\times \\frac{5}{28}a=\\frac{45}{784}{{a}^{2}}$$，  $${{S}_{{甲}}}:{{S}_{{乙}}}=\\frac{5}{98}{{a}^{2}}:\\frac{45}{784}{{a}^{2}}=8:9$$．  故选$$\\text{C}$$． ", "<p>解析：甲的长宽比为$$5:2=10:4$$</p>\n<p>乙的长宽比为$$9:5$$</p>\n<p>甲乙面积比$$=10\\times4:9\\times5$$</p>\n<p>$$=40:45$$</p>\n<p>$$=8:9$$．</p>"], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3095", "queId": "e6780c4b96dd428597fb4e9b927c0ec5", "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->循环小数比大小"], "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": "1453", "queId": "440b5a0176fd4fb8b2a47910f8c0397c", "competition_source_list": ["2012年IMAS小学中年级竞赛第一轮检测试题第14题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$3$$根吸管连在一块，好像一根魔术棒，每根吸管长$$20\\text{cm}$$，两根之间的接头重叠部分是$$2\\text{cm}$$，请问这根``魔术棒''的长度为多少$$\\text{cm}$$？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$54$$ "}], [{"aoVal": "B", "content": "$$55$$ "}], [{"aoVal": "C", "content": "$$56$$ "}], [{"aoVal": "D", "content": "$$58$$ "}], [{"aoVal": "E", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->加减法应用->重叠问题->重叠直线型", "Overseas Competition->知识点->应用题模块->加减法应用->重叠问题"], "answer_analysis": ["若没有接头，$$3$$根吸管共长$$20\\times 3=60(\\text{cm})$$，一个接头重叠部分需$$2\\text{cm}$$，两个接头重叠部分需$$4\\text{cm}$$，则``魔术棒''长$$60-4=56(\\text{cm})$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2704", "queId": "ba227cb64a274939875333080f36e71c", "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$$a$$和$$b$$是确定的数，$$x$$是待求的量，有如下四个结论：  ①~~ $$ab=ba$$不是一个方程；  ②~~ $$\\left( {{a}^{2}}+1 \\right)x=2bx+7$$和$${{a}^{2}}x-7=\\left( 2b-1 \\right)x$$是同解方程；  ③~~ $$\\frac{1}{a-2}+2a$$中的字母$$a$$不能取$$2$$；  ④~~ $$2x+6y=29$$有整数解．  其中正确的结论是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "①，②和③ "}], [{"aoVal": "B", "content": "①，②和④ "}], [{"aoVal": "C", "content": "①，③和④ "}], [{"aoVal": "D", "content": "②，③和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->整数系数方程"], "answer_analysis": ["①是乘法交换律，不是方程，结论正确；②两个方程是同解方程；③结论正确；④没有整数解，否则，左端是偶数，右端是奇数． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2382", "queId": "0b4450e0b8174c9db964b500fbd2e4a2", "competition_source_list": ["2009年五年级竞赛创新杯", "2009年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "将长为15的木棒截成长度为整数的三段，使它们构成一个三角形的三边，则得到不同的三角形的个数为 (  )个. ", "answer_option_list": [[{"aoVal": "A", "content": "8 "}], [{"aoVal": "B", "content": "7 "}], [{"aoVal": "C", "content": "6 "}], [{"aoVal": "D", "content": "5 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->不等式->不等式求解"], "answer_analysis": ["设能构成三角形的边长为整数$$a$$,$$b$$,$$c$$($$a\\leqslant b\\leqslant c$$),则$$a+b+c=15$$,且$$a+b\\textgreater c$$. 从而$$\\left( a,b,c \\right)=\\left( 1,7,7\\right),\\left( 2,6,7 \\right),\\left( 3,5,7 \\right),\\left( 3,6,6 \\right),\\left( 4,4,7 \\right),\\left( 4,5,6 \\right),\\left( 5,5,5 \\right)$$, 其中$$\\left( a,b,c \\right)=\\left( 1,7,7 \\right)$$,表示$$a=1$$,$$b=7$$,$$c=7$$. 故选B "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1161", "queId": "113cb027aa9e4ef888535b536eb2af7a", "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$$4$$个，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->逻辑分析"], "answer_analysis": ["一个数的一半的一半是$$4$$，则这个数的一半为$$8$$，则这个数为$$16$$，故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2743", "queId": "ff8080814518d52401451b56b37a06a1", "competition_source_list": ["2014年全国迎春杯三年级竞赛复赛第12题"], "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->能力->运算求解"], "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": "3454", "queId": "cb1443dd93bc40e7afa9be690d5faf94", "competition_source_list": ["2018年第23届华杯赛小学中年级竞赛初赛第3题", "2018年华杯赛小学中年级竞赛初赛第3题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "连接正方形$$ABCD$$的对角线，并将四个顶点分别染成红色或黄色，将顶点颜色全相同的三角形称为同色三角形，则图中有同色三角形染色方法共有（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$22$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["每个顶点有两种染色方法，则一共有$$2\\times 2\\times 2\\times 2=16$$（种）染色方法．  要存在同色三角形，则不能有``两个顶点染红，两个顶点染黄''这种情况有$$C_{4}^{2}=6$$（种），故存在同色三角形的染色方法有$$16-6=10$$（种）． ", "<p>要存在同色三角形，则应排除&ldquo;对角线两侧不同色&rdquo;的情况，这种情况有$$2\\times 3=6$$（种），故存在同色三角形的染色方法有$$16&minus;6=10$$（种）．</p>"], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "274", "queId": "a7cfa20c9a9c402f8af1784e506a4eae", "competition_source_list": ["2017年第17届湖北武汉世奥赛五年级竞赛决赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "奶奶家的后院有$$6$$缸自家酿造的酱，这$$6$$个缸里分别装有$$4$$缸大酱和$$2$$缸酱油；这$$6$$缸酱都是不同年代的：这些酱中最长的年份是$$15$$年，最短的年份是$$9$$年年份最长的酱油比年份最短的大酱多$$4$$年；年份最长的大酱比年份最短的酱油多$$4$$年．年代第二久远的酱油的年份是（~ ）年．（注：$$6$$缸酱的年份都是整数年） ", "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": ["设大酱从大到小分别是：$$A$$、$$B$$、$$C$$、$$D$$，酱油从大到小依次是$$E$$、$$F$$．那么有$$E-D=4$$，$$A-F=4$$．如果$$E=15$$，有$$D=11$$，$$F=9$$，$$A=13$$，不满足有$$4$$个不同的整数$$A$$、$$B$$、$$C$$、$$D$$；那么$$A=15$$，$$F=11$$．所以是$$11$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2298", "queId": "9c2463a85e8642a2b808be856eb83a54", "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4题5分", "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3题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "客车和货车分别从甲、乙两站同时相向开出，第一次相遇在离甲站$$40$$千米的地方，相遇后两车仍以原速度继续前进，客车到达乙站、货车到达甲站后均立即返回，结果它们又在离乙站$$20$$千米的地方相遇．则甲、乙两站之间的距离为千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$80$$ "}], [{"aoVal": "B", "content": "$$90$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$110$$ "}]], "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->两次相遇"], "answer_analysis": ["两次相遇客车行驶的总路程：$$40\\times3=120$$（千米），  所以甲、乙两站之间的距离：$$120-20=100$$（千米）．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1684", "queId": "536276fe53c142359163d4ce44e436bd", "competition_source_list": ["2008年第6届创新杯六年级竞赛初赛B卷第2题5分", "2008年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "某种商品若按标价的八折出售，可获利$$20 \\%$$，若按原标价出售，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": ["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": ["设原标价为$$x$$元，进价为$$y$$元，则$$x\\times 80 \\%=y\\times\\left( 1+20 \\% \\right)$$，从而$$x=y\\times\\left( 1+20 \\% \\right)\\div 80 \\%$$，$$x-y=y\\times \\left( 1+20 \\% \\right)\\div 80 \\%-y$$，$$\\frac{x-y}{y}=\\left( 1+20 \\% \\right)\\div 80 \\%-1=0.5=50 \\%$$，所以，按原标价出售，可获利$$50 \\%$$. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3480", "queId": "fe70b13011774a4896f28dd78ad9efcc", "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3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在所有的两位数中，个位数字比十位数字大的两位数有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$45$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$38$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["这道题要求找出两位数中，个位数字比十位数字大的两位数，  如果十位是$$1$$，那么个位可以是$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$这$$8$$个数，  如果十位是$$2$$，那么个位可以是$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$这$$7$$个数，  如果十位是$$3$$，这样的数有$$6$$个，  如果十位是$$4$$，这样的数有$$5$$个，  如果十位是$$5$$，这样的数有$$4$$个，  如果十位是$$6$$，这样的数有$$3$$个，  如果十位是$$7$$，这样的数有$$2$$个，  如果十位是$$8$$，那么个位可以是$$9$$这$$1$$个数，  $$8+7+6+5+4+3+2+1=36$$，所以一共有$$36$$个． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "498", "queId": "c6381abfa0f44ffcbd01039e9e7b600c", "competition_source_list": ["2018年北京迎春杯五年级竞赛初赛A卷"], "difficulty": "3", "qtype": "single_choice", "problem": "在下列横式中，相同的汉字代表相同的数字，不同的汉字代表不同的数字，且没有汉字代表$$7$$，``迎''、``春''、``杯''均不等于$$1$$，那么``迎''、``春''、``杯''所代表三个数字的和是~\\uline{~~~~~~~~~~}~．  $$\\overline{学数学}\\times 7\\times迎 \\times 春\\times$$$$杯$$$$=\\overline{加油加油}\\times 吧$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$16$$ "}], [{"aoVal": "E", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理", "Overseas Competition->知识点->组合模块->数字谜->横式数字谜"], "answer_analysis": ["若含$$5$$，则必为``加''此时$$\\overline{加油}=56$$，此时$$3$$和$$9$$各剩一个，无法满足，所以不含$$5$$．$$\\overline{加油}$$为$$7$$的倍数，且不为$$49$$，考虑$$3$$、$$6$$、$$9$$的分配．  第一种情况，吧$$=9$$，则$$3$$、$$6$$在左侧，且$$\\overline{加油}$$不是$$3$$的倍数，则$$\\overline{加油}=14$$或$$28$$，无解；  第二种情况，$$9$$在左侧，则$$3$$、$$6$$在右侧，可得$$1\\times 2\\times 4\\times 9\\times 7=63\\times 8$$．  ``迎''、``春''、``杯''的和为$$15$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "400", "queId": "ada0d69adf8844d89191d0e386656c0c", "competition_source_list": ["2017年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "小明行李箱锁的密码是数字组成只包含$$8$$和$$5$$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->拓展思维->组合模块->抽屉原理->最不利原则"], "answer_analysis": ["枚举有$$5$$、$$8$$构成的三位数有$$558$$、$$588$$、$$585$$、$$855$$、$$858$$、$$885$$，共$$6$$个，根据最不利原则，必须要试$$6$$次才可以保证打开。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3003", "queId": "bc13aa0e65ba463a9cb508ff4c5fad50", "competition_source_list": ["2009年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "已知下面A，B，C，D四个算式中，得数最大的是(  ) ", "answer_option_list": [[{"aoVal": "A", "content": "$$2008+\\frac{2008}{2009}$$ "}], [{"aoVal": "B", "content": "$$2008-\\frac{2008}{2009}$$ "}], [{"aoVal": "C", "content": "$$2008\\div \\frac{2008}{2009}$$ "}], [{"aoVal": "D", "content": "$$2008\\times \\frac{2008}{2009}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之凑整法->整数基准数法"], "answer_analysis": ["$$2008\\div \\frac{2008}{2009}=2009$$，而A、B、D的值均小于2009，所以得数最大的算式是C. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "933", "queId": "eec2aa96246f4944b06b747759a92d47", "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1题"], "difficulty": "1", "qtype": "single_choice", "problem": "若三个连续偶数的和是$$162$$，则它们的乘积是（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$157248$$ "}], [{"aoVal": "B", "content": "$$125748$$ "}], [{"aoVal": "C", "content": "$$157284$$ "}], [{"aoVal": "D", "content": "$$172584$$ "}]], "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", "课内体系->知识模块->数与代数"], "answer_analysis": ["这三个连续偶数的中间的那个数为$$162\\div 3=54$$，那么这三个偶数分别为$$52$$、$$54$$、$$56$$，它们的乘积是$$52\\times 54\\times 56=157248$$，故选A。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "911", "queId": "a5641a69b2e146498eb1ac5dde8a4d16", "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2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知数$$a$$和数$$b$$的最小公倍数是$$a\\times b$$，那么它们的最大公因数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$a$$ "}], [{"aoVal": "C", "content": "$$b$$ "}], [{"aoVal": "D", "content": "$$ab$$ "}]], "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": ["由题干可知：$$a$$和$$b$$的最小公倍数是$$a\\times b$$，  那么可以推出$$a$$和$$b$$是互质数，  所以它们的最大公因数是$$1$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2574", "queId": "70958a0b13604d0ea8ec84377cd0f7ec", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（一）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列说法正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "计算小数加减法，得数的小数部分有$$0$$的要把$$0$$去掉 "}], [{"aoVal": "B", "content": "小数加减法混合运算一定要先算加法，后算减法 "}], [{"aoVal": "C", "content": "在$$2.3$$的末尾添上$$2$$个$$0$$，这个数的大小不变 "}], [{"aoVal": "D", "content": "比$$0.1$$大而又比$$0.5$$小的数只有$$3$$个 "}]], "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": ["$$A$$项，不是末尾的$$0$$不能去掉；  $$B$$项，小数加减法混合运算应从左往右计算；  $$D$$项，比$$0.1$$大而又比$$0.5$$小的数有无数个，比如$$0.11$$、$$0.111$$、$$\\cdots \\cdots $$ "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1913", "queId": "d2ce8ef841bc4ec299ec0e9d6d01f443", "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8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "小红将$$99$$个糖果放入两种规格的盒子中，每个大盒可以放$$12$$颗糖果，每个小盒可以放$$5$$颗糖，如果要求装$$10$$个盒子以上，且每个盒子都装满，那么这些糖果正好装完，需要个小盒． ", "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->思想->方程思想"], "answer_analysis": ["设大盒子的个数为$$x$$，小盒子的个数为$$y$$，得$$12x+5y=99$$，且$$x+y\\geqslant 10$$，由于$$12x+5y=99$$得$$99-12x=5y$$，所以$$99-12x$$的差必须是$$5$$的倍数，经实验，只有当$$x=2$$，$$y=15$$和$$x=7$$，$$y=3$$时，$$2+15\\textgreater10$$符合题意，$$7+3=10$$不符合题意，所以小盒子的个数是$$15$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2236", "queId": "c7e9e3874a9340b38f0fa1ce645354f3", "competition_source_list": ["2008年第6届创新杯四年级竞赛复赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个车队以$$5$$米/秒的速度缓缓通过一座长$$200$$米的大桥，共用$$145$$秒．已知每辆车长$$5$$米，两车间隔$$8$$米，那么这个车队共有车（ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$39$$辆 "}], [{"aoVal": "B", "content": "$$40$$辆 "}], [{"aoVal": "C", "content": "$$41$$辆 "}], [{"aoVal": "D", "content": "$$42$$辆 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["车队走的路程$$145\\times 5=725$$米，车队的长度$$725-200=525$$米，  由``车队长$$=$$车辆数$$\\times 5+$$（车辆数$$-1$$）$$\\times 8=5+$$（车辆数$$-1$$）$$\\times \\left( 5+8 \\right)$$''得  $$\\left( 525-5 \\right)\\div \\left( 5+8 \\right)=40$$（辆），$$40+1=41$$（辆）. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "489", "queId": "dc868072d3b54578b0e67ba5d9bb3301", "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7题"], "difficulty": "1", "qtype": "single_choice", "problem": "设x和y是选自1-500这500个自然数中的两个不同的数，那么$$\\left( x+y \\right)\\div \\left( x-y \\right)$$的最大值是(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "1000 "}], [{"aoVal": "B", "content": "990 "}], [{"aoVal": "C", "content": "999 "}], [{"aoVal": "D", "content": "998 "}]], "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": ["要求$$\\left( x+y \\right)\\div \\left( x-y \\right)$$的最大值，则$$x+y$$应尽量最大，而$$x-y$$则应最小.  当$$x+y$$最大，$$x-y$$最小时，$$x=500$$ ，$$y=499$$满足本题要求，于是$$x+y=999，x-y=1$$，因此$$\\left( x+y \\right)\\div \\left( x-y \\right)=999$$，故选C "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2526", "queId": "8fc70f5f576f432ab6031c5ffa34ccc2", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛A卷第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "下面各题中，两种量成正比例关系的是． ", "answer_option_list": [[{"aoVal": "A", "content": "速度一定，路程和时间 "}], [{"aoVal": "B", "content": "圆的半径和面积 "}], [{"aoVal": "C", "content": "圆锥的体积一定，底面积和它的高 "}], [{"aoVal": "D", "content": "一袋水泥，用去的部分和剩下的部分． "}]], "knowledge_point_routes": ["拓展思维->能力->图形认知"], "answer_analysis": ["正比例是指两种相关联的量．  一种量变化，另一种量也变化．  并且这两种量中对应的两个数比值一定．  选项$$\\text{A}$$：路程$$=$$速度$$\\times $$时间．  速度一点，路程和时间成正比例关系；  选项$$\\text{B}$$：圆的面积$$=\\pi \\times $$半径$$\\times $$半径，  半径和面积不成正比例，错误；  选项$$\\text{C}$$：圆锥体积$$=$$底面积$$\\times $$高$$\\times \\frac{1}{3}$$，  体积一定，底面积和高成反比例．错误；  选项$$\\text{D}$$：水泥总量$$=$$用去的部分$$+$$剩下的部分；  用去的部分和剩下的部分不成比例，错误；  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": "3393", "queId": "a97399739c014d8e805dd4d40cae6032", "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": "$$106$$ "}], [{"aoVal": "E", "content": "$$108$$ "}]], "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->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": "1097", "queId": "9d5a5bea7a3c4a718f2346ef8a109184", "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卷第4题10分"], "difficulty": "1", "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->能力->构造模型->模型思想"], "answer_analysis": ["设田径队员为$$a$$人，则合唱队员$$2a$$人，舞蹈队员$$(2a+10)$$人，$$2a+a+2a+10=100$$，则$$a=18$$，所以舞蹈队员$$18\\times 2+10=46$$人． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "689", "queId": "831ed292e5f64b668d183ad3f2e26cb4", "competition_source_list": ["2016年河南郑州联合杯六年级竞赛初赛第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$1-9$$这九个自然数中任取一个，是$$2$$的倍数的概率是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{4}{9}$$~~~~~~ "}], [{"aoVal": "B", "content": "$$\\frac{2}{9}$$~~~~~~ "}], [{"aoVal": "C", "content": "$$\\frac{5}{9}$$~~~~~~ "}], [{"aoVal": "D", "content": "$$\\frac{2}{3}$$~~~~~~ "}]], "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": ["概率与可能性；$$1-9$$中$$2$$的倍数有$$4$$个，所以概率为$$\\frac{4}{9}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2093", "queId": "e26b7255c32c411c9990b9d2476108c3", "competition_source_list": ["2017年湖北武汉中环杯五年级竞赛初赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "小明在黑板上写了一个数，小红把这个数先乘$$2$$，再加上$$10$$，再除以$$4$$，得到的结果为$$8$$．那么，小明在黑板上写的数为~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["倒推$$(8\\times 4-10)\\div 2=11$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "853", "queId": "736438decc744de59383f96ff9a64b44", "competition_source_list": ["2018年华杯赛六年级竞赛模拟卷"], "difficulty": "2", "qtype": "single_choice", "problem": "在序列$$20181$$\\ldots 中，从第$$5$$个数字开始，每个数字都是前面$$4$$个数字和的个位数，这样的序列可以一直写下去，那么这样的数列中，一定不会出现的数组是~\\uline{~~~~~~~~}~。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$7838$$ "}], [{"aoVal": "B", "content": "$$0721$$ "}], [{"aoVal": "C", "content": "$$3474$$ "}], [{"aoVal": "D", "content": "$$2017$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的应用"], "answer_analysis": ["观察数列奇偶性，多写几个就会发现数列满足``偶偶奇偶奇''的周期顺序，不会出现两个奇数相邻的情况，所以$$2017$$一定不会出现。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3146", "queId": "03b6769c9dbc4d39879f2607709c5776", "competition_source_list": ["2016年创新杯五年级竞赛训练题（三）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "在``和谐幸福家庭''报告会上，小红一家五口坐成一排为同学们签名留念，从左往右依次坐得顺序为爸爸、奶奶、小红、爷爷、妈妈．已知每位同学都恰好找座位相邻的三人签名，其中有$$24$$位同学同时找爷爷和奶奶签名．小红一共签名$$45$$次，爸爸比妈妈多签$$5$$次名，那么奶奶一共签名（~ ）次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$37$$ "}], [{"aoVal": "B", "content": "$$39$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$45$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["奶奶一共签名$$37$$次． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "112", "queId": "b4b954802f0441118b6a634952a2d4cf", "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->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": "1943", "queId": "d795fd17a1fd4f069a20862f3f10b116", "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7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$3$$套茶具的价格相当于$$6$$个水瓶的价格．买$$1$$套茶具与$$2$$个水瓶要付$$58$$元．问$$1$$套茶具多少元？下面列式正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$58\\div (6\\div3+2)$$ "}], [{"aoVal": "B", "content": "$$58\\div[1+2\\times (6\\div3)]$$ "}], [{"aoVal": "C", "content": "$$58\\div[(1+2)\\times2]$$ "}], [{"aoVal": "D", "content": "$$58\\div [1+2\\div (6\\div3)]$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["$$3$$套茶具的价格相当于$$6$$个水瓶的价格，$$6\\div 3=2$$，  所以$$1$$套茶具的价格相当于$$2$$个水瓶的价格，买$$1$$套茶具与$$2$$个水瓶，可以把$$2$$个水瓶的价格看作$$1$$套茶具的价格，即$$2\\div (6\\div 3)$$，再加上$$1$$套茶具，相当于买了$$\\left[ 1+2\\div (6\\div 3) \\right]$$套茶具，共花了$$58$$元，  所以$$1$$套茶具的价格为$$58\\div \\left[ 1+2\\div (6\\div 3) \\right]$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1676", "queId": "5c19f892b0f94376bd0ee059e6fad67d", "competition_source_list": ["2012年天津陈省身杯三年级竞赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "被减数、减数与差的和是$$280$$，减数是$$100$$，那么差是~\\uline{~~~~~~~~~~}~． ", "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->思想->对应思想", "Overseas Competition->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": ["由题意，被减数与差的和是$$280-100=180$$，差是$$100$$，由和差问题的公式，差是较小数，差为$$\\left( 180-100 \\right)\\div 2=40$$. "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1296", "queId": "59715768e4534b2b962e792c824f0b0b", "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11题"], "difficulty": "3", "qtype": "single_choice", "problem": "某地水费的收费标准为：如果每月用水量不超过$$15$$吨，那么按每吨$$2.5$$元收费；如果超过$$15$$吨，那么超过部分每吨再多收$$0.5$$元，如果某个月李老师家比洪老师家多交了$$10.5$$元的水费，且两家的用水量均为整数吨，那么李老师家这个月的用水量为~\\uline{~~~~~~~~~~}~吨． ", "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->思想->对应思想"], "answer_analysis": ["设李老师家用水$$x$$吨，洪老师家用水$$y$$吨，由$$10.5\\div 2.5=4.2$$，$$10.5\\div 3=3.5$$可以知道$$x$$，$$y$$必有$$x\\textgreater15$$，$$y\\textless{}15$$，那么$$15\\times 2.5+\\left( x-15 \\right)\\times 3-y\\times 2.5=10.5$$，$$6x-5y=36$$．$$x=16$$，$$y=12$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "211", "queId": "3f01e70282854bc286137841b4486eb3", "competition_source_list": ["2017年全国迎春杯小学中年级四年级竞赛初赛学而思线上模拟第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "一次活动中有中国人、美国人、英国人、日本人$$4$$个国家的人共$$35$$人，且每个国家的人数互不相同．其中中国人和英国人共有$$16$$人，美国人和英国人共有$$17$$人．有一个国家的人数恰好是$$9$$人， 这个国家是~\\uline{~~~~~~~~~~}~. ", "answer_option_list": [[{"aoVal": "A", "content": "中国 "}], [{"aoVal": "B", "content": "美国 "}], [{"aoVal": "C", "content": "英国 "}], [{"aoVal": "D", "content": "日本 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["由美国人和英国人共有$$17$$人可知，中国人和日本人共有$$18$$人，因为人数不同，所以排除中国和日本．  如果美国人有$$9$$人，则英国和中国人都是$$8$$人，所以只能是英国人$$9$$人． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "503", "queId": "f859b184360c4b74aa6336b8f52926ba", "competition_source_list": ["2013年华杯赛四年级竞赛初赛", "2013年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2013$$年的钟声敲响了，小明哥哥感慨地说：这是我有生以来第一次将要渡过一个没有重复数字的年份.已知小明哥哥出生的年份是$$19$$的倍数，那么$$2013$$年小明哥哥的年龄是（  ）岁。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$22$$ "}]], "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": ["有生以来遇到的第一个没有重复数字的年份，说明他生于$$1987$$年之后.年份又是$$19$$的倍数，$$1987$$到$$2013$$之间，只有$$1995$$.所以哥哥生于$$1995$$年，$$2013$$年他$$18$$岁。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2275", "queId": "660e42250825431aabba05632fa08b95", "competition_source_list": ["2014年迎春杯六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$地在$$B$$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->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": ["$$n$$小时后$${S}_{乙}+{S}_{丙}=60$$千米  $$2n$$小时$${S}_{乙}+{S}_{丙}=120$$千米  设甲乙、丙丁$$2n$$小时内的路程差为$${S}_{0}$$，  则$${S}_乙={S}_甲-{S}_0,{S}_{丙}={S}_{丁}+{S}_{0}$$，  则$${{S}_{甲}}+{{S}_{丁}}=120$$（千米）①  $$2n$$小时甲追上丁：$${{S}_{甲}}-{{S}_{丁}}=60$$（千米）②  将①式$$+$$②式得$${{S}_{甲}}=90$$千米，$${{S}_{丁}}=30$$千米  $$BC$$的距离正好是$${{\\text{S}}_{丁}}$$，  答：$$BC$$两地距离$$30$$千米． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2365", "queId": "1795f699221e4f149dc024ca2b38ed67", "competition_source_list": ["2008年第6届创新杯六年级竞赛复赛第6题4分", "2008年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$\\frac{1}{46}$$写成两个分母不同的单位分数(分子为$$1$$的真分数)之和，写法有(  ) ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$种 "}], [{"aoVal": "B", "content": "$$3$$种 "}], [{"aoVal": "C", "content": "$$4$$种 "}], [{"aoVal": "D", "content": "$$5$$种 "}]], "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": ["解法一：设$$\\frac{1}{46}=\\frac{1}{a}+\\frac{1}{b}$$，$$a\\textgreater b$$，则$$ab=46\\left( a+b \\right)$$. 令$$\\left( a,b \\right)=d$$，从而$$a={{a}_{1}}d$$，$$b={{b}_{1}}d$$ 且$$\\left( {{a}_{1}},{{b}_{1}} \\right)=1$$，得$${{a}_{1}}{{b}_{1}}d=46\\left( {{a}_{1}},{{b}_{1}} \\right)$$ ①。 又$$\\left( {{a}_{1}},{{a}_{1}}+{{b}_{1}} \\right)=\\left( {{a}_{1}},{{b}_{1}} \\right)=1$$，同理$$\\left( {{b}_{1}},{{a}_{1}}+{{b}_{1}} \\right)=1$$，所以$$\\left( {{a}_{1}}{{b}_{1}},{{a}_{1}}+{{b}_{1}} \\right)=1$$，因此$${{a}_{1}}\\textbar46$$，$${{{b}}_{1}}\\textbar46$$，$${{a}_{1}}{{b}_{1}}\\left\\textbar{} 46 \\right.$$，$$\\left( {{a}_{1}},{{b}_{1}} \\right)=1$$，$${{a}_{1}}\\textgreater{{b}_{1}}$$，故 $$\\left( {{a}_{1}},{{b}_{1}} \\right)=\\left( 2,1 \\right)$$，$$\\left( 23,1 \\right)$$，$$\\left( 23,2 \\right)$$或$$\\left( 46,1 \\right)$$，其中$$\\left( {{a}_{1}},{{b}_{1}} \\right)=\\left( 2,1 \\right)$$表示$${{a}_{1}}=2$$，$${{b}_{1}}=1$$，其余类推。 将$${{a}_{1}},{{b}_{1}}$$之值代入①依次得$$d=69$$，$$48$$，$$25$$，$$47$$则得， $$\\left( a,b \\right)=\\left( 138,69 \\right)$$，$$\\left( 1104,48 \\right)$$，$$\\left( 575,50 \\right)$$，$$\\left( 2162,47 \\right)$$， 从而$$\\frac{1}{46}=\\frac{1}{138}+\\frac{1}{69}$$，$$\\frac{1}{46}=\\frac{1}{1104}+\\frac{1}{48}$$，$$\\frac{1}{46}=\\frac{1}{575}+\\frac{1}{50}$$，$$\\frac{1}{46}=\\frac{1}{2162}+\\frac{1}{47}$$。  解法二：设$$\\frac{1}{46}=\\frac{1}{a}+\\frac{1}{b}$$，$$a\\textgreater b$$，则$$\\frac{1}{b} \\textless{} \\frac{1}{46}=\\frac{1}{a}+\\frac{1}{b} \\textless{} \\frac{2}{b}$$，从而$$46 \\textless{} b \\textless{} 92$$。 又由$$\\frac{1}{46}=\\frac{1}{a}+\\frac{1}{b}$$化简得$$ab=46\\left( a+b \\right)$$ ①， 即$$b\\left( a-46 \\right)=46a$$，所以$$b=\\frac{46a}{a-46}=46+\\frac{{{46}^{2}}}{a-46}$$，因此$$\\left( a-46 \\right)\\textbar{{46}^{2}}$$ ②， 且$$0 \\textless{} \\frac{{{46}^{2}}}{a-46} \\textless{} 46$$，即$$a-46\\textgreater46$$ ③， 由②，③知$$a-46=4\\cdot 23$$，$${{23}^{2}}$$，$$2\\cdot {{23}^{2}}$$或$$4\\cdot {{23}^{2}}$$， 依次可得$$a=138$$，$$575$$，$$1104$$或$$2162$$， 依次将$$a$$的值代入$$b=46+\\frac{{{46}^{2}}}{a-46}$$，得$$b=69$$，$$50$$，$$48$$或$$47$$。  因此$$\\frac{1}{46}=\\frac{1}{138}+\\frac{1}{69}$$，$$\\frac{1}{46}=\\frac{1}{575}+\\frac{1}{50}$$，$$\\frac{1}{46}=\\frac{1}{1104}+\\frac{1}{48}$$，$$\\frac{1}{46}=\\frac{1}{2162}+\\frac{1}{47}$$。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2514", "queId": "2bab57525fc741d2866b618d95725572", "competition_source_list": ["2019年第24届YMO二年级竞赛决赛第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "$\\bigcirc$，$$★$$，$\\triangle$代表三个数，$\\bigcirc+\\bigcirc=12$，$$★+★+★=15$$，$$△+△+△+△=24$$，$\\bigcirc+\\bigstar+\\triangle=$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意分析可知，  $\\bigcirc+\\bigcirc=12$，即$2\\bigcirc=12$，$\\bigcirc=12\\div2=6$，  $$★+★+★=15$$，$$3★=15$$，$\\bigstar=15\\div3=5$，  $$△+△+△+△=24$$，$$4△=24$$，$\\triangle=24\\div4=6$，  由此可知，$\\bigcirc+\\bigstar+\\triangle=6+5+6=17$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1491", "queId": "7f305cb9733d41b49209feb6ea33bce3", "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，今年的国庆节是星期． ", "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": "1118", "queId": "145590362b8f4adc8eb73de5e1066409", "competition_source_list": ["2018年IMAS小学高年级竞赛（第一轮）第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "餐厅有面粉$$240\\text{kg}$$，原计划使用$$8$$天．改变食谱后，每天比原计划少用$$6\\text{kg}$$．请问这批面粉实际使用了多少天？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"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->运算求解"], "answer_analysis": ["按照原计划，每天使用$$240\\div 8=30\\text{kg}$$．  改变食谱后，每天使用$$30-6=24\\text{kg}$$．  得知这批面粉实际使用了$$240\\div 24=10$$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": "934", "queId": "eec360863e0249058b72f0f71106a28f", "competition_source_list": ["2015年第4届广东广州羊排赛六年级竞赛第6题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "五百多年前的今天------$$1445$$年$$3$$月$$1$$日，佛罗伦萨画派艺术家桑德罗·波提切利诞生．他以画作《春》、《维纳斯的诞生》而声名大噪．将他的诞生日对应数字$$14450301$$，该数字可以读出个零． ", "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->数据处理"], "answer_analysis": ["$$14450301$$读作一千四百四十五万零三百零一，可以读出$$2$$个零．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "646", "queId": "46c54075bec5471dadcfa20bcf5de499", "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38题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$A$$为一个可整除$$11$$但不可整除$$10$$的正整数，若$$B$$为一个颠倒$$A$$数位的整数，那么$$B$$必须被几整除? ", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$33$$ "}], [{"aoVal": "D", "content": "$$A$$．$$B$$．$$C$$都不选 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->整除->整除性质的应用"], "answer_analysis": ["整除特征，反过来奇数位和偶数位的差值不变，所以还是被$$11$$整除． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "9", "queId": "054308464ac844ca9eb0f664c9ca558d", "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．它们所有等候时间的总和~ ~ ~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": "1102", "queId": "58989d014d13449c900aed2e2802541d", "competition_source_list": ["2018年第6届湖北长江杯六年级竞赛初赛A卷第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列说法正确的有．  ①液氮变成氮气，体积增大$$a \\%$$，氮气压缩成液氮，体积减小$$a \\%$$；  ②唐朝$$666$$年的二月有$$29$$天；  ③$$\\frac{a}{b}$$可能比$$\\frac{a+b}{2b}$$大（$$a$$，$$b$$均为非零自然数）；  ④$$1.88\\div 0.03=62\\cdots \\cdots 2$$，所以$$1.88$$除以$$0.03$$的余数是$$2$$；  ⑤六年级的美术小组学生人数占六年级学生总人数的$$35 \\%$$，六年级创客小组学生人数占六年级男生总人数的$$75 \\%$$，六年级的美术小组学生人数比六年级的创客小组学生人数要少． ", "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": ["拓展思维->能力->数据处理"], "answer_analysis": ["①单位$$1$$不同，对应的分率也不同，故①说法错误；  ②唐朝$$666$$年是平年，所以$$2$$月有$$28$$天，所以②说法错误；  ③当$$a=2$$，$$b=1$$时，$$\\frac{a}{b}=2$$，$$\\frac{a+b}{2b}=\\frac{2+1}{2\\times 1}=\\frac{3}{2}$$，因为$$2\\textgreater\\frac{3}{2}$$，所以$$\\frac{a}{b}$$，可能比$$\\frac{a+b}{2b}$$大，故③说法正确；  ④在除法算式中，余数小于除数，所以④说法错误；  ⑤因为不知道男生与总人数的数量关系，所以美术小组人数与创客小组人数无法进行比较，故⑤说法错误．一共有$$1$$个正确．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1567", "queId": "5b449e052d5e458d9b85285467becaf3", "competition_source_list": ["2015年第27届广东广州五羊杯小学高年级竞赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2015$$年的中国股市大起大落，使很多新入市的股民损失惨重．一位中学生高考结束后，在$$7$$月中旬开户炒股票．他买的第一只股票的价格第一天就下跌了$$10 \\%$$，后又下跌了$$15 \\%$$．现在市场形势转好，股市大涨，这只股票又上涨了$$20 \\%$$．那么，这只股票的现在价格与原价相比． ", "answer_option_list": [[{"aoVal": "A", "content": "下跌了$$5 \\%$$ "}], [{"aoVal": "B", "content": "下跌了$$4.1 \\%$$ "}], [{"aoVal": "C", "content": "下跌了$$8.2 \\%$$ "}], [{"aoVal": "D", "content": "下跌了$$7.6 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["本题主要考察分数单位$$1$$的灵活应用，可以用设数法进行求解．  若这只股票原来的价格为$$100$$元，则下跌$$10 \\%$$后的价格为$$90$$元；  在此基础上，再下跌$$15 \\%$$的价格则为$$76.5$$元；  这时若上涨$$20 \\%$$，  则现在的价格为$$91.8$$元．  所以同原来$$100$$元价格相比，下跌了$$8.2 \\%$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2703", "queId": "88797c46b4d44a73aa7dea14ad14608b", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一串分数：$$\\frac{1}{4}$$，$$\\frac{2}{4}$$，$$\\frac{3}{4}$$，$$\\frac{2}{4}$$，$$\\frac{1}{4}$$，$$\\frac{1}{7}$$，$$\\frac{2}{7}$$，$$\\frac{3}{7}$$，$$\\frac{4}{7}$$，$$\\frac{5}{7}$$，$$\\frac{6}{7}$$，$$\\frac{5}{7}$$，$$\\frac{4}{7}$$，$$\\frac{3}{7}$$，$$\\frac{2}{7}$$，$$\\frac{1}{7}$$，$$\\frac{1}{10}$$，$$\\frac{2}{10}$$，$$\\frac{3}{10}$$，\\ldots，$$\\frac{9}{10}$$，$$\\frac{8}{10}$$，\\ldots，$$\\frac{1}{10}$$，$$\\frac{1}{13}$$，$$\\frac{2}{13}$$．求第$$2016$$个分数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2016}$$ "}], [{"aoVal": "B", "content": "$$\\frac{64}{79}$$ "}], [{"aoVal": "C", "content": "$$\\frac{65}{79}$$ "}], [{"aoVal": "D", "content": "$$\\frac{21}{82}$$ "}]], "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": ["观察可知，分母为$$4$$的分数有$$5$$个，分母为$$7$$的分数有$$11$$个，分母为$$10$$的分数有$$17$$个，\\ldots{} 由上面的规律可知，分母分别为：$$4$$，$$7$$，$$10$$，\\ldots 成公差为$$3$$的等差数列，相应的分数的个数分别 为 ：$$5$$，$$11$$，$$17$$， \\ldots 成公差为$$6$$的等差数列， 因为$$5+11+17+...+149=1925$$，$$5+11+17+...+155=2080$$，所以分母是第$$26$$个数，为$$4+\\left( 26-1 \\right)\\times 3=79$$，而前面$$25$$个分母共计$$1925$$个分数，少$$2016-1925=91$$（个）分数，又$$91-78=13$$，$$78-13=65$$，第$$2016$$个分数的分子是$$65$$， 故这串分数中第$$2016$$个分数是$$\\frac{65}{79}$$ "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3309", "queId": "ff8080814502fa2401450753aa7e09ea", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第12题"], "difficulty": "2", "qtype": "single_choice", "problem": "在``爸爸去哪儿''的节目中有一个任务，五个参加任务的孩子（天天、石头、Kimi、Cindy、Angela）需要换爸爸（每个小朋友可以选择除了自己爸爸之外其他四位父亲中的任何一位），那么最终五人有（~~~~~~~ ）种不同的选择结果． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$44$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "knowledge_point_routes": ["拓展思维->能力->抽象概括"], "answer_analysis": ["设五个爸爸分别是$$ABCDE$$，五个孩子分别是$$abcde$$，$$a$$ 有4种选择，假设$$a$$选择$$B$$ ，接着让$$b$$选择，有两种可能，选择$$A$$和不选择$$A$$，（1）选择$$A$$，$$c$$、$$d$$、$$e$$ 选择三个人错排，（2）不选择$A$，则$$bcde$$ 选择情况同$$4$$人错排．所以$${{S}_{5}}=4\\times ({{S}_{4}}+{{S}_{3}})$$ 同理$${{S}_{4}}=3\\times({{S}_{3}}+{{S}_{2}})$$ ，$${{S}_{3}}=2\\times ({{S}_{2}}+{{S}_{1}})$$，而$${{S}_{1}}=0$$（不可能排错），$${{S}_{2}}=1$$，所以$${{S}_{3}}=2$$，$${{S}_{4}}=9$$，$${{S}_{5}}=44$$．  方法二：  $$5$$个人错排，可以拆出$$5$$人全部错排，与$$3+2$$人错排，  $$5$$人全部错排有$$5!=4\\times3\\times2\\times1=24$$，  $$3+2$$人错排有$$\\text{C}_{5}^{2}\\times 2=20$$  $$24+20=44$$. "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2197", "queId": "47589997d42247099c3e0cb613edb3b3", "competition_source_list": ["2017年河南郑州联合杯竞赛第5题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "中午$$12$$点整时，钟面上时针与分针重合，那么到当晚$$12$$点整时，时针与分针又重合了（~ ）次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$次 "}], [{"aoVal": "B", "content": "$$11$$次 "}], [{"aoVal": "C", "content": "$$12$$次 "}], [{"aoVal": "D", "content": "$$13$$次 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->行程模块->时钟问题->时钟上的相遇问题->时钟上的相遇问题对称"], "answer_analysis": ["时钟问题；$$12\\div \\left[ 60\\div \\left( 60-5 \\right) \\right]=12\\div \\left( 60\\div 55 \\right)=11$$（次），则时针与分针重合了$$11$$次． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1794", "queId": "b69b395df8354ccd87d36f5e5101b711", "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2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "兄弟俩今年的年龄和是$$25$$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": "$$10$$ "}], [{"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->年龄问题->年龄问题基本关系->年龄问题基本关系和差问题"], "answer_analysis": ["年龄差是个不变量，本题的关键是求出他们的年龄差，然后再根据和差公式进一步解答．  根据题意，假设他们的年龄差是$$1$$份，由``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''可知弟弟的年龄是$$2$$份，哥哥的年龄是$$3$$份，再根据兄弟俩今年的年龄和是$$25$$岁，可以求出年龄差，然后再进一步解答．  他们的年龄差是：$$25\\div \\left( 2+3 \\right)=5$$（岁），  哥哥的年龄是：$$\\left( 25+5 \\right)\\div 2=15$$（岁），  弟弟的年龄是：$$25-15=10$$（岁）．  答：弟弟今年$$10$$岁．  故答案为：$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2929", "queId": "a962c38fbfb54ca7828c504df7b3139b", "competition_source_list": ["2018年IMAS小学高年级竞赛（第一轮）第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把分数$$\\frac{1}{4}$$、$$\\frac{5}{6}$$与$$\\frac{3}{8}$$全部通分，请问通分后分母最小为多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$48$$ "}], [{"aoVal": "E", "content": "$$96$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["三个分数的分母$$4$$、$$6$$与$$8$$之最小公倍数为$$24$$，  通分后分别为$$\\frac{6}{24}$$、$$\\frac{20}{24}$$与$$\\frac{9}{24}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2387", "queId": "1393c38468d244e0ba6c7d9a1b414468", "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$$2016-7=2009$$，$$2009=7\\times7\\times41$$，所以$$2009$$中比$$7$$大的因数有$$41$$，$$7\\times7$$，$$7\\times41$$，$$7\\times7\\times41$$，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": "830", "queId": "b18f5ac6368e49a0aae0da06834133eb", "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": "小明在做一道加法算式题，由于粗心，将个位上的$$5$$看作$$9$$，把十位上的$$8$$看作$$3$$，结果所得的和是$$123$$，正确的结果应该是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$85$$ "}], [{"aoVal": "B", "content": "$$39$$ "}], [{"aoVal": "C", "content": "$$169$$ "}], [{"aoVal": "D", "content": "$$195$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["看错的加数是$$39$$，因此得到错误的和是$$123$$，则另一个没看错的加数是$$123-39=84$$，所以正确结果应该是$$84+85=169$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1611", "queId": "6d8b9ae8a44d445cbbba84296a99ddc8", "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2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲班和乙班共83人，乙班和丙班共86人，丙班和丁班共88人，那么甲班和丁班人数之和为(  )人. ", "answer_option_list": [[{"aoVal": "A", "content": "85 "}], [{"aoVal": "B", "content": "86 "}], [{"aoVal": "C", "content": "87 "}], [{"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": ["$$83\\text{+}88-86\\text{=}85$$（人） "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "329", "queId": "60a69c8d909c4a7c93b24cd9781cba4a", "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她放在冰箱里的整个蛋糕都吃光了．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:卡洛斯吃了蛋糕．  本尼:奥斯汀吃了蛋糕．  卡洛斯:本尼在撒谎．  丹妮拉:吃蛋糕的不是卡洛斯．  艾拉:我没有吃蛋糕．  如果只有两个孩子说的是真话，那么是谁吃了蛋糕? ", "answer_option_list": [[{"aoVal": "A", "content": "本尼 "}], [{"aoVal": "B", "content": "卡洛斯 "}], [{"aoVal": "C", "content": "丹妮拉 "}], [{"aoVal": "D", "content": "艾拉 "}]], "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->找一致（同伙）"], "answer_analysis": ["因为奥斯汀和丹妮拉说的话是相矛盾的，本尼和卡洛斯说的话是相矛盾的，所以四句话中必有两句真话和两句假话，因此艾拉说的这能是假话，因此真实情况是艾拉吃了蛋糕. "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3417", "queId": "aa05c383449440b395761198fa7e6046", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（四）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "辉辉小学五年级有$$58$$人参加钢琴兴趣小组，$$62$$人参加美术兴趣小组，$$25$$人既参加钢  琴兴趣小组又参加美术兴趣小组，那么参加钢琴或美术小组的同学一共有人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$95$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$145$$ "}]], "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": ["根据容斥原理可知，参加两个小组的同学一共有$$58+62-25=95$$（人）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2341", "queId": "0569da6c4823478399a3fb3ba8a4896f", "competition_source_list": ["2012年IMAS小学高年级竞赛第一轮检测试题第3题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "请问下列哪组分数的总和大于$$1$$？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$、$$\\frac{1}{4}$$、$$\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$、$$\\frac{1}{4}$$、$$\\frac{1}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$、$$\\frac{1}{3}$$、$$\\frac{1}{6}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$、$$\\frac{1}{4}$$、$$\\frac{1}{5}$$ "}], [{"aoVal": "E", "content": "$$\\frac{1}{2}$$、$$\\frac{1}{3}$$、$$\\frac{1}{5}$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数运算->分数加减"], "answer_analysis": ["$$\\text{A}$$选项三个分数之和是$$\\frac{47}{60}$$小于$$1$$，$$\\text{B}$$选项三个分数之和是$$\\frac{11}{12}$$小于$$1$$，$$\\text{C}$$选项三个分数之等于$$1$$，$$\\text{D}$$选项三个分数之和是$$\\frac{19}{20}$$小于$$1$$，$$\\text{E}$$选项三个分数之和（等于$$\\frac{31}{30}$$）大于$$\\text{C}$$选项三个分数之和，故大于$$1$$，所以答案是$$\\text{E}$$．  我们可以估算每组的分数之和是否大于$$1$$，不一定要精确算出三个分数之和．在$$\\text{A}$$选项中，$$\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}\\textless{}\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{3}=1$$．在$$\\text{B}$$选项中，$$\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{6}\\textless{}\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{4}=1$$．在$$\\text{D}$$选项中，$$\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{5}\\textless{}\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{4}=1$$．而在$$\\text{E}$$选项中，$$\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{5}\\textgreater\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}=1$$．所以选$$\\text{E}$$． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1916", "queId": "aa21ef88cf75427786d075ceb528c273", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第16题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$A$$、$$B$$、$$C$$、$$D$$四人各拿出一样多的钱合买了一批笔记本．由于每人需求不同，分完笔记本时，$$A$$、$$B$$、$$C$$分别比$$D$$多拿了$$6$$、$$7$$、$$11$$本．为了公平起见，$$A$$、$$B$$、$$C$$三人一共拿$$48$$元给$$D$$，请问每本笔记本的价格是多少元？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$12$$ "}], [{"aoVal": "E", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["可知$$A$$、$$B$$、$$C$$总共比$$D$$多拿了$$6+7+11=24$$本，  故$$D$$拿的笔记本数会比四人平均笔记本数少了$$24\\div4=6$$本，  因此每本笔记本的价格为$$48\\div6=8$$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": "792", "queId": "6d96d66850d041e7833f153c7387e318", "competition_source_list": ["2017年新希望杯六年级竞赛训练题（三）第2题", "2018年湖北武汉新希望杯六年级竞赛训练题（三）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列选项中，是完全平方数的是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7567$$ "}], [{"aoVal": "B", "content": "$$8645$$ "}], [{"aoVal": "C", "content": "$$7744$$ "}], [{"aoVal": "D", "content": "$$8286$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["平方数的个位只能是$$0$$，$$1$$,$$4$$,$$5$$,$$6$$, $$9$$，所以排除$$\\text{A}$$；如果平方数的个位是$$5$$，那么十位一定是$$2$$，排除$$\\text{B}$$；  如果平方数是的$$3$$倍数，那么平方数一定是$$9$$的倍数，排除$$\\text{D}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2653", "queId": "3f7c4a26337747359501cf36879da538", "competition_source_list": ["2007年走美杯三年级竞赛初赛", "2007年走美杯四年级竞赛初赛"], "difficulty": "0", "qtype": "single_choice", "problem": "$$113\\times 5-37\\times 15$$，这个题目中隐藏的公因数是（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$37$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数构造提取->整数倍数关系"], "answer_analysis": ["略 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2748", "queId": "68dd314417d74e43af897b20f78be20c", "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$$20$$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$$1\\div 5=\\frac{1}{5}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3398", "queId": "92f708dacd5e447884580fdfa7435b66", "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": "$$98$$ "}], [{"aoVal": "C", "content": "$$480$$ "}], [{"aoVal": "D", "content": "$$720$$ "}]], "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": "3001", "queId": "a5907ead6b44453b81113cc8eaa7d2ed", "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． ", "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->拓展思维->计算模块->小数->循环小数->循环小数的概念"], "answer_analysis": ["要使小点数最小，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": "1894", "queId": "8aac50a74ff4b16201500e92dc792af7", "competition_source_list": ["2009年全国迎春杯小学中年级四年级竞赛初赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "老师在黑板上写了三个不同的整数，小明每次先擦掉第一个数，然后在最后写上另两个数的平均数，如此做了$$7$$次，这时黑板上三个数的和为$$159$$．如果开始时老师在黑板上写的三个数之和为$$2008$$，且所有写过的数都是整数．那么开始时老师在黑板上写的第一个数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2008$$ "}], [{"aoVal": "B", "content": "$$1860$$ "}], [{"aoVal": "C", "content": "$$159$$ "}], [{"aoVal": "D", "content": "$$53$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["设老师写的第二个数和第三个数分别为$$a$$和$$b$$，那么小明写的第一个数是$$\\frac{a+b}{2}$$，第二个数是$$\\frac{a+3b}{4}$$，第三个数是$$\\frac{3a+5b}{8}$$，第四个数是$$\\frac{5a+11b}{16}$$，第五个数是$$\\frac{11a+21b}{32}$$，第六个数是$$\\frac{21a+43b}{64}$$，第七个数是$$\\frac{43a+85b}{128}$$．所以有$$\\frac{11a+21b}{32}+\\frac{21a+43b}{64}+\\frac{43a+85b}{128}=159$$，化简为$$\\frac{43a+85b}{128}=53$$，显然$$a=53$$，$$b=53$$是不定方程的一组正整数解，但题目要求的$$a\\ne b$$，所以$$a=53+85=138$$，$$b=53-43=10$$是满足条件的唯一组正整数解．老师写的第一个数是$$2008-138-10=1860$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1523", "queId": "491e2b4559254568af9b32d413ca2780", "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->能力->运算求解"], "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$$10.21-8.56=1.65$$，所以这个一位小数是$$1.65\\times 10=16.5$$．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": "622", "queId": "11314bda28fe47178bc5246dcfaf963e", "competition_source_list": ["2010年全国华杯赛竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "恰有$$20$$个因数的最小自然数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$240$$ "}], [{"aoVal": "C", "content": "$$360$$ "}], [{"aoVal": "D", "content": "$$432$$ "}]], "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": ["$$20=1\\times20=2\\times 10=4\\times 5=2\\times 2\\times 5$$ 四种情况下的最小自然数分别为：$${{2}^{19}}$$、$${{2}^{9}}\\times 3$$、$${{2}^{4}}\\times {{3}^{3}}$$、$${{2}^{4}}\\times 3\\times 5$$，其中最小的是最后一个，为$$240$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "645", "queId": "6254a62815a54841863cf5a2822c8223", "competition_source_list": ["2021年华杯赛小学高年级竞赛初赛（华数之星）第8题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$0$$至$$9$$这十个数字中，组成三个三位数，使得这三个三位数的和等于$$2020$$，那么未被选中的数字是=． ", "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->运算求解"], "answer_analysis": ["复杂问题利用余数简化！  $$2020$$除以$$9$$余$$4$$．  $$0+1+2+\\cdots +9=45$$，$$45$$是$$9$$的倍数．  要想让$$9$$个数的和除以$$9$$余$$4$$，  则未被选中的数字为$$9-4=5$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1545", "queId": "df156d840956478c8eb5b71ce991bb1f", "competition_source_list": ["2015年第5届广东广州羊排赛六年级竞赛第6题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2015$$年的元旦是星期四，那么$$2015$$年的圣诞节（$$12$$月$$25$$日）是星期． ", "answer_option_list": [[{"aoVal": "A", "content": "四 "}], [{"aoVal": "B", "content": "五 "}], [{"aoVal": "C", "content": "六 "}], [{"aoVal": "D", "content": "日 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["从$$2015$$年$$1$$月$$1$$日（元旦）到$$2015$$年$$12$$月$$25$$日（圣诞节）共经历了$$359$$天，$$359\\div 7=51\\cdots \\cdots 2$$，对应星期五．故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "517", "queId": "d43e33988d494f9d83777dcc90bd1eed", "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） ", "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": "1197", "queId": "345c9a43c0e6428da5db69c9a137d513", "competition_source_list": ["2019年广东广州学而思综合能力诊断五年级竞赛第13题12分"], "difficulty": "2", "qtype": "single_choice", "problem": "有$$2019$$枚棋子，毛毛和伦伦轮流取走一些棋子，每次能取走的棋子数可以是$$1$$、$$2$$、$$4$$、8$$\\cdots \\cdots $$，谁让对方取走最后一枚棋子则获胜．毛毛先取，若要必胜，则第一次取走棋子数的所有可能值之和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$660$$ "}], [{"aoVal": "B", "content": "$$682$$ "}], [{"aoVal": "C", "content": "$$703$$ "}], [{"aoVal": "D", "content": "$$724$$ "}], [{"aoVal": "E", "content": "$$746$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->应用题模块->分百应用题->认识单位1", "拓展思维->思想->逆向思想"], "answer_analysis": ["倒推必胜点和必败点，毛毛要获胜，那么得想办法挤进必胜点，同时逼着伦伦进入必败点，必败点可以走的点中有必胜点即可，必胜点所有能走的都得是必败点．  那下面我们用剩几个来表示必胜点和必败点：首先第一个必胜点是$$1$$，能走到这个点的都是必败点，有：$$2$$、$$3$$、$$5$$、$$9$$、$$17$$、$$33$$、$$65$$、$$129\\cdots \\cdots $$，能逼着对方走进这些点的是下一个必胜点，目前只有$$4$$是能保证对方走进$$2$$或$$3$$的，那么能走进$$4$$的都是必败点，有$$5$$、$$6$$、$$8$$、$$12$$、$$20$$、$$36$$、$$68$$、$$132\\cdots \\cdots $$，整理目前所有的必败点：$$2$$、$$3$$、$$5$$、$$6$$、$$8$$、$$9$$、$$12$$、$$17$$、$$20$$、$$33$$、$$36$$、$$65$$、$$68$$、$$129$$、$$132\\cdots \\cdots $$，必胜点有$$1$$和$$4$$，那下一个必胜点目前只有$$7$$，能走进$$7$$的都是必败点，有：$$8$$、$$9$$、$$11$$、$$15$$、$$23$$、$$39$$、$$71$$、$$135\\cdots \\cdots $$，整理目前所有的必败点：$$2$$、$$3$$、$$5$$、$$6$$、$$8$$、$$9$$、$$11$$、$$12$$、$$15$$、$$17$$、$$20$$、$$23$$、$$33$$、$$36$$、$$39$$、$$65$$、$$68$$、$$71$$、$$129$$、$$132$$、$$135\\cdots \\cdots $$，必胜点目前有$$1$$、$$4$$、$$7$$，那么下一个必胜点目前只有$$10$$和$$13\\cdots \\cdots $$其实此时已经能发现必胜点的规律：$$1$$、$$4$$、$$7$$、$$10$$、$$13$$全是$$3n+1$$的数，  所以毛毛取完之后，必须剩下$$\\left( 3n+1 \\right)$$枚棋子，$$2019\\div 3\\cdots \\cdots 0$$，  所以第一次得取$$\\left( 3k+2 \\right)$$枚棋子，符合要求的数量有$$2$$、$$8$$、$$32$$、$$128$$、$$512$$，加起来有$$2+8+32+128+512=682$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1089", "queId": "6fc9ae23a0fe4425b74c314d924d26aa", "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": ["Overseas Competition->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"], "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": "2032", "queId": "bd5f02200fbf4ea29538fc977612faae", "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10题"], "difficulty": "1", "qtype": "single_choice", "problem": "某城市按以下规定收取每月煤气费:用煤气如果不超过60立方米，按每立方米0.8元收费;如果超过60立方米，超过部分按每立方米1.2元收费，已知某用户4月份的煤气费平均每立方米0.88元，那么，4月份这个用户应交煤气费(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "60元 "}], [{"aoVal": "B", "content": "66元 "}], [{"aoVal": "C", "content": "75元 "}], [{"aoVal": "D", "content": "78元 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->经济问题->复杂的经济问题"], "answer_analysis": ["由这个用户4月份的煤气费平均每立方米0.88元可知这个用户这个月的用量肯定超过60立方米，这60立方米的实际费用比平均费用少$$60\\times \\left( 0.88-0.8 \\right)=4.8$$(元)，那么超过60立方米的实际费用比平均费用每立方米多$$1.2-0.88=0.32 $$元，则超过60立方米的部分的用量为$$4.8\\div \\left( 1.2-0.88 \\right)=15$$(立方米)，所以这个用户4月份应交煤气费$$0.88\\times \\left( 60+15 \\right)=66$$元，选B "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1207", "queId": "62479f725eb44d75801eaf1754877c94", "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": ["Overseas Competition->知识点->应用题模块->分百应用题->认识单位1", "拓展思维->思想->逐步调整思想"], "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": "2896", "queId": "a905e4f9c8ec42848d65d7d09450bd29", "competition_source_list": ["2015年IMAS小学高年级竞赛第一轮检测试题第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$39$$位小朋友在操场上排成若干排，第一排有$$4$$位，后面每排都比前一排多一位，请问最后一排有多少位小朋友？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$39$$ "}], [{"aoVal": "D", "content": "$$26$$ "}], [{"aoVal": "E", "content": "$$35$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$4+5+6+7+8+9=39$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2670", "queId": "7155e18a2a4f479c81fb488148af7d0d", "competition_source_list": ["2020年新希望杯四年级竞赛决赛（8月）第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "规定$$a\\otimes b=(a+b-1)\\times (a-b+2)$$，那么$$(5\\otimes 3)\\otimes 3=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$800$$ "}], [{"aoVal": "B", "content": "$$830$$ "}], [{"aoVal": "C", "content": "$$820$$ "}], [{"aoVal": "D", "content": "$$810$$ "}]], "knowledge_point_routes": ["拓展思维->能力->符号代换->代数运算"], "answer_analysis": ["根据定义可知$$(5\\otimes 3)\\otimes 3$$，先算$$5\\otimes 3$$部分为：  $$5\\otimes 3$$  $$=(5+3-1)\\times (5-3+2)$$  $$=7\\times 4$$  $$=28$$．  再算$$28\\otimes 3$$  $$=(28+3-1)\\times (28-3+2)$$  $$=30\\times 27$$  $$=810$$．  故答案为：$$810$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1650", "queId": "4a3ec25918e84b3d83763f1619e6a58f", "competition_source_list": ["2017年IMAS小学高年级竞赛（第一轮）第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有浓度为$$3.2 \\%$$的食盐水$$500\\text{g}$$，当水全部蒸发掉后，请问剩下多少$$\\text{g}$$食盐？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$32$$ "}], [{"aoVal": "C", "content": "$$64$$ "}], [{"aoVal": "D", "content": "$$100$$ "}], [{"aoVal": "E", "content": "$$128$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["由题意可知盐水中含有盐$$500\\times 3.2 \\%=16\\text{g}$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1346", "queId": "246395590d814e1b96a944ca38101e29", "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4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一年六月份中的星期五比星期四多$$1$$个，那么这年七月一日为． ", "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": ["拓展思维->能力->运算求解"], "answer_analysis": ["星期五比星期四多$$1$$个，说明$$6$$月$$1$$号是星期五，那么七月一日是$$30$$天之后，$$30\\div 7=4$$（组）$$\\cdots \\cdots 2$$（天），说明七月一日是星期日． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "624", "queId": "b4a8ca17327a4acbbfedd0227c7b4933", "competition_source_list": ["2018年美国数学大联盟杯三年级竞赛初赛第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "两个奇数之间的差不可能是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["本题题意为``两个奇数之间的差不可能是多少''，两个奇数之间的差结果是偶数，因此选项中$$\\text{C}$$选项不可能是两个奇数相减得来，因此本题选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1734", "queId": "73024699c1474ff69068e73afdcc62a9", "competition_source_list": ["2016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": "一种盐水的浓度是$$20  \\% $$，加入$$800$$克水后，它的浓度变为$$12  \\% $$，这种盐水溶液原来有克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$300$$ "}], [{"aoVal": "B", "content": "$$800$$ "}], [{"aoVal": "C", "content": "$$1200$$ "}], [{"aoVal": "D", "content": "$$2400$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设原来有$$x$$克，  ∵盐水的浓度为$$20  \\% $$，  ∴盐水溶液中，  有水：$$x\\cdot \\left( 1-20 \\% \\right)=0.8x$$克，  有盐：$$x\\cdot 20 \\%=0.2x$$克，  加入$$800$$克的水后浓度变为$$12  \\% $$，  ∴$$\\left( x+800 \\right)\\cdot 12 \\%=0.2x$$，  解得：$$x=1200$$，  ∴盐水溶液原来有$$1200$$克． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1267", "queId": "5deddb6a01b044d59dc8a7e9e0d9193d", "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->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$$2019$$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": "1252", "queId": "74e2af09ef0c47dea1363e7d2f3211c5", "competition_source_list": ["2011年全国迎春杯五年级竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "猪八戒的糖是巧克力的$$3$$倍，若吃了$$24$$颗，则巧克力是糖的$$3$$倍，那么猪八戒原巧克力和糖一共~\\uline{~~~~~~~~~~}~颗． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->逐步调整思想"], "answer_analysis": ["列方程，设原来巧克力为$$x$$，则糖原来有$$3x$$，吃了 后为$$\\left( 3x-24 \\right)$$，  $$3\\left( 3x-24 \\right)=x$$，求得$$x=9$$人，原来糖为$$9\\times 3=27$$，共$$9+27=36$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "798", "queId": "ff80808146dc29ee0146e1142409079c", "competition_source_list": ["2014年全国华杯赛小学高年级竞赛初赛B卷第2题", "2016年吉林长春二道区吉大附中力旺实验小学小升初第5题2分"], "difficulty": "3", "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$$0$ $9$$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->思想->逐步调整思想"], "answer_analysis": ["首先可以确定的是$$E\\times F=0$$，$$\\Rightarrow$$ $$E$$，$$F$$中必有一个是$$0$$．那么$$I+J=4$$，$$\\Rightarrow I$$，$$J$$只能为$$1$$，$$3$$；此时剩下的数字还有$$2$$，$$4$$，$$5$$，$$6$$，$$7$$，$$8$$，$$9$$．$$G-H=1$$，$$\\Rightarrow$$ $$G$$，$$H$$相差$$1$$；讨论如下  若$$\\overline{AB}=54\\Rightarrow \\overline{CD}=27$$，那么$$G$$，$$H$$为$$8$$，$$9$$  若$$\\overline{AB}=58\\Rightarrow \\overline{CD}=29$$，那么$$G$$，$$H$$为$$6$$，$$7$$  若$$\\overline{AB}=92\\Rightarrow \\overline{CD}=46$$，那么$$G$$，$$H$$为$$7$$，$$8$$  若$$\\overline{AB}=96\\Rightarrow \\overline{CD}=48$$，此时$$G$$，$$H$$无法取值．所以$$\\overline{AB}\\ne96$$，选D． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1112", "queId": "385697a9bbe8413ca971b6e904103e79", "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$$42$$美元．请问约翰有美元. ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$31$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->实践应用"], "answer_analysis": ["$$(42+20)\\div 2=31$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "269", "queId": "5203591737a34bb5be97b00ca7534cd9", "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": "$$A$$、$$B$$、$$C$$三人进行乒乓球赛，两人比赛一人观战，每赛一场后胜者继续打，负者换另一个人上场，一直这样打下去，结果$$A$$打胜了$$11$$场，$$B$$胜了$$9$$场，$$C$$胜了$$7$$场，那么$$C$$总共打了（~ ）场比赛． ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$27$$ "}]], "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": ["假设从$$A$$、$$B$$、$$C$$三人各自获胜次数中分别扣除$$1$$场构造，$$A$$胜$$B$$，$$B$$胜$$C$$，$$C$$胜$$A$$，此时三人分别已赛$$2$$场．那么$$A$$、$$B$$、$$C$$三人分别获胜次数还剩下$$10$$场、$$8$$场、$$6$$场．可知$$B$$与$$C$$分别输给$$A:10\\div 2=5$$场，$$A$$与$$C$$分别输给$$\\text{B:}8\\div 2=4$$场．$$A$$与$$B$$分别输给$$\\text{C:}6\\div 2=3$$场．综上可知：  $$A$$共赛：$$2+10+4+3=19$$场  $$B$$共赛：$$2+8+5+3=18$$场  $$C$$共赛：$$2+6+5+4=17$$场 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1133", "queId": "263aa81760d24597a97df0ec9934ae32", "competition_source_list": ["2020年第1届广东深圳超常思维竞赛六年级竞赛初赛第24题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "新建成的芜宣机场停有$$10$$架飞机，第一架飞机起飞后，每隔$$40$$分钟，有一架飞机起飞，在第一架飞机起飞$$20$$分钟后，有一架飞机降落该机场，以后每隔$$1$$个小时有一架飞机降落，降落的飞机在原有的$$10$$架飞机起飞后，又依次每隔$$40$$分钟起飞一架，那么从第一架飞机起飞后，经过小时，飞机场第一次出现没有飞机的现象． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$16\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$17\\frac{1}{3}$$ "}], [{"aoVal": "E", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["当第一架飞机起飞后，后面只要出现起飞的飞机数比降落的飞机数多$$9$$，  则机场会第一次出现没有飞机的情况，  设$$x$$分钟后达到要求，  $$\\left( 1+\\frac{x}{40} \\right)-\\left( \\frac{x-20}{60}+1 \\right)=9$$，  所以：$$x=1040$$，  $$1040$$分钟$$=17 \\frac{1}{3}$$小时，  则在第$$18$$小时的时候，刚好满足题目要求．  故选$$\\text{E}$$． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "163", "queId": "239e13e528674e708128af78b763dd19", "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，但比甲、丁多．甲没有丁多．按苹果由少到多的顺序排列为． ", "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": ["由题意可以知道：乙\\textgreater 甲、丁，但乙不是最多的，那么最多的只能是丙；  再有丁\\textgreater 甲，可以得到由少到多的排序：甲、丁、乙、丙. "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2123", "queId": "f0a1187fcbe04a61a4783c47b132ed93", "competition_source_list": ["2017年河南郑州豫才杯六年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "在书写日期时，有人将$$3$$月$$8$$日写成$$3$$/$$8$$，有些人则写成$$8$$/$$3$$，这样会造成些混淆，因为当我们看到$$3$$/$$8$$时，不知到底是指$$8$$月$$3$$日，还是指$$3$$月$$8$$日，但是$$13$$/$$6$$及$$6$$/$$13$$则很容易区分，因为一年中只有$$12$$个月．使用这种写法，一年中（~ ）天会造成混淆． ", "answer_option_list": [[{"aoVal": "A", "content": "$$132$$ "}], [{"aoVal": "B", "content": "$$130$$ "}], [{"aoVal": "C", "content": "$$131$$ "}], [{"aoVal": "D", "content": "$$133$$ "}]], "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": ["$$1$$月会造成混淆的日期有：$$1$$月$$2$$日、$$1$$月$$3$$日、$$\\cdot \\cdot \\cdot $$、$$1$$月$$11$$日、$$1$$月$$12$$日，一共有$$11$$天；$$2$$月、$$3$$月、$$4$$月、$$\\cdot \\cdot \\cdot $$、$$12$$月会造成混淆的日期各有$$11$$天，所以一年中有$$11\\times 12=132$$（天）会造成混淆． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3015", "queId": "aa601abe63cc49a8b9bf2f78e859c9f9", "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18题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "观察$$1$$，$$2$$，$$3$$，$$6$$，$$12$$，$$23$$，$$44$$，☆，$$164$$的规律，☆表示的数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$67$$ "}], [{"aoVal": "B", "content": "$$85$$ "}], [{"aoVal": "C", "content": "$$89$$ "}], [{"aoVal": "D", "content": "$$92$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["暂无 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2548", "queId": "947d983fa39142d38c75dc47586c98e4", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第11题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "规定$$*$$是一种运算符号，若$$4*2=82$$、$$6*3=183$$、$$8*4=324$$、$$9*3=276$$、$$9*5=454$$，请问$$10*2$$是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$55$$ "}], [{"aoVal": "B", "content": "$$125$$ "}], [{"aoVal": "C", "content": "$$202$$ "}], [{"aoVal": "D", "content": "$$208$$ "}], [{"aoVal": "E", "content": "$$2002$$ "}]], "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": ["由题意典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差．  因$$10\\times 2=20$$、$$10-2=8$$，  所以$$10*2=208$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2169", "queId": "342f97a1daf8455d9929288f242da49c", "competition_source_list": ["2016年第14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": "甲、乙二人分别从$$A$$、$$B$$两地同时出发，相向而行、往返跑步．以每分钟跑$$300$$米，甲每分钟跑$$240$$米．如果他们的第$$12$$次迎面相撞点与第$$13$$次迎面相遇点相距为$$300$$米，则$$A$$、$$B$$两点间的距离是（~ ）米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$400$$ "}], [{"aoVal": "B", "content": "$$450$$ "}], [{"aoVal": "C", "content": "$$500$$ "}], [{"aoVal": "D", "content": "$$550$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["由于二者速度比为$$300:240=5:4$$，则两地相遇$$300\\div 6\\times 9=450$$米． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1572", "queId": "ff8080814518d5240145190918fb030d", "competition_source_list": ["2014年全国迎春杯五年级竞赛初赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "为了减少城市交通拥堵的情况，某城市拟定从$$2014$$年$$1$$月$$1$$日起开始试行新的限行规则，规定尾号为$$1$$、$$6$$的车辆周一、周二限行，尾号$$2$$、$$7$$的车辆周二、周三限行，尾号$$3$$、$$8$$的车辆周三、周四限行，尾号$$4$$、$$9$$的车辆周四、周五限行，尾号$$5$$、$$0$$的车辆周五、周一限行，周六、周日不限行．由于$$1$$月$$31$$日是春节，因此，$$1$$月$$30$$日和$$1$$月$$31$$日两天不限行．已知$$2014$$年$$1$$月$$1$$日是周三并且限行，那么$$2014$$年$$1$$月份（~~~~~~~ ）组尾号可出行的天数最少． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$、$$6$$ "}], [{"aoVal": "B", "content": "$$2$$、$$7$$ "}], [{"aoVal": "C", "content": "$$4$$、$$9$$ "}], [{"aoVal": "D", "content": "$$5$$、$$0$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["$$1$$月份共$$31$$天，由于$$1$$月$$1$$日是周三，所以$$1$$月份周三、周四、周五共$$5$$天，周一、周二共$$4$$天．其中$$1$$月$$30$$日周四、$$1$$月$$31$$日周五．所以只看周三即可．周三$$2$$、$$7$$以及$$3$$、$$8$$限行．所以此题B组尾号可出行的天数最少． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1666", "queId": "4eda6ff8edc34e658a8c7075d9aaaf58", "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4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "小丽有$$17$$朵花，她给小明$$5$$朵后，两人的花就同样多．小明原来有朵花． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$12$$ "}]], "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": ["小明接受小丽的$$5$$朵花之后有：$$17-5=12$$（朵）．  所以小明原来有：$$12-5=7$$（朵）．  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": "662", "queId": "6711f482e8374424929588a0c663eeb1", "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复赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "把数字1、2、3、6、7分别写在5张卡片上，从中任意取出两张拼成两位数，其中写着6的一张可以当9用.那么在这些拼成的两位数中，质数的个数是(  ) ", "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->数论模块->质数与合数->特殊质数运用->常用质数"], "answer_analysis": ["质数有13、17、19、23、29、31、37、61、67、71、73、79、97共13个。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "639", "queId": "74bac41bb4734c4c9149485008d6d93a", "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2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "两个数的最大公因数是$$5$$，最小公倍数是$$75$$，那么这两个数的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$80$$ "}], [{"aoVal": "C", "content": "$$40$$或$$50$$ "}], [{"aoVal": "D", "content": "$$40$$或$$80$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["枚举：$$5$$、$$75$$；$$15$$、$$25$$满足和为$$80$$或40 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "395", "queId": "7cc854c45f5e4ec2aa0abbe4f6e3d991", "competition_source_list": ["2014年全国迎春杯小学中年级四年级竞赛初赛第9题"], "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，从第一间房间里的人开始，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": "$$0$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$11$$ "}], [{"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": ["第一个人看别的房间，开灯的$$9$$间，关灯的$$10$$间，所以会关灯．  第二个人看别的房间关灯的至少$$10$$间，开灯的至多$$9$$间，所以会关灯．  第三个人看别的房间，关灯的至少$$10$$间，所以会关灯．  第四个人看别的房间，关灯的至少$$10$$间，所以也会关灯．  $$\\cdots\\cdots$$  所以最后所有房间均为关灯． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2293", "queId": "e951e978b3ed4871807c36d3d432b540", "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": "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->单人简单行程问题"], "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": "687", "queId": "d567e459c1f6405783c1ed8d435c7a7b", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "今有物不知其数，三三数之剩二，五五数之剩三，七七数之剩二，问物最少几何？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$26$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->数论模块->余数问题->余数的性质"], "answer_analysis": ["此数除以$$3$$余$$2$$，除以$$5$$余$$3$$，除以$$7$$余$$2$$，满足条件最小数是$$23$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "627", "queId": "220f842994154ce78269e298ab9358ed", "competition_source_list": ["2010年全国华杯赛竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "恰有$$20$$个因数的最小自然数是（~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$240$$ "}], [{"aoVal": "C", "content": "$$360$$ "}], [{"aoVal": "D", "content": "$$432$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$20=1\\times20=2\\times 10=4\\times 5=2\\times 2\\times 5$$ 四种情况下的最小自然数分别为：$${{2}^{19}}$$、$${{2}^{9}}\\times 3$$、$${{2}^{4}}\\times {{3}^{3}}$$、$${{2}^{4}}\\times 3\\times 5$$，其中最小的是最后一个，为$$240$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "938", "queId": "f377e08dcb4c43db8f2033d94f0407fe", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（二）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$1\\sim 200$$这$$200$$个自然数中既不是$$3$$的倍数，又不是$$5$$的倍数的数从小到大排成一排，那么第$$100$$个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$188$$ "}], [{"aoVal": "B", "content": "$$187$$ "}], [{"aoVal": "C", "content": "$$184$$ "}], [{"aoVal": "D", "content": "$$182$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$1$$至$$200$$的自然数中，是$$3$$的倍数的有$$66$$个，是$$5$$的倍数的数有$$40$$人，既是$$3$$又是$$5$$的倍数的数有$$13$$个．  所以从$$1$$至$$200$$的自然数中是$$3$$或$$5$$的倍数的数有$$66+4013=93$$个，所以从$$1$$至$$200$$的这$$200$$的自然数中，既不是$$3$$又不是$$5$$的倍数的数有$$20093=107$$个．  从小到大第$$100$$个即倒数第$$8$$个，将这些数从大到小列出：$$199$$、$$197$$、$$196$$、$$194$$、$$193$$、$$191$$、$$188$$、$$187\\ldots \\ldots $$故从小到大排列第$$100$$个数是$$187$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "804", "queId": "ed208f240a4c43ff8d3913f268707ff5", "competition_source_list": ["2009年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2008200920$$$$\\square \\square $$这个$$12$$位数能被$$36$$整除，这个数的个位数最小是(  )。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "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"], "answer_analysis": ["由于$$12$$位数$$2008200920$$□□能被$$36$$整除，且$$36=4\\times 9$$，所以这个数能被$$9$$和$$4$$整除，设此数的最后两个数字为$$a$$、$$b$$，则$$\\overline{ab}$$是$$4$$的倍数，且此$$12$$位数的数字和$$23+a+b$$为$$9$$的倍数，从而$$a+b=4$$或$$13$$，且$$\\overline{ab}$$为$$4$$的倍数。符合这些条件的个位数$$b$$最小值为$$0$$ $$\\left(a=4,b=0\\right)$$。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "935", "queId": "bcf22945d87d4bdcb87f458b3e27fea1", "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$$6$$个正整数的最小公倍数有多少个正因数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "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": ["前$$6$$分正整数的最小公倍数是：$$2\\times 3\\times 2\\times 5$$，因数个数:$$3\\times 2\\times 2=12$$个． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3385", "queId": "85e6f250cb404e80b28284632d297c8f", "competition_source_list": ["2020年新希望杯六年级竞赛（2月）第22题", "2020年希望杯六年级竞赛模拟第22题"], "difficulty": "1", "qtype": "single_choice", "problem": "疯狂动物城的雨林学校、沙漠学校、冰原学校组织春节联欢晚会，共演出$$15$$个节目，如果每校至少演出$$3$$个节目，那么这三所学校演出节目数的不同情况共有~\\uline{~~~~~~~~~~}~种． ", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用", "Overseas Competition->知识点->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": ["$$15=3+3+9\\to3$$种，  $$=3+4+8\\to6$$种，  $$=3+5+7\\to6$$种，  $$=3+6+6\\to3$$种，  $$=4+4+7\\to3$$种，  $$=4+5+6\\to6$$种，  $$=5+5+5\\to1$$种，  共计：$$3+6+6+3+3+6+1=28$$（种）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "963", "queId": "fde87b01e4ac4dcdbeb52365b4800fc5", "competition_source_list": ["2017年迎春杯五年级竞赛", "2017年迎春杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "村长让喜羊羊买东西，他把若干个羊币分成$$6$$份，分别包在$$6$$个小纸包里，村长告诉喜羊羊，从$$1$$羊币到$$63$$羊币，不管是多少，都能从这$$6$$包中挑出一包或几包来支付并且不用找钱；那么这$$6$$包中最大的那一包装了（  ）个羊币。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$63$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "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": ["$${{2}^{6}}\\text{=64}$$，可以转化成二进制来解决，那么这$$6$$包所包装的羊币分别为$$1,2,4,8,16,32,$$则可以满足$$1$$个羊币到$$63$$个羊币都可取出，最大的那一包装了$$32$$个羊币。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "100", "queId": "22178b8691b241168650d4fe553bab4b", "competition_source_list": ["2016年全国迎春杯小学中年级四年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "某班$$43$$名同学围成一圈．由班长起从$$1$$开始连续报数，谁报到$$100$$，谁就表演一个节目；然后再 由这个同学起从$$1$$开始连续报数，结果第一个表演节目的是小明，第二个演节目的是小强．那么小明和小强之间有~\\uline{~~~~~~~~~~}~名同学． ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$或$$30$$ "}], [{"aoVal": "B", "content": "$$12$$或$$29$$ "}], [{"aoVal": "C", "content": "$$13$$或$$30$$ "}], [{"aoVal": "D", "content": "$$13$$或$$29$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["显然小明是$$100$$号，小强是$$199$$号（本题的陷阱）  $$100\\div 43=2\\cdots \\cdots 14$$  $$199\\div 43=4\\cdots \\cdots 27$$  所以，小明和小强间有$$15$$号至$$26$$号共$$12$$人，或者$$43-12-2=29$$人． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2639", "queId": "4cbc882df3e74b5298b6d7b423c06cc3", "competition_source_list": ["2016年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "算式$$\\underbrace{999\\cdots 9}_{2016\\text{个}9}\\times \\underbrace{999\\cdots 9}_{2016\\text{个}9}$$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": ["课内体系->知识模块->数与代数", "拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数构造提取->整数拆补构造"], "answer_analysis": ["把$$\\underbrace{999\\cdots 9}_{2016\\text{个}9}$$变形为$$1\\underbrace{000\\cdots 0}_{2016\\text{个}0}-1$$，然后根据乘法的分配律拆分，再进一步解答即可。  解：$$\\underbrace{999\\cdots 9}_{2016\\text{个}9}\\times \\underbrace{999\\cdots 9}_{2016\\text{个}9}$$  $$=(1\\underbrace{000\\cdots 0}_{2016\\text{个}0}-1)\\times \\underbrace{999\\cdots 9}_{2016\\text{个}9}$$  $$=1\\underbrace{000\\cdots 0}_{2016\\text{个}0}\\times \\underbrace{999\\cdots 9}_{2016\\text{个}9}-\\underbrace{999\\cdots 9}_{2016\\text{个}9}$$  $$=\\underbrace{999\\cdots 000}_{2016\\text{个}9\\text{和}0}-\\underbrace{999\\cdots 9}_{2016\\text{个}9}$$  个位$$0$$减$$9$$不够减，需要连续退位，个位数得$$1$$，所以数字$$0$$的个数是：  $$2016-1=2015$$（个）。  故选：C "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3299", "queId": "c7f3ba78b34e489aa2e87a1a57711930", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛B卷第9题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "随机投掷一枚均匀的硬币两次，两次都正面朝上的概率是~\\uline{~~~~~~~~~~}~. ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "E", "content": "$$\\frac25$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想", "Overseas Competition->知识点->计数模块->统计与概率->概率"], "answer_analysis": ["因为由题干可知，现随机投掷一枚均匀的硬币两次，  则可能的情况为：``正正''、``正负''、``负正''、``负负''，  所以两次都正面朝上的概率是$$\\frac{1}{4}$$，  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1881", "queId": "f742af557ffb452a8ff20237f785eb20", "competition_source_list": ["2013年第11届创新杯三年级竞赛初赛第1题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "在做一道加法算式时，小芳把一个加数个位上的$$6$$看成了$$9$$．把另一个加数十位上的$$3$$看成了$$5$$，结果算成$$120$$，正确答案应该是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$115$$ "}], [{"aoVal": "B", "content": "$$97$$ "}], [{"aoVal": "C", "content": "$$125$$ "}], [{"aoVal": "D", "content": "$$143$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->还原问题->小马虎题型"], "answer_analysis": ["个位多加了︰$$9-6=3$$，  十位多加了︰$$50-30=20$$，  正确结果是：$$120-20-3=97$$．  答︰正确结果是$$97$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2911", "queId": "c907ea7f9b5a44c4981b906a3ada0fb9", "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": "$$T\\textgreater R\\textgreater S\\textgreater P\\textgreater Q$$ "}], [{"aoVal": "B", "content": "$$P\\textgreater S\\textgreater T\\textgreater Q\\textgreater R$$ "}], [{"aoVal": "C", "content": "$$S\\textgreater R\\textgreater T\\textgreater Q\\textgreater P$$ "}], [{"aoVal": "D", "content": "$$S\\textgreater Q\\textgreater P\\textgreater T\\textgreater R$$ "}]], "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": "2988", "queId": "c9a21ab35f9e4dc794d6e787b1d20a03", "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": ["Overseas Competition->知识点->计算模块", "拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left(2^{2} \\times 2^{4} \\times 2^{6}\\times \\ldots ~\\times 2^{98} \\times 2^{100}\\right) \\div \\left(2^{1} \\times 2^{3} \\times 2^{5}\\times \\ldots ~\\times 2^{97} \\times 2^{99}\\right)$$  $$=\\left( {{2}^{2}}\\div {{2}^{1}} \\right)\\times \\left( {{2}^{4}}\\div {{2}^{3}} \\right)\\times \\left( {{2}^{6}}\\div {{2}^{5}} \\right)\\times \\ldots \\times \\left( {{2}^{98}}\\div {{2}^{97}} \\right)\\times \\left( {{2}^{100}}\\div {{2}^{99}} \\right)$$  $$=\\underbrace{2\\times 2\\times 2\\times \\cdots \\times 2}_{\\text{fifty 2s}}$$  $$={{2}^{50}}$$．  The answer is $$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1134", "queId": "3d16b66bbb9249b9898777fe8018003c", "competition_source_list": ["2016年IMAS小学高年级竞赛第一轮检测试题第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知汽油的价格为每公升$$6$$元，小汽车每行驶$$100\\text{km}$$需耗油$$8$$公升，请问小汽车加油$$200$$元最多能行驶多少$$\\text{km}$$？（结果只取整数部分） ", "answer_option_list": [[{"aoVal": "A", "content": "$$416$$ "}], [{"aoVal": "B", "content": "$$417$$ "}], [{"aoVal": "C", "content": "$$418$$ "}], [{"aoVal": "D", "content": "$$419$$ "}], [{"aoVal": "E", "content": "$$420$$ "}]], "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->多量和倍问题->多量和倍"], "answer_analysis": ["$$100\\div 8=12.5$$（千米/公升）  $$200\\div 6\\times 12.5= 416.\\overset{\\cdot }{\\mathop{6}} ,$$（千米），用去尾法最多可行驶$$416$$千米． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2366", "queId": "130d90e37aac4a5c95742179fb82a192", "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复赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在自然数1，2，3，$$\\cdots$$，2008中，末位是3的所有数的和是(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "201603 "}], [{"aoVal": "B", "content": "201703 "}], [{"aoVal": "C", "content": "201803 "}], [{"aoVal": "D", "content": "201903 "}]], "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": ["$$3+13+23+\\cdots +2003=\\left( 1+2+\\cdots +200 \\right)\\times 10+3\\times 201=201603$$. "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "463", "queId": "b7b979daf316487d86e13039a9ab8c1b", "competition_source_list": ["2008年六年级竞赛学而思杯"], "difficulty": "1", "qtype": "single_choice", "problem": "古时候的一场大地震几乎毁灭了整个人类，只有两个部落死里逃生．最初在这两个部落中，神帝部落所有的人都坚信人性本恶，圣地部落所有的人都坚信人性本善，并且没有既相信人性本善又相信人性本恶的人存在．后来两个部落繁衍生息，信仰追随和部落划分也遵循着一定的规律．部落内通婚，所生的孩子追随父母的信仰，归属原来的部落；部落间通婚，所生孩子追随母亲的信仰，归属母亲的部落．我们发现神圣子是相信人性本善的．在以下各项对神圣子身份的判断中，不可能为真的是． ", "answer_option_list": [[{"aoVal": "A", "content": "神圣子的父亲是神帝部落的人 "}], [{"aoVal": "B", "content": "神圣子的母亲是神帝部落的人 "}], [{"aoVal": "C", "content": "神圣子的父母都是圣地部落的人 "}], [{"aoVal": "D", "content": "神圣子的母亲是圣地部落的人 "}], [{"aoVal": "E", "content": "神圣子的姥姥是圣地部落的人 "}]], "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": ["不管是部落间还是部落内通婚，生下的孩子都会追随母亲的信仰，由于神圣子坚信人性本善，所以他的母亲一定是圣地部落的． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2320", "queId": "d8a032cfbef84fa3aba2dc3c32d996e2", "competition_source_list": ["2020年新希望杯四年级竞赛初赛（个人战）第7题", "2020年新希望杯四年级竞赛决赛（8月）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": "$$20$$ "}], [{"aoVal": "B", "content": "$$17.5$$ "}], [{"aoVal": "C", "content": "$$22$$ "}], [{"aoVal": "D", "content": "$$25$$ "}], [{"aoVal": "E", "content": "$$27.5$$ "}]], "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": ["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": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "693", "queId": "70c48fa563e74e6eae7d5f7ebe163322", "competition_source_list": ["2006年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "某加油站有两位员工，从今年规定，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（  ）天要聘临时工。 ", "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->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": ["甲每工作$$3$$天后休息$$1$$天，共$$3+1=4$$天，乙每工作$$5$$天后休息$$2$$天，共$$5+2=7$$天，$$4\\times 7=28$$天，在$$28$$天中甲和乙有$$2$$天同休，又$$365\\div 28=13\\cdots\\cdots 1$$，（如果是$$366$$天就余$$2$$天），所以一年中有$$2\\times 13=26$$天要聘临时工。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1404", "queId": "31e99edb3ec44b15b5b46f1329733a05", "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1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "把一根粗细均匀的木料锯成$$7$$段，每锯一次需要$$4$$分钟，一共要分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$20$$ "}]], "knowledge_point_routes": ["知识标签->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$$7$$段需要锯$$6$$次，所以一共需要的是时间为：$$6\\times 4=24$$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": "2003", "queId": "d39971e4ac9142bfb018c41802cf8901", "competition_source_list": ["2009年全国迎春杯五年级竞赛初赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "某班女同学人数是男同学的$$2$$倍，如果女同学的平均身高是$$150$$厘米，男同学的平均身高是$$162$$厘米．那么全班同学的平均身高是~\\uline{~~~~~~~~~~}~厘米． ", "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->七大能力->运算求解"], "answer_analysis": ["由于女同学人数是男同学的$$2$$倍，所以将男同学人数看做一份，则女同学就有$$2$$份，男女生共$$3$$份．$$(150\\times 2+162)\\div 3=154$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "618", "queId": "1d7358037c2d4bda9a78dc52fccea8e6", "competition_source_list": ["2014年华杯赛四年级竞赛初赛"], "difficulty": "2", "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->数论模块->整除->整除特征->尾数系整除特征"], "answer_analysis": ["能被$$4$$整除的数末两位必能被$$4$$整除，则除以$$4$$余$$1$$的数末两位必为$$33$$，这样的九位数前七位由$$6$$个$$3$$和$$1$$个$$0$$组成，共有$$6$$种不同组合方式。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1574", "queId": "ff8080814518d5240145190945380314", "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． ", "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": ["$$2000\\div (4+3+2+1)=200$$（组）． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "223", "queId": "67e9d71a4a194a34b102220ba55c8301", "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9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个不透明的盒子里装着一些同样大小的玻璃球，其中有$$4$$个黄玻璃球，$$5$$个白玻璃球，$$6$$个红玻璃球，至少摸出个才能确保摸到的有黄玻璃球． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["从题干中可以知道，一个不透明的盒子里装着一些同样大小的玻璃球，  其中有$$4$$个黄玻璃球，$$5$$个白玻璃球，$$6$$个红玻璃球，  根据最不利原则，有其他颜色球存在的情况下，都不会摸到黄玻璃球，  其他颜色球一共：$$5+6=11$$（个），  所以摸到第$$12$$个球时一定是黄玻璃球．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "371", "queId": "c8be181dd0634c86a0ef5da864772141", "competition_source_list": ["2008年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "电视台要播放一部$$32$$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（  ）天。 ", "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->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$$4$$集未播出，由于已有过一天播出$$4$$集的情况，因此，这余下4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$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$11$$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": "2912", "queId": "7406988fbd8c49808ede9babed772f8d", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$2.020\\times2.0212021-2.021\\times2.0202020=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2.020$$ "}], [{"aoVal": "D", "content": "$$2.021$$ "}], [{"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": ["原式$$=2.020\\times2.021\\times10001-2.021\\times2.020\\times10001=0$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2278", "queId": "85af9e066aa5452d89d3768412455d91", "competition_source_list": ["2015年美国数学大联盟杯六年级竞赛初赛第35题5分(每题5分)"], "difficulty": "1", "qtype": "single_choice", "problem": "船长们划船$$5$$个小时．从第二个小时开始，每个小时行驶的距离比前一个小时少了$$2$$千米．如果$$5$$个小时的平均行驶速度是$$6$$千米$$/$$小时，那么前$$2$$个小时行驶了多少千米？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["设第$$1$$个小时走$$x\\text{km}$$，  则可列方程：  $$\\begin{eqnarray} x+x-2+x-4+x-6+x-8\\&=\\&5\\times 6   5x-20\\&=\\&30   5x\\&=\\&50   x\\&=\\&10．\\end{eqnarray}$$  前两个小时走了$$10+10-2=18$$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": "2044", "queId": "e646c1e2e147477e9843848c8dbcad64", "competition_source_list": ["2004年希望杯五年级竞赛复赛", "2004年第2届希望杯五年级竞赛复赛第3题", "2004年希望杯一年级竞赛复赛", "2004年希望杯二年级竞赛复赛", "2004年希望杯三年级竞赛复赛", "2004年希望杯六年级竞赛复赛", "2004年希望杯四年级竞赛复赛"], "difficulty": "1", "qtype": "single_choice", "problem": "在一列数$$9$$、$$2$$、$$1$$、$$3$$、$$4$$、$$7$$、$$1$$\\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": "$$11$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}]], "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": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2429", "queId": "3cef69a301df4ce2baa92e49ee63f636", "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"], "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": "3337", "queId": "ff80808147342b7e01474868ebad2592", "competition_source_list": ["2019年第24届YMO五年级竞赛决赛第8题3分", "2019年四川绵阳涪城区绵阳东辰国际学校小升初（十三）第9题3分", "2013年全国华杯赛小学高年级竞赛初赛A卷第4题"], "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->计数模块->统计与概率->概率->典型问题->摸小球"], "answer_analysis": ["方法一：  取出一粒黑子，总数少一粒棋子，黑子数占全部的$$\\frac{9}{16}$$，取出一粒白子，总数少一粒棋子，黑子数占全部的$$\\frac{7}{12}$$，$$1\\div (\\frac{7}{12}-\\frac{9}{16})=48$$（粒）盒子里原有黑子数比白子数多$$48\\times (\\frac{9}{16}-\\frac{7}{16})+1=7$$（粒）  方法二：方程法  设白棋有$$7x$$粒，黑棋有$$9x+1$$粒，列方程为$$\\frac{7x-1}{9x+1}=\\frac{7}{5}$$；$$x=3$$，白棋有$$21$$粒，黑棋有$$28$$粒，盒子里原有黑子数比白子数多$$28-21=7$$（粒）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3287", "queId": "561971a1081749fb8949591d3b214140", "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，那么不能作为小龙$$3$$个月读书数量的总和． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["利用$$10$$以内质数$$2$$、$$3$$、$$5$$、$$7$$进行枚举分析：$$10=2+3+5$$，$$12=2+3+7$$，$$15=3+5+7$$，只有$$13$$不可以． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2204", "queId": "35c41a4873e54e37900b0843309cf2cd", "competition_source_list": ["2007年第5届创新杯五年级竞赛第10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$，$$B$$两地相距$$950$$米，甲、乙两人同时由$$A$$出发，在$$A$$，$$B$$两地间往返．甲步行每分钟$$40$$米，乙跑步每分钟$$150$$米，甲、乙两人第二次相遇时相距$$B$$地有米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$120$$ "}], [{"aoVal": "C", "content": "$$140$$ "}], [{"aoVal": "D", "content": "$$150$$ "}]], "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->往返相遇"], "answer_analysis": ["$$950\\times2=1900$$（米），  $$1900\\div (40+150)=10$$（分钟），  $$10\\times2=20$$（分钟），  $$20\\times40=800$$（米），  $$950-800=150$$（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": "972", "queId": "fe7e8d3d2a354176914e8137b28ecabf", "competition_source_list": ["2005年第3届创新杯五年级竞赛初赛第1题"], "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->思想->转化与化归的思想"], "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$$a，b$$互质)，那么$$15\\times a\\times b=360$$，即$$a\\times b=24$$，又$$a，b$$互质，则$$( a$$，$$b )=( 1$$，$$24 )$$或$$( 3$$，$$8 )$$.当$$( a$$，$$b )=( 1$$，$$24 )$$时，这两个整数的和为$$15\\times 1+15\\times 24=15\\times 25=375$$；当$$( a$$，$$b )=( 3$$，$$8 )$$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$$\\rm B$$符合结论，故选$$\\rm B$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1405", "queId": "a2b3895812a44e8997d7219f7820f08c", "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->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": "3158", "queId": "337624fc47c04539983ed90f06e28719", "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6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "用数字$$0$$、$$3$$、$$4$$、$$5$$、$$6$$最多可以组成个无重复数字的三位数． ", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$64$$ "}], [{"aoVal": "D", "content": "$$100$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["∵首先百位数不能选$$0$$有$$4$$种选择，  当百位数确定后十位数有$$4$$种选择，  当十位数确定后个位数有$$3$$种选择，  ∴根据分步计数原理所组成的三位数为$$4\\times4\\times3=48$$种．  故选：$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2182", "queId": "303d3cc8af124d37b514cd7a13fa8475", "competition_source_list": ["2010年第8届创新杯六年级竞赛初赛第7题4分", "2010年六年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "一架小飞机，在静止的空气中飞行速度是$$320$$千米/时。现在有风，风速为$$40$$千米/时(风速不变)，逆风飞行全程需要时间$$135$$分钟，顺风返回时需要(  )分钟。(飞机起飞和着陆的时间略去不计) ", "answer_option_list": [[{"aoVal": "A", "content": "94.5 "}], [{"aoVal": "B", "content": "105 "}], [{"aoVal": "C", "content": "112.5 "}], [{"aoVal": "D", "content": "120 "}]], "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->基本关系"], "answer_analysis": ["设顺风的速度为$${{v}_{1}}$$千米/时，逆风的速度为$${{v}_{2}}$$千米/时，则$${{v}_{1}}$$:$$320+40=360$$(千米/时)，$${{v}_{2}}$$:$$320-40=280$$(千米/时)。 $${{v}_{1}}:{{v}_{2}}=360:280=9:7$$，所以$${{t}_{1}}:{{t}_{2}}=7:9$$，顺风返回要:$$135\\times \\frac{7}{9}=105$$(分钟)。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2889", "queId": "858f9422fa3246d2a63eb54e55c79433", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（一）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a\\times 15\\times 1=b\\times \\frac{2}{3}\\div \\frac{3}{4}\\times 15=c\\times 15.2\\div \\frac{4}{5}=d\\times 14.8\\times \\frac{73}{74}$$，$$a$$、$$b$$、$$c$$、$$d$$四个数由  大到小排列为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$b\\textgreater d\\textgreater a\\textgreater c$$ "}], [{"aoVal": "B", "content": "$$b\\textgreater a\\textgreater c\\textgreater d$$ "}], [{"aoVal": "C", "content": "$$c\\textgreater a\\textgreater b\\textgreater d$$ "}], [{"aoVal": "D", "content": "$$c\\textgreater a\\textgreater d\\textgreater b$$ "}]], "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->两数相除法"], "answer_analysis": ["设它们的值均为$$1$$，由题设条件知：$$15a=\\frac{40}{3}\\times b=19\\times c=14.6\\times d=1$$，又$$19\\textgreater15\\textgreater14.6\\textgreater\\frac{40}{3}$$，所以$$b\\textgreater d\\textgreater a\\textgreater c$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3361", "queId": "966389d075a74364b0b26ab740353773", "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9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙三人都在读同一本故事书，书中有$$100$$个故事．已知甲读了$$85$$个故事，乙读了$$70$$个故事，丙读了$$62$$个故事．甲、乙、丙三人都读过的故事最少有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$62$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["要求三人都读过的故事最少有多少个，  则三人读的故事错开越多越好，不妨令这$$100$$个故事编号为$$1$$，$$2$$，$$3$$$$\\cdots \\cdots $$$$100$$，  则甲读$$1\\sim 85$$号，乙读$$31\\sim 100$$号，从$$31\\sim 85$$是甲乙重复的．  那么丙读$$1\\sim 30$$号，$$86\\sim 100$$号，共计$$45$$个故事，此时没有三人都重复的，  但丙读的$$62$$个故事还缺：$$62-45=17$$（个）．这$$17$$个故事只能从$$31\\sim 85$$中选取，  故三人会最少重复$$17$$个故事．  即选择$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "454", "queId": "a58614ba69e44464b1c00c009b4ba7d1", "competition_source_list": ["2008年六年级竞赛学而思杯"], "difficulty": "1", "qtype": "single_choice", "problem": "古时候的一场大地震几乎毁灭了整个人类，只有两个部落死里逃生。最初在这两个部落中，神帝部落所有的人都坚信人性本恶，圣地部落所有的人都坚信人性本善，并且没有既相信人性本善又相信人性本恶的人存在。后来两个部落繁衍生息，信仰追随和部落划分也遵循着一定的规律。部落内通婚，所生的孩子追随父母的信仰，归属原来的部落；部落间通婚，所生孩子追随母亲的信仰，归属母亲的部落。 我们发现神圣子是相信人性本善的。 在以下各项对神圣子身份的判断中，不可能为真的是(  ) ", "answer_option_list": [[{"aoVal": "A", "content": "神圣子的父亲是神帝部落的人。 "}], [{"aoVal": "B", "content": "神圣子的母亲是神帝部落的人。 "}], [{"aoVal": "C", "content": "神圣子的父母都是圣地部落的人。 "}], [{"aoVal": "D", "content": "神圣子的母亲是圣地部落的人。 "}], [{"aoVal": "E", "content": "神圣子的姥姥是圣地部落的人。 "}]], "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": ["不管是部落间还是部落内通婚，生下的孩子都会追随母亲的信仰，由于神圣子坚信人性本善，所以他的母亲一定是圣地部落的。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "323", "queId": "800d2885668145eb997b0bc02de16636", "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9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$15$$瓶未开封的口香糖，其中$$14$$瓶的质量相同，只有$$1$$瓶的质量比其他几瓶稍轻一点．如果要确保找出轻的那一瓶口香糖，至少需要用天平称次．（只有天平，没有砝码） ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["先将$$15$$瓶盖达木糖醇口香糖分成$$7$$、$$7$$、$$1$$组，第一次两边各放$$7$$个，留$$1$$个，如果两边一样重，留出的那个为轻的；若不一样重，再把轻的那个分成$$3$$、$$3$$、$$1$$，称$$3$$、$$3$$的两组；进而再称轻的$$3$$个，这样只需$$3$$次就可以找出那件次品．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3300", "queId": "63fa6ae25d534898a8acbad73f099f2c", "competition_source_list": ["2019年第24届YMO四年级竞赛决赛第7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7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7题3分", "2020年第24届YMO四年级竞赛决赛第7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7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "由$$1$$、$$2$$、$$3$$、$$4$$、$$5$$这五个数字组成的各个数位互不相同的五位数中，从大到小的第$$96$$个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21345$$ "}], [{"aoVal": "B", "content": "$$21354$$ "}], [{"aoVal": "C", "content": "$$21435$$ "}], [{"aoVal": "D", "content": "$$31245$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["由$$1$$、$$2$$、$$3$$、$$4$$、$$5$$这五个数字组成的各个数位互不相同的五位数，共有：$$5\\times4\\times3\\times2\\times1=120$$（个），  其中$$1$$、$$2$$、$$3$$、$$4$$、$$5$$开头各有：$$4\\times3\\times2\\times1=24$$个，  又由$$96\\div 24=4$$，  可知第$$96$$个数为从$$2$$开头的倒数第$$1$$小的数，即$$21345$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2183", "queId": "304afd9518664702a68fe865bebf3340", "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10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在一条马路上，小明骑车与小光同向而行，小明骑车速度是小光速度的$$3$$倍，每隔$$7$$分有一辆公共汽车超过小光，每隔$$14$$分有一辆公共汽车超过小明．已知公共汽车从始发站每次间隔同样的时间发一辆车，问：相邻两车间隔是分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$5.6$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["根据每隔$$7$$分钟就有一辆公交车超过小光可知：  车距$$=7\\times ($$车速$$-$$小光速度$$)$$①；  再根据每隔$$14$$分钟就有一辆公共汽车超过小明可知：  车距$$=14\\times ($$车速$$-3$$小光速度$$)$$②；  ①$$\\times 6-$$②得：$$5$$车距$$=28$$车速，  所以车距$$=5.6$$车速．  所以发车时间间隔为$$5.6$$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": "3181", "queId": "b94c54f4e8bd4751848f2b4dcbfc4394", "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，红黄绿交替且没有间隔。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->能力->公式记忆->符号化数学原理"], "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": "2431", "queId": "0ccd52d3177143c1b434d9da63ee1c94", "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": "458", "queId": "fffa6fb4b3f54170a604cbd4751a481b", "competition_source_list": ["2010年陈省身杯五年级竞赛国际青少年数学解题能力展示活动"], "difficulty": "2", "qtype": "single_choice", "problem": "在下面的算式中，不同的字母代表不同的数字，则$$A+B+C+D+E+F+G=$$~\\uline{~~~~~~~~~~}~．  $$\\overline{ABCD}+\\overline{EFG}=2010$$ ", "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->数学思想->转化与化归的思想"], "answer_analysis": ["题中的式子不好看，我们把它列成竖式观察，$$D+G$$的和个位是$$0$$，所以$$D+G=10$$，我们需要进一位，$$C+F$$下面对应是$$1$$，由于还有前面的进位$$1$$，所以$$C+F=10$$，还要进位$$1$$，$$B+E=9$$，加上进位的$$1$$，也要向前进位$$1$$，所以$$A=1$$，所以$$A+B+C+D+E+F+G=10+10+9+1=30$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1704", "queId": "bf47456bd65a45f9bad6bd10d8101150", "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$$42$$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": "$$31$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$(42+20)\\div 2=31$$（美元）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2900", "queId": "c467d018330d44cb93f174b04cdea3d8", "competition_source_list": ["2019年华夏杯一年级竞赛复赛（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": "2290", "queId": "8e3cf51a935440dea9acff7205cc9e73", "competition_source_list": ["2016年迎春杯六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "变形金刚擎天柱和大黄蜂分别从$$A$$、$$B$$两地同时出发，相向而行，以机器人身份出发时，他们的速度比是$$4:3$$，相遇后变形为汽车人，擎天柱速度提高$$25 \\%$$，大黄蜂提高$$30 \\%$$，当擎天柱到达$$B$$地时，大黄蜂离$$A$$地还有$$83$$千米。那么$$A$$、$$B$$两地相距  千米。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$350$$ "}], [{"aoVal": "B", "content": "$$450$$ "}], [{"aoVal": "C", "content": "$$330$$ "}], [{"aoVal": "D", "content": "$$420$$ "}]], "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": ["解：$$4\\times (1+25 \\%):3\\times (1+30 \\%)$$  $$=5:\\frac{39}{10}$$  $$=50:39$$  $$83\\div \\left( \\frac{4}{4+3}-\\frac{3}{4+3}\\times \\frac{39}{50} \\right)$$  $$=83\\div \\left( \\frac{4}{7}-\\frac{117}{350} \\right)$$  $$=83\\div \\frac{83}{350}$$  $$=350$$（千米）  答：$$A$$、$$B$$两地相距$$350$$千米。  故答案为：$$350$$。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "672", "queId": "d5543ea617a5400a819386742edd8028", "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": "数列$$20160$$、$$20161$$、$$20162$$、$$\\dots\\dots20169$$中共有~\\uline{~~~~~~~~~~}~个质数． ", "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->逻辑分析"], "answer_analysis": ["大于$$2$$的偶数一定是合数，所以需要判断的只有$$20161$$、$$20163$$、$$20165$$、$$20167$$、$$20169$$．其中，$$20163$$、$$20169$$是$$3$$的倍数；$$20165$$是$$5$$的倍数；利用三位截断法可判断出$$20167$$是$$7$$的倍数；用试除法得到$$20161$$是质数． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2607", "queId": "47c38c1fe65a46fe85b396307a21dac9", "competition_source_list": ["2014年IMAS小学中年级竞赛第一轮检测试题第18题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "小朋友玩游戏，老师让小朋友们站成一排，并从第一位开始依照$$1$$、$$2$$、$$3$$循环报数，最后一位小朋友报的数是$$2$$，请问这一排可能共有多少位小朋友？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$26$$ "}], [{"aoVal": "D", "content": "$$27$$ "}], [{"aoVal": "E", "content": "$$28$$ "}]], "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": ["由题意可知这一派同学的人数除以$$3$$后所得的余数是$$2$$，在各个选项中只有选项$$\\text{C}$$符合要求．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1243", "queId": "67018a016d514639995663432a35038c", "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8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多$$1$$个，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": "$$36$$ "}], [{"aoVal": "B", "content": "$$38$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$42$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["根据题意分析可知，第二天摘后剩下桃子$$9$$个，  那么第一天摘后剩下桃子$$\\left( 9+1 \\right)\\times 2=20$$（个），  由此可知，原来树上有桃子$$\\left( 20+1 \\right)\\times 2=21\\times 2=42$$（个）．  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": "3092", "queId": "f41af76d955949a0b90b79b99193d894", "competition_source_list": ["2021年第8届鹏程杯四年级竞赛初赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$，$$b$$都表示数，规定：若$$a\\textgreater b$$， 则$$\\textbar a -b\\textbar=a -b$$；若$$a=b$$， 则$$\\textbar a - b\\textbar{} =0$$；若$$a\\textless{}b$$，则$$\\textbar a -b\\textbar=b-a$$．  现在请你从$$1\\sim 10$$这$$10$$个自然数中任选$$5$$个数，从小到大记为$$a_1$$，$$a_2$$，$$a_3$$，$$a_4$$，$$a_5$$，将剩下的$$5$$个数从大到小记为$$b_1$$，$$b_2$$，$$b_3$$，$$b_4$$，$$b_5$$，计算：$$\\textbar a _1-b_1\\textbar+\\textbar a_2-b_2\\textbar+\\textbar a_3-b_3\\textbar+\\textbar a_4-b_4\\textbar+\\textbar a _5-b_5\\textbar=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$25$$ "}], [{"aoVal": "D", "content": "$$27$$ "}], [{"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": ["构造特例：$$a_1$$，$$a_2$$，$$a_3$$，$$a_4$$，$$a_5$$分别为$$6\\sim 10$$，  $$b_1$$，$$b_2$$，$$b_3$$，$$b_4$$，$$b_5$$分别为$$5$$、$$4$$、$$3$$、$$2$$、$$1$$．   $$\\textbar a _1-b_1\\textbar+\\textbar a_2-b_2\\textbar+\\textbar a_3-b_3\\textbar+\\textbar a_4-b_4\\textbar+\\textbar a _5-b_5\\textbar$$  $$=\\textbar6-5\\textbar+\\textbar7-4\\textbar+\\textbar8-3\\textbar+\\textbar9-2\\textbar+\\textbar10-1\\textbar$$  $$=1+3+5+7+9$$  $$=25$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "589", "queId": "0bcad56b1ca04227b13512edf0b102a6", "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（~ ）． ", "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->拓展思维->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$$B$$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$$A$$选项．估算$${{48}^{2}}=2304$$，$${{49}^{2}}=2401$$，排除  $$C$$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": "718", "queId": "3229034863c54bcf99fa8ab1bb781f3f", "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16题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一些两位数，加$$47$$，就变成三位数；减$$47$$，就变成一位数，这样的两位数有个． ", "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->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["$$100-47=53$$，  有一些两位数，加$$47$$，就变成三位数，那么这个两位数一定不小于$$53$$，  $$47+9=56$$，  减$$47$$，就变成一位数，那么这个两位数一定不大于$$56$$，  故有$$53$$，$$54$$，$$55$$，$$56$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1198", "queId": "15d5f5b5bdb7444baf2d4223ea85ef77", "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": "~某商品八折出售，仍能获得$$20 \\%$$的利润，那么定价时期望的利润率是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25 \\%$$ "}], [{"aoVal": "B", "content": "$$30 \\%$$ "}], [{"aoVal": "C", "content": "$$40 \\%$$ "}], [{"aoVal": "D", "content": "$$50 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["设进价为$$1$$份，售价为$$1\\times \\left( 1+20  \\% ~\\right)=1.2$$（份），定价为$$1.2\\div 80 \\%=1.5$$（份），期望利润率为$$\\left( 1.51 \\right)+1\\times 100 \\%=50 \\%$$. "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2194", "queId": "281cf54d2dfa4c8b8a89c65b2de08813", "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": ["简单计算可得 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "807", "queId": "ff80808147248448014724de1a560131", "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$$A$$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->分解质因数->分解质因数（式）"], "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": "1981", "queId": "bccebc83b38742b3b8f8a78b1bf13932", "competition_source_list": ["2013年小机灵杯三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "小明妈妈花$$8$$元买了一条鱼，以$$9$$元价格卖掉，然后觉得不合算，又花$$10$$元买回来，以$$11$$元卖给另一个人，那么小明妈妈赚了（  ）元。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->加减法应用"], "answer_analysis": ["解：$$11+9-(8+10)$$  $$=20-18$$  $$=2$$（元）  答：小明妈妈赚了$$2$$元；  故选:B。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1311", "queId": "c775f53362f04013bbf44a3434ee6b86", "competition_source_list": ["2004年第2届创新杯六年级竞赛初赛第5题", "2004年六年级竞赛创新杯", "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": ["设甲、乙两根绳子的长度都为$$9x$$米，则：  甲剩下：  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde9x\\times \\left( 1-\\frac{1}{3} \\right)-\\frac{1}{3}$$  $$=6x-\\frac{1}{3}$$（米），  乙剩下：  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left( 9x-\\frac{1}{3} \\right)\\times \\left( 1-\\frac{1}{3} \\right)$$  $$=6x-\\frac{2}{9}$$（米），  $$6x-\\frac{1}{3}$$米$$ ~\\textless{} ~6x-\\frac{2}{9}$$米．  答：乙绳剩下部分长．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "940", "queId": "d3d6c0ce6be442e0907edc16079efd6a", "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和是多少岁？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["知识标签->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": ["$$3080=2^{3}\\times5\\times7\\times11$$,7、$$11$$不可以跟其他质因数组合，$$5$$如果跟$$2$$组合，那么剩余的两个$$2$$只能凑成$$4$$，到不了$$5$$岁，所以$$5$$只能单独存在，剩下的三个$$3$$凑成8.  $$5+7+8+11=31$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1873", "queId": "8ea0c87592bc4c6fb81956e1a63a2417", "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": "【补充4】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": "965", "queId": "de20369acc224b8c92027b912e490cda", "competition_source_list": ["2014年第12届全国创新杯六年级竞赛第5题", "小学高年级六年级其它2014年数学思维能力等级测试第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "写有数字$$6$$、$$10$$、$$18$$的卡片各$$10$$张，现在从这$$30$$张中适当选出$$9$$张计算出它们的和，可能的是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$93$$ "}], [{"aoVal": "B", "content": "$$98$$ "}], [{"aoVal": "C", "content": "$$104$$ "}], [{"aoVal": "D", "content": "$$107$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理"], "answer_analysis": ["考虑到$$6$$、$$10$$、$$18$$都除以$$4$$余$$2$$，而$$9$$个数和除以$$4$$也余$$2$$，选项中只有$$98$$除以$$4$$也余$$2$$，故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2403", "queId": "13cf0c8d9a2e416586d08e4640ea2519", "competition_source_list": ["2014年IMAS小学高年级竞赛第二轮检测试题第5题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$A=15984$$，$$B=48951$$，且正整数$$n$$的平方等于$$A$$与$$B$$的乘积，请问$$n$$的各位数码之和为多少？（~ ~ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$21$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$27$$ "}], [{"aoVal": "E", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$15987={{2}^{4}}\\times {{3}^{3}}\\times 37$$  $$48951={{3}^{3}}\\times {{7}^{2}}\\times 37$$  $$A\\times B={{2}^{4}}\\times {{3}^{3}}\\times 37\\times {{3}^{3}}\\times {{7}^{2}}\\times 37$$  $$={{2}^{4}}\\times {{3}^{6}}\\times {{7}^{2}}\\times {{37}^{2}}$$  ∴$$n={{2}^{2}}\\times {{3}^{3}}\\times 7\\times 37=27972$$  $$2+7+9+7+2=27$$  $$n$$的各位数码之和为$$27$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1885", "queId": "97d71f7c62714f0a8cef6370c00b29aa", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛初赛模拟"], "difficulty": "2", "qtype": "single_choice", "problem": "学校商店购进了进价为每千克$$4$$元的桔子$$1000$$千克，开始售价为$$8$$元，由于销售状况不佳，几天后决定按售价的八五折出售桔子，又卖了几天后，发现开始按$$8$$元所销售的重量比此时余下的桔子的重量多$$100$$千克，决定再调整售价，直到商店卖完后一共获利$$2920$$元．则余下的桔子按每千克~\\uline{~~~~~~~~~~}~元销售． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5.6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$6.2$$ "}], [{"aoVal": "D", "content": "$$2.6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["设余下的桔子按每千克$$x$$元销售，余下$$y$$千克，  $$4\\times\\left(y+100\\right)+2.8\\times\\left(900-2y\\right)+\\left(x-4\\right)y=2920$$  解得$$x=5.6$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "580", "queId": "4a93680325334f6ebcc261093dfa742e", "competition_source_list": ["2016年创新杯五年级竞赛训练题（二）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$M$$、$$N$$为非零自然数，且$${{2014}^{M}}+{{2015}^{N}}$$被$$7$$整除，$$M+N$$的最小值是（~ ）． ", "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->思想->逐步调整思想"], "answer_analysis": ["$$2014$$除以$$7$$的余数为$$5$$，$$2015$$除以$$7$$的余数为$$6$$，相当于要求$${{5}^{M}}+{{6}^{N}}$$是$$7$$的倍数．经尝试，当$$M=3$$，$$N=2$$时满足要求，此时的和是最小的，故$$M+N$$的最小值为$$5$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "322", "queId": "579804477877457eae86e2a8edc882eb", "competition_source_list": ["2011年全国创新杯小学高年级五年级竞赛5分", "2016年创新杯小学高年级五年级竞赛训练题（四）第2题"], "difficulty": "1", "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$1+2+\\cdot \\cdot \\cdot +x$这个和代替，例如$2$被$3(3=1+2)$代替，$5$被$15(15=1+2+3+4+5)$代替，计算机其他功能都正常，如果你计算$1+3+5$，计算机结果是（~ ~ ~ ~）． ", "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->能力->运算求解"], "answer_analysis": ["$$3$$被$$1+2+3=6$$代替  $$5$$被$$1+2+3+4+5=15$$代替  ∴$$1+3+5$$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": "2437", "queId": "1063f1c375f54abe9837ab250a9aca7b", "competition_source_list": ["2017年IMAS小学中年级竞赛（第二轮）第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$A\\times B\\times C=30$$、$$B\\times C\\times D=90$$、$$C\\times D\\times E=120$$﹐请问$$A\\times C\\times E$$等于多少． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$50$$ "}], [{"aoVal": "E", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->能力->符号代换->代数运算"], "answer_analysis": ["由于$$A\\times C\\times E=(A\\times B\\times C)\\times (C\\times D\\times E)\\div(B\\times C\\times D)$$，  因此$$A\\times C\\times E=30\\times120\\div90=40$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1466", "queId": "99bae8bef6b947fda1fafe9b9118e9a0", "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8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小王买$$2$$支毛笔和$$3$$支钢笔，用去$$74$$元；小李买同样的毛笔$$4$$支和钢笔$$2$$支，用去$$68$$元．求每支钢笔售价多少元？下面列式正确的是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 74\\times 2-68 \\right)\\div \\left( 3\\times 2-2 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 74\\times 3-68 \\right)\\div \\left( 2\\times 2-2 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( 74\\times 2+68 \\right)\\div \\left( 3\\times 2-2 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( 74\\times 3+68 \\right)\\div \\left( 2\\times 2-2 \\right)$$ "}]], "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": ["两人对比，当小王再多买一份后，用去钱数为之前的$$2$$倍，就有$$4$$支毛笔和$$6$$支钢笔，此时，小王比小李多出$$4$$支钢笔，算出此时小王比小李多用的钱，就是$$4$$支钢笔的价格，可求出一支钢笔的价格：$$(74\\times 2-68)\\div (3\\times 2-2)$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1354", "queId": "2066bc02799e4a08bb321537ed227f04", "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": "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": "3256", "queId": "7578a0eebbe14c82b49a006be0e8a1db", "competition_source_list": ["2017年河南郑州豫才杯竞赛第12题", "2017年河南郑州小升初豫才杯第二场第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "将若干个$$1$$立方米的正方形木块，摆成一个最小的正方体（不包括一块）至少需要（~ ）块． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$块 "}], [{"aoVal": "B", "content": "$$8$$块 "}], [{"aoVal": "C", "content": "$$16$$块 "}], [{"aoVal": "D", "content": "$$27$$块 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->几何计数->立体图形计数->立体小方块计数"], "answer_analysis": ["符合题意最小的正方体的棱长为$$2$$米，即每个棱长上都有$$2$$个$$1$$立方米的正方体木块，所以需要$$2\\times 2\\times 2=8$$个． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "266", "queId": "90bfd866ba16423282f1a17f9a402a42", "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（~ ~ ~ ）． ", "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": ["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": "2039", "queId": "dd1bf592d1d444dfa0a2cd857be0321c", "competition_source_list": ["2012年全国迎春杯四年级竞赛初赛第9题", "2019年湖南长沙雨花区中雅培粹中学小升初（2）第10题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "甲乙二人相距$$30$$米面对面站好．两人玩``石头、剪子、布''．胜者向前走$$3$$米，负者向后退$$2$$米．平局两人各向前走$$1$$米．玩了$$15$$局后，甲距出发点$$17$$米，乙距出发点$$2$$米．甲胜了~\\uline{~~~~~~~~~~}~次． ", "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->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", "Overseas Competition->知识点->应用题模块->鸡兔同笼问题"], "answer_analysis": ["（$$1$$）有胜有负的局，两人距离缩短$$1$$米；平局两人距离缩短$$2$$米．$$15$$局后两人之间的距离缩短$$15$$ \\textasciitilde{} $$30$$米．（$$2$$）如果两人最后的效果都是后退，两人之间的距离会变大，与（$$1$$）矛盾．（$$3$$）如果两人最后的效果是``一人前进，另一人后退''，两人距离会缩短$$15$$米．但如果两人距离缩短$$15$$米，只能是$$15$$局都是``胜负局''．假设甲$$15$$局都是胜者，他会前进$$45$$米，每把一次``胜者''换成一次``负者''，他会少前进$$5$$米．$$45$$减去多少个$$5$$都不可能等于$$17$$．这种情况不成立．（$$4$$）如果两人最后的效果是都向前进，两人的距离缩短$$19$$米．假设$$15$$局都是``胜负局''，两人之间距离缩短$$15$$米，每把一局``胜负局''换成平局，两人之间距离多缩短$$1$$米．由``鸡兔同笼''法求出，``胜负局''有$$11$$局，平局有$$4$$局．（$$5$$）$$4$$局平局中甲前进了$$4$$米．假设甲其余$$11$$局都是胜者，他一共前进$$33+4=37$$米．每把一局胜局改为败局，他会后退$$5$$米，改$$4$$局，他一共前进$$37-20=17$$米．（$$6$$）验算：甲$$7$$胜$$4$$平$$4$$败，前进$$21+4-8=17$$米；乙$$4$$胜$$7$$败$$4$$平，前进$$12+4-14=2$$米． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "529", "queId": "e6bf9b8e64f3459a9f4ff49df16be5e9", "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10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小强、小刚、小东、小平四个人比高矮．小强不是最矮的；小平不是最高的，但比小强高；小东不比大家高．最高的人是． ", "answer_option_list": [[{"aoVal": "A", "content": "小强 "}], [{"aoVal": "B", "content": "小刚 "}], [{"aoVal": "C", "content": "小东 "}], [{"aoVal": "D", "content": "小平 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["根据题意分析可知，已知小平不是最高的，且小平又比小强高，所以最高的不是小平和小强；小东不比大家高所以小东也不是最高的，那么剩下的小刚就是最高的．  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1715", "queId": "69cd85aaa8d144989311089130f60ff7", "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（$$159$$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->拓展思维->应用题模块->加减法应用->加减法应用顺口溜", "课内体系->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  $$ = \\left( 1250 + 150 \\right) + \\left( 370 + 230 \\right) + \\left( 8 + 9 - 5 \\right) = 1400 + 600 + 12 = 2012$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "432", "queId": "93261d73f9334519845067ab1318acb1", "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": "在一次美术比赛中，$$A$$，$$B$$，$$C$$，$$D$$，$$E$$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，请问$$D$$是第~\\uline{~~~~~~~~~~}~名． ", "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->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": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1549", "queId": "95753f5997ab4cdcb63f4b3c67098c36", "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": ["属龙的人共有25-8=17（人），当全部男生都属龙的时候，女生中属龙的人数最少，至少有17-11=6（人）. "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2363", "queId": "0478eda45b524c7496bd3b59b21d8af2", "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->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": ["$$\\frac{99}{100}=1- \\frac{1}{100}=1- \\frac{2}{200}$$，  $$\\frac{100}{101}=1- \\frac{1}{101}=1- \\frac{2}{202}$$，  $$\\frac{199}{201}=1- \\frac{2}{201}$$，  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$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{99}{100}$$．  所以选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2752", "queId": "767fa1ecae5c4674b5143d6dcb71fa0d", "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2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$2$$，$$a$$，$$b$$，$$23$$这四个数成等差数列，则$$a+b=$$ ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$21$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "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，$$2$$，$$a$$，$$b$$，$$23$$这四个数成等差数列，  等差数列有个特点：若$$a$$，$$b$$，$$c$$三个数按这个顺序排列成等差数列，那么$$b$$叫$$a$$，$$c$$的等差中项，$$a$$，$$b$$，$$c$$满足$$b-a=c-b$$；  也就是$$a-2=b-a=23-b$$；  整理可得：$$a+b=2+23=25$$，所以$$a+b=25$$．  故选：$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2556", "queId": "30be2eaa68644829aaa03dc4262659f6", "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1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$1$$，$$a$$，$$b$$，$$2018$$这四个数成等差数列，则$$a+b=$$． ", "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->对应思想"], "answer_analysis": ["若$$a$$，$$b$$，$$c$$三个数按这个顺序排列成等差数列，那么$$b$$叫$$a$$，$$c$$的等差中项．$$a$$，$$b$$，$$c$$满足$$b-a=c-b$$，$$2b=a+c$$．对于这道题，$$1$$、$$a$$、$$b$$、$$2018$$这四个数成等差数列，根据等差数列等差中项的含义，$$a+b=1+2018=2019$$．  故选：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1654", "queId": "76e32f867bd7448882d34514575f8e40", "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几种情况． ", "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": "1986", "queId": "ea29f1301c0d47ea8fdb6102ba59a5ab", "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10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$1991991\\cdots\\cdots$$这串数中，从第三个数开始，每个数都是前两个数相乘后积的尾数（个位数字），那么把这串数字写到第$$40$$位时，第$$40$$位的数字是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$91$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["观察此数字串是按照如下规律排列：$$199199199199\\cdots $$以$$199$$为数字循环节，  第$$40$$位为：$$40\\div 3=13$$余$$1$$，也就是循环节里的第一位，也就是$$1$$，故选$$\\text{A}$$ "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2276", "queId": "a45bc588521c447f93c8669c28875ab6", "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1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明早上$$6:30$$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$$11:30$$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->单人简单行程问题"], "answer_analysis": ["小明一共走了：$$11-6-1=4$$（小时），  所以甲、乙两地之间的距离是：$$4\\times 8=32$$（千米），  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1824", "queId": "92922d7b711b44b8bf4af6c7a1cfbaf0", "competition_source_list": ["2019年陕西西安灞桥区铁一中滨河中学小升初入学真卷5第6题3分", "2019年湖南长沙雨花区中雅培粹中学小升初第9题3分", "2018年陕西西安灞桥区铁一中滨河中学小升初入学真卷2第3题3分", "2018~2019学年四川成都高新区成都外国语学校附属小学六年级上学期月考第20题2分", "2013年全国世奥赛竞赛第7题", "2011年北京北大附属实验学校小升初分班测试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "一次考试，参加的学生中有$$\\frac{1}{7}$$得优，$$\\frac{1}{3}$$得良，$$\\frac{1}{2}$$得中，其余的得差，已知参加考试的学生不满$$50$$人，那么得差的学生有~\\uline{~~~~~~~~~~}~人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->对应思想"], "answer_analysis": ["选$$\\text{C}$$．由题意``参加的学生中有$$\\frac{1}{7}$$ 得优，$$\\frac{1}{3}$$得良，$$\\frac{1}{2}$$ 得中''，  可知参加考试的学生人数是$$7$$，$$3$$，$$2$$ 的倍数，  因为$$7$$，$$3$$，$$2$$ 的最小公倍数为$$42$$ ，$$42 \\times 2 = 84 \\textgreater{} 50$$ ~，  所以参加的学生总数为$$42$$ 人．那么得差的学生有：$$42 \\times (1 - \\frac{1}{7} - \\frac{1}{3} - \\frac{1}{2}) = 1$$ ~人． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3459", "queId": "ef865526383148dea33c10687f9ca944", "competition_source_list": ["2003年第1届创新杯六年级竞赛复赛第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "在自然数中，有些相邻的两个自然数相加时不会出现进位的情况，如$$1003$$和$$1004$$、$$1004$$和$$1005$$、$$1399$$和$$1400$$就是$$3$$对这样的自然数．那么在自然数$$1000$$、$$1001$$、$$1002$$、$$1003$$、$$\\cdots\\cdots$$、$$2002$$、$$2003$$中，像这样的两个相邻自然数的对数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$128$$对 "}], [{"aoVal": "B", "content": "$$153$$对 "}], [{"aoVal": "C", "content": "$$158$$对 "}], [{"aoVal": "D", "content": "$$159$$对 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["根据题意，相邻的两个自然数相加时不会出现进位的情况，能够确定出各个数位上的数字的特点，然后计算出满足条件的这样的两个相邻的自然数的对数即可，  根据题意，符合条件的数应满足条件：  个位上的数要么$$\\leqslant 4$$，要么等于$$9$$，  十位上的数要么$$\\leqslant 4$$，要么和个位一样，等于$$99$$，  百位上的数要么$$\\leqslant 4$$，要么和十位以及个位一样，等于$$999$$，  千位上的数相加肯定不会出现进位的情况，  因此从$$1000$$到$$2000$$满足条件的两个相邻自然数的对数共有：  $$\\textasciitilde\\textasciitilde\\textasciitilde5\\times5\\times5+5\\times5\\times1+5\\times1+1$$  $$=125+25+5+1$$  $$=156$$（对），  $$2000$$和$$2001$$、$$2001$$和$$2002$$、$$2002$$和$$2003$$共有$$3$$对，  $$156+3=159$$（对）．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2354", "queId": "0332d9ff28ff421b819204e731d91325", "competition_source_list": ["2011年全国华杯赛竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "若连续的四个自然数都为合数，那么这四个数之和的最小值为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$101$$ "}], [{"aoVal": "C", "content": "$$102$$ "}], [{"aoVal": "D", "content": "$$103$$ "}]], "knowledge_point_routes": ["知识标签->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": ["任何四个连续自然数之和一定被$$4$$除余$$2$$，所以只有$$102$$满足条件． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1644", "queId": "ff80808147248448014724de1bef0137", "competition_source_list": ["2016年全国华杯赛小学中年级竞赛在线模拟第4题", "2013年全国华杯赛小学中年级竞赛初赛A卷第4题", "2013年全国华杯赛小学高年级竞赛初赛A卷第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2013$$年的钟声敲响了，小明的哥哥感慨地说：这是我有生以来遇到的第一个没有重复数字的年份．已知小明哥哥出生的年份是$$19$$的倍数，那么$$2013$$年小明哥哥的年龄是（~ ~ ~ ~）岁． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$22$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据``小明哥哥出生的年份是$$19$$的倍数，第一个没有重复数字的年份．''可以考虑之前的年份是$$19-\\/-$$年，所以考虑到重复数字$$199-$$年，$$1995$$年是$$19$$的倍数．$$2013-1995=18$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1732", "queId": "bacd85fe4e9e456884485f9802304ed1", "competition_source_list": ["2019年华夏杯二年级竞赛决赛第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$2019$$年$$4$$月$$14$$日是星期日，问$$24$$日后是星期几？ ", "answer_option_list": [[{"aoVal": "A", "content": "星期二 "}], [{"aoVal": "B", "content": "星期三 "}], [{"aoVal": "C", "content": "星期四 "}], [{"aoVal": "D", "content": "星期一 "}], [{"aoVal": "E", "content": "都不对 "}]], "knowledge_point_routes": ["Overseas Competition->知识点->组合模块->时间问题", "拓展思维->能力->逻辑分析"], "answer_analysis": ["$$24=7+7+7+3$$．  $$24$$日后是星期三． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "952", "queId": "cb34f2160c604c6cbeef54c9ff44c9a4", "competition_source_list": ["2019年第7届湖北长江杯五年级竞赛复赛A卷第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "把所有三位数的质数相乘，它们的积是一个． ", "answer_option_list": [[{"aoVal": "A", "content": "质数 "}], [{"aoVal": "B", "content": "偶数 "}], [{"aoVal": "C", "content": "奇数 "}], [{"aoVal": "D", "content": "不确定 "}]], "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": ["在质数中$$2$$是唯一的偶数，其它质数都是奇数，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，既然是若干个质数（$$2$$除外）相乘，这些质数自然是它们积的因数，所以，乘积必定是合数，不会是质数，也不会是偶数，据此进行选择．因为，所有的三位质数都是奇数，而奇数$$\\times $$奇$$=$$奇数，所以，把所有三位数的质数相乘，它们的积是一个奇数．  故选：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1719", "queId": "daddc3c52fd2411fa8e462e0bea1962b", "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1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "时钟$$6$$点敲$$6$$下，$$10$$秒钟敲完，$$12$$点敲$$12$$下需要秒钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$22$$ "}]], "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->敲钟问题"], "answer_analysis": ["敲$$6$$下，则中间停顿$$5$$次，所以每次停顿时长：$$10\\div 5=2$$秒．敲$$12$$下，则停顿$$11$$次，所以时长：$$11\\times 2=22$$秒．  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": "280", "queId": "ff8080814502fa240145065892570315", "competition_source_list": ["2014年全国迎春杯四年级竞赛复赛第12题", "2017年全国小升初八中入学备考课程"], "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->能力->运算求解"], "answer_analysis": ["由于$$25=9+7+5+3+1$$ ，所以我们可以猜测$$0～50$$之间的所有偶数都有可能得到，那么总和就是$$(0+50)\\times 26\\div 2=650$$，但是我们认真思考下，$$4$$和$$46$$是无法凑出来的．所以答案是$$650-4-46=600$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2211", "queId": "31abe6db538e40b386602ad7271b5eeb", "competition_source_list": ["其它改编自2013年全国希望杯六年级竞赛初赛第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$12$$点开始，经过~\\uline{~~~~~~~~~~}~分钟，时针与分针第一次成$$90°$$角，$$12$$点之后，时针与分针第二次成$$90°$$角的时刻是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16\\frac{4}{11}$$；$$45$$分 "}], [{"aoVal": "B", "content": "$$15$$；$$49\\frac{1}{11}$$分 "}], [{"aoVal": "C", "content": "$$16\\frac{1}{11}$$；$$49\\frac{4}{11}$$分 "}], [{"aoVal": "D", "content": "$$16\\frac{4}{11}$$；$$49\\frac{1}{11}$$分 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["转化为钟面上的行程问题，时针和分针的速度差为$$5.5$$度每分钟，第一次和第二次的路程差分别为$$90$$度和$$270$$度，所以$$90\\div 5.5=16\\frac{4}{11}$$分钟第一次成$$90$$度，而第二次时刻是$$270\\div5.5=49\\frac{1}{11}$$分． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "767", "queId": "d5e6f3bb2b6948a88c7dd7edd898e141", "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": "$$14!$$或$$15!$$ "}], [{"aoVal": "B", "content": "$$15!$$ "}], [{"aoVal": "C", "content": "$$16!$$或$$17!$$ "}], [{"aoVal": "D", "content": "$$15!$$或$$16!$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "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": "2340", "queId": "0566070df41f4d199d7276610f40bee7", "competition_source_list": ["2013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$$3\\times 11\\times 61+3+11+61$$的值等于什么？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2013$$ "}], [{"aoVal": "B", "content": "$$2088$$ "}], [{"aoVal": "C", "content": "$$2113$$ "}], [{"aoVal": "D", "content": "$$4026$$ "}], [{"aoVal": "E", "content": "$$4052169$$ "}]], "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": ["解法$$1$$：因为$$3\\times 11\\times 61+3+11+61=2013+3+11+61=2088$$，故选$$\\text{B}$$．  解法$$2$$：虽然$$3\\times 11\\times 61+3+11+61$$为四个奇数相加，故其和为偶数，所以$$\\text{A}$$、$$\\text{C}$$、$$\\text{E}$$不合；又$$3\\times 11\\times 61+3+11+61\\leqslant 40\\times 61+100\\leqslant 2540$$，所以$$\\text{D}$$不合．故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "11", "queId": "054a7499b4fc4887bf6ac1e167efc38f", "competition_source_list": ["2013年全国华杯赛小学中年级竞赛初赛A卷第3题", "2016年全国华杯赛小学中年级竞赛在线模拟第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（$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$）． ", "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", "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": "1309", "queId": "79bc8d6b3480495d84f820170d0c4812", "competition_source_list": ["2008年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "在一条长$$2008$$米的公路的两侧栽树（两端都栽），每隔$$8$$米栽一棵，一共要栽（  ）棵。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$251$$ "}], [{"aoVal": "B", "content": "$$253$$ "}], [{"aoVal": "C", "content": "$$502$$ "}], [{"aoVal": "D", "content": "$$504$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->间隔问题"], "answer_analysis": ["公路每侧栽树$$2008\\div 8+1=252$$（棵），两侧栽树共$$252\\times 2=504$$（棵）。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1857", "queId": "ae05664690394d6e87551977e39a64b3", "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": "一台新上市的苹果$$\\text{iphone6}$$，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->思想->对应思想"], "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": "2855", "queId": "612c39f677e54c2a92b220f437743f2d", "competition_source_list": ["2006年第11届全国华杯赛竞赛复赛第5题", "2017年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": "先写出一个两位数$$62$$，接着在$$62$$右端写这两个数字的和$$8$$，得到$$628$$，再写末两位数字$$2$$和$$8$$的和$$10$$，得到$$62810$$，用上述方法得到一个有$$60$$位的整数：$$628101123$$$\\cdots$，则这个整数的个位是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["此题要首先找出数字出现的规律：$$6281011235813471123581347$$\\ldots，从第六个开始，每十个数字一个循环，然后按周期问题解答即可．  $$\\left( 60-5 \\right)\\div 10=5\\cdots 5$$；个位数字是周期中的第$$5$$个，即为$$5$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2333", "queId": "0513871ebda246429f546912ffa2e157", "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$$315$$道题目需要多少小时？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$4.5$$ "}], [{"aoVal": "D", "content": "$$270$$ "}]], "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": ["$$35$$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": "3321", "queId": "910bc657cc3a43b9abfd7f9422d865c5", "competition_source_list": ["2006年全国迎春杯小学中年级竞赛复赛第2题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a$$、$$b$$、$$c$$、$$d$$、$$e$$这五个数各不相同，它们两两相乘后的积从小到大排列依次为：$$3$$，$$6$$，$$15$$，$$18$$，$$20$$，$$50$$，$$60$$，$$100$$，$$120$$，$$300$$．那么，这五个数中从小到大排列第$$2$$个数的平方是( ~ ~ ~ ~ ~)． ", "answer_option_list": [[{"aoVal": "A", "content": "1 "}], [{"aoVal": "B", "content": "3 "}], [{"aoVal": "C", "content": "5 "}], [{"aoVal": "D", "content": "10 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["解设$$a\\textless{}b\\textless{}c\\textless{}d\\textless{}e$$，由$$ab=3$$,$$ac=6$$.推知$$c=2b$$，由$$ce=120$$,$$de=300$$,推知$$d=\\frac{5}{2}c=5b$$．  $$bc=b\\times 2b=2{{b}^{2}}$$,$$bd=b\\times 5b=5{{b}^{2}}$$,$$cd=2b\\times 5b=10{{b}^{2}}$$.  在$$15$$、$$18$$、$$20$$、$$50$$、$$60$$、$$100$$中，满足$$2:5:10$$的三个数是$$20:50:100$$，所以$${{b}^{2}}=100\\div 10=10$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "970", "queId": "f5264e062ce047b882a65e786b552c07", "competition_source_list": ["2020年广东深圳鹏程杯六年级竞赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "若三个质数$$x$$，$$y$$，$$z$$使得等式$$x\\times y\\times z+7=2020$$成立，则$$x+y+z=$$ ． ", "answer_option_list": [[{"aoVal": "A", "content": "$$75$$ "}], [{"aoVal": "B", "content": "$$78$$ "}], [{"aoVal": "C", "content": "$$81$$ "}], [{"aoVal": "D", "content": "$$82$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->运算求解"], "answer_analysis": ["分解质因数，  $$x\\times y\\times z+7=2020$$，$$x\\times y\\times z=2013$$，  $$2013=3\\times 11\\times 61$$，  $$x+y+z=3+11+61=75$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3029", "queId": "ca20d8ed78744f4c8ece29c85122f776", "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，总要进行自查自检，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": "3343", "queId": "ff808081477bd88b0147859206210e04", "competition_source_list": ["2014年全国华杯赛小学中年级竞赛初赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁、戊围坐在圆形桌子边玩扑克，甲有自己的固定座位．如果乙和丁的座位不能相邻，那么共有（~ ~ ~ ~）种不同的围坐方法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["乙、丙、丁、戊四个人的位置有$$4\\times 3\\times 2\\times 1=24$$种方法；乙丁相邻可以把他们看成一个人，三个人的排法有$$2\\times 3\\times 2\\times 1=12$$，则乙和丁的座位不能相邻的排法有$$24-12=12$$种． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2128", "queId": "12bd0cea9195484e964299a7753bda8b", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛B卷第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有快、中、慢三辆车同时从同一地点出发，沿同一条公路追赶路上一个骑车人．这三辆车分别用$$6$$小时、$$10$$小时、$$12$$小时追上骑车人．现在知道快车每小时走$$24$$千米，中速车每小时走$$20$$千米，那么，慢速车每小时走千米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$19$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由题意可知在，快车走了$$6$$个小时，走了$$24\\times 6=144$$千米．  中车走了$$10$$个小时，走了$$20\\times 10=200$$千米，  两车时间上相差$$4$$小时，而路程的差正是人在这$$4$$小时内走的距离，  所以人的速度是（$$\\left( 200-144 \\right)\\div 4=14$$，  行人开始距起点的距离就是$$\\left( 24-14 \\right)\\times 6=60$$千米，  慢车追上行人的时候，行人向前走了$$14\\times12=168$$千米，  慢车一共走了$$168+60=228$$千米，  慢车速度$$228\\div12=19$$千米$$/$$小时，就可以解答本题．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1303", "queId": "d5518ff3771a48eca3c77089f5b2b602", "competition_source_list": ["2017年河南郑州K6联赛竞赛模拟第七套第4题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "哥哥和弟弟进行$$100$$米赛跑，当哥哥到达终点时，弟弟才到达$$95$$米处．如果让弟弟在原点起跑，哥哥后退$$5$$米，兄弟俩的速度不变，先到达终点的是（ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "哥哥 "}], [{"aoVal": "B", "content": "弟弟 "}], [{"aoVal": "C", "content": "同时达到 "}], [{"aoVal": "D", "content": "无法判断 "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["哥哥与弟弟的速度比为$$100:95=20:19$$，如果哥哥后退$$5$$米，相当于哥哥跑$$105$$米，则弟弟跑$$105\\div 20\\times 19=99.75\\textless{}100$$，所以先到达终点的是哥哥． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3108", "queId": "e6cfb72144954b50bda1ac2dfb78ba71", "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->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": "2142", "queId": "084aa0aa52dc4f749dd4491fed59a482", "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->能力->运算求解"], "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": "2504", "queId": "7e10902e6da14f9aaa2a0ad015a9635f", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第3题", "2011年全国创新杯五年级竞赛第5题5分", "2016年创新杯五年级竞赛训练题（二）第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$a$是一个满足下列条件的最大的正整数：使得$a$除$64$的余数是$4$，用$a$除$155$的余数是$5$，用$a$除$187$的余数是$7$，则$a=$（~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$\\text{10}$ "}], [{"aoVal": "B", "content": "$\\text{15}$ "}], [{"aoVal": "C", "content": "$\\text{30}$ "}], [{"aoVal": "D", "content": "$60$ "}]], "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": ["$$a\\textbar60$$，$$a\\textbar150$$，$$a\\textbar180$$，$$(60,150,180)=30$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1851", "queId": "d25fc49c150441a9a3395f4066747325", "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2题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "学校画廊展出了每个年级学生的图画作品，其中有$$27$$幅不是三年级的，有$$29$$幅不是四年级的，三、四两个年级参展的画共有$$12$$幅．除三四年级外，其他年级参展的画共有幅． ", "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": ["其中有$$27$$幅不是三年级的，有$$29$$幅不是四年级的，三、四两个年级参展的画共有$$12$$幅，则所有一共有$$(27+29+12)\\div2=34$$（幅），则除三四年级外，其他年级参展的画共有$$34-12=22$$（幅）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "39", "queId": "06ec1b86eb7444ff9d66f29785daf1ae", "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，第一名是谁？ ", "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": ["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->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": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1087", "queId": "8abd9e4dd60f4ca1a31b0429c1617e5d", "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3题"], "difficulty": "1", "qtype": "single_choice", "problem": "在一个停车场上共停了$24$辆车，其中汽车有$4$个轮子，摩托车有$3$个轮子，这些车共有$86$个轮子，那么三轮摩托车有辆． ", "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->鸡兔同笼问题"], "answer_analysis": ["假设这$$24$$辆车都为汽车，那么共有轮子$$24\\times 4=96$$（个），每把$$1$$辆汽车换成$$1$$辆摩托车，轮子的总个数就少$$4-3=1$$（个），那么当轮子的总个数少到$$86$$个时，一共换了$$\\left( 96-86 \\right)\\div 1=10$$（辆）汽车，即摩托车的数量为$$10$$辆，选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2090", "queId": "d93fca6c98e4405da57d8ddbb5020de8", "competition_source_list": ["2015年IMAS小学中年级竞赛第二轮检测试题第11题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "某次联谊活动共有$$20$$位学生，其中第一位女学生和$$7$$位男学生握过手；第二位女学生和$$8$$位男学生握过手；第三位女学生和$$9$$位男学生握过手；以此类推，最后一位女学生和全体男学生都握过手，请问这$$20$$位学生中，有~\\uline{~~~~~~~~~~}~位男学生。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "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", "课内体系->能力->运算求解"], "answer_analysis": ["不妨设最后一名女学生为第$$n$$位，则第$$n$$位女生与$$(n+6)$$位男生握过手，所以全体女生有$$n$$位，全体男生有$$(n+6)$$位，则$$n+(n+6)=20$$，解得$$n=7$$，男生有$$7+6=13$$（位）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "173", "queId": "282b6ed85cd5488c90013746ef207372", "competition_source_list": ["2017年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "两个有限小数的整数部分分别是$$7$$和$$11$$，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$$7\\times 11 \\textless{} a\\times b \\textless{} 8\\times 12$$，即$$77 \\textless{} a\\times b \\textless{} 96$$，则乘积的整数部分$$M$$满足$$M\\leqslant a\\times b$$，则$$77\\leqslant M \\textless{} 96$$，因此可得整数部分可以取$$77$$到$$95$$的所有整数，共有$$95-77+1=19$$（个），因此选D。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "245", "queId": "951ee334d9e941afae0a76a97b422431", "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": "1248", "queId": "1ebe86a834aa4b6aaa33cf768b4e4d22", "competition_source_list": ["2006年四年级竞赛创新杯"], "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->拓展思维->数论模块->余数问题->余数的性质->余数的可加性"], "answer_analysis": ["从1\\textasciitilde10，10个自然数的平方的末位数字依次是1，4，9，6，5，6，9，4，1，0;  从11\\textasciitilde20，10个自然数的平方的末位数字依次是1，4，9，6，5，6，9，4，1，0;  从21\\textasciitilde30，10个自然数的平方的末位数字依次是1，4，9，6，5，6，9，4，1，0;  $$\\cdots\\cdots$$  由上可知，从1开始，每10个自然数的平方的末位数以1，4，9，6，5，6，9，4，1，0为一个周期循环.  $$2006\\div 10=200$$周期$$\\cdots\\cdots 6$$.因为一个周期内的十个数的个位数之和为45，所以题中2006个数的个位数之和为$$45\\times 200+\\left( 1+4+9+6+5+6 \\right)=45\\times 200+31$$，因此所求的个位数字为1，故选A. "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "86", "queId": "21c3f9dfad534553bc3212ecfbbf6bc5", "competition_source_list": ["2014年全国中环杯五年级竞赛决赛第7题", "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"], "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~\\uline{~~~~~~~~~~}~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": "653", "queId": "82b6e95bcc5447b6918cce70ea1ce605", "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5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$、$$b$$、$$c$$都是质数，并且$$a+b=55$$，$$b+c=66$$，则$$a+b+c=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$68$$ "}], [{"aoVal": "B", "content": "$$69$$ "}], [{"aoVal": "C", "content": "$$70$$ "}], [{"aoVal": "D", "content": "$$71$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "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．  从题干可以知道$$a$$、$$b$$、$$c$$都是质数，所有质数中，只有$$2$$是偶数，其他都是奇数；  只有一个奇数和一个偶数的和才是奇数；两个奇数的和一定是偶数；  $$a+b=55$$；$$b+c=66$$；可以推出$$a=2$$；那么$$a+b+c=2+66=68$$．  故选：$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1230", "queId": "3daa5797c99e40ad87b257f5008183c7", "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": "$$2020$$年元旦是星期二，今年的国庆节是星期． ", "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": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3415", "queId": "8aac50a7508d5d410150d5348736773a", "competition_source_list": ["2015年北京华杯赛小学高年级竞赛初赛A卷第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "桌上有编号$$1$$至$$20$$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~\\uline{~~~~~~~~~~}~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->能力->实践应用"], "answer_analysis": ["由于有$$2$$倍多$$2$$的关系，所以$$1$$、$$4$$、$$10$$只能取其中两个，$$2$$、$$6$$、$$14$$只能取其中两个，$$3$$、$$8$$、$$18$$只能取其中两个．即这里至少有$$3$$个数取不到，而$$11$$、$$13$$、$$15$$、$$17$$、$$19$$不满足$$2$$倍多$$2$$的关系，也无法取到．合计至少有$$8$$个数取不到，取$$12$$个数为最多的情况．列举最多的一种情况：$$1$$、$$4$$；$$2$$、$$6$$；$$3$$、$$8$$；$$5$$，$$12$$；$$7$$，$$16$$；$$9$$，$$20$$．取到了最多的$$12$$个数的情况． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2219", "queId": "4cac8b3a67ae45c7a50f686bcdcf3d1c", "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": "$$12$$ "}]], "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": ["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": "1312", "queId": "5986db8b47574607a2e7a0a9820a99fa", "competition_source_list": ["2017年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": "1", "qtype": "single_choice", "problem": "有$$A$$、$$B$$两个仓库有同样多的货物，甲单独搬完一个仓库需要$$10$$小时，乙单独搬完一个仓库需要$$12$$小时，丙单独搬完一个仓库需要$$15$$小时，现在甲搬$$A$$仓库，乙搬$$B$$仓库，丙一会帮甲，一会帮乙．最后两个仓库同时搬完，用了小时． ", "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->思想->对应思想"], "answer_analysis": ["看成甲乙丙三个人一起合作搬两个仓库的货物  $$\\left( 1+1 \\right)\\div \\left( \\frac{1}{10}+\\frac{1}{12}+\\frac{1}{15} \\right)=8$$小时. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2922", "queId": "81a48661044e4f288ca5fabfe867a90d", "competition_source_list": ["2017年全国亚太杯五年级竞赛初赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "分母为$$48$$的最简真分数有~\\uline{~~~~~~~~~~}~个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$47$$ "}], [{"aoVal": "C", "content": "$$30$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["分解质因数；  $$48={{2}^{4}}\\times 3$$，所以分子不能是$$2$$或$$3$$的倍数，因为是最简真分数，所以分子小于分母，且为正整数即分子范围为$$1\\sim 47$$ ．  $$1\\sim 47$$ 里$$2$$的倍数有$$48\\div 2-1=23$$个．  $$3$$的倍数有$$48\\div 3-1=15$$个．  $$6$$的倍数有$$48\\div 6-1=7$$个．（因为$$2$$的倍数和$$3$$的倍数都包括了$$6$$的倍数，所以要减去）$$47-23-15+7=16$$（个）．  所以分母为$$48$$的最简真分数有$$16$$个． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1058", "queId": "07d979dda21b401aa970beaa548250ec", "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4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "笑笑去上学，前$$4$$分钟走了全程的$$\\frac{1}{2}$$，接着$$4$$分钟又行了剩下路程的$$\\frac{1}{2}$$，这时距离学校还有全程的． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{4}$$ "}]], "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": ["因为由题干可知，前$$4$$分钟走了全程的$$\\frac{1}{2}$$，则还剩下全程的$$1-\\frac{1}{2}=\\frac{1}{2}$$，接着$$4$$分钟又行了剩下路程的$$\\frac{1}{2}$$，所以这时距离学校还有全程的$$\\frac{1}{2}-\\frac{1}{2}\\times \\frac{1}{2}=\\frac{1}{4}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1429", "queId": "636de7bd903c49138b8817341981054c", "competition_source_list": ["2009年第14届全国华杯赛竞赛初赛第2题10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->能力->实践应用"], "answer_analysis": ["由于开学前$$6$$填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$$6$$填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$$60\\div（3-1）\\times 3=90$$道题，他平均每天做$$90\\div 6=15$$道题．正确答案为$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3322", "queId": "76975059cf524957a99e36c55ad70ef9", "competition_source_list": ["2020年第24届YMO六年级竞赛决赛第6题3分", "2019年第24届YMO六年级竞赛决赛第6题3分"], "difficulty": "3", "qtype": "single_choice", "problem": "一条道路上有$$20$$盏路灯，为了节省用电，又不影响照明，现关闭其中的$$8$$盏路灯，要求道路两端的路灯不能关闭，相邻的路灯不能都关闭．那么关闭路灯有种不同的方式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$150$$ "}], [{"aoVal": "C", "content": "$$165$$ "}], [{"aoVal": "D", "content": "$$185$$ "}]], "knowledge_point_routes": ["拓展思维->能力->公式记忆->符号化数学原理"], "answer_analysis": ["由题干可得出有：$$20-8=12$$盏路灯是打开状态，  $$12$$盏开着的路灯中间有$$11$$个空档（两端不能放关闭的路灯），  把$$8$$盏关闭的路灯分别放入这$$11$$个空档中（每个空档只能放$$1$$盏）的方法有：  $$\\left( 11\\times 10\\times 9\\times 8\\times 7\\times 6\\times 5\\times 4 \\right)\\div \\left( 8\\times 7\\times 6\\times 5\\times 4\\times 3\\times 2 \\right)=165$$（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": "3458", "queId": "ef7a8df1dec24d29b7ee90c885f88714", "competition_source_list": ["2017年全国华杯赛小学中年级竞赛初赛模拟第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "刘老师在某一个星期中要去$$3$$次健身馆，但是为了防止运动过量，不能连续两天都去，刘老师一共有~\\uline{~~~~~~~~~~}~种满足条件的时间安排． ", "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->枚举思想"], "answer_analysis": ["可以用枚举：一、三、五；一、三、六；一、三、七；一、四、六；一、四、七；一、五、七；二、四、六；二、四、七；二、五、七；三、五、七；共$$10$$种 ~ "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "411", "queId": "8a8a792b59b14112acd90c5cb84798c9", "competition_source_list": ["2016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": "将$$3$$颗相同的小球放到$$A$$、$$B$$、$$C$$三个不同的盒子中，允许某些盒子是空的，请问共有多少种不同的放法？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$18$$ "}], [{"aoVal": "E", "content": "$$27$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["①放在$$3$$个盒子里，$$1$$种；  ②放在$$2$$个盒子里，$$\\text{A}_{3}^{2}=6$$种；  ③放在$$1$$个盒子里，$$3$$种；  $$1+6+3=10$$（种）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "995", "queId": "4a6cbee14b0e4c44bfce2b429f7a5e58", "competition_source_list": ["2015年第13届全国创新杯六年级竞赛第2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2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "两根电线都是$$2$$米长，第一根用去$$\\frac{1}{4}$$米，第二根用去$$\\frac{1}{4}$$，剩下的电线相比，（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "第二根剩下的长 "}], [{"aoVal": "B", "content": "第一根剩下的长 "}], [{"aoVal": "C", "content": "两根剩下的一样长 "}], [{"aoVal": "D", "content": "无法比较 "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["第一根剩$$2-\\frac{1}{4}=1\\frac{3}{4}$$米，第二根剩$$2\\times \\frac{3}{4}=1\\frac{1}{2}$$米，第一根剩的长． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2227", "queId": "a77ccc00ebd542ef8f3e4e98d7fc7c01", "competition_source_list": ["2017年第15届全国希望杯六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "小明放学回家，休息了一会儿开始做作业，此时他看到钟面上分针略超过时针．完成作业时，小明发现分针与时针恰好互换了位置．小明做家庭作业用了~\\uline{~~~~~~~~~~}~分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{720}{11}$$ "}], [{"aoVal": "B", "content": "$$\\frac{720}{13}$$ "}], [{"aoVal": "C", "content": "$$\\frac{360}{11}$$ "}], [{"aoVal": "D", "content": "$$\\frac{360}{13}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设小明做作业用了$$x$$分钟．根据题意：分针每走一小格转动$$6$$°，时针每走一大格（$$1$$小时） 转动$$30$$°，即一分钟走$$0.5$$°．于是有$$6x+0.5x=360$$，解得：$$x=\\frac{720}{13}$$．所以小明做作业用 了$$\\frac{720}{13}$$分钟． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "630", "queId": "f5a5effb27104a1bb399c306fd829219", "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1题"], "difficulty": "1", "qtype": "single_choice", "problem": "两个两位数的积是3927.这两个数的和是（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "124 "}], [{"aoVal": "B", "content": "126 "}], [{"aoVal": "C", "content": "128 "}], [{"aoVal": "D", "content": "130 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->数论模块->分解质因数"], "answer_analysis": ["$$3927=3\\times 7\\times 11\\times 17$$，这两个两位数只能为$$7\\times 11=77$$和$$3\\times 17=51$$，这两个数的和为$$77+51=128$$，选C. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1242", "queId": "5001e5d2821d4b18a990af932dad25ef", "competition_source_list": ["2016年河南郑州联合杯六年级竞赛初赛第11题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一次小测验，甲的成绩是$$85$$分，比乙的成绩低$$9$$分，比丙的成绩高$$3$$分．那么他们三人的平均成绩是（~ ）分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$94$$~~~ "}], [{"aoVal": "B", "content": "$$91$$~~~ "}], [{"aoVal": "C", "content": "$$87$$~~~ "}], [{"aoVal": "D", "content": "$$82$$~~~ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["平均数；甲：$$85$$分，乙：$$94$$，丙：$$82$$分，则三人的平均成绩是$$\\left( 85+94+82 \\right)\\div 3=87$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1062", "queId": "259befe324de4cba8aa55a2dc99302df", "competition_source_list": ["2016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": "0", "qtype": "single_choice", "problem": "一种纺织品的合格率是$$98 \\%$$，$$300$$件产品中大约会有件不合格． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$294$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["合格率为$$98 \\%$$，  ∴不合格率为：$$100 \\%-98 \\%=2 \\%$$，  ∴$$300$$件产品中大约会有不合格产品：$$300\\times 2 \\%=6$$（件）．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2893", "queId": "b6a0aee66e954700887a0ee4cb7cc831", "competition_source_list": ["2011年第7届全国新希望杯五年级竞赛A卷第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$20.11-\\square \\times 0.03+0.27=19.9$$，则$$\\square $$=（~ ）． ", "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->思想->方程思想"], "answer_analysis": ["将当成未知数，解方程即可得到=$$16$$，故选$$B$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1630", "queId": "ff80808146ec1d88014701e886711545", "competition_source_list": ["2019年北京四年级上学期单元测试《数学百花园》（北京版）（2）第2题", "2012年全国华杯赛小学中年级竞赛初赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "小虎在$$19\\times 19$$的围棋盘的格点上摆棋子，先摆成了一个长方形的实心点阵．然后再加上$$45$$枚棋子，就正好摆成一边不变的较大的长方形的实心点阵．那么小虎最多用了（ ~ ~ ~ ）枚棋子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$285$$ "}], [{"aoVal": "B", "content": "$$171$$ "}], [{"aoVal": "C", "content": "$$95$$ "}], [{"aoVal": "D", "content": "$$57$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$45=3\\times 3\\times 5$$，它小于$$19$$的最大约数为$$15$$，所以不变的边长应为$$15$$，另一边最长为$$19$$，所以小虎最多用了$$15\\times19=285$$（枚）棋子． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1980", "queId": "eea8715fd1084d47875bebb16fa215c0", "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33题"], "difficulty": "2", "qtype": "single_choice", "problem": "四个人每人有一美元（＄$$1$$），再加上一个人，则五个人平均每人$$10$$美元（＄$$10$$）．求第五人有多少钱？ ", "answer_option_list": [[{"aoVal": "A", "content": "＄$$9$$ "}], [{"aoVal": "B", "content": "＄$$19$$ "}], [{"aoVal": "C", "content": "＄$$46$$ "}], [{"aoVal": "D", "content": "＄$$49$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$5\\times10-4\\times1=46$$（美元）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3109", "queId": "d026a163717e4dd299c0b6a1c61929f9", "competition_source_list": ["2003年六年级竞赛创新杯", "2003年五年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$a\\times \\frac{5}{4}=b\\times \\frac{6}{7}=c\\times \\frac{4}{9}=d\\times \\frac{8}{8}$$，并且$$a$$、$$b$$、$$c$$、$$d$$都不等于零，那么$$a$$、$$b$$、$$c$$、$$d$$四个数的大小关系是(  ) ", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater d\\textgreater b\\textgreater c$$ "}], [{"aoVal": "B", "content": "$$b\\textgreater c\\textgreater d\\textgreater a$$ "}], [{"aoVal": "C", "content": "$$c\\textgreater b\\textgreater d\\textgreater a$$ "}], [{"aoVal": "D", "content": "$$c\\textgreater d\\textgreater b\\textgreater a$$ "}]], "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": ["若$$a\\times \\frac{5}{4}=b\\times \\frac{6}{7}=c\\times \\frac{4}{9}=d\\times \\frac{8}{8}=1$$，  那么$$a=\\frac{4}{5}$$，$$b=\\frac{7}{6}$$，$$c=\\frac{9}{4}$$，$$d=1$$，  显然$$c\\textgreater b\\textgreater d\\textgreater a$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "521", "queId": "cb63199c4b4c483b935b740a99f37b1c", "competition_source_list": ["2013年中环杯三年级竞赛决赛"], "difficulty": "0", "qtype": "single_choice", "problem": "星期天，许多帮助妈妈做一些家务。已知做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。为了节约时间，上面哪些事情可以同时做？（  ） ", "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": "用洗衣机洗衣服、洗碗 "}]], "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：合理利用空闲时间。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "82", "queId": "2608785033c3488e956b9d2535788f0b", "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": "橙子 "}]], "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": "2174", "queId": "98e9a9f26ea44322b6462920404eac1d", "competition_source_list": ["2017年全国华杯赛小学高年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$3$$点$$15$$分到$$3$$点$$45$$分，钟面的分针转动了一个（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "锐角 "}], [{"aoVal": "B", "content": "直角 "}], [{"aoVal": "C", "content": "平角 "}], [{"aoVal": "D", "content": "周角 "}]], "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": ["分针走了$$30$$分钟，$$60$$分钟为一个周角，$$30$$分钟则是一个平角． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2576", "queId": "39e8ab81daf0467d95529ddc4517982f", "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": "469", "queId": "bc6b1472941549bab1df3bace6be2aa0", "competition_source_list": ["2016年全国华杯赛小学中年级竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙、丁四支足球队进行比赛．  懒羊羊说：甲第一，丁第四；  喜羊羊说：丁第二，丙第三；  沸羊羊说：丙第二，乙第一，每个的预测都只对了一半，那么，实际的第一名至第四名的球队依次是． ", "answer_option_list": [[{"aoVal": "A", "content": "甲乙丁丙 "}], [{"aoVal": "B", "content": "甲丁乙丙 "}], [{"aoVal": "C", "content": "乙甲丙丁 "}], [{"aoVal": "D", "content": "丙甲乙丁 "}]], "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->半真半假"], "answer_analysis": ["假设懒羊羊所说的``甲第一''正确，则``丁第四''错误；沸羊羊所说的``乙第一''错误，则``丙第二''正确；此时喜羊羊的两句话都错误，与题意不符，故假设不成立．  假设懒羊羊所说的``甲第一''错误，则``丁第四''正确；喜羊羊所说的``丁第二''错误，则``丙第三''正确；沸羊羊所说的``丙第二''错误，则``乙第一''正确．故实际的第一名至第四为：乙甲丙丁．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3125", "queId": "ebfce811c4a54631bea9279d5bd1c55f", "competition_source_list": ["2019年第7届湖北长江杯六年级竞赛复赛A卷第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一袋面粉的质量标识为``$$25\\pm 0.25$$千克''，则下列面粉中合格的有． ", "answer_option_list": [[{"aoVal": "A", "content": "$$24.70$$千克 "}], [{"aoVal": "B", "content": "$$25.30$$千克 "}], [{"aoVal": "C", "content": "$$25.51$$千克 "}], [{"aoVal": "D", "content": "$$24.80$$千克 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["因为由题干可知，这袋面粉的质量标识为``$$25\\pm 0.25$$千克''，  则合格的范围是：$$24.75$$到$$25.25$$之间，  所以下列面粉中合格的有$$24.80$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "696", "queId": "755f88efb72f4f8891844d79e9ffcab8", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛决赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "三个质数的倒数和为$$\\frac{311}{1001}$$，那么这三个质数的和为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$311$$ "}], [{"aoVal": "B", "content": "$$35$$ "}], [{"aoVal": "C", "content": "$$31$$ "}], [{"aoVal": "D", "content": "$$29$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$1001=7\\times 11\\times 13$$，$$\\frac{311}{1001}=\\frac{7\\times 11+7\\times 13+11\\times 13}{1001}=\\frac{1}{7}+\\frac{1}{11}+\\frac{1}{13}$$，$$7+11+13=31$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "791", "queId": "6491b2d5ced04522ac9d66b595f676e2", "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": "$$135$$ "}], [{"aoVal": "E", "content": "$$136$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "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": "678", "queId": "7e721e4bcc4443e8968e6cee3d07c691", "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": "【闯关1】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->赋值思想"], "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": "254", "queId": "e395458a7f3945a087dd25d869360976", "competition_source_list": ["2019年第24届YMO三年级竞赛决赛第6题3分", "2020年第24届YMO三年级竞赛决赛第6题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "桌上放着$$100$$张卡片，二人轮流每次取卡片，每次最少取$$1$$张、最多取$$5$$张，规定谁取走最后一张卡片谁获胜．如果第一个取的人想获胜，第一次他应该取张卡片． ", "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": ["由题干可知：每次每人最少取$$1$$张卡片，最多取$$5$$张卡片，  $$100\\div \\left( 5+1 \\right)=100\\div 6=16$$（组）$$\\cdots \\cdots 4$$（张），  所以首先他先拿$$4$$张，剩下的$$6$$张卡片为一组，  保证每次与对方共取一组，这样最后一张卡片必是他的．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "320", "queId": "848827c2654e41468925e2fd7bf4cdf7", "competition_source_list": ["2016年全国小学生数学学习能力测评五年级竞赛复赛第7题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "音乐课上，聪聪坐在音乐教室的第$$4$$列、第$$2$$行，用数对表示是$$(3,1)$$，明明坐在聪聪正后方的第一个位置上，明明的位置用数对表示是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 3,2 \\right)$$ "}], [{"aoVal": "B", "content": "$$(4,1)$$ "}], [{"aoVal": "C", "content": "$$(4,2)$$ "}], [{"aoVal": "D", "content": "$$(5,2)$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->方向与坐标->坐标"], "answer_analysis": ["聪聪的正后方的第一个位置和聪聪的列数相同，行数加一，所以表示为$$\\left( 3,2 \\right)$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "879", "queId": "d6b9abb4084d48bc8eb5d9f4a2ba8521", "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": "$$53$$ "}], [{"aoVal": "C", "content": "$$158$$ "}], [{"aoVal": "D", "content": "$$22$$ "}]], "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": ["设这个数为$$a$$，则$$\\begin{cases}a\\div 3\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 2    a\\div 5\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 3    a\\div 7\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 4 \\end{cases}$$，整理得$$\\begin{cases}a\\div 3\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 11    a\\div 5\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 3    a\\div 7\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 11 \\end{cases}$$，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": "2496", "queId": "8b07a0da056143d586a33aa39791822e", "competition_source_list": ["2020年希望杯二年级竞赛模拟第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "根据数串的规律，括号中应填的数是．  1，$$4$$，$$7$$，$$10$$，，16 ", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->思想->数形结合思想"], "answer_analysis": ["等差数串，每两个数之间的差是3 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3170", "queId": "5419be99998b4f57ac4a239ade3772e9", "competition_source_list": ["2016年陕西西安小升初交大附中入学真卷5第11题", "2016年河南郑州K6联赛六年级竞赛第12题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一次外语小测验只有两道题，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$$10$$人是全部做对，则有$$25-10=15$$人是只做对第一题，而做错第二题的．已知第二题总共有$$18$$人做错，则多余的$$3$$人就是全错的． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2636", "queId": "5a3a654aea174b6dabdb761e8e6ce18c", "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": "725", "queId": "516e742a37bf40c286c49818c5ca1e88", "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5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "两个连续奇数的乘积是$$783$$，那么这两个奇数的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$52$$ "}], [{"aoVal": "B", "content": "$$56$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$64$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$783=27\\times 29$$，  则这两个连续奇数的和为：$$27+29=56$$．  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": "2679", "queId": "68479852782b43f5bfcda02cdc8adfd9", "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4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "把$$150$$块糖分成若干份，每份最少$$1$$块，任意两份数量都不相同，则最多可以分成份． ", "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": ["根据等差数列求和公式（首项$$+$$末项）$$\\times $$项数$$\\div 2$$可得$$\\left( 1+n \\right)\\times n\\div 2\\leqslant 150$$，则$$n=16$$，所以答案选择$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "735", "queId": "da556eb4ba3a4ecd9565736d0009b7e1", "competition_source_list": ["2009年第14届全国华杯赛竞赛复赛第11题", "2016年浙江杭州五年级模拟考试卓学一周一练12第3题", "2009年全国华杯赛竞赛复赛第11题"], "difficulty": "3", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$，$$c$$是三个自然数，且$$a$$与$$b$$的最小公倍数是$$60$$，$$a$$与$$c$$的最小公倍数是$$270$$．求$$b$$与$$c$$的最小公倍数是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$540$$ "}], [{"aoVal": "B", "content": "$$108$$ "}], [{"aoVal": "C", "content": "$$A$$,$$B$$都对 "}], [{"aoVal": "D", "content": "$$A$$,$$B$$都不对 "}]], "knowledge_point_routes": ["知识标签->课内知识点->数的认识->数的特征->倍数->最小公倍数"], "answer_analysis": ["如果$$b$$不是$${{2}^{2}}$$的倍数，因为$$\\left[ \\text{a,b} \\right]={{2}^{2}}\\times 3\\times 5$$，则$$a$$一定是$${{2}^{2}}$$的倍数．由此可知$$\\left[ \\text{a,c} \\right]$$一定是$${{2}^{2}}$$的倍数，但是$$\\left[ \\text{a,c} \\right]={{2}^{2}}\\times 3\\times 5$$不是$${{2}^{2}}$$的倍数．所以$$b$$是$${{2}^{2}}$$的倍数．同理可得$$c$$是$${{3}^{3}}$$的倍数，所以$$\\left[ \\text{b,c} \\right]={{2}^{2}}\\cdot {{3}^{3}}$$整除．  因为$$\\left[ \\text{a,b} \\right]=60$$，$$\\left[ \\text{a,c} \\right]=270$$，所以$$60$$是$$b$$的倍数，$$270$$是$$c$$的倍数，所以$$b$$，$$c$$的最小公倍数$$\\left[ \\text{b,c} \\right]$$是$$\\left[ 60,270 \\right]$$的约数．因为$$\\left[ 60,270 \\right]={{2}^{2}}\\cdot 3\\cdot 5$$，所以$$\\left[ \\text{b,c} \\right]={{2}^{2}}\\cdot 3\\cdot 5$$=540，或$$\\left[ \\text{b,c} \\right]={{2}^{2}}\\cdot 3$$=108．  当$$a=1$$,b=60,c=270时，$$\\left[ \\text{a,c} \\right]=60$$，$$\\left[ \\text{a,c} \\right]=270$$，$$\\left[ \\text{b,c} \\right]=540$$；  当$$a=5$$,b=12,c=54时，$$\\left[ \\text{a,c} \\right]=60$$，$$\\left[ \\text{a,c} \\right]=270$$，$$\\left[ \\text{b,c} \\right]=108$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3319", "queId": "9a245445bf9a4e89a064a7e0e3848e4d", "competition_source_list": ["2009年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "由数字$$1$$、$$2$$、$$3$$、$$4$$、$$5$$组成没有重复数字的五位数，由小到大排列，则$$35214$$是第（ ）个数。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$72$$ "}], [{"aoVal": "B", "content": "$$70$$ "}], [{"aoVal": "C", "content": "$$71$$ "}], [{"aoVal": "D", "content": "$$69$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->排列组合->排列"], "answer_analysis": ["比$$35214$$小的五位数（由数字$$1$$、$$2$$、$$3$$、$$4$$、$$5$$组成的没有重复数字的五位数）有以下三类：  （I）以$$1$$或$$2$$开头的五位数：形如$$12345$$的有$$4\\times 3\\times 2\\times 1=24$$（个）；如$$21345$$的有$$4\\times 3\\times 2\\times 1=24$$（个）；  （II）以$$31$$、$$32$$或$$34$$开头的五位数：形如$$31245$$的有$$3\\times 2\\times 1=6$$（个）；同理，形如$$32145$$，$$34512$$的各有$$6$$个；  （III）以$$351$$开头的五位数有$$2\\times 1=2$$（个）。  所以$$35214$$是第$$24\\times 2+6\\times 3+2+1=69$$（个）数。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1890", "queId": "8aac50a74ff4b16201500e7a7e832aa4", "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$$3$$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": "3202", "queId": "2b8f395758624ce98f3b2b4e275e03ba", "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3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两数的最大公因数是$$50$$，最小公倍数是$$600$$，这两个数的差最小是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$200$$ "}]], "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": ["因为$$600\\div 50=12$$，$$12=1\\times 12=2\\times 6=3\\times 4$$，所以这两个数有如下情况：  $$50\\times 1=50$$；$$50\\times 12=600$$；$$50\\times 2=100$$；$$50\\times 6=300$$；  $$50\\times 3=150$$；$$50\\times 4=200$$；  差依次是：$$600-50=550$$；$$300-100=200$$；$$200-150=50$$，最小是$$50$$，  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": "251", "queId": "a2f584ccc1ad4b20b08e78bd9b718ee2", "competition_source_list": ["2017年全国华杯赛小学中年级竞赛初赛模拟第4题", "其它改编题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁四人对$$A$$先生的藏书数目作了一个估计．甲说：``$$A$$先生有$$500$$本书．''乙说：``$$A$$先生至少有$$1000$$本书．''丙说：``$$A$$先生的书不到$$2000$$本．''丁说：``$$A$$先生最少有$$1$$本书．''实际上这四个人的估计中只有一句是对的．请问：$$A$$先生究竟有~\\uline{~~~~~~~~~~}~本书． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$500$$ "}], [{"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->真假话->找矛盾"], "answer_analysis": ["假设甲对：则丙也是对的，矛盾，假设不成立；假设乙对：则丁也是对的，矛盾，假设不成立；假设丙对：则其他三人的话可以全错，假设可以成立，此时，$$A$$先生有$$0$$本书；假设丁对：则其他三人必须全错，看甲、$$A$$先生藏书不是$$500$$本，看乙、$$A$$先生藏书不够$$1000$$本，看丙、$$A$$先生藏书至少$$2000$$本，出现矛盾，所以假设不成立．所以，丙说的对，$$A$$先生实际上没有书，$$0$$本． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3403", "queId": "cde6b7fa226043ffae1718ced33a44cc", "competition_source_list": ["2008年第6届创新杯五年级竞赛初赛A卷第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "从$$1$$到$$50$$这五十个自然数中，取两个数相加，要使它们的和大于$$50$$，共有种不同取法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$560$$ "}], [{"aoVal": "B", "content": "$$605$$ "}], [{"aoVal": "C", "content": "$$625$$ "}], [{"aoVal": "D", "content": "$$750$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["从$$1$$到$$50$$这$$50$$个自然数中，取两个数相加，  要使它们的和大于$$50$$，这样的数太多了，  我们计数时可以首先取$$50$$，  它与$$1$$、$$2$$、$$3\\cdots 49$$的和都大于$$50$$，  有$$49$$个数；然后再取$$49$$，  和它前面的数字$$2$$、$$3$$、$$4\\cdots 48$$的和都大于$$50$$，  有$$47$$个；  依次向前取$$48$$，和它的和大于$$50$$的有$$45$$个数；  $$\\cdots $$直到取数字$$26$$，  只有$$25+26=51$$大于$$50$$，$$1$$个数字；  这样分步完成，符合加法原理，是一组等差为$$2$$的数字，因此得解．  $$49+47+45+43+\\cdots +1$$  $$=(1+49)\\times 25\\div 2$$  $$=25\\times 25$$  $$=625$$（种）．  答：从$$1$$到$$50$$这$$50$$个自然数中，取两个数相加，要使它们的和大于$$50$$，共有$$625$$种不同的取法．  故答案为：$$625$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1370", "queId": "6c24efad627a486d8e51da9f130998e1", "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": "$$90$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$110$$ "}], [{"aoVal": "D", "content": "$$120$$ "}], [{"aoVal": "E", "content": "$$132$$ "}]], "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": "147", "queId": "34f2d687bc0b4fc4a3e44e72c16a9c28", "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7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在下面的``$$\\bigcirc $$''中选出相同的数，使等式成立．  $$4.3\\times \\bigcirc -1.1=1.3\\times \\bigcirc +3.7$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.6$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$1.6$$ "}], [{"aoVal": "D", "content": "$$2.3$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意，本题属于算式谜，解题关键是先化简；  首先把$$3.7$$和$$1.1$$相加，把$$\\bigcirc $$想成未知数；  得到算式$$4.3\\times \\bigcirc -1.3\\times \\bigcirc =3.7+1.1$$，进一步解答，求出$$\\bigcirc $$所代表的数即可．  $$4.3\\times \\bigcirc -1.1=1.3\\times \\bigcirc +3.7$$  $$4.3\\times \\bigcirc -1.3\\times \\bigcirc =3.7+1.1$$  $$3\\times \\bigcirc =4.8$$  $$\\bigcirc =1.6$$  因此$$\\bigcirc $$代表的数字是$$1.6$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1909", "queId": "ffea06f28a574d27af16cea014c02ee5", "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10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$1991991\\cdots\\cdots$$这串数中，从第三个数开始，每个数都是前两个数相乘后积的尾数（个位数字），那么把这串数字写到第$$40$$位时，各个数位上的数字和是．（某年全国小学生数学学习能力测评四年级竞赛初赛） ", "answer_option_list": [[{"aoVal": "A", "content": "$$290$$ "}], [{"aoVal": "B", "content": "$$248$$ "}], [{"aoVal": "C", "content": "$$250$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["观察此数字串是按照如下规律排列：$$199199199199\\cdots $$以$$199$$为数字循环节，  第$$40$$位时共有$$13$$个$$199$$，然后是$$1$$，  所以总和为$$13\\times (1+9+9)+1=248$$．  不难发现，这串数有周期性的规律；$$1$$，$$9$$，$$9$$循环出现．  $$40\\div3=13$$（组）$$\\cdots\\cdots 1$$（个），  $$13\\times (1+9+9)+1=248$$，  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "38", "queId": "06c23fd9456144ee8ab864c1d9c1b946", "competition_source_list": ["2023年袋鼠数学竞赛(Math Kangaroo)小学高年级竞赛第14题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$Maria$$、$$Peter$$、$$Richard$$ 和$$Tina$$ 在教室里踢足球，其中一个人打碎了一扇窗户．校长询问是谁干的，得到以下回答：  $$Maria$$说：``是$$Peter$$干的．\"  $$Peter$$说： `` 是$$Richard$$干的．  $$Richard$$ 说：``不是我．\"  $$Tina$$说：``不是我．''只有一个孩子说了实话．请问是谁打碎了窗户 ", "answer_option_list": [[{"aoVal": "A", "content": "$$Maria$$ "}], [{"aoVal": "B", "content": "$$Tina$$ "}], [{"aoVal": "C", "content": "$$Peter$$ "}], [{"aoVal": "D", "content": "$$Richard$$ "}], [{"aoVal": "E", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->能力->推理推导->言语逻辑推理"], "answer_analysis": ["Peter said that Richard did it, while Richard said that he didn't do it. Among these two sentences, one must be true and the other must be false. Since only one child among the four told the truth, the remaining two people, Maria and Tina, must have both told lies. Therefore, Tina's statement that she didn't do it is a lie. The one who broke the window is Tina. Choose B.  $$Peter$$说是$$Richard$$干的，而$$Richard$$说不是他干的，这两句话当中必定有一句真一句假，而四个人当中只有一个孩子说了真话，因此剩下的两个人$$Maria$$和$$Tina$$说的一定都是假话，因此$$Tina$$说不是她干的是假话，打碎窗户的就是$$Tina$$，选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2624", "queId": "295648acfa6845b7ac7e1feac56f6089", "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（五）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$A=\\underbrace{111\\cdots 1}_{2018个1}$$，那么$$A$$除以$$7$$的余数是（~ ）． ", "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->能力->运算求解"], "answer_analysis": ["$$111111$$可被$$7$$整除，$$2018\\div 6=336\\cdots \\cdots 2$$，最后剩下两个$$1$$，$$11\\div 7=1\\cdots \\cdots 4$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "709", "queId": "2944c4dee83c4b14abcbfc5db12cb9cb", "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->能力->逻辑分析", "Overseas Competition->知识点->数论模块->分解质因数"], "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": "777", "queId": "7b0bb563f95d48b2b341df44ff07439a", "competition_source_list": ["2014年第26届广东广州五羊杯六年级竞赛第6题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "乘积$$1\\times 2\\times 3\\times 4\\times 5\\times \\cdots \\cdots \\times 49\\times 50$$可以简记为$$50$$！，读作$$50$$的阶乘，这个乘积能被$$3$$整除，如果把这个乘积去反复除以$$3$$，直到不能被$$3$$整除为止，从第一次除以$$3$$开始算共可除以$$3$$次． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "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": ["这是分解质因数的方法（一直除到商比质因数$$3$$小为止）：  $$50\\div 3=16\\cdots 2$$；$$16\\div 3=5\\cdots 1$$；$$5\\div 3=1\\cdots 2$$．  把商求和得$$16+5+1=22$$．  其他质因数的指数也可用这种方法．常见的题目就是求阶乘的尾零个数． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3397", "queId": "cdcd64f2e2b8450cab7cd34bf0cf23aa", "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8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某阅览室有不同的文科类图书$$60$$本，不同的理科类图书$$80$$本，如果每类图书都最多只能借一本，则共有种不同的借法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$80$$ "}], [{"aoVal": "B", "content": "$$140$$ "}], [{"aoVal": "C", "content": "$$4800$$ "}], [{"aoVal": "D", "content": "$$4940$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["文科书有$$60$$本，所以单借一本文科书有$$60$$种借法；  理科书有$$80$$本，所以单借一本理科书有$$80$$种借法；  同时借一本文科书和一本理科书有$$60\\times80=4800$$种借法．  则共有$$60+80+4800=4940$$种借法．  故选：$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2606", "queId": "c796b0fff44e44d5b9529e695411da42", "competition_source_list": ["其它改编自2013年全国希望杯六年级竞赛初赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "计算：$$30 \\%\\div 1\\frac{2}{5}\\times(\\frac{1}{3}+\\frac{1}{7})=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{21}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{21}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{49}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{49}$$ "}]], "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": ["$$30 \\%\\div 1\\frac{2}{5}\\times(\\frac{1}{3}+\\frac{1}{7})=\\frac{3}{10}\\times \\frac{5}{7}\\times \\frac{10}{21}=\\frac{5}{49}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2730", "queId": "ff8080814502fa2401450bc9eefe16a2", "competition_source_list": ["2008年第13届全国迎春杯竞赛初赛第5题10分", "2008年全国华杯赛竞赛初赛第5题", "小学高年级六年级其它小高著名杯赛拉分题第4题"], "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$$a$$的所有数位上的数字和等于（ ~ ~ ~ ）． ", "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->拓展思维->计算模块->整数->整数分配律->整数乘法巧算之分配律->拆数法构造分配律"], "answer_analysis": ["$${a = \\underbrace {1515 \\cdots 15}_{1004 个 15}~ \\times ~\\underbrace {333 \\cdots 3}_{2008个3}}$$  $${ = \\underbrace {505050 \\cdots 5}_{1004个 5 和 1003 个0}\\times ~ \\underbrace {999 \\cdots 9}_{2008个9}}$$  $${ = \\underbrace {505050 \\cdots 5}_{1004个 5 和 1003 个0}\\times~ （1 \\underbrace {0000 \\cdots 0}_{2008个0} - 1}）$$  $${ = \\underbrace {505050 \\cdots 50}_{1004个50} \\underbrace {0000 \\cdots 0}_{2007个0} ~ - \\underbrace {505050 \\cdots 5}_{1004个 5 和 1003 个0}}$$  $${ = \\underbrace {505050 \\cdots 50}_{1003个50}\\underbrace {494949 \\cdots 49}_{1004个49}{5}}$$  所以整数$$a$$的所有数位上的数字和$$ = 1003 \\times 5 + 1004 \\times 4 + 9 + 5 = 18072$$ ． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1315", "queId": "2831f6a594824631ad5f9dd569ab534c", "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": "$$10$$，$$10$$，$$10$$，$$20$$ "}], [{"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->能力->运算求解"], "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": "2807", "queId": "4ec72c3bb8984873bb7a33f84ef4bfe0", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第5题4分", "2021年第8届鹏程杯四年级竞赛初赛第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->能力->运算求解->运算符号应用"], "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": "2432", "queId": "98b95285948644958615d25c287aa8a8", "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6题5分", "2007年第5届创新杯五年级竞赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\frac{6}{7}$$化成小数后，小数点后面前2007位数字之和等于（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "9038 "}], [{"aoVal": "B", "content": "9040 "}], [{"aoVal": "C", "content": "9041 "}], [{"aoVal": "D", "content": "9044 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->循环小数->分数化循环小数"], "answer_analysis": ["因为$$\\frac{6}{7}=0.857142857142\\cdots $$，所以(一个循环的数字和)$$8+5+7+1+4+2=27$$，又$$2007=6\\times 334+3$$，因此，小数点后面前$$2007$$位数字之和为$$27\\times 334+\\left( 8+5+7 \\right)=9038$$. "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "263", "queId": "beb62c88e5f94b599af19cdcffc43f98", "competition_source_list": ["2017年天津陈省身杯六年级竞赛国际青少年数学邀请赛第16题", "2009年全国迎春杯小学中年级竞赛复赛第5题", "六年级其它", "2017年天津陈省身杯六年级竞赛第16题", "2009年全国迎春杯小学中年级竞赛复赛第5题8分"], "difficulty": "4", "qtype": "single_choice", "problem": "$$A$$、$$B$$、$$C$$、$$D$$、$$E$$、$$F$$六个足球队进行单循环比赛，每两个队之间都要赛一场，且只赛一场．胜者得$$3$$分，负者得$$0$$分，平局每队各得$$1$$分．比赛结果，各队得分由高到低恰好为一个等差数列，获得第$$3$$名的队得了$$8$$分，那么这次比赛中共有~\\uline{~~~~~~~~~~}~场平局． ", "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->组合模块->逻辑推理->体育比赛->3-1-0 积分制"], "answer_analysis": ["六个足球队进行单循环比赛，总共有$$5+4+3+2+1=15$$（场）比赛．  平局的两队总分为$$1+1=2$$（分），非平局总分为$$0+3=3$$（分），因此，如果全是非平局总分有$$15\\times 3=45$$（分），否则多一场平局总分减少$$1$$分．  由于第$$3$$名得了$$8$$分，最后一名至少$$0$$分，所以各队得分的构成的等差数列的公差不超过$$8\\div 3=2\\frac{2}{3}$$（分），只可能为$$1$$分或$$2$$分．  如果各队得分的构成的等差数列公差为$$1$$，则这六个队的总分为$$（8+7）\\times 3=45$$（分），则有$$0$$场平局，每场比赛每队都得$$0$$分或$$3$$分，则每支队的得分都应是$$3$$的倍数，与第$$3$$名得$$8$$分不符．  如果各队得分的构成的等差数列公差为$$2$$，则这六个队的总分为$$（8+6）\\times 3=42$$（分），有$$45-42=3$$（场）平局，符合题意．所以这次比赛中共有$$3$$场平局． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "142", "queId": "1ad41aa41a0940d6814c0c4098a513f6", "competition_source_list": ["2011年四年级竞赛明心奥数挑战赛"], "difficulty": "1", "qtype": "single_choice", "problem": "算式：$$\\text{猫}\\times\\text{鼠} \\times \\overline{\\text{猫鼠}}=\\overline{\\text{鼠鼠鼠}}$$，其中猫和鼠各代表一个一位数，则猫$$+$$鼠$$=$$（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜"], "answer_analysis": ["设猫$$=a$$，鼠$$=b$$，猫$$\\times $$鼠$$\\times \\overline{\\text{猫鼠}}=\\overline{\\text{鼠鼠鼠}}=a\\cdot b\\cdot \\overline{ab}=\\overline{bbb}=b\\times 111=b\\times 3\\times 37$$，可知$$a\\times \\overline{ab}=3\\times 37$$，所以$$a=3$$，$$b=7$$，则猫$$+$$鼠$$=3+7=10$$。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "430", "queId": "b729ac2743cc45fcb9129b66778c4bb4", "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": "$$22:05$$ "}]], "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": "560", "queId": "412cc7a038ba4d57978c2df68d72482c", "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决赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个自然数被$$3$$、$$5$$、$$7$$除的余数分别为$$1$$、$$2$$、$$4$$，三个商的整数部分之和是$$257$$，那么这个自然数除以$$11$$的余数是． ", "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->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": ["余数问题．假设这个数是$$N$$，根据$$N\\div 3\\cdots 1$$、$$N\\div 5\\cdots 2$$、$$N\\div 7\\cdots 4$$，  由逐级满足法可得$$N=105k+67$$，  $$[\\frac{67}{3}]+[\\frac{67}{5}]+[\\frac{67}{7}]=44$$，  $$(257-44)\\div ([\\frac{105}{3}]+[\\frac{105}{5}]+[\\frac{105}{7}])=3$$，  所以$$N=105\\times 3+67=382$$，$$382\\div 11=34\\cdots 8$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "844", "queId": "a8a71e5b7115479094ab3d30bd4719c8", "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40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": "将$$1$$，$$2$$，$$\\cdots $$，$$13$$，这$$13$$个不同的自然数重新排成一行，记作$${{a}_{1}}=13$$，$${{a}_{2}}=1$$．并使得$${{a}_{1}}+{{a}_{2}}+\\cdots +{{a}_{k}}$$能被$${{a}_{k+1}}$$整除（$$k=1$$，$$2$$，$$\\cdots $$，$$12$$），求$${{a}_{3}}+{{a}_{13}}$$的值． ", "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->能力->运算求解"], "answer_analysis": ["$$1$$到$$13$$的加和一定是$${{a}_{13}}$$的倍数，$$1+2+3\\cdots \\cdots +13=7\\times 13$$，所以$${{a}_{13}}$$是$$7$$，由于$${{a}_{1}}+{{a}_{2}}=14$$，所以$${{a}_{3}}$$只能是$$2$$，所以$${{a}_{3}}+{{a}_{13}}=9$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2065", "queId": "e2017c1922764fd5ab1e9e3e414d436a", "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": "2010", "queId": "e13c9107a0724936b8af778a84873bbe", "competition_source_list": ["2014年全国华杯赛小学高年级竞赛初赛B卷第5题", "2017年全国小升初八中入学备考课程"], "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，乙 $$13$$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": "175", "queId": "3e560a8c75334f6ea9dff671f495ec3c", "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、亮亮和刚刚三个人是好朋友，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，判断是工人．  （$$1$$）明明不是工人．  （$$2$$）亮亮不是医生．  （$$3$$）明明和亮亮正在听解放军讲战斗故事． ", "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，明明和亮亮都不是解放军，所以刚刚是解放军．那么亮亮既不是解放军也不是医生，所以亮亮是工人  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "880", "queId": "cd93804611e442408ac82d51e56230a4", "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$${{2}^{77232917}}-1$$．以下各数中，能被$$15$$整除的是． ", "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->思想->对应思想"], "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": "654", "queId": "273d221e080b43c2a011193e62bbaf76", "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2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "两个数的最大公因数是$$5$$，最小公倍数是$$75$$，那么这两个数的和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$80$$ "}], [{"aoVal": "C", "content": "$$40$$或$$50$$ "}], [{"aoVal": "D", "content": "$$40$$或$$80$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["公因数：两个或多个数都有的因数叫\\textbar 做公因数公倍数，  两个或多个数都有的倍数叫做公倍数最大公因数，  两个或多个数都有的因数里最大的叫做最大公因数最小公倍数，  两个或多个数都有的倍数里最小的叫做最小公倍数（没有最大公倍数），  两个数的最大公因数是$$5$$，最小公倍数是$$75$$，那么这两个数可以是$$5$$和$$75$$，它们的和就是$$80$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "878", "queId": "96edf656fde945c6b631556a549d47d3", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第12题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "一串数：$$1$$，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，$$\\cdots $$即从第三个数起，每个数都等于前面相邻两个数的和．则第$$2021$$个数被$$6$$除的余数是． ", "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->数论模块->余数问题->余数的性质->余数找规律"], "answer_analysis": ["把每一个数除以$$6$$的余数依次写出来，观察余数的变化规律，  这串数除以$$6$$的余数依次是$$1$$、$$1$$、$$2$$、$$3$$、$$5$$、$$2$$、$$1$$、$$3$$、$$4$$、$$1$$、$$5$$、$$0$$、$$5$$、$$5$$、$$4$$、$$3$$、$$1$$、$$4$$、$$5$$、$$3$$、$$2$$、$$5$$、$$1$$、$$0$$、$$1$$、$$1\\cdots \\cdots $$，  可以看出来周期是$$1$$、$$1$$、$$2$$、$$3$$、$$5$$、$$2$$、$$1$$、$$3$$、$$4$$、$$1$$、$$5$$、$$0$$、$$5$$、$$5$$、$$4$$、$$3$$、$$1$$、$$4$$、$$5$$、$$3$$、$$2$$、$$5$$、$$1$$、$$0$$，一共有$$24$$个，  $$2021\\div 24=84\\cdots \\cdots 5$$，  所以第$$2021$$个数除以$$6$$的余数和第五个数除以$$6$$的余数相同，第五个除以$$6$$的余数为$$5$$．  故选$$\\text{E}$$． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2820", "queId": "80164d6d748643b8ae3bf80b77f5833a", "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$$\\square +5=13-6$$中，请问$$\\square $$中应填入什么数才能使算式正确？ ", "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->整数->整数加减->整数减法运算->减法横式"], "answer_analysis": ["$$2+5=13-6$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1435", "queId": "2e2dd9a16c6a408194ee697140141b6f", "competition_source_list": ["2014年IMAS小学中年级竞赛第二轮检测试题第5题4分"], "difficulty": "2", "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->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": ["$$27=3\\times 3\\times 3$$，$$27$$分成一些连续的正整数，$$27$$一定是中间数的整数倍，所以小朋友最多有$$3$$位． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2231", "queId": "e82f5aaed4a14efabd70a8b76dfb82a5", "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": "$$6:22$$ "}], [{"aoVal": "E", "content": "7:33 "}]], "knowledge_point_routes": ["拓展思维->能力->图形认知"], "answer_analysis": ["略． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1108", "queId": "0c7a8afb4b694e89817ab48e0942e9ec", "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2题3分", "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", "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2题3分", "2014年迎春杯四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有四个数，它们的和是$$45$$，把第一个数加$$2$$，第二个数减$$2$$，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": "$$8$$，$$12$$，$$5$$，$$20$$ "}], [{"aoVal": "D", "content": "$$9$$，$$11$$，$$12$$，$$13$$ "}]], "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": ["解：设相同的结果为$$2x$$，  根据题意有：$$2x-2+2x+2+x+4x=45$$，  解得$$x=5$$，  所以原来的$$4$$个数依次是$$8$$，$$12$$，$$5$$，$$20$$。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "584", "queId": "2a401b12ec2b428da31cc7ab67a4779f", "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36题"], "difficulty": "2", "qtype": "single_choice", "problem": "满足以下条件的最小整数是多少：``除以$$3$$余$$2$$．除以$$5$$余$$4$$，除以$$7$$余$$6$$．'' ", "answer_option_list": [[{"aoVal": "A", "content": "$$59$$ "}], [{"aoVal": "B", "content": "$$61$$ "}], [{"aoVal": "C", "content": "$$104$$ "}], [{"aoVal": "D", "content": "$$106$$ "}]], "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": ["余数定理应用． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2205", "queId": "47a8086e28e04be3839ce16e4c04b0d0", "competition_source_list": ["2017年第15届湖北武汉创新杯六年级竞赛初赛第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个时钟每小时快$$3$$分钟，上午$$8:00$$对准标准时间，当这个钟在次日上午指示着$$8:30$$时，标准时间是（上午）~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8:20$$ "}], [{"aoVal": "B", "content": "$$7:20$$ "}], [{"aoVal": "C", "content": "$$7:22$$ "}]], "knowledge_point_routes": ["拓展思维->知识点->行程模块->时钟问题->坏钟问题"], "answer_analysis": ["该钟与标准时间的比为$$63:60=21:20$$，从上午$$8:00$$到次日上午$$8:30$$，共$$24\\times 60+30=1470$$分钟，标准时间经过$$1470\\div 21\\times 20=1400$$分钟，即$$23$$小时$$20$$分，所以标准时间是$$7:20$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2654", "queId": "4ce8294e868849e5882f8c1a3754d23d", "competition_source_list": ["2017年第22届全国华杯赛小学高年级竞赛初赛第5题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "在序列$$20170\\cdots $$中，从第$$5$$个数字开始，每个数字都是前面$$4$$个数字和的个位数．这样的序列可以一直写下去．那么从第$$5$$个数字开始，该序列中一定不会出现的数组是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8615$$ "}], [{"aoVal": "B", "content": "$$2016$$ "}], [{"aoVal": "C", "content": "$$4023$$ "}], [{"aoVal": "D", "content": "$$2017$$ "}]], "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": ["观察这列数的奇偶性，从第一项起前四个数为``偶偶奇奇''，下一项为``偶''，再下一项为``偶''，再下一项为``奇''，再下一项为``奇''，重复出现了``偶偶奇奇''，发生循环，所以这列数的奇偶性为:``偶偶奇奇偶偶偶奇奇偶偶偶奇奇偶$$\\cdots $$''．中间没有``偶偶奇偶''这种情况，所以不可能出砚$$2016$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1625", "queId": "649ed845d3414a97bfc347a5a8975225", "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5题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "小马虎在计算$$\\left( \\square +15 \\right)\\times 4$$时，忘掉了小括号，最后算得结果是$$90$$，你知道正确的答案是多少吗？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$90$$ "}], [{"aoVal": "C", "content": "$$180$$ "}], [{"aoVal": "D", "content": "以上都错 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->还原问题->小马虎题型"], "answer_analysis": ["根据题意，小马虎忘掉了括号结果是$$90$$，也就是$$\\square +15\\times 4=90$$，  那么$$\\square =90-60=30$$，  把$$\\square =30$$代入原式子，也就是：$$\\left( 30+15 \\right)\\times 4=45\\times 4=180$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3274", "queId": "9e39e60f45be47cc98f38e408e8c3510", "competition_source_list": ["2017年世界少年奥林匹克数学竞赛五年级竞赛初赛A卷第14题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "某班学生有$$40$$人参加篮球社，$$30$$人参加足球社，两个社团都参加的有$$20$$人，两个社团都没参加的一个也没有，这个班的学生共有人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$70$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$50$$ "}], [{"aoVal": "D", "content": "$$40$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["这个班有学生$$50$$人．  $$40+30=70$$（人），  $$70-20=50$$（人）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1722", "queId": "cd1af5d3b4274e2fb3bfc07231fd49ff", "competition_source_list": ["2015年全国迎春杯六年级竞赛网校模拟第7题"], "difficulty": "3", "qtype": "single_choice", "problem": "运动会连续开了$$n$$天，一共发了$$m$$枚奖章．第一天发一枚以及剩下$$(m-1)$$枚的$$\\frac{1}{7}$$，第二天发$$2$$枚以及发后剩下的$$\\frac{1}{7}$$，以后每天都按此规律发奖章，在最后一天即第$$n$$天发了剩下的$$n$$枚奖章，那么运动会一共发了枚奖章． ", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$49$$ "}], [{"aoVal": "D", "content": "$$64$$ "}]], "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": ["倒数第二天发了$$n-1$$枚和余下的$$\\frac{1}{7}$$，那么最后一天发的$$n$$枚就是余下的$$\\frac{6}{7}$$，可以看出$$n$$是$$6$$的倍数，进一步尝试发现，当$$n=6$$时，依次逆推可得，每天都发$$6$$枚奖章，运动会$$6$$天开完，恰好符合题目要求．所以一共发了$$36$$枚奖章． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "663", "queId": "1ef8f7c606e343a5ac372458d0008294", "competition_source_list": ["2011年全国学而思杯五年级竞赛第5题", "2011年全国学而思杯四年级竞赛第5题", "2011年北京学而思综合能力诊断六年级竞赛第3题"], "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~\\uline{~~~~~~~~~~}~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": "509", "queId": "e1ac535afddb471da569347349934379", "competition_source_list": ["2010年六年级竞赛明心奥数挑战赛"], "difficulty": "1", "qtype": "single_choice", "problem": "有一群体育爱好者，他们之中所有的桥牌爱好者都爱好围棋，有围棋爱好者爱好武术；所有的武术爱好者都不爱好健身操，有桥牌爱好者同时爱好健身操，如果上述结论都是真实的，则以下说法不对的是(  ) ", "answer_option_list": [[{"aoVal": "A", "content": "所有的围棋爱好者也都爱好桥牌； "}], [{"aoVal": "B", "content": "有的桥牌爱好者爱好武术； "}], [{"aoVal": "C", "content": "健身操爱好者都爱好围棋； "}], [{"aoVal": "D", "content": "围棋爱好者都爱好健身操。 "}]], "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 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2108", "queId": "ebcd787d47ff4e1c92f9f3db1550cac4", "competition_source_list": ["2018年第8届北京学而思综合能力诊断三年级竞赛年度教学质量监测第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "绿化队$$3$$小时种树$$21$$棵，还要种$$77$$棵，照这样的工作效率，完成任务还需要~\\uline{~~~~~~~~~~}~小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$21$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->归一归总问题->归总问题"], "answer_analysis": ["先计算$$1$$小时的种树量$$21\\div 3=7$$棵，$$77\\div 7=11$$小时 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3094", "queId": "f8c85ae7100c4e289d41fa683942581d", "competition_source_list": ["2016年全国世奥赛五年级竞赛A卷第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2003$$年九月六日，即将完成的价值$$2.9$$亿美元的$$NOAA-N-Prime$$高级气象卫星在工作时不慎落下．当时，一名技术人员把卫星底部的螺丝松掉了，但忘记记录在案．随后上来的一组技术人员在一侧转动承载着卫星的支架，却不知道底部的螺丝已经松掉了，也忽略了在卫星移动前需要查看所有螺丝是否固定的步骤．结果导致$$NOAA-N-Prime$$气象卫星摔倒在离地面大约$$1\\text{m}$$的混凝土地板上，历史上各国同样因为粗心，将螺丝钉遗忘或未拧紧，使得卫星爆炸导致机毁人亡的事件也有许多，惨痛的教训告诉我们，哪怕是一个螺丝钉一个小数点，因为粗心也会造成不可挽回的损失，下面看看，在数学学习过程中，由于粗心导致的错误吧！胡图图在计算四个分数$$\\frac{5}{3}$$，$$\\frac{3}{2}$$，$$\\frac{13}{8}$$，$$\\frac{5}{8}$$的平均值时由于粗心，把其中一个分数的分子和分母抄颠倒了，抄错后的答案和正确的答案最大相差（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{105}{416}$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{24}$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{15}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["要使抄错后的答案和正确的答案差最大，则分子和分母抄颠倒的分数最大，四个分数中最大的是$$\\frac{5}{3}$$，所以抄错后的答案和正确的答案最大相差$$(\\frac{5}{3}-\\frac{3}{5})\\div 4=\\frac{4}{15}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2106", "queId": "e2aad192f1134525a4e5a616c1449d64", "competition_source_list": ["2022年第9届广东深圳鹏程杯四年级竞赛初赛第25题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "今年父亲的年龄比儿子年龄的$$6$$倍大$$1$$岁，$$10$$年后父亲的年龄比儿子年龄的$$3$$倍小$$1$$ 岁，则今年儿子和父亲的年龄之和是岁. ", "answer_option_list": [[{"aoVal": "A", "content": "$$43$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$47$$ "}], [{"aoVal": "D", "content": "$$49$$ "}], [{"aoVal": "E", "content": "$$50$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之变倍型"], "answer_analysis": ["无 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1925", "queId": "f77786671d154e84b1a0f77913ebf316", "competition_source_list": ["2020年长江杯六年级竞赛复赛B卷第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "新星小学开展广播操比赛，六年级学生站成一个实心方阵（正方形队列）时，还多$$10$$人，如果站成一个每边多$$1$$人的实心方阵，则还缺少$$15$$人．该校六年级有人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$154$$ "}], [{"aoVal": "B", "content": "$$144$$ "}], [{"aoVal": "C", "content": "$$169$$ "}], [{"aoVal": "D", "content": "$$174$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["此题解答的关键是扩大的方阵共有多少人，要求扩大的方阵共有多少人，就要求出扩大的方阵每边上的人数．  扩大的方阵每边上有：  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left( 10+15+1 \\right)\\div 2$$  $$=16\\div 2$$  $$=13$$（人），  原来人数：  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde13\\times 13-15$$  $$=169-15$$  $$=154$$（人）．  答：原来有$$154$$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": "2938", "queId": "7d6c1960b0b146ad87150051539b698f", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第3题"], "difficulty": "2", "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->计算模块->整数->整数乘除->整数除法运算->带余除法"], "answer_analysis": ["除数$$=(472-427)\\div 5=9$$，$$472\\div 9$$余$$4$$，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": "1442", "queId": "2e40ff835fa747aa8854bca572d76823", "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": "【破冰行动3】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->对应思想"], "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": "2334", "queId": "1bec8b5bb6e549deaa27b12564eae88a", "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3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$\\frac{\\dfrac{2+3+6}{\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{6}}-\\dfrac{\\dfrac{1}{2}+\\dfrac{1}{3}+\\dfrac{1}{6}}{2+3+6}}{2\\times 3\\times 6}=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{11}{36}$$ "}], [{"aoVal": "B", "content": "$$\\frac{10}{36}$$ "}], [{"aoVal": "C", "content": "$$\\frac{10}{33}$$ "}], [{"aoVal": "D", "content": "$$-\\frac{10}{33}$$ "}], [{"aoVal": "E", "content": "$$-\\frac{11}{36}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$2+3+6=11$$，$$\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}=1$$，  则原式$$=\\frac{\\dfrac{11}{1}-\\dfrac{1}{11}}{2\\times 3\\times 6}=\\frac{\\dfrac{120}{11}}{36}=\\frac{120}{11}\\times \\frac{1}{36}=\\frac{10}{33}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3323", "queId": "7fb23a38bd264d97b871b52df5b63a18", "competition_source_list": ["2007年四年级竞赛创新杯", "2007年六年级竞赛创新杯", "2007年五年级竞赛创新杯", "2007年第5届创新杯六年级竞赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一部电视剧共$$8$$集，要在$$3$$天里播完，每天至少播一集，则安排播出的方法共有(  )种可能。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$22$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"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": ["设第一天、第二天、第三天依次播出$$x$$，$$y$$，$$z$$集，则$$x+y+z=8$$， 其中整数$$x$$，$$y$$，$$z$$均大于零，显然$$\\left( x\\text{，}y\\text{，}z \\right)=\\left( 6\\text{，}1\\text{，}1 \\right)$$；$$\\left( 5\\text{，}2\\text{，}1 \\right)$$；$$\\left( 5\\text{，}1\\text{，}2 \\right)$$；$$\\left( 4\\text{，}3\\text{，}1 \\right)$$；$$\\left( 4\\text{，}2\\text{，}2 \\right)$$；$$\\left( 4\\text{，}1\\text{，}3 \\right)$$；$$\\left( 3\\text{，}4\\text{，}1 \\right)$$；$$\\left( 3\\text{，}3\\text{，}2 \\right)$$；$$\\left( 3\\text{，}2\\text{，}3 \\right)$$；$$\\left( 3\\text{，}1\\text{，}4 \\right)$$； $$\\left( 2\\text{，}5\\text{，}1 \\right)$$；$$\\left( 2\\text{，}4\\text{，}2 \\right)$$；$$\\left( 2\\text{，}3\\text{，}3 \\right)$$；$$\\left( 2\\text{，}2\\text{，}4 \\right)$$；$$\\left( 2\\text{，}1\\text{，}5 \\right)$$；$$\\left( 1\\text{，}6\\text{，}1 \\right)$$；$$\\left( 1\\text{，}5\\text{，}2 \\right)$$；$$\\left( 1\\text{，}4\\text{，}3 \\right)$$；$$\\left( 1\\text{，}3\\text{，}4 \\right)$$；$$\\left( 1\\text{，}2\\text{，}5 \\right)$$；$$\\left( 1\\text{，}1\\text{，}6 \\right)$$，其中 $$\\left( x\\text{，}y\\text{，}z \\right)=\\left( 4\\text{，}3\\text{，}1 \\right)$$表示$$x=4$$，$$y=3$$，$$z=1$$，同理$$\\left( x\\text{，}y\\text{，}z \\right)=\\left( a\\text{，}b\\text{，}c \\right)$$表示$$x=a$$，$$y=b$$，$$z=c$$。因此，播出的方法有$$1+2+3+4+5+6=\\frac{1+6}{2}\\times 6=21$$（种）。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2145", "queId": "ab46d4d05d1b4d6486a659af9ce9600a", "competition_source_list": ["六年级上学期其它北师大版53天天练", "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"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两地相距$$1500$$千米．飞机从甲地到乙地是顺风，需$$2$$小时；从乙地返回甲地是逆风，需$$2.5$$小时，则飞机往返的平均速度是（~ ）千米∕时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$700$$~ "}], [{"aoVal": "B", "content": "$$666\\frac{2}{3}$$~ "}], [{"aoVal": "C", "content": "$$675$$~ "}], [{"aoVal": "D", "content": "$$650$$~ "}]], "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": ["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": "1187", "queId": "3d68bfe04a2f400ab35a79386447f9fa", "competition_source_list": ["2018年第17届春蕾杯一年级竞赛初赛第13题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "把一根粗细均匀的竹竿锯成$$2$$段需要$$3$$分钟，照这样计算把这根竹竿锯成$$5$$段需要几分钟? ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["两端不种树类型，次数$$=$$段数$$-1$$．锯成$$5$$段，需要锯$$4$$次，每一次$$3$$分钟；那么一共$$4+4+4+4=12$$（分钟）．  答：把这根竹竿锯成$$5$$段需要$$12$$分钟． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3196", "queId": "227c621faa4348b1845d45dc35297e24", "competition_source_list": ["2020年福建河仁杯六年级竞赛初赛A卷第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "若自然数$$n$$可以使得$$n+\\left( n+1 \\right)+\\left( n+2 \\right)$$不产生进位，则称$$n$$为有序数，如因为$$11+12+13$$不产生进位，所以``$$11$$''是有序数；因为$$40+41+42$$会产生进位，所以``$$40$$''不是有序数，则小于$$200$$的``有序数''共有个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$21$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["要想不发生任何进位，则不管三个数中的最小数为几位数，其个位一定不能超过$$2$$，十位及后面的数位都不能超过$$3$$，再利用加乘原理即可．  （$$1$$）个位数最小数只能是$$0$$、$$1$$、$$2$$共$$3$$种，  （$$2$$）两位数最小数十位$$1$$、$$2$$、$$3$$三种选择，个位$$0$$、$$1$$、$$2$$三种选择，  则共有：$$3\\times 3=9$$种，  （$$3$$）三位数最小数百位只能是$$1$$，十位$$0$$、$$1$$、$$2$$、$$3$$共四种选择，  个位$$0$$、$$1$$、$$2$$三种选择，共有$$1\\times 4\\times 3=12$$种，  综上，$$3+9+12=24$$种．  故本题正确答案为$$\\text{C}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1944", "queId": "e9d59cd68b0d48508905f2c733b3fc91", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（三）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "一项工程，小王单独干$$6$$天后，小刘接着单独干$$9$$天，可以完成总任务量的$$\\frac{2}{5}$$，如果小王单独干$$9$$天后，小刘接着干$$6$$天，可以完成总任务量的$$\\frac{7}{20}$$．则小王和小刘一起完成这项工程需要（~ ）天． ", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["小王单独干$$15$$天，小刘接着单独干$$15$$天，可以完成总任务量的$$\\left( \\frac{2}{5}+\\frac{7}{20} \\right)$$．  每天完成$$\\left( \\frac{2}{5}+\\frac{7}{20} \\right)\\div 15=\\frac{1}{20}$$，$$1\\div \\frac{1}{20}=20$$(天)． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "961", "queId": "ddba988391cd468497360c7af525579b", "competition_source_list": ["2016年创新杯六年级竞赛训练题（一）第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个自然数，用它分别去除$$63$$，$$90$$，$$130$$都有余数，三个余数的和为$$25$$，这三个余数中最小的一个是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$43$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$\\left( 63+90+130 \\right)-25=258$$是除数的倍数，$$258=2\\times 3\\times 43$$，验证只有除数是$$43$$的时候满足，此时余数中最小的是$$1$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2426", "queId": "14212be105324aba9bc5e6d59bf933ab", "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$$219$$，从第$$4$$个数字开始，每个数字都是前面$$3$$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": ["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": ["按照题目规则，写出数串$$21922372215847906512811\\cdots \\cdots $$发现数字的奇偶性为``偶奇奇偶偶奇奇偶偶奇奇偶偶$$\\cdots \\cdots$$''不可能出现选项$$A$$、$$B$$、$$C$$中三个偶数相连、三个奇数相连或奇偶奇偶的情况，且$$2215$$在数串中出现，则选$$C$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1047", "queId": "2ec2217457b3439db4be5efb2cdb82de", "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10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$1991991\\cdots\\cdots$$这串数中，从第三个数开始，每个数都是前两个数相乘后积的尾数（个位数字），那么把这串数字写到第$$40$$位时，各个数位上的数字和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$290$$ "}], [{"aoVal": "B", "content": "$$248$$ "}], [{"aoVal": "C", "content": "$$250$$ "}]], "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": ["观察此数字串是按照如下规律排列：$$199199199199\\cdots $$以$$199$$为数字循环节，  第$$40$$位时共有$$13$$个$$199$$，然后是$$1$$，  所以总和为$$13\\times (1+9+9)+1=248$$．  不难发现，这串数有周期性的规律；$$1$$，$$9$$，$$9$$循环出现．  $$40\\div3=13$$（组）$$\\cdots\\cdots 1$$（个），  $$13\\times (1+9+9)+1=248$$，  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2177", "queId": "6b8d36ed2e0c44149c087084dbda0f82", "competition_source_list": ["2022年第9届广东深圳鹏程杯四年级竞赛初赛第22题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙二人练习跑步，若甲让乙先跑$$20$$米，则甲跑$$20$$秒钟可追上乙；若乙比甲先跑$$7$$秒钟，则甲跑$$35$$秒钟能追上乙．则两人每秒钟共跑~\\uline{~~~~~~~~~~}~米． ", "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": ["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": ["无 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "540", "queId": "f07d6b5fd2fa463da1d888f05258c909", "competition_source_list": ["2020年新希望杯三年级竞赛初赛（巅峰对决）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "宝宝、星星、乐乐、强强四人参加长跑比赛．他们对名次分别做了预测.  星星说：\"乐乐是第一名.\"  宝宝说：\"星星是第一名.\"  乐乐说：\"我不是第一名.\"  强强说：\"我不是第一名.\"  比赛结束后发现，只有一人预测正确，那么． ", "answer_option_list": [[{"aoVal": "A", "content": "宝宝预测正确，星星是第一名 "}], [{"aoVal": "B", "content": "星星预测正确，乐乐是第一名 "}], [{"aoVal": "C", "content": "强强预测正确，乐乐是第一名 "}], [{"aoVal": "D", "content": "强强预测正确，宝宝是第一名 "}], [{"aoVal": "E", "content": "乐乐预测正确，强强是第一名 "}]], "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": "1051", "queId": "079d29f71c754927a258de8179ce5755", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（四）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "有一群大雁向南飞，迎面飞来一群雀，两雀对一雁，还多$$1$$雁；两雁对三雀还多$$2$$雀，雁有只． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设有$$x$$只雁，$$y$$只雀  $$x=\\frac{1}{2}y+1$$  $$y=\\frac{1}{2}x\\centerdot 3+2$$  可得$$x=8$$  $$y=14 $$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "393", "queId": "d218fa3f0f5244f89d100cae93777132", "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": "已知$$\\square \\square \\square +\\square \\square \\square =1949$$，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": ["知识标签->学习能力->七大能力->运算求解"], "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，所以各位的数字之和是$$9$$，十位数字之和是$$14$$，百位数字之和是$$18$$，共$$9+14+18=41$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3023", "queId": "b7fb3072396b4cbca2bb8466bf88c574", "competition_source_list": ["2014年IMAS小学中年级竞赛第一轮检测试题第18题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "小朋友玩游戏，老师让小朋友们站成一排，并从第一位开始依照$$1$$、$$2$$、$$3$$循环报数，最后一位小朋友报的数是$$2$$，请问这一派可能共有多少位小朋友？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$26$$ "}], [{"aoVal": "D", "content": "$$27$$ "}], [{"aoVal": "E", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由题意可知这一派同学的人数除以$$3$$后所得的余数是$$2$$，在各个选项中只有选项$$\\text{C}$$符合要求．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2640", "queId": "9e2f432807eb4f519dcf8a954963618e", "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35题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$9$$个连续整数之和是$$99$$．这$$9$$个数中最大的数是？ ", "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->对应思想"], "answer_analysis": ["$$99 \\div9+4=15$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1911", "queId": "e99f7a271e9c4fe29859ffff6c8787d5", "competition_source_list": ["2013年全国迎春杯小学高年级竞赛复赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "迎春小学六年级同学在某次体育达标测试中，达标的有$$900$$人，达标率为$$75 \\%$$，因故没有参加体育达标测试的占该年级全体同学人数的$$4 \\%$$．该年级共有~\\uline{~~~~~~~~~~}~人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1250$$ "}], [{"aoVal": "B", "content": "$$1200$$ "}], [{"aoVal": "C", "content": "$$1000$$ "}], [{"aoVal": "D", "content": "$$900$$ "}]], "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": ["参加测试的有$$900\\div 75 \\% =1200$$（人），所以共有$$1200\\div (1-4 \\% )=1250$$（人）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "997", "queId": "61a5d203edc8481d8e9179adfbd4aa52", "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": "$$145$$ "}], [{"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": "475", "queId": "af116cc958b54798b4f5d5067547ce2f", "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7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "用一个平底锅煎饼，每次可以放$$3$$张饼，每面要煎$$1$$分钟．如果有$$4$$张饼，两面都要煎，至少要用分钟． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["需要煎饼的总面数：$$4\\times 2=8$$（面）．  需要煎饼的次数：$$8\\div 3=2$$（次）$$\\cdots \\cdots 2$$（面），$$2+1=3$$（次）．  需要的总时间：$$3\\times 1=3$$（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": "2288", "queId": "db65fa9b18c7460bba788042ea1ed2c2", "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7题"], "difficulty": "0", "qtype": "single_choice", "problem": "我们都知道，灭霸和雷神都是外国人，所以他们其实是说英语的，灭霸说：``Leishen''~flies at $$800$$ km/hour and have $$4400$$ km to travel． It will take him  hours to complete his trip． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$4.5$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$5.5$$ "}]], "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->单人简单行程问题"], "answer_analysis": ["雷神每小时飞行$$800$$千米，它还有$$4400$$千米要飞，它还需要花多久？  $$4400\\div 800=5.5$$（小时）． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2040", "queId": "ef5216f106b54db08a04494ab2f7b5d6", "competition_source_list": ["2018年湖北武汉新希望杯五年级竞赛训练题（二）第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$56$$名学生给甲、乙、丙三人投票，开票中途累计，甲得$$17$$票，乙得$$14$$票，丙得$$11$$票，若得票最多的人当选，甲至少再得票才能保证当选． ", "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": ["$$(56-11)\\div 2=22.5$$，$$23-17=6$$（票）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "766", "queId": "ff8080814502fa240145171bd04a398e", "competition_source_list": ["2014年全国迎春杯六年级竞赛初赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "老师把一个三位完全平方数的百位告诉了甲，十位告诉了乙，个位告诉了丙，并且告诉三人他们的数字互不相同．三人都不知道其他两人的数是多少，他们展开了如下对话：  甲：我不知道这个完全平方数是多少．  乙：不用你说，我也知道你一定不知道．  丙：我已经知道这个数是多少了．  甲：听了丙的话，我也知道这个数是多少了．  乙：听了甲的话，我也知道这个数是多少了．  请问这个数是（~~~~~~~ ）的平方． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$28$$ "}], [{"aoVal": "D", "content": "$$29$$ "}]], "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": ["通过枚举不难发现，百位是$$6$$，$$8$$，$$9$$ 的满足条件的平方数分别只有$$625$$，$$841$$，$$961$$，  因此第一句说明百位不是$$6$$，$$8$$，$$9$$；  进而得知第二句说明十位不是$$2$$，$$4$$，$$6$$；  第三句说明这个数的个位在剩下所有可能中是唯一的，而只有当个位是$$4$$或$$9$$，$${{28}^{2}}=784$$，$${{17}^{2}}=289$$是唯一满足之前所有条件的数;  第四句说明甲在丙说话之前还不知道结果，而若百位是$$7$$，而$${{28}^{2}}=784$$，$${{17}^{2}}=289$$，  于是甲听完乙说话后已经知道结果了，因此百位只能是$$2$$．  从而这个数为$${{17}^{2}}=289$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "429", "queId": "c4c1a7e8eea14d77825dc62ba30bbbd0", "competition_source_list": ["2014年华杯赛六年级竞赛初赛", "2014年华杯赛五年级竞赛初赛"], "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->拓展思维->组合模块->逻辑推理->假设型逻辑推理"], "answer_analysis": ["如果甲是真的，则甲的卡片为$$5$$和$$2$$；乙无论是哪张卡片，一定是真的；  如果丙是真的，则丙的卡片为$$5$$和$$0$$；如果丁是真的，则丁的卡片为$$0$$和$$3$$；  通过观察可以发现，丙与甲和丁两个人均矛盾，所以说谎的一定是丙。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2052", "queId": "e65ca280bfb64ee4babc3c7ec15474ef", "competition_source_list": ["2014年全国学而思杯一年级竞赛第7题", "2017年全国小学低年级二年级其它思维创新大通关第6讲巧填算符第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "在下面算式的``$$\\square$$''中填入合适的``$$+$$''或``$$-$$''符号，使得结果尽可能大，那么，结果最大是~\\uline{~~~~~~~~~~}~．  $$12-(4 \\square3）\\square(2\\square1)$$ ", "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->思想->转化与化归的思想", "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": ["保证结果最大，就要尽量减小的数，后面加的尽量是大数，$$12-(4-3)+(2+1)=14$$． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3096", "queId": "e68470c66f884483a199d54cd2483f84", "competition_source_list": ["2016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": "若规定运算$$2*6=26-2-6=18$$、$$7*3=73-7-3=63$$，已知$$a*b=36$$，其中$$a$$、$$b$$都是不为零的一个数码，请问$$a$$等于多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}], [{"aoVal": "E", "content": "$$9$$ "}]], "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": ["有上面的可知$$a\\cdot b=10a+b-a-b=9a=36$$，∴$$a=4$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1357", "queId": "35b8c658925e4310829362a840a4daee", "competition_source_list": ["2017年新希望杯小学高年级六年级竞赛训练题（二）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（二）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "体育用品商店有篮球和排球共$$50$$个，其中篮球占$$60 \\%$$，当卖出一批篮球后，篮球占现存总数的$$37.5 \\%$$，卖出了（ ）个篮球． ", "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->思想->方程思想"], "answer_analysis": ["$$50\\times 60 \\%=30$$，设卖出$$x$$个篮球，则$$\\frac{30-x}{50-x}=37.5 \\%$$，解得$$x=18$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "125", "queId": "1668605c5f7f45c19657f04e4d67790a", "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9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$8$$个球编号是①至⑧，其中有$$6$$个球一样重，另外两个球都轻$$1$$克．为了找出这两个轻球，用天平称了$$3$$次．结果如下：  第一次：①$$+$$②比③$$+$$④重；  第二次：⑤$$+$$⑥比⑦$$+$$⑧轻；  第三次：①$$+$$③$$+$$⑤与②$$+$$④$$+$$⑧一样重．  那么两个轻球分别是． ", "answer_option_list": [[{"aoVal": "A", "content": "①，④ "}], [{"aoVal": "B", "content": "③，⑧ "}], [{"aoVal": "C", "content": "④，⑤ "}], [{"aoVal": "D", "content": "②，⑤ "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["第一次：①$$+$$②$$\\textgreater$$③$$+$$④，则③与④有轻球；  第二次：⑤$$+$$⑥$$ ~\\textless{} ~$$⑦$$+$$⑧，则⑤与⑥有轻球；  第三次：①$$+$$③$$+$$⑤$$=$$②$$+$$④$$+$$⑧，则轻球①、③、⑤占$$1$$个，②、④、⑧占$$1$$个．  则②、④、⑧中的轻球为④号，则③排除，所以①、③、⑤中轻球为⑤，  故④、⑤为轻球．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1337", "queId": "a282d67386bf4d30a7c61db9bbeaf95b", "competition_source_list": ["2017年第15届湖北武汉创新杯五年级竞赛初赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "某幼儿园的小朋友比老师多$$40$$人，每位老师管理$$4$$个还剩$$1$$个，那么这个幼儿园的老师人数是人 ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "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": ["$$(40-1)\\div (4-1)=39\\div 3=13$$人． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "45", "queId": "4f30969234994fef85c1965775200334", "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": ["拓展思维->思想->整体思想"], "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": "710", "queId": "9e13cb7b26b34312bdff87c69c3330ae", "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": "1281", "queId": "c76ad4aa6a014eacb338df7c4b0b31d2", "competition_source_list": ["2020年第36届AMC竞赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "2020年第$$36$$届$$AMC$$竞赛第$$3$$题 Carrie has a rectangular garden that measures $$6$$ feet by $$8$$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $$4$$ strawberry plants per square foot, and she harvests an average of $$10$$ straw-berries per plant. How many strawberries can she expect to harvest? ", "answer_option_list": [[{"aoVal": "A", "content": "$$560$$ "}], [{"aoVal": "B", "content": "$$960$$ "}], [{"aoVal": "C", "content": "$$1120$$ "}], [{"aoVal": "D", "content": "$$1920$$ "}], [{"aoVal": "E", "content": "$$3840$$ "}]], "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", "Overseas Competition->知识点->应用题模块->平均数问题"], "answer_analysis": ["There are $$6\\times 8=48$$ square foot, and hence $$4\\times 6\\times 8=192$$ plants.  Consquently $$192\\times 10=1920$$ strawberrics does Carrie harvest. "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3306", "queId": "642e8b964f1c4155bbbd63734410951f", "competition_source_list": ["2008年四年级竞赛创新杯"], "difficulty": "2", "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，砝码可以放在天平的两边，能称出不同的重量的种数是（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"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": ["每次从中任意拿出$$2$$个砝码使用，若保证不重复，共可拿$$6$$次，每次拿出的两个砝码同时使用能称出两种不同的重量，两砝码重量之和及两砝码重量之差。同时使用$$1$$克、$$2$$克的砝码可称$$3$$克、$$1$$克的重量，同时使用$$1$$克、$$4$$克的砝码可称$$5$$克、$$3$$克的重量，同时使用$$1$$克、$$8$$克的砝码可称$$9$$克、$$7$$克的重量，同时使用$$2$$克、$$4$$克的砝码可称$$6$$克、$$2$$克的重量，同时使用$$2$$克、$$8$$克的砝码可称$$10$$克、$$6$$克的重量，同时使用$$4$$克、$$8$$克的砝码可称$$12$$克、$$4$$克的重量，去掉重复的重量，可称$$10$$种不同重量：$$1$$克，$$2$$克，$$3$$克，$$4$$克，$$5$$克，$$6$$克，$$7$$克，$$9$$克，$$10$$克，$$12$$克。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3255", "queId": "5e8be4d9ec8a42be8e21a82fb97a2e1e", "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4题"], "difficulty": "1", "qtype": "single_choice", "problem": "由3，4，5，6排成没有重复数字的四位数，从小到大排起来，6345是第(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "16个 "}], [{"aoVal": "B", "content": "17个 "}], [{"aoVal": "C", "content": "18个 "}], [{"aoVal": "D", "content": "19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": ["以3、4、5为最高位的四位数共有$$3\\times 2\\times 3=18$$个，而6345是以6为最高位的最小四位数，所以从小到大排起来6345是第19个数 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1893", "queId": "8aac50a74ff4b16201500e88c73a2ae2", "competition_source_list": ["2009年全国迎春杯小学中年级四年级竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "一些奇异的动物在草坪上聚会．有独脚兽（$$1$$个头、$$1$$只脚）、双头龙（$$2$$个头、$$4$$只脚）、三脚猫（$$1$$个头、$$3$$只脚）和四脚蛇（$$1$$个头、$$4$$只脚）．如果草坪上的动物共有$$58$$个头、$$160$$只脚，且四脚蛇的数量恰好是双头龙的$$2$$倍，那么其中独脚兽有~\\uline{~~~~~~~~~~}~只． ", "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->实践应用"], "answer_analysis": ["鸡兔同笼问题．把$$2$$个四脚蛇和$$1$$个双头龙捆绑在一起，则是$$4$$头$$12$$脚，即$$1$$头$$3$$脚，同三脚猫是一样的，所以可以假设都是$$1$$头$$3$$脚，则有$$3\\times58=174$$只脚，但只有$$160$$只脚，差了$$174-160=14$$只脚，替换：$$14÷2=7$$只，故有$$7$$只独脚兽． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2874", "queId": "8d6e4e1dd16a48728bb8f98856366bf3", "competition_source_list": ["2009年希望杯五年级竞赛初赛", "希望杯五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "计算： $$100\\div 1.2\\times 3\\div \\frac{5}{6}\\times 1\\frac{4}{15}=$$~\\uline{~~~~~~~~~~}~。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$230$$ "}], [{"aoVal": "B", "content": "$$280$$ "}], [{"aoVal": "C", "content": "$$380$$ "}], [{"aoVal": "D", "content": "$$330$$ "}]], "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": ["原式$$=100\\times \\frac{5}{6}\\times 3\\times \\frac{6}{5}\\times \\frac{19}{15}=380$$。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "470", "queId": "c57e2d7e9d8c4e5c8bc7a2b94d066597", "competition_source_list": ["2006年第4届创新杯四年级竞赛初赛B卷第9题"], "difficulty": "2", "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": "$$6$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"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->24点游戏"], "answer_analysis": ["因为$$1\\times2\\times3\\times4=24$$，所以$$4$$可用；  因为$$(5-1)\\times2\\times3=24$$，所以$$5$$可用；  因为$$(3+2-1)\\times6=24$$，所以$$6$$可用；  因为$$3\\times7+1+2=24$$，所以$$7$$可用；  因为$$3\\times8\\times (2-1)=24$$，所以$$8$$可用；  因为$$3\\times9-2-1=24$$，所以$$9$$可用；  因为$$10\\times2+1+3=24$$，所以$$10$$可用．  答：可用的数字是$$7$$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": "22", "queId": "027dfab0cc714bdfbd956d810ad15d1e", "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4题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "布袋中有$$27$$个同样大小的小球，其中白球$$9$$个，红球$$12$$个，蓝球$$6$$个．从布袋中摸出若干个球，为保证剩下的球中至少有$$8$$个球同色，摸出的球数最多为（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$个 "}], [{"aoVal": "B", "content": "$$7$$个 "}], [{"aoVal": "C", "content": "$$8$$个 "}], [{"aoVal": "D", "content": "$$21$$个 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["逆向思考至少留几个才能保证有$$8$$个球同色显然$$6+7+7+1=21$$，因此最多取$$21-15=6$$个． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1194", "queId": "a6ceec85ff84491a9c6a0a7fd13764c9", "competition_source_list": ["其它改编自2012年全国希望杯六年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙两家商店出售同一款兔宝宝玩具，每只原售价都是$$25$$元，为了促销，甲店先提价$$10 \\%$$，再降价$$20 \\%$$；乙店则直接降价$$10 \\%$$．那么，调价后对于这款兔宝宝玩具，~\\uline{~~~~~~~~~~}~店的售价更便宜，便宜~\\uline{~~~~~~~~~~}~元． ", "answer_option_list": [[{"aoVal": "A", "content": "甲；$$0.5$$ "}], [{"aoVal": "B", "content": "甲；$$1$$ "}], [{"aoVal": "C", "content": "乙；$$0.5$$ "}], [{"aoVal": "D", "content": "乙；$$1$$ "}]], "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->已知成本利润率求利润"], "answer_analysis": ["甲：$$25\\times (1+10 \\%)\\times (1-20 \\%)=22$$（元）；乙：$$25\\times(1-10 \\%)=22.5$$（元），$$22.5-22=0.5$$（元）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2711", "queId": "b106549781ba46018e3eabc730a2430b", "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$$99 \\%$$，运抵武昌后测得含水量为$$98 \\%$$，运抵武昌后葡萄重(  )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->应用题模块->列方程解应用题"], "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$$x\\times \\left( 1-98 \\% \\right)=4\\times \\left( 1-99 \\% \\right)$$，解得$$x=2$$，所以运抵武昌后葡萄剩下2吨. "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1636", "queId": "bf11a40d5ecd4a7bb742bd49cb87fb0a", "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->思想->整体思想"], "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": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "408", "queId": "bff953b73c9141938df9657935e90a22", "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1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "小明和小东拿着同样的钱去买铅笔，小明买了$$5$$支，小东买了$$7$$支，买的铅笔便宜． ", "answer_option_list": [[{"aoVal": "A", "content": "小明 "}], [{"aoVal": "B", "content": "小东 "}], [{"aoVal": "C", "content": "一样 "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["从题干中可以知道：小明和小东拿着同样的钱去买铅笔，也就是二者钱数相同；  问谁买的铅笔便宜一些，也就是看谁可以用相同的钱买到更多的笔；  小明买了$$5$$支，小东买了$$7$$支，所以小东买的铅笔便宜．  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1796", "queId": "bb257f2d2e724ef5900f241ef950eaf6", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "小明参加有奖竞猜，共有$$30$$道选择题，评分标准是：自己答对一题得$$4$$分；现场求助答对得$$2$$分；不答不得分；答错一题倒扣$$3$$分（现场求助的题打错也扣$$3$$分）．小明最后得分为$$50$$分，而且他自己答对的和不打的题一样多；现场求助答对的题比不答的多$$1$$题，那么他现场求助的题有（~ ）题． ", "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->逻辑分析"], "answer_analysis": ["设自己答对的和答不对的共有$$x$$道，则求助答对有$$x+1$$道，则答错的为$$30-(3x+1)$$道． 根据题目可列$$4x+2(x+1)-3(29-3x)=50$$解得$$x=9$$，则求助的有$$10$$道题． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2726", "queId": "ff8080814502fa2401450bc67b2f1596", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "算式$$\\frac{25\\times 8+1\\div \\frac{5}{7}}{2014-201.4\\times 2}$$的计算结果是（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{7}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{8}$$ "}]], "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": ["$$\\frac{25\\times 8+1\\div\\frac{5}{7}}{2014  -201.4\\times 2}=\\frac{200+1.4}{201.4\\times 10-201.4\\times2}=\\frac{201.4}{201.4\\times 8}=\\frac{1}{8}$$ "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1334", "queId": "3e72dafe44d943eeb707808fa4fffb8d", "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39题"], "difficulty": "2", "qtype": "single_choice", "problem": "珍妮花$$50$$美金买了一条围巾．她一开始准备以$$60$$美金卖出，然后把价格增加$$25 \\%$$，再把新的价格减少$$20 \\%$$并且出售．在这笔交易之中，她（ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "赚了$$7$$美金 "}], [{"aoVal": "B", "content": "赚了$$10$$美金 "}], [{"aoVal": "C", "content": "亏了$$7$$美金 "}], [{"aoVal": "D", "content": "亏了$$10$$美金 "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["算出实际花的钱，还有实际收入的钱，最终的结果是赚了$$10$$美元． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2028", "queId": "cf55c2289e9848fa8c90e4437e1eaea3", "competition_source_list": ["2019年美国数学大联盟杯三年级竞赛初赛第11题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "（2019 Math League, Grade 3, Question\\# 11）  Helen wrote out all the numbers from $$1$$ to $$60$$ on a piece of paper. How many times did she write the digit $5$?  海伦在纸上写下了从$1$到$60$的所有数，请问她一共写了多少个数字$5$? ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["海伦把从$$1$$到$$60$$的所有数字写在一张纸上．她写了多少次数字$$5$$？  在$$1\\sim 10$$中，$$5$$出现$$1$$次，  在$$11\\sim 60$$中，$$5$$在十位上出现$$10$$次，在个位上出现$$5$$次，  所以一共出现了$$1+10+5=16$$（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": "70", "queId": "0c03971e8ad24ddfa993e9d3d45b1e9c", "competition_source_list": ["2015年全国迎春杯四年级竞赛网络版第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "甲乙丙三人参加比赛，他们得了前$$3$$名（无并列）．  甲说：``我是第一．''  乙说：``我不是第一．''  丙说：``我不是第三．''  三人中只有一个人说真话，则第一名是~\\uline{~~~~~~~~~~}~. ", "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": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["逻辑推理．假设法：假设甲说真话，那么乙说的肯定是真话，不符题意；假设乙说真话那丙说了假话，丙是第三，乙只能是第二，甲只能是第一，与前提矛盾；那么只能是丙说了真话，甲乙说了假话．推断出甲第三，乙第一，丙第二．故选$$\\text{B}$$. "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2000", "queId": "b40b81e1dc8a426a90a0fcc49d9d3b3e", "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": "价格相同的一种商品，甲店：买四赠一．乙店：优惠$$\\frac{1}{4}$$．如果只从经济折扣方面考虑，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": "942", "queId": "c65aec288d424187ae937e114b3f7407", "competition_source_list": ["2005年第3届创新杯五年级竞赛复赛第2题"], "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->思想->对应思想"], "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$$\\rm A$$. "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2356", "queId": "0e6558655d78415a822ced0404ae124c", "competition_source_list": ["2006年走美杯四年级竞赛", "2006年走美杯五年级竞赛", "2006年走美杯六年级竞赛"], "difficulty": "0", "qtype": "single_choice", "problem": "$$a\\vee b$$表示$$a$$、$$b$$两个数中取较大的一个，$$a\\wedge b$$表示$$a$$、$$b$$两个数中取较小的一个，则$$\\left( 2006\\vee 2008 \\right)\\wedge \\left( 2007\\vee 2009 \\right)$$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": ["$$2006\\vee 2008$$两个数中取较大的是$$2008$$，$$2007\\vee 2009$$两个数中取较大的是$$2009$$，$$2008\\wedge 2009$$两个数中取较小的是$$2008$$。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "811", "queId": "9f0515c9da1c4959b705e6bf391f7db9", "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": "$$32$$ "}]], "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": ["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": "1219", "queId": "4232a605153f4bce8632a19bc113a070", "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．【本题欣赏题】答案：C ", "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->拓展思维->应用题模块->浓度问题->抓不变量"], "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$$400$$克糖水．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": "1383", "queId": "24e4bf43bb40452e8b24e82b244a8e42", "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4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$10$$个小朋友排一队，从前面数小红排在第$$2$$个，小军排在小红后面第$$4$$个，那么小军从后往前数排第个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->能力->构造模型->模型思想"], "answer_analysis": ["从前往后数，小军排第$$2+4=6$$个，则从后往前数，小军排：$$10-6+1=5$$个．  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": "926", "queId": "dc52894c0a2d44ba81f4bd0bdbfc79dd", "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7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个两位奇数除$$1477$$，余数是$$49$$，那么这个两位奇数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$57$$ "}], [{"aoVal": "B", "content": "$$59$$ "}], [{"aoVal": "C", "content": "$$51$$ "}], [{"aoVal": "D", "content": "$$61$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["本题是利用了除法算式中各部分的关系求解．  根据被除数$$=$$除数$$\\times $$商$$+$$余数，  除数应是被除数$$-$$余数的因数，除数大于余数，由此解决问题．  $$1477-49=1428$$，  $$1428=7\\times 2\\times 2\\times 3\\times 17$$，  所以$$1428$$大于$$49$$的两位数因数有：  $$17\\times 3=51$$，  $$2\\times 2\\times 17=68$$，  $$2\\times 2\\times 3\\times 7=84$$，  所以这个两位奇数是$$51$$、$$68$$、$$84$$．  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": "594", "queId": "0faeeff329f64705bac538b3d7ce9aa3", "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1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知一个长方体的长宽高分别为$$3$$个连续自然数，并且这三个自然数均为合数，那么，这个长方体的体积最小是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$210$$ "}], [{"aoVal": "B", "content": "$$720$$ "}], [{"aoVal": "C", "content": "$$504$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$3$$个连续自然数最小为$$8$$，$$9$$，$$10$$，  则体积为：$$8\\times 9\\times 10=720$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1322", "queId": "79cab9972c1a480d9ec80b36345b3148", "competition_source_list": ["2006年第4届创新杯六年级竞赛初赛B卷第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "在一家三口人中，每两个人的平均年龄加上余下一人的年龄得到$$47$$，$$61$$，$$60$$，那么这三个人中最大年龄与最小年龄的差是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$ "}], [{"aoVal": "B", "content": "$$27$$ "}], [{"aoVal": "C", "content": "$$26$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设三人的年龄为$$X$$、$$Y$$、$$Z$$，  则有$$\\frac{X+Y}{2}+Z=47$$，  $$\\frac{X+Z}{2}+Y=61$$，  $$\\frac{Y+Z}{2}+X=60$$，  可将上三式变化为：  $$X+Y+3Z=94(1)$$  $$X+Z+2Y=122(2)$$  $$Y+Z+2X=120(3)$$  $$(2)-(3)Y-X=2(4)$$，  $$2\\times (3)-(1)Y+3X=146(5)$$，  $$(5)-(4)4X=144$$，  ∴$$X=36$$，  ∴由（$$4$$）可得$$Y=38$$，  把$$X$$、$$Y$$代入（$$1$$）中得$$Z=10$$，  ∴极差为$$38-10=28$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "56", "queId": "0b720a53e90a4301b3e1d6ee3bb5731d", "competition_source_list": ["2016年创新杯五年级竞赛训练题（四）第13题", "2014年全国世奥赛五年级竞赛第11题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某班军训时进行射击比赛，几名教官在讨论该班的成绩  ~ ~ ~ ~王教官说：``这次军训时间太短，这个班没有人射击成绩会是优秀．''  ~ ~ ~ ~孙教官说：``不会吧，有几个人以前训练过，他们的射击成绩会是优秀．''  ~ ~ ~ ~周教官说：``我看班长或是体育委员能打出优秀成绩．''  结果发现三位教官中只有一人说对了．  由此可以推出以下~\\uline{~~~~~~~~~~}~选项肯定为真． ", "answer_option_list": [[{"aoVal": "A", "content": "全班所有人的射击成绩都不是优秀 "}], [{"aoVal": "B", "content": "班里所有人的射击成绩都是优秀 "}], [{"aoVal": "C", "content": "班长的射击成绩是优秀 "}], [{"aoVal": "D", "content": "体育委员的射击成绩不是优秀 "}]], "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": ["王教官和孙教官的说法矛盾，二人必有一对一错，因为只有一个人对，故周教官是错的，因此班长和体育委员都不优秀，选$$\\rm D$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1166", "queId": "543900f3912a4d479a05f856d4f3b4f8", "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23题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "五年级举行安全知识竞赛，共有$$20$$道试题．做对一道得$$5$$分，做错或没做一道都要扣$$3$$分．笑笑得了$$60$$分，那么她做对了道题． ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["利用方程法来解：设笑笑做对了$$x$$道题，  $$5x-3\\times (20-x)=60$$，  解：$$\\textasciitilde5x-60+3x=60$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde8x=120$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x=15$$，  所以：笑笑做对了$$15$$道题．  答案选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "588", "queId": "181f7481a5a44eab87a1465ef471476c", "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": "0", "qtype": "single_choice", "problem": "以下哪个数是质数？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$2013$$ "}], [{"aoVal": "B", "content": "$$2015$$ "}], [{"aoVal": "C", "content": "$$2017$$ "}], [{"aoVal": "D", "content": "$$2019$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$2013$$是$$3$$的倍数，$$2015$$是$$5$$的倍数，$$2019$$是$$3$$的倍数． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "442", "queId": "97b95831a49e4df3ad37232022356c84", "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->逻辑分析"], "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": "506", "queId": "c1e6564899b646068ec18dcfc5a94e34", "competition_source_list": ["2017年世界少年奥林匹克数学竞赛四年级竞赛初赛A卷第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "选择``$$+$$、$$-$$、$$\\times $$、$$\\div $$''填入圆圈中，使得等式$$21\\bigcirc 10\\bigcirc 2\\bigcirc 8=24$$成立． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\times $$；$$\\div $$；$$+$$． "}], [{"aoVal": "B", "content": "$$\\times $$；$$\\div $$；$$-$$． "}], [{"aoVal": "C", "content": "$$-$$；$$\\div $$；$$-$$． "}], [{"aoVal": "D", "content": "$$-$$；$$\\div $$；$$+$$． "}]], "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": ["$$21-10\\div 2+8=24$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2206", "queId": "31715628b69a4205b521c89d1a9ae80c", "competition_source_list": ["2006年第4届创新杯四年级竞赛复赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "从山下到山上的路程是$$1200$$米，小华上山时平均速度为每分钟走$$60$$米，下山时平均每分钟走$$120$$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->平均速度->公式法"], "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），下山时间为$$1200\\div 120=10$$（分），总时间为$$20+10=30$$（分），总路程为$$1200\\times 2=2400$$（米） ，平均速度:$$2400\\div 30=80$$（米/分），故选：$\\text{B}$． ", "<p>$$\\quad\\left(1+1\\right)\\div\\left(\\frac{1}{60}+ \\frac{1}{120}\\right)$$</p>\n<p>$$=2 \\div \\frac{1}{40}$$</p>\n<p>$$=80$$（米），</p>\n<p>或$$\\left(1200+1200\\right)\\div\\left(\\frac{1200}{60}+ \\frac{1200}{120} \\right)$$</p>\n<p>$$=2400\\div30$$</p>\n<p>$$=80$$（米）．</p>\n<p>所以$$\\text{B}$$选项是正确的．</p>"], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1996", "queId": "dc909f2acedc49e2b2d16ad51a46564f", "competition_source_list": ["2020年世界少年奥林匹克数学竞赛六年级竞赛复赛第1题12分"], "difficulty": "2", "qtype": "single_choice", "problem": "奶茶店用$$2$$份红茶、$$3$$份水和$$5$$份牛奶调配制作成奶茶，如果这三种材料都是$$5$$千克，当红茶全部用完时，水需要增加多少千克？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1.5$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$2.5$$ "}], [{"aoVal": "D", "content": "$$3$$ "}], [{"aoVal": "E", "content": "$$4$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->应用题模块", "拓展思维->思想->对应思想"], "answer_analysis": ["茶$$:$$水$$:$$牛奶$$=2:3:5$$，  红茶为$$2$$份，用了$$5$$千克，则$$1$$份为$$5\\div 2=2.5$$（千克），  则水需要：$$2.5\\times 3=7.5$$（千克），  水需要增加：$$7.5-5=2.5$$（千克），  则牛奶需要：$$2.5\\times 5=12.5$$（千克），  牛奶需要增加：$$12.5-5=7.5$$（千克）．  答：水需要增加$$2.5$$千克，牛奶需要增加$$7.5$$千克． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1945", "queId": "e5434bed51774609879618fffac25364", "competition_source_list": ["2014年北京学而思杯六年级竞赛B卷第6题", "2014年全国学而思杯六年级竞赛B卷第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两所小学，甲校的人数是乙校人数的$$\\frac{2}{5}$$，甲校的女生人数占全校人数的$$40 \\%$$，乙校男生人数占全校人数的$$60 \\%$$．如果将甲、乙两校合并，女生人数占总人数的$$ \\%$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$50$$ "}], [{"aoVal": "E", "content": "$$60$$ "}]], "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": ["设甲乙两校人数分别为$$2$$份和$$5$$份，则女生共$$2\\times 40 \\%+5\\times (1-60 \\%)=2.8$$，占$$2.8\\div (5+2)=40 \\%$$，实际上，从甲乙两校女生都占各自的$$40 \\%$$即可得出结论． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "932", "queId": "e11337edf5964b719288b164864a64a4", "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->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．  或者使用公式法：$120=2^{3}\\times3\\times5$，$$120$$的因数个数是：$\\left( 3+1\\right)\\times\\left( 1+1\\right)\\times\\left( 1+1\\right)=16$（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": "2074", "queId": "f45579d3383c4298bcfc3a076c94a4dd", "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1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "小明今年$$7$$岁，小刚今年$$4$$岁，$$2$$年后，小明比小刚大岁． ", "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": ["根据题意分析可知，今年小明比小刚大：$$7-4=3$$（岁），年龄差是一直不变的，所以$$2$$年后小明也比小刚大$$3$$岁，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": "1844", "queId": "a968fe78c23f492da539332b6b6a4153", "competition_source_list": ["2012年第10届全国创新杯小学高年级六年级竞赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "某校师生到郊外植树，已知老师人数是学生人数的$$\\frac{1}{3}$$，若每位男生种$$13$$棵树，女生每人种$$10$$棵树，每个老师种$$15$$棵树，他们一共种了$$201$$棵树，那么老师有（~ ）人． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "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": ["设男生有$$a$$人，女生有$$b$$人，则老师人数为$$\\frac{a+b}{3}$$，有  $$13a+10b+5\\left( a+b \\right)=201$$，化简得$$6a+5b=67$$，  考虑到$$a+b$$为整数，只有一组解$$a=2$$，$$b=10$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1634", "queId": "52e79ced04a74b999f4282e6d83b8023", "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->能力->运算求解"], "answer_analysis": ["倒推．$$44$$ 对应的是$$44-2=42$$，颠倒后是$$24$$，平均分后为$$12$$．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": "2859", "queId": "614395e9093940fab17563e6a2dba8b5", "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": "7.5：x=24:12，则x=~\\uline{~~~~~~~~~~}~ ", "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": ["$$\\frac{8}{9}\\times \\left[ \\frac{3}{4}-\\left( \\frac{7}{16}-\\frac{1}{4} \\right) \\right]=0.5$$．  $$\\frac{8}{9}\\times \\left[ \\frac{3}{4}-\\left( \\frac{7}{16}-\\frac{1}{4} \\right) \\right]$$  $$=\\frac{8}{9}\\times \\left( \\frac{3}{4}-\\frac{7}{16}+\\frac{1}{4} \\right)$$  $$=\\frac{8}{9}\\times \\frac{9}{16}$$  $$=\\frac{1}{2}$$  $$=0.5$$ "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2262", "queId": "9a9475fceb994833a5ea022c140d888a", "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15题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "上午$$8$$点整，甲从$$A$$地出发匀速去$$B$$地，$$8$$点$$20$$分甲与从$$B$$地出发匀速去$$A$$地的乙相遇；相遇后甲将速度提高到原来的$$3$$倍，乙速度不变；$$8$$点$$30$$分，甲，乙两人同时到达各自的目的地．那么，乙从$$B$$地出发时是$$8$$点分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$05$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["本题考查比的应用和路程问题；首先，根据题意可知甲在相遇前的路程即为乙在相遇后的路程，甲在相遇后的路程即为乙在相遇前的路程；  相遇后，甲从$$A$$地出发到与乙相遇的$$20$$分钟路程乙用了$$10$$分钟，可知乙的速度是甲原速度的$$2$$倍；遇后甲将速度提高到原来的$$3$$倍，可知乙的速度：甲相遇后的速度$$=2:3$$，因此，甲在相遇后花了十分钟，因此乙相遇前所花时间$$:$$甲相遇后所花时间$$=3:2$$，乙在相遇前花了$$(30-20)\\div \\frac{2}{3}=15$$分钟，  $$8$$时$$20$$分$$-8$$时$$15$$分$$=8$$时$$5$$分，  所以乙从$$B$$地出发时是$$8$$点$$5$$分．  故选Ａ． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "101", "queId": "342e1014c38e419780110bee1e174d48", "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": "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"], "answer_analysis": ["小东与小南说的话对立，所以两人中一定一对一错．已知只有一个人说对了，则小西和小北说的是错的．\"小北说：小南说的也不对．\"这句话是错的，可知小南说的是对的．  故选：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2300", "queId": "8aac50a7508d5d410150d53608d47758", "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年北京华杯赛小学高年级竞赛初赛A卷第5题"], "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->能力->实践应用"], "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），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），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），即慢了$$12$$分钟． ", "<p>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．</p>"], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "89", "queId": "41bad802484e42ea817ab9c5f1a095b6", "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": "蛋君举办了一年一度的汽车大赛，本次有$$4$$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->拓展思维->组合模块->逻辑推理->比较型逻辑推理"], "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": "485", "queId": "e0fe25156f9b4b86b001f8d1257051d5", "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：小南比我高；小南说：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": "小南 "}], [{"aoVal": "C", "content": "小朱 "}], [{"aoVal": "D", "content": "园园 "}]], "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": ["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": "2913", "queId": "9b934af1dc844008be23c9eab46655e5", "competition_source_list": ["2013年IMAS小学高年级竞赛第一轮检测试题第20题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$P$$、$$Q$$、$$R$$、$$S$$四位小朋友在计算$$\\frac{4}{5}+\\frac{5}{6}+\\frac{7}{9}+\\frac{9}{11}$$时，$$P$$不小心把$$\\frac{4}{5}$$的分子、分母颠倒了，$$Q$$不小心把$$\\frac{5}{6}$$的分子、分母颠倒了，$$R$$不小心把$$\\frac{7}{9}$$的分子、分母颠倒了，$$S$$不小心把$$\\frac{9}{11}$$的分子、分母颠倒了．请问哪一位小朋友计算出来的结果与正确答案的值相差最小？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$P$$ "}], [{"aoVal": "B", "content": "$$Q$$ "}], [{"aoVal": "C", "content": "$$R$$ "}], [{"aoVal": "D", "content": "$$S$$ "}], [{"aoVal": "E", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->分数->分数运算->分数加减"], "answer_analysis": ["解法$$1$$：$$P$$不小心把$$\\frac{4}{5}$$的分子、分母颠倒了，所以他算出来的和比正确的值大$$\\frac{5}{4}-\\frac{4}{5}=\\frac{9}{20}$$；  $$Q$$不小心把$$\\frac{5}{6}$$的分子、分母颠倒了，所以他算出来的和比正确的值大$$\\frac{6}{5}-\\frac{5}{6}=\\frac{11}{30}$$；  $$R$$不小心把$$\\frac{7}{9}$$的分子、分母颠倒了，所以他算出来的和比正确的值大$$\\frac{9}{7}-\\frac{7}{9}=\\frac{32}{63}$$；  $$S$$不小心把$$\\frac{9}{11}$$的分子、分母颠倒了，所以他算出来的和比正确的之大$$\\frac{11}{9}-\\frac{9}{11}=\\frac{40}{99}$$．  可知$$\\frac{11}{30}\\textless{}\\frac{12}{30}=\\frac{2}{5}$$而$$\\frac{9}{20}\\textgreater\\frac{8}{20}=\\frac{2}{5}$$、$$\\frac{32}{63}\\textgreater\\frac{1}{2}\\textgreater\\frac{2}{5}$$、$$\\frac{40}{99}\\textgreater\\frac{40}{100}=\\frac{2}{5}$$，所以$$Q$$所求出的和与正确值相差最小．故选$$\\text{B}$$．  解法$$2$$：两个小于$$1$$的分数，如果$$a\\textgreater b$$，则$$\\frac{1}{a}\\textless{}\\frac{1}{b}$$，从而$$\\frac{1}{a}-a\\textless{}\\frac{1}{b}-b$$，所以分子、分母写颠倒的分数应该是四个分数中最大的那个，这样才能使得所求的和与正确值相差最小．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1460", "queId": "2e748fdb61a7440e854211b9e1bd946d", "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": "$$8$$岁 "}], [{"aoVal": "B", "content": "$$9$$岁 "}], [{"aoVal": "C", "content": "$$10$$岁 "}], [{"aoVal": "D", "content": "$$11$$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": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1446", "queId": "faaf5db0b43443d8b5d027480ce33472", "competition_source_list": ["2014年迎春杯三年级竞赛复赛"], "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->两量还原问题"], "answer_analysis": ["解：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，  一共有：$$\\left( 12-1 \\right)\\times 2$$  $$=11\\times 2$$  $$=22$$（瓶）  答：小明家一共买了$$22$$瓶果汁。  故选：B。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2485", "queId": "26e57cd2cfe44d6e89e0b755574db441", "competition_source_list": ["2021年新希望杯五年级竞赛模拟（考前培训100题）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$3+33+333+\\cdots+\\underbrace{33\\cdots3}_{8个3}$$的结果个位数字是~\\uline{~~~~~~~~~~}~。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->整数->多位数计算->多位数的加减运算"], "answer_analysis": ["原式的末三位和每个数的末三位有关系，有$$2007$$个$$3$$，$$2006$$个$$30$$，$$2005$$个$$300$$，则$$2007\\times 3+2006\\times 30+2005\\times 300=6021+60180+601500=667701$$，原式末三位数字为$$701$$。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1392", "queId": "ff2fe5eb874646c5937ee970e88a925d", "competition_source_list": ["2018年IMAS小学中年级竞赛（第一轮）第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "将数码和为$$3$$且不含数码$$2$$的所有正整数从小到大排列，请问最小三个数的和是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$63$$ "}], [{"aoVal": "C", "content": "$$144$$ "}], [{"aoVal": "D", "content": "$$206$$ "}], [{"aoVal": "E", "content": "$$414$$ "}]], "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": ["可知$$3=1+1+1$$，故这样的数由小至大依序为$$3$$、$$30$$、$$111$$、$$300$$、$$\\cdots $$．  所以最小三个数的和是$$3+30+111=144$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1078", "queId": "18152259d67844eeb4e09e469570acf1", "competition_source_list": ["2014年迎春杯四年级竞赛复赛"], "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$$10$$个字，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->多人合作"], "answer_analysis": ["解：依题意可知小元每打$$10$$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，打错$$1$$份。  小元和小芳每错一份正确的字数比为$$9:4$$。因为总的正确数量比为$$2:1$$。那么小元和小芳的答错题的份数比为$$\\frac{2}{9}:\\frac{1}{4}=8:9$$，  小元和小芳答正确的份数为$$(8\\times 9):(4\\times 9)=72:36$$。  小元答题总数为：$$72\\div 9\\times 10=80$$（份）  小芳答题总数为：$$36\\div 4\\times 5=45$$（份）  总字数份数为：$$80+45=125$$（份）  每份字数为：$$10000\\div 125=80$$（字）  小元和小芳共正确的字数为：$$(72+36)\\times 80=8640$$（字）  故选：D。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2966", "queId": "9c0fa379257046879de94b9d17c220f5", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "把奇数按以下规律进行分组：$$\\left( 1 \\right)$$，$$\\left( 3,5 \\right)$$，$$\\left( 7,9,11 \\right)$$，$$\\cdots $$其中第$$1$$组有$$1$$个数，第$$2$$组有$$2$$个数，第$$3$$组有$$3$$个数，第$$4$$组有$$4$$个数\\ldots\\ldots 那么第$$100$$组中的$$100$$个数的和为． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10000$$ "}], [{"aoVal": "B", "content": "$$100000$$ "}], [{"aoVal": "C", "content": "$$1000000$$ "}], [{"aoVal": "D", "content": "$$10000000$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->公式类运算->连续自然数立方和公式"], "answer_analysis": ["第一组和为$$1={{1}^{3}}$$，第二组和为$$8={{2}^{3}}$$，第三组和为$$27={{3}^{3}}$$，则第$$n$$组和$$={{n}^{3}}$$，  所以第$$100$$组和$$={{100}^{3}}=1000000$$．  答案：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1169", "queId": "872be11b967a415594ed33db569e7935", "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4题"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$1$$千克甲种盐水与$$2$$千克浓度为$$20 \\%$$的乙种盐水混合后，浓度变为$$25 \\%$$，甲种盐水的浓度为（~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$30 \\%$$~~~~~~~ "}], [{"aoVal": "B", "content": "$$35 \\%$$~~~~~~~ "}], [{"aoVal": "C", "content": "$$40 \\%$$~~~~~~~ "}], [{"aoVal": "D", "content": "$$45 \\%$$ "}]], "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": ["混合溶液为$$1+2=3$$（千克），其中盐有$$3\\times 25 \\%=0.75$$（千克），甲中含盐：$$0.75-2\\times 20 \\%=0.35$$（千克），甲浓度：$$\\frac{0.35}{1}\\times 100 \\%=35 \\%$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1743", "queId": "77a110332a084a38addcc1482411d107", "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8题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "小林家住九楼，放学回家时，从$$1$$楼走到$$4$$楼共用了$$48$$秒；照这样，小林回到家还要秒． ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$80$$ "}], [{"aoVal": "C", "content": "$$144$$ "}]], "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": ["从题干可以知道，小林家住九楼，放学回家时，从$$1$$楼走到$$4$$楼共用了$$48$$秒，  也就是爬了三层楼用了$$48$$秒，爬一层楼需要：$$48\\div3=16$$（秒），  小林还需要爬$$5$$层楼，$$16\\times5=80$$（秒），也就是还需要$$80$$秒．  故选$$\\text{B}$$； "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2115", "queId": "fe4db4bd53f140829a475987fba39d57", "competition_source_list": ["2017年全国小学低年级二年级其它思维创新大通关第8讲周期问题第15题", "2017年第11届北京学而思综合能力诊断二年级竞赛第14题", "2017年第11届北京学而思杯小学低年级二年级竞赛第14题12分"], "difficulty": "3", "qtype": "single_choice", "problem": "艾迪同学在$$2017$$年$$4$$月$$1$$日养了一盆仙人掌，但是粗心的艾迪总是会忘了给仙人掌浇水，每周日和周一才记得给仙人掌浇水．如果一个月内浇仙人掌的天数少于$$9$$天，仙人掌就会退化变成仙人球．已知$$2017$$年$$4$$月$$1$$日是周六，那么在~\\uline{~~~~~~~~~~}~月，仙人掌会因为浇水天数不够而退化成仙人球． ", "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": ["Overseas Competition->知识点->应用题模块->周期问题", "拓展思维->思想->逐步调整思想"], "answer_analysis": ["$$4$$月份有$$30$$天，$$30=7\\times 4+2$$，$$4$$月$$1$$日是周六，则$$4$$月$$30$$日是周日，所以$$4$$月份有$$5$$个周日，$$4$$个周一，浇水的天数等于$$9$$天，不会退化成仙人球；$$5$$月份有$$31$$天，$$31=7\\times 4+3$$，$$5$$月$$1$$日是周一，则$$5$$月$$31$$日是周三，所以$$5$$月份有$$4$$个周日，$$5$$个周一，浇水的天数等于$$9$$天，不会退化成仙人球；$$6$$月份有$$30$$天，$$30=7\\times 4+2$$，$$6$$月$$1$$日是周四，则$$6$$月$$30$$日是周五，所以$$6$$月份有$$4$$个周日，$$4$$个周一，浇水的天数小于$$9$$天，所以会退化成仙人球．  $$4$$月份有$$30$$天，每个星期有$$7$$天，  又因为$$4$$月$$1$$日是周六，  所以$$4$$月的每一个星期周期都是以周六开始，周五结束，  这样的周期共有$$30\\div7=4$$（个）$$\\cdots\\cdots2$$（天），  则$$4$$月$$30$$日是周五往后数两天，则为周日，  所以$$4$$月份有$$5$$个周日，$$4$$个周一，浇水的天数等于$$9$$天，不会退化成仙人球．  $$5$$月份有$$31$$天，$$5$$月$$1$$日是周一，每个周期从周一开始，周日结束，  这样的周期有$$31\\div7=4$$（个）$$\\cdots\\cdots3$$（天），  则$$5$$月$$31$$日是周日往后数$$3$$天，为周三，  所以$$5$$月份有$$4$$个周日，$$5$$个周一，浇水的天数等于$$9$$天，不会退化成仙人球．  $$6$$月份有$$30$$天，$$6$$月$$1$$日是周四，每个周期从周四开始周三结束，  这样的周期有$$30\\div7=4$$（个）$$\\cdots\\cdots2$$（天），  则$$6$$月$$30$$日是周三往后数$$2$$天，是周五，  所以$$6$$月份有$$4$$个周日，$$4$$个周一，浇水的天数小于$$9$$天，所以会退化成仙人球． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "27", "queId": "8aac50a7519fa10a01519ff186a000b8", "competition_source_list": ["2016年全国华杯赛小学中年级竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙、丁四支足球队进行比赛．懒羊羊说：甲第一，丁第四；喜羊羊说：丁第二，丙第三；沸羊羊说：丙第二，乙第一，每个的预测都只对了一半，那么，实际的第一名至第四名的球队依次是（~ ~ ~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "甲乙丁丙 "}], [{"aoVal": "B", "content": "甲丁乙丙 "}], [{"aoVal": "C", "content": "乙甲丙丁 "}], [{"aoVal": "D", "content": "丙甲乙丁 "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["假设懒羊羊所说的``甲第一''正确，则``丁第四''错误；沸羊羊所说的``乙第一''错误，则``丙第二''正确；此时喜羊羊的两句话都错误，与题意不符，故假设不成立．  假设懒羊羊所说的``甲第一''错误，则``丁第四''正确；喜羊羊所说的``丁第二''错误，则``丙第三''正确；沸羊羊所说的``丙第二''错误，则``乙第一''正确．故实际的第一名至第四为：乙甲丙丁．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3165", "queId": "14184379e93746be8d0d4aef7d614ea3", "competition_source_list": ["2011年六年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "连接武汉与广州的高铁，中途只停六个车站，那么应该印制（  ）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->拓展思维->计数模块->排列组合->组合->组合的基本运算"], "answer_analysis": ["加上首尾两站共$$8$$个站，共有$$\\text{C}_{8}^{2}=28$$（种）组合，考虑车票的次序共有$$28\\times 2=56$$（种）。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1291", "queId": "9dc7150bc56e41b68bd97c1c2f121b9d", "competition_source_list": ["2020年新希望杯五年级竞赛初赛（个人战）第17题", "2020年新希望杯五年级竞赛决赛（8月）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->思想->对应思想"], "answer_analysis": ["因为$$2020$$年是闰年．  从$$2019$$年国庆节经过$$366$$天是$$2020$$年国庆节．  闰年往后推两天，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": "2721", "queId": "ff8080814502fa24014507b670c60ba0", "competition_source_list": ["2014年全国迎春杯三年级竞赛初赛第14题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2$$个樱桃的价钱与$$3$$个苹果价钱一样，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->物品代换"], "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": "2779", "queId": "b145836967ac489a88d6c96b51fda063", "competition_source_list": ["2004年第2届创新杯六年级竞赛初赛第3题", "2004年六年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "在甲、乙、丙三个数中，如果甲数是乙数的$$\\frac{3}{5}$$，乙数是丙数的$$\\frac{2}{3}$$，那么甲数与丙数的关系(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "甲数是丙数的$$\\frac{2}{5}$$ "}], [{"aoVal": "B", "content": "甲数是丙数的$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "甲数是丙数的$$\\frac{9}{10}$$ "}], [{"aoVal": "D", "content": "甲数是丙数的$$\\frac{10}{9}$$ "}]], "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": ["由题干可得出：甲$$=\\frac{3}{5}$$乙，乙$$=\\frac{2}{3}$$丙，  所以甲$$=\\frac{3}{5}\\times \\frac{2}{3}$$丙，即甲$$=\\frac{2}{5}$$丙．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "568", "queId": "0e76abb88d954109bf986d861acd6ba3", "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$$\\left[ \\frac{3x+7}{7} \\right]=3$$，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->能力->运算求解"], "answer_analysis": ["$$\\frac{14}{3}\\leqslant x ~\\textless{} ~7$$，对应的整数共$$2$$个数．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1941", "queId": "ee5f306e495f4510a7d61fe1e843e224", "competition_source_list": ["2019年广东广州学而思综合能力诊断五年级竞赛第15题12分"], "difficulty": "3", "qtype": "single_choice", "problem": "老师将$$21$$个两位质数写在卡片上，发给四个同学，每人五张，老师手里留下一张，进行五轮游戏，每轮规则如下：  ①每轮每人打出$$1$$张卡片，如果个位与别人重复则出局；  ②在留下的人中，数最大的人获胜．  （例如，四个同学分别打出$$17$$、$$41$$、$$73$$、$$83$$．那么，打出$$73$$与$$83$$的两个人出局，打出$$41$$的人获得胜利）  已知，小明前三轮游戏打出的是$$11$$、$$13$$、$$17$$均获得了胜利；第$$4$$轮小明打出的是$$83$$，出局了，打出$$41$$的人获得了胜利：第$$5$$轮小明打出的是$$97$$，也出局了，打出$$59$$的人获得了胜利．如果$$59$$是第五轮第二大的卡片，那么，老师手里留下的卡片写的质数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$41$$ "}], [{"aoVal": "B", "content": "$$47$$ "}], [{"aoVal": "C", "content": "$$59$$ "}], [{"aoVal": "D", "content": "$$67$$ "}], [{"aoVal": "E", "content": "$$83$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->游戏策略->数字游戏", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["先分析每种个位题目中未出现过的数：个位$$1$$的有$$41$$、$$61$$、$$71$$；个位$$3$$的有$$23$$、$$43$$、$$53$$、$$73$$；个位$$7$$的有$$37$$、$$47$$、$$67$$；个位$$9$$的有$$19$$、$$29$$、$$79$$、$$89$$，  第一轮$$11$$能赢，$$11$$身为最小的两位质数，说明其他三个人的个位应该是一样的，且不为$$1$$；第二轮$$13$$能赢，比$$13$$小的$$11$$出现过了，所以也是其他三人个位都一样，且不为$$3$$；第三轮$$17$$也能赢，考虑到$$11$$和$$13$$都出现过了，那么其他三人个位也一样，且不为$$7$$；第四轮打出$$83$$出局了，并且赢的人是$$41$$，说明第四轮应该出现了另一个个位是$$3$$的质数；第五轮$$97$$也出局，那应该是出现了另一个个位是$$7$$的质数；  考虑到这为止，前三轮个位$$1$$、$$3$$、$$7$$、$$9$$能用到的质数个数应该分别为$$3$$个、$$3$$个、$$2$$个、$$4$$个，因此第一轮到第三轮其他三个数的个位应该是$$1$$、$$3$$、$$9$$（顺序暂时未确定），此时后两轮能出现的数只剩一个个位$$3$$的质数、三个个位是$$7$$的质数，以及一个个位是$$9$$的质数，其中个位$$3$$的质数必定出现在第四轮，和$$83$$同归于尽，第五轮会出现至少一个个位是$$7$$的质数；  考虑到最后一轮打出$$59$$能赢，那么最后一轮不能出现个位是$$9$$的，那么除了$$97$$和$$59$$外，另外两个质数个位只能是$$7$$，而题目说$$59$$第二大，鉴于$$97$$比$$59$$大，那么这两个个位是$$7$$的数只可能是$$37$$和$$47$$；  第四轮除了$$83$$、$$41$$和一个个位$$3$$的质数外，还有$$67$$和个位是$$9$$的质数能选择，由于没有直接出局，所以必须得比$$41$$小，因此只可能是后者，所以唯一没有出现过的质数是$$67$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1976", "queId": "ca515d9ac28640b6af07ce70ab8b9bc3", "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7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "同学们做早操，$$37$$名同学排成一排，每$$2$$名女生中间是$$2$$名男生．如果第一名是女生，这列队伍共有名男生． ", "answer_option_list": [[{"aoVal": "A", "content": "$$18$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$22$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["拓展思维->能力->归纳总结->归纳推理"], "answer_analysis": ["根据题意发现排队规律是``女男男女男男女男男$$\\cdots \\cdots $$''  $$37\\div 3=12$$（组）$$\\cdots \\cdots 1$$（名），$$2\\times 12=24$$（名）．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2149", "queId": "a205554e968b4f17acf829abc56dd994", "competition_source_list": ["2014年迎春杯六年级竞赛初赛"], "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$$B$$、$$C$$两地相距多少千米？ ", "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->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": ["由$$n$$小时乙丙相遇，知$$n$$小时内$${{s}_{\\text{乙}}}+{{s}_{\\text{丙}}}\\text{=}60$$千米，因此在$$2n$$小时内$${{s}_{\\text{乙}}}+{{s}_{\\text{丙}}}\\text{=120}$$千米。由$$2n$$小时甲追上丁，知$$2n$$小时内$${{s}_{\\text{甲}}}-{{s}_{\\text{丁}}}\\text{=60}$$。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$${{s}_{\\text{甲}}}-{{s}_{\\text{丁}}}\\text{=60}$$知路程的公差为$$\\text{60}\\div \\text{3=20}$$千米。由$${{s}_{\\text{乙}}}+{{s}_{\\text{丙}}}\\text{=120}$$千米容易解出$${{s}_{\\text{乙}}}\\text{=70}$$，$${{s}_{\\text{丙}}}\\text{=70}$$，进而求出$${{s}_{\\text{丁}}}\\text{=30}$$，而$${{s}_{\\text{丁}}}$$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": "2786", "queId": "bf136a67929a4ebd8ac291c7f4b63647", "competition_source_list": ["2015年湖北武汉世奥赛小学高年级五年级竞赛模拟训练题（一）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "李老师买$$4$$个足球和$$3$$个篮球共用去$$388$$元，王老师买$$8$$个排球和$$2$$个篮球共用去$$444$$元，那么，买同样的$$1$$个篮球、$$1$$个足球和$$1$$个排球需要元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$152$$ "}], [{"aoVal": "B", "content": "$$152.5$$ "}], [{"aoVal": "C", "content": "$$153$$ "}], [{"aoVal": "D", "content": "$$153.5$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["分解题意，可得如下关系式：  $$4$$个足球$$+3$$个篮球$$=388$$元 ①  $$8$$个排球$$+2$$个篮球$$=444$$元 ②  $$\\times 2+$$②，得：$$8$$个足球$$+8$$个篮球$$+8$$个排球$$=1220$$元，  所以，买$$1$$个篮球、$$1$$个足球和$$1$$个排球需要$$1220\\div 8=152.5$$（元）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3207", "queId": "a245fd8babb3496da98d282f2b5de82c", "competition_source_list": ["2017年IMAS小学中年级竞赛（第一轮）第15题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$2$$颗相同的黑球与$$2$$颗相同的白球放入$$A$$、$$B$$、$$C$$三个不同的盒子中，要求每个盒子至少放$$1$$颗球，请问共有多少种不同的放置方法？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$15$$ "}], [{"aoVal": "E", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由题意可判断出$$A$$、$$B$$、$$C$$三个不同的盒子中有可能为  （$$1$$）黑黑、白、白；（$$2$$）白白、黑、黑；（$$3$$）黑白、黑、白；  （$$1$$）中同一盒内放置两颗黑球的情况有$$3$$种、  （$$2$$）中同一盒内放置两颗白球的情况有$$3$$种、  （$$3$$）中同一盒内放置一颗黑球与一颗白球的情况有$$3$$种且另二颗球放到另二个盒子的情况有$$2$$种，因此总共有$$3+3+3\\times2=12$$种不同情况．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1165", "queId": "151e2d913aa743558e52ba0a6278cf0a", "competition_source_list": ["2018年第8届北京学而思综合能力诊断四年级竞赛年度教学质量监测第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "今年小王子$$8$$岁，大王子$$14$$岁，当两人的年龄之和是$$40$$岁时，应该是~\\uline{~~~~~~~~~~}~年之后的事了． ", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["方法一：利用年龄和的变化特点来解决：  今年两人年龄和：$$8+14=22$$（岁），  几年之后两人共长：$$40-22=18$$（岁），  几年之后小王子长几岁，大王子就长几岁，所以分别长了$$18\\div 2=9$$（岁），  所以当两人年龄和是$$40$$岁时，应该是$$9$$年之后的事了．  方法二：根据年龄差不变，可以用和差问题的方法来解决：  大王子小王子两人年龄差：$$14-8=6$$（岁），  当两人年龄和是$$40$$岁时小王子的年龄：$$(40-6)\\div 2=17$$（岁），  小王子长大了：$$17-8=9$$（岁），  所以当两人年龄和是$$40$$岁时，应该是$$9$$年之后的事了． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2529", "queId": "16f31754d12c4619b71b98afb7d85c43", "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：$$\\frac{5}{14}$$、$$\\frac{10}{27}$$、$$\\frac{20}{53}$$、$$\\frac{12}{31}$$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->运算求解"], "answer_analysis": ["通分子$$\\frac{5}{14}=\\frac{60}{168}$$  $$\\frac{10}{27}=\\frac{60}{162}$$  $$\\frac{20}{53}=\\frac{60}{159}$$  $$\\frac{12}{31}=\\frac{60}{155}$$ "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "376", "queId": "db20f6e5320742c3a143cd248ca0cb0b", "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，但比甲、丁多．甲没有丁多．按苹果由少到多的顺序排列为． ", "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->拓展思维->组合模块->逻辑推理->比较型逻辑推理"], "answer_analysis": ["暂无 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3348", "queId": "77231028a9b946fc8d95523bacd8abe8", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第19题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "地图上有$$A$$、$$B$$、$$C$$、$$D$$、$$E$$五个国家，有五种不同的颜色去染色，要求相邻国家染不同的颜色，则共有种不同的染色方法． ", "answer_option_list": [[{"aoVal": "A", "content": "$$120$$ "}], [{"aoVal": "B", "content": "$$240$$ "}], [{"aoVal": "C", "content": "$$420$$ "}], [{"aoVal": "D", "content": "$$480$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->加乘原理->染色计数问题"], "answer_analysis": ["从中间入手，$$E$$有$$5$$种涂法，再考虑$$AC$$同色，则有$$4\\times 1$$种，$$BD$$各有$$3$$种，  所以是$$4\\times 1\\times 3\\times 3=36$$种，若$$AC$$不同色，有$$4\\times 3$$，$$BD$$各有$$2$$种，  所以是$$4\\times 3\\times 2\\times 2=48$$种，一共是$$5\\times \\left( 36+48 \\right)=420$$种．  答案：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2359", "queId": "6657d0431a744ee3b7e6c3a259ffacee", "competition_source_list": ["2017年第1届重庆华杯赛竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$3\\frac{3}{5}\\times 2345+5555\\div \\frac{25}{256}+654.3\\times 36$$ ", "answer_option_list": [[{"aoVal": "A", "content": "$$8888$$ "}], [{"aoVal": "B", "content": "$$88880$$ "}], [{"aoVal": "C", "content": "$$88888$$ "}], [{"aoVal": "D", "content": "$$8880$$ "}]], "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": ["原式$$=3.6\\times 2345+1111\\times \\frac{256}{5}+6543\\times 3.6$$  $$=3.6\\times (2345+6543)+1111\\times \\frac{256}{5}$$  $$=3.6\\times 8888+1111\\times \\frac{256}{5}$$  $$=3.6\\times 8888+8888\\times \\frac{32}{5}$$  $$=8888\\times (3.6+\\frac{32}{5})$$  $$=88880$$ "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "816", "queId": "ed352489bb744ecdb39db6f495982376", "competition_source_list": ["2013年第9届全国新希望杯五年级竞赛复赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "在有$$10$$个不同因数的自然数中，最小的两个数的和是（~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$92$$ "}], [{"aoVal": "B", "content": "$$128$$ "}], [{"aoVal": "C", "content": "$$210$$ "}], [{"aoVal": "D", "content": "$$560$$ "}]], "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->因数和"], "answer_analysis": ["$$A$$有$$10$$个因数，$$10=2\\times 5$$，故$$A={{m}^{9}}$$或$$A={{m}^{2}}\\times {{n}^{4}}$$若为第一种$$A$$最小为$$512$$舍去，  $$\\therefore A={{m}^{2}}\\times {{n}^{4}}$$最小为$${{2}^{4}}\\times 3=48$$次小为$${{2}^{4}}\\times 5=80$$则和为$$48+80=128$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2558", "queId": "597c2747f26a4272a72a7e7a62d20aa8", "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B卷第7题5分"], "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 $$\\begin{cases} x+y+z=100    3x+2y+z=180   \\end{cases}$$ 解这个方程组可以得到$$z-x=20$$，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": "692", "queId": "6c262160590b430ca363d63c2c95df74", "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6题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "对于两个整数$$a$$、$$b$$，若它们除以整数$$m$$所得的余数相等，则称$$a$$、$$b$$对于模$$m$$同余，记作$$a\\equiv b\\left( modm \\right)$$，如：$$11\\equiv 3\\left( mod4 \\right)$$，$$14\\equiv 19\\left( mod5 \\right)$$．已知$$p$$是小于$$2013$$的自然数，且$$p\\equiv 2013\\left( mod9 \\right)$$，则$$p$$的取值共有（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$222$$个 "}], [{"aoVal": "B", "content": "$$223$$个 "}], [{"aoVal": "C", "content": "$$224$$个 "}], [{"aoVal": "D", "content": "$$225$$个 "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "answer_analysis": ["$$p\\equiv 2013\\equiv 6\\left( \\bmod 9 \\right)$$ $$2013\\div 9=223\\cdots\\cdots 6$$ 即$$2013=223\\times 9+6$$ ∴ $$ p=a\\times 9+6$$ 且$$a\\textless{}223$$，因此共有$$223$$种情况（$$a$$可以为$$0$$）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3334", "queId": "ff80808147248448014724de18cb012b", "competition_source_list": ["2012年全国华杯赛小学高年级竞赛初赛网络版第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$39$$，$$41$$，$$44$$，$$45$$，$$47$$，$$52$$，$$55$$这$$7$$个数重新排成一列，使得其中任意相邻的三个数的和都为$$3$$的倍数．在所有这样的排列中，第四个数的最大值是（~ ~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$44$$ "}], [{"aoVal": "B", "content": "$$45$$ "}], [{"aoVal": "C", "content": "$$47$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["除以$$3$$分别于余：$$0$$，$$2$$，$$2$$，$$0$$，$$2$$，$$1$$，$$1$$，使得其中任意相邻的三个数的和都为$$3$$的倍数重新排成一列，余数顺序为$$2$$，$$0$$，$$1$$，$$2$$，$$0$$，$$1$$，$$2$$．第四个数就是余$$2$$的数，则余$$2$$的数分别是$$41$$，$$44$$，$$47$$．最大是$$47$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3262", "queId": "5111aecd12974b3cb9462e5a24555d7b", "competition_source_list": ["五年级其它", "2017年第13届湖北武汉新希望杯五年级竞赛决赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "图书馆阅览室有$$100$$本读物，星星借阅过$$80$$本，希希借阅过$$78$$本，望望借阅过$$76$$本，则三人共同借阅过的读物最少有（~ ）本． ", "answer_option_list": [[{"aoVal": "A", "content": "$$33$$~~~ "}], [{"aoVal": "B", "content": "$$34$$~~~ "}], [{"aoVal": "C", "content": "$$35$$~~~ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["$$(80+78+76)-2\\times 100=34$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2324", "queId": "ddc14f56bb3940c19207f679045f8fad", "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": ["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->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": "832", "queId": "656817c7e2d24e62ba282187304d9c82", "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． ", "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->能力->逻辑分析"], "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": "2806", "queId": "72703e3c96174992a04946a0a1fd338f", "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": "2505", "queId": "1e78f1368aa24db1b80279a13d5c2f8b", "competition_source_list": ["2020年长江杯五年级竞赛复赛A卷第2题5分", "2020年长江杯五年级竞赛复赛A卷"], "difficulty": "1", "qtype": "single_choice", "problem": "两个数相除，如果除数不变，被除数增加$$10.8$$，商就增加$$0.4$$；如果被除数减少$$4.8$$，除数减少$$0.3$$，商不变．原来的被除数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$32.4$$ "}], [{"aoVal": "B", "content": "$$432$$ "}], [{"aoVal": "C", "content": "$$324$$ "}], [{"aoVal": "D", "content": "$$43.2$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["两个数相除，如果除数不变，被除数增加$$10.8$$，则商增加了$$0.4$$．  因此我们可以得到除数是：$$10.8\\div 0.4=27$$．  如果被除数减少$$4.8$$，除数减少$$0.3$$．商不变，  因此商是$$4.8\\div 0.3=16$$．  则原来的被除数是$$27\\times 16=432$$．  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1554", "queId": "496c41c9892941618953d5509281c6d9", "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，他发现他的前面有$$3$$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->拓展思维->应用题模块->应用题模块排队问题->单主角求总数"], "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": "73", "queId": "1cc6fa102189456c9185929666c48c20", "competition_source_list": ["2010年第8届创新杯四年级竞赛初赛第1题6分", "2010年四年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "2010个连续自然数的和的最小值是(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "2021055 "}], [{"aoVal": "B", "content": "2019045 "}], [{"aoVal": "C", "content": "2020050 "}], [{"aoVal": "D", "content": "2009045 "}]], "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": ["最小的自然数是0，要使和最小，则这2010个连续自然数是0\\textasciitilde2009，$$\\left( 1+2009 \\right)\\times 2009\\div 2=1005\\times 2009=2019045$$ "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2369", "queId": "6aff4e5acef742788fe6fbeebea3b92a", "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9题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "有10个表面涂满红色的立方体，它们的棱长分别为1，2，3...10厘米，如果把这些立方体全部分割成棱长为1厘米的小立方体，在这些小立方体中至少有一面是红色的块数是(  )块. ", "answer_option_list": [[{"aoVal": "A", "content": "1000 "}], [{"aoVal": "B", "content": "1250 "}], [{"aoVal": "C", "content": "1729 "}], [{"aoVal": "D", "content": "2007 "}]], "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": ["$$\\left( {{10}^{3}}-{{8}^{3}} \\right)+\\left( {{9}^{3}}-{{7}^{3}} \\right)+\\cdots +\\left( {{3}^{3}}-{{1}^{3}} \\right)+{{2}^{3}}+{{1}^{3}}={{10}^{3}}+{{9}^{3}}=1729$$（块） "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "172", "queId": "62baef99400346a2b92e5fde42fa9b39", "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": ["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": ["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": "1263", "queId": "396b08a0ce5d4abca6da26e85235ff3e", "competition_source_list": ["2019年亚洲国际数学奥林匹克公开赛（AIMO）六年级竞赛决赛第27题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一个圆形报时机器【钟$$K$$】，特性如下:  与一般时钟相同﹐分针每$$60$$分钟走一圈．不过，时针每$$1440$$分钟走一圈．钟面顺时针方向等距分布【$$1$$】至【$$24$$】(正上方是【$$24$$】)．在【$$5$$时$$x$$分】时，钟面上的【$$10$$】位于钟$$K$$的时针和分针的正中间，求$$x$$的值．  Below is the descriptions of a circular time-reporting machines, namely clock $$K$$.  Similar to normal clock, it takes only $$60$$ minutes for the minute-hand to complete one cycle, but it takes $$1440$$ minutes for the hour-hand to complete one cycle. Numbers to $$24$$ are placed equidistantly on its edge clockwise. (\"$$24$$\"is at the top of the clock.) At \"$$x$$ minutes past $$5$$\", the number \"$$10$$\"lies at the right middle of the two hands of CLOCK $$K$$, find the value of $$x$$. ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$42$$ "}], [{"aoVal": "E", "content": "$$48$$ "}]], "knowledge_point_routes": ["Overseas Competition->知识点->应用题模块", "拓展思维->能力->运算求解"], "answer_analysis": ["时针每分钟走$$0.25{}^{}\\circ $$，分针每分钟走$$6{}^{}\\circ $$．  每两个数字之间的角度为$$15{}^{}\\circ $$．  由此可得：  $$\\begin{eqnarray}150-(5\\times 15+0.25x)\\&=\\&6x-150   225\\&=\\&6.25x   x\\&=\\&36 \\end{eqnarray}$$ "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "907", "queId": "d713c294f1144df19c15851086030724", "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$${{(1011)}_{2}}=1\\times {{2}^{3}}+0\\times {{2}^{2}}+1\\times 2+1=11$$，那么二进制数$${{(11011)}_{2}}$$，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": ["知识标签->课内知识点->数学广角->数与形->数与形数形结合思想"], "answer_analysis": ["$${{(11011)}_{2}}=1\\times {{2}^{4}}+1\\times {{2}^{3}}+0\\times {{2}^{2}}+1\\times 2+1=27$$，因此选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3148", "queId": "0462c881d12e40aebd841e7d410536b5", "competition_source_list": ["2017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": "1", "qtype": "single_choice", "problem": "给一个正方体的表面涂上红、黄、蓝三种颜色，每个面只涂一种颜色，任意抛一次，红色朝上的可能性最大．黄色和蓝色朝上的可能性相同，这个正方体可能有个面涂红色． ", "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->统计与概率->概率->基本概率->可能性"], "answer_analysis": ["在正方体的六个面，分别涂上红、黄、蓝三种颜色，要使哪种颜色朝上的可能性大，必须使这种颜色涂的面数多；要使两种颜色面朝上的可能性相同，就必须使这两种颜色的数量相同；正方体有$$6$$个面，要使黄色和蓝色朝上的可能性相同，可以把一个面涂成红色，一个面涂成蓝色，剩下的面都涂成红色，据此分析解答．  $$6-1-1=4$$．  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": "1597", "queId": "455a047536884873a69cfae0a645c4a8", "competition_source_list": ["其它改编题", "2016年创新杯六年级竞赛训练题（二）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "~某俱乐部男、女会员的人数之比是$$3：2$$，分为甲、乙、丙三组．已知甲、乙、丙三组人数比是$$10：8：7$$，甲组中男、女会员的人数之比是$$3：1$$，乙组中男、女会同的人数之比是$$5：3$$．求丙组中男、女会员人数之比． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2：9$$ "}], [{"aoVal": "B", "content": "$$5：9$$ "}], [{"aoVal": "C", "content": "$$7：9$$ "}], [{"aoVal": "D", "content": "$$4：9$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["设总人数为$$1$$  甲组男会员为$$\\frac{10}{10+8+7} \\times \\frac{3}{3+1}=\\frac{3}{10}$$，女会员为$$\\frac{3}{10} \\times \\frac{1}{3}=\\frac{1}{10}$$；  乙组男会员为$$\\frac{8}{10+8+7} \\times \\frac{5}{5+3}=\\frac{1}{5}$$，女会员为 $$\\frac{1}{5} \\times \\frac{3}{5}=\\frac{3}{25}$$；  丙组男会员为$$\\frac{3}{3+2} -(\\frac{3}{10}+\\frac{1}{5})=\\frac{1}{10}$$，女会员为$$\\frac{2}{3+2} -(\\frac{1}{10}+\\frac{3}{25})=\\frac{9}{50}$$；  丙组中男女会员之比为：$$5：9$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1264", "queId": "705c5425322447c5a8e30305de444c2a", "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：周长是$$48$$厘米，那么长$$+$$宽：$$48\\div$$ 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": "3231", "queId": "42d019b6c2de46de902b43700f0dfc0d", "competition_source_list": ["2006年迎春杯四年级竞赛复赛", "2006年迎春杯三年级竞赛复赛"], "difficulty": "3", "qtype": "single_choice", "problem": "动物园里猩猩比狒狒多，猴子比猩猩多.一天，饲养员拿了10箱香蕉分给它们.每只猩猩比每只狒狒多分1根，每只猴子比每只猩猩多分1根.分完后，只剩下2根香蕉.如果每箱香蕉数量相同，都是40多个，而且猴子比狒狒多6只，猩猩有16只.那么，动物园里有（ ）只猴子. ", "answer_option_list": [[{"aoVal": "A", "content": "18 "}], [{"aoVal": "B", "content": "19 "}], [{"aoVal": "C", "content": "20 "}], [{"aoVal": "D", "content": "17 "}]], "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": ["由题意知，共吃了香蕉$$400\\tilde{  }500$$根，且个位数是$$8$$.设有猴子$$n$$只，则有狒狒$$\\left( n-6 \\right)$$只.因为猩猩的数量介于猴子与狒狒之间，所以$$17\\leqslant n\\leqslant 21$$.再设每只猴子吃了$$a$$根香蕉，则每只猩猩吃了$$\\left( a-1 \\right)$$根，每只狒狒吃了$$\\left( a-2 \\right)$$根.共吃香蕉 $$an+\\left( a-1 \\right)\\times 16+\\left( a-2 \\right)\\left( n-6 \\right)$$ $$=an+16a-16+an-2n-6a+12$$ $$=2an+10a-2n-4$$ $$=2a\\left( n+5 \\right)-2n-4$$ $$=X$$.$$X$$的个位数是8，且$$17$$≤$$n$$≤$$21$$ 若$$n=17$$，则$$X=44a-38$$.$$a$$的个位只能是$$4$$或$$9$$，不满足$$400 \\textless{} X \\textless{} 500$$; 若$$n=18$$，则$$X=46a-40$$.$$a$$的个位只能是$$3$$或$$8$$，不满足$$400 \\textless{} X \\textless{} 500$$; 若$$n=19$$，则$$X=48a-42$$.$$a$$的个位只能是$$0$$或$$5$$，当$$a=10$$时，$$X=438$$，符合题意; 同理，$$n=20$$或$$21$$都不符合题意. 所以有猴子$$19$$只. "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1841", "queId": "8632595493ce4ad594aeaba20b286803", "competition_source_list": ["2017年全国华杯赛竞赛初赛模拟试卷2第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "两个盒子$$A$$和 $$B$$分别装着不同数量的两个小球，两盒中小球的重量是一样的，同时从$$A，B$$盒子中拿出一个小球放入$$C$$盒子,记为一次操作，$$C$$盒子开始为空盒子，经过四十次操作后，$$C$$盒子的总重量与$$B$$盒子一样，又经过十次操作，$$C$$盒子总重量与$$A$$盒子一样，那么$$A，B$$盒子中单个小球的重量比为（ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1:1$$ "}], [{"aoVal": "B", "content": "$$1:2$$ "}], [{"aoVal": "C", "content": "$$1:3$$ "}], [{"aoVal": "D", "content": "$$1:4$$ "}]], "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$$A$$盒子中单个小球重量为$$1$$，$$B$$盒子中单个小球重量为$$k$$．根据题目，可以判断$$k\\geqslant1$$($$A$$盒子小球比$$B$$盒子小球轻)，则有 $$40k-50=10k+10$$， 解方程得$$k=2$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2433", "queId": "0cd6943042cd434b9cf760a468144785", "competition_source_list": ["2018年第6届湖北长江杯五年级竞赛初赛B卷第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$10$$个互不相同的两位奇数之和为$$898$$，那么这是个奇数中最大和最小的奇数之和是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$174$$ "}], [{"aoVal": "B", "content": "$$176$$ "}], [{"aoVal": "C", "content": "$$178$$ "}], [{"aoVal": "D", "content": "$$180$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["$$898\\div 10=89.8$$  这$$10$$个互不相同的两位奇数中间的应该是$$89$$．  据此可以推算出这$$10$$个奇数分别是：  $$79$$、$$83$$、$$85$$、$$87$$、$$89$$、$$91$$、$$93$$、$$95$$、$$97$$、$$99$$．  $$79+99=178$$  故选：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1457", "queId": "51785c7533374eebbf55fe4ce320ffaf", "competition_source_list": ["2016年第14届全国创新杯五年级竞赛初赛第4题", "2016年全国创新杯竞赛A卷第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "将某数的$$3$$倍减$$5$$，计算出答案：将这个答案的$$3$$倍减$$5$$，计算出答案，$$\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot $$，这样反复$$4$$次，最后得出的结果是$$1177$$，那么原数是（~ ）． ", "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->逻辑分析->代数逻辑推理"], "answer_analysis": ["逆推：$$(1177+5)\\div 3=394$$；$$(394+5)\\div 3=133$$；$$(133+5)\\div 3=46$$；$$(46+5)\\div 3=17$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1149", "queId": "265c0520a1314ccda875221cec835b0d", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（三）第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个最简分数，如果分子加$$a$$，等于$$\\frac{1}{2}$$，分母加$$a$$，则等于$$\\frac{1}{4}$$．原来的分数是，$$a$$是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{10}$$，$$3$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{10}$$，$$1$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{10}$$，$$2$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{10}$$，$$2$$ "}]], "knowledge_point_routes": ["知识标签->拓展思维->应用题模块->还原问题->逆运算"], "answer_analysis": ["设原来的分数为$$\\frac{\\text{n}}{\\text{m}}$$，则有$$\\frac{\\text{n+a}}{\\text{m}}=\\frac{1}{2}$$，$$\\frac{\\text{n}}{\\text{m+a}}=\\frac{1}{4}$$，$$\\text{2n=3a}$$，解得$$\\text{m}=10$$，$$\\text{n=3}$$，$$\\text{a}=2$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1724", "queId": "df8062747d0242dda2539f34f7ad1ccb", "competition_source_list": ["2017年四川成都六年级竞赛“全能明星”选拔赛第6题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "一双鞋子如果卖$$140$$元，可赚$$40 \\%$$；如果卖$$120$$元，可赚． ", "answer_option_list": [[{"aoVal": "A", "content": "$$20 \\%$$ "}], [{"aoVal": "B", "content": "$$22 \\%$$ "}], [{"aoVal": "C", "content": "$$25 \\%$$ "}], [{"aoVal": "D", "content": "$$28 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["这双鞋子的成本价：$$140\\div (1+40 \\%)=100$$（元）；  如果卖$$120$$元，可赚$$(120-100)\\div 100=0.2=20 \\%$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1269", "queId": "628e9d4d27f745e7bd6f21117785ec15", "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": "超市中出售的糖果有每包$$11$$粒装和每包$$15$$粒装两种．斌斌一共购买了$$89$$粒糖果．那么，斌斌一共购买了（~ ）包糖果． ", "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->方程思想"], "answer_analysis": ["设分别买了$$x$$和$$y$$包糖果，$$11x+11y=89$$，得$$x=4$$，$$y=3$$，所以一共买了$$7$$包． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2527", "queId": "c2b814c07d0b48bb85dc48c47c6ba497", "competition_source_list": ["2015年全国美国数学大联盟杯五年级竞赛初赛第37题"], "difficulty": "2", "qtype": "single_choice", "problem": "飞人以每小时$$360$$千米的速度飞行，这相当于每秒多少米? ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$64$$ "}], [{"aoVal": "D", "content": "$$100$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->单位换算->复合单位的认识换算"], "answer_analysis": ["单位转化． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "202", "queId": "d0d7eb0f07de4a5b8c0e25fc4ac50418", "competition_source_list": ["2010年第9届小机灵杯四年级竞赛复赛第12题9分", "2021年新希望杯五年级竞赛模拟（考前培训100题）第88题"], "difficulty": "1", "qtype": "single_choice", "problem": "将$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$这$$8$$个数分成三组，分别计算各组数的和，已知这三个和互不相同，且最大的和是最小的和的$$2$$倍，最小的和是~\\uline{~~~~~~~~~~}~． ", "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->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": ["$$1+2+3+4+5+6+7+8=36$$，最大的和是最小的和的$$2$$倍，所以三个和之和应该大于最小的和的$$4$$倍，得到最小的和小于$$36\\div 4=9$$；然后三个和之和应该小于最小的和的$$5$$倍，得到最小的和大于$$36\\div 5=7.2$$，所以最小的和是$$8$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "451", "queId": "a56c25b5a51c4e7d9dd8ea215e45ee30", "competition_source_list": ["2017年河南郑州K6联赛竞赛模拟第八套第10题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "小明的家和学校相隔$$8$$条街，学校早上$$8$$时开始上课．他$$7$$时$$42$$分离开家，$$7$$时$$54$$分到达学校．他从家到学校花了多长时间？解答这个问题需要用到的信息是（ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$条，$$8$$时，$$7$$时$$42$$分，$$7$$时$$54$$分 "}], [{"aoVal": "B", "content": "$$8$$时，$$7$$时$$42$$分，$$7$$时$$54$$分 "}], [{"aoVal": "C", "content": "$$8$$条，$$8$$时，$$7$$时$$42$$分 "}], [{"aoVal": "D", "content": "$$7$$时$$42$$分，$$7$$时$$54$$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": ["$$7$$时$$54$$分$$-7$$时$$42$$分$$=12$$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": "1494", "queId": "5f4ba313eb68409e9c6a9d11aa255fce", "competition_source_list": ["2020年新希望杯三年级竞赛决赛（8月）第12题", "2020年新希望杯三年级竞赛初赛（个人战）第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$4$$ years ago, Fang Fang and Jia Jia\\textquotesingle s ages were $$20$$. This year Fang Fang is $$15$$ years old. How many years later will Jia Jia be $$15$$ years old too?  $$4$$年前，方方和佳佳的年龄和是$$20$$岁．今年方方$$15$$岁，再过~\\uline{~~~~~~~~~~}~年佳佳也是$$15$$岁． ", "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->公式记忆->符号化数学原理", "Overseas Competition->知识点->应用题模块->年龄问题"], "answer_analysis": ["今年年龄和为：$$20+4+4=28$$ （岁），所以今年佳佳：$$28-15=13$$ （岁）．$$15-13=2$$ （年）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2452", "queId": "18fbb2b09b864101bc37de6446b7022f", "competition_source_list": ["2011年六年级竞赛创新杯"], "difficulty": "0", "qtype": "single_choice", "problem": "两个一位小数相乘，如果先把它们四舍五入到个位再相乘，其积是51，那么不四舍五入直接用原数相乘，则积最大值与最小值依次为（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "59.16， 25.25 "}], [{"aoVal": "B", "content": "59.16， 41.25 "}], [{"aoVal": "C", "content": "71.96， 25.05 "}], [{"aoVal": "D", "content": "71.96， 25.25 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->小数->小数四则运算"], "answer_analysis": ["$$51=3\\times 17=1\\times 51$$ 两种情况对应的最大值分别为$$3.4\\times 17.4=59.16$$，$$1.4\\times 51.4=71.96$$;对应的最小值分别为$$2.5\\times 16.5=41.25$$，$$0.5\\times 50.5=25.25$$，对比选项，答案为D. "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3053", "queId": "e5b85f25758e4727a335f454f437ea99", "competition_source_list": ["2009年六年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "美是一种感觉，本应没有客观的标准，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照，这个比例称为黄金分割.在人体下躯体(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，若此比例越接近0.618，就越给人一种美的感觉.如果某女士身高为1.60m，下躯干与身高的比为0.6，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": "2.5cm "}], [{"aoVal": "B", "content": "5.1cm "}], [{"aoVal": "C", "content": "7.5cm "}], [{"aoVal": "D", "content": "8.2cm "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->方程基础->比例方程"], "answer_analysis": ["该女士的下躯干高$$1.6\\times 0.6=0.96(m)$$，设高跟鞋高为$$xm$$，则有$$\\frac{0.96+x}{1.60+x}=0.618$$，解得$$x\\approx 0.075$$，$$0.075m=7.5cm$$. "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3338", "queId": "ff8080814767495f01476b21736f0631", "competition_source_list": ["2014年全国华杯赛小学中年级竞赛初赛第3题"], "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->转化与化归的思想"], "answer_analysis": ["判断能否被$$4$$整除，要看末两位．用$$8$$个$$3$$和$$1$$个$$0$$组成的九位数，被$$4$$除余$$1$$的，末尾只能是$$33$$，所以有$$6$$种． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1590", "queId": "842368a9b2e04bd9877beb58ddae129f", "competition_source_list": ["2018年广东广州学而思综合能力诊断五年级竞赛第13题"], "difficulty": "2", "qtype": "single_choice", "problem": "小伦打算步行去小招家，一路上小伦注意到，每隔$$8$$分钟就有一辆公交车从后方超越他．小伦走到半路，体力不支走不动了，只好打出租车去小招家，这时小伦又发现出租车每隔$$6$$分钟超越一辆公交车．已知出租车的速度是小伦步行速度的$$8$$倍，如果公交车的发车时间间隔固定，且公交车、小伦步行和出租车的速度固定的话，公交车的发车时间间隔为分钟． ", "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->思想->方程思想", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "answer_analysis": ["本题考的是发车问题，解题的关键是公交车与人的追及路程等于出租车与公交车的追及路程等于两辆公交车之间的距离，根据追及问题中的关系式列方程即可解决．  设出租车的速度为$$8$$，则人的速度为$$1$$，  解：设公交车的速度为$$x$$，  $$\\begin{eqnarray} 8\\times \\left( x-1 \\right)\\&=\\&6\\left( 8-x \\right)   8x-8\\&=\\&48-6x   14x\\&=\\&56   x\\&=\\&4\\end{eqnarray}$$  $$8\\times \\left( 4-1 \\right)=24$$，  $$24\\div 4=6$$（分）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2647", "queId": "3b0eb2b57c4542aa8ad46681fae40f02", "competition_source_list": ["2016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": "小亮看一本故事书的正文，他从第二天开始每天比前一天多看一页，他已经连续看了$$9$$天，这本书的正文还剩$$48$$页没看．已知他第四天看了$$39$$页，请问这本数的正文总共有多少页？~\\uline{~~~~~~~~~~}~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$351$$ "}], [{"aoVal": "B", "content": "$$399$$ "}], [{"aoVal": "C", "content": "$$360$$ "}], [{"aoVal": "D", "content": "$$408$$ "}], [{"aoVal": "E", "content": "$$432$$ "}]], "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": ["由等差数列的知识：末项$$=$$首项$$+$$（项数$$-1$$）$$\\times $$公差  可知，第一天看了$$36$$页，第九天看了：$$36+(9-1)\\times 1=44$$（页）  $$9$$天看的总和$$=$$（首项$$+$$末项）$$\\times $$项数$$\\div 2$$  $$=(36+44)\\times 9\\div 2$$  $$=360$$（页）．  书共有：$$360+48=408$$（页）． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "428", "queId": "ede137cd6220423abb4a9a9d9759daec", "competition_source_list": ["2017年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "两个小三角形不重叠放置可以拼成一个大三角形，那么这个大三角形不可能由（  ）拼成。 ", "answer_option_list": [[{"aoVal": "A", "content": "两个锐角三角形 "}], [{"aoVal": "B", "content": "两个直角三角形 "}], [{"aoVal": "C", "content": "两个钝角三角形 "}], [{"aoVal": "D", "content": "一个锐角三角形和一个钝角三角形 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->逻辑推理"], "answer_analysis": ["此题可以采用目标倒推法，从三角形的一个顶点出发对三角形进行分割，看看可以得到的情况，  下面是BCD三种情况举例  （$$1$$）一个钝角三角形、一个锐角三角形（$$2$$）两个直角三角形（$$3$$）两个钝角三角形。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "748", "queId": "404c4959362945b28068d5d54a6cbb23", "competition_source_list": ["小学高年级六年级其它小高著名杯赛拉分题第11题", "2008年全国华杯赛小学高年级五年级竞赛复赛"], "difficulty": "2", "qtype": "single_choice", "problem": "黑板上写着$$1$$至$$2008$$共$$2008$$个自然数，小明每次擦去两个奇偶性相同的数，再写上它们的平均数，最后黑板上只剩下一个自然数，这个数可能的最大值和最小值的差是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2007$$ "}], [{"aoVal": "B", "content": "$$2005$$ "}], [{"aoVal": "C", "content": "$$2006$$ "}], [{"aoVal": "D", "content": "$$2004$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->逐步调整思想"], "answer_analysis": ["先求剩下数的最大值，那么擦去的数应该尽量小，找到规律：  首先擦去$$1$$，$$3$$，写上$$2$$．  擦去$$2$$，$$2$$，写上$$2$$．  擦去$$2$$，$$4$$，写上$$3$$．  擦去$$3$$，$$5$$，写上$$4$$．  擦去$$4$$，$$6$$，写上$$5$$．  $$ \\cdots ~\\cdots $$  擦去$$2006$$，$$2008$$，写上$$2007$$．  所以剩下数的最大值为$$2007$$．  同理可知剩下数的最小值为$$2$$．  所以最大值和最小值的差是$$2005$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1668", "queId": "d18e76bfbb164163bf3f00eb2adf6d31", "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2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "学校一共购买了$$300$$本书，其中有五分之二是数学书，三分之一是语文书，其余是英语书．那么，英语书共有本． ", "answer_option_list": [[{"aoVal": "A", "content": "$$80$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$120$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["数学书：$$300\\times \\frac{2}{5}=120$$（本），  语文书：$$300\\times \\frac{1}{3}=100$$（本），  英语书：$$300-120-100=80$$（本）．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "704", "queId": "c2fdb5b021624685b060acdba8f9c142", "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": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "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．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1997", "queId": "eec6e0b216774e0f853878707e2292f1", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（四）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "《庄子·天下篇》中有一句话：``一尺之锤，日取其半，万世不竭．''意思是：一尺长的木棍，每天截去它的一半，永远也截不完．那么第$$10$$天取的长度是第$$7$$天的（~ ）．~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{16}$$ "}]], "knowledge_point_routes": ["拓展思维->思想->整体思想"], "answer_analysis": ["第$$10$$天是第$$9$$天的$$\\frac{1}{2}$$，第$$9$$天是第$$8$$天的$$\\frac{1}{2}$$，第$$8$$天是第$$7$$天的$$\\frac{1}{2}$$．  第$$10$$天是第$$7$$天的$$\\frac{1}{2}\\times \\frac{1}{2}\\times \\frac{1}{2}=\\frac{1}{8}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2134", "queId": "06e8861f411f4c83896336c9844f09a7", "competition_source_list": ["2019年广东深圳华杯赛小学高年级竞赛初赛第5题10分", "2019年华杯赛小学高年级竞赛第5题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "一艘小船逆水而上，突然，船上一只小鸭落入水中顺流往下漂，船行$$5$$分钟后船夫才发现，立马一边呼唤鸭子，一边掉头行驶．小鸭听到呼唤后，也往船的方向游过来．已知鸭子的游速是船速的$$\\frac{1}{4}$$，船掉头后~\\uline{~~~~~~~~~~}~分钟追上鸭子． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$2.5$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$3.5$$ "}], [{"aoVal": "E", "content": "$$4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["$$5$$分钟后，船与鸭子相距$$5\\left( {{V}_{船}}-{{V}_{水}}+{{V}_{水}} \\right)=5{{V}_{船}}$$，  则再用$$5{{V}_{船}}\\div \\left( {{V}_{船}}+{{V}_{水}}+{{V}_{鸭}}-{{V}_{水}} \\right)$$  $$          =5{{V}_{船}}\\div \\left( {{V}_{船}}+{{V}_{鸭}} \\right)$$  $$          =5{{V}_{船}}\\div \\left( {{V}_{船}}+\\frac{1}{4}{{V}_{船}} \\right)$$  $$          =5{{V}_{船}}\\div \\frac{5}{4}{{V}_{船}}$$  $$          =4$$分钟追上鸭子． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3007", "queId": "c53dc39ee25647c5835e96402a3547f7", "competition_source_list": ["其它改编自2014年全国希望杯六年级竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "计算$$\\dfrac{1}{1+\\dfrac{2}{1+\\dfrac{3}{1+\\dfrac{4}{5}+1}+1}+1}=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{13}{14}$$ "}], [{"aoVal": "B", "content": "$$\\frac{13}{42}$$ "}], [{"aoVal": "C", "content": "$$\\frac{23}{114}$$ "}], [{"aoVal": "D", "content": "$$\\frac{43}{114}$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["原式$$=\\dfrac{1}{1+\\dfrac{2}{1+\\dfrac{3}{\\dfrac{14}{5}}+1}+1}=\\dfrac{1}{1+\\dfrac{2}{1+\\dfrac{15}{14}+1}+1}=\\dfrac{1}{1+\\dfrac{2}{\\dfrac{43}{14}}+1}=\\dfrac{1}{1+\\dfrac{28}{43}+1}=\\dfrac{1}{\\dfrac{114}{43}}=\\dfrac{43}{114}$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "776", "queId": "40f04cd536fc45e4bafb14cc2e68d6a6", "competition_source_list": ["2011年全国希望杯五年级竞赛初赛第13题"], "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$$0-9$$这$$10$$个数字全都用上了，不重也不漏．''那么，维纳这一年~\\uline{~~~~~~~~~~}~岁． ", "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->思想->对应思想"], "answer_analysis": ["此题的关键是他的年龄的立方是一个四位数，年龄的四次方是一个六位数．由于立方是四位数，四次方是六位数，所以年龄的范围大致应在$$17$$到$$22$$之间，所以，符合题意的只有$$18$$． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2463", "queId": "268037b0447044e285bad99d097d11c0", "competition_source_list": ["2013年四川成都小升初", "2009年全国迎春杯三年级竞赛初赛第5题", "2019年四川成都小升初集训题第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\triangle$$ represents a number.$$\\triangle$$代表一个数.  $$\\triangle +\\triangle =a, \\triangle -\\triangle =b, \\triangle \\times \\triangle =c, \\triangle \\div \\triangle =d,$$ and $$a+b+c+d=121$$. $$\\triangle =$$~\\uline{~~~~~~~~~~}~. ", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}], [{"aoVal": "E", "content": "$$12$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析->代数逻辑推理", "Overseas Competition->知识点->计算模块->方程基础->等量代换->字母表示数"], "answer_analysis": ["$$b=\\triangle - \\triangle =0$$  $$d=\\triangle \\div \\triangle =1$$  $$a+b+c+d$$  $$=\\triangle +\\triangle +0+\\triangle \\times \\triangle +1=121$$  $$\\triangle +\\triangle +\\triangle \\times \\triangle =120$$  $$\\triangle \\times \\left( \\triangle +2 \\right)=120=10\\times12$$  $$\\triangle =10$$ "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3275", "queId": "4869dc01ce5945c3b176131a33c63638", "competition_source_list": ["2016年第3届广东深圳鹏程杯四年级竞赛第6题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "在数列$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$，$$8$$，$$9$$，$$\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot $$，$$2015$$，$$2016$$中，去掉带有数字$$1$$和$$9$$的数，把剩下的数从小大大排成一列：$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$，$$8$$，$$20$$，$$22$$，$$23$$，$$24$$，$$25$$，$$26$$，$$27$$，$$28$$，$$30$$，$$\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot $$，那么在列数从左到右第$$319$$个数是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$321$$ "}], [{"aoVal": "B", "content": "$$376$$ "}], [{"aoVal": "C", "content": "$$439$$ "}], [{"aoVal": "D", "content": "$$506$$ "}], [{"aoVal": "E", "content": "$$588$$ "}]], "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": ["$$1$$到$$99$$以内去掉$$1$$，$$9$$，$$10-\\/-19$$，$$21$$，$$29$$，$$31$$，$$39$$，$$\\cdot \\cdot \\cdot $$，$$90-\\/-99$$，还剩$$99-2-10-2\\times 7-10=63$$，$$100-\\/-199$$，$$200-\\/-299$$，$$100-2-10-2\\times 7-10=64$$，$$300-399(64)$$，$$400-\\/-499(64)$$，$$500-599(64)$$，$$64\\times 4+63=319$$，第$$319$$个数是：$$588$$． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1195", "queId": "19c9e58a46874718b4b9a45867ba84af", "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": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->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": "1228", "queId": "4b5bfcb4f8fc48758e59b9b8dc6a68a2", "competition_source_list": ["2020年新希望杯一年级竞赛初赛（团战）第31题"], "difficulty": "1", "qtype": "single_choice", "problem": "欣欣一家$$5$$月$$30$$日白天到达海边（五月有$$31$$天)，并入住附近宾馆，一直住到$$6$$月$$4$$日白天才离开．欣欣一家在海边住了个晚上． ", "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->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": "1626", "queId": "f65e907ce7ae4ba8843fd8e6d045eaf4", "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5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$30$$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": "$$18$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->能力->实践应用"], "answer_analysis": ["涛涛前面有十个人，涛涛前面的人加上涛涛后面的人加上涛涛共有$$30$$个，  所以$$30-10-1=19$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2849", "queId": "69fd1fac559844ddba945f8b17c0e963", "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9题3分", "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"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$5$$，$$a$$，$$b$$，$$c$$，$$4035$$这五个数成等差数列，则$$b=$$．  Given an arithemetic sequence $$5$$, $$a$$, $$b$$, $$c$$, and $$4035$$ ， 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->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": "1297", "queId": "c76ee9943eac4ad7984cec03fc1696bd", "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": "羊村储备了一些草，其中青草是黄草的$$3$$倍多$$2$$千克．每天吃$$15$$千克青草，$$6$$千克黄草；吃了若干天后，青草还剩下$$74$$千克，黄草剩下$$4$$千克．羊村的青草和黄草共（~ ）千克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$250$$ "}], [{"aoVal": "B", "content": "$$374$$ "}], [{"aoVal": "C", "content": "$$474$$ "}], [{"aoVal": "D", "content": "$$498$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["设吃了$$x$$天，则原有青草$$\\left( 15x+74 \\right)$$千克，原有黄草$$\\left( 6x+4 \\right)$$千克，依题意可得方程：$$15x+74=3\\left( 6x+4 \\right)+2$$，解之得$$x=20$$，羊村原有青草和黄草共$$15\\times 20+74+6\\times 20+4=498$$（千克）． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "507", "queId": "dd0f161c3a6f46d28276bdfa454c313e", "competition_source_list": ["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": "现在是早上$$8$$时，喜羊羊准备$$1$$小时后到草原上割青草，请问喜羊羊几点去割青草？ ", "answer_option_list": [[{"aoVal": "A", "content": "早上$$7$$时 "}], [{"aoVal": "B", "content": "早上$$8$$时 "}], [{"aoVal": "C", "content": "早上$$9$$时 "}], [{"aoVal": "D", "content": "晚上$$8$$时 "}]], "knowledge_point_routes": ["知识标签->课内知识点->常见的量->钟表->认识钟表：结构、时针、分针"], "answer_analysis": ["早上$$8$$时，$$1$$小时后是早上$$9$$时 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1821", "queId": "85dc889898394dc385771804a6880996", "competition_source_list": ["2020年第1届广东深圳超常思维竞赛六年级竞赛初赛第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "某竞赛考试有$$30$$个选择题，计分规则是：基础分$$30$$分，答对一题得$$4$$分，答错一题扣$$1$$分，不答不给分．用公式表示就是$$S=30+4c-w$$，其中$$S$$为分数，$$c$$是答对的题数，$$w$$是答错的题数，允许不答．小马在这次考试中得分在$$80$$以上，他把分数告诉了小姜，小姜据此能推算出小马做出了几道题．如果小马的得分少一些，但仍在$$80$$分以上，小姜就无法推算了．那么小马得了分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$110$$ "}], [{"aoVal": "C", "content": "$$119$$ "}], [{"aoVal": "D", "content": "$$120$$ "}], [{"aoVal": "E", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["假设小明全对，得分为$$30+30\\times4=150$$分，  每错一道，损失$$5$$分，没答一道损失$$4$$分，  设错$$x$$道，没答$$y$$道，  得分$$\\begin{cases}100    110    119    120   \\end{cases}\\to 失分\\begin{cases}50    40    31    30   \\end{cases}$$，  ①$$5x+4y=50$$，  $$\\begin{matrix}\\begin{cases}x=10    y=0  \\end{cases}, \\begin{cases}x=6    y=5   \\end{cases}, \\begin{cases}x=2    y=10   \\end{cases}\\times      \\end{matrix}$$，  ②$$5x+4y=40$$，  $$\\begin{cases}x=8    y=0   \\end{cases}, \\begin{cases}x=4    y=5   \\end{cases}, \\begin{cases}x=0    y=10   \\end{cases}\\times $$，  ③$$5x+4y=31$$，  $$\\begin{cases}x=3    y=4   \\end{cases}\\surd $$，  ④$$5x+4y=30$$  $$\\begin{cases}x=6    y=0   \\end{cases}, \\begin{cases}x=2    y=5   \\end{cases}\\times $$，  根据不定方程组的答案，  ①②④这三个不定方程均有多个解，  不符合小姜推算出了小马答题的情况，  根据③号的答案扣分$$31$$分，则得分$$119$$分．  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": "631", "queId": "15704a4f387f4a498a68f1c658d23cc0", "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": "1", "qtype": "single_choice", "problem": "如果将一周的时间表示为分钟，这个分钟数的最大质因数是多少？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$21$$ "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["如果将一周的时间表示为分钟，这个分钟数的最大质因数是$$7$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3151", "queId": "074114e037154df8bd728a3821973200", "competition_source_list": ["2017年湖南长沙雨花区湖南广益实验中学小升初（五）第7题6分", "2017年全国小升初师达入学备考", "2019年湖南长沙开福区青竹湖湘一外国语学校小升初（一）第5题3分", "2015年湖北武汉世奥赛六年级竞赛模拟训练题（一）第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "有写着数字$$2、$$5$$、8$$的卡片各$$10$$张，现在从中任意抽出$$7$$张，这$$7$$张卡片的和可能等于． ", "answer_option_list": [[{"aoVal": "A", "content": "$$21$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$29$$ "}], [{"aoVal": "D", "content": "$$58$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["各数被$$3$$除余$$2$$，故$$7$$张卡片的和$$\\equiv 2\\times 7\\equiv 2  (\\bmod   3)$$，只有$$C$$满足要求． ", "<p>因为在写着数字$$2$$、$$5$$、$$8$$的卡片各$$10$$张中任意抽出$$7$$张，可以组成的数有$$14$$、$$17$$、$$20$$、$$23$$、$$26$$、$$29$$、$$32$$、$$35$$、$$38$$、$$41$$、$$44$$、$$47$$、$$0$$、$$53$$、$$56$$，所以$$A$$、$$B$$、$$D$$是不可能的．</p>\n", "<p>$$2$$、$$5$$、$$8$$被$$3$$除，余数都是$$2$$，同余．</p>\n<p>所以取出$$7$$张卡片求和，余数变成了$$14$$．</p>\n<p>因为减去$$14$$，剩下的数可以被$$3$$整除（$$7$$张$$2$$的情况，和为$$14$$，减去$$14$$为$$0$$）．</p>\n<p>或者$$14$$被$$3$$除，余数是$$2$$，即$$7$$张卡片求和，被$$3$$除，余数为$$2$$．</p>\n<p>只有$$29$$符合题意．</p>\n<p>故选$$\\text{C}$$．</p>"], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3279", "queId": "882a21e2cbb54c42b5ee6aaa424fc341", "competition_source_list": ["2011年北京五年级竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "气象台预报``本市明天降雨概率是$$80 \\%''$$．对此信息，下列说法中正确的是： ", "answer_option_list": [[{"aoVal": "A", "content": "本市明天将有$$80 \\%$$的地区降水． "}], [{"aoVal": "B", "content": "本市明天将有$$80 \\%$$的时间降水． "}], [{"aoVal": "C", "content": "明天肯定下雨． "}], [{"aoVal": "D", "content": "明天降水的可能性比较大． "}]], "knowledge_point_routes": ["课内体系->知识点->随机现象->随机现象的可能性", "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": ["降水概率指的是可能性的大小，并不是降水覆盖的地区或者降水的时间．$$80 \\%$$的概率也不是指肯定下雨，$$100 \\%$$的概率才是肯定下雨．$$80 \\%$$的概率是说明有比较大的可能性下雨． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3195", "queId": "2b5ab8cf9902471c9c3c3a8244d3d1f2", "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": "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->拓展思维->计数模块->排列组合->排列->捆绑法"], "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）；总的排法有$$5\\times 4\\times 3\\times 2\\times 1=120$$（种）。  所以贝贝和妮妮不相邻的排法有$$120-2\\times 24=72$$（种）。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3137", "queId": "208ea734253943a087e35011117c1951", "competition_source_list": ["2013年美国数学大联盟杯小学高年级竞赛初赛第35题5分", "2012年全国美国数学大联盟杯小学高年级竞赛初赛第35题"], "difficulty": "1", "qtype": "single_choice", "problem": "小威的中药需要分三步进行熬制，现有蝾螈、苍蝇、甲壳虫、蛇和蜗牛五种中药原料，要求熬制过程中每步只能选择一种原料，且每次选择的原料必是不相同的．问：小威制成的中药有（~ ~ ~）种．（注：对于相同的中药原料，若熬制的顺序不同，则最终制成的中药也必然不同．） ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$60$$ "}]], "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"], "answer_analysis": ["$$5\\times4\\times3=60$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1686", "queId": "4f083f67a21148549e19301761a1cd71", "competition_source_list": ["2004年第2届创新杯五年级竞赛初赛第8题", "2004年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "刘老师给钱让小明去买每本1.4元的练习本，正好可以找回1元，但是小明在途中不小心弄丢了5元，又来不及回家取钱，小明就买回了同样本数每本1.2元的练习本，并找回3元，刘老师当时给小明的钱数是（ ）. ", "answer_option_list": [[{"aoVal": "A", "content": "50元 "}], [{"aoVal": "B", "content": "36元 "}], [{"aoVal": "C", "content": "22元 "}], [{"aoVal": "D", "content": "15元 "}]], "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": ["小明买每本1.4元的练习本可以找回1元，买每本1.2元的练习本不但丢了5元还可以找回3元，那么买每本1.2元的练习本一共少花$$5+3-1=7$$(元)，则练习本共有$$7\\div \\left( 1.4-1.2 \\right)=35$$(本)，刘老师当时给了小明$$35\\times 1.4+1=50$$(元)，选A "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "162", "queId": "2c3cd6f23e9e493483eafa7181b6672f", "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13题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙三人同时说了三句话．甲说：``乙正在说谎．''乙说：``丙正在说谎．''丙说：``他俩正在说谎．''那么，说谎的人是 ． ", "answer_option_list": [[{"aoVal": "A", "content": "甲 "}], [{"aoVal": "B", "content": "乙 "}], [{"aoVal": "C", "content": "甲和丙 "}], [{"aoVal": "D", "content": "乙和丙 "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->逻辑分析"], "answer_analysis": ["甲和丙 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3222", "queId": "23ac701fab8344269b7bfdeb4157749c", "competition_source_list": ["2013年IMAS小学高年级竞赛第一轮检测试题第14题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "一个箱子里面装有十颗白球和十颗黑球，小明按照以下规则玩游戏：每一次他可选择两种方式之一进行操作：（$$a$$）从箱子里拿出一个白球和一个黑球放置在箱子外；（$$b$$）如果箱子外有黑球，从箱子里拿出一个白球放置在箱子外并从箱子外已取出的球中拿出一个黑球放回箱子里．游戏刚开始时，箱子外面没有球，请问经过$$6$$次操作后，小明算了以下箱子外面的球数之总数．请问小明所算的球数共有多少种可能的值？ ", "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->思想->整体思想"], "answer_analysis": ["因为白球每次都拿出一个，所以箱子外面不同球数的情况由黑球决定．  由游戏的规则可知，如果进行六次（$$a$$）操作则箱子外面有$$6$$个黑球；  如果进行五次（$$a$$）操作一次（$$b$$）操作则箱子外面有$$4$$个黑球；  如果进行四次（$$a$$）操作二次（$$b$$）操作则箱子外面有$$2$$个黑球；  如果进行三次（$$a$$）操作三次（$$b$$）操作则箱子外面有$$0$$个黑球．  不可能进行三次以上（$$b$$）操作，否则箱子外没有足够的黑球，所以箱子外面的球数有$$6$$、$$8$$、$$10$$、$$12$$四种可能，故选$$C$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "528", "queId": "dda18fa27fcf4e57a48a1d33cd1ee3dc", "competition_source_list": ["2006年四年级竞赛创新杯"], "difficulty": "2", "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->拓展思维->组合模块->逻辑推理->条件型逻辑推理->神推理"], "answer_analysis": ["选项A:12题，按第一种给分方法，最多得$$12\\times 5=60$$分，排除A; 选项B:15题，按第一种给分方法，最多得$$15\\times 5=75$$分，排除B; 选项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D.故选C "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2308", "queId": "a18fda1e727b4ec3aab390fa07940b1a", "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": ["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": ["速度和$$\\times $$相遇时间$$=$$相遇路程，第一次相遇和第二次相遇速度和相等，相遇时间和相遇路程成正比例关系，两次相遇时间之比为$$1:3$$，则甲第二次相遇共走$$8\\times 3=24$$千米，则$$A$$、$$B$$两地相距$$24-4=20$$千米． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1301", "queId": "82eb63f7d3314904b104395d17045ca6", "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->能力->构造模型->模型思想"], "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": "2877", "queId": "bf991bb55c0145ddbc6d4d959b2a9729", "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": "1", "qtype": "single_choice", "problem": "算式$$\\textasciitilde4+5\\times 6$$的正确结果是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$54$$ "}], [{"aoVal": "B", "content": "$$34$$ "}], [{"aoVal": "C", "content": "$$25$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->运算求解"], "answer_analysis": ["先算乘除，再算加减，$$4+5\\times 6=4+30=34$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "53", "queId": "0f28a4cf8b4142ab89d75789f7270a2c", "competition_source_list": ["2004年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "$$A$$，$$B$$，$$C$$三家超市在同一条南北大街上。$$A$$超市在$$B$$超市的南边$$40$$米处，$$C$$超市在$$B$$超市的北边$$100$$米处。小明从$$B$$超市出发沿街向北走了$$50$$米，接着又向南走了$$60$$米，此时他的位置在(  )。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$B$$超市 "}], [{"aoVal": "B", "content": "$$C$$超市北边$$30$$米处 "}], [{"aoVal": "C", "content": "$$A$$超市北边$$30$$米处 "}], [{"aoVal": "D", "content": "$$B$$超市北边$$10$$米处 "}]], "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": ["$$A$$，$$B$$，$$C$$三家超市的位置依次由南向北，$$AB=40$$米，$$BC=100$$米，小明从$$B$$超市出发沿街向北走$$50$$米然后再向南走$$60$$米，实际上的位置是$$B$$超市南边$$10$$米，即$$A$$超市北边$$30$$米，故选C。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2652", "queId": "ac16463d9f264c59832633a762f97d86", "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": "$$1.01\\times 0.99\\textless0.99$$ "}], [{"aoVal": "B", "content": "$$1.1\\times 1.1\\textgreater1.1$$ "}], [{"aoVal": "C", "content": "$$10.4\\times 0.1\\textless{}1.04$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}-\\frac{1}{3}\\textless{}\\frac{1}{3}-\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["知识标签->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": ["$$\\text{A}$$：$$1.01\\times 0.99\\textgreater{}0.99$$，故该选项错误；  $$\\text{C}$$：$$10.4\\times 0.1=1.04$$，故该选项错误；  $$\\text{D}$$：$$\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$$，$$\\frac{1}{3}-\\frac{1}{4}=\\frac{1}{12}$$，$$\\frac{1}{6}\\textgreater\\frac{1}{12}$$，故该选项错误；  所以答案为$$\\text{B}$$ "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3013", "queId": "c558102269a9495ba475ff32bf1cb03c", "competition_source_list": ["2017年河南郑州K6联赛竞赛第三轮第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "定义新运算：已知$$A\\bigcirc B=A+B-1$$，$$A△B=A\\times B-1$$，$$x\\bigcirc \\left( x△4 \\right)=30$$，则$$x=$$． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$6.2$$ "}], [{"aoVal": "C", "content": "$$6.4$$ "}], [{"aoVal": "D", "content": "$$14$$ "}]], "knowledge_point_routes": ["知识标签->数学思想->赋值思想"], "answer_analysis": ["$$x?4=4x-1$$，$$x\\bigcirc \\left( 4x-1 \\right)=x+\\left( 4x-1 \\right)-1=5x-2$$，即$$5x-2=30$$，解得$$x=6.4$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2720", "queId": "ff8080814502fa24014507533c8809e2", "competition_source_list": ["2014年全国迎春杯五年级竞赛复赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "整数除法算式：$$a\\div b=c\\cdots \\cdots r$$，若$$a$$和$$b$$同时扩大$$3$$倍，则（~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$r$$不变 "}], [{"aoVal": "B", "content": "$$c$$扩大$$3$$倍 "}], [{"aoVal": "C", "content": "$$c$$和$$r$$都扩大$$3$$倍 "}], [{"aoVal": "D", "content": "$$r$$扩大$$3$$倍 "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析"], "answer_analysis": ["被除数和除数同时乘或除以相同的数（$$0$$除外），商不变，但是余数相应的扩大或缩小相同的倍数． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1576", "queId": "ff8080814518d524014519095fde0319", "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": "2537", "queId": "d54312a21eaa463aa0bf71444fdebcb8", "competition_source_list": ["2014年全国创新杯五年级竞赛第3题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "用$$\\min \\left( a,b \\right)$$表示$$a$$、$$b$$两数中的较小者，用$$\\max \\left( a,b \\right)$$表示$$a$$、$$b$$两数中的较大者，例如$$\\min \\left( 3,5 \\right)=3$$，$$\\max \\left( 3,5 \\right)=5$$，$$\\min \\left( 3,3 \\right)=3$$，$$\\max \\left( 5,5 \\right)=5$$．设$$a$$、$$b$$、$$c$$、$$d$$是互不相等的自然数，$$\\min \\left( a,b \\right)=p$$，$$\\min \\left( c,d \\right)=q$$，$$\\max \\left( p,q \\right)=x$$，$$\\max \\left( a,b \\right)=m$$，$$\\max \\left( c,d \\right)=n$$，$$\\min \\left( m,n \\right)=y$$，则（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textgreater y$$ "}], [{"aoVal": "B", "content": "$$x\\textless{}y$$ "}], [{"aoVal": "C", "content": "$$x=y$$ "}], [{"aoVal": "D", "content": "$$x\\textgreater y$$和$$x\\textless{}y$$都有可能 "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["因为$$a$$、$$b$$、$$c$$、$$d$$大小关系不定，取$$a$$、$$b$$、$$c$$、$$d$$为$$4$$、$$3$$、$$2$$、$$1$$，$$p=3$$，$$q=1$$，$$x=3$$，$$m=4$$，$$n=2$$，$$y=2$$，故$$x\\textgreater y$$，若取$$a$$、$$b$$、$$c$$、$$d$$为$$4$$、$$3$$、$$2$$、$$1$$，得$$x=2$$，$$y=3$$．故$$x\\textless{}y$$都有可能． ", "<p>若$$\\begin{cases}a=1 \\\\ b=2 \\\\ c=3 \\\\ d=4 \\\\ \\end{cases}$$，那么$$\\begin{cases}x=\\max \\left[ \\min \\left( 1,2 \\right),\\min \\left( 3,4 \\right) \\right]=\\max \\left( 1,3 \\right)=3 \\\\ y=\\min \\left[ \\max \\left( 1,2 \\right),\\max \\left( 3,4 \\right) \\right]=\\min \\left( 2,4 \\right)=2 \\\\ \\end{cases}$$，那么$$x&gt;y$$；</p>\n<p>若$$\\begin{cases}a=1 \\\\ b=4 \\\\ c=3 \\\\ d=2 \\\\ \\end{cases}$$，那么$$\\begin{cases}x=\\max \\left[ \\min \\left( 1,4 \\right),\\min \\left( 3,2 \\right) \\right]=\\max \\left( 1,2 \\right)=2 \\\\ y=\\min \\left[ \\max \\left( 1,4 \\right),\\max \\left( 3,2 \\right) \\right]=\\min \\left( 4,3 \\right)=3 \\\\ \\end{cases}$$，那么$$x&lt;{}y$$；</p>\n<p>即$$x&gt;y$$和$$x&lt;{}y$$都有可能，故选$$\\text{D}$$．</p>\n"], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1964", "queId": "e566e74a689b46cba9ace49453e7d101", "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": "三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．三个同学得到的结果分别是$$2274$$、$$2003$$、$$2320$$，老师看后指出．这三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": "$$2320$$ "}]], "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$$\\text{C}$$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．  $$\\left( 2310-2003 \\right)=307$$非偶，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": "1436", "queId": "a2d800888981426693aa7fa6c52d07c1", "competition_source_list": ["2016年新希望杯六年级竞赛训练题（五）第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "现有$$2000$$克浓度为$$15  \\% $$的盐水，要配制浓度为$$20  \\% $$的盐水，需要加入（~ ）克浓度为$$24  \\% $$的盐水． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1800$$ "}], [{"aoVal": "B", "content": "$$2000$$ "}], [{"aoVal": "C", "content": "$$2400$$ "}], [{"aoVal": "D", "content": "$$2500$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["假设需要的重量为$$x$$克，列方程$$2000\\times 15  \\% +24  \\% \\times x=20  \\% \\times \\left( 2000+x \\right)$$，解得：$$x=2500$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "874", "queId": "73dbc03a8f974d489949f4c4fa35c9a3", "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4题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "$$a$$、$$b$$、$$c$$三个数字共可组成六个无重复数字的三位数，这六个三位数的和为$$3774$$，那么这六个三位数中最大数与最小数的差最小是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$99$$ "}], [{"aoVal": "B", "content": "$$297$$ "}], [{"aoVal": "C", "content": "$$369$$ "}], [{"aoVal": "D", "content": "$$396$$ "}]], "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": ["这六个三位数之和为$$222\\times \\left( a+b+c \\right)=3774$$，即$$a+b+c=17$$．要使最大和最小的差最小，即  让三个数尽可能接近，即为$$4$$、$$6$$、$$7$$，此时最大减最小为$$764-467=297$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3303", "queId": "ac63482ba1e64a248334f25dbf11ca7e", "competition_source_list": ["2017年第17届世奥赛六年级竞赛决赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "过年包饺子时，在饺子里放一枚硬币，吃到硬币的人在新的一年会运气好，事事顺意，是一种好运的象征．妈妈包了$$50$$个饺子，其中有$$1$$个饺子包了硬币，煮好后平均分成$$5$$碗，你从这$$5$$碗中随便选$$1$$碗，能吃到包有硬币的饺子的概率是（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2 \\%$$ "}], [{"aoVal": "B", "content": "$$10 \\%$$ "}], [{"aoVal": "C", "content": "$$20 \\%$$ "}], [{"aoVal": "D", "content": "$$50 \\%$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["五碗中一定有$$1$$碗是的，那么概率即为$$\\frac{1}{5}=20 \\%$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "948", "queId": "c67d7e83341b47ee89d5f9ab10ffa8a3", "competition_source_list": ["2016年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有一种数，是以法国数学家梅森的名字命名的，形如$${{2}^{n}}-1$$（$$n$$为质数）的数为梅森数，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。那么第一个梅森合数是（  ） ", "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->拓展思维->数论模块->质数与合数"], "answer_analysis": ["解：选项A：$${{2}^{n}}-1=4$$，$$n$$无整数解；  选项B：$${{2}^{n}}-1=15$$，$$n$$为$$4$$，$$n$$不是质数，故舍去；  选项C：$${{2}^{n}}-1=127$$，$$n$$为$$7$$，$$127$$不是合数，故舍去；  选项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": "1117", "queId": "0cbd59d9044746f08390b22a3de2762b", "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，龙嘴里不断吐出不同颜色的龙珠，按$$4$$颗红珠，$$3$$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": ["$$2000\\div (4+3+2+1)=200$$（组）． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1278", "queId": "5960dcaaecc54792ae9d2db5c02ce73e", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（四）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "有一圆形花坛，绕它走一圈是$$120$$米，如果在花坛周围每隔$$6$$米栽一株丁香花，再在每相邻的两株丁香之间栽$$2$$株月季花，共栽月季花朵． ", "answer_option_list": [[{"aoVal": "A", "content": "$$38$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$44$$ "}]], "knowledge_point_routes": ["拓展思维->七大能力->实践应用"], "answer_analysis": ["这是一个环形植树问题，棵树$$=$$段数，故丁香花的数量为$$120\\div 6=20$$（株），月季花的数量为$$20\\times 2=40$$（株）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3120", "queId": "f9a772aebbe14fc6a338b8282dc6aab0", "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", "2007年华杯赛四年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "对于所有的数$$a$$、$$b$$，把运算$$a*b$$定义为$$a*b=ab-a+b$$，则方程$$5*x=17$$的解是$$x=$$(  ). ", "answer_option_list": [[{"aoVal": "A", "content": "$$3\\frac{2}{5}$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3\\frac{2}{3}$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "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": ["$$5*x=17$$  $$5x-5+x{=}17$$  $$6x=22$$  $$x=3\\frac{2}{3}$$ "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1298", "queId": "878acad4c29e495eb61fe6b3fba961af", "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年全国迎春杯三年级竞赛初赛第$$8$$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": ["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": ["$$2000\\div (4+3+2+1)=200$$（组）． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1810", "queId": "9ff37d0ff6fc417e8566523c78482d1f", "competition_source_list": ["2012年第10届全国创新杯小学高年级六年级竞赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "~某工程，可由若干台机器在规定时间内完成，如果增加两台机器，则节省了八分之一的时间，如果减少两台机器，就要推迟$$\\frac{2}{3}$$小时做完，那么一台机器完成这个工程需要（~ ）小时． ", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$58$$ "}], [{"aoVal": "D", "content": "$$42$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["增加两台机器后，时间比为$$8:7$$，可以得到机器数比为$$7:8$$，那么原来应该有$$14$$台机器；若减少两台机器，推迟$$\\frac{2}{3}$$小时，可以得到时间比为$$6:7$$，原来时间为$$4$$小时，因共有$$14$$台机器，所以每台机器完成需$$14\\times 4=56$$小时． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1128", "queId": "2aaff3e845ab4135a39666cd57af239c", "competition_source_list": ["2016年新希望杯小学高年级六年级竞赛训练题（二）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲商品打$$8$$折后的价格是乙商品原价的$$4$$倍，小华分别以$$8$$折和$$7$$折的价格买下甲、乙两种商品，支出总额比甲商品原价少$$6$$元，乙商品实际售价是（~ ）元． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$21$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["设乙原价为$$x$$元．  $$4x+0.7x=\\frac{4x}{0.8}-6$$，$$x=20$$，$$20\\times 0.7=14$$（元）． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2017", "queId": "cf2c6ec75c2b4f62a7ed8fcb09db27e2", "competition_source_list": ["2013年华杯赛六年级竞赛初赛", "2020年第24届YMO五年级竞赛决赛第8题3分", "2013年华杯赛五年级竞赛初赛"], "difficulty": "1", "qtype": "single_choice", "problem": "盒子里有黑棋子和白棋子若干，若取出一枚黑子，则余下的黑子数与白子数之比为$$9:7$$，若取出一粒白子，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比白子多个 ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$6$$ "}], [{"aoVal": "E", "content": "$$7$$ "}]], "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": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2336", "queId": "01367556e4f7435396eb1a8b3b3cefa4", "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": "一根$$5$$米长的绳子，平均分成$$3$$段，每段是这根绳子的$$\\frac{\\left( {          } \\right)}{\\left( {          } \\right)}$$，每段长$$\\frac{\\left( {          } \\right)}{\\left( {          } \\right)}$$米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$，$$\\frac{5}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{6}$$，$$\\frac{5}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{6}$$，$$\\frac{1}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{3}$$，$$\\frac{1}{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数的认识->认、读、写数->分数->分数的意义", "拓展思维->拓展思维->计算模块->分数->分数基础->分数的意义"], "answer_analysis": ["绳子总共分为$$3$$段，每段是全长的$$\\frac{1}{3}$$；每段长：$$5\\div 3=\\frac{5}{3}$$（米）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1307", "queId": "42c2b90b933e45899b5d3172672e1099", "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7题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "现有大、小油瓶共$$50$$个，每个大瓶可装油$$4$$千克，每个小瓶可装油$$2$$千克，大瓶比小瓶共多装$$20$$千克，问：大瓶有个． ", "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->思想->对应思想"], "answer_analysis": ["（头和脚差型；分组法）分组时，$$1$$个大瓶配$$2$$个小瓶，这样每组大瓶小瓶装油重量相等；  由于大瓶比小瓶共多装$$20$$千克，所以共有$$20\\div 4=5$$ （个）多余的大瓶．  组数：$$(50-5)\\div 3=15$$ （组）；  大瓶有：$$15+5=20$$（个）；  小瓶有：$$15\\times 2=30$$（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": "761", "queId": "ff8080814502fa24014503a7819e01a9", "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$$\\frac{2014}{x+1}$$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->能力->运算求解"], "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": "2516", "queId": "425137604faa4bc8a27dec50b961afb4", "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->思想->数形结合思想"], "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": "3239", "queId": "2895344fdd4742529fff388af4f68993", "competition_source_list": ["2022年第九届鹏程杯四年级竞赛初赛第27题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$19$$，$$20$$，$$21$$，$$22$$，···，$$73$$，$$74$$这些数中，任取两个数，使其和为偶数的选法共有~\\uline{~~~~~~~~~~}~种． ", "answer_option_list": [[{"aoVal": "A", "content": "$$656$$ "}], [{"aoVal": "B", "content": "$$756$$ "}], [{"aoVal": "C", "content": "$$856$$ "}], [{"aoVal": "D", "content": "$$956$$ "}], [{"aoVal": "E", "content": "$$1056$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->排列组合"], "answer_analysis": ["无 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "107", "queId": "7014c1ca75334576a59b360cac6f0bd9", "competition_source_list": ["2009年六年级竞赛创新杯"], "difficulty": "1", "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": "$$23\\leqslant M\\leqslant 69$$ "}], [{"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->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": ["当其余四个数为$$1$$、$$3$$、$$5$$、$$7$$时，$$M$$有最大值$$69$$。设最大的一个等于$$x$$，其余四个数为$$x-2$$， $$x-4$$，$$x-6$$，$$x-8$$时，$$M$$有最小值，由$$x+\\left( x-2 \\right)+\\left( x-4 \\right)+\\left( x-6 \\right)+\\left( x-8 \\right)=85$$，解得$$x=21$$，所以$$21\\leqslant M\\leqslant 69$$。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "785", "queId": "646f88e3229a489a859ff9cc314bcb29", "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": ["知识标签->学习能力->七大能力->逻辑分析"], "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": "2816", "queId": "727d969e22414ac48efa2db2752250c0", "competition_source_list": ["2014年全国学而思杯一年级竞赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "计算：$$18+9-8=$$~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$19$$ "}], [{"aoVal": "E", "content": "$$21$$ "}]], "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->拓展思维->计算模块->整数->整数加减->整数加减巧算之凑整法->整数同尾相减"], "answer_analysis": ["$$18+9-8=18-8+9=10+9=19$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2467", "queId": "7deace7598d74749a287051f96b15763", "competition_source_list": ["2014年全港小学数学挑战赛四年级竞赛初赛A卷第11题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算：$$9+99+999+9999+99999+999999$$．  Find the value of $$9+99+999+9999+99999+999999$$. ", "answer_option_list": [[{"aoVal": "A", "content": "$$1111104$$． "}], [{"aoVal": "B", "content": "$$1111114$$． "}], [{"aoVal": "C", "content": "$$1111105$$． "}]], "knowledge_point_routes": ["拓展思维->思想->转化与化归的思想"], "answer_analysis": ["$$9+99+999+9999+99999+999999$$$$=10-1+100-1+1000-1+10000-1+100000-1+1000000-1$$$$=1111110-6=1111104‬$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1500", "queId": "fac886a0b2854f76b59a3ecdaf08d637", "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$$13$$个图形是什么？ $$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\cdots$$ ", "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->课内知识点->数学广角->简单的周期->简单的周期规律"], "answer_analysis": ["根据规律，「$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$」为一个周期  他的顺序是：  $$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$  故答案为：$$\\blacksquare$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "215", "queId": "4c7ef97d72bc441b9a87282608af6ae5", "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．已知：  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 "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "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": "2127", "queId": "009d777b14654bea8ac04d05433d8125", "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->能力->实践应用"], "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": "235", "queId": "36e92373b40941d6a450472f7d98c338", "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~\\uline{~~~~~~~~~~}~分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$13$$ "}]], "knowledge_point_routes": ["知识标签->学习能力->七大能力->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": "623", "queId": "1932fdcadd8448f582065efdc1e07489", "competition_source_list": ["2019年广东深圳华杯赛小学高年级竞赛第6题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$1\\times 2\\times 3\\times \\cdots \\times 99\\times 100={{12}^{n}}M$$，其中$$M$$为自然数，$$n$$为使得等式成立的最大的自然数．下面有$$4$$个答案：其中正确． ", "answer_option_list": [[{"aoVal": "A", "content": "$$M$$能被$$2$$整除，但不能被$$3$$整除 "}], [{"aoVal": "B", "content": "$$M$$能被$$3$$整除，但不能被$$2$$整除 "}], [{"aoVal": "C", "content": "$$M$$能被$$4$$整除，但不能被$$3$$整除 "}], [{"aoVal": "D", "content": "$$M$$不能被$$3$$整除，也不能被$$2$$整除 "}]], "knowledge_point_routes": ["拓展思维->思想->逐步调整思想"], "answer_analysis": ["$$1$$至$$100$$中，  $$3$$的$$1$$次方的倍数共有$$100\\div 3$$，整商$$=33$$个，  $$3$$的$$2$$次方的倍数共有$$100\\div \\left( 3\\times 3 \\right)$$，整商$$=11$$个，  $$3$$的$$3$$次方的倍数共有$$100\\div \\left( 3\\times 3\\times 3 \\right)$$，整商$$=3$$个，  $$3$$的$$4$$次方的倍数共有$$100\\div \\left( 3\\times 3\\times 3\\times 3 \\right)$$，整商$$=1$$个，  所以$$1\\times 2\\times 3\\cdots \\times 99\\times 100$$的结果包含质因数$$3$$的次数是$$33+11+3+1=48$$，  $$2$$的$$1$$次方的倍数共有$$100\\div 2$$，整商$$=50$$个，  $$2$$的$$2$$次方的倍数共有$$100\\div \\left( 2\\times 2 \\right)$$，整商$$=25$$个，  $$2$$的$$3$$次方的倍数共有$$100\\div \\left( 2\\times 2\\times 2 \\right)$$，整商$$=12$$个，  $$2$$的$$4$$次方的倍数共有$$100\\div \\left( 2\\times 2\\times 2\\times 2 \\right)$$，整商$$=6$$个，  $$2$$的$$5$$次方的倍数共有$$100\\div \\left( 2\\times 2\\times 2\\times 2\\times 2 \\right)$$，整商$$=3$$个，  $$2$$的$$6$$次方的倍数共有$$100\\div \\left( 2\\times 2\\times 2\\times 2\\times 2\\times 2 \\right)$$，整商$$=1$$个，  所以$$1\\times 2\\times 3\\cdots \\times 99\\times 100$$的结果包含质因数$$2$$的次数是$$50+25+12+6+3+1=97$$，  $${{2}^{97}}={{4}^{48}}\\times 2$$，  $${{3}^{48}}\\times {{4}^{48}}={{12}^{48}}$$，  ∴$$1\\times 2\\times 3\\cdots \\times 99\\times 100$$的结果包含因数$$12$$的最大次数是$$48$$，  ∴$$M$$能被$$2$$整除，但不能被$$3$$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": "1444", "queId": "75c6d824c64a415c9b733d92a728d3a7", "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": "$$56$$ "}], [{"aoVal": "E", "content": "$$60$$ "}]], "knowledge_point_routes": ["拓展思维->能力->逻辑分析", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1"], "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": "3391", "queId": "862bc940f48c4c6aa581a4e88a548425", "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": "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->能力->公式记忆->符号化数学原理"], "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": "1769", "queId": "cd54202aaf914c94837e871d09246335", "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": "$$3$$ "}], [{"aoVal": "D", "content": "$$0$$ "}], [{"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": ["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": "1646", "queId": "ff80808147342b05014738dd13d203ee", "competition_source_list": ["2013年全国华杯赛小学中年级竞赛初赛B卷第2题"], "difficulty": "1", "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->思想->逐步调整思想"], "answer_analysis": ["$$A+B+C=111$$，$$A+B=31\\times 2=62\\Rightarrow C=111-62=49$$，  $$A+C=37\\times 2=74\\Rightarrow B=111-74=37$$．  所以，$$(B+C)\\div 2=(49+37)\\div 2=43$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "7", "queId": "015542c1796f47adb043267187c6d3d2", "competition_source_list": ["2017年华杯赛六年级竞赛初赛"], "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->拓展思维->组合模块->抽屉原理->构造型抽屉原理"], "answer_analysis": ["解：构造抽屉，把这$$20$$个数分组，看成$$10$$个抽屉：$$\\left { 1,20 \\right }$$，$$\\left { 2,19 \\right }$$，$$\\cdots $$，$$\\left { 10,11 \\right }$$。  从这$$10$$个数组的$$20$$个数中任取$$11$$个数，根据抽屉原理可得，其中必有两个数的和等于$$21$$，  故选：C。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "340", "queId": "72bdf91e724e424f8ce957add3ef066c", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第25题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "下列说法中，不能实现的个数为．  ①桌子上有$$7$$只杯子，$$3$$只杯口向上，$$4$$只杯口向下．每次任意翻转其中$$2$$只，经过若干次翻转后，可将$$7$$只杯子全变成杯口向下；  ②有两堆石子，一堆有$$1234$$颗，另一堆有$$4321$$颗．现有两种操作，一种操作是从两堆石子中拿走相同数量的石子（每次拿的数量可以不同），另一种操作是从其中一堆中拿出若干石子放入另一堆，则若干次操作后可以将两堆的石子同时取光；  ③有$$777$$个孩子，依次编为$$1\\sim 777$$号，能够将这些孩子分成若干组，使每组中都有一个孩子的号码数等于本组其余孩子的号码数的和；  ④走廊上六盏全部都关着的灯，依次编号为$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，经过$$9988$$次循环往复的拉动（从$$1$$号到$$6$$号拉动，再从$$1$$号到$$6$$号拉动），最后可以将$$4$$号灯拉亮． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "以上都不对 "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->操作问题->翻杯子问题"], "answer_analysis": ["①每个杯子翻动奇数次才能改变杯子的状态，开始时有$$3$$个杯口向上，$$4$$个杯口向下，要使所有的杯子都杯口向下，则开始时向上的杯子每个需要翻动奇数次，$$4$$个杯口向下的需要翻动偶数次，总共需要翻动奇$$+$$奇$$+$$奇$$+$$偶$$+$$偶$$+$$偶$$+$$偶$$=$$奇数次，但是每次任意翻动$$2$$只，只翻动了偶数次，不论翻动多少次，总翻动的次数都是偶数，偶数不等于奇数，所以不能实现；  ②每次操作要么从两堆中拿走相同数量的石子，即石子的总数减少偶数个，要么从其中一堆中拿出若干个石子放入另一堆，则石子的总数不变，所以每次操作时，两堆石子的总数奇偶性不变，原来总共有$$1234+4321=5555$$颗石子，为奇数，那每次操作后石子的数量还是奇数，所以不可能若干次操作后可以将两堆石子同时取光，所以②不能实现；  ③若每组中都有一个孩子的号码数等于本组其余孩子的号码数的和，则每组的号码和都是偶数，那么总共的号码和也是偶数，但是$$1+2+3+\\cdots \\cdots +777=$$奇数，所以③不能实现；  ④要改变灯的状态必须拉动奇数次，$$9988\\div 6=1664\\cdots \\cdots 4$$，说明$$1$$、$$2$$、$$3$$、$$4$$都拉动了$$1664+1=1665$$次，为奇数，则$$1$$、$$2$$、$$3$$、$$4$$的状态都改变了，原来是关，最后就是亮，$$5$$、$$6$$拉动了$$1664$$次，为偶数，状态没变，还是关，所以④可以实现．  所以不能实现的是①②③，共$$3$$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": "3316", "queId": "90fcb64478ef49fdb8a9fe7106d80cd8", "competition_source_list": ["2014年走美杯五年级竞赛初赛"], "difficulty": "0", "qtype": "single_choice", "problem": "$$5$$个人围坐在一张圆桌旁就餐，有（  ）种不同的坐法。（如果某种排法可以通过旋转得到另一种排法，那么这两种排法算作同一种） ", "answer_option_list": [[{"aoVal": "A", "content": "$$22$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$26$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计数模块->排列组合->排列->排列的基本应用"], "answer_analysis": ["圆桌可以旋转，先固定$$1$$人，剩下的$$4$$人进行全排列即可，所以有$$1\\times\\text{A}_{\\text{4}}^{\\text{4}}=1\\times4\\times3\\times2\\times1=24$$（种）。  故选：B。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1637", "queId": "8ca9744a6b594e7d84865445de7e039a", "competition_source_list": ["2013年华杯赛六年级竞赛", "2013年华杯赛六年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一只青蛙$$8$$点从深为$$12$$米的井底向上爬，它每向上爬$$3$$米，因为井壁打滑，就会下滑$$1$$米，下滑$$1$$米的时间是向上爬$$3$$米所用时间的三分之一。$$8$$点$$17$$分时，青蛙第二次爬至离井口$$3$$米处，那么青蛙从井底爬到井口时所花的时间为(  )分钟。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$22$$ "}], [{"aoVal": "B", "content": "$$20$$ "}], [{"aoVal": "C", "content": "$$17$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "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"], "answer_analysis": ["解：以爬$$3$$米，滑一米为一个周期；$$\\left( 3-1 \\right)\\times 3+3=9\\text{（m）}$$，青蛙第一次爬至离井口$$3$$米之处，$$\\left( 3-1 \\right)\\times 4+1=9\\text{（m）}$$，青蛙第二次爬至离井口$$3$$米之处，此时，青蛙爬了$$4$$个周期加$$1$$米，用时$$17$$分钟，所以青蛙每爬$$1\\text{m}$$或滑$$1\\text{m}$$所用时间为$$1$$分钟； $$\\left( 12-3 \\right)\\div \\left( 3-1 \\right)=4\\cdots \\cdots 1$$，青蛙从井底爬到井口经过$$5$$个周期，再爬$$2\\text{m}$$，用时$$5\\times \\left( 3+1 \\right)+2=22$$（分钟）；  故选：A。 "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "23", "queId": "029684cd15364b2da493bdaec06b423d", "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 "}]], "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": "1744", "queId": "91e1f278995a4151ad00893c4ef97149", "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4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "有盐水若干克，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->拓展思维->应用题模块->浓度问题->抓不变量"], "answer_analysis": ["根据题意分析可知，设第一次加入一定量的水后，盐水为$$100$$克，  则第二次加水后盐水有$$100\\times 4 \\%\\div 3 \\%\\approx 133$$（克），  $$133-100=33$$（克），说明每次加的水为$$33$$克；  只含盐$$100\\times 3 \\%=3$$（克），第三次又加入水$$33$$克，  所以第三次加入同样的水后盐水浓度为：  $$3\\div (133+33)=3\\div 166\\approx 2 \\%$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "826", "queId": "5c55caef74104b9584cdfba194ab5184", "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": "~判定自然数$$2017$$是质数还是合数，答案为（~ ~ ~）． ", "answer_option_list": [[{"aoVal": "A", "content": "是合数 "}], [{"aoVal": "B", "content": "是质数 "}], [{"aoVal": "C", "content": "都不是 "}], [{"aoVal": "D", "content": "无法确定 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["用筛法去判定，因为$$2017\\textless4{{7}^{2}}$$，用$$47$$以内的质数去试除都无法整除，所以$$2017$$为质数，选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1731", "queId": "777eac6b99544f0d98787316fd863175", "competition_source_list": ["2005年第3届创新杯六年级竞赛初赛第6题"], "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$$\\frac{1}{3}$$，那么全班人数是(  )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->思想->对应思想"], "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$\\text{B}$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1836", "queId": "92b6972fc4684cf7a991d42326afa925", "competition_source_list": ["2016年全国世奥赛竞赛A卷第5题", "2016年第16届世奥赛六年级竞赛决赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "阿基米德是古希腊伟大的数学家，在物理学、工程学上也有卓越贡献，传授阿基米德为判金王冠中是否掺银而冥思苦想，最后在洗澡时悟出了浮力定律，并利用它解决王冠的难题．由于水对物体的浮力作用，纯金和纯银完全浸入水中称重时，分别减轻约$$\\frac{1}{20}$$和$$\\frac{1}{10}$$．假定一顶王冠重$$0.9$$千克，浸入水中称重，减轻了$$\\frac{1}{18}$$，则可断定其中掺入了白银，那么这顶王冠中含银（~ ）千克． ", "answer_option_list": [[{"aoVal": "A", "content": "$$0.1$$ "}], [{"aoVal": "B", "content": "$$0.2$$ "}], [{"aoVal": "C", "content": "$$0.3$$ "}], [{"aoVal": "D", "content": "$$0.4$$ "}]], "knowledge_point_routes": ["拓展思维->思想->方程思想"], "answer_analysis": ["方法一：方程法  设王冠中含有银千克，则金的质量为千克  则可列方程为，解得  故含有银的量为千克  方法二：十字交叉（两种物体混合的问题都可以转化成类浓度问题，用十字交叉解决，如行程、浓度、工程、降级问题都可以）  根据题目可知，剩余金的和银的是整个王冠的，即  金：银的质量比，则银的质量为千克． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "954", "queId": "dd5e7ccc5fd646358fca7303d49413e2", "competition_source_list": ["2013年全国学而思杯五年级竞赛A卷第20题"], "difficulty": "4", "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": ["拓展思维->思想->逐步调整思想", "课内体系->思想->逐步调整思想"], "answer_analysis": ["$$2222$$与$$0000$$的每一位数字均不相同，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": "1792", "queId": "6ac42139386043998ce7361330d1a73a", "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10题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$6$$年前，母亲的年龄是儿子的$$5$$倍。$$6$$年后母子年龄总和是$$78$$岁。问：母亲今年岁。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$66$$ "}], [{"aoVal": "B", "content": "$$54$$ "}], [{"aoVal": "C", "content": "$$51$$ "}], [{"aoVal": "D", "content": "$$45$$ "}]], "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->年龄问题"], "answer_analysis": ["$$6$$年前，母子的年龄和为$$78-\\left( 6+6 \\right)\\times 2=54$$岁，儿子那时$$54\\div \\left( 5+1 \\right)=9$$岁，母亲那时$$9\\times 5=45$$岁，母亲今年$$45+6=51$$岁 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1389", "queId": "435a9a5389a04eef93eab99126d73172", "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10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "有四位小朋友，其中每三位小朋友的岁数之和分别为$$22$$、$$20$$、$$17$$、$$25$$，请问这四位小朋友中年龄最大的比年龄最小的大岁． ", "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->能力->构造模型->模型思想"], "answer_analysis": ["设这$$4$$个小朋友的年龄分别为$$a$$岁，$$b$$岁，$$c$$岁，$$d$$岁，  根据题意，可列出$$\\begin{cases}a+b+c=22 ①    a+b+d=20 ②    b+c+d=17  ③   a+c+d=25 ④   \\end{cases}$$，  ①$$+$$②$$+$$③$$+$$④，  $$3a+3b+3c+3d=22+20+17+25$$，  $$3(a+b+c+d)=84$$，  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde a+b+c+d$$  $$=84\\div 3$$  $$=28$$，  所以$$4$$个人年龄和为$$28$$岁，  又因为$$17 ~\\textless{} ~20 ~\\textless{} ~22 ~\\textless{} ~25$$，  所以年龄最大的比年龄最小的大：  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde(28-17)-(28-25)$$  $$=11-3$$  $$=8$$（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": "2013", "queId": "b8a045a0fc1c41c380e0c101a608b37e", "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，正好赶上新冠疫情，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->思想->对应思想"], "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": "3412", "queId": "ae581a583c684dac8abc25530462747a", "competition_source_list": ["2016年全国世奥赛五年级竞赛A卷第7题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "在$$1829$$年法国盲人路易布莱尔发明了点字，他是由一个框内的$$6$$个凸或者不凸的点组成，以凸点数的多少和位置的不同，代表不同的符号，因此可以变化组成（~ ）个不同的符号（没有任何凸点就不计数），这已成为现代国际上普遍采用的盲文形式． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$63$$ "}], [{"aoVal": "D", "content": "$$64$$ "}]], "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"], "answer_analysis": ["由于没有任何凸点就不计数，则不同的符号有$${{2}^{6}}-1=63$$个． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2629", "queId": "da2d3bde8f374b34ba151b30947ff5d2", "competition_source_list": ["2006年第11届全国华杯赛竞赛复赛第5题", "2017年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": "先写出一个两位数$$62$$，接着在$$62$$右端写这两个数字的和$$8$$，得到$$628$$，再写末两位数字$$2$$和$$8$$的和$$10$$，得到$$62810$$，用上述方法得到一个有$$123$$位的整数：$$628101123$$$\\cdots$，则这个整数的个位是~\\uline{~~~~~~~~~~}~． ", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->计算模块->数列与数表->数列规律->数形结合找规律"], "answer_analysis": ["此题要首先找出数字出现的规律：$$6281011235813471123581347$$\\ldots，从第六个开始，每十个数字一个循环，然后按周期问题解答即可．  $$\\left( 123-5 \\right)\\div 10=11\\cdots 8$$；个位数字是周期中的第$$8$$个，即为$$3$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1093", "queId": "b92f4d703e454b26822fadb8bc1ed0de", "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": "$$11$$ "}], [{"aoVal": "E", "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->等量代换替换型"], "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": "288", "queId": "4dedb4d2c519461580f0c49c109077e8", "competition_source_list": ["2006年第4届创新杯五年级竞赛初赛A卷第5题", "2006年五年级竞赛创新杯"], "difficulty": "2", "qtype": "single_choice", "problem": "有4名棋手进行循环比赛，每两人赛一局，胜一局得2分，平一局得1分，负一局得0分，如果每人得分互不相同，第一名不是全胜，那么至少有(  )局是平局. ", "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->体育比赛->单循环赛"], "answer_analysis": ["因为比赛共进行$$4\\times \\left( 4-1 \\right)\\div 2=6$$局，所以4名选手得分之和为$$2\\times 6=12$$分.如果没有平局，那么4个选手得分是4个互不相同的偶数，所以4个选手得分依次为6分，4分，2分和0分，从而得知第一名三局得6分，即第一名三局全胜与题设条件矛盾，当第一名有一局为平局时，4个选手得分可以依次为5分，4分，2分和1分，符合题意，因此，至少有一局是平局. "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1610", "queId": "ff8080814613dae901461c2ec57a05e1", "competition_source_list": ["2011年北京五年级竞赛", "小学高年级六年级其它小高著名杯赛拉分题第9题", "2009年第14届全国华杯赛竞赛初赛第6题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$A$$、$$B$$、$$C$$、$$D$$、$$E$$五个小朋友做游戏，每轮游戏都按照下面的箭头方向把原来手里的玩具传给另外一个小朋友：$$A\\to C$$，$$B\\to E$$，$$C\\to A$$，$$D\\to B$$，$$E\\to D$$．开始$$A$$、$$B$$拿着福娃，$$C$$、$$D$$、$$E$$拿着福牛，传递完$$5$$轮时，拿着福娃的小朋友是（ ~ ~ ~~）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$C$$与$$D$$ "}], [{"aoVal": "B", "content": "$$A$$与$$D$$ "}], [{"aoVal": "C", "content": "$$C$$与$$E$$ "}], [{"aoVal": "D", "content": "$$A$$与$$B$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["根据题意，$$A$$与$$C$$互相传，$$B$$、$$D$$、$$E$$之间则按$$B$$→$$E$$→$$D$$→$$B$$→\\ldots 的顺序轮流传．开始时，两个福娃分别在$$A$$、$$B$$手上，其中$$A$$手上的福娃经过$$5$$轮的传递将到$$C$$的手里，$$B$$手上的福娃经过$$5$$轮的传递将到$$D$$的手里．所以传递完$$5$$轮时，拿着福娃的小朋友是$$C$$和$$D$$．正确答案为$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1598", "queId": "c8256fad14ef497d8bd645c928326ea4", "competition_source_list": ["2014年迎春杯三年级竞赛初赛"], "difficulty": "2", "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": ["解：$$10+31=41$$（天）  $$41\\div 7=5$$（周）$$\\cdots \\cdots6$$（天）  余数是$$6$$，从星期六再过$$6$$天就是星期五。  答：$$2014$$年$$1$$月$$31$$日是星期五。  故选：C。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "900", "queId": "86594577b1414289a5f092b619f473b3", "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）． ", "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": "635", "queId": "58ef207bdc0041a388c88d7ebd42cc24", "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$$\\text{n}$$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． ", "<p>枚举$$n=2$$、$$3$$、$$5$$、$$7$$、$11\\cdots$ 当$$n=11$$时首次满足要求，$2^{11}-1=2047=23\\times89$，是合数.</p>"], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "743", "queId": "3795262d8e3e46d6b28069e97915586a", "competition_source_list": ["2013年第11届创新杯三年级竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个儿童用棱长为$$1$$厘米的$$42$$个正方体黏合成一个各面为长方形的立体砖．如果其底面的周长是$$18$$厘米，则这块砖的高是厘米． ", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["因为棱长为$$1\\text{cm}$$的$$42$$个正方体黏合成一个各面为长方形的立体砖，  其底面的周长是$$18\\text{cm}$$，  所以底面立方体个数一定小于$$18$$大于$$8$$，  因为$$42$$个正方体黏合成一个各面为长方形的立体砖，  即可得出组合图形整除底面个数，  所以底面个数一定是整除$$42$$，  所以符合要求的只有：$$14$$，  所以底面立方体个数为$$14$$，  所以这个立体砖的高为：$$42\\div14=3$$（厘米）．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3008", "queId": "f7774e9b906445cd8d3e096db68766d0", "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": "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->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"], "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": "3479", "queId": "ec08c0b5aeb04af09c11aad3714763d7", "competition_source_list": ["2020年希望杯一年级竞赛模拟第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$4$$人握手，每两人之间都握一次手，一共要握次．（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->七大能力->实践应用"], "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": "1696", "queId": "9188fdf8a4364039ae170acb4a7aa94f", "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）共有多少辆缆车？ ", "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->思想->整体思想"], "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": "2126", "queId": "004cfc42c1c346a3ba0d632dcfa4bde1", "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": ["知识标签->数学思想->转化与化归的思想"], "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": "313", "queId": "ff808081472482f50147325d98630ac2", "competition_source_list": ["2013年全国华杯赛小学高年级竞赛初赛C卷第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$～$$11$$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->能力->抽象概括"], "answer_analysis": ["对于``任意$$\\cdots$$必有''这样的语句，应该考虑到``抽屉原理''．如果需要证明结论正确的话，那就要构造抽屉，而抽屉的个数，应该是小于$$6$$的  第一句说必有两个数互质，如果这是正确的话，那么就要构造出小于$$6$$的抽屉，且每一组抽屉中的数一定是两两互质的，而很容易想到，每相邻的两个数都是互质的，所以可以这样构造（$$1$$，$$2$$，$$3$$）（$$4$$，$$5$$）（$$6$$，$$7$$）（$$8$$，$$9$$）（$$10$$，$$11$$）其实构造的方法不是唯一的，还有很多构造方法如：【（$$1$$，$$3$$，$$4$$）（$$2$$，$$9$$）（$$8$$，$$11$$）（$$5$$，$$6$$）（$$7$$，$$10$$）】；  第二句很容易举出反例，（$$6$$、$$7$$、$$8$$、$$9$$、$$10$$、$$11$$）最大的六个数就没有倍数关系，同样的还有：（$$4$$、$$5$$、$$6$$、$$7$$、$$9$$、$$11$$）；  第三句，根据第二句话的反例，可以看出，第三句话是成立的，那么就要严谨的证明一下，还是要构造抽屉，按照什么构造呢，可以按照$$1$$到$$11$$中的$$5$$个质因数来构造$$5$$个抽屉，$$1$$放在哪个抽屉里都可以，（$$1$$，$$2$$，$$4$$，$$8$$）（$$3$$，$$6$$，$$9$$）（$$5$$，$$10$$）（$$7$$）（$$11$$）这五个抽屉中，要任意取$$6$$个数，必有两个数在同一个抽屉中，就必满足其中一个数的$$2$$倍是另一个数的倍数． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1234", "queId": "591ffa675d5f48058df674787a809560", "competition_source_list": ["2015年湖北武汉世奥赛五年级竞赛模拟训练题（一）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "某个游戏满分为$$100$$分，每人可以做$$4$$次，以平均分为游戏的成绩．小乐的分数是$$85$$分，则他任何一次游戏的得分都不低于分． ", "answer_option_list": [[{"aoVal": "A", "content": "$$40$$ "}], [{"aoVal": "B", "content": "$$50$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$70$$ "}]], "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": ["小乐$$4$$次游戏的部分是$$85\\times 4=340$$（分），当他有$$3$$次游戏的分数最高时，剩余$$1$$次的分数最低．  因为游戏满分为$$100$$分，故最低分为$$3403\\times 100=40$$（分）． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3314", "queId": "ff8080814518d5240145192bf4320578", "competition_source_list": ["2014年全国迎春杯六年级竞赛复赛第15题"], "difficulty": "3", "qtype": "single_choice", "problem": "老师把某个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：``听了他们两人的话，我也知道这个两位数是多少了．''  $$E$$：``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": "20", "queId": "86ddbd59ec044e24bca5f6d1acfba4b4", "competition_source_list": ["2016年创新杯小学高年级六年级竞赛训练题（一）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$32$$块石头，重量各不相同，无法从外观及手感区分它们的重量，现有一架无砝码的天平用来比较它们的重量．最少需要称（~ ~ ~ ）次，一定能称出最重的和第二重的石块． ", "answer_option_list": [[{"aoVal": "A", "content": "$$31$$ "}], [{"aoVal": "B", "content": "$$35$$ "}], [{"aoVal": "C", "content": "$$61$$ "}], [{"aoVal": "D", "content": "$$62$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->操作与策略->操作问题->称重找假物->不知轻重"], "answer_analysis": ["秤出最重的需要$$16+8+4+2+1=31$$（次），第二重的只用把和最重比较过的$$5$$块石头取出继续比较，还需要$$2+1+1=4$$（次），一共是$$35$$次． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3325", "queId": "6009b0bba4964a67995c10e2acff49bf", "competition_source_list": ["2015年第13届全国创新杯五年级竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "甲、乙、丙、丁做竞猜游戏，在一个箱子中装有大小一样的二十个球，编号分别为$$1$$、$$2$$、$$3$$、$$\\cdot \\cdot \\cdot $$、$$20$$，从中任意取出一个球，猜测球的号码的情况．甲猜：奇数；乙猜：偶数；丙猜：$$3$$的倍数；丁猜：含有数字$$1$$．赢得可能性最大的是（~ ）． ", "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->拓展思维->计数模块->统计与概率->概率->基本概率->可能性"], "answer_analysis": ["甲：$$\\frac{10}{20}$$；乙：$$\\frac{10}{20}$$；丙：$$\\frac{6}{20}$$；丁：$$\\frac{11}{20}$$；丁最大． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1582", "queId": "ff8080814518d5240145201ac0000a6e", "competition_source_list": ["2014年全国迎春杯三年级竞赛复赛第10题"], "difficulty": "1", "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->思想->转化与化归的思想"], "answer_analysis": ["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$$(23-2)\\div (4-1)=7$$（个），蓝泳帽有$$7+23=30$$（个），共有$$30+7=37$$（位）同学． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3081", "queId": "ef466b661ee34f4da9a153c34cf08b71", "competition_source_list": ["2021年第8届鹏程杯五年级竞赛初赛第5题4分", "2021年第8届鹏程杯四年级竞赛初赛第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，$$1$$和$$2$$一组，$$3$$和$$4$$一组，$$\\cdots \\cdots 2019$$和$$2020$$一组，每组结果都是$$-1$$，一共有$$2020\\div2=1010$$组，所以前面的和是$$-1010$$，$$2021-1010=1011$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2413", "queId": "3cdfded0821f4df0888bdc4474bb1a17", "competition_source_list": ["2011年全国世奥赛三年级竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "$$100 - 99 + 98 - 97~ ~\\cdots ~+4 - 3 + 2 - 1$$的结果是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$150$$ "}], [{"aoVal": "C", "content": "$$100$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["知识标签->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": ["$$50$$ "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "40", "queId": "20d68350f7444e2daae69580ca98e0fe", "competition_source_list": ["2008年三年级竞赛学而思杯"], "difficulty": "1", "qtype": "single_choice", "problem": "美国前总统林肯说过：``最高明的骗子，可能在某个时刻欺骗所有人，也可能在所有时刻欺骗某些人，但不可能在所有时刻欺骗所有人。''如果林肯上述判定为真，则以下（  ）为假。 ", "answer_option_list": [[{"aoVal": "A", "content": "林肯可能在某个时刻受骗    "}], [{"aoVal": "B", "content": "林肯可能在所有时刻不受骗 "}], [{"aoVal": "C", "content": "骗子可能在某个时刻受骗     "}], [{"aoVal": "D", "content": "不存在某一时刻有人可能不受骗 "}], [{"aoVal": "E", "content": "存在某一时刻，有人可能不受骗 "}]], "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": ["A项，林肯可能遇到骗子并欺骗了他；B项，林肯可能一直遇不到骗子，从而不受骗；C项，骗子可能遇到骗子；D项，如果某一时刻有人未遇到骗子，则他可能不受骗，从而D为假；E项，骗子不可能在所有时刻欺骗所有人，故存在某一时刻，有人可能不受骗，选D。 "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1470", "queId": "5a951a83bfb34f05aaf007f22dc77e63", "competition_source_list": ["2015年湖北武汉世奥赛六年级竞赛模拟训练题（三）第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个最简分数，如果分子加$$a$$，则等于$$\\frac{1}{2}$$．原来的分数是，$$a$$是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{10}$$，$$3$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{10}$$，$$1$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{10}$$，$$2$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{10}$$，$$2$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["设原来的分数为$$\\frac{\\text{n}}{\\text{m}}$$，则有$$\\frac{\\text{n+a}}{\\text{m}}=\\frac{1}{2}$$，$$\\frac{\\text{n}}{\\text{m+a}}=\\frac{1}{4}$$，$$\\text{2n=3a}$$，解得$$\\text{m}=10$$，$$\\text{n=3}$$，$$\\text{a}=2$$． "], "answer_value": "D"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "835", "queId": "d65731f3028f4fb697f5b68270300f4e", "competition_source_list": ["2014年全国迎春杯六年级竞赛初赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "老师把一个三位完全平方数的百位告诉了甲，十位告诉了乙，个位告诉了丙，并且告诉三人他们的数字互不相同．三人都不知道其他两人的数是多少，他们展开了如下对话：  甲：我不知道这个完全平方数是多少．  乙：不用你说，我也知道你一定不知道．  丙：我已经知道这个数是多少了．  甲：听了丙的话，我也知道这个数是多少了．  乙：听了甲的话，我也知道这个数是多少了．  请问这个数是的平方． ", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$28$$ "}], [{"aoVal": "D", "content": "$$29$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["通过枚举不难发现，百位是$$6$$，$$8$$，$$9$$ 的满足条件的平方数分别只有$$625$$，$$841$$，$$961$$，  因此第一句说明百位不是$$6$$，$$8$$，$$9$$；  进而得知第二句说明十位不是$$2$$，$$4$$，$$6$$；  第三句说明这个数的个位在剩下所有可能中是唯一的，而只有当个位是$$4$$或$$9$$，$${{28}^{2}}=784$$，$${{17}^{2}}=289$$是唯一满足之前所有条件的数;  第四句说明甲在丙说话之前还不知道结果，而若百位是$$7$$，而$${{28}^{2}}=784$$，$${{17}^{2}}=289$$，  于是甲听完乙说话后已经知道结果了，因此百位只能是$$2$$．  从而这个数为$${{17}^{2}}=289$$． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2872", "queId": "91f2cd99aa8740a8a644b31ee7a157c1", "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": "$$208$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算求解", "拓展思维->思想->方程思想"], "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": "3119", "queId": "ebe64fadbebb41aaad24c993d99c6e06", "competition_source_list": ["2014年IMAS小学中年级竞赛第一轮检测试题第18题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "小朋友玩游戏，老师让小朋友们站成一排，并从第一位开始依照$$1$$、$$2$$、$$3$$循环报数，最后一位小朋友报的数是$$2$$，请问这一排可能共有多少位小朋友？ ", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$25$$ "}], [{"aoVal": "C", "content": "$$26$$ "}], [{"aoVal": "D", "content": "$$27$$ "}], [{"aoVal": "E", "content": "$$28$$ "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["由题意可知这一派同学的人数除以$$3$$后所得的余数是$$2$$，在各个选项中只有选项$$\\text{C}$$符合要求．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3353", "queId": "df83562d6c8e4ec9a34994f472a8786b", "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": "$$160$$ "}], [{"aoVal": "C", "content": "$$480$$ "}], [{"aoVal": "D", "content": "$$720$$ "}]], "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": ["$$2\\times 2\\times 3=12$$（种）． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2974", "queId": "9c2bdedc87b449a7a23d6928bd164435", "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．  $$124+129+106+141+237-500+113=$$~\\uline{~~~~~~~~~~}~ ", "answer_option_list": [[{"aoVal": "A", "content": "$$370$$ "}], [{"aoVal": "B", "content": "$$360$$ "}], [{"aoVal": "C", "content": "$$350$$ "}]], "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“好朋友数”的应用"], "answer_analysis": ["$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde$$$$124+129+106+141+237-500+113$$  $$=\\left( 124+106 \\right)+\\left( 129+141 \\right)+\\left( 237+113 \\right)-500$$  $$=230+270+350-500$$  $$=500-500+350$$  $$=350$$ "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2216", "queId": "321bd94459564da786e8ee5f60ddd8d4", "competition_source_list": ["2004年第2届创新杯五年级竞赛初赛第5题", "2004年五年级竞赛创新杯"], "difficulty": "1", "qtype": "single_choice", "problem": "甲乙两地相距$$120$$千米，一辆汽车从甲地开往乙地，每小时行$$40$$千米，返回时每小时行$$60$$千米，这辆汽车往返两地的平均速度是（  ）。 ", "answer_option_list": [[{"aoVal": "A", "content": "每小时$$100$$千米 "}], [{"aoVal": "B", "content": "每小时$$50$$千米 "}], [{"aoVal": "C", "content": "每小时$$48$$千米 "}], [{"aoVal": "D", "content": "每小时$$24$$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": ["求平均速度的公式为：平均速度$$=$$总路程$$\\div $$总时间  这辆汽车去时用了$$120\\div 40=3$$(小时)，返回时用了$$120\\div 60=2$$(小时)，那么这辆汽车的平均速度为：$$120\\times 2\\div \\left( 3+2 \\right)=48$$(千米/时)，选C 。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2200", "queId": "509a3dc4a7f24736b16160aaab63ba63", "competition_source_list": ["2018年华杯赛六年级竞赛模拟卷"], "difficulty": "2", "qtype": "single_choice", "problem": "甲乙两人从相距$$600$$米的地方同向而行，甲走了$$1000$$米追上乙，如果甲每秒多走一米，那么走了$$900$$米就可以追上乙。如果一开始甲每秒少走一米，甲乙两人相向而行~\\uline{~~~~~~~~}~秒相遇。 ", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$120$$ "}], [{"aoVal": "D", "content": "$$150$$ "}]], "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": ["甲速度不变时，他走$$1000$$米所用时间和追上乙的时间相同；速度提升时，甲走$$900$$的时间和追上乙所用时间相同。我们可以列出关系：$$\\begin{cases}\\frac{600}{{{V}_{\\text{甲}}}-{{V}_{\\text{乙}}}}\\text{=}\\frac{1000}{{{V}_{\\text{甲}}}}    \\frac{600}{\\left( {{V}_{\\text{甲}}}\\text{+1} \\right)-{{V}_{\\text{乙}}}}\\text{=}\\frac{900}{{{V}_{\\text{甲}}}\\text{+}1}   \\end{cases}$$，解得$$\\begin{cases}{{V}_{\\text{甲}}}\\text{=}5    {{V}_{\\text{乙}}}\\text{=}2   \\end{cases}$$。甲的速度减慢，两人相遇的时间为$$600\\div \\left( 4+2 \\right)=100$$秒。 "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1823", "queId": "8e0641bf28e447e88836935bf2be2c52", "competition_source_list": ["2017年第8届广东广州羊排赛六年级竞赛第5题1分"], "difficulty": "1", "qtype": "single_choice", "problem": "苗苗有一杯$$20 \\%$$的盐水$$100$$克．向杯中再加入$$100$$克纯净水，此时盐水的浓度． ", "answer_option_list": [[{"aoVal": "A", "content": "等于$$10 \\%$$ "}], [{"aoVal": "B", "content": "大于$$10 \\%$$ "}], [{"aoVal": "C", "content": "小于$$10 \\%$$ "}], [{"aoVal": "D", "content": "不能确定 "}]], "knowledge_point_routes": ["拓展思维->能力->数据处理"], "answer_analysis": ["盐水浓度为$$\\frac{100\\times 20 \\%}{100+100}\\times 100 \\%=10 \\%$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "119", "queId": "30040540731b4f21b219e583cfa3e194", "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": "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"], "answer_analysis": ["小东与小南说的话对立，所以两人中一定一对一错．已知只有一个人说对了，则小西和小北说的是错的．\"小北说：小南说的也不对．\"这句话是错的，可知小南说的是对的．  故选：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1985", "queId": "b85e44e132b04854bfd327c9a98ec69a", "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->拓展思维->应用题模块->比例应用题->复杂的比例问题", "课内体系->思想->逐步调整思想"], "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．为了让得到的钱数相同，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": "192", "queId": "5e622a1b65734244b28b7a4a6c4e2372", "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->思想->逐步调整思想"], "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": "2095", "queId": "e700390fefab4fcc99d1507a06633d4e", "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->思想->对应思想"], "answer_analysis": ["线段图 详见课程回放 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1059", "queId": "20f383278e6b4fd1bf390446a82c9660", "competition_source_list": ["2008年陈省身杯小学高年级六年级竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "一根绳子分成甲乙两段，甲占全长的$$\\frac{3}{5}$$，乙长$$\\frac{3}{5}$$米，两段绳子相比较（~~~~~~~~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "甲长 "}], [{"aoVal": "B", "content": "乙长 "}], [{"aoVal": "C", "content": "一样长 "}], [{"aoVal": "D", "content": "无法确定 "}]], "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": ["根据题意得，一根绳子分成甲乙两段，甲占全长的$$\\frac{3}{5}$$，乙长$$\\frac{3}{5}$$米，两段绳子相比较甲长，故答案为$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "1529", "queId": "4d9a4b8184e9425581d4921a9eafcc8e", "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2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个数，减去$$15$$之后除以$$13$$，其结果加上$$3$$之后乘$$2$$，得到的结果是$$314$$，那么原来的这个数是． ", "answer_option_list": [[{"aoVal": "A", "content": "$$2015$$ "}], [{"aoVal": "B", "content": "$$2016$$ "}], [{"aoVal": "C", "content": "$$2017$$ "}], [{"aoVal": "D", "content": "$$2018$$ "}]], "knowledge_point_routes": ["拓展思维->思想->逆向思想"], "answer_analysis": ["倒推还原$$\\left( 314\\div 2-3 \\right)\\times 13+15=2017$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "214", "queId": "363d880fe1754e25add48047147b6a91", "competition_source_list": ["2016年华杯赛四年级竞赛初赛"], "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（如果初练也可只用$$1\\sim10$$这$$40$$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，如抽出的牌是$$3$$、$$8$$、$$8$$、$$9$$，那么算式为$$\\left( 9-8 \\right)\\times 8\\times 3$$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": "$$1$$、$$2$$、$$3$$、$$3$$ "}], [{"aoVal": "B", "content": "$$1$$、$$5$$、$$5$$、$$5$$ "}], [{"aoVal": "C", "content": "$$2$$、$$2$$、$$2$$、$$2$$ "}], [{"aoVal": "D", "content": "$$3$$、$$3$$、$$3$$、$$3$$ "}]], "knowledge_point_routes": ["拓展思维->拓展思维->组合模块->数字谜->巧填算符"], "answer_analysis": ["可以通过计算它们能达到的最大值，找到不能凑出$$24$$点的选项。  解：A、$$(1+2)\\times 3\\times 3=27\\textgreater24$$，有可能凑出$$24$$点；  B、$$(1+5)\\times 5\\times 5=150\\textgreater24,$$有可能凑出$$24$$点；  C、$$2\\times 2\\times 2\\times 2=16 \\textless{} 24$$，不可能凑出$$24$$点；  D、$$3\\times 3\\times 3\\times 3=81\\textgreater24$$，有可能凑出$$24$$点；  故选：C。 "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "505", "queId": "f3c12a4c48254c60bc2cb5b85f6ef5b9", "competition_source_list": ["2012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": "奥运会足球小组赛分为上、下两个半长，每半长$$45$$分钟，中间休息$$15$$分钟．伦敦奥运会期间，某场足球小组赛于$$13:00$$开始，并且此赛事没有伤停与延长赛时间，请问这场足球小组赛在什么时刻结束？（~ ~ ） ", "answer_option_list": [[{"aoVal": "A", "content": "$$13:45$$ "}], [{"aoVal": "B", "content": "$$14:00$$ "}], [{"aoVal": "C", "content": "$$14:30$$ "}], [{"aoVal": "D", "content": "$$14:35$$ "}], [{"aoVal": "E", "content": "$$14:45$$ "}]], "knowledge_point_routes": ["拓展思维->思想->对应思想"], "answer_analysis": ["此场足球赛时长$$45+45+15=105$$分钟，所以结束时间是$$14:45$$． "], "answer_value": "E"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "2122", "queId": "ec131654762a4210ac1ec71949f4dfdd", "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2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲要完成一批零件，原计划$$10$$天完成．实际上甲每天比原计划多做$$16$$个，结果$$8$$天完成．这批零件共个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$200$$ "}], [{"aoVal": "B", "content": "$$100$$ "}], [{"aoVal": "C", "content": "$$640$$ "}]], "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->具体量工程问题"], "answer_analysis": ["实际每天比原来多做$$16$$个，  则$$8$$天比原来多做：$$8\\times 16=128$$（个），  多做的部分相当于原来$$10-8=2$$天的量．  所以原来每天做：$$128\\div 2=64$$（个），  则这批零件共$$64\\times 10=640$$（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": "3083", "queId": "fd241d105f5343c282a2b701ebb8d110", "competition_source_list": ["2014年全国创新杯五年级竞赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "把足够大的一张厚度为$$0.1mm$$纸连续对折，要使对折后的整叠纸总厚度超过$$12mm$$，至少要对折扣（~ ）． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$次 "}], [{"aoVal": "B", "content": "$$7$$次 "}], [{"aoVal": "C", "content": "$$8$$次 "}], [{"aoVal": "D", "content": "$$9$$次 "}]], "knowledge_point_routes": ["拓展思维->能力->运算求解"], "answer_analysis": ["折$$n$$次变为原来的$${{2}^{n}}$$倍，$${{2}^{7}}=128$$，$$128\\times 0.1=12.8mm$$，至少要折$$7$$次． "], "answer_value": "B"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "468", "queId": "d2fe9e7bc09e424a9af68cda9d1b8b32", "competition_source_list": ["2005年第3届创新杯五年级竞赛复赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a_1$$，$$a_2$$，$$a_3$$，$$a_4$$，$$a_5$$是任意五个奇数，且$$a_1\\textless{}a_2\\textless{}a_3\\textless{}a_4\\textless{}a_5$$，$$a_1+a_2+a_3+a_4+a_5=85$$，符合这些条件的五个奇数显然有很多，如$$a_1=1$$，$$a_2=3$$，$$a_3=7$$，$$a_4=11$$，$$a_5=63$$，或$$a_1=5$$，$$a_2=7$$，$$a_3=13$$，$$a_4=21$$，$$a_5=39$$，等等，在这些答案中，记$$a_5$$的最大值和最小值分别为$$M$$和$$m$$，则． ", "answer_option_list": [[{"aoVal": "A", "content": "$$M=67$$，$$m=23$$ "}], [{"aoVal": "B", "content": "$$M=67$$，$$m=19$$ "}], [{"aoVal": "C", "content": "$$M=69$$，$$m=21$$ "}], [{"aoVal": "D", "content": "$$M=69$$，$$m=17$$ "}]], "knowledge_point_routes": ["拓展思维->思想->分类讨论思想"], "answer_analysis": ["当$$a_1=1$$，$$a_2=3$$，$$a_3=5$$，$$a_4=7$$时，$$a_5$$最大，  所以$$M=85-(1+3+5+7)=69$$，  因为$$85\\div5=17$$，  所以$$a_5$$最小时$$a_1=13$$，$$a_2=15$$，$$a_3=17$$，$$a_4=19$$，$$a_5=21$$，  所以$$m=21$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "prime_math_competition_ch_single_choice_3.2K_dev", "dataset_version": "2023-07-07", "qid": "3254", "queId": "59f6adf3c5e747a29b8a8c51f8d75bc6", "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共有（~ ~ ~ ）个． ", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$30$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$42$$ "}]], "knowledge_point_routes": ["拓展思维->思想->枚举思想"], "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": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "539", "queId": "b0842f505b9240b48452693cde5c2a10", "competition_source_list": ["2018年全国全国初中数学联赛初一竞赛A卷"], "difficulty": "2", "qtype": "single_choice", "problem": "设二次函数$$y={{x}^{2}}+2ax+\\frac{{{a}^{2}}}{2}$$的图象的顶点为$$A$$，与$$x$$轴的交点为$$B$$，$$C$$．当$$\\triangle ABC$$为等边三角形时，其边长为 (~ ~ ) .", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{6}$$ "}], [{"aoVal": "B", "content": "$$2\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$2\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$3\\sqrt{2}$$ "}]], "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->二次函数与等边三角形结合"], "answer_analysis": ["由题设知$$A\\left( -a,-\\frac{{{a}^{2}}}{2} \\right)$$．设$$B({{x}_{1}},0)$$，$$C({{x}_{2}},0)$$，  二次函数的图象的对称轴与$$x$$轴的交点为$$D$$，  则$$BC=\\left\\textbar{} {{x}_{1}}-{{x}_{ }}_{2} \\right\\textbar=\\sqrt{{{({{x}_{1}}+{{x}_{2}})}^{2}}-4{{x}_{1}}{{x}_{2}}}=\\sqrt{4{{a}^{2}}-4\\times \\frac{{{a}^{2}}}{2}}=\\sqrt{2{{a}^{2}}}$$.  又$$AD=\\frac{\\sqrt{3}}{2}BC$$，则$$\\left\\textbar{} -\\frac{{{a}^{2}}}{2} \\right\\textbar=\\frac{\\sqrt{3}}{2}\\cdot \\sqrt{2{{a}^{2}}}$$，解得$${{a}^{2}}=6$$或$${{a}^{2}}=0$$ (舍去)  所以，$$\\triangle ABC$$的边长$$BC=\\sqrt{2{{a}^{2}}}=2\\sqrt{3}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1433", "queId": "dc4e601351f74197bc8074c6c6be35bc", "competition_source_list": ["2000年第11届希望杯初一竞赛第4题"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$a=2$$，$$b=3$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$a{{x}^{3}}{{y}^{2}}$$和$$b{{m}^{3}}{{n}^{2}}$$是同类项 "}], [{"aoVal": "B", "content": "$$3{{x}^{a}}{{y}^{3}}$$和$$b{{x}^{3}}{{y}^{3}}$$是同类项 "}], [{"aoVal": "C", "content": "$$b{{x}^{2a+1}}{{y}^{4}}$$和$$a{{x}^{5}}{{y}^{b+1}}$$是同类项 "}], [{"aoVal": "D", "content": "$$5{{m}^{2b}}{{n}^{5a}}$$和$$6{{n}^{2b}}{{m}^{5a}}$$是同类项 "}]], "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->同类项的定义"], "answer_analysis": ["由同类项的定义可知，当$$a=2$$，$$b=3$$时，  $$\\text{A}$$为：$$2{{x}^{3}}{{y}^{2}}$$和$$3{{m}^{2}}{{n}^{2}}$$，显然不是同类项．  $$\\text{B}$$为$$3{{x}^{2}}{{y}^{3}}$$和$$3{{x}^{3}}{{y}^{3}}$$，$${{x}^{2}}$$与$${{x}^{3}}$$不同，所以也不是同类项．  $$\\text{C}$$为$$3{{x}^{2\\times 2+1}}{{y}^{4}}$$和$$3{{x}^{5}}{{y}^{3+1}}$$，即$$3{{x}^{5}}{{y}^{4}}$$和$$3{{x}^{5}}{{y}^{4}}$$，是同类项．  $$\\text{D}$$为$$5{{m}^{2}}\\times 3{{n}^{5\\times 2}}=5{{m}^{6}}{{n}^{10}}$$和$$6{{n}^{2\\times 3}}{{m}^{5\\times 2}}=6{{n}^{6}}{{m}^{10}}$$，显然不是．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1238", "queId": "8aac4907507fb884015089ba06891ad1", "competition_source_list": ["1991年第8届全国初中数学联赛竞赛（第一试）第6题", "初一下学期单元测试《二元一次方程组》三元一次方程组第8题", "2022年浙江温州瓯海区浙江省温州中学初三自主招生第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a$$、$$c$$、$$d$$是整数，$$b$$是正整数，且满足$$a+b=c$$，$$b+c=d$$，$$c+d=a$$，那么$$a+b+c+d$$的最大值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$-5$$ "}], [{"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->知识点->式->整式的加减->整式的加减运算->整式加减的综合"], "answer_analysis": ["∵$$a+b=c$$，∴$$a=c-b$$．  又∵$$b+c=d$$，$$c+d=a$$，  ∴$$c=-2b$$，$$a=-3b$$，$$d=-b$$，  ∴$$a+b+c+d=-5b$$．  ∵$$b$$是正整数，其最小值为$$1$$，  ∴$$a+b+c+d=-5b$$的最大值是$$-5$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "209", "queId": "33f11597b54e43b687374d27bf32deb7", "competition_source_list": ["2009年第20届希望杯初一竞赛第1试第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$${{x}^{2}}+x-2=0$$，则$${{x}^{3}}+2{{x}^{2}}-x+2019$$=．", "answer_option_list": [[{"aoVal": "A", "content": "$$2020$$ "}], [{"aoVal": "B", "content": "$$2021$$ "}], [{"aoVal": "C", "content": "$$-2020$$ "}], [{"aoVal": "D", "content": "$$-2021$$ "}]], "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->整式加减化简求值->整式加减化简求值-降次化简求值"], "answer_analysis": ["$${{x}^{2}}+x-2=0$$，则$${{x}^{2}}+x=2$$，  ∴$${{x}^{3}}+2{{x}^{2}}-x+2019$$  $$={{x}^{3}}+{{x}^{2}}+{{x}^{2}}+x-2x+2019$$  $$=x\\left( {{x}^{2}}+x \\right)+\\left( {{x}^{2}}+x \\right)-2x+2019$$  $$=2x+2-2x+2019$$  $$=2021$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "277", "queId": "795194e55f8d4179a21ae2055bb78874", "competition_source_list": ["2009年第20届希望杯初一竞赛第1试第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$${{x}^{2}}+x-2=0$$，则$${{x}^{3}}+2{{x}^{2}}-x+2007$$=．", "answer_option_list": [[{"aoVal": "A", "content": "$$2009$$ "}], [{"aoVal": "B", "content": "$$2008$$ "}], [{"aoVal": "C", "content": "$$-2008$$ "}], [{"aoVal": "D", "content": "$$-2009$$ "}]], "knowledge_point_routes": ["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-降次化简求值", "课内体系->能力->运算能力"], "answer_analysis": ["$${{x}^{2}}+x-2=0$$，则$${{x}^{2}}+x=2$$，  ∴$${{x}^{3}}+2{{x}^{2}}-x+2007$$  $$={{x}^{3}}+{{x}^{2}}+{{x}^{2}}+x-2x+2007$$  $$=x\\left( {{x}^{2}}+x \\right)+\\left( {{x}^{2}}+x \\right)-2x+2007$$  $$=2x+2-2x+2007$$  $$=2009$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "858", "queId": "57dcc4f7b1e74aa49d12b690ea4ddaab", "competition_source_list": ["2010年第15届华杯赛初一竞赛决赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "互不相等的有理数$$a$$，$$b$$，$$c$$在数轴上的对应点分别为$$A$$，$$B$$，$$C$$．如果$$\\left\\textbar{} a-b \\right\\textbar+\\left\\textbar{} b-c \\right\\textbar=\\left\\textbar{} c-a \\right\\textbar$$，那么在点$$A$$，$$B$$，$$C$$中，居中的是点~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$A$$ "}], [{"aoVal": "B", "content": "$$B$$ "}], [{"aoVal": "C", "content": "$$C$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->绝对值->绝对值的几何意义", "课内体系->知识点->数->有理数->数轴"], "answer_analysis": ["当$$a\\textless{}b\\textless{}c$$时，$$\\left\\textbar{} a-b \\right\\textbar+\\left\\textbar{} b-c \\right\\textbar=b-a+c-b=c-a=\\left\\textbar{} c-a \\right\\textbar$$；  当$$a\\textgreater b\\textgreater c$$时，$$\\left\\textbar{} a-b \\right\\textbar+\\left\\textbar{} b-c \\right\\textbar=a-b+b-c=a-c=\\left\\textbar{} c-a \\right\\textbar$$；  所以点$$B$$在点$$A$$与点$$C$$之间．  当点$$B$$不在$$A$$，$$C$$两个点之间时，$$\\left\\textbar{} a-b \\right\\textbar+\\left\\textbar{} b-c \\right\\textbar=\\left\\textbar{} c-a \\right\\textbar$$不成立．  事实上，当$$b\\textless{}a\\textless{}c$$时，$$\\left\\textbar{} a-b \\right\\textbar+\\left\\textbar{} b-c \\right\\textbar=a-b+c-b=a+c-2b$$，$$\\left\\textbar{} c-a \\right\\textbar=c-a$$．  这时不可能有$$\\left\\textbar{} a-b \\right\\textbar+\\left\\textbar{} b-c \\right\\textbar=\\left\\textbar{} c-a \\right\\textbar$$，  否则，$$a+c-2b=c-a$$，即$$2a=2b$$，  得出$$a$$和$$b$$相等，与题设条件矛盾．类似地可以讨论其他情形． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1076", "queId": "bfdb2b2df4164804837e1d1e09f10600", "competition_source_list": ["2005年第22届全国初中数学联赛竞赛初赛第5题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "在直角坐标系中，横纵坐标都是整数的点称为整点，设$$k$$ 为整数，当直线$$y=x-3$$ 与$$y=kx+k$$ 的交点为整数时，$$k$$ 的值可以取．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ 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->函数->一次函数->一次函数与方程、不等式->一次函数与二元一次方程组", "课内体系->知识点->方程与不等式->二元一次方程（组）->含参二元一次方程组->二元一次方程组的含参整数解问题", "课内体系->能力->运算能力"], "answer_analysis": ["由题意得：$$\\begin{cases}y=kx+k    y=x-3   \\end{cases}$$ ，  解得：$$\\begin{cases}x=\\dfrac{k+3}{1-k}    y=\\dfrac{4k}{1-k}   \\end{cases}$$ ，  ∴$$\\begin{cases}x=-1+\\dfrac{4}{1-k}    y=-4+\\dfrac{4}{1-k}   \\end{cases}$$ ，  ∵交点为整数，  ∴$$k$$ 可取的整数解有$$0$$ ，$$2$$ ，$$3$$ ，$$5$$ ，$$-1$$ ，$$-3$$ 共$$6$$ 个．  故选$$\\text{C}$$ ． ", "<p>由题意可知，$$x-3=k\\left( x+1 \\right)$$，</p>\n<p>所以$$k=\\frac{x-3}{x+1}=1-\\frac{4}{x+1}$$，</p>\n<p>但$$k$$为整数，</p>\n<p>于是$$x+1=\\pm 1$$，$$\\pm 2$$，$$\\pm 4$$，$$k$$可取$$6$$个值，</p>\n<p>故选$$\\text{C}$$．</p>"], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "921", "queId": "e8c9a2c46f6f4c42bad1d5d35438fdef", "competition_source_list": ["其它", "2007年第18届希望杯初二竞赛第1试第4题"], "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": "329", "queId": "22ca2d3c500e4e41812b1c6fedadfdf0", "competition_source_list": ["2014年第25届全国希望杯初二竞赛复赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "方程组$$\\left { \\begin{matrix}\\left\\textbar{} x \\right\\textbar+y=10    x+\\left\\textbar{} y \\right\\textbar=4   \\end{matrix} \\right.$$的解有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$组 "}], [{"aoVal": "B", "content": "$$2$$组 "}], [{"aoVal": "C", "content": "$$3$$组 "}], [{"aoVal": "D", "content": "$$4$$组 "}]], "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": ["当$$x\\geqslant 0$$，$$y\\geqslant 0$$时，方程组是$$\\left { \\begin{matrix}x+y=10    x+y=4   \\end{matrix} \\right.$$，两个方程互相矛盾，无解；  当$$x\\textless{}0$$，$$y\\textless{}0$$时，方程组是$$\\left { \\begin{matrix}-x+y=10    x-y=4   \\end{matrix} \\right.$$，两个方程互相矛盾，无解；  当$$x\\geqslant 0$$，$$y\\textless{}0$$时，方程组是$$\\left { \\begin{matrix}x+y=10    x-y=4   \\end{matrix} \\right.$$，解得$$\\left { \\begin{matrix}x=7    y=3   \\end{matrix} \\right.$$，与条件矛盾，舍去；  当$$x\\textless{}0$$，$$y\\geqslant 0$$时，方程组是$$\\left { \\begin{matrix}-x+y=10    x+y=4   \\end{matrix} \\right.$$，解得$$\\left { \\begin{matrix}x=-3    y=7   \\end{matrix} \\right.$$，符合条件．  综上所述，原方程组只有一组解． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "319", "queId": "9456d720820e4c00b9c694f8ac89d685", "competition_source_list": ["2012年第23届全国希望杯初一竞赛初赛第8题4分", "初一单元测试《列方程解应用题—设元的技巧》第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "某研究所全体员工的月平均工资为$$5500$$元，男员工月平均工资为$$6500$$元，女员工月平均工资为$$5000$$元，则该研究所男、女员工人数之比是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2:3$$ "}], [{"aoVal": "B", "content": "$$3:2$$ "}], [{"aoVal": "C", "content": "$$1:2$$ "}], [{"aoVal": "D", "content": "$$2:1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->分式方程->分式方程与实际问题->分式方程的其他实际问题", "课内体系->能力->分析和解决问题能力"], "answer_analysis": ["设该研究所有$$x$$名男员工，$$y$$名女员工，  由题意，得$$\\frac{6500x+5000y}{x+y}=5500$$，  整理，得$$y=2x$$，  所以$$x:y=1:2$$．  即该研究所男、女员工人数之比是$$1:2$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "629", "queId": "f61552dcad47489fb23e77c11e141390", "competition_source_list": ["1999年第4届华杯赛初一竞赛复赛第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "有一批长度分别为$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$，$$8$$，$$9$$，$$10$$和$$11$$厘米的细木条，它们的数量都足够多，从中适当选取$$3$$根木条作为$$3$$条边，可围成一个三角形．如果规定底边是$$11$$厘米长，你能围成个不同的三角形?", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$48$$ "}]], "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->三角形的三边关系", "竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["规定底边长是$$11$$厘米，其余两边长分别是$$a$$及$$b$$时，不妨设$$a\\leqslant b$$．任意一个三角形，它的任意两条边的长度之和，总比余下的一条边要长，即必有$$11\\textless{}a+b$$．于是$$\\left( a,b \\right)$$的可能的值便有：$$\\left( 11,11 \\right)$$；$$\\left( 10,10 \\right)$$，$$\\left( 10,11 \\right)$$；$$\\left( 9,9 \\right)$$，$$\\left( 9,10 \\right)$$，$$\\left( 9,11 \\right)$$；$$\\left( 8,8 \\right)$$，$$\\left( 8,9 \\right)$$，$$\\left( 8,10 \\right)$$，$$\\left( 8,11 \\right)$$；$$\\left( 7,7 \\right)$$，$$\\left( 7,8 \\right)$$，$$\\left( 7,9 \\right)$$，$$\\left( 7,10 \\right)$$，$$\\left( 7,11 \\right)$$；$$\\left( 6,6 \\right)$$，$$\\left( 6,7 \\right)$$，$$\\left( 6,8 \\right)$$，$$\\left( 6,9 \\right)$$，$$\\left( 6,10 \\right)$$，$$\\left( 6,11 \\right)$$；$$\\left( 5,7 \\right)$$，$$\\left( 5,8 \\right)$$，$$\\left( 5,9 \\right)$$，$$\\left( 5,10 \\right)$$，$$\\left( 5,11 \\right)$$；$$\\left( 4,8 \\right)$$，$$\\left( 4,9 \\right)$$，$$\\left( 4,10 \\right)$$，$$\\left( 4,11 \\right)$$；$$\\left( 3,9 \\right)$$，$$\\left( 3,10 \\right)$$，$$\\left( 3,11 \\right)$$；$$\\left( 2,10 \\right)$$，$$\\left( 2,11 \\right)$$；$$\\left( 1,11 \\right)$$．  可见，总数等于$$2\\times \\left( 1+2+3+4+5 \\right)+6=36$$ （个）．  答：能围成$$36$$个不同的三角形． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1326", "queId": "8aac50a751148307015114d8c704039f", "competition_source_list": ["2015年第26届全国希望杯初二竞赛复赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "若代数式$${{x}^{2}}-6x+b$$可化为$${{(x-a)}^{2}}-1$$，则$$b-a$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "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", "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": ["由题意$$\\left { \\begin{matrix}2a=6    b={{a}^{2}}-1   \\end{matrix} \\right.$$，解得$$\\left { \\begin{matrix}a=3    b=8   \\end{matrix} \\right.$$，于是$$b-a=5$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1589", "queId": "f9c51645f7a646c6b7318f384274fb37", "competition_source_list": ["2012年四川成都青羊区成都树德实验中学初三自主招生第8题5分", "2017~2018学年福建泉州鲤城区泉州市第六中学初二下学期期中第10题4分", "1998年竞赛（全国初中数学竞赛）第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$abc\\ne 0$$，并且$$\\frac{a+b}{c}=\\frac{b+c}{a}=\\frac{c+a}{b}=p$$，  那么直线$$y=px+p$$一定通过第象限．", "answer_option_list": [[{"aoVal": "A", "content": "一、二 "}], [{"aoVal": "B", "content": "二、三 "}], [{"aoVal": "C", "content": "三、四 "}], [{"aoVal": "D", "content": "一、四 "}]], "knowledge_point_routes": ["知识标签->知识点->函数->一次函数->一次函数的图象与性质->一次函数的图象与系数的关系", "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->学习能力->运算能力", "知识标签->题型->函数->一次函数->一次函数基础->题型：一次函数图象与性质", "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->数学思想->数形结合思想"], "answer_analysis": ["由条件得：①$$a+b=pc$$，②$$b+c=pa$$，③$$a+c=pb$$，  三式相加得: $$2\\left( a+b+c \\right)=p\\left( a+b+c \\right)$$．  ∴$$p=2$$或$$a+b+c=0$$．  当$$p=2$$时，$$y=2x+2$$．则直线通过第一、二、三象限．  当$$a+b+c=0$$时，不妨取$$a+b=-c$$，于是$$p=\\frac{a+b}{c}=-1$$，  ∴$$y=-x-1$$，则直线通过第二、三、四象限．  综合上述两种情况，直线一定通过第二、三象限．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1083", "queId": "ff8080814d9539f1014d9891b326023d", "competition_source_list": ["2008年第19届希望杯初一竞赛第2试第10题4分", "初一其它"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$1$$，$$2$$，$$3$$，$$4$$，\\ldots，$$12$$，$$13$$这$$13$$个整数分为两组，使得一组中所有数的和比另一组中所有数的和大$$10$$，这样的分组方法（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "只有一种 "}], [{"aoVal": "B", "content": "恰有两种 "}], [{"aoVal": "C", "content": "多于三种 "}], [{"aoVal": "D", "content": "不存在 "}]], "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->二元一次方程（组）->二元一次方程组应用题", "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": ["设一组数的和为$$x$$，另一组数的和为$$y$$，由$$\\left { \\begin{matrix}x+y=91    x-y=10   \\end{matrix} \\right.$$，得$$2y=81$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "390", "queId": "6295eaf93a864be2a85c336d0b4c6107", "competition_source_list": ["2001年第12届希望杯初一竞赛第2试第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "满足$${{(a-b)}^{2}}+(b-a)\\left\\textbar{} a-b \\right\\textbar=ab\\left( ab\\ne 0 \\right)$$的有理数$$a$$、$$b$$，一定不满足的关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$ab \\textless{} 0$$ "}], [{"aoVal": "B", "content": "$$ab\\textgreater0$$ "}], [{"aoVal": "C", "content": "$$a+b\\textgreater0$$ "}], [{"aoVal": "D", "content": "$$a+b \\textless{} 0$$ "}]], "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": ["若$$a\\geqslant b$$，$${{(a-b)}^{2}}+(b-a)\\left\\textbar{} a-b \\right\\textbar={{(a-b)}^{2}}-{{(a-b)}^{2}}=0\\ne ab$$，  若$$a \\textless{} b$$，$${{(a-b)}^{2}}+(b-a)\\left\\textbar{} a-b \\right\\textbar={{(a-b)}^{2}}+{{(b-a)}^{2}}=2{{(a-b)}^{2}}=ab$$，  从平方的非负性我们知道$$ab\\geqslant 0$$，且$$ab\\ne 0$$，所以$$ab\\textgreater0$$，则选项$$\\text A$$一定不满足． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1128", "queId": "e914cd86d5a74200ac5b9d5b6c7f3cd5", "competition_source_list": ["2014年第25届全国希望杯初二竞赛初赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有以下$$5$$个命题：  ①两组邻角互补的四边形是平行四边形  ②有一条对角线平分一个内角的平行四边形是菱形  ③一边上的两个角相等的梯形是等腰梯形  ④对角线相等的四边形是矩形  ⑤菱形的面积等于两条对角线的乘积  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->四边形->特殊平行四边形->矩形->矩形的判定及应用->矩形的判定-从平行四边形", "课内体系->知识点->四边形->特殊平行四边形->菱形->菱形的周长与面积"], "answer_analysis": ["两组邻角互补的四边形可能是梯形，命题①不正确；  有一条对角线平分一个内角的平行四边形是菱形，命题②正确；  一边上的两个角相等的梯形可能是直角梯形，命题③不正确；  对角线相等的四边形不一定是矩形，对角线相等的平行四边形才是矩形，命题④不正确；  菱形的面积等于两条对角线的乘积的一半，命题⑤不正确．  故正确的命题的个数是$$1$$个． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "360", "queId": "5939daa9431b40be9bb625925e34e7cf", "competition_source_list": ["2012年第23届全国希望杯初二竞赛复赛第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "实数$$a$$，$$b$$，$$c$$，$$d$$满足：①$$a+b=c+d$$；②$$a+d\\textless{}b+c$$；③$$c\\textless{}d$$，则$$a$$，$$b$$，$$c$$，$$d$$的大小关系是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless{}c\\textless{}d\\textless{}b$$ "}], [{"aoVal": "B", "content": "$$b\\textless{}c\\textless{}d\\textless{}a$$ "}], [{"aoVal": "C", "content": "$$c\\textless{}d\\textless{}a\\textless{}b$$ "}], [{"aoVal": "D", "content": "$$c\\textless{}d\\textless{}b\\textless{}a$$ "}]], "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": ["由①，知$$a-d=c-b$$，④  ②$$+$$④，得$$2a\\textless{}2c$$，即$$a\\textless{}c$$，⑤  因为$$c\\textless{}d$$，  所以$$a\\textless{}c\\textless{}d$$，  又由①，知$$b-d=c-a\\textgreater0$$，  所以$$b\\textgreater d$$，  所以$$a\\textless{}c\\textless{}d\\textless{}b$$． ", "<p>①$$+$$②得：$$2a+b+d &nbsp;&lt; &nbsp;2c+b+d$$，故$$a &nbsp;&lt; &nbsp;c$$，</p>\n<p>①$$-$$②得：$$b-d&gt;d-b$$，故$$b&gt;d$$，</p>\n<p>又∵$$c &nbsp;&lt; &nbsp;d$$，&there4;$$a &nbsp;&lt; &nbsp;c &nbsp;&lt; &nbsp;d &nbsp;&lt; &nbsp;b$$．</p>", "<p>设$$a=1$$，$$b=6$$，$$c=3$$，$$d=4$$，满足条件，故$$a &nbsp;&lt; &nbsp;c &nbsp;&lt; &nbsp;d &nbsp;&lt; &nbsp;b$$．</p>\n<p>故选$$\\text{A}$$．</p>"], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1561", "queId": "fda88fd4ad104ba69b7626fbb084f389", "competition_source_list": ["1993年第10届全国初中数学联赛竞赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$x$$是实数，$$y=\\left\\textbar{} x-1 \\right\\textbar+\\left\\textbar{} x+1 \\right\\textbar$$．下列四个结论：  Ⅰ．$$y$$没有最小值~  Ⅱ．只有一个$$x$$使$$y$$取到最小值~  Ⅲ．有限多个$$x$$(不止一个)使$$y$$取到最大值~  Ⅳ．有无穷多个$$x$$使$$y$$取到最小值~  其中正确的是（~ ）.", "answer_option_list": [[{"aoVal": "A", "content": "I "}], [{"aoVal": "B", "content": "Ⅱ "}], [{"aoVal": "C", "content": "Ⅲ "}], [{"aoVal": "D", "content": "Ⅳ "}]], "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->利用绝对值求最值", "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", "课内体系->思想->分类讨论思想"], "answer_analysis": ["因为$$\\left\\textbar{} x-1 \\right\\textbar$$，$$\\left\\textbar{} x+1 \\right\\textbar$$分别表示数轴上点$$x$$到点$$1$$和点$$-1$$的距离，  因此，当$$-1\\leqslant x\\leqslant 1$$时，$$y=\\left\\textbar{} x-1 \\right\\textbar+\\left\\textbar{} x+1 \\right\\textbar=2$$，  当$$x ~\\textless{} ~-1$$时，$$y=\\left\\textbar{} x-1 \\right\\textbar+\\left\\textbar{} x+1 \\right\\textbar=2+2\\left\\textbar{} x+1 \\right\\textbar\\textgreater2$$，  当$$x\\textgreater1$$时，$$y=\\left\\textbar{} x-1 \\right\\textbar+\\left\\textbar{} x+1 \\right\\textbar=2+2\\left\\textbar{} x+1 \\right\\textbar\\textgreater2$$，  而在$$-1$$与$$1$$之间有无穷个实数$$x$$，故有无穷多个$$x$$使$$y$$取到最小值. "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "786", "queId": "b13bfbcd3f624b8d87cb7f75df57ee1d", "competition_source_list": ["2013年第24届全国希望杯初一竞赛复赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\triangle ABC$$的内角分别为$$\\angle A$$，$$\\angle B$$，$$\\angle C$$，若$$\\angle 1=\\angle A+\\angle B$$，$$\\angle 2=\\angle B+\\angle C$$，$$\\angle 3=\\angle A+\\angle C$$，则$$\\angle 1$$，$$\\angle 2$$，$$\\angle 3$$中（~ ~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "至少有一个锐角 "}], [{"aoVal": "B", "content": "三个都是钝角 "}], [{"aoVal": "C", "content": "至少有两个钝角 "}], [{"aoVal": "D", "content": "可以有两个直角 "}]], "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->知识点->三角形->三角形及多边形->与三角形有关的角->三角形的外角定义及性质"], "answer_analysis": ["因为三角形的内角中至少有$$2$$个锐角，  而$$\\angle 1$$，$$\\angle 2$$，$$\\angle 3$$是三角形的三个外角，  所以$$\\angle 1$$，$$\\angle 2$$，$$\\angle 3$$中至少有两个钝角． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "648", "queId": "373ff9afcf304d1caf47c82c569aefd0", "competition_source_list": ["1994年第11届全国初中数学联赛竞赛第10题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "当$$\\left\\textbar{} x+1 \\right\\textbar\\leqslant 6$$时，函数$$y=x\\left\\textbar{} x \\right\\textbar-2x+1$$的最大值是 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$24$$ "}], [{"aoVal": "D", "content": "$$34$$ "}]], "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->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->能力->运算能力"], "answer_analysis": ["由$$\\left\\textbar{} x+1 \\right\\textbar\\leqslant 6$$解得$$-7\\leqslant x\\leqslant 5$$．  当$$0\\leqslant x\\leqslant 5$$时，$$y={{x}^{2}}-2x+1={{(x-1)}^{2}}$$，此时$${{y}_{最大}}={{(5-1)}^{2}}=16$$．  当$$-7\\leqslant x\\leqslant 0$$时，$$y=-{{x}^{2}}-2x+1=2-{{(x+1)}^{2}}$$，此时$${{y}_{最大}}=2$$．  因此，当$$-7\\leqslant x\\leqslant 5$$时，$$y$$的最大值是$$16$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "653", "queId": "7a8c6842feba4eedba64e52075db3b05", "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": ["知识标签->题型->式->因式分解->因式分解综合应用", "知识标签->知识点->式->因式分解->因式分解：其他方法"], "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": "900", "queId": "a88672040a19443f8d31a06316ec7a91", "competition_source_list": ["1995年第6届希望杯初二竞赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "某位同学在代数变形中，得到下列四个式子．  ①$$^{3}\\sqrt{-{{(1-x)}^{3}}}=1-x$$．  ②当$$x$$$$=\\pm 2$$时，分式$$\\frac{\\left\\textbar{} x\\left. {} \\right\\textbar-2 \\right.}{{{x}^{2}}-x-6}$$的值均为$$0$$．  ③分解因式:$$$$$${{x}^{n+1}}-3{{x}^{n}}+2{{x}^{n-1}}={{x}^{n}}.x-3{{x}^{n}}+{{x}^{n}}\\frac{2}{x}={{x}^{n}}(x-3+\\frac{2}{x})$$．  ④$${{9997}^{2}}=({{9997}^{2}}-{{3}^{2}})+9=(9997+3)(9997-3)+9=99940009$$．  其中正确的个数是 ．", "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->因式分解->因式分解：公式法", "竞赛->知识点->数与式->因式分解->因式分解：提公因式法", "竞赛->知识点->数与式->绝对值->给定范围绝对值化简"], "answer_analysis": ["显然①式不成立；在②当中，当$$x=-2$$时，分母为$$0$$，故②式不成立；当$$x=0$$时，③式不成立；只有④式成立．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1043", "queId": "ff8080814d7978b9014d866b715d243c", "competition_source_list": ["1990年第1届全国希望杯初一竞赛初赛第10题", "2019年浙江杭州拱墅区杭州市文澜中学中考模拟（一）第10题3分"], "difficulty": "4", "qtype": "single_choice", "problem": "轮船往返于一条河的两码头之间，如果船本身在静水中的速度是固定的，那么，当这条河的水流速度增大时，船往返一次所用的时间将（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "增多 "}], [{"aoVal": "B", "content": "减少 "}], [{"aoVal": "C", "content": "不变 "}], [{"aoVal": "D", "content": "增多、减少都有可能 "}]], "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->分式加减混合运算"], "answer_analysis": ["设两码头之间距离为$$s$$，船在静水中速度为$$a$$，水速为$${{v}_{0}}$$，则往返一次所用时间为$${{t}_{0}}=\\frac{s}{a+{{v}_{0}}}+\\frac{s}{a-{{v}_{0}}}$$，  设河水速度增大后为$$v(v\\textgreater{{v}_{0}})$$，则往返一次所用时间为$$t=\\frac{s}{a+v}+\\frac{s}{a-v}$$．  计算：$${{t}_{0}}-t=\\frac{s}{a+{{v}_{0}}}+\\frac{s}{a-{{v}_{0}}}-\\frac{s}{a+v}-\\frac{s}{a-v}$$  $$=s[(\\frac{1}{a+{{v}_{0}}}-\\frac{1}{a+v})+(\\frac{1}{a-{{v}_{0}}}-\\frac{1}{a-v})]$$  $$=s[\\frac{v-{{v}_{0}}}{(a+{{v}_{0}})(a+v)}+\\frac{{{v}_{0}}-v}{(a-{{v}_{0}})(a-v)}]$$  $$=s(v-{{v}_{0}})[\\frac{1}{(a+{{v}_{0}})(a+v)}-\\frac{1}{(a-{{v}_{0}})(a-v)}]$$．  由于$$v-{{v}_{0}}\\textgreater0$$，$$a+{{v}_{0}}\\textgreater a-{{v}_{0}}$$，$$a+v\\textgreater a-v$$  所以$$(a+{{v}_{0}})(a+v)\\textgreater(a-{{v}_{0}})(a-v)$$，  因此$$\\frac{1}{(a+{{v}_{0}})(a+v)}\\textless\\frac{1}{(a-{{v}_{0}})(a-v)}$$，  即$$\\frac{1}{(a+{{v}_{0}})(a+v)}-\\frac{1}{(a-{{v}_{0}})(a-v)}\\textless0$$．  ∴$${{t}_{0}}-t\\textless0$$，即$${{t}_{0}}\\textless t$$．因此河水速增大所用时间将增多，选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1393", "queId": "98761c859cf64df8b7824fe809cedc4d", "competition_source_list": ["2014年第25届全国希望杯初二竞赛复赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "中国古代``五行''学说认为：``物质分金、木、水、火、土五种属性，金克木，木克土，土克水，水克火，火克金．''从五种属性互不相同的物质中随机抽取两种，则抽取的两种物质不相克的概率是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{10}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{5}$$ "}]], "knowledge_point_routes": ["知识标签->知识点->统计与概率->概率->概率公式", "知识标签->题型->统计与概率->概率->题型：概率计算公式"], "answer_analysis": ["基本事件为：金、木；金、水；金、火；金、土；木、水；木、火；木、土；水、火；水、土；火、土．  共计$$10$$种，  其中不相克的事件数为$$m=10-5=5$$（种），  所以抽取的两种物质不相克的概率是$$\\frac{5}{10}=\\frac{1}{2}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "704", "queId": "a7c39b31a67f43b69ed3fe5547051265", "competition_source_list": ["2017年湖南长沙天心区初二竞赛长郡教育集团抵达杯第3题5分", "2004年第21届全国初中数学联赛竞赛第1题", "初一下学期其它", "2018~2019学年湖北襄阳市樊城区襄阳市第五中学初三上学期期末（培优班）第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$abc\\ne 0$$，且$$a+b+c=0$$，则代数式$$\\frac{{{a}^{2}}}{bc}+\\frac{{{b}^{2}}}{ca}+\\frac{{{c}^{2}}}{ab}$$的值是．", "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->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->能力->运算能力"], "answer_analysis": ["$$\\frac{{{a}^{2}}}{bc}+\\frac{{{b}^{2}}}{ca}+\\frac{{{c}^{2}}}{ab}$$  $$=\\frac{{{\\left( b+c \\right)}^{2}}}{bc}+\\frac{{{\\left( a+c \\right)}^{2}}}{ac}+\\frac{{{\\left( a+b \\right)}^{2}}}{ab}$$  $$=\\frac{b}{c}+2+\\frac{c}{b}+\\frac{a}{c}+2+\\frac{c}{a}+\\frac{a}{b}+2+\\frac{b}{a}$$  $$=\\frac{a+b}{c}+\\frac{a+c}{b}+\\frac{b+c}{a}+6$$  $$=\\frac{a+b+c}{a}+\\frac{a+b+c}{b}+\\frac{a+b+c}{c}-1-1-1+6$$  $$=3$$． ", "<p>把$$a=-\\left( b+c \\right)$$，$$b=-\\left( a+c \\right)$$，$$c=-\\left( a+b \\right)$$代入，</p>\n<p>原式$$=\\frac{-\\left( b+c \\right)\\cdot a}{ab}+\\frac{-\\left( a+c \\right)\\cdot b}{ac}+\\frac{-\\left( a+b \\right)\\cdot c}{ab}$$</p>\n<p>$$~~~~~=-\\left( \\frac{ba+ca}{bc} \\right)-\\left( \\frac{ab+cb}{ac} \\right)-\\left( \\frac{ac+bc}{ab} \\right)$$</p>\n<p>$$~~~~~=-\\left( \\frac{a}{b}+\\frac{a}{c} \\right)-\\left( \\frac{b}{a}+\\frac{b}{c} \\right)-\\left( \\frac{c}{a}+\\frac{c}{b} \\right)$$</p>\n<p>$$~~~~~=\\frac{-\\left( b+c \\right)}{a}+\\frac{-\\left( a+c \\right)}{b}+\\frac{-\\left( a+b \\right)}{c}$$</p>\n<p>$$~~~~~=\\frac{a}{a}+\\frac{b}{b}+\\frac{c}{c}$$</p>\n<p>$$~~~~~=3$$，</p>", "<p>$$\\frac{{{a}^{2}}}{bc}+\\frac{{{b}^{2}}}{ca}+\\frac{{{c}^{2}}}{ab}$$</p>\n<p>$$=\\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{abc}$$</p>\n<p>$$=\\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc+3abc}{abc}$$</p>\n<p>$$=\\frac{\\left( a+b+c \\right)\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \\right)+3abc}{abc}$$</p>\n<p>$$=\\frac{3abc}{abc}$$</p>\n<p>$$=3$$．</p>"], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1254", "queId": "c056f81e11584e3581664c433cccebea", "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": "934", "queId": "b1cb418aba7f43cf86ac3a4451b60270", "competition_source_list": ["2008年竞赛第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$1$$，$$2$$，$$3$$，$$4$$，$$5$$这五个数字排成一排，最后一个数是奇数，且使得其中任意连续三个数之和都能被这三个数中的第一个数整除，那么满足要求的排法有．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$种 "}], [{"aoVal": "B", "content": "$$3$$种 "}], [{"aoVal": "C", "content": "$$4$$种 "}], [{"aoVal": "D", "content": "$$5$$种 "}]], "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$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$${{a}_{4}}$$，$${{a}_{5}}$$是$$1$$，$$2$$，$$3$$，$$4$$，$$5$$的一个满足要求的排列．  首先，对于$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$${{a}_{4}}$$，不能有连续的两个都是偶数，否则，这两个之后都是偶数，与已知条件矛盾．  又如果$${{a}_{i}}\\left( 1\\leqslant i\\leqslant 3 \\right)$$是偶数，$${{a}_{i+1}}$$是奇数，则$${{a}_{i+2}}$$是奇数，这说明一个偶数后面一定要接两个或两个以上的奇数，除非接的这个奇数是最后一个数．  所以$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$${{a}_{4}}$$，$${{a}_{5}}$$只能是：偶，奇，奇，偶，奇，有如下$$5$$种情形满足条件：  $$2$$，$$1$$，$$3$$，$$4$$，$$5$$；  $$2$$，$$3$$，$$5$$，$$4$$，$$1$$；  $$2$$，$$5$$，$$1$$，$$4$$，$$3$$；  $$4$$，$$3$$，$$1$$，$$2$$，$$5$$；  $$4$$，$$5$$，$$3$$，$$2$$，$$1$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1560", "queId": "cb8a1179858247eb8fc81d552b314112", "competition_source_list": ["2004年竞赛第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$$x$$和$$y$$是非零实数，使得$$\\left\\textbar{} x \\right\\textbar+y=3$$和$$\\left\\textbar{} x \\right\\textbar y+{{x}^{3}}=0$$，那么$$x+y$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{13}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1-\\sqrt{13}}{2}$$ "}], [{"aoVal": "D", "content": "$$4-\\sqrt{13}$$ "}]], "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->绝对值综合"], "answer_analysis": ["将$$y=3-\\left\\textbar{} x \\right\\textbar$$代入$$\\textbar x\\textbar y+{{x}^{3}}=0$$，得$${{x}^{3}}-{{x}^{2}}+3\\textbar x\\textbar=0$$．  （$$1$$）当$$x\\textgreater0$$时，$${{x}^{3}}-{{x}^{2}}+3x=0$$，方程$${{x}^{2}}-x+3=0$$无实根；  （$$2$$）当$$x\\textless{}0$$时，$${{x}^{3}}-{{x}^{2}}-3x=0$$，得方程$${{x}^{2}}-x-3=0$$．  解得$$x=\\frac{1\\pm \\sqrt{13}}{2}$$，正根舍去，从而$$x=\\frac{1-\\sqrt{13}}{2}$$．  于是$$y=3-\\left\\textbar{} x \\right\\textbar=3+\\frac{1-\\sqrt{13}}{2}=\\frac{7-\\sqrt{13}}{2}$$，  故$$x+y=4-\\sqrt{13}$$．  因此，结论$$\\text{D}$$是在正确的． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1527", "queId": "cf8ccd2e6ebf4a6084a57d4879434f24", "competition_source_list": ["1998年第9届希望杯初一竞赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "方程组$$\\begin{cases}3x+y=7    5x-8y=31   \\end{cases}$$的解$$(x,y)$$是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(3,-2)$$ "}], [{"aoVal": "B", "content": "$$(2,1)$$ "}], [{"aoVal": "C", "content": "$$(4,-5)$$ "}], [{"aoVal": "D", "content": "$$(0,7)$$ "}]], "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）->解二元一次方程组->加减消元法解二元一次方程组"], "answer_analysis": ["以$$(3,-2)$$，$$(2,1)$$，$$(4,-5)$$，$$(0,7)$$代入方程组检验，只有$$(3,-2)$$满足方程组，故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "490", "queId": "24970eaca8f24dec88b934ec0d830c68", "competition_source_list": ["2008年第13届华杯赛初一竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "关于数$$a$$有下面四个命题：  ①若$${{a}^{2}}=\\left\\textbar{} a \\right\\textbar$$，则$$a$$必为$$0$$．  ②若$${{a}^{2}}=\\left\\textbar{} a \\right\\textbar$$，则$$a$$，$$a+1$$，$$a-1$$中至少有一个为$$0$$．  ③若$${{a}^{2}}=\\left\\textbar{} a \\right\\textbar$$，则$$a=0$$，或$$a=1$$．  ④若$${{a}^{2}}=\\left\\textbar{} a \\right\\textbar$$，则$${{a}^{3}}-a$$的值必为$$0$$．  四个命题中正确的个数为．", "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->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", "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$${{a}^{2}}=\\left\\textbar{} a \\right\\textbar$$，所以$$a=0$$，或$$a=1$$，或$$a=1$$．即命题``$$a$$必为$$0$$''是错误的．因为$$a=0$$，或$$a=1$$，或$$a=1$$，所以$$a$$，$$a+1$$，$$a-1$$中至少有一个为零．即命题②是正确的．由此可知命题③是错误的．因为$${{a}^{3}}-a=a\\left( a+1 \\right)\\left( a-1 \\right)$$，所以$${{a}^{3}}-a=0$$，即命题④正确．所以正确的命题个数为$$2$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "513", "queId": "31924e70bed1432db670d75fcd234673", "competition_source_list": ["1996年第7届全国希望杯初一竞赛复赛第9题"], "difficulty": "3", "qtype": "single_choice", "problem": "如果关于$$x$$的方程$$3\\left( x+4 \\right)=2a+5$$的解大于关于$$x$$的方程$$\\frac{(4a+1)x}{4}=\\frac{a(3x-4)}{3}$$的解，那么．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater2$$ "}], [{"aoVal": "B", "content": "$$a\\textless{}2$$ "}], [{"aoVal": "C", "content": "$$a\\textless{}\\frac{7}{18}$$ "}], [{"aoVal": "D", "content": "$$a\\textgreater\\frac{7}{18}$$ "}]], "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->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->不等式（组）解的情况确定参数的范围"], "answer_analysis": ["关于$$x$$的方程$$3\\left( x+4 \\right)=2a+5$$的解为$$x=\\frac{2a-7}{3}$$，  关于$$x$$的方程$$\\frac{(4a+1)x}{4}=\\frac{a(3x-4)}{3}$$的解为$$x=-\\frac{16}{3}a$$，  由题意$$\\frac{2a-7}{3}\\textgreater-\\frac{16}{3}a$$，解得$$a\\textgreater\\frac{7}{18}$$，选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "42", "queId": "05ebf9851ac84527ad05a6b3e861c3a1", "competition_source_list": ["1993年第4届希望杯初二竞赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个三角形的三边长分别为$$2$$，$$4$$，$$a$$，如果$$a$$的数恰是方程$$4\\textbar x-2{{\\textbar}^{2}}-4\\textbar x-2\\textbar+1=0$$的根，那么三角形的周长为．", "answer_option_list": [[{"aoVal": "A", "content": "$$7\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$8\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "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->三角形的三边关系", "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->与三角形三边关系结合"], "answer_analysis": ["原方程为$${{(2\\textbar x-2\\textbar-1)}^{2}}=0$$，可得$$\\textbar x-2\\textbar=\\frac{1}{2}$$，  所以$${{x}_{1}}=1\\frac{1}{2}$$，$${{x}_{2}}=2\\frac{1}{2}$$，若$$a=1\\frac{1}{2}$$，则$$2+1\\frac{1}{2}=3\\frac{1}{2}\\textless{}4$$，  故$$2$$，$$4$$，$$1\\frac{1}{2}$$不能是三角形的三边长，又若$$a=2\\frac{1}{2}$$，则  $$2+2\\frac{1}{2}=4\\frac{1}{2}\\textgreater4$$，  故$$2$$，$$4$$，$$2\\frac{1}{2}$$是三角形的三边长，周长为$$8\\frac{1}{2}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "26", "queId": "01b1868bab7f41e5a813fe9900dba905", "competition_source_list": ["2012年第29届全国全国初中数学联赛竞赛第5题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "若方程$${{x}^{2}}+2px-3p-2=0$$的两个不相等的实数根$${{x}_{1}}$$，$${{x}_{2}}$$满足$$x_{1}^{2}+x_{1}^{3}=4-(x_{2}^{2}+x_{2}^{3})$$，则实数$$p$$的所有可能的值之和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$-\\frac{3}{4}$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "$$-\\frac{5}{4}$$ "}]], "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": ["因为方程$${{x}^{2}}+2px-3p-2=0$$两个不相等的实数根，  故$$\\Delta ={{(2p)}^{2}}+4(3p+2)=4{{p}^{2}}+12p+8\\textgreater0$$，  解得$$p\\textless{}-2$$或$$p\\textgreater-1$$．  因为$${{x}_{1}}$$，$${{x}_{2}}$$满足$${{x}_{1}}+{{x}_{2}}=-2p$$，$${{x}_{1}}\\cdot {{x}_{2}}=-3p-2$$，  故$$x_{1}^{2}+x_{2}^{2}={{({{x}_{1}}+{{x}_{2}})}^{2}}-2{{x}_{1}}{{x}_{2}}$$  $$=4{{p}^{2}}+2(3p+2)$$  $$=4{{p}^{2}}+6p+4$$，  $$x_{1}^{3}+x_{2}^{3}=({{x}_{1}}+{{x}_{2}})(x_{1}^{2}-{{x}_{1}}{{x}_{2}}+x_{2}^{2})$$  $$=-2p(4{{p}^{2}}+6p+4+3p+2)$$  $$=-8{{p}^{3}}-18{{p}^{2}}-12p$$，  由$$x_{1}^{2}+x_{1}^{3}=4-(x_{2}^{2}+x_{2}^{3})$$得$$x_{1}^{2}+x_{2}^{2}+x_{1}^{3}+x_{2}^{3}=4$$  即$$4{{p}^{2}}+6p+4-8{{p}^{3}}-18{{p}^{2}}-12p=4$$，  整理得$$p(4p^{2}+7p+3)=0$$，  解得$${{p}_{1}}=-\\frac{3}{4}$$，$${{p}_{2}}=-1$$，$${{p}_{3}}=0$$．  ∵$$p\\textless{}-2$$或$$p\\textgreater-1$$，  ∴$$p=-\\frac{3}{4}$$，  ∴实数$$p$$的所有可能的值之和为$$-\\frac{3}{4}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "807", "queId": "daacea01f56d45adade81be796e9c961", "competition_source_list": ["2017~2018学年北京海淀区中国人民大学附属中学初一上学期单元测试《数形结合谈数轴》第1题", "1998年第9届希望杯初一竞赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知有理数$$a$$在数轴上原点的右方，有理数$$b$$在原点的左方，那么（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$ab\\textless{}b$$ "}], [{"aoVal": "B", "content": "$$ab\\textgreater b$$ "}], [{"aoVal": "C", "content": "$$a+b\\textgreater0$$ "}], [{"aoVal": "D", "content": "$$a-b\\textgreater0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->数轴->数轴上的规律探究", "课内体系->能力->运算能力", "课内体系->思想->数形结合思想"], "answer_analysis": ["∵$$a$$在数轴上原点右方，$$b$$在原点左方，  ∴$$a\\textgreater0$$，$$b\\textless{}0$$，  当$$a=1$$，$$ab=b$$，显然应排除$$\\text{A}$$、$$\\text{B}$$选项．  当$$a=1$$，$$b=-2$$时，$$a+b=-1\\textless{}0$$，排除$$\\text{C}$$选项．  当$$a\\textgreater0$$，$$b\\textless{}0$$时，$$a-b\\textgreater0$$总成立，故$$\\text{D}$$选项正确． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "19", "queId": "097ea9a4eb794bb3930c7c1dad30837c", "competition_source_list": ["2012年全国全国初中数学联赛初一竞赛", "2012年第29届全国全国初中数学联赛竞赛A卷第11题20分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知直角三角形的边长均为整数，周长为$$30$$，则它的外接圆的面积为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{121}{4} \\pi $$ "}], [{"aoVal": "B", "content": "$$\\frac{169}{4} \\pi $$ "}], [{"aoVal": "C", "content": "$$\\frac{225}{4} \\pi $$ "}], [{"aoVal": "D", "content": "以上答案都不对 "}]], "knowledge_point_routes": ["课内体系->知识点->圆->圆与多边形->正多边形与圆", "课内体系->知识点->三角形->勾股定理及应用->勾股定理基础->勾股定理的证明", "课内体系->知识点->三角形->勾股定理及应用->勾股定理应用->勾股定理的综合应用"], "answer_analysis": ["设直角三角形的三边长分别为$$a$$，$$b$$，$$c$$（$$a\\leqslant b\\textless{}c$$），则$$a+b+c=30$$．  显然，三角形的外接圆的直径即为斜边长$$c$$，下面先求$$c$$的值．  由$$a\\leqslant b\\textless{}c$$及$$a+b+c=30$$得$$30=a+b+c\\textless{}3c$$，  所以$$c\\textgreater10$$．  由$$a+b\\textgreater c$$及$$a+b+c=30$$得$$30=a+b+c\\textgreater2c$$，  所以$$c\\textless{}15$$．  又因为$$c$$为整数，  所以$$11\\leqslant c\\leqslant 14$$．  根据勾股定理可得$${{a}^{2}}+{{b}^{2}}={{c}^{2}}$$，  把$$c=30-a-b$$代入，化简得$$ab-30(a+b)+450=0$$，  所以$$(30-a)(30-b)=450=2\\times {{3}^{2}}\\times {{5}^{2}}$$，  因为$$a$$，$$b$$均为整数且$$a\\leqslant b$$，  所以只可能是$$\\begin{cases}30-a={{5}^{2}}    30-b=2\\times {{3}^{2}} \\end{cases}$$，解得$$\\begin{cases}a=5    b=12 \\end{cases}$$，  所以，直角三角形的斜边长$$c=13$$，三角形的外接圆的面积为$$\\frac{169}{4} \\pi $$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "204", "queId": "263de7ee70e84be296ecbd478a501533", "competition_source_list": ["2012年第23届全国希望杯初二竞赛复赛第10题4分"], "difficulty": "3", "qtype": "single_choice", "problem": "循环节长度是$$4$$的纯循环小数化成最简分数后，分母是三位数，这样的循环小数有（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$798$$个 "}], [{"aoVal": "B", "content": "$$898$$个 "}], [{"aoVal": "C", "content": "$$900$$个 "}], [{"aoVal": "D", "content": "$$998$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->算术基本定理"], "answer_analysis": ["设$$0.\\dot{a}bc\\dot{d}=\\frac{\\overline{abcd}}{9999}=\\frac{q}{p}$$，$$(p,q)=1$$，$$q\\textless{}p$$，$$p$$是$$9999$$的约数，且是三位数．  由于$$9999={{3}^{2}}\\times 11\\times 101$$，  所以，$$p$$的值只可能是$$101$$，$$303$$，$$909$$．  分母$$p=101$$时，因为$$101$$是质数，  所以，分子可以是$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$100$$，共$$100$$个数满足条件；  分母$$p=303$$时，在$$1$$～$$302$$中，$$3$$的倍数有$$100$$个，$$101$$的倍数有$$2$$个，  所以，共有$$302-102=200$$个数满足条件；  分母$$p=909$$时，在$$1$$～$$908$$中，$$3$$的倍数有$$302$$个，$$101$$的倍数有$$8$$个，既是$$3$$的倍数又是$$101$$的倍数的数有$$2$$个：$$303$$，$$606$$，  所以，共有$$908-302-8+2=600$$个数满足条件．  综上，满足条件的数共有$$100+200+600=900$$个． ", "\n<p>，故$$9999$$的三位数约数只有$$101$$，$$303$$，$$909$$，</p>\n<p>即对于分数$$\\frac{x}{909}$$，必有$$101|x$$，故共有$$900$$个．</p>", "\n<p>即与$$9999$$约分后得$$3$$位数．</p>\n<p></p>\n<p>&there4;约分后为</p>\n<p>①，与$$101$$互质的有$$100$$个，（$$x&gt;1$$，$$2$$，$$\\cdots $$，$$100$$），</p>\n<p>②，与$$303$$互质，（$$x=1$$，$$2$$，$$\\cdots 302$$），</p>\n<p>其中$$3$$的倍数有$$100$$个，加上$$101$$，$$202$$，共$$102$$个，</p>\n<p>&there4;有$$200$$个符合的．</p>\n<p>③，与$$909$$互质（$$x=1$$，$$2$$，$$\\cdots 908$$），</p>\n<p>其中$$3$$的倍数有$$302$$个，加上$$101$$，$$202$$，$$404$$，$$505$$，$$707$$，$$808$$，共$$308$$个．</p>\n<p>&there4;有$$600$$个符合的．</p>\n<p>综上，共$$900$$个这样的循环小数．</p>\n<p>故选$$\\text{C}$$．</p>"], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "362", "queId": "e3289c79a3454edcad85ad2c9a623f9c", "competition_source_list": ["其它", "2011年第28届全国全国初中数学联赛竞赛第4题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "今有长度分别为$$1$$，$$2$$，$$\\cdots $$，$$9$$的线段各一条，现从中选出若干条线段组成``线段组''，由这一组线段恰好可以拼接成一个正方形，则这样的``线段组''的组数有（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$组 "}], [{"aoVal": "B", "content": "$$7$$组 "}], [{"aoVal": "C", "content": "$$9$$组 "}], [{"aoVal": "D", "content": "$$11$$组 "}]], "knowledge_point_routes": ["竞赛->知识点->组合->计数问题-枚举法"], "answer_analysis": ["显然用这些线段去拼接成正方形，至少要$$7$$条．  当用$$7$$条线段去拼接成正方形时，有$$3$$条边每边都用$$2$$条线段连接，而另一条边只用$$1$$条线段，其长度恰好等于其它$$3$$条边中每两条线段的长度之和．  当用$$8$$条线段去拼接成正方形时，则每边用两条线段相接，其长度和相等．  又因为$$1+2+\\cdots +9=45$$，  所以正方形的边长不大于$$\\left[ \\frac{45}{4} \\right]=11$$．  由于$$7=1+6=2+5=3+4$$；  $$8=1+7=2+6=3+5$$；  $$9=1+8=2+7=3+6=4+5$$；  $$1+9=2+8=3+7=4+6$$；  $$2+9=3+8=4+7=5+6$$．  所以，组成边长为$$7$$、$$8$$、$$10$$、$$11$$的正方形，各有一种方法；  组成边长为$$9$$的正方形，有$$5$$种方法．  故满足条件的``线段组''的组数为$$1\\times 45=9$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "231", "queId": "7ddfbddac38e4b64892bbb5b28dea406", "competition_source_list": ["2018年全国初中数学联赛竞赛A卷"], "difficulty": "4", "qtype": "single_choice", "problem": "设$$p$$，$$q$$均为大于$$3$$的素数，则使$${{p}^{2}}+5pq+4{{q}^{2}}$$为完全平方数的素数对$$(p,q)$$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->数论->整除->素数与合数", "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->因式分解->分组分解法", "课内体系->知识点->式->因式分解->因式分解的应用"], "answer_analysis": ["设$${{p}^{2}}+5pq+4{{q}^{2}}={{m}^{2}}$$ ($$m$$为自然数)，则$${{(p+2q)}^{2}}+pq={{m}^{2}}$$，即  $$(m-p-2q)(m+p+2q)=pq$$．  由于$$pq$$为素数，且$$m+p+2q\\textgreater p$$，$$m+p+2q\\textgreater q$$，  所以$$m-p-2q=1$$，$$m+p+2q=pq$$，  从而$$pq-2p-4q-1=0$$，即$$(p-4)(q-2)=9$$，  所以，$$(p,q)=(5,11)$$或$$(7,5)$$．  所以，满足条件的素数对$$(p,q)$$的个数为$$2$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "444", "queId": "476b2adc9c82422cab275d9706644810", "competition_source_list": ["2016年第27届全国希望杯初一竞赛复赛第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "将一张$$1\\text{m}\\times 1\\text{m}$$的正方形白纸对折$$8$$次（每一次都沿平行于正方形边的方向对折），那么所有折痕的长度的和的最小是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$32\\text{m}$$ "}], [{"aoVal": "B", "content": "$$30\\text{m}$$ "}], [{"aoVal": "C", "content": "$$16\\text{m}$$ "}], [{"aoVal": "D", "content": "$$14\\text{m}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->相交线与平行线->平行线的判定与性质->平行线的性质", "课内体系->知识点->几何图形初步->相交线与平行线->平行线的判定与性质->平行线相关折叠问题"], "answer_analysis": ["第一次折叠，折痕长度为$$1\\text{m}$$，  第二次折叠，折痕最小为$$\\frac{1}{2}\\times 2=1\\text{m}$$，  第三次折叠，折痕最小为$$\\frac{1}{2}\\times 4=2\\text{m}$$，  第四次折叠，折痕最小为$$\\frac{1}{4}\\times 8=2\\text{m}$$，  第五次折叠，折痕最小为$$\\frac{1}{4}\\times 16=4\\text{m}$$，  第六次折叠，折痕最小为$$\\frac{1}{8}\\times 32=4\\text{m}$$，  第七次折叠，折痕最小为$$\\frac{1}{8}\\times 64=8\\text{m}$$，  第八次折叠，折痕最小为$$\\frac{1}{16}\\times 128=8\\text{m}$$，  ∴所有折痕的长度的和的最小是$$1+1+2+2+4+4+8+8=30\\text{m}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "794", "queId": "c83f9434014442bdad886bf1fee57e1d", "competition_source_list": ["2001年第12届希望杯初二竞赛第2试第1题", "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"], "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": "828", "queId": "915eaac17d8443faac9cb70d401aaf72", "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12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12题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知：$$243=3\\times 3\\times 3\\times 3\\times 3$$，表示成$$g\\left( 243 \\right)=5$$，那么$$g\\left( 81 \\right)$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$27$$ "}], [{"aoVal": "D", "content": "$$81$$ "}]], "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->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": ["∵$${{3}^{4}}=81$$，  ∴$$g\\left( 81 \\right)=4$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "396", "queId": "a260c20401ef44caa7600df9f0addeab", "competition_source_list": ["2016年第27届全国希望杯初一竞赛初赛第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若定义$$a*b=ab+a+b$$，从左到右依次计算$$x=1*2*3*\\cdots *(n-1)*n$$，则满足$$x\\textgreater2016$$的最小正整数$$n$$是（~ ~ ~）．", "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->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->整式的加减->整式的加减运算", "课内体系->能力->运算能力"], "answer_analysis": ["$$1*2=1\\times 2+1+2=5$$，  $$1*2*3=5*3=5\\times 3+5+3=23$$，  $$1*2*3*4=23*4=23\\times 4+23+4=119$$，  $$1*2*3*4*5=119*5=119\\times 5+119+5=719$$，  $$1*2*3*4*5*6=719*6=719\\times 6+719+6=5039\\textgreater2016$$，  ∴$$n=6$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1119", "queId": "ff8080814d9efd56014daa7ae6940acc", "competition_source_list": ["1993年第4届全国希望杯初一竞赛初赛第11题"], "difficulty": "2", "qtype": "single_choice", "problem": "有理数$$a$$、$$b$$小于零，并且使$${{(a-b)}^{3}}\\textless{}0$$，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": "$${{a}^{2}}\\textgreater{{b}^{4}}$$ "}]], "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": ["由$${{(a-b)}^{3}}\\textless{}0$$，得出$$a-b\\textless{}0$$．即$$a\\textless{}b$$．  ∵$$a$$，$$b\\textless{}0$$，∴$$\\left\\textbar{} a \\right\\textbar\\textless{}\\left\\textbar{} b \\right\\textbar$$，选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1471", "queId": "fca3eacda1784f74aa107d4f36ca0b1b", "competition_source_list": ["1990年第1届希望杯初二竞赛第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$${{f}_{1}}=\\dfrac{1}{1-\\dfrac{1}{x}}$$， $${{f}_{2}}=\\dfrac{1}{2个\\left { 1-\\dfrac{1}{1-\\dfrac{1}{x}} \\right.}$$ ，\\ldots，$${{f}_{1990}}=\\dfrac{1}{1990个\\left { 1-\\dfrac{1}{1-\\dfrac{1}{1-\\ddots 1-\\dfrac{1}{x}}} \\right.}$$，将$${{f}_{1990}}$$化简，应等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{x}{x-1}$$ "}], [{"aoVal": "B", "content": "$$1-x$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{x}$$ "}], [{"aoVal": "D", "content": "$$x$$ "}]], "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->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": ["计算：$${{f}_{1}}=\\frac{x}{x-1}$$，$${{f}_{2}}=1-x$$，$${{f}_{3}}=\\frac{1}{x}$$，$${{f}_{4}}=\\frac{x}{x-1}$$，\\ldots{}  可见$${{f}_{3n+1}}=\\frac{x}{x-1}$$，而$$1990=3\\times 663+1$$．  所以$${{f}_{1990}}=\\frac{x}{x-1}$$．  故选（$$\\mathbf{A}$$）． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "986", "queId": "73a51d5d314e4a3f9363e2ef9ec300cc", "competition_source_list": ["2016年第27届全国希望杯初二竞赛初赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$A$$、$$B$$两班共有$$112$$名学生，且各班男生、女生的人数比分别是$$5:6$$，$$12:11$$，那么两班混合后男生、女生的人数比是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1:1$$ "}], [{"aoVal": "B", "content": "$$12:11$$ "}], [{"aoVal": "C", "content": "$$27:29$$ "}], [{"aoVal": "D", "content": "$$25:31$$ "}]], "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）->二元一次方程组应用题->二元一次方程组的数字问题"], "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": "1334", "queId": "8aac50a7511483070151192950b80f25", "competition_source_list": ["2015年第26届全国希望杯初三竞赛初赛（特）第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$\\alpha $$为锐角，且$$\\cos \\alpha +2\\sin \\alpha =2$$，则$$\\tan \\alpha $$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["知识标签->题型->三角形->锐角三角函数及解直角三角形->锐角三角函数->正弦定理与余弦定理", "知识标签->知识点->三角形->锐角三角函数及解直角三角形->锐角三角函数->锐角三角函数的定义"], "answer_analysis": ["∵$$\\cos \\alpha +2\\sin \\alpha =2$$，  ∴$$\\cos \\alpha =2-2\\sin \\alpha $$．  又$${{\\sin }^{2}}\\alpha +{{\\cos }^{2}}\\alpha =1$$，  ∴$${{\\sin }^{2}}\\alpha +{{(2-2\\sin \\alpha )}^{2}}=1$$，  整理得：$$5{{\\sin }^{2}}\\alpha -8\\sin \\alpha +3=0$$，  解得$$\\sin \\alpha =\\frac{3}{5}$$或$$\\sin \\alpha =1$$，  ∵$$\\alpha $$为锐角，  ∴$$\\sin \\alpha =\\frac{3}{5}$$．  ∴$$\\cos \\alpha =2-2\\sin \\alpha =\\frac{4}{5}$$，  ∴$$\\tan \\alpha =\\frac{\\sin \\alpha }{\\cos \\alpha }=\\frac{3}{4}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1351", "queId": "c9a76894bebd47dcadbb5d91e8de571f", "competition_source_list": ["2003年第14届希望杯初二竞赛第2试第8题"], "difficulty": "0", "qtype": "single_choice", "problem": "将长为$$12$$的线段截成长度为整数的三段，使它们成为一个三角形的三边，则构成的三角形．", "answer_option_list": [[{"aoVal": "A", "content": "不可能是等腰三角形 "}], [{"aoVal": "B", "content": "不可能是直角三角形 "}], [{"aoVal": "C", "content": "不可能是等边三角形 "}], [{"aoVal": "D", "content": "不可能是钝角三角形 "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->特殊三角形->等腰三角形"], "answer_analysis": ["可以构成的三角形的三条边长分别为$$3$$，$$4$$，$$5$$（直角三角形）；$$4$$，$$4$$，$$4$$（等边三角形）；$$5$$，$$5$$，$$2$$（等腰三角形）．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "525", "queId": "2d559ed50e5c4527a4c87e8cf40665ee", "competition_source_list": ["2008年第19届希望杯初二竞赛第1试第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "使方程$$3x+2y=200$$成立的正整数对$$(x,y)$$有．", "answer_option_list": [[{"aoVal": "A", "content": "$$66$$个 "}], [{"aoVal": "B", "content": "$$33$$个 "}], [{"aoVal": "C", "content": "$$30$$个 "}], [{"aoVal": "D", "content": "$$18$$个 "}]], "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->二元一次方程的整数解", "竞赛->知识点->数论->不定方程"], "answer_analysis": ["因为$$3x+2y=200$$，$$x$$，$$y$$为正整数，所以  $$y=\\frac{200-3x}{2}\\textgreater0$$，  因此$$x\\textless{}66\\frac{2}{3}$$，  又因为$$x$$为正偶数，所以$$x$$可取$$2$$，$$4$$，$$6$$，$$\\cdots $$，$$66$$，共$$33$$个数，  故原方程的正整数解有$$33$$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": "737", "queId": "56c49781599e46d08402c10945958098", "competition_source_list": ["1998年第9届希望杯初一竞赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "一条直线上距离相等地立有$$10$$根标杆，一名学生匀速地从第$$1$$杆向第$$10$$杆行走，当他走到第$$6$$杆时用了$$6.6$$秒，则当他走到第$$10$$杆时所用时间是．", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$秒 "}], [{"aoVal": "B", "content": "$$13.2$$秒 "}], [{"aoVal": "C", "content": "$$11.88$$秒 "}], [{"aoVal": "D", "content": "$$9.9$$秒 "}]], "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": ["从第$$1$$根标杆到第$$6$$根标杆有$$5$$个间隔．  因而，每个间隔行进$$6.6\\div 5=1.32$$（秒）．  而从第$$1$$根标杆到第$$10$$根标杆共有$$9$$个间隔．  所以行进$$9$$个间隔共用$$1.32\\times 9=11.88$$（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": "1231", "queId": "c0437e6e48d24248b61903753c292c2e", "competition_source_list": ["2014年第25届全国希望杯初二竞赛初赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "在平面直角坐标系$$xOy$$中，函数$$\\sqrt{xy}+\\left\\textbar{} x-y+1 \\right\\textbar=0$$的图象是（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "直线$$x=0$$、$$y=0$$和$$x-y+1=0$$ "}], [{"aoVal": "B", "content": "直线$$x=0$$和$$x-y+1=0$$ "}], [{"aoVal": "C", "content": "点$$(0,0)$$和直线$$x-y+1=0$$ "}], [{"aoVal": "D", "content": "点$$(0,1)$$和$$(-1,0)$$ "}]], "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->知识点->函数->函数概念和图象->函数图象->函数的图象", "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$$\\sqrt{xy}\\geqslant 0$$，$$\\left\\textbar{} x-y+1 \\right\\textbar\\geqslant 0$$，  由题设$$\\sqrt{xy}+\\left\\textbar{} x-y+1 \\right\\textbar=0$$，  所以$$\\left { \\begin{matrix}xy=0    x-y+1=0   \\end{matrix} \\right.$$，  即$$\\left { \\begin{matrix}x=0    x-y+1=0   \\end{matrix} \\right.$$或$$\\left { \\begin{matrix}y=0    x-y+1=0   \\end{matrix} \\right.$$，  解得$$\\left { \\begin{matrix}x=0    y=1   \\end{matrix} \\right.$$或$$\\left { \\begin{matrix}x=-1    y=0   \\end{matrix} \\right.$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "336", "queId": "34a7a8bdc8d34e43bad694b5f6381e01", "competition_source_list": ["2011年第22届全国希望杯初二竞赛复赛第5题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$n+1={{2010}^{2}}+{{2011}^{2}}$$，则$$\\sqrt{2n+1}=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2011$$ "}], [{"aoVal": "B", "content": "$$2010$$ "}], [{"aoVal": "C", "content": "$$4022$$ "}], [{"aoVal": "D", "content": "$$4021$$ "}]], "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": ["由题设，得$$n={{2010}^{2}}+{{2011}^{2}}-1$$，  $$2n+1=2\\times {{2010}^{2}}+2\\times {{2011}^{2}}-2\\times 1+1$$  $$=2\\times {{2010}^{2}}+2\\times {{(2010+1)}^{2}}-1$$  $$=2\\times {{2010}^{2}}+2\\times {{2010}^{2}}+4\\times 2010+1$$  $$={{4020}^{2}}+2\\times 4020+1$$  $$={{(4020+1)}^{2}}$$  $$={{4021}^{2}}$$．  所以$$\\sqrt{2n+1}=4021$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "810", "queId": "64d04678e17648f79c18c08b46edae7a", "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7题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "一个三角形木架三边长分别是$$75\\text{cm}$$，$$100\\text{cm}$$，$$125\\text{cm}$$，现要做一个与其相似的三角形木架，而只有长为$$40\\text{cm}$$和$$120\\text{cm}$$的两根木条．要求以其中一根为一边，从另一根截下两段作为另两边（允许有余料），则不同的截法有．", "answer_option_list": [[{"aoVal": "A", "content": "一种 "}], [{"aoVal": "B", "content": "两种 "}], [{"aoVal": "C", "content": "三种 "}], [{"aoVal": "D", "content": "四种 "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->相似三角形->相似三角形的判定->相似三角形的判定-三边对应成比例"], "answer_analysis": ["由题知新形成的三角形钢架与原三角形钢架相似，则它们的对应边成比例．  （$$1$$）若以$$40\\text{cm}$$的钢筋为三角形一边，去截$$120\\text{cm}$$长的钢筋，且截成两段分别为$$x\\text{cm}$$，$$y\\text{cm}$$，  则有三种情况：  ①以$$40\\text{cm}$$为最长边，则可列得$$\\frac{40}{125}=\\frac{x}{100}=\\frac{y}{75}$$，  解得：$$x=32$$，$$y=24$$，且$$32+24=56\\textless{}120$$，  所以此法可行；  ②以$$40\\text{cm}$$为较长边时，则$$\\frac{x}{125}=\\frac{40}{100}=\\frac{y}{75}$$，  解得：$$x=50$$，$$y=30$$，且$$50+30=80\\textless{}120$$，  所以此法可行；  ③以$$40\\text{cm}$$为最短边，则$$\\frac{x}{125}=\\frac{y}{100}=\\frac{40}{75}$$，  解得：$$x=\\frac{200}{3}$$，$$y=\\frac{160}{3}$$，且$$\\frac{200}{3}+\\frac{160}{3}=120$$，  所以此法可行；  （$$2$$）若以$$120\\text{cm}$$的钢筋为三角形一边，去截$$40\\text{cm}$$长的钢筋，且截成的两段分别为$$x\\text{cm}$$，$$y\\text{cm}$$，  很容易的值$$120$$只能作为最长边，  则$$\\frac{120}{125}=\\frac{x}{100}=\\frac{y}{75}$$，  解得：$$x=96$$，$$y=72$$，  但$$96\\textgreater40$$，$$72\\textgreater40$$，  所以此法不可行．  综上可知只有$$3$$种截法．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1396", "queId": "f2f41bd71d1f4674b9191823c2e028b5", "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10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知：反比例函数$$y=\\frac{2}{x}$$与正比例函数$$y=x$$的图象交于$$A$$，$$B$$两点（$$A$$在第一象限），点$${{F}_{1}}\\left( -2,-2 \\right)$$，$${{F}_{2}}\\left( 2,2 \\right)$$在直线$$y=x$$上．设点$$P\\left( {{x}_{0}},{{y}_{0}} \\right)$$是反比例函数$$y=\\frac{2}{x}$$图象上的任意一点，记点$$P$$与$${{F}_{1}}$$，$${{F}_{2}}$$两点之间的距离之差$$d=\\left\\textbar{} P{{F}_{1}}-P{{F}_{2}} \\right\\textbar$$，由反比例函数的定义可知$$d=\\left\\textbar{} P{{F}_{1}}-P{{F}_{2}} \\right\\textbar$$为常数，值为线段$$AB$$的长．现请你在反比例函数$$y=\\frac{2}{x}$$第一象限内的分支上找一点$$P$$，使点$$P$$到$${{F}_{2}}\\left( 2,2 \\right)$$和点$$C\\left( 6,4 \\right)$$的距离之和最小，则点$$P$$的坐标为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 2,1 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 1,1 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( 3,2 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( 2,3 \\right)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->反比例函数->反比例函数与代数综合->反比例函数与一次函数综合", "课内体系->知识点->综合与实践->新定义->与函数有关的新定义", "课内体系->特色题型->新定义", "课内体系->能力->抽象概括能力"], "answer_analysis": ["解由$$y=\\frac{2}{x}$$和$$y=x$$组成的方程组可得$$A$$、$$B$$两点的坐标分别为，$$\\left( \\sqrt{2},\\sqrt{2} \\right)$$、$$\\left( -\\sqrt{2},-\\sqrt{2} \\right)$$，线段$$AB$$的长度$$=4$$，  ∵点$$P\\left( {{x}_{0}},{{y}_{0}} \\right)$$是反比例函数$$y=\\frac{2}{x}$$图象上一点，  ∴$${{y}_{0}}=\\frac{2}{{{x}_{0}}}$$，  ∴$$P{{F}_{1}}=\\sqrt{{{\\left( {{x}_{0}}+2 \\right)}^{2}}+{{\\left( \\frac{2}{{{x}_{0}}}+2 \\right)}^{2}}}=\\left\\textbar{} \\frac{{{\\left( {{x}_{0}}+1 \\right)}^{2}}+1}{{{x}_{0}}} \\right\\textbar$$，  $$P{{F}_{2}}=\\sqrt{{{\\left( {{x}_{0}}-2 \\right)}^{2}}+{{\\left( \\frac{2}{{{x}_{0}}}-2 \\right)}^{2}}}=\\left\\textbar{} \\frac{{{\\left( {{x}_{0}}-1 \\right)}^{2}}+1}{{{x}_{0}}} \\right\\textbar$$，  ∴$$d=\\left\\textbar{} P{{F}_{1}}-P{{F}_{2}} \\right\\textbar=\\left\\textbar{} \\left\\textbar{} \\frac{{{\\left( {{x}_{0}}+1 \\right)}^{2}}+1}{{{x}_{0}}} \\right\\textbar-\\left\\textbar{} \\frac{{{\\left( {{x}_{0}}-1 \\right)}^{2}}+1}{{{x}_{0}}} \\right\\textbar{} \\right\\textbar$$，  当$${{x}_{0}}\\textgreater0$$时，$$d=4$$；当$${{x}_{0}}\\textless{}0$$时，$$d=4$$，  因此，无论点$$P$$的位置如何，线段$$AB$$的长度与$$d$$一定相等．  由条件$$P{{F}_{2}}=P{{F}_{1}}-4$$，知$$P{{F}_{2}}+PC=P{{F}_{1}}+PC-4$$，  由$${{F}_{1}}$$，$$-P$$，$$C$$三点共线时最小，  此时可解得$$P\\left( 2,1 \\right)$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1220", "queId": "db8f7cc91d4449a5b58868f9d82b7bbe", "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4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$${{y}_{1}}=2a\\left( a\\ne 1 \\right)$$，$${{y}_{2}}=\\frac{2}{{{y}_{1}}}$$，$${{y}_{3}}=\\frac{2}{{{y}_{2}}}$$，$$\\cdot \\cdot \\cdot $$，$${{y}_{2017}}=\\frac{2}{{{y}_{2016}}}$$ ，$${{y}_{2018}}=\\frac{2}{{{y}_{2017}}}$$，则$${{y}_{1}}{{y}_{2018}}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["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"], "answer_analysis": ["将$${{y}_{1}}=2a$$代入$${{y}_{2}}=\\frac{2}{{{y}_{1}}}$$中得$${{y}_{2}}=\\frac{1}{a}$$，同样的方法得出$${{y}_{3}}=2a$$，$$\\cdot \\cdot \\cdot $$，  则对于$${{y}_{n}}$$，当$$n$$为奇数时，$${{y}_{n}}=2a$$，当$$n$$为偶数时，$${{y}_{n}}=\\frac{1}{a}$$，  所以$${{y}_{2018}}=\\frac{1}{a}$$，  故$${{y}_{1}}\\cdot {{y}_{2018}}=2a\\cdot \\frac{1}{a}=2$$，  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1017", "queId": "ff8080814d4b1928014d4c13d008058b", "competition_source_list": ["2015年第26届全国希望杯初一竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$是整数，且$$\\left\\textbar{} a-1 \\right\\textbar+\\left\\textbar{} b+2 \\right\\textbar=1$$，则$${{(a-1)}^{2}}\\times {{(b+2)}^{4}}=$$（~ ）．", "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->知识点->数->有理数->绝对值->绝对值的非负性"], "answer_analysis": ["∵$$a$$，$$b$$是整数，且$$\\left\\textbar{} a-1 \\right\\textbar+\\left\\textbar{} b+2 \\right\\textbar=1$$，  ∴$$\\left { \\begin{matrix}\\left\\textbar{} a-1 \\right\\textbar=1    \\left\\textbar{} b+2 \\right\\textbar=0   \\end{matrix} \\right.$$或$$\\left { \\begin{matrix}\\left\\textbar{} a-1 \\right\\textbar=0    \\left\\textbar{} b+2 \\right\\textbar=1   \\end{matrix} \\right.$$，  ∴$${{(a-1)}^{2}}\\times {{(b+2)}^{4}}=0$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1377", "queId": "983dc809620b47a3aeff4c3efbc81701", "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4题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "把抛物线$$y=-(x-1)^{2}+1$$向右平移$$1$$个单位，再向下平移$$1$$个单位，得到新函数的解析式为．", "answer_option_list": [[{"aoVal": "A", "content": "$$y=-x^{2}$$ "}], [{"aoVal": "B", "content": "$$y=-x^{2}+2$$ "}], [{"aoVal": "C", "content": "$$y=-(x-2)^{2}$$ "}], [{"aoVal": "D", "content": "$$y=-(x-2)^{2}+2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->二次函数->二次函数的几何变换->二次函数平移变换"], "answer_analysis": ["∵抛物线$$y=-(x-1)^{2}+1$$的顶点坐标为$$(1,1)$$，  ∴向右平移$$1$$个单位，再向下平移$$1$$个单位后的顶点坐标是$$(2,0)$$，  ∴所得抛物线解析式是$$y=- (x-2)^{2}$$，  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1558", "queId": "c6fc3459362246efb3885d72ab0dde52", "competition_source_list": ["2010年第15届华杯赛初一竞赛初赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "乘积为$$-240$$的不同的五个整数的平均值最大是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{17}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{18}{5}$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"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"], "answer_analysis": ["假设$$-240=-1\\times b\\times c\\times d\\times e$$，$$-1 ~\\textless{} ~b ~\\textless{} ~c ~\\textless{} ~d ~\\textless{} ~e$$，$$-240=A\\times B\\times C\\times D\\times E$$，$$A ~\\textless{} ~B ~\\textless{} ~C ~\\textless{} ~D ~\\textless{} ~E$$，必有$$-1\\geqslant A$$，且可以取$$b=B$$，$$c=C$$，$$d=D$$，$$e=\\left\\textbar{} A \\right\\textbar\\times E$$，总有$$-1+b+c+d+e\\geqslant A+B+C+D+E$$，故只需求出$$\\left( b+c+d+e \\right)$$的最大值即可，其中$$b\\times c\\times d\\times e=240$$，取$$b=1$$，朋$$1\\times {{c}^{\\prime }}\\times {{d}^{\\prime }}\\times {{e}^{\\prime }}=240$$，且可以取$${{c}^{\\prime }}=c$$，$${{d}^{\\prime }}=d$$，$${{e}^{\\prime }}=b\\times e$$，则得到：$$1+{{c}^{\\prime }}+{{d}^{\\prime }}+{{e}^{\\prime }}-b-c-d-e=1+be-b-e=\\left( b-1 \\right)\\times \\left( e-1 \\right)\\textgreater0$$，故只需求出$$1+{{c}^{\\prime }}+{{d}^{\\prime }}+{{e}^{\\prime }}$$的最大值即可．依次类推可得：$$-240=-1\\times 1\\times 2\\times 3\\times 40$$，因此乘积为$$-240$$的不同的五个整数的平均值最大是$$9$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "906", "queId": "5cbd99a693d947abaf4ad5ff30cfbf98", "competition_source_list": ["2006年第17届希望杯初二竞赛第1试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$m$$，$$n$$是实数，且满足 $${{m}^{2}}+2{{n}^{2}}+m-\\frac{4}{3}n+\\frac{17}{36}=0$$，则$$-m{{n}^{2}}$$的平方根是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{2}}{6}$$ "}], [{"aoVal": "B", "content": "$$\\pm \\frac{\\sqrt{2}}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{6}$$ "}], [{"aoVal": "D", "content": "$$\\pm \\frac{1}{6}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->乘法公式"], "answer_analysis": ["由题设，得：  $$\\left( {{m}^{2}}+m+\\frac{1}{4} \\right)+2\\left( {{n}^{2}}-\\frac{2}{3}n+\\frac{1}{9} \\right)+\\frac{17}{36}-\\frac{1}{4}-\\frac{2}{9}=0$$，  即$${{\\left( m+\\frac{1}{2} \\right)}^{2}}+2{{\\left( n-\\frac{1}{3} \\right)}^{2}}=0$$，  因为$${{\\left( m+\\frac{1}{2} \\right)}^{2}}\\geqslant 0$$，$$2{{\\left( n-\\frac{1}{3} \\right)}^{2}}\\geqslant 0$$，  所以$$m=-\\frac{1}{2}$$，$$n=\\frac{1}{3}$$，  $$-m{{n}^{2}}=\\frac{1}{18}$$，它的平方根是$$\\pm \\frac{\\sqrt{2}}{6}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "13", "queId": "173752e241004b1dac8e788a62428962", "competition_source_list": ["2013年第18届华杯赛初一竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "将乘积$$0.\\dot{2}4\\dot{3}\\times 0.3\\dot{2}523\\dot{3}$$化为小数，小数点后第$$2014$$位数字是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["$$0.\\dot{2}4\\dot{3}\\times 0.\\dot{3}2523\\dot{3}$$  $$=\\frac{243}{999}\\times \\frac{325230}{999990}$$  $$=\\frac{27\\times 9}{27\\times 37}\\times \\frac{32523}{99999}$$  $$=\\frac{9}{37}\\times \\frac{879\\times 37}{99999}$$  $$=\\frac{7911}{99999}$$  $$=0.0\\dot{7}91\\dot{1}$$．  因为$$2014\\div 5=402\\cdots \\cdots 4$$，所以，小数点后第$$2014$$位数字是$$1$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "720", "queId": "49505e179da54eed92dedd99a27f1726", "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": "例题3：  1、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": "1177", "queId": "8aac49074e023206014e20e3fab1660f", "competition_source_list": ["1994年第5届全国希望杯初一竞赛复赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果等式$$1992+1994+1996+1998=5000-\\square $$成立，则$$\\square $$中应当填的数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$-980$$ "}], [{"aoVal": "C", "content": "$$-1990$$ "}], [{"aoVal": "D", "content": "$$-2980$$ "}]], "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->常规一元一次方程解法"], "answer_analysis": ["设$$\\square $$的数是$$x$$，则$$1992+1994+1996+1998=5000-x$$，即$$7980=5000-x$$，  ∴$$x=5000-7980=-2980$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1495", "queId": "b8af4d093d444869a3348bcf1a4a5aa8", "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$$a+b$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$-5$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$-1$$ "}]], "knowledge_point_routes": ["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", "知识标签->知识点->方程与不等式->一元二次方程->一元二次方程的解", "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"], "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": "70", "queId": "1c368f7dd9514054a5326a3aa18b0616", "competition_source_list": ["2016年第27届全国希望杯初一竞赛初赛第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "下列计算中，正确的是（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{x}^{2}}+{{x}^{3}}={{x}^{5}}$$ "}], [{"aoVal": "B", "content": "$${{x}^{4}}-{{x}^{2}}={{x}^{2}}$$ "}], [{"aoVal": "C", "content": "$${{x}^{2}}\\cdot {{x}^{3}}={{x}^{6}}$$ "}], [{"aoVal": "D", "content": "$${{x}^{3}}\\div {{x}^{2}}=x$$ "}]], "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->式->整式的乘除->整式的乘除运算->单项式除以单项式", "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": ["略． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1460", "queId": "dc8b511965fd400f80e6081f8d649d9a", "competition_source_list": ["2018~2019学年10月天津南开区天津市第二十五中学初一上学期月考第2题3分", "2019~2020学年10月天津河北区天津外国语大学附属外国语学校初一上学期月考第7题3分", "2011年第22届全国希望杯初一竞赛初赛第1题4分", "2019~2020学年10月天津河北区天津外国语大学附属外国语学校初一上学期月考第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a$$的负倒数的相反数是$$8$$，$$b$$的相反数的负倒数也是$$8$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a=b$$ "}], [{"aoVal": "B", "content": "$$a\\textless{}b$$ "}], [{"aoVal": "C", "content": "$$a\\textgreater b$$ "}], [{"aoVal": "D", "content": "$$ab=1$$ "}]], "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", "课内体系->知识点->数->有理数->相反数->相反数的定义"], "answer_analysis": ["由$$-\\left( -\\frac{1}{a} \\right)=8$$，得$$a=\\frac{1}{8}$$；  由$$-\\left( \\frac{1}{-b} \\right)=8$$，得$$b=\\frac{1}{8}$$；  所以$$a=b$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1154", "queId": "f70c2a730bd647b698eef2fb12a1dd61", "competition_source_list": ["2012年第23届全国希望杯初一竞赛复赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$x+y=3$$，$$xy=1$$．则$${{x}^{5}}+{{y}^{5}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$33$$ "}], [{"aoVal": "B", "content": "$$231$$ "}], [{"aoVal": "C", "content": "$$123$$ "}], [{"aoVal": "D", "content": "$$312$$ "}]], "knowledge_point_routes": ["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": ["因为$$x+y=3$$，  所以$${{(x+y)}^{2}}=9$$，  即$${{x}^{2}}+{{y}^{2}}+2xy=9$$．  因为$$xy=1$$，  所以$${{x}^{2}}+{{y}^{2}}=7$$．  所以$${{x}^{3}}+{{y}^{3}}=(x+y)({{x}^{2}}+{{y}^{2}}-xy)=3\\times (7-1)=18$$，  所以$${{x}^{5}}+{{y}^{5}}=({{x}^{3}}+{{y}^{3}})({{x}^{2}}+{{y}^{2}})-{{(xy)}^{2}}(x+y)=18\\times 7-{{1}^{2}}\\times 3=123$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "160", "queId": "0c78d5b15aea42d9977c8de3ee6a0d05", "competition_source_list": ["2018~2019学年湖北宜昌初三下学期期中（东部）第7题", "2019年广东惠州惠城区光正实验学校初二竞赛第10题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "甲、乙、丙、丁四人进行射击测试，每人$$10$$次射击的平均成绩恰好都是$$9.2$$环，方差分别是$$S_{甲}^{2}=0.56$$，$$S_{乙}^{2}=0.45$$，$$S_{丙}^{2}=0.50$$，$$S_{丁}^{2}=0.60$$，则成绩最稳定的是．", "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->能力->数据处理能力", "课内体系->知识点->统计与概率->数据的分析->方差"], "answer_analysis": ["由题意得：  $$S_{乙}^{2}\\textless{}S_{丙}^{2}\\textless{}S_{甲}^{2}\\textless{}S_{丁}^{2}$$，  ∴最稳定的是乙．  选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "145", "queId": "2f0ea5d9e0984e1b83830ac17cb8b706", "competition_source_list": ["2016年第27届全国希望杯初一竞赛初赛第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$99$$，$$100$$这$$100$$个自然数中，不是$$2$$的倍数，不是$$3$$的倍数，且不是$$5$$的倍数的数共有$$k$$个，则$$k=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$26$$ "}], [{"aoVal": "C", "content": "$$27$$ "}], [{"aoVal": "D", "content": "$$28$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->整除的概念与基本性质"], "answer_analysis": ["在$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$99$$，$$100$$这$$100$$个自然数中，  $$2$$的倍数有$$50$$个，  $$3$$的倍数有$$33$$个，  $$5$$的倍数有$$20$$个，  $$2$$和$$3$$的公倍数有$$16$$个，  $$2$$和$$5$$的公倍数有$$10$$个，  $$3$$和$$5$$的公倍数有$$6$$个，  $$2$$、$$3$$和$$5$$的公倍数有$$3$$个．  共计$$50+33+20-16-10-6+3=74$$个，  ∴$$k=100-74=26$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "571", "queId": "be791bfe44534e80b13afb7b4c1c660b", "competition_source_list": ["2010年第21届希望杯初二竞赛第1试第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "顺次连结一个凸四边形各边的中点，得到一个菱形，则这个四边形一定是．", "answer_option_list": [[{"aoVal": "A", "content": "任意的四边形 "}], [{"aoVal": "B", "content": "两条对角线等长的四边形 "}], [{"aoVal": "C", "content": "矩形 "}], [{"aoVal": "D", "content": "平行四边形 "}]], "knowledge_point_routes": ["课内体系->知识点->四边形->特殊平行四边形->菱形->菱形的性质"], "answer_analysis": ["顺次连结任意四边形的中点，得到的四边形的两组对边分别平行于原四边形的两条对角线，并且分别等于原四边形的两条对角线长的一半，所以得到的四边形是平行四边形，当原四边形的两条对角线长度相等时，顺次连结该四边形各边中点所得到的平行四边形的两组对边等长，即为菱形．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "423", "queId": "8b4ba8b7d31a41bd8dfc4b4740fbd1c9", "competition_source_list": ["1992年第3届希望杯初二竞赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "在式子$$\\left\\textbar{} x+1 \\right\\textbar+\\left\\textbar{} x+2 \\right\\textbar+\\left\\textbar{} x+3 \\right\\textbar+\\left\\textbar{} x+4 \\right\\textbar$$中，用不同的$$x$$值代入，得到对应的值，在这些对应值中，最小的值是（~ ）．", "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->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->利用绝对值求最值", "课内体系->思想->分类讨论思想"], "answer_analysis": ["令$$a=\\left\\textbar{} x+1 \\right\\textbar+\\left\\textbar{} x+4 \\right\\textbar$$，$$b=\\left\\textbar{} x+2 \\right\\textbar+\\left\\textbar{} x+3 \\right\\textbar$$，  $$t=\\left\\textbar{} x+1 \\right\\textbar+\\left\\textbar{} x+2 \\right\\textbar+\\left\\textbar{} x+3 \\right\\textbar+\\left\\textbar{} x+4 \\right\\textbar=a+b$$，  根据绝对值的几何意义，$$a$$表示点$$x$$到$$-1$$与$$-4$$两点的距离之和，  分析可得当$$-4\\leqslant x\\leqslant -1$$时，$$a$$最小，其值为$$3$$，  $$b$$表示点$$x$$到$$-2$$与$$-3$$两点的距离之和，  分析可得当$$-3\\leqslant x\\leqslant -2$$时，$$b$$最小，其值为$$1$$，  综合可得，当$$-3\\leqslant x\\leqslant -2$$，$$a$$、$$b$$均取得最小值，  故此时$$t$$取得最小值，且$$t$$的最小值为$$3+1=4$$，  故答案为$$4$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "316", "queId": "1a436a8790154a37803880881d04e919", "competition_source_list": ["2015年第32届全国全国初中数学联赛竞赛第2题7分", "2017~2018学年12月天津天津市益中学校初三上学期月考第11题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "若抛物线$$y={{x}^{2}}+bx+c$$与$$x$$轴只有一个公共点，且过点$$A(m,n)$$，$$B(m-8,n)$$，则$$n=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "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->二次函数与一元二次方程的关系", "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 的对称轴->已知抛物线上对称的两点求对称轴"], "answer_analysis": ["∵抛物线过点$$A(m,n)$$，$$B(m-8,n)$$，  ∴抛物线的对称轴为$$x=-\\frac{b}{2}=\\frac{m+m-8}{2}$$，  ∴$$b=8-2m$$．  ∵抛物线$$y={{x}^{2}}+bx+c$$与$$x$$轴只有一个公共点，  ∴$$\\Delta ={{b}^{2}}-4c=0$$，  ∴$$c=\\frac{1}{4}{{b}^{2}}={{(4-m)}^{2}}$$，  ∴$$n={{m}^{2}}+(8-2m)m+{{(4-m)}^{2}}=16$$． ", "<p>由题意得：$${{b}^{2}}-4c=0$$，$$b=8-2m$$，</p>\n<p>&there4;$$n={{m}^{2}}+bm+c={{m}^{2}}+\\left( 8-2m \\right)m+{{\\left( 4-m \\right)}^{2}}=16$$．</p>"], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "180", "queId": "3cfc3560503c44f29a259a26da387e1c", "competition_source_list": ["1990年第1届希望杯初二竞赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "等腰三角形周长是$$\\text{24cm}$$，一腰的中线将周长分成$$5:3$$的两部分，那么这个三角形的底边长是．", "answer_option_list": [[{"aoVal": "A", "content": "$$7\\text{.5cm}$$ "}], [{"aoVal": "B", "content": "$$\\text{12cm}$$ "}], [{"aoVal": "C", "content": "$$\\text{4cm}$$ "}], [{"aoVal": "D", "content": "$$\\text{12cm}$$或$$\\text{4cm}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["若底边长为$$12$$．则其他二边之和也是$$12$$，矛盾．故不可能是$$\\text{B}$$或$$\\text{D}$$．  又底为$$4$$时，腰长是$$10$$．符合题意．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1452", "queId": "f348d176ae394d2189ad5befb79016e8", "competition_source_list": ["1993年第4届希望杯初二竞赛第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "若一元二次方程$$2x(kx-4)-{{x}^{2}}+6=0$$有实数根，则$$k$$的最大整数值是．", "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->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": ["原方程整理为$$(2k-1){{x}^{2}}-8x+6=0$$，  当$$\\Delta \\geqslant 0$$时，方程有实数根，则$${{(-8)}^{2}}-4\\cdot (2k-1)\\cdot 6\\geqslant 0$$，  即$$48k\\leqslant 88$$，  所以$$k\\leqslant \\frac{11}{6}$$，  故$$k$$的最大整数值是$$1$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "285", "queId": "a22ee33ab00e4881abb3f6383bd3fb36", "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": "小伟、小胜、小臣、小琳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：  小伟：苹果不在我这里；  小胜：苹果在小琳那里；  小臣：苹果在小胜那里；  小琳：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": "小伟 "}], [{"aoVal": "B", "content": "小胜 "}], [{"aoVal": "C", "content": "小臣 "}], [{"aoVal": "D", "content": "小琳 "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->命题与证明"], "answer_analysis": ["用假设法：  （$$1$$）假设小伟说了假话，那么苹果在小伟那里，小胜和小臣就说的是假话，与题设冲突；  （$$2$$）假设小胜说的是假话，其他人说的都是真话，那么苹果在小胜那里；  （$$3$$）假设小臣说了假话，则小胜说``苹果在小琳那里''与小琳说``苹果不在我这里''都是真话，互相矛盾；  （$$4$$）假设小琳说的是假话，那么小臣说的就是假话，这样小琳、小臣都说假话，与``其中只有一人说的是假话''的题设不符．  因此苹果在小胜那里． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1039", "queId": "ff8080814d7978b9014d865e712323f4", "competition_source_list": ["1990年第1届全国希望杯初一竞赛初赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "有四种说法：  甲：正数的平方不一定大于它本身；  乙：正数的立方不一定大于它本身；  丙：负数的平方不一定大于它本身；  丁：负数的立方不一定大于它本身．  这四种说法中，不正确的说法的个数是（~ ）．", "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->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->有理数乘法->有理数乘法运算", "课内体系->能力->运算能力"], "answer_analysis": ["由$${{1}^{2}}=1$$，$${{1}^{3}}=1$$可知甲、乙两种说法是正确的．  由$${{(-1)}^{3}}=-1$$，可知丁也是正确的说法．  而负数的平方均为正数，即负数的平方一定大于它本身，  所以``负数平方不一定大于它本身''的说法不正确．即丙不正确．  在甲、乙、丙、丁四个说法中，只有丙$$1$$个说法不正确．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": "1243", "queId": "8aac4907508d5d3d0150a48f552a2599", "competition_source_list": ["1997年第8届全国希望杯初一竞赛复赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$1997$$个不全相等的有理数之和为零，则这$$1997$$个有理数中（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "至少有一个是零 "}], [{"aoVal": "B", "content": "至少有$$998$$个正数 "}], [{"aoVal": "C", "content": "至少有一个是负数 "}], [{"aoVal": "D", "content": "至多有$$1995$$个是负数 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["由题意，这$$1997$$个有理数可以有零，也可以没有零，则排除$$\\text{A}$$．  这$$1997$$个有理数中，必须有正数和负数．  例如，$$1996$$个$$-1$$和一个$$1996$$相加为零，则否定了$$\\text{B}$$和$$\\text{D}$$．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "107", "queId": "1c794bf05aaa44b78d783e97283d9cf0", "competition_source_list": ["2014年第25届全国希望杯初二竞赛初赛第2题4分", "2017~2018学年河南洛阳涧西区洛阳市东升第二中学初二上学期期中第5题3分", "2017~2018学年湖南益阳桃江县源嘉桥镇中学初二上学期期中第9题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "分式$$\\frac{{{b}^{2}}-1}{{{b}^{2}}-2b-3}$$的值为$$0$$，则$$b$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$\\pm 1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的基础->分式值为0", "课内体系->知识点->式->分式->分式的基础->分式有意义的条件"], "answer_analysis": ["因为$$\\frac{{{b}^{2}}-1}{{{b}^{2}}-2b-3}=0$$，  所以$${{b}^{2}}-1=0$$且$${{b}^{2}}-2b-3\\ne 0$$，  解得$$b=1$$或$$b=-1$$，且$$b\\ne -1$$且$$b\\ne 3$$，  所以$$b=1$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1468", "queId": "ab219d8315384f39b404babe6433cb54", "competition_source_list": ["2010年第21届希望杯初二竞赛第2试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$${{2}^{12}}\\times {{5}^{9}}$$，得数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$位数 "}], [{"aoVal": "B", "content": "$$10$$位数 "}], [{"aoVal": "C", "content": "$$11$$位数 "}], [{"aoVal": "D", "content": "$$12$$位数 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["$${{2}^{12}}\\times {{5}^{9}}={{2}^{3}}\\times {{2}^{9}}\\times {{5}^{9}}=8\\times {{\\left( 2\\times 5 \\right)}^{9}}=8\\times {{10}^{9}}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "324", "queId": "7e119b1fd37d4f8c8aa783ebc7af9dae", "competition_source_list": ["2011年第22届全国希望杯初一竞赛初赛第19题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$a$$与$$b$$是互为相反数，且$$\\left\\textbar{} a-2b \\right\\textbar=\\frac{3}{2}$$，则$$\\frac{2a-ab-{{b}^{2}}+2}{{{a}^{2}}+ab+b-1}$$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->数->有理数->绝对值->绝对值的定义", "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->式->分式->分式化简求值->分式化简求值-直接代入数值求值", "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"], "answer_analysis": ["因为$$a$$与$$b$$是互为相反数，  所以$$a+b=0$$，即$$b=-a$$．  所以$$\\left\\textbar{} a-2b \\right\\textbar=\\left\\textbar{} 3a \\right\\textbar=\\frac{3}{2}$$，  所以$$\\left\\textbar{} a \\right\\textbar=\\left\\textbar{} b \\right\\textbar=\\frac{1}{2}$$．  所以$${{a}^{2}}+ab+b-1\\ne 0$$．  所以$$\\frac{2a-ab-{{b}^{2}}+2}{{{a}^{2}}+ab+b-1}$$  $$=\\frac{2a+{{a}^{2}}-{{a}^{2}}+2}{{{a}^{2}}-{{a}^{2}}-a-1}$$  $$=\\frac{2a+2}{-a-1}$$  $$=-2$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "918", "queId": "73185d6b80954dc18756e832ab04e199", "competition_source_list": ["2005年第16届希望杯初二竞赛复赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\triangle ABC$$的三个内角的比是$$m:\\left( m+1 \\right):\\left( m+2 \\right)$$，其中$$m$$是大于$$1$$的正整数，那么$$\\triangle ABC$$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": ["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": ["设一个角为$$m$$，则两外两个角为$$m+1$$，$$m+2$$，  则：$$3m+3=180{}^{}\\circ $$，  解得：$$m=59{}^{}\\circ $$，即可得三角分别为$$59{}^{}\\circ $$，$$60{}^{}\\circ $$，$$61{}^{}\\circ $$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1149", "queId": "9bb3985d8cd841189b132a913cf044f8", "competition_source_list": ["2012年第23届全国希望杯初一竞赛初赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$3$$枚面值$$3$$元的硬币和$$5$$枚面值$$5$$元的硬币中任意取出$$1$$枚或多于$$1$$枚，可以得到$$n$$种不同的面值，则$$n$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$23$$ "}], [{"aoVal": "D", "content": "$$26$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->计数问题-分步处理与乘法原理"], "answer_analysis": ["分两步取硬币：  第一步，取面值为$$3$$元的硬币，可以是$$0$$枚，$$1$$枚，$$2$$枚．  第二步，取面值为$$5$$元的硬币，可以是$$0$$枚，$$1$$枚，$$2$$枚，$$3$$枚，$$4$$枚，$$5$$枚．  于是，共有取法$$4\\times 6=24$$（种）．  因为要求意取出$$1$$枚或多于$$1$$枚，  所以共有取法$$24-1=23$$（种）． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "559", "queId": "481314bbebd64df6be25b816a47f25c2", "competition_source_list": ["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", "初一下学期其它第18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": "919", "queId": "65b8740d9f904498af31da703c28285b", "competition_source_list": ["2021年福建竞赛（\"大梦杯”青少年水平测试）第1~1题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知二次函数$$y=a{{x}^{2}}+bx+c$$的图象交\\emph{x}轴于\\emph{A}(\\emph{x}\\textsubscript{1}，0)，\\emph{B}(\\emph{x}\\textsubscript{2}，0)两点，交\\emph{y}轴于点\\emph{C}(0，3)，若$${{x}_{1}}+{{x}_{2}}=4$$，且△\\emph{ABC}的面积为3，则\\emph{a+b}（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "3 "}], [{"aoVal": "B", "content": "-5 "}], [{"aoVal": "C", "content": "3 "}], [{"aoVal": "D", "content": "5 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 方法一：由根与系数的关系可得$${{x}_{1}}+{{x}_{2}}=-\\frac{b}{a}=4$$，$${{x}_{1}}{{x}_{2}}=\\frac{c}{a}=\\frac{3}{a}$$，再利用$$AB=\\left\\textbar{} {{x}_{1}}-{{x}_{2}} \\right\\textbar=\\sqrt{{{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-4{{x}_{1}}{{x}_{2}}}=\\sqrt{16-4\\times \\frac{3}{a}}=2\\sqrt{4-\\frac{3}{a}}$$列方程求解$$a,b$$，再检验即可得到答案；方法二：不妨设$${{x}_{1}}\\textless{} {{x}_{2}}$$，由三角形的面积先求解$$AB={{x}_{2}}-{{x}_{1}}=2$$，结合$${{x}_{1}}+{{x}_{2}}=4$$，再求解$${{x}_{1}},{{x}_{2}},$$再利用待定系数法求解$$a,b,$$ 从而可得答案.\\\\ 【详解】\\\\ 解：方法一：依题意$${{x}_{1}},{{x}_{2}}$$为方程$$a{{x}^{2}}+bx+c=0$$的两根，且$$c=3$$．\\\\ 所以$${{x}_{1}}+{{x}_{2}}=-\\frac{b}{a}=4$$，$${{x}_{1}}{{x}_{2}}=\\frac{c}{a}=\\frac{3}{a}$$．\\\\ 所以$$AB=\\left\\textbar{} {{x}_{1}}-{{x}_{2}} \\right\\textbar=\\sqrt{{{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-4{{x}_{1}}{{x}_{2}}}=\\sqrt{16-4\\times \\frac{3}{a}}=2\\sqrt{4-\\frac{3}{a}}$$，\\\\ 所以$$\\vartriangle ABC$$面积$$S=\\frac{1}{2}AB\\times 3=\\frac{1}{2}\\times 2\\sqrt{4-\\frac{3}{a}}\\times 3=3$$．\\\\ 解得$$a=1$$，经检验符合题意，\\\\ $$\\therefore $$ $$b=-4a=-4$$．\\\\ 因为函数$$y={{x}^{2}}-4x+3$$的图象与$$x$$轴有两个不同交点，因此$$a=1$$，$$b=-4$$，$$c=3$$符合要求．\\\\ 所以$$a+b=-3$$．\\\\ 方法二：不妨设$${{x}_{1}}\\textless{} {{x}_{2}}$$，则$$AB={{x}_{2}}-{{x}_{1}}$$，由$$\\vartriangle ABC$$的面积为3，且$$C\\left( 0,3 \\right)$$，得$$AB=2$$．\\\\ 所以$$AB={{x}_{2}}-{{x}_{1}}=2$$，又$${{x}_{1}}+{{x}_{2}}=4$$，\\\\ 解得：$${{x}_{1}}=1$$，$${{x}_{2}}=3$$．\\\\ 因此$$y=a{{x}^{2}}+bx+c=a\\left( x-1 \\right)\\left( x-3 \\right)$$．\\\\ 将$$x=0$$代入，得$$y=3a=3$$，所以$$a=1$$．\\\\ 所以$$y=a{{x}^{2}}+bx+c=\\left( x-1 \\right)\\left( x-3 \\right)={{x}^{2}}-4x+3$$，\\\\ 因此$$a+b=1+\\left( -4 \\right)=-3$$．\\\\ 故选C\\\\ 【点睛】\\\\ 本题考查的是二次函数的性质，一元二次方程根与系数的关系，掌握二次函数与\\emph{x}轴的交点坐标的含义是解本题的关键. "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "55", "queId": "93f65b4c05fe4570b336b3d61666ed82", "competition_source_list": ["2013年竞赛第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$，$$c$$是实常数，关于$$x$$的一元二次方程$$a{{x}^{2}}+bx+c=0$$有两个非零实根$${{x}_{1}}$$，$${{x}_{2}}$$，则下列关于$$x$$的一元二次方程中，以$$\\frac{1}{x_{1}^{2}}$$，$$\\frac{1}{x_{2}^{2}}$$为两个实根的是(  )．", "answer_option_list": [[{"aoVal": "A", "content": "$${{c}^{2}}{{x}^{2}}+\\left( {{b}^{2}}-2ac \\right)x+{{a}^{2}}=0$$ "}], [{"aoVal": "B", "content": "$${{c}^{2}}{{x}^{2}}-\\left( {{b}^{2}}-2ac \\right)x+{{a}^{2}}=0$$ "}], [{"aoVal": "C", "content": "$${{c}^{2}}{{x}^{2}}+\\left( {{b}^{2}}-2ac \\right)x-{{a}^{2}}=0$$ "}], [{"aoVal": "D", "content": "$${{c}^{2}}{{x}^{2}}-\\left( {{b}^{2}}-2ac \\right)x-{{a}^{2}}=0$$ "}]], "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": ["由于$$a{{x}^{2}}+bx+c=0$$是关于$$x$$的一元二次方程，则$$a\\ne 0$$．  因为$${{x}_{1}}+{{x}_{2}}=-\\frac{b}{a}$$，$${{x}_{1}}{{x}_{2}}=\\frac{c}{a}$$，且$${{x}_{1}}{{x}_{2}}\\ne 0$$，所以$$c\\ne 0$$，且$$\\frac{1}{x_{1}^{2}}+\\frac{1}{x_{2}^{2}}=\\frac{{{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}}}{x_{1}^{2}x_{2}^{2}}=\\frac{{{b}^{2}}-2ac}{{{c}^{2}}}$$，$$\\frac{1}{x_{1}^{2}}\\cdot \\frac{1}{x_{2}^{2}}=\\frac{{{a}^{2}}}{{{c}^{2}}}$$，  于是根据方程根与系数的关系，以$$\\frac{1}{x_{1}^{2}}$$，$$\\frac{1}{x_{2}^{2}}$$为两个实根的一元二次方程是$${{x}^{2}}-\\frac{{{b}^{2}}-2ac}{{{c}^{2}}}x+\\frac{{{a}^{2}}}{c}=0$$，即$${{c}^{2}}{{x}^{2}}-\\left( {{b}^{2}}-2ac \\right)x+{{a}^{2}}=0$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "201", "queId": "0d522f459bad4b04838eb48993449c51", "competition_source_list": ["2016年上海浦东新区进才中学自主招生第8题10分", "2016年第33届全国全国初中数学联赛竞赛A卷第3题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果一个正整数可以表示为两个连续奇数的立方差，则称这个正整数为``和谐数''．如：$$2={{1}^{3}}-{{(-1)}^{3}}$$，$$26={{3}^{3}}-{{1}^{3}}$$，$$2$$和$$26$$均为``和谐数''．那么，不超过$$2016$$的正整数中，所有的``和谐数''之和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$6858$$ "}], [{"aoVal": "B", "content": "$$6860$$ "}], [{"aoVal": "C", "content": "$$9260$$ "}], [{"aoVal": "D", "content": "$$9262$$ "}]], "knowledge_point_routes": ["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->整式的加减->整式有关的概念->整式->整式的定义", "课内体系->知识点->综合与实践->新定义->新定义综合其它", "课内体系->能力->运算能力"], "answer_analysis": ["方法一：$${{(2k+1)}^{3}}-{{(2k-1)}^{3}}$$  $$=\\left[ (2k+1)-(2k-1) \\right]\\left[ {{(2k+1)}^{2}}+(2k+1)(2k-1)+{{(2k-1)}^{2}} \\right]$$  $$=2(12{{k}^{2}}+1)$$（其中$$k$$为非负整数），  由$$2(12{{k}^{2}}+1)\\leqslant 2016$$得，$$k\\leqslant 9$$，  ∴$$k=0$$，$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$9$$，  即得所有不超过$$2016$$的``和谐数''的和为  $$\\left[ {{1}^{3}}-{{(-1)}^{3}} \\right]+({{3}^{3}}-{{1}^{3}})+({{5}^{3}}-{{3}^{3}})+\\cdots +({{17}^{3}}-{{15}^{3}})+({{19}^{3}}-{{17}^{3}})={{19}^{3}}+1=6860$$．  方法二：设``和谐数''为$$m$$，一定可以表示为$$m={{n}^{3}}-{{(n-2)}^{3}}$$，其中$$n$$为正整数．题目要求$$m\\leqslant 2016$$，通过计算可得$$n$$最大为$$19$$，于是有  $$2={{1}^{3}}-{{(-1)}^{3}}$$，  $$26={{3}^{3}}-{{1}^{3}}$$，  $$\\cdots $$  $$1946={{19}^{3}}-{{17}^{3}}$$，  等式两边分别相加，可得``和谐数''总和$$={{19}^{3}}-{{(-1)}^{3}}=6860$$，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": "108", "queId": "0874c8c988924b92a562ac5dd29d076f", "competition_source_list": ["1994年第5届希望杯初二竞赛第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$n$$为大于$$1$$的自然数，则下列四个式子的值一定不是完全平方数的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3{{n}^{2}}-3n+3$$ "}], [{"aoVal": "B", "content": "$$5{{n}^{2}}-5n-5$$ "}], [{"aoVal": "C", "content": "$$9{{n}^{2}}-9n+9$$ "}], [{"aoVal": "D", "content": "$$11{{n}^{2}}-11n-11$$ "}]], "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": ["欲使$$9{{n}^{2}}-9n+9$$为某自然数的平方，因为$$9{{n}^{2}}-9n+9=9\\left( {{n}^{2}}-n+1 \\right)$$，  故必须使$${{n}^{2}}-n+1$$ 为某自然数的平方，  而$$n\\textgreater1$$时有$${{n}^{2}}-2n+1\\textless{}{{n}^{2}}-n+1\\textless{}{{n}^{2}}$$，  即$${{n}^{2}}-n+1$$不可能为某自然数的完全平方，  故选$$\\text{C}$$． ", "<p>当$$n=2$$时，$$3{{n}^{2}}-3n+3=9$$，</p>\n<p>当$$n=3$$时，$$5{{n}^{2}}-5n-5=25$$，</p>\n<p>当$$n=4$$时，$$11{{n}^{2}}-11n-11=121$$均为完全平方数，所以排除$$\\text{ABD}$$，</p>\n<p>故选$$\\text{C}$$．</p>"], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "485", "queId": "315c52655c7b465887f1a19eb3cefb70", "competition_source_list": ["2002年第13届希望杯初一竞赛第2试第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面四个命题中一定不正确的命题是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3{{a}^{2}}{{b}^{7}}$$和$$7{{b}^{7}}{{a}^{2}}$$是同类项 "}], [{"aoVal": "B", "content": "$$3x-1=0$$和$$3+\\frac{2}{x-1}=0$$是同解方程 "}], [{"aoVal": "C", "content": "$$a-3$$和$$3-a$$互为倒数 "}], [{"aoVal": "D", "content": "$${{x}^{3}}-b$$和$$-{{x}^{3}}-b$$互为相反数 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["$$\\text{A}$$．显然正确；$$\\text{B}$$．中两个方程的解 $$x=\\frac{1}{3}$$，是同解方程；  当$$b=0$$时，$$\\text{D}$$正确；若存在$$a$$，使得$$\\left( a-3 \\right)$$与$$\\left( 3-a \\right)$$互为倒数，则  $${{\\left( a-3 \\right)}^{2}}=-1\\textless{}0$$，矛盾．  所以对任何数$$a$$，$$a-3$$与$$3-a$$都不能互为倒数．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "373", "queId": "1f01edaaf5dd43dd8f64a9cb967e48bf", "competition_source_list": ["辽宁大连竞赛", "初一单元测试《分解方法的延拓（1）》第14题", "2016~2017学年广东深圳福田区深圳实验学校中学部初一下学期单元测试《整式乘除》第27题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a+b=-\\frac{1}{5}$$，$$a+3b=1$$，则$$3{{a}^{2}}+12ab+9{{b}^{2}}+\\frac{3}{5}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2}{9}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{4}{5}$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "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": ["$$3{{a}^{2}}+12ab+9{{b}^{2}}+\\frac{3}{5}$$  $$=3({{a}^{2}}+4ab+3{{b}^{2}})+\\frac{3}{5}$$  $$=3(a+b)(a+3b)+\\frac{3}{5}$$  $$=3\\times \\left( -\\frac{1}{5} \\right)\\times 1+\\frac{3}{5}$$  $$=0$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "694", "queId": "56757307f756491fbb4a9441780fadba", "competition_source_list": ["2016年第27届全国希望杯初二竞赛复赛第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "三条边长都是质数的三角形可能是（~ ）．  ①锐角三角形；②直角三角形；③钝角三角形；④等腰三角形；⑤等边三角形．", "answer_option_list": [[{"aoVal": "A", "content": "①②③④ "}], [{"aoVal": "B", "content": "②③④⑤ "}], [{"aoVal": "C", "content": "①③④⑤ "}], [{"aoVal": "D", "content": "①②③④⑤ "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->直角三角形->直角三角形的定义", "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", "课内体系->知识点->三角形->等腰三角形->等腰三角形基础->等腰三角形的定义", "课内体系->知识点->三角形->等腰三角形->等边三角形->等边三角形的定义"], "answer_analysis": ["例如边长为$$2$$，$$2$$，$$2$$的三角形，满足题意，  该三角形为锐角三角形，等腰三角形，等边三角形．  例如边长为$$3$$，$$3$$，$$5$$的三角形，满足题意，  该三角形为钝角三角形．  若该三角形为直角三角形，设三边长为$$a$$，$$b$$，$$c$$，$$a\\textgreater b\\geqslant c$$且均为质数．  ∴$${{a}^{2}}={{b}^{2}}+{{c}^{2}}$$．  由于偶质数只有$$2$$，且奇数的平方仍然数奇数，  ∴一定有$$c=2$$，  ∴$${{a}^{2}}-{{b}^{2}}=4$$．  ∵最小的两个奇质数的平方差为$${{5}^{2}}-{{3}^{2}}=16\\ne 4$$，  ∴不存在三边长都是质数的直角三角形．  综上，满足条件的三角形可能是锐角三角形，钝角三角形，等腰三角形和等边三角形． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "761", "queId": "646e1cca4d7a42268f6498582cdc2862", "competition_source_list": ["1985年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "如果一个七位数的最高位不是$$0$$和$$1$$，那么这个七位数以$$4$$开头，以$$0$$结束的概率是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac 1{63}$$ "}], [{"aoVal": "B", "content": "$$\\frac 1{80}$$ "}], [{"aoVal": "C", "content": "$$\\frac 1{81}$$ "}], [{"aoVal": "D", "content": "$$\\frac 1{90}$$ "}], [{"aoVal": "E", "content": "$$\\frac 1{100}$$ "}]], "knowledge_point_routes": ["美国amc8->知识点->组合->排列组合->排列问题"], "answer_analysis": ["对于第一个数字总共有$$10-2=8$$可能性，所以数字以$$4$$开头的概率是$$\\frac{1}{8}$$对于最后一个数字总共有$$10$$种肯能性，所以数字$$0$$开头的概率是$$\\frac{1}{10}$$，因为这是独立事件，所以两种情况都发生的概率是$$\\frac{1}{8}\\cdot \\frac{1}{10}=\\frac{1}{80}$$，故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "166", "queId": "1cf259d0e9a7431f922e778ad17ce1e3", "competition_source_list": ["1993年第4届希望杯初二竞赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$x=\\frac{\\sqrt{n+1}-\\sqrt{n}}{\\sqrt{n+1}+\\sqrt{n}}$$，$$y=\\frac{\\sqrt{n+1}+\\sqrt{n}}{\\sqrt{n+1}-\\sqrt{n}}$$，（$$n$$为自然数）．如果$$2{{x}^{2}}+197xy+2{{y}^{2}}=1993$$成立，那么$$n$$的值为．", "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->数与式->二次根式->二次根式的性质与运算"], "answer_analysis": ["由题设得：$$xy=1$$，$$x+y=4n+2$$由$$2{{x}^{2}}+197xy+2{{y}^{2}}=1993$$，  得$$2{{(x+y)}^{2}}+193xy=1993$$．  将$$xy=1$$，$$x+y=4n+2$$代入上式得$${{(4n+2)}^{2}}=900$$，即  $$4n+2=30$$．  ∴$$n=7$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1410", "queId": "f3083b169e6241f2a51c55f404d29678", "competition_source_list": ["2002年第13届希望杯初二竞赛第1试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "使分式$$\\frac{-x-\\textbar x\\textbar}{{{x}^{2}}+x}$$的值为零的$$x$$的一个值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "$$-2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->绝对值->绝对值的代数意义"], "answer_analysis": ["符合题意的$$x$$满足下列条件  $$\\begin{cases}-x-\\textbar x\\textbar=0    {{x}^{2}}+x\\ne 0    \\end{cases}$$，即$$\\begin{cases}x=-\\textbar x\\textbar{}    x(x+1)\\ne 0    \\end{cases}$$．  所以$$\\begin{cases}x\\leqslant 0    x\\ne 0且x\\ne -1    \\end{cases}$$，即$$x\\textless{}0$$且$$x\\ne -1$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1364", "queId": "93b11ceada4843c98d5af50a871577e5", "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->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": "1004", "queId": "6f4b9391e3314665802a5caf380deb1e", "competition_source_list": ["1993年第4届希望杯初二竞赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "等腰三角形的某个内角的外角是$$130{}^{}\\circ $$，那么这个三角形的三个内角的大小是．", "answer_option_list": [[{"aoVal": "A", "content": "$$50{}^{}\\circ $$，$$50{}^{}\\circ $$，$$80{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$50{}^{}\\circ $$，$$50{}^{}\\circ $$，$$80{}^{}\\circ $$或$$130{}^{}\\circ $$，$$25{}^{}\\circ $$，$$25{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$50{}^{}\\circ $$，$$65{}^{}\\circ $$，$$65{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$50{}^{}\\circ $$，$$50{}^{}\\circ $$，$$80{}^{}\\circ $$或$$50{}^{}\\circ $$，$$65{}^{}\\circ $$，$$65{}^{}\\circ $$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->等腰三角形->等腰三角形的性质->等腰三角形的性质-等边对等角", "课内体系->知识点->三角形->等腰三角形->等腰三角形的性质->已知一角求其余两角", "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": ["由已知得等腰三角形的某个内角是$$50{}^{}\\circ $$．若它是底角，则三个内角是$$50{}^{}\\circ $$，$$50{}^{}\\circ $$，$$80{}^{}\\circ $$；若它是顶角，则三个内角是$$50{}^{}\\circ $$，$$65{}^{}\\circ $$，$$65{}^{}\\circ $$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "80", "queId": "61be04e5bf5f442ca143af9630c12d4e", "competition_source_list": ["2015年山东青岛黄岛区山东省青岛第九中学自主招生", "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": ["知识标签->知识点->方程与不等式->不定方程", "知识标签->题型->式->整式的乘除->幂的运算->题型：幂的综合运算", "知识标签->数学思想->分类讨论思想"], "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$$m=-1$$，  故满足等式$${{\\left( 2-m \\right)}^{{{m}^{2}}-m-2}}=1$$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": "1581", "queId": "f0543b32d9ed4057a6ce1460624f58f5", "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": ["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$$\\frac{2}{x}+\\frac{3}{y}+\\frac{4}{z}=\\frac{2}{x}+\\frac{3}{\\dfrac{5}{3}x}+\\dfrac{4}{5x}=\\frac{23}{5x}=2$$． ", "<p>由通分然后取倒数得，</p>\n<p>$$\\frac{xy+xz}{x+y+z}=2$$，$$\\frac{xy+yz}{x+y+z}=3$$，$$\\frac{zx+yz}{x+y+z}=4$$，</p>\n<p>$$\\therefore \\frac{2}{x}+\\frac{3}{y}+\\frac{4}{z}=\\frac{xy+zx}{x+y+z}\\cdot \\frac{1}{x}+\\frac{xy+yz}{x+y+z}\\cdot \\frac{1}{y}+\\frac{zx+zy}{x+y+z}\\cdot \\frac{1}{z}$$</p>\n<p>$$=\\frac{x+y+x+z+y+z}{x+y+z}=2$$．</p>"], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1179", "queId": "8aac49074e023206014e20e890056623", "competition_source_list": ["2016~2017学年湖北天门市岳口初级中学初一上学期期中第17题3分", "1994年第5届全国希望杯初一竞赛复赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$-0.1428$$中用数字$$3$$替换其中一个非$$0$$数码后，使所得的数最大，则替换的数字是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "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": ["实际上是比较$$-0.3428$$（$$3$$换$$1$$）、$$-0.1328$$（$$3$$换$$4$$）、$$-0.1438$$（$$3$$换$$2$$）、$$-0.1423$$（$$3$$换$$8$$）哪个最大，  即比较$$0.3428$$、$$0.1328$$、$$0.1438$$、$$0.1423$$哪个最小．  易知$$0.1328$$最小．  所以在$$-0.1428$$中用数字$$3$$换$$4$$，所得之数最大． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1246", "queId": "97966b5f6cfa4d16bbd30c1461ff5b3b", "competition_source_list": ["2019~2020学年5月四川成都天府新区师大一中（麓山校区）初二下学期周测C卷第5题3分", "1993年第4届希望杯初二竞赛第1题", "2019~2020学年吉林白山长白朝鲜族自治县初一上学期期中第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$a\\textless{}b\\textless{}0$$，那么在下列结论中正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a+b\\textless{}-1$$ "}], [{"aoVal": "B", "content": "$$ab\\textless{}1$$ "}], [{"aoVal": "C", "content": "$$\\frac{a}{b}\\textless{}1$$ "}], [{"aoVal": "D", "content": "$$\\frac{a}{b}\\textgreater1$$ "}]], "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": ["由$$a\\textless{}b\\textless{}0$$，知$$\\frac{a}{b}\\textgreater0$$，  又由$$a\\textless{}b$$，知$$\\textbar a\\textbar\\textgreater\\textbar b\\textbar$$，  所以得$$\\frac{a}{b}\\textgreater1$$，  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "862", "queId": "538359ec6f9b4d2cb5e7dc02d6e2d0f4", "competition_source_list": ["2016年上海浦东新区进才中学自主招生第8题10分", "2016年第33届全国全国初中数学联赛竞赛A卷第3题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果一个正整数可以表示为两个连续奇数的立方差，则称这个正整数为``和谐数''．如：$$2={{1}^{3}}-{{(-1)}^{3}}$$，$$26={{3}^{3}}-{{1}^{3}}$$，$$2$$和$$26$$均为``和谐数''．那么，不超过$$2016$$的正整数中，所有的``和谐数''之和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$6858$$ "}], [{"aoVal": "B", "content": "$$6860$$ "}], [{"aoVal": "C", "content": "$$9260$$ "}], [{"aoVal": "D", "content": "$$9262$$ "}]], "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->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->综合与实践->新定义->新定义综合其它", "课内体系->能力->运算能力"], "answer_analysis": ["方法一：$${{(2k+1)}^{3}}-{{(2k-1)}^{3}}$$  $$=\\left[ (2k+1)-(2k-1) \\right]\\left[ {{(2k+1)}^{2}}+(2k+1)(2k-1)+{{(2k-1)}^{2}} \\right]$$  $$=2(12{{k}^{2}}+1)$$（其中$$k$$为非负整数），  由$$2(12{{k}^{2}}+1)\\leqslant 2016$$得，$$k\\leqslant 9$$，  ∴$$k=0$$，$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$9$$，  即得所有不超过$$2016$$的``和谐数''，它们的和为  $$\\left[ {{1}^{3}}-{{(-1)}^{3}} \\right]+({{3}^{3}}-{{1}^{3}})+({{5}^{3}}-{{3}^{3}})+\\cdots +({{17}^{3}}-{{15}^{3}})+({{19}^{3}}-{{17}^{3}})={{19}^{3}}+1=6860$$．  方法二：``和谐数''设为$$m$$，它一定可以表示为$$m={{n}^{3}}-{{(n-2)}^{3}}$$，其中$$n$$为正整数．题目要求$$m\\leqslant 2016$$，大致计算可得$$n$$最大为$$19$$，于是有  $$2={{1}^{3}}-{{(-1)}^{3}}$$，  $$26={{3}^{3}}-{{1}^{3}}$$，  $$\\cdots $$  $$1946={{19}^{3}}-{{17}^{3}}$$，  等式右边两两相消是有  ``和谐数''总和$$={{19}^{3}}-{{(-1)}^{3}}=6860$$，选$$\\text{B}$$. "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "878", "queId": "60e0a6b0d8284298a456093ebf6b3327", "competition_source_list": ["初二其它", "2018~2019学年11月浙江杭州滨江区杭州滨兰实验学校初中部初二上学期月考第8题3分", "2008年第19届希望杯初二竞赛第1试第10题", "2016~2017学年浙江杭州拱墅区杭州市文澜中学初二上学期期末第6题3分", "2016~2017学年浙江杭州拱墅区杭州市文澜中学初二上学期期末第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "一次函数$$y=kx+b$$的图象经过点$$\\left( 0,5 \\right)$$和$$B\\left( 4,0 \\right)$$，则在该图象和坐标轴围成的三角形内，横坐标和纵坐标都是正整数的点有．", "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": ["课内体系->能力->运算能力", "课内体系->知识点->函数->一次函数->一次函数与几何综合->一次函数与整点问题"], "answer_analysis": ["把点$$\\left( 0, 5 \\right)$$和$$B\\left( 4,0 \\right)$$代入$$y=kx+b$$得$$y=-\\dfrac{5}{4}x+5$$．  当$$x=1$$时，$$y=\\dfrac{15}{4}$$，所以横坐标和纵坐标都是正整数的点有$$\\left( 1,1 \\right)$$，$$\\left( 1,2 \\right)$$，$$\\left( 1,3 \\right)$$；  当$$x=2$$时，$$y=\\dfrac{5}{2}$$，所以横坐标和纵坐标都是正整数的点有$$\\left( 2,1 \\right)$$，$$\\left( 2,2 \\right)$$；  当$$x=3$$时，$$y=\\dfrac{5}{4}$$，所以横坐标和纵坐标都是正整数的点有$$\\left( 3, 1 \\right)$$．  综上，共有$$6$$个． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "95", "queId": "53d4e597e0624f2da2803c996e09ed80", "competition_source_list": ["2015年第26届全国希望杯初二竞赛复赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$x=\\frac{1}{\\sqrt{3}-\\sqrt{2}}$$，则$${{x}^{6}}-2\\sqrt{2}{{x}^{5}}-{{x}^{4}}+{{x}^{3}}-2\\sqrt{3}{{x}^{2}}+2x-\\sqrt{2}$$的值是  ．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$-\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{3}$$ "}]], "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->二次根式直接化简求值"], "answer_analysis": ["$$x=\\frac{1}{\\sqrt{3}-\\sqrt{2}}=\\sqrt{3}+\\sqrt{2}$$，则$${{x}^{2}}-2\\sqrt{2}x+2=3$$，$${{x}^{2}}-2\\sqrt{3}x+3=2$$，  所求式$$={{x}^{4}}({{x}^{2}}-2\\sqrt{2}x-1)+x({{x}^{2}}-2\\sqrt{3}x+1)+x-\\sqrt{2}=x-\\sqrt{2}=\\sqrt{3}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "855", "queId": "6e20767be63a49d6a216d1ebf0eda47a", "competition_source_list": ["2002年第13届希望杯初一竞赛第1试第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$2a+b=0$$，则$$\\left\\textbar{} \\frac{a}{\\left\\textbar{} b \\right\\textbar}-1 \\right\\textbar+\\left\\textbar{} \\frac{\\left\\textbar{} a \\right\\textbar}{b}-2 \\right\\textbar$$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->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$$2a+b=0$$得$$b=-2a$$，进而有$$\\frac{a}{\\left\\textbar{} b \\right\\textbar}=\\frac{a}{\\left\\textbar{} -2a \\right\\textbar}=\\frac{a}{\\left\\textbar{} -2 \\right\\textbar\\cdot \\left\\textbar{} a \\right\\textbar}=\\frac{1}{2}\\cdot \\frac{a}{\\left\\textbar{} a \\right\\textbar}$$，$$\\frac{\\left\\textbar{} a \\right\\textbar}{b}=\\frac{\\left\\textbar{} a \\right\\textbar}{-2a}=-\\frac{1}{2}\\cdot \\frac{\\left\\textbar{} a \\right\\textbar}{a}$$  若$$a\\textgreater0$$，则$$\\left\\textbar{} \\frac{a}{\\left\\textbar{} b \\right\\textbar}-1 \\right\\textbar+\\left\\textbar{} \\frac{\\left\\textbar{} a \\right\\textbar}{b}-2 \\right\\textbar=\\left\\textbar{} \\frac{1}{2}-1 \\right\\textbar+\\left\\textbar{} -\\frac{1}{2}-2 \\right\\textbar=3$$，  若$$a \\textless{} 0$$，则$$\\left\\textbar{} \\frac{a}{\\left\\textbar{} b \\right\\textbar}-1 \\right\\textbar+\\left\\textbar{} \\frac{\\left\\textbar{} a \\right\\textbar}{b}-2 \\right\\textbar=\\left\\textbar{} -\\frac{1}{2}-1 \\right\\textbar+\\left\\textbar{} \\frac{1}{2}-2 \\right\\textbar=3$$ ． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "612", "queId": "e381a41bfbcb4d6490eeea75da66ea51", "competition_source_list": ["2013年第24届全国希望杯初一竞赛初赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，$$\\angle A+\\angle C=2\\angle B$$，$$2\\angle A+\\angle B=2\\angle C$$，则$$\\triangle ABC$$是．", "answer_option_list": [[{"aoVal": "A", "content": "锐角且不等边三角形 "}], [{"aoVal": "B", "content": "直角三角形 "}], [{"aoVal": "C", "content": "钝角三角形 "}], [{"aoVal": "D", "content": "等边三角形 "}]], "knowledge_point_routes": ["知识标签->题型->三角形->三角形及多边形->三角形的基础->题型：三角形的分类", "知识标签->知识点->三角形->三角形及多边形->与三角形有关的角->三角形的内角、内角和", "知识标签->知识点->三角形->三角形及多边形->三角形的概念及分类"], "answer_analysis": ["∵$$\\angle A+\\angle C=2\\angle B$$，  ∴$$\\angle A+\\angle B+\\angle C=3\\angle B$$．  ∵$$\\angle A+\\angle B+\\angle C=180{}^{}\\circ $$，  ∴$$\\angle B=60{}^{}\\circ $$．  ∵$$2\\angle A+\\angle B=2\\angle C$$，  ∴$$2\\angle A+60{}^{}\\circ =2\\angle C$$，  ∴$$2\\angle A+60{}^{}\\circ =2(120{}^{}\\circ -\\angle A)$$，  解得$$\\angle A=45{}^{}\\circ $$，  ∴$$\\angle C=180{}^{}\\circ -45{}^{}\\circ -60{}^{}\\circ =75{}^{}\\circ $$，  ∴$$\\triangle ABC$$是锐角且不等边三角形． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "755", "queId": "5b629433ed0841cdb3e5d03cbe318480", "competition_source_list": ["2016年全国全国初中数学联赛初一竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$为实数，关于$$x$$，$$y$$的方程组$$\\left { \\begin{array}{*{35}{l}} ax+2y=24    2x+2y=a   \\end{array} \\right.$$有整数解，则$$a$$的个数为（~ ）", "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->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）->含参二元一次方程组->二元一次方程组的含参整数解问题", "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": ["由$$\\left { \\begin{array}{*{35}{l}} ax+2y=24    2x+2y=a   \\end{array} \\right.$$得$$\\left( a-2 \\right)x=24-a$$，∴$$x=\\frac{24-a}{a-2}=-1+\\frac{22}{a-2}$$；  由$$2x+2y=a$$，可知$$a$$必为偶数，又$$-1+\\frac{22}{a-2}$$为整数，所以$$a=0,4,24,-20$$ "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1466", "queId": "ca7d1c1a0ef640c4b9600775d412a0a4", "competition_source_list": ["2001年第12届希望杯初二竞赛第1试第5题"], "difficulty": "0", "qtype": "single_choice", "problem": "已知点$$D$$在线段$$EF$$上，下列四个等式：①$$DE=2DF$$，②$$DE=\\frac{1}{3}EF$$，③$$EF=2DF$$，④$$DF=\\frac{1}{2}DE$$，其中能表示：点$$D$$是线段$$EF$$的一个三等分点的表达式是．", "answer_option_list": [[{"aoVal": "A", "content": "①②③ "}], [{"aoVal": "B", "content": "②③④ "}], [{"aoVal": "C", "content": "①②④ "}], [{"aoVal": "D", "content": "①③④ "}]], "knowledge_point_routes": ["竞赛->知识点->几何图形初步->线"], "answer_analysis": ["点$$D$$在线段$$EF$$上．  ①若$$DE=2DF$$，则$$D$$是线段$$EF$$的一个三等分点．  所以①正确．  ②若$$DE=\\frac{1}{3}EF$$，同样点$$D$$是线段$$EF$$的一个三等分点．  所以②正确．  ④若$$DF=\\frac{1}{2}DE$$，点$$D$$也是线段$$EF$$的一个三等分点．  所以④正确．  综上可知①②④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": "1000", "queId": "b6979c93f01f4fc98b4d90db11420609", "competition_source_list": ["2012年第23届全国希望杯初一竞赛初赛第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "计算：$$1+{{(-2)}^{2}}-\\frac{-4\\times {{(-1)}^{2}}}{4}=$$（~ ~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"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", "课内体系->能力->运算能力"], "answer_analysis": ["$$1+{{(-2)}^{2}}-\\frac{-4\\times {{(-1)}^{2}}}{4}$$  $$=1+4+1$$  $$=6$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1539", "queId": "bdb887ab87bb49eaa60eeeb00897fe14", "competition_source_list": ["2008年第19届希望杯初一竞赛第2试第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "``嫦娥一号''卫星在未打开太阳翼时，外形是长$$222$$厘米、宽$$172$$厘米、高$$220$$厘米的长方体．若在表面包裹$$1$$厘米厚的防震材料层，在这外面还有$$1$$厘米厚的木板包装箱．则木板包装箱所需木材的体积至少是．", "answer_option_list": [[{"aoVal": "A", "content": "$$224\\times 174\\times 222-222\\times 172\\times 220$$立方厘米 "}], [{"aoVal": "B", "content": "$$223\\times 173\\times 221-221\\times 171\\times 219$$立方厘米 "}], [{"aoVal": "C", "content": "$$225\\times 175\\times 223-224\\times 174\\times 222$$立方厘米 "}], [{"aoVal": "D", "content": "$$226\\times 176\\times 224-224\\times 174\\times 222$$立方厘米 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数与实际问题"], "answer_analysis": ["因包装箱最外边的尺寸每个方向向外增加$$1$$厘米，长，宽，高就各增加$$2$$厘米，整个体积减去包裹了防震材料层的体积即是最少的木材用料．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1550", "queId": "dd7e0cf4e9b94c198d7f8ddf1b541ad3", "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$$\\left\\textbar{} a \\right\\textbar$$，$$-a$$，$${{a}^{2009}}$$，$${{a}^{2010}}$$，$$\\left\\textbar{} -a \\right\\textbar$$，$$\\left( \\frac{{{a}^{2}}}{a}+a \\right)$$，$$\\left( \\frac{{{a}^{2}}}{a}-a \\right)$$中负数个数．", "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->数->有理数->数轴与有理数有关的概念->题型：区分正负数", "知识标签->知识点->数->有理数->正数和负数->正数和负数的定义"], "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": "618", "queId": "3b4269c06d594f60ae810ed5ef026dce", "competition_source_list": ["2017~2018学年江苏南通崇川区南通市启秀中学初一下学期期中第8题3分", "1997年第8届希望杯初二竞赛第1试第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$m$$、$$n$$是整数，$$3m+2=5n+3$$，且$$3m+2\\textgreater30$$，$$5n+3\\textless{}40$$，则$$mn$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$70$$ "}], [{"aoVal": "B", "content": "$$72$$ "}], [{"aoVal": "C", "content": "$$77$$ "}], [{"aoVal": "D", "content": "$$84$$ "}]], "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）->一元一次不等式组->一元一次不等式组的整数解", "课内体系->能力->运算能力"], "answer_analysis": ["由题意得：$$\\begin{cases}3m+2\\textgreater30    3m+2\\textless{}40    \\end{cases}$$，  解得：$$\\frac{28}{3}\\textless{}m\\textless{}\\frac{38}{3}$$，  ∵$$m$$是整数，  ∴$$m=10$$或$$11$$或$$12$$，  由$$\\begin{cases}5n+3\\textgreater30    5n+3\\textless{}40    \\end{cases}$$，解得：$$\\frac{27}{5}\\textless{}n\\textless{}\\frac{37}{5}$$，  ∵$$n$$是整数，  ∴$$n=6$$或$$7$$，  根据$$3m+2=5n+3$$成立时，$$m=12$$，$$n=7$$，  则$$mn=12\\times 7=84$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "527", "queId": "70dc2ee6710b42e2bd8c12654fb48ac7", "competition_source_list": ["2016年浙江宁波鄞州区初三中考一模第6题4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1题4分", "2017~2018学年陕西宝鸡凤翔县初三上学期期末第4题3分", "2018~2019学年浙江宁波鄞州区宁波兴宁中学初三上学期期中第8题3分", "2019年陕西西安鄠邑区初三中考一模第4题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "从长度分别为$$2,4,6,7$$的四条线段中随机取三条，能构成三角形的概率是(  )", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{4}$$ "}]], "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->统计与概率->概率->概率的计算方法->列举法求概率"], "answer_analysis": ["共有$$4$$种等可能的结果数，它们为$$2$$、$$4$$、$$6,2$$、$$4$$、$$7,2$$、$$6$$、$$7,4$$、$$6$$、$$7$$，其中能构成三角形的结果数为$$2$$种，  所以能构成三角形的概率$$=\\frac{2}{4}=\\frac{1}{2}$$．  故选$$\\rm C$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1365", "queId": "9c96f4c0363047718cc19e9a96345db1", "competition_source_list": ["1999年第10届希望杯初二竞赛第1试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "某工厂到车站的路程为$$m$$公里，现有一辆汽车从工厂到车站拉货，去时的速度为$$3a$$公里$$/$$小时，返回时的速度为$$2a$$公里$$/$$小时，那么这辆车往返一次的平均速度为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{2}a$$公里$$/$$小时 "}], [{"aoVal": "B", "content": "$$\\frac{2}{5}ma$$公里$$/$$小时 "}], [{"aoVal": "C", "content": "$$\\frac{7}{3}a$$公里$$/$$小时 "}], [{"aoVal": "D", "content": "$$\\frac{12}{5}a$$公里$$/$$小时 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->分式->分式的基本运算"], "answer_analysis": ["该车从工厂到车站所用时间$${{t}_{1}}=\\frac{m}{3a}$$（小时），  从车站返回工厂所用时间为$${{t}_{2}}=\\frac{m}{2a}$$（小时）．  所以往返一次的平均速度为$$v=\\frac{2m}{{{t}_{1}}+{{t}_{2}}}=\\frac{2m}{\\dfrac{m}{3a}+\\dfrac{m}{2a}}=\\frac{12}{5}a$$（公里$$/$$小时）．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "441", "queId": "42e5d61da2a14133a26dd618bd5ff19a", "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$$\\left\\textbar{} \\left\\textbar{} x-1 \\right\\textbar-\\left\\textbar{} x \\right\\textbar{} \\right\\textbar-\\left\\textbar{} x-1 \\right\\textbar+\\left\\textbar{} x \\right\\textbar=1$$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": ["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": "338", "queId": "1a81714631974f0694f6e413dc38ca39", "competition_source_list": ["1993年第4届希望杯初二竞赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列各组数可以成为三角形的三边长度的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$，$$2$$，$$3$$． "}], [{"aoVal": "B", "content": "$$a+1$$，$$a+2$$，$$a+3$$，并且$$a\\textgreater0$$ "}], [{"aoVal": "C", "content": "$$a$$，$$b$$，$$c$$，并且$$a+b\\textgreater c$$． "}], [{"aoVal": "D", "content": "$$1$$，$$m$$，$$n$$，并且$$1-m\\textless{}n$$． "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["$$\\text{A}$$选项：由$$(a+1)+(a+2)=2a+3\\textgreater a+3$$（因为$$a\\textgreater0$$)，所以$$a+1$$，$$a+2$$，$$a+3$$可以成为三角形的三边，而$$1+2=3$$，故排除$$\\text{A}$$选项；  $$\\text{C}$$选项：$$1+3\\textgreater2$$，$$1$$、$$2$$、$$3$$不能构成三角形，故排除$$\\text{C}$$选项；  $$\\text{D}$$选项：$$1-2\\textless{}1$$，$$1$$、$$1$$、$$2$$不能构成三角形，故排除$$\\text{D}$$选项；  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "917", "queId": "e8c65d83d1d74d32b3cc2e2eacfa22a9", "competition_source_list": ["1996年第13届全国初中数学联赛竞赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "实数$$a$$，$$b$$满足$$ab=1$$，记$$M=\\frac{1}{1+a}+\\frac{1}{1+b}$$，$$N=\\frac{a}{1+a}+\\frac{b}{1+b}$$，则$$M$$与$$N$$的关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$M\\textgreater N$$ "}], [{"aoVal": "B", "content": "$$M=N$$ "}], [{"aoVal": "C", "content": "$$M\\textless{}N$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的运算->分式加法运算"], "answer_analysis": ["【方法一】将$$a=\\frac{1}{b}$$分别代入$$M$$和$$N$$，  则$$M=\\frac{1}{1+\\frac{1}{b}}+\\frac{1}{1+b}=\\frac{b}{1+b}+\\frac{1}{1+b}=1$$，  $$N=\\frac{\\frac{1}{b}}{1+\\frac{1}{b}}+\\frac{b}{1+b}=\\frac{1}{1+b}+\\frac{b}{1+b}=1$$，  所以$$M=N$$．选$$\\text{B}$$，  【方法二】$$M=\\frac{1}{1+a}+\\frac{1}{1+b}=\\frac{b}{b+ab}+\\frac{a}{a+ab}$$，  又由$$ab=1$$，  得到$$M=\\frac{b}{b+1}+\\frac{a}{a+1}=N$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1473", "queId": "eed2d19bb519497ca1900194c1c380ae", "competition_source_list": ["2012年第23届全国希望杯初二竞赛复赛第9题4分", "初一单元测试《有条件的分式化解与求值》第23题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$，$$b$$是实数，且$$\\frac{1}{1+a}-\\frac{1}{1+b}=\\frac{1}{b-a}$$，则$$\\frac{1+b}{1+a}+\\frac{1+a}{1+b}$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$-3$$ "}], [{"aoVal": "C", "content": "$$3(b-a)$$ "}], [{"aoVal": "D", "content": "无法确定的 "}]], "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->思想->整体思想"], "answer_analysis": ["因为$$\\frac{1}{1+a}-\\frac{1}{1+b}=\\frac{1}{b-a}$$，  所以$$\\frac{1}{1+a}=\\frac{1}{b-a}+\\frac{1}{1+b}$$，  即$$\\frac{1+b}{1+a}=\\frac{1+b}{b-a}+1$$．  同理$$\\frac{1}{1+b}=\\frac{1}{1+a}-\\frac{1}{b-a}$$，  即$$\\frac{1+a}{1+b}=1-\\frac{1+a}{b-a}$$．  两式相加，得$$\\frac{1+b}{1+a}+\\frac{1+a}{1+b}=\\frac{1+b}{b-a}+1+1-\\frac{1+a}{b-a}=3$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "183", "queId": "0cd28ea4436d4ba9b830184af1640f9c", "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12题3分", "2019~2020学年四川宜宾叙州区初一上学期期中", "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12题3分", "2019~2020学年4月辽宁沈阳铁西区沈阳雨田实验中学初一下学期月考第5题3分", "2019~2020学年四川宜宾叙州区四川省宜宾县第二中学校初二上学期期末模拟（一）第10题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "我们规定这样一种运算：如果$${{a}^{b}}=N(a\\textgreater0,N\\textgreater0)$$，那么$$b$$就叫做以$$a$$为底$$N$$的对数，记作$$b={{\\log }_{a}}{N}$$．例如：因为$${{2}^{3}}=8$$，所以$${{\\log }_{2}}{8}=3$$，那么$${{\\log }_{3}}{81}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$27$$ "}], [{"aoVal": "D", "content": "$$81$$ "}]], "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->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->抽象概括能力"], "answer_analysis": ["∵$${{3}^{4}}=81$$，  ∴$${{\\log }_{3}}81=4$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "799", "queId": "4a295e2a39b14943b3f5bc83215524df", "competition_source_list": ["1994年第5届希望杯初二竞赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知关于$$x$$的二次方程$$2{{x}^{2}}+ax-2a+1=0$$的两个实根的平方和为$$7\\frac{1}{4}$$，则$$a$$的值．", "answer_option_list": [[{"aoVal": "A", "content": "$$-11$$或$$3$$ "}], [{"aoVal": "B", "content": "$$-11$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "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->方程与不等式->二次方程->一元二次方程根的判别式"], "answer_analysis": ["设方程的两个实根为$${{x}_{1}}$$，$${{x}_{2}}$$，  则$$\\begin{cases}\\Delta ={{a}^{2}}-8\\left( -2a+1 \\right)={{a}^{2}}+16a-8\\geqslant 0①    x_{1}^{2}+x_{2}^{2}={{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-2{{x}_{1}}{{x}_{2}}                              ={{\\left( -\\dfrac{a}{2} \\right)}^{2}}-2\\cdot \\left( \\dfrac{-2a+1}{2} \\right)=\\dfrac{29}{4}②    \\end{cases}$$，  整理②式得，$${{a}^{2}}+8a-33=0$$，  解得$$a=3$$或$$a=-11$$．  将$$a=3$$代入①式得$${{3}^{2}}+16\\times 3-8\\textgreater0$$．  将$$a=-11$$代入①式得$${{\\left( -11 \\right)}^{2}}+16\\times \\left( -11 \\right)-8\\textless{}0$$矛盾．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "576", "queId": "6c74d180f1344c3a99d4c15ec064c8af", "competition_source_list": ["2018年浙江宁波余姚市余姚市实验学校初二竞赛第7题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知由小到大的$$10$$个正整数$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$$\\cdots \\cdots $$，$${{a}_{10}}$$的和是$$2018$$（$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$$\\cdots \\cdots $$，$${{a}_{10}}$$中任何两个数都不相等），那么$${{a}_{5}}$$的最大值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$329$$ "}], [{"aoVal": "B", "content": "$$330$$ "}], [{"aoVal": "C", "content": "$$331$$ "}], [{"aoVal": "D", "content": "$$332$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->整除的概念与基本性质"], "answer_analysis": ["∵$${{a}_{5}}$$最大，就要$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$${{a}_{4}}$$最小，还要$${{a}_{6}}$$，$${{a}_{7}}$$，$${{a}_{8}}$$，$${{a}_{9}}$$，$${{a}_{10}}$$尽可能接近$${{a}_{5}}$$，  设$${{a}_{1}}=1$$，$${{a}_{2}}=2$$，$${{a}_{3}}=3$$，$${{a}_{4}}=4$$，  即剩下$$6$$个数的和为$$2018-\\left( 1+2+3+4 \\right)=2008$$，  设$${{a}_{6}}={{a}_{5}}+1$$，$${{a}_{7}}={{a}_{5}}+2$$，$${{a}_{8}}={{a}_{5}}+3$$，$${{a}_{9}}={{a}_{5}}+4$$，$${{a}_{10}}={{a}_{5}}+5$$，  ∴$${{a}_{5}}+{{a}_{5}}+1+{{a}_{5}}+2+{{a}_{5}}+3+{{a}_{5}}+4+{{a}_{5}}+5=2008$$，  ∴$$6{{a}_{5}}+15=2008$$，  ∴$$6{{a}_{5}}=1993$$，  ∵$$1993$$无法整除$$6$$，  ∵$$1992$$可以整除$$6$$，  ∴令$${{a}_{4}}=5$$，  即$$6{{a}_{5}}=1992$$，  ∴$${{a}_{5}}=332$$，  即当$${{a}_{1}}=1$$，$${{a}_{2}}=2$$，$${{a}_{3}}=3$$，$${{a}_{4}}=5$$时，  $${{a}_{5}}$$取得最大值为$$332$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1329", "queId": "8aac50a751148307015114dbd56403bd", "competition_source_list": ["2015年第26届全国希望杯初二竞赛复赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$${{a}^{2}}={{b}^{4}}\\ne 0$$，则$$\\frac{a}{{{b}^{2}}}$$的值（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "$$1$$或$$-1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->平方根->利用平方根解方程", "课内体系->知识点->式->二次根式->二次根式的基础->二次根式的定义"], "answer_analysis": ["$${{a}^{2}}={{b}^{4}}\\ne 0$$，则$$a={{b}^{2}}$$或$$-a={{b}^{2}}$$，∴$$\\frac{a}{{{b}^{2}}}=1$$或$$-1$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1135", "queId": "78909f5f86e14ff192cfcc58f1e67c78", "competition_source_list": ["2017年第1届重庆全国初中数学联赛竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "化简$$\\frac{b-c}{\\left( a-b \\right)\\left( a-c \\right)}+\\frac{c-a}{\\left( b-c \\right)\\left( b-a \\right)}+\\frac{a-b}{\\left( c-a \\right)\\left( c-b \\right)}+\\frac{2}{b-a}-\\frac{2}{c-a}$$", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2}{a-c}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{b-c}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{b-c}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{b-c}$$ "}]], "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-降次化简求值"], "answer_analysis": ["原式$$=\\frac{1}{a-b}-\\frac{1}{a-c}+\\frac{1}{b-c}-\\frac{1}{b-a}+\\frac{1}{c-a}-\\frac{1}{c-b}+\\frac{2}{b-a}-\\frac{2}{c-a}$$ $$=\\frac{2}{b-c}$$ "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "722", "queId": "4092e876da5a44c082f67b84ceca0fe0", "competition_source_list": ["初一上学期单元测试《一元一次方程》第20题", "2001年第12届希望杯初一竞赛第10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": "1559", "queId": "e21e42a124d845e49a989cf72df8a32f", "competition_source_list": ["2017年第1届重庆全国初中数学联赛竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "化简$$\\frac{b-c}{\\left( a-b \\right)\\left( a-c \\right)}+\\frac{c-a}{\\left( b-c \\right)\\left( b-a \\right)}+\\frac{a-b}{\\left( c-a \\right)\\left( c-b \\right)}+\\frac{2}{b-a}-\\frac{2}{c-a}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2}{a-c}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{b-c}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{b-c}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{b-c}$$ "}]], "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-降次化简求值"], "answer_analysis": ["原式$$=\\frac{1}{a-b}-\\frac{1}{a-c}+\\frac{1}{b-c}-\\frac{1}{b-a}+\\frac{1}{c-a}-\\frac{1}{c-b}+\\frac{2}{b-a}-\\frac{2}{c-a}$$ $$=\\frac{2}{b-c}$$ "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "977", "queId": "6aab8d0fc4cc4402b5493adc969e1ec6", "competition_source_list": ["2019~2020学年浙江宁波鄞州区宁波市鄞州蓝青学校初一上学期期末第6题", "初一上学期单元测试《物以类聚》第19题", "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2题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果对于某一特定范围内的$$x$$的任一允许值，$$p=\\left\\textbar{} 1-2x \\right\\textbar+\\left\\textbar{} 1-3x \\right\\textbar+\\cdots +\\left\\textbar{} 1-9x \\right\\textbar+\\left\\textbar{} 1-10x \\right\\textbar$$为定值，则此定值为．", "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->知识点->数->有理数->绝对值->零点分段法", "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->能力->运算能力"], "answer_analysis": ["∵$$P$$为定值，  ∴$$P$$的表达式化简后与$$x$$无关，  即$$x$$的系数为$$0$$，  观察发现：$$2+3+4+5+6+7=8+9+10$$，  ∴只需$$1-7x\\geqslant 0$$且$$1-8x\\leqslant 0$$，  ∴$$\\frac{1}{8}\\leqslant x\\leqslant \\frac{1}{7}$$，  ∴$$P=(1-2x)+(1-3x)+(1-4x)+(1-5x)+(1-6x)+(1-7x)-(1-8x)-(1-9x)-(1-10x)$$  $$=3$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "739", "queId": "49807abb4fb64022ae52bb8599e455a9", "competition_source_list": ["2012年竞赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$a$$，$$b$$为给定的实数，且$$1 ~\\textless{} ~a ~\\textless{} ~b$$，那么$$1$$，$$a+1$$，$$2a+b$$，$$a+b+1$$这四个数据的平均数与中位数之差的绝对值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\frac{2a-1}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->数据的分析->中位数", "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", "课内体系->能力->数据处理能力"], "answer_analysis": ["由绝对值的知识点可知$$1$$，$$a+1$$，$$2a+b$$，$$a+b+1$$这四个数据的平均数与中位数之差的绝对值是$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "24", "queId": "12aba4c8474448748c83b40a21896cf2", "competition_source_list": ["2002年第13届希望杯初二竞赛第1试第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "若三角形的三个内角$$A$$，$$B$$、$$C$$的关系满足$$A\\textgreater3B$$，$$C\\textless2B$$，那么这个三角形是．", "answer_option_list": [[{"aoVal": "A", "content": "钝角三角形 "}], [{"aoVal": "B", "content": "直角三角形 "}], [{"aoVal": "C", "content": "等边三角形 "}], [{"aoVal": "D", "content": "不等边的锐角三角形 "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["∵$$A\\textgreater3B$$，$$C\\textless2B$$ ，$$2B\\textless3B$$ ．  ∴$$A\\textgreater B$$，$$A\\textgreater C$$，即$$A$$为最大的内角．  若:$$A≤90°$$，则$$B \\textless{} 30°$$，$$C \\textless{} 60°$$．  $$A+B+C ~\\textless{} ~90{}^{}\\circ +30{}^{}\\circ +60{}^{}\\circ =180{}^{}\\circ $$与$$A+B+C=180{}^{}\\circ $$矛盾．  故：可排除$$\\text{B}$$，$$\\text{C}$$，$$\\text{D}$$，选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "853", "queId": "d6364fa140ac46ed83f742a96572528c", "competition_source_list": ["1999年第10届希望杯初二竞赛第1试第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "代数式$$\\frac{\\textbar\\pi-3 \\cdot 1416\\textbar}{3 \\cdot 1415-\\pi}$$的值．", "answer_option_list": [[{"aoVal": "A", "content": "是零 "}], [{"aoVal": "B", "content": "在$$0$$与$$1$$之间 "}], [{"aoVal": "C", "content": "在$$-1$$与$$0$$之间 "}], [{"aoVal": "D", "content": "等于$$1$$或$$-1$$ "}]], "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$$\\pi=3.141592 \\cdots\\textless3.1416$$，  所以$$\\frac{\\textbar\\pi-3.1416\\textbar}{3.1415-\\pi}$$  $$=\\frac{3.1416-\\pi}{3.1415-\\pi}$$  $$=1+\\frac{0.0001}{3.1415-\\pi}$$  $$=1-\\frac{0.0001}{0.00009 \\cdots}$$  $$\\approx-\\frac{1}{9}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1193", "queId": "974db426be43405b969a2b9fdb083692", "competition_source_list": ["2011年第22届全国希望杯初二竞赛初赛第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若关于$$x$$，$$y$$的方程组$$\\begin{cases}x+ay+1=0    bx-2y+a=0 \\end{cases}$$无实数解，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$ab=-2$$ "}], [{"aoVal": "B", "content": "$$ab=-2$$且$$a\\ne 1$$ "}], [{"aoVal": "C", "content": "$$ab\\ne -2$$ "}], [{"aoVal": "D", "content": "$$ab=-2$$且$$a\\ne 2$$ "}]], "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", "知识标签->题型->方程与不等式->二元一次方程（组）->含参二元一次方程组->题型：二元一次方程组解的情况", "知识标签->学习能力->运算能力"], "answer_analysis": ["题设方程组没有实数解得条件是$$\\frac{1}{b}=\\frac{a}{-2}\\ne \\frac{1}{a}$$，  即$$\\begin{cases}ab=-2    {{a}^{2}}\\ne-2    a\\ne b \\end{cases}$$．  当$$ab=-2$$时，$$ab\\ne 0$$，$$a$$，$$b$$异号，则$$a\\ne b$$，  又$${{a}^{2}}\\ne -2$$在实数范围内成立，  所以$$ab=-2$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "223", "queId": "4f914d930f3c4b3aacc1cbda569bd381", "competition_source_list": ["2013年第24届全国希望杯初二竞赛复赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知直线$$y=kx+b$$（$$k\\ne 0$$）与$$x$$轴的交点在$$x$$轴的正半轴上，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$k\\textgreater0$$，$$b\\textgreater0$$ "}], [{"aoVal": "B", "content": "$$k\\textless{}0$$，$$b\\textless{}0$$ "}], [{"aoVal": "C", "content": "$$kb\\textgreater0$$ "}], [{"aoVal": "D", "content": "$$kb\\textless{}0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->一次函数->一次函数基础"], "answer_analysis": ["令$$y=0$$，$$x=-\\frac{b}{k}\\textgreater0$$，  所以$$\\frac{b}{k}\\textless{}0$$，即$$kb\\textless{}0$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "825", "queId": "accac0b2f1ed468b899cdb1b95c7af53", "competition_source_list": ["2011年竞赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "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->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->方法->代入法", "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": "424", "queId": "4bd4baf1ede54b679c655d2928e1cfb7", "competition_source_list": ["2012年第23届全国希望杯初一竞赛初赛第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "计算：$$1+{{(-2)}^{2}}-\\frac{-4\\times {{(-1)}^{2}}}{4}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["知识标签->知识点->数->有理数->有理数的运算->有理数的混合运算", "知识标签->题型->数->有理数->有理数基础运算->题型：有理数四则综合运算"], "answer_analysis": ["解：原式=$$1+4-\\frac{-4\\times {1}}{4}$$  $$=1+4+1$$  $$=6$$ "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "887", "queId": "72dccbfb3d8b4b8f999417448c27a050", "competition_source_list": ["2016年第33届全国全国初中数学联赛竞赛B卷第3题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知二次函数$$y=a{{x}^{2}}+bx+1$$（$$a\\ne 0$$）的图象的顶点在第二象限，且过点$$(1,0)$$．当$$a-b$$为整数时，$$ab=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$-\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$-2$$ "}]], "knowledge_point_routes": ["知识标签->知识点->函数->二次函数->二次函数的图象及性质->二次函数的图象和系数的关系", "知识标签->题型->函数->二次函数->二次函数图象与性质->图象与系数的关系"], "answer_analysis": ["依题意知$$a\\textless{}0$$，$$-\\frac{b}{2a}\\textless{}0$$，$$a+b+1=0$$，  故$$b\\textless{}0$$且$$b=-a-1$$，  ∴$$a-b=a-(-a-1)=2a+1$$，  于是$$-1\\textless{}a\\textless{}0$$，  ∴$$-1\\textless{}2a+1\\textless{}1$$，  又$$a-b$$为整数，  ∴$$2a+1=0$$，  故$$a=-\\frac{1}{2}=b$$，  ∴$$ab=\\frac{1}{4}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1369", "queId": "a120770057c14180ac3370fe1e504dce", "competition_source_list": ["2000年第11届希望杯初二竞赛第2试第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个三角形的三边长分别是$$a$$，$$b$$，$$c$$（$$a$$，$$b$$，$$c$$都是质数），且$$a+b+c=16$$，则这个三角形是（ ~ ~）．", "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": "994", "queId": "ff8080814d043c29014d0a5c42d12d9c", "competition_source_list": ["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", "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", "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->能力->运算能力", "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"], "answer_analysis": ["由条件得$$a\\geqslant 3$$，原等式为$$\\left\\textbar{} b+2 \\right\\textbar+\\sqrt{\\left( a-3 \\right){{b}^{2}}}=0$$，$$a+b=1$$． ", "<p>由题意$$(a-3){{b}^{2}}\\geqslant 0$$，∵$${{b}^{2}}\\geqslant 0$$，&there4;$$a-3\\geqslant 0$$，&there4;$$2a-4&gt;0$$，</p>\n<p>&there4;原式可化简为$$2a-4+|b+2|+\\sqrt{(a-3){{b}^{2}}}+4=2a$$，$$|b+2|+\\sqrt{(a-3){{b}^{2}}}=0$$，</p>\n<p>&there4;$$b+2=(a-3){{b}^{2}}=0$$，&there4;$$a=3$$，$$b=-2$$，选$$\\text{C}$$．</p>"], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "814", "queId": "bf1d42d150c1445794a2c2d3d3c328b7", "competition_source_list": ["2017年湖南长沙天心区初二竞赛长郡教育集团抵达杯第3题5分", "2004年第21届全国初中数学联赛竞赛第1题", "初一下学期其它", "2018~2019学年湖北襄阳市樊城区襄阳市第五中学初三上学期期末（培优班）第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$abc\\ne 0$$，且$$a+b+c=0$$，则代数式$$\\frac{{{a}^{2}}}{bc}+\\frac{{{b}^{2}}}{ca}+\\frac{{{c}^{2}}}{ab}$$的值是．", "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->式->分式->分式的运算->分式加减、乘除、乘方混合运算"], "answer_analysis": ["方法一：$$\\frac{{{a}^{2}}}{bc}+\\frac{{{b}^{2}}}{ca}+\\frac{{{c}^{2}}}{ab}$$  $$=\\frac{{{\\left( b+c \\right)}^{2}}}{bc}+\\frac{{{\\left( a+c \\right)}^{2}}}{ac}+\\frac{{{\\left( a+b \\right)}^{2}}}{ab}$$  $$=\\frac{b}{c}+2+\\frac{c}{b}+\\frac{a}{c}+2+\\frac{c}{a}+\\frac{a}{b}+2+\\frac{b}{a}$$  $$=\\frac{a+b}{c}+\\frac{a+c}{b}+\\frac{b+c}{a}+6$$  $$=\\frac{a+b+c}{a}+\\frac{a+b+c}{b}+\\frac{a+b+c}{c}-1-1-1+6$$  $$=3$$．  方法二：把$$a=-\\left( b+c \\right)$$，$$b=-\\left( a+c \\right)$$，$$c=-\\left( a+b \\right)$$代入，  原式$$=\\frac{-\\left( b+c \\right)\\cdot a}{ab}+\\frac{-\\left( a+c \\right)\\cdot b}{ac}+\\frac{-\\left( a+b \\right)\\cdot c}{ab}$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=-\\left( \\frac{ba+ca}{bc} \\right)-\\left( \\frac{ab+cb}{ac} \\right)-\\left( \\frac{ac+bc}{ab} \\right)$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=-\\left( \\frac{a}{b}+\\frac{a}{c} \\right)-\\left( \\frac{b}{a}+\\frac{b}{c} \\right)-\\left( \\frac{c}{a}+\\frac{c}{b} \\right)$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\frac{-\\left( b+c \\right)}{a}+\\frac{-\\left( a+c \\right)}{b}+\\frac{-\\left( a+b \\right)}{c}$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\frac{a}{a}+\\frac{b}{b}+\\frac{c}{c}$$  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=3$$，  方法三：$$\\frac{{{a}^{2}}}{bc}+\\frac{{{b}^{2}}}{ca}+\\frac{{{c}^{2}}}{ab}$$  $$=\\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{abc}$$  $$=\\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc+3abc}{abc}$$  $$=\\frac{\\left( a+b+c \\right)\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \\right)+3abc}{abc}$$  $$=\\frac{3abc}{abc}$$  $$=3$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "17", "queId": "05123aa1329b42309438dcba25494bc5", "competition_source_list": ["2011年竞赛第4题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知正整数$${{a}_{1}}$$，$${{a}_{2}}$$，$$\\cdots $$，$${{a}_{10}}$$满足$${{a}_{1}}\\leqslant {{a}_{2}}\\leqslant \\cdots \\leqslant {{a}_{10}}$$，且其中任意三个都不能成为三角形的三边长，则$$\\frac{{{a}_{10}}}{{{a}_{1}}}$$的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$34$$ "}], [{"aoVal": "B", "content": "$$55$$ "}], [{"aoVal": "C", "content": "$$89$$ "}], [{"aoVal": "D", "content": "$$144$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["因为$${{a}_{1}}+{{a}_{2}}\\leqslant {{a}_{3}}$$，$${{a}_{2}}+{{a}_{3}}\\leqslant {{a}_{4}}$$，$$\\cdots $$，$${{a}_{8}}+{{a}_{9}}\\leqslant {{a}_{10}}$$，  所以$${{a}_{1}}+2{{a}_{2}}\\leqslant {{a}_{4}}$$，$$2{{a}_{1}}+3{{a}_{2}}\\leqslant {{a}_{5}}$$，$$3{{a}_{1}}+5{{a}_{2}}\\leqslant {{a}_{6}}$$，  $$5{{a}_{1}}+8{{a}_{2}}\\leqslant {{a}_{7}}$$，$$8{{a}_{1}}+13{{a}_{2}}\\leqslant {{a}_{8}}$$，$$13{{a}_{1}}+21{{a}_{2}}\\leqslant {{a}_{9}}$$，$$21{{a}_{1}}+34{{a}_{2}}\\leqslant {{a}_{10}}$$．  所以$$55{{a}_{1}}\\leqslant {{a}_{10}}$$，即 $$\\frac{{{a}_{10}}}{{{a}_{1}}}\\geqslant 55$$．  另一方面，当$${{a}_{1}}={{a}_{2}}=1$$，$${{a}_{n+2}}={{a}_{n+1}}+{{a}_{n}}$$，$$n=1$$，$$2$$，$$\\cdots $$，$$8$$时， $$\\frac{{{a}_{10}}}{{{a}_{1}}}=55$$．  所以，$$\\frac{{{a}_{10}}}{{{a}_{1}}}$$的最小值是$$55$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "886", "queId": "b62ab2d5eca04a03abbcee2bde4e9288", "competition_source_list": ["2011年第22届全国希望杯初一竞赛复赛第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "有理数$$a$$，$$b$$满足$$20a+11\\left\\textbar{} b \\right\\textbar=0$$（$$b\\textgreater0$$），则$$\\frac{a}{{{b}^{2}}}$$是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "正数 "}], [{"aoVal": "B", "content": "负数 "}], [{"aoVal": "C", "content": "非正数 "}], [{"aoVal": "D", "content": "非负数 "}]], "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->知识点->数->有理数->绝对值->绝对值的非负性"], "answer_analysis": ["由$$20a+11\\left\\textbar{} b \\right\\textbar=0$$（$$b\\textgreater0$$），  得$$a=-\\frac{11}{20}\\left\\textbar{} b \\right\\textbar\\textless{}0$$，  所以$$\\frac{a}{{{b}^{2}}}\\textless{}0$$，为负数． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "421", "queId": "751b26c3bdf74dd49481f1a72818e965", "competition_source_list": ["2016年第33届全国全国初中数学联赛竞赛A卷第3题7分", "2016年上海浦东新区进才中学自主招生第8题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果一个正整数可以表示为两个连续奇数的立方差，则称这个正整数为``和谐数''．如：$$2={{1}^{3}}-{{(-1)}^{3}}$$，$$26={{3}^{3}}-{{1}^{3}}$$，$$2$$和$$26$$均为``和谐数''．那么，不超过$$2020$$的正整数中，所有的``和谐数''之和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$6858$$ "}], [{"aoVal": "B", "content": "$$6860$$ "}], [{"aoVal": "C", "content": "$$9260$$ "}], [{"aoVal": "D", "content": "$$9262$$ "}]], "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->利用立方和与立方差公式因式分解", "课内体系->知识点->式->整式的加减->整式有关的概念->整式->整式的定义", "课内体系->知识点->综合与实践->新定义->新定义综合其它"], "answer_analysis": ["$${{(2k+1)}^{3}}-{{(2k-1)}^{3}}$$  $$=\\left[ (2k+1)-(2k-1) \\right]\\left[ {{(2k+1)}^{2}}+(2k+1)(2k-1)+{{(2k-1)}^{2}} \\right]$$  $$=2(12{{k}^{2}}+1)$$（其中$$k$$为非负整数），  由$$2(12{{k}^{2}}+1)\\leqslant 2020$$得，$$k\\leqslant 9$$，  ∴$$k=0$$，$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$9$$，  即得所有不超过$$2020$$的``和谐数''，它们的和为  $$\\left[ {{1}^{3}}-{{(-1)}^{3}} \\right]+({{3}^{3}}-{{1}^{3}})+({{5}^{3}}-{{3}^{3}})+\\cdots +({{17}^{3}}-{{15}^{3}})+({{19}^{3}}-{{17}^{3}})={{19}^{3}}+1=6860$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "369", "queId": "1aeb4e2c47544c6f8279d7090121fcbd", "competition_source_list": ["2013年第24届全国希望杯初一竞赛初赛第9题4分", "初一单元测试《列方程解应用题-有趣的行程问题》第21题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$A$$，$$B$$两地相距$$60$$千米，甲、乙两人驾车（匀速）从$$A$$驶向$$B$$，甲的时速为$$120$$千米，乙的时速为$$90$$千米，如果乙比甲早出发$$6$$分钟，则当甲追上乙以后，乙再经过（~ ）分钟可以到达$$B$$．", "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->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": ["由题意可知，甲每分钟行驶$$2$$千米，乙每分钟行驶$$1.5$$千米．  设甲$$t$$分钟追上乙，则$$2t=1.5(6+t)$$，  解得$$t=18$$．  这时乙离$$B$$地还有$$60-1.5\\times \\left( 18+6 \\right)=24$$（千米），  所以，乙还要走$$24\\div 1.5=16$$（分钟）．  答：当甲追上乙以后，乙再经过$$16$$分钟可以到达$$B$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1111", "queId": "ff8080814d9efd56014daa7265f70a85", "competition_source_list": ["1993年第4届全国希望杯初一竞赛初赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$n$$是正整数，并且有理数$$a$$，$$b$$满足$$a+\\frac{1}{b}=0$$，则必有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{a}^{n}}+{(\\frac{1}{b})^{2n}}=0$$ "}], [{"aoVal": "B", "content": "$${{a}^{2n}}+{(\\frac{1}{b})^{2n+1}}=0$$ "}], [{"aoVal": "C", "content": "$${{a}^{2n}}+{(\\frac{1}{b})^{3n}}=0$$ "}], [{"aoVal": "D", "content": "$${{a}^{2n+1}}+{(\\frac{1}{b})^{2n+1}}=0$$ "}]], "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"], "answer_analysis": ["因为$$b\\ne 0$$，$$a+\\frac{1}{b}=0$$，所以$$a=-\\frac{1}{b}$$，$$a\\ne 0$$．  易知，当$$n$$为自然数时，$${{a}^{2n+1}}={(-\\frac{1}{b})^{2n+1}}=-{(\\frac{1}{b})^{2n+1}}$$，  所以$${{a}^{2n+1}}+{(\\frac{1}{b})^{2n+1}}=0$$，选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1522", "queId": "ef45ae7734fc4415b4b2bf8fb2634602", "competition_source_list": ["1987年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "瓶子里有$$10$$个标号为$$1$$到$$10$$的小球． Jack 把手伸进去，然后随机拿走了一个小球．然后 Jill 也把手伸进去随机拿走了一个不同的小球．那有多大概率小球上两个数字的和为偶数？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{4}{9}$$ "}], [{"aoVal": "B", "content": "$$\\frac{9}{19}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{10}{9}$$ "}], [{"aoVal": "E", "content": "$$\\frac{5}{9}$$ "}]], "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": ["瓶子里有$$10$$个标号为$$1$$到$$10$$的小球． Jack 把手伸进去，然后随机拿走了一个小球．然后 Jill 也把手伸进去随机拿走了一个不同的小球．那有多大概率小球上两个数字的和为偶数？  为了使两个数的和为偶数，它们必须具有相同的奇偶性．有五个偶数和五个奇数．  不管杰克选择什么，具有相同奇偶性的数字的数目是$$4$$．一共有九个数字，所以吉尔选择一个与杰克的相同的奇偶性的数字的概率是$$\\frac{4}{9}$$．  故选$$\\text{A}$$．  我们发现，只有当所加的数都是偶数或奇数时，和才可能是偶数．当我们加一个偶数和一个奇数时，我们会得到一个奇数．我们可以使用互补计数来帮助解决这个问题．因为杰克可以选择$$10$$个数字，吉尔可以选择$$9$$个，所以总共有$$90$$种可能性．  两个数字不同的情况有$$50$$种，因为杰克可以选择$$10$$个数字中的任何一个，吉尔必须从集合中的$$5$$个数字中选择与杰克选择的具有不同奇偶性的数字．所以和为奇数的概率是$$\\frac{5}{9}$$，$$1$$减去这个数，得到答案$$\\frac{4}{9}$$．  故选$$\\text{A}$$．  For the sum of the two numbers removed to be even, they must be of the same \\uline{parity}. There are five even values and five \\uline{odd} values.  No matter what Jack chooses, the number of numbers with the same parity is four. There are nine numbers total, so the probability Jill chooses a number with the same parity as Jack\\textquotesingle s is $$\\frac{4}{9}$$.  We find that it is only possible for the sum to be even if the numbers added are both even or odd. We will get an odd number when we add an even and an odd. We can use \\uline{complementary counting} to help solve the problem. There are a total of $$90$$ possibilities since Jack can chose $$10$$ numbers and Jill can pick $$9$$. There are $$50$$ possibilities for the two numbers to be different since Jack can pick any of the $$10$$ numbers and Jill has to pick from $$5$$ numbers in the set with a different \\uline{parity} than the one that Jack picks. So the probability that the sum will be odd is $$\\frac{50}{90}=\\frac{5}{9}$$. Subtracting this by one gets the answer $$\\frac{4}{9}$$. "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1087", "queId": "8a4c83710caa43b49dbb10f6c570910a", "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": "$$\\frac{12}{25}$$ "}]], "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": ["$$t=2xy+yz+2zx$$  $$=2x(y+z)+yz$$  $$\\leqslant 2x(y+z)+\\frac{1}{4}{{(y+z)}^{2}}$$  $$=2x(1-x)+\\frac{1}{4}{{(1-x)}^{2}}$$  $$=-\\frac{7}{4}{{x}^{2}}+\\frac{3}{2}x+\\frac{1}{4}$$  $$=-\\frac{7}{4}{{\\left( x-\\frac{3}{7} \\right)}^{2}}+\\frac{4}{7}$$，  易知：当$$x=\\frac{3}{7}$$，$$y=z=\\frac{2}{7}$$时，  $$t=2xy+yz+2zx$$取得最大值$$\\frac{4}{7}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "246", "queId": "41d9a49977e744fdad395d794d0c35a5", "competition_source_list": ["1996年第7届全国希望杯初一竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$$a\\textgreater b$$，且$$c\\textless{}0$$，那么在下面不等式中：  ①$$a+c\\textgreater b+c$$；②$$ac\\textgreater bc$$；③$$-\\frac{a}{c}\\textgreater-\\frac{b}{c}$$．成立的个数是．", "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->方程与不等式->不等式（组）->不等式->判断不等式的变形是否正确", "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": ["已知$$a\\textgreater b$$，$$c\\textless{}0$$，由不等式的基本性质，得  $$a+c\\textgreater b+c$$，成立，$$ac\\textgreater bc$$不成立，$$-\\frac{a}{c}\\textgreater-\\frac{b}{c}$$成立，$$a{{c}^{2}}\\textless{}b{{c}^{2}}$$不成立．  所以只有①③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": "544", "queId": "9e134b79a0df437cb88a02ce9e0ab1c2", "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->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", "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->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->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->方法->作差法"], "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": "703", "queId": "f18f5997596d410e9b42bd62a2326c9e", "competition_source_list": ["2018年第29届希望杯初一竞赛初赛第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在黑板上按下面的方案写数：在第一行写数$$1$$；在第二行写两个数$$2$$和$$3$$；在第三行写三个数$$3$$，$$4$$和$$5$$；以此类推（在第$$n$$行写由$$n$$开始的$$n$$个连续的自然数），一直写完$$2000$$行，这时在黑板上共出现$$2018$$次．", "answer_option_list": [[{"aoVal": "A", "content": "$$991$$ "}], [{"aoVal": "B", "content": "$$993$$ "}], [{"aoVal": "C", "content": "$$995$$ "}], [{"aoVal": "D", "content": "$$997$$ "}]], "knowledge_point_routes": ["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->数列找规律->数列找规律-其他数列规律"], "answer_analysis": ["根据题意，第$$n$$行从$$n$$开始，写到$$n+n-1$$，  令$$n+n-1=2018$$，解得$$n=1009.5$$，  所以从第$$1010$$行，$$2018$$开始出现，  所以$$2018$$一共出现$$2000-1010+1=991$$（次）． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1300", "queId": "8aac50a74f25f755014f29940c3c0fb9", "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（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$时$$30$$分 "}], [{"aoVal": "B", "content": "$$10$$时$$5$$分 "}], [{"aoVal": "C", "content": "$$10$$时$$5\\frac{5}{11}$$分 "}], [{"aoVal": "D", "content": "$$9$$时$$32\\frac{8}{11}$$分 "}]], "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->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->一元一次方程的钟表问题"], "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": "662", "queId": "40094784b8a140748fb88c946976ea2e", "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": "479", "queId": "62ea73a6cb6c4a148ddb972d02f2c178", "competition_source_list": ["2014年第25届全国希望杯初二竞赛初赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$x=\\sqrt{5}$$，$$y$$是不大于$$x$$的最大整数，则$$\\frac{1}{x-y}$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{5}-2$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{5}+2$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{5}-1$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{5}+1$$ "}]], "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", "课内体系->知识点->式->二次根式->二次根式的运算"], "answer_analysis": ["因为$$2\\textless{}x\\textless{}3$$，  所以$$y=2$$，  则$$\\frac{1}{x-y}=\\frac{1}{\\sqrt{5}-2}=\\sqrt{5}+2$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1138", "queId": "9b95d410617b405cba980178a2463a62", "competition_source_list": ["2014年第25届全国希望杯初一竞赛初赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "不等式$$(x-7)(x+2)\\textless{}0$$的整数解的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->其它不等式->高次不等式", "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": ["原不等式可化为$$\\left { \\begin{matrix}x-7\\textgreater0    x+2\\textless{}0   \\end{matrix} \\right.$$或$$\\left { \\begin{matrix}x-7\\textless{}0    x+2\\textgreater0   \\end{matrix} \\right.$$，  解$$\\left { \\begin{matrix}x-7\\textgreater0    x+2\\textless{}0   \\end{matrix} \\right.$$得$$x\\textgreater7$$或$$x\\textless{}-2$$，不成立，舍去；  解$$\\left { \\begin{matrix}x-7\\textless{}0    x+2\\textgreater0   \\end{matrix} \\right.$$得$$-2\\textless{}x\\textless{}7$$，  ∴$$x$$的整数值有$$-1$$，$$0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，共$$8$$个． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1321", "queId": "8aac50a7511483070151148e530e0062", "competition_source_list": ["2015年第26届全国希望杯初二竞赛初赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "若一次函数$$y=x+5$$的图象经过点$$P(a,b)$$和点$$Q(c,d)$$，则$$ad+bc-ac-bd$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$25$$ "}], [{"aoVal": "D", "content": "$$-25$$ "}]], "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->一次函数与二元一次方程组"], "answer_analysis": ["∵一次函数$$y=x+5$$的图象经过点$$P(a,b)$$和点$$Q(c,d)$$，  ∴$$b=a+5$$，$$d=c+5$$．  ∴$$b-a=5$$，$$d-c=5$$．  ∴$$ad+bc-ac-bd$$  $$=a(d-c)-b(d-c)$$  $$=(a-b)(d-c)$$  $$=-25$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1101", "queId": "ff8080814d9efd56014da556dbc00757", "competition_source_list": ["1992年第3届全国希望杯初一竞赛复赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a\\textgreater0$$，$$b\\textless{}0$$且$$a\\textless{}\\left\\textbar{} b \\right\\textbar$$，则下列关系式中正确的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-b\\textgreater a\\textgreater-a\\textgreater b$$ "}], [{"aoVal": "B", "content": "$$b\\textgreater a\\textgreater-b\\textgreater-a$$ "}], [{"aoVal": "C", "content": "$$-b\\textgreater a\\textgreater b\\textgreater-a$$ "}], [{"aoVal": "D", "content": "$$a\\textgreater b\\textgreater-a\\textgreater-b$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->数轴", "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->已知范围化简绝对值"], "answer_analysis": ["已知$$a\\textgreater0$$，$$b\\textless{}0$$且$$a\\textless{}\\left\\textbar{} b \\right\\textbar$$．  在数轴上直观表示出来，$$b$$到原点的距离大于$$a$$到原点的距离，  所以$$-b\\textgreater a\\textgreater-a\\textgreater b$$，选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "295", "queId": "19ed5806e8cc4d35af40fddbcb81f49e", "competition_source_list": ["初三上学期其它", "2005年第22届全国初中数学联赛竞赛初赛第2题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a+b+c=3$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=3$$，则$${{a}^{2005}}+{{b}^{2005}}+{{c}^{2005}}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$${{2}^{2005}}$$ "}], [{"aoVal": "D", "content": "$$3\\times {{2}^{2005}}$$ "}]], "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->能力->运算能力"], "answer_analysis": ["解法一：  由$$a+b+c=3$$可知$$c=3-a-b$$，将其代入$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=3$$并化简得  $${{a}^{2}}+ab+{{b}^{2}}-3a-3b+3=0$$．以下同例$$14$$第（$$2$$）问．  解得$$a=1$$，$$b=1$$，此时$$c=3-a-b=1$$，故$${{a}^{2005}}+{{b}^{2005}}+{{c}^{2005}}=3$$．  解法二：  由$$a+b+c=3$$可得$${{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2(ab+bc+ac)=9$$，  又因$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=3$$，解得$$ab+bc+ca=3={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$，可知$$a=b=c$$，  代入$$a+b+c=3$$中得$$a=b=c=1$$，故$${{a}^{2005}}+{{b}^{2005}}+{{c}^{2005}}=3$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "395", "queId": "1f5c03b8a0ef49f8810ab8d4e3df307b", "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->三角形->三角形基础", "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", "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": ["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": "905", "queId": "91c54f3855434c57958fac0b1f8e4810", "competition_source_list": ["2006年第17届希望杯初一竞赛复赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$和$$b$$是满足$$ab\\ne 0$$的有理数，现有四个命题：  ①$$\\frac{a-2}{{{b}^{2}}+4}$$的相反数是$$\\frac{2-a}{{{b}^{2}}+4}$$；  ②$$a-b$$的相反数是$$a$$的相反数与$$b$$的相反数的差；  ③$$ab$$的相反数是$$a$$的相反数和$$b$$的相反数的乘积；  ④$$ab$$的倒数是$$a$$的倒数和$$b$$的倒数的乘积．  其中真命题有．", "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->能力->运算能力", "课内体系->知识点->数->有理数->相反数", "课内体系->知识点->数->有理数->倒数与负倒数"], "answer_analysis": ["因为$$-\\frac{a-2}{{{b}^{2}}+4}=\\frac{2-a}{{{b}^{2}}+4}$$，  所以命题①是真命题；  因为$$a-b$$的相反数为$$-(a-b)=-a-(-b)$$，  所以命题②是真命题；  因为$$ab$$的相反数为$$-ab$$，$$(-a)(-b)=ab$$，又$$ab\\ne 0$$，  所以$$-ab\\ne ab$$，因此，③不是真命题；  因为$$ab\\ne 0$$，  所以$$ab$$的倒数为$$\\frac{1}{ab}=\\frac{1}{a}\\cdot \\frac{1}{b}$$，因此，④是真命题．  故选($$\\text{C}$$)． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "620", "queId": "4cfd57eb0ff5459398db11515fdada52", "competition_source_list": ["2001年第12届希望杯初二竞赛第2试第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "小张上周工作$$a$$小时，每小时的工资为$$b$$元，本周他的工作时间比上周减少$$10 \\%$$，而每小时的工资数额增加$$10 \\%$$，则他本周的工资总额与上周的工资总额相比．", "answer_option_list": [[{"aoVal": "A", "content": "增加$$1 \\%$$ "}], [{"aoVal": "B", "content": "减少$$1 \\%$$ "}], [{"aoVal": "C", "content": "增加$$1.5 \\%$$ "}], [{"aoVal": "D", "content": "减少$$1.5 \\%$$ "}]], "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->整式的乘除运算->整式乘除与实际问题", "课内体系->能力->运算能力"], "answer_analysis": ["小张上周的工资总额为$$ab$$元，他本周的工作时间为$$a(1-10 \\%)$$小时，每小时工资金额为$$b(1+10 \\%)$$．  所以$$a(1-10 \\%)\\cdot b(1+10 \\%)=0.99ab$$．  所以本周与上周工资相比，减少$$1 \\%$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1201", "queId": "8aac49074e724b45014e87c0242f50e9", "competition_source_list": ["1996年第7届全国希望杯初一竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$\\frac{a}{b}=0$$，那么有理数$$a$$，$$b$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "都是零 "}], [{"aoVal": "B", "content": "互为相反数 "}], [{"aoVal": "C", "content": "互为倒数 "}], [{"aoVal": "D", "content": "不都是零 "}]], "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->分式->分式的基础->分式值为0"], "answer_analysis": ["当$$b=0$$时，$$\\frac{a}{b}$$无意义，故排除$$\\text{A}$$；  当$$a=0$$且$$b=1$$时，$$\\frac{a}{b}=0$$，易知$$1$$与$$0$$不互为相反数，也不互为倒数，排除$$\\text{B}$$、$$\\text{C}$$．  因此选$$\\text{D}$$．  事实上，$$\\frac{a}{b}=0$$，只能$$a=0$$，$$b\\ne 0$$，此时$$a$$，$$b$$不都是零． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1251", "queId": "c0518e3870e3490b91aeabd46c177261", "competition_source_list": ["1996年第7届希望杯初二竞赛第6题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "三角形的三边长分别为$$2{{n}^{2}}+2n$$，$$2n+1$$，$$2{{n}^{2}}+2n+1$$ （$$n$$是自然数），这样的三角形是．", "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": "锐角三角形或直角三角形 "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->特殊三角形->直角三角形"], "answer_analysis": ["因为$$n$$是自然数，  所以  $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde{{(2{{n}^{2}}+2n+1)}^{2}}-{{(2{{n}^{2}}+2n)}^{2}}$$  $$=\\left[ (2{{n}^{2}}+2n+1)+(2{{n}^{2}}+2n) \\right]\\left[ (2{{n}^{2}}+2n+1)-(2{{n}^{2}}+2n) \\right]$$  $$=4{{n}^{2}}+4n+1$$  $$={{(2n+1)}^{2}}$$，  所以三角形的三边长满足勾股定理，  所以，该三角形是直角三角形．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "809", "queId": "52fe492916d44d71803cef8451880c69", "competition_source_list": ["2002年第13届希望杯初二竞赛第2试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a\\ne b$$，$$a$$，$$b$$，$$\\sqrt{a}-\\sqrt{b}$$都是有理数，那么$$\\sqrt{a}$$和$$\\sqrt{b}$$（~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "都是有理数 "}], [{"aoVal": "B", "content": "一个是有理数，另一个是无理数 "}], [{"aoVal": "C", "content": "都是无理数 "}], [{"aoVal": "D", "content": "是有理数还是无理数不能确定 "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->无理数有关的计算->两个数运算后判断是否为无理数"], "answer_analysis": ["因为$$a$$，$$b$$，$$\\sqrt{a}-\\sqrt{b}$$都是有理数，  所以$$\\sqrt{a}+\\sqrt{b}=\\frac{a-b}{\\sqrt{a}-\\sqrt{b}}$$是有理数．  所以$$\\sqrt{a}=\\frac{1}{2}(\\sqrt{a}+\\sqrt{b})+\\frac{1}{2}(\\sqrt{a}-\\sqrt{b})$$是有理数，$$\\sqrt{b}=\\frac{1}{2}(\\sqrt{a}+\\sqrt{b})-\\frac{1}{2}(\\sqrt{a}-\\sqrt{b})$$也是有理数．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1143", "queId": "d6ce4cf534c346ba9afc60127de6b638", "competition_source_list": ["2016年第27届全国希望杯初一竞赛复赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "三个内角的度数都是质数的三角形的种数（三个内角的度数对应相等的两个三角形视为一种）是（~ ~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "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"], "answer_analysis": ["设三角形的三个内角的度数分别为$$x{}^{}\\circ $$，$$y{}^{}\\circ $$，$$z{}^{}\\circ $$．  ∴$$x+y+z=180$$．  ∵$$x$$，$$y$$，$$z$$均为质数，$$180$$为偶数，  而三个奇质数的和为奇数，  ∴其中必然有一个是偶质数$$2$$．  不妨令$$z=2$$，则$$x+y=178$$．  将$$178$$分成两个质数之和如下：  $$178=5+173$$；  $$178=11+167$$；  $$178=29+149$$；  $$178=41+137$$；  $$178=47+131$$；  $$178=71+107$$；  $$178=89+89$$．  ∴满足条件的三角形有$$7$$个． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1018", "queId": "ff8080814d4b1928014d4c17770605ea", "competition_source_list": ["2015年第26届全国希望杯初一竞赛初赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$x$$，$$y$$，$$m$$，$$n$$为有理数，若$${{x}^{2}}+{{y}^{2}}={{m}^{2}}+{{n}^{2}}=8$$，则$$xy+mn$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "有最小值$$4$$ "}], [{"aoVal": "B", "content": "有最大值$$4$$ "}], [{"aoVal": "C", "content": "有最小值$$8$$ "}], [{"aoVal": "D", "content": "有最大值$$8$$ "}]], "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->整式的加减->整式的加减运算->整式加减", "课内体系->方法->配方法"], "answer_analysis": ["方法一：∵$$2xy\\leqslant {{x}^{2}}+{{y}^{2}}=8$$，  ∴$$xy\\leqslant 4$$；  同理，$$mn\\leqslant 4$$．  故$$xy+mn$$有最大值$$8$$．  方法二：  因为$${{(x-y)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy\\geqslant 0$$，所以$${{x}^{2}}+{{y}^{2}}\\geqslant 2xy$$，①  因为$${{(m-n)}^{2}}={{m}^{2}}+{{n}^{2}}-2mn\\geqslant 0$$，所以$${{m}^{2}}+{{n}^{2}}\\geqslant 2mn$$，②  ①$$+$$②，得$${{x}^{2}}+{{y}^{2}}+{{m}^{2}}+{{n}^{2}}\\geqslant 2mn+2xy$$，③  把$${{x}^{2}}+{{y}^{2}}={{m}^{2}}+{{n}^{2}}=8$$代入③，得$$16\\geqslant 2(xy+mn)$$，即$$xy+mn\\leqslant 8$$．  当$$x=m=y=n=2$$时，$$2\\times 2+2\\times 2=8$$，即可达到$$8$$．即$$xy+mn$$有最大值为$$8$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "409", "queId": "39a5ccc63d3c4250aaa9b68d40031bb2", "competition_source_list": ["2009年第20届希望杯初一竞赛第1试第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$${{x}^{2}}+x-2=0$$，则$${{x}^{3}}+2{{x}^{2}}-x+2020$$=．", "answer_option_list": [[{"aoVal": "A", "content": "$$2022$$ "}], [{"aoVal": "B", "content": "$$2020$$ "}], [{"aoVal": "C", "content": "$$-2020$$ "}], [{"aoVal": "D", "content": "$$-2022$$ "}]], "knowledge_point_routes": ["知识标签->题型->式->整式加减->整式加减化简求值->题型：整式加减条件化简求值", "知识标签->知识点->式->整式的加减->整式的加减运算", "知识标签->学习能力->运算能力"], "answer_analysis": ["$${{x}^{2}}+x-2=0$$，则$${{x}^{2}}+x=2$$，  ∴$${{x}^{3}}+2{{x}^{2}}-x+2007$$  $$={{x}^{3}}+{{x}^{2}}+{{x}^{2}}+x-2x+2020$$  $$=x\\left( {{x}^{2}}+x \\right)+\\left( {{x}^{2}}+x \\right)-2x+2020$$  $$=2x+2-2x+2020$$  $$=2021$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1455", "queId": "ab0478d0d28f46dabbb22248274042a8", "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->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->方法->整体法", "课内体系->思想->整体思想"], "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．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "480", "queId": "4316a999b41e41de8976f2e096d8265f", "competition_source_list": ["初三上学期其它", "2007年竞赛第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知三个关于$$x$$的一元二次方程$$a{{x}^{2}}+bx+c=0$$，$$b{{x}^{2}}+cx+a=0$$，$$c{{x}^{2}}+ax+b=0$$恰有一个公共实数根，则$$\\frac{{{a}^{2}}}{bc}+\\frac{{{b}^{2}}}{ca}+\\frac{{{c}^{2}}}{ab}$$的值为（ ~ ~）．", "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->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->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": ["~设$${{x}_{0}}$$是它们的一个公共实数根，  则$$ax_{0}^{2}+b{{x}_{0}}+c=0$$，$$bx_{0}^{2}+c{{x}_{0}}+a=0$$，$$cx_{0}^{2}+a{{x}_{0}}+b=0$$．  把上面三个式子相加，并整理得  $$(a+b+c)(x_{0}^{2}+{{x}_{0}}+1)=0$$．  因为$$x_{0}^{2}+{{x}_{0}}+1={{\\left( {{x}_{0}}+\\frac{1}{2} \\right)}^{2}}+\\frac{3}{4}\\textgreater0$$，  所以$$a+b+c=0$$．  于是$$\\frac{{{a}^{2}}}{bc}+\\frac{{{b}^{2}}}{ca}+\\frac{{{c}^{2}}}{ab}$$  $$=\\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{abc}$$  $$=\\frac{{{a}^{3}}+{{b}^{3}}-{{(a+b)}^{3}}}{abc}$$  $$=\\frac{-3ab(a+b)}{abc}$$  $$=3$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "47", "queId": "7d946633b73b42ce89847bda85fb2bf2", "competition_source_list": ["2017~2018学年北京海淀区清华大学附属中学初一上学期期中第8题3分", "1996年第7届希望杯初二竞赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个角的补角的一半比这个角的余角的$$2$$倍小$$3{}^{}\\circ$$，那么这个角等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$58{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$59{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$60{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$61{}^{}\\circ $$ "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->几何图形", "课内体系->知识点->几何图形初步->角->角的定义和分类->余角和补角"], "answer_analysis": ["设这个角为$$x{}^{}\\circ $$，列方程得$$\\frac{180-x}{2}=2(90-x)-3$$，  解得$$x=58$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "37", "queId": "02d5d5db1542432b9e3c76e7a6dafc4d", "competition_source_list": ["初三上学期其它", "2018年全国竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "关于$$x$$的方程$$\\left\\textbar{} \\frac{{{x}^{2}}}{x-1} \\right\\textbar=a$$仅有两个不同的实根．则实数$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater0$$ "}], [{"aoVal": "B", "content": "$$a\\geqslant 4$$ "}], [{"aoVal": "C", "content": "$$2\\textless{}a\\textless{}4$$ "}], [{"aoVal": "D", "content": "$$0\\textless{}a\\textless{}4$$ "}]], "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->由分式方程的解确定参数", "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", "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": ["当$$a\\textless{}0$$时，无解；  当$$a=0$$时，$$x=0$$只有一解，不合题意；  当$$a\\textgreater0$$时，原方程化为$$\\frac{{{x}^{2}}}{x-1}=\\pm a$$，整理得  $${{x}^{2}}-ax+a=0$$ ①  或$${{x}^{2}}+ax-a=0$$．②  因为方程②的判别式大于$$0$$，所以方程②有两个不同实根；  又因为原方程仅有两个不同的实根，故必有方程①的判别式小于$$0$$，从而$$0\\textless{}a\\textless{}4$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "361", "queId": "c2b8227a2951460ca9c1c52fcbf9f29a", "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->整式的乘除->幂的运算->幂的综合运算", "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": ["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": "796", "queId": "ff80808146ec1f920146ffc18b7815a4", "competition_source_list": ["2006年第17届希望杯初一竞赛初赛第4题4分", "北京其它"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$m=\\frac{a+2}{a+3}$$，$$n=\\frac{a+1}{a+2}$$，$$p=\\frac{a}{a+1}$$，若$$a\\textless-3$$，则（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$m\\textless{n}\\textless{p}$$ "}], [{"aoVal": "B", "content": "$$n\\textless{p}\\textless{m}$$ "}], [{"aoVal": "C", "content": "$${p}\\textless{n}\\textless{m}$$ "}], [{"aoVal": "D", "content": "$$p\\textless{m}\\textless{n}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的运算->分式比较大小"], "answer_analysis": ["∵$$a\\textless-3$$，令$$a=-4$$，  ∴$$m=2$$，$$n=\\frac{3}{2}$$，$$p=\\frac{4}{3}$$，  ∵$$\\frac{4}{3}\\textless\\frac{3}{2}\\textless2$$ ，  ∴$${p}\\textless{n}\\textless{m}$$． ", "<p>因为$$\\frac{a+2}{a+3}=1-\\frac{1}{a+3}$$，$$\\frac{a+1}{a+2}=1-\\frac{1}{a+2}$$，$$\\frac{a}{a+1}=1-\\frac{1}{a+1}$$．</p>\n<p>又$$a+1&lt;{}a+2&lt;{}a+3&lt;{}0$$，</p>\n<p>可得$$0&lt;{}-\\left( a+3 \\right)&lt;{}-\\left( a+2 \\right)&lt;{}-\\left( a+1 \\right)$$，</p>\n<p>所以$$-\\frac{1}{a+1}&lt;{}-\\frac{1}{a+2}&lt;{}-\\frac{1}{a+3}$$，</p>\n<p>所以$$\\frac{a}{a+1}&lt;{}\\frac{a+1}{a+2}&lt;{}\\frac{a+2}{a+3}$$．</p>"], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1194", "queId": "d6ea348bbe464519806dd61d962020ba", "competition_source_list": ["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4题4分", "2019~2020学年广东深圳福田区深圳外国语学校初二上学期期末第6题3分", "2018~2019学年四川内江初一下学期期末第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知关于$$x$$，$$y$$的方程组$$\\begin{cases}mx+ny=7    2mx-3ny=4    \\end{cases}$$的解为$$\\begin{cases}x=1    y=2    \\end{cases}$$，则$$m$$，$$n$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\begin{cases}m=5    n=1    \\end{cases}$$ "}], [{"aoVal": "B", "content": "$$\\begin{cases}m=1    n=5    \\end{cases}$$ "}], [{"aoVal": "C", "content": "$$\\begin{cases}m=3    n=2    \\end{cases}$$ "}], [{"aoVal": "D", "content": "$$\\begin{cases}m=2    n=3    \\end{cases}$$ "}]], "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->由二元一次方程组的解求参数的值", "课内体系->能力->运算能力"], "answer_analysis": ["关于$$x$$，$$y$$的方程组$$\\begin{cases}mx+ny=7    2mx-3ny=4    \\end{cases}$$的解为$$\\begin{cases}x=1    y=2    \\end{cases}$$，  ∴$$\\begin{cases}m+2n=7    2m-6n=4    \\end{cases}$$，  解得$$\\begin{cases}m=5    n=1    \\end{cases}$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1267", "queId": "8aac50a74e442d83014e4cccc70a1d43", "competition_source_list": ["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", "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"], "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（~ ）．", "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": "$$x-{{y}^{2}}\\textless{}0$$ "}]], "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": "1037", "queId": "ff8080814d7978b9014d86591d2123cc", "competition_source_list": ["1990年第1届全国希望杯初一竞赛初赛第4题", "2018~2019学年10月吉林长春南关区东北师范大学附属中学（明珠校区）初一上学期周测C卷第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$a$$，$$b$$代表有理数，并且$$a+b$$的值大于$$a-b$$的值，那么（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$，$$b$$同号 "}], [{"aoVal": "B", "content": "$$a$$，$$b$$异号 "}], [{"aoVal": "C", "content": "$$a\\textgreater0$$ "}], [{"aoVal": "D", "content": "$$b\\textgreater0$$ "}]], "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": ["由于$$a+b\\textgreater{}a-b$$，则$$b\\textgreater-b$$，所以$$2b\\textgreater0$$，所以$$b\\textgreater0$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "494", "queId": "3a3deb370e75434a8d0c5da8067a1b12", "competition_source_list": ["2017年河北石家庄裕华区石家庄外国语学校初三竞赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\frac{a}{3}=\\frac{b}{4}=\\frac{c}{5}\\ne 0$$，则$$\\frac{c-a}{b}$$的值为（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{3}$$ "}]], "knowledge_point_routes": ["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": ["设$$k=\\frac{a}{3}=\\frac{b}{4}=\\frac{c}{5}\\ne 0$$，由此得到$$a=3k$$，$$b=4k$$，$$c=5k$$  ∴$$\\frac{c-a}{b}$$  $$=\\frac{5k-3k}{4k}$$  $$=\\frac{1}{2}$$，故$$\\text{B}$$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": "745", "queId": "da865e2fcd9147d1bea8788994527780", "competition_source_list": ["1997年第8届希望杯初二竞赛第2试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知实数$$a$$，$$b$$满足条件：$${{a}^{2}}+4{{b}^{2}}-a+4b+\\frac{5}{4}=0$$，那么$$-ab$$的平方根是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\pm 2$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$\\pm \\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->整式的乘除运算"], "answer_analysis": ["因为$${{a}^{2}}+4{{b}^{2}}-a+4b+\\frac{5}{4}=0$$，  所以$${{\\left( a-\\frac{1}{2} \\right)}^{2}}+{{(2b+1)}^{2}}=0$$，  因为$${{\\left( a-\\frac{1}{2} \\right)}^{2}}\\geqslant 0$$，$${{(2b+1)}^{2}}\\geqslant 0$$，  所以$$a=\\frac{1}{2}$$，$$b=-\\frac{1}{2}$$，  所以$$-ab=\\frac{1}{4}$$，  所以$$\\frac{1}{4}$$的平方根是$$\\pm \\frac{1}{2}$$．  故选：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1584", "queId": "f0681b0979894e03889c9bd44f0ddd6b", "competition_source_list": ["2022~2023学年福建漳州龙海市初一上学期期中（阶段性评价）第6题", "2022~2023学年浙江宁波初三月考（六校强基竞赛）第3题", "2021~2022学年福建厦门思明区厦门外国语学校初一上学期期中第6题", "2022~2023学年广东广州增城区初一上学期期中（测试）第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$a\\textless{} 0$，$b\\textgreater0$，则\\emph{b}、$b+a$、$b-a$、\\emph{ab}中最大的一个数是", "answer_option_list": [[{"aoVal": "A", "content": "\\emph{b} "}], [{"aoVal": "B", "content": "$b+a$ "}], [{"aoVal": "C", "content": "$b-a$ "}], [{"aoVal": "D", "content": "\\emph{ab} "}]], "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：∵$a\\textless0，b\\textgreater0，$  $\\therefore$$ab\\textless0\\textless b﹣a，$  故$b+a\\textless b，b﹣a\\textgreater b，$  ∴$b+a\\textless b\\textless b﹣a．$  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1358", "queId": "9c8c1d7e52d64fe5ad5de9714b22548b", "competition_source_list": ["2017年第1届重庆全国初中数学联赛初一竞赛初赛"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$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$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式", "课内体系->能力->运算能力"], "answer_analysis": ["$$\\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}$$ "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "393", "queId": "237e3ff68dca479d9183ac3680daf9ce", "competition_source_list": ["2005年第16届希望杯初二竞赛初赛第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "某人月初用$$x$$元人民币投资股票，由于行情较好，他的资金每月都增加$$\\frac{1}{3}$$，即使他每月末都取出$$1000$$元用于日常开销，他的资金仍然在$$3$$个月后增长了一倍，那么$$x$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$9000$$ "}], [{"aoVal": "B", "content": "$$10000$$ "}], [{"aoVal": "C", "content": "$$11000$$ "}], [{"aoVal": "D", "content": "$$11100$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的经济问题->一元一次方程的经济问题-其它", "课内体系->能力->运算能力", "课内体系->思想->方程思想"], "answer_analysis": ["一个月后，他的资金为$$\\frac{4}{3}x-1000$$，  两个月后，他的资金为$$\\frac{4}{3}\\left( \\frac{4}{3}x-1000 \\right)-1000=\\frac{16}{9}x-\\frac{7000}{3}$$，  三个月后，他的资金为$$\\frac{4}{3}\\left( \\frac{16}{9}x-\\frac{7000}{3} \\right)-1000=\\frac{64}{27}x-\\frac{37000}{9}$$，  由题设，得$$\\frac{64}{27}x-\\frac{37000}{9}=2x$$，  解得$$x=11100$$．  选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "67", "queId": "06ffd4bbe1944003a08238b564631d7c", "competition_source_list": ["2013年第24届全国希望杯初一竞赛复赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\triangle ABC$$的内角分别为$$\\angle A$$，$$\\angle B$$，$$\\angle C$$，若$$\\angle 1=\\angle A+\\angle B$$，$$\\angle 2=\\angle B+\\angle C$$，$$\\angle 3=\\angle A+\\angle C$$，则$$\\angle 1$$，$$\\angle 2$$，$$\\angle 3$$中．", "answer_option_list": [[{"aoVal": "A", "content": "至少有一个锐角 "}], [{"aoVal": "B", "content": "三个都是钝角 "}], [{"aoVal": "C", "content": "至少有两个钝角 "}], [{"aoVal": "D", "content": "可以有两个直角 "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->三角形及多边形->与三角形有关的角->三角形的外角定义及性质", "课内体系->知识点->三角形->三角形及多边形->与三角形有关的角->三角形内角和的应用"], "answer_analysis": ["因为三角形的内角中至少有$$2$$个锐角，  而$$\\angle 1$$，$$\\angle 2$$，$$\\angle 3$$是三角形的三个外角，  所以$$\\angle 1$$，$$\\angle 2$$，$$\\angle 3$$中至少有两个钝角． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "38", "queId": "17651554e8374101a614a0a6ba69a019", "competition_source_list": ["1979年竞赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "计算：$$41.7-32.8=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$74.5$$ "}], [{"aoVal": "B", "content": "$$9.1$$ "}], [{"aoVal": "C", "content": "$$11.1$$ "}], [{"aoVal": "D", "content": "$$8.9$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["$$41.7-32.8=41.7-32.7-0.1=9.0-0.1=8.9$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "604", "queId": "7ee5c3736a8a42e8a3b60506756f6aca", "competition_source_list": ["2019~2020学年10月北京朝阳区北京十七中初一上学期月考第9题3分", "2019~2020学年9月浙江杭州下城区杭州观成中学初一上学期月考第20题4分", "2000年第11届希望杯初一竞赛第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "$$a$$是有理数，则$$\\frac{11}{a+2000}$$的值不能是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$-2000$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["∵$$a$$是有理数，  ∴不论$$a$$取任何有理数$$\\frac{11}{a+2000}$$的值永远不会是$$0$$．  ∴选$$\\text{C}$$．  但要注意当选$$\\text{D}$$时，$$\\frac{11}{a+2000}$$这个式子本身无意义，  ∴不能选$$\\text{D}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1375", "queId": "bc273cf47a9749b494a6c1867877a7cc", "competition_source_list": ["2012年竞赛第8题4分", "2019~2020学年陕西西安雁塔区陕西师范大学附属中学初三上学期开学考试第17题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果关于$$x$$的方程$${{x}^{2}}+kx+\\frac{3}{4}{{k}^{2}}-3k+\\frac{9}{2}=0$$的两个实数根分别为$${{x}_{1}}$$，$${{x}_{2}}$$，那么$$\\frac{{{x}_{1}}^{2011}}{{{x}_{2}}^{2012}}$$的值为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$-\\frac{1}{3}$ "}], [{"aoVal": "B", "content": "$-\\frac{2}{3}$ "}], [{"aoVal": "C", "content": "$-\\frac{4}{3}$ "}], [{"aoVal": "D", "content": "$-\\frac{5}{3}$ "}]], "knowledge_point_routes": ["知识标签->题型->方程与不等式->一元二次方程->根的判别式->题型：判断一元二次方程根的情况", "知识标签->题型->方程与不等式->一元二次方程->根的判别式->题型：由一元二次方程根的情况确定参数", "知识标签->知识点->方程与不等式->一元二次方程->一元二次方程的解", "知识标签->知识点->方程与不等式->一元二次方程->一元二次方程根的判别式", "知识标签->学习能力->运算能力"], "answer_analysis": ["根据题意可得  ∵方程有实数根，  ∴$$\\Delta ={{b}^{2}}-4ac\\geqslant 0$$，  即$${{k}^{2}}-4\\left( \\frac{3}{4}{{k}^{2}}-3k+\\frac{9}{2} \\right)\\geqslant 0$$，  ∴$$-2{{(k-3)}^{2}}\\geqslant 0$$，  ∵$${{(k-3)}^{2}}\\leqslant 0$$，  ∴$$k-3=0$$，  即$$k=3$$，  ∴原方程为：$${{x}^{2}}+3x+\\frac{9}{4}=0$$，  ∴$${{x}_{1}}={{x}_{2}}=-\\frac{3}{2}$$，  ∴$$\\frac{x_{1}^{2011}}{x_{2}^{2012}}={{\\left( \\frac{{{x}_{1}}}{{{x}_{2}}} \\right)}^{2011}}\\cdot \\frac{1}{{{x}_{2}}}=\\frac{1}{{{x}_{2}}}=-\\frac{2}{3}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "946", "queId": "d67e99e5554f4a0887a909da34ce68f4", "competition_source_list": ["1997年第8届希望杯初二竞赛第1试第10题", "2018~2019学年3月江苏扬州广陵区扬州市树人中学初一下学期月考（2+4学制）第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙两种茶叶，以$$x:y$$ （重量比）相混合制成一种混合茶，甲种茶叶的价格每公斤$$50$$元，乙种茶叶的价格每公斤$$40$$元，现在甲种茶叶的价格上调了$$10 \\%$$，乙种茶叶的价格下调了$$10 \\%$$，但混合茶的价格不变，则$$x:y$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$1:1$$ "}], [{"aoVal": "B", "content": "$$5:4$$ "}], [{"aoVal": "C", "content": "$$4:5$$ "}], [{"aoVal": "D", "content": "$$5:6$$ "}]], "knowledge_point_routes": ["课内体系->思想->方程思想", "课内体系->知识点->方程与不等式->二元一次方程（组）->二元一次方程组应用题->二元一次方程组的经济问题", "课内体系->能力->运算能力"], "answer_analysis": ["$$50x+40y=50(1+10 \\%)x+40(1-10 \\%)y$$，  化简后得到$$\\frac{x}{y}=\\frac{4}{5}$$，  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "293", "queId": "15de66d28b5b44a387a188569dcda7bf", "competition_source_list": ["2012年第23届全国希望杯初二竞赛复赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "化简：$$\\sqrt{4+\\sqrt{7}}-\\sqrt{4-\\sqrt{7}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的运算->多重二次根式", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算"], "answer_analysis": ["设$$a=\\sqrt{4+\\sqrt{7}}-\\sqrt{4-\\sqrt{7}}$$，  则$${{a}^{2}}=4+\\sqrt{7}-2\\sqrt{(4+\\sqrt{7})(4-\\sqrt{7})}+4-\\sqrt{7}=2$$，  于是$$\\sqrt{4+\\sqrt{7}}-\\sqrt{4-\\sqrt{7}}=$\\sqrt{2}$$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "162", "queId": "f5936be9e1ae44febc90138fb37a8451", "competition_source_list": ["1987年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "吉米在前两场篮球赛中$$20$$投$$10$$进，从而让他的赛季命中率为$$50 \\%$$．在他的下场比赛中，他投了$$30$$次篮，使得赛季命中率提高到了$$70 \\%$$．这$$30$$次投篮中他命中了几次？", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$23$$ "}], [{"aoVal": "E", "content": "$$25$$ "}]], "knowledge_point_routes": ["美国amc8->知识点->数与运算->比和比例->比例"], "answer_analysis": ["在三场比赛中，他投了$$50$$次，其中他的赛季得分是$$70 \\%$$，  所以他投中了$$50\\times 0.7=35$$球．  前两场中投了$$10$$球，所以在最后一场中他投了$$35-10=25$$球．  故选$$\\text{E}$$． "], "answer_value": "E"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "775", "queId": "4e46ff68558849b2a210d8cce6737169", "competition_source_list": ["2019年湖南长沙雨花区湖南广益实验中学初一竞赛（广益杯）第10题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "现定义一种新运算：$$a\\otimes b={{b}^{2}}-ab$$，如：$$1\\otimes 2={{2}^{2}}-1\\times 2$$，则$$(-2\\otimes 1)\\otimes 3$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->综合与实践->规律探究与程序框图->定义新运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->含乘方的有理数混合运算"], "answer_analysis": ["∵$$a\\otimes b={{b}^{2}}-ab$$，  $$-2\\otimes 1={{1}^{2}}-\\left( -2 \\right)\\times 1=1-\\left( -2 \\right)=3$$，  $$3\\otimes 3={{3}^{2}}-3\\times 3=9-9=0$$，  故$$(-2\\otimes 1)\\otimes 3=0$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1003", "queId": "9fda92681a3941ea8c8b20fca62228db", "competition_source_list": ["2005年第16届希望杯初二竞赛复赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知两位数$$\\overline{ab}$$能够被$$3$$整除，它的十位数字与个位数字的乘积等于它的个位数字，且它的任意次幂的个位数字等于它的个位数字，这样的两位数共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$个 "}], [{"aoVal": "B", "content": "$$3$$个 "}], [{"aoVal": "C", "content": "$$4$$个 "}], [{"aoVal": "D", "content": "$$5$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->整除的概念与基本性质", "课内体系->能力->运算能力"], "answer_analysis": ["由题意知$$ab=b$$，  所以$$b(a-1)=0$$，  解得$$a=1$$或$$b=0$$，  当$$a=1$$时，只有$$b=5$$符合题意，  当$$b=0$$时，$$a=3$$，$$6$$或$$9$$均符合题意，  所以，符合题意的两位数有$$4$$个，即$$15$$，$$30$$，$$60$$，$$90$$．  故选($$\\text{C}$$)． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1034", "queId": "ff8080814d7978b9014d865117102393", "competition_source_list": ["1990年第1届全国希望杯初一竞赛初赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "如果$$a$$，$$b$$都代表有理数，并且$$a+b=0$$，那么（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$，$$b$$都是$$0$$ "}], [{"aoVal": "B", "content": "$$a$$，$$b$$之一是$$0$$ "}], [{"aoVal": "C", "content": "$$a$$，$$b$$互为相反数 "}], [{"aoVal": "D", "content": "$$a$$，$$b$$互为倒数 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->相反数->相反数的性质"], "answer_analysis": ["令$$a=2$$，$$b=-2$$，满足$$2+(-2)=0$$，由此可排出$$\\text{A}$$，$$\\text{B}$$，$$\\text{D}$$．  事实上，$$a+b=0\\Rightarrow a=-b$$，表明$$a$$，$$b$$互为相反数． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "844", "queId": "6512d8d120be43ceb028386da34561ae", "competition_source_list": ["2002年竞赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设关于$$x$$的方程$$a{{x}^{2}}+\\left( a+2 \\right)x+9a=0$$有两个不相等的实数根$${{x}_{1}}$$、$${{x}_{2}}$$，且$${{x}_{1}}\\textless1\\textless{{x}_{2}}$$，则$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{2}{7}\\textless{}a\\textless\\frac{2}{5}$$ "}], [{"aoVal": "B", "content": "$$a\\textgreater\\frac{2}{5}$$ "}], [{"aoVal": "C", "content": "$$a\\textless-\\frac{2}{7}$$ "}], [{"aoVal": "D", "content": "$$-\\frac{2}{11}\\textless{}a\\textless0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->由一元二次方程根的情况确定参数", "课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->一元二次方程根的判别式"], "answer_analysis": ["由于方程有两个不相等的实根，故$$a\\ne 0$$，  原方程可变形为$${{x}^{2}}+\\left( 1+\\frac{2}{a} \\right)x+9=0$$，  记$$y={{x}^{2}}+\\left( 1+\\frac{2}{a} \\right)x+9$$，  则这个抛物线开口向上，  因$${{x}_{1}}\\textless{}1\\textless{}{{x}_{2}}$$，  故当$$x=1$$时，$$y\\textless{}0$$，即$$1+\\left( 1+\\frac{2}{a} \\right)+9\\textless{}0$$，  解得$$-\\frac{2}{11}\\textless{}a\\textless{}0$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1470", "queId": "ca8161edc4634467b6f7e98eabf9b56b", "competition_source_list": ["2018年全国初中数学联赛竞赛A卷"], "difficulty": "4", "qtype": "single_choice", "problem": "设$$p$$，$$q$$均为大于$$3$$的素数，则使$${{p}^{2}}+5pq+4{{q}^{2}}$$为完全平方数的素数对$$(p,q)$$的个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->能力->抽象概括能力", "课内体系->能力->运算能力", "课内体系->知识点->式->因式分解->因式分解的基础->已知因式分解结果求参数", "课内体系->知识点->式->因式分解->因式分解的应用", "竞赛->知识点->数论->整除->素数与合数"], "answer_analysis": ["设$${{p}^{2}}+5pq+4{{q}^{2}}={{m}^{2}}$$ ($$m$$为自然数)，则$${{(p+2q)}^{2}}+pq={{m}^{2}}$$，即  $$(m-p-2q)(m+p+2q)=pq$$．  由于$$pq$$为素数，且$$m+p+2q\\textgreater p$$，$$m+p+2q\\textgreater q$$，  所以$$m-p-2q=1$$，$$m+p+2q=pq$$，  从而$$pq-2p-4q-1=0$$，即$$(p-4)(q-2)=9$$，  所以，$$(p,q)=(5,11)$$或$$(7,5)$$．  所以，满足条件的素数对$$(p,q)$$的个数为$$2$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "670", "queId": "5acb0d64de0d4721a9d070110cf500fd", "competition_source_list": ["1992年第3届希望杯初二竞赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$a\\textgreater b\\textgreater0$$，则有．", "answer_option_list": [[{"aoVal": "A", "content": "$$a+b\\textgreater1$$ "}], [{"aoVal": "B", "content": "$$ab\\textgreater1$$ "}], [{"aoVal": "C", "content": "$$\\frac{a}{b}\\textgreater1$$ "}], [{"aoVal": "D", "content": "$$a-b\\textgreater1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["用特殊值法，不妨设$$a=0.4$$，$$b=0.2$$，则$$a+b=0.6 ~\\textless{} ~1$$，可排除（$$\\text{A}$$）；  $$ab=0.08 ~\\textless{} ~1$$，可排除（$$\\text{B}$$）；  $$a-b=0.4-0.2=0.2 ~\\textless{} ~1$$，可排除（$$\\text{D}$$），所以应选（$$\\text{C}$$）．  事实上，$$a\\textgreater0$$，$$b\\textgreater0$$，且$$a\\textgreater b$$，所以$$\\frac{a}{b}\\textgreater1$$成立．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1478", "queId": "fca8c86bbf1b43c5a6644218f9f9f093", "competition_source_list": ["2019年湖南长沙雨花区湖南广益实验中学初一竞赛（广益杯）第2题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "在$$-\\left( -8 \\right)$$，$${{\\left( -1 \\right)}^{2000}}$$，$$-{{3}^{2}}$$，$$\\left\\textbar{} -1 \\right\\textbar$$，$$-\\left\\textbar{} -3 \\right\\textbar$$中，负数共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$个 "}], [{"aoVal": "B", "content": "$$3$$个 "}], [{"aoVal": "C", "content": "$$2$$个 "}], [{"aoVal": "D", "content": "$$1$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->有理数乘方运算", "课内体系->知识点->数->有理数->正数和负数->正数、负数定义"], "answer_analysis": ["$$-\\left( -8 \\right)=8$$，$${{\\left( -1 \\right)}^{2000}}=1$$，$$-{{3}^{2}}=-9$$，$$\\left\\textbar{} -1 \\right\\textbar=1$$，$$-\\left\\textbar{} -3 \\right\\textbar=-3$$，  所以负数有$$-{{3}^{2}}$$和$$-\\left\\textbar{} -3 \\right\\textbar$$，共$$2$$个，  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "913", "queId": "5ccef22943b545ca967e34dddf0f3610", "competition_source_list": ["2018年山东淄博临淄区初三一模第6题4分", "2018年陕西西安灞桥区西安铁一中东城滨河中学初三四模第6题3分", "2017~2018学年广东东莞市万江区翰林实验学校初三上学期期中第6题3分", "2018年浙江宁波余姚市余姚市实验学校初二竞赛第2题5分", "2018~2019学年山东济宁曲阜市初三上学期期中第7题3分", "2018~2019学年山东济宁曲阜市曲阜市实验中学初三上学期期中第7题3分", "2017~2018学年山东临沂蒙阴县初三上学期期末第6题3分", "2019~2020学年9月四川达州渠县四川省渠县中学初三上学期月考第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$${{x}_{1}},{{x}_{2}}$$是方程$${{x}^{2}}-2mx+{{m}^{2}}-m-1=0$$的两个根，且$${{x}_{1}}+{{x}_{2}}=1-{{x}_{1}}{{x}_{2}}$$，则$$m$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$或$$2$$ "}], [{"aoVal": "B", "content": "$$1$$或$$-2$$ "}], [{"aoVal": "C", "content": "$$-2$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->一元二次方程->根与系数的关系->一元二次方程根与系数的关系"], "answer_analysis": ["根据韦达定理，$${{x}_{1}}+{{x}_{2}}=2m$$，$${{x}_{1}}{{x}_{2}}={{m}^{2}}-m-1$$．  $${{x}_{1}}+{{x}_{2}}=1-{{x}_{1}}{{x}_{2}}$$，  ∴$$2m=1-({{m}^{2}}-m-1)$$，  解得$$m=-2$$或$$1$$，  而$$\\Delta =4{{m}^{2}}-4({{m}^{2}}-m-1)\\geqslant0$$，  ∴$$m\\textgreater-1$$．  ∵$${{x}_{1}}$$，$${{x}_{2}}$$是方程$${{x}^{2}}-2mx+{{m}^{2}}-m-1=0$$两个根，  ∴$${{x}_{1}}+{{x}_{2}}=2m$$，$${{x}_{1}}\\cdot {{x}_{2}}={{m}^{2}}-m-1$$．  ∵$${{x}_{1}}+{{x}_{2}}=1-{{x}_{1}}{{x}_{2}}$$，  ∴$$2m=1-({{m}^{2}}-m-1)$$，即$${{m}^{2}}+m-2=0$$，  解得：$${{m}_{1}}=-2$$，$${{m}_{2}}=1$$．  ∵方程$${{x}^{2}}-2mx+{{m}^{2}}-m-1=0$$有实数根，  ∴$$\\Delta ={{(-2m)}^{2}}-4({{m}^{2}}-m-1)=4m+4\\geqslant 0$$，  解得：$$m\\geqslant -1$$．  ∴$$m=1$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "343", "queId": "98fd1084c4c143ab805f2ae08678b068", "competition_source_list": ["2017年第19届浙江宁波余姚市余姚市实验学校初三竞赛（实验杯）第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "仔细观察一列数中的前五个数：$$0$$，$$9$$，$$26$$，$$65$$，$$124$$，则根据规律，第六个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$186$$ "}], [{"aoVal": "B", "content": "$$215$$ "}], [{"aoVal": "C", "content": "$$216$$ "}], [{"aoVal": "D", "content": "$$217$$ "}]], "knowledge_point_routes": ["课内体系->知识点->综合与实践->规律探究与程序框图->规律探究->数列找规律->数列找规律-其他数列规律"], "answer_analysis": ["规律为$${{1}^{3}}-1$$，$${{2}^{3}}+1$$，$${{3}^{3}}-1$$，$${{4}^{3}}+1$$，$${{5}^{3}}-1$$，$${{6}^{3}}+1$$，  ∴$${{6}^{3}}+1=217$$．  故$$\\text{ABC}$$错误，$$\\text{D}$$正确．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "33", "queId": "02b66527a4e948e1b3e11a7d31653947", "competition_source_list": ["2013年第24届全国希望杯初二竞赛复赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列命题中，正确的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "如果三角形三个内角的度数比是$$3:4:5$$，那么这个三角形是直角三角形 "}], [{"aoVal": "B", "content": "如果直角三角形的两条直角边的长分别是$$a$$和$$b$$，那么斜边的长是$${{a}^{2}}+{{b}^{2}}$$ "}], [{"aoVal": "C", "content": "如果三角形三条边长的比是$$1:2:3$$，那么这个三角形是直角三角形 "}], [{"aoVal": "D", "content": "如果直角三角形的两条直角边的长分别是$$a$$和$$b$$，斜边长是$$c$$，那么斜边上的高的长是$$\\frac{ab}{c}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->勾股定理及应用->勾股定理基础->勾股定理的逆定理"], "answer_analysis": ["如果三角形三个内角的度数比是$$3:4:5$$，可得这个三角形三个内角的度数分别是$$45{}^{}\\circ $$，$$60{}^{}\\circ $$，$$75{}^{}\\circ $$，因此这个三角形不是直角三角形，$$\\text{A}$$选项错误；  如果直角三角形的两条直角边的长分别是$$a$$和$$b$$，那么斜边的长是$$\\sqrt{{{a}^{2}}+{{b}^{2}}}$$，$$\\text{B}$$选项错误；  长度比是$$1:2:3$$的三条线段不能构成三角形，$$\\text{C}$$选项错误；  如果直角三角形的两条直角边的长分别是$$a$$和$$b$$，斜边长是$$c$$，  设斜边上的高的长是$$h$$，  由三角形的面积公式知$$\\frac{1}{2}ab=\\frac{1}{2}ch$$，得$$h=\\frac{ab}{c}$$，$$\\text{D}$$选项正确． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "911", "queId": "850bdfd19cc74944b0ff8f9bf061786b", "competition_source_list": ["1997年第8届希望杯初二竞赛第2试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\triangle ABC$$的三边长为$$a$$，$$b$$，$$c$$，满足条件$$\\frac{2}{b}=\\frac{1}{a}+\\frac{1}{c}$$，则$$b$$边所对的角$$B$$的大小是．", "answer_option_list": [[{"aoVal": "A", "content": "锐角 "}], [{"aoVal": "B", "content": "直角 "}], [{"aoVal": "C", "content": "钝角 "}], [{"aoVal": "D", "content": "锐角、直角、钝角都有可能 "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["若$$a\\geqslant c$$，则$$\\frac{1}{a}\\leqslant \\frac{1}{c}$$，  所以$$\\frac{2}{b}=\\frac{1}{a}+\\frac{1}{c}\\geqslant \\frac{2}{a}$$，  所以$$b\\leqslant a$$，  所以$$\\angle B\\leqslant \\angle A$$，$$\\angle B$$为锐角，  同理，若$$a\\leqslant c$$，可知$$b\\leqslant c$$，  所以$$\\angle B\\leqslant \\angle C$$，$$\\angle B$$为锐角，  所以选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "112", "queId": "86f53e57b46149ca8c9a4c7e8a2824ed", "competition_source_list": ["2018年第29届希望杯初一竞赛初赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$\\textbar a\\textbar\\textless{}1$$，$$\\textbar b\\textbar\\textless{}1$$，$$\\textbar c\\textbar\\textless{}1$$时，给出下列判断：  ①$$\\textbar abc\\textbar\\textless{}1$$；②$$\\textbar a+b+c\\textbar\\textless{}3$$；③$$ab+bc+ca\\textless{}1$$；④$$ab+bc+ca\\textgreater-1$$．  其中，正确判断的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->绝对值的非负性", "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质", "课内体系->能力->分析和解决问题能力"], "answer_analysis": ["①因为$$\\textbar a\\textbar\\textless{}1$$，$$\\textbar b\\textbar\\textless{}1$$，$$\\textbar c\\textbar\\textless{}1$$，  所以$$\\textbar abc\\textbar=\\textbar a\\textbar\\cdot\\textbar b\\textbar\\cdot\\textbar c\\textbar\\textless1$$，正确．  ②$$\\textbar a+b+c\\textbar\\leqslant \\textbar a\\textbar+\\textbar b\\textbar+\\textbar c\\textbar\\textless{}3$$，正确．  ③令$$a=b=c=\\frac{2}{3}$$，  可得$$ab+bc+ca=\\frac{4}{3}$$，说法错误．  ④将$$ab+bc+ac+1$$看成$$a$$的函数，  则$$f(a)=ab+bc+ac+1=(b+c)a+bc+1$$是一次函数或常值函数，  在$$-1\\textless a\\textless1$$上的图象是一条线段．  因为$$f(1)=b+c+bc+1=(b+1)(c+1)$$，且$$\\textbar b\\textbar\\textless1$$，$$\\textbar c\\textbar\\textless1$$，  所以$$f(1)\\textgreater0$$．  因为$$f(-1)=-b-c+bc+1=(b-1)(c-1)$$，且$$\\textbar b\\textbar\\textless1$$，$$\\textbar c\\textbar\\textless1$$，  所以$$f(-1)\\textgreater0$$．  所以对任何$$-1\\textless a\\textless1$$，都有$$f(a)\\textgreater0$$，  即$$ab+bc+ac+1\\textgreater0$$，  即$$ab+bc+ca\\textgreater-1$$，正确．  故正确的有$$3$$个． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1508", "queId": "b8d56a037b644462bb7217dbf381c374", "competition_source_list": ["2014年第31届全国全国初中数学联赛竞赛第2题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知非负实数$$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}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->因式分解->因式分解：添项、拆项", "课内体系->知识点->式->整式的乘除->乘法公式->配方求最值"], "answer_analysis": ["\\textbf{（知识点：配方法求最值问题）}  $$t=2xy+yz+2zx$$  $$=2x(y+z)+yz$$  $$\\leqslant 2x(y+z)+\\frac{1}{4}{{(y+z)}^{2}}$$  $$=2x(1-x)+\\frac{1}{4}{{(1-x)}^{2}}$$  $$=-\\frac{7}{4}{{x}^{2}}+\\frac{3}{2}x+\\frac{1}{4}$$  $$=-\\frac{7}{4}{{\\left( x-\\frac{3}{7} \\right)}^{2}}+\\frac{4}{7}$$，  易知：当$$x=\\frac{3}{7}$$，$$y=z=\\frac{2}{7}$$时，  $$t=2xy+yz+2zx$$取得最大值$$\\frac{4}{7}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "406", "queId": "54dcdbbf2cf44b169b3821ea01552dbc", "competition_source_list": ["2001年第12届希望杯初一竞赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "下面四个命题中，正确的命题是．", "answer_option_list": [[{"aoVal": "A", "content": "两个不同的整数之间必定有一个正数 "}], [{"aoVal": "B", "content": "两个不同的整数之间必定有一个整数 "}], [{"aoVal": "C", "content": "两个不同的整数之间必定有一个有理数 "}], [{"aoVal": "D", "content": "两个不同的整数之间必定有一个负数 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["$$\\text{A}$$、$$\\text{B}$$．$$-1$$与$$0$$是两个相邻的整数，它们之间既没有整数，也没在正数，故$$\\text{A}$$、$$\\text{B}$$错误；  $$\\text{C}$$．事实上，$$a$$，$$b$$是两个不同的整数（设$$a\\textless{}b$$），则$$\\frac{a+b}{2}$$是有理数且成立$$a\\textless{}\\frac{a+b}{2}\\textless{}b$$，即$$\\frac{a+b}{2}$$在$$a$$，$$b$$之间，故$$\\text{C}$$正确；  $$\\text{D}$$．而两个正整数之间的数都是正数，故$$\\text{D}$$错误；  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1311", "queId": "8aac50a74fb1a33e014feb23fd571cbc", "competition_source_list": ["2012年浙江宁波余姚中学自主招生第5题7分", "2011年竞赛第5题3分", "初一下学期单元测试《放缩法》第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$S=\\frac{1}{{{1}^{3}}}+\\frac{1}{{{2}^{3}}}+\\cdots +\\frac{1}{{{99}^{3}}}$$．则$$4S$$的整数部分等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->综合与实践->规律探究与程序框图->规律探究->算式找规律", "课内体系->知识点->式->整式的加减->整式的加减运算"], "answer_analysis": ["当$$n=2$$，$$3$$，$$\\cdot \\cdot \\cdot $$，$$99$$时，  ∵$$\\frac{1}{{{n}^{3}}}\\textless{}\\frac{1}{n({{n}^{2}}-1)}=\\frac{1}{2}\\left[ \\frac{1}{(n-1)n}-\\frac{1}{n(n+1)} \\right]$$，  ∴$$1\\textless{}S=\\frac{1}{{{1}^{3}}}+\\frac{1}{{{2}^{3}}}+\\cdot \\cdot \\cdot +\\frac{1}{{{99}^{3}}}\\textless{}1+\\frac{1}{2}\\left( \\frac{1}{2}-\\frac{1}{99\\times 100} \\right)\\textless{}\\frac{5}{4}$$，  于是有$$4\\textless{}4S\\textless{}5$$，得$$4S$$的正数部分为$$4$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1308", "queId": "a9e2005e086942779a12ed95d8c6bf90", "competition_source_list": ["2018年第29届希望杯初一竞赛初赛第10题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "满足$$a+c=2b$$的三位数$$\\overline{abc}$$共有个．", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$49$$ "}], [{"aoVal": "E", "content": "$$52$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->计数问题-枚举法", "课内体系->知识点->方程与不等式->其他方程->不定方程"], "answer_analysis": ["若$$b=0$$，则$$a+c=0$$，不可能  若$$b=1$$，则$$a+c=2$$，有$$2$$种情况；  若$$b=2$$，则$$a+c=4$$，有$$4$$种情况；  若$$b=3$$，则$$a+c=6$$，有$$6$$种情况；  若$$b=4$$，则$$a+c=8$$，有$$8$$种情况；  若$$b=5$$，则$$a+c=10$$，有$$9$$种情况；  若$$b=6$$，则$$a+c=12$$，有$$7$$种情况；  若$$b=7$$，则$$a+c=14$$，有$$5$$种情况；  若$$b=8$$，则$$a+c=16$$，有$$3$$种情况；  若$$b=9$$，则$$a+c=18$$，有$$1$$种情况；  综上，共有$$2+4+6+8+9+7+5+3+1=45$$（种）情况． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "705", "queId": "f18ff06e44b542ecb9fe71fdaa40cdbf", "competition_source_list": ["2012年第23届全国希望杯初一竞赛复赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "(★★)若$$x+y=3$$，$$xy=1$$．则$${{x}^{5}}+{{y}^{5}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$33$$ "}], [{"aoVal": "B", "content": "$$231$$ "}], [{"aoVal": "C", "content": "$$123$$ "}], [{"aoVal": "D", "content": "$$312$$ "}]], "knowledge_point_routes": ["知识标签->知识点->式->整式的乘除->乘法公式->完全平方公式", "知识标签->知识点->式->整式的乘除->乘法公式->立方和与立方差公式", "知识标签->题型->式->整式的乘除->乘法公式->题型：和与差的立方公式"], "answer_analysis": ["因为$$x+y=3$$，  所以$${{(x+y)}^{2}}=9$$，  即$${{x}^{2}}+{{y}^{2}}+2xy=9$$．  因为$$xy=1$$，  所以$${{x}^{2}}+{{y}^{2}}=7$$．  所以$${{x}^{3}}+{{y}^{3}}=(x+y)({{x}^{2}}+{{y}^{2}}-xy)=3\\times (7-1)=18$$，  所以$${{x}^{5}}+{{y}^{5}}=({{x}^{3}}+{{y}^{3}})({{x}^{2}}+{{y}^{2}})-{{(xy)}^{2}}(x+y)=18\\times 7-{{1}^{2}}\\times 3=123$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1173", "queId": "a96715b6edb141758f2dc2ad232070d3", "competition_source_list": ["1999年第16届全国初中数学联赛竞赛第1题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "计算$$\\frac{1}{1-\\sqrt[4]{3}}+\\frac{1}{1+\\sqrt[4]{3}}+\\frac{2}{1+\\sqrt{3}}$$的值是（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$-2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->二次根式->二次根式的运算"], "answer_analysis": ["通分得  原式$$=\\frac{2}{1-\\sqrt{3}}+\\frac{2}{1+\\sqrt{3}}=\\frac{4}{1-3}=-2$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1049", "queId": "ff8080814d7978b9014d86d48d84257a", "competition_source_list": ["1991年第2届全国希望杯初一竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$为有理数，则一定成立的关系式是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$7a\\textgreater{}a$$ "}], [{"aoVal": "B", "content": "$$7+a\\textgreater{}a$$ "}], [{"aoVal": "C", "content": "$$7+a\\textgreater7$$ "}], [{"aoVal": "D", "content": "$$\\left\\textbar{} a \\right\\textbar\\geqslant 7$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["若$$a=0$$，$$7\\times 0=0$$，排除$$\\text{A}$$；  $$7+0=7$$，排除$$\\text{C}$$；  $$\\left\\textbar{} 0 \\right\\textbar\\textless7$$，排除$$\\text{D}$$；  事实上因为$$7\\textgreater0$$，必有$$7+a\\textgreater0+a=a$$．选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1310", "queId": "a55882b1bab94d14a2f0777b24a46e67", "competition_source_list": ["2006年竞赛第1题6分", "2018~2019学年浙江绍兴绍兴县绍兴市元培中学初一下学期期末第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "在高速公路上，从$$3$$千米处开始，每隔$$4$$千米经过一个限速标志牌；并且从$$10$$千米处开始，每隔$$9$$千米经过一个速度监控仪．刚好在$$19$$千米处第一次同时经过这两种设施，那么第二次同时经过这两种设施的千米数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$ "}], [{"aoVal": "B", "content": "$$37$$ "}], [{"aoVal": "C", "content": "$$55$$ "}], [{"aoVal": "D", "content": "$$90$$ "}]], "knowledge_point_routes": ["课内体系->能力->分析和解决问题能力", "课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数与实际问题->有理数加减法与实际问题", "竞赛->知识点->数论->整除->因数与倍数"], "answer_analysis": ["因为$$4$$和$$9$$的最小公倍数为$$36$$，$$19+36=55$$，所以第二次同时经过这两种设施是在$$55$$千米处．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "552", "queId": "b530da51ec03480cb744689de181de57", "competition_source_list": ["2009年第20届希望杯初一竞赛第1试第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在用数字$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$组成的没有重复数字的三位数中，是$$9$$的倍数的数有（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$个 "}], [{"aoVal": "B", "content": "$$18$$个 "}], [{"aoVal": "C", "content": "$$20$$个 "}], [{"aoVal": "D", "content": "$$30$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->因数与倍数", "竞赛->知识点->数论->整除->整除的概念与基本性质", "课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数除法运算"], "answer_analysis": ["因为能被$$9$$整除的数的各位数字之和能被$$9$$整除，  所以从数字$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$中随机抽取$$3$$个数组成的没有重复数字的三位数，  且是$$9$$的倍数的数有以下三种情况：  （$$1$$）由数字$$1$$，$$3$$，$$5$$可组成$$6$$个三位数；  （$$2$$）由数字$$2$$，$$3$$，$$4$$可组成$$6$$个三位数；  （$$3$$）由数字$$1$$，$$2$$，$$5$$可组成$$6$$个三位数．  则共可组合$$18$$个满足题意的三位数，  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1293", "queId": "dbbc492136f844d3ad9bba73f4e4c133", "competition_source_list": ["2002年第19届全国初中数学联赛竞赛第2题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$${{m}^{2}}=n+2$$，$${{n}^{2}}=m+2(m\\ne n)$$，则$${{m}^{3}}-2mn+{{n}^{3}}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "$$-2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->整式的乘除运算"], "answer_analysis": ["$${{m}^{2}}=n+2$$，$${{n}^{2}}=m+2$$，  两式相减得到$${{m}^{2}}-{{n}^{2}}=n-m\\Rightarrow (m-n)(m+n+1)=0$$．  又$$m\\ne n$$，所以$$m+n=-1$$，  由立方和公式$${{m}^{3}}-2mn+{{n}^{3}}=(m+n)({{m}^{2}}-mn+{{n}^{2}})-2mn$$  $$=-({{m}^{2}}-mn+{{n}^{2}})-2mn$$  $$=-{{m}^{2}}-mn-{{n}^{2}}$$  $$=-m(m+n)-{{n}^{2}}$$  $$=m-{{n}^{2}}$$．  又由已知条件$${{n}^{2}}=m+2$$，所以$${{m}^{3}}-2mn+{{n}^{3}}=m-{{n}^{2}}=-2$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "298", "queId": "7021c4c21b514a78ae32968c931827ba", "competition_source_list": ["2018年湖南长沙初二竞赛长郡教育集团（觉园杯）第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$xyz\\ne 0$$，且$$x+2y+z=0$$，$$5x+y-4z=0$$，则$$\\frac{{{x}^{2}}+6{{y}^{2}}-10{{z}^{2}}}{3{{x}^{2}}-4yz+5{{z}^{2}}}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{7}{46}$$ "}], [{"aoVal": "B", "content": "$$\\frac{15}{46}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式化简求值->分式化简求值-条件化简求值"], "answer_analysis": ["由$$\\begin{cases}x+2y+z=0    5x+4y-4z=0   \\end{cases}$$得$$\\left { \\begin{array}{*{35}{l}} x=2z    y=-\\dfrac{3}{2}z   \\end{array} \\right.$$，  ∴原式$$=\\frac{4{{z}^{2}}+6\\times \\dfrac{9}{4}{{z}^{2}}-10{{z}^{2}}}{3\\times 4{{z}^{2}}+6{{z}^{2}}+5{{z}^{2}}}=\\frac{15}{46}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1579", "queId": "d98f78911a56449c96fd3fbd73675c7d", "competition_source_list": ["1978年竞赛第15题", "1978年Math League竞赛(7年级[.org])第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "集合$$S$$ 只有一个子集，则集合$$S$$中有几个元素?.", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}], [{"aoVal": "E", "content": "none of these "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Combinatorics->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets", "美国AMC8->Knowledge Point->Combinatorics->Inclusion-Exclusion Principle->Counting Number of Figures using Multi-Set Inclusion-Exclusion Principle", "课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->一元一次不等式的解集"], "answer_analysis": ["The only set with exactly one sunset is the empty set.  集合 $$S$$只有一个子集 ， 集合 $$S$$中有多少个元素 ?  $$\\text{A.}$$$$0$$~ ~ ~ ~ ~ $$\\text{B.}$$$$1$$~ ~ ~ ~ ~ $$\\text{C.}$$$$2$$~ ~ ~ ~ ~ $$\\text{D.}$$没有选项  只有一个子集的集合是空集．故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "530", "queId": "31b4f6f6d67f43629b32d9da88b0dadb", "competition_source_list": ["2005年第16届希望杯初一竞赛复赛第1题4分", "2018~2019学年3月江苏苏州苏州工业园区东沙湖学校初一下学期月考第8题2分", "2019年河北保定莲池区初三中考一模第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$${{(a+b)}^{2}}-{{(a-b)}^{2}}=4$$，则一定成立的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$是$$b$$的相反数 "}], [{"aoVal": "B", "content": "$$a$$是$$-b$$的相反数 "}], [{"aoVal": "C", "content": "$$a$$是$$b$$的倒数 "}], [{"aoVal": "D", "content": "$$a$$是$$-b$$的倒数 "}]], "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": ["化简得$${{(a+b)}^{2}}-{{(a-b)}^{2}}=4ab=4$$，  故$$ab=1$$，$$a$$是$$b$$的倒数． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "599", "queId": "5a4cf3a5b9ed4d3cbd5fa4153bba7d2c", "competition_source_list": ["2013年第24届全国希望杯初二竞赛复赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知有理数$$a$$，$$b$$，$$x$$，$$y$$满足$$ax+by=3$$，$$ax-by=5$$，那么$$({{a}^{2}}+{{b}^{2}})({{x}^{2}}+{{y}^{2}})$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$225$$ "}], [{"aoVal": "B", "content": "$$75$$ "}], [{"aoVal": "C", "content": "$$54$$ "}], [{"aoVal": "D", "content": "$$34$$ "}]], "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"], "answer_analysis": ["$$({{a}^{2}}+{{b}^{2}})({{x}^{2}}+{{y}^{2}})$$  $$={{(ax+by)}^{2}}+{{(ax-by)}^{2}}$$  $$={{3}^{2}}+{{5}^{2}}$$  $$=34$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "725", "queId": "3c4d16aa82244e30a4c9694c41c6ac2c", "competition_source_list": ["1984年竞赛第36题"], "difficulty": "1", "qtype": "single_choice", "problem": "1984年$$AMC8$$竞赛第$$36$$题  If $$3x=21$$, then $$21x=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$63$$ "}], [{"aoVal": "D", "content": "$$147$$ "}]], "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", "美国AMC8->Knowledge Point->Algebra and Sequences->Equations->Linear Equations"], "answer_analysis": ["If $$3x=21$$, then $$7\\times 3x=7\\times 21$$ or $$21x=147$$.  如果$$3x=21$$，那么$$21x=$$．  $$\\text{A}$$．$$3$$~ ~ ~ $$\\text{B}$$．$$7$$~ ~ ~ $$\\text{C}$$．$$63$$~ ~ ~ $$\\text{D}$$．$$147$$  如果$$3x=21$$，那么$$21x=7\\times 21=147$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1178", "queId": "8aac49074e023206014e20e7205b6619", "competition_source_list": ["1994年第5届全国希望杯初一竞赛复赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "据报道目前用超级计算机找到的最大质数是$${{2}^{859433}}-1$$，这个质数的末尾数字是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "knowledge_point_routes": ["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", "课内体系->能力->推理论证能力"], "answer_analysis": ["$${{2}^{n}}$$的末位数字对指数以$$4$$为周期而变化，  $${{2}^{1}}=2$$，$${{2}^{2}}=4$$，$${{2}^{3}}=8$$，$${{2}^{4}}$$末位数是$$6$$．  一般地$${{2}^{4k+1}}$$末位数字为$$2$$，$${{2}^{4k+2}}$$末位数字为$$4$$，$${{2}^{4k+3}}$$末位数字为$$8$$，$${{2}^{4k+4}}$$末位数字是$$6$$（其中k是非负整数）．  $$859433=214858\\times 4+1$$，$${{2}^{859433}}={{2}^{4\\times 214858+1}}$$  ∴$${{2}^{859433}}$$末位数字为$$2$$．  ∴$${{2}^{859433}}-1$$末位数字为$$1$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1419", "queId": "f30e3a5d7ba0471d9aaaaedb6b9da5d0", "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": "258", "queId": "873098b1c11c4ce0a8f52b782ee3c437", "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", "2019~2020学年河南郑州金水区郑州八中初一上学期期中第8~8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$a$$，$$b$$，$$c$$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": ["知识标签->知识点->数->有理数->绝对值->绝对值的性质", "知识标签->题型->式->分式->分式化简求值->题型：含绝对的分式化简", "知识标签->学习能力->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$$a\\leqslant b\\leqslant c$$，  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": "110", "queId": "45fb70dc19a847278a64cb4f76944c24", "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": "在下列$$2021$$个数中：$$1^{2}$$，$$2^{2}$$，$$3^{2}$$，$$\\cdots $$，$$2020^{2}$$，$$2021^{2}$$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->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，$$1^{2}$$，$$2^{2}$$，$$3^{2}$$，$$\\cdots $$，$$2020^{2}$$，$$2021^{2}$$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": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "487", "queId": "2d031b05a15847088936ef98e464824f", "competition_source_list": ["1998年第9届希望杯初二竞赛第1试第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$，$$c$$，$$d$$为正实数，且$${{a}^{2}}=2$$，$${{b}^{3}}=3$$，$${{c}^{4}}=4$$，$${{d}^{5}}=5$$，则$$a$$，$$b$$，$$c$$，$$d$$中最大的数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$ "}], [{"aoVal": "B", "content": "$$b$$ "}], [{"aoVal": "C", "content": "$$c$$ "}], [{"aoVal": "D", "content": "$$d$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->幂的运算->幂的乘方", "课内体系->能力->运算能力"], "answer_analysis": ["因为$${{a}^{2}}=2$$，$${{c}^{4}}=4$$，  所以$${{c}^{2}}=2={{a}^{2}}$$，$$a=c$$．  又因为$${{a}^{6}}={{({{a}^{2}})}^{3}}=8$$，$${{b}^{6}}={{({{b}^{3}})}^{2}}=9$$．  所以$$b\\textgreater a=c$$．  最后比较$$b$$与$$d$$的大小．  因为$${{b}^{15}}={{({{b}^{3}})}^{5}}=243$$，$${{d}^{15}}={{({{d}^{5}})}^{3}}=125$$．  所以$$b\\textgreater d$$．  所以$$a$$，$$b$$，$$c$$，$$d$$中$$b$$最大．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1481", "queId": "c6172be053ad450689cf3e890189f697", "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": "$$1 ~\\textless{} ~m ~\\textless{} ~2$$ "}]], "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": "241", "queId": "ec52a08130654aeda55fdc1abb2ee5e9", "competition_source_list": ["2014年第31届全国全国初中数学联赛竞赛第1题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x$$，$$y$$为整数，且满足$$\\left( \\frac{1}{x}+\\frac{1}{y} \\right)\\left( \\frac{1}{{{x}^{2}}}+\\frac{1}{{{y}^{2}}} \\right)=-\\frac{2}{3}\\left( \\frac{1}{{{x}^{4}}}-\\frac{1}{{{y}^{4}}} \\right)$$，则$$x+y$$的可能的值有．", "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->方程与不等式->分式方程->解复杂分式方程"], "answer_analysis": ["由已知等式得$$\\frac{x+y}{xy}\\cdot \\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}{{y}^{2}}}=\\frac{2}{3}\\cdot \\frac{{{x}^{4}}-{{y}^{4}}}{{{x}^{4}}{{y}^{4}}}$$，  显然$$x$$，$$y$$均不为$$0$$，  所以$$x+y=0$$或$$3xy=2(x-y)$$．  若$$3xy=2(x-y)$$，则$$(3x+2)(3y-2)=-4$$．  又$$x$$，$$y$$为整数，可求得$$\\left { \\begin{matrix}x=-1    y=2   \\end{matrix} \\right.$$或$$\\left { \\begin{matrix}x=-2    y=1   \\end{matrix} \\right.$$  所以$$x+y=1$$或$$x+y=-1$$．  因此，$$x+y$$的可能的值有$$3$$个． ", "<p><span><span><span>&nbsp;显然</span>$$x\\ne y$$<span>（否则原式不成立），故原式两边可同时乘上</span></span></span>$$\\left( \\frac{1}{x}-\\frac{1}{y} \\right)$$<span><span><span>，化为：</span>$$3k+1$$，</span></span></p>\n<p><span><span><span>（$$1$$）若</span>$$\\frac{1}{{{x}^{4}}}-\\frac{1}{{{y}^{4}}}=0$$<span>，则</span>$$x=-y$$<span>，故</span>$$x+y=0$$<span>；</span></span></span></p>\n<p><span><span><span>（$$2$$）若</span>$$f(x,y,z)=9(z+1)$$<span>，</span></span></span></p>\n<p><span><span><span>则</span></span></span>$$-\\frac{2}{3}\\left( \\frac{1}{x}-\\frac{1}{y} \\right)$$<span><span>$$=1\\Leftrightarrow x=\\frac{2y}{2-3y}$$，</span></span></p>\n<p style=\"text-align:justify\"><span><span><span>因为</span>$$x$$，$$y$$<span>为整数，所以</span>$$3x=\\frac{6y}{2-3y}=\\frac{4}{2-3y}-2$$<span>为整数</span>．</span></span></p>\n<p><span><span><span>易得：当</span></span></span>$$y=1$$<span><span><span>时，</span></span></span>$$x=-2$$<span><span><span>，符合题意，此时</span>$$x=z+2,y=z+1$$<span>；</span></span></span></p>\n<p><span><span><span>当</span></span></span>$$y=2$$<span><span><span>时，</span></span></span>$$x=-1$$<span><span><span>，符合题意，此时</span>$$x+y=1$$．</span></span></p>\n<p><span><span><span>所以</span>$$x+y$$<span>的可能的值有</span></span></span>$$3$$<span><span><span>个．</span></span></span></p>"], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1061", "queId": "ff8080814d7978b9014d88ba1c5d2a3d", "competition_source_list": ["1991年第2届全国希望杯初一竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个分数的分子与分母都是正整数，且分子比分母小$$1$$，若分子和分母都减去$$1$$，则所得分数为小于$$\\frac{6}{7}$$的正数，则满足上述条件的分数共有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$个 "}], [{"aoVal": "B", "content": "$$6$$个 "}], [{"aoVal": "C", "content": "$$7$$个 "}], [{"aoVal": "D", "content": "$$8$$个 "}]], "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": ["设$$a$$为正整数，该分数表示为$$\\frac{a}{a+1}$$．  依题意$$\\frac{a-1}{a}\\textless{}\\frac{6}{7}$$，解得$$0\\textless{}a\\textless{}7$$，  所以$$a$$可取$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，六个值．  因此满足条件的分数共有$$5$$个：$$\\frac{2}{3}$$，$$\\frac{3}{4}$$，$$\\frac{4}{5}$$，$$\\frac{5}{6}$$，$$\\frac{6}{7}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "574", "queId": "367ab737b4ac4f7ca3c996246a859599", "competition_source_list": ["2016年第27届全国希望杯初二竞赛初赛第2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．若一次函数$$y=x-3$$与$$y=kx-k$$（$$k$$为整数）的图象的交点是整点，则$$k$$的不同取值的个数是．", "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->能力->推理论证能力", "课内体系->知识点->函数->一次函数->一次函数与方程、不等式->一次函数与二元一次方程组"], "answer_analysis": ["由题意得$$\\begin{cases}y=x-3    y=kx-k \\end{cases}$$，  解得$$\\begin{cases}x=\\dfrac{k-3}{k-1}    y=\\dfrac{-2k}{k-1} \\end{cases}$$，  ∴$$x=1-\\frac{2}{k-1}$$，$$y=-2-\\frac{2}{k-1}$$，  ∵交点是整点，  ∴$$k-1=\\pm 1$$或$$\\pm 2$$，且$$k-1\\ne 0$$，  解得$$k=0$$，$$2$$，$$-1$$或$$3$$．  ∵$$k\\ne 0$$，  ∴$$k$$的不同取值的个数是$$3$$．  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": "198", "queId": "a6b43f976edf488785454744ca5d69e0", "competition_source_list": ["2000年第11届希望杯初二竞赛第1试第3题", "2019~2020学年12月广东深圳福田区红岭中学园岭校区初一上学期月考第12题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "适合$$\\textbar2a+7\\textbar+\\textbar2a-1\\textbar=8$$的整数$$a$$的值的个数有．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->绝对值->绝对值的几何意义"], "answer_analysis": ["$$a$$是整数，$$2a$$表示偶数，  由已知$$\\textbar2a+7\\textbar+\\textbar2a-1\\textbar=8$$，即在数轴上表示$$2a$$的点到$$-7$$与$$+1$$的距离的和等于$$8$$，所以$$2a$$表示从$$-7$$到$$1$$之间的偶数，  所以$$2a=-6$$或$$2a=-4$$，或$$2a=-2$$或$$2a=0$$，  即$$a=-3$$或$$a=-2$$或$$a=-1$$或$$a=0$$，共有$$4$$个数值．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "937", "queId": "65e5f3af08514a2992bc39531c2dc98a", "competition_source_list": ["1993年第4届希望杯初二竞赛第10题"], "difficulty": "0", "qtype": "single_choice", "problem": "如果方程$$\\left\\textbar3x\\right\\textbar-ax-1=0$$的根是负数，那么$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater-3$$ "}], [{"aoVal": "B", "content": "$$a\\geqslant 3$$ "}], [{"aoVal": "C", "content": "$$a \\textless{} 3$$ "}], [{"aoVal": "D", "content": "$$a\\leqslant -3$$ "}]], "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->不等式（组）->含参不等式->不等式（组）解的情况确定参数的范围", "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->解一元一次方程"], "answer_analysis": ["原方程即$$\\left\\textbar3x\\right\\textbar=ax+1$$．  $$\\left(1\\right)$$若$$a=3$$时，则$$\\left\\textbar3x\\right\\textbar=3x+1$$．  当$$x \\textless{} 0$$时，得$$-3x=3x+1$$．所以$$x=-\\frac{1}{6}$$；  当$$x\\geqslant 0$$时，得$$3x=3x+1$$，不成立．  所以当$$a=3$$时，原方程的根为$$x=-\\frac{1}{6}$$．  $$\\left(2\\right)$$若$$a\\textgreater3$$．  当$$x \\textless{} 0$$时，得$$-3x=ax+1$$，所以$$x=-\\frac{1}{a+3} \\textless{} 0$$；  当$$x\\geqslant 0$$时，得$$3x=ax+1$$，所以$$x=\\frac{1}{3-a} \\textless{} 0$$，矛盾．  所以当$$a\\textgreater3$$时，原方程的解为$$x=-\\frac{1}{a+3} \\textless{} 0$$．  $$\\left(3\\right)$$若$$a \\textless{} 3$$  当$$x\\geqslant 0$$时，得$$3x=ax+1$$，所以$$x=\\frac{1}{3-a}\\textgreater0$$，原方程有正根，与题设矛盾．  综上所述，$$a\\geqslant 3$$时，原方程的根是负数．  所以应选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1244", "queId": "8aac4907508d5d3d0150a493762e25bf", "competition_source_list": ["1997年第8届全国希望杯初一竞赛复赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "平面上三条直线相互间的交点个数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$1$$或$$3$$ "}], [{"aoVal": "C", "content": "$$1$$或$$2$$或$$3$$ "}], [{"aoVal": "D", "content": "不一定是$$1$$，$$2$$，$$3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->几何图形初步->线"], "answer_analysis": ["当平面上三条直线互相平行时，没有交点．故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "445", "queId": "be3fd5fae76f4534b7306f8b5a29ff9d", "competition_source_list": ["2011年竞赛第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x$$，$$y$$，$$z$$为实数，满足$$x+2y-5z=3$$，$$x-2y-z=-5$$，则$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}$$的最小值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{11}$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$\\frac{54}{11}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的加减->整式的加减运算", "课内体系->知识点->式->整式的乘除->乘法公式->配方求最值", "课内体系->能力->运算能力", "课内体系->方法->配方法"], "answer_analysis": ["由$$\\begin{cases}x+2y-5z=3    x-2y-z=-5    \\end{cases}$$，  可得$$\\begin{cases}x=3z-1    y=z+2    \\end{cases}$$，  于是$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=11{{z}^{2}}-2z+5=11\\left( {{z}^{2}}-2\\cdot z\\cdot \\frac{1}{11}+\\frac{1}{121} \\right)+5-\\frac{1}{11}$$，  $$=11{{\\left( z-\\frac{1}{11} \\right)}^{2}}+\\frac{54}{11}\\geqslant \\frac{54}{11}$$，  因此，当$$z-\\frac{1}{11}=0$$即$$z=\\frac{1}{11}$$时，$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}$$最小，最小值为$$\\frac{54}{11}$$，  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "199", "queId": "2ab7b4f741e249938ad59ae5c39a17c7", "competition_source_list": ["2004年第15届希望杯初二竞赛第1试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$x$$是实数，且$$\\left( {{x}^{2}}-9x+20 \\right)\\sqrt{3-x}=0$$，那么$${{x}^{2}}+x+1=$$(  ).", "answer_option_list": [[{"aoVal": "A", "content": "$$31$$ "}], [{"aoVal": "B", "content": "$$21$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$13$$或$$21$$或$$31$$ "}]], "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"], "answer_analysis": ["提示：$$3-x\\geqslant 0$$，$$\\therefore x\\leqslant 3$$．  又$$\\because {{x}^{2}}-9x+20=\\left( x-4 \\right)\\left( x-5 \\right)$$，  当$$x\\leqslant 3$$时，$$\\left( x-4 \\right)\\left( x-5 \\right)\\ne 0$$，  $$\\therefore \\sqrt{3-x}=0$$，$$\\therefore x=3$$．  故$${{x}^{2}}+x+1={{3}^{2}}+3+1=13$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1029", "queId": "ff8080814d56502a014d6effb5d642fe", "competition_source_list": ["初一上学期单元测试《走进美妙的数学世界》第15题", "2018~2019学年湖南长沙岳麓区长郡双语实验中学初一上学期分班考试第1题5分", "2010年第21届全国希望杯初一竞赛复赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若一个两位数恰等于它的各位数字之和的$$4$$倍，则这个两位数称为``巧数''，则不是``巧数''的两位数的个数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$82$$ "}], [{"aoVal": "B", "content": "$$84$$ "}], [{"aoVal": "C", "content": "$$86$$ "}], [{"aoVal": "D", "content": "$$88$$ "}]], "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->能力->运算能力"], "answer_analysis": ["设两位数为$$10a+b$$，其中$$a$$，$$b$$是自然数，且$$1\\leqslant a\\leqslant 9$$，$$0\\leqslant b\\leqslant 9$$．  由题意，得$$10a+b=4(a+b)$$，  即$$b=2a$$．  因为满足$$b=2a$$的两位数只有$$12$$，$$24$$，$$36$$，$$48$$这$$4$$个，  所以``巧数''只有$$4$$个，  因此不是``巧数''的两位数共有$$90-4=86$$（个）． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "573", "queId": "fa9fcfdda56342aa8f6f1fdf51270a26", "competition_source_list": ["初一下学期其它第17题", "2005年第16届希望杯初二竞赛复赛第10题4分"], "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$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=2$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1320", "queId": "8aac50a7511483070151148d18e30048", "competition_source_list": ["2015年第26届全国希望杯初二竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a+b=10$$，$$ab=24$$，则$${{a}^{2}}+{{b}^{2}}$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$76$$ "}], [{"aoVal": "C", "content": "$$58$$ "}], [{"aoVal": "D", "content": "$$52$$ "}]], "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": ["$${{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab=100-48=52$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1124", "queId": "cd93a27afeef430aa60d7ad0261cad98", "competition_source_list": ["2007年第12届华杯赛初一竞赛初赛第4题", "2019~2020学年北京海淀区北京一零一中学初一上学期单元测试《有理数》第20题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\textbar x+y\\textbar=y-x$$，则有．", "answer_option_list": [[{"aoVal": "A", "content": "$$y\\textgreater0$$，$$x\\textless{}0$$ "}], [{"aoVal": "B", "content": "$$y\\textless{}0$$，$$x\\textgreater0$$ "}], [{"aoVal": "C", "content": "$$y\\textless{}0$$，$$x\\textless{}0$$ "}], [{"aoVal": "D", "content": "$$x=0$$，$$y\\geqslant 0$$或$$y=0$$，$$x\\leqslant 0$$ "}]], "knowledge_point_routes": ["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->数->有理数->绝对值->绝对值的非负性"], "answer_analysis": ["∵$$\\textbar x+y\\textbar=y-x$$，  ∴①当$$x+y\\geqslant 0$$时，  $$x+y=y-x$$，即$$x=0$$；  ②当$$x+y\\leqslant 0$$时，  $$-x-y=y-x$$，即$$y=0$$，  综上所述，$$x=0$$，$$y\\geqslant 0$$或$$y=0$$，$$x\\geqslant 0$$．  故$$\\text{D}$$选项正确． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "706", "queId": "5212dc2bbcd94c4e96bb5e769f7489bb", "competition_source_list": ["2012年第17届华杯赛初一竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知甲瓶盐水浓度为$$8 \\%$$，乙瓶盐水浓度为$$5 \\%$$，混合后浓度为$$6.2 \\%$$．那么四分之一的甲瓶盐水与六分之一的乙瓶盐水混合后的浓度为．", "answer_option_list": [[{"aoVal": "A", "content": "$$5.5 \\%$$ "}], [{"aoVal": "B", "content": "$$6 \\%$$ "}], [{"aoVal": "C", "content": "$$6.5 \\%$$ "}], [{"aoVal": "D", "content": "$$7.5 \\%$$ "}]], "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->分式方程->分式方程与实际问题->分式方程的其他实际问题"], "answer_analysis": ["设甲瓶盐水重$$x$$，乙瓶盐水重$$y$$，所以$$8 \\%x+5 \\%y=6.2 \\%\\times \\left( x+y \\right)$$，  解之得$$\\frac{x}{y}=\\frac{2}{3}$$，若从甲瓶取$$\\frac{1}{4}$$盐水，从乙瓶取$$\\frac{1}{6}$$盐水，则混合后的浓度为$$6.5 \\%$$，运算如下：$$\\frac{8 \\%\\times \\dfrac{1}{4}x+5 \\%\\times \\dfrac{1}{6}y}{\\dfrac{1}{4}x+\\dfrac{1}{6}y}=\\frac{2x+\\dfrac{5}{6}y}{\\dfrac{1}{4}x+\\dfrac{1}{6}y} \\%=\\frac{2\\times \\dfrac{x}{y}+\\dfrac{5}{6}}{\\dfrac{1}{4}\\times \\dfrac{x}{y}+\\dfrac{1}{6}} \\%=\\frac{2\\times \\dfrac{2}{3}+\\dfrac{5}{6}}{\\dfrac{1}{4}\\times \\dfrac{2}{3}+\\dfrac{1}{6}} \\%=\\frac{13}{2} \\%=6.5 \\%$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1139", "queId": "a4a832cb749349b5907130fc18d3915a", "competition_source_list": ["2017~2018学年4月安徽滁州明光市鲁山中学初一下学期月考第2题4分", "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3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若实数$$a$$，$$b$$，$$c$$满足$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=9$$，代数式$${{\\left( a-b \\right)}^{2}}+{{\\left( b-c \\right)}^{2}}+{{\\left( c-a \\right)}^{2}}$$的最大值是（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$27$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["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", "课内体系->能力->运算能力"], "answer_analysis": ["$${{\\left( a-b \\right)}^{2}}+{{\\left( b-c \\right)}^{2}}+{{\\left( c-a \\right)}^{2}}=3({{a}^{2}}+{{b}^{2}}+{{c}^{2}})-{{(a+b+c)}^{2}}\\leqslant 3({{a}^{2}}+{{b}^{2}}+{{c}^{2}})=27$$．  当且仅当$$a+b+c=0$$时成立． ", "<p>$${{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}=2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})-2(ab+bc+ca)$$，</p>\n<p>考虑到等式$${{\\left( a+b+c \\right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\\left( ab+bc+ca \\right)$$，</p>\n<p>将两个等式相加得到$${{\\left( a-b \\right)}^{2}}+{{\\left( b-c \\right)}^{2}}+{{\\left( c-a \\right)}^{2}}+{{(a+b+c)}^{2}}=3\\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\right)=27$$，</p>\n<p>当$$a+b+c=0$$时，$${{\\left( a-b \\right)}^{2}}+{{\\left( b-c \\right)}^{2}}+{{\\left( c-a \\right)}^{2}}$$有最大值$$27$$．</p>\n<p>故选$$\\text{A}$$．</p>\n", "<p>原式$$=2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ab-2bc-2ca=18-2ab-2bc-2ca=18-\\left[ {{\\left( a+b+c \\right)}^{2}}-{{a}^{2}}-{{b}^{2}}-{{c}^{2}} \\right]=27-{{\\left( a+b+c \\right)}^{2}}\\leqslant 27$$，</p>\n<p>&there4;原式的最大值为$$27$$，当$$a+b+c=0$$时取到．</p>\n<p>故选$$\\text{A}$$．</p>\n"], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "515", "queId": "50d3e28c2cbd41a184f54f1f33a88b7d", "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{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$257$$ "}], [{"aoVal": "B", "content": "$$506$$ "}], [{"aoVal": "C", "content": "$$1007$$ "}], [{"aoVal": "D", "content": "$$2010$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->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": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1212", "queId": "8aac49074ed448ae014ee9baf3e44d17", "competition_source_list": ["初一上学期单元测试《有理数》绝对值的应用第24题", "2019~2020学年浙江宁波鄞州区宁波市鄞州蓝青学校初一上学期期末第4题", "1992年第9届全国初中数学联赛竞赛第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "满足$$\\left\\textbar{} a-b \\right\\textbar+ab=1$$的非负整数（$$a$$，$$b$$）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->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": ["由条件得$$\\left { \\begin{matrix}\\left\\textbar{} a-b \\right\\textbar=1    ab=0   \\end{matrix} \\right.$$或$$\\left { \\begin{matrix}\\left\\textbar{} a-b \\right\\textbar=0    ab=1   \\end{matrix} \\right.$$，则满足条件的数对有$$(0  ,  1)$$，$$(1  ,  0)$$，$$(1  ,  1)$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "205", "queId": "14b21eb04ba24f7ba50c84760e589d27", "competition_source_list": ["2002年第19届全国初中数学联赛竞赛第4题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "直角三角形$$ABC$$的面积为$$240$$，且$$\\angle BAC=90{}^{}\\circ $$，$$AD$$是斜边上的中线，过$$D$$作$$DE\\bot AB$$于$$E$$，连$$CE$$交$$AD$$于$$F$$，则$$\\triangle AFE$$的面积为．", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$44$$ "}], [{"aoVal": "C", "content": "$$40$$ "}], [{"aoVal": "D", "content": "$$36$$ "}]], "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->与中线或等分线有关的等积变换", "课内体系->知识点->三角形->直角三角形->直角三角形的性质->直角三角形斜边中线性质以及应用"], "answer_analysis": ["$$C$$ "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1284", "queId": "8aac50a74e724b3f014e832aefe83c60", "competition_source_list": ["1996年第7届全国希望杯初一竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$$a\\textgreater b$$，且$$c\\textless{}0$$，那么在下面不等式中：  ①$$a+c\\textgreater b+c$$；②$$ac\\textgreater bc$$；③$$-\\frac{a}{c}\\textgreater-\\frac{b}{c}$$；④$$a{{c}^{2}}\\textless{}b{{c}^{2}}$$．成立的个数是（~ ）．", "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->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->方程与不等式->不等式（组）->不等式->判断不等式的变形是否正确"], "answer_analysis": ["已知$$a\\textgreater b$$，$$c\\textless{}0$$，由不等式的基本性质，得  $$a+c\\textgreater b+c$$，成立，$$ac\\textgreater bc$$不成立，$$-\\frac{a}{c}\\textgreater-\\frac{b}{c}$$成立，$$a{{c}^{2}}\\textless{}b{{c}^{2}}$$不成立．  所以只有①③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": "1272", "queId": "8aac50a74e442d83014e4cd2ab8e1d9b", "competition_source_list": ["1995年第6届全国希望杯初一竞赛复赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$$1995x+6y=420000$$的一组整数解$$\\left( x,y \\right)$$是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 61,48723 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 62,48725 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( 63,48726 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( 64,48720 \\right)$$ "}]], "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->二元一次方程的整数解"], "answer_analysis": ["设$$x$$，$$y$$均为整数，且满足$$1995x+6y=420000$$．  则$$5\\textbar1995x$$，$$5\\textbar420000$$，所以$$5\\textbar6y$$．  但$$\\left( 5,6 \\right)=1$$，因此$$5\\textbar y$$．所以排除$$\\text{A}$$，$$\\text{C}$$．  对$$\\text{B}$$，$$\\left( 62,48725 \\right)$$不满足方程．  事实上，$$1995\\times 64+6\\times 48720=420000$$成立．选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1340", "queId": "8aac50a7511483070151196c51ff1158", "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": "$$-2$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{4}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->配方思想的运用", "课内体系->能力->运算能力"], "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$$4{{x}^{2}}-5x+4=0$$，两边同时除以$$x$$，得$$4x-5+\\frac {4}{x}=0$$，  ∴$$4\\left (x+\\frac {1}{x}\\right )=5$$，  ∴$$x+\\frac {1}{x}=\\frac {5}{4}$$. ", "<p>原方程化为$$4\\left( {{x}^{2}}+\\frac{1}{{{x}^{2}}} \\right)+3\\left( x+\\frac{1}{x} \\right)-2=0$$．</p>\n<p>故选$$\\text{B}$$．</p>"], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "597", "queId": "6c89d0a1c49f4609ad5a7c0e178d2cd2", "competition_source_list": ["2003年第14届希望杯初二竞赛第2试第3题", "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": "整数$$x$$、$$y$$满足不等式$${{x}^{2}}+{{y}^{2}}+1\\leqslant 2x+2y$$，则$$x+y$$的值有．", "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->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["由题意，有$${{x}^{2}}-2x+1+{{y}^{2}}-2y+1\\leqslant 1$$，即$${{\\left( x-1 \\right)}^{2}}+{{\\left( y-1 \\right)}^{2}}\\leqslant 1$$.  $$\\because {{\\left( x-1 \\right)}^{2}}\\geqslant 0$$，$${{\\left( y-1 \\right)}^{2}}\\geqslant 0$$，$$x$$、$$y$$为整数，  $$\\therefore \\begin{cases}x-1=0    y-1=0 \\end{cases}$$或$$\\begin{cases}x-1=\\pm 1    y-1=0 \\end{cases}$$或$$\\begin{cases}x-1=0    y-1=\\pm 1 \\end{cases}$$，  解得$$\\left( x,y \\right)=\\left( 1,1 \\right)\\left( 2,1 \\right)\\left( 0,1 \\right)\\left( 1,2 \\right)\\left( 1,0 \\right)$$，故$$x+y$$的不同值是$$1$$、$$2$$、$$3$$. "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "830", "queId": "9161db4a18524ea1b8269473009413a4", "competition_source_list": ["2016年第27届全国希望杯初一竞赛初赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "关于多项式$$\\frac{1}{2}{{x}^{3}}y+5{{y}^{4}}{{x}^{2}}-2{{y}^{7}}+4$$，有以下叙述：  ①该多项式是六次三项式；  ②该多项式是七次四项式；  ③该多项式是七次三项式；  ④该多项式最高次项的系数是$$-2$$；  ⑤该多项式常数项是$$-4$$．  其中，正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "①④ "}], [{"aoVal": "B", "content": "③⑤ "}], [{"aoVal": "C", "content": "②④ "}], [{"aoVal": "D", "content": "②⑤ "}]], "knowledge_point_routes": ["知识标签->题型->式->整式加减->整式有关的概念->求多项式的项和次数", "知识标签->知识点->式->整式的加减->多项式->多项式的项和次数"], "answer_analysis": ["该多项式为七次四项式，最高次项的系数为$$-2$$，常数项为$$4$$．故正确的是②④． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1420", "queId": "a613b935892640e6885129ca5467d271", "competition_source_list": ["1995年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "一支队伍在前$50$场比赛中赢了$40$场比赛，这支队伍在剩下的$$40$$场比赛中至少赢多少场才能保证自己整个赛季的胜率能超过$$70 \\%$$？", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$23$$ "}], [{"aoVal": "C", "content": "$$28$$ "}], [{"aoVal": "D", "content": "$$30$$ "}], [{"aoVal": "E", "content": "$$35$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率", "美国AMC8->Knowledge Point->Counting, Probability and Statistics->Classical Probability"], "answer_analysis": ["一个队伍赢了开始$$50$$场比赛中的$$40$$场．这只队在剩下来的$$40$$场比赛中必须最少赢多少场才能让它恰好整个赛季胜率为$$70 \\%$$？  注意到$$70 \\%$$与$$\\frac{70}{100}=\\frac{7}{10}$$相等，当$$x$$是$$40$$场比赛中获胜的数量时，获胜的分数是$$\\frac{40+x}{50+40}=\\frac{40+x}{90}$$，我们所要做的就是把它们相等：$$\\frac{40+x}{90}=\\frac{7}{10}$$，$$40+x=63$$，$$x=(\\text{B})23$$．  另外，我们可以注意到他们总共会打$$40+50=90$$场比赛，必须赢得$$0.7(90)=63$$场比赛．因为他们赢了$$40$$场比赛，他们还需要$$63-40=(\\text{B})23$$场比赛．  Noting that $$70 \\%$$ is the same as $$\\frac{70}{100}= \\frac{7}{10}$$, and that, when x is the amount of wins in the last $$40$$ games, the fraction of games won is $$\\frac{40+x}{50+40}= \\frac{40+x}{90}$$, all we have to do is set them equal:  $$\\frac{40+x}{90}= \\frac{7}{10}$$, $$40+x=63$$, $$x=23$$.  Alternatively we can note that they will play a total of $$40+50 =90$$ games and must win $$0.7(90) = 63$$ games. Since they won $$40$$ games already they need $$63-40=23$$ more games. "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "142", "queId": "25d8eca34fbb44599e9282cf439ded04", "competition_source_list": ["2006年第11届华杯赛初一竞赛初赛第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "有如下四个命题：①最大的负数是$$-1$$；②最大的负整数是$$-1$$；③最小的整数是$$1$$；④最小的正整数是$$1$$；其中真命题有个．", "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->数->有理数->有理数的概念及分类->有理数的分类"], "answer_analysis": ["$$-0.1$$大于$$-1$$，所以①不正确；整数包括负整数，显然整数没有下界，所以③不正确．正整数自$$1$$开始，$$1$$，$$2$$，$$3$$，\\ldots，所以④正确；同样，②正确． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "717", "queId": "7ad6c34d49244b7e8c5a60d6af2c06b7", "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$$M$$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": ["知识标签->题型->数->有理数->有理数基础运算->题型：有理数乘法的计算题", "知识标签->知识点->数->有理数->数轴", "知识标签->知识点->数->有理数->有理数的运算->有理数加法的运算", "知识标签->知识点->数->有理数->绝对值->绝对值的性质"], "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": "52", "queId": "066a7eea619c499f94013d064afea558", "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": ["课内体系->知识点->数->有理数->分数", "美国AMC8->Knowledge Point->Number and Operations->Fractions->Fraction Operations"], "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": "935", "queId": "6a4ae453015b4ffd8d0b71d579f91897", "competition_source_list": ["2000年第11届希望杯初二竞赛第2试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "若一个等腰三角形的三条边长均为整数，且周长为$$10$$，则底边的长为．", "answer_option_list": [[{"aoVal": "A", "content": "一切偶数 "}], [{"aoVal": "B", "content": "$$2$$或$$4$$或$$6$$或$$8$$ "}], [{"aoVal": "C", "content": "$$2$$或$$4$$或$$6$$ "}], [{"aoVal": "D", "content": "$$2$$或$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->特殊三角形->等腰三角形"], "answer_analysis": ["设等腰三角形的腰长为$$a$$，底边为$$b$$，  所以$$2a+b=10$$，$$b=10-2a$$．  所以$$b$$是偶数．  又$$b ~\\textless{} ~2a$$，  所以$$2b ~\\textless{} ~10$$，  解得$$b ~\\textless{} ~5$$，  所以$$b$$为小于$$5$$的偶数．  所以$$b=2$$或$$b=4$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1041", "queId": "ff8080814d7978b9014d8661f11f2418", "competition_source_list": ["1990年第1届全国希望杯初一竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "在解方程的过程中，为了使得到的方程和原方程同解，可以在原方程的两边．", "answer_option_list": [[{"aoVal": "A", "content": "乘同一个数 "}], [{"aoVal": "B", "content": "乘同一个整式 "}], [{"aoVal": "C", "content": "加上同一个代数式 "}], [{"aoVal": "D", "content": "都加上$$1$$ "}]], "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->等式与方程->等式->等式的性质->等式性质1"], "answer_analysis": ["对方程同解变形，要求方程两边同乘不等于$$0$$的数，所以排除$$\\text{A}$$．  我们考察方程$$x-2=0$$，易知其根为$$x=2$$．若该方程两边同乘一个整式$$x-1$$，得$$(x-1)(x-2)=0$$，其根为$$x=1$$及$$x=2$$，不与原方程同解，排除$$\\text{B}$$．  若在方程$$x-2=0$$两边加上同一个代数式$$\\frac{1}{x-2}$$得$$\\frac{1}{x-2}+(x-2)=\\frac{1}{x-2}$$，此方程误解，失去了原方程$$x=2$$的根，所以应排除$$\\text{C}$$．  事实上方程两边同时加上一个常数，新方程与原方程同解，对$$\\text{D}$$，这里所加常数为$$1$$，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": "1413", "queId": "a60c6b6a35754802857b0c2c0cf55625", "competition_source_list": ["1997年全美数学竞赛（AMC）竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "有一组五个正整数，它们的平均值是$$5$$，中位数是$$5$$，唯一的众数是$$8$$．这组数中的最大整数和最小整数之间的差是什么?", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$7$$ "}], [{"aoVal": "E", "content": "$$8$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Combinatorics->Fun Problems in Math->Fun Math Problems", "课内体系->知识点->统计与概率->数据的分析->中位数"], "answer_analysis": ["有一组五个正整数，它们的平均值是$$5$$，中位数是$$5$$，唯一的众数是$$8$$．集合中最大整数和最小整数之间的差是什么?  当这些数字按升序排列时，中间值$$5$$就在中间，这是左边的第三个整数．由于$$8$$是唯一的已知数，$$5$$右边的两个整数都必须是$$8$$．由于平均值是$$5$$，整数的和是$$25$$，这意味着$$2$$个最左边的整数必须和是$$4$$．$$2$$和$$2$$无效，因为这将导致两种模式．但是，$$1$$和$$3$$可以，所以我们的答案是$$8-1=7$$．  When these numbers are ordered in ascending order, $$5$$, the median, falls right in the middle, which is the third integer from the left. Since there is a unique mode of $$8$$, both integers to the right of $$5$$ must be $$8\\rm s$$. Since the mean is $$5$$, the sum of the integers is $$25$$, which means the $$2$$ leftmost integers have to sum to $$4$$. $$2$$ and $$2$$ does not work because that would result in two modes. However, $$1$$ and $$3$$ does, and so our answer is $$8-1=7$$. "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1451", "queId": "bcd2d215f57545d79efc7a151b223fb3", "competition_source_list": ["2016年第27届全国希望杯初二竞赛复赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$${{3}^{2016}}+5$$除以$${{3}^{2013}}-1$$，所得的余数是（~ ~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{3}^{15}}-1$$ "}], [{"aoVal": "B", "content": "$${{3}^{12}}-1$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "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->有理数乘方运算"], "answer_analysis": ["$${{3}^{2016}}+5$$  $$={{3}^{3}}\\times {{3}^{2013}}+5$$  $$={{3}^{3}}\\times ({{3}^{2013}}-1)+{{3}^{3}}+5$$  $$={{3}^{3}}\\times ({{3}^{2013}}-1)+32$$，  ∴$${{3}^{2016}}+5$$除以$${{3}^{2013}}-1$$，所得的余数是$$32$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1315", "queId": "a5668a8ff34e46f9997599b237ec8b61", "competition_source_list": ["2008年第19届希望杯初二竞赛第1试第1题", "2018~2019学年10月四川成都锦江区四川师范大学附属第一实验中学初二上学期周测A卷第6题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "下列说法中正确的是（~ ~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$的平方根和$$1$$的立方根相同 "}], [{"aoVal": "B", "content": "$$0$$的平方根和$$0$$的立方根相同 "}], [{"aoVal": "C", "content": "$$\\sqrt{4}$$的平方根是$$\\pm 2$$ "}], [{"aoVal": "D", "content": "$$8$$的立方根是$$\\pm 2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->平方根->开平方", "课内体系->知识点->数->实数->立方根->开立方"], "answer_analysis": ["$$\\text{A}$$、$$1$$的平方根为$$\\pm 1$$，$$1$$的立方根为$$1$$，二者不相等，故$$\\text{A}$$选项错误；  $$\\text{B}$$、$$0$$的平方根和立方根均为$$0$$，正确；  $$\\text{C}$$、可知$$\\sqrt{4}=2$$，所以$$2$$的平方根为$$\\pm \\sqrt{2}$$，故$$\\text{C}$$选项错误；  $$\\text{D}$$、$$8$$的立方根为$$2$$，故$$\\text{D}$$选项错误；  故答案选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "171", "queId": "3cf490ae084b49ad8c7f2534f93a8e9c", "competition_source_list": ["2018年四川成都武侯区成都七中初三自主招生第7题5分", "初二上学期单元测试《等腰三角形的判定》第24题", "2008年第19届希望杯初二竞赛第1试第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\triangle ABC$$的三边的长分别为$$a$$、$$b$$、$$c$$，且$$\\frac{a}{b}+\\frac{a}{c}=\\frac{b+c}{b+c-a}$$，则$$\\triangle ABC$$一定是（ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "等边三角形 "}], [{"aoVal": "B", "content": "腰长为$$a$$的等腰三角形 "}], [{"aoVal": "C", "content": "底边长为$$a$$的等腰三角形 "}], [{"aoVal": "D", "content": "等腰直角三角形 "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->等腰三角形->等腰三角形的判定->等腰三角形的判定-两边相等"], "answer_analysis": ["∵$$\\frac{a}{b}+\\frac{a}{c}=\\frac{b+c}{b+c-a}$$．  ∴$$a\\left( \\frac{1}{b}+\\frac{1}{c} \\right)=\\frac{b+c}{b+c-a}$$．  $$\\frac{a(b+c)}{bc}=\\frac{b+c}{b+c-a}$$．  ∵$$b+c\\textgreater0$$．  ∴$$\\frac{a}{bc}=\\frac{1}{b+c-a}$$．  ∴$$ab+ac-{{a}^{2}}-bc=0$$．  ∴$$(b-a)(c-a)=0$$．  ∴$$a=b$$或$$c=a$$．  ∴$$\\triangle ABC$$为以$$a$$为腰长的等腰三角形． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1006", "queId": "a46927ae1afd40589df530d5178f74bb", "competition_source_list": ["2010年第21届希望杯初二竞赛第2试第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$xy\\ne 0$$，$$x+y\\ne 0$$，$$\\frac{1}{x}+\\frac{1}{y}$$与$$x+y$$成反比，则$${{(x+y)}^{2}}$$与$${{x}^{2}}+{{y}^{2}}$$．", "answer_option_list": [[{"aoVal": "A", "content": "成正比 "}], [{"aoVal": "B", "content": "成反比 "}], [{"aoVal": "C", "content": "既不成正也不成反比 "}], [{"aoVal": "D", "content": "的关系不确定 "}]], "knowledge_point_routes": ["课内体系->知识点->函数->反比例函数->反比例函数基础->反比例函数的定义", "课内体系->知识点->函数->反比例函数->反比例函数基础->根据反比例函数定义解题"], "answer_analysis": ["题设$$\\frac{1}{x}+\\frac{1}{y}$$和$$x+y$$成反比，可令  $$\\frac{1}{x}+\\frac{1}{y}=\\frac{k}{x+y}$$（$$k$$为常数，且$$k\\ne 0$$），  则$${{(x+y)}^{2}}=kxy$$，  所以$${{x}^{2}}+{{y}^{2}}={{(x+y)}^{2}}-2xy=(k-2)xy$$，  $$\\frac{{{(x+y)}^{2}}}{{{x}^{2}}+{{y}^{2}}}=\\frac{kxy}{(k-2)xy}=\\frac{k}{k-2}$$，  $${{(x+y)}^{2}}=\\frac{k}{k-2}({{x}^{2}}+{{y}^{2}})$$，  因为$$xy\\ne 0$$，  那么$${{x}^{2}}+{{y}^{2}}=(k-2)xy\\ne 0$$，  故有$$k\\ne 2$$．  即$$k$$为常数，且$$k\\ne 0$$，$$k\\ne 2$$，所以$$\\frac{k}{k-2}$$是不等于$$0$$的常数，可知$${{(x+y)}^{2}}$$与$${{x}^{2}}+{{y}^{2}}$$成正比．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1366", "queId": "a596bf40bf924ad2a9fc6ecb0da11a56", "competition_source_list": ["1991年第2届希望杯初二竞赛第14题"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$a\\textless{}-1$$时，方程$$\\left( {{a}^{3}}+1 \\right){{x}^{2}}+\\left( {{a}^{2}}+1 \\right)x-\\left( a+1 \\right)=0$$的根的情况是．", "answer_option_list": [[{"aoVal": "A", "content": "两负根 "}], [{"aoVal": "B", "content": "一正根、一负根且负根的绝对值大 "}], [{"aoVal": "C", "content": "一正根、一负根且负根的绝对值小 "}], [{"aoVal": "D", "content": "没有实数根 "}]], "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": ["$$a\\textless{}-1$$时，$${{a}^{3}}+1\\textless{}0$$，$${{a}^{2}}+1\\textgreater0$$，$$a+1\\textless{}0$$．而若方程的两根为$${{x}_{1}}$$，$${{x}_{2}}$$，则有  $${{x}_{1}}\\cdot {{x}_{2}}=-\\frac{a+1}{{{a}^{3}}+1}\\textless{}0$$，表示有一正根一负根．又$${{x}_{1}}+{{x}_{2}}=-\\frac{{{a}^{2}}+1}{{{a}^{3}}+1}\\textgreater0$$，表示负根的绝对值小于正的绝对值．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1318", "queId": "93894e50d01e4877972ae48a39e8dd6f", "competition_source_list": ["2013年竞赛第1题4分", "初一下学期其它第16题"], "difficulty": "2", "qtype": "single_choice", "problem": "设非零实数$$a$$，$$b$$，$$c$$，满足$$\\begin{cases}a+2b+3c=0    2a+3b+4c=0    \\end{cases}$$，则$$\\frac{ab+bc+ca}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "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->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->整式的加减->整式有关的概念->整式->整式的定义"], "answer_analysis": ["由已知得$$a+b+c=(2a+3b+4c)-(a+2b+3c)=0$$，  故$${{(a+b+c)}^{2}}=0$$，  于是$$ab+bc+ca=-\\frac{1}{2}({{a}^{2}}+{{b}^{2}}+{{c}^{2}})$$，  ∴$$\\frac{ab+bc+ca}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=-\\frac{1}{2}$$，  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "729", "queId": "56ad385809a44d7081a991a0f60c4b61", "competition_source_list": ["2014年第31届全国全国初中数学联赛竞赛第8题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "使得不等式$$\\frac{9}{17}\\textless{}\\frac{n}{n+k}\\textless{}\\frac{8}{15}$$对唯一的整数$$k$$成立的最大正整数$$n$$为．", "answer_option_list": [[{"aoVal": "A", "content": "$$142$$ "}], [{"aoVal": "B", "content": "$$143$$ "}], [{"aoVal": "C", "content": "$$144$$ "}], [{"aoVal": "D", "content": "$$145$$ "}]], "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": ["由条件得$$\\frac{7}{8}\\textless{}\\frac{k}{n}\\textless{}\\frac{8}{9}$$，  由$$k$$的唯一性，得$$\\frac{k-1}{n}\\leqslant \\frac{7}{8}$$且$$\\frac{k+1}{n}\\geqslant \\frac{8}{9}$$，  所以$$\\frac{2}{n}=\\frac{k+1}{n}-\\frac{k-1}{n}\\geqslant \\frac{8}{9}-\\frac{7}{8}=\\frac{1}{72}$$，  所以$$n\\leqslant 144$$．  当$$n=144$$时，由$$\\frac{7}{8}\\textless{}\\frac{k}{n}\\textless{}\\frac{8}{9}$$可得$$126\\textless{}k\\textless{}128$$，$$k$$可取唯一整数值$$127$$．  故满足条件的正整数$$n$$的最大值为$$144$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "457", "queId": "62d6a1d376a04ecfb054d372a5f5dfba", "competition_source_list": ["2008年第19届希望杯初一竞赛第1试第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\triangle ABC$$的三个内角$$A$$，$$B$$，$$C$$的外角依次记为$$\\alpha $$，$$\\beta $$，$$\\gamma $$，若$$\\beta =2B$$，$$\\alpha -\\gamma =40{}^{}\\circ $$，则三个内角$$A$$，$$B$$，$$C$$的度数依次是．", "answer_option_list": [[{"aoVal": "A", "content": "$$60{}^{}\\circ $$，$$60{}^{}\\circ $$，$$60{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$30{}^{}\\circ $$，$$60{}^{}\\circ $$，$$90{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$40{}^{}\\circ $$，$$60{}^{}\\circ $$，$$80{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$50{}^{}\\circ $$，$$60{}^{}\\circ $$，$$70{}^{}\\circ $$ "}]], "knowledge_point_routes": ["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->与三角形有关的角->三角形的外角定义及性质", "课内体系->知识点->三角形->三角形及多边形->与三角形有关的角->三角形内角和的应用", "课内体系->思想->方程思想"], "answer_analysis": ["由题意得$$\\begin{cases}\\beta +B=180{}^{}\\circ    \\beta =2B   \\end{cases}$$$$\\Rightarrow B=60{}^{}\\circ $$，$$A+C=120{}^{}\\circ $$，  因此 $$\\alpha +\\gamma =2\\times 180{}^{}\\circ -120{}^{}\\circ =240{}^{}\\circ $$，  又知$$\\alpha -\\gamma =40{}^{}\\circ $$，解得 $$\\alpha =140{}^{}\\circ $$，$$\\gamma =100{}^{}\\circ $$，  所以 $$A=40{}^{}\\circ $$，$$C=80{}^{}\\circ $$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "678", "queId": "5adb5116f06f4f929143320e5aa918e6", "competition_source_list": ["初一下学期其它", "2000年第11届希望杯初二竞赛第2试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$-\\frac{1999}{2000}$$，$$-\\frac{1998}{1999}$$，$$-\\frac{998}{999}$$，$$-\\frac{999}{1000}$$这四个数从小到大的排列顺序是（  ）", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{1999}{2000}\\textless{}-\\frac{1998}{1999}\\textless{}-\\frac{999}{1000}\\textless{}-\\frac{998}{999}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{998}{999}\\textless{}-\\frac{999}{1000}\\textless{}-\\frac{1998}{1999}\\textless{}-\\frac{1999}{2000}$$ "}], [{"aoVal": "C", "content": "$$-\\frac{1998}{1999}\\textless{}-\\frac{1999}{2000}\\textless{}-\\frac{999}{1000}\\textless{}-\\frac{998}{999}$$ "}], [{"aoVal": "D", "content": "$$-\\frac{999}{1000}\\textless{}-\\frac{998}{999}\\textless{}-\\frac{1999}{2000}\\textless{}-\\frac{1998}{1999}$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性"], "answer_analysis": ["$$-\\frac{1999}{2000}=-1+\\frac{1}{2000}$$，$$-\\frac{1998}{1999}=-1+\\frac{1}{1999}$$，  $$-\\frac{998}{999}=-1+\\frac{1}{999}$$， $$-\\frac{999}{1000}=-1+\\frac{1}{1000}$$．  因为$$\\frac{1}{2000}\\textless{}\\frac{1}{1999}\\textless{}\\frac{1}{1000}\\textless{}\\frac{1}{999}$$，  所以$$-\\frac{1999}{2000}\\textless{}-\\frac{1998}{1999}\\textless{}-\\frac{999}{1000}\\textless{}-\\frac{998}{999}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "488", "queId": "62f7239a9f0d49688bd40db10ea93b2a", "competition_source_list": ["2007年第12届华杯赛初一竞赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "全班$$43$$名同学参加校庆，每人不是持彩旗，就是举鲜花．现有$$10$$位男同学持彩旗，$$30$$人举鲜花．如果全班有$$15$$名女同学，问：她们中有多少人举鲜花?", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$10$$ "}], [{"aoVal": "E", "content": "$$9$$ "}]], "knowledge_point_routes": ["课内体系->知识点->四边形->平面向量->向量的分解与线性组合", "竞赛->知识点->组合->容斥原理"], "answer_analysis": ["全班持彩旗的人数为$$43-30=13$$（人），其中持彩旗的女同学有$$13-10=3$$（人）．由于全班女同学共$$15$$人，所以女同学中举鲜花的人数为$$15-3=12$$（人）． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "117", "queId": "08ba196acac846a69a16110f7cd9e49e", "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1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知数据$${{x}_{1}}$$，$${{x}_{2}}$$的平均数是$$a$$，方差是$$b$$，则数据$$x_{1}^{2}$$，$$x_{2}^{2}$$的平均数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{b+2{{a}^{2}}}{2}$$ "}], [{"aoVal": "B", "content": "$$b+2{{a}^{2}}$$ "}], [{"aoVal": "C", "content": "$$\\frac{b+{{a}^{2}}}{2}$$ "}], [{"aoVal": "D", "content": "$$b+{{a}^{2}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->数据的分析->统计综合", "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", "课内体系->知识点->统计与概率->数据的分析->方差"], "answer_analysis": ["∵$${{x}_{1}}$$、$${{x}_{2}}$$平均数为$$a$$，  ∴$$\\frac{{{x}_{1}}+{{x}_{2}}}{2}=a$$，  又$${{x}_{1}}{{x}_{2}}$$方差为$$b$$，  ∴$$\\frac{{{\\left( {{x}_{1}}-a \\right)}^{2}}+{{\\left( {{x}_{2}}-a \\right)}^{2}}}{2}=b$$  $$\\frac{x_{1}^{2}-2a{{x}_{1}}+{{a}^{2}}+x_{2}^{2}-2a{{x}_{2}}+{{a}^{2}}}{2}=b$$  $$\\frac{x_{1}^{2}+x_{2}^{2}}{2}-a{{x}_{1}}-a{{x}_{2}}+{{a}^{2}}=b$$  $$\\frac{x_{1}^{2}+x_{2}^{2}}{2}=a\\left( {{x}_{1}}+{{x}_{2}} \\right)+b-{{a}^{2}}$$，  又$${{x}_{1}}+{{x}_{2}}=2a$$，  ∴$$\\frac{x_{1}^{2}+x_{2}^{2}}{2}-2{{a}^{2}}-{{a}^{2}}+b$$  $$={{a}^{2}}+b$$，  即$$x_{1}^{2}$$，$$x_{2}^{2}$$平均数为$${{a}^{2}}+b$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "833", "queId": "4edafe9242294656a47af2b9bb946595", "competition_source_list": ["2019~2020学年9月湖北武汉武昌区武汉市武珞路实验初级中学初三上学期周测A卷第8题3分", "2004年第21届全国初中数学联赛竞赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$${{b}^{2}}-4ac$$是一元二次方程$$a{{x}^{2}}+bx+c=0\\left( a\\ne 0 \\right)$$的一个实数根，则$$ab$$的取值范围为．", "answer_option_list": [[{"aoVal": "A", "content": "$$ab\\geqslant \\frac{1}{8}$$ "}], [{"aoVal": "B", "content": "$$ab\\leqslant \\frac{1}{8}$$ "}], [{"aoVal": "C", "content": "$$ab\\geqslant \\frac{1}{4}$$ "}], [{"aoVal": "D", "content": "$$ab\\leqslant \\frac{1}{4}$$ "}]], "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->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", "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": ["\\textbf{（知识点：判别式的应用）}  方法一：由求根公式得到方程的根为$$x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}$$，设$$\\sqrt{{{b}^{2}}-4ac}=y$$，又因为$${{y}^{2}}={{b}^{2}}-4ac$$是原方程的根，所以$${{y}^{2}}=\\frac{-b+y}{2a}$$或者$${{y}^{2}}=\\frac{-b-y}{2a}$$，因为$$a\\ne 0$$，这两个方程可化成关于$$y$$的一元二次方程$$2a{{y}^{2}}\\pm y+b=0$$，考虑这个方程的判别式得到$$1-8ab\\geqslant 0$$，所以$$ab{\\leqslant } \\frac{1}{8}$$，选择$$\\text{B}$$．  方法二：因为方程有实数解，所以$${{b}^{2}}-4ac{\\geqslant }0$$，由题意，  有$$\\frac{-b+\\sqrt{{{b}^{2}}-4ac}}{2a}={{b}^{2}}-4ac$$或$$\\frac{-b-\\sqrt{{{b}^{2}}-4ac}}{2a}={{b}^{2}}-4ac$$．  令$$u=\\sqrt{{{b}^{2}}-4ac}$$，得方程$$2a{{u}^{2}}-u+b=0$$或$$2a{{u}^{2}}+u+b=0$$，  因为$$u=\\sqrt{{{b}^{2}}-4ac}$$是该方程的解，所以方程的判别式非负，即$$1-8ab{\\geqslant }0$$，所以$$ab{\\leqslant }\\frac{1}{8}$$．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": "354", "queId": "b96f70ab626c4da999f6ce2607881013", "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。假设每个孩子都有可能是男孩或女孩。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": ["课内体系->知识点->统计与概率->概率->可能性的大小", "美国AMC8->Knowledge Point->Counting, Probability and Statistics->Statistical Distribution"], "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": "1234", "queId": "8aac4907507fb8840150803ea036011b", "competition_source_list": ["1997年第8届全国希望杯初一竞赛复赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "文具店、书店和玩具店依次坐落在一条东西走向的大街上，文具店在书店西边$$20$$米处，玩具店位于书店东边$$100$$米处，小明从书店沿街向东走了$$40$$米，接着又向东走了$$-60$$米，此时小明的位置在（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "文具店 "}], [{"aoVal": "B", "content": "玩具店 "}], [{"aoVal": "C", "content": "文具店西边$$40$$米 "}], [{"aoVal": "D", "content": "玩具店东$$-60$$米 "}]], "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": ["向东走了$$-60$$米就是向西走了$$60$$米．  所以，小明从书店向东走了$$40$$米，再向西走$$60$$米，  结果是小明的位置在书店西边$$20$$米，也就是文具店的位置． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "914", "queId": "6a192fc993ee4620b0fe8fa3e73ea546", "competition_source_list": ["2019~2020学年5月四川绵阳涪城区绵阳南山中学双语学校初一下学期月考第12题3分", "2017~2018学年湖南长沙开福区青竹湖湘一外国语学校初一下学期期中第12题3分", "2019~2020学年天津和平区天津市益中学校初一下学期单元测试《不等式》第8题", "2007年第18届希望杯初二竞赛第2试第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知关于$$x$$的不等式组$$\\begin{cases}2a+3x\\textgreater0    3a-2x\\geqslant 0   \\end{cases}$$恰有$$3$$个整数解，则$$a$$的取值范围是（~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2}{3}\\leqslant a\\leqslant \\frac{3}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{3}\\leqslant a\\leqslant \\frac{3}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{4}{3} ~\\textless{} ~a\\leqslant \\frac{3}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{3}\\leqslant a ~\\textless{} ~\\frac{3}{2}$$ "}]], "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）的整数解情况求参数范围", "课内体系->能力->运算能力"], "answer_analysis": ["由于不等式组有解，则$$-\\frac{2a}{3} ~\\textless{} ~x\\leqslant \\frac{3a}{2}$$，必定有整数解$$0$$，  ∵$$\\left\\textbar{} \\frac{3a}{2} \\right\\textbar\\textgreater\\left\\textbar{} -\\frac{2a}{3} \\right\\textbar$$，  ∴三个整数解不可能是$$-2$$，$$-1$$，$$0$$．  若三个整数解为$$-1$$，$$0$$，$$1$$，则不等式组$$\\begin{cases}1\\leqslant \\frac{3a}{2} ~\\textless~ 2   -3\\leqslant -\\frac{2a}{3} ~\\textless~ -2   \\end{cases}$$无解；  若三个整数解为$$0$$，$$1$$，$$2$$，则$$\\begin{cases}2\\leqslant \\frac{3a}{2} ~\\textless{} ~3    -1\\leqslant -\\frac{2a}{3} ~\\textless{} ~0   \\end{cases}$$，  解得$$\\frac{4}{3}\\leqslant a\\leqslant \\frac{3}{2}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1136", "queId": "bff13c924c6042c1ba47848174f1e8a4", "competition_source_list": ["2013年第24届全国希望杯初二竞赛初赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "小王与小李约定下午$$3$$点在学校门口见面，为此，他们在早上$$8$$点将自己的手表对准，小王于下午$$3$$点到达学校门口，可是小李还没到，原来小李的手表比正确时间每小时慢$$4$$分钟．如果小李按他自己的手表在$$3$$点到达，则小王还需要等（~ ）．（正确时间）", "answer_option_list": [[{"aoVal": "A", "content": "$$26$$分钟 "}], [{"aoVal": "B", "content": "$$28$$分钟 "}], [{"aoVal": "C", "content": "$$30$$分钟 "}], [{"aoVal": "D", "content": "$$32$$分钟 "}]], "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", "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": ["小李的表在$$7$$小时内比正确时间慢$$7\\times 4=28$$分．  下午$$3$$时的时候，小李的表才$$2$$时$$32$$分，  小李的表走$$28$$分，正确的时间是$$28\\times \\frac{60}{56}=30$$（分）． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1339", "queId": "8aac50a7511483070151196750ac1139", "competition_source_list": ["2015年第26届全国希望杯初三竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "掷$$2$$次骰子，第一次掷出的点数（骰子落地后，朝上一面的点数）记为$$a$$，第二次掷出的点数记为$$b$$，则点$$(a  ,  b)$$在直线$$y=2x$$上的概率为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{12}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{9}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{9}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率->概率的定义", "课内体系->思想->数形结合思想"], "answer_analysis": ["总共有$$36$$种可能，其中$$a=1$$，$$b=2$$或$$a=2$$，$$b=4$$或$$a=3$$，$$b=6$$这三种结果能使点$$(a  ,  b)$$在直线$$y=2x$$上，  ∴点$$(a  ,  b)$$在直线$$y=2x$$上的概率为$$\\frac{3}{36}=\\frac{1}{12}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "560", "queId": "3204a483f38c4af2baf2913bfaf9ab53", "competition_source_list": ["1996年第7届希望杯初二竞赛第10题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "有一架不准确的天平（左臂长为$$a$$厘米，右臂长为$$b$$厘米，$$a\\ne b$$）某人用它来计量某件重物．先将重物放在左盘，砝码放在右盘，需用$${{m}_{1}}$$千克使天平平衡；然后再将重物放在右盘，砝码放在左盘，需用$${{m}_{2}}$$千克使天平平衡，于是用$$Q=\\frac{{{m}_{1}}+{{m}_{2}}}{2}$$千克估算重物的实际重量，若重物的实际重量为$$P$$千克，那么．", "answer_option_list": [[{"aoVal": "A", "content": "$$Q\\textgreater P$$ "}], [{"aoVal": "B", "content": "$$Q=P$$ "}], [{"aoVal": "C", "content": "$$Q\\leqslant P$$ "}], [{"aoVal": "D", "content": "$$Q ~\\textless{} ~P$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->乘法公式"], "answer_analysis": ["重物的实际重量为$$P$$，  ∴$$Pa={{m}_{1}}b$$，  ∴$${{m}_{1}}=\\frac{a}{b}P$$，  同理$${{m}_{2}}=\\frac{b}{a}P$$，  $$Q=\\frac{1}{2}({{m}_{1}}+{{m}_{2}})$$  $$=\\frac{1}{2}\\left( \\frac{a}{b}+\\frac{b}{a} \\right)P$$  $$=\\frac{{{a}^{2}}+{{b}^{2}}}{2ab}P$$  $$=\\frac{{{(a-b)}^{2}}+2ab}{2ab}P$$  $$=\\left[ \\frac{{{(a-b)}^{2}}}{2ab}+1 \\right]P$$，  ∵$$a\\ne b$$，  ∴$${{(a-b)}^{2}}\\textgreater0$$，  ∴$$Q\\textgreater P$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "645", "queId": "444c09087fbf48bf9615dfaf066585e8", "competition_source_list": ["2012年第29届全国全国初中数学联赛竞赛A卷第11题20分", "2012年全国全国初中数学联赛初一竞赛"], "difficulty": "3", "qtype": "single_choice", "problem": "已知直角三角形的边长均为整数，周长为$$30$$，则它的外接圆的面积为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{121}{4} \\pi $$ "}], [{"aoVal": "B", "content": "$$\\frac{169}{4} \\pi $$ "}], [{"aoVal": "C", "content": "$$\\frac{225}{4} \\pi $$ "}], [{"aoVal": "D", "content": "以上答案都不对 "}]], "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->勾股定理的证明", "课内体系->知识点->三角形->勾股定理及应用->勾股定理应用->勾股定理的综合应用", "课内体系->知识点->圆->圆与多边形->正多边形与圆"], "answer_analysis": ["设直角三角形的三边长分别为$$a$$，$$b$$，$$c$$（$$a\\leqslant b\\textless{}c$$），则$$a+b+c=30$$．  显然，三角形的外接圆的直径即为斜边长$$c$$，下面先求$$c$$的值．  由$$a\\leqslant b\\textless{}c$$及$$a+b+c=30$$得$$30=a+b+c\\textless{}3c$$，  所以$$c\\textgreater10$$．  由$$a+b\\textgreater c$$及$$a+b+c=30$$得$$30=a+b+c\\textgreater2c$$，  所以$$c\\textless{}15$$．  又因为$$c$$为整数，  所以$$11\\leqslant c\\leqslant 14$$．  根据勾股定理可得$${{a}^{2}}+{{b}^{2}}={{c}^{2}}$$，  把$$c=30-a-b$$代入，化简得$$ab-30(a+b)+450=0$$，  所以$$(30-a)(30-b)=450=2\\times {{3}^{2}}\\times {{5}^{2}}$$，  因为$$a$$，$$b$$均为整数且$$a\\leqslant b$$，  所以只可能是$$\\begin{cases}30-a={{5}^{2}}    30-b=2\\times {{3}^{2}} \\end{cases}$$，解得$$\\begin{cases}a=5    b=12 \\end{cases}$$，  所以，直角三角形的斜边长$$c=13$$，三角形的外接圆的面积为$$\\frac{169}{4} \\pi $$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "952", "queId": "6a7ff93572b14236946c4b53cad9b4ac", "competition_source_list": ["1994年第11届全国初中数学联赛竞赛第8题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$${{a}^{x}}={{b}^{y}}={{1994}^{z}}$$（其中$$a$$，$$b$$是自然数），且有$$\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{z}$$，则$$2a+b$$的一切可能的取值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1001$$ "}], [{"aoVal": "B", "content": "$$1001$$，$$3989$$ "}], [{"aoVal": "C", "content": "$$1001$$，$$1996$$ "}], [{"aoVal": "D", "content": "$$1001$$，$$1996$$，$$3989$$ "}]], "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->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->幂的运算->幂的乘方", "课内体系->知识点->式->整式的乘除->幂的运算->同底数幂的乘法", "课内体系->知识点->式->整式的乘除->幂的运算->积的乘方"], "answer_analysis": ["由$${{a}^{x}}={{1994}^{x}}$$及$${{b}^{y}}={{1994}^{z}}$$可得$${{a}^{\\frac{1}{z}}}={{1994}^{\\frac{1}{x}}}$$及$${{b}^{\\frac{1}{z}}}={{1994}^{\\frac{1}{y}}}$$．  故$${{(ab)}^{\\frac{1}{z}}}={{1994}^{\\frac{1}{z}+\\frac{1}{y}}}={{1994}^{\\frac{1}{z}}}$$．  因此$$ab=1994=1\\times 1994=2\\times 997$$．  但$$a$$，$$b$$均不为$$1$$，故有$$a=2$$，$$b=997$$，或$$a=997$$，$$b=2$$．  于是，$$2a+b=1001$$或$$1996$$．  故选 C． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "193", "queId": "d5119074654e4472b1214130cb279fcb", "competition_source_list": ["1992年第3届希望杯初二竞赛第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "方程$${{x}^{2}}+667x+1992=0$$的较大的那个实根的负倒数等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{664}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{1}{667}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{1992}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{3}$$ "}]], "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->一元二次方程->一元二次方程的基础->一元二次方程的根"], "answer_analysis": ["设较大的那个实根为$$x$$，  则当$$-\\frac{1}{x}=\\frac{1}{664}$$时，$$x=-664$$，  当$$-\\frac{1}{x}=-\\frac{1}{667}$$时，$$x=667$$，  当$$-\\frac{1}{x}=\\frac{1}{1992}$$时，$$x=-1992$$，  当$$-\\frac{1}{x}=\\frac{1}{3}$$时，$$x=-3$$．  由于方程$${{x}^{2}}+667x+1992=0$$不能有非负根，所以$$x=667$$排除，  剩下的$$-664$$，$$-1992$$，$$-3$$这三个数中，最大者为$$-3$$，  以$$-3$$代入原方程，恰好满足方程．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "848", "queId": "c3ce460c765a450cbbd0ebb61405c2e2", "competition_source_list": ["2018~2019学年10月天津河西区天津市海河中学初三上学期月考第12题3分", "2017~2018学年1月陕西西安雁塔区西安市第八十五中学初三上学期月考第10题3分", "2019~2020学年12月河北石家庄裕华区石家庄外国语学校初三上学期周测B卷第24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第8题4分", "2017年天津中考真题第12题3分", "2017~2018学年山东济宁汶上县初三上学期期末第9题3分", "2018年陕西西安新城区西光中学初三中考二模第10题3分", "2018年山东德州夏津县初三中考二模第11题4分", "2017~2018学年3月陕西咸阳秦都区陕西科技大学附属中学初三下学期月考第1次（四模）第10题3分", "2019~2020学年陕西西安雁塔区西安高新逸翠园学校初三上学期期末（西安高新第三中学联考）第8题3分", "2017~2018学年浙江杭州拱墅区杭州市文澜中学初三上学期期中第8题3分", "2019~2020学年山东威海文登区初三上学期期末（五四制）第8题3分", "2017~2018学年2月天津天津市第二南开中学初三下学期月考第11题3分", "2017~2018学年陕西西安高新区西安高新第一中学初三上学期期末第8题3分", "2019~2020学年天津河北区天津外国语大学附属外国语学校初三上学期期末第10题3分", "2017~2018学年9月浙江金华义乌市义乌市绣湖中学初一上学期月考第6题", "2019年广东广州荔湾区广雅实验学校初三中考二模第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知抛物线$$y={{x}^{2}}-4x+3$$与$$x$$轴相交于点$$A$$，$$B$$（点$$A$$在点$$B$$左侧），顶点为$$M$$．平移该抛物线，使点$$M$$平移后的对应点$${M}'$$落在$$x$$轴上，点$$B$$平移后的对应点$${B}'$$落在$$y$$轴上．则平移后的抛物线解析式为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$y={{x}^{2}}+2x+1$$ "}], [{"aoVal": "B", "content": "$$y={{x}^{2}}+2x-1$$ "}], [{"aoVal": "C", "content": "$$y={{x}^{2}}-2x+1$$ "}], [{"aoVal": "D", "content": "$$y={{x}^{2}}-2x-1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->二次函数->二次函数的几何变换->二次函数平移变换"], "answer_analysis": ["令$$y=0={{x}^{2}}-4x+3$$，$${{x}_{1}}=1$$，$${{x}_{2}}=3$$，  ∵$$A\\left( 1,0 \\right)$$，$$B\\left( 3,0 \\right)$$，$$y={{x}^{2}}-4x+3={{\\left( x-2 \\right)}^{2}}-1$$，  ∴$$M\\left( 2,-1 \\right)$$，  ∵平移后$${{M}^{\\prime }}$$落在$$x$$轴上，  ∴图像向上平移$$1$$个单位，  ∵平移后$${{B}^{\\prime }}$$落在$$y$$轴上，  ∴图像向左平移$$3$$个单位，  ∴平移后的解析式为$$y=(x-2+3){{}^{2}}-1+1=(x+1){{}^{2}}={{x}^{2}}+2x+1$$，  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "553", "queId": "2da9465436cb40deb81816073e0a4d46", "competition_source_list": ["2009年第14届华杯赛初一竞赛初赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "数学课上，全班同学每人各报一个数．如果男生所报的数之和与女生所报的数之和相等，且男生所报数的平均值是$$-\\frac{3}{8}$$，女生所报数的平均值是$$-\\frac{1}{4}$$，那么全班同学所报数的平均值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{3}{12}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{5}{8}$$ "}], [{"aoVal": "C", "content": "$$-\\frac{3}{10}$$ "}], [{"aoVal": "D", "content": "$$-\\frac{5}{12}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->方程应用"], "answer_analysis": ["设男生人数为$$x$$，女生人数为$$y$$，所有男生所报数的和为$$S$$．根据题意，$$S=-\\frac{3}{8}x=-\\frac{1}{4}y$$，该班男生与女生的人数之比，与它们的平均数之比成反比，则女生人数是男生人数的$$\\frac{y}{x}=\\left( -\\frac{3}{8} \\right)\\div \\left( -\\frac{1}{4} \\right)=\\frac{3}{2}$$．则女生人数为$$y=\\frac{3}{2}x$$．于是，全班所报数的平均数是：$$\\frac{-\\dfrac{3}{8}\\times x+\\left( -\\dfrac{1}{4} \\right)\\times \\dfrac{3}{2}x}{x+\\dfrac{3}{2}x}=\\left( -\\frac{3}{4} \\right)\\div \\left( 1+\\frac{3}{2} \\right)=-\\frac{3}{10}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1", "queId": "00308ac6d1db428ebf5694c42e902530", "competition_source_list": ["1998年第9届希望杯初二竞赛第2试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a+b+c=0$$，则$${{a}^{3}}+{{a}^{2}}c-abc+{{b}^{2}}c+{{b}^{3}}$$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->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": ["$${{a}^{3}}+{{a}^{2}}c-abc+{{b}^{2}}c+{{b}^{3}}$$  $$=\\left( {{a}^{3}}+{{b}^{3}} \\right)+\\left( {{a}^{2}}+{{b}^{2}} \\right)c-abc$$  $$=\\left( a+b \\right)\\left( {{a}^{2}}-ab+{{b}^{2}} \\right)+\\left( {{a}^{2}}+{{b}^{2}} \\right)c-abc$$  $$=\\left( a+b \\right)\\left( {{a}^{2}}+{{b}^{2}} \\right)-ab\\left( a+b \\right)+\\left( {{a}^{2}}+{{b}^{2}} \\right)c-abc$$  因为$$a+b+c=0$$，所以$$a+b=-c$$．  所以原式$$=-c\\left( {{a}^{2}}+{{b}^{2}} \\right)+abc+\\left( {{a}^{2}}+{{b}^{2}} \\right)c-abc=0$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "30", "queId": "05a16f5dca57460ebfe38c224d4935ea", "competition_source_list": ["2018年全美数学竞赛（AMC）竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "贝拉开始从她的房子走向她的朋友艾拉的房子．与此同时，艾拉开始骑自行车向贝拉的家走去．他们每个人都保持恒定的速度，$$Ella$$骑的速度是$$Bella$$走路速度的$5$倍．他们家之间的距离是$2$英里，也就是$10560$英尺，贝拉每走一步就走$2.5$英尺．当$$Bella$$遇到$$Ella$$的时候她会走多少步?", "answer_option_list": [[{"aoVal": "A", "content": "$$704$$ "}], [{"aoVal": "B", "content": "$$845$$ "}], [{"aoVal": "C", "content": "$$1056$$ "}], [{"aoVal": "D", "content": "$$1760$$ "}], [{"aoVal": "E", "content": "$$3520$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Word Problem->Travel Word Problems->Travel Word Problems on Straight Road", "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-相遇问题"], "answer_analysis": ["Since Ella rides $$5$$ times as fast as Bella, Ella rides at a rate of $$2.5\\times5=12.5$$. Together,they move $$15$$ feet towards each other every \"step\". Dividing $$10560$$ by $$15$$ to find the number of steps Ella takes results in the answer or $$\\boxed{ (\\rm A)704}$$. "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1289", "queId": "a9cfca4c8b5a4ffe97247c01a523cf71", "competition_source_list": ["1998年第9届希望杯初一竞赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "下面的四个判断中，不正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$34{{x}^{3}}{{y}^{6}}$$与$$34{{a}^{3}}{{b}^{6}}$$不是同类项 "}], [{"aoVal": "B", "content": "$$3x$$和$$-3x+1$$不能互为相反数 "}], [{"aoVal": "C", "content": "$$4(x-7)=6(5-27x)$$和$$6(5-27y)=4(y-7)$$不是同解方程 "}], [{"aoVal": "D", "content": "$$3$$和$$\\frac{1}{a}+\\frac{1}{3}$$不能互为倒数 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["①$$34{{x}^{3}}y6$$与$$34{{a}^{3}}{{b}^{6}}$$，因字母不同，不是同类项，所以$$\\text{A}$$是正确的，排除$$\\text{A}$$．  ②若$$3x$$与$$-3x+1$$互为相反数，则$$-(3x)=-3x+1$$得出$$0=1$$的矛盾．所以``$$3x$$和$$-3x+1$$不能互为相反数''这句话正确，排除$$\\text{B}$$．  ③$$4(x-7)=6(5-27x)$$的解为$$\\frac{29}{83}$$．  $$6(5-27y)=4(y-7)$$的解也是$$\\frac{29}{83}$$．  因为这两个方程的解相同，因此，它们是同解方程．即$$\\text{C}$$``$$4(x-7)=6(5-27x)$$和$$6(5-27y)=4(y-7)$$不是同解方程''这句话不正确．  ④若$$3$$与$$\\frac{1}{a}+\\frac{1}{3}$$互为倒数，则$$\\frac{1}{3}=\\frac{1}{a}+\\frac{1}{3}\\Rightarrow 0=\\frac{1}{a}$$，矛盾．所以``$$3$$和$$\\frac{1}{a}+\\frac{1}{3}$$不能互为倒数''这句话是正确的，所以排除$$\\text{D}$$．  综上所述，应选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1592", "queId": "fe98cfd839234c4a94c2231f98915f7b", "competition_source_list": ["2001年第12届希望杯初二竞赛第1试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$b\\ne c$$，且满足$$(\\sqrt{3}+1)(a-b)+\\sqrt{2}(b-c)=a-c$$，则$$\\frac{a-b}{b-c}$$的值．", "answer_option_list": [[{"aoVal": "A", "content": "大于零 "}], [{"aoVal": "B", "content": "等于零 "}], [{"aoVal": "C", "content": "小于零 "}], [{"aoVal": "D", "content": "正负号不确定 "}]], "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->式->二次根式->二次根式化简求值"], "answer_analysis": ["因为$$(\\sqrt{3}+1)(a-b)+\\sqrt{2}(b-c)=a-c$$，  所以$$(\\sqrt{3}+1)(a-b)+\\sqrt{2}(b-c)-(a-b)-(b-c)=0$$，  所以$$\\sqrt{3}(a-b)+(\\sqrt{2}-1)\\cdot (b-c)=0$$，  即$$\\sqrt{3}(a-b)=(1-\\sqrt{2})(b-c)$$，  因为$$b\\ne c$$，$$b-c\\ne 0$$，  所以$$\\frac{a-b}{b-c}=\\frac{1-\\sqrt{2}}{\\sqrt{3}}\\textless{}0$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "196", "queId": "4642eeadf54f4a9e9bc9081999213318", "competition_source_list": ["1999年第10届希望杯初一竞赛第7题", "初一上学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "$$-\\frac{1997}{1998}$$，$$-\\frac{97}{98}$$，$$-\\frac{1998}{1999}$$，$$-\\frac{98}{99}$$这四个数由小到大的排列顺序是（ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{1997}{1998}\\textless{}-\\frac{97}{98}\\textless{}-\\frac{1998}{1999}\\textless{}-\\frac{98}{99}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{1998}{1999}\\textless{}-\\frac{1997}{1998}\\textless{}-\\frac{98}{99}\\textless{}-\\frac{97}{98}$$ "}], [{"aoVal": "C", "content": "$$-\\frac{97}{98}\\textless{}-\\frac{98}{99}\\textless{}-\\frac{1997}{1998}\\textless{}-\\frac{1998}{1999}$$ "}], [{"aoVal": "D", "content": "$$-\\frac{98}{99}\\textless{}-\\frac{1998}{1999}\\textless{}-\\frac{97}{98}\\textless{}-\\frac{1997}{1998}$$ "}]], "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": ["~将每个数都加$$1$$，大小次序不变．  $$-\\frac{1997}{1998}+1=\\frac{1}{1998}$$，$$-\\frac{97}{98}+1=\\frac{1}{98}$$．  $$-\\frac{1998}{1999}+1=\\frac{1}{1999}$$，$$-\\frac{98}{99}+1=\\frac{1}{99}$$．  因为$$\\frac{1}{1999}\\textless{}\\frac{1}{1998}\\textless{}\\frac{1}{99}\\textless{}\\frac{1}{98}$$．  所以$$-\\frac{1998}{1999}\\textless{}-\\frac{1997}{1998}\\textless{}-\\frac{98}{99}\\textless{}-\\frac{97}{98}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "774", "queId": "648d0bca5eab4cf398a6f5648c3cbe96", "competition_source_list": ["2009年第20届希望杯初二竞赛第1试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$a+\\sqrt{2}ab+b=\\sqrt{2}$$，且$$b$$是有理数，那么．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$是整数 "}], [{"aoVal": "B", "content": "$$a$$是有理数 "}], [{"aoVal": "C", "content": "$$a$$是无理数 "}], [{"aoVal": "D", "content": "$$a$$可能是有理数，也可能是无理数 "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->无理数有关的计算->无理数运算与参数问题"], "answer_analysis": ["若$$a$$是有理数，由$$a+\\sqrt{2}ab+b=\\sqrt{2}$$，可知$$a+b=0$$，故$$-{{a}^{2}}\\sqrt{2}=\\sqrt{2}$$，$$-{{a}^{2}}=1$$，矛盾，$$a$$是无理数． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "827", "queId": "dabb2a0296774ceb8819c58b234b3b6b", "competition_source_list": ["2014年第25届全国希望杯初二竞赛复赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$2014$$中选出不能表示成两个整数的平方差的数，这些数的和是（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$507024$$ "}], [{"aoVal": "B", "content": "$$508032$$ "}], [{"aoVal": "C", "content": "$$1014049$$ "}], [{"aoVal": "D", "content": "$$1015056$$ "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "竞赛->知识点->数论->同余->完全平方数"], "answer_analysis": ["若整数$$a$$是奇数，  则$$a={{\\left( \\frac{a+1}{2} \\right)}^{2}}-{{\\left( \\frac{a-1}{2} \\right)}^{2}}$$．  若整数$$a$$是$$4$$的倍数，  则$$a={{\\left( \\frac{a}{4}+1 \\right)}^{2}}-{{\\left( \\frac{a}{4}-1 \\right)}^{2}}$$．  若整数$$a$$是偶数，并且不是$$4$$的倍数，则$$a=4k\\pm 2$$，  此情形下$$a$$不能表示成两个整数的平方差，理由如下：  设整数$$a=4k\\pm 2$$可表示为$$a={{p}^{2}}-{{q}^{2}}$$，则$$p$$，$$q$$有相同的奇偶性．  当$$p$$，$$q$$都是奇数时，  有$$a={{p}^{2}}-{{q}^{2}}$$  $$={{(2m+1)}^{2}}-{{(2n+1)}^{2}}$$  $$=4({{m}^{2}}-{{n}^{2}}+m-n)$$，  则$$a$$是$$4$$的倍数，与$$a=4k\\pm 2$$矛盾；  当$$p$$，$$q$$都是偶数时，  有$$a={{p}^{2}}-{{q}^{2}}$$  $$={{(2m)}^{2}}-{{(2n)}^{2}}$$  $$=4({{m}^{2}}-{{n}^{2}})$$，  则$$a$$也是$$4$$的倍数，与$$a=4k\\pm 2$$矛盾．  因此，只有$$4k+2$$（$$k=0$$，$$1$$，$$2$$，$$\\cdots $$，$$503$$）不能表示成两个整数的平方差．  它们的和是  $$2+6+10+\\cdots +(4k+2)+\\cdots +2006+2010+2014$$  $$=\\frac{(2014+2)\\times 504}{2}$$  $$=508032$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1011", "queId": "dfccb8639838479f913305364a1132d9", "competition_source_list": ["1992年第3届希望杯初二竞赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "等腰三角形的周长为$$a\\text{cm}$$．一腰的中线将周长分成$$5:3$$，则三角形的底边长为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{a}{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{5}a$$ "}], [{"aoVal": "C", "content": "$$\\frac{a}{6}$$或$$\\frac{3}{5}a$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{5}a$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->特殊三角形->等腰三角形"], "answer_analysis": ["设等腰三角形的腰长为$$b\\text{cm}$$，底边长为$$c\\text{cm}$$，$$c=a-2b$$，分两种情形讨论：  （$$1$$）若$$\\dfrac{b+\\dfrac{b}{2}}{c+\\dfrac{b}{2}}=\\frac{5}{3}$$，即$$\\frac{\\dfrac{3}{2}b}{a-\\dfrac{3}{2}b}=\\frac{5}{3}$$，  解得$$b=\\frac{5}{12}a$$，所以$$c=a-2\\times \\frac{5}{12}a=\\frac{a}{6}$$．  （$$2$$）若$$\\frac{c+\\dfrac{b}{2}}{b+\\dfrac{b}{2}}=\\frac{5}{3}$$，即$$\\frac{a-\\dfrac{3}{2}b}{\\dfrac{3}{2}b}=\\frac{5}{3}$$，  解得$$b=\\frac{a}{4}$$，所以$$c=a-2\\times \\frac{a}{4}=\\frac{a}{2}$$．  这是不可能的，因此选$$\\text{A}$$．  事实上，$$\\frac{3}{5}a$$已超过$$a$$的一半，即三角形有一边大于其他两边之和，这是不可能的，所以排除$$B$$，$$C$$，$$D$$，．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1189", "queId": "8aac49074e023206014e25c047c572b4", "competition_source_list": ["1997年第8届全国希望杯初一竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "有理数$$a$$、$$b$$满足$$\\left\\textbar{} a+b \\right\\textbar\\textless{}\\left\\textbar{} a-b \\right\\textbar$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a+b\\geqslant 0$$ "}], [{"aoVal": "B", "content": "$$a+b\\textless{}0$$ "}], [{"aoVal": "C", "content": "$$ab\\textless{}0$$ "}], [{"aoVal": "D", "content": "$$ab\\geqslant 0$$ "}]], "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->知识点->数->有理数->绝对值"], "answer_analysis": ["由$$\\left\\textbar{} a+b \\right\\textbar\\textless{}\\left\\textbar{} a-b \\right\\textbar$$有$${{\\left( a+b \\right)}^{2}}\\textless{}{{\\left( a-b \\right)}^{2}}$$，  即$${{a}^{2}}+2ab+{{b}^{2}}\\textless{}{{a}^{2}}-2ab+{{b}^{2}}$$．  不等式两边都减去$${{a}^{2}}+{{b}^{2}}$$，然后除以$$2$$，则有$$ab\\textless{}-ab$$，  只有$$ab\\textless{}0$$时才能成立，选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "735", "queId": "e3bb0fcc669245879927eef57ffd07a0", "competition_source_list": ["2018年湖南长沙初二竞赛长郡教育集团（觉园杯）第4题4分", "2012年竞赛第4题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "小倩和小玲每人都有若干面值为整数元的人民币，小倩对小玲说：``你若给我$$2$$元，我的钱数将是你的$$n$$倍''，小玲对小倩说：``你若给我$$n$$元，我的钱数将是你的$$2$$倍''，其中$$n$$为正整数，则$$n$$的可能值的个数是．", "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->方程与不等式->二元一次方程（组）->二元一次方程组应用题", "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->二元一次方程的整数解", "课内体系->能力->推理论证能力"], "answer_analysis": ["设小倩所有的钱数为$$x$$元，小玲所有的钱数为$$y$$元，$$x$$，$$y$$均为非负整数，由题设可得$$\\begin{cases}x+2=n(y-2)    y+n=2\\left( x-n \\right)   \\end{cases}$$，  消去$$x$$得$$\\left( 2y-7 \\right)n=y+4$$，  $$2n=\\frac{\\left( 2y-7 \\right)+15}{2y-7}=1+\\frac{15}{2y-7}$$，  因为$$\\frac{15}{2y-7}$$为正整数，所以$$2y-7$$的值分别为$$1$$、$$3$$、$$5$$、$$15$$，所以$$y$$的值只能为$$4$$、$$5$$、$$6$$、$$11$$，从而$$n$$的值分别为$$8$$、$$3$$、$$2$$、$$1$$，$$x$$的值分别为$$14$$、$$7$$、$$6$$、$$7$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "637", "queId": "9e5a10d1bff24a1d9d837ebb10f440f3", "competition_source_list": ["2016年第33届全国全国初中数学联赛竞赛第2题7分", "2016年上海浦东新区进才中学自主招生第6题10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（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$种 "}], [{"aoVal": "B", "content": "$$10$$种 "}], [{"aoVal": "C", "content": "$$11$$种 "}], [{"aoVal": "D", "content": "$$12$$种 "}]], "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$$\\left { \\begin{matrix}y+z=30-x    3y+4z=100-2x   \\end{matrix} \\right.$$，  解得$$\\left { \\begin{matrix}y=20-2x    z=10+x   \\end{matrix} \\right.$$，  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": "181", "queId": "104dc22addc74cd684c45c3444f54fcc", "competition_source_list": ["1993年第4届希望杯初二竞赛第10题"], "difficulty": "0", "qtype": "single_choice", "problem": "如果方程$$\\left\\textbar3x\\right\\textbar-ax-1=0$$的根是负数，那么$$a$$的取值范围是(  )", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater-3$$ "}], [{"aoVal": "B", "content": "$$a\\geqslant 3$$ "}], [{"aoVal": "C", "content": "$$a \\textless{} 3$$ "}], [{"aoVal": "D", "content": "$$a\\leqslant -3$$ "}]], "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", "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->不等式（组）->含参不等式->不等式（组）解的情况确定参数的范围"], "answer_analysis": ["原方程即$$\\left\\textbar3x\\right\\textbar=ax+1$$．  $$\\left(1\\right)$$若$$a=3$$时，则$$\\left\\textbar3x\\right\\textbar=3x+1$$．  当$$x \\textless{} 0$$时，得$$-3x=3x+1$$．所以$$x=-\\frac{1}{6}$$；  当$$x\\geqslant 0$$时，得$$3x=3x+1$$，不成立．  所以当$$a=3$$时，原方程的根为$$x=-\\frac{1}{6}$$．  $$\\left(2\\right)$$若$$a\\textgreater3$$．  当$$x \\textless{} 0$$时，得$$-3x=ax+1$$，所以$$x=-\\frac{1}{a+3} \\textless{} 0$$；  当$$x\\geqslant 0$$时，得$$3x=ax+1$$，所以$$x=\\frac{1}{3-a} \\textless{} 0$$，矛盾．  所以当$$a\\textgreater3$$时，原方程的解为$$x=-\\frac{1}{a+3} \\textless{} 0$$．  $$\\left(3\\right)$$若$$a \\textless{} 3$$  当$$x\\geqslant 0$$时，得$$3x=ax+1$$，所以$$x=\\frac{1}{3-a}\\textgreater0$$，原方程有正根，与题设矛盾．  综上所述，$$a\\geqslant 3$$时，原方程的根是负数．  所以应选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "184", "queId": "ab562bb014e2404eae215c1d07c18ee5", "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"], "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": "936", "queId": "6ec6b84b90bc4b47955e385037aae5cd", "competition_source_list": ["2006年第17届希望杯初二竞赛第1试第7题", "2018~2019学年浙江杭州下城区杭州观成中学初二下学期单元测试《平行四边形》第8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个凸多边形剪去一个角后形成的多边形的内角和是$$2520{}^{}\\circ $$，则原多边形的边数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$15$$或$$16$$ "}], [{"aoVal": "D", "content": "$$15$$或$$16$$或$$17$$ "}]], "knowledge_point_routes": ["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", "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"], "answer_analysis": ["设这个多边形的边数是$$n$$．  由题意得：$$(n-2)\\times 180{}^{}\\circ =2520{}^{}\\circ $$，  解得$$n=16$$．  所以这个多边形的边数是$$16$$．  故原多边形的边数为$$15$$或$$16$$或$$17$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1400", "queId": "ce7f512a181d4cb9a5178553c1a68eb5", "competition_source_list": ["2009年第20届希望杯初二竞赛第2试第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "每只玩具熊的售价为$$250$$元．熊的四条腿上各有两个饰物，标号依次为$$1$$，$$2$$，$$3$$，$$\\cdots $$， $$8$$．卖家说：``第$$1$$，$$2$$，$$3$$，$$4$$，$$\\cdots $$，$$8$$号饰物依次要收$$1$$，$$2$$，$$4$$，$$8$$，$$\\cdots $$，$$128$$元．如果购买全部饰物，那么玩具熊就免费赠送．''若按这样的付费办法，这只熊比原售价便宜了．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$元 "}], [{"aoVal": "B", "content": "$$-5$$元 "}], [{"aoVal": "C", "content": "$$6$$元 "}], [{"aoVal": "D", "content": "$$-6$$元 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数与实际问题->有理数加减法与实际问题"], "answer_analysis": ["$$1+2+4+8+16+32+64+128=255$$（元）．  $$250-255=-5$$（元）．  则这只熊比原售价便宜了$$-5$$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": "1144", "queId": "8a7cfe5740a745d2a1f5601debcdb179", "competition_source_list": ["2013年全美数学竞赛（AMC）竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "这个算式的值是多少$$4\\cdot (-1+2-3+4-5+6-7+\\cdots +1000)$$?．", "answer_option_list": [[{"aoVal": "A", "content": "$$-10$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$500$$ "}], [{"aoVal": "E", "content": "$$2000$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Number and Operations->Integers->Addition and subtraction of integers", "课内体系->知识点->数->有理数->有理数运算巧解->分组求和"], "answer_analysis": ["求$$4\\cdot (-1+2-3+4-5+6-7+\\cdots +1000)$$．  注意，我们可以每两个数字配对，得出$$\\left( -1+2-3+4-\\cdot \\cdot \\cdot +1000 \\right)=\\left( \\left( -1+2 \\right)+\\left( -3+4 \\right)+\\cdot \\cdot \\cdot +\\left( -999+1000 \\right) \\right)$$$$=\\left( 1+1+\\cdot \\cdot \\cdot 1 \\right)=500$$，  因此答案是$$4\\times 500=2000$$．  故选$$\\text{E}$$．  Notice that we can pair up every two numbers to make a sum of $$1$$:  $$(-1+2-3+4-\\cdots+1000)$$  $$=((-1+2)+(-3+4)+\\cdots+(-999+1000))$$  $$=(1+1+\\cdots+1)$$  $$= 500$$  Therefore, the answer is $$4\\cdot500=\\boxed{(\\text{E})2000}$$. "], "answer_value": "E"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "965", "queId": "96a723388f614616902b5cb3b141ee21", "competition_source_list": ["2016年全国全国初中数学联赛初一竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$为实数，关于$$x$$，$$y$$的方程组$$\\left { \\begin{array}{*{35}{l}} ax+2y=22    2x+2y=a   \\end{array} \\right.$$有整数解，则$$a$$的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "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->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）->含参二元一次方程组->题型：含参整数解问题", "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：加减消元法", "知识标签->学习能力->运算能力"], "answer_analysis": ["由$$\\left { \\begin{array}{*{35}{l}} ax+2y=22    2x+2y=a   \\end{array} \\right.$$得$$\\left( a-2 \\right)x=22-a$$，∴$$x=\\frac{22-a}{a-2}=-1+\\frac{20}{a-2}$$；  由$$2x+2y=a$$，可知$$a$$必为偶数，又$$-1+\\frac{20}{a-2}$$为整数，所以$$a=0,4,6,-2,12,-8,22,-18$$ "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "301", "queId": "46aef1dd088640408629627e9aecafb0", "competition_source_list": ["2012年第29届全国全国初中数学联赛竞赛第5题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "若方程$${{x}^{2}}+2px-3p-2=0$$的两个不相等的实数根$${{x}_{1}}$$，$${{x}_{2}}$$满足$$x_{1}^{2}+x_{1}^{3}=4-(x_{2}^{2}+x_{2}^{3})$$，则实数$$p$$的所有可能的值之和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$-\\frac{3}{4}$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "$$-\\frac{5}{4}$$ "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "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": ["因为方程$${{x}^{2}}+2px-3p-2=0$$两个不相等的实数根，  故$$\\Delta ={{(2p)}^{2}}+4(3p+2)=4{{p}^{2}}+12p+8\\textgreater0$$，  解得$$p\\textless{}-2$$或$$p\\textgreater-1$$．  因为$${{x}_{1}}$$，$${{x}_{2}}$$满足$${{x}_{1}}+{{x}_{2}}=-2p$$，$${{x}_{1}}\\cdot {{x}_{2}}=-3p-2$$，  故$$x_{1}^{2}+x_{2}^{2}={{({{x}_{1}}+{{x}_{2}})}^{2}}-2{{x}_{1}}{{x}_{2}}$$  $$=4{{p}^{2}}+2(3p+2)$$  $$=4{{p}^{2}}+6p+4$$，  $$x_{1}^{3}+x_{2}^{3}=({{x}_{1}}+{{x}_{2}})(x_{1}^{2}-{{x}_{1}}{{x}_{2}}+x_{2}^{2})$$  $$=-2p(4{{p}^{2}}+6p+4+3p+2)$$  $$=-8{{p}^{3}}-18{{p}^{2}}-12p$$，  由$$x_{1}^{2}+x_{1}^{3}=4-(x_{2}^{2}+x_{2}^{3})$$得$$x_{1}^{2}+x_{2}^{2}+x_{1}^{3}+x_{2}^{3}=4$$  即$$4{{p}^{2}}+6p+4-8{{p}^{3}}-18{{p}^{2}}-12p=4$$，  整理得$$p(4p^{2}-7p-3)=0$$，  解得$${{p}_{1}}=\\frac{7+\\sqrt{97}}{8}$$，$${{p}_{2}}=\\frac{7-\\sqrt{97}}{8}$$，$${{p}_{3}}=0$$．  ∵$$p\\textless{}-2$$或$$p\\textgreater-1$$，  ∴$$p=\\frac{7\\pm\\sqrt{97}}{8}$$，  ∴实数$$p$$的所有可能的值之和为$$\\frac{7}{4}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1166", "queId": "8aac49074e023206014e1e29ac455b89", "competition_source_list": ["1993年第4届全国希望杯初一竞赛复赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$x$$是正数，$$\\textless{}x\\textgreater$$表示不超过$$x$$的质数的个数，如$$\\textless{}5.1\\textgreater=3$$．即不超过$$5.1$$的质数有$$2$$，$$3$$，$$5$$共$$3$$个．那么$$\\textless{}\\textless{}19\\textgreater+\\textless{}93\\textgreater+\\textless{}4\\textgreater\\times \\textless{}1\\textgreater\\times \\textless{}8\\textgreater\\textgreater$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$11$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "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->式->整式的加减->整式的加减运算"], "answer_analysis": ["$$\\textless{}19\\textgreater$$为不超过$$19$$的质数，有$$2$$，$$3$$，$$5$$，$$7$$，$$11$$，$$13$$，$$17$$，$$19$$共$$8$$个．  $$\\textless{}93\\textgreater$$为不超过$$93$$的质数，共$$24$$个，易知$$\\textless{}1\\textgreater=0$$．  所以$$\\textless{}\\textless{}19\\textgreater+\\textless{}93\\textgreater+\\textless{}4\\textgreater\\times \\textless{}1\\textgreater\\times \\textless{}8\\textgreater\\textgreater=\\textless{}\\textless{}19\\textgreater+\\textless{}93\\textgreater\\textgreater=\\textless{}8+24\\textgreater=\\textless{}32\\textgreater=11$$，选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "105", "queId": "0b32296d905a4be1a203a6587a1dd8e6", "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->方法->作差法", "知识标签->学习能力->运算能力", "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->式->整式的乘除->乘法公式->平方差公式", "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", "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->因式分解->提公因式法与公式法->题型：提公因式法", "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"], "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": "109", "queId": "8f5afabfd39a4d08b3cf5c8ff4f5809f", "competition_source_list": ["2011年第16届华杯赛初一竞赛初赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "对四堆石子进行如下``操作''：每次允许从每堆中各拿掉相同个数的石子，或从任一堆中取出一些石子放入另一堆中．若四堆石子的个数分别为$$2011$$，$$2010$$，$$2009$$，$$2008$$，则按上述方式进行若干次``操作''后，四堆石子的个数可能是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$，$$0$$，$$0$$，$$1$$ "}], [{"aoVal": "B", "content": "$$0$$，$$0$$，$$0$$，$$2$$ "}], [{"aoVal": "C", "content": "$$0$$，$$0$$，$$0$$，$$3$$ "}], [{"aoVal": "D", "content": "$$0$$，$$0$$，$$0$$，$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->整除的概念与基本性质"], "answer_analysis": ["因为，每操作一次四堆石子总数的减少是$$4$$的倍数，  而$$2011+2010+2009+2008=2009\\times 4+2$$．所以，最后四堆石子的个数分别是$$0$$，$$0$$，$$0$$，$$2$$. "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "642", "queId": "da4fcdf9892e448ba515c9864a33762f", "competition_source_list": ["1993年第4届希望杯初二竞赛第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "设二次方程$$a{{x}^{2}}+bx+c=0$$的两根为$${{x}_{1}}$$，$${{x}_{2}}$$，记$${{S}_{1}}={{x}_{1}}+1993{{x}_{2}}$$，$${{S}_{2}}=x_{1}^{2}+1993x_{2}^{2}$$，$$\\cdots $$，$${{S}_{n}}=x_{1}^{n}+1993x_{2}^{n}$$，则$$a{{S}_{1993}}+b{{S}_{1992}}+c{{S}_{1991}}=$$  ．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$2010$$ "}], [{"aoVal": "C", "content": "$$2011$$ "}], [{"aoVal": "D", "content": "$$2012$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->二次方程->方程的构造"], "answer_analysis": ["由于方程两根为$${{x}_{1}}$$，$${{x}_{2}}$$，  则$$ax_{1}^{2}+b{{x}_{1}}+c=0$$，$$ax_{2}^{2}+b{{x}_{2}}+c=0$$．  所以$$a{{S}_{1993}}+b{{S}_{1992}}+c{{S}_{1991}}$$  $$=a\\left( {{x}_{1}}^{1993}+1993{{x}_{2}}^{1993} \\right)+b\\left( {{x}_{1}}^{1992}+1993{{x}_{2}}^{1992} \\right)+c\\left( {{x}_{1}}^{1991}+1993{{x}_{2}}^{1991} \\right)$$  $$={{x}_{1}}^{1991}\\left( a{{x}_{1}}^{2}+b{{x}_{1}}+c \\right)+1993{{x}_{2}}^{1991}\\left( ax_{2}^{2}+b{{x}_{2}}+c \\right)$$  $$=0$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "695", "queId": "3c05f72d1c5d4b56b3c357b8db36662f", "competition_source_list": ["初一其它", "北京竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "把$$2009$$表示成两个整数的平方差的形式，则不同的表示法有．", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$种 "}], [{"aoVal": "B", "content": "$$14$$种 "}], [{"aoVal": "C", "content": "$$12$$种 "}], [{"aoVal": "D", "content": "$$10$$种 "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->公式法->利用平方差公式因式分解", "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": ["$$\\left( x+y \\right)\\left( x-y \\right)=2009={{7}^{2}}\\times 41$$有$$6$$个正因数，分别是$$1$$，$$7$$，$$41$$，$$49$$，$$287$$和$$2009$$，因此对应的方程组为$$\\begin{cases}x+y=-1,-7,-41,-49,-287,    -2009,1,7,41,49,287,2009;    x-y=-2009,-287,-49,-41,-7,-1,    2009,287,49,41,7,1 \\end{cases}$$．  故$$\\left( x,y \\right)$$共有$$12$$组不同的表示． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1316", "queId": "8effba2d57b841f89b27235729705519", "competition_source_list": ["2008年第19届希望杯初二竞赛第1试第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "初二（$$1$$）班有$$48$$名同学，其中有男同学$$n$$名，将他们编成$$1$$号，$$2$$号，$$\\cdots $$，$$n$$号，在寒假期间，$$1$$号给$$3$$名同学打过电话，$$2$$号给$$4$$名同学打过电话，$$3$$号给$$5$$名同学打过电话，$$\\cdots \\cdots $$，$$n$$号给全班一半同学打过电话，由此可知该班女同学的人数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$22$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$25$$ "}], [{"aoVal": "D", "content": "$$26$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数与实际问题"], "answer_analysis": ["由题意，可知$$n$$号同学给$$(n+2)$$名同学打过电话，  所以$$n+2=48\\div 2$$，$$n=22$$，  所以女同学有$$48-22=26$$（名）．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1133", "queId": "b23ea5bae7154c5ba9e3bebf7841b511", "competition_source_list": ["2001年第12届希望杯初二竞赛第1试第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "在正常情况下，一个司机每天驾车行驶$$t$$小时，且平均速度为$$v$$千米/小时，若他一天内多行驶$$1$$小时，平均速度比平时快$$5$$千米/小时，则比平时多行驶$$70$$千米，若他一天内少行驶$$1$$小时，平均速度比平时慢$$5$$千米/小时，他将比平时少行驶．", "answer_option_list": [[{"aoVal": "A", "content": "$$60$$千米 "}], [{"aoVal": "B", "content": "$$70$$千米 "}], [{"aoVal": "C", "content": "$$75$$千米 "}], [{"aoVal": "D", "content": "$$80$$千米 "}]], "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": ["由题意知$$(t+1)\\cdot (v+5)-vt=70$$，  即$$5t+v+5=70$$，  所以$$5t+v=65$$．  若每天少行驶$$1$$小时，且速度比平时慢$$5$$千米/小时，则  $$vt-(t-1)\\cdot (v-5)=vt-vt+5t+v-5=60$$（千米）．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "163", "queId": "4f65a69f008742f28a488b12fc6c9343", "competition_source_list": ["山西太原竞赛", "2020~2021学年广东深圳龙岗区深圳市百合外国语学校初一下学期单元测试《整式的乘除》第9题3分", "2019~2020学年四川成都金牛区成都外国语学校初二下学期单元测试《因式分解》第8题2分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$ 满足等式$$x={{a}^{2}}+{{b}^{2}}+20$$，$$y=4\\left( 2b-a \\right)$$ ，则$$x$$，$$y$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\leqslant y$$ "}], [{"aoVal": "B", "content": "$$x\\geqslant y$$ "}], [{"aoVal": "C", "content": "$$x\\textless{}y$$ "}], [{"aoVal": "D", "content": "$$x\\textgreater{}y$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->配方思想的运用", "课内体系->能力->运算能力"], "answer_analysis": ["∵$$x={{a}^{2}}+{{b}^{2}}+20$$，$$y=4\\left( 2b-a \\right)=8b-4a$$，  ∴$$x-y={{a}^{2}}+{{b}^{2}}+20-8b+4a$$  $$=\\left( {{a}^{2}}+4a+4 \\right)+\\left( {{b}^{2}}-8b+16 \\right)$$  $$={{\\left( a+2 \\right)}^{2}}+{{\\left( b-4 \\right)}^{2}}$$，  ∵$${{\\left( a+2 \\right)}^{2}}\\geqslant 0$$，$${{\\left( b-4 \\right)}^{2}}\\geqslant 0$$，  ∴$${{\\left( a+2 \\right)}^{2}}+{{\\left( b-4 \\right)}^{2}}\\geqslant 0$$，  ∴$$x-y\\geqslant 0$$，即$$x\\geqslant y$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "422", "queId": "42c1e6440696497995488bc0d1d412ec", "competition_source_list": ["2014年第25届全国希望杯初二竞赛初赛第6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若关于$$x$$的不等式组$$\\left { \\begin{matrix} 2-3x\\geqslant 0    2x+m\\textgreater0   \\end{matrix} \\right.$$没有实数解，则实数$$m$$的取值范围是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$m\\textless{}-\\frac{4}{3}$$ "}], [{"aoVal": "B", "content": "$$m\\leqslant -\\frac{4}{3}$$ "}], [{"aoVal": "C", "content": "$$m\\textgreater-\\frac{4}{3}$$ "}], [{"aoVal": "D", "content": "$$m\\geqslant -\\frac{4}{3}$$ "}]], "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）->一元一次不等式组->解一元一次不等式组"], "answer_analysis": ["解不等式组$$\\left { \\begin{matrix} 2-3x\\geqslant 0    2x+m\\textgreater0   \\end{matrix} \\right.$$可得$$\\left { \\begin{matrix}x\\leqslant \\dfrac{2}{3}    x\\textgreater-\\dfrac{m}{2}   \\end{matrix} \\right.$$．  因为原不等式组没有实数解，  所以$$-\\frac{m}{2}\\geqslant \\frac{2}{3}$$，  解得$$m\\leqslant -\\frac{4}{3}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1493", "queId": "d3b4ea10265d4dbf8b9d9a46316796b9", "competition_source_list": ["初一上学期单元测试《有理数》数轴的概念及应用第10题", "1995年第6届全国希望杯初一竞赛复赛第5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": ["知识标签->题型->数->有理数->数轴与有理数有关的概念->题型：数轴上的规律探究", "知识标签->知识点->数->有理数->数轴", "知识标签->学习能力->空间想象能力", "知识标签->学习能力->运算能力"], "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": "1065", "queId": "ff8080814d7978b9014d88c5ae7d2ab6", "competition_source_list": ["1991年第2届全国希望杯初一竞赛复赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$均为有理数．在下列两个结论中，（~ ）．  甲：若$$a\\textgreater b$$，则$$a{{c}^{2}}\\textgreater b{{c}^{2}}$$．  乙：若$$a{{c}^{2}}\\textgreater b{{c}^{2}}$$，则$$a\\textgreater b$$．", "answer_option_list": [[{"aoVal": "A", "content": "甲、乙都真 "}], [{"aoVal": "B", "content": "甲真，乙不真 "}], [{"aoVal": "C", "content": "甲不真，乙真 "}], [{"aoVal": "D", "content": "甲、乙都不真 "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->命题与证明", "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": ["若$$c=0$$，甲不正确．  对于乙，若$$a{{c}^{2}}\\textgreater b{{c}^{2}}$$，可推出$$c\\ne 0$$，∴$${{c}^{2}}\\textgreater0$$，进而推出$$a\\textgreater b$$，乙正确．选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1582", "queId": "f98e1a6a62f14df9a1d9923d38671475", "competition_source_list": ["2016年第27届全国希望杯初二竞赛初赛第2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．若一次函数$$y=x-3$$与$$y=kx-k$$（$$k$$为整数）的图象的交点是整点，则$$k$$的不同取值的个数是．", "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->知识点->函数->一次函数->一次函数与方程、不等式->一次函数与二元一次方程组", "课内体系->能力->推理论证能力"], "answer_analysis": ["由题意得$$\\begin{cases}y=x-3    y=kx-k \\end{cases}$$，  解得$$\\begin{cases}x=\\dfrac{k-3}{k-1}    y=\\dfrac{-2k}{k-1} \\end{cases}$$，  ∴$$x=1-\\frac{2}{k-1}$$，$$y=-2-\\frac{2}{k-1}$$，  ∵交点是整点，  ∴$$k-1=\\pm 1$$或$$\\pm 2$$，且$$k-1\\ne 0$$，  解得$$k=0$$，$$2$$，$$-1$$或$$3$$．  ∵$$k\\ne 0$$，  ∴$$k$$的不同取值的个数是$$3$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "983", "queId": "ff8080814cfa9b24014cfe948e250a26", "competition_source_list": ["初二其它", "2020~2021学年北京海淀区清华大学附属初级中学初一上学期期中第9题2分", "1997年第8届希望杯初二竞赛第2试第2题", "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8题"], "difficulty": "2", "qtype": "single_choice", "problem": "把多项式$${{x}^{2}}-{{y}^{2}}-2x-4y-3$$因式分解之后，正确的结果是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$(x+y+3)(x-y-1)$$ "}], [{"aoVal": "B", "content": "$$(x+y-1)(x-y+3)$$ "}], [{"aoVal": "C", "content": "$$(x+y-3)(x-y+1)$$ "}], [{"aoVal": "D", "content": "$$(x+y+1)(x-y-3)$$ "}]], "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->拆添项法"], "answer_analysis": ["$${{x}^{2}}-{{y}^{2}}-2x-4y-3$$  $$={{x}^{2}}-2x+1-({{y}^{2}}+4y+4)$$  $$={{(x-1)}^{2}}-{{(y+2)}^{2}}$$  $$=(x-1+y+2)(x-1-y-2)$$  $$=(x+y+1)(x-y-3)$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "9", "queId": "009577e8481c4333b912dbe924e56bbd", "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": ["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->同余->剩余系及其应用", "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": ["∵$$5{{n}^{2}}+3n-5$$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": "1302", "queId": "fbe9517b7c254564b9709e43b551a2d9", "competition_source_list": ["2001年第12届希望杯初二竞赛第1试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\triangle ABC$$中，$$\\angle B=60{}^{}\\circ $$，$$\\angle C\\textgreater\\angle A$$，且$${{(\\angle C)}^{2}}={{(\\angle A)}^{2}}+{{(\\angle B)}^{2}}$$，则$$\\triangle ABC$$的形状是．", "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": "直角或钝角三角形 "}]], "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"], "answer_analysis": ["$$\\triangle ABC$$中，$$\\angle B=60{}^{}\\circ $$，$$\\angle C\\textgreater\\angle A$$，设$$\\angle C=60{}^{}\\circ +x{}^{}\\circ $$，则$$\\angle A=60{}^{}\\circ -x{}^{}\\circ $$，  因为$${{(\\angle C)}^{2}}={{(\\angle A)}^{2}}+{{(\\angle B)}^{2}}$$，  所以$${{(\\angle B)}^{2}}={{(\\angle C)}^{2}}-{{(\\angle A)}^{2}}$$，  $$=(\\angle C+\\angle A)\\cdot (\\angle C-\\angle A)$$，  所以$$3600=120\\times 2x$$，所以$$x{}^{}\\circ =15{}^{}\\circ $$，  所以$$\\angle C=75{}^{}\\circ $$，$$\\angle A=45{}^{}\\circ $$．  故$$\\triangle ABC$$是锐角三角形．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1217", "queId": "8aac49074f25f754014f3ebcc878486a", "competition_source_list": ["2012年第23届全国希望杯初二竞赛初赛第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "在$${{2}^{77}}$$，$${{3}^{55}}$$，$${{5}^{44}}$$，$${{6}^{33}}$$这四个数中，最大的数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{2}^{77}}$$ "}], [{"aoVal": "B", "content": "$${{3}^{55}}$$ "}], [{"aoVal": "C", "content": "$${{5}^{44}}$$ "}], [{"aoVal": "D", "content": "$${{6}^{33}}$$ "}]], "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->式->整式的乘除->幂的运算->利用幂的乘方比较大小", "课内体系->能力->运算能力"], "answer_analysis": ["将题设的四个数化成同指数的幂，得  $${{2}^{77}}={{({{2}^{7}})}^{11}}={{128}^{11}}$$，  $${{3}^{55}}={{({{3}^{5}})}^{11}}={{243}^{11}}$$，  $${{5}^{44}}={{({{5}^{4}})}^{11}}={{625}^{11}}$$，  $${{6}^{33}}={{({{6}^{3}})}^{11}}={{216}^{11}}$$，  所以最大的数是$${{5}^{44}}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "21", "queId": "0523dddad54b46dcbd91a6080961c740", "competition_source_list": ["2012年竞赛第2题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "在平面直角坐标系$$xOy$$中，满足不等式$${{x}^{2}}+{{y}^{2}}\\leqslant 2x+2y$$的整数点坐标$$\\left( x,y \\right)$$的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->平面直角坐标系->坐标系综合->坐标与距离"], "answer_analysis": ["由题设$${{x}^{2}}+{{y}^{2}}\\leqslant 2x+2y$$，得$$0\\leqslant {{\\left( x-1 \\right)}^{2}}{{\\left( y-1 \\right)}^{2}}\\leqslant 2$$，  因为$$x$$，$$y$$均为整数，所以有  $$\\begin{cases}{{(x-1)}^{2}}=0    {{\\left( y-1 \\right)}^{2}}=0   \\end{cases}$$或$$\\begin{cases}{{\\left( x-1 \\right)}^{2}}=0    {{\\left( y-1 \\right)}^{2}}=1   \\end{cases}$$或 $$\\begin{cases}{{\\left( x-1 \\right)}^{2}}=1    {{\\left( y-1 \\right)}^{2}}=0   \\end{cases}$$或$$\\begin{cases}{{\\left( x-1 \\right)}^{2}}=1    {{\\left( y-1 \\right)}^{2}}=1   \\end{cases}$$，  解得：$$\\begin{cases}x=1    y=1   \\end{cases}$$；$$\\begin{cases}x=1    y=2   \\end{cases}$$；$$\\begin{cases}x=1    y=0   \\end{cases}$$；$$\\begin{cases}x=0    y=1   \\end{cases}$$；$$\\begin{cases}x=0    y=0   \\end{cases}$$；$$\\begin{cases}x=0    y=2   \\end{cases}$$；$$\\begin{cases}x=2    y=1   \\end{cases}$$；$$\\begin{cases}x=2    y=0   \\end{cases}$$；$$\\begin{cases}x=2    y=2   \\end{cases}$$，  以上共计$$9$$对$$(x,y)$$．  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "213", "queId": "0d933d6733e941738c4cc59ca8f5298f", "competition_source_list": ["2009年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "一个长方形长变为原来的$$1.5$$倍，宽减少20\\%，则原面积占新面积的几分之几？ ．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{7}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "E", "content": "$$\\frac{1}{3}$$ "}]], "knowledge_point_routes": ["美国amc8->知识点->几何->直线型->几何模型"], "answer_analysis": ["在维度为$$10\\times 10$$的矩形中，新的矩形将具有维度为$$15\\times 8$$，新区域与旧区域比例为$$\\frac{100}{120}=\\frac{5}{6}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1346", "queId": "8aac50a75139269a0151626dc2d352ed", "competition_source_list": ["1992年第9届全国初中数学联赛竞赛第2题", "初三上学期其它"], "difficulty": "0", "qtype": "single_choice", "problem": "若$${{x}_{0}}$$是一元二次方程$$a{{x}^{2}}+bx+c=0(a\\ne 0)$$的根，则判别式$$\\Delta ={{b}^{2}}-4ac$$与平方式$$M={{(2a{{x}_{0}}+b)}^{2}}$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\Delta \\textgreater M$$ "}], [{"aoVal": "B", "content": "$$\\Delta =M$$ "}], [{"aoVal": "C", "content": "$$\\Delta \\textless{}M$$ "}], [{"aoVal": "D", "content": "不能确定 "}]], "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"], "answer_analysis": ["把$${{x}_{0}}$$代入方程$$a{{x}^{2}}+bx+c=0$$中得$$ax_{0}^{2}+b{{x}_{0}}=-c$$，  ∵$${{(2a{{x}_{0}}+b)}^{2}}=4{{a}^{2}}x_{0}^{2}+4ab{{x}_{0}}+{{b}^{2}}$$，  ∴$${{(2a{{x}_{0}}+b)}^{2}}=4a(a{{x}_{0}}^{2}+b{{x}_{0}})+{{b}^{2}}=-4ac+{{b}^{2}}=\\Delta $$，  ∴$$M=\\Delta $$．  故选$$\\text{B}$$． ", "<p>∵$$x_0$$是一元二次方程$$a{{x}^{2}}+bx+c=0(a\\ne 0)$$的根，</p>\n<p>&there4;$${{x}_{0}}=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}$$，</p>\n<p>&there4;$${{b}^{2}}-4ac={{(2a{{x}_{0}}+b)}^{2}}$$，</p>\n<p>&there4;$$\\Delta =M$$．</p>\n<p>故选$$\\text{B}$$．</p>"], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1461", "queId": "af7bbc1900274af6901c1f20eb4fe384", "competition_source_list": ["2006年第17届希望杯初一竞赛复赛第3题4分", "2018~2019学年湖南常德澧县初一上学期期中第8题3分", "2015~2016学年上海长宁区上海市复旦初级中学初一上学期期中第18题2分", "2012~2013学年广东广州海珠区中山大学附属中学初一上学期期中第7题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在代数式$$x{{y}^{2}}$$中，$$x$$和$$y$$的值各减少$$25 \\%$$，则该代数式的值减少了（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$50 \\%$$ "}], [{"aoVal": "B", "content": "$$75 \\%$$ "}], [{"aoVal": "C", "content": "$$\\frac{37}{64}$$ "}], [{"aoVal": "D", "content": "$$\\frac{27}{64}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->整式的乘除运算->整式乘除与实际问题", "课内体系->能力->运算能力"], "answer_analysis": ["∵在代数式$$x{{y}^{2}}$$中，$$x$$和$$y$$的值各减少$$25 \\%$$，  ∴知$${{x}^{\\prime }}=\\frac{3}{4}x$$，$${{y}^{\\prime }}=\\frac{3}{4}y$$，  ∴$${{x}^{\\prime }}{{({{y}^{\\prime }})}^{2}}=\\left( \\frac{3}{4}x \\right)\\times {{\\left( \\frac{3}{4}y \\right)}^{2}}=\\frac{27}{64}x{{y}^{2}}$$，  ∴该代数式的值减少了$$\\frac{37}{64}$$．  故选$$\\text{C}$$． ", "<p>因为$$x(1-25\\%)\\cdot {{\\left[ y(1-25\\%) \\right]}^{2}}=\\frac{27}{64}x{{y}^{2}}$$</p>\n<p>所以代数式的值减少了$$1-\\frac{27}{64}=\\frac{37}{64}$$</p>\n<p>故选$$\\text{C}$$．</p>"], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "71", "queId": "131cdd1a36dc453ea36896741414fa7e", "competition_source_list": ["2016年第27届全国希望杯初二竞赛复赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a$$和$$b$$都是个位数字和十位数字相同的两位数，$$c$$是各位数字都相同的四位数，且$${{a}^{2}}+b=c$$，则$$a+b+c$$的最大值和最小值的差是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$6732$$ "}], [{"aoVal": "B", "content": "$$3179$$ "}], [{"aoVal": "C", "content": "$$6723$$ "}], [{"aoVal": "D", "content": "$$3187$$ "}]], "knowledge_point_routes": ["课内体系->方法->枚举法", "课内体系->知识点->方程与不等式->其他方程->不定方程", "课内体系->思想->分类讨论思想"], "answer_analysis": ["由分析可知：  若$$a=11$$，$${{a}^{2}}=121$$，$$c=1111$$，则$$b=990$$，不符合题意；  若$$a=22$$，$${{a}^{2}}=484$$，$$c=1111$$，则$$b=627$$，不符合题意；  若$$a=33$$，$${{a}^{2}}=1089$$，$$c=1111$$，则$$b=22$$，符合题意；  若$$a=44$$，$${{a}^{2}}=1936$$，$$c=2222$$，则$$b=296$$，不符合题意；  若$$a=55$$，$${{a}^{2}}=3025$$，$$c=3333$$，则$$b=308$$，不符合题意；  若$$a=66$$，$${{a}^{2}}=4356$$，$$c=4444$$，则$$b=88$$，符合题意；  若$$a=77$$，$${{a}^{2}}=5929$$，$$c=6666$$，则$$b=737$$，不符合题意；  若$$a=88$$，$${{a}^{2}}=7744$$，$$c=7777$$，则$$b=33$$，符合题意；  若$$a=99$$，$${{a}^{2}}=9801$$，$$c=9999$$，则$$b=198$$，不符合题意．  综上，满足条件的解有三组，即$$\\left { \\begin{matrix}a=33    b=22    c=1111   \\end{matrix} \\right.$$，$$\\left { \\begin{matrix}a=66    b=88    c=4444   \\end{matrix} \\right.$$，$$\\left { \\begin{matrix}a=88    b=33    c=7777   \\end{matrix} \\right.$$，  ∴$$a+b+c$$的值最大为$$7898$$，最小为$$1166$$，  ∴$$a+b+c$$的最大值和最小值的差是$$7898-1166=6732$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "0", "queId": "0005bdfa24104b999c91bf403480acf4", "competition_source_list": ["1993年第4届希望杯初二竞赛第10题"], "difficulty": "0", "qtype": "single_choice", "problem": "如果方程$$\\left\\textbar3x\\right\\textbar-ax-1=0$$的根是负数，那么$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater-3$$ "}], [{"aoVal": "B", "content": "$$a\\geqslant 3$$ "}], [{"aoVal": "C", "content": "$$a \\textless{} 3$$ "}], [{"aoVal": "D", "content": "$$a\\leqslant -3$$ "}]], "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->不等式（组）->含参不等式->不等式（组）解的情况确定参数的范围", "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->解一元一次方程"], "answer_analysis": ["原方程即$$\\left\\textbar3x\\right\\textbar=ax+1$$．  $$\\left(1\\right)$$若$$a=3$$，则$$\\left\\textbar3x\\right\\textbar=3x+1$$．  当$$x \\textless{} 0$$时，得$$-3x=3x+1$$．所以$$x=-\\frac{1}{6}$$；  当$$x\\geqslant 0$$时，得$$3x=3x+1$$，不成立．  所以当$$a=3$$时，原方程的根为$$x=-\\frac{1}{6}$$．  $$\\left(2\\right)$$若$$a\\textgreater3$$．  当$$x \\textless{} 0$$时，得$$-3x=ax+1$$，所以$$x=-\\frac{1}{a+3} \\textless{} 0$$；  当$$x\\geqslant 0$$时，得$$3x=ax+1$$，所以$$x=\\frac{1}{3-a} \\textless{} 0$$，矛盾．  所以当$$a\\textgreater3$$时，原方程的解为$$x=-\\frac{1}{a+3} \\textless{} 0$$．  $$\\left(3\\right)$$若$$a \\textless{} 3$$．  当$$x\\geqslant 0$$时，得$$3x=ax+1$$，所以$$x=\\frac{1}{3-a}\\textgreater0$$，原方程有正根，与题设矛盾．  综上所述，$$a\\geqslant 3$$时，原方程的根是负数．  所以应选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1432", "queId": "aab7866c18c749248ace27dd855b36e4", "competition_source_list": ["1996年全美数学竞赛（AMC）竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "在一个圆形区域内随机选择一个点．此点距离圆心比距离圆周更近的概率是多少?", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac 14$$ "}], [{"aoVal": "B", "content": "$$\\frac 13$$ "}], [{"aoVal": "C", "content": "$$\\frac 12$$ "}], [{"aoVal": "D", "content": "$$\\frac 23$$ "}], [{"aoVal": "E", "content": "$$\\frac 34$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Counting, Probability and Statistics->Geometric Probability", "课内体系->知识点->统计与概率"], "answer_analysis": ["在一个圆形区域内随机选择一个点．此点圆心比离圆周更近的概率是多少?  画一个半径为$$2$$的圆．画一个半径为$$1$$的同心圆．这个内圆的边缘是所有点的集合，到中心和外圆分别为$$1$$和$$1$$．换句话说，它是到圆心与大圆等距的所有点的集合．半径为$$1$$的圆的内侧是所有点的集合，这些点比外圆的边界更靠近区域的中心．``垫圈''区域在半径$$1$$的圆外，但在半径$$2$$的圆内，是所有靠近边界而不是圆心的点的集合．  如果在区域$$B$$中选择一个随机点，则该点位于较小子区域$$A$$中的概率为$$\\frac{A}{B}$$的比率．在这种情况下，$$B= \\pi \\cdot {{2}^{2}}=4 \\pi $$，$$A= \\pi \\cdot {{1}^{2}}= \\pi $$，面积比是$$\\frac{ \\pi }{4 \\pi }=\\frac{1}{4}$$．  故选$$\\text{A}$$．  Draw a circle with a radius of $$2$$ . Draw a concentric circle with radius $$1$$ . The edge of this nner circle is the set of all points that are $$1$$ from the center, and $$1$$ from the outer circle. In other words, it is the set of all points that are equidistan t from the center of the circles to the outside of the big circle.  The inside of this circle of radius $$1$$ is the set of all points that are closer to the center of the region than to the boundary of the outer circle. The \"washer\" region that is outside the circle of radius $$1$$, but inside the circle of radius $$2$$, is the set of all points that are closer to the boundary than to the center of the circle.  If you select a random point in a region of area $$B$$, the probability that the point is in a smaller subregion $$A$$ is the ratio$$\\frac{A}{B}$$. In this case, $$B=\\pi \\cdot {{2}^{2}}=4\\pi $$,~ and $$A=\\pi \\cdot {{1}^{2}}=\\pi $$, and the ratio of areas is $$\\frac{\\pi }{4\\pi }=\\frac{1}{4}$$, and the answer is $$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1059", "queId": "ff8080814d7978b9014d86e6be4d25f4", "competition_source_list": ["1991年第2届全国希望杯初一竞赛初赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "浓度为$$p \\%$$的盐水$$m$$公斤与浓度为$$q \\%$$的盐水$$n$$公斤混合后的溶液浓度是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{p+q}{2} \\%$$ "}], [{"aoVal": "B", "content": "$$(mp+nq) \\%$$ "}], [{"aoVal": "C", "content": "$$\\frac{(mp+nq)}{p+q} \\%$$ "}], [{"aoVal": "D", "content": "$$\\frac{(mp+nq)}{m+n} \\%$$ "}]], "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": ["设混合溶液浓度为$$x$$，则$$m\\times p \\%+n\\times q \\%=(m+n)x$$．  $$x=\\frac{mp+nq}{m+n} \\%$$．选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "947", "queId": "a42e0d3365f24a859cf4a0d78672d01a", "competition_source_list": ["2001年第12届希望杯初二竞赛第2试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$\\left\\textbar{} a \\right\\textbar{} ~\\textless{} ~\\left\\textbar{} c \\right\\textbar$$，$$b=\\frac{a+c}{2}$$，$$\\left\\textbar{} b \\right\\textbar{} ~\\textless{} ~2\\left\\textbar{} a \\right\\textbar$$，$${{S}_{1}}=\\left\\textbar{} \\frac{a-b}{c} \\right\\textbar$$，$${{S}_{2}}=\\left\\textbar{} \\frac{b-c}{a} \\right\\textbar$$，$${{S}_{3}}=\\left\\textbar{} \\frac{a-c}{b} \\right\\textbar$$，则$${{S}_{1}}$$，$${{S}_{2}}$$，$${{S}_{3}}$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{S}_{1}}\\textless{{S}_{2}}\\textless{{S}_{3}}$$ "}], [{"aoVal": "B", "content": "$${{S}_{1}}\\textgreater{{S}_{2}}\\textgreater{{S}_{3}}$$ "}], [{"aoVal": "C", "content": "$${{S}_{1}}\\textless{{S}_{3}}\\textless{{S}_{2}}$$ "}], [{"aoVal": "D", "content": "$${{S}_{1}}\\textgreater{{S}_{3}}\\textgreater{{S}_{2}}$$ "}]], "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$$b=\\frac{a+c}{2}$$，  所以$${{S}_{1}}=\\left\\textbar{} \\frac{a-b}{c} \\right\\textbar=\\left\\textbar{} \\frac{a-\\dfrac{a+c}{2}}{c} \\right\\textbar=\\left\\textbar{} \\frac{a-c}{2c} \\right\\textbar$$.  $${{S}_{2}}=\\left\\textbar{} \\frac{b-c}{a} \\right\\textbar=\\left\\textbar{} \\frac{\\dfrac{a+c}{2}-c}{a} \\right\\textbar=\\left\\textbar{} \\frac{a-c}{2a} \\right\\textbar$$，  因为$$\\left\\textbar{} a \\right\\textbar{} ~\\textless{} ~\\left\\textbar{} c \\right\\textbar$$，  所以$${{S}_{1}} ~\\textless{} ~{{S}_{2}}$$，  又$$\\left\\textbar{} b \\right\\textbar{} ~\\textless{} ~2\\left\\textbar{} a \\right\\textbar$$，  所以$${{S}_{3}}=\\textbar\\frac{a-c}{b}\\textbar\\textgreater\\textbar\\frac{a-c}{2a}\\textbar={{S}_{2}}$$  所以$${{S}_{1}} ~\\textless{} ~{{S}_{2}} ~\\textless{} ~{{S}_{3}}$$．选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1192", "queId": "a4d9032099e342a8977b0283dbc22b9b", "competition_source_list": ["2019年第1届广东深圳罗湖区深圳中学初一竞赛（凤凰木杯）第9题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a{{x}^{\\left\\textbar{} a+2 \\right\\textbar}}+3x+1$$为一次多项式，则$$a$$的所有取值之和为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$-3$$ "}], [{"aoVal": "D", "content": "$$-6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->对称多项式 ", "课内体系->知识点->式->整式的加减->整式有关的概念->多项式->由多项式的项和次数求参数的值"], "answer_analysis": ["∵$$a{{x}^{\\left\\textbar{} a+2 \\right\\textbar}}+3x+1$$为一次多项式，  ∴①$$a=0$$时，原式为$$3x+1$$满足条件，  ②$$\\left\\textbar{} a+2 \\right\\textbar=1$$时，$$a=-1$$或$$-3$$，  $$a=-1$$时原式为$$-x+3x+1=2x+1$$，满足条件，  $$a=-3$$时原式为$$-3x+3x+1=0$$，不满足条件，  ③$$\\left\\textbar{} a+2 \\right\\textbar=0$$时，$$a=-2$$，  原式$$=-2{{x}^{0}}+3x+1=3x-1$$满足条件，  综上$$a=0$$或$$-1$$或$$-2$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "437", "queId": "55097d0a7b9f46a3849a6ffada509ab7", "competition_source_list": ["2016年第27届全国希望杯初二竞赛初赛第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\triangle ABC$$和$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$，$$D$$和$${{D}_{1}}$$分别是$$BC$$和$${{B}_{1}}{{C}_{1}}$$的中点，下列条件中能判定$$\\triangle ABC$$≌$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\angle C=\\angle {{C}_{1}}$$，$$AC={{A}_{1}}{{C}_{1}}$$，$$AD={{A}_{1}}{{D}_{1}}$$ "}], [{"aoVal": "B", "content": "$$\\angle B=\\angle {{B}_{1}}$$，$$AB={{A}_{1}}{{B}_{1}}$$，$$AC={{A}_{1}}{{C}_{1}}$$ "}], [{"aoVal": "C", "content": "$$AD={{A}_{1}}{{D}_{1}}$$，$$BD={{B}_{1}}{{D}_{1}}$$，$$CD={{C}_{1}}{{D}_{1}}$$ "}], [{"aoVal": "D", "content": "$$\\angle BAD=\\angle {{B}_{1}}{{A}_{1}}{{D}_{1}}$$，$$AB={{A}_{1}}{{B}_{1}}$$，$$AD={{A}_{1}}{{D}_{1}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->全等三角形->全等三角形的判定->SAS"], "answer_analysis": ["$$\\text{A}$$中，$$\\text{SSA}$$不能证明$$\\triangle ACD$$≌$$\\triangle {{A}_{1}}{{C}_{1}}{{D}_{1}}$$，进而也不能证明$$\\triangle ABC$$≌$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$；  $$\\text{B}$$中，$$\\text{SSA}$$不能证明$$\\triangle ABC$$≌$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$；  $$\\text{C}$$中，任意两个斜边相等的直角三角形，都满足，不能证明$$\\triangle ABC$$≌$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$；  $$\\text{D}$$中，通过$$\\text{SAS}$$可证明$$\\triangle ABD$$≌$$\\triangle {{A}_{1}}{{B}_{1}}{{D}_{1}}$$，  ∴$$\\angle B=\\angle {{B}_{1}}$$，$$BD={{B}_{1}}{{D}_{1}}$$，  ∴$$BC={{B}_{1}}{{C}_{1}}$$，  ∴$$\\triangle ABC$$≌$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "365", "queId": "3dded09eaf70426e9258c234f92a4ac5", "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": "$$uncertain$$ "}]], "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": "1110", "queId": "ff8080814d9efd56014daa6f01150a76", "competition_source_list": ["1993年第4届全国希望杯初一竞赛初赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a\\textless{}b$$，则$$(a-b)\\left\\textbar{} a-b \\right\\textbar$$等于（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{(a-b)}^{2}}$$ "}], [{"aoVal": "B", "content": "$${{b}^{2}}-{{a}^{2}}$$ "}], [{"aoVal": "C", "content": "$${{a}^{2}}-{{b}^{2}}$$ "}], [{"aoVal": "D", "content": "$$-{{(a-b)}^{2}}$$ "}]], "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->知识点->数->有理数->绝对值->绝对值的非负性"], "answer_analysis": ["因为$$a\\textless{}b$$，所以$$a-b\\textless{}0$$，此时$$\\left\\textbar{} a-b \\right\\textbar=b-a$$．  所以$$(a-b)\\left\\textbar{} a-b \\right\\textbar=(a-b)(b-a)=-(a-b)(a-b)=-{{(a-b)}^{2}}$$，选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "999", "queId": "cd73d781180049fa9ed04a6861d114b6", "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3题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "定义一种运算：$${\\text{C}_{m}}^{n}=\\frac{m\\times \\left( m-1 \\right)\\times \\left( m-2 \\right)\\times \\cdots~ \\times \\left( m-n+1 \\right)}{n\\times \\left( n-1 \\right)\\times \\left( n-2 \\right)\\times \\cdots~ \\times 1}$$，则$${\\text{C}_{5}}^{3}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{10}{3}$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->排列与组合"], "answer_analysis": ["∵$$\\text{C}_{m}^{n}=\\frac{m\\times \\left( m-1 \\right)\\times \\left( m-2 \\right)\\times \\cdots~ \\times \\left( m-n+1 \\right)}{n\\times \\left( n-1 \\right)\\times \\left( n-2 \\right)\\times \\cdots \\times 1}$$，  ∴$$\\text{C}_{5}^{3}=\\frac{5\\times 4\\times \\left( 5-3+1 \\right)}{3\\times 2\\times 1}=\\frac{5\\times 4\\times 3}{6}=10$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "364", "queId": "d5409cf0a46d4d57a3352b9a23c7c750", "competition_source_list": ["1996年第13届全国初中数学联赛竞赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "实数$$a$$，$$b$$满足$$ab=1$$，记$$M=\\frac{1}{1+a}+\\frac{1}{1+b}$$，$$N=\\frac{a}{1+a}+\\frac{b}{1+b}$$，则$$M$$与$$N$$的关系是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$M\\textgreater N$$ "}], [{"aoVal": "B", "content": "$$M=N$$ "}], [{"aoVal": "C", "content": "$$M\\textless{}N$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的运算->分式加法运算"], "answer_analysis": ["方法一：将$$a=\\frac{1}{b}$$分别代入$$M$$和$$N$$，  则$$M=\\frac{1}{1+\\frac{1}{b}}+\\frac{1}{1+b}=\\frac{b}{1+b}+\\frac{1}{1+b}=1$$，  $$N=\\frac{\\frac{1}{b}}{1+\\frac{1}{b}}+\\frac{b}{1+b}=\\frac{1}{1+b}+\\frac{b}{1+b}=1$$，  所以$$M=N$$．选$$\\text{B}$$，  方法二：$$M=\\frac{1}{1+a}+\\frac{1}{1+b}=\\frac{b}{b+ab}+\\frac{a}{a+ab}$$，  又由$$ab=1$$，  得到$$M=\\frac{b}{b+1}+\\frac{a}{a+1}=N$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "11", "queId": "1bdf8eb25e4941868ae98b552b609cf8", "competition_source_list": ["2018年山东淄博临淄区初三中考一模第6题4分", "2019~2020学年9月四川达州渠县四川省渠县中学初三上学期月考第8题3分", "2018年浙江宁波余姚市余姚市实验学校初二竞赛第2题5分", "2018~2019学年山东济宁曲阜市初三上学期期中第7题3分", "2017~2018学年山东临沂蒙阴县初三上学期期末第6题3分", "2018~2019学年山东济宁曲阜市曲阜市实验中学初三上学期期中第7题3分", "2017~2018学年广东东莞市万江区东莞市北师大翰林实验学校初三上学期期中第6题3分", "2018年陕西西安灞桥区西安铁一中滨河学校初三中考四模第6题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$${{x}_{1}},{{x}_{2}}$$是方程$${{x}^{2}}-2mx+{{m}^{2}}-m-1=0$$的两个根，且$${{x}_{1}}+{{x}_{2}}=1-{{x}_{1}}{{x}_{2}}$$，则$$m$$的值为（  ）", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$或$$2$$ "}], [{"aoVal": "B", "content": "$$1$$或$$-2$$ "}], [{"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", "课内体系->能力->运算能力"], "answer_analysis": ["根据韦达定理，$${{x}_{1}}+{{x}_{2}}=2m$$，$${{x}_{1}}{{x}_{2}}={{m}^{2}}-m-1$$．  $${{x}_{1}}+{{x}_{2}}=1-{{x}_{1}}{{x}_{2}}$$，  ∴$$2m=1-({{m}^{2}}-m-1)$$，  解得$$m=-2$$或$$1$$，  而$$\\Delta =4{{m}^{2}}-4({{m}^{2}}-m-1)\\geqslant0$$，  ∴$$m\\textgreater-1$$． ", "<p>∵$${{x}_{1}}$$，$${{x}_{2}}$$是方程$${{x}^{2}}-2mx+{{m}^{2}}-m-1=0$$两个根，</p>\n<p>&there4;$${{x}_{1}}+{{x}_{2}}=2m$$，$${{x}_{1}}\\cdot {{x}_{2}}={{m}^{2}}-m-1$$．</p>\n<p>∵$${{x}_{1}}+{{x}_{2}}=1-{{x}_{1}}{{x}_{2}}$$，</p>\n<p>&there4;$$2m=1-({{m}^{2}}-m-1)$$，即$${{m}^{2}}+m-2=0$$，</p>\n<p>解得：$${{m}_{1}}=-2$$，$${{m}_{2}}=1$$．</p>\n<p>∵方程$${{x}^{2}}-2mx+{{m}^{2}}-m-1=0$$有实数根，</p>\n<p>&there4;$$\\Delta ={{(-2m)}^{2}}-4({{m}^{2}}-m-1)=4m+4\\geqslant 0$$，</p>\n<p>解得：$$m\\geqslant -1$$．</p>\n<p>&there4;$$m=1$$．</p>\n<p>故选$$\\text{D}$$．</p>"], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1543", "queId": "eaf40c34c77e4dcdb69c286795415c99", "competition_source_list": ["2006年第17届希望杯初二竞赛第1试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$m$$，$$n$$是实数，且满足 $${{m}^{2}}+2{{n}^{2}}+m-\\frac{4}{3}n+\\frac{17}{36}=0$$，则$$-m{{n}^{2}}$$的平方根是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{2}}{6}$$ "}], [{"aoVal": "B", "content": "$$\\pm \\frac{\\sqrt{2}}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{6}$$ "}], [{"aoVal": "D", "content": "$$\\pm \\frac{1}{6}$$ "}]], "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->整式的乘除->乘法公式->配方思想的运用"], "answer_analysis": ["\\textbf{（知识点：配方法）}  由题设，得：  $$\\left( {{m}^{2}}+m+\\frac{1}{4} \\right)+2\\left( {{n}^{2}}-\\frac{2}{3}n+\\frac{1}{9} \\right)+\\frac{17}{36}-\\frac{1}{4}-\\frac{2}{9}=0$$，  即$${{\\left( m+\\frac{1}{2} \\right)}^{2}}+2{{\\left( n-\\frac{1}{3} \\right)}^{2}}=0$$，  因为$${{\\left( m+\\frac{1}{2} \\right)}^{2}}\\geqslant 0$$，$$2{{\\left( n-\\frac{1}{3} \\right)}^{2}}\\geqslant 0$$，  所以$$m=-\\frac{1}{2}$$，$$n=\\frac{1}{3}$$，  $$-m{{n}^{2}}=\\frac{1}{18}$$，它的平方根是$$\\pm \\frac{\\sqrt{2}}{6}$$．  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": "1548", "queId": "d8e73f5059e646ad8c5efc78106eed7f", "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": "在下列$$2021$$个数中：$$1^{2}$$，$$2^{2}$$，$$3^{2}$$，$$\\cdots $$，$$2020^{2}$$，$$2021^{2}$$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->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，$$1^{2}$$，$$2^{2}$$，$$3^{2}$$，$$\\cdots $$，$$2020^{2}$$，$$2021^{2}$$的每一个数前面任意添上``$$+$$''号或``$$-$$''号，其代数和的奇偶性与$$1^{2}+2^{2}+3^{2}+4^{2}+5^{2}-\\cdots -2014^{2}+2015^{2}+2016^{2}-2017^{2}+2018^{2}-2019^{2}-2020^{2}+2021^{2}=55$$（八项归零）的奇偶性相同，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": "690", "queId": "404faeffc0a4464aa90246c21d543ba1", "competition_source_list": ["1998年第9届希望杯初一竞赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "有以下两个数串：$$1$$，$$3$$，$$5$$，$$7$$，\\ldots，$$1991$$，$$1993$$，$$1995$$，$$1997$$，$$1999$$和$$1$$，$$4$$，$$7$$，$$10$$，\\ldots，$$1990$$，$$1993$$，$$1996$$，$$1999$$．同时出现在这两个数串中的数共有（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$333$$ "}], [{"aoVal": "B", "content": "$$334$$ "}], [{"aoVal": "C", "content": "$$335$$ "}], [{"aoVal": "D", "content": "$$336$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->同余->同余的概念与进制性质", "竞赛->知识点->数论->同余->剩余系及其应用", "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->有理数基础运算->有理数除法->有理数四则混合运算"], "answer_analysis": ["第一个数串是$$1 \\tilde{  } 1999$$的整数中被$$2$$除余$$1$$的数，共有$$1000$$个，第二个数串是$$1 \\tilde{  } 1999$$的整数中被$$3$$除余$$1$$的数，共有$$667$$个，同时出现在这两个数串中数是$$1 \\tilde{  } 1999$$的整数中被$$6$$除余$$1$$的数，它们是：$$1$$，$$7$$，$$19$$，$$25$$，\\ldots，$$1993$$，$$1999$$，共$$334$$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": "470", "queId": "59b1ac0eefb8418c80d7f4015b57d185", "competition_source_list": ["2010年第21届全国希望杯初一竞赛复赛第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "若$$a-b$$的相反数是$$2b-a$$，则$$b=$$（~ ）．", "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->相反数->相反数的定义", "课内体系->知识点->数->有理数->相反数->相反数的性质"], "answer_analysis": ["因为$$a-b$$的相反数是$$2b-a$$，  所以$$a-b+2b-a=0$$，  解得$$b=0$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1557", "queId": "c6f8fe2aa9d14315ab1f838d92e35108", "competition_source_list": ["2018年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\rm Bella$$從自己家向$$\\rm Ella$$的房子前進．與此同時 $$\\rm Ella$$ 騎車朝 $$Bella$$ 的家行進．她們各自保持恒定速度，且$$\\rm Ella$$的速度是$$\\rm Bella$$的$$5$$倍．她們的房子相距$$2$$英裏，即$$10560$$英尺，又知$$\\rm Bella$$每步前进 $$2.5$$英尺．在她們相遇時$$B$$走了多少步？  Bella begins to walk from her house toward her friend Ella\\textquotesingle s house. At the same time, Ella begins to ride her bicycle toward Bella\\textquotesingle s house. They each maintain a constant speed, and Ella rides $$5$$ times as fast as Bella walks. The distance between their houses is $$2$$ miles, which is $$10,560$$ feet, and Bella covers $$2\\frac 12$$ feet with each step. How many steps will Bella take by the time she meets Ella?．", "answer_option_list": [[{"aoVal": "A", "content": "$$704$$ "}], [{"aoVal": "B", "content": "$$845$$ "}], [{"aoVal": "C", "content": "$$1056$$ "}], [{"aoVal": "D", "content": "$$1760$$ "}], [{"aoVal": "E", "content": "$$3520$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["因為艾拉騎得比貝拉快$$5$$倍，艾拉騎的速度是$$\\frac{25}{2}$$或$$12\\frac{1}{2}$$．它們一起移動$$15$$英尺每個單位．將$$10560$$除以$$15$$，得出的結果是$$704$$．  故選$$\\text{A}$$． ", "<p>由于 ella 骑得比 bella 快$$5$$倍，它们的速度比是$$5:1$$．这意味着贝拉行驶了$$\\frac{1}{6}$$的路程，$$\\frac{1}{6}$$的$$10560$$英尺是$$1760$$英尺．贝拉也一步行走$$2.5$$英尺，$$1760$$除以$$2.5$$就是$$704$$．</p>\n<p>故选$$\\text{A}$$．</p>", "<p>Since Ella rides $$5$$ times as fast as Bella, Ella rides at a rate of $$\\frac{25}{2}$$ or $${12 \\frac{1}{2}}$$. Together,they move $$15$$ feet towards each other every unit. Dividing $$10560$$ by $$15$$ to find the number of steps Ella takes results in the answer or $$\\boxed{ (\\rm A)704}$$.</p>", "<p>Since Ella rides $$5$$ times faster than Bella, the ratio of their speeds is $$5:1$$. This means that Bella travels $$\\frac{1}{6}$$ of the way, and $$\\frac{1}{6}$$ of $$10560$$ feet is $$1760$$ feet. Bella also walks $$2.5$$ feet in a step, and $$1760$$ divided by $$2.5$$ is $$704$$.</p>"], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1022", "queId": "ed9d694da9bf438aa015f7e17583809a", "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$$a \\%$$，则所用的时间减少$$b \\%$$，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->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": "1064", "queId": "ff8080814d7978b9014d88c4711c2aa2", "competition_source_list": ["1991年第2届全国希望杯初一竞赛复赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$$1\\textless{}x\\textless{}2$$，则代数式$$\\frac{\\left\\textbar{} x-2 \\right\\textbar}{x-2}-\\frac{\\left\\textbar{} x-1 \\right\\textbar}{x-1}+\\frac{\\left\\textbar{} x \\right\\textbar}{x}$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->绝对值->|a|/a的化简", "课内体系->知识点->数->有理数->绝对值->绝对值的非负性"], "answer_analysis": ["当$$1\\textless{}x\\textless{}2$$时，$$x\\textgreater0$$，$$x-1\\textgreater0$$，$$x-2\\textless{}0$$．  ∴$$\\left\\textbar{} x \\right\\textbar=x$$，$$\\left\\textbar{} x-1 \\right\\textbar=x-1$$，$$\\left\\textbar{} x-2 \\right\\textbar=2-x$$．  ∴$$\\frac{\\left\\textbar{} x-2 \\right\\textbar}{x-2}-\\frac{\\left\\textbar{} x-1 \\right\\textbar}{x-1}+\\frac{\\left\\textbar{} x \\right\\textbar}{x}=\\frac{2-x}{x-2}-\\frac{x-1}{x-1}+\\frac{x}{x}=-1-(-1)+1=1$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "570", "queId": "3f2fd9a69fc54813bd004f006ba79ce0", "competition_source_list": ["初三上学期其它", "2018年全国竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "~关于$$x$$的方程$$\\left\\textbar{} \\frac{{{x}^{2}}}{x-1} \\right\\textbar=a$$仅有两个不同的实根．则实数$$a$$的取值范围是（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater0$$ "}], [{"aoVal": "B", "content": "$$a\\geqslant 4$$ "}], [{"aoVal": "C", "content": "$$2\\textless{}a\\textless{}4$$ "}], [{"aoVal": "D", "content": "$$0\\textless{}a\\textless{}4$$ "}]], "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->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", "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": ["当$$a\\textless{}0$$时，无解；  当$$a=0$$时，$$x=0$$只有一解，不合题意；  当$$a\\textgreater0$$时，原方程化为$$\\frac{{{x}^{2}}}{x-1}=\\pm a$$，整理得  $${{x}^{2}}-ax+a=0$$ ①  或$${{x}^{2}}+ax-a=0$$．②  因为方程②的判别式大于$$0$$，所以方程②有两个不同实根；  又因为原方程仅有两个不同的实根，故必有方程①的判别式小于0，从而$$0\\textless{}a\\textless{}4$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "20", "queId": "014270bea300409e9b3514fb0ce54ab4", "competition_source_list": ["2016年第33届全国全国初中数学联赛竞赛A卷第3题7分", "2016年上海浦东新区进才中学自主招生第8题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果一个正整数可以表示为两个连续奇数的立方差，则称这个正整数为``和谐数''．如：$$2={{1}^{3}}-{{(-1)}^{3}}$$，$$26={{3}^{3}}-{{1}^{3}}$$，$$2$$和$$26$$均为``和谐数''．那么，不超过$$2016$$的正整数中，所有的``和谐数''之和为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$6858$$ "}], [{"aoVal": "B", "content": "$$6860$$ "}], [{"aoVal": "C", "content": "$$9260$$ "}], [{"aoVal": "D", "content": "$$9262$$ "}]], "knowledge_point_routes": ["课内体系->知识点->综合与实践->新定义->新定义综合其它", "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": ["$${{(2k+1)}^{3}}-{{(2k-1)}^{3}}$$  $$=\\left[ (2k+1)-(2k-1) \\right]\\left[ {{(2k+1)}^{2}}+(2k+1)(2k-1)+{{(2k-1)}^{2}} \\right]$$  $$=2(12{{k}^{2}}+1)$$（其中$$k$$为非负整数），  由$$2(12{{k}^{2}}+1)\\leqslant 2016$$得，$$k\\leqslant 9$$，  ∴$$k=0$$，$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$9$$，  即得所有不超过$$2016$$的``和谐数''，它们的和为  $$\\left[ {{1}^{3}}-{{(-1)}^{3}} \\right]+({{3}^{3}}-{{1}^{3}})+({{5}^{3}}-{{3}^{3}})+\\cdots +({{17}^{3}}-{{15}^{3}})+({{19}^{3}}-{{17}^{3}})={{19}^{3}}+1=6860$$． ", "<p>&ldquo;和谐数&rdquo;设为$$m$$，它一定可以表示为$$m={{n}^{3}}-{{(n-2)}^{3}}$$，其中$$n$$为正整数．题目要求$$m\\leqslant 2016$$，大致计算可得$$n$$最大为$$19$$，于是有</p>\n<p>$$2={{1}^{3}}-{{(-1)}^{3}}$$，</p>\n<p>$$26={{3}^{3}}-{{1}^{3}}$$，</p>\n<p>$$\\cdots $$</p>\n<p>$$1946={{19}^{3}}-{{17}^{3}}$$，</p>\n<p>等式右边两两相消是有</p>\n<p>&ldquo;和谐数&rdquo;总和$$={{19}^{3}}-{{(-1)}^{3}}=6860$$，选$$\\text{B}$$.</p>\n"], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "968", "queId": "89f839b38eb942f5818b7656a077d0a4", "competition_source_list": ["2010年第21届全国希望杯初一竞赛初赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$2009$$年$$8$$月，台风``莫拉克''给台湾海峡两岸人民带来了严重灾难，台湾当局领导人马英九在追悼``八八水灾''罹难民众和救灾殉职人员的大会的致辞中说到，大陆同胞购款金额约$$50$$亿新台币，是台湾接到的最大一笔捐款，展现了两岸人民血浓于水的情感．$$50$$亿新台币折合人民币约$$11$$亿多元．若设$$1.1=m$$，则$$11$$亿这个数可表示成（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$9m$$ "}], [{"aoVal": "B", "content": "$${{m}^{9}}$$ "}], [{"aoVal": "C", "content": "$$m\\times {{10}^{9}}$$ "}], [{"aoVal": "D", "content": "$$m\\times {{10}^{10}}$$ "}]], "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": ["$$11$$亿$$=1100000000=1.1\\times {{10}^{9}}$$，  因为$$1.1=m$$，  所以$$11$$亿这个数可表示成$$m\\times {{10}^{9}}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1363", "queId": "b30f3b7e345c4b9dae036a6d2844f8c2", "competition_source_list": ["1998年第9届希望杯初二竞赛第2试第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\triangle ABC$$的一个内角的大小是$$40{}^{}\\circ $$，且$$\\angle A=\\angle B$$，那么$$\\angle C$$的外角的大小是．", "answer_option_list": [[{"aoVal": "A", "content": "$$140{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$80{}^{}\\circ $$或$$100{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$100{}^{}\\circ $$或$$140{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$80{}^{}\\circ $$或$$140{}^{}\\circ $$ "}]], "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->与三角形有关的角->三角形内角和的应用", "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->三角形的外角定义及性质"], "answer_analysis": ["①若$$\\angle A=40{}^{}\\circ $$，则$$\\angle B=40{}^{}\\circ $$，$$\\angle C=100{}^{}\\circ $$，$$\\angle C$$的外角为$$80{}^{}\\circ $$．  ②若$$\\angle C=40{}^{}\\circ $$，则$$\\angle C$$的外角为$$140{}^{}\\circ $$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "973", "queId": "922cbbffd2f7479b8ca185d10af91084", "competition_source_list": ["2011年第22届全国希望杯初二竞赛初赛第9题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若关于$$x$$，$$y$$的方程组$$\\begin{cases}x+ay+1=0    bx-2y+a=0 \\end{cases}$$没有实数解，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$ab=-2$$ "}], [{"aoVal": "B", "content": "$$ab=-2$$且$$a\\ne 1$$ "}], [{"aoVal": "C", "content": "$$ab\\ne -2$$ "}], [{"aoVal": "D", "content": "$$ab=-2$$且$$a\\ne 2$$ "}]], "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->二元一次方程组解的情况", "课内体系->能力->运算能力"], "answer_analysis": ["整理方程得$$\\begin{cases}bx+aby+b=0    bx-2y+a=0    \\end{cases}$$，  则$$\\left( ab+2 \\right)y=a-b$$，  要使方程组没有实数解，则$$\\begin{cases}ab=-2    a-b\\ne 0    \\end{cases}$$，  ∵$$ab=-2\\textless{}0$$，  ∴$$a$$不可能等于$$b$$，  ∴满足$$ab=-2$$即可． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1580", "queId": "e2aa5efa235d4c8b97a0357c7d472f27", "competition_source_list": ["1998年第9届希望杯初一竞赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$，$$b$$都是有理数，现有$$4$$个判断：  ①如果$$a+b ~\\textless{} ~a$$，则$$b ~\\textless{} ~0$$．②如果$$ab ~\\textless{} ~a$$，那么$$b ~\\textless{} ~0$$．  ③如果$$a-b ~\\textless{} ~a$$，则$$b\\textgreater0$$．④如果$$a\\textgreater b$$，那么$$\\frac{a}{b}\\textgreater1$$．  其中正确的判断是．", "answer_option_list": [[{"aoVal": "A", "content": "①② "}], [{"aoVal": "B", "content": "②③ "}], [{"aoVal": "C", "content": "①④ "}], [{"aoVal": "D", "content": "①③ "}]], "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": ["由$$a+b ~\\textless{} ~a\\Rightarrow b ~\\textless{} ~0$$，所以①判断正确．  当$$a=2$$，$$b=\\frac{1}{2}$$时，$$ab=1 ~\\textless{} ~2=a$$．而$$b=\\frac{1}{2}\\textgreater0$$，所以判断②不正确．  由$$a-b ~\\textless{} ~a\\Rightarrow -b ~\\textless{} ~0\\Rightarrow b\\textgreater0$$．③判断正确．  当$$a=2$$，$$b=-1$$时，显然$$a\\textgreater b$$，但$$\\frac{a}{b}=-2 ~\\textless{} ~1$$．所以④判断不正确．  综上分析可知，选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "518", "queId": "31a21953db97424fa35b2bf801f14b92", "competition_source_list": ["2011年第22届全国希望杯初一竞赛复赛第7题4分"], "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乙，获利$$10 \\%$$，而后乙又将这些股票反卖给甲，但乙损失了$$10 \\%$$．最后甲按乙卖给甲的价格的九折将这些股票卖给了乙，若上述股票交易中的其它费用忽略不计，则甲（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "盈亏平衡 "}], [{"aoVal": "B", "content": "盈利$$1$$元 "}], [{"aoVal": "C", "content": "盈利$$9$$元 "}], [{"aoVal": "D", "content": "亏损$$1.1$$元 "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->有理数与实际问题"], "answer_analysis": ["甲将股票卖给乙的价格是$$1000\\times (1+10 \\%)=1100$$（元），  乙将股票卖给甲的价格是$$1000\\times (1-10 \\%)=990$$（元），  甲再卖给乙的价格是$$990\\times (1-10 \\%)=891$$（元），  因此甲先赚了$$1100-1000=100$$（元），接着又亏了$$990-891=99$$（元），  故甲盈利$$1$$元． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "940", "queId": "6a582e78e915481cb928bbaf67ace3ed", "competition_source_list": ["2019~2020学年河北邯郸复兴区邯郸市锦玉中学初一上学期期末第15题2分", "2006年第17届希望杯初一竞赛复赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a$$是有理数，用$$[a]$$表示不超过$$a$$的最大整数，如$$[1.7]=1$$，$$[-1]=-1$$，$$[0]=0$$，$$[-1.2]=-2$$，则在以下四个结论中，正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$[a]+[-a]=0$$ "}], [{"aoVal": "B", "content": "$$[a]+[-a]=0$$或$$1$$ "}], [{"aoVal": "C", "content": "$$[a]+[-a]\\ne 0$$ "}], [{"aoVal": "D", "content": "$$[a]+[-a]=0$$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": ["当$$a=1.1$$时，$$[a]=1$$，$$[-a]=-2$$，所以$$\\text{A}$$、$$\\text{B}$$不成立．  当$$a=1$$时，$$[a]=1$$，$$[-a]=-1$$，所以$$\\text{C}$$不成立．  当$$a\\geqslant 0$$时，$$a$$可以写成$$a=[a]+ {a }$$，而 $$0\\leqslant  {a } ~\\textless{} ~1$$，$$-a=-[a]- {a }$$．  如果$$ {a }=0$$，即$$a$$是正整数，则$$[-a]=-[a]$$，所以 $$[a]+[-a]=0$$．  如果$$ {a }\\textgreater0$$，则$$[-a]=-[a]=-1$$，所以$$[a]+[-a]=-1$$．  当$$a ~\\textless{} ~0$$时，令$$b=-a\\textgreater0$$，将上面讨论中的$$a$$换成$$b$$，仍可以得到$$[a]+[-a]=0$$或$$-1$$．  故选$$\\text{D}$$. "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "380", "queId": "5de85f67e44e42c1b8b5df4835bea5ef", "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：$$a+b+c=3$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=4$$，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": "$$12$$ "}], [{"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": ["∵$$a+b+c=3$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=4$$，  ∴$$\\frac{{{a}^{2}}+{{b}^{2}}}{2-c}+\\frac{{{b}^{2}}+{{c}^{2}}}{2-a}+\\frac{{{c}^{2}}+{{a}^{2}}}{2-b}$$  $$=\\frac{4-{{c}^{2}}}{2-c}+\\frac{4-{{a}^{2}}}{2-a}+\\frac{4-{{b}^{2}}}{2-b}$$  $$=2+c+2+a+2+b$$  $$=a+b+c+6$$  $$=3+6$$  $$=9$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "283", "queId": "d0865eef21b44055b44271eb76c3bc57", "competition_source_list": ["2000年第11届希望杯初二竞赛第1试第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "在一个凸八边形中，每三个顶点形成三个角（如由$$A$$、$$B$$、$$C$$三个顶点形成$$\\angle ABC$$，$$\\angle BAC$$，$$\\angle ACB$$），一共可以作出$$168$$个角，那么这些角中最小的一个一定．", "answer_option_list": [[{"aoVal": "A", "content": "小于或等于$$20^{\\circ}$$ "}], [{"aoVal": "B", "content": "小于或等于$$22.5^{\\circ}$$ "}], [{"aoVal": "C", "content": "小于或等于$$25^{\\circ}$$ "}], [{"aoVal": "D", "content": "小于或等于$$27.5^{\\circ}$$ "}]], "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->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": ["凸八边形的$$8$$个内角的和为$$6\\times180^{\\circ}=1080^{\\circ}$$，其中至少有一个内角小于等于$$1080^{\\circ}\\div8=135^{\\circ}$$．  不妨设$$\\angle A \\leqslant 135^{\\circ}$$，在八边形中除$$A$$点及它的两个相邻顶点外，还有$$5$$个顶点，它们与$$A$$连$$5$$条对角线，把$$\\angle A$$分成$$6$$个角，在这$$6$$个角中必有一个角小于等于$$135^{\\circ}\\div6=22.5^{\\circ}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1549", "queId": "f440c73d944a428fafa4e7eb8536e058", "competition_source_list": ["2001年第12届希望杯初一竞赛第2试第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "古人用天干和地支记次序，其中天干有$$10$$个：甲乙丙丁度己庚辛壬癸．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$$12$$个汉字分别循环排列成如下两行：  甲乙丙丁戊己庚辛壬癸甲乙丙丁戊己庚辛壬癸$$\\cdots \\cdots $$  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$$\\cdots \\cdots $$，  则当第$$2$$次甲和子在同一列时，该列的序号是．", "answer_option_list": [[{"aoVal": "A", "content": "$$31$$ "}], [{"aoVal": "B", "content": "$$61$$ "}], [{"aoVal": "C", "content": "$$91$$ "}], [{"aoVal": "D", "content": "$$121$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->排列与组合"], "answer_analysis": ["``甲''在第一行出现的位置是$$10m+1$$，$$m=0$$，$$1$$，$$2$$，$$\\cdots $$，``子''在第二行出现的位置是$$12n+1$$，$$n=0$$，$$1$$，$$2$$，$$\\cdots $$．  所以``甲''和``子''在同一列时应有  $$10m+1=12n+1$$，  即$$10m=12n$$，  当$$m=n=0$$时是第一次``甲''\\,``子''同列，第二次``甲''\\,``子''同列时应是使得$$10m=12n$$成立的最小正整数$$m$$和$$n$$，即$$m=6$$，$$n=5$$．  所以，应是第$$61$$号位置．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1407", "queId": "f303def9023a44a28db5f689ac2aa7d9", "competition_source_list": ["2001年第12届希望杯初一竞赛第2试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "数$$a$$的任意正奇数次幂都等于$$a$$的相反数，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$a=0$$ "}], [{"aoVal": "B", "content": "$$a=-1$$ "}], [{"aoVal": "C", "content": "$$a=1$$ "}], [{"aoVal": "D", "content": "不存在这样的$$a$$值 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["根据题意，对任意正奇数$$n$$，$${{a}^{n}}=-a$$，  如果$$a\\textless{}0$$，则$$-a\\textgreater0$$，  而$${{a}^{n}}\\textless{}0$$，$${{a}^{n}}\\ne -a$$，  因此$$a$$不能是负数，  如果$$a\\textgreater0$$，则$$-a\\textless{}0$$，  而$${{a}^{n}}\\textgreater0$$，$${{a}^{n}}\\ne -a$$，  因此$$a$$不能是正数，  由于$$0$$的相反数是$$0$$，  所以$$a=0$$时，$${{a}^{n}}={{0}^{n}}=0=-a$$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": "1162", "queId": "8aac49074e023206014e1e1155915aa4", "competition_source_list": ["1993年第4届全国希望杯初一竞赛复赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "$${{19}^{93}}+{{93}^{19}}$$的末位数字是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["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->有理数乘方运算"], "answer_analysis": ["$${{19}^{93}}={{19}^{2\\times 46+1}}$$，$${{93}^{19}}={{93}^{4\\times 4+3}}$$  所以$${{19}^{93}}$$与$${{19}^{1}}$$的末位数相同是$$9$$、$${{93}^{19}}$$与$${{93}^{3}}$$末位数字相同是$$7$$．  因此$${{19}^{93}}+{{93}^{10}}$$末位数字是$$9+7=16$$的末位数字$$6$$，选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1403", "queId": "dc256085dafd4914adb5435f3af65c90", "competition_source_list": ["2013年第24届全国希望杯初二竞赛复赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$q=mn$$，$$p=\\sqrt{q+n}+\\sqrt{q-m}$$，其中$$m$$，$$n$$是两个连续的自然数（$$m\\textless{}n$$）．则$$p$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "总是奇数 "}], [{"aoVal": "B", "content": "总是偶数 "}], [{"aoVal": "C", "content": "有时是奇数，有时是偶数 "}], [{"aoVal": "D", "content": "有时是有理数，有时是无理数 "}]], "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": ["由$$m$$，$$n$$是两个连续的自然数（$$m\\textless{}n$$），得$$n=m+1$$，  所以$$q=mn=m(m+1)={{m}^{2}}+m$$．  从而$$p=\\sqrt{q+n}+\\sqrt{q-m}$$  $$=\\sqrt{{{m}^{2}}+m+m+1}+\\sqrt{{{m}^{2}}+m-m}$$  $$=m+1+m$$  $$=2m+1$$．  可知，$$p$$总是奇数． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1499", "queId": "ef07e0b46c81455d8b5adbffd5c556d4", "competition_source_list": ["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", "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", "初一上学期单元测试《解读绝对值》第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$是非零有理数，且$$a+b+c=0$$，那么$$\\frac{a}{\\textbar a\\textbar}+\\frac{b}{\\textbar b\\textbar}+\\frac{c}{\\textbar c\\textbar}+\\frac{abc}{\\textbar abc\\textbar}$$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": "956", "queId": "7c6a9d1b9b87458ca3aaa9fea6530be5", "competition_source_list": ["2019~2020学年9月湖北武汉武昌区武汉市武珞路实验初级中学初三上学期周测A卷第8题3分", "2004年第21届全国初中数学联赛竞赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$${{b}^{2}}-4ac$$是一元二次方程$$a{{x}^{2}}+bx+c=0\\left( a\\ne 0 \\right)$$的一个实数根，则$$ab$$的取值范围为．", "answer_option_list": [[{"aoVal": "A", "content": "$$ab\\geqslant \\frac{1}{8}$$ "}], [{"aoVal": "B", "content": "$$ab\\leqslant \\frac{1}{8}$$ "}], [{"aoVal": "C", "content": "$$ab\\geqslant \\frac{1}{4}$$ "}], [{"aoVal": "D", "content": "$$ab\\leqslant \\frac{1}{4}$$ "}]], "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->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->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", "课内体系->能力->推理论证能力", "课内体系->能力->运算能力"], "answer_analysis": ["由求根公式得到方程的根为$$x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}$$，设$$\\sqrt{{{b}^{2}}-4ac}=y$$，又因为$${{y}^{2}}={{b}^{2}}-4ac$$是原方程的根，所以$${{y}^{2}}=\\frac{-b+y}{2a}$$或者$${{y}^{2}}=\\frac{-b-y}{2a}$$，因为$$a\\ne 0$$，这两个方程可化成关于$$y$$的一元二次方程$$2a{{y}^{2}}\\pm y+b=0$$，考虑这个方程的判别式得到$$1-8ab\\geqslant 0$$，所以$$ab{\\leqslant } \\frac{1}{8}$$，选择$$\\text{B}$$． ", "<p>因为方程有实数解，故$${{b}^{2}}-4ac{\\geqslant }0$$，由题意，</p>\n<p>有$$\\frac{-b+\\sqrt{{{b}^{2}}-4ac}}{2a}={{b}^{2}}-4ac$$或$$\\frac{-b-\\sqrt{{{b}^{2}}-4ac}}{2a}={{b}^{2}}-4ac$$．</p>\n<p>令$$u=\\sqrt{{{b}^{2}}-4ac}$$，得方程$$2a{{u}^{2}}-u+b=0$$或$$2a{{u}^{2}}+u+b=0$$，</p>\n<p>因为$$u=\\sqrt{{{b}^{2}}-4ac}$$是其解，所以方程的判别式非负，即$$1-8ab{\\geqslant }0$$，故$$ab{\\leqslant }\\frac{1}{8}$$．</p>\n<p>故选$$\\text{B}$$．</p>"], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "172", "queId": "2a8f8ad78b724878987487a070960690", "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", "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"], "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$${{v}_{1}}\\textgreater{{v}_{2}}\\textgreater{{v}_{3}}\\textgreater{{v}_{4}}\\textgreater0$$，  所以有$${{t}_{1}}\\textgreater{{t}_{2}}\\textgreater{{t}_{3}}$$，  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": "1188", "queId": "8aac49074e023206014e25beb5e972a9", "competition_source_list": ["1997年第8届全国希望杯初一竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "有理数$$a$$、$$b$$满足$$a=1997b$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\geqslant b$$ "}], [{"aoVal": "B", "content": "$$\\left\\textbar{} a \\right\\textbar\\leqslant b$$ "}], [{"aoVal": "C", "content": "$$a\\geqslant \\left\\textbar{} b \\right\\textbar$$ "}], [{"aoVal": "D", "content": "$$\\left\\textbar{} a \\right\\textbar\\geqslant \\left\\textbar{} b \\right\\textbar$$ "}]], "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": ["∵$$1997\\textgreater0$$，可以确定有理数$$a$$、$$b$$同是正数，或同是负数，或同是$$0$$．  又∵$$1997\\textgreater1$$，所以必须$$\\left\\textbar{} a \\right\\textbar\\geqslant \\left\\textbar{} b \\right\\textbar$$，选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "305", "queId": "66e491a48aeb4d82a5b24edf363072ff", "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11题3分", "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11题3分", "2020~2021学年10月湖南长沙岳麓区长郡集团郡维学校初一上学期月考第10题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "计算机是将信息转换成二进制进行处理的，二进制即``逢二进一''，如$${{(1101)}_{2}}$$表示二进制数，将它转换成十进制形式是$$1\\times {{2}^{3}}+1\\times {{2}^{2}}+0\\times {{2}^{1}}+1\\times {{2}^{0}}=13$$，那么将二进制数$${{(11111)}_{2}}$$转换成十进制形式的数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["课内体系->能力->抽象概括能力", "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->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": ["∵$${{(1101)}_{2}}=1\\times {{2}^{3}}+1\\times {{2}^{2}}+0\\times {{2}^{1}}+1\\times {{2}^{0}}=13$$，  ∴$${{(11111)}_{2}}=1\\times {{2}^{4}}+1\\times ~{{2}^{3}}+1\\times {{2}^{2}}+1\\times ~{{2}^{1}}+1\\times {{2}^{0}}$$，  $$=16+8+4+2+1$$  $$=31$$．  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "985", "queId": "ff8080814cfa9b24014cff35a27e0e65", "competition_source_list": ["2011年第22届全国希望杯初二竞赛初赛第8题4分", "初一单元测试《分解方法的延拓（1）》第22题"], "difficulty": "2", "qtype": "single_choice", "problem": "$${{2}^{16}}-1$$能分解成$$n$$个质因数的乘积，$$n$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->因式分解->公式法->公式法综合应用", "课内体系->能力->运算能力"], "answer_analysis": ["$${{2}^{16}}-1$$  $$=({{2}^{8}}+1)({{2}^{8}}-1)$$  $$=\\cdots $$  $$=({{2}^{8}}+1)({{2}^{4}}+1)({{2}^{2}}+1)({{2}^{1}}+1)({{2}^{1}}-1)$$，  其中，除了$${{2}^{1}}-1=1$$外，其余$$4$$个括号内的数都是质数，  所以$${{2}^{16}}-1$$可以分解为$$4$$个质因数的乘积． ", "<p>首先利用平方差公式分解因式，即可求得：</p>\n<p>$$216-1=\\left( 28+1 \\right)\\left( 24+1 \\right)\\left( 22+1 \\right)\\left( 2+1 \\right)\\left( 2-1 \\right)$$，</p>\n<p>则问题得解．</p>\n<p>∵$$216-1=\\left( 28+1 \\right)\\left( 28-1 \\right)=\\left( 28+1 \\right)\\left( 24+1 \\right)\\left( 24-1 \\right)$$</p>\n<p>$$=\\left( 28+1 \\right)\\left( 24+1 \\right)\\left( 22+1 \\right)\\left( 22-1 \\right)$$</p>\n<p>$$=\\left( 28+1 \\right)\\left( 24+1 \\right)\\left( 22+1 \\right)\\left( 2+1 \\right)\\left( 2-1 \\right)=257\\times 17\\times 5\\times 3$$．</p>\n<p>&there4;$$n=4$$．</p>\n<p>故选$$\\text{C}$$．</p>\n"], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1210", "queId": "d6f0b834ba394b9a9053e886809e6c0a", "competition_source_list": ["2010年全国全国初中数学联赛初一竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "对于自然数$$n$$，将其各位数字之和记为$${{a}_{n}}$$，如$${{a}_{2009}}=2+0+0+9=11$$，$${{a}_{2010}}=2+0+1+0=3$$，则$${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+\\cdots +{{a}_{2009}}+{{a}_{2010}}=$$ （~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$28062$$ "}], [{"aoVal": "B", "content": "$$28065$$ "}], [{"aoVal": "C", "content": "$$28067$$ "}], [{"aoVal": "D", "content": "$$28068$$ "}]], "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->有理数乘除混合运算"], "answer_analysis": ["根据弃九法，它和$$1$$到$$2010$$的和被$$9$$除的余数相等．每连续$$9$$个自然数之和被$$9$$整除，$$2010$$被$$9$$除余$$3$$，$$1+2+3=6$$，所以只有$$\\text{D}$$符合． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1296", "queId": "d29ba67ff2a34c22901b0424872bd0af", "competition_source_list": ["2003年第20届全国初中数学联赛竞赛第2题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "在凸$$10$$边形的所有内角中，锐角的个数最多是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["由于任何凸多边形的外角之和都是$$360{}^{}\\circ $$，故外角中钝角的个数不超过$$3$$个，即内角中锐角最多不超过$$3$$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": "514", "queId": "7e99d3d7d278457b8ae5364479f5bd45", "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$$2020\\times20212021-2021\\times20202020$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2020$$ "}], [{"aoVal": "D", "content": "$$2021$$ "}]], "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": ["$$2020\\times20212021-2021\\times20202020$$  $$=2020\\times2021\\times10001-2021\\times2020\\times10001$$  $$=0$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1203", "queId": "8aac49074e724b45014e87c423295111", "competition_source_list": ["1996年第7届全国希望杯初一竞赛复赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$A$$、$$B$$两地相距$$s$$千米．甲、乙的速度分别是$$a$$千米/小时，$$b$$千米/小时$$\\left( a\\textgreater b \\right)$$．甲、乙都从$$A$$到$$B$$去开会，如果甲比乙先出发$$1$$小时，那么乙比甲晚到$$B$$地的小时数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{s}{a}-\\left( \\frac{s}{b}+1 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\frac{s}{b}-\\left( \\frac{s}{a}+1 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\frac{s}{a}-\\left( \\frac{s}{b}-1 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\frac{s}{b}-\\left( \\frac{s}{a}-1 \\right)$$ "}]], "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": ["走完全程乙用$$\\frac{s}{b}$$小时，甲用$$\\frac{s}{a}$$小时，  所以乙比甲晚到$$\\frac{s}{b}-\\left( \\frac{s}{a}-1 \\right)$$小时，选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "575", "queId": "5126d2468d534964a677437012c9b81d", "competition_source_list": ["初三上学期其它", "1999年第16届全国初中数学联赛竞赛第6题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "有下列三个命题：  （甲）若$$\\alpha $$、$$\\beta $$是不相等的无理数，则$$\\alpha \\beta +\\alpha -\\beta $$是无理数；  （乙）若$$\\alpha $$、$$\\beta $$是不相等的无理数，则$$\\frac{\\alpha -\\beta }{\\alpha +\\beta }$$是无理数；  （丙）若$$\\alpha $$、$$\\beta $$是不相等的无理数，则$$\\sqrt{\\alpha }+\\sqrt[3]{\\beta }$$是无理数．  其中正确命题的个数是（~ ~ ）", "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->数->实数->无理数有关的计算->无理数的定义"], "answer_analysis": ["因为$$\\alpha \\beta +\\alpha -\\beta =\\alpha \\beta +\\alpha -\\beta -1+1=\\left( \\alpha -1 \\right)\\left( \\beta +1 \\right)+1$$，  只要令$$\\alpha =1+\\sqrt{2}$$，$$\\beta =-1+\\sqrt{2}$$，则$$\\alpha \\beta +\\alpha -\\beta =3$$为有理数，故（甲）不对；  又若令$$\\alpha =2\\sqrt{2}$$，$$\\beta =\\sqrt{2}$$，则$$\\frac{\\alpha -\\beta }{\\alpha +\\beta }=\\frac{1}{3}$$为有理数，故（乙）不对；  又若令$$\\alpha =\\sqrt[3]{2}$$，$$\\beta =-\\sqrt{2}$$，则$$\\sqrt{\\alpha }+\\sqrt[3]{\\beta }=0$$为有理数，故（丙）不对．  因此正确命题个数是$$0$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "126", "queId": "7daf6c264b564f7380ff8b805b54bc24", "competition_source_list": ["1999年竞赛（全国初中数学竞赛）第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果抛物线$$y={{x}^{2}}-\\left( k-1 \\right)x-k-1$$与$$x$$轴的交点分别为$$A$$，$$B$$，顶点为$$C$$，那么$$\\triangle ABC$$面积的最小值是．", "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->推理论证能力", "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->知识点->函数->二次函数->二次函数与几何综合->二次函数与面积"], "answer_analysis": ["首先，$$\\Delta ={{\\left( k-1 \\right)}^{2}}+4\\left( k+1 \\right)={{k}^{2}}+2k+5\\textgreater0$$，  所以抛物线与$$x$$轴总有两个交点，设抛物线与$$x$$轴交点的横坐标分别为$${{x}_{1}}$$，$${{x}_{2}}$$，那么  $$\\left\\textbar{} AB \\right\\textbar=\\sqrt{{{\\left( {{x}_{1}}-{{x}_{2}} \\right)}^{2}}}=\\sqrt{{{\\left( {{x}_{1}}+{{x}_{2}} \\right)}^{2}}-4{{x}_{1}}{{x}_{2}}}=\\sqrt{{{k}^{2}}+2k+5}$$．  又抛物线的顶点坐标是$$C\\left( \\frac{k-1}{2},-\\frac{{{k}^{2}}+2k+5}{4} \\right)$$，  所以  $${{S}_{\\triangle ABC}}=\\frac{1}{2}\\sqrt{{{k}^{2}}+2k+5}\\cdot \\frac{{{k}^{2}}+2k+5}{4}$$  $$=\\frac{1}{8}{{\\left[ {{\\left( k+1 \\right)}^{2}}+4 \\right]}^{\\dfrac{3}{2}}}\\geqslant \\frac{1}{8}\\cdot {{4}^{\\frac{3}{2}}}=1$$，  当$$k=-1$$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": "170", "queId": "0c99e9a2bd9f4549b79bb29d2c296a93", "competition_source_list": ["2018年第29届希望杯初一竞赛初赛第10题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "满足$$a+c=2b$$的三位数$$\\overline{abc}$$共有个．", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$36$$ "}], [{"aoVal": "C", "content": "$$45$$ "}], [{"aoVal": "D", "content": "$$49$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->计数问题-枚举法"], "answer_analysis": ["若$$b=0$$，则$$a+c=0$$，不可能  若$$b=1$$，则$$a+c=2$$，有$$2$$种情况；  若$$b=2$$，则$$a+c=4$$，有$$4$$种情况；  若$$b=3$$，则$$a+c=6$$，有$$6$$种情况；  若$$b=4$$，则$$a+c=8$$，有$$8$$种情况；  若$$b=5$$，则$$a+c=10$$，有$$9$$种情况；  若$$b=6$$，则$$a+c=12$$，有$$7$$种情况；  若$$b=7$$，则$$a+c=14$$，有$$5$$种情况；  若$$b=8$$，则$$a+c=16$$，有$$3$$种情况；  若$$b=9$$，则$$a+c=18$$，有$$1$$种情况；  综上，共有$$2+4+6+8+9+7+5+3+1=45$$（种）情况． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "819", "queId": "726bf8602fd04854bd489db763fa304a", "competition_source_list": ["2011年竞赛第5题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$S=\\frac{1}{{{1}^{3}}}+\\frac{1}{{{2}^{3}}}+\\frac{1}{{{3}^{3}}}+\\cdots +\\frac{1}{{{2011}^{3}}} $$，则$$4S$$的整数部分等于．", "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->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", "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": ["当$$k=2$$，$$3$$，$$\\cdots $$，$$2011$$时，因为  $$\\frac{1}{{{k}^{3}}} ~\\textless{} ~\\frac{1}{k({{k}^{2}}-1)}=\\frac{1}{2}\\left[ \\frac{1}{\\left( k-1 \\right)k}-\\frac{1}{k\\left( k+1 \\right)} \\right]$$，  所以$$1 ~\\textless{} ~S=1+\\frac{1}{{{2}^{3}}}+\\frac{1}{{{3}^{3}}}+\\cdots +\\frac{1}{{{2011}^{3}}} ~\\textless{} ~1+\\frac{1}{2}\\left( \\frac{1}{2}-\\frac{1}{2011\\times 2012} \\right) ~\\textless{} ~\\frac{5}{4}$$．  于是有$$4\\leqslant 4S\\leqslant 5$$，故$$4S$$的整数部分等于$$4$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "520", "queId": "996376c924584965bbb0dbd0ff6936c3", "competition_source_list": ["2019~2020学年4月山东潍坊奎文区潍坊德润国际双语学校初一下学期月考第6题3分", "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2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2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$2x+5y+4z=0$$，$$3x+y-7z=0$$，则$$x+y-z$$的值等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "不能求出 "}]], "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->方法->代入法", "课内体系->方法->整体法"], "answer_analysis": ["根据题意得：$$\\begin{cases}2x+5y+4z=0①    3x+y-7z=0②   \\end{cases}$$，  把②变形为：$$y=7z-3x$$③，  代入①得：$$x=3z$$，  代入③得：$$y=-2z$$，  则$$x+y-z=3z-2z-z=0$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1530", "queId": "c69a0885c84b4e8bae26531d9aaec972", "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": "题5：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": ["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-顺风、顺水问题", "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": "1297", "queId": "a0c7047130cf44049e30446c6a19c5fb", "competition_source_list": ["2013年第24届全国希望杯初一竞赛初赛第8题4分", "初一上学期单元测试《解读绝对值》第6题"], "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": ["解法一：  ∵$$b\\textless{}0\\textless{}c$$，$$\\left\\textbar{} b \\right\\textbar\\textless{}c$$，  ∴$$-b\\textless{}c$$，  ∴$$b+c\\textgreater0$$．  ∵$$a\\textless{}0\\textless{}d$$，$$\\left\\textbar{} a \\right\\textbar\\textless{}d$$，  ∴$$-a\\textless{}d$$，  ∴$$a+d\\textgreater0$$．  ∴$$a+b+c+d\\textgreater0$$．  解法二：  取特殊值，依题设，可假定$$b=-2$$，$$c=3$$，$$a=-4$$，$$d=5$$，  则$$a+b+c+d=2\\textgreater0$$，  故$$a+b+c+d$$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": "840", "queId": "c3c3ddcf00ef4f52b867554c21e484fb", "competition_source_list": ["2018年重庆江津区四川省江津市第二中学初三中考一模（八校联考）第12题4分", "2017年湖南长沙天心区初二竞赛长郡教育集团抵达杯第6题5分", "2017年四川达州渠县初三中考二模九校联合中考第9题3分", "2016年重庆中考真题B卷第12题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果关于$$x$$的分式方程$$\\dfrac{a}{x+1}-3=\\dfrac{1-x}{x+1}$$有负分数解，且关于$$x$$的不等式组$$\\left { \\begin{matrix}2(a-x)\\geqslant -x-4    \\dfrac{3x+4}{2}\\textless{}x+1   \\end{matrix} \\right.$$的解集为$$x\\textless{}-2$$，那么符合条件的所有整数$$a$$的积是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-3$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "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->由分式方程的解确定参数", "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）的解集求参数的范围", "课内体系->能力->分析和解决问题能力"], "answer_analysis": ["$$\\left { \\begin{matrix}2(a-x)\\geqslant -x-4\\&①    \\dfrac{3x+4}{2}\\textless{}x+1\\&②   \\end{matrix} \\right.$$，  由①得：$$x\\leqslant 2a+4$$，  由②得：$$x\\textless{}-2$$，  由不等式组的解集为$$x\\textless{}-2$$，得到$$2a+4\\geqslant -2$$，即$$a\\geqslant -3$$，  分式方程去分母得：$$a-3x-3=1-x$$，  把$$a=-3$$代入整式方程得：$$-3x-6=1-x$$，即$$x=-\\dfrac{7}{2}$$，符合题意；  把$$a=-2$$代入整式方程得：$$-3x-5=1-x$$，即$$x=-3$$，不合题意；  把$$a=-1$$代入整式方程得：$$-3x-4=1-x$$，即$$x=-\\dfrac{5}{2}$$，符合题意；  把$$a=0$$代入整式方程得：$$-3x-3=1-x$$，即$$x=-2$$，不合题意；  把$$a=1$$代入整式方程得：$$-3x-2=1-x$$，即$$x=-\\dfrac{3}{2}$$，符合题意；  把$$a=2$$代入整式方程得：$$-3x-1=1-x$$，即$$x=1$$，不合题意；  把$$a=3$$代入整式方程得：$$-3x=1-x$$，即$$x=-\\dfrac{1}{2}$$，符合题意；  把$$a=4$$代入整式方程得：$$-3x+1=1-x$$，即$$x=0$$，不合题意，  ∴符合条件的整数$$a$$取值为$$-3$$；$$-1$$；$$1$$；$$3$$，之积为$$9$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1183", "queId": "8aac49074e023206014e25b52cb9726b", "competition_source_list": ["1997年第8届全国希望杯初一竞赛初赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面说法中，不正确的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "小于$$-1$$的有理数比它的倒数小 "}], [{"aoVal": "B", "content": "非负数的相反数不一定比它本身小 "}], [{"aoVal": "C", "content": "小于$$0$$的有理数的二次幂大于原数 "}], [{"aoVal": "D", "content": "小于$$0$$的有理数的立方小于原数 "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->数->有理数->倒数与负倒数->倒数的定义", "课内体系->知识点->数->有理数->相反数->相反数的定义", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->有理数乘方运算"], "answer_analysis": ["设$$a$$为有理数，当$$-1\\textless{}a\\textless{}0$$时，$${{a}^{3}}\\textgreater a$$，∴$$\\text{D}$$的说法不正确． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "732", "queId": "4dd6fce1123241cdbdb79df0008a1885", "competition_source_list": ["2016年第27届全国希望杯初二竞赛复赛第1题4分", "2020~2021学年江西南昌东湖区南昌市第三中学初一下学期期末第6题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "在平面直角坐标系中，点$$A(a-2,a-1)$$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": ["∵$$a-2\\textless{}a-1$$，  ∴当$$a-2\\textgreater0$$时，不可能有$$a-1\\textless{}0$$，  ∴点$$A(a-2,a-1)$$不可能在第四象限． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "715", "queId": "494346db7cee4017acb304ec3c4172d8", "competition_source_list": ["2005年第16届希望杯初一竞赛初赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "公园里准备修五条直的甬道，并在甬道交叉口处设一个报亭，这样的报亭最多可设．", "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": ["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->几何图形初步->相交线与平行线->相交线->相交的定义"], "answer_analysis": ["因为五条直线相交最多有$$10$$个交点，所以最多可设$$10$$个报亭．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1151", "queId": "d24931c63e6644bfa6ac741e9cbca9a5", "competition_source_list": ["2005年第16届希望杯初一竞赛复赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$x=-\\frac{7}{12}$$时，式子$${{(x-2)}^{2}}-2(2-2x)-(1+x)(1-x)$$的值等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{23}{72}$$ "}], [{"aoVal": "B", "content": "$$\\frac{23}{72}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$\\frac{49}{72}$$ "}]], "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-直接代入化简求值"], "answer_analysis": ["原式$$={{x}^{2}}-4x+4-4+4x-1+{{x}^{2}}$$  $$=2{{x}^{2}}-1$$，  当$$x=-\\frac{7}{12}$$时，  $$2{{x}^{2}}-1=2\\times {{\\left( -\\frac{7}{12} \\right)}^{2}}-1=-\\frac{23}{72}$$．  故选($$\\text{A}$$)． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "280", "queId": "159e916ad39842deb4a09d4036d69635", "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": "768", "queId": "5707c426ec144faf875ec6aa6cdddd07", "competition_source_list": ["2018年全国初中数学联赛初一竞赛B卷"], "difficulty": "2", "qtype": "single_choice", "problem": "已知实数$$a$$，$$b$$满足$${{a}^{3}}-3{{a}^{2}}+5a=1$$，$${{b}^{3}}-3{{b}^{2}}+5b=5$$，则$$a+b=$$（~ ）．", "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->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": ["有已知条件可得$${{\\left( a-1 \\right)}^{3}}+2\\left( a-1 \\right)=-2$$，$${{\\left( b-1 \\right)}^{3}}+2\\left( b-1 \\right)=2$$，两式相加得$${{\\left( a-1 \\right)}^{3}}+2\\left( a-1 \\right)+{{\\left( b-1 \\right)}^{3}}+2\\left( b-1 \\right)=0$$，  因式分解得$$\\left( a+b-2 \\right)\\left[ {{\\left( a-1 \\right)}^{2}}-\\left( a-1 \\right)\\left( b-1 \\right)+\\left( b-{{1}^{2}}+2 \\right) \\right]=0$$，  因为$${{\\left( a-1 \\right)}^{2}}-\\left( a-1 \\right)\\left( b-1 \\right)+{{\\left( b-1 \\right)}^{2}}+2={{\\left[ \\left( a-1 \\right)-\\frac{1}{2}\\left( b-1 \\right) \\right]}^{2}}+\\frac{3}{4}{{\\left( b-1 \\right)}^{2}}+2\\textgreater0$$，  所以$$a+b-2=0$$，因此$$a+b=2$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1570", "queId": "ddebcf3711ea4c6da616f6dd741f0514", "competition_source_list": ["2018年亚洲国际数学奥林匹克公开赛（AIMO）初二竞赛决赛第23题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知一个正整数有$$48$$个正因数，它最多可以有多少个质因数？  It is known that a positive integer has $$48$$ positive factors, at most how many prime factors does it have?", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$12$$ "}], [{"aoVal": "E", "content": "$$48$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数", "竞赛->知识点->数论->整除->因数与倍数"], "answer_analysis": ["$$48=2\\times 2\\times 2\\times 2\\times 3$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1120", "queId": "ff8080814d9efd56014daa7da15f0ad6", "competition_source_list": ["1993年第4届全国希望杯初一竞赛初赛第12题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$M$$表示$$a$$与$$b$$的和的平方，$$N$$表示$$a$$与$$b$$的平方的和，则当$$a=7$$，$$b=-5$$时，$$M-N$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-28$$ "}], [{"aoVal": "B", "content": "$$70$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "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->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": ["$$M={{(a+b)}^{2}}$$，$$N=a+{{b}^{2}}$$．  $$M-N={{(a+b)}^{2}}-(a+{{b}^{2}})={{a}^{2}}+2ab+{{b}^{2}}-a-{{b}^{2}}={{a}^{2}}+2ab-a$$．  ∵$$a=7$$，$$b=-5$$，  ∴$$M-N={{7}^{2}}+2\\times 7\\times (-5)-7=-28$$．选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "400", "queId": "1f6c4b48ef6d40468e05a8135ea15dd5", "competition_source_list": ["1992年第9届全国初中数学联赛竞赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "在半径为$$1$$的圆中有一内接多边形，若它的边长皆大于$$1$$且小于$$\\sqrt{2}$$，则这个多边形的边数必为（~ ）", "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->知识点->圆->圆与多边形->正多边形与圆", "课内体系->能力->运算能力"], "answer_analysis": ["若满足条件的多边形的边数大于或等于$$6$$，则至少有一边所对的圆心角不大于$$60{}^{}\\circ $$，由余弦定理知该边长必不大于$$1$$；同理，若存在满足条件的四边形，则它至少有一边长不小于$$\\sqrt{2}$$．  故选$$\\text C$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "634", "queId": "b0c06daa1aa74cf6855fd2543237ff6d", "competition_source_list": ["1993年第10届全国初中数学联赛竞赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "不等式$$x-1\\textless{}{{(x-1)}^{2}}\\textless{}3x+7$$的整数解的个数．", "answer_option_list": [[{"aoVal": "A", "content": "等于$$4$$ "}], [{"aoVal": "B", "content": "小于$$4$$ "}], [{"aoVal": "C", "content": "大于$$5$$ "}], [{"aoVal": "D", "content": "等于$$5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->方程与不等式->不等式（组）->其它不等式->高次不等式", "课内体系->能力->运算能力"], "answer_analysis": ["注意到$$x-1\\textless{}{{(x-1)}^{2}}\\textless{}3x+7\\Rightarrow \\begin{cases}(x-2)(x-1)\\textgreater0    (x+1)(x-6)\\textless{}0    \\end{cases}$$，  $$(x+1)(x-6)\\textless{}0\\Leftrightarrow -1\\textless{}x\\textless{}6$$．  于是原不等式的整数解是介于$$-1$$与$$6$$之间且不等于$$1$$，$$2$$的整数，即$$0$$，$$3$$，$$4$$，$$5$$四个整数． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "61", "queId": "8aac50a7519fa10a0151b81f734b4ebe", "competition_source_list": ["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7题2分", "2000年第11届希望杯初二竞赛第1试第10题", "2016年初一下学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$、$$b$$、$$c$$均为正数，若$$\\frac{c}{a+b}\\textless{}\\frac{a}{b+c}\\textless{}\\frac{b}{c+a}$$，则$$a$$、$$b$$、$$c$$三个数的大小关系是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$c\\textless{}a\\textless{}b$$ "}], [{"aoVal": "B", "content": "$$b\\textless{}c\\textless{}a$$ "}], [{"aoVal": "C", "content": "$$a\\textless{}b\\textless{}c$$ "}], [{"aoVal": "D", "content": "$$c\\textless{}b\\textless{}a$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的运算->分式比较大小"], "answer_analysis": ["因为$$\\frac{c}{a+b}\\textless\\frac{a}{b+c}\\textless\\frac{b}{c+a}$$，  所以$$\\frac{a+b}{c}\\textgreater\\frac{b+c}{a}\\textgreater\\frac{c+a}{b}$$，  所以$$\\frac{a+b}{c}+1\\textgreater\\frac{b+c}{a}+1\\textgreater\\frac{c+a}{b}+1$$，  所以$$\\frac{a+b+c}{c}\\textgreater\\frac{a+b+c}{a}\\textgreater\\frac{a+b+c}{b}$$，  所以$$c\\textless a\\textless b$$．  故选$$\\text{A}$$． ", "<p>由$$\\frac{c}{a+b}&lt;\\frac{a}{b+c}$$得$$c(b+c)&lt;a(a+b)$$，</p>\n\n<p>$$b c+c^{2}-a^{2}-a b&lt;0$$，</p>\n<p>$$(c-a)(a+b+c)&lt;0$$．</p>\n<p>所以$$c&lt;a$$．</p>\n<p>同理，由$$\\frac{a}{b+c}&lt;\\frac{b}{c+a}$$得$$a&lt;b$$．</p>\n<p>所以$$c&lt;a&lt;b$$．</p>\n<p>故选$$\\text{A}$$．</p>\n"], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "791", "queId": "4e6d90748b9345699b89119af82e06ed", "competition_source_list": ["2013年第30届全国全国初中数学联赛竞赛第4题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "不定方程$$3{{x}^{2}}+7xy-2x-5y-17=0$$的全部正整数解$$(x,y)$$的组数为（~ ）．", "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->其他方程->不定方程", "课内体系->能力->运算能力"], "answer_analysis": ["由$$3{{x}^{2}}+7xy-2x-5y-17=0$$，得$$y=\\frac{-3{{x}^{2}}+2x+17}{7x-5}$$，  因为$$x$$，$$y$$为正整数，  故$$x\\geqslant 1$$，$$y\\geqslant 1$$，  从而$$7x-5\\textgreater0$$，  于是$$-3{{x}^{2}}+2x+17\\geqslant 7x-5$$，$$3{{x}^{2}}+5x-22\\leqslant 0$$，  即$$(x-2)(3x+11)\\leqslant 0$$，  由$$x\\geqslant 1$$，知$$3x+11\\textgreater0$$，  故$$x-2\\leqslant 0$$，$$x\\leqslant 2$$，  故$$x=1$$或$$x=2$$．  当$$x=1$$时，$$y=8$$；  当$$x=2$$时，$$y=1$$．  故原不定方程的全部正整数解$$(x,y)$$有两组：$$(1,8)$$，$$(2,1)$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "287", "queId": "469d8c2f71c34f68b9aee25d9d516eba", "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 side, 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->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->运算能力", "课内体系->能力->分析和解决问题能力"], "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": "1488", "queId": "b8a309dcf54642a58978ab8faaa3a4ed", "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": "下列说法中，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": "191", "queId": "146fc342b2254abeb5a35eccda0a8416", "competition_source_list": ["2011年第22届全国希望杯初二竞赛初赛第8题4分", "初一单元测试《分解方法的延拓（1）》第22题"], "difficulty": "2", "qtype": "single_choice", "problem": "$${{2}^{16}}-1$$能分解成$$n$$个质因数的乘积，$$n$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"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->因式分解->公式法->公式法综合应用"], "answer_analysis": ["方法一：$${{2}^{16}}-1$$  $$=({{2}^{8}}+1)({{2}^{8}}-1)$$  $$=\\cdots $$  $$=({{2}^{8}}+1)({{2}^{4}}+1)({{2}^{2}}+1)({{2}^{1}}+1)({{2}^{1}}-1)$$，  其中，除了$${{2}^{1}}-1=1$$外，其余$$4$$个括号内的数都是质数，  所以$${{2}^{16}}-1$$可以分解为$$4$$个质因数的乘积．  方法二：首先利用平方差公式分解因式，即可求得：  $$216-1=\\left( 28+1 \\right)\\left( 24+1 \\right)\\left( 22+1 \\right)\\left( 2+1 \\right)\\left( 2-1 \\right)$$，  则问题得解．  ∵$$216-1=\\left( 28+1 \\right)\\left( 28-1 \\right)=\\left( 28+1 \\right)\\left( 24+1 \\right)\\left( 24-1 \\right)$$  $$=\\left( 28+1 \\right)\\left( 24+1 \\right)\\left( 22+1 \\right)\\left( 22-1 \\right)$$  $$=\\left( 28+1 \\right)\\left( 24+1 \\right)\\left( 22+1 \\right)\\left( 2+1 \\right)\\left( 2-1 \\right)=257\\times 17\\times 5\\times 3$$．  ∴$$n=4$$．  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": "698", "queId": "a323af94355b4ea48d0732fdd5c229e3", "competition_source_list": ["2018~2019学年广东深圳福田区深圳实验学校中学部初二下学期单元测试《图形的平移与旋转》第4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2题2分", "2019~2020学年10月安徽马鞍山金家庄区马鞍山市二中初二上学期月考第2题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "在直角坐标系中，点$$M(-3,-4)$$)先右移$$3$$个单位，再下移$$2$$个单位，则点$$M$$的坐标变为．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-6,-6)$$ "}], [{"aoVal": "B", "content": "$$(0,-6)$$ "}], [{"aoVal": "C", "content": "$$(0,-2)$$ "}], [{"aoVal": "D", "content": "$$(-6,-2)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->平面直角坐标系->坐标系综合->坐标系中的平移->坐标系中的平移-点的平移", "课内体系->能力->运算能力"], "answer_analysis": ["点$$M(-3,-4)$$，先右移$$3$$个单位，再下移$$2$$个单位后点的坐标为$$(-3+3,-4-2)$$，即$$(0,-6)$$，故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1207", "queId": "8aac49074e724b45014e87caced5514b", "competition_source_list": ["1996年第7届全国希望杯初一竞赛复赛第9题"], "difficulty": "3", "qtype": "single_choice", "problem": "如果关于$$x$$的方程$$3\\left( x+4 \\right)=2a+5$$的解大于关于$$x$$的方程$$\\frac{(4a+1)x}{4}=\\frac{a(3x-4)}{3}$$的解，  那么．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater2$$ "}], [{"aoVal": "B", "content": "$$a\\textless{}2$$ "}], [{"aoVal": "C", "content": "$$a\\textless{}\\frac{7}{18}$$ "}], [{"aoVal": "D", "content": "$$a\\textgreater\\frac{7}{18}$$ "}]], "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->不等式（组）解的情况确定参数的范围", "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"], "answer_analysis": ["关于$$x$$的方程$$3\\left( x+4 \\right)=2a+5$$的解为$$x=\\frac{2a-7}{3}$$，  关于$$x$$的方程$$\\frac{(4a+1)x}{4}=\\frac{a(3x-4)}{3}$$的解为$$x=-\\frac{16}{3}a$$，  由题意$$\\frac{2a-7}{3}\\textgreater-\\frac{16}{3}a$$，解得$$a\\textgreater\\frac{7}{18}$$，选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "966", "queId": "663b3b63448e41b39676867cde581451", "competition_source_list": ["2009年第20届希望杯初一竞赛第1试第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$${{x}^{2}}+x-2=0$$，则$${{x}^{3}}+2{{x}^{2}}-x+2020$$=．", "answer_option_list": [[{"aoVal": "A", "content": "$$2021$$ "}], [{"aoVal": "B", "content": "$$2020$$ "}], [{"aoVal": "C", "content": "$$-2020$$ "}], [{"aoVal": "D", "content": "$$-2021$$ "}]], "knowledge_point_routes": ["知识标签->题型->式->整式加减->整式加减化简求值->题型：整式加减条件化简求值", "知识标签->知识点->式->整式的加减->整式的加减运算", "知识标签->学习能力->运算能力"], "answer_analysis": ["$${{x}^{2}}+x-2=0$$，则$${{x}^{2}}+x=2$$，  ∴$${{x}^{3}}+2{{x}^{2}}-x+2007$$  $$={{x}^{3}}+{{x}^{2}}+{{x}^{2}}+x-2x+2020$$  $$=x\\left( {{x}^{2}}+x \\right)+\\left( {{x}^{2}}+x \\right)-2x+2020$$  $$=2x+2-2x+2020$$  $$=2021$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1067", "queId": "ff8080814d7978b9014d88d515d42bce", "competition_source_list": ["1991年第2届全国希望杯初一竞赛复赛第9题", "2018~2019学年10月重庆北碚区重庆市西南大学附属中学初一上学期月考第7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$abc=1$$，则$$\\frac{a}{ab+a+1}+\\frac{b}{bc+b+1}+\\frac{c}{ca+c+1}$$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": ["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": ["$$abc=1$$，则$$a$$，$$b$$，$$c$$均不为$$0$$．  $$\\frac{a}{ab+a+1}+\\frac{b}{bc+b+1}+\\frac{c}{ca+c+1}$$  $$=\\frac{a}{ab+a+abc}+\\frac{b}{bc+b+abc}+\\frac{c}{ca+c+abc}$$  $$=\\frac{1}{b+1+bc}+\\frac{b}{bc+b+1}+\\frac{1}{a+1+ab}$$  $$=\\frac{b+1}{b+1+bc}+\\frac{abc}{a+abc+ab}$$  $$=\\frac{b+1}{b+1+bc}+\\frac{bc}{1+bc+b}$$  $$=1$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "268", "queId": "5d843a9f886d4e559497d1fe23b0407a", "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5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$y=\\begin{cases}-x,x\\leqslant 0    {{x}^{2}},x\\textgreater0    \\end{cases}$$，若$$y=4$$，则数$$x$$值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-4$$或$$-2$$ "}], [{"aoVal": "B", "content": "$$-4$$或$$2$$ "}], [{"aoVal": "C", "content": "$$-2$$或$$4$$ "}], [{"aoVal": "D", "content": "$$-2$$或$$2$$ "}]], "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->函数值"], "answer_analysis": ["∵$$y=\\begin{cases}-x,x\\leqslant 0    {{x}^{2}},x\\textgreater0    \\end{cases}$$，  ∴当$$x\\leqslant 0$$时，$$y=4$$，即$$-x=4$$，  解得$$x=-4$$，  当$$x\\textgreater0$$时，$$y=4$$，即$${{x}^{2}}=4$$，  解得$$x=2$$或$$x=-2$$（舍去）．  综上所述，$$x$$的值为$$-4$$或$$2$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "69", "queId": "1311733d5ee74e1d80877de34ca8e794", "competition_source_list": ["2011年第22届全国希望杯初一竞赛初赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$x\\ne -\\frac{5}{b}$$时，$$\\frac{a+x}{-bx-5}=2$$成立，则$${{a}^{2}}-{{b}^{2}}=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$99.25$$ "}], [{"aoVal": "D", "content": "$$99.75$$ "}]], "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": ["因为$$\\frac{a+x}{-bx-5}=2$$，  所以$$a+x=2(-bx-5)$$，  即$$a+x=-2bx-10$$，  于是$$a=-10$$，$$b=-\\frac{1}{2}$$，  所以$${{a}^{2}}-{{b}^{2}}=99.75$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "823", "queId": "bf261e682f8442f395e9f6da22564c04", "competition_source_list": ["2001年第12届希望杯初二竞赛第2试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\triangle ABC$$的三边长是$$a$$，$$b$$，$$c$$，且满足$${{a}^{4}}={{b}^{4}}+{{c}^{4}}-{{b}^{2}}{{c}^{2}}$$，$${{b}^{4}}={{c}^{4}}+{{a}^{4}}-{{a}^{2}}{{c}^{2}}$$，$${{c}^{4}}={{a}^{4}}+{{b}^{4}}-{{a}^{2}}{{b}^{2}}$$，则$$\\triangle ABC$$是．", "answer_option_list": [[{"aoVal": "A", "content": "钝角三角形 "}], [{"aoVal": "B", "content": "直角三角形 "}], [{"aoVal": "C", "content": "等腰直角三角形 "}], [{"aoVal": "D", "content": "等边三角形 "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["在$$\\triangle ABC$$中，$${{a}^{4}}={{b}^{4}}+{{c}^{4}}-{{b}^{2}}{{c}^{2}}$$，且$${{b}^{4}}={{c}^{4}}+{{a}^{4}}-{{a}^{2}}{{c}^{2}}$$，$${{c}^{4}}={{a}^{4}}+{{b}^{4}}-{{a}^{2}}{{b}^{2}}$$，  三式相加得  $${{a}^{4}}+{{b}^{4}}+{{c}^{4}}=2{{a}^{4}}+2{{b}^{4}}+2{{c}^{4}}-{{b}^{2}}{{c}^{2}}-{{a}^{2}}{{b}^{2}}-{{a}^{2}}{{c}^{2}}$$，  所以$${{a}^{4}}+{{b}^{4}}+{{c}^{4}}-{{a}^{2}}{{b}^{2}}-{{b}^{2}}{{c}^{2}}-{{c}^{2}}{{a}^{2}}=0$$．  所以$$2{{a}^{4}}+2{{b}^{4}}+2{{c}^{4}}-2{{a}^{2}}{{b}^{2}}-2{{b}^{2}}{{c}^{2}}-2{{c}^{2}}{{a}^{2}}=0$$.  所以$${{({{a}^{2}}-{{b}^{2}})}^{2}}+{{({{b}^{2}}-{{c}^{2}})}^{2}}+{{({{c}^{2}}-{{a}^{2}})}^{2}}=0$$，  所以$${{a}^{2}}={{b}^{2}}={{c}^{2}}$$，即$$a=b=c$$．  所以$$\\triangle ABC$$为等边三角形．选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "88", "queId": "4a8e7fc1e55c47d5b73a12d0aa706ebf", "competition_source_list": ["2006年第17届希望杯初一竞赛复赛第4题4分", "2019~2020学年10月河南郑州中原区朗悦慧外国语中学初一上学期月考第8题3分", "2019~2020学年10月河南郑州高新区郑州市枫杨外国语（西分）初一上学期月考第8题3分", "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7题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "若$$a＜b＜0＜c＜d$$，则以下四个结论中，正确的是（ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a+b+c+d$$一定是正数 "}], [{"aoVal": "B", "content": "$$d+c-a-b$$可能是负数 "}], [{"aoVal": "C", "content": "$$d-c-b-a$$一定是正数 "}], [{"aoVal": "D", "content": "$$c-d-b-a$$一定是正数 "}]], "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->正数和负数->正数、负数定义"], "answer_analysis": ["$$a+b+c+d$$不能确定正负；$$d+c-a-b$$一定为正；$$d-c-b-a$$一定是正数；$$c-d$$为负，$$-b-a$$为正，$$c-d-b-a$$不能确定正负． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "672", "queId": "b57b80ac70d74c8fbea03dc388ba8ecf", "competition_source_list": ["1998年第9届希望杯初一竞赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "``$$a$$与$$b$$的和的立方''的代数式表示是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{a}^{3}}+{{b}^{3}}$$ "}], [{"aoVal": "B", "content": "$$a+{{b}^{3}}$$ "}], [{"aoVal": "C", "content": "$${{a}^{3}}+b$$ "}], [{"aoVal": "D", "content": "$${{(a+b)}^{3}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->立方根->开立方"], "answer_analysis": ["$${{a}^{3}}+{{b}^{3}}$$的意义是$$a$$立方与$$b$$立方之和；  $$a+{{b}^{3}}$$的意义是$$a$$与$$b$$立方之和；  $${{a}^{3}}+b$$的意义是$$a$$立方与$$b$$之和；  $${{(a+b)}^{3}}$$的意义是$$a$$与$$b$$的和的立方．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "959", "queId": "8d8dd0ec4be74461aa16679a595cc27f", "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->能力->运算能力", "课内体系->思想->分类讨论思想", "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->幂的运算->零指数幂"], "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": "418", "queId": "a71057bdee3145ef909aef751fb679a4", "competition_source_list": ["2018~2019学年湖北武汉汉阳区初一上学期期中第7题3分", "2016~2017学年湖北武汉武昌区武汉市武珞路中学初一上学期期中第10题3分", "2005年第16届希望杯初一竞赛初赛第6题4分", "2018~2019学年10月广东东莞市虎门镇东莞市虎门外国语学校初一上学期月考第10题2分", "2020~2021学年广东广州花都区黄冈中学广州学校初一上学期期中第10题3分", "2018~2019学年湖北武汉汉阳区初一上学期期中"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$a+b+c=0$$，且$$\\left\\textbar{} a \\right\\textbar\\textgreater\\left\\textbar{} b \\right\\textbar\\textgreater\\left\\textbar{} c \\right\\textbar$$，则下列说法中可能成立的是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$、$$b$$为正数，$$c$$为负数 "}], [{"aoVal": "B", "content": "$$a$$、$$c$$为正数，$$b$$为负数 "}], [{"aoVal": "C", "content": "$$b$$、$$c$$为正数，$$a$$为负数 "}], [{"aoVal": "D", "content": "$$a$$、$$c$$为负数，$$b$$为正数 "}]], "knowledge_point_routes": ["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$$\\left\\textbar{} a \\right\\textbar\\textgreater\\left\\textbar{} b \\right\\textbar\\textgreater\\left\\textbar{} c \\right\\textbar$$，  ∴$$a$$、$$b$$、$$c$$三个数中只有两负一正或者两正一负两种情况．  如果两负一正情况成立，则$$b\\textless{}0$$、$$c\\textless{}0$$、$$a\\textgreater0$$．  于是应该在两正一负的答案中寻找正确答案．  若$$a$$、$$b$$为正数，$$c$$为负数时， 则：$$\\left\\textbar{} a \\right\\textbar+\\left\\textbar{} b \\right\\textbar\\textgreater\\left\\textbar{} c \\right\\textbar$$，  ∴$$a+b+c\\ne 0$$， $$\\text{A}$$不成立．  若$$a$$、$$c$$为正数，$$b$$为负数时， 则：$$\\left\\textbar{} a \\right\\textbar+\\left\\textbar{} c \\right\\textbar\\textgreater\\left\\textbar{} b \\right\\textbar$$，  ∴$$a+b+c\\ne 0$$， $$\\text{B}$$不成立．  只有$$\\text{C}$$符合题意． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "824", "queId": "e3efbb95b7c5421b811365280405dd77", "competition_source_list": ["2008年第19届希望杯初二竞赛第2试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "给出两列数:（$$1$$）$$1$$，$$3$$，$$5$$，$$7$$，$$\\cdots $$，$$2007$$；（$$2$$） $$1$$，$$6$$，$$11$$，$$16$$，$$\\cdots $$，$$2006$$，则同时出现在两列数中的数的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$201$$ "}], [{"aoVal": "B", "content": "$$200$$ "}], [{"aoVal": "C", "content": "$$199$$ "}], [{"aoVal": "D", "content": "$$198$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["由观察可知，同时出现在两列数中的数是$$1$$，$$11$$，$$21$$，$$\\cdots $$，$$2001$$，即每相邻两个数之间相差$$10$$，所以总数是  $$\\frac{2001-1}{10}+1=201$$，  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "498", "queId": "316f391afadb4baba18997e4cc37cb92", "competition_source_list": ["2000年第11届希望杯初一竞赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知代数式$$\\frac{{{x}^{2}}(a{{x}^{5}}+b{{x}^{3}}+cx)}{{{x}^{4}}+d{{x}^{2}}}$$，当$$x=1$$时，值为$$1$$，那么该代数式当$$x=-1$$时的值是．", "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->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": ["将$$x=1$$代入代数式得值为$$1$$，有$$\\frac{(a+b+c)}{1+d}=1$$．  即$$\\frac{a+b+c}{1+d}=1$$．  将$$x=-1$$代入代数式中得：  $$\\frac{{{(-1)}^{2}}\\left[ a{{(-1)}^{5}}+b{{(-1)}^{3}}+c(-1) \\right]}{{{(-1)}^{4}}+d{{(-1)}^{2}}}=-\\frac{a+b+c}{1+d}=-1$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1045", "queId": "ff8080814d7978b9014d86a0d9ff24de", "competition_source_list": ["1990年第1届全国希望杯初一竞赛复赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "甲杯中盛有$$2m$$毫升红墨水，乙杯中盛有$$m$$毫升蓝墨水，从甲杯倒出$$a$$毫升到乙杯里，$$0\\textless{a}\\textless{m}$$，搅匀后，又从乙杯倒出$$a$$毫升到甲杯里，则这时（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "甲杯中混入的蓝墨水比乙杯中混入的红墨水少 "}], [{"aoVal": "B", "content": "甲杯中混入的蓝墨水比乙杯中混入的红墨水多 "}], [{"aoVal": "C", "content": "甲杯中混入的蓝墨水和乙杯中混入的红墨水相同 "}], [{"aoVal": "D", "content": "甲杯中混入的蓝墨水与乙杯中混入的红墨水多少关系不定 "}]], "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": ["从甲杯倒出$$a$$毫升红墨水到乙杯中以后：  乙杯中含红墨水的比例是$$\\frac{a}{m+a}$$，  乙杯中含蓝墨水的比例是$$\\frac{m}{m+a}$$．  再从乙杯倒出$$a$$毫升混合墨水到甲杯中以后：  乙杯中含有的红墨水的数量是$$a-a\\cdot \\frac{a}{m+a}=\\frac{ma}{m+a}$$毫升，  乙杯中减少的蓝墨水的数量是$$a\\cdot \\frac{m}{m+a}=\\frac{ma}{m+a}$$毫升．  两者相等． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "468", "queId": "830ff63a66174d5194087508a718c114", "competition_source_list": ["2014年第25届全国希望杯初二竞赛初赛第1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "化简$${{\\left[ {{(-1)}^{n+1}}{{p}^{3}} \\right]}^{n}}$$（$$n$$是自然数），得（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{p}^{3n}}$$ "}], [{"aoVal": "B", "content": "$$-{{p}^{3n}}$$ "}], [{"aoVal": "C", "content": "$$-{{p}^{n+3}}$$ "}], [{"aoVal": "D", "content": "$${{p}^{n+3}}$$ "}]], "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->式->整式的乘除->幂的运算->幂的乘方"], "answer_analysis": ["$${{\\left[ {{(-1)}^{n+1}}{{p}^{3}} \\right]}^{n}}={{(-1)}^{n(n+1)}}{{p}^{3n}}$$．  因为$$n$$和$$n+1$$是连续的自然数，  所以$$n(n+1)$$是偶数，  所以$${{\\left[ {{(-1)}^{n+1}}{{p}^{3}} \\right]}^{n}}={{(-1)}^{n(n+1)}}{{p}^{3n}}={{p}^{3n}}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "950", "queId": "8d7ae982c570468e98fec67b21a91805", "competition_source_list": ["2004年第15届希望杯初二竞赛第1试第10题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$[x]$$表示不大于$$x$$的最大整数，如$$[3.15]=3$$，$$[-2.7]=-3$$，$$[4]=4$$，则$$\\frac{[\\sqrt{1\\times 2}]+[\\sqrt{2\\times 3}]+\\cdots +[\\sqrt{2003\\times 2004}]}{1002}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$1001$$ "}], [{"aoVal": "B", "content": "$$2003$$ "}], [{"aoVal": "C", "content": "$$2004$$ "}], [{"aoVal": "D", "content": "$$1002$$ "}]], "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->数->实数->平方根->算术平方根的定义", "课内体系->能力->抽象概括能力"], "answer_analysis": ["因为$${{n}^{2}}\\textless{}n(n+1)\\textless{}{{(n+1)}^{2}}$$，  $$n\\textless{}\\sqrt{n(n+1)}\\textless{}n+1$$，  所以$$[\\sqrt{n(n+1)}]=n$$，  $$[\\sqrt{1\\times 2}]=1$$，$$[\\sqrt{2\\times 3}]=2$$，$$\\cdots $$，$$[\\sqrt{2003\\times 2004}]=2003$$，  所以，原式$$=\\frac{1+2+3+\\cdots +2003}{1002}$$  $$=\\frac{(1+2003)\\times 2003}{2\\times 1002}$$  $$=2003$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "958", "queId": "d68b68b59f35484386cdb91bb271b382", "competition_source_list": ["2002年第13届希望杯初一竞赛第1试第5题"], "difficulty": "0", "qtype": "single_choice", "problem": "如果$$x=-1$$是方程$${{x}^{2}}+mx+n=0$$的一个根，那么$$m$$，$$n$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$m\\textgreater n$$ "}], [{"aoVal": "B", "content": "$$m=n$$ "}], [{"aoVal": "C", "content": "$$m\\textless{}n$$ "}], [{"aoVal": "D", "content": "不确定的 "}]], "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->由一元二次方程的解求参数的值", "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": ["因为$$x=-1$$是$${{x}^{2}}+mx+n=0$$的根，  则有$${{(-1)}^{2}}+m\\cdot (-1)+n=0$$，  所以$$m-n=1$$，即$$m\\textgreater n$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "224", "queId": "14e8197825f1481b97de64eff9974cd3", "competition_source_list": ["2002年第13届希望杯初一竞赛第1试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$${{\\left( a+b \\right)}^{2001}}=-1$$，$${{\\left( a-b \\right)}^{2002}}=1$$，则$${{a}^{2003}}+{{b}^{2003}}$$的值是．", "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->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$${{\\left( a+b \\right)}^{2001}}=-1$$，$${{\\left( a-b \\right)}^{2002}}=1$$，得  $$\\begin{cases} a+b=-1    a-b=-1   \\end{cases}$$或$$\\begin{cases} a+b=-1    a-b=1   \\end{cases}$$，  解得$$\\begin{cases} a=-1    b=0   \\end{cases}$$或$$\\begin{cases} a=0    b=-1   \\end{cases}$$，  所以$${{a}^{2003}}+{{b}^{2003}}=-1$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "238", "queId": "15185b79cea247a28cc39c3deb0312d1", "competition_source_list": ["2016年第27届全国希望杯初二竞赛初赛第9题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "从长度分别是$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$14$$的$$14$$条线段中任取出$$n$$条，其中必有$$3$$条线段能构成一个三角形，则$$n$$的最小值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "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", "课内体系->能力->推理论证能力"], "answer_analysis": ["取极端情况，若取出$$6$$条，$$1$$，$$2$$，$$3$$，$$5$$，$$8$$，$$13$$，其中任意$$3$$条都不能构成一个三角形，再多取出一条，则有$$3$$条线段能构成一个三角形，故$$n$$的最小值为$$7$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1525", "queId": "b9033b0913584964b07b707f74de3c26", "competition_source_list": ["2019~2020学年广东深圳南山区深圳市南山第二外国语学校集团海德学校初二上学期期中（南山二外教育集团）第11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4题4分", "2017~2018学年4月陕西西安高新区西安高新第一中学初三下学期月考（六模）第7题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知直线$${{l}_{1}}:y=kx+b\\left( k\\ne 0 \\right)$$与直线$${{l}_{2}}:y={{k}_{1}}x-6\\left( {{k}_{1}}\\textless{}0 \\right)$$在第三象限交于点$$M$$，若直线$${{l}_{1}}$$与$$x$$轴的交点为$$B\\left( 3,0 \\right)$$，则$$k$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2\\textless{}k\\textless{}2$$ "}], [{"aoVal": "B", "content": "$$-2\\textless{}k\\textless{}0$$ "}], [{"aoVal": "C", "content": "$$0\\textless{}k\\textless{}4$$ "}], [{"aoVal": "D", "content": "$$0\\textless{}k\\textless{}2$$ "}]], "knowledge_point_routes": ["课内体系->思想->数形结合思想", "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", "课内体系->知识点->函数->一次函数->一次函数与方程、不等式->一次函数与二元一次方程组"], "answer_analysis": ["$$B\\left( 3,0 \\right)$$代入$${{l}_{1}}:0=3k+b$$，  ∴$$b=-3k$$，  ∴$${{l}_{1}}:y=kx-3k$$，  将$$\\left( 0,-6 \\right)$$代入：$$k=2$$，  由图象知：$$0\\textless{}k\\textless{}2$$时，  第三象限有交点．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1528", "queId": "f88e9b7ba1674b15b22df73daeb0c8d0", "competition_source_list": ["2013年第24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$$x+\\frac{1}{x}=6$$（$$0\\textless{}x\\textless{}1$$），则$$\\sqrt{x}-\\frac{1}{\\sqrt{x}}$$的值是（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\sqrt{5}$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{5}$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "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": ["因为$${{\\left( \\sqrt{x}-\\frac{1}{\\sqrt{x}} \\right)}^{2}}=x+\\frac{1}{x}-2=4$$，  所以$$\\sqrt{x}-\\frac{1}{\\sqrt{x}}=\\pm 2$$．  又$$0\\textless{}x\\textless{}1$$，  所以$$\\sqrt{x}-\\frac{1}{\\sqrt{x}}=-2$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1402", "queId": "c9fc5370df7e4ba6af5df0634b0d40f5", "competition_source_list": ["2002年第19届全国初中数学联赛竞赛第1题7分", "初二下学期单元测试《二次根式》二次根式的运算第43题"], "difficulty": "1", "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": "$$b\\textless{}a\\textless{}c$$ "}], [{"aoVal": "C", "content": "$$c\\textless{}b\\textless{}a$$ "}], [{"aoVal": "D", "content": "$$c\\textless{}a\\textless{}b$$ "}]], "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->二次根式比较大小"], "answer_analysis": ["\\textbf{（知识点：二次根式比较大小）}  ∵$$a-b$$$$=\\sqrt{2}-1-\\left( 2\\sqrt{2}-\\sqrt{6} \\right)$$$$=\\sqrt{6}-\\left( 1+\\sqrt{2} \\right)$$$$\\approx 2.449-2.414\\textgreater0$$，  ∴$$a\\textgreater b$$；  ∵$$a-c$$$$=\\sqrt{2}-1-\\left( \\sqrt{6}-2 \\right)$$$$=\\sqrt{2}+1-\\sqrt{6}$$$$\\approx 2.414-2.449\\textless{}0$$，  ∴$$a\\textless{}c$$；于是$$b\\textless{}a\\textless{}c$$．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": "1572", "queId": "f02d660b54914fa086a8250627136091", "competition_source_list": ["2014年第25届全国希望杯初二竞赛复赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "中国古代``五行''学说认为：``物质分金、木、水、火、土五种属性，金克木，木克土，土克水，水克火，火克金．''从五种属性互不相同的物质中随机抽取两种，则抽取的两种物质不相克的概率是（~ ）．  （注：概率等于时间所包含的各种可能的结果数在全部可能的结果数中所占的比）", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{10}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{5}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率->概率的计算方法"], "answer_analysis": ["基本事件为：金、木；金、水；金、火；金、土；木、水；木、火；木、土；水、火；水、土；火、土．  共计$$10$$种，  其中不相克的事件数为$$m=10-5=5$$（种），  所以抽取的两种物质不相克的概率是$$\\frac{5}{10}=\\frac{1}{2}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1038", "queId": "ff8080814d7978b9014d865bbc2d23db", "competition_source_list": ["1990年第1届全国希望杯初一竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "大于$$-\\pi$$并且不是自然数的整数有（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$个 "}], [{"aoVal": "B", "content": "$$3$$个 "}], [{"aoVal": "C", "content": "$$4$$个 "}], [{"aoVal": "D", "content": "无数个 "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->实数基础概念->实数与数轴"], "answer_analysis": ["在数轴上容易看出：在$$-\\pi$$右边$$0$$的左边的整数只有$$-3$$，$$-2$$，$$-1$$，共$$3$$个． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1303", "queId": "ffceed6e567f4b51b78b75c8aa8cc7de", "competition_source_list": ["2017~2018学年4月广东深圳福田区深圳实验学校中学部初三下学期月考第3题3分", "2010年竞赛第1题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\frac{a}{b}=20$$，$$\\frac{b}{c}=10$$，则$$\\frac{a+b}{b+c}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{11}{21}$$ "}], [{"aoVal": "B", "content": "$$\\frac{21}{11}$$ "}], [{"aoVal": "C", "content": "$$\\frac{110}{21}$$ "}], [{"aoVal": "D", "content": "$$\\frac{210}{11}$$ "}]], "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-条件化简求值"], "answer_analysis": ["由$$\\begin{cases}\\dfrac{a}{b}=20    \\dfrac{b}{c}=10 \\end{cases}$$，得$$\\begin{cases}a=20b    b=10c \\end{cases}$$，  则$$\\frac{a+b}{b+c}=\\frac{20b+b}{10c+c}=\\frac{21b}{11c}=\\frac{21\\cdot 10c}{11c}=\\frac{210}{11}$$． ", "<p>$$\\frac{a+b}{b+c}=\\frac{\\dfrac{a}{b}+1}{1+\\dfrac{c}{b}}=\\frac{20+1}{1+\\dfrac{1}{10}}=\\frac{210}{11}$$．</p>\n<p>故选$$\\text{D}$$．</p>"], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "230", "queId": "5433691d02e34792859003caa89f1902", "competition_source_list": ["2017年河北石家庄裕华区石家庄外国语学校初三竞赛第1题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\frac{a}{3}=\\frac{b}{4}=\\frac{c}{5}\\ne 0$$，则$$\\frac{c-a}{b}$$的值为（ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{3}$$ "}]], "knowledge_point_routes": ["知识标签->题型->三角形->相似三角形->比例线段->题型：比例的综合应用", "知识标签->知识点->三角形->相似->比例线段->比例线段的性质"], "answer_analysis": ["设$$k=\\frac{a}{3}=\\frac{b}{4}=\\frac{c}{5}\\ne 0$$，由此得到$$a=3k$$，$$b=4k$$，$$c=5k$$  ∴$$\\frac{c-a}{b}$$  $$=\\frac{5k-3k}{4k}$$  $$=\\frac{1}{2}$$，故$$\\text{B}$$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": "1187", "queId": "8aac49074e023206014e25bb7d43729f", "competition_source_list": ["1997年第8届全国希望杯初一竞赛初赛第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$a=\\frac{19951995}{19961996}$$，$$b=\\frac{19961996}{19971997}$$，$$c=\\frac{19971997}{19981998}$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless{}b\\textless{}c$$ "}], [{"aoVal": "B", "content": "$$b\\textless{}c\\textless{}a$$ "}], [{"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-作商法", "课内体系->能力->运算能力"], "answer_analysis": ["设$$A=19951995$$，$$B=19961996$$，$$C=19971997$$，$$D=19981998$$，  则有$$B=A+10001$$，$$C=B+10001$$，$$D=C+10001$$．  ∵$$\\left( B+10001 \\right)\\left( B-10001 \\right)={{B}^{2}}-{{10001}^{2}}$$  亦即，$$C\\cdot A={{B}^{2}}-{{10001}^{2}}$$，∴$$C\\cdot A\\textless{}{{B}^{2}}$$．  由于$$B$$、$$C$$均为正数，不等式两边同时除以$$B\\cdot C$$，得到$$\\frac{A}{B}\\textless{}\\frac{B}{C}$$，  所以$$\\frac{19951995}{19961996}\\textless{}\\frac{19961996}{19971997}$$，同理$$\\frac{19961996}{19971997}\\textless{}\\frac{19971997}{19981998}$$，  即$$a\\textless{}b\\textless{}c$$．故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "471", "queId": "5098043bf2654dc8b35bddb811b46dc4", "competition_source_list": ["2006年第17届希望杯初二竞赛第2试第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$n(n\\geqslant2)$$个正整数$$a_1$$，$$a_2$$，$$a_3$$，$$\\cdots $$，$$a_n$$，任意改变它们的顺序后，记作$$b_1$$，$$b_2$$，$$b_3$$，$$\\cdots $$，$$b_n$$，若$$P=(a_1-b_1)(a_2-b_2)(a_3-b_3)\\cdots(a_n-b_n)$$，则（ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$P$$一定是奇数 "}], [{"aoVal": "B", "content": "$$P$$一定是偶数 "}], [{"aoVal": "C", "content": "当$$n$$是奇数时，$$P$$是偶数 "}], [{"aoVal": "D", "content": "当$$n$$是偶数时，$$P$$是奇数 "}]], "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->奇数与偶数"], "answer_analysis": ["无论$$n$$是奇数偶数，可以假设$$a_n=b_n$$，$$P=0$$为偶数，$$\\rm A$$、$$\\rm D$$不能选，  现在在$$\\rm B$$和$$\\rm C$$中选择，要让$$P$$为奇数，那么必须它的$$n$$个因式都是奇数，  也就是每个因式都是一个奇数与一个偶数的差，  因为$$b_1$$，$$b_2$$，$$\\cdots$$，$$b_n$$都是$$a_n$$变来的，  所以原来如果是$$x$$个奇数与$$n-x$$个偶数的话，奇数与偶数的数目必须也是一样的，即$$x=n-x$$，$$n=2x$$为偶数，  也就是说，$$P$$若为奇数，$$n$$必须是偶数，可以推出，$$n$$为奇数，$$P$$必须为偶数．  所以$$\\rm B$$错，$$\\rm C$$正确．  故选$$\\rm C$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "29", "queId": "0592a49b55534702a9ea332c48bf962c", "competition_source_list": ["2022~2023学年浙江宁波初三月考（六校强基竞赛）第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "若关于\\emph{x}的不等式组$\\left {\\begin{array}{c} 3-2x\\geq 1   x\\geq m+1 \\end{array}\\right.$共有2个整数解，则\\emph{m}的取值范围是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$m=-1$ "}], [{"aoVal": "B", "content": "$-2\\textless{} m\\leq -1$ "}], [{"aoVal": "C", "content": "$-2\\leq m\\leq -1$ "}], [{"aoVal": "D", "content": "$m\\textless{} -1$ "}]], "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"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 先解不等式$3-2x\\geq 1$，得$x\\leq 1$，结合不等式组的整数解的情况，得出关于\\emph{m}的不等式组，求解即可．\\\\ 【详解】\\\\ 解不等式$3-2x\\geq 1$，得$x\\leq 1$，\\\\ ∵关于\\emph{x}的不等式组$\\left {\\begin{array}{c} 3-2x\\geq 1   x\\geq m+1 \\end{array}\\right.$共有2个整数解，\\\\ ∴这两个整数解为$0,1$，\\\\ ∴$-1\\textless{} m+1\\leq 0$，\\\\ 解得$-2\\textless{} m\\leq -1$，\\\\ 故选：B．\\\\ 【点睛】\\\\ 本题考查一元一次不等式组的整数解，解答本题的关键是明确题意，得出关于\\emph{m}的不等式组． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "616", "queId": "55eeaf4a6fbf4643a4e3b4ccfd6de3d4", "competition_source_list": ["2009年第20届希望杯初二竞赛第2试第17题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$12$$人参加象棋单循环赛，规定胜$$1$$局得$$2$$分，平$$1$$局得$$1$$分，败者不得分，比赛结果是第二名的得分与最后$$5$$名的得分之和相同，那么第二名得分为．", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$19$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率->游戏公平性", "竞赛->知识点->组合->极端原理"], "answer_analysis": ["因为是循环赛，所以最后$$5$$名棋手相互之间要赛$$10$$场，  不论胜负情况如何，每场比赛两人得分的和都是$$2$$分，  即最后$$5$$名棋手得分的和至少是$$20$$分，  所以第二名至少得$$20$$分，  又因为第二名要进行$$1$$场比赛，他不能全胜得$$22$$分，否则他是第一名，  同样他也不能得$$21$$分，否则他的成绩是$$10$$胜$$1$$平，  其中平局一定是与第一名对阵的结果，这样他们两人将并列第一名，不是第二名，  综上可得第二名只能得$$20$$分．（检验构造该种情况） "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1449", "queId": "af5c9036c3b947299b0ab8cb73f31491", "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": "812", "queId": "8cae81b0ccff45ed96d194c9a9b514d9", "competition_source_list": ["2006年第17届希望杯初二竞赛第1试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "实数$$m={{2005}^{3}}-2005$$，下列各数中不能整除$$m$$的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$2006$$ "}], [{"aoVal": "B", "content": "$$2005$$ "}], [{"aoVal": "C", "content": "$$2004$$ "}], [{"aoVal": "D", "content": "$$2003$$ "}]], "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->知识点->式->整式的乘除->乘法公式->平方差公式的计算"], "answer_analysis": ["因为$$m={{2005}^{3}}-2005$$  $$=2005({{2005}^{2}}-1)$$  $$=2005(2005+1)(2005-1)$$  $$=2005\\times 2006\\times 2004$$．  所以，$$m$$不能被$$2003$$整除．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "464", "queId": "830d30a407ee40fdb41a1a434a8bdb57", "competition_source_list": ["1999年第10届希望杯初一竞赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$三个整数满足$$a \\textless{} b \\textless{} c$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$a+c \\textless{} b+c$$ "}], [{"aoVal": "B", "content": "$$\\left\\textbar{} a \\right\\textbar+\\left\\textbar{} c \\right\\textbar{} \\textless{} \\left\\textbar{} b \\right\\textbar+\\left\\textbar{} c \\right\\textbar$$ "}], [{"aoVal": "C", "content": "$$ab \\textless{} ac$$ "}], [{"aoVal": "D", "content": "$$\\left\\textbar{} a \\right\\textbar\\left\\textbar{} b \\right\\textbar{} \\textless{} \\left\\textbar{} a \\right\\textbar\\left\\textbar{} c \\right\\textbar$$ "}]], "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->绝对值综合", "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->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["用特殊值法，取$$a=0$$，易知$$\\text{C}$$、$$\\text{D}$$不成立，  而当$$a=-1$$，$$b=0$$，$$c=1$$时，$$\\text{B}$$不成立．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "656", "queId": "4d3e28cfff8e4ac785ddb12a2437390d", "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->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-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": "1553", "queId": "f8f15b768ff24a65a8baa63d33f6f083", "competition_source_list": ["2013年第24届全国希望杯初二竞赛初赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若定义``$$\\odot $$''：$$a\\odot b={{b}^{a}}$$，如$$3\\odot 2={{2}^{3}}=8$$，则$$3\\odot \\frac{1}{2}$$等于（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{6}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{2}$$ "}]], "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"], "answer_analysis": ["由运算定义，知$$3\\odot \\frac{1}{2}={{\\left( \\frac{1}{2} \\right)}^{3}}=\\frac{1}{8}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "806", "queId": "5762a2c528d444feb48d0a7836ada0b6", "competition_source_list": ["2012年第23届全国希望杯初一竞赛复赛第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x$$、$$y$$是非负整数，且使$$\\frac{x-1}{2}=\\frac{4-y}{3}$$是整数，那么这样的数对$$(x,y)$$有（~ ）个．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$2012$$ "}]], "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": ["设$$\\frac{x-1}{2}=\\frac{4-y}{3}=k$$，$$k$$为整数．  所以$$\\left { \\begin{matrix}x=2k+1\\geqslant 0    y=4-3k\\geqslant 0   \\end{matrix} \\right.$$，  解得$$-\\frac{1}{2}\\leqslant k\\leqslant \\frac{4}{3}$$．  因为$$k$$是整数，  所以$$k=0$$或$$1$$．  当$$k=0$$时，$$x=1$$，$$y=4$$，产生数对$$(1,4)$$；  当$$k=1$$时，$$x=3$$，$$y=1$$，产生数对$$(3,1)$$．  所以，这样的数对$$(x,y)$$有$$2$$个． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "432", "queId": "4be7755dcf2948c68aa88eff1f81099b", "competition_source_list": ["2018~2019学年5月天津南开区天津市南开翔宇学校初二下学期月考第5题", "2018~2019学年5月天津和平区天津市益中学校初二下学期月考翔宇益中初二联考第5题", "2000年第11届希望杯初二竞赛第1试第7题"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$\\sqrt{7}=a$$，$$\\sqrt{70}=b$$，则$$\\sqrt{4.9}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{a+b}{10}$$ "}], [{"aoVal": "B", "content": "$$\\frac{b-a}{10}$$ "}], [{"aoVal": "C", "content": "$$\\frac{b}{a}$$ "}], [{"aoVal": "D", "content": "$$\\frac{ab}{10}$$ "}]], "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->二次根式的乘法"], "answer_analysis": ["$$\\sqrt{4.9}=\\sqrt{\\frac{49}{10}}=\\frac{\\sqrt{490}}{10}=\\frac{\\sqrt{7}\\cdot \\sqrt{70}}{10}=\\frac{ab}{10}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "83", "queId": "0abc39b996d3466fbeed18e09c9df330", "competition_source_list": ["2004年竞赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若直角三角形的两条直角边长为$$a$$，$$b$$，斜边长为$$c$$，斜边上的高为$$h$$，则有(  ).", "answer_option_list": [[{"aoVal": "A", "content": "$$ab={{h}^{2}}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{a}+\\frac{1}{b}=\\frac{1}{h}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{{{a}^{2}}}+\\frac{1}{{{b}^{2}}}=\\frac{1}{{{h}^{2}}}$$ "}], [{"aoVal": "D", "content": "$${{a}^{2}}+{{b}^{2}}=2{{h}^{2}}$$ "}]], "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->三角形的周长与面积问题", "课内体系->知识点->三角形->勾股定理及应用->勾股定理基础->勾股定理"], "answer_analysis": ["$$\\because \\frac{1}{2}ab=\\frac{1}{2}ch$$  $$\\therefore h=\\frac{ab}{c}$$  $$\\therefore \\frac{1}{h}=\\frac{c}{ab}$$  $$\\therefore \\frac{1}{{{a}^{2}}}+\\frac{1}{{{b}^{2}}}=\\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}{{b}^{2}}}=\\frac{{{c}^{2}}}{{{a}^{2}}{{b}^{2}}}=\\frac{1}{{{h}^{2}}}$$．故选$$\\text{C}$$ "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "710", "queId": "8c31c7b1998e45e59785f46e6d5e6a81", "competition_source_list": ["2000年第11届希望杯初二竞赛第1试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$x\\ne 0$$，则$$\\frac{x-\\left\\textbar{} x \\right\\textbar}{\\sqrt{{{x}^{2}}}}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$0$$或$$-2$$ "}], [{"aoVal": "D", "content": "$$0$$或$$2$$ "}]], "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": ["当$$x\\textgreater0$$时，$$\\frac{x-\\left\\textbar{} x \\right\\textbar}{\\sqrt{{{x}^{2}}}}=0$$，  当$$x\\textless{}0$$时，$$\\frac{x-\\left\\textbar{} x \\right\\textbar}{\\sqrt{{{x}^{2}}}}=\\frac{2x}{-x}=-2$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1028", "queId": "9fedb58afd504c459265205fe5b91bc1", "competition_source_list": ["2000年第11届希望杯初一竞赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "有三个正整数$$a$$，$$b$$，$$c$$，其中$$a$$与$$b$$互质且$$b$$与$$c$$也互质，给出下面四个判断：  ①$${{\\left( a+c \\right)}^{2}}$$不能被$$b$$整除  ②$${{a}^{2}}+{{c}^{2}}$$不能被$$b$$整除  ③$${{\\left( a+b \\right)}^{2}}$$不能被$$c$$整除  ④$${{a}^{2}}+{{b}^{2}}$$不能被$$c$$整除  其中，不正确的判断有(  ) .", "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->式->整式的乘除->整式的乘除运算->多项式除以单项式", "竞赛->知识点->数论->整除->整除的概念与基本性质"], "answer_analysis": ["如果$$a=3$$，$$b=5$$，$$c=2$$，则$$a$$，$$b$$互质，$$b$$，$$c$$互质，但$${{\\left( a+c \\right)}^{2}}=25$$能被$$5$$整除，$${{\\left( a+b \\right)}^{2}}=64$$及$${{a}^{2}}+{{b}^{2}}=34$$能被$$2$$整除，所以①、③、④都是假命题．当$$a=3$$，$$b=2$$，$$c=5$$时，$${{a}^{2}}+{{c}^{2}}=34$$能被$$2$$整除，所以②也是假命题．题中所给$$4$$个命题均为假命题． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "139", "queId": "0bd2084de8b5445db75f308ec7c98d49", "competition_source_list": ["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6题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$x$$、$$y$$、$$z$$是三个非负实数，满足$$3x+2y+z=5$$，$$x+y-z=2$$，若$$S=2x+y-z$$，则$$S$$的最大值与最小值的和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "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": ["联立得方程组$$\\begin{cases}3x+2y+z=5(1)    x+y-z=2(2)~   \\end{cases}$$，  $$\\left( 1 \\right)+\\left( 2 \\right)$$得$$4x+3y=7$$，$$y=\\frac{7-4x}{3}$$，  $$\\left( 1 \\right)-\\left( 2 \\right)\\times 2$$得，$$x+3z=1$$，$$z=\\frac{1-x}{3}$$，  把$$y=\\frac{7-4x}{3}$$，$$z=\\frac{1-x}{3}$$代入$$S=2x+y-z$$，整理得，$$S=x+2$$，当$$x$$取最小值时，$$S$$有最小值，  ∵$$x$$、$$y$$、$$z$$是三个非负实数，  ∴$$x$$的最小值是$$0$$，  ∴$${{S}_{{最小}}}=2$$，  $$\\left( 1 \\right)-\\left( 2 \\right)$$得到：$$2x+y=3-2z$$，  ∴$$S=3-z$$，  ∵$$z$$是非负数，  ∴$$z=0$$时，$$S$$有最大值$$3$$，  ∴$$S$$的最大值与最小值的和$$3+2=5$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "189", "queId": "6fdfb12b14914133afe1df6ba0382275", "competition_source_list": ["2008年第19届希望杯初一竞赛第2试第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "一片牧场上的草长得一样快，已知$$60$$头牛$$24$$天可将草吃完，而$$30$$头牛$$60$$天也可将草吃完．那么，若在$$120$$天里将草吃完，则需要牛．", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$头 "}], [{"aoVal": "B", "content": "$$18$$头 "}], [{"aoVal": "C", "content": "$$20$$头 "}], [{"aoVal": "D", "content": "$$22$$头 "}]], "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->方法->整体法", "课内体系->思想->转化与化归思想", "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", "课内体系->知识点->函数->一次函数->一次函数与实际问题"], "answer_analysis": ["设牧场原有青草量为$$a$$，每天长出的青草量为$$x$$，$$y$$头牛在$$120$$天里恰可把牧场上的草吃完，则依题意列方程组$$\\frac{a+24x}{24\\times 60}=\\frac{a+60x}{60\\times 30}=\\frac{a+120x}{120y}$$，  由$$\\frac{a+24x}{24\\times 60}=\\frac{a+60x}{60\\times 30}$$，解得$$a=120x$$，  代入$$\\frac{a+60x}{60\\times 30}=\\frac{a+120x}{120y}$$，得  $$\\frac{120x+60x}{60\\times 30}=\\frac{120x+120x}{120y}$$，  消去$$x$$，解得$$y=20$$（头）．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1274", "queId": "a9c371420a694870b9dc626e4cba44c7", "competition_source_list": ["1996年第7届希望杯初二竞赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知一个三角形中两条边的长分别为$$a$$，$$b$$，且$$a\\textgreater b$$，那么这个三角形的周长$$l$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3a\\textgreater l\\textgreater3b$$ "}], [{"aoVal": "B", "content": "$$2\\left( a+b \\right)\\textgreater l\\textgreater2a$$ "}], [{"aoVal": "C", "content": "$$2a+b\\textgreater l\\textgreater2b+a$$ "}], [{"aoVal": "D", "content": "$$3a-b\\textgreater l\\textgreater a+2b$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["三角形中两边长为$$a$$，$$b$$，且$$a\\textgreater b$$，则第三边为$$c$$，满足条件$$a-b ~\\textless{} ~c ~\\textless{} ~a+b$$，  所以$$a+b+\\left( a-b \\right) ~\\textless{} ~a+b+c ~\\textless{} ~a+b+\\left( a+b \\right)$$，  即$$2a ~\\textless{} ~a+b+c ~\\textless{} ~2\\left( a+b \\right)$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1093", "queId": "ff8080814d9539f1014d9b5da62d0a1f", "competition_source_list": ["1992年第3届全国希望杯初一竞赛初赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a=-123.4-(-123.5)$$，$$b=123.4-123.5$$，$$c=123.4-(-123.5)$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$c\\textgreater b\\textgreater a$$ "}], [{"aoVal": "B", "content": "$$c\\textgreater a\\textgreater b$$ "}], [{"aoVal": "C", "content": "$$a\\textgreater b\\textgreater c$$ "}], [{"aoVal": "D", "content": "$$b\\textgreater c\\textgreater a$$ "}]], "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"], "answer_analysis": ["易见$$a=-123.4+123.5=0.1$$，$$b=123.4-123.5\\textless{}0$$，$$c=123.4-(-123.5)\\textgreater123.4\\textgreater a$$，  所以$$b\\textless{}a\\textless{}c$$，选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "535", "queId": "94d16cf8df1d43e49c3a72f9351ce423", "competition_source_list": ["1998年第9届希望杯初一竞赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "有理数$$a$$等于它的倒数，有理数$$b$$等于它的相反数，则$${{a}^{1998}}+{{b}^{1998}}$$等于．", "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->数->有理数->倒数与负倒数", "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", "课内体系->能力->运算能力"], "answer_analysis": ["有理数$$a$$等于它的倒数，即$$a=\\frac{1}{a}\\Rightarrow {{a}^{2}}=1$$，所以$$a=\\pm 1$$，因为$${{a}^{1998}}=1$$．  有理数$$b$$等于它的相反数，则$$b=-b$$．即$$2b=0\\Rightarrow b=0$$，因此$${{b}^{1998}}=0$$．  所以$${{a}^{1998}}+{{b}^{1998}}=1+0=1$$，故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1057", "queId": "ff8080814d7978b9014d86e42b9b25e0", "competition_source_list": ["1991年第2届全国希望杯初一竞赛初赛第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$-4$$，$$-1$$，$$-2.5$$，$$-0.01$$与$$-15$$这五个数中，最大的数与绝对值最大的那个数的乘积是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$225$$ "}], [{"aoVal": "B", "content": "$$0.15$$ "}], [{"aoVal": "C", "content": "$$0.0001$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["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->有理数比较大小->有理数比较大小-利用有理数正负性"], "answer_analysis": ["$$-4$$，$$-1$$，$$-2.5$$，$$-0.01$$与$$-15$$中最大的数是$$-0.01$$，绝对值最大的数是$$-15$$，$$(-001)\\times (-15)=0.15$$．选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "186", "queId": "e2fe84c682254446b80dc814f1afaf2e", "competition_source_list": ["2003年竞赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "在本埠投寄平信，每封信质量不超过$$20\\text{g}$$时付邮费$$0.80$$元，超过$$20\\text{g}$$而不超过$$40\\text{g}$$时付邮费$$1.60$$元，依次类推，每增加$$20\\text{g}$$需增加邮费$$0.80$$元（信的质量在$$100\\text{g}$$以内)．如果某人所寄一封信的质量为$$72.5\\text{g}$$，那么他应付邮费．", "answer_option_list": [[{"aoVal": "A", "content": "$$2.4$$元 "}], [{"aoVal": "B", "content": "$$2.8$$元 "}], [{"aoVal": "C", "content": "$$3$$元 "}], [{"aoVal": "D", "content": "$$3.2$$元 "}]], "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$$20\\times 3\\textless{}72.5\\textless{}20\\times 4$$，所以根据题意，可知需付邮费$$0.8\\times 4=3.2$$（元）．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "949", "queId": "73627cc6310b491bad4e82a83aa7d8d8", "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2题3分", "初一上学期单元测试《物以类聚》第19题", "2019~2020学年浙江宁波鄞州区宁波市鄞州蓝青学校初一上学期期末第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果对于某一特定范围内$$x$$的任意允许值, $$p=\\left\\textbar{} 1-2x \\right\\textbar+\\left\\textbar{} 1-3x \\right\\textbar+\\cdots +\\left\\textbar{} 1-9x \\right\\textbar+\\left\\textbar{} 1-10x \\right\\textbar$$ 的值恒为一常数，则此值为．", "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->知识点->数->有理数->绝对值->零点分段法", "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->能力->运算能力"], "answer_analysis": ["∵$$P$$为定值，  ∴$$P$$的表达式化简后与$$x$$无关，  即$$x$$的系数为$$0$$，  观察发现：$$2+3+4+5+6+7=8+9+10$$，  ∴只需$$1-7x\\geqslant 0$$且$$1-8x\\leqslant 0$$，  ∴$$\\frac{1}{8}\\leqslant x\\leqslant \\frac{1}{7}$$，  ∴$$P=(1-2x)+(1-3x)+(1-4x)+(1-5x)+(1-6x)+(1-7x)-(1-8x)-(1-9x)-(1-10x)$$  $$=3$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "220", "queId": "2ad883229e92441c9956c6682dae2ba6", "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": "（2002年全国初中联赛题）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->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->整式的乘除->乘法公式->完全平方公式"], "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": "1500", "queId": "d3c999c0a7df4c45a8dae604a5bc57e0", "competition_source_list": ["2013年第24届全国希望杯初二竞赛初赛第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x=\\sqrt{2}+\\sqrt{3}$$，且$${{x}^{8}}+1={{x}^{4}}(6y+8)$$，则$$y$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$30$$ "}]], "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->二次根式->二次根式的运算->二次根式的四则混合运算"], "answer_analysis": ["因为$$x=\\sqrt{2}+\\sqrt{3}$$，  所以$${{x}^{2}}=5+2\\sqrt{6}$$，  所以$${{x}^{2}}+{{x}^{-2}}=5+2\\sqrt{6}+\\frac{1}{5+2\\sqrt{6}}$$  $$=5+2\\sqrt{6}+\\frac{5-2\\sqrt{6}}{(5+2\\sqrt{6})(5-2\\sqrt{6})}$$  $$=5+2\\sqrt{6}+5-2\\sqrt{6}$$  $$=10$$，  于是$${{x}^{4}}+{{x}^{-4}}={{({{x}^{2}}+{{x}^{-2}})}^{2}}-2={{10}^{2}}-2=98$$．  因为$${{x}^{8}}+1={{x}^{4}}(6y+8)$$，  所以$$y=\\frac{{{x}^{4}}+{{x}^{-4}}-8}{6}=15$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "998", "queId": "db31b1d0e05c4cbb91c99f0700ffb628", "competition_source_list": ["2014年第25届全国希望杯初一竞赛复赛第2题4分", "2018年贵州铜仁地区中考真题第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果一个凸多边形的内角和等于外角和的$$3$$倍，那么，这个多边形的边数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "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->求多边形的内角和", "课内体系->知识点->三角形->三角形及多边形->多边形->多边形的外角和定理"], "answer_analysis": ["设多边形的边数为$$n$$，  则$$(n-2)\\times 180{}^{}\\circ =3\\times 360{}^{}\\circ $$，  解得$$n=8$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "675", "queId": "7f3e770d2b78464086f893c76de006e4", "competition_source_list": ["2003年竞赛第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "某校初三两个毕业班的学生和教师共$$100$$人一起在台阶上拍毕业照留念，摄影师要将其排列成前多后少的梯形队阵（排数$$≥3$$），且要求各行的人数必须是连续的自然数，这样才能使后一排的人均站在前一排两人间的空挡处，那么，满足上述要求的排法的方案种数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "knowledge_point_routes": ["竞赛->知识点->组合->排列与组合"], "answer_analysis": ["设最后一排有$$k$$个人，共有$$n$$排，那么从后往前各排的人数分别为$$k$$，$$k+1$$，$$k+2$$，\\ldots，$$k+\\left( n-1 \\right)$$，由题意可知：  $$kn+\\frac{n\\left( n-1 \\right)}{2}=100$$，  即$$n\\left[ 2k+\\left( n-1 \\right) \\right]=200$$．因为$$k$$，$$n$$都是正整数，且$$n\\geqslant 3$$，所以$$n\\leqslant 2k+\\left( n-1 \\right)$$，且$$n$$与$$2k+\\left( n-1 \\right)$$的奇偶性不同．将$$200$$分解质因数，可知$$n=5$$或$$8$$．当$$n=5$$时，$$k=18$$；当$$n=8$$时，$$k=9$$．共有两种不同方案．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "404", "queId": "5e0796b418fa4851a90591a3ba76e1d6", "competition_source_list": ["2013年第24届全国希望杯初一竞赛初赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$${{x}^{2}}-3x+2=0$$，则$${{x}^{3}}-{{x}^{2}}-4x+10$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "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": ["$${{x}^{3}}-{{x}^{2}}-4x+10$$  $$=x({{x}^{2}}-3x+2)+2({{x}^{2}}-3x+2)+6$$  $$=0+0+6$$  $$=6$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "76", "queId": "1324cc6673ba449595e047e04173106d", "competition_source_list": ["2008年第19届希望杯初一竞赛第1试第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "正整数$$x$$，$$y$$满足$$(2x-5)(2y-5)=25$$，则$$x+y$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$26$$ "}], [{"aoVal": "D", "content": "$$10$$或$$18$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->一次方程->特殊方程"], "answer_analysis": ["$$(2x-5)(2y-5)=25$$，$$x$$及$$y$$都是正整数．  所以$$\\begin{cases}2x-5=1    2y-5=25   \\end{cases}$$，或$$\\begin{cases}2x-5=5    2y-5=5   \\end{cases}$$，或$$\\begin{cases}2x-5=25    2y-5=1   \\end{cases}$$．  所以$$\\begin{cases}x=3    y=15   \\end{cases}$$，或$$\\begin{cases}x=5    y=5   \\end{cases}$$，或$$\\begin{cases}x=15    y=3   \\end{cases}$$．  所以$$x+y=18$$或$$x+y=10$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "392", "queId": "27d0ae9b44b143bf9d3409f8a6455e08", "competition_source_list": ["2017年全国初中数学联赛竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "已知实数$$a$$，$$b$$，$$c$$满足$$2a+13b+3c=90$$，$$3a+9b+c=72$$，则$$\\frac{3b+c}{a+2b}=$$（~ ）．", "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->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": ["已知等式可变形为$$2(a+2b)+3(3b+c)=90$$，$$3(a+2b)+(3b+c)=72$$，解得$$a+2b=18$$，$$3b+c=18$$．所以$$\\frac{3b+c}{a+2b}=1$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "169", "queId": "98b77c5dbad34460851c3ce3a17b9eb4", "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": ["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->式->因式分解->因式分解的应用"], "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": "928", "queId": "7c250c094247495ea3dfdd4a3d1e8ee0", "competition_source_list": ["1999年第10届希望杯初二竞赛第1试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\frac{1}{\\sqrt{30}-\\sqrt{31}}$$与$$\\sqrt{30}+\\sqrt{31}$$的关系是．", "answer_option_list": [[{"aoVal": "A", "content": "相等 "}], [{"aoVal": "B", "content": "互为相反数 "}], [{"aoVal": "C", "content": "互为倒数 "}], [{"aoVal": "D", "content": "互为负倒数 "}]], "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": ["因为$$\\frac{1}{\\sqrt{30}-\\sqrt{31}}=\\frac{\\sqrt{30}+\\sqrt{31}}{(\\sqrt{30}-\\sqrt{31})(\\sqrt{30}+\\sqrt{31})}=-(\\sqrt{30}+\\sqrt{31})$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "700", "queId": "405c85955ae04d6eae0b515fd1391b87", "competition_source_list": ["2013年第24届全国希望杯初一竞赛复赛第2题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "某天，黑河凌晨的温度比上午$$9$$点的温度低$$12$$℃，中午$$12$$点的温度比凌晨的温度高$$20$$℃，晚上$$9$$点的温度比中午$$12$$点的温度低$$19$$℃，若当天上午$$9$$点的温度记为$$a$$℃，则当天晚上$$9$$点的温度应记为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( a-32 \\right)$$℃ "}], [{"aoVal": "B", "content": "$$\\left( a-11 \\right)$$℃ "}], [{"aoVal": "C", "content": "$$\\left( 32-a \\right)$$℃ "}], [{"aoVal": "D", "content": "$$\\left( 11-a \\right)$$℃ "}]], "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": ["由题意知，上午$$9$$点的温度是$$a$$℃，  凌晨的温度是$$(a-12)$$℃，  中午$$12$$点的温度是$$(a-12)+20=(a+8)$$℃，  晚上$$9$$点的温度是$$(a+8)-19=(a-11)$$℃． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1092", "queId": "ff8080814d9539f1014d9b5c53a50a15", "competition_source_list": ["1992年第3届全国希望杯初一竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a+1\\textless{}0$$，则在下列每组四个数中，按从小到大的顺序排列的一组是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$，$$-1$$，$$1$$，$$-a$$ "}], [{"aoVal": "B", "content": "$$-a$$，$$-1$$，$$1$$，$$a$$ "}], [{"aoVal": "C", "content": "$$-1$$，$$-a$$，$$a$$，$$1$$ "}], [{"aoVal": "D", "content": "$$-1$$，$$a$$，$$1$$，$$-a$$ "}]], "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"], "answer_analysis": ["由$$a+1\\textless{}0$$，知$$a\\textless{}-1$$，所以$$-a\\textgreater1$$．  于是由小到大的排列次序应是$$a\\textless{}-1\\textless{}1\\textless{}-a$$，选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1044", "queId": "ff8080814d7978b9014d869e991024c5", "competition_source_list": ["1990年第1届全国希望杯初一竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "某工厂去年的生产总值比前年增长$$a \\%$$，则前年比去年少的百分数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a \\%$$ "}], [{"aoVal": "B", "content": "$$(1+a) \\%$$ "}], [{"aoVal": "C", "content": "$$\\frac{a+1}{100a}$$ "}], [{"aoVal": "D", "content": "$$\\frac{a}{100+a}$$ "}]], "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": ["设前年的生产总值是$$m$$，则去年的生产总值是$$m(1+\\frac{a}{100})$$，  前年比去年少$$m(1+\\frac{a}{100})-m=\\frac{ma}{100}$$，  这个产值差占去年的$$\\frac{\\frac{ma}{100}}{m(1+\\frac{a}{100})}=\\frac{a}{100+a}$$．应选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1480", "queId": "bd14190627cc4ac19646a2ef68672171", "competition_source_list": ["初一上学期单元测试《一元一次方程》第8题", "2017~2018学年湖南长沙雨花区湖南广益实验中学初二上学期开学考试第11题3分", "1998年第9届希望杯初一竞赛第4题", "2019~2020学年浙江温州鹿城区温州市实验中学初一下学期期中模拟第2题", "2019~2020学年9月浙江杭州下城区杭州观成中学初一上学期月考第22题4分", "2018~2019学年湖北武汉江夏区武汉外国语学校初一上学期期末第8题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知关于$$x$$的方程$$(3a+8b)x+7=0$$无解，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->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->能力->运算能力"], "answer_analysis": ["整理原方程得：$$(3a+8b)x=-7$$，  ∵原方程无解，  ∴$$3a+8b=0$$，  ∴$$a=-\\frac{8}{3}b$$，故$$a$$、$$b$$异号或都为$$0$$，  ∴$$ab\\leqslant 0$$，  ∴$$ab$$是非正数 . "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "168", "queId": "1435a526f1c146d0b12be724b89a1b93", "competition_source_list": ["1997年第8届希望杯初二竞赛第1试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "直角三角形的两个锐角的外角平分线所夹的锐角的大小是．", "answer_option_list": [[{"aoVal": "A", "content": "$$30{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$45{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$60{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$15{}^{}\\circ $$或$$75{}^{}\\circ $$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["两个外角分别等于其不相邻的锐角与直角之和，因此两个外角之和等于$$270{}^{}\\circ $$．  两个外角的角平分线的夹角$$=180{}^{}\\circ -\\frac{270{}^{}\\circ }{2}=45{}^{}\\circ $$，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": "290", "queId": "1e184a33ef304d79a3b5e271b4bbe280", "competition_source_list": ["2009年第20届希望杯初二竞赛第1试第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "计算：$$(11+4 \\sqrt{7})^{\\frac{3}{2}}+(11-4 \\sqrt{7})^{\\frac{3}{2}}$$，结果等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$58$$ "}], [{"aoVal": "B", "content": "$$38 \\sqrt{7}$$ "}], [{"aoVal": "C", "content": "$$24 \\sqrt{7}$$ "}], [{"aoVal": "D", "content": "$$32 \\sqrt{7}$$ "}]], "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": ["原式$$=(7+2\\times 2\\sqrt{7}+2^{2})^{\\dfrac{3}{2}}+(7-2\\times 2\\sqrt{7}+2^{2})^{\\dfrac{3}{2}}$$  $$=[(\\sqrt{7}+2)]^{\\dfrac{3}{2}}+[(\\sqrt{7}-2)^{2}]^{\\dfrac{3}{2}}$$  $$=(\\sqrt{7}+2)^{3}+(\\sqrt{7}-2)^{3}$$  $$=(\\sqrt{7})^{3}+3\\times (\\sqrt{7})^{2}\\times 2+3\\times\\sqrt{7}\\times 2^{2}+2^{3}+(\\sqrt{7})^{3}-3\\times(\\sqrt{7})^{2}\\times 2+3\\times \\sqrt{7}\\times 2^{2}-2^{3}$$  $$=2\\times 7\\sqrt{7}+6\\sqrt{7}\\times 4=38\\sqrt{7}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1176", "queId": "8aac49074e023206014e20e11bb365f9", "competition_source_list": ["1994年第5届全国希望杯初一竞赛复赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\frac{3+4+5+6+7}{5}=\\frac{1993+1994+1995+1996+1997}{N}$$，则$$N=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1991$$ "}], [{"aoVal": "B", "content": "$$1993$$ "}], [{"aoVal": "C", "content": "$$1995$$ "}], [{"aoVal": "D", "content": "$$1997$$ "}]], "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"], "answer_analysis": ["∵$$\\frac{3+4+5+6+7}{5}=\\frac{1993+1994+1995+1996+1997}{N}$$，  ∴$$\\frac{5\\times 5}{5}=\\frac{5\\times 1995}{N}\\Rightarrow N=1995$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "872", "queId": "60da1557d6be4313aebc721f4b549af5", "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，则不同的排法一共有．", "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": "200", "queId": "26377fb9fbca44c6a8ba6339f33a6afa", "competition_source_list": ["2004年第21届全国初中数学联赛竞赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$p$$，$$q$$均为质数，且满足$$5{{p}^{2}}+3q=59$$，则以$$p+3$$，$$1-p+q$$，$$2p+q-4$$为边长的三角形是（~ ~）．", "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": ["课内体系->能力->推理论证能力", "课内体系->知识点->三角形->勾股定理及应用->勾股定理基础->勾股定理", "课内体系->知识点->三角形->勾股定理及应用->勾股定理基础->勾股定理的逆定理", "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"], "answer_analysis": ["由方程知$$5{{p}^{2}} ~\\textless{} ~59$$，所以$$p ~\\textless{} ~4$$，又$$p$$为质数，所以$$p$$只可能是$$2$$或$$3$$，代回方程$$5{{p}^{2}}+3q=59$$中，因为$$q$$也是质数，只有当$$p=2$$时$$q=13$$为质数．  所以$$p=2$$，$$q=13$$．  由$$p$$，$$q$$的值得到三角形三边长为$$5$$，$$12$$，$$13$$，由勾股定理逆定理得到这是个直角三角形．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "893", "queId": "dae54674203f4731897cea8def84cac1", "competition_source_list": ["1994年第11届全国初中数学联赛竞赛第2题6分", "初一上学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$，$$b$$，$$c$$是不全相等的任意实数，若$$x={{a}^{2}}-bc$$，$$y={{b}^{2}}-ca$$，$$z={{c}^{2}}-ab$$，则$$x$$，$$y$$，$$z$$．", "answer_option_list": [[{"aoVal": "A", "content": "都不小于$$0$$ "}], [{"aoVal": "B", "content": "都不大于$$0$$ "}], [{"aoVal": "C", "content": "至少有一个小于$$0$$ "}], [{"aoVal": "D", "content": "至少有一个大于$$0$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数运算巧解->完全平方非负性", "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": ["$$x+y+z={{a}^{2}}-bc+{{b}^{2}}-ca+{{c}^{2}}-ab=\\frac{1}{2}[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}]\\textgreater0$$，  ∴$$x$$，$$y$$，$$z$$中至少有一个大于$$0$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1307", "queId": "8aac50a74fb1a33e014fb624e0f7193c", "competition_source_list": ["初一上学期其它", "1994年第11届全国初中数学联赛竞赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$，$$b$$，$$c$$是不全相等的任意实数，若$$x={{a}^{2}}-bc$$，$$y={{b}^{2}}-ca$$，$$z={{c}^{2}}-ab$$，则$$x$$，$$y$$，$$z$$（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "都不小于$$0$$ "}], [{"aoVal": "B", "content": "都不大于$$0$$ "}], [{"aoVal": "C", "content": "至少有一个小于$$0$$ "}], [{"aoVal": "D", "content": "至少有一个大于$$0$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->配方思想的运用", "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", "课内体系->能力->运算能力"], "answer_analysis": ["$$x+y+z={{a}^{2}}-bc+{{b}^{2}}-ca+{{c}^{2}}-ab=\\frac{1}{2}[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}]\\textgreater0$$，  ∴$$x$$，$$y$$，$$z$$中至少有一个大于$$0$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "577", "queId": "29bc5b0ade534d539a50476add626117", "competition_source_list": ["2022~2023学年浙江宁波初三月考（六校强基竞赛）第1题", "2022~2023学年浙江宁波初三月考（六校强基竞赛）第1题", "2021~2022学年黑龙江哈尔滨南岗区哈尔滨市松雷中学校初二上学期期中第4题"], "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": "$$(-2x)\\cdot \\left( -5{{x}^{2}}y \\right)=-10{{x}^{3}}y$$ "}], [{"aoVal": "D", "content": "$$(-2xy)\\cdot \\left( -3{{x}^{2}}y \\right)=6{{x}^{3}}y$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->整式的乘除运算", "课内体系->能力->运算能力"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据单项式乘单项式、同底数幂的运算法则逐一计算即可得出答案．\\\\ 【详解】\\\\ 解：A.$$3{{x}^{3}}\\cdot 2{{x}^{2}}y=6{{x}^{5}}y$$，计算错误，不符合题意；\\\\ B.$$2{{a}^{2}}\\cdot 3{{a}^{3}}=6{{a}^{5}}$$，计算正确，符合题意；\\\\ C.$$(-2x)\\cdot \\left( -5{{x}^{2}}y \\right)=10{{x}^{3}}y$$，计算错误，不符合题意；\\\\ D.$$(-2xy)\\cdot \\left( -3{{x}^{2}}y \\right)=6{{x}^{3}}{{y}^{2}}$$ ，计算错误，不符合题意；\\\\ 故选B．\\\\ 【点睛】\\\\ 本题考查了单项式乘单项式、同底数幂，熟练掌握运算法则是解题的关键． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "124", "queId": "2a4c10e5e7dd4dd9903377c9afc5512c", "competition_source_list": ["2019~2020学年4月四川绵阳游仙区绵阳中学英才学校初一下学期月考A卷第12题3分", "1999年第10届希望杯初二竞赛第1试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$x$$，$$y$$为实数，设$$a=\\dfrac{\\sqrt{{{\\left( -x \\right)}^{2}}}}{\\left\\textbar{} -x \\right\\textbar}$$，$$b=\\dfrac{y}{{{\\left( \\sqrt{-y} \\right)}^{2}}}$$，$$c=\\dfrac{3+\\dfrac{1}{4}}{4-\\dfrac{1}{2}}$$，则$$a$$，$$b$$，$$c$$的大小关系为．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless{}b\\textless{}c$$ "}], [{"aoVal": "B", "content": "$$b\\textless{}a\\textless{}c$$ "}], [{"aoVal": "C", "content": "$$b\\textless{}c\\textless{}a$$ "}], [{"aoVal": "D", "content": "$$a=b\\textgreater c$$ "}]], "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->数->实数->实数运算->含二次根式的实数的运算", "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": ["$$a=\\frac{\\sqrt{{{\\left( -x \\right)}^{2}}}}{\\left\\textbar{} -x \\right\\textbar}=\\left\\textbar{} \\frac{-x}{-x} \\right\\textbar=1$$，  $$b=\\frac{y}{{{\\left( \\sqrt{-y} \\right)}^{2}}}=\\frac{y}{-y}=-1$$，  $$c=\\dfrac{3+\\dfrac{1}{4}}{4-\\dfrac{1}{2}}=\\frac{13}{14}$$．  所以$$b\\textless{}c\\textless{}a$$，选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "281", "queId": "b4b47decafc74624a4a3944690b37dd9", "competition_source_list": ["2010年第21届全国希望杯初一竞赛复赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$m=\\frac{2009\\times 2010+2010\\times 2011}{2}$$，则$$m$$是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "奇数，且是完全平方数 "}], [{"aoVal": "B", "content": "偶数，且是完全平方数 "}], [{"aoVal": "C", "content": "奇数，但不是完全平方数 "}], [{"aoVal": "D", "content": "偶数，但不是完全平方数 "}]], "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"], "answer_analysis": ["$$m=\\frac{2009\\times 2010+2010\\times 2011}{2}$$  $$=\\frac{2010\\times (2009+2011)}{2}$$  $$=\\frac{2010\\times 2010\\times 2}{2}$$  $$={{2010}^{2}}$$．  故$$m$$是偶数，且是完全平方数． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "532", "queId": "59f6d03c61574aa58745e63081b0b43f", "competition_source_list": ["1998年第9届希望杯初二竞赛第1试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "设实数$$m$$，$$n$$满足$${{m}^{2}}{{n}^{2}}+{{m}^{2}}+{{n}^{2}}+10mn+16=0$$，则有．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\begin{cases}m=2,    n=2,    \\end{cases}$$或$$\\begin{cases}m=-2,    n=-2.    \\end{cases}$$ "}], [{"aoVal": "B", "content": "$$\\begin{cases}m=2,    n=2,    \\end{cases}$$或$$\\begin{cases}m=2,    n=-2.    \\end{cases}$$ "}], [{"aoVal": "C", "content": "$$\\begin{cases}m=2,    n=-2,    \\end{cases}$$或$$\\begin{cases}m=-2,    n=2.    \\end{cases}$$ "}], [{"aoVal": "D", "content": "$$\\begin{cases}m=-2,    n=-2,    \\end{cases}$$或$$\\begin{cases}m=-2,    n=2.    \\end{cases}$$ "}]], "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": ["因为$${{m}^{2}}{{n}^{2}}+{{m}^{2}}+{{n}^{2}}+10mn+16=0$$，  所以$$({{m}^{2}}{{n}^{2}}+8mn+16)+({{m}^{2}}+2mn+{{n}^{2}})=0$$，  所以$${{(mn+4)}^{2}}+{{(m+n)}^{2}}=0$$．  又因为$${{(mn+4)}^{2}}\\geqslant 0$$，$${{(m+n)}^{2}}\\geqslant 0$$，  所以$${{(mn+4)}^{2}}=0$$，$${{(m+n)}^{2}}=0$$，  即$$\\begin{cases}mn+4=0    m+n=0    \\end{cases}$$，  解得$$\\begin{cases}m=2,    n=-2,    \\end{cases}$$或$$\\begin{cases}m=-2,    n=2.    \\end{cases}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "341", "queId": "169a59f1e45c4d0bb42278b8803dded8", "competition_source_list": ["2005年第16届希望杯初一竞赛初赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$105$$的负约数的和等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$-105$$ "}], [{"aoVal": "B", "content": "$$-87$$ "}], [{"aoVal": "C", "content": "$$-86$$ "}], [{"aoVal": "D", "content": "$$-192$$ "}]], "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->有理数除法->有理数除法运算", "竞赛->知识点->数论->整除->因数与倍数"], "answer_analysis": ["因为$$105=1\\times 3\\times 5\\times 7$$，  所以$$105$$的负约数为$$-1$$，$$-3$$，$$-5$$，$$-7$$，$$-15$$，$$-21$$，$$-35$$，$$-105$$，  它们的和为$$-\\left( 1+3+5+7+15+21+35+105 \\right)=-192$$．  故选 $$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "749", "queId": "4e0a0a67d9a444f4ad3842fba7f314dc", "competition_source_list": ["2007年第18届希望杯初一竞赛复赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "对于彼此互质的三个正整数$$a$$，$$b$$，$$c$$，有以下判断：①$$a$$，$$b$$，$$c$$均为奇数；②$$a$$，$$b$$，$$c$$中必有一个偶数；③$$a$$，$$b$$，$$c$$没有公因数；④$$a$$，$$b$$，$$c$$必有公因数．  其中，不正确的判断的个数为．", "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->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->思想->分类讨论思想", "课内体系->能力->抽象概括能力", "竞赛->知识点->数论->整除->因数与倍数"], "answer_analysis": ["令$$a=2$$，$$b=3$$，$$c=5$$，知①不正确；  令$$a=7$$，$$b=3$$，$$c=5$$，知②不正确；  因为$$1$$显然是$$a$$，$$b$$，$$c$$的公因数，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": "94", "queId": "8aaeae6d6e8945e9a68a3b51b27a6d4f", "competition_source_list": ["1995年第12届全国初中数学联赛竞赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "设实数$$a$$，$$b$$满足不等式$$\\textbar\\textbar a\\textbar-(a+b)\\textbar{} ~\\textless{} ~\\textbar a-\\textbar a+b\\textbar\\textbar$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater0$$且$$b\\textgreater0$$ "}], [{"aoVal": "B", "content": "$$a ~\\textless{} ~0$$且$$b\\textgreater0$$ "}], [{"aoVal": "C", "content": "$$a\\textgreater0$$且$$b ~\\textless{} ~0$$ "}], [{"aoVal": "D", "content": "$$a ~\\textless{} ~0$$且$$b ~\\textless{} ~0$$ "}]], "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->解含绝对值的一元一次不等式（组）", "课内体系->能力->运算能力"], "answer_analysis": ["取$$a=-1$$，$$b=2$$可否定$$\\text{A}$$，$$\\text{C}$$，$$\\text{D}$$，选$$\\text{B}$$．  一般地，对已知不等式平方，有$$\\textbar a\\textbar(a+b)\\textgreater a\\textbar a+b\\textbar$$．  显然$$\\textbar a\\textbar\\textbar(a+b)\\textbar\\textgreater0$$（若等于$$0$$，则与上式矛盾），有$$\\frac{a+b}{\\textbar a+b\\textbar}\\textgreater\\frac{a}{\\textbar a\\textbar}$$．  两边都只能取$$1$$或$$-1$$，故只有$$1\\textgreater-1$$，即有$$a ~\\textless{} ~0$$且$$a+b\\textgreater0$$，从而$$b\\textgreater-a\\textgreater0$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "311", "queId": "74c8e2e6f76e4ec088be667e33863dbd", "competition_source_list": ["1991年第2届希望杯初二竞赛第9题"], "difficulty": "1", "qtype": "single_choice", "problem": "在整数$$0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$7$$，$$8$$，$$9$$中，设质数的个数为$$x$$，偶数的个数为$$y$$，完全平方数的个数为$$z$$，合数的个数为$$u$$．则$$x+y+z+u$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$17$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$11$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->整除->素数与合数"], "answer_analysis": ["因为$$x=4$$，$$y=5$$，$$z=4$$，$$u=4$$．  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": "922", "queId": "5ce49925bacf4bb498198760846ea5b4", "competition_source_list": ["2017年湖南长沙天心区初二竞赛长郡教育集团抵达杯第5题5分", "2020~2021学年浙江宁波鄞州区宁波市鄞州蓝青学校初三上学期期中（加试）第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一副扑克牌有$$4$$种花色，每种花色有$$13$$张，从中任意抽牌，最小要抽张才能保证有$$4$$张牌是同一花色的．", "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->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->不等式（组）->不等式组应用", "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->一元一次方程的其他实际问题"], "answer_analysis": ["$$4$$种花色相当于$$4$$个抽屉，  设最少要抽$$x$$张扑克，  问题相当于把$$x$$张扑克放进$$4$$个抽屉，至少有$$4$$张牌在同一个抽屉，有  $$x=3\\times 3+4=13$$，  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "607", "queId": "6806e7a7dca44f42aed91ef9f96bdfa9", "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->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->因式分解->提公因式法与公式法->题型：提公因式法", "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->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", "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->学习能力->运算能力", "知识标签->方法->作差法"], "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": "1345", "queId": "8aac50a75139269a01515e115b4d4186", "competition_source_list": ["2015年初一上学期其它", "1999年第10届希望杯初一竞赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "一个角的补角的$$\\frac{1}{17}$$是$$6{}^{}\\circ $$，则这个角是( ~ )．", "answer_option_list": [[{"aoVal": "A", "content": "$$68{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$78{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$88{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$98{}^{}\\circ $$ "}]], "knowledge_point_routes": ["课内体系->知识点->几何图形初步->角->角度的运算->角的和差的计算与证明->角的和差的计算与证明-不需要分类讨论", "课内体系->知识点->几何图形初步->角->角的定义和分类->余角和补角"], "answer_analysis": ["设这个角为$$\\alpha $$，$$\\alpha $$的补角等于$$180{}^{}\\circ -\\alpha $$，其$$\\frac{1}{17}$$为$$\\frac{180{}^{}\\circ -\\alpha }{17}$$，所以$$\\frac{180{}^{}\\circ -\\alpha }{17}=6{}^{}\\circ $$．解得$$\\alpha =78{}^{}\\circ $$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1099", "queId": "ff8080814d9efd56014da55422170742", "competition_source_list": ["1992年第3届全国希望杯初一竞赛复赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$${{8.047}^{3}}=521.077119823$$，则$${{0.8047}^{3}}$$等于（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0.521077119823$$ "}], [{"aoVal": "B", "content": "$$52.1077119823$$ "}], [{"aoVal": "C", "content": "$$571077.119823$$ "}], [{"aoVal": "D", "content": "$$0.00521077119823$$ "}]], "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": ["∵$${{(\\frac{a}{10})}^{3}}=\\frac{{{a}^{3}}}{{{10}^{3}}}$$，  ∴将$${{8.047}^{3}}=512.077119823$$的小数点向前移三位得$$0.512077119823$$，即为$${{0.8047}^{3}}$$的值，选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "701", "queId": "7f53c5256f5f421ab2ed0a8dde10f63d", "competition_source_list": ["2007年第18届希望杯初一竞赛初赛第9题4分", "初一上学期单元测试《一元一次方程》第17题"], "difficulty": "3", "qtype": "single_choice", "problem": "以$$x$$为未知数的方程$$2007x+2007a+2008b=0$$（$$a$$，$$b$$为有理数，且$$b\\textgreater0$$）有正整数解，则$$ab$$是．", "answer_option_list": [[{"aoVal": "A", "content": "负数 "}], [{"aoVal": "B", "content": "非负数 "}], [{"aoVal": "C", "content": "正数 "}], [{"aoVal": "D", "content": "零 "}]], "knowledge_point_routes": ["课内体系->能力->推理论证能力", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程的相关概念->由一元一次方程的解求参数的值", "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": ["方程两边同乘$$b$$得：  $$2007bx+2007ab+2008{{b}^{2}}=0$$，  ∴$$ab=-\\frac{2007bx+2008{{b}^{2}}}{2007}$$，  ∵$$b\\textgreater0$$，是$$x$$为正整数，  ∴$$ab\\textless{}0$$，为负数．  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": "1496", "queId": "afc1d0f1607f4e0eae5f0c082b67bcde", "competition_source_list": ["1997年第14届全国初中数学联赛竞赛第3题", "2020~2021学年浙江宁波奉化区奉化实验中学初二上学期期中第29题5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3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若实数$$a$$，$$b$$，$$c$$满足$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=9$$，代数式$${{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}$$的最大值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$27$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->不等式->柯西不等式"], "answer_analysis": ["$${{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}=3({{a}^{2}}+{{b}^{2}}+{{c}^{2}})-{{(a+b+c)}^{2}}\\leqslant 3({{a}^{2}}+{{b}^{2}}+{{c}^{2}})=27$$．  当且仅当$$a+b+c=0$$时成立．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1148", "queId": "a95113d43aba43fa965ce0e88a2b3676", "competition_source_list": ["2012年第23届全国希望杯初一竞赛初赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知有理数$$x$$满足方程$$\\frac{1}{2012-\\dfrac{x}{x-1}}=\\frac{1}{2012}$$，则$$\\frac{{{x}^{4}}-2009}{{{x}^{9}}+49}=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-41$$ "}], [{"aoVal": "B", "content": "$$-49$$ "}], [{"aoVal": "C", "content": "$$41$$ "}], [{"aoVal": "D", "content": "$$49$$ "}]], "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"], "answer_analysis": ["由有理数$$x$$满足方程$$\\frac{1}{2012-\\dfrac{x}{x-1}}=\\frac{1}{2012}$$，  易知$$x\\ne 1$$，$$\\frac{x}{x-1}\\ne 2012$$，即$$x\\ne \\frac{2012}{2011}$$，  整理，得$$\\frac{x}{x-1}=0$$，  因为$$x-1\\ne 0$$，  所以$$x=0$$．  所以$$\\frac{{{x}^{4}}-2009}{{{x}^{9}}+49}=\\frac{-2009}{49}=-41$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "930", "queId": "7330adf8ce694d1da128292d70c26059", "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->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": "971", "queId": "738fdebd35c847ec9e393b0390b06ca2", "competition_source_list": ["2012年第23届全国希望杯初一竞赛复赛第2题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若两位自然数$$\\overline{ab}$$是质数，且交换数字后的两位数$$\\overline{ba}$$也是质数，则称$$\\overline{ab}$$为绝对质数．于是两位数中的所有绝对质数的乘积的个位数字是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "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->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"], "answer_analysis": ["两位数中的绝对质数有$$11$$，$$13$$，$$17$$，$$31$$，$$37$$，$$71$$，$$73$$，$$79$$，$$97$$．  其乘积是$$11\\times 13\\times 17\\times 31\\times 37\\times 71\\times 73\\times 79\\times 97$$，  要求其个位数字，就是求$$1\\times 3\\times 7\\times 1\\times 7\\times 1\\times 3\\times 9\\times 7$$的个位数字，  易得，个位数字是$$3$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1430", "queId": "d334c7206746411da38f71bf3e32dc4f", "competition_source_list": ["2009年第20届希望杯初一竞赛第1试第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$${{x}^{2}}+x-2=0$$，则$${{x}^{3}}+2{{x}^{2}}-x+2007$$=．", "answer_option_list": [[{"aoVal": "A", "content": "$$2009$$ "}], [{"aoVal": "B", "content": "$$2008$$ "}], [{"aoVal": "C", "content": "$$-2008$$ "}], [{"aoVal": "D", "content": "$$-2009$$ "}]], "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->整式加减化简求值->整式加减化简求值-降次化简求值"], "answer_analysis": ["$${{x}^{2}}+x-2=0$$，则$${{x}^{2}}+x=2$$，  ∴$${{x}^{3}}+2{{x}^{2}}-x+2007$$  $$={{x}^{3}}+{{x}^{2}}+{{x}^{2}}+x-2x+2007$$  $$=x\\left( {{x}^{2}}+x \\right)+\\left( {{x}^{2}}+x \\right)-2x+2007$$  $$=2x+2-2x+2007$$  $$=2009$$．  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": "753", "queId": "49a1f3d1c237445386368e3a7e6b4784", "competition_source_list": ["2014年第25届全国希望杯初二竞赛复赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "无论$$m$$为何实数，直线$$y=x-m$$与直线$$y=-2x+3$$的交点都不可能在（~ ）．", "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->推理论证能力", "课内体系->知识点->函数->一次函数->一次函数基础->一次函数的图象特征与性质"], "answer_analysis": ["直线$$y=x-m$$与直线$$y=-2x+3$$的交点必在直线$$y=-2x+3$$上，  而直线$$y=-2x+3$$不经过第三象限，  所以直线$$y=x-m$$与直线$$y=-2x+3$$的交点不可能在第三象限． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1336", "queId": "8aac50a7511483070151192cbc3a0f52", "competition_source_list": ["2015年第26届全国希望杯初三竞赛初赛（特）第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "两个标准的骰子各被投掷一次，以骰子向上一面的数字和作为圆的直径．则圆的面积数值小于圆的周长数值的概率是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{36}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{12}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{6}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率->概率的计算方法"], "answer_analysis": ["圆的面积数值小于圆的周长数值，  即$$\\pi{{r}^{2}}\\textless{}2\\pi r$$，  即$$0\\textless{}r\\textless{}2$$．  ∴直径$$0\\textless{}d\\textless{}4$$．  骰子向上一面的数字和作为圆的直径，  ∴直径$$d\\textless{}4$$的情况有$$1$$，$$2$$或$$1$$，$$1$$或$$2$$，$$1$$三种，  故概率$$=\\frac{3}{36}=\\frac{1}{12}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "866", "queId": "69b5044a6bf9408097de3494935a4fd1", "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->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", "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->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$${{M}_{\\max }}=\\frac{3}{4}$$． ", "<p>令$$y=1-x-z$$，代入$$M$$，则：</p>\n<p>$$M=x\\left( 1-x-z \\right)+2\\left( 1-x-z \\right)z+3xz=x-{{x}^{2}}-xz+2z-2xz-2{{z}^{2}}+3xz$$</p>\n<p>$$=-{{x}^{2}}+x-2{{z}^{2}}+2z=-{{\\left( x-\\frac{1}{2} \\right)}^{2}}-2{{\\left( z-\\frac{1}{2} \\right)}^{2}}+\\frac{3}{4}$$，$${{M}_{\\max }}=\\frac{3}{4}$$．</p>\n<p>故选$$\\text{C}$$．</p>"], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "133", "queId": "f0e3d3fd27954a229ce48ae7c80f1e32", "competition_source_list": ["2018~2019学年山东济南市中区初二上学期期末第28题5分", "2001年第12届希望杯初二竞赛第2试第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "平面内有$$n$$条直线$$(n\\geqslant 2)$$，这$$n$$条直线两两相交，最多可以得到$$a$$个交点，最少可以得到$$b$$个交点，则$$a+b$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$n(n-1)$$ "}], [{"aoVal": "B", "content": "$${{n}^{2}}-n+1$$ "}], [{"aoVal": "C", "content": "$$\\frac{{{n}^{2}}-n}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{{{n}^{2}}-n+2}{2}$$ "}]], "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->相交线的交点问题找规律"], "answer_analysis": ["$$a=1+2+\\cdots +n-1$$  $$=(1+n-1)\\times \\frac{n-1-1+1}{2}$$  $$=\\frac{n(n-1)}{2}$$，$$b=1$$，  $$a+b=\\frac{n(n-1)}{2}+1=\\frac{{{n}^{2}}-n+2}{2}$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "636", "queId": "6cc3c4c00f534560b4720a6abefa537a", "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": ["\\textbf{（知识点：二次根式化简求值）}  $$1+{{x}^{2}}=1+\\frac{1}{4}\\left( {{1991}^{\\frac{2}{n}}}-2+{{1991}^{-\\frac{2}{n}}} \\right)=\\frac{1}{4}{{\\left( {{1991}^{\\frac{1}{n}}}+{{1991}^{-\\frac{1}{n}}} \\right)}^{2}}$$，  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": "1372", "queId": "fc12344875164cfc97d2a7fd090c5ade", "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？  Arrange prime numbers within $$100$$ in a string of numbers from smallest to largest, and finishing the following five tasks in order is called one operation:  ($$1$$) move the first digit on the left to the rightmost of the number string;  ($$2$$) form several two-digit numbers from the left to the right with two digits in one section;  ($$3$$) cross out the composite numbers from these two-digit numbers;  ($$4$$) if there are identical two-digit prime numbers left, keep the one on the left and cross out the rest;  ($$5$$) the remaining two-digit prime numbers are kept in the numerical order and formed into a new number string.  Q: After $$1997$$ operations, what is the final string of numbers?", "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$$1731$$．  The $$1$$st operation yields the number string $$711131737$$; the $$2$$nd operation yields the number string $$11133173$$; the $$3$$th operation yields the number string $$111731$$; the $$4$$th operation yields the number string $$1173$$; the $$5$$th operation yields the number string $$1731$$; the $$6$$th operation yields the number string $$7311$$; the $$7$$th operation yields the number string $$3117$$; the $$8$$ operation yields the number string $$1173$$. The following is a cycle of $$4$$, that is, the string of numbers obtained in $$4k$$ operations is $$1173$$. $$1996=4\\times 499$$, so the $$1996$$th operation yields the number string $$1173$$, so the $$1997$$th operation yields the number string $$1731$$. "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "297", "queId": "5d9f2878bebc4ba3ab0f5c6a141e8d1e", "competition_source_list": ["1991年第2届希望杯初二竞赛第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$11$$个连续奇数的和是$$1991$$，把这些数按大小顺序排列来，第六个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$179$$ "}], [{"aoVal": "B", "content": "$$181$$ "}], [{"aoVal": "C", "content": "$$183$$ "}], [{"aoVal": "D", "content": "$$185$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论->同余->奇数与偶数"], "answer_analysis": ["设这$$11$$个连续奇数为：$$2n+1$$，$$2n+3$$，$$2n+5$$，\\ldots，$$2n+21$$．  则$$\\left( {2n + 1} \\right) + \\left( {2n + 3} \\right) + \\left( {2n + 5} \\right) + ~\\cdots ~+ \\left( {2n + 21} \\right) = 1991$$，  即$$11 \\times \\left( {2n + 11} \\right) = 1991$$，解得$$n=85$$．  所以第六个数是$$2\\times 85+11=181$$，  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "669", "queId": "d123d5f0da924f708a770573beee626e", "competition_source_list": ["2000年竞赛第24题"], "difficulty": "1", "qtype": "single_choice", "problem": "2000年$$AMC8$$竞赛第$$24$$题  If $$a$$ and $$b$$ are integers, what is the least integer $$x$$ for which the inequality $$a ~\\textless{} ~b ~\\textless{} ~a+x$$ could be true．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "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）->不等式组应用->不等式组的其他实际问题"], "answer_analysis": ["Since $$a$$ and $$b$$ are integers, the least possible $$b=a$$$$+1$$.~ Since $$x$$ is also an integer, the least possible $$a+x=$$$$b+1$$ $$=a$$$$+2$$.  如果$$a$$，$$b$$是整数 ， 那么满足不等式$$\\textasciitilde a ~\\textless{} ~b ~\\textless{} ~a+x$$的最小的整数$$x$$是多少．  $$\\text{A}$$．$$-1$$~ ~ $$\\text{B}$$． $$0$$~ ~ $$\\text{C}$$．$$1$$~ ~$$\\text{D}$$．$$2$$  因为$$a$$，$$b$$是整数 ， 那么最小的可能就是$$b=a+1$$．因为$$x$$是一个整数 ， 那么最小值就是$$\\textasciitilde a+x=b+1=a+2$$， 也就是 $$x=0$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "879", "queId": "580c3519f2d44275a1650b85986e2081", "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": "992", "queId": "9b44657317774ec7a7b37ea6435d0b19", "competition_source_list": ["2019~2020学年广东广州越秀区广州市第二中学初一下学期单元测试《相交线与平行线》第7题4分", "2012年第23届全国希望杯初一竞赛初赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "某人在练车场上练习驾驶汽车，两次拐弯后的行驶方向与原来的方向相反，则这两次拐弯的角度可能是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "第一次向左拐$$40{}^{}\\circ $$，第二次向右拐$$40{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "第一次向右拐$$50{}^{}\\circ $$，第二次向左拐$$130{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "第一次向右拐$$70{}^{}\\circ $$，第二次向左拐$$110{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "第一次向左拐$$70{}^{}\\circ $$，第二次向左拐$$110{}^{}\\circ $$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->有理数->正数和负数->相反意义的量"], "answer_analysis": ["记向左为``$$+$$''，向右为``$$-$$''，则题中四个选项依次为  $$+40{}^{}\\circ -40{}^{}\\circ =0{}^{}\\circ $$，方向不变；  $$-50{}^{}\\circ +130{}^{}\\circ =80{}^{}\\circ $$；  $$-70{}^{}\\circ +110{}^{}\\circ =40{}^{}\\circ $$；  $$+70{}^{}\\circ +110{}^{}\\circ =180{}^{}\\circ $$，方向相反．  故$$\\text{D}$$正确． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "587", "queId": "5a402e2af2e6421d8a3128c8f3fc96d7", "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->思想->整体思想", "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->方法->代入法", "课内体系->能力->运算能力"], "answer_analysis": ["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$$\\left[ x \\right]=-2$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1412", "queId": "b7f9a75ab7df49278fb864503e177da9", "competition_source_list": ["2011年竞赛第3题4分", "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"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$A$$，$$B$$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": ["知识标签->学习能力->推理论证能力", "知识标签->题型->三角形->锐角三角函数及解直角三角形->锐角三角函数->题型：正弦定理与余弦定理", "知识标签->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", "知识标签->知识点->方程与不等式->一元二次方程->一元二次方程的解", "知识标签->知识点->方程与不等式->一元二次方程->解一元二次方程->因式分解法解一元二次方程", "知识标签->数学思想->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": "483", "queId": "28bf6cd14f4e460785683a68b1e35d60", "competition_source_list": ["2002年第13届希望杯初二竞赛第1试第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$m=\\frac{4-x}{3}$$，$$n=\\frac{x+3}{4}$$，$$p=\\frac{2-3x}{5}$$，且$$m\\textgreater n\\textgreater p$$，那么$$x$$的取值范围是（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$x ~\\textless{} ~1$$ "}], [{"aoVal": "B", "content": "$$-\\frac{14}{5} ~\\textless{} ~x ~\\textless{} ~1$$ "}], [{"aoVal": "C", "content": "$$-\\frac{7}{17} ~\\textless{} ~x ~\\textless{} ~1$$ "}], [{"aoVal": "D", "content": "$$-\\frac{14}{5} ~\\textless{} ~x ~\\textless{} ~-\\frac{7}{17}$$ "}]], "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"], "answer_analysis": ["$$m\\textgreater n\\textgreater p$$，即$$\\frac{4-x}{3}\\textgreater\\frac{x+3}{4}\\textgreater\\frac{2-3x}{5}$$．  所以$$\\begin{cases}\\frac{4-x}{3}\\textgreater\\frac{x+3}{4}    \\frac{x+3}{4}\\textgreater\\frac{2-3x}{5}   \\end{cases}$$，  即$$\\begin{cases}16-4x\\textgreater3x+9    5x+15\\textgreater8-12x   \\end{cases}$$，  解得$$\\begin{cases}x ~\\textless{} ~1    x\\textgreater-\\frac{7}{17}   \\end{cases}$$，即$$-\\frac{7}{17} ~\\textless{} ~x ~\\textless{} ~1$$，  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "75", "queId": "fa2bca3a5a9b4c659764955978d9d111", "competition_source_list": ["1991年第2届全国希望杯初一竞赛复赛第17题", "2019~2020学年北京西城区北京四中初一上学期开学考试第9题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "【补10】  计算：$98^{2}+2^{2}$~=~\\uline{~~~~~~~~~~}~  马马虎的做法如下：  原式=$\\left( 98+2\\right)^{2}+2\\times98\\times2$  马马虎的解法正确吗？请你帮他挑挑错吧！", "answer_option_list": [[{"aoVal": "A", "content": "$$9608$$ "}], [{"aoVal": "B", "content": "$$9610$$ "}], [{"aoVal": "C", "content": "$$9708$$ "}], [{"aoVal": "D", "content": "$$9710$$ "}]], "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": ["符号错了，应为减法：  原式=$\\left( 98+2\\right)^{2}-2\\times98\\times2$ "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "303", "queId": "228e973c3976483986a0ba0150303329", "competition_source_list": ["2001年竞赛（全国初中数学竞赛）第2题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$$a$$，$$b$$，$$c$$是三个任意整数，那么$$\\frac{a+b}{2}$$，$$\\frac{a+c}{2}$$，$$\\frac{b+c}{2}$$．", "answer_option_list": [[{"aoVal": "A", "content": "都不是整数 "}], [{"aoVal": "B", "content": "至少有两个整数 "}], [{"aoVal": "C", "content": "至少有一个整数 "}], [{"aoVal": "D", "content": "都是整数 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题", "课内体系->知识点->数->有理数"], "answer_analysis": ["因为$$a$$，$$b$$，$$c$$是任意整数，所以$$a$$，$$b$$，$$c$$中至少有两个同为奇数或同为偶数，而两个奇数或两个偶数的和是偶数，所以$$\\frac{a+b}{2}$$，$$\\frac{a+c}{2}$$，$$\\frac{b+c}{2}$$中至少有一个是整数，如当$$a=1$$，$$b=2$$ ，$$c=3$$时，$$\\frac{a+c}{2}=2$$，$$\\frac{a+b}{2}=\\frac{3}{2}$$，$$\\frac{b+c}{2}=\\frac{5}{2}$$．此时，恰有一个整数．  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": "1236", "queId": "8aac4907507fb8840150804b77e5018d", "competition_source_list": ["1997年第8届全国希望杯初一竞赛复赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a\\textless{}-b$$，且$$\\frac{a}{b}\\textgreater0$$，则$$\\left\\textbar{} a \\right\\textbar-\\left\\textbar{} b \\right\\textbar+\\left\\textbar{} a+b \\right\\textbar+\\left\\textbar{} ab \\right\\textbar$$等于（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2a+2b+ab$$ "}], [{"aoVal": "B", "content": "$$-ab$$ "}], [{"aoVal": "C", "content": "$$-2a-2b+ab$$ "}], [{"aoVal": "D", "content": "$$-2a+ab$$ "}]], "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->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["∵$$\\frac{a}{b}\\textgreater0$$，  ∴$$a$$和$$b$$同号，  又∵$$a\\textless{}-b$$，  ∴$$a$$和$$b$$均为负数．  ∴$$\\left\\textbar{} a \\right\\textbar-\\left\\textbar{} b \\right\\textbar+\\left\\textbar{} a+b \\right\\textbar+\\left\\textbar{} ab \\right\\textbar$$  $$=-a-\\left( -b \\right)+\\left[ -\\left( a+b \\right) \\right]+ab$$  $$=-a+b-a-b+ab$$  $$=-2a+ab$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1348", "queId": "e989734a6bc346a09cc1146be97bda29", "competition_source_list": ["2012年第23届全国希望杯初二竞赛初赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "方程组$$\\left { \\begin{matrix}x+y+z=10    3x+y-z=50    2x+y=40   \\end{matrix} \\right.$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "无解 "}], [{"aoVal": "B", "content": "有$$1$$组解 "}], [{"aoVal": "C", "content": "有$$2$$组解 "}], [{"aoVal": "D", "content": "有无穷多组解 "}]], "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->能力->推理论证能力"], "answer_analysis": ["由$$x+y+z=10$$，$$3x+y-z=50$$两式相加，  得$$4x+2y=60$$，即$$2x+y=30$$，  与原方程组的$$2x+y=40$$矛盾，  所以原方程组无解． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "434", "queId": "d9fc02a7196c47bba01aa74235d7c23e", "competition_source_list": ["1991年第2届希望杯初二竞赛第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$D$$是等腰锐角三角形$$ABC$$的底边$$BC$$上的一点，则$$AD$$，$$BD$$，$$CD$$ 满足关系式．", "answer_option_list": [[{"aoVal": "A", "content": "$$A{{D}^{2}}=B{{D}^{2}}+C{{D}^{2}}$$ "}], [{"aoVal": "B", "content": "$$A{{D}^{2}}\\textgreater B{{D}^{2}}+C{{D}^{2}}$$ "}], [{"aoVal": "C", "content": "$$2A{{D}^{2}}=B{{D}^{2}}+C{{D}^{2}}$$ "}], [{"aoVal": "D", "content": "$$2A{{D}^{2}}\\textgreater B{{D}^{2}}+C{{D}^{2}}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->等腰三角形->等腰三角形的性质->等腰三角形的性质-等边对等角"], "answer_analysis": ["以等边三角形为例，当$$D$$为$$BC$$边上的中点时，有$$A{{D}^{2}}\\textgreater B{{D}^{2}}+C{{D}^{2}}$$，当$$D$$为$$BC$$边的端点时，有$$A{{D}^{2}}=B{{D}^{2}}+C{{D}^{2}}$$，故有$$2A{{D}^{2}}\\textgreater B{{D}^{2}}+C{{D}^{2}}$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "689", "queId": "3bfce26405cb419686abb36d7fcfee22", "competition_source_list": ["2015~2016学年山东济南商河县胡集中学初二上学期期中第7题3分", "1992年第3届希望杯初二竞赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$a=\\frac{1}{2}(\\sqrt{5}+\\sqrt{3})$$，$$b=\\frac{1}{2}(\\sqrt{5}-\\sqrt{3})$$，那么$${{a}^{2}}-ab+{{b}^{2}}$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{7}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{9}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{11}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{15}}{2}-1$$ "}]], "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->整式乘除化简求值->整式乘除化简求值-整体代入化简求值"], "answer_analysis": ["由已知得$$a+b=\\sqrt{5}$$，$$ab=\\frac{1}{2}$$，  ∴$${{a}^{2}}-ab+{{b}^{2}}={{(a+b)}^{2}}-3ab$$，  $$={{(\\sqrt{5})}^{2}}-3\\times \\frac{1}{2}=\\frac{7}{2}$$，  故选：$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1422", "queId": "a19d43a3ef3b458c86a7735a129cdf18", "competition_source_list": ["2008年第19届希望杯初一竞赛第2试第1题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$a$$是最大的负整数，$$b$$是绝对值最小的有理数，则$${{a}^{2007}}+\\frac{{{b}^{2009}}}{2008}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2008}$$ "}], [{"aoVal": "D", "content": "$$2007$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["依题意知$$a=-1$$，$$b=0$$，所以$${{a}^{2007}}+\\frac{{{b}^{2009}}}{2008}={{\\left( -1 \\right)}^{2007}}+\\frac{0}{2008}=-1$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1578", "queId": "ebb2e22c60064229b10f77114b08f7bc", "competition_source_list": ["2007年第18届希望杯初一竞赛复赛第18题4分"], "difficulty": "4", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$都是质数，且$$a+b+c+abc=99$$，则$$\\left\\textbar{} \\frac{1}{a}-\\frac{1}{b} \\right\\textbar+\\left\\textbar{} \\frac{1}{b}-\\frac{1}{c} \\right\\textbar+\\left\\textbar{} \\frac{1}{c}-\\frac{1}{a} \\right\\textbar=$$ ．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{15}{17}$$ "}], [{"aoVal": "B", "content": "$$\\frac{15}{34}$$ "}], [{"aoVal": "C", "content": "$$\\frac{17}{19}$$ "}], [{"aoVal": "D", "content": "$$\\frac{17}{38}$$ "}]], "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->绝对值综合"], "answer_analysis": ["若$$a$$，$$b$$，$$c$$都是奇数，那么$$abc$$也为奇数，则$$a+b+c+abc$$为偶数，与$$a+b+c+abc=99$$矛盾．  ∴$$a$$，$$b$$，$$c$$中必有一个偶数，  又∵$$a$$，$$b$$，$$c$$都是质数，  ∴$$a$$，$$b$$，$$c$$中必有一个偶数是$$2$$，  令$$a=2$$，则$$b+c+2bc=97$$，  同理，若$$b$$，$$c$$都是奇数，则$$bc$$为奇数，则$$b+c+2bc$$为偶数，与$$b+c+2bc=97$$矛盾，  ∴$$b$$，$$c$$中也必有一个偶数，则偶数必是$$2$$，  令$$b=2$$，可得$$c=19$$，  ∴$$\\left\\textbar{} \\frac{1}{a}-\\frac{1}{b} \\right\\textbar+\\left\\textbar{} \\frac{1}{b}-\\frac{1}{c} \\right\\textbar+\\left\\textbar{} \\frac{1}{c}-\\frac{1}{a} \\right\\textbar=\\frac{17}{19}$$．  故答案为：$$\\frac{17}{19}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1247", "queId": "a9a9573a44a44102afdc020f7a7a213c", "competition_source_list": ["2004年第15届希望杯初二竞赛第1试第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\triangle ABC$$的三个内角$$\\angle A$$、$$\\angle B$$、$$\\angle C$$，满足$$3\\angle A\\textgreater5\\angle B$$，$$3\\angle C\\leqslant 2\\angle B$$，则这个三角形是．", "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": "等边三角形 "}]], "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->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"], "answer_analysis": ["由已知得$$\\angle B ~\\textless{} ~\\frac{3}{5}\\angle A$$，$$\\angle C\\leqslant \\frac{2}{3}\\angle B ~\\textless{} ~\\frac{2}{5}\\angle A$$，所以$$\\angle B+\\angle C ~\\textless{} ~\\angle A$$，$$\\triangle ABC$$是钝角三角形，选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1237", "queId": "8aac4907507fb8840150897542fb17b7", "competition_source_list": ["1999年第10届希望杯初一竞赛第9题", "2017~2018学年四川凉山西昌市西昌一中俊波外国语学校初一上学期期中第14题2分", "初一下学期单元测试《二元一次方程组》二元一次方程（组）的相关概念第24题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$\\left\\textbar{} a+b+1 \\right\\textbar$$与$${{(a-b+1)}^{2}}$$互为相反数，则$$a$$与$$b$$的大小关系是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b$$ "}], [{"aoVal": "B", "content": "$$a=b$$ "}], [{"aoVal": "C", "content": "$$a\\textless{}b$$ "}], [{"aoVal": "D", "content": "$$a\\geqslant b$$ "}]], "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"], "answer_analysis": ["$$\\left\\textbar{} a+b+1 \\right\\textbar$$与$${{(a-b+1)}^{2}}$$互为相反数，  由绝对值和偶次方的非负性知，  $$a+b+1=0$$，$$a-b+1=0$$，  解得$$a=-1$$，$$b=0$$，  ∴$$a\\textless{}b$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "547", "queId": "4c82cc756ac94db2b690c0193962ef0b", "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", "课内体系->能力->运算能力"], "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": "686", "queId": "4d81cf7b3dca453bbe9f34ec13fe44ae", "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": "1020", "queId": "ff8080814d4b1928014d4c1a10670630", "competition_source_list": ["2015年第26届全国希望杯初一竞赛初赛第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$x=2$$时，多项式$$m{{x}^{3}}-x+5m+3$$的值是$$118$$，则多项式$${{m}^{2}}-6m-7$$的值为（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-16$$ "}], [{"aoVal": "B", "content": "$$-7$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$93$$ "}]], "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->常规一元一次方程解法", "课内体系->知识点->式->整式的加减->整式的加减运算"], "answer_analysis": ["∵当$$x=2$$时，多项式$$m{{x}^{3}}-x+5m+3$$的值是$$118$$，  ∴$$8m-2+5m+3=118$$，解得$$m=9$$．  ∴$${{m}^{2}}-6m-7=20$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "875", "queId": "5c6cdcd6ae184447b9e2058ad63c3078", "competition_source_list": ["初三上学期其它", "初一单元测试《分解方法的延拓（2）》第21题", "2005年竞赛第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$M=3{{x}^{2}}-8xy+9{{y}^{2}}-4x+6y+13$$（$$x$$，$$y$$是实数），则$$M$$的值一定是．", "answer_option_list": [[{"aoVal": "A", "content": "正数 "}], [{"aoVal": "B", "content": "负数 "}], [{"aoVal": "C", "content": "零 "}], [{"aoVal": "D", "content": "非负数 "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->配方求最值", "课内体系->能力->运算能力"], "answer_analysis": ["$$M=3{{x}^{2}}-8xy+9{{y}^{2}}-4x+6y+13$$  $$=(2{{x}^{2}}-8xy+8{{y}^{2}})+({{x}^{2}}-4x+4)+({{y}^{2}}+6y+9)$$  $$=2{{(x-2y)}^{2}}+{{(x-2)}^{2}}+{{(y+3)}^{2}}$$，  ∵$${{(x-2y)}^{2}}\\geqslant 0$$，$${{(x-2)}^{2}}\\geqslant 0$$，$${{(y+3)}^{2}}\\geqslant 0$$，  又∵$$x-2y$$，$$x-2$$，$$y+3$$三项不能同时取$$0$$，  ∴$$M=2{{(x-2y)}^{2}}+{{(x-2)}^{2}}+{{(y+3)}^{2}}\\textgreater0$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "439", "queId": "79ca107a5f144a03a0aa8f7fa3f08af7", "competition_source_list": ["2013年第24届全国希望杯初二竞赛复赛第17题4分", "初一下学期单元测试《坐标平面上的直线》第24题"], "difficulty": "2", "qtype": "single_choice", "problem": "直线$$y=x-1$$与$$x$$轴、$$y$$轴分别交于$$A$$，$$B$$两点，点$$C$$在坐标轴上，$$\\triangle ABC$$是等腰三角形，则满足条件的点$$C$$有个．", "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->知识点->函数->一次函数->一次函数与几何综合->一次函数与等腰三角形结合", "课内体系->思想->分类讨论思想"], "answer_analysis": ["$$C\\left( 0,0 \\right)$$、$$C\\left( 0,1 \\right)$$、$$C\\left( 0,-1-\\sqrt{2} \\right)$$、$$C\\left( 1+\\sqrt{2},0 \\right)$$、$$C\\left( 1-\\sqrt{2},0 \\right)$$、$$C\\left( -1,0 \\right)$$、$$C\\left( 0,\\sqrt{2}-1 \\right)$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "261", "queId": "4fb2df3194bb460a948dc2a4c964d3f1", "competition_source_list": ["2004年第15届希望杯初二竞赛第1试第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "代数式$$\\frac{1}{x-1}+\\frac{1}{x+1}+\\frac{2x}{{{x}^{2}}+1}+\\frac{4{{x}^{3}}}{{{x}^{4}}+1}$$的化简结果是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{8{{x}^{5}}}{{{x}^{8}}-1}$$ "}], [{"aoVal": "B", "content": "$$\\frac{8{{x}^{4}}}{{{x}^{8}}-1}$$ "}], [{"aoVal": "C", "content": "$$\\frac{4{{x}^{7}}}{{{x}^{8}}-1}$$ "}], [{"aoVal": "D", "content": "$$\\frac{8{{x}^{7}}}{{{x}^{8}}-1}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->分式->分式的基本运算"], "answer_analysis": ["$$\\frac{1}{x-1}+\\frac{1}{x+1}+\\frac{2x}{{{x}^{2}}+1}+\\frac{4{{x}^{3}}}{{{x}^{4}}+1}$$  $$=\\frac{2x}{{{x}^{2}}-1}+\\frac{2x}{{{x}^{2}}+1}+\\frac{4{{x}^{3}}}{{{x}^{4}}+1}$$  $$=\\frac{4{{x}^{3}}}{{{x}^{4}}-1}+\\frac{4{{x}^{3}}}{{{x}^{4}}+1}$$  $$=\\frac{8{{x}^{7}}}{{{x}^{8}}-1}$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1418", "queId": "e0beacc5af76481585c79436ecd2a9d0", "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2题3分", "2018~2019学年浙江嘉兴初一上学期期末第7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->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->同类项的定义", "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"], "answer_analysis": ["由题意，得：  $$2m=2$$，$$n-1=1$$．  ∴$$m=1$$，$$n=2$$．  ∴$$m-2n=1-2\\times 2=-3$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "368", "queId": "54af56bb73774292a990f051d1f628c1", "competition_source_list": ["1996年第13届全国初中数学联赛竞赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果一个三角形的面积和周长都被一直线平分，那么该直线必通过这个三角形的（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "内心 "}], [{"aoVal": "B", "content": "外心 "}], [{"aoVal": "C", "content": "重心 "}], [{"aoVal": "D", "content": "垂心 "}]], "knowledge_point_routes": ["课内体系->知识点->圆->圆与多边形->三角形五心", "课内体系->知识点->圆->与圆有关的位置关系->直线与圆->三角形内切圆"], "answer_analysis": ["设直线$$DE$$平分$${\\triangle }ABC$$的周长和面积，$$D$$，$$E$$分别在边$$AB$$和$$AC$$上，作$$\\angle A$$的平分线交$$DE$$于$$P$$，记$$P$$到$$AB$$，$$AC$$的距离为$$r$$，$$P$$到$$BC$$的距离为$${{r}_{1}}$$，于是依题意有：  $$\\begin{cases}AD+AE=DB+BC+CE    \\dfrac{r}{2}(AD+AE)=\\dfrac{r}{2}(DB+CE)+\\dfrac{{{r}_{1}}}{2}BC \\end{cases}$$，  解得$$r={{r}_{1}}$$，即$$P$$为$${\\triangle }ABC$$的内心，  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1287", "queId": "8ecb03ec4e884f6ea75064cb4298f84f", "competition_source_list": ["2011年竞赛第2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第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "对于任意实数$$a$$，$$b$$，$$c$$，$$d$$，定义有序实数对$$\\left( a,b \\right)$$与$$\\left( c,d \\right)$$之间的运算``$$\\Delta $$''为：$$\\left( a,b \\right)\\Delta \\left( c,d \\right)=\\left( ac+bd,ad+bc \\right)$$，如果对于任意实数$$u$$，$$v$$，都有$$\\left( u,v \\right)\\Delta \\left( x,y \\right)=\\left( u,v \\right)$$，那么$$\\left( x,y \\right)$$为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 0,1 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 1,0 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( -1,0 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( 0,-1 \\right)$$ "}]], "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->整式的加减->整式的加减运算", "课内体系->能力->运算能力"], "answer_analysis": ["依照定义的运算法则，有$$\\begin{cases}ux+vy=u    vx+uy=v    \\end{cases}$$，即$$\\begin{cases}u(x-1)+vy=0    v(x-1)+uy=0    \\end{cases}$$，对任何实数$$u$$，$$v$$都成立．由于实数$$u$$，$$v$$的任意性，得$$(x,y)=(1,0)$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1342", "queId": "b2fa3bc4924743beab96eed388b014ee", "competition_source_list": ["2018~2019学年湖南株洲荷塘区初一下学期期末第10题4分", "初一单元测试《一次方程组的应用》第22题", "2019~2020学年浙江杭州杭州第十三中学初一下学期单元测试《二元一次方程组》（杭州市十三中教育集团）第17题3分", "2005年第16届希望杯初二竞赛复赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "购买铅笔$$7$$支，作业本$$3$$个，圆珠笔$$1$$支共需$$3$$元；购买铅笔$$10$$支，作业本$$4$$个，圆珠笔$$1$$支共需$$4$$元，则购买铅笔$$11$$支，作业本$$5$$个，圆珠笔$$2$$支共需（  ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4.5$$元 "}], [{"aoVal": "B", "content": "$$5$$元 "}], [{"aoVal": "C", "content": "$$6$$元 "}], [{"aoVal": "D", "content": "$$6.5$$元 "}]], "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->整式的加减->整式有关的概念->代数式->列代数式", "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": ["铅笔、作业本、圆珠笔的单价分别为$$a$$，$$b$$，$$c$$，  则$$7a+3b+c=3$$，$$10a+4b+c=4$$，  于是$$11a+5b+2c$$$$=3\\left( 7a+3b+c \\right)-\\left( 10a+4b+c \\right)$$$$=5$$．  ~ "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1112", "queId": "ff8080814d9efd56014daa73e3160a8f", "competition_source_list": ["1993年第4届全国希望杯初一竞赛初赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "如果有理数$$a$$，$$b$$满足$$\\frac{1}{a}+\\frac{1}{b}=0$$，则下列说法中不正确的一个是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$与$$b$$的和是$$0$$ "}], [{"aoVal": "B", "content": "$$a$$与$$b$$的差是正数 "}], [{"aoVal": "C", "content": "$$a$$与$$b$$的积是负数 "}], [{"aoVal": "D", "content": "$$a$$除以$$b$$，得到的商是$$-1$$ "}]], "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": ["有理数$$a$$，$$b$$满足$$\\frac{1}{a}+\\frac{1}{b}=0$$，所以$$a\\ne 0$$，$$b\\ne 0$$，$$\\frac{1}{a}=-\\frac{1}{b}$$，  所以$$a=-b$$，所以一定有$$a+b=0$$，$$ab\\textless{}0$$，$$\\frac{a}{b}=-1$$，  所以排除$$\\text{A}$$，$$\\text{C}$$，$$\\text{D}$$．  事实上，如$$a=-2$$，$$b=2$$，但$$a-b=-2-2=-4\\textless{}0$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1080", "queId": "ada2fda653fe4b74bdf6cfdc5a36146b", "competition_source_list": ["2008年第19届希望杯初一竞赛第1试第3题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "有以下两个结论：  ①任何一个有理数和它的相反数之间至少有一个有理数；  ②如果一个有理数有倒数，则这个有理数与它的倒数之间至少有一个有理数．  则．", "answer_option_list": [[{"aoVal": "A", "content": "①、②都不对 "}], [{"aoVal": "B", "content": "①对，②不对 "}], [{"aoVal": "C", "content": "①、②都对 "}], [{"aoVal": "D", "content": "①不对，②对 "}]], "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$$0$$的相反数是$$0$$，所以①是假命题；因为$$1$$的倒数是$$1$$，故命题②也是假命题．  故选：$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "910", "queId": "8087034d02c840e88b6dd6f5f5ebbc94", "competition_source_list": ["2009年第14届华杯赛初一竞赛初赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a$$、$$b$$为两个负数，且$$a ~\\textless{} ~b$$，则下面四个数中一定为大于$$a$$且小于$$b$$的数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3a+2b}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}a$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{3}b$$ "}], [{"aoVal": "D", "content": "$$\\frac{3a+2b}{4}$$ "}]], "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->正数和负数->正数、负数定义"], "answer_analysis": ["因为$$a ~\\textless{} ~b$$，所以$$5a ~\\textless{} ~3a+2b ~\\textless{} ~5b$$，因此，$$a ~\\textless{} ~\\frac{3a+2b}{5} ~\\textless{} ~b$$，故$$\\text{A}$$正确；  $$a=-2$$，$$b=-1$$，$$a ~\\textless{} ~b$$，$$\\frac{1}{2}a=-1$$，故$$\\text{B}$$错误；  $$a=-2$$，$$b=-1$$，$$\\frac{1}{3}b=-\\frac{1}{3}$$，故$$\\text{C}$$错误；  $$a=-2$$，$$b=-1$$，$$\\frac{3a+2b}{4}=\\frac{-6-2}{4}=-2$$．其实对于$$\\text{A}$$，也可取$$a=-2$$，$$b=-1$$，$$\\frac{3a+2b}{5}=\\frac{-6-2}{5}=-\\frac{8}{5}$$，满足$$-2 ~\\textless{} ~-\\frac{8}{5} ~\\textless{} ~-1$$，故$$\\text{D}$$错误． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "882", "queId": "6e4faa20754a42de92adc0601405900c", "competition_source_list": ["2010年第21届希望杯初二竞赛第2试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知三角形三个内角的度数之比为$$x:y:z$$，且$$x+y ~\\textless{} ~z$$，则这个三角形是．", "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": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["这个三角形的最大内角为：$$\\frac{180{}^{}\\circ }{x+y+z}\\cdot z\\textgreater\\frac{180{}^{}\\circ }{2z}\\cdot z=90{}^{}\\circ $$．则它是钝角三角形．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "894", "queId": "583749ebd89e4313aeebe183919af71b", "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": ["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->题型->式->二次根式->二次根式化简求值->题型：二次根式的化简求值", "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": ["$$1+{{x}^{2}}=1+\\frac{1}{4}\\left( {{1991}^{\\frac{2}{n}}}-2+{{1991}^{-\\frac{2}{n}}} \\right)=\\frac{1}{4}{{\\left( {{1991}^{\\frac{1}{n}}}+{{1991}^{-\\frac{1}{n}}} \\right)}^{2}}$$，  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": "344", "queId": "22e9eb875b184369a229bb8e1300236f", "competition_source_list": ["2002年第19届全国初中数学联赛竞赛第2题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "$${{m}^{2}}=n+2$$，$${{n}^{2}}=m+2(m\\ne n)$$, find $${{m}^{3}}-2mn+{{n}^{3}}$$．  若$${{m}^{2}}=n+2$$，$${{n}^{2}}=m+2(m\\ne n)$$，则$${{m}^{3}}-2mn+{{n}^{3}}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$-1$$ "}], [{"aoVal": "E", "content": "$$-2$$ "}]], "knowledge_point_routes": ["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", "竞赛->知识点->数与式->整式->整式的乘除运算"], "answer_analysis": ["$${{m}^{2}}=n+2$$，$${{n}^{2}}=m+2$$，  两式相减得到$${{m}^{2}}-{{n}^{2}}=n-m\\Rightarrow (m-n)(m+n+1)=0$$．  又$$m\\ne n$$，所以$$m+n=-1$$，  由立方和公式$${{m}^{3}}-2mn+{{n}^{3}}=(m+n)({{m}^{2}}-mn+{{n}^{2}})-2mn$$  $$=-({{m}^{2}}-mn+{{n}^{2}})-2mn$$  $$=-{{m}^{2}}-mn-{{n}^{2}}$$  $$=-m(m+n)-{{n}^{2}}$$  $$=m-{{n}^{2}}$$．  又由已知条件$${{n}^{2}}=m+2$$，所以$${{m}^{3}}-2mn+{{n}^{3}}=m-{{n}^{2}}=-2$$． "], "answer_value": "E"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "659", "queId": "d11c6cc0b1304458a91df80944a74ece", "competition_source_list": ["1992年第3届希望杯初二竞赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知实数$$a$$满足$$\\left\\textbar{} 1992-a \\right\\textbar+\\sqrt{a-1993}=a$$，那么$$a-{{1992}^{2}}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1991$$ "}], [{"aoVal": "B", "content": "$$1992$$ "}], [{"aoVal": "C", "content": "$$1993$$ "}], [{"aoVal": "D", "content": "$$1994$$ "}]], "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": ["由题意知$$a-1993\\geqslant 0$$，因而$$a\\geqslant 1993$$，  于是$$\\left\\textbar{} 1992-a \\right\\textbar=a-1992$$，  由$$\\left\\textbar{} 1992-a \\right\\textbar+\\sqrt{a-1993}=a$$，  可知，$$a-1992+\\sqrt{a-1993}=a$$，  即$$\\sqrt{a-1993}=1992$$，  从而$$a-1993={{1992}^{2}}$$，  故$$a-{{1992}^{2}}=1993$$．  所以应选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "646", "queId": "da5262f0062441e5aaf168581d877cda", "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": "874", "queId": "6552df3ab87b4aa8abd706f09fb3a826", "competition_source_list": ["1991年第2届希望杯初二竞赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$\\left( x+a \\right)\\left( x+b \\right)+\\left( x+b \\right)\\left( x+c \\right)-\\left( x+c \\right)\\left( x+a \\right)$$是完全平方式．则$$a$$，$$b$$，$$c$$的大小关系可以写成．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless{}b\\textless{}c$$ "}], [{"aoVal": "B", "content": "$${{\\left( a-b \\right)}^{2}}+{{\\left( b-c \\right)}^{2}}=0$$ "}], [{"aoVal": "C", "content": "$$c\\textless{}a\\textless{}b$$ "}], [{"aoVal": "D", "content": "$$a=b\\ne c$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->利用完全平方式求参数的值"], "answer_analysis": ["容易看到$$a=b=c$$时，原式成为$$3{{\\left( x+a \\right)}^{2}}$$，是完全平方式．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1333", "queId": "8aac50a7511483070151192885130f1a", "competition_source_list": ["2015年第26届全国希望杯初三竞赛初赛（特）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$a={{(2+\\sqrt{3})}^{-1}}$$，$$b={{(2-\\sqrt{3})}^{-1}}$$，$${{(a-2)}^{-3}}+{{(b-2)}^{-3}}$$则的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{2\\sqrt{3}}{9}$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$\\frac{2\\sqrt{3}}{9}$$ "}], [{"aoVal": "D", "content": "$$\\pm \\frac{2\\sqrt{3}}{9}$$ "}]], "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": ["$$a={{(2+\\sqrt{3})}^{-1}}$$，$$b={{(2-\\sqrt{3})}^{-1}}$$，  ∴$$a=\\frac{1}{2+\\sqrt{3}}=2-\\sqrt{3}$$，$$b=\\frac{1}{2-\\sqrt{3}}=2+\\sqrt{3}$$，  ∴$${{(a-2)}^{-3}}+{{(b-2)}^{-3}}=\\frac{1}{{{(a-2)}^{3}}}+\\frac{1}{{{(b-2)}^{3}}}=\\frac{1}{{{(-\\sqrt{3})}^{3}}}+\\frac{1}{{{(\\sqrt{3})}^{3}}}=0$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1467", "queId": "ca7e7e18abf84f938112b34c7d0531da", "competition_source_list": ["2011年第22届全国希望杯初一竞赛初赛第5题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$6$$个人用$$35$$天完成了某项工程的$$\\frac{1}{3}$$，如果再增加工作效率相同的$$8$$个人，那么完成这项工程，前后共用的天数是（~ ~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$65$$ "}]], "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"], "answer_analysis": ["总工程量的$$\\frac{1}{3}$$为$$6\\times 35=210$$（人$$\\cdot $$天），  还有$$\\frac{2}{3}$$，即$$2\\times 210=420$$（人$$\\cdot $$天），  还需要$$\\frac{420}{6+8}=30$$（天），  则完成这项工程共用的天数是$$35+30=65$$（天）． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "192", "queId": "33d6abf9da5f4634b483b5b6a85b71fe", "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": "有一组 "}], [{"aoVal": "B", "content": "有二组 "}], [{"aoVal": "C", "content": "多于二组 "}], [{"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．故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "792", "queId": "4e6fd36e58414a6c8d5feb76ef16f96d", "competition_source_list": ["1994年第5届希望杯初二竞赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "若三角形的三边长度均为整数，其中两边长的差是$$7$$，且三角形的周长是奇数，则第三边长可能是．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "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->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->求三角形第三边的取值范围"], "answer_analysis": ["不妨设三角形三边长的度为$$a$$，$$b$$，$$c$$．  且$$a-b=7$$，则$$a$$与$$b$$为一奇一偶，又题设知$$a+b+c$$为奇数，  所以$$c$$一定是偶数，又三角形两边之差小于第三边，即$$c\\textgreater a-b=7$$，  所以第三边的长度可能是$$8$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "43", "queId": "05f70a78426549bf85a3263b2bc022b6", "competition_source_list": ["2019~2020学年5月四川成都金牛区成都外国语学校初一下学期周测A卷第4题3分", "2014~2015学年广东深圳罗湖区初一下学期期中第12题3分", "2002年竞赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "~已知$$a=1999x+2000    ， b=1999x+2001    ，c=1999x+2002$$，则多项式$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca$$的值为（ ~ ~）．", "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->知识点->式->整式的乘除->乘法公式->配方思想的运用", "课内体系->能力->运算能力", "课内体系->方法->配方法"], "answer_analysis": ["原式$$=\\frac{1}{2}\\left[ {{\\left( a-b \\right)}^{2}}+{{\\left( a-c \\right)}^{2}}+{{\\left( b-c \\right)}^{2}} \\right]=\\frac{1}{2}\\times \\left[ 1+4+1 \\right]=3$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "489", "queId": "d56df669e2194a8fa6a227bd23f85c15", "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": "93", "queId": "17c9e601f0ba4c4a9c130a89e67e672e", "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": "$$\\sqrt{\\text{3}}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{\\text{6}}$$ "}], [{"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": "300", "queId": "66e2a0b66095432281b07235954d95d9", "competition_source_list": ["2002年第13届希望杯初一竞赛第2试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "爸爸给女儿园园买了一个（圆柱形的）生日蛋糕，园园想把蛋糕切成大小不定相等的若干块（不少于$$10$$块），分给$$10$$个小朋友，若沿竖直方向切分这块蛋糕，至少需要切刀．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$9$$ "}]], "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"], "answer_analysis": ["易知，$$2$$条直线至多可以分圆面为$$4$$块；$$3$$条直线至多分圆面为$$7$$块；$$4$$条直线至多分圆面为$$11$$块，因此要把蛋糕分成不少于$$10$$块，至少需要切$$4$$刀．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "989", "queId": "ff8080814cfa9b24014d02c290ed138b", "competition_source_list": ["2012年第29届全国全国初中数学联赛竞赛第2题7分", "2019年浙江温州瓯海区浙江省温州中学初三自主招生第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "方程$${{x}^{2}}+2xy+3{{y}^{2}}=34$$的整数解$$(x,y)$$的组数为（~ ）．", "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->知识点->方程与不等式->其他方程->二元二次方程（组）"], "answer_analysis": ["方程变形得：$${{(x+y)}^{2}}+2{{y}^{2}}=34$$，  ∵$$34$$与$$2{{y}^{2}}$$是偶数，  ∴$$x+y$$必须是偶数，  设$$x+y=2t$$，  则原方程变为：$${{(2t)}^{2}}+2{{y}^{2}}=34$$，  ∴$$2{{t}^{2}}+{{y}^{2}}=17$$，  它的整数解为$$\\begin{cases}t=\\pm 2    y=\\pm 3 \\end{cases}$$，  则当$$y=3$$，$$t=2$$时，$$x=1$$；  当$$y=3$$，$$t=-2$$时，$$x=-7$$；  当$$y=-3$$，$$t=2$$时，$$x=7$$；  当$$y=-3$$，$$t=-2$$时，$$x=-1$$．  ∴原方程的整数解为：$$(1,3)$$，$$(-7,3)$$，$$(7,-3)$$，$$(-1,-3)$$共$$4$$组． ", "<p>$${{\\left( x+y \\right)}^{2}}+2{{y}^{2}}=34$$ 尝试可知，$${{\\left( x+y \\right)}^{2}}=16$$，$$ {{y}^{2}}=9$$，</p>\n<p>&there4;$$\\begin{cases}x=1 \\\\ y=3 \\\\\\end{cases}$$，$$\\begin{cases}x=-1 \\\\ y=-3 \\\\\\end{cases}$$，$$\\begin{cases}x=7 \\\\ y=-3 \\\\\\end{cases}$$，$$\\begin{cases}x=-7 \\\\ y=3 \\\\\\end{cases}$$．</p>\n<p>故选$$\\text{B}$$．</p>"], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "727", "queId": "df1714354de8407eb3130b7d5f711f4a", "competition_source_list": ["2014年第25届全国希望杯初一竞赛复赛第4题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$({{a}^{m}}{{b}^{n+2}})\\cdot ({{a}^{2n}}{{b}^{2m}})={{a}^{5}}{{b}^{3}}$$，则$$m+n$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$-3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->整式的乘除运算->单项式乘单项式", "课内体系->能力->运算能力"], "answer_analysis": ["∵$$({{a}^{m}}{{b}^{n+2}})\\cdot ({{a}^{2n}}{{b}^{2m}})={{a}^{m+2n}}{{b}^{2m+n+2}}={{a}^{5}}{{b}^{3}}$$，  ∴$$m+2n=5$$，$$2m+n+2=3$$，  两式相加，得$$m+n=2$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1066", "queId": "ff8080814d7978b9014d88d3b5ce2bb9", "competition_source_list": ["1991年第2届全国希望杯初一竞赛复赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "①若$$a=0$$，$$b\\ne 0$$，则方程$$ax=b$$无解；  ②若$$a=0$$，$$b\\ne 0$$，则不等式$$ax\\textgreater b$$无解；  ③若$$a\\ne 0$$，则方程$$ax=b$$有唯一解$$x=\\frac{b}{a}$$；  ④若$$a\\ne 0$$，则不等式$$ax\\textgreater b$$的解为$$x\\textgreater\\frac{b}{a}$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "①、②、③、④都正确 "}], [{"aoVal": "B", "content": "①、③正确，②、④不正确 "}], [{"aoVal": "C", "content": "①、③不正确，②、④正确 "}], [{"aoVal": "D", "content": "①、②、③、④都不正确 "}]], "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"], "answer_analysis": ["若$$a=0$$，$$b=-1$$，则$$0x\\textgreater-1$$，可见②无解不正确．  若$$a\\ne 0$$，比如$$a=-1$$，$$-x\\textgreater b\\Rightarrow x\\textless{}-b$$，④不正确．  只有①③正确． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "323", "queId": "6b91289fe0b642eaa73e6a9e798c769b", "competition_source_list": ["2011年第22届全国希望杯初二竞赛初赛第1题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$a$$千克含盐$$10 \\%$$的盐水配制成含盐$$15 \\%$$的盐水，需加盐$$x$$千克，则由此可列出方程为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a(1-10 \\%)=(a+x)(1-15 \\%)$$ "}], [{"aoVal": "B", "content": "$$a\\times 10 \\%=(a+x)\\times 15 \\%$$ "}], [{"aoVal": "C", "content": "$$a\\times 10 \\%+x=a\\times 15 \\%$$ "}], [{"aoVal": "D", "content": "$$a(1-10 \\%)=x(1-15 \\%)$$ "}]], "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->按要求写出一元一次方程", "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->一元一次方程的浓度问题"], "answer_analysis": ["加盐前后，水的含量不变，  加盐前含水$$a(1-10 \\%)$$，加盐后含水$$(a+x)(1-15 \\%)$$，  可列出方程为$$a(1-10 \\%)=(a+x)(1-15 \\%)$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "101", "queId": "afdd044b67204c40b64839f883887acd", "competition_source_list": ["2017年第1届重庆全国初中数学联赛竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$\\frac{5x+11}{2{{x}^{2}}+7x-4}=\\frac{A}{2x-1}+\\frac{B}{x+4}$$，其中$$A$$、$$B$$为常数，求$$4A-3B$$的值（~ ~ ~）.", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$0$$ "}]], "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->分式通分", "课内体系->知识点->式->分式->分式的基础->分式为特殊值"], "answer_analysis": ["将已知等式右边通分得$$\\frac{5x+11}{2{{x}^{2}}+7x-4}=\\frac{\\left( A+2B \\right)x+\\left( 4A-B \\right)}{2{{x}^{2}}+7x-4}$$  于是$$\\begin{cases}A+2B=5    4A-B=11 \\end{cases}$$，解得$$A=3,B=1$$  故$$4A-3B=4\\times 3-3\\times 1=9$$ "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1077", "queId": "73ee5828a1a74702917560b917d41dcb", "competition_source_list": ["2012年第29届全国全国初中数学联赛竞赛第6题7分"], "difficulty": "2", "qtype": "single_choice", "problem": "由$$1$$，$$2$$，$$3$$，$$4$$这四个数字组成四位数$$\\overline{abcd}$$（数字可重复使用），要求满足$$a+c=b+d$$．这样的四位数共有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$36$$个 "}], [{"aoVal": "B", "content": "$$40$$个 "}], [{"aoVal": "C", "content": "$$44$$个 "}], [{"aoVal": "D", "content": "$$48$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->组合->排列与组合"], "answer_analysis": ["分为$$4$$类：  ①由同一个数字组成如$$1111$$共$$4$$个数；  ②由两个不同数字组成如$$1221$$，$$1122$$，$$2112$$，$$2211$$，而从$$4$$个数里面取$$2$$个，  共六种取法故此类可构成$$4\\times 6=24$$个数；  ③由三个不同数字组成如$$1232$$，$$3212$$，$$2123$$，$$2321$$，  此类只有两种组合，即$$1+3=2+2$$和$$2+4=3+3$$，  故可构成$$2\\times 4=8$$个数；  ④由四个不同数字组成如$$1243$$，$$1342$$，$$4213$$，$$4312$$，$$2134$$，$$2431$$，$$3124$$，$$3421$$，共$$8$$个．  综上所有的数共$$4+24+8+8=44$$个． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1013", "queId": "ff8080814d4b1928014d4c0b132f04f3", "competition_source_list": ["2015年第26届全国希望杯初一竞赛初赛第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "若$$\\frac{x}{2}+\\frac{x}{3}+\\frac{x}{6}=-2015$$，则$$x=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2015$$ "}], [{"aoVal": "B", "content": "$$-403$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "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->常规一元一次方程解法"], "answer_analysis": ["$$\\frac{x}{2}+\\frac{x}{3}+\\frac{x}{6}=-2015$$，  ∴$$\\frac{3+2+1}{6}x=-2015$$，  ∴$$x=-2015$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1082", "queId": "ff8080814d9539f1014d986ce14001c3", "competition_source_list": ["2005年第16届希望杯初二竞赛初赛第5题4分", "初一其它"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$A=\\frac{1}{{{2}^{3}}}+\\frac{1}{{{2}^{3}}+1}+\\frac{1}{{{2}^{3}}+2}+\\cdots +\\frac{1}{{{2}^{4}}-1}$$，则$$A$$与$$1$$的大小关系是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$A\\textgreater1$$ "}], [{"aoVal": "B", "content": "$$A=1$$ "}], [{"aoVal": "C", "content": "$$A\\textless{}1$$ "}], [{"aoVal": "D", "content": "无法确定 "}]], "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-利用有理数正负性", "课内体系->能力->运算能力"], "answer_analysis": ["从$$\\frac{1}{{{2}^{3}}}$$到$$\\frac{1}{{{2}^{4}}-1}$$共有$$\\left( {{2}^{4}}-1 \\right)-{{2}^{3}}+1={{2}^{3}}$$个数，故$$A\\textless{}\\underbrace{\\frac{1}{{{2}^{3}}}+\\frac{1}{{{2}^{3}}}+\\frac{1}{{{2}^{3}}}+\\cdot \\cdot \\cdot +\\frac{1}{{{2}^{3}}}}_{{{2}^{3}}}=1$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "623", "queId": "32b4e6ded4cd4b1babf788505dcc35de", "competition_source_list": ["2015年四川成都金牛区成都外国语学校初三自主招生第14题3分", "2000年竞赛（全国初中数学竞赛）第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$\\triangle ABC$$三边长分别为$$a$$，$$b$$，$$c$$，面积为$$s$$．$$\\triangle {{A}^{\\prime }}{{B}^{\\prime }}{{C}^{\\prime }}$$的三边长分别为$${{a}^{\\prime }}$$、$${{b}^{\\prime }}$$、$${{c}^{\\prime }}$$，面积为$${{s}^{\\prime }}$$，且$$a\\textgreater{{a}^{\\prime }}$$，$$b\\textgreater{{b}^{\\prime }}$$，$$c\\textgreater{{c}^{\\prime }}$$，则$$s$$与$${{s}^{\\prime }}$$的大小关系一定是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$s\\textgreater{{s}^{\\prime }}$$ "}], [{"aoVal": "B", "content": "$$s\\textless{}{{s}^{\\prime }}$$ "}], [{"aoVal": "C", "content": "$$s={{s}^{\\prime }}$$ "}], [{"aoVal": "D", "content": "不确定 "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->相似三角形->相似三角形基础->相似三角形的性质", "课内体系->知识点->三角形->相似三角形->相似三角形的判定->相似三角形的性质与判定综合", "课内体系->思想->分类讨论思想"], "answer_analysis": ["分别构造$$\\triangle ABC$$与$$\\triangle {{A}^{\\prime }}{{B}^{\\prime }}{{C}^{\\prime }}$$如下：  ①作$$\\triangle ABC\\backsim \\triangle {{A}^{\\prime }}{{B}^{\\prime }}{{C}^{\\prime }}$$，显然$$\\frac{s}{{{s}^{\\prime }}}={{\\left( \\frac{a}{{{a}^{\\prime }}} \\right)}^{2}}\\textgreater1$$，  ∴$$s\\textgreater{{s}^{\\prime }}$$．  ②设$$a=b=\\sqrt{101}$$，$$c=20$$，则$${{h}_{c}}=1$$，$$s=10$$，  设$${{a}^{\\prime }}={{b}^{\\prime }}={{c}^{\\prime }}=10$$，则$${{s}^{\\prime }}=\\frac{\\sqrt{3}}{4}\\times 100\\textgreater10$$，  ∴$$s\\textless{}{{s}^{\\prime }}$$．  ③设$$a=b=\\sqrt{101}$$，$$c=20$$，则$${{h}_{c}}=1$$，$$s=10$$，  $${{a}^{\\prime }}={{b}^{\\prime }}=\\sqrt{29}$$，$${{c}^{\\prime }}=10$$，则$${{h}_{c}}=2$$，$${{s}^{\\prime }}=10$$，  ∴$$s={{s}^{\\prime }}$$，  ∴$$s$$与$${{s}^{\\prime }}$$的大小关系不确定． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "57", "queId": "0a2b2fa13c514815abd6d8f5ddcc2314", "competition_source_list": ["2001年竞赛（全国初中数学竞赛）第1题5分", "2019~2020学年5月浙江杭州萧山区杭州萧山金山初中初一下学期周测C卷第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "化简：$$\\frac{{{2}^{n+4}}-2\\times {{2}^{n}}}{2\\times {{2}^{n+3}}}$$得（  ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{2}^{n+1}}-\\frac{1}{8}$$ "}], [{"aoVal": "B", "content": "$$-{{2}^{n+1}}$$ "}], [{"aoVal": "C", "content": "$$\\frac{7}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{7}{4}$$ "}]], "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": ["原式$$=\\frac{{{2}^{n+4}}- {{2}^{n+1}}}{{{2}^{n+4}}}=\\frac{{2}^{n+1}(8-1)}{{{2}^{n+4}}}=\\frac{7}{8}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "486", "queId": "83202b51721f43f89e1273e16bcd8a00", "competition_source_list": ["2005年竞赛第2题6分", "初一单元测试《分解方法的延拓（2）》第21题", "初三上学期其它"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$M=3{{x}^{2}}-8xy+9{{y}^{2}}-4x+6y+13$$（$$x$$，$$y$$是实数），则$$M$$的值一定是（  ）．", "answer_option_list": [[{"aoVal": "A", "content": "正数 "}], [{"aoVal": "B", "content": "负数 "}], [{"aoVal": "C", "content": "零 "}], [{"aoVal": "D", "content": "整数 "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->配方求最值"], "answer_analysis": ["$$M=3{{x}^{2}}-8xy+9{{y}^{2}}-4x+6y+13$$  $$=(2{{x}^{2}}-8xy+8{{y}^{2}})+({{x}^{2}}-4x+4)+({{y}^{2}}+6y+9)$$  $$=2{{(x-2y)}^{2}}+{{(x-2)}^{2}}+{{(y+3)}^{2}}$$，  ∵$${{(x-2y)}^{2}}\\geqslant 0$$，$${{(x-2)}^{2}}\\geqslant 0$$，$${{(y+3)}^{2}}\\geqslant 0$$，  又∵$$x-2y$$，$$x-2$$，$$y+3$$三项不能同时取$$0$$，  ∴$$M=2{{(x-2y)}^{2}}+{{(x-2)}^{2}}+{{(y+3)}^{2}}\\textgreater0$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1073", "queId": "ff8080814d7978b9014d8977cc8c353a", "competition_source_list": ["初一其它", "2010年第21届希望杯初二竞赛第2试第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "某次足球比赛的计分规则是:胜一场得$$3$$分，平一场得$$1$$分，负一场得$$0$$分．某球队参赛$$15$$场，积$$33$$分，若不考虑比赛顺序，则该队胜、平、负的情况可能有（~ ~ ~）．", "answer_option_list": [[{"aoVal": "A", "content": "$$15$$种 "}], [{"aoVal": "B", "content": "$$11$$种 "}], [{"aoVal": "C", "content": "$$5$$种 "}], [{"aoVal": "D", "content": "$$3$$种 "}]], "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", "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": ["设该队胜$$x$$场，平$$y$$场，负$$z$$场，$$x$$，$$y$$，$$z$$都为整数，且$$0\\leqslant x,y,z\\leqslant 15$$，  则$$\\left { \\begin{matrix}x+y+z=15    3x+y=33   \\end{matrix} \\right.$$，  解得$$x=11-\\frac{y}{3}$$，则$$y=0$$，$$3$$，$$6$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "895", "queId": "776ddede051a4c98a91c3af01f6dd50c", "competition_source_list": ["2012年竞赛第2题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "在平面直角坐标系$$xOy$$中，满足不等式$${{x}^{2}}+{{y}^{2}}\\leqslant 2x+2y$$的整数$$\\left( x,y \\right)$$的对数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数->平面直角坐标系->坐标系综合->坐标与距离"], "answer_analysis": ["由题设$${{x}^{2}}+{{y}^{2}}\\leqslant 2x+2y$$，得$$0\\leqslant {{\\left( x-1 \\right)}^{2}}{{\\left( y-1 \\right)}^{2}}\\leqslant 2$$，  因为$$x$$，$$y$$均为整数，所以有  $$\\begin{cases}{{(x-1)}^{2}}=0    {{\\left( y-1 \\right)}^{2}}=0   \\end{cases}$$或$$\\begin{cases}{{\\left( x-1 \\right)}^{2}}=0    {{\\left( y-1 \\right)}^{2}}=1   \\end{cases}$$或 $$\\begin{cases}{{\\left( x-1 \\right)}^{2}}=1    {{\\left( y-1 \\right)}^{2}}=0   \\end{cases}$$或$$\\begin{cases}{{\\left( x-1 \\right)}^{2}}=1    {{\\left( y-1 \\right)}^{2}}=1   \\end{cases}$$，  解得：$$\\begin{cases}x=1    y=1   \\end{cases}$$；$$\\begin{cases}x=1    y=2   \\end{cases}$$；$$\\begin{cases}x=1    y=0   \\end{cases}$$；$$\\begin{cases}x=0    y=1   \\end{cases}$$；$$\\begin{cases}x=0    y=0   \\end{cases}$$；$$\\begin{cases}x=0    y=2   \\end{cases}$$；$$\\begin{cases}x=2    y=1   \\end{cases}$$；$$\\begin{cases}x=2    y=0   \\end{cases}$$；$$\\begin{cases}x=2    y=2   \\end{cases}$$，  以上共计$$9$$对$$(x,y)$$．  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "48", "queId": "09f9428290ba4848bb7200a9d9eb03ad", "competition_source_list": ["2012年四川成都青羊区成都树德实验中学初三自主招生第8题5分", "2017~2018学年福建泉州鲤城区泉州市第六中学初二下学期期中第10题4分", "1998年竞赛（全国初中数学竞赛）第4题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$abc\\ne 0$$，并且$$\\frac{a+b}{c}=\\frac{b+c}{a}=\\frac{c+a}{b}=p$$，那么直线$$y=px-p$$一定通过第象限．", "answer_option_list": [[{"aoVal": "A", "content": "一、二 "}], [{"aoVal": "B", "content": "二、三 "}], [{"aoVal": "C", "content": "三、四 "}], [{"aoVal": "D", "content": "一、四 "}]], "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->题型->函数->一次函数->一次函数基础->题型：一次函数图象与性质", "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", "知识标签->知识点->函数->一次函数->一次函数的图象与性质->一次函数的图象与系数的关系", "知识标签->数学思想->数形结合思想"], "answer_analysis": ["由条件得：①$$a+b=pc$$，②$$b+c=pa$$，③$$a+c=pb$$，  三式相加得: $$2\\left( a+b+c \\right)=p\\left( a+b+c \\right)$$．  ∴$$p=2$$或$$a+b+c=0$$．  当$$p=2$$时，$$y=2x-2$$．则直线通过第一、三、四象限．  当$$a+b+c=0$$时，不妨取$$a+b=-c$$，于是$$p=\\frac{a+b}{c}=-1$$，  ∴$$y=-x+1$$，则直线通过第一、二、四象限．  综合上述两种情况，直线一定通过第一、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": "1014", "queId": "a907506f464e4d40abaa1bdd45534937", "competition_source_list": ["2006年第17届希望杯初二竞赛第2试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "要使代数式$$\\frac{\\sqrt{\\textbar x-3\\textbar-2}}{{{x}^{2}}-4x+3}$$有意义，那么实数$$x$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1 \\textless{} x\\leqslant 5$$ "}], [{"aoVal": "B", "content": "$$x \\textless{} 1$$或$$x\\geqslant 5$$ "}], [{"aoVal": "C", "content": "$$x\\leqslant 1$$或$$x\\geqslant 5$$ "}], [{"aoVal": "D", "content": "$$x \\textless{} 1$$或$$x\\textgreater5$$ "}]], "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->能力->运算能力"], "answer_analysis": ["由题意，必须使分式和二次根式同时有意义，得$$\\begin{cases}{{x}^{2}}-4x+3\\ne 0    \\textbar x-3\\textbar-2\\geqslant 0   \\end{cases}$$  $${{x}^{2}}-4x+3\\ne 0$$得$$x\\ne 1$$且$$x\\ne 3$$，由$$\\textbar x-3\\textbar-2\\geqslant 0,$$得$$x\\leqslant 1$$或$$x\\geqslant 5$$  从而$$x$$的取值范围是$$x \\textless{} 1$$或$$x\\geqslant 5$$，故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1575", "queId": "f4d2de109f6c4ffbafca1c3cd77406e0", "competition_source_list": ["2007年第18届希望杯初一竞赛初赛第1题4分"], "difficulty": "0", "qtype": "single_choice", "problem": "在$${{(-1)}^{2007}}$$，$${{\\left\\textbar{} -1 \\right\\textbar}^{3}}$$，$$-{{(-1)}^{18}}$$，$$18$$这四个有理数中，负数共有．", "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->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": ["$${{(-1)}^{2007}}=-1\\textless{}0$$，$${{\\left\\textbar{} -1 \\right\\textbar}^{3}}=+1\\textless{}0$$，$$-{{(-1)}^{18}}=-1\\textless{}0$$，$$18\\textgreater0$$，所以这四个有理数中，共有两个负数．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "592", "queId": "c7be93027e1349bdb32d56a20b1e23c2", "competition_source_list": ["1991年第8届全国初中数学联赛竞赛（第一试）第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$1\\times 2\\times 3\\times \\ldots \\times 99\\times 100={{12}^{n}}M$$，其中$$M$$为自然数，$$n$$为使得等式成立的最大的自然数，则$$M$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "能被$$2$$整除，但不能被$$3$$整除 "}], [{"aoVal": "B", "content": "能被$$3$$整除，但不能被$$2$$整除 "}], [{"aoVal": "C", "content": "能被$$4$$整除，但不能被$$3$$整除 "}], [{"aoVal": "D", "content": "不能被$$3$$整除，也不能被$$2$$整除 "}]], "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->知识点->数->有理数->数的特征->整除的条件"], "answer_analysis": ["在$$1\\times 2\\times 3\\times \\ldots \\times 100$$的质因数分解中，$$2$$的因子有$$\\left[ \\frac{100}{2} \\right]+\\left[ \\frac{100}{{{2}^{2}}} \\right]+\\left[ \\frac{100}{{{2}^{3}}} \\right]+\\ldots +\\left[ \\frac{100}{{{2}^{6}}} \\right]$$  $$=50+25+12+6+3+1=97($$个$$)$$，  $$3$$的因子有  $$\\left[ \\frac{100}{3} \\right]+\\left[ \\frac{100}{{{3}^{2}}} \\right]+\\left[ \\frac{100}{{{3}^{3}}} \\right]+\\left[ \\frac{100}{{{3}^{4}}} \\right]$$  $$=33+11+3+1=48($$个$$)$$  所以$$1\\times 2\\times 3\\times \\ldots \\times 100$$  $$={{2}^{97}}\\times {{3}^{48}}\\times P$$  $$={{12}^{48}}\\times 2P$$，其中$$2$$不整除$$P$$，$$3$$不整除$$P$$，因而$$M=2P$$，故选$$A$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1096", "queId": "ff8080814d9539f1014d9b6355d50a5a", "competition_source_list": ["1992年第3届全国希望杯初一竞赛初赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$a$$，$$b$$，$$c$$，$$m$$都是有理数，并且$$a+2b+3c=m$$，$$a+b+2c=m$$，那么$$b$$与$$c$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "互为相反数 "}], [{"aoVal": "B", "content": "互为倒数 "}], [{"aoVal": "C", "content": "互为负倒数 "}], [{"aoVal": "D", "content": "相等 "}]], "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->数->有理数->相反数->相反数的性质"], "answer_analysis": ["因为$$a+2b+3c=m=a+b+2c$$，  所以$$b+c=0$$，即$$b$$，$$c$$互为相反数，选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "652", "queId": "99c92b9c2b054af6b8baa065ee6fc5a1", "competition_source_list": ["北京初二上学期单元测试《三角形的边与角》第24题", "2011年第28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$$\\triangle ABC$$的两条高线的长分别为$$5$$和$$20$$，若第三条高线的长也是整数，则第三条高线长的最大值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "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->整数边问题"], "answer_analysis": ["设$$\\triangle ABC$$的面积为$$S$$，  所求的第三条高线的长为$$h$$，  则三边长分别为$$\\frac{2S}{5}$$，$$\\frac{2S}{20}$$，$$\\frac{2S}{h}$$．  显然$$\\frac{2S}{5}\\textgreater\\frac{2S}{20}$$，  于是由三边关系，得$$\\left { \\begin{matrix} \\dfrac{2S}{20}+\\dfrac{2S}{h}\\textgreater\\dfrac{2S}{5}    \\dfrac{2S}{20}+\\dfrac{2S}{5}\\textgreater\\dfrac{2S}{h}   \\end{matrix} \\right.$$，  解得$$4\\textless{}h\\textless{}\\frac{20}{3}$$．  所以$$h$$的最大整数值为$$6$$，即第三条高线的长的最大值为$$6$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "60", "queId": "06b189e2c7384dd5a9b18455bcfc6bd6", "competition_source_list": ["1996年第7届希望杯初二竞赛第9题"], "difficulty": "2", "qtype": "single_choice", "problem": "在一家三口人中，每两人的平均年龄加上余下一人的年龄分别得到$$47$$、$$61$$、$$60$$，那么这三个人中最大年龄与最小年龄的差是(  )．", "answer_option_list": [[{"aoVal": "A", "content": "$$28$$ "}], [{"aoVal": "B", "content": "$$27$$ "}], [{"aoVal": "C", "content": "$$26$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "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->其他方程->三元一次方程组->三元一次方程组的实际应用", "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": ["设三人年龄各为$$x$$、$$y$$、$$z$$，则：  $$\\begin{cases}\\dfrac{x+y}{2}+z=47    \\dfrac{y+z}{2}+x=61    \\dfrac{x+z}{2}+y=60   \\end{cases}$$得$$\\begin{cases}x+y+2z=94    2x+y+z=122    x+2y+z=120   \\end{cases}$$．  三个方程相加得$$x+y+z=84$$，$$\\therefore x=38$$，$$y=36$$，$$z=10$$．  $$\\therefore x-z=38-10=28$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1165", "queId": "8aac49074e023206014e1e28098f5b65", "competition_source_list": ["1993年第4届全国希望杯初一竞赛复赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "绝对值小于$$100$$的所有被$$3$$除余$$1$$的整数之和等于（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$-32$$ "}], [{"aoVal": "C", "content": "$$33$$ "}], [{"aoVal": "D", "content": "$$-33$$ "}]], "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->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"], "answer_analysis": ["即求$$-100$$与$$100$$之间被$$3$$除余$$1$$的整数之和，  在$$0$$到$$100$$之间被$$3$$除余$$1$$的整数是$$1$$，$$4$$，$$7$$，$$\\cdots $$，$$91$$，$$94$$，$$97$$共计$$33$$个．  在$$-100$$到$$0$$之间被$$3$$除余$$1$$的整数是$$-98$$，$$-95$$，$$-92$$，$$-89$$，$$\\cdots $$，$$-8$$，$$-5$$，$$-2$$共$$33$$个．  其总和为$$-33$$．选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1260", "queId": "8aac50a74e023208014e3f5885c81920", "competition_source_list": ["1995年第6届全国希望杯初一竞赛初赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "在绝对值小于$$1000$$的整数中，完全平方数的个数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$62$$ "}], [{"aoVal": "B", "content": "$$63$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$31$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数->实数->平方根->算术平方根的定义", "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"], "answer_analysis": ["在绝对值小于$$1000$$的整数中，共计$$1999$$个整数，  其中$$-1999$$，$$-1998$$，$$\\cdots $$，$$-2$$，$$-1$$，这$$999$$个负整数都不能写成整数的平方．  因此可以写成整数的平方的数只能在$$0$$，$$1$$，$$2$$，$$\\cdots $$，$$998$$，$$999$$这一千个整数中去找．  $$0={{0}^{2}}$$，$$1={{1}^{2}}$$，$$4={{2}^{2}}$$，$$\\cdots $$，$$961={{31}^{2}}$$．共计$$32$$个，选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1521", "queId": "caf5c40334e941eb874902cc7271dbfd", "competition_source_list": ["2009年竞赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "2009年$$AMC8$$竞赛第$$5$$题  $$(x+1)(x-1)-(x+2)(x-2)=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$-5$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$2{{x}^{2}}-5$$ "}], [{"aoVal": "D", "content": "$$2{{x}^{2}}+3$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Algebra and Sequences->Algebraic expression", "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"], "answer_analysis": ["$$(x+1)(x-1)-(x+2)(x-2)=({{x}^{2}}-1)-({{x}^{2}}-4)={{x}^{2}}-1-{{x}^{2}}+4=3$$，  故选Ｂ． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "177", "queId": "463008d9edd14f6c967efecb1c0e41ae", "competition_source_list": ["2006年第17届希望杯初一竞赛复赛第7题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "三角形三边的长$$a$$，$$b$$，$$c$$都是整数，且$$[a,b,c]=60$$，$$(a,b)=4$$，$$(b,c)=3$$．（注：$$[a,b,c]$$表示$$a$$, $$b$$, $$c$$ 的最小公倍数；$$(a,b)$$表示$$a$$，$$b$$的最大公约数），则$$a+b+c$$的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$31$$ "}], [{"aoVal": "C", "content": "$$32$$ "}], [{"aoVal": "D", "content": "$$33$$ "}]], "knowledge_point_routes": ["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->与三角形有关的线段->求三角形第三边的取值范围", "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": ["由题意知$$b$$既能被$$4$$整除，又能被$$3$$整除，  所以$$b$$能被$$12$$整除．  又$$60$$能被$$b$$整除，  所以$$b=12$$或$$60$$．  若$$b=12$$，则$$60\\div b=5$$，  因为$$(5,4)=1$$，$$(5,3)=1$$，  所以，$$a$$，$$c$$中至少有一个含因数$$5$$．  若$$a$$含因数$$5$$，则$$a\\geqslant 20$$，  又$$c\\geqslant 3$$，  所以$$a+b+c\\geqslant 20+12+3=35$$；  若$$c$$含因数$$5$$，则$$c\\geqslant 15$$，  又$$a\\geqslant 4$$，  所以$$a+b+c\\geqslant 4+12+15=31$$，  取$$a=4$$，$$b=12$$，$$c=15$$能构成三角形．  若$$b=60$$，则$$a+b+c\\textgreater60\\textgreater31$$．  综上知：$$a+b+c$$的最小值为$$31$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "718", "queId": "5b1ada8c9b2f40d380e29968bdd1bfb8", "competition_source_list": ["2012年第23届全国希望杯初一竞赛复赛第10题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$x+y=3$$，$$xy=1$$．则$${{x}^{5}}+{{y}^{5}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$33$$ "}], [{"aoVal": "B", "content": "$$231$$ "}], [{"aoVal": "C", "content": "$$123$$ "}], [{"aoVal": "D", "content": "$$312$$ "}]], "knowledge_point_routes": ["课内体系->知识点->式->整式的乘除->乘法公式->高次方公式", "课内体系->能力->运算能力"], "answer_analysis": ["因为$$x+y=3$$，  所以$${{(x+y)}^{2}}=9$$，  即$${{x}^{2}}+{{y}^{2}}+2xy=9$$．  因为$$xy=1$$，  所以$${{x}^{2}}+{{y}^{2}}=7$$．  所以$${{x}^{3}}+{{y}^{3}}=(x+y)({{x}^{2}}+{{y}^{2}}-xy)=3\\times (7-1)=18$$，  所以$${{x}^{5}}+{{y}^{5}}=({{x}^{3}}+{{y}^{3}})({{x}^{2}}+{{y}^{2}})-{{(xy)}^{2}}(x+y)=18\\times 7-{{1}^{2}}\\times 3=123$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1062", "queId": "ff8080814d7978b9014d88c127872a72", "competition_source_list": ["2018~2019学年10月重庆北碚区重庆市西南大学附属中学初一上学期月考第3题3分", "1991年第2届全国希望杯初一竞赛复赛第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "下列四个等式：$$\\frac{a}{b}=0$$，$$ab=0$$，$${{a}^{2}}=0$$，$${{a}^{2}}+{{b}^{2}}=0$$中，可以断定$$a$$必等于$$0$$的式子共有（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$个 "}], [{"aoVal": "B", "content": "$$2$$个 "}], [{"aoVal": "C", "content": "$$1$$个 "}], [{"aoVal": "D", "content": "$$0$$个 "}]], "knowledge_point_routes": ["课内体系->知识点->式->分式->分式的基础->分式值为0", "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->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->有理数基础运算->有理数乘法->有理数乘法运算"], "answer_analysis": ["$$\\frac{a}{b}=0$$，由于分母$$b\\ne 0$$，所以$$a=0$$；  $$ab=0$$，可得$$a=0$$或$$b=0$$，$$a$$未必为$$0$$；  $${{a}^{2}}=0$$，可得$$a=0$$；  $${{a}^{2}}+{{b}^{2}}=0$$，则$$a=0$$且$$b=0$$．  所以$$a$$必为$$0$$的式子共有$$3$$个． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "469", "queId": "cc26de9ceaa647e384e12f7605bf4c78", "competition_source_list": ["2007年第18届希望杯初二竞赛第1试第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "让$$k$$依次取$$1$$，$$2$$，$$3$$，$$\\cdots $$等自然数，当取到某一个数之后，以下四个代数式：①$$k+2$$，②$${{k}^{2}}$$，③$$2k$$，④$${{2}^{k}}$$就排成一个不变的大小顺序，这个顺序是．", "answer_option_list": [[{"aoVal": "A", "content": "①$$ ~\\textless{} ~$$②$$ ~\\textless{} ~$$③$$ ~\\textless{} ~$$④ "}], [{"aoVal": "B", "content": "②$$ ~\\textless{} ~$$①$$ ~\\textless{} ~$$③$$ ~\\textless{} ~$$④ "}], [{"aoVal": "C", "content": "①$$ ~\\textless{} ~$$③$$ ~\\textless{} ~$$②$$ ~\\textless{} ~$$④ "}], [{"aoVal": "D", "content": "③$$ ~\\textless{} ~$$②$$ ~\\textless{} ~$$①$$ ~\\textless{} ~$$④ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->整式->整式的乘除运算"], "answer_analysis": ["可逐个试验：  当$$k=1$$时，②$$ ~\\textless{} ~$$③$$=$$④$$ ~\\textless{} ~$$①；  当$$k=2$$时，①$$=$$②$$=$$③$$=$$④；  当$$k=3$$时，①$$ ~\\textless{} ~$$③$$ ~\\textless{} ~$$④$$ ~\\textless{} ~$$②；  当$$k=4$$时，①$$ ~\\textless{} ~$$③$$ ~\\textless{} ~$$②$$=$$④；  当$$k\\geqslant 5$$时，①$$ ~\\textless{} ~$$③$$ ~\\textless{} ~$$②$$ ~\\textless{} ~$$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": "317", "queId": "5914587f674a41d8a58faa39f192ff5c", "competition_source_list": ["2016年第33届全国全国初中数学联赛竞赛第2题7分", "2016年上海浦东新区进才中学自主招生第6题10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．", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$种 "}], [{"aoVal": "B", "content": "$$10$$种 "}], [{"aoVal": "C", "content": "$$11$$种 "}], [{"aoVal": "D", "content": "$$12$$种 "}]], "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->二元一次方程组的配套问题", "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->二元一次方程组的解"], "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": "41", "queId": "0309623c0ffc4809bd5be97f2897d153", "competition_source_list": ["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": "785", "queId": "45a5a4408cdf40b8ac7a7347a09f3a08", "competition_source_list": ["2002年第13届希望杯初二竞赛第1试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$P$$是等边$$\\triangle ABC$$内任意一点，从点$$P$$作三边的垂线$$PD$$，$$PE$$，$$PF$$，点$$D$$，$$E$$，$$F$$是垂足，则$$\\frac{PD+PE+PF}{AB+BC+CA}$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{3}}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{3}}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2\\sqrt{3}}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角形->三角形基础"], "answer_analysis": ["取特殊点法．  当$$P$$点为$$\\triangle ABC$$三条垂直平分线交点时，因为$$\\triangle ABC$$为等边三角形．  故$$P$$点既是$$\\triangle ABC$$内心又是重心．  $$PD=\\frac{1}{3}AD=\\frac{1}{3}\\times \\frac{\\sqrt{3}}{2}AB=\\frac{\\sqrt{3}}{6}AB$$．  又$$PD=PE=PF$$，$$AB=AC=BC$$．  ∴$$\\frac{PD+PE+PF}{AB+BC+CA}=\\frac{\\sqrt{3}}{6}$$，  选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "428", "queId": "2c740543a88648f0ac91c985bd0d56d0", "competition_source_list": ["2016年上海浦东新区进才中学自主招生第8题10分", "2016年第33届全国全国初中数学联赛竞赛A卷第3题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果一个正整数可以表示为两个连续奇数的立方差，则称这个正整数为``和谐数''．如：$$2={{1}^{3}}-{{(-1)}^{3}}$$，$$26={{3}^{3}}-{{1}^{3}}$$，$$2$$和$$26$$均为``和谐数''．那么，不超过$$2016$$的正整数中，所有的``和谐数''之和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$6858$$ "}], [{"aoVal": "B", "content": "$$6860$$ "}], [{"aoVal": "C", "content": "$$9260$$ "}], [{"aoVal": "D", "content": "$$9262$$ "}]], "knowledge_point_routes": ["课内体系->知识点->综合与实践->新定义->新定义综合其它", "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->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->能力->运算能力"], "answer_analysis": ["$${{(2k+1)}^{3}}-{{(2k-1)}^{3}}$$  $$=\\left[ (2k+1)-(2k-1) \\right]\\left[ {{(2k+1)}^{2}}+(2k+1)(2k-1)+{{(2k-1)}^{2}} \\right]$$  $$=2(12{{k}^{2}}+1)$$（其中$$k$$为非负整数），  由$$2(12{{k}^{2}}+1)\\leqslant 2016$$得，$$k\\leqslant 9$$，  ∴$$k=0$$，$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$9$$，  即得所有不超过$$2016$$的``和谐数''，它们的和为  $$\\left[ {{1}^{3}}-{{(-1)}^{3}} \\right]+({{3}^{3}}-{{1}^{3}})+({{5}^{3}}-{{3}^{3}})+\\cdots +({{17}^{3}}-{{15}^{3}})+({{19}^{3}}-{{17}^{3}})={{19}^{3}}+1=6860$$．  ``和谐数''设为$$m$$，它一定可以表示为$$m={{n}^{3}}-{{(n-2)}^{3}}$$，其中$$n$$为正整数．题目要求$$m\\leqslant 2016$$，大致计算可得$$n$$最大为$$19$$，于是有  $$2={{1}^{3}}-{{(-1)}^{3}}$$，  $$26={{3}^{3}}-{{1}^{3}}$$，  $$\\cdots $$  $$1946={{19}^{3}}-{{17}^{3}}$$，  等式右边两两相消是有  ``和谐数''总和$$={{19}^{3}}-{{(-1)}^{3}}=6860$$，选$$\\text{B}$$. "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "87", "queId": "07b09358cfbb4667befb958aa2709799", "competition_source_list": ["1994年第5届希望杯初二竞赛第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "对于三角形的三个外角，下面结论中正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "可能有两个直角 "}], [{"aoVal": "B", "content": "最少有一个锐角 "}], [{"aoVal": "C", "content": "不可能有三个钝角 "}], [{"aoVal": "D", "content": "最多有一个锐角 "}]], "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"], "answer_analysis": ["由于三角形的三个内角最多只能有一个钝角或者直角，  所以它的三个外角中，不可能有两个直角，  可能有三个钝角（此时三角形的三个内角均为锐角）．  否定了$$\\text{A}$$，$$\\text{B}$$，$$\\text{C}$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "5", "queId": "04b873553d0b494badcaac8811ffd7f8", "competition_source_list": ["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", "初一上学期单元测试《解读绝对值》第3题", "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": "如果$$a$$，$$b$$，$$c$$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->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": "602", "queId": "8bc7a43e17dd4e7487df692cfc344e07", "competition_source_list": ["1999年第10届希望杯初一竞赛第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a ~\\textless{} ~0$$，则下述命题中正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$的偶次方的偶次方是负数 "}], [{"aoVal": "B", "content": "$$a$$的奇次方的偶次方是负数 "}], [{"aoVal": "C", "content": "$$a$$的奇次方的奇次方是负数 "}], [{"aoVal": "D", "content": "$$a$$的偶次方的奇次方是负数 "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["有理数的偶次方得非负数．即可排除$$\\text{A}$$、$$\\text{B}$$．  $$a ~\\textless{} ~0$$而$$a$$的偶次方是正数，正数的奇次方是正数，所以排除$$\\text{D}$$，因此应选$$\\text{C}$$．  事实上$$a ~\\textless{} ~0$$，$$a$$的奇次方是负数，负数的奇次方仍是负数，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": "529", "queId": "be6236895488497b8622dfa8d86df4a9", "competition_source_list": ["2001年竞赛（全国初中数学竞赛）第2题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$$a$$，$$b$$，$$c$$是三个任意整数，那么$$\\frac{a+b}{2}$$，$$\\frac{b+c}{2}$$，$$\\frac{c+a}{2}$$．", "answer_option_list": [[{"aoVal": "A", "content": "都不是整数 "}], [{"aoVal": "B", "content": "至少有两个整数 "}], [{"aoVal": "C", "content": "至少有一个整数 "}], [{"aoVal": "D", "content": "都是整数 "}]], "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$$a$$，$$b$$，$$c$$是任意整数，所以$$a$$，$$b$$，$$c$$中至少有两个同为奇数或同为偶数，而两个奇数或两个偶数的和是偶数，故$$\\frac{a+b}{2}$$，$$\\frac{b+c}{2}$$，$$\\frac{c+a}{2}$$中至少有一个是整数，如当$$a=1$$，$$b=2$$ ，$$c=3$$时，$$\\frac{a+c}{2}=2$$，$$\\frac{a+b}{2}=\\frac{3}{2}$$，$$\\frac{b+c}{2}=\\frac{5}{2}$$．此时，恰有一个整数．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "942", "queId": "a8b83fb0c1df49f0a2e2631d28605adc", "competition_source_list": ["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", "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"], "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": "352", "queId": "8fc35bb8821b4610838a701dd246f87d", "competition_source_list": ["2009年第20届希望杯初二竞赛第1试第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "在一次视力检查中，初二$$(1)$$班的$$50$$人中只有$$8$$人的视力达标．用扇形图表示视力检查结果，则表示视力达标的扇形的圆心角是．", "answer_option_list": [[{"aoVal": "A", "content": "$$64.8{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$57.6{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$48{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$16{}^{}\\circ $$ "}]], "knowledge_point_routes": ["竞赛->知识点->数与式->数的运算->有理数运算问题"], "answer_analysis": ["周角为$$360{}^{}\\circ $$，对应全体学生，则表示视力达标的扇形的圆心角为  $$\\frac{8}{50}\\times 360{}^{}\\circ =57.6{}^{}\\circ $$，故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "793", "queId": "9ee6a854080e40a999031c122d7fdc09", "competition_source_list": ["1994年全美数学竞赛（AMC）竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "1994年全美数学竞赛（AMC）竞赛  The number halfway between $$\\frac{1}{6}$$ and $$\\frac{1}{4}$$ is．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{10}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{24}$$ "}], [{"aoVal": "D", "content": "$$\\frac{7}{24}$$ "}], [{"aoVal": "E", "content": "$$\\frac{5}{12}$$ "}]], "knowledge_point_routes": ["美国AMC8->Knowledge Point->Number and Operations->Fractions->Fraction Basics", "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": ["$$\\frac{1}{6}$$和$$\\frac{1}{4}$$中间的数是．  中间的数字是平均数，  $$\\frac{\\dfrac{1}{6}+\\dfrac{1}{4}}{2}=\\frac{\\dfrac{2}{12}+\\dfrac{3}{12}}{2}=\\text{C}\\frac{5}{24}$$．  The number halfway between is the average.  $$\\dfrac{\\dfrac{1}{6}+\\dfrac{1}{4}}{2}=\\dfrac{\\dfrac{2}{12}+\\dfrac{3}{12}}{2}=\\dfrac{5}{24}$$. "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "926", "queId": "6a337fb8ff5c41588e54b80b74b0e018", "competition_source_list": ["2004年第21届全国初中数学联赛竞赛第1题", "2004年第21届全国初中数学联赛竞赛（第二试A）第1题"], "difficulty": "3", "qtype": "single_choice", "problem": "已知关于$$x$$的方程$${{x}^{2}}-6x-4{{n}^{2}}-32n=0$$的根都是整数，则整数$$n$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$或$$-18$$ "}], [{"aoVal": "B", "content": "$$0$$或$$-8$$ "}], [{"aoVal": "C", "content": "$$10$$或$$-18$$或$$0$$或$$-8$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "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->根与系数的关系->一元二次方程根与系数的关系"], "answer_analysis": ["利用求根公式得到$$x=3\\pm \\sqrt{4{{n}^{2}}+32n+9}$$，  为使$$x$$为整数，只要求$$4{{n}^{2}}+32n+9={{k}^{2}}$$$$\\left( k\\geqslant 0 \\right)$$，（这种方法在处理完全平方数的问题中很常用）  配方得到$${{\\left( 2n+8 \\right)}^{2}}-{{k}^{2}}=55$$，  即$$\\left( 2n+8+k \\right)\\left( 2n+8-k \\right)=55$$，  考虑到$$k$$是非负的，所以$$2n+8+k=55$$或$$11$$或$$-1$$或$$-5$$，  对应的$$2n+8-k=1$$或$$5$$或$$-55$$或$$-11$$．  得到$$n=10$$或$$0$$或$$-18$$或$$-8$$．  由题方程$${{x}^{2}}-6x-4{{n}^{2}}-32n=0$$的根都是整数，  则$${{b}^{2}}-4ac=36-4(-4{{n}^{2}}-32n)=4(4{{n}^{2}}+32n+9)$$为一个完全平方数，  所以$$4{{n}^{2}}+32n+9$$为一个完全平方数，  不妨设$$4{{n}^{2}}+32n+9={{k}^{2}}$$$$(k\\geqslant 0)$$，（这种方法在处理完全平方数的问题中很常用）  配方得到$${{(2n+8)}^{2}}-{{k}^{2}}=55$$，  即$$(2n+8+k)(2n+8-k)=55$$，考虑到$$k$$是非负的，  所以$$2n+8+k=55$$或$$11$$或$$-1$$或$$-5$$，  对应的$$2n+8-k=1$$或$$5$$或$$-55$$或$$-11$$．得到$$n=10$$或$$0$$或$$-18$$或$$-8$$． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1429", "queId": "e9ec7c4a483e49c2947d86fd642c2265", "competition_source_list": ["2005年第16届希望杯初二竞赛复赛第2题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$b\\textless{}a\\textless{}0$$，$${{a}^{2}}+{{b}^{2}}=\\frac{5}{2}ab$$，则$$\\frac{a+b}{a-b}$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$-3$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->能力->运算能力", "课内体系->知识点->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"], "answer_analysis": ["由$${{a}^{2}}+{{b}^{2}}=\\frac{5}{2}ab$$，得  $${{(a+b)}^{2}}=\\frac{9}{2}ab$$，$${{(a-b)}^{2}}=\\frac{1}{2}ab$$，  即$${{\\left( \\frac{a+b}{a-b} \\right)}^{2}}=9$$．  又$$b\\textless{}a\\textless{}0$$，  所以$$a+b\\textless{}0$$，$$a-b\\textgreater0$$，  所以$$\\frac{a+b}{a-b}=-3$$．  故选($$\\text{C}$$）． "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "757", "queId": "7f9f630300de4f84b7ea61d3d4b0a555", "competition_source_list": ["2003年第14届希望杯初二竞赛第2试第3题", "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": "整数$$x$$、$$y$$满足不等式$${{x}^{2}}+{{y}^{2}}+1\\leqslant 2x+2y$$，则$$x+y$$的值有．", "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->方程与不等式->不等式（组）->不等式->不等式的性质"], "answer_analysis": ["由题意，有$${{x}^{2}}-2x+1+{{y}^{2}}-2y+1\\leqslant 1$$，即$${{\\left( x-1 \\right)}^{2}}+{{\\left( y-1 \\right)}^{2}}\\leqslant 1$$.  $$\\because {{\\left( x-1 \\right)}^{2}}\\geqslant 0$$，$${{\\left( y-1 \\right)}^{2}}\\geqslant 0$$，$$x$$、$$y$$为整数，  $$\\therefore \\begin{cases}x-1=0    y-1=0 \\end{cases}$$或$$\\begin{cases}x-1=\\pm 1    y-1=0 \\end{cases}$$或$$\\begin{cases}x-1=0    y-1=\\pm 1 \\end{cases}$$，  解得$$\\left( x,y \\right)=\\left( 1,1 \\right)\\left( 2,1 \\right)\\left( 0,1 \\right)\\left( 1,2 \\right)\\left( 1,0 \\right)$$，故$$x+y$$的不同值是$$1$$、$$2$$、$$3$$. "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "593", "queId": "48446516a4b146fc8ba13b0aa0f1a6b2", "competition_source_list": ["2019~2020学年9月四川成都青羊区成都市树德中学初一上学期周测A卷第4题3分", "1999年第10届希望杯初一竞赛第2题"], "difficulty": "0", "qtype": "single_choice", "problem": "下列命题正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$是最小的正有理数 "}], [{"aoVal": "B", "content": "$$-1$$是最大的负有理数 "}], [{"aoVal": "C", "content": "$$0$$是最小的正整数 "}], [{"aoVal": "D", "content": "$$0$$是最大的非正整数 "}]], "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"], "answer_analysis": ["$$\\text{A}$$、$$\\text{B}$$．没有最小的正有理数，也没有最大的负有理数，故$$\\text{A}$$、$$\\text{B}$$错误；  $$\\text{C}$$．$$0$$既不是正数也不是负数，故$$\\text{C}$$错误；  $$\\text{D}$$．$$0$$是最大的非正整数，故$$\\text{D}$$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": "1417", "queId": "c10914c031444269993a0e0a22226876", "competition_source_list": ["2017年第1届重庆全国初中数学联赛竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "分式$$\\frac{6{{x}^{2}}+12x+10}{{{x}^{2}}+2x+2}$$可取得的最小值为~\\uline{~~~~~~~~~~}~.", "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->知识点->式->分式->分式的基础->分式的值", "知识标签->题型->式->分式->分式的运算->题型：分式约分"], "answer_analysis": ["$$\\frac{6{{x}^{2}}+12x+10}{{{x}^{2}}+2x+2}=\\frac{6{{x}^{2}}+12x+12-2}{{{x}^{2}}+2x+2}=6-\\frac{2}{{{x}^{2}}+2x+2}=6-\\frac{2}{{{\\left( x+1 \\right)}^{2}}+1}\\geqslant 6-\\frac{2}{1}=4$$ "], "answer_value": "C"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1567", "queId": "eff9747b32ab4ae09dd7099c8651d9d8", "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": ["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": "684", "queId": "762412d28a044417834dcd8c45055c68", "competition_source_list": ["2007年第12届华杯赛初一竞赛初赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$\\frac{a}{b}\\textgreater0$$，$$\\frac{c}{d}\\textgreater0$$．有如下四个结论：  ①如果$$ad\\textgreater bc$$，则必定有$$\\frac{a}{b}\\textgreater\\frac{c}{d}$$  ②如果$$ad\\textgreater bc$$，则必定有$$\\frac{a}{b} ~\\textless{} ~\\frac{c}{d}$$  ③如果$$ad ~\\textless{} ~bc$$，则必定有$$\\frac{a}{b} ~\\textless{} ~\\frac{c}{d}$$  ④如果$$ad ~\\textless{} ~bc$$，则必定有$$\\frac{a}{b}\\textgreater\\frac{c}{d}$$  其中正确结论的个数是．", "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->数->有理数->有理数基础运算->有理数乘法->有理数的乘法法则"], "answer_analysis": ["取$$a=-2$$，$$b=-3$$，$$c=4$$，$$d=5$$，$$\\frac{a}{b}=\\frac{-2}{-3}=\\frac{2}{3}$$，$$\\frac{c}{d}=\\frac{4}{5}$$即$$\\frac{a}{b}$$和$$\\frac{c}{d}$$是正数，且$$ad\\textgreater bc$$，此时结论①不成立；  取$$a=4$$，$$b=5$$，$$c=2$$，$$d=3$$，即$$\\frac{a}{b}$$和$$\\frac{c}{d}$$是正数，并且$$ad\\textgreater bc$$，此时结论②不成立；  若③正确，则①正确；  若④正确，则②正确．所以，$$4$$个结论无一正确．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "766", "queId": "df2f848a1ec2416889b4de89d1c9b4b2", "competition_source_list": ["1992年第3届希望杯初二竞赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知三角形的三个内角度数之比为$$1:2:3$$，若这个三角形的最短边长为$$\\sqrt{2}$$，那么它的最长边等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$2\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$3\\sqrt{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角形->直角三角形->直角三角形的性质->30°特殊角的性质应用", "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->解直角三角形的综合应用", "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": ["根据三角形内角和为180°及三个内角度数之比为$$1:2:3$$，容易得出三个内角为$$30{}^{}\\circ $$，$$60{}^{}\\circ $$，$$90{}^{}\\circ $$，$$30{}^{}\\circ $$角对边为最短边，由题设知，它的边长为$$\\sqrt{2}\\cdot 90{}^{}\\circ $$角所对边为最长边，即直角三角形的斜边，其长应为$$30{}^{}\\circ $$角对边的$$2$$倍，即为$$2\\sqrt{2}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "1222", "queId": "a07edc14d52f42ac90134b09ccfdac8e", "competition_source_list": ["1999年第16届全国初中数学联赛竞赛第6题7分", "初三上学期其它"], "difficulty": "2", "qtype": "single_choice", "problem": "有下列三个命题：  （甲）若$$\\alpha $$、$$\\beta $$是不相等的无理数，则$$\\alpha \\beta +\\alpha -\\beta $$是无理数；  （乙）若$$\\alpha $$、$$\\beta $$是不相等的无理数，则$$\\frac{\\alpha -\\beta }{\\alpha +\\beta }$$是无理数；  （丙）若$$\\alpha $$、$$\\beta $$是不相等的无理数，则$$\\sqrt{\\alpha }+\\sqrt[3]{\\beta }$$是无理数．  其中正确命题的个数是．", "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->数->实数->无理数有关的计算->无理数的定义"], "answer_analysis": ["因为$$\\alpha \\beta +\\alpha -\\beta =\\alpha \\beta +\\alpha -\\beta -1+1=\\left( \\alpha -1 \\right)\\left( \\beta +1 \\right)+1$$，  只要令$$\\alpha =1+\\sqrt{2}$$，$$\\beta =-1+\\sqrt{2}$$，则$$\\alpha \\beta +\\alpha -\\beta =3$$为有理数，所以（甲）不对；  又若令$$\\alpha =2\\sqrt{2}$$，$$\\beta =\\sqrt{2}$$，则$$\\frac{\\alpha -\\beta }{\\alpha +\\beta }=\\frac{1}{3}$$为有理数，所以（乙）不对；  又若令$$\\alpha =\\sqrt[3]{2}$$，$$\\beta =-\\sqrt{2}$$，则$$\\sqrt{\\alpha }+\\sqrt[3]{\\beta }=0$$为有理数，所以（丙）不对．  因此正确命题个数是$$0$$．  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": "248", "queId": "828668852a494088a946a438c8823971", "competition_source_list": ["2010年第21届希望杯初二竞赛第1试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$${{a}^{2}}\\geqslant {{a}^{3}}\\geqslant 0$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{a}\\geqslant \\sqrt[3]{a}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{a}\\leqslant \\sqrt[3]{a}$$ "}], [{"aoVal": "C", "content": "$$a\\geqslant 1$$ "}], [{"aoVal": "D", "content": "$$0\\textless{}a\\textless{}1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->方程与不等式->不等式->二次不等式"], "answer_analysis": ["由$${{a}^{3}}\\geqslant 0$$，得$$a\\geqslant 0$$，  由$${{a}^{2}}\\geqslant {{a}^{3}}$$得，$${{a}^{2}}(1-a)\\geqslant 0$$，  又$${{a}^{2}}\\geqslant 0$$得，$$1-a\\geqslant 0$$，$$a\\leqslant 1$$，  即$$0\\leqslant a\\leqslant 1$$，  所以$$\\sqrt{a}\\leqslant \\sqrt[3]{a}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "mid_math_competition_ch_single_choice_1.5K_dev", "dataset_version": "2023-07-07", "qid": "244", "queId": "194fe942217040aa950db13ce4bcde66", "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$$\\angle BOC$$互为邻补角，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": ["课内体系->知识点->几何图形初步->角->角度的运算->角的和差的计算与证明->角的和差的计算与证明-不需要分类讨论"], "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": "628", "queId": "6cb6f211c8d1431295a0a18d754f28ef", "competition_source_list": ["2010年第21届全国希望杯初一竞赛初赛第8题4分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$是三个连续的正整数，则（~ ）．  （说明：$$a$$可被$$b$$整除，记作$$b\\textbar a$$．）", "answer_option_list": [[{"aoVal": "A", "content": "$$a_{1}^{3}\\textbar({{a}_{1}}{{a}_{2}}{{a}_{3}}+{{a}_{2}})$$ "}], [{"aoVal": "B", "content": "$$a_{2}^{3}\\textbar({{a}_{1}}{{a}_{2}}{{a}_{3}}+{{a}_{2}})$$ "}], [{"aoVal": "C", "content": "$$a_{3}^{3}\\textbar({{a}_{1}}{{a}_{2}}{{a}_{3}}+{{a}_{2}})$$ "}], [{"aoVal": "D", "content": "$${{a}_{1}}{{a}_{2}}{{a}_{3}}\\textbar({{a}_{1}}{{a}_{2}}{{a}_{3}}+{{a}_{2}})$$ "}]], "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", "竞赛->知识点->数论->整除->整除的概念与基本性质"], "answer_analysis": ["依题意，设$${{a}_{1}}=n$$（$$n$$为正整数），  则$${{a}_{2}}=n+1$$，$${{a}_{3}}=n+2$$．  所以$${{a}_{1}}{{a}_{2}}{{a}_{3}}+{{a}_{2}}=n(n+1)(n+2)+(n+1)=(n+1)\\left[ n(n+2)+1 \\right]={{(n+1)}^{3}}=a_{2}^{3}$$，  所以$$a_{2}^{3}\\textbar({{a}_{1}}{{a}_{2}}{{a}_{3}}+{{a}_{2}})$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "885", "queId": "c8dd5ea0c5284cfebf16f6b096ef1d4d", "competition_source_list": ["2020~2021学年广东广州天河区广州中学高二上学期期中第7题5分", "2016年北京西城区高三二模理科第5题5分", "2017~2018学年山东泰安高三上学期期中理科第2题5分", "2015年吉林全国高中数学联赛竞赛初赛第2题6分", "2016~2017学年北京西城区高二下学期期末文科第5题5分", "2016~2017学年北京丰台区北京市第十二中学高二上学期期末理科第2题5分", "2017~2018学年山东泰安高三上学期期中文科第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "``$$a$$，$$b$$，$$c$$，$$d$$成等差数列''是``$$a+d=b+c$$''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": "充分必要条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["课内体系->知识点->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->素养->逻辑推理"], "answer_analysis": ["由等差数列的性质可知命题的充分性，取$$a=1$$，$$b=2$$，$$c=4$$，$$d=5$$可知不满足必要性，即为充分不必要条件． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "780", "queId": "8962b7a621384c8086a7e0b3124e4594", "competition_source_list": ["1986年全国高中数学联赛竞赛一试第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "设$$-1\\textless{}a\\textless{}0$$，$$\\theta =\\arcsin a$$，那么不等式$$\\sin x\\textless{}a$$的解集为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left { x\\textbar2n \\pi +\\theta \\textless{}x\\textless{}\\left( 2n+1 \\right) \\pi -\\theta ,n\\in \\mathbf{Z} \\right }$$ "}], [{"aoVal": "B", "content": "$$\\left { x\\textbar2n \\pi -\\theta \\textless{}x\\textless{}\\left( 2n+1 \\right) \\pi +\\theta ,n\\in \\mathbf{Z} \\right }$$ "}], [{"aoVal": "C", "content": "$$\\left { x\\textbar\\left( 2n-1 \\right) \\pi +\\theta \\textless{}x\\textless{}2n \\pi -\\theta ,n\\in \\mathbf{Z} \\right }$$ "}], [{"aoVal": "D", "content": "$$\\left { x\\textbar\\left( 2n-1 \\right) \\pi- \\theta \\textless{}x\\textless{}2n \\pi +\\theta ,n\\in \\mathbf{Z} \\right }$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->反三角函数", "竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["∵ $$-1\\textless{}a\\textless{}0$$，  ∴$$\\theta =\\arcsin a\\in \\left( -\\frac{ \\pi }{2},0 \\right)$$．  在区间$$(- \\pi , \\pi )$$中，  满足方程$$\\sin x=a$$的解为：  $$x=- \\pi -\\arcsin a$$或$$x=\\arcsin a$$．  因此，不等式$$\\sin x\\textless{}a$$的解集为：  $$\\left { x\\textbar\\left( 2n-1 \\right) \\pi -\\theta \\textless{}x\\textless{}2n \\pi +\\theta ,n\\in \\mathbf{Z} \\right }$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "250", "queId": "2fce74d281234bb99f5f8057a6b4459d", "competition_source_list": ["2009年竞赛珠海市第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某种放射性元素，$$100$$年后只剩原来质量的一半，现有这种元素$$1$$克，$$3$$年后剩下．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3\\times 0.5}{100}$$克 "}], [{"aoVal": "B", "content": "$${{(1-0.5 \\%)}^{3}}$$克 "}], [{"aoVal": "C", "content": "$$0.925$$克 "}], [{"aoVal": "D", "content": "$$\\sqrt[100]{0.125}$$克 "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->解方程（组）"], "answer_analysis": ["列方程，$${{x}^{100}}=0.5$$，则$${{x}^{3}}=\\sqrt[100]{0.125}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "184", "queId": "1d4e3d4519f6443d9ac1e4a40fe1eaf3", "competition_source_list": ["2006年全国高中数学联赛竞赛一试第11题9分"], "difficulty": "2", "qtype": "single_choice", "problem": "方程$$\\left( {{x}^{2006}}+1 \\right)\\left( 1+{{x}^{2}}+{{x}^{4}}+\\cdots +{{x}^{2004}} \\right)=2006\\cdot {{x}^{2005}}$$的实数解的个数为~\\uline{~~~~~~~~~~}~．", "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）", "课内体系->知识点->等式与不等式->不等式->基本不等式->多元均值不等式的应用"], "answer_analysis": ["$$\\left( {{x}^{2006}}+1 \\right)\\left( 1+{{x}^{2}}+{{x}^{4}}+\\cdots +{{x}^{2004}} \\right)=2006\\cdot {{x}^{2005}}$$，  $$\\Leftrightarrow \\left( x+\\frac{1}{{{x}^{2005}}} \\right)\\left( 1+{{x}^{2}}+{{x}^{4}}+\\cdots +{{x}^{2004}} \\right)=2006$$，  $$\\Leftrightarrow x+{{x}^{3}}+{{x}^{5}}+\\cdots +{{x}^{2005}}+\\frac{1}{{{x}^{2005}}}+\\frac{1}{{{x}^{2003}}}+\\frac{1}{{{x}^{2001}}}+\\cdots +\\frac{1}{x}=2006$$．  当$$x\\textgreater0$$时，方程左边$$\\geqslant 2\\cdot 1003=2006$$，等号当$$x=\\frac{1}{x},{{x}^{3}}=\\frac{1}{{{x}^{3}}},\\cdots ,{{x}^{2005}}=\\frac{1}{{{x}^{2005}}}$$时，即$$x=1$$时取得；  当$$x\\textless{}0$$时，方程左边$$\\textless{}0$$，显然不满足方程．  综上，$$x=1$$是原方程的全部解．因此原方程的实数解个数为$$1$$．  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": "54", "queId": "06edbfe2e9b94ee38e0cb9a0b1becc3e", "competition_source_list": ["2014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": "若关于$$x$$的不等式$$\\frac{4x}{a}+\\frac{1}{x}\\geqslant 4$$在区间$$\\left[ 1,  2 \\right]$$上恒成立，则实数$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 0,\\frac{4}{3} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left( 1,\\frac{4}{3} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left[ 1,\\frac{4}{3} \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left[ \\frac{16}{7},\\frac{4}{3} \\right]$$ "}]], "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->函数的概念"], "answer_analysis": ["取$$x=1$$，得$$\\frac{4}{a}\\geqslant 3$$，可知$$0\\textless{}a\\leqslant \\frac{4}{3}$$，$$a$$是正数．这样原不等式变为  $$a\\leqslant \\frac{4x}{4-\\frac{1}{x}}$$，$$\\forall x\\in \\left[ 1,  2 \\right]$$．  从而只需求函数$$f\\left( x \\right)=\\frac{4x}{4-\\frac{1}{x}}$$在区间$$\\left[ 1,  2 \\right]$$的最小值．易知$${f}'\\left( x \\right)=\\frac{8x\\left( 2x-1 \\right)}{{{\\left( 4x-1 \\right)}^{2}}}$$在$$\\left[ 1,  2 \\right]$$上恒为正数，即$$f\\left( x \\right)$$在$$\\left[ 1,  2 \\right]$$上单调增，故$$f\\left( x \\right)$$的最小值为$$f\\left( 1 \\right)=\\frac{4}{3}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "27", "queId": "0268c5c2ca4f43799f63743c663023b3", "competition_source_list": ["2011年江西全国高中数学联赛竞赛初赛第7题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "满足$${{x}^{2}}+7{{y}^{2}}=2011$$的一组正整数$$(x,  y)$$有组.", "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->数论模块->不定方程->同余分析"], "answer_analysis": ["由于$$2011$$是$$4N+3$$形状的数，所以$$y$$必为奇数，而$$x$$为偶数， 设$$x=2m$$，$$y=2n+1$$，代人得  $$4{{m}^{2}}+28n(n+1)=2004$$，  即$${{m}^{2}}+7n(n+1)=501$$. ①  而$$n(n+1)$$为偶数，则$${{m}^{2}}$$为奇数，设$$m=2k+1$$，则  $${{m}^{2}}=4k(k+1)+1$$，  由①得，$$k(k+1)+7\\cdot \\frac{n(n+1)}{4}=125$$， ②  则$$\\frac{n(n+1)}{4}$$为奇数，且$$n,  n+1$$中恰有一个是$$4$$的倍数，当$$n=4r$$，为使$$7\\cdot \\frac{n(n+1)}{4}=7r(4r+1)$$为奇数，且$$7r(4r+1)\\textless{}125$$，只有$$r=1$$，②成为$$k(k+1)+35=125$$，  即$$k(k+1)=90$$，于是$$n=4,  k=9,  x=38,  y=9$$；  若$$n+1=4r$$，为使$$7\\cdot \\frac{n(n+1)}{4}=7r(4r-1)$$为奇数，且$$7r(4r-1)\\textless{}125$$，只有$$r=1$$，②成为$$k(k+1)+21=125$$，即$$k(k+1)=104$$，它无整解；  于是$$(x,  y)=(38,  9)$$是唯一解：$${{38}^{2}}+7\\cdot {{9}^{2}}=2011$$．  （另外，也可由$$x$$为偶数出发，使$$2011-{{x}^{2}}=2009-({{x}^{2}}-2)=7\\times 287-({{x}^{2}}-2)$$  为$$7$$的倍数，那么$${{x}^{2}}-2$$是$$7$$的倍数，故$$x$$是$$7k\\pm 3$$形状的偶数，依次取$$k=1,  3,  5$$，检验相应的六个数即可．） "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "598", "queId": "63fbf683ce144c3aacdcf4fb1f633c23", "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8题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知椭圆$$C:\\frac{{{x}^{2}}}{4}+\\frac{{{y}^{2}}}{2}=1$$，过点$$M(1,1)$$的直线与椭面$$C$$相交于$$P$$，$$Q$$两点，若弦$$PQ$$恰被点$$M$$平分，则直线$$l$$的斜率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2$$ "}], [{"aoVal": "B", "content": "$$-\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$-1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "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->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->圆锥曲线->直线与圆锥曲线问题->中点弦问题"], "answer_analysis": ["设$$P({{x}_{1}},{{y}_{1}})$$，$$Q({{x}_{2}},{{y}_{2}})$$，  代入$$\\frac{{{x}^{2}}}{4}+\\frac{{{y}^{2}}}{2}=1$$得$$\\frac{x_{1}^{2}}{4}+\\frac{y_{1}^{2}}{2}=1$$，$$\\frac{x_{2}^{2}}{4}+\\frac{y_{2}^{2}}{2}=1$$，  两式相减得$$\\frac{({{x}_{1}}+{{x}_{2}})({{x}_{1}}-{{x}_{2}})}{4}+\\frac{({{y}_{1}}+{{y}_{2}})({{y}_{1}}-{{y}_{2}})}{2}=0$$，  整理得$$\\frac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}=-\\frac{2}{4}\\cdot \\frac{{{x}_{1}}+{{x}_{2}}}{{{y}_{1}}+{{y}_{2}}}$$，  又因为$$PQ$$恰被点$$M$$平分，  所以$$1=\\frac{{{x}_{1}}+{{x}_{2}}}{2}$$，$$1=\\frac{{{y}_{1}}+{{y}_{2}}}{2}$$，  代入上式得$$\\frac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}}=-\\frac{1}{2}$$，  即直线$$l$$的斜率为$$-\\frac{1}{2}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "59", "queId": "0709a642ddbe4b858fd4a9c09c8d8553", "competition_source_list": ["2012年AMC10竞赛A第7题", "2012年AMC12竞赛A第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2012-AMC10A-7$$  In a bag of marbles, $$\\frac{3}{5}$$ of the marbles are blue and the rest are red. If the number of red marbles is doubled and the number of blue marbles stays the same, what fraction of the marbles will be red?  在一袋弹珠中，$$\\frac{3}{5}$$ 的弹珠是蓝色的，其余的是红色的。 如果红色弹珠的数量翻倍而蓝色弹珠的数量保持不变，那么红色弹珠的比例是多少？（", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{7}$$ "}], [{"aoVal": "C", "content": "$$\\frac{4}{7}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{5}$$ "}], [{"aoVal": "E", "content": "$$\\frac{4}{5}$$ "}]], "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": ["Assume that there are $$5$$ total marbles in the bag. The actual number does not matter, since all we care about is the ratios, and the only operation performed on the marbles in the bag is doubling.  There are $$3$$ blue marbles in the bag and $$2$$ red marbles. If you double the amount of red marbles, there will still be $$3$$ blue marbles but now there will be $$4$$ red marbles. Thus, the answer is $$\\boxed{\\rm(C)\\textasciitilde\\frac{4}{7}}$$.  Let us say that there are $$x$$ marbles in the bag. Therefore, $$\\frac{3x}{5x}$$ are blue, and $$\\frac{2x}{5x}$$ are red.~ When the red marbles are doubled, we now have $$\\frac{2\\times 2x}{5x+2x}= \\frac{4x}{7x}= \\frac{4}{7}\\Rightarrow \\boxed{\\rm(C)}$$. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "576", "queId": "5f318f2c0b074d529808650f8ba3bb1f", "competition_source_list": ["2017~2018学年福建漳州东山县福建省东山第二中学高二上学期期中文科第1题5分", "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9题3分", "2017~2018学年5月江西南昌东湖区南昌市第十中学高一下学期月考文科第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "有下列调查方式：  ①学校为了解高一学生的数学学习情况，从每班抽$$2$$人进行座谈；  ②一次数学竞赛中，某班有$$15$$人在$$100$$分以上，$$35$$人在$$90\\sim 100$$分，$$10$$人低于$$90$$分．现在从中抽取$$12$$人座谈了解情况；  ③运动会中工作人员为参加$$400\\text{m}$$比赛的$$6$$名同学公平安排跑道．  就这三个调查方式，最合适的抽样方法依次为．", "answer_option_list": [[{"aoVal": "A", "content": "分层抽样，系统抽样，简单随机抽样 "}], [{"aoVal": "B", "content": "系统抽样，系统抽样，简单随机抽样 "}], [{"aoVal": "C", "content": "分层抽样，简单随机抽样，简单随机抽样 "}], [{"aoVal": "D", "content": "系统抽样，分层抽样，简单随机抽样 "}]], "knowledge_point_routes": ["课内体系->素养->数据分析", "课内体系->知识点->统计与概率->统计->随机抽样->系统抽样", "课内体系->知识点->统计与概率->统计->随机抽样->分层随机抽样", "课内体系->知识点->统计与概率->统计->随机抽样->简单随机抽样"], "answer_analysis": ["在①中，因为总体已经按班级进行分组，故适合于系统抽样；  在②中，因为总体形成差异明显的三个层次，故适合于分层抽样；  在③中，因为总体单元数较少，故适合于简单随机抽样．  故选：$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1089", "queId": "c5e3de606c7541c490162e4467cbd3f8", "competition_source_list": ["2021年贵州全国高中数学联赛竞赛初赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$y=\\sin x$$与$$y={{\\log }_{2021}}\\textbar x\\textbar$$的图象的交点个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1284$$ "}], [{"aoVal": "B", "content": "$$1285$$ "}], [{"aoVal": "C", "content": "$$1286$$ "}], [{"aoVal": "D", "content": "$$1287$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数方程"], "answer_analysis": ["$$643\\pi \\textless{}2021\\textless{}643.5\\pi $$，故绘制两个函数图象可知两侧交点个数均为$$\\frac{643-1}{2}\\times 2+1=643$$个，  故共有$$1286$$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": "716", "queId": "e889f89591954fde9d35b1a95e125669", "competition_source_list": ["2003年AMC10竞赛A第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "从顶点为$(0,0)$、$(4,0)$、$(4,1)$和$(0,1)$的矩形内随机选取一个点$(x,y)$.$x\\textless~y$的概率是多少?", "answer_option_list": [[{"aoVal": "A", "content": "$$\\dfrac{1}{8}$$ "}], [{"aoVal": "B", "content": "$$\\dfrac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$\\dfrac{3}{8}$$ "}], [{"aoVal": "D", "content": "$$\\dfrac{1}{2}$$ "}], [{"aoVal": "E", "content": "$$\\dfrac{3}{4}$$ "}]], "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->与面积有关的几何概率的计算", "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities"], "answer_analysis": ["The rectangle has a width of $$4$$ and a height of $$1$$.  The area of this rectangle is $$4\\cdot1=4$$.  The line $$x=y$$ intersects the rectangle at $$(0,0)$$ and $$(1,1)$$.  The area which $$x\\lt y$$ is the right isosceles triangle with side length $$1$$ that has vertices at $$(0,0)$$, $$(1,1)$$, and $$(0,1)$$.  The area of this triangle is $$\\dfrac{1}{2}\\cdot1^{2}=\\dfrac{1}{2}$$  Therefore, the probability that $$x\\lt y$$ is $$\\dfrac{\\dfrac{1}{2}}{4}=\\dfrac{1}{8}$$. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "606", "queId": "71a2b5a0bb1f43d785dfd46d80a04edf", "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2~2题"], "difficulty": "0", "qtype": "single_choice", "problem": "已知点$$P(\\sin \\alpha ,\\tan \\alpha )$$在第二象限，则角$$\\alpha $$的终边在（~~~~~~~）", "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->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】\\\\ 利用任意角的三角函数的定义，三角函数在各个象限中的负号，求得角α所在的象限．\\\\ 【详解】\\\\ 解：∵点\\emph{P}（sinα，tanα）在第二象限，\\\\ ∴sinα＜0，tanα＞0，\\\\ 若角α顶点为坐标原点，始边为\\emph{x}轴的非负半轴，则α的终边落在第三象限，\\\\ 故选：C． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "500", "queId": "2de085a027d4423caf878dd2ba6d2821", "competition_source_list": ["竞赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "当$a\\textgreater1$时，若不等式$\\frac{1}{n+1}+\\frac{1}{n+2}+\\cdots +\\frac{1}{2n}\\textgreater\\frac{7}{12}\\left( {{\\log }_{a+1}}x-{{\\log }_{a}}x+1 \\right)$对于不小于2的正整数n恒成立.则x的范围是.", "answer_option_list": [[{"aoVal": "A", "content": "$2\\textless{} x\\textless{} \\frac{29}{17}$ "}], [{"aoVal": "B", "content": "$0\\textless{} x\\textless{} 1$ "}], [{"aoVal": "C", "content": "$0\\textless{} x\\textless{} 4$ "}], [{"aoVal": "D", "content": "$x\\textgreater1$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【详解】\\\\ 设$f\\left( n \\right)=\\frac{1}{n+1}+\\frac{1}{n+2}+\\cdots +\\frac{1}{n+n}$.\\\\ 则$f\\left( n+1 \\right)=f\\left( n \\right)+\\frac{1}{2n+1}+\\frac{1}{2n+2}-\\frac{1}{n+1}=f\\left( n \\right)+\\frac{1}{2n+1}-\\frac{1}{2n+2}\\textgreater f\\left( n \\right)$.\\\\ 所以，$f\\left( n \\right)$为增函数，故$f\\left( n \\right)\\ge f\\left( 2 \\right)$.\\\\ 又已知不等式恒成立，所以，$\\text{ } ! !\\textasciitilde ! !\\text{ }\\left( \\text{lo}{{\\text{g}}_{a+1}}x-\\text{lo}{{\\text{g}}_{a}}x+1 \\right)\\textless{} \\frac{7}{12}$，即$\\text{lo}{{\\text{g}}_{a+1}}x\\textless{} \\text{lo}{{\\text{g}}_{a}}x$.\\\\ 由于$a\\textgreater1$，则$x\\textgreater1$，选D. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "268", "queId": "98ecb0de98b14aa4b825b418312b0477", "competition_source_list": ["2008年甘肃全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$${{\\left( \\frac{\\sqrt{3}}{2}+\\frac{x}{2}\\text{i} \\right)}^{2008}}=f(x)+\\text{ig}(x)$$，其中$$f(x)$$、$$g(x)$$均为实系数多项式，则$$f(x)$$的系数之和是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{\\sqrt{3}}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{3}}{2}$$ "}], [{"aoVal": "C", "content": "$$-\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->多项式与复数综合"], "answer_analysis": ["取$$x=1$$，则$${{\\left( \\frac{\\sqrt{3}}{2}+\\frac{1}{2}\\text{i} \\right)}^{2008}}=f(1)+\\text{ig}(1)$$，  即有$${{\\left( \\cos \\frac{ \\pi }{6}+\\text{i}\\sin \\frac{ \\pi }{6} \\right)}^{2008}}=f(1)+\\text{i}g(1)$$，  从而$${{\\left( \\cos \\frac{ \\pi }{6}+\\text{i}\\sin \\frac{ \\pi }{6} \\right)}^{4}}=f(1)+\\text{i}g(1)$$，  即$$\\cos \\frac{2 \\pi }{3}+\\text{i}\\sin \\frac{2 \\pi }{3}=f(1)+\\text{i}g(1)$$  $$=-\\frac{1}{2}+\\frac{\\sqrt{3}}{2}\\text{i}$$，  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "562", "queId": "3729064e5e054f48baaea401bb18ae2d", "competition_source_list": ["2012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": "如果对一切正实数$$x$$，$$y$$，不等式$$\\frac{y}{4}-{{\\cos }^{2}}x\\geqslant a\\sin x-\\frac{9}{y}$$恒成立，则实数$$a$$的取值范围是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$【-3，3】$ "}], [{"aoVal": "B", "content": "$【-2，2】$ "}], [{"aoVal": "C", "content": "$【-3\\sqrt{2}，3\\sqrt{2}】$ "}], [{"aoVal": "D", "content": "$【-2\\sqrt{2}，2\\sqrt{2}】$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->基本不等式->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->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->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->素养->数学抽象", "课内体系->素养->数学建模", "课内体系->思想->转化化归思想"], "answer_analysis": ["原式$$\\Rightarrow \\dfrac{y}{4}+\\dfrac{9}{y}\\geqslant {{\\cos }^{2}}x+a\\sin x\\Rightarrow {{\\left( \\dfrac{y}{4}+\\dfrac{9}{y} \\right)}_{\\min }}\\geqslant {{({{\\cos }^{2}}x+a\\sin x)}_{\\max }}$$  ∵$$\\dfrac{y}{4}+\\dfrac{9}{y}\\geqslant 2\\sqrt{\\dfrac{y}{4}\\cdot \\dfrac{9}{y}}=3$$，  ∴$${{\\left( \\dfrac{y}{4}+\\dfrac{9}{y} \\right)}_{\\min }}=3$$  ∵$${{\\cos }^{2}}x+a\\sin x=-{{\\sin }^{2}}x+a\\sin x+1$$，设$$t=\\sin x$$，$$t\\in [-1,1]$$  ∴$${{y}_{\\max }}=\\begin{cases}-a,\\dfrac{a}{2}\\leqslant -1    \\dfrac{{{a}^{2}}}{4}+1,-1\\textless{}\\dfrac{a}{2}\\textless{}1    a,\\dfrac{a}{2}\\geqslant 1    \\end{cases}$$  ∴问题转化为$$\\begin{cases}\\dfrac{a}{2}\\leqslant -1    -a\\leqslant 3    \\end{cases}$$或$$\\begin{cases}-1\\textless{}\\dfrac{a}{2}\\textless{}1    \\dfrac{{{a}^{2}}}{4}+1\\leqslant 3    \\end{cases}$$或$$\\begin{cases}\\dfrac{a}{2}\\geqslant 1    a\\leqslant 3    \\end{cases}\\Rightarrow -3\\leqslant a\\leqslant 3$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "984", "queId": "8aac50a74e023208014e28c462045043", "competition_source_list": ["2015~2016学年湖北武汉武昌区华中师大一附中高二下学期期中理科第7题5分", "高一上学期单元测试《导数的应用》竞赛第15题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$f\\left( x \\right)$$是定义在$$\\left( 0,+\\infty \\right)$$上的非负可导函数，且满足$$x{f}'\\left( x \\right)-f\\left( x \\right)\\leqslant 0$$，对任意正数$$a$$、$$b$$，若$$a\\textless b$$，则必有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$af(b)\\leqslant bf(a)$$ "}], [{"aoVal": "B", "content": "$$bf(a)\\leqslant af(b)$$ "}], [{"aoVal": "C", "content": "$$af(a)\\leqslant f(b)$$ "}], [{"aoVal": "D", "content": "$$bf(b)\\leqslant f(a)$$ "}]], "knowledge_point_routes": ["课内体系->方法->构造法", "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", "课内体系->知识点->导数->导数的应用->导数与单调性->利用导数求函数的单调性、单调区间", "课内体系->素养->数学运算"], "answer_analysis": ["设$$g(x)=\\frac{f(x)}{x}$$，  ∴ $${{g}^{\\prime }}(x)=\\frac{{{f}^{\\prime }}(x)x-f(x)}{{{x}^{2}}}\\leqslant 0$$  ∴$$g(x)$$在$$(0,+\\infty )$$上单调递减或为常函数，  ∵$$0\\textless{}a\\textless{}b$$，∴$$g(a)\\geqslant g(b)$$，  ∴$$\\frac{f(a)}{a}\\geqslant \\frac{f(b)}{b}\\Rightarrow bf(a)\\geqslant af(b)$$，  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "579", "queId": "3313cbfda15444d1828514b1769b6b95", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第16题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$p:\\left\\textbar{} 1-\\frac{x-2}{3} \\right\\textbar\\leqslant 2,  q:{{x}^{2}}-2x+1-{{m}^{2}}\\leqslant 0(m\\textgreater0)$$，而$$\\neg p$$是$$\\neg q$$的必要不充分条件，则实数$$m$$的取值范围为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$[10,+\\infty)$$ "}], [{"aoVal": "B", "content": "$$(-\\infty,10]$$ "}], [{"aoVal": "C", "content": "$$[5,+\\infty)$$ "}], [{"aoVal": "D", "content": "$$(-\\infty,5]$$ "}]], "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->充要条件与不等式结合", "课内体系->知识点->常用逻辑用语->命题->四种命题", "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->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->方法->定义法", "课内体系->素养->数学运算"], "answer_analysis": ["$$-2\\leqslant 1-\\frac{x-2}{3}\\leqslant 2$$，解得$$1\\leqslant x\\leqslant 13$$，所以$$\\neg p$$是$$x\\textless{}1$$或$$x\\textgreater11$$．  $${{\\left( x-1 \\right)}^{2}}\\leqslant {{m}^{2}}$$，解得$$1-m\\leqslant x\\leqslant 1+m$$，所以$$\\neg q$$是$$x\\textless{}1-m$$或$$x\\textgreater1+m$$．  因为$$\\neg p$$是$$\\neg q$$的必要不充分条件，且$$x\\textless{}1-m\\textless{}1$$成立，所以$$1+m\\geqslant 11\\Rightarrow m\\geqslant 10$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "332", "queId": "7e3e1d77918a4155a4028ab4ad6f60a1", "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": ["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", "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": "235", "queId": "d9c68ecccf5d42e29562b320b1364300", "competition_source_list": ["2008年安徽全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "下列$$4$$个数中与$$\\cos (1{}^{}\\circ )+\\cos (2{}^{}\\circ )+\\ldots +\\cos (2008{}^{}\\circ )$$最接近的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2008$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2008$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["因为$$\\cos 1{}^{}\\circ +\\cos 2{}^{}\\circ +\\cdots +\\cos 360{}^{}\\circ =0$$，$$2008=360\\cdot 5+208$$，  所以$$\\cos 1{}^{}\\circ +\\cos 2{}^{}\\circ +\\cdots +\\cos 2008{}^{}\\circ =\\cos 1{}^{}\\circ +\\cos 2{}^{}\\circ +\\cdots +\\cos 208{}^{}\\circ $$．  再由$$\\cos x{}^{}\\circ +\\cos \\left( x{}^{}\\circ +180{}^{}\\circ \\right)=0$$，知  $$\\cos 1{}^{}\\circ +\\cos 2{}^{}\\circ +\\cdots +\\cos 208{}^{}\\circ =\\cos 29{}^{}\\circ +\\cos 30{}^{}\\circ +\\cdots +\\cos 180{}^{}\\circ $$．  又$$\\cos x{}^{}\\circ +\\cos \\left( 180{}^{}\\circ -x{}^{}\\circ \\right)=0$$，则  $$\\cos 29{}^{}\\circ +\\cos 30{}^{}\\circ +\\cdots +\\cos 180{}^{}\\circ =\\cos 152{}^{}\\circ +\\cos 153{}^{}\\circ +\\cdots +\\cos 180{}^{}\\circ $$．  显然$$-29\\textless{}\\cos 152{}^{}\\circ +\\cos 153{}^{}\\circ +\\cdots +\\cos 180{}^{}\\circ \\textless{}-1$$，选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "737", "queId": "84a54ab84a8b49ea9da8f6e1d5f7dd9b", "competition_source_list": ["2016年AMC10竞赛第18题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在正立方體的各頂點標上數字$$1$$至$$8$$，每個數字只用一次，且使得各個面上四個頂點的數字和都相等．各頂點安排的數字能由旋轉正立方體而得到相同狀況都視爲一樣的安排．試間有多少種不同安排頂點數字的方法？  Each vertex of a cube is to be labeled with an integer $$1$$ through $$8$$, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangenents that can be obtained from each other through rotations of the cube are considered to be the sane. How nany different arrangements are possible?", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$12$$ "}], [{"aoVal": "E", "content": "$$24$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["Note that the sum of the numbers on each face must be $$18$$, because $$\\frac {1+2+\\cdots +8}2=18$$. So now consider the opposite edges (two edges which are parallel but not on same face of the cube); they must have the same sum value too. Now think about the points $$1$$ and $$8$$. If they are not on the same edge, they must be endpoints of opposite edges}, we should have $$1+x=8+Y$$, but no solution for $$[2,7]$$, contradiction.  The points $$1$$ and $$8$$ are therefore on the same side and all edges parallel must also sum to $$9$$. Now we have $$4$$ parallel sides $$1-8$$, $$2-7$$, $$3-6$$, $$4-5$$. thinking about $$4$$ endpoints number need to have a sum of $$8$$. It is easy to notice only $$1-7-6-4$$ and $$8-2-3-5$$ would work.  So if we fix one direction $$1-8$$ or $$(8-1)$$ all other $$3$$ parallel sides must lay in one particular direction. $$(1-8,7-2,6-3, 4-5)$$ or$$(8-1,2-7,3-6,5-4)$$  Now, the problem is same as the problem to arrange $$4$$ points in a two-dimensional square. which is $$\\frac {4!}4=6$$.  Again, all faces sum to $$18$$. If $$x$$, $$y$$, $$z$$ are the vertices next to $$1$$, then the remaining vertices are $$17-x-y$$, $$17-y-x$$, $$17-x-z$$, $$x+y+z-16$$. Now it remains to test possibilities. Note that we must have $$x+y+z\\textgreater17$$. Without loss of generality, let $$x\\textless y\\textless z$$. $$3$$, $$7$$, $$8$$: Does not work. $$4$$, $$6$$, $$8$$： Works $$4$$, $$7$$, $$8$$: Works. $$5$$, $$6$$, $$7$$: Does not work. $$5$$, $$6$$, $$8$$: Does not work. $$5$$, $$7$$, $$8$$: Does not work. $$6$$, $$7$$, $$8$$: Works.  Keeping in mind that a) solutions that can be obtained by rotating each other count as one solution and b) a cyclic sequence is always asymmetrical, we can see that there are two solutions (one with $$[x,y,z]$$ and one with $$[z,y,x]$$) for each combination of $$x$$, $$y$$, and $$z$$ from above. So, our answer is $$3\\cdot 2=6$$.  We know the sum of each face is $$18$$. If we look at an edge of the cube whose numbers sum to $$x$$, it must be possible to achieve the sum $$18-x$$ in two distinct ways, looking at the two faces which contain the edge. If $$8$$ and $$6$$ were on the same edge, it is possible to achieve the desired sum only with the numbers $$1$$ and $$3$$ since the values must be distinct. Similarly, if $$8$$ and $$7$$ were on the same edge, the only way to get the sum is with $$1$$ and $$2$$. This means that $$6$$ and $$7$$ are not on the same edge as $$8$$, or in other words they are diagonally across from it on the same face, or on the other end of the cube. Now we look at three cases, each yielding two solutions which are reflections of each other: $$1$$) $$6$$ and $$7$$ are diagonally opposite $$8$$ on the same face. $$2$$) $$6$$ is diagonally across the cube from $$8$$, while $$7$$ is diagonally across from $$8$$ on the same face. $$3$$) $$7$$ is diagonally across the cube from $$8$$, while $$6$$ is diagonally across from $$8$$ on the same face.  This means the answer is $$3\\cdot 2=6$$. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "107", "queId": "0bbd8b322dc44081a3028dc82b639df4", "competition_source_list": ["2014年黑龙江全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，角$$A$$、$$B$$、$$C$$所对的边分别为$$a$$、$$b$$、$$c$$，且$$1+\\frac{\\tan A}{\\tan B}=\\frac{2c}{b}$$，则角$$A$$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": "$$\\frac{2 \\pi }{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角形中的问题->解三角形"], "answer_analysis": ["$$1+\\frac{\\tan A}{\\tan B}=\\frac{2c}{b}\\Leftrightarrow 1+\\frac{\\tan A}{\\tan B}=\\frac{2\\sin C}{\\sin B}\\Leftrightarrow \\frac{\\sin B\\cos A+\\sin A\\cos B}{\\sin B\\cos A}=\\frac{2\\sin C}{\\sin B}$$  $$\\Leftrightarrow \\frac{\\sin C}{\\sin B\\cos A}=\\frac{2\\sin C}{\\sin B}\\Leftrightarrow \\cos A=\\frac{1}{2}$$，故角$$A$$为$$\\frac{ \\pi }{3}$$. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "867", "queId": "dfb607d8bf8840eba259ee008f9cb8dd", "competition_source_list": ["1985年全国高中数学联赛竞赛一试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知方程$$\\arccos \\frac{4}{5}-\\arccos \\left( -\\frac{4}{5} \\right)=\\arcsin x$$，则（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$x=\\frac{24}{25}$$ "}], [{"aoVal": "B", "content": "$$x=-\\frac{24}{25}$$ "}], [{"aoVal": "C", "content": "$$x=0$$ "}], [{"aoVal": "D", "content": "这样的$$x$$不存在 "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->反三角函数"], "answer_analysis": ["∵$$\\text{arc}\\cos \\left( -\\frac{4}{5} \\right)= \\pi -\\arccos \\frac{4}{5}$$  ∴ 原方程为$$2\\arccos \\frac{4}{5}-\\arcsin x= \\pi $$  又∵$$\\arccos x$$是减函数，  ∴$$\\arccos \\frac{4}{5}\\textless{}\\arccos \\frac{\\sqrt{2}}{2}=\\frac{ \\pi }{4}$$  又 $$-\\arcsin x\\mathsf{\\leqslant }\\frac{ \\pi }{2}$$，  ∴$$2\\arccos \\frac{4}{5}-\\arcsin x\\textless{} \\pi $$  此式表明没有能满足方程的$$x$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "546", "queId": "3fad6f13c6c44fe7a99b326b84ceb350", "competition_source_list": ["2014年吉林全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知函数$$f\\left( x \\right)=\\left { \\begin{matrix}0,  x\\textless{}0,   \\pi ,  x=0, \\x+1,  x\\textgreater0\\end{matrix} \\right }$$则$$f\\left { f\\left[ f\\left( -1 \\right) \\right] \\right }$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$ \\pi +1$$ "}], [{"aoVal": "B", "content": "0 "}], [{"aoVal": "C", "content": "1 "}], [{"aoVal": "D", "content": "$$ \\pi $$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的概念"], "answer_analysis": ["$$f\\left { f\\left[ f\\left( -1 \\right) \\right] \\right }=f\\left( f\\left( 0 \\right) \\right)=f\\left( ~\\pi ~\\right)= \\pi +1$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1185", "queId": "eb8e9e69bd8a481d897b569f9fbddc0b", "competition_source_list": ["2013年天津全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$B$$，$$C$$是定点且都不在平面$$\\pi $$上，动点$$A$$在平面$$ \\pi $$上且$$\\sin \\angle ABC{=}\\frac{1}{2}$$．那么，$$A$$点的轨迹是．", "answer_option_list": [[{"aoVal": "A", "content": "椭圆 "}], [{"aoVal": "B", "content": "抛物线 "}], [{"aoVal": "C", "content": "双曲线 "}], [{"aoVal": "D", "content": "以上皆有可能 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->椭圆", "竞赛->知识点->解析几何->轨迹方程（二试）", "竞赛->知识点->解析几何->双曲线", "竞赛->知识点->解析几何->抛物线"], "answer_analysis": ["满足$$\\sin \\angle ABC=\\frac{1}{2}$$的$$A$$点轨迹是以$$B$$为锥顶，以$$BC$$为轴线的一个圆锥面（母线与轴线的夹角为$$30{}^{}\\circ $$）．现在，$$A$$点还需落在平面$$ \\pi $$上，从而其轨迹是圆锥面与平面$$ \\pi $$的交线，可能是三种圆锥曲线中的任何一种． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "326", "queId": "8fd53b68b70140939f9f9b9c31b1607e", "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学年江苏盐城高一上学期期末第5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", "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", "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"], "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": ["【方法点睛】  换元法！例：设$$\\frac{\\pi }{4}-x=t$$，则$$x=\\frac{\\pi }{4}-t$$，$$2x=\\frac{\\pi }{2}-2t$$.  。  。  $$\\sin \\left( \\frac{ \\pi }{4}-x \\right)=\\frac{\\sqrt{2}}{2}(\\cos x-\\sin x)=\\frac{3}{5}$$，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": "483", "queId": "3f10f8e442cc4bb1945dac20a854f207", "competition_source_list": ["2015~2016学年湖北武汉武昌区高二下学期期末理科第10题5分", "2014年天津全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知圆柱的底面半径为$$r$$，高为$$h$$，体积为$$2$$，表面积为$$24$$，则$$\\frac{1}{r}+\\frac{1}{h}=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "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->空间几何体的体积", "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->空间几何体的体积、表面积->空间几何体的表面积", "课内体系->素养->数学运算"], "answer_analysis": ["由题意可知$$\\begin{cases} \\pi {{r}^{2}}h=2    2 \\pi {{r}^{2}}+2 \\pi rh=24 \\end{cases}$$，  两式相比可得$$\\frac{ \\pi {{r}^{2}}h}{2 \\pi {{r}^{2}}+2 \\pi rh}=\\frac{1}{12}$$，  ∴$$\\frac{rh}{r+h}=\\frac{1}{6}$$，  ∴$$\\frac{1}{r}+\\frac{1}{h}=\\frac{r+h}{rh}=6$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "428", "queId": "94b476f109914d5497d8e0bc76504cff", "competition_source_list": ["2017年天津全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知等差数列$$\\left { {{a}_{n}} \\right }$$的公差不为零，且$${{a}_{2}},{{a}_{3}},{{a}_{9}}$$构成等比数列，则$$\\frac{{{a}_{4}}+{{a}_{5}}+{{a}_{6}}}{{{a}_{2}}+{{a}_{3}}+{{a}_{4}}}$$的值．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{8}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{7}{3}$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{3}$$ "}]], "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": ["设公差为$$d$$，由题意，$$\\left( {{a}_{3}}-d \\right)\\left( {{a}_{3}}+6d \\right)=a_{3}^{2}$$，化简得$${{a}_{3}}=\\frac{6}{5}d$$．  于是$$\\frac{{{a}_{4}}+{{a}_{5}}+{{a}_{6}}}{{{a}_{2}}+{{a}_{3}}+{{a}_{4}}}=\\frac{\\left( {{a}_{3}}+d \\right)+\\left( {{a}_{3}}+2d \\right)+\\left( {{a}_{3}}+3d \\right)}{3{{a}_{3}}}=\\frac{8}{3}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "278", "queId": "3493725e55ac4d74b784bfd69328f924", "competition_source_list": ["2010年AMC10竞赛B第21题"], "difficulty": "3", "qtype": "single_choice", "problem": "我们随机选择了一个介于$1000$和$10,000$之间的回文数，它能被$7$整除的概率是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\dfrac{1}{10}$$ "}], [{"aoVal": "B", "content": "$$\\dfrac{1}{9}$$ "}], [{"aoVal": "C", "content": "$$\\dfrac{1}{7}$$ "}], [{"aoVal": "D", "content": "$$\\dfrac{1}{6}$$ "}], [{"aoVal": "E", "content": "$$\\dfrac{1}{5}$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities", "课内体系->知识点->等式与不等式->等式->方程组的解集"], "answer_analysis": ["The palindromes can be expressed as: $$1000 x+100 y+10 y+x$$ (since it is a four digit palindrome, it must be of the form $$xyyx$$, where $$x$$ and $$y$$ are integers from $$ [1, 9] $$ and $$[0, 9]$$, respectively.) We simplify this to: $$1001 x+110 y$$. Because the question asks for it to be divisible by $$7$$, We express it as $$1001 x+110 y≡0 (\\bmod 7)$$. Because $$1001≡0 (\\bmod 7)$$, We can substitute $$1001$$ for $$0$$, We are left with $$110y≡0(\\bmod 7)$$. Since $$110≡5(\\bmod 7)$$ we can simplfy the $$110$$ in the expression to $$5y≡0(\\bmod 7)$$. In order for this to be true, $$y≡0(\\bmod 7)$$ must also be true.~ Thus we solve:~ $$y≡0(\\bmod 7)$$. Which has two soutions: $$0$$ and $$7$$. There are thus two options for $$y$$ out of the $$10$$, so the answer is $$2/10= (\\rm E)\\frac{1}{5}$$. "], "answer_value": "E"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "21", "queId": "022ab20adcb6421dba6f3de25f25333f", "competition_source_list": ["2002年全国全国高中数学联赛竞赛一试第1题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)={{\\log }_{\\frac{1}{2}}}({{x}^{2}}-2x-3)$$的单调递增区间是（~ ~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\infty ,-1)$$ "}], [{"aoVal": "B", "content": "$$(-\\infty ,1)$$ "}], [{"aoVal": "C", "content": "$$(1,+\\infty )$$ "}], [{"aoVal": "D", "content": "$$(3,+\\infty )$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学抽象", "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->函数的概念与性质->函数的性质->单调性->函数单调性的判定->判断复合函数单调性", "课内体系->知识点->函数的概念与性质->函数的性质->单调性->函数单调性的判定->求单调区间", "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"], "answer_analysis": ["由$${{x}^{2}}-2x-3\\textgreater0\\Rightarrow x\\textless{}-1$$或$$x\\textgreater3$$，令$$f\\left( x \\right)={{\\log }_{\\frac{1}{2}}}u$$，$$u= {{x}^{2}}-2x-3$$，$$u$$在$$\\left( -\\infty ,1 \\right)$$上单调减，结合$$f\\left( x \\right)$$的定义域．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "521", "queId": "da3bb047cdbc4e59b4705e8d04b96d31", "competition_source_list": ["2000年全国高中数学联赛竞赛一试第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设全集$$U$$是实数，若$$A=\\left { x\\textbar\\sqrt{x-2}\\leqslant 0 \\right }$$，$$B=\\left { x\\textbar{{10}^{{{x}^{2}}-2}}={{10}^{x}} \\right }$$，则$$A\\bigcap\\left( \\complement_{U}B\\right)$$是．", "answer_option_list": [[{"aoVal": "A", "content": "$$ {2 }$$ "}], [{"aoVal": "B", "content": "$$ {-1 }$$ "}], [{"aoVal": "C", "content": "$$\\left { x\\textbar x\\leqslant 2 \\right }$$ "}], [{"aoVal": "D", "content": "$$\\varnothing $$ "}]], "knowledge_point_routes": ["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", "竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["由$$\\sqrt{x-2}\\leqslant 0$$，得$$x=2$$，故$$A=\\left { 2 \\right }$$；由$${{10}^{{{x}^{2}}-2}}={{10}^{x}}$$，得$${{x}^{2}}-x-2=0$$，故$$B= {-1,2 }$$．所以$$A\\bigcap\\left( \\complement_{U}B\\right)=\\varnothing$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1003", "queId": "a55ed4518bb14af1a6ab161fd965391e", "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": "924", "queId": "8a7563cf44a44acdbd133b175ce6475e", "competition_source_list": ["竞赛第15题"], "difficulty": "0", "qtype": "single_choice", "problem": "某乡间小路的一侧共有12盏路灯，因行人较少，为节约用电，每晚仅开4盏灯，要求道路两头的路灯都不开，任何两盏开的路灯都不相邻，则不同的开灯方法共有（~~~~~~~）种．", "answer_option_list": [[{"aoVal": "A", "content": "32 "}], [{"aoVal": "B", "content": "35 "}], [{"aoVal": "C", "content": "42 "}], [{"aoVal": "D", "content": "46 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 利用组合可求不同的开灯方法.\\\\ 【详解】\\\\ 根据题意，不同的开灯方法数共有${\\mathrm{C}}_{7}^{4}=35$种．\\\\ 故选：B. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "46", "queId": "041785172372460cac3eee5cd7ca9140", "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文科第11题5分", "2009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": "已知关于$$x$$、$$y$$的方程组$$\\begin{cases}{{x}^{2}}+{{y}^{2}}=2{{k}^{2}}   kx-y=2k \\end{cases}$$仅有一组实数解，则符合条件的实数$$k$$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->多项式与方程->解方程（组）"], "answer_analysis": ["若$$k=0$$，显然方程组仅有一组解$$(0,0)$$，故$$k=0$$符合条件；若$$k\\ne 0$$，则$${{x}^{2}}+{{y}^{2}}=2{{k}^{2}}$$的图象是一个以$$(0,0)$$为圆心，以$$r=\\sqrt{2}\\textbar k\\textbar$$为半径的圆，而$$kx-y=2k$$表示直线．由题设条件知$$\\frac{\\textbar2k\\textbar}{\\sqrt{{{k}^{2}}+1}}=\\sqrt{2}\\textbar k\\textbar$$，即$$\\frac{4{{k}^{2}}}{1+{{k}^{2}}}=2{{k}^{2}}$$，解得$$k=\\pm 1$$．  综上所述，符合条件的实数$$k$$共有$$3$$个． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "0", "queId": "000f5e135125437792426c998a270190", "competition_source_list": ["2016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": "在$$\\triangle ABC$$中，设内角$$A$$、$$B$$、$$C$$的对边长分别为$$a$$，$$b$$，$$c$$．命题$$p$$：$$B+C=2A$$，且$$b+c=2a$$；命题$$q$$：$$\\triangle ABC$$是正三角形，则命题$$p$$是命题$$q$$的（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "充要条件 "}], [{"aoVal": "B", "content": "充分不必要条件 "}], [{"aoVal": "C", "content": "必要不充分条件 "}], [{"aoVal": "D", "content": "既不是充分条件也不是必要条件 "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->常用逻辑用语->充分条件与必要条件", "课内体系->知识点->解三角形->正余弦定理的综合应用->正余弦定理综合判断三角形形状", "课内体系->知识点->解三角形->正弦定理", "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->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->和差角公式化简求值综合运用"], "answer_analysis": ["当$$B+C=2A$$时，则$$A=60{}^{}\\circ $$．  由余弦定理，$${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\\cos 60{}^{}\\circ ={{b}^{2}}+{{c}^{2}}-bc$$．  又$$b+c=2a$$，则$${{b}^{2}}+2bc+{{c}^{2}}=4{{a}^{2}}=4\\left( {{b}^{2}}+{{c}^{2}}-bc \\right)\\Rightarrow b=c$$，所以$$\\triangle ABC$$是正三角形．  反之，若$$\\triangle ABC$$是正三角形，显然有$$B+C=2A$$和$$b+c=2a$$成立．  因此，$$p$$是$$q$$的充要条件． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "911", "queId": "bb47876b47ae47e786ee0f387ef984ae", "competition_source_list": ["2008年吉林全国高中数学联赛竞赛初赛第3题6分", "2019~2020学年3月天津南开区天津市南开中学高一下学期周测B卷第12题2分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$ABC$$是平面上不共线的三点，$$O$$是三角形$$ABC$$的重心，动点$$P$$满足$$\\overrightarrow{OP}=\\frac{1}{3}\\left( \\frac{1}{2}\\overrightarrow{OA}+\\frac{1}{2}\\overrightarrow{OB}+2\\overrightarrow{OC} \\right)$$，则点$$P$$一定为三角形$$ABC$$的（  ）", "answer_option_list": [[{"aoVal": "A", "content": "$$AB$$边中线的中点 "}], [{"aoVal": "B", "content": "$$AB$$边中线的三等分点（非重心） "}], [{"aoVal": "C", "content": "重心 "}], [{"aoVal": "D", "content": "$$AB$$边的中点 "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->数学建模", "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->平面向量与三角形的“心”", "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->平面向量的数乘运算及运算规则"], "answer_analysis": ["由选项知，本题应该选择$$C$$为起点，  于是得：$$6\\overrightarrow{CP}-6\\overrightarrow{CO}=\\overrightarrow{CA}-\\overrightarrow{CO}+\\overrightarrow{CB}-\\overrightarrow{CO}-4\\overrightarrow{CO}$$，  从而$$6\\overrightarrow{CP}=\\overrightarrow{CA}+\\overrightarrow{CB}$$，即$$\\overrightarrow{CP}=\\frac{1}{6}(\\overrightarrow{CA}+\\overrightarrow{CB})$$，  取$$AB$$边的中点$$M$$，则$$\\overrightarrow{CA}+\\overrightarrow{CB}=2\\overrightarrow{CM}$$，从而$$\\overrightarrow{CP}=\\frac{1}{3}\\overrightarrow{CM}$$，  即点$$P$$为三角形中$$AB$$边上的中线的一个三等分点，且点$$P$$不是重心．  点评：由$$\\overrightarrow{CP}=\\frac{1}{6}(\\overrightarrow{CA}+\\overrightarrow{CB})$$，$$\\frac{1}{6}:\\frac{1}{6}=1$$知点$$P$$在$$BC$$边的中线上；由$$\\frac{1}{6}+\\frac{1}{6}=\\frac{1}{3}$$知，是中线上靠近$$C$$的一个三等分点，且不是重心，  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "980", "queId": "ae346d014a404c6c8d636a5887e0eb01", "competition_source_list": ["2011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": "已知函数$$y=f(x)$$有反函数$$y={{f}^{-1}}(x)$$，将$$y=f(x)$$的图象绕$$(1,  -1)$$逆时针旋转$$90{}^{}\\circ $$，所得曲线的方程是．", "answer_option_list": [[{"aoVal": "A", "content": "$$y={{f}^{-1}}(-x)-2$$ "}], [{"aoVal": "B", "content": "$$y=-{{f}^{-1}}(-x)-2$$ "}], [{"aoVal": "C", "content": "$$y={{f}^{-1}}(-x+1)-1$$ "}], [{"aoVal": "D", "content": "$$y={{f}^{-1}}(-x-1)+1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数综合", "竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["点$$(t,  f(t))$$绕$$(1,  -1)$$旋转$$90{}^{}\\circ $$，得到$$(-f(t),  t-2)$$．令$$x=-f(t)$$，则$$t={{f}^{-1}}(-x)$$，因此$$y=t-2={{f}^{-1}}(-x)-2$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "426", "queId": "f5e13a23613c4ad1ac523de992aa3a45", "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": "465", "queId": "3a7747dfacce408fa554b6f72c28d6f1", "competition_source_list": ["2005年AMC12竞赛A第19题"], "difficulty": "3", "qtype": "single_choice", "problem": "2005AMC12A, 19  A faulty car odometer proceeds from digit $$3$$ to digit $$5$$, always skipping the digit $$4$$, regardless of position. If the odometer now reads $$002005$$, how many miles has the car actually traveled?  一个有故障的汽车里程表，它会从数字$$3$$ ，越过数字$$4$$，直接跳到数字$$5$$，在每一位上都是如此。 如果里程表现在显示$$002005$$，汽车实际行驶了多少英里？", "answer_option_list": [[{"aoVal": "A", "content": "$$1404$$ "}], [{"aoVal": "B", "content": "$$1462$$ "}], [{"aoVal": "C", "content": "$$1604$$ "}], [{"aoVal": "D", "content": "$$1605$$ "}], [{"aoVal": "E", "content": "$$1804$$ "}]], "knowledge_point_routes": ["课内体系->知识点->算法与框图->算法案例->进位制", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Simple Logical Reasoning"], "answer_analysis": ["这实际上是一个考察进制的题。  在这个错误的里程表上，每位上可取的数字从$0 9$变成了$0, 1, 2, 3, 5, 6, 7, 8, 9$，也就是从$$10$$个变成了$$9$$个。因此这是一个$$9$$进制里程表，它每一位上的$0, 1, 2, 3, 5, 6, 7, 8, 9$，分别对应$$9$$进制中的$0, 1, 2, 3, 4, 5, 6, 7, 8$。  那么，实际里程数$=2004_9=4+2\\times 9^{3}=1462$（注意里程表上的$$5$$对应的是$$9$$进制的4），选$$\\text{B}$$。 "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "543", "queId": "dee39d21357141729241365d8855db1b", "competition_source_list": ["2002年AMC10竞赛A第19题"], "difficulty": "3", "qtype": "single_choice", "problem": "2002年$$AMC10$$竞赛$$A$$第$$19$$题  Spot\\textquotesingle s doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside of the doghouse that Spot can reach?  Spot 的狗窝有一个规则的六边形底座，每边长一码。 他被一根两码长的绳子拴在一个顶点上。 Spot 可以到达的狗窝外区域的面积是多少平方码？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac {2\\pi }3$$ "}], [{"aoVal": "B", "content": "$$2\\pi $$ "}], [{"aoVal": "C", "content": "$$\\frac {5\\pi }2$$ "}], [{"aoVal": "D", "content": "$$\\frac {8\\pi} 3$$ "}], [{"aoVal": "E", "content": "$$3\\pi $$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Regular Polygon", "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": ["Part of what Spot can reach is $$\\frac {240}{360}=\\frac 23$$ of a circle with radius $$2$$, which gives him $$\\frac {8\\pi }3$$. He can also reach two $$\\frac {60}{360}$$ parts of a unit circle, which combines to give $$\\frac {\\pi }3$$. The total area is then $$3\\pi $$, which gives $$\\rm E$$.  Note   We can clearly see that the area must be more than $$\\frac {8\\pi }3$$, and the only such answer is $$\\rm E$$. "], "answer_value": "E"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "545", "queId": "5f0755fd9e6e4b3caca98be0926c1963", "competition_source_list": ["2012年甘肃全国高中数学联赛竞赛初赛第8题10分"], "difficulty": "0", "qtype": "single_choice", "problem": "实数$$x,  y,  z$$满足$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1$$，则$$xy+yz$$的最大值为~\\uline{~~~~~~~~~~}~.", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{2}}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{3}}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{5}}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->均值", "课内体系->知识点->等式与不等式->不等式"], "answer_analysis": ["因为$$1=\\left( {{x}^{2}}+\\frac{1}{2}{{y}^{2}} \\right)+\\left( \\frac{1}{2}{{y}^{2}}+{{z}^{2}} \\right)$$ $$\\geqslant 2\\sqrt{\\frac{1}{2}{{x}^{2}}{{y}^{2}}}+2\\sqrt{\\frac{1}{2}{{y}^{2}}{{z}^{2}}}$$ $$=\\sqrt{2}\\left( xy+yz \\right)$$，  所以$$xy+yz\\leqslant \\frac{\\sqrt{2}}{2}$$．  又当$$x=1,  y=\\frac{\\sqrt{2}}{2},  z=1$$时取到等号，故最大值为$$\\frac{\\sqrt{2}}{2}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "128", "queId": "1cd567b04e34452b8865da618e331563", "competition_source_list": ["2018年湖北全国高中数学联赛竞赛初赛第2题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "设数列$$ {{{a}_{n}} }$$满足：$${{a}_{1}}=1$$，$$4{{a}_{n+1}}-{{a}_{n+1}}{{a}_{n}}+4{{a}_{n}}=9$$．则$${{a}_{2018}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{3}$$ "}]], "knowledge_point_routes": ["知识标签->思想->转化化归思想", "知识标签->思想->函数思想", "知识标签->题型->数列->数列的概念->数列的函数特性问题->数列的周期性问题", "知识标签->题型->数列->数列的概念-> 数列求通项问题->常见数列的通项公式", "知识标签->素养->数学运算", "知识标签->素养->逻辑推理", "知识标签->知识点->数列->数列的概念->数列的函数特性", "知识标签->知识点->数列->数列的概念->递推数列与递推方法", "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": ["由题设得$$(4-{{a}_{n}})(4-{{a}_{n+1}})=7$$，  设$${{b}_{n}}=4-{{a}_{n}}$$，则$${{b}_{n}}{{b}_{n+1}}=7$$，  又$${{b}_{1}}=4-{{a}_{1}}=3$$，故$${{b}_{2}}=\\frac{7}{3}$$，  一般地，有$${{b}_{2k-1}}=3$$，$${{b}_{2k}}=\\frac{7}{3}$$，  于是，$${{a}_{2k}}=4-\\frac{7}{3}=\\frac{5}{3}$$，$${{a}_{2018}}=\\frac{5}{3}$$．  故答案为：$$\\frac{5}{3}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1017", "queId": "a58872bf91364f59ad7d8c16516c36f3", "competition_source_list": ["2014年湖南全国高中数学联赛竞赛初赛第6题5分", "2010年上海复旦大学自主招生千分考第26题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$C$$是以$$O$$为圆心、$$r$$为半径的圆周，两点$$P$$、$${P}'$$在以$$O$$为起点的射线上，并且满足$$\\left\\textbar{} OP \\right\\textbar\\cdot \\left\\textbar{} O{P}' \\right\\textbar={{r}^{2}}$$，则称$$P$$、$${P}'$$关于圆周$$C$$对称．那么，双曲线$${{x}^{2}}-{{y}^{2}}=1$$上的点$$P\\left( x,y \\right)$$关于单位圆周$$C$$：$${{x}^{2}}+{{y}^{2}}=1$$的对称点$${P}'$$所满足的方程为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{x}^{2}}-{{y}^{2}}={{x}^{4}}+{{y}^{4}}$$ "}], [{"aoVal": "B", "content": "$${{x}^{2}}-{{y}^{2}}={{\\left( {{x}^{2}}+{{y}^{2}} \\right)}^{2}}$$ "}], [{"aoVal": "C", "content": "$${{x}^{2}}-{{y}^{2}}=2\\left( {{x}^{4}}+{{y}^{4}} \\right)$$ "}], [{"aoVal": "D", "content": "$${{x}^{2}}-{{y}^{2}}=2{{\\left( {{x}^{2}}+{{y}^{2}} \\right)}^{2}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->极坐标系与直角坐标系"], "answer_analysis": ["化为极坐标系，则$$P\\left( \\rho ,\\theta \\right)$$关于圆周$$C$$对称的点为$${P}'\\left( \\frac{{{r}^{2}}}{\\rho },\\theta \\right)$$，  ∴双曲线$${{x}^{2}}-{{y}^{2}}=1$$，即$${{\\rho }^{2}}{{\\cos }^{2}}\\theta -{{\\rho }^{2}}{{\\sin }^{2}}\\theta =1$$，  关于单位圆周$$C$$对称的方程为$$\\frac{1}{{{\\rho }^{2}}}{{\\cos }^{2}}\\theta -\\frac{1}{{{\\rho }^{2}}}{{\\sin }^{2}}\\theta =1$$，即$${{\\rho }^{2}}{{\\cos }^{2}}\\theta -{{\\rho }^{2}}{{\\sin }^{2}}\\theta ={{\\rho }^{4}}$$，  也即$${{x}^{2}}-{{y}^{2}}={{\\left( {{x}^{2}}+{{y}^{2}} \\right)}^{2}}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "870", "queId": "cd643ab7f07d45a4b716256a8a1d374f", "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->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", "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": "1110", "queId": "bd2ad4057e794dbc8b940d4237d11342", "competition_source_list": ["2017年山西全国高中数学联赛竞赛初赛第7题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "$${{7}^{1}}+{{17}^{2}}+{{27}^{3}}+{{37}^{4}}+\\cdots +{{2017}^{202}}$$的末位数字是．", "answer_option_list": [[{"aoVal": "A", "content": "$5$ "}], [{"aoVal": "B", "content": "$6$ "}], [{"aoVal": "C", "content": "$7$ "}], [{"aoVal": "D", "content": "$8$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->（模）周期数列"], "answer_analysis": ["$${{7}^{k}}$$的末位数字依次是$$7,9,3,1,7\\cdots $$，成周期为$$4$$的周期数列，所以  $${{7}^{1}}+{{17}^{2}}+{{27}^{3}}+{{37}^{4}}+\\cdots +{{2017}^{202}}$$的末位数字为$$\\frac{202}{4}\\left( 7+9+3+1 \\right)+7+9=1026$$的末位数字． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "792", "queId": "a87af65da19c4a96a08fb69b10882eb4", "competition_source_list": ["2015年山西全国高中数学联赛竞赛初赛第7题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a=\\sqrt{3x+1}+\\sqrt{3y+1}+\\sqrt{3z+1}$$，其中$$x+y+z=1$$，$$x,y,z\\geqslant 0$$，则$$\\left[ a \\right]=$$ ．", "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->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，$${{\\left( \\frac{a}{3} \\right)}^{2}}\\leqslant \\frac{\\left( 3x+1 \\right)+\\left( 3y+1 \\right)+\\left( 3z+1 \\right)}{3}=2$$，所以$$a\\leqslant 3\\sqrt{2}$$；  另外，当$$0\\leqslant x\\leqslant 1$$时，$$\\sqrt{3x+1}\\geqslant \\sqrt{{{x}^{2}}+2x+1}=x+1$$，  于是$$a\\geqslant \\left( x+1 \\right)+\\left( y+1 \\right)+\\left( z+1 \\right)=4$$．  因此$$\\left[ a \\right]=4$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "419", "queId": "4311f632409641c790f5cca3decba15e", "competition_source_list": ["2017年AMC10竞赛B第23题", "2017年AMC12竞赛B第19题"], "difficulty": "3", "qtype": "single_choice", "problem": "2017 AMC10B P23  Let $$N=123456789101112\\ldots 4344$$ be the $$79$$-digit number that is formed by writing the integers from $$1$$ to $$44$$ in order, one after the other. What is the remainder when $$N$$ is divided by $$45$$?  设$N=123456789101112\\ldots 4344$$是$79$位数，由$$1$$至$$44$$的整数依次写成。当$N$除以$$45$$时，余数是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$18$$ "}], [{"aoVal": "E", "content": "$$44$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Number Theory->Aliquot Theory->Divisibility Rules of Certain Numbers (2/3/5/9/11)"], "answer_analysis": ["We only need to find the remainders of N when divided by $$5$$and $$9$$ to determine the answer. By inspection，$$N=4$$ (mod$$5$$)，$$The$$ remainder when Nis divided  by $$9$$is$$1+2+3+4+\\ldots +1+0+1+1+1+2+\\ldots +4+3+4+4$$，but  $$sin c{{e}^{10=1(mod9)}}$$，$$we$$ can also write this  $$a{{s}^{1+2+3+\\cdots +10+11+12+\\cdots 43+44=\\frac{44\\cdot 45}{2}=22\\cdot 45}}$$，$$which$$ has a remairder of 0 rod$$9$$. Therefoer，$$by$$ inspection，$$the$$ answer is ($$\\text{C}$$)$$9$$  Note：$$the$$ sum of the digits or N is$$270$$.  Noting the solution above，$$we$$ try to find the sum of the digits to figure out its remainder when divided by$$9$$. From1thru$$9$$，$$the$$ sum is$$45$$，$$10$$ thru$$19$$，$$the$$ sum is$$55$$，$$20$$ thru$$29$$is$$65$$，and$$30$$thru$$39$$is$$75$$．$$Thus$$ the sum of the digits is$$45+55+65+75+4+5+6+7+8=240+30=270$$，$$and$$ thus N is divisible by$$9$$. Now.refer to the above solution $$^{N=4\\left( mod5 \\right)}an{{d}^{ N=0\\left( mod9 \\right)}}$$From this information，$$we$$ can conclude this information, we can conclude that $$^{N\\equiv 54 \\left( mod 5 \\right)}and{{ }^{N\\equiv 54\\left( mod 9 \\right)}}$$.  Therefore $$^{N=54\\left( md 45 \\right)}an{{d}^{N=9\\left( md 45 \\right)}}$$ so the remainder is ($$\\text{C}$$)$$9$$.  Because a number is equivalent to the sum of its digits modulo$$9$$，$$we$$ have that $$^{N\\equiv 1+2+3+4+5+\\cdots +44\\equiv \\frac{44\\times 45}{2}\\equiv 0(mod9)}$$Furthemore，$$we$$ see that $$N-9$$ends in the digt$$5$$and thus is divisible by$$5$$，$$so$$ $$N-9$$is divisible by$$45$$，$$meaning$$ the remainder is($$\\text{C}$$)$$9$$. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "880", "queId": "781a7724d7e14ecfb94ae5a51f76acc3", "competition_source_list": ["2008年贵州全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在直三棱柱$$ABC-{{A}_{1}}{{B}_{1}}{{C}_{1}}$$中，$$A{{A}_{1}}=AB=AC$$，$$AB\\bot AC$$，$$M$$是$$C{{C}_{1}}$$的中点，$$Q$$是$$BC$$的中点，点$$P$$在$${{A}_{1}}{{B}_{1}}$$上，则直线$$PQ$$与直线$$AM$$所成的角等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$30{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$45{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$60{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$90{}^{}\\circ $$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的角与距离"], "answer_analysis": ["可取$$AC$$的中点$$N$$，连接$${{A}_{1}}N$$、$$QN$$，  可得$$\\left. \\begin{matrix}AB\\bot AM    AB//NQ   \\end{matrix} \\right }\\Rightarrow \\left. \\begin{matrix}AM\\bot NQ    AM\\bot {{A}_{1}}N   \\end{matrix} \\right }$$，  所以有$$AM\\bot $$面$${{A}_{1}}{{B}_{1}}QN$$，  又$$PQ\\subset $$面$${{A}_{1}}{{B}_{1}}QN$$，  可得$$AM\\bot PQ$$，故选$$\\text{D}$$. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "701", "queId": "4e8b8379c426467f9a7de93dcc5b89a2", "competition_source_list": ["2008年全国高中数学联赛竞赛一试第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "方程组$$\\begin{cases}x+y+z=0    xyz+z=0    xy+yz+xz+y=0    \\end{cases}$$的有理数解$$(x,y,z)$$的个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->解方程（组）"], "answer_analysis": ["若$$z=0$$，则$$\\begin{cases} x+y=0    xy+y=0 \\end{cases}$$解得$$\\begin{cases} x=0    y=0 \\end{cases}$$或$$\\begin{cases} x=-1    y=1 \\end{cases}$$，  若$$z\\ne 0$$，则由$$xyz+z=0$$得$$xy=-1$$①，  由$$x+y+z=0$$得$$z=-x-y$$②，  将②代入$$xy+yz+xz+y=0$$得$${{x}^{2}}+{{y}^{2}}+xy-y=0$$③，  由①得$$x=-\\frac{1}{y}$$，代入③化简得$$(y-1)({{y}^{3}}-y-1)=0$$，  易知$${{y}^{3}}-y-1=0$$无有理数根，故$$y=1$$，由①得$$x=-1$$，由②得$$z=0$$，与$$z\\ne 0$$矛盾，故该方程组共有两组有理数解$$\\begin{cases} x=0    y=0    z=0 \\end{cases}$$或$$\\begin{cases} x=-1    y=1    z=0 \\end{cases}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "206", "queId": "e7aa46314f1f4eb1a050059d32c8566c", "competition_source_list": ["2013年浙江全国高中数学联赛竞赛初赛第10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$f\\left( x \\right),g\\left( x \\right)$$，$$h\\left( x \\right)$$为一次函数，若对实数$$x$$满足  $$\\left\\textbar{} f\\left( x \\right) \\right\\textbar-\\left\\textbar{} g\\left( x \\right) \\right\\textbar+h\\left( x \\right)=\\begin{cases}-1,x\\textless{}-1    3x+2,-1{\\leqslant }x\\textless{}0    -2x+2,x{\\geqslant }0 \\end{cases}$$．  则$$h\\left( x \\right)$$的表达式为．", "answer_option_list": [[{"aoVal": "A", "content": "$$h\\left( x \\right)=x-\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$h\\left( x \\right)=-x-\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$h\\left( x \\right)=-x+\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$h\\left( x \\right)=x+\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["$$h\\left( x \\right)=\\frac{-2x+2+\\left( -1 \\right)}{2}=-x+\\frac{1}{2}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "19", "queId": "0557de50cf324bf59c5563ff65ad6f20", "competition_source_list": ["2010年四川全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$${{A}_{1}},  {{A}_{2}}$$为椭圆$$\\frac{{{x}^{2}}}{{{a}^{2}}}+\\frac{{{y}^{2}}}{{{b}^{2}}}=1(ab0)$$的左、右顶点，若在椭圆上存在异于$${{A}_{1}},  {{A}_{2}}$$的  点$$P$$，使得$$\\overrightarrow{PO}\\cdot \\overrightarrow{P{{A}_{2}}}=0$$，其中$$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)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算", "竞赛->知识点->解析几何->椭圆"], "answer_analysis": ["由题设知$$\\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}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "720", "queId": "b5f27a849de843018c4c2fe03e95c8ad", "competition_source_list": ["2020~2021学年河北石家庄新华区石家庄市第二中学高二上学期期中（竞赛班）第6题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知命题$$p:x\\textgreater1$$，命题$$q:{{x}^{2}}\\textgreater x$$，则$$\\neg q$$是$$\\neg p$$的．", "answer_option_list": [[{"aoVal": "A", "content": "充分不必要条件 "}], [{"aoVal": "B", "content": "必要不充分条件 "}], [{"aoVal": "C", "content": "充要条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与不等式结合", "课内体系->知识点->等式与不等式->不等式->解不等式->一元二次不等式", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理"], "answer_analysis": ["∵$$p:x\\textgreater1$$，  命题$$q:{{x}^{2}}\\textgreater x$$，得$$x\\textless{}0$$或$$x\\textgreater1$$，  ∴$$\\neg p:x\\leqslant 1$$，$$\\neg q:0\\leqslant x\\leqslant 1$$，  根据充分必要条件的定义可判断：  $$\\neg q$$是$$\\neg p$$的充分不必要条件．  故选：$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "122", "queId": "1cc52f1449d04806b3c196db3dd10ca2", "competition_source_list": ["2017年AMC10竞赛第18题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "阿米莉亞有一枚硬幣有$$ \\frac{1}{3} $$概率落在正面，布萊恩有一枚硬幣有$$ \\frac{2}{5} $$ 概率落在正面．阿米莉亞和布萊恩交替投擲硬幣，直到有人是正面；第一個是正面的人獲勝．所有拋硬幣的事件都是獨立的．阿米莉亞先拋，她獲勝的概率為$$ \\frac{p}{q} $$，其中$$p$$和$$q$$是互質的正整數．$$q−p$$是多少？  Amelia has a coin that lands heads with probability$$ \\frac{1}{3} $$ , and Blaine has a coin that lands on heads with probability$$ \\frac{2}{5} $$ melia and Blaine alternately toss their coins untill someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is$$ \\frac{p}{q} $$ , where p and q are relatively prime positive integers. What is $$q---p$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$4$$ "}], [{"aoVal": "E", "content": "$$5$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["Let $$P$$ be the probability Amelia wins. Note that $$P=$$ chance she wins on her first turn $$+$$ chance she gets to her turn again$$·P$$, as if she gets to her turn again, she is back where she started with probability of winning $$P$$. The chance she wins on her first turn is $$\\dfrac{1}{3}$$. The chance she makes it to her turn again is a combination of her failing to win the first turn$$-\\dfrac{2}{3}$$ and Blaine failing to win $$-\\dfrac{3}{5}$$. Multiplying gives us $$\\dfrac{2}{5}$$. Thus,~ $$P= \\dfrac{1}{3}+ \\dfrac{2}{5}P$$. Therefore, $$p=\\dfrac{9}{5}$$, so the answer is $$9-5=(\\rm D)4$$.  Let $$P$$ be the probability Amelia wins. Note that $$P=$$ chance she wins on her frst turn $$+$$ chance she gets to her second turn $$\\cdot\\dfrac{1}{3}+$$ chance she gets to her third turn $$\\cdot\\dfrac{1}{3}\\ldots$$. This can be represented by an infinite geometric series. $$p=\\dfrac{\\dfrac{1}{3}}{1-\\dfrac{2}{3}\\cdot\\dfrac{3}{6}}=\\dfrac{\\dfrac{1}{3}}{1-\\dfrac{2}{5}}=\\dfrac{\\dfrac{1}{3}}{\\dfrac{3}{5}}=\\dfrac{1}{3}\\cdot\\dfrac{5}{3}=\\dfrac{5}{9}$$. Therefre $$P= \\dfrac{5}{9}$$, so the answer is $$9-5=(\\rm D)4$$. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "548", "queId": "3b598794b5e44138ab026807160ca741", "competition_source_list": ["2009年高二竞赛广州市第4题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$[x]$$表示不超过$$x$$的最大整数，则$$\\left[ {{\\log }_{2}}1 \\right]+\\left[ {{\\log }_{2}}2 \\right]+\\left[ {{\\log }_{2}}3 \\right]+\\cdots +\\left[ {{\\log }_{2}}2009 \\right]$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$18054$$ "}], [{"aoVal": "B", "content": "$$18044$$ "}], [{"aoVal": "C", "content": "$$17954$$ "}], [{"aoVal": "D", "content": "$$17944$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->取整函数->取整函数的定义与性质"], "answer_analysis": ["根据题意，当$${{2}^{k}}\\leqslant n\\textless{}{{2}^{k+1}}$$时，$$[{{\\log }_{2}}n]=k\\left( k\\in N \\right)$$，于是  $$M=0\\cdot \\left( {{2}^{1}}-{{2}^{0}} \\right)+1\\cdot \\left( {{2}^{2}}-{{2}^{1}} \\right)+2\\cdot \\left( {{2}^{3}}-{{2}^{2}} \\right)+\\cdots +9\\cdot \\left( {{2}^{10}}-{{2}^{9}} \\right)+10\\cdot \\left( 2009-{{2}^{10}}+1 \\right)=18054$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "290", "queId": "22dcd68a7e654050aefd74275a677507", "competition_source_list": ["2014年浙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知复数$${{z}_{1}}$$，$${{z}_{2}}$$，且$$\\left\\textbar{} {{z}_{1}} \\right\\textbar=2\\left\\textbar{} {{z}_{2}} \\right\\textbar=2$$，$$\\left\\textbar{} {{z}_{1}}+{{z}_{2}} \\right\\textbar=\\sqrt{7}$$，则$$\\left\\textbar{} {{z}_{1}}-{{z}_{2}} \\right\\textbar$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{5}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{7}$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["因为  $$\\left\\textbar{} {{z}_{1}}-{{z}_{2}} \\right\\textbar=\\sqrt{2{{\\left\\textbar{} {{z}_{1}} \\right\\textbar}^{2}}+2{{\\left\\textbar{} {{z}_{2}} \\right\\textbar}^{2}}-{{\\left\\textbar{} {{z}_{1}}+{{z}_{1}} \\right\\textbar}^{2}}}=\\sqrt{3}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "283", "queId": "1e92fb7e8c8e404aa6837af5475c5912", "competition_source_list": ["2009年四川全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$、$$b$$、$$c$$均为大于$$0$$的实数，设命题$$P:$$以$$a$$、$$b$$、$$c$$为长度的线段可以构成三角形的三边，命题$$Q:{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\\textless{}2(ab+bc+ca)$$，则$$P$$是$$Q$$的．", "answer_option_list": [[{"aoVal": "A", "content": "充分但不必要条件 "}], [{"aoVal": "B", "content": "必要但不充分条件 "}], [{"aoVal": "C", "content": "充要条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的证明"], "answer_analysis": ["一方面，若$$P$$成立，则$$b+c\\textgreater a$$，故$$a(b+c)\\textgreater{{a}^{2}}$$，即$$ab+ac\\textgreater{{a}^{2}}$$．  同理$$ba+bc\\textgreater{{b}^{2}},ca+cb\\textgreater{{c}^{2}}$$．所以，  $${{a}^{2}}+{{b}^{2}}+{{c}^{2}}\\textless{}2(ab+bc+ca)$$，  即$$Q$$成立．  另一方面，若$$Q$$成立，取$$b=c=1,a=2$$，这时以$$a$$、$$b$$、$$c$$为长度的线段不能构成三角形的三边，即$$P$$不成立．  综上所述，$$P$$是$$Q$$的充分但不必要条件． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "192", "queId": "871ef90dcf014d368d97b59ff249a58f", "competition_source_list": ["2016年天津全国高中数学联赛竞赛初赛第7题9分", "2016年高考真题天津卷"], "difficulty": "1", "qtype": "single_choice", "problem": "椭圆$${{x}^{2}}+k{{y}^{2}}=1$$与双曲线$$\\frac{{{x}^{2}}}{4}-\\frac{{{y}^{2}}}{5}=1$$有相同的准线，则$$k$$等于 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{7}{16}$$ "}], [{"aoVal": "B", "content": "$$\\frac{16}{7}$$ "}], [{"aoVal": "C", "content": "$$\\frac{4}{7}$$ "}], [{"aoVal": "D", "content": "$$\\frac{7}{4}$$ "}]], "knowledge_point_routes": ["课内体系->素养->直观想象", "课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->圆锥曲线->圆锥曲线的第二定义->椭圆的第二定义", "课内体系->知识点->圆锥曲线->圆锥曲线的第二定义->双曲线的第二定义"], "answer_analysis": ["双曲线的准线方程是  $$x=\\pm \\frac{4}{\\sqrt{4+5}}=\\pm \\frac{4}{3}$$，  椭圆的准线方程是$$x=\\pm \\frac{1}{\\sqrt{1-\\frac{1}{k}}}$$，因此由方程$$\\frac{1}{\\sqrt{1-\\frac{1}{k}}}=\\frac{4}{3}$$得$$1-\\frac{1}{k}=\\frac{9}{16}$$，解得$$k=\\frac{16}{7}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "921", "queId": "97138fdbc19b4003bb62f78bfcc517b1", "competition_source_list": ["2012年浙江全国高中数学联赛竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$${{F}_{1}}$$、$${{F}_{2}}$$分别为双曲线$$C:\\frac{{{x}^{2}}}{9}-\\frac{{{y}^{2}}}{27}=1$$的左、右焦点，点$$A$$的坐标为$$\\left( \\frac{9}{2},\\frac{\\sqrt{135}}{2} \\right)$$，则$$\\angle {{F}_{1}}A{{F}_{2}}$$的平分线与$$x$$轴交点$$M$$的坐标为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 2,  0 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( -2,  0 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( 4,  0 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( -4,  0 \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->双曲线"], "answer_analysis": ["$${{F}_{1}}\\left( -6,  0 \\right),  {{F}_{2}}\\left( 6,  0 \\right)\\Rightarrow \\frac{\\left\\textbar{} A{{F}_{1}} \\right\\textbar}{\\left\\textbar{} A{{F}_{2}} \\right\\textbar}=2\\Rightarrow \\frac{\\left\\textbar{} M{{F}_{1}} \\right\\textbar}{\\left\\textbar{} M{{F}_{2}} \\right\\textbar}=2\\Rightarrow M\\left( 2,  0 \\right)$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "837", "queId": "ff8080814acd9dc6014adbd75bf62310", "competition_source_list": ["2014年黑龙江全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知三棱锥$$S-ABC$$，在三棱锥内任取一点$$P$$，使得$${{V}_{p-ABC}}\\textless\\frac{1}{2}{{V}_{S-ABC}}$$的概率是（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{7}{8}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学抽象", "课内体系->素养->数学运算", "课内体系->知识点->统计与概率->概率->事件与概率->几何概型->与体积有关的几何概率的计算"], "answer_analysis": ["当$$P$$在三棱锥的中截面与下底面构成的三棱台内时符合要求，由几何概型知，$$P=1-\\frac{1}{8}=\\frac{7}{8}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "673", "queId": "529e47692dd544f7bbd5b7820fd619fc", "competition_source_list": ["2012年山东全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知平面内三点$$A,  B,  C$$满足$$\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar=3$$，$$\\left\\textbar{} \\overrightarrow{BC} \\right\\textbar=5$$，$$\\left\\textbar{} \\overrightarrow{CA} \\right\\textbar=6$$，则$$\\overrightarrow{AB}\\cdot \\overrightarrow{BC}+\\overrightarrow{BC}\\cdot \\overrightarrow{CA}+\\overrightarrow{CA}\\cdot \\overrightarrow{AB}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-55$$ "}], [{"aoVal": "B", "content": "$$-45$$ "}], [{"aoVal": "C", "content": "$$-35$$ "}], [{"aoVal": "D", "content": "$$-25$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角形中的问题->解三角形", "竞赛->知识点->复数与平面向量->平面向量的应用"], "answer_analysis": ["由余弦定理，知$$\\overrightarrow{AB}\\cdot \\overrightarrow{BC}=-\\overrightarrow{BA}\\cdot \\overrightarrow{BC}=-\\frac{{{\\left\\textbar{} \\overrightarrow{BA} \\right\\textbar}^{2}}+{{\\left\\textbar{} \\overrightarrow{BC} \\right\\textbar}^{2}}-{{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}^{2}}}{2}$$  同理  $$\\overrightarrow{BC}\\cdot \\overrightarrow{CA}=\\frac{{{\\left\\textbar{} \\overrightarrow{CA} \\right\\textbar}^{2}}+{{\\left\\textbar{} \\overrightarrow{CB} \\right\\textbar}^{2}}-{{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}^{2}}}{2}$$  $$\\overrightarrow{CA}\\cdot \\overrightarrow{AB}=\\frac{{{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}^{2}}+{{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}^{2}}-{{\\left\\textbar{} \\overrightarrow{BC} \\right\\textbar}^{2}}}{2}$$  从而得  $$\\overrightarrow{AB}\\cdot \\overrightarrow{BC}+\\overrightarrow{BC}\\cdot \\overrightarrow{CA}+\\overrightarrow{CA}\\cdot \\overrightarrow{AB}$$  $$=-\\frac{{{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}^{2}}+{{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}^{2}}+{{\\left\\textbar{} \\overrightarrow{BC} \\right\\textbar}^{2}}}{2}=-35$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "201", "queId": "21dda1fb0e1f4df5b0162e74cdf4921f", "competition_source_list": ["竞赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "羅拉把兩個三位正整數相加．若這兩個正整款的六個位數的數字都不相同，且羅拉計算出來之和是一個三位數$$S$$，則$$S$$的所有各位救字之和的最小可能値為何？  Laura added two three---digit positive integers. All six digits in these nunbers are different. Laura\\textquotesingle s sun is a three---digit number $$S$$. What is the smallest possible value for the sun of the digits of $$S$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$15$$ "}], [{"aoVal": "E", "content": "$$20$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["Let the two three$$-$$digit numbers she added be $$a$$ and $$b$$ with $$a+b=S$$ and $$a\\textless b$$. The hundreds digits of these numbers must be at least $$1$$ and $$2$$, so $$a\\geqslant 100$$ and $$b\\geqslant 200$$.  Say $$a=100+p$$ and $$b=200+q$$; then we just need $$p+q=100$$ with $$p$$ and $$q$$ having different digits which aren\\textquotesingle t $$1$$ or $$2$$. There are many solutions,  but $$p=3$$ and $$q=97$$ give $$103+297=400$$ which proves that $$\\left(\\text{B}\\right) 4$$ is attainable.  For this problem, to find the $$3-$$digit integer with the smallest sum of digits, one should make the units and tens digit add to $$0$$ . To do that, we need to make sure the digits are all distinct. For the units digit, we can have a variety of digits that work. $$7$$ works best for the top number which makes the bottom digit $$3$$. The tens digits need to add to $$9$$ because of the $$1$$ that needs to be carried from the addition of the units digits. We see that $$5$$ and~ $$4$$ work the best as we can\\textquotesingle t use $$6$$ and $$3$$. Finally, we use $$2$$ and $$1$$ for our hundreds place digits.  Adding the numbers $$257$$ and $$143$$, we get $$400$$ which means our answer is $$\\left(\\text{B}\\right) 4$$. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "8", "queId": "04f79d07fceb4665b78172c3cc3d619e", "competition_source_list": ["2018年四川全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知多项式$$f\\left( x \\right)={{x}^{12}}-{{x}^{6}}+1$$除以$${{x}^{2}}+1$$的商式为$$q\\left( x \\right)$$，余式$$r\\left( x \\right)=ax+b$$，其中，$$a$$，$$b$$为实数．则$$b$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->多项式与复数综合"], "answer_analysis": ["注意到，$${{\\text{i}}^{2}}+1=0$$．  令$$x=\\text{i}$$，得$${{\\text{i}}^{12}}-{{\\text{i}}^{6}}+1=a\\text{i}+b$$  $$\\Rightarrow a\\text{i}+b=3\\Rightarrow a=0$$，$$b=3$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "860", "queId": "9faf9fa9074f42d59c5af001d9d461c7", "competition_source_list": ["2017年AMC10竞赛第14题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "羅傑每週從零用錢中支付一張電影票和一杯汽水． 上周，羅傑的零用錢是$$A$$美元． 他的電影票的價格是$$A$$和他的汽水價格差的$$20 \\%$$，而他的汽水價格是$$A$$和他的電影票價格差的$$5 \\%$$． 按最接近整數的百分比计算，羅傑為他的電影票和汽水支付$$A$$的百分之多少？  Every week Roger pays for a novie ticket and a soda out of his allowance. Last week, Roger\\textquotesingle s allowance was $$A$$ dollars. The cost of his movie ticket was $$20$$\\% of the difference between $$A$$ and the cost of his soda, while the cost of his soda was $$5$$\\% of the difference betreen $$A$$ and the cost of his movie ticket. To the nearest whole percent, what fraction of $$A$$ did Roger pay for his movie ticket and soda?", "answer_option_list": [[{"aoVal": "A", "content": "$$9$$\\% "}], [{"aoVal": "B", "content": "$$19$$\\% "}], [{"aoVal": "C", "content": "$$22$$\\% "}], [{"aoVal": "D", "content": "$$23$$\\% "}], [{"aoVal": "E", "content": "$$25$$\\% "}]], "knowledge_point_routes": [], "answer_analysis": ["A patten starts to emerge as the function is continued. The repeating patten is $$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$$. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "135", "queId": "33b48054181d42448cfafff789cb4fde", "competition_source_list": ["2013年吉林全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f\\left( x \\right)=2\\sin \\left( \\omega x+\\frac{ \\pi }{3} \\right)+\\cos \\left( \\omega x-\\frac{ \\pi }{6} \\right)\\left( \\omega \\textgreater0 \\right)$$的最小正周期为$$ \\pi $$，则$$\\omega $$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["$$f\\left( x \\right)=2\\sin \\left( \\omega x+\\frac{ \\pi }{3} \\right)+\\cos \\left( \\omega x-\\frac{ \\pi }{6} \\right)$$  $$=2\\sin \\left( \\omega x+\\frac{ \\pi }{3} \\right)+\\cos \\left( \\omega x+\\frac{ \\pi }{3}-\\frac{ \\pi }{2} \\right)$$  $$=2\\sin \\left( \\omega x+\\frac{ \\pi }{3} \\right)+\\sin \\left( \\omega x+\\frac{ \\pi }{3} \\right)=3\\sin \\left( \\omega x+\\frac{ \\pi }{3} \\right).$$  因为$$T=\\frac{2 \\pi }{\\omega }$$，所以$$\\frac{2 \\pi }{\\omega }= \\pi $$，即$$\\omega =2$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "78", "queId": "0f0de950a67c415f9546d7838d86a781", "competition_source_list": ["2017~2018学年广东湛江高三上学期期中文科第9题5分", "2018年黑龙江全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$y={{\\cos }^{2}}x+\\sqrt{3}\\sin x\\cos x$$在区间$$\\left[ -\\frac{\\pi }{6},\\frac{\\pi }{4} \\right]$$上的值域是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ -\\frac{1}{2},1 \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left[ -\\frac{1}{2},\\frac{\\sqrt{3}}{2} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left[ 0,\\frac{3}{2} \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left[ 0,\\frac{\\sqrt{3}+1}{2} \\right]$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->三角函数->三角恒等变换->倍角、和差角公式综合", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的余弦", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->半角公式", "课内体系->知识点->三角函数->三角恒等变换->二倍角公式->二倍角的正弦", "课内体系->知识点->三角函数->三角恒等变换->和差角公式->辅助角公式", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦函数的图象和性质", "课内体系->知识点->三角函数->三角函数的图象与性质->正弦型函数的图象与性质"], "answer_analysis": ["由题意得：$$y={{\\cos }^{2}}x+\\sqrt{3}\\sin x\\cdot \\cos x$$  $$=\\frac{1}{2}\\cos 2x+\\frac{1}{2}+\\frac{\\sqrt{3}}{2}\\sin 2x$$  $$=\\sin \\left( 2x+\\frac{ \\pi }{6} \\right)+\\frac{1}{2}$$，  由$$x\\in \\left[ -\\frac{ \\pi }{6},\\frac{ \\pi }{4} \\right]$$，  且据$$\\sin \\left( 2x+\\frac{ \\pi }{6} \\right)+\\frac{1}{2}$$的单调性可知，  当$$x=-\\frac{ \\pi }{6}$$时，$$y$$取最小值$$0$$，  当$$x=\\frac{ \\pi }{6}$$时，$$y$$取最大值$$\\frac{3}{2}$$．  因此$$y\\in \\left[ 0,\\frac{3}{2} \\right]$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "551", "queId": "8835b831055f44b099735aea89d0b766", "competition_source_list": ["2013年天津全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果曲线$$y=2\\sin \\frac{x}{2}$$两条互相垂直的切线交于$$P$$点，则$$P$$点的坐标不可能是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( ~\\pi , \\pi ~\\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( 3 \\pi ,- \\pi ~\\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( 5 \\pi ,- \\pi ~\\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( 7 \\pi ,- \\pi ~\\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["函数$$y=2\\sin \\frac{x}{2}$$的图象在$$x={{x}_{i}}$$处的切线斜率为$$\\cos \\frac{{{x}_{i}}}{2}$$，因此，若$${{x}_{1}}$$和$${{x}_{2}}$$处的切线互相垂直，则$$\\cos \\frac{{{x}_{1}}}{2}\\cos \\frac{{{x}_{2}}}{2}=-1$$不妨设$$\\cos \\frac{{{x}_{1}}}{2}{\\leqslant }\\cos \\frac{{{x}_{2}}}{2}$$，则从上式可得$$\\cos \\frac{{{x}_{1}}}{2}=-1$$，$$\\cos \\frac{{{x}_{2}}}{2}=1$$，即$$\\frac{{{x}_{1}}}{2}=\\left( 2a+1 \\right) \\pi $$，$$\\frac{{{x}_{2}}}{2}=2b \\pi $$，$$a$$，$$b\\in \\mathbf{Z}$$．进一步，可求出两条切线的交点$$P$$的坐标为$$\\left( {{x}_{0}}{,}{{y}_{0}} \\right)=\\left( \\frac{\\left( {{x}_{1}}+{{x}_{2}} \\right)}{2},\\frac{\\left( {{x}_{1}}-{{x}_{2}} \\right)}{2} \\right)$$．可见，$$P$$的横坐标都是$$ \\pi $$的奇数倍，且两者之差是$$4 \\pi $$的整数倍．四个选项中只有$$C$$．不符合这一特征． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "958", "queId": "a4e6a12c5d6f4f23b0b7c76d1371a0d9", "competition_source_list": ["2011年浙江全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$A$$ ，$$B$$为两个互不相同的集合，命题P：$$x\\in A\\cap B$$， 命题q：$$x\\in A$$或$$x\\in B$$，则$$p$$是$$q$$的．", "answer_option_list": [[{"aoVal": "A", "content": "充分且必要条件 "}], [{"aoVal": "B", "content": "充分非必要条件 "}], [{"aoVal": "C", "content": "必要非充分条件 "}], [{"aoVal": "D", "content": "非充分且非必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->常用逻辑用语", "竞赛->知识点->逻辑->逻辑推理"], "answer_analysis": ["$$p$$是$$q$$的充分非必要条件． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "256", "queId": "2267458ac9ac467aa96fc053819e201c", "competition_source_list": ["2018年陕西全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$0 \\textless{} x \\textless{} \\frac{\\pi }{2}$$，且$$\\frac{{{\\sin }^{4}}x}{9}+\\frac{{{\\cos }^{4}}x}{4}=\\frac{1}{13}$$，则$$\\tan x$$的值是(  )", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式", "竞赛->知识点->三角函数"], "answer_analysis": ["解：由柯西不等式$$\\left( 9+4 \\right)\\left( \\frac{{{\\sin }^{4}}x}{9}+\\frac{{{\\cos }^{4}}x}{4} \\right)\\geqslant {{\\left( {{\\sin }^{2}}x+{{\\cos }^{2}}x \\right)}^{2}}$$，由取等条件知$$\\frac{{{\\sin }^{4}}x}{81}=\\frac{{{\\cos }^{4}}x}{16}\\Rightarrow \\tan x=\\frac{3}{2}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1055", "queId": "aa77ee285680452f9ecf10d97f057e35", "competition_source_list": ["1981年全国高中数学联赛竞赛一试第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "下面四个图形中，哪一个面积最大？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\mathsf{\\triangle }ABC:\\angle A=60{}^{}\\circ ，\\angle B=45{}^{}\\circ ，AC=\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "梯形：两对角线长度分别为$$\\sqrt{2}$$和$$\\sqrt{3}$$，夹角为$$75°$$ "}], [{"aoVal": "C", "content": "圆：半径为$$1$$ "}], [{"aoVal": "D", "content": "正方形：对角线的长度为$$2.5$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角形中的问题->面积问题"], "answer_analysis": ["先计算出△$$ABC$$的外接圆半径$$R$$，有  $$R=\\frac{AC}{2\\sin B}=\\frac{\\sqrt{2}}{2\\sin 45{}^{}\\circ }=1$$  现在把$$A$$与$$C$$比较，当然$$C$$的面积大，再由凸四边形面积等于对角线乘积再乘以它们夹角的正弦的一半，可得  B的面积与$$D$$的面积相比较  $$\\frac{1}{2}\\cdot \\sqrt{3}\\cdot \\sqrt{2}\\sin 75{}^{}\\circ ~~\\textless{} ~\\frac{1}{2}\\cdot 2.5\\cdot 2.5\\sin 90{}^{}\\circ $$．  所以$$C$$大于$$B$$，再比较$$C$$与$$D$$．  C的面积为$$\\pi $$．$$D$$的面积为$$\\frac{6.25}{2}=3.125$$，$$C$$的面积最大． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "338", "queId": "503ca957cdeb4cc1981d92a2242c66fa", "competition_source_list": ["2008年江苏全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知等比数列$$ {{{a}_{n}} }$$的公比$$q\\textless{}0$$，其前$$n$$项和为$${{S}_{n}}$$，则$${{a}_{9}}{{S}_{8}}$$与$${{a}_{8}}{{S}_{9}}$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{a}_{9}}{{S}_{8}}\\textgreater{{a}_{8}}{{S}_{9}}$$ "}], [{"aoVal": "B", "content": "$${{a}_{9}}{{S}_{8}}\\textless{}{{a}_{8}}{{S}_{9}}$$ "}], [{"aoVal": "C", "content": "$${{a}_{9}}{{S}_{8}}={{a}_{8}}{{S}_{9}}$$ "}], [{"aoVal": "D", "content": "与$${{a}_{1}}$$的值有关 "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["$${{a}_{9}}{{S}_{8}}-{{a}_{8}}{{S}_{9}}=\\frac{a_{1}^{2}{{q}^{7}}}{1-q}[q(1-{{q}^{8}})-(1-{{q}^{9}})]=-a_{1}^{2}{{q}^{7}}\\textgreater0$$，  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1031", "queId": "a5a49d86e9da41619c6fb12b718b9d66", "competition_source_list": ["2008年安徽全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "四面体的$$6$$个二面角中钝角个数最多为．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->平面几何->完全四边形（二试）"], "answer_analysis": ["假设四边形的四个内角都是钝角，那么这四个内角的和$$\\textgreater360{}^{}\\circ $$，  与四边形的内角和定理矛盾，  所以四边形的四个内角不能都是钝角，  即钝角个数最多有$$3$$个，故选$$\\text{A}$$. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "163", "queId": "58b118c51c3e4d27965b3ab8c49c4d94", "competition_source_list": ["2010年浙江全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$\\overrightarrow{a}$$，$$\\overrightarrow{b}$$为两个相互垂直的单位向量．已知$$\\overrightarrow{OP}=\\overrightarrow{a}$$，$$\\overrightarrow{OQ}=\\overrightarrow{b}$$，$$\\overrightarrow{OR}=r\\overrightarrow{a}+k\\overrightarrow{b}$$，若$$\\triangle PQR$$为等边三角形，则$$k,   r$$的取值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$k=r=\\frac{-1\\pm \\sqrt{3}}{2}$$ "}], [{"aoVal": "B", "content": "$$k=\\frac{1\\pm \\sqrt{3}}{2},  r=\\frac{1\\pm \\sqrt{3}}{2}$$ "}], [{"aoVal": "C", "content": "$$k=r=\\frac{1\\pm \\sqrt{3}}{2}$$ "}], [{"aoVal": "D", "content": "$$k=\\frac{-1\\pm \\sqrt{3}}{2},  r=\\frac{-1\\pm \\sqrt{3}}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算"], "answer_analysis": ["$$\\left\\textbar{} PQ \\right\\textbar=\\left\\textbar{} QR \\right\\textbar=\\left\\textbar{} PR \\right\\textbar$$，  即$$\\sqrt{{{r}^{2}}+{{(k-1)}^{2}}}=\\sqrt{{{(r-1)}^{2}}+{{k}^{2}}}=\\sqrt{2}$$,  解得$$r=k=\\frac{1\\pm \\sqrt{3}}{2}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1144", "queId": "c693bee8ef6c443dba73a29c7a13bb95", "competition_source_list": ["2009年第二十届全国希望杯高二竞赛复赛第5题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$a,b,m,n,x,y$$都是正数，且$$a\\textless{}b$$，又知$$a,m,b,x$$成等差数列，$$a,n,b,y$$成等比数列，则有．", "answer_option_list": [[{"aoVal": "A", "content": "$$m\\textgreater n,x\\textgreater y$$ "}], [{"aoVal": "B", "content": "$$m\\textgreater n,x\\textless{}y$$ "}], [{"aoVal": "C", "content": "$$m\\textless{}n,x\\textgreater y$$ "}], [{"aoVal": "D", "content": "$$m\\textless{}n,x\\textless{}y$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["$$m=\\frac{a+b}{2}\\textgreater\\sqrt{ab}=n$$，由$$b=\\sqrt{ny}=\\frac{m+x}{2}\\textgreater\\sqrt{mx}$$及$$m\\textgreater n$$可知$$y\\textgreater x$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "660", "queId": "f64f4bd7ce3d4c3b94a8ebca9c96cfea", "competition_source_list": ["2022年江苏徐州贾汪区江苏省贾汪中学高一竞赛（下学期春季）第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "当$0\\textless{} x\\textless{} \\frac{\\pi }{4}$时,函数$f(x)=\\frac{{{\\cos }^{2}}x}{\\cos x\\sin x-{{\\sin }^{2}}x}$的最小值是", "answer_option_list": [[{"aoVal": "A", "content": "$\\frac{1}{4}$ "}], [{"aoVal": "B", "content": "$\\frac{1}{2}$ "}], [{"aoVal": "C", "content": "$2$ "}], [{"aoVal": "D", "content": "4 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 分子与分母同除以${{\\cos }^{2}}x$，得$f(x)=\\frac{1}{\\tan x-{{\\tan }^{2}}x}$利用二次函数求最值即可解答\\\\ 【详解】\\\\ 分子与分母同除以${{\\cos }^{2}}x$，得$f(x)=\\frac{1}{\\tan x-{{\\tan }^{2}}x}$，$\\begin{array}{*{35}{l}}\\textless{} br/\\textgreater{} \\because 0\\textless{} x\\textless{} \\frac{\\pi }{4},\\therefore 0\\textless{} \\tan x\\textless{} 1,\\therefore \\tan x-{{\\tan }^{2}}x=-{{\\left( \\tan x-\\frac{1}{2} \\right)}^{2}}+\\frac{1}{4}   \\textless{} br/\\textgreater{} {}   \\textless{} br/\\textgreater\\end{array}$\\\\ $\\therefore \\tan x=\\frac{1}{2}$时，$\\tan x-{{\\tan }^{2}}x$的最大值为$\\frac{1}{4}$\\\\ 综上，$f(x)=\\frac{{{\\cos }^{2}}x}{\\cos x\\sin x-{{\\sin }^{2}}x}$的最小值为4\\\\ 故选D\\\\ 【点睛】\\\\ 本题考查同角三角函数基本关系，考查二次函数求最值，注意公式的合理运用，是基础题 "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1181", "queId": "d945ca3e228a4efa9780c5163f6ed465", "competition_source_list": ["2018~2019学年浙江宁波北仑区浙江省北仑中学高一下学期期中A卷第5题4分", "2015年浙江全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若实数$$a,b$$满足$$\\begin{cases}a+b-2\\geqslant 0    b-a-1\\leqslant 0    a\\leqslant 1 \\end{cases}$$，则$$\\frac{a+2b}{2a+b}$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{7}{5}$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->线性规划"], "answer_analysis": ["由$$a,b$$满足的条件知$$1\\leqslant \\frac{b}{a}\\leqslant 3$$，所以$$\\frac{a+2b}{2a+b}=2-\\frac{3}{2+\\frac{b}{a}}\\leqslant \\frac{7}{5}$$，当$$(a,b)=(\\frac{1}{2},\\frac{3}{2})$$取等号． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1025", "queId": "e504e97eb7fb46c89c6c8acdde5ebfb8", "competition_source_list": ["2009年第二十届全国希望杯高二竞赛复赛第9题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$0\\textless{}\\theta \\textless{}2 \\pi $$，并且$${{\\sin }^{5}}\\theta +3\\sin \\theta \\textgreater{{\\cos }^{5}}\\theta +3\\cos \\theta $$，则$$\\theta $$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( \\frac{ \\pi }{4},\\frac{5}{4} \\pi ~\\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( \\frac{ \\pi }{4},\\frac{3}{4} \\pi ~\\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{ \\pi }{2},\\frac{5}{4} \\pi ~\\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{3}{4} \\pi, \\frac{5}{4} \\pi ~\\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["$${{\\sin }^{5}}\\theta +3\\sin \\theta -({{\\cos }^{5}}\\theta +3\\cos \\theta )$$  $$=(\\sin \\theta -\\cos \\theta )({{\\sin }^{4}}\\theta +{{\\sin }^{3}}\\theta \\cos \\theta +{{\\sin }^{2}}\\theta {{\\cos }^{2}}\\theta +\\sin \\theta {{\\cos }^{3}}\\theta +{{\\cos }^{4}}\\theta +3)$$  而$${{\\sin }^{4}}\\theta +{{\\sin }^{3}}\\theta \\cos \\theta +{{\\sin }^{2}}\\theta {{\\cos }^{2}}\\theta +\\sin \\theta {{\\cos }^{3}}\\theta +{{\\cos }^{4}}\\theta +3$$  $$={{({{\\sin }^{2}}\\theta +{{\\cos }^{2}}\\theta )}^{2}}+{{\\sin }^{3}}\\theta \\cos \\theta +\\sin \\theta {{\\cos }^{3}}\\theta -{{\\sin }^{2}}\\theta {{\\cos }^{2}}\\theta +3$$  $$={{\\sin }^{3}}\\theta \\cos \\theta +\\sin \\theta {{\\cos }^{3}}\\theta -{{\\sin }^{2}}\\theta {{\\cos }^{2}}\\theta +4\\textgreater0$$，  所以原不等式等价于$$\\sin \\theta -\\cos \\theta \\textgreater0$$，因此$$\\frac{ \\pi }{4}\\textless{}\\theta \\textless{}\\frac{5}{4} \\pi $$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "683", "queId": "bf03c34c55c146889b174af0d77e3dda", "competition_source_list": ["2021年北京海淀区清华大学竞赛（科学领军人才培养计划）第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$，$$c$$，$$d$$都是正整数，且$${{a}^{3}}={{b}^{2}}$$，$${{c}^{5}}={{d}^{4}}$$，$$c-a=77$$，求$$d-b$$．", "answer_option_list": [[{"aoVal": "A", "content": "$231$ "}], [{"aoVal": "B", "content": "$$240$$ "}], [{"aoVal": "C", "content": "$233$ "}], [{"aoVal": "D", "content": "$$235$$ "}], [{"aoVal": "E", "content": "$$240$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质", "竞赛->知识点->多项式与方程->因式定理"], "answer_analysis": ["由题意可设$$a={{x}^{2}}$$，$$b={{x}^{3}}$$，$$c={{y}^{4}}$$，$$d={{y}^{5}}$$，从而$${{y}^{4}}-{{x}^{2}}=77$$，因式分解得$$({{y}^{2}}-x)({{y}^{2}}+x)=77$$，从而$${{y}^{2}}-x=7$$，$${{y}^{2}}+x=11$$，解得$$x=2$$，$$y=3$$，因此$$d-b={{y}^{5}}-{{x}^{3}}=235$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "723", "queId": "c3baf90b2bb44e31893cc0d5e0c6327e", "competition_source_list": ["2017年湖南全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知点$$P$$为正三棱柱$$ABC-{{A}_{1}}{{B}_{1}}{{C}_{1}}$$上底面$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$的中心，作平面$$BCD\\bot AP$$，与棱$$A{{A}_{1}}$$交于$$D$$，若$$A{{A}_{1}}=2AB=2$$，则三棱锥$$D-ABC$$的体积为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{3}}{48}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{3}}{24}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{3}}{16}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{3}}{12}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["过$$A{{A}_{1}}$$和$$BC$$中点$$M$$作截面，$$AP$$在这个截面内．  由$$AP\\bot MD$$，可得$$AD=\\frac{1}{4}$$．  所以，$${{V}_{D-ABC}}=\\frac{1}{3}\\cdot \\frac{\\sqrt{3}}{4}\\cdot \\frac{1}{4}=\\frac{\\sqrt{3}}{48}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1053", "queId": "c56fb68a913143bbb29dc0c2527b24d6", "competition_source_list": ["2010年浙江全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "化简三角有理式$$\\frac{{{\\cos }^{4}}x+{{\\sin }^{4}}x+{{\\sin }^{2}}x{{\\cos }^{2}}x}{{{\\sin }^{6}}x+{{\\cos }^{6}}x+2{{\\sin }^{2}}x{{\\cos }^{2}}x}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\sin x+\\cos x$$ "}], [{"aoVal": "C", "content": "$$\\sin x\\cos x$$ "}], [{"aoVal": "D", "content": "1+$$\\sin x\\cos x$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["分母$$=({{\\sin }^{2}}x+{{\\cos }^{2}}x)({{\\sin }^{4}}x+{{\\cos }^{4}}x-{{\\sin }^{2}}x{{\\cos }^{2}}x)+2{{\\sin }^{2}}x{{\\cos }^{2}}x$$ $$={{\\sin }^{4}}x+{{\\cos }^{4}}x+{{\\sin }^{2}}x{{\\cos }^{2}}x$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "990", "queId": "c97bcb8dba16439c9b33132af1c2a7ed", "competition_source_list": ["2019~2020学年辽宁高一上学期期中（辽南协作体）第11题5分", "2008年浙江全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$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)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->函数的概念"], "answer_analysis": ["函数$$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}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1136", "queId": "d89b91aa16fa498a961bec9eb5def425", "competition_source_list": ["2009年四川全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "下列函数中，以$$\\frac{\\pi }{2}$$为最小正周期的偶函数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$y=\\sin 2x+\\cos 2x$$ "}], [{"aoVal": "B", "content": "$$y=\\sin 2x\\cos 2x$$ "}], [{"aoVal": "C", "content": "$$y={{\\sin }^{2}}x+\\cos 2x$$ "}], [{"aoVal": "D", "content": "$$y={{\\sin }^{2}}2x-{{\\cos }^{2}}2x$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["在$$\\text{A}$$选项中，取$$x=\\frac{\\pi }{4}$$，则$$y=1$$；取$$x=-\\frac{\\pi }{4}$$，则$$y=-1$$，从而$$y$$不是偶函数；在$$\\text{B}$$选项中，$$y=\\frac{1}{2}\\sin 4x$$，它不是偶函数；在$$\\text{C}$$选项中，$$y=\\frac{1+\\cos 2x}{2}$$，它的最小正周期为$$\\pi $$；在$$\\text{D}$$选项中，$$y=-\\cos 4x$$，符合条件． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "320", "queId": "be223a5254084ded948c13fcf8222361", "competition_source_list": ["2011年四川全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "甲、乙、丙三人一起玩``剪刀、石头、布''的游戏．每一局甲、乙、丙同时出``剪刀、石头、布''中的一种手势，且是相互独立的．设在一局中甲赢的人数为$$\\xi $$，则随机变量$$\\xi $$的数学期望$$E\\xi $$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{9}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->概率初步"], "answer_analysis": ["$$P(\\xi =0)=\\frac{3\\times 4}{27}=\\frac{4}{9}$$,$$P(\\xi =1)=\\frac{3\\times 4}{27}=\\frac{4}{9}$$,$$P(\\xi =2)=\\frac{3\\times 1}{27}=\\frac{1}{9}$$,  于是$$E\\xi =\\frac{4}{9}\\times 0+\\frac{4}{9}\\times 1+\\frac{1}{9}\\times 2=\\frac{2}{3}$$. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "198", "queId": "1103e15424174072957a3236b6147ead", "competition_source_list": ["2013年浙江全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知直线$$AB$$与抛物线$${{y}^{2}}=4x$$交于$$A$$，$$B$$两点，$$M$$为$$AB$$的中点，$$C$$为抛物线上一个动点，若$${{C}_{0}}$$满足$$\\overrightarrow{{{C}_{0}}A}\\cdot \\overrightarrow{{{C}_{0}}B}=\\min \\left { \\overrightarrow{CA}\\cdot \\overrightarrow{CB} \\right }$$，则下列一定成立的是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{C}_{0}}M\\bot AB$$ "}], [{"aoVal": "B", "content": "$${{C}_{0}}M\\bot l$$，其中$$l$$是抛物线过$${{C}_{0}}$$的切线 "}], [{"aoVal": "C", "content": "$${{C}_{0}}A\\bot {{C}_{0}}B$$ "}], [{"aoVal": "D", "content": "$${{C}_{0}}M=\\frac{1}{2}AB$$ "}]], "knowledge_point_routes": ["课内体系->知识点->平面向量->平面向量基本定理及其坐标表示->坐标表示平面向量的垂直", "课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积的运算律", "课内体系->知识点->平面向量->平面向量的运算->数量积->向量的数量积的定义", "课内体系->知识点->平面向量->平面向量的运算->数量积->平面向量数量积运算（非坐标）", "课内体系->知识点->圆锥曲线->直线与圆锥曲线问题->直线与圆锥曲线的位置关系判断", "课内体系->知识点->圆锥曲线->抛物线->直线和抛物线的位置关系", "课内体系->素养->直观想象"], "answer_analysis": ["$$\\overrightarrow{CA}\\cdot \\overrightarrow{CB}=(\\overrightarrow{CM}\\cdot \\overrightarrow{AM})\\cdot (\\overrightarrow{CM}\\cdot \\overrightarrow{BM})$$  $$={{\\left\\textbar{} \\overrightarrow{CM} \\right\\textbar}^{2}}-\\overrightarrow{CM}(\\overrightarrow{AM}+\\overrightarrow{BM})+\\overrightarrow{AM}\\cdot \\overrightarrow{BM}$$  $$={{\\left\\textbar{} \\overrightarrow{CM} \\right\\textbar}^{2}}-{{\\left\\textbar{} \\overrightarrow{AM} \\right\\textbar}^{2}}$$  $$\\Rightarrow \\min \\left { \\overrightarrow{CA}\\cdot \\overrightarrow{CB} \\right }$$  $$={{\\left\\textbar{} \\overrightarrow{CM} \\right\\textbar}^{2}}_{\\min }$$  $$\\Leftrightarrow CM\\text{//}l$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "345", "queId": "54dbaa7382644b59ba6e1e5baf4ebe7d", "competition_source_list": ["2005年全国高中数学联赛竞赛一试第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "空间四点$$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}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "927", "queId": "971f5707f43a4599942e3215212186c7", "competition_source_list": ["第二十届全国希望杯高一竞赛复赛邀请赛第8题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\text{e}$$是自然对数的底，那么$$\\sqrt{\\text{e}},\\frac{1}{\\sqrt{\\text{e}}},\\sin \\sqrt{\\text{e}},\\tan \\sqrt{\\text{e}}$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{\\sqrt{\\text{e}}}\\textless{}\\sin \\sqrt{\\text{e}}\\textless{}\\sqrt{\\text{e}}\\textless{}\\tan \\sqrt{\\text{e}}$$ "}], [{"aoVal": "B", "content": "$$\\sin \\sqrt{\\text{e}}\\textless{}\\frac{1}{\\sqrt{\\text{e}}}\\textless{}\\sqrt{\\text{e}}\\textless{}\\tan \\sqrt{\\text{e}}$$ "}], [{"aoVal": "C", "content": "$$\\tan \\sqrt{\\text{e}}\\textless{}\\frac{1}{\\sqrt{\\text{e}}}\\textless{}\\sin \\sqrt{\\text{e}}\\textless{}\\sqrt{\\text{e}}$$ "}], [{"aoVal": "D", "content": "$$\\tan \\sqrt{\\text{e}}\\textless{}\\sin \\sqrt{\\text{e}}\\textless{}\\frac{1}{\\sqrt{\\text{e}}}\\textless{}\\sqrt{\\text{e}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["首先$$ \\pi \\textgreater\\sqrt{\\text{e}}\\textgreater\\frac{ \\pi }{2}\\textgreater1$$，$$\\tan \\sqrt{\\text{e}}\\textless{}0$$，$$\\sin \\sqrt{\\text{e}}\\textgreater\\sin \\frac{ \\pi }{4}=\\frac{1}{\\sqrt{2}}\\textgreater\\frac{1}{\\sqrt{\\text{e}}}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "29", "queId": "3339df915f7140f9b8a2a5fb1cf84e84", "competition_source_list": ["2011年江西全国高中数学联赛竞赛初赛第7题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "满足$${{x}^{2}}+7{{y}^{2}}=2011$$的正整数$$(x,  y)$$的组数为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数", "竞赛->知识点->数论模块->不定方程->同余分析"], "answer_analysis": ["由于$$2011$$是$$4N+3$$形状的数，所以$$y$$必为奇数，而$$x$$为偶数， 设$$x=2m$$，$$y=2n+1$$，代人得  $$4{{m}^{2}}+28n(n+1)=2004$$，  即$${{m}^{2}}+7n(n+1)=501$$. ①  而$$n(n+1)$$为偶数，则$${{m}^{2}}$$为奇数，设$$m=2k+1$$，则  $${{m}^{2}}=4k(k+1)+1$$，  由①得，$$k(k+1)+7\\cdot \\frac{n(n+1)}{4}=125$$， ②  则$$\\frac{n(n+1)}{4}$$为奇数，且$$n,  n+1$$中恰有一个是$$4$$的倍数，当$$n=4r$$，为使$$7\\cdot \\frac{n(n+1)}{4}=7r(4r+1)$$为奇数，且$$7r(4r+1)\\textless{}125$$，只有$$r=1$$，②成为$$k(k+1)+35=125$$，  即$$k(k+1)=90$$，于是$$n=4,  k=9,  x=38,  y=9$$；  若$$n+1=4r$$，为使$$7\\cdot \\frac{n(n+1)}{4}=7r(4r-1)$$为奇数，且$$7r(4r-1)\\textless{}125$$，只有$$r=1$$，②成为$$k(k+1)+21=125$$，即$$k(k+1)=104$$，它无整解；  于是$$(x,  y)=(38,  9)$$是唯一解：$${{38}^{2}}+7\\cdot {{9}^{2}}=2011$$．  （另外，也可由$$x$$为偶数出发，使$$2011-{{x}^{2}}=2009-({{x}^{2}}-2)=7\\times 287-({{x}^{2}}-2)$$  为$$7$$的倍数，那么$${{x}^{2}}-2$$是$$7$$的倍数，故$$x$$是$$7k\\pm 3$$形状的偶数，依次取$$k=1,  3,  5$$，检验相应的六个数即可．） "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "863", "queId": "ff8080814bfd15e4014c2af4652835f7", "competition_source_list": ["2012年黑龙江全国高中数学联赛竞赛初赛第1题5分", "2011年高考真题湖北卷理科第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "$$\\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$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->复数->复数的运算->复数的乘法和除法", "课内体系->知识点->复数->复数的运算->复数中的周期问题"], "answer_analysis": ["因为$$\\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}$$，  故选A． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1106", "queId": "d83308ab16c64a09a8c1eef6e6a4c101", "competition_source_list": ["2010年山东全国高中数学联赛竞赛初赛第7题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$z$$为复数，$$\\text{i}$$为虚数单位．若$$\\left\\textbar{} z \\right\\textbar=1,\\left\\textbar{} \\overline{z}+\\text{i} \\right\\textbar=1$$，则当$${{(z+\\text{i})}^{n}}$$（$$n$$为正整数）为实数时，$${{\\left\\textbar{} z+\\text{i} \\right\\textbar}^{n}}$$的最小值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$2\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$3\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["设$$z=x+y\\text{i}(x,y\\in \\mathbf{R})$$，  由$$\\left\\textbar{} z \\right\\textbar=1,\\left\\textbar{} \\overline{z}+\\text{i} \\right\\textbar=1$$，得$$\\begin{cases}{{x}^{2}}+{{y}^{2}}=1    {{x}^{2}}+{{(1-y)}^{2}}=1 \\end{cases}$$  解得$$\\begin{cases}x=\\pm \\frac{\\sqrt{3}}{2}    y=\\frac{1}{2} \\end{cases}$$所以$$z=\\pm \\frac{\\sqrt{3}}{2}+\\frac{1}{2}\\text{i}$$，  $$z+\\text{i}=\\pm \\frac{\\sqrt{3}}{2}\\text{i}+\\frac{3}{2}\\text{i}=\\sqrt{3}\\left( \\pm \\frac{1}{2}+\\frac{\\sqrt{3}}{2}\\text{i} \\right)$$．  $${{\\left\\textbar{} z+\\text{i} \\right\\textbar}^{n}}={{\\left\\textbar{} \\sqrt{3}\\left( \\pm \\frac{1}{2}+\\frac{\\sqrt{3}}{2}\\text{i} \\right) \\right\\textbar}^{n}}$$  $$={{(\\sqrt{3})}^{n}}{{\\left\\textbar{} \\pm \\frac{1}{2}+\\frac{\\sqrt{3}}{2}\\text{i} \\right\\textbar}^{n}}={{(\\sqrt{3})}^{n}}.$$  由此可见，当$$n$$取最小值$${{n}_{0}}$$时，$${{\\left\\textbar{} z+\\text{i} \\right\\textbar}^{n}}$$取最小值$${{(\\sqrt{3})}^{{{n}_{0}}}}$$．  当$$z=\\frac{\\sqrt{3}}{2}+\\frac{1}{2}\\text{i}$$时，  $$z+\\text{i}=\\sqrt{3}\\left( \\frac{1}{2}+\\frac{\\sqrt{3}}{2}\\text{i} \\right)=\\sqrt{3}\\left( \\cos \\frac{ \\pi }{3}+\\text{i}\\sin \\frac{ \\pi }{3} \\right)$$．  当$$z=-\\frac{\\sqrt{3}}{2}+\\frac{1}{2}\\text{i}$$时，  $$z+\\text{i=}\\sqrt{3}\\left( -\\frac{1}{2}+\\frac{\\sqrt{3}}{2}\\text{i} \\right)=\\sqrt{3}\\left( \\cos \\frac{2 \\pi }{3}+\\text{i}\\sin \\frac{2 \\pi }{3} \\right)$$．  可见使$${{(z+\\text{i})}^{n}}$$为实数时$$n$$的最小值为$$3$$，  此时$${{\\left\\textbar{} z+\\text{i} \\right\\textbar}^{n}}$$的最小值为$${{(\\sqrt{3})}^{3}}=3\\sqrt{3}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1113", "queId": "eef6e78b871343debacb079bf82dff6e", "competition_source_list": ["2016年AMC12竞赛A第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "對於所有的實數定義餘數$$\\text{rem}\\left(x,y\\right)=x-y\\left\\lfloor {\\frac{x}{y}} \\right\\rfloor$$：，這裡$$\\left\\lfloor {\\frac{x}{y}} \\right\\rfloor$$ 定義為小於等於$$\\frac{x}{y}$$的最大整数．试求$$\\text{rem}\\left(\\frac{3}{8},-\\frac{2}{5}\\right)$$之値爲何？  The remainder can be defined for all real numbers $$x$$ and $$y$$ with $$y\\ne 0$$ by $$\\text{rem}\\left(x,y\\right)=x-y\\left\\lfloor {\\frac{x}{y}} \\right\\rfloor$$ where $$\\left\\lfloor {\\frac{x}{y}} \\right\\rfloor$$ denotes the greatest integer less than or equal to $$\\frac{x}{y}$$. What is the value of $$\\text{rem}\\left(\\frac{3}{8},-\\frac{2}{5}\\right)$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{3}{8}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{1}{40}$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{8}$$ "}], [{"aoVal": "E", "content": "$$\\frac{31}{40}$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["The value, by definition, is  $$\\text {rem}\\left(\\frac 38,-\\frac 25\\right)=\\frac 38-\\left(-\\frac 25\\right)\\left[\\frac {\\frac 38}{-\\frac 25}\\right]$$  $$=\\frac 38-\\left(-\\frac 25\\right)\\left[\\frac 38\\times \\frac {-5}2\\right]$$  $$=\\frac 38-\\left(-\\frac 25\\right)\\left[\\frac {-15}{16}\\right]$$  $$=\\frac 38-\\left(-\\frac 25\\right)(-1)$$  $$=\\frac 38-\\frac 25$$  $$=-\\frac 1{40}$$.  Do note that the denominator of the answer will be a multiple of $$5$$ and $$8(40)$$ and that the answer will also be negative. The only answer choice that satisfies this is $$\\rm B$$. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "653", "queId": "7677610fff6a4117a5bc9ac0580bbf29", "competition_source_list": ["2008年陕西全国高中数学联赛竞赛初赛第10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$k$$、$$m$$、$$n$$均为整数，过圆$${{x}^{2}}+{{y}^{2}}={{(3k+1)}^{2}}$$外一点$$P({{m}^{3}}-m,{{n}^{3}}-n)$$向该圆引两条切线，切点分别为$$A$$、$$B$$，则直线$$AB$$上整点（横坐标和纵坐标都是整数的点）有．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$个 "}], [{"aoVal": "B", "content": "$$1$$个 "}], [{"aoVal": "C", "content": "$$0$$个 "}], [{"aoVal": "D", "content": "无数个 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆综合"], "answer_analysis": ["易知，切点弦所在直线$$AB$$的方程为$$({{m}^{3}}-m)x+({{n}^{3}}-n)y={{(3k+1)}^{2}}$$．  若直线$$AB$$上存在整点$$({{x}_{0}},{{y}_{0}})$$，  则有$$(m-1)m(m+1){{x}_{0}}+(n-1)n(n+1){{y}_{0}}={{(3k+1)}^{2}}$$．  因为$$(m-1)m(m+1)$$和$$(n-1)n(n+1)$$都是$$3$$的倍数，  所以上式左边能被$$3$$整除．而右边被$$3$$除余$$1$$，矛盾．  故直线$$AB$$上不存在整点．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "505", "queId": "29d621ba1ac948bdafe89937fcead0f6", "competition_source_list": ["2013年黑龙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "$$\\text{i}$$为虚数单位，则$$\\text{i}+{{\\text{i}}^{2}}+{{\\text{i}}^{3}}+{{\\text{i}}^{4}}+{\\ldots }+{{\\text{i}}^{2013}}$$=．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\text{i}$$ "}], [{"aoVal": "B", "content": "$$-\\text{i}$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算"], "answer_analysis": ["由复数性质可知选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "9", "queId": "0504985f7b29402eaad0f5b948c5ee07", "competition_source_list": ["2010年黑龙江全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "与圆$${{(x-2)}^{2}}+{{y}^{2}}=1$$相切，且在两坐标轴上截距相等的直线共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$条 "}], [{"aoVal": "B", "content": "$$3$$条 "}], [{"aoVal": "C", "content": "$$4$$条 "}], [{"aoVal": "D", "content": "$$6$$条 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆综合", "竞赛->知识点->解析几何->圆与方程", "竞赛->知识点->解析几何->直线与方程"], "answer_analysis": ["由数形结合可知，与圆相切且在两坐标轴上截距相等的直线分成两类：一类是过原点满足条件的直线有$$2$$条，另一类是斜率为$$-1$$满足条件的直线也有$$2$$条，所以共$$4$$条． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "807", "queId": "bf6b249da6e54af8962312e5459cc5a6", "competition_source_list": ["2013年辽宁全国高中数学联赛竞赛初赛第3题6分", "高二上学期单元测试《代数变形1》自招第9题", "1991年全国高中数学联赛竞赛一试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$，$$b$$，$$c$$均为非零复数，且$$\\frac{a}{b}=\\frac{b}{c}=\\frac{c}{a}$$，则$$\\frac{a+b-c}{a-b+c}$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\pm \\omega $$ "}], [{"aoVal": "C", "content": "$$1,\\omega ,{{\\omega }^{2}}$$ "}], [{"aoVal": "D", "content": "$$1,\\omega ,{{\\omega }^{2}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->换元技巧->代数换元"], "answer_analysis": ["令$$\\frac{a}{b}=\\frac{b}{c}=\\frac{c}{a}=t$$，则  $$a=bt=c{{t}^{2}}=a{{t}^{3}}$$．  由$$a\\mathsf{\\ne }0$$，知  $${{t}^{3}}=1$$，  因此$$t=1,\\omega ,{{\\omega }^{2}}$$．  利用比例性质，知原式等于$$\\frac{1}{t}$$．  故当$$t=1,\\omega ,{{\\omega }^{2}}$$时，原式分别取$$1,\\omega ,{{\\omega }^{2}}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "340", "queId": "54cf7e8bc38c462f9658987c92870bea", "competition_source_list": ["2018年湖北全国高中数学联赛竞赛初赛第3题10分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$f(x)$$为定义在区间$$(0,+\\infty )$$上的单调函数．若对任意的$$x\\in (0,+\\infty )$$，均有$$f(f(x)-2{{\\log }_{2}}x)=4$$，则不等式$$f(x)\\textless{}6$$的解集为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\infty,2)$$ "}], [{"aoVal": "B", "content": "$$(\\infty,4)$$ "}], [{"aoVal": "C", "content": "$$(0,2)$$ "}], [{"aoVal": "D", "content": "$$(0,4)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["由题设，知存在正常数$$c$$，使得$$f(c)=4$$，且对任意的$$x\\in (0,+\\infty )$$，  均有$$f(x)-2{{\\log }_{x}}x=c$$．  当$$x=c$$时，有$$f(c)=2{{\\log }_{2}}c+c=4$$，  由单调性，知此方程只有唯一解$$c=2$$．  于是，$$f(x)=2{{\\log }_{x}}x+2$$．  由$$f(x)\\textless{}6\\Rightarrow 2{{\\log }_{2}}x+2\\textless{}6\\Rightarrow 0\\textless{}x\\textless{}4$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "289", "queId": "1a7bfb74af954c07840d8af10e080d0c", "competition_source_list": ["2019~2020学年4月山西太原迎泽区太原市第五中学校高二下学期周测C卷理科第6题6分", "2012年黑龙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设集合$$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}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1130", "queId": "d3f782f1144c453d974135fbbd6e81c9", "competition_source_list": ["2016年天津全国高中数学联赛竞赛初赛第6题6分", "2016年高考真题天津卷"], "difficulty": "1", "qtype": "single_choice", "problem": "设函数$$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)$$是单调递增函数；  （$$4$$）若方程$$f\\left( f\\left( x \\right) \\right)=x$$有实根，则方程$$f\\left( x \\right)=x$$也有实根．  正确的命题共有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$个 "}], [{"aoVal": "B", "content": "$$2$$个 "}], [{"aoVal": "C", "content": "$$3$$个 "}], [{"aoVal": "D", "content": "$$4$$个 "}]], "knowledge_point_routes": ["课内体系->素养->数据分析", "课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->函数的应用->函数与方程", "课内体系->知识点->函数的概念与性质->函数的性质->周期性->函数周期性判断", "课内体系->知识点->函数的概念与性质->函数的性质->奇偶性->函数奇偶性的判定->利用定义判断函数奇偶性", "课内体系->知识点->函数的概念与性质->函数的性质->单调性->函数单调性的判定->用定义法证明函数的单调性", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->复合函数"], "answer_analysis": ["若$$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}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "230", "queId": "cbe1a086ffca4d21a2c67ffe28935c1f", "competition_source_list": ["2018~2019学年北京东城区北京市第二中学高一上学期期中第6题5分", "2013年黑龙江全国高中数学联赛竞赛初赛第1题5分", "2017~2018学年10月北京朝阳区北京陈经纶中学高一上学期月考理科第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知集合$$A=\\left { x\\left\\textbar{} {{\\left( \\frac{1}{2} \\right)}^{x}}\\leqslant 1 \\right. \\right }$$，$$B=\\left { x\\textbar{{x}^{2}}-6x+8\\leqslant 0 \\right }$$，则$A\\cap\\overline{B}$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left { x\\textbar x\\leqslant 0 \\right }$$ "}], [{"aoVal": "B", "content": "$$\\left { x\\textbar2\\leqslant x\\leqslant 4 \\right }$$ "}], [{"aoVal": "C", "content": "$$\\left { x\\textbar0\\leqslant x\\textless{}2 \\right.$$或$$x\\textgreater4 }$$ "}], [{"aoVal": "D", "content": "$$ {x\\textbar0\\textless{}x\\leqslant 2$$或$$x\\geqslant 4 }$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->等式与不等式->不等式->解不等式->一元二次不等式", "课内体系->知识点->基本初等函数->指数函数->指数方程和指数不等式", "课内体系->知识点->集合->集合的基本运算->交集", "课内体系->知识点->集合->集合的基本运算->交、并、补集混合运算", "课内体系->知识点->集合->集合的基本运算->补集"], "answer_analysis": ["∵$${{\\left( \\frac{1}{2} \\right)}^{x}}\\leqslant 1$$，  解出$$x\\geqslant 0$$，即$$A=\\left { x\\textbar x\\geqslant 0 \\right }$$．  ∵$${{x}^{2}}-6x+8=(x-2)(x-4)\\leqslant 0$$，  解出$$2\\leqslant x\\leqslant 4$$，即$$B=\\left { x\\textbar2\\leqslant x\\leqslant 4 \\right }$$，  ∴$${{\\complement }_{\\mathbf{R}}}B= {x\\textbar x\\textless{}2$$或$$x\\textgreater4 }$$，  ∴$$A\\cap ({{\\complement }_{\\mathbf{R}}}B)=\\left { x\\textbar0\\leqslant x\\textless{}2 \\right.$$或$$x\\textgreater4 }$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "120", "queId": "0bf5efa003104ff5ae9df152dfc4ecc5", "competition_source_list": ["2016年AMC12竞赛B第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "當$$a=\\frac{1}{2}$$时，$$\\frac{2a^{-1}+\\dfrac{a^{-1}}{2}}{a}$$ 的值為何？  What is the value of $$\\frac{2a^{-1}+\\dfrac{a^{-1}}{2}}{a}$$ when $$a=\\frac{1}{2}$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{2}$$ "}], [{"aoVal": "D", "content": "$$10$$ "}], [{"aoVal": "E", "content": "$$20$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["Factorizing the numerator, $$\\dfrac{\\dfrac{1}{a}\\cdot\\left( 2+\\dfrac{1}{2}\\right)}{a}$$ then becomes $$\\dfrac{\\dfrac{5}{2}}{a^{2}}$$ which is equal to $$\\dfrac{5}{2}\\cdot2^{2}$$ which is $$\\left(\\text{D}\\right)10$$. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "824", "queId": "809feba8139f4270a861e952b9584c18", "competition_source_list": ["2008年山东全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "若函数$$f(x)={{\\log }_{a}}x(a\\textless0$$且$$a\\ne 1)$$，满足$$f\\left( \\frac{2}{a} \\right)\\textgreater f\\left( \\frac{3}{a} \\right)$$，则$$f\\left( 1-\\frac{1}{x} \\right)\\textgreater1$$的解集是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0\\textless{}x\\textless{}\\frac{1}{a}$$ "}], [{"aoVal": "B", "content": "$$0\\textless{}x\\textless{}\\frac{1}{1-a}$$ "}], [{"aoVal": "C", "content": "$$1\\textless{}x\\textless{}\\frac{1}{a}$$ "}], [{"aoVal": "D", "content": "$$1\\textless{}x\\textless{}\\frac{1}{1-a}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["由$$f\\left( \\frac{2}{a} \\right)={{\\log }_{a}}\\frac{2}{a}\\textgreater f\\left( \\frac{3}{a} \\right)={{\\log }_{a}}\\frac{3}{a}$$，  得$$0\\textless{}a\\textless{}1$$．  所以，原不等式等价于不等式$$0\\textless{}1-\\frac{1}{x}\\textless{}a$$．  解此不等式得$$1\\textless{}x\\textless{}\\frac{1}{1-a}$$．故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "303", "queId": "2320a8032d39438594cc9a8f2338cfe5", "competition_source_list": ["2017年辽宁全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\triangle ABC$$的三个内角为$$A,B,C$$，若$${{\\sin }^{2}}A+{{\\sin }^{2}}B+{{\\sin }^{2}}C=2$$，则$$\\cos A+\\cos B+2\\cos C$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{2}\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{4}{3}\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{6}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->三角函数->三角形中的问题->解三角形"], "answer_analysis": ["$$2={{\\sin }^{2}}A+{{\\sin }^{2}}B+{{\\sin }^{2}}C=1-{{\\cos }^{2}}A+\\left( 1-{{\\cos }^{2}}B \\right)+\\left( 1-{{\\cos }^{2}}C \\right)$$，  则$$1-{{\\cos }^{2}}C={{\\cos }^{2}}A+{{\\cos }^{2}}B=\\frac{1+\\cos 2A}{2}+\\frac{1+\\cos 2B}{2}$$  $$=1+\\cos \\left( A+B \\right)\\cos \\left( A-B \\right)=1-\\cos C\\cos \\left( A-B \\right)$$  所以，$${{\\cos }^{2}}C=\\cos C\\cos \\left( A-B \\right)$$．  若$$\\cos C=0$$，则$$C=90{}^{}\\circ $$，$$\\cos A+\\cos B+2\\cos C=\\cos A+\\sin A\\leqslant \\sqrt{2}$$；  若$$\\cos C\\ne 0$$，则$$\\cos C=\\cos \\left( A-B \\right)$$，$$C=A-B$$或$$C=B-A$$，得$$A=90{}^{}\\circ $$或$$B=90{}^{}\\circ $$．  于是，$$\\cos A+\\cos B+2\\cos C=\\sin C+2\\cos C\\leqslant \\sqrt{5}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "688", "queId": "7fc5b7d7679347da9a37f1baa5f960b5", "competition_source_list": ["2019年北京高一竞赛初赛（中学生数学竞赛）第1题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知集合$$M= {2,0,1,9 }$$，$$A$$为$$M$$的子集，且$$A$$中各元素之和为$$3$$的倍数．则满足条件的子集$$A$$的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$5$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["经枚举，各元素之和为$$3$$的倍数的子集有：  $$ {0 }$$、$$ {9 }$$、$$ {2,1 }$$、$$ {0,9 }$$、$$ {2,0,1 }$$、$$ {2,1,9 }$$、$$ {2,0,1,9 }$$，共$$7$$个． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1049", "queId": "fc2e6f7c7c25496e96617cdbf2e11651", "competition_source_list": ["1986年全国高中数学联赛竞赛一试第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "设实数$$a$$、$$b$$、$$c$$满足$$\\begin{cases}{{a}^{2}}-bc-8a+7=0    {{b}^{2}}+{{c}^{2}}+bc-6a+6=0    \\end{cases}$$，那么$$a$$的取值范围是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\infty ,+\\infty )$$ "}], [{"aoVal": "B", "content": "$$(-\\infty ,1)\\cup [9,+\\infty ]$$ "}], [{"aoVal": "C", "content": "$$(0,7)$$ "}], [{"aoVal": "D", "content": "$$[1,9]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的综合应用（数列与不等式）", "竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["由$$\\begin{cases}{{a}^{2}}-bc-8a+7=0    {{b}^{2}}+{{c}^{2}}+bc-6a+6=0    \\end{cases}$$，  得$$\\begin{cases}bc={{a}^{2}}-8a+7    {{b}^{2}}+{{c}^{2}}+bc=6a-6    \\end{cases}$$，  ∴$${{b}^{2}}+{{c}^{2}}-2bc=-3({{a}^{2}}-10a+9)$$，  即$${{(b+c)}^{2}}=-3(a-1)(a-9)\\mathsf{\\geqslant 0}$$，  ∴$$1\\mathsf{\\leqslant }a\\mathsf{\\leqslant }9$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "877", "queId": "9fc83e0f0aa346b897f1ebd2c1812822", "competition_source_list": ["2017年辽宁全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "对于三位数$$\\overline{abc}$$，满足$$\\overline{abc}=37\\left( a+b+c \\right)$$的三位数的个数共有个", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$15$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->整除->最大公约数和最小公倍数"], "answer_analysis": ["由题设，$$100a+10b+c=37\\left( a+b+c \\right)$$，即$$63a=27b+36c$$，则$$7a=3b+4c$$．  $$3\\left( a-b \\right)=4\\left( c-a \\right)$$，则$$3\\left\\textbar{} c-a4\\left\\textbar{} a-b \\right. \\right.$$．  ①$$a-b=4$$，$$c-a=3$$，则$$\\left( abc \\right)=\\left( 407 \\right)\\left( 518 \\right)\\left( 629 \\right)$$；  ②$$a-b=8c-a=6$$，无解；  ③$$a-b=-4c-a=-3$$，则$$\\left( abc \\right)=\\left( 370 \\right)\\left( 481 \\right)\\left( 592 \\right)$$；  ④$$a-b=0=c-a$$，则$$\\left( abc \\right)$$有$$9$$种可能．  综上，三位数的个数共有$$3+3+9=15$$个． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "763", "queId": "c3daf03d147b40dba11f5181287b1971", "competition_source_list": ["2017年辽宁全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "数列$${{a}_{0}}=134,{{a}_{1}}=150$$，$${{a}_{k+1}}={{a}_{k-1}}-\\frac{k}{{{a}_{k}}}\\left( k=1,2,\\cdots ,n-1 \\right)$$，若$${{a}_{n}}=0$$，则$$n$$为．", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$201$$ "}], [{"aoVal": "C", "content": "$$2017$$ "}], [{"aoVal": "D", "content": "$$20101$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["由题设，有$${{a}_{n+1}}{{a}_{n}}={{a}_{n}}{{a}_{n-1}}-n$$，  于是，$${{a}_{n}}{{a}_{n-1}}={{a}_{n-1}}{{a}_{n-2}}-\\left( n-1 \\right)$$，  \\ldots\\ldots{}  $${{a}_{2}}{{a}_{1}}={{a}_{1}}{{a}_{0}}-1$$，  以上式子累加，得$${{a}_{n+1}}{{a}_{n}}={{a}_{1}}{{a}_{0}}-\\frac{n\\left( n+1 \\right)}{2}=20100-\\frac{n\\left( n+1 \\right)}{2}$$．  若$${{a}_{n}}=0$$，则$${{a}_{n}}{{a}_{n-1}}=0$$，解得$$n=201$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "341", "queId": "3e0e8ee1bc794b648e61e40632d08701", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第8题5分", "2005年高考真题山东卷文科第7题5分", "2005年高考真题山东卷理科第6题5分", "2015~2016学年广东广州越秀区广州市执信中学高三上学期期末理科第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f(x)=\\begin{cases}\\sin (\\pi{{x}^{2}}),-1\\textless{}x\\textless{}0    {{\\text{e}}^{x-1}},x\\geqslant 0 \\end{cases}$$若$$f(1)+f(a)=2$$，则$$a$$的所有可能值为（ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-\\frac{\\sqrt{2}}{2}$$ "}], [{"aoVal": "C", "content": "$$1$$，$$-\\frac{\\sqrt{2}}{2}$$ "}], [{"aoVal": "D", "content": "$$1$$，$$\\frac{\\sqrt{2}}{2}$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->函数求值问题", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->分段函数"], "answer_analysis": ["由题意知，当$$-1\\textless{}x\\textless{}0$$时，$$f(x)=\\sin (\\pi{{x}^{2}})$$；  当$$x\\geqslant 0$$时，$$f(x)={{\\text{e}}^{x-1}}$$；  ∴$$f(1)={{\\text{e}}^{1-1}}=1$$．  若$$f(1)+f(a)=2$$，则$$f(a)=1$$；  当$$a\\geqslant 0$$时，$${{\\text{e}}^{a-1}}=1$$，  ∴$$a=1$$；  当$$-1\\textless{}a\\textless{}0$$时，$$\\sin (\\pi{{x}^{2}})=1$$，  ∴$${{x}^{2}}=\\frac{1}{2}$$，$$x=\\frac{\\sqrt{2}}{2}$$（不满足条件，舍去），或$$x=-\\frac{\\sqrt{2}}{2}$$．  所以$$a$$的所有可能值为：$$1$$，$$-\\frac{\\sqrt{2}}{2}$$．  故答案为：$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "32", "queId": "09e327c1b12b4917b19a5a4a1e595f2f", "competition_source_list": ["2010年山东全国高中数学联赛竞赛初赛第10题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "某人从一层上到二层需跨$$10$$个台阶．他一步可能跨$$1$$个台阶，称为一阶步，也可能跨$$2$$个台阶，称为二阶步，最多能跨$$3$$个台阶，称为三阶步．从一层上到二层他总共跨了$$6$$步，而且任何相邻两步均不同阶，那么他从一层到二层可能的不同过程共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$个 "}], [{"aoVal": "B", "content": "$$8$$个 "}], [{"aoVal": "C", "content": "$$10$$个 "}], [{"aoVal": "D", "content": "$$12$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->逻辑推理"], "answer_analysis": ["按题意要求，不难验证这$$6$$步中不可能没有三阶步，也不可能有多于$$1$$个的三阶步，所以只能是$$1$$个三阶步，$$2$$个二阶步，$$3$$个一阶步．  为形象起见，我们以白、黑、红三种颜色的球来记录从一层到二层跨越$$10$$个台阶的过程：白球表示一阶步，黑球表示二阶步，红球表示三阶步．每一过程可表示为$$3$$个白球，$$2$$个黑球，$$1$$个红球的一个同色球不相邻的排列．  情况1：第$$1$$、第$$6$$球均为白球，则两黑球分别位于中间白球的两侧．此时共有$$4$$个黑白球之间的空位放置红球．所以此种情况共有$$4$$种可能的不同排列．  情况2：第$$1$$球不是白球．  （$$1$$）第$$1$$球为红球，则余下$$5$$球只有一种可能的排列；  （$$2$$）第$$1$$球为黑球，则余下$$5$$球因红、黑球的位置不同有两种不同的排列．  因此此种情况共有$$3$$种不同排列．  情况3：第$$6$$球不是白球，显然这种情况与情况$$2$$对称共有$$3$$种不同排列．  总之，按题意要求从一层到二层共有4+3+3=10（种）可能的不同过程． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1171", "queId": "d4959bc9ead144ed9dfbb5ed914752a0", "competition_source_list": ["2008年湖南全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "连接球面上两点的线段称为球的弦．半径为$$4$$的球的两条弦$$AB$$、$$CD$$的长度分别等于$$2\\sqrt{7}$$和$$4\\sqrt{3}$$，$$M$$、$$N$$分别为$$AB$$、$$CD$$的中点，两条弦的两端都在球面上运动，有下面四个命题：  ①弦$$AB$$、$$CD$$可能相交于点$$M$$；②弦$$AB$$、$$CD$$可能相交于点$$N$$；  ③$$MN$$的最大值为$$5$$；④$$MN$$的最小值为$$1$$．  其中真命题为．", "answer_option_list": [[{"aoVal": "A", "content": "①③④ "}], [{"aoVal": "B", "content": "①②③ "}], [{"aoVal": "C", "content": "①②④ "}], [{"aoVal": "D", "content": "②③④ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的角与距离", "竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["假设$$AB$$、$$CD$$相交于点$$N$$，则$$AB$$、$$CD$$共面，  所以$$A$$、$$B$$、$$C$$、$$D$$四点共圆，  而过圆的弦$$CD$$的中点$$N$$的弦$$AB$$的长度显然有$$AB\\geqslant CD$$，  所以②是错的．  容易证明，当以$$AB$$为直径的圆面与以$$CD$$为直径的圆面平行且在球心两侧时，  $$MN$$最大为５，故③对．  当以$$AB$$为直径的圆面与以$$CD$$为直径的圆面平行且在球心同侧时，  $$MN$$最小为$$1$$，故④对．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1094", "queId": "bcec07fd16f9424d846de784fe489882", "competition_source_list": ["1998年全国高中数学联赛竞赛一试第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "在正方体的$$8$$个顶点，$$12$$条棱的中点，$$6$$个面的中心及正方体的中心共$$27$$个点中，共线的三点组的个数是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$57$$ "}], [{"aoVal": "B", "content": "$$49$$ "}], [{"aoVal": "C", "content": "$$43$$ "}], [{"aoVal": "D", "content": "$$37$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->两个基本计数原理"], "answer_analysis": ["$$8$$个顶点中无$$3$$点共线，故共线的三点组中至少有一个是棱中点或面中心或体中心．  （$$1$$）体中心为中点：$$4$$对顶点，$$6$$对棱中点，$$3$$对面中心；共$$13$$组；  （$$2$$）面中心为中点：$$4\\times 6=24$$组；  （$$3$$）棱中点为中点：$$12$$个．  共$$13+24+12=49$$个．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "666", "queId": "768b1848297142f69aee5bcca83be4dd", "competition_source_list": ["2015年甘肃全国高中数学联赛竞赛初赛第4题7分"], "difficulty": "1", "qtype": "single_choice", "problem": "在数列$$\\left { {{a}_{n}} \\right }$$中，$${{a}_{1}}=2,{{a}_{2}}=10$$，对所有的正整数$$n$$都有$${{a}_{n+2}}={{a}_{n+1}}-{{a}_{n}}$$，则$${{a}_{2021}}=$$~\\uline{~~~~~~~~~~}~", "answer_option_list": [[{"aoVal": "A", "content": "$$-10$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$2$$ "}], [{"aoVal": "E", "content": "$$10$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->（模）周期数列", "课内体系->知识点->数列->数列的概念->递推数列与递推方法"], "answer_analysis": ["$${{a}_{n+3}}={{a}_{n+2}}-{{a}_{n+1}}=\\left( {{a}_{n+1}}-{{a}_{n}} \\right)-{{a}_{n+1}}=-{{a}_{n}}$$，所以$${{a}_{n+6}}=-{{a}_{n+3}}={{a}_{n}}$$，即$$\\left { {{a}_{n}} \\right }$$是周期为$$6$$的周期数列．故$${{a}_{2021}}={{a}_{5}}=-{{a}_{2}}=-10$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "28", "queId": "12c4f23ac14d43bba90c0f4923556bee", "competition_source_list": ["2020~2021学年河北石家庄新华区石家庄市第二中学高二上学期期中（竞赛班）第12题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$A$$，$$B$$是抛物线$${{y}^{2}}=8x$$上的两个动点且$$\\textbar AB\\textbar=16$$，则线段$$AB$$中点$$M$$到直线$$x=-3$$距离的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$10$$ "}]], "knowledge_point_routes": ["课内体系->知识点->圆锥曲线->抛物线->抛物线的定义、标准方程->抛物线的标准方程", "课内体系->知识点->圆锥曲线->抛物线->抛物线的定义、标准方程->抛物线中其他最值问题", "课内体系->素养->数学运算"], "answer_analysis": ["设$$A({{x}_{1}},{{y}_{1}})$$，$$B({{x}_{2}},{{y}_{2}})$$，  抛物线$${{y}^{2}}=8x$$的线准线$$x=-2$$，  线段$$AB$$中点$$M$$到直线$$x=-3$$距离：  $$S=\\left\\textbar{} \\frac{{{x}_{1}}+{{x}_{2}}}{2} \\right\\textbar+3$$  $$=\\frac{{{x}_{1}}+2+{{x}_{2}}+2}{2}+1$$  $$=\\frac{\\textbar AF\\textbar+\\textbar BF\\textbar}{2}+1$$．  （两边之和大于第三边且$$A$$，$$B$$，$$F$$三点共线时取等号）  ∴$$\\frac{\\textbar AF\\textbar+\\textbar BF\\textbar}{2}+1\\geqslant \\frac{\\textbar AB\\textbar}{2}+1=8+1=9$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "62", "queId": "53c8fdb5547f48059951ea1e434c9c5e", "competition_source_list": ["2022年浙江宁波竞赛第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知正方形$ABCD$的边长为1，则$\\left\\textbar{} \\overrightarrow{AB}+2\\overrightarrow{BC}+\\overrightarrow{AC} \\right\\textbar=$（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "1 "}], [{"aoVal": "B", "content": "$\\sqrt{6}$ "}], [{"aoVal": "C", "content": "$\\sqrt{7}$ "}], [{"aoVal": "D", "content": "$\\sqrt{13}$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 利用向量的线性运算和垂直向量的数量积为0可求题设中向量的模.\\\\ 【详解】\\\\ $\\left\\textbar{} \\overrightarrow{AB}+2\\overrightarrow{BC}+\\overrightarrow{AC}\\left\\textbar{} = \\right\\textbar2\\overrightarrow{AB}+3\\overrightarrow{BC} \\right\\textbar=\\sqrt{4{{\\overrightarrow{AB}}^{2}}+9{{\\overrightarrow{BC}}^{2}}}=\\sqrt{13}$，\\\\ 故选：D． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "802", "queId": "8979e65063b54556a697d49f49a9bb23", "competition_source_list": ["2020~2021学年河北石家庄新华区石家庄市第二中学高二上学期期中（竞赛班）第5题3分"], "difficulty": "1", "qtype": "single_choice", "problem": "从一箱产品中随机地抽取一件，设事件$$A=$$``抽到一等品''，事件$$B=$$``抽到二等品''，事件$$C=$$``抽到三等品''，且已知$$P\\left( A \\right)=0.65$$，$$P\\left( B \\right)=0.2$$， $$P\\left( C \\right)=0.1$$，则事件``抽到的不是一等品''的概率为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$0.65$$ "}], [{"aoVal": "B", "content": "$$0.35$$ "}], [{"aoVal": "C", "content": "$$0.3$$ "}], [{"aoVal": "D", "content": "$$0.005$$ "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->知识点->统计与概率->概率->事件与概率->互斥事件->对立事件的概率和为1", "课内体系->知识点->统计与概率->概率->事件与概率->互斥事件->互斥事件与对立事件", "课内体系->知识点->统计与概率->概率->事件与概率->概率的概念->概率的基本性质"], "answer_analysis": ["由题意知本题是求一个对立事件的概率，因为抽到的不是一等品的对立事件是抽到一等品，$$P\\left( A \\right)=0.65$$，所以抽到的不是一等品的概率是$$1-0.65=0.35$$．  故答案为：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "125", "queId": "417c1be393e8440cb27aa28edf60be6f", "competition_source_list": ["1988年全国高中数学联赛竞赛一试第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "设有三个函数，第一个是$$y=\\varphi \\left( x \\right)$$，它的反函数就是第二个函数，而第三个函数的图象与第二个函数的图象关于直线$$x+y=0$$对称．那么第三个函数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$y=-\\varphi \\left( x \\right)$$ "}], [{"aoVal": "B", "content": "$$y=-\\varphi \\left( -x \\right)$$ "}], [{"aoVal": "C", "content": "$$y=-{{\\varphi }^{-1}}\\left( x \\right)$$ "}], [{"aoVal": "D", "content": "$$y=-{{\\varphi }^{-1}}\\left( -x \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的概念", "竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["第一个函数的图象与第二个函数的图象关于$$x-y=0$$对称，第二个函数的图象与第三个函数的图象关于$$x+y=0$$对称，所以第一个函数的图象与第三个函数的图象关于原点对称． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1030", "queId": "d763efa8290b49de886146ec6afe56ba", "competition_source_list": ["2016年湖南郴州高三三模理科第2题5分", "2013年黑龙江全国高中数学联赛竞赛初赛第3题5分", "2016年湖南郴州高三三模文科第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "命题``所有实数的平方都是正数''的否定为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "所有实数的平方都不是正数 "}], [{"aoVal": "B", "content": "有的实数的平方是正数 "}], [{"aoVal": "C", "content": "至少有一个实数的平方是正数 "}], [{"aoVal": "D", "content": "至少有一个实数的平方不是正数 "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->知识点->常用逻辑用语->命题->逻辑联结词", "课内体系->知识点->常用逻辑用语->命题->全称量词命题与存在量词命题的否定"], "answer_analysis": ["∵``全称命题''的否定一定是``存在性命题''，  ∴命题``所有实数的平方都是正数''的否定是：``至少有一个实数的平方不是正数''．  故选：$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1177", "queId": "e6d211ff9981423ab6c62153299d02fa", "competition_source_list": ["1996年全国高中数学联赛竞赛一试第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "如果在区间$$[1,2]$$上，函数$$f(x)={{x}^{2}}+px+q$$与$$g(x)=x+\\frac{1}{{{x}^{2}}}$$在同一点取相同的最小值，那么$$f(x)$$在该区间上的最大值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4+\\frac{11}{2}\\sqrt[3]{2}+\\sqrt[3]{4}$$ "}], [{"aoVal": "B", "content": "$$4-\\frac{5}{2}\\sqrt[3]{2}+\\sqrt[3]{4}$$ "}], [{"aoVal": "C", "content": "$$1-\\frac{1}{2}\\sqrt[3]{2}+\\sqrt[3]{4}$$ "}], [{"aoVal": "D", "content": "以上答案都对 "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->均值", "竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->函数的概念"], "answer_analysis": ["在$$[1,2]$$上$$g(x)=x+\\frac{1}{{{x}^{2}}}=\\frac{x}{2}+\\frac{x}{2}+\\frac{1}{{{x}^{2}}}\\geqslant 3\\sqrt[3]{\\frac{1}{4}}=\\frac{3}{2}\\sqrt[3]{2}$$，  ∵$$f(x)$$与$$g(x)$$在同一点上取相等的最小值，  当$$\\frac{x}{2}=\\frac{1}{{{x}^{2}}}$$，即$$x=\\sqrt[2]{3}{\\in }\\left[ 1    2 \\right]$$时，$$g{{(x)}_{最小}}=\\frac{3}{2}\\sqrt[3]{2}$$，  ∴$$-\\frac{p}{2}=\\sqrt[3]{2}$$，$$\\frac{4q-{{p}^{2}}}{4}=\\frac{3}{2}\\sqrt[3]{2}$$，  解得：$$p=-2\\sqrt[3]{2}$$，$$q=\\frac{3}{2}\\sqrt[3]{2}+\\sqrt[3]{4}$$，  于是$$f(x)={{x}^{2}}-2\\sqrt[2]{2x}+\\frac{3}{2}\\sqrt[3]{2}+\\sqrt[3]{4}$$，  由于$$\\sqrt[3]{2}-1\\textless{}2-\\sqrt[3]{2}$$，  ∴$$f{{(x)}_{最大}}=f(2)=4-\\frac{5}{2}\\sqrt[3]{2}+\\sqrt[3]{4}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "170", "queId": "2f487043e9bb4b429be74f1116b0bb99", "competition_source_list": ["2012年黑龙江全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "函数$$y={{\\text{e}}^{x}}$$的图象与直线$$x=0$$、直线$$y=\\text{e}$$所围成的区域面积是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\text{e}-1$$ "}], [{"aoVal": "C", "content": "$$\\text{e}$$ "}], [{"aoVal": "D", "content": "$$2\\text{e}-1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->积分"], "answer_analysis": ["就是$$\\int_{0}^{1}{\\left( \\text{e}-{{\\text{e}}^{x}} \\right)\\text{dx}=1}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1152", "queId": "e1cf08db3e2b4b65a11c6481fe007268", "competition_source_list": ["2012年湖南全国高中数学联赛竞赛初赛第4题6分", "2009年上海复旦大学自主招生千分考第23题", "高三上学期单元测试《复数》自招第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "设实数$$r\\textgreater1$$，如果复平面上的动点$$z$$满足$$\\textbar z\\textbar=r$$，则动点$$\\omega =z+\\frac{1}{z}$$的轨迹是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "焦距为$$4$$的椭圆 "}], [{"aoVal": "B", "content": "焦距为$$\\frac{4}{r}$$的椭圆 "}], [{"aoVal": "C", "content": "焦距为$$2$$的椭圆 "}], [{"aoVal": "D", "content": "焦距为$$\\frac{2}{r}$$的椭圆 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->曲线与方程", "竞赛->知识点->复数与平面向量->模、辐角与单位根", "竞赛->知识点->复数与平面向量->复数的应用"], "answer_analysis": ["令$$z=r\\left( \\cos \\theta +\\text{i}\\sin \\theta \\right)$$，则  $$\\omega =r\\left( \\cos \\theta +\\text{i}\\sin \\theta \\right)+\\frac{1}{r}\\left( \\cos \\theta -\\text{i}\\sin \\theta \\right)=\\left( r+\\frac{1}{r} \\right)\\cos \\theta +\\text{i}\\cdot \\left( r-\\frac{1}{r} \\right)\\sin \\theta $$，  设$$\\omega =x+y\\text{i}$$，则$$\\frac{{{x}^{2}}}{{{\\left( r+\\frac{1}{r} \\right)}^{2}}}+\\frac{{{y}^{2}}}{{{\\left( r-\\frac{1}{r} \\right)}^{2}}}=1$$，  这是一个椭圆．$${{c}^{2}}={{\\left( r+\\frac{1}{r} \\right)}^{2}}-{{\\left( r-\\frac{1}{r} \\right)}^{2}}=4$$，焦距为$$4$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "976", "queId": "97a04486e0e546e08444f4c3970ed029", "competition_source_list": ["2008年贵州全国高中数学联赛竞赛初赛第9题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "顶点在同一球面上的正四棱柱$$ABCD-{A}'{B}'{C}'{D}'$$中，$$AB=1$$，$$A{A}'=\\sqrt{2}$$，则$$A$$，$$C$$两点间的球面距离为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\pi }{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\pi }{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{2}}{4}\\pi $$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{2}}{2}\\pi $$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["设球心为$$O$$，则$$O$$为正四棱柱对角线的交点，  因为$$AC=\\sqrt{2}$$，$$A{C}'=2$$，  故$$OA=OC=1$$，$$\\triangle OAC$$为直角三角形，  $$\\angle AOC=\\frac{ \\pi }{2}$$，故选$$\\text{B}$$. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1139", "queId": "fd25eb3deee54f7ab72dcb715de6e665", "competition_source_list": ["2017~2018学年10月北京海淀区清华大学附属中学高一上学期月考理科第5题", "2008年高考真题辽宁卷理科第12题5分", "2010年黑龙江全国高中数学联赛竞赛初赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$f(x)$$是连续的偶函数，且当$$x\\textgreater0$$时$$f(x)$$是单调函数，则满足$$f(x)=f\\left( \\frac{x+3}{x+4} \\right)$$的所有$$x$$之和为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-3$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$-8$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->素养->逻辑推理", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->抽象函数", "课内体系->知识点->函数的概念与性质->函数的性质->奇偶性->函数奇偶性的判定->函数奇偶性的运算"], "answer_analysis": ["由$$f(x)$$的性质知$$f(x)=f\\left( \\frac{x+3}{x+4} \\right)\\Rightarrow \\left\\textbar{} x \\right\\textbar=\\left\\textbar{} \\frac{x+3}{x+4} \\right\\textbar$$，若$$x=\\frac{x+3}{x+4}$$，即$${{x}^{2}}+3x-3=0$$，有解，且两根之和为$$-3$$；  若$$-x=\\frac{x+3}{x+4}$$，即$${{x}^{2}}+5x+3=0$$，有解，且两根之和为$$-5$$，故总和为$$-8$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "231", "queId": "8af059d43e654b6f9019af9b18dd3812", "competition_source_list": ["2010年四川全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$p:\\sqrt{1+\\sin 2\\alpha }=\\frac{4}{3}$$和$$q:\\left\\textbar{} \\sin \\alpha +\\cos \\alpha \\right\\textbar=\\frac{4}{3}$$．则$$p$$是$$q$$的．", "answer_option_list": [[{"aoVal": "A", "content": "充分但不必要条件 "}], [{"aoVal": "B", "content": "必要但不充分条件 "}], [{"aoVal": "C", "content": "充要条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换", "竞赛->知识点->逻辑->常用逻辑用语"], "answer_analysis": ["因为$$\\sqrt{1+\\sin 2\\alpha }=\\sqrt{{{(\\sin \\alpha +\\cos \\alpha )}^{2}}}=\\left\\textbar{} \\sin \\alpha +\\cos \\alpha \\right\\textbar=\\frac{4}{3}$$，故$$p$$是$$q$$的充要条件．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1197", "queId": "f4fbc08aa432421dbe5593a0e80c25cd", "competition_source_list": ["2022年江苏徐州贾汪区江苏省贾汪中学高一竞赛（下学期春季）第1题", "2022年江苏徐州贾汪区江苏省贾汪中学高一竞赛（下学期春季）第1题", "2022年江苏徐州贾汪区江苏省贾汪中学高一竞赛（下学期春季）第1题", "2022年江苏徐州贾汪区江苏省贾汪中学高一竞赛（下学期春季）第1题", "2021~2022学年四川南充顺庆区四川省南充高级中学高二上学期开学考试理科第5~5题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\text{sin}\\left( \\frac{\\pi }{6}-\\theta \\right)=\\text{cos}\\left( \\frac{\\pi }{6}+\\theta \\right)$$，则$$\\cos 2\\theta =$$（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "1 "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "0 "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角恒等变换->二倍角公式"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据两角差的正弦公式和两角和的余弦公式，可得$$\\tan \\theta =-1$$，即可得到$$\\theta =-\\frac{\\pi }{4}+k\\pi ,k\\in \\mathbf{Z}$$，再根据诱导公式即可求出$$\\cos 2\\theta $$.\\\\ 【详解】\\\\ 因为$$\\text{sin}\\left( \\frac{\\pi }{6}-\\theta \\right)=\\text{cos}\\left( \\frac{\\pi }{6}+\\theta \\right)$$，所以$$\\frac{1}{2}\\cos \\theta -\\frac{\\sqrt{3}}{2}\\text{sin}\\theta =\\frac{\\sqrt{3}}{2}\\cos \\theta -\\frac{1}{2}\\text{sin}\\theta $$，\\\\ 所以$$\\frac{1-\\sqrt{3}}{2}\\text{sin}\\theta =\\frac{\\sqrt{3}-1}{2}\\cos \\theta $$，所以$$\\tan \\theta =-1$$，\\\\ 所以$$\\theta =-\\frac{\\pi }{4}+k\\pi ,k\\in \\mathbf{Z}$$，\\\\ 所以$$\\cos 2\\theta =\\cos \\left[ 2\\times \\left( -\\frac{\\pi }{4}+k\\pi \\right) \\right]=\\cos \\left( -\\frac{\\pi }{2}+2k\\pi \\right)=\\cos \\left( -\\frac{\\pi }{2} \\right)=0,k\\in \\mathbf{Z}$$.\\\\ 故选：D. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "496", "queId": "b536d9872be44036b10bd9ad59e84255", "competition_source_list": ["2015年AMC10竞赛B第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "对于多少个整数$$x$$来说，点$$(x,-x)$$在以$$(5.5)$$为中心的半径为$$10$$的圆内或圆上？", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$14$$ "}], [{"aoVal": "E", "content": "$$15$$ "}]], "knowledge_point_routes": ["课内体系->知识点->直线和圆的方程->圆与方程->圆的标准方程与一般方程->圆的标准方程问题", "美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Circle"], "answer_analysis": ["The equation of the circle is $$(x-5)^{2}+(y-5)^{2}=100$$. Plugging in the given conditions we have $$(x-5)^{2}+(-x-5)^{2}\\leqslant 100$$. Expanding gives: $$x^{2}-10x+25+x^{2}+10x+25\\leqslant 100$$, which simpifies to $$x^{2}\\leqslant 25$$ and therefore $$x\\leqslant 5$$ and $$x\\geqslant -5$$. So $$x$$ ranges from $$-5$$ to $$5$$, for a total of $$\\boxed{\\rm (A)\\textasciitilde11}$$ values. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "747", "queId": "ace7d0ecedcd4ef78f7e886f585bdd45", "competition_source_list": ["2022~2023学年吉林长春朝阳区长春市第五中学高一上学期期末第7题", "2022~2023学年吉林长春朝阳区长春市第五中学高一上学期期末第7题", "2022~2023学年吉林长春朝阳区长春市第五中学高一上学期期末第7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8题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$\\cos \\alpha =\\frac{1}{3}$，$\\cos \\left( \\beta -\\alpha \\right)=\\frac{\\sqrt{3}}{3}$，且$0\\textless{} \\beta \\textless{} \\alpha \\textless{} \\pi $，则$\\cos \\beta =$", "answer_option_list": [[{"aoVal": "A", "content": "$-\\frac{5\\sqrt{3}}{9}$ "}], [{"aoVal": "B", "content": "$-\\frac{\\sqrt{3}}{3}$ "}], [{"aoVal": "C", "content": "$\\frac{2\\sqrt{3}}{9}$ "}], [{"aoVal": "D", "content": "$\\frac{5\\sqrt{3}}{9}$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 利用同角三角函数之间的关系求出$\\sin\\alpha =\\frac{2\\sqrt{2}}{3},\\sin\\left( \\beta -\\alpha \\right)=-\\frac{\\sqrt{6}}{3}$，再利用$\\cos \\beta =\\cos \\left[ \\left( \\beta -\\alpha \\right)+\\alpha \\right]$求解即可.\\\\ 【详解】\\\\ $\\because $ $\\text{cos}\\alpha =\\frac{1}{3}$，$\\text{cos}\\left( \\beta -\\alpha \\right)=\\frac{\\sqrt{3}}{3}$，且$0\\textless{} \\beta \\textless{} \\alpha \\textless{} \\pi $，\\\\ $\\therefore -\\pi \\textless{} \\beta -\\alpha \\textless{} 0$，\\\\ $\\therefore \\sin\\alpha =\\sqrt{1-\\frac{1}{9}}=\\frac{2\\sqrt{2}}{3},\\sin\\left( \\beta -\\alpha \\right)=-\\sqrt{1-\\frac{1}{3}}=-\\frac{\\sqrt{6}}{3}$，\\\\ $\\therefore \\cos \\beta =\\cos \\left[ \\left( \\beta -\\alpha \\right)+\\alpha \\right]$\\\\ $=\\cos \\left( \\beta -\\alpha \\right)\\cos \\alpha -\\sin\\left( \\beta -\\alpha \\right)\\sin\\alpha $\\\\ $=\\frac{1}{3}\\times \\frac{\\sqrt{3}}{3}-\\frac{2\\sqrt{2}}{3}\\times \\left( -\\frac{\\sqrt{6}}{3} \\right)=\\frac{5\\sqrt{3}}{9}$，故选D.\\\\ 【点睛】\\\\ 三角函数求值有三类，(1)``给角求值''：一般所给出的角都是非特殊角，从表面上来看是很难的，但仔细观察非特殊角与特殊角总有一定关系，解题时，要利用观察得到的关系，结合公式转化为特殊角并且消除非特殊角的三角函数而得解．(2)``给值求值''：给出某些角的三角函数式的值，求另外一些角的三角函数值，解题关键在于``变角''，使其角相同或具有某种关系．(3)``给值求角''：实质是转化为``给值求值''，先求角的某一函数值，再求角的范围，确定角． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "416", "queId": "79e061263ef7483da3b1b4f24efb06bd", "competition_source_list": ["2011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$$n$$为正整数，$$x={{\\left( 1+\\frac{1}{n} \\right)}^{n}},  y={{\\left( 1+\\frac{1}{n} \\right)}^{n+1}}$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$${{x}^{y}}\\textgreater{{y}^{x}}$$ "}], [{"aoVal": "B", "content": "$${{x}^{y}}={{y}^{x}}$$ "}], [{"aoVal": "C", "content": "$${{x}^{y}}\\textless{}{{y}^{x}}$$ "}], [{"aoVal": "D", "content": "以上都有可能 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["由于$$x=\\frac{{{(n+1)}^{n}}}{{{n}^{n}}},  y=\\frac{{{(n+1)}^{n+1}}}{{{n}^{n+1}}}$$，取对数易得$${{x}^{y}}={{y}^{x}}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1145", "queId": "e63f0dd37e404eb5b8880fc44fbc3768", "competition_source_list": ["竞赛第14题"], "difficulty": "3", "qtype": "single_choice", "problem": "甲、乙两人参加一次选举，已知结果：甲7票，乙5票，则计票过程中，甲的票数始终领先乙的概率以及甲的票数始终不落后于乙的概率分别为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\frac{1}{6},\\frac{3}{8}$ "}], [{"aoVal": "B", "content": "$\\frac{1}{12},\\frac{3}{8}$ "}], [{"aoVal": "C", "content": "$\\frac{1}{6},\\frac{3}{13}$ "}], [{"aoVal": "D", "content": "$\\frac{1}{12},\\frac{3}{13}$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据卡特兰数的模型可求甲的票数始终领先乙的概率以及甲的票数始终不落后于乙的概率.\\\\ 【详解】\\\\ 根据卡特兰数的模型，可得甲的票数不落后于乙的概率为$\\frac{{\\mathrm{C}}_{12}^{5}-{\\mathrm{C}}_{12}^{4}}{{\\mathrm{C}}_{12}^{5}}=\\frac{297}{792}=\\frac{3}{8}$．\\\\ 甲的票数领先乙的概率为（第一票必然为甲获得，相等于甲6票，乙5票的卡特兰模型）\\\\ $\\frac{{\\mathrm{C}}_{11}^{5}-{\\mathrm{C}}_{11}^{4}}{{\\mathrm{C}}_{12}^{5}}=\\frac{132}{792}=\\frac{1}{6}$．\\\\ 故选：A. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "366", "queId": "1bb3df3f506f4f0c92c28b28da2c0167", "competition_source_list": ["2015年黑龙江全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设正项等比数列$$ {{{a}_{n}} }$$的前$$n$$项和为$${{S}_{n}}$$，且$${{2}^{\\^{10}}}{{S}_{30}}+{{S}_{10}}=({{2}^{10}}+1){{S}_{20}}$$，则数列$$ {{{a}_{n}} }$$的公比为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{8}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列", "竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["$${{2}^{10}}\\left( 1+{{q}^{10}}+{{q}^{20}} \\right)+1=\\left( {{2}^{10}}+1 \\right)\\left( 1+{{q}^{10}} \\right)$$，解得$${{q}^{10}}=\\frac{1}{{{2}^{10}}}$$，所以$$q=\\frac{1}{2}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "160", "queId": "1d13701fbeba4b2e957e015efe2536cf", "competition_source_list": ["2008年吉林全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$ABC$$是平面上不共线的三点，$$G$$是三角形$$ABC$$的重心，动点$$P$$满足$$\\overrightarrow{GP}=\\frac{1}{3}\\left( \\frac{1}{2}\\overrightarrow{GA}+\\frac{1}{2}\\overrightarrow{GB}+2\\overrightarrow{GC} \\right)$$，则点$$P$$一定为三角形$$ABC$$的．", "answer_option_list": [[{"aoVal": "A", "content": "$$AB$$边中线的中点 "}], [{"aoVal": "B", "content": "$$AB$$边中线的三等分点（非重心） "}], [{"aoVal": "C", "content": "重心 "}], [{"aoVal": "D", "content": "$$AB$$边的中点 "}]], "knowledge_point_routes": ["知识标签->素养->数学运算", "知识标签->素养->数学建模", "知识标签->题型->平面向量->平面向量的运算->平面向量的线性运算->平面向量与三角形的“心”", "知识标签->题型->平面向量->平面向量的运算->平面向量的线性运算->几何、代数、坐标进行的线性运算", "知识标签->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量的数乘运算及运算规则", "知识标签->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量的减法运算及运算规则"], "answer_analysis": ["由选项知，本题应该选择$$C$$为起点，  于是得：$$6\\overrightarrow{CP}-6\\overrightarrow{CG}=\\overrightarrow{CA}-\\overrightarrow{CG}+\\overrightarrow{CB}-\\overrightarrow{CG}-4\\overrightarrow{CG}$$，  从而$$6\\overrightarrow{CP}=\\overrightarrow{CA}+\\overrightarrow{CB}$$，即$$\\overrightarrow{CP}=\\frac{1}{6}(\\overrightarrow{CA}+\\overrightarrow{CB})$$，  取$$AB$$边的中点$$M$$，则$$\\overrightarrow{CA}+\\overrightarrow{CB}=2\\overrightarrow{CM}$$，从而$$\\overrightarrow{CP}=\\frac{1}{3}\\overrightarrow{CM}$$，  即点$$P$$为三角形中$$AB$$边上的中线的一个三等分点，且点$$P$$不是重心．  点评：由$$\\overrightarrow{CP}=\\frac{1}{6}(\\overrightarrow{CA}+\\overrightarrow{CB})$$，$$\\frac{1}{6}:\\frac{1}{6}=1$$知点$$P$$在$$BC$$边的中线上；由$$\\frac{1}{6}+\\frac{1}{6}=\\frac{1}{3}$$知，是中线上靠近$$C$$的一个三等分点，且不是重心，  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "694", "queId": "76b53c7fa14742ebab70ea8244d8c52e", "competition_source_list": ["竞赛第16题"], "difficulty": "4", "qtype": "single_choice", "problem": "对$n$个正整数用\\emph{k}种颜色染色，使得无法从中选出三个不同色的正整数构成等差数列，设\\emph{k}的最大值为$f(n)$，则（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\log _{3}n\\leq f(n)\\leq \\log _{2}n+1$ "}], [{"aoVal": "B", "content": "$\\log _{2}n\\leq f(n)\\leq \\log _{2}n+1$ "}], [{"aoVal": "C", "content": "$\\log _{3}n\\leq f(n)\\leq \\log _{2}n$ "}], [{"aoVal": "D", "content": "$\\log _{3}n\\leq f(n)\\leq \\log _{3}n+1$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 利用特例可判断BCD的正误，利用整除性和数学归纳法可判断A的正误.\\\\ 【详解】\\\\ 取$n=4$，则$f(4)=3$，将1234染成$ABBC$即可．因此选项CD错误．\\\\ 下面证明$\\log _{3}n\\leq f(n)\\leq \\log _{2}n+1$．\\\\ 设$v(n,p)$为正整数\\emph{n}中素数\\emph{p}的幂次，即$p^{v(n,p)}\\parallel n$．\\\\ 对于任意的正整数$m(1\\leq m\\leq n)$，将\\emph{m}染成第$v(m,3)+1$种颜色，这样就用了$\\left[\\log _{3}n\\right]+1$种颜色．\\\\ 此时对于任意不同色的正整数\\emph{x}，\\emph{y}，有$v(2x-y,3)=v(2y-x,3)=\\min  {v(x,3),v(y,3) }$，\\\\ 因此$2x-y$和$2y-x$均与\\emph{x}，\\emph{y}之一同色，符合题意．这样就证明了$f(n)\\geq \\left[\\log _{3}n\\right]+1\\textgreater{} \\log _{3}n$．\\\\ 接下来设$2^{k}\\leq n\\textless{} 2^{k+1}\\left(k\\in \\mathbf{N}^{\\ast }\\right)$，则右侧不等式即$f(n)\\leq k+1$．\\\\ 对\\emph{k}进行归纳，应用数学归纳法证明．\\\\ 当$k=1$时，$n=2,3$，此时$f(n)=2$，命题成立．\\\\ 假设命题对$k-1$成立，对已有$1,2,\\cdots ,2^{k}-1$的染色，考虑1，2，\\ldots，\\emph{n}的染色，若其中有两种或两种以上新增的颜色，不妨设为红色和蓝色，且最小的红色数为\\emph{x}，最小的蓝色数为\\emph{y}，且$x\\textless{} y$，则$2^{k}\\leq x\\textless{} y\\leq n\\textless{} 2^{k+1}\\Rightarrow 2x-y\\textgreater{} 0$，\\\\ 考虑正整数$2x-y$，它与异色数对$(x,y)$成等差数列，于是必然与\\emph{x}，\\emph{y}之一同色，但$2x-y\\textless{} x\\textless{} y$，这与之前假设的\\emph{x}，\\emph{y}的最小性矛盾．\\\\ 因此1，2，\\ldots，\\emph{n}的染色至多在之前的基础上增加1种，从而命题得证．\\\\ 综上所述，有$\\log _{3}n\\leq f(n)\\leq \\log _{2}n+1$．\\\\ 接下来考虑$n=9$的情形，此时$f(n)=3$，排除选项B．\\\\ 故选：A. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1090", "queId": "f7f09030d6e4476994fd5f94766475cd", "competition_source_list": ["2017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": "已知正数$$a,b,c,d$$满足$$a+2b=1c+2d=1$$，则$$\\frac{1}{a}+\\frac{1}{bcd}$$的最小值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{16}$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{5}$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->均值"], "answer_analysis": ["因为$$c+2d=1\\geqslant 2\\sqrt{2cd}\\Rightarrow cd\\leqslant \\frac{1}{8}$$，所以  $$\\frac{1}{a}+\\frac{1}{bcd}\\geqslant \\frac{1}{a}+\\frac{8}{b}=\\left( \\frac{1}{a}+\\frac{8}{b} \\right)\\left( a+2b \\right)=17+\\frac{2b}{a}+\\frac{8a}{b}\\geqslant 17+8=25$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1064", "queId": "c12273f35c014bbb9301daaa1d82dda1", "competition_source_list": ["2019年江西全国高中数学联赛竞赛初赛第1题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$1,2,\\cdots , 19 $$中任意两个互异的数作乘积．则所有这种乘积的和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$15815$$ "}], [{"aoVal": "B", "content": "$$16815$$ "}], [{"aoVal": "C", "content": "$$17815$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的综合应用", "课内体系->知识点->等式与不等式->等式->代数式化简求值"], "answer_analysis": ["无 "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "693", "queId": "64a74b1a9bff45d4a15bfdef1636cf19", "competition_source_list": ["2009年河北全国高中数学联赛竞赛初赛第10题9分"], "difficulty": "1", "qtype": "single_choice", "problem": "甲、乙、丙、丁四个人做传球练习，球首先由甲传出，每个人得到球后都等可能地传给其余三个人之一，设$${{p}_{n}}$$表示经过$$n$$次传递后球回到甲手中的概率，则$${{p}_{6}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{19}{81}$$ "}], [{"aoVal": "B", "content": "$$\\frac{20}{81}$$ "}], [{"aoVal": "C", "content": "$$\\frac{59}{243}$$ "}], [{"aoVal": "D", "content": "$$\\frac{61}{243}$$ "}]], "knowledge_point_routes": ["知识标签->知识点->概率->条件概率与事件的独立性->事件的独立性->相互独立事件的概率乘法公式", "知识标签->知识点->概率->事件与概率->概率的概念->概率的基本性质", "知识标签->题型->概率->事件的独立性与条件概率->概率乘法公式问题", "知识标签->素养->逻辑推理"], "answer_analysis": ["经过一次传递后，球落在乙、丙、丁手中的概率分别为$$\\frac{1}{3}$$，而落在甲手中的概率为$$0$$，因此$${{p}_{1}}=0$$．两次传递后球落在甲手中的概率为  $${{p}_{2}}=\\frac{1}{3}\\times \\frac{1}{3}+\\frac{1}{3}\\times \\frac{1}{3}+\\frac{1}{3}\\times \\frac{1}{3}=\\frac{1}{3}$$．  下面考虑递推．要想经过$$n$$次传递后球落在甲的手中，那么在$$n-1$$次传递后球一定不在甲手中，所以$${{p}_{n}}=\\frac{1}{3}(1-{{p}_{n-1}})$$，$$n=1,2,3,4,\\cdots $$因此  $${{p}_{3}}=\\frac{1}{3}(1-{{p}_{2}})=\\frac{1}{3}\\times \\frac{2}{3}=\\frac{2}{9}$$，  $${{p}_{4}}=\\frac{1}{3}(1-{{p}_{3}})=\\frac{1}{3}\\times \\frac{7}{9}=\\frac{7}{27}$$，  $${{p}_{5}}=\\frac{1}{3}(1-{{p}_{4}})=\\frac{1}{3}\\times \\frac{20}{27}=\\frac{20}{81}$$，  $${{p}_{6}}=\\frac{1}{3}(1-{{p}_{5}})=\\frac{1}{3}\\times \\frac{61}{81}=\\frac{61}{243}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1133", "queId": "dd1bb00194024a4992098511f332f314", "competition_source_list": ["1996年全国全国高中数学联赛竞赛一试第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "把圆$${{x}^{2}}+{{(y-1)}^{2}}=1$$与椭圆$$9{{x}^{2}}+{{(y+1)}^{2}}=9$$的公共点，用线段连接起来所得到的图形为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "线段 "}], [{"aoVal": "B", "content": "不等边三角形 "}], [{"aoVal": "C", "content": "等边三角形 "}], [{"aoVal": "D", "content": "四边形 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->圆与方程"], "answer_analysis": ["圆$${{x}^{2}}+(y-1)=1$$与椭圆$$9{{x}^{2}}+{{(y+1)}^{2}}=9$$的交点$$A(xy)$$的坐标应满足方程，$${{x}^{2}}+\\frac{{{(y+1)}^{2}}}{9}={{x}^{2}}+{{(y-1)}^{2}}$$，  解得，$$y=2$$，或$$\\frac{1}{2}$$．于是相应地求得$$x=0$$，或$$x=\\pm \\frac{\\sqrt{3}}{2}$$．这样．得到给定圆与椭圆的三个交点：$${{A}_{1}}(0,2)$$，$${{A}_{2}}\\left( \\frac{\\sqrt{3}}{2}    \\frac{1}{2} \\right)$$，$${{A}_{3}}\\left( -\\frac{\\sqrt{3}}{2}    \\frac{1}{2} \\right)$$易求得，$$\\left\\textbar{} {{A}_{1}}{{A}_{2}} \\right\\textbar=\\left\\textbar{} {{A}_{1}}{{A}_{3}} \\right\\textbar=\\left\\textbar{} {{A}_{2}}{{A}_{3}} \\right\\textbar=\\sqrt{3}$$，  因此所得到的图形为正三角形．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "626", "queId": "ff766b71be264da08e3cfd41df5d8018", "competition_source_list": ["2017年天津全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "将曲线$$y={{\\log }_{2}}x$$沿$$x$$轴正方向移动$$1$$个单位，再沿$$y$$轴负方向移动$$2$$个单位，得到曲线$$C$$．在下列曲线中，与$$C$$关于直线$$x+y=0$$对称的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$y={{2}^{x+2}}-1$$ "}], [{"aoVal": "B", "content": "$$y=-{{2}^{x+2}}-1$$ "}], [{"aoVal": "C", "content": "$$y=-{{2}^{2-x}}-1$$ "}], [{"aoVal": "D", "content": "$$y={{2}^{2-x}}-1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["曲线$$C$$：$$y={{\\log }_{2}}\\left( x-1 \\right)-2$$，与$$C$$关于$$x+y=0$$对称：$$-x={{\\log }_{2}}\\left( -y-1 \\right)-2$$，  化简得$$y=-{{2}^{2-x}}-1$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "570", "queId": "8843d2e8e0fe42f397dd8a735a1924b4", "competition_source_list": ["2019~2020学年上海浦东新区华东师范大学第二附属中学高二上学期单元测试第1题", "2000年全国高中数学联赛竞赛一试第3题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知点$$A$$为双曲线$${{x}^{2}}-{{y}^{2}}=1$$的左顶点，点$$B$$和点$$C$$在双曲线的右分支上，$$\\triangle ABC$$是等边三角形，则$$\\triangle ABC$$的面积是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{3}}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3\\sqrt{3}}{2}$$ "}], [{"aoVal": "C", "content": "$$3\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$6\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->双曲线"], "answer_analysis": ["设点$$B\\left( {{x}_{1}},{{y}_{1}} \\right)$$和点$$C\\left( {{x}_{2}},{{y}_{2}} \\right)$$，  则$$\\left { \\begin{array}{*{35}{l}} {{\\left( {{x}_{1}}+1 \\right)}^{2}}+y_{1}^{2}={{\\left( {{x}_{2}}+1 \\right)}^{2}}+y_{2}^{2}    x_{1}^{2}-y_{1}^{2}=x_{2}^{2}-y_{2}^{2}=1   \\end{array} \\right.$$$$\\Rightarrow \\left( {{x}_{1}}-{{x}_{2}} \\right)\\left( {{x}_{1}}+{{x}_{2}}+1 \\right)=0$$，  因为点$$B$$和点$$C$$在双曲线的右分支上，  所以$${{x}_{1}}={{x}_{2}}$$，$${{k}_{AB}}=\\frac{\\sqrt{3}}{3}$$，  所以$$AB$$直线方程为$$y=\\frac{\\sqrt{3}}{3}\\left( x+1 \\right)$$，  联立$$\\left { \\begin{array}{*{35}{l}} y=\\dfrac{\\sqrt{3}}{3}\\left( x+1 \\right)    {{x}^{2}}-{{y}^{2}}=1   \\end{array} \\right.$$，  得$$\\left\\textbar{} AB \\right\\textbar=2\\sqrt{3}$$，  $${{S}_{\\triangle ABC}}=3\\sqrt{3}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1096", "queId": "dc9178e66ecd4b258cb7640533af365b", "competition_source_list": ["2005年AMC12竞赛B第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个信封里有八张钞票：$$2$$张$$1$$元，$$2$$张$$5$$元，$$2$$张$$10$$元，$$2$$张$$20$$元$$.$$ 不放回地随机抽出两张钞票，这两张钞票的总和大于等于$20$的概率是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{7}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{7}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "E", "content": "$$\\frac{2}{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率", "美国AMC10/12->Knowledge Point->Combination->Probability Calculation->Classical Models of Probabilities"], "answer_analysis": ["The only way to get a total of $$$20$$ or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of $$\\left(\\begin{array}{l} 8    2 \\end{array}\\right)=\\frac{8 \\times 7}{2 \\times 1}=28$$ ways to choose $2$ bills out of $8$. There are $12$ ways to choose a twenty and some other non-twenty bill. There is $1$ way to choose both twenties, and also $1$ way to choose both tens. Adding these up, we find that there are a total of $14$ ways to attain a sum of $20$ or greater, so there is a total probability of $\\frac{14}{28}=\\frac{1}{2}$. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "357", "queId": "2c50825fb2334212a8a169b02dc41711", "competition_source_list": ["2016年湖北全国高中数学联赛高二竞赛初赛第3题9分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\triangle ABC$$是等边三角形，椭圆$$\\Gamma $$的一个焦点为$$A$$，另一个焦点$$F$$在线段$$BC$$上，如果椭圆$$\\Gamma $$恰好经过$$B$$，$$C$$两点，则它的离心率为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2\\sqrt{5}}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2\\sqrt{3}}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{3}}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{5}}{5}$$ "}]], "knowledge_point_routes": ["知识标签->素养->直观想象", "知识标签->素养->数学运算", "知识标签->知识点->平面解析几何->椭圆->椭圆的定义、标准方程->椭圆的标准方程", "知识标签->知识点->平面解析几何->椭圆->椭圆的简单几何性质->椭圆的离心率", "知识标签->题型->平面解析几何->椭圆与双曲线中确定离心率的问题->求椭圆的离心率", "知识标签->题型->平面解析几何->圆锥曲线中确定基本量的问题->椭圆、双曲线、抛物线的基本量求解"], "answer_analysis": ["设$$\\triangle ABC$$边长为$$2$$，则$$4a=2\\times 3=6$$，$$a=\\frac{3}{2}$$，且$$F$$只能是$$BC$$中点，$$2c=\\sqrt{3}$$，$$\\frac{c}{a}=\\frac{\\sqrt{3}}{3}$$．  故答案为：$$\\frac{\\sqrt{3}}{3}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "139", "queId": "10105c7cad614cee89ab8a90d80c7132", "competition_source_list": ["2015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": "已知向量$$\\overrightarrow{a},\\overrightarrow{b}$$的夹角为$$60{}^{}\\circ $$，且$$\\left\\textbar{} \\overrightarrow{a} \\right\\textbar=1$$，$$\\left\\textbar{} \\overrightarrow{a}-2\\overrightarrow{b} \\right\\textbar=\\sqrt{13}$$，则$$\\left\\textbar{} \\overrightarrow{b} \\right\\textbar=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "C", "content": "$$2\\sqrt{2}$$ "}], [{"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"], "answer_analysis": ["$${{\\left\\textbar{} \\overrightarrow{a} \\right\\textbar}^{2}}-4\\overrightarrow{a}\\cdot \\overrightarrow{b}+4{{\\left\\textbar{} \\overrightarrow{b} \\right\\textbar}^{2}}=13$$，即$$1-2\\left\\textbar{} \\overrightarrow{b} \\right\\textbar+4{{\\left\\textbar{} \\overrightarrow{b} \\right\\textbar}^{2}}=13$$，解得$$\\left\\textbar{} \\overrightarrow{b} \\right\\textbar=2$$ "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "243", "queId": "19a4c6c21fa641edbba71ee830d23b50", "competition_source_list": ["2009年黑龙江全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知复数$$z=1-\\text{i}$$，则$$\\frac{{{z}^{2}}-2z}{z-1}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2\\text{i}$$ "}], [{"aoVal": "B", "content": "$$2\\text{i}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$-2$$ "}]], "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": ["$$\\frac{{{z}^{2}}-2z}{z-1}=\\frac{{{\\left( z-1 \\right)}^{2}}-1}{z-1}=z-1-\\frac{1}{z-1}=-\\text{i}-\\frac{1}{-\\text{i}}=-\\text{i}-\\text{i}=-2\\text{i}$$ "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "439", "queId": "8328ca2242ad4730aebeb47e7725c162", "competition_source_list": ["1982年全国高中数学联赛竞赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "当$$ab$$是两个不相等的正数时，下列三个代数式：  甲：$$(a+\\frac{1}{a})(b+\\frac{1}{b})$$，  乙：$${{(\\sqrt{ab}+\\frac{1}{\\sqrt{ab}})}^{2}}$$，  丙：$${{(\\frac{a+b}{2}+\\frac{2}{a+b})}^{2}}$$中间，值最大的一个（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "必定是甲； "}], [{"aoVal": "B", "content": "必定是乙； "}], [{"aoVal": "C", "content": "必定是丙； "}], [{"aoVal": "D", "content": "一般并不确定，而与$$a$$、$$b$$的取值有关 "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的证明"], "answer_analysis": ["甲$$\\mathsf{=}(a+\\frac{1}{a})(b+\\frac{1}{b})=ab+\\frac{1}{ab}+\\frac{b}{a}+\\frac{a}{b}$$  $$\\textgreater ab+\\frac{1}{ab}+2$$  $$={{(\\sqrt{ab}+\\frac{1}{\\sqrt{ab}})}^{2}}=$$乙．  $$\\therefore $$ 甲\\textgreater 乙．因此，最大者不会是乙，排除$$B$$只须比较甲、丙的大小．  取特殊值$$a=1\\mathsf{，}b=2$$得  甲$$=(1+1)(2+\\frac{1}{2})  =5$$，丙$$={{(\\frac{3}{2}+\\frac{2}{3})}^{2}}=4\\frac{25}{36}$$  $$\\therefore $$ 甲\\textgreater 丙，因此，丙不会取最大的，排除$$C$$．  再令 $$a=1\\mathsf{，}b=5$$，  甲$$=(1+1)(5+\\frac{1}{5})=10\\frac{2}{5}$$．  丙$$={{(\\frac{6}{2}\\mathsf{+}\\frac{2}{6})}^{2}}=11\\frac{1}{9}$$．  所以此时甲 ~\\textless{} ~丙，所以甲也不会取最大值，排除$$A$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "429", "queId": "9df44c9ec5e44632bab3968c4a28ef9e", "competition_source_list": ["2006年AMC12竞赛B第11题"], "difficulty": "2", "qtype": "single_choice", "problem": "$$2006-AMC12A-11$$  Joe and JoAnn each bought $$12$$ ounces of coffee in a $$16$$-ounce cup. Joe drank $$2$$ ounces of his coffee and then added $$2$$ ounces of cream. JoAnn added $$2$$ ounces of cream, stirred the coffee well, and then drank $$2$$ ounces. What is the resulting ratio of the amount of cream in Joe\\textquotesingle s coffee to that in JoAnn\\textquotesingle s coffee?.  乔和乔安每人买了 12 盎司的咖啡，装在一个 16 盎司的杯子里。 乔喝了 2 盎司咖啡，然后加了 2 盎司奶油。 乔安加了 2 盎司奶油，将咖啡搅拌均匀，然后喝了 2 盎司。 乔的咖啡中的奶油量与乔安的咖啡中的奶油量的最终比是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{6}{7}$$ "}], [{"aoVal": "B", "content": "$$\\frac{13}{14}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$\\frac{14}{13}$$ "}], [{"aoVal": "E", "content": "$$\\frac{7}{6}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->集合->容斥原理", "美国AMC10/12->Knowledge Point->Algebra->Application->Proportion Word Problems"], "answer_analysis": ["$\\dfrac{2}{2-\\dfrac{2}{14}\\times2}=\\dfrac{7}{6}$ "], "answer_value": "E"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "57", "queId": "3c98f6df61244b51bf0cee835c57af19", "competition_source_list": ["2011年山东全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "直线$$y=5$$与$$y=-1$$在区间$$\\left[ 0,\\left. \\frac{4 \\pi }{\\omega } \\right] \\right.$$上截曲线$$y=m\\sin \\frac{\\omega }{2}x+n (m\\textgreater0,  n\\textgreater0)$$所得的弦长相等且不为零，则下列描述正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$m\\leqslant \\frac{3}{2},n=\\frac{5}{2}$$ "}], [{"aoVal": "B", "content": "$$m\\leqslant 3,  n=2$$ "}], [{"aoVal": "C", "content": "$$m\\textgreater\\frac{3}{2},n=\\frac{5}{2}$$ "}], [{"aoVal": "D", "content": "$$m\\textgreater3,  n=2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["函数$${{y}_{1}}=m\\sin \\frac{\\omega x}{2}$$，$$x\\in \\left[ 0,\\left. \\frac{4 \\pi }{\\omega } \\right] \\right.$$的图象只有被$$y=a$$及$$ y=-a,  0\\leqslant a\\textless{}m$$这样的两直线所截，截得的弦长才能相等，且不为零．所以截取函数  $$y=m\\sin \\frac{\\omega x}{2}+n,\\quad x\\in \\left[ 0,\\left. \\frac{4 \\pi }{\\omega } \\right] \\right.$$  的图象所得弦长相等且不为零的两直线应为$$y=n+a,  y=n-a,  0\\leqslant a\\textless{}m$$，即有$$n+a=5,  n-a=-1 $$解得$$n=2$$，$$a=3$$．进而$$m\\textgreater3$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1134", "queId": "f3ea9627402e4af5b6b554f650ba112a", "competition_source_list": ["2015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": "设质数$$p$$，满足存在正整数$$xy$$使得$$p-1=2{{x}^{2}},{{p}^{2}}-1=2{{y}^{2}}$$，则符合条件的质数$$p$$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->多项式与方程->解方程（组）"], "answer_analysis": ["由已知，得$$2{{y}^{2}}=\\left( p-1 \\right)\\left( p+1 \\right)=2{{x}^{2}}\\left( p+1 \\right)\\Rightarrow p+1={{\\left( \\frac{y}{x} \\right)}^{2}}$$，只能$$\\frac{y}{x}\\in {{\\mathbf{N}}^{*}}$$，设$$\\frac{y}{x}=k$$，则$$p={{k}^{2}}-1=\\left( k+1 \\right)\\left( k-1 \\right)$$，于是$$k=2$$，$$p=3$$．经检验$$p=3$$满足要求． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1002", "queId": "a5596f10b819411286d534271ee0ffcd", "competition_source_list": ["2010年四川全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "长方体$$ABCD-{{A}_{1}}{{B}_{1}}{{C}_{1}}{{D}_{1}}$$的八个顶点都在球$$O$$的球面上，其中$$A{{A}_{1}}=1,  AB=2\\sqrt{2  },  AD=3\\sqrt{3}$$，则经过$$B,  C$$两点的球面距离是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2 \\pi }{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4 \\pi }{3}$$ "}], [{"aoVal": "C", "content": "2$$ \\pi $$ "}], [{"aoVal": "D", "content": "$$4 \\pi $$ "}]], "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": ["球$$O$$的半径$$R=\\frac{1}{2}\\sqrt{1+{{(2\\sqrt{2})}^{2}}+{{(3\\sqrt{3})}^{2}}}=3$$，在$$\\triangle OBC$$中$$OB=OC=3$$，$$BC=AD=3\\sqrt{3}$$，则$$\\cos \\angle BOC=-\\frac{1}{2}$$，从而$$\\angle BOC=\\frac{2 \\pi }{3}$$．  所以，经过$$BC$$两点的球面距离是$$2 \\pi \\times 3\\times \\frac{1}{3}=2 \\pi $$．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "648", "queId": "84209363065e489681e3dafbefce5742", "competition_source_list": ["2018年天津全国高中数学联赛竞赛初赛第8题9分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$$a$$为正实数，且$$f(x)=\\log _{2}(a x+\\sqrt{2 x^{2}+1})$$为奇函数，则$$f(x)\\textgreater\\frac{3}{2}$$的解集为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left(\\frac{7}{8},+\\infty\\right)$$ "}], [{"aoVal": "B", "content": "$$\\left(\\frac{5}{8},+\\infty\\right)$$ "}], [{"aoVal": "C", "content": "$$\\left(\\frac{3}{8},+\\infty\\right)$$ "}], [{"aoVal": "D", "content": "$$\\left(\\frac{1}{8},+\\infty\\right)$$ "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->素养->数学抽象", "课内体系->素养->数学运算", "课内体系->知识点->基本初等函数->对数函数->对数方程和对数不等式->解对数不等式", "课内体系->知识点->基本初等函数->对数函数->对数型复合函数的性质->对数型函数的其它类型"], "answer_analysis": ["由$$f(x)+f(-x)=0$$$$\\Rightarrow(a x+\\sqrt{2 x^{2}+1})(-a x+\\sqrt{2 x^{2}+1})=1$$$$\\Rightarrow\\left(2-a^{2}\\right) x^{2}=0 \\Rightarrow a=\\sqrt{2}$$，  由于$$\\sqrt{2} x+\\sqrt{2 x^{2}+1}$$在区间$$(0,+\\infty)$$上为增函数，于是，$$f(x)$$在区间$$(0,+\\infty)$$上为增函数．  结合$$f(x)$$为奇函数，知$$f(x)$$在$$\\mathbf{R}$$上为增函数．  则方程$$f(x)=\\frac{3}{2}$$$$\\Leftrightarrow \\sqrt{2} x+\\sqrt{2 x^{2}+1}=2^{\\frac{3}{2}}=2 \\sqrt{2}$$$$\\Leftrightarrow \\sqrt{2 x^{2}+1}=\\sqrt{2}(2-x)$$$$\\Rightarrow x=\\frac{7}{8}$$．  故$$f(x)\\textgreater\\frac{3}{2}$$的解集为$$\\left(\\frac{7}{8},+\\infty\\right)$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "916", "queId": "85e54350a13141a8a2da058bc8696e96", "competition_source_list": ["2019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": "函数$$f\\left( x \\right)=\\left( \\sqrt{3}\\sin x+\\cos x \\right)\\left( \\sqrt{3}\\cos x-\\sin x \\right)$$的最小正周期为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\pi }{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2\\pi }{3}$$ "}], [{"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"], "answer_analysis": ["由题给条件知：  $$f\\left( x \\right)=2\\sin \\left( 2x+\\frac{\\pi }{2} \\right)$$，  故$$f\\left( x \\right)$$的最小正周期为$$\\pi $$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "595", "queId": "ff6b7bef030a4e36911b1064769c22ea", "competition_source_list": ["2009年湖南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$$f(x)$$的定义域为$$R$$，若$$f(x+1)$$与$$f(x-1)$$都是奇函数，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$f(x)$$是偶函数 "}], [{"aoVal": "B", "content": "$$f(x)$$是奇函数 "}], [{"aoVal": "C", "content": "$$f(x+3)$$是奇函数 "}], [{"aoVal": "D", "content": "$$f(x+3)$$是偶函数 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["因为$$f(x+1)$$是奇函数，所以  $$f(x+1)=-f(-x+1), f(x)=-f(-x+2)$$．①  同理，由$$f(x-1)$$是奇函数，可得  $$f(x)=-f(-x-2)$$．②  由①、②可得，$$f(-x+2)=f(-x-2)$$，所以$$f(x+2)=f(x-2), f(x+4)=f(x)$$，即$$f(x)$$是周期为$$4$$的函数．  又因为$$f(x-1)$$是奇函数，所以$$f(x+3)$$是奇函数，故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "183", "queId": "0d7371a5e0354658aa3162ef8d0ddcd1", "competition_source_list": ["2013年AMC10竞赛A第8题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\dfrac{2^{2022}+2^{2020}}{2^{2022}-2^{2020}}$$ 等于多少? (改编自 2013 AMC10 A 第$$8$$题)", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$\\dfrac{5}{3}$$ "}], [{"aoVal": "D", "content": "$$2013$$ "}], [{"aoVal": "E", "content": "$$2^{4024}$$ "}]], "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": ["Factoring out, we get: $$\\dfrac{2^{2020}\\left( 2^{2}+1\\right)}{2^{2020}\\left( 2^{2}-1\\right)}$$.  Cancelling out the $$2^{2020}$$ from the numerator and denominator, we see that it simplifies to $$\\left( \\text{C}\\right)\\dfrac{5}{3}$$. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "246", "queId": "9443c89c9a5b41999aa5de62f592cc46", "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6题"], "difficulty": "1", "qtype": "single_choice", "problem": "若函数$y={{a}^{\\left\\textbar{} x \\right\\textbar}}+m-1\\left( 0\\textless{} a\\textless{} 1 \\right)$的图象和$x$轴有交点，则实数$m$的取值范围是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\left[ 1,+\\infty \\right)$ "}], [{"aoVal": "B", "content": "$\\left( 0,1 \\right)$ "}], [{"aoVal": "C", "content": "$\\left( -\\infty ,1 \\right)$ "}], [{"aoVal": "D", "content": "$\\left[ 0,1 \\right)$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的应用->函数的零点->零点的个数问题", "课内体系->知识点->基本初等函数->指数函数"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 将题意等价转化为函数$y={{a}^{\\left\\textbar{} x \\right\\textbar}}$的图象与$y=1-m$的图象有交点，根据指数函数的性质可得$0\\textless{} {{a}^{\\textbar x\\textbar}}\\le 1$，列出关于$m$的不等式，解出即可.\\\\ 【详解】\\\\ 函数$y={{a}^{\\left\\textbar{} x \\right\\textbar}}+m-1\\left( 0\\textless{} a\\textless{} 1 \\right)$的图象和$x$轴有交点，即方程${{a}^{\\left\\textbar{} x \\right\\textbar}}=1-m\\left( 0\\textless{} a\\textless{} 1 \\right)$，\\\\ 等价于函数$y={{a}^{\\left\\textbar{} x \\right\\textbar}}$的图象与$y=1-m$的图象有交点，\\\\ $0\\textless{} a\\textless{} 1$时，$0\\textless{} {{a}^{\\textbar x\\textbar}}\\le 1$，\\\\ 即$0\\textless{} 1-m\\le 1$，解得$0\\le m\\textless{} 1$，\\\\ 即实数$m$的取值范围是$\\left[ 0,1 \\right)$，\\\\ 故选：\\emph{D}.\\\\ 【点睛】\\\\ 本题主要考查了指数函数的图象与性质的应用问题，将问题转化为函数$y={{a}^{\\left\\textbar{} x \\right\\textbar}}$的图象与$y=1-m$的图象有交点是解题的关键，属于基础题. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "601", "queId": "9e8692cd7c084df3975adb10a93783ca", "competition_source_list": ["2021年吉林全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，$$\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar=2$$ ，$$\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar=3$$，$$\\overrightarrow{AB}\\cdot \\overrightarrow{CA}\\textgreater0$$ ，且$$\\triangle ABC$$的面积为$$\\frac{3}{2}$$，则$$\\angle BAC$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$${{150}^{\\circ }}$$ "}], [{"aoVal": "B", "content": "$${{30}^{\\circ }}$$ "}], [{"aoVal": "C", "content": "$${{120}^{\\circ }}$$ "}], [{"aoVal": "D", "content": "$${{60}^{\\circ }}$$ "}]], "knowledge_point_routes": ["课内体系->方法->定义法", "课内体系->特色题型->选择压轴", "课内体系->知识点->解三角形->三角形面积公式", "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->思想->数形结合思想", "课内体系->素养->数学运算"], "answer_analysis": ["$$\\triangle ABC$$的面积$${{S}_{\\triangle ABC}}$$$$=\\frac{1}{2}\\times AB\\times AC\\times \\sin \\angle BAC$$  $$=\\frac{1}{2}\\times 2\\times 3\\times \\sin \\angle BAC$$$$=\\frac{3}{2}$$，  所以$$\\sin \\angle BAC=\\frac{1}{2}$$，  又由已知$$\\angle BAC$$为纯角，  从而$$\\angle BAC={{150}^{\\circ }}$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1202", "queId": "e2c498c5a0754866aae84bb0b2a96058", "competition_source_list": ["竞赛第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知实数\\emph{a}，\\emph{b}，\\emph{c}，\\emph{d}，\\emph{e}满足$\\begin{cases} a+b+c+d+e=8,   a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16, \\end{cases}$则\\emph{e}的取值范围是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$[-2,2]$ "}], [{"aoVal": "B", "content": "$[0,1]$ "}], [{"aoVal": "C", "content": "$[0,2)$ "}], [{"aoVal": "D", "content": "以上答案都不对 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据柯西不等式可求\\emph{e}的取值范围.\\\\ 【详解】\\\\ 根据柯西不等式，有$-\\sqrt{4}\\cdot \\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}\\leq a+b+c+d\\leq \\sqrt{4}\\cdot \\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}$，\\\\ 从而$\\textbar{} 8-e\\textbar{} \\leq 2\\sqrt{16-e^{2}}\\Rightarrow 0\\leq e\\leq \\frac{16}{5}$，\\\\ 因此\\emph{e}的取值范围是$\\left[0,\\frac{16}{5}\\right]$．\\\\ 故选：D. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "771", "queId": "804738ee337040308734bf73a86c788a", "competition_source_list": ["2023年陕西榆林绥德县实验中学高三上学期高考模拟（第四次模拟）第12题", "2022~2023学年9月辽宁高三上学期月考（名校联盟）第8题", "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8题", "2022年四川成都青羊区四川省成都市石室中学高三上学期高考模拟理科（一诊模拟）第12题"], "difficulty": "4", "qtype": "single_choice", "problem": "已知$a={{\\text{e}}^{0.2}}-1,b=\\ln 1.2,c=\\tan 0.2$，其中$\\text{e}=2.71828\\cdots $为自然对数的底数，则（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$c\\textgreater a\\textgreater b$ "}], [{"aoVal": "B", "content": "$a\\textgreater c\\textgreater b$ "}], [{"aoVal": "C", "content": "$b\\textgreater a\\textgreater c$ "}], [{"aoVal": "D", "content": "$a\\textgreater b\\textgreater c$ "}]], "knowledge_point_routes": ["课内体系->知识点->基本初等函数", "课内体系->知识点->三角函数->三角函数的图象与性质"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 观察$a={{\\text{e}}^{0.2}}-1,b=\\ln 1.2,c=\\tan 0.2$，发现都含有$0.2$，把$0.2$换成$x$，自变量在$(0,1)$或其子集范围内构造函数，利用导数证明其单调性，比较$a,b,c$的大小.\\\\ 【详解】\\\\ 令$f(x)={{\\text{e}}^{x}}-1-\\tan x=\\frac{\\cos x{{\\text{e}}^{x}}-\\cos x-\\sin x}{\\cos x}$，$0\\textless{} x\\textless{} \\frac{\\pi }{4}$，\\\\ 令$g(x)=\\cos x{{\\text{e}}^{x}}-\\cos x-\\sin x$，${g}\\textquotesingle(x)=(-\\sin x+\\cos x){{\\text{e}}^{x}}+\\sin x-\\cos x=({{\\text{e}}^{x}}-1)\\cdot (\\cos x-\\sin x)$，\\\\ 当$0\\textless{} x\\textless{} \\frac{\\pi }{4}$时，${g}\\textquotesingle(x)\\textgreater0$，$g(x)$单调递增，\\\\ 又$g(0)=1-1=0$，所以$g(x)\\textgreater0$，又$\\cos x\\textgreater0$，\\\\ 所以$f(x)\\textgreater0$，在$(0,\\frac{\\pi }{4})$成立，所以$f(0.2)\\textgreater0$即$a\\textgreater c$，\\\\ 令$h(x)=\\ln (x+1)-x$，${h}\\textquotesingle(x)=\\frac{1}{x+1}-1=\\frac{-x}{x+1}$，$h(x)$在$x\\in (0,\\frac{\\pi }{2})$为减函数，所以$h(x)\\textless{} h(0)=0$，即$\\ln (x+1)\\textless{} x$，\\\\ 令$m(x)=x-\\tan x$，${m}\\textquotesingle(x)=1-\\frac{1}{{{\\cos }^{2}}x}$，$m(x)$在$x\\in (0,\\frac{\\pi }{2})$为减函数，所以$m(x)\\textless{} m(0)=0$，即$x\\textless{} \\tan x$，\\\\ 所以$\\ln (x+1)\\textless{} x\\textless{} \\tan x$，$x\\in (0,\\frac{\\pi }{2})$成立，\\\\ 令$x=0.2$，则上式变为$\\ln (0.2+1)\\textless{} 0.2\\textless{} \\tan 0.2$，所以$b\\textless{} 0.2\\textless{} c$\\\\ 所以$b\\textless{} c$，\\\\ 所以$b\\textless{} c\\textless{} a$.\\\\ 故答案为：B.\\\\ 【点睛】\\\\ 比较大小题目，是高考的热点，也是难点，通过观察和构造函数是基本的解题要求，难点在于构造后的证明，需要平时多积累常见的结论，达到深入理解，举一反三，融会贯通. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "948", "queId": "86401cc8d36d4b0cb8f306ea838e78c5", "competition_source_list": ["2012年四川全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设抛物线$${{y}^{2}}=4x$$的焦点为$$F$$，顶点为$$O,  M$$是抛物线上的动点，则$$\\frac{\\left\\textbar{} MO \\right\\textbar}{\\left\\textbar{} MF \\right\\textbar}$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{3}}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2\\sqrt{3}}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{4}{3}$$ "}], [{"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": ["由条件知$$F\\left( 1,  0 \\right)$$，$$O\\left( 0,  0 \\right)$$，点$$M$$坐标为$$\\left( x,  y \\right)$$，显然$$x\\geqslant 0$$，则  $${{\\left( \\frac{\\left\\textbar{} MO \\right\\textbar}{\\left\\textbar{} MF \\right\\textbar} \\right)}^{2}}=\\frac{{{x}^{2}}+{{y}^{2}}}{{{\\left( x-1 \\right)}^{2}}+{{y}^{2}}}=\\frac{{{x}^{2}}+4x}{{{x}^{2}}+2x+1}$$ $$=\\frac{4}{3}-3{{\\left( \\frac{1}{x+1}-\\frac{1}{3} \\right)}^{2}}\\leqslant \\frac{4}{3}$$．  当$$x=2$$时等号成立，故$$\\frac{\\left\\textbar{} MO \\right\\textbar}{\\left\\textbar{} MF \\right\\textbar}$$的最大值为$$\\frac{2\\sqrt{3}}{3}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1095", "queId": "ca7254401dd145a4bd530b2d437bd076", "competition_source_list": ["2014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$$a$$、$$b$$、$$c$$、$$d$$均为实数，函数$$f\\left( x \\right)=\\frac{a}{3}{{x}^{3}}+\\frac{b}{2}{{x}^{2}}+cx+d$$（$$a\\textless{}0$$）有两个极值点$${{x}_{1}}$$、$${{x}_{2}}$$（$${{x}_{1}}\\textless{}{{x}_{2}}$$），满足$$f\\left( {{x}_{2}} \\right)={{x}_{1}}$$，则方程$$a{{\\left[ f\\left( x \\right) \\right]}^{2}}+b\\left[ f\\left( x \\right) \\right]+c=0$$的实根个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$个 "}], [{"aoVal": "B", "content": "$$2$$个 "}], [{"aoVal": "C", "content": "$$3$$个 "}], [{"aoVal": "D", "content": "$$4$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->导数", "竞赛->知识点->函数->二次函数"], "answer_analysis": ["$${f}'\\left( x \\right)=a{{x}^{2}}+bx+c$$，由于$$a\\textless{}0$$，$${{x}_{1}}\\textless{}{{x}_{2}}$$，则$$f\\left( x \\right)$$在$$\\left( -\\infty ,  {{x}_{1}} \\right)$$单调递减，在$$\\left( {{x}_{1}},  {{x}_{2}} \\right)$$单调递增，在$$\\left( {{x}_{2}},  +\\infty \\right)$$单调递减．  且$$ax_{1}^{2}+b{{x}_{1}}+c=0$$，$$ax_{2}^{2}+b{{x}_{2}}+c=0$$．  若$$f\\left( x \\right)={{x}_{1}}$$，此时方程有两个根，若$$f\\left( x \\right)={{x}_{2}}$$，此时方程有一个根．  所以，方程应有三个实根． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "14", "queId": "253acf98c5ce43de909f75718b68a46c", "competition_source_list": ["2011年江苏全国高中数学联赛竞赛复赛第2题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知函数$$f(x)=\\frac{x}{\\sqrt{{{x}^{2}}+2x+4}}$$，则$$f(x)$$的值域为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ -\\sqrt{3},  1 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left[ -\\frac{2\\sqrt{3}}{3},  1 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left[ -\\frac{\\sqrt{3}}{2},  1 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left[ 0,  1 \\right)$$ "}]], "knowledge_point_routes": ["课内体系->方法->定义法", "课内体系->思想->分类讨论思想", "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->求复合函数的值域", "课内体系->素养->数学运算"], "answer_analysis": ["$$x\\textgreater0$$时，$$f\\left( x \\right)=\\frac{1}{\\sqrt{\\frac{4}{{{x}^{2}}}+\\frac{2}{x}+1}}\\textless{}\\frac{1}{\\sqrt{1}}=1$$；  $$x=0$$时，$$f\\left( x \\right)=0$$；  $$x\\textless{}0$$时，$$f\\left( x \\right)=-\\frac{1}{\\sqrt{\\frac{4}{{{x}^{2}}}+\\frac{2}{x}+1}}=-\\frac{1}{\\sqrt{{{\\left( \\frac{2}{x}+\\frac{1}{2} \\right)}^{2}}+\\frac{3}{4}}}\\geqslant -\\frac{1}{\\sqrt{\\frac{3}{4}}}=-\\frac{2}{3}\\sqrt{3}$$．  所以，$$f\\left( x \\right)$$的值域为$$\\left[ -\\frac{2\\sqrt{3}}{3},  1 \\right)$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "407", "queId": "3e79a005afef4ce7aa4404ea5980d4e9", "competition_source_list": ["2013年天津全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "3", "qtype": "single_choice", "problem": "在$${\\triangle }ABC$$中，$$\\overrightarrow{BC}\\cdot \\overrightarrow{BA}=\\overrightarrow{CB}\\cdot \\overrightarrow{CA}$$，则$${\\triangle }ABC$$是．", "answer_option_list": [[{"aoVal": "A", "content": "等腰三角形 "}], [{"aoVal": "B", "content": "直角三角形 "}], [{"aoVal": "C", "content": "等腰直角三角形 "}], [{"aoVal": "D", "content": "以上均不对 "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的应用", "竞赛->知识点->复数与平面向量->平面向量的概念与运算"], "answer_analysis": ["所给数量积等式可转化为$$\\overrightarrow{BC}\\cdot \\left( \\overrightarrow{AB}+\\overrightarrow{AC} \\right)=0$$．若记$$BC$$的中点为$$D$$，$$\\overrightarrow{AB}+\\overrightarrow{AC}=2\\overrightarrow{AD}$$，从而$$\\overrightarrow{BC}\\cdot \\overrightarrow{AD}=0$$．这表明$$BC$$与中线$$AD$$垂直，因此$$\\triangle ABC$$是等腰三角形． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "855", "queId": "80da1f3142b045e68edbf26e14c4a569", "competition_source_list": ["2015~2016学年3月陕西西安雁塔区唐南中学高一下学期月考理科第10题4分", "2013~2014学年10月辽宁沈阳沈阳铁路实验中学高二上学期月考第6题5分", "2012年吉林全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知等差数列$$\\left { {{a}_{n}} \\right }$$满足，$${{a}_{1}}\\textgreater0$$，$$5{{a}_{8}}=8{{a}_{13}}$$ ，则前$$n$$项和$${{S}_{n}}$$取最大值时，$$n$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$20$$ "}], [{"aoVal": "B", "content": "$$21$$ "}], [{"aoVal": "C", "content": "$$22$$ "}], [{"aoVal": "D", "content": "$$23$$ "}]], "knowledge_point_routes": ["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  ", "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": ["设数列的公差为$$d$$，由$$5{{a}_{8}}=8{{a}_{13}}$$得$$5\\left( {{a}_{1}}+7d \\right)=8\\left( {{a}_{1}}+12d \\right)$$，解得$$d=-\\frac{3}{61}{{a}_{1}}$$，  由$${{a}_{n}}={{a}_{1}}+\\left( n-1 \\right)d={{a}_{1}}+\\left( n-1 \\right)\\left( -\\frac{3}{61}{{a}_{1}} \\right)\\geqslant 0$$，可得$$n\\leqslant \\frac{64}{3}=21\\frac{1}{3}$$，  所以数列$$\\left { {{a}_{n}} \\right }$$前$$21$$项都是正数，以后各项都是负数，  故$${{S}_{n}}$$取最大值时，$$n$$的值为$$21$$.  故选：$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "389", "queId": "5e2ec7fabe71497d83bfa505663cd3fa", "competition_source_list": ["竞赛第16题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "若一無窮等比級數之和是一正數$$S$$，且級數的第$$2$$项是$$1$$，則$$S$$的最小可能値爲何？  The sum of an infinite geometric series is a positive number S, and the second tern in the series is $$1$$. What is  the smallest possible value of $$S$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$ \\frac{1+ \\sqrt{5}}{2} $$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$ \\sqrt{5} $$ "}], [{"aoVal": "D", "content": "$$3$$ "}], [{"aoVal": "E", "content": "$$4$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["The sum of an infinite geometric series is of the form: $$S=\\dfrac{a_{1}}{1-r}$$ where $$a_{1}$$ is the first term and $$r$$ is the ratio whose absolute value is less than $$1$$.  We know that the second term is the first term multiplied by the ratio. In other words: $$\\begin{array}{l} {a_{1}\\cdot r=1}  {a_{1}=\\dfrac{1}{r}} \\end{array}$$ Thus, the sum is the following: $$\\begin{array}{l} {S=\\dfrac{\\dfrac{1}{r}}{1-r}}  {S=\\dfrac{1}{r-r^{2}}} \\end{array}$$.  Since we want the minimum value of this expression, we want the maximum value for the denominator, $$-r^{2}+r$$. The maximum x$$-$$value of a quadratic with negative $$a$$ is $$\\dfrac{-b}{2a}$$. $$\\begin{array}{l} {r=\\dfrac{-\\left( 1\\right)}{2\\left( -1\\right)}}  {r=\\dfrac{1}{2}} \\end{array}$$ Plugging $${r=\\dfrac{1}{2}}$$ into the quadratic yields: $$\\begin{array}{l} {S=\\dfrac{1}{\\dfrac{1}{2}-\\left( \\dfrac{1}{2}\\right)^{2}}}  {S=\\dfrac{1}{\\dfrac{1}{4}}} \\end{array}$$.  Therefore, the minimum sum of our infinite geometric sequence is $$\\left( \\text{E}\\right)4$$. $$\\left({}\\right.$$Solution by akaashp$$11\\left.{}\\right)$$ As an extension to find the maximum value for the denominator we can find the derivative of $$-r^{2}+r$$ to get $$1-2r$$. we know that this changes sign when $$r =\\dfrac{1}{2}$$ so plugging it in into the original equation we find the answer is $$\\left( \\text{E}\\right)4$$.  After observation we realize that in order to minimize our sum $$\\dfrac{a}{1-r}$$ with $$a$$ being the reciprocal of $$r$$.  The common ratio $$r$$ has to be in the form of $$\\dfrac{1}{x}$$ with $$x$$ being an integer as anything more than $$1$$ divided by $$x$$ would give a larger sum than a ratio in the form of $$\\dfrac{1}{x}$$.  The first term has to be $$x$$, so then in order to minimize the sum, we have minimize $$x$$.  The smallest possible value for $$x$$ such that it is an integer that\\textquotesingle s greater than $$1$$ is $$2$$. So our first term is $$2$$ and our common ratio is $$\\dfrac{1}{2}$$. Thus the sum is $$\\dfrac{2}{\\dfrac{1}{2}}$$ or $$\\left( \\text{E}\\right)4$$. Solution $$2$$ by No_One.  We can see that if $$a$$ is the first term, and $$r$$ is the common ratio between each of the terms, then we can get~ $$S=\\dfrac{a}{1-r}\\Rightarrow S-Sr=a$$ Also, we know that the second term can be expressed as $$a\\cdot r$$ notice if we multiply $$S-Sr=a$$ by $$r$$, we would get $$r\\left( S-Sr\\right)=ar\\Rightarrow Sr-Sr^{2}=1\\Rightarrow Sr^{2}-Sr+1=0$$ This quadratic has real solutions if the discriminant is greater than or equal to zero, or $$S^{2}-4\\cdot S\\cdot1\\geqslant0$$ This yields that $$S\\leqslant0$$ or $$S\\geqslant4$$. However, since we know that $$S$$ has to be positive, we can safely conclude that the minimum possible value of $$S$$ is $$\\left( \\text{E}\\right)4$$. "], "answer_value": "E"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "972", "queId": "9c163da49d8e488cb3129496e838f315", "competition_source_list": ["2006年第24届美国数学邀请赛（AIME）竞赛第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$M=1!2!3\\cdots 99!100!$$，$$N$$是$$M$$末位连续零的个数，试求$$N$$被$$1000$$除所得的余数．", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$104$$ "}], [{"aoVal": "C", "content": "$$124$$ "}], [{"aoVal": "D", "content": "$$224$$ "}], [{"aoVal": "E", "content": "$$634$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->整除->质数（算数基本定理）", "课内体系->知识点->计数原理"], "answer_analysis": ["设$$M=1!2!3!\\cdots 99!100!$$．  那么$$N$$等于$$M$$含有的因子$$5$$的个数．  注意到$$M={{1}^{100}}\\cdot {{2}^{99}}\\cdot \\cdots \\cdot {{99}^{2}}\\cdot {{100}^{1}}$$，  其中每出现一个$$5$$的倍数产生一个$$5$$的因子，  每出现一个$$25$$的倍数就额外加增加一个$$5$$的因子．  $$5$$的倍数一共有$$96+91+86+\\cdots +1=970$$个，  $$25$$的倍数一共有$$76+51+26+1=154$$个，  所以最终$$M$$末尾零的个数为$$970+154=1124$$，  故所得的余数为$$124$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "745", "queId": "ba9f4fed1e894b259499add3f3e9fef6", "competition_source_list": ["2000年全国高中数学联赛竞赛一试第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$\\text{sin }\\alpha \\textgreater0$$，$$\\cos \\alpha \\textless{}0$$，且$$\\sin \\frac{\\alpha }{3}\\textgreater\\cos \\frac{\\alpha }{3}$$，则$$\\frac{\\alpha }{3}$$的取值范围是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 2k \\pi +\\frac{ \\pi }{6},2k \\pi +\\frac{ \\pi }{3} \\right)$$，$$k\\in \\mathbf{Z}$$ "}], [{"aoVal": "B", "content": "$$\\left( \\frac{2k \\pi }{3}+\\frac{ \\pi }{6},\\frac{2k \\pi }{3}+\\frac{ \\pi }{3} \\right)$$，$$k\\in \\mathbf{Z}$$ "}], [{"aoVal": "C", "content": "$$\\left( 2k \\pi +\\frac{5 \\pi }{6},2k \\pi + \\pi ~\\right)$$，$$k\\in \\mathbf{Z}$$ "}], [{"aoVal": "D", "content": "$$\\left( 2k \\pi +\\frac{ \\pi }{4},2k \\pi +\\frac{ \\pi }{3} \\right)\\cup \\left( 2k \\pi +\\frac{5 \\pi }{6},2k \\pi + \\pi ~\\right)$$，$$k\\in \\mathbf{Z}$$ "}]], "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->素养->直观想象", "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->余弦型三角函数的图象与性质", "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->余弦函数的图象和性质"], "answer_analysis": ["由$$\\sin \\alpha \\textgreater0$$，$$\\cos \\alpha \\textless{}0$$，得$$\\alpha \\in \\left( 2k \\pi +\\frac{ \\pi }{2},2k \\pi + \\pi ~\\right),k\\in \\mathbf{Z}$$  从而有$$\\frac{\\alpha }{3}\\in \\left( \\frac{2k \\pi }{3}+\\frac{ \\pi }{6},\\frac{2k \\pi }{3}+\\frac{ \\pi }{3} \\right),k\\in \\mathbf{Z}$$  又因为$$\\sin \\frac{\\alpha }{3}\\textgreater\\cos \\frac{\\alpha }{3}$$，所以又有$$\\frac{\\alpha }{3}\\in \\left( 2k \\pi +\\frac{ \\pi }{4},2k \\pi +\\frac{5 \\pi }{4} \\right),k\\in \\mathbf{Z}$$  于是$$\\frac{\\alpha }{3}$$的范围为$$\\left( 2k \\pi +\\frac{ \\pi }{4},2k \\pi +\\frac{ \\pi }{3} \\right)\\cup \\left( 2k \\pi +\\frac{5 \\pi }{6},2k \\pi + \\pi ~\\right),k\\in \\mathbf{Z}$$．  故选$$\\text{D}$$． ", "<p>$$\\sin \\alpha &gt;0$$，$$\\cos \\alpha &nbsp;&nbsp;&lt; &nbsp;0$$，$$\\frac{\\pi }{6}+\\frac{2k\\pi }{3} &nbsp;&lt; &nbsp;\\frac{\\alpha }{3} &nbsp;&lt; &nbsp;\\frac{\\pi }{3}+\\frac{2k\\pi }{3}$$，</p>\n<p>$$k=3n$$，$$\\frac{\\pi }{6}+2n\\pi &nbsp;&nbsp;&lt; &nbsp;\\frac{\\alpha }{3} &nbsp;&lt; &nbsp;\\frac{\\pi }{3}+2n\\pi $$，$$\\sin \\frac{\\alpha }{3}&gt;\\cos \\frac{\\alpha }{3}$$，</p>\n<p>$$\\frac{\\pi }{4}+2n\\pi &nbsp;&nbsp;&lt; &nbsp;\\frac{\\alpha }{3} &nbsp;&lt; &nbsp;\\frac{5\\pi }{4}+2n\\pi $$，从而$$\\frac{\\pi }{4}+2n\\pi &nbsp;&nbsp;&lt; &nbsp;\\frac{\\alpha }{3} &nbsp;&lt; &nbsp;\\frac{\\pi }{3}+2n\\pi $$，</p>\n<p>同理$$k=3n+1$$，$$k=3n+2$$，可得答案．</p>\n<p>故选：$$\\text{D}$$．</p>"], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "583", "queId": "51c4b46bbedc48c39d4ff33ae40fa3e6", "competition_source_list": ["2000年全国高中数学联赛竞赛一试第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "平面上整点（纵、横坐标都是整数的点）到直线$$y=\\frac{5}{3}x+\\frac{4}{5}$$的距离中的最小值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{34}}{170}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{34}}{85}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{20}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{30}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与方程"], "answer_analysis": ["设整点坐标$$(m,n)$$，则它到直线$$25x-15y+12=0$$的距离为  $$d=\\frac{\\left\\textbar{} 25m-15n+12 \\right\\textbar}{\\sqrt{{{25}^{2}}+{{(-15)}^{2}}}}=\\frac{\\left\\textbar{} 5(5m-3n)+12 \\right\\textbar}{5\\sqrt{34}}$$，  由于$$m,n\\in \\mathbf{Z}$$，故$$5\\left( 5m-3n \\right)$$是$$5$$的倍数，只有当$$m=n=-1$$，时$$5\\left( 5m-3n \\right)=-10$$与$$12$$的和的绝对值最小，其值为$$2$$，从而所求的最小值为$$\\frac{\\sqrt{34}}{85}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1158", "queId": "d44d2f5a7c7045bfab9c3d69f4e9ea30", "competition_source_list": ["2011年四川全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "双曲线$$\\frac{{{x}^{2}}}{{{a}^{2}}}-\\frac{{{y}^{2}}}{{{b}^{2}}}=1$$的左、右准线$${{l}_{1}},{{l}_{2}}$$将线段$${{F}_{1}}{{F}_{2}}$$三等分（其中$${{F}_{1}}$$、$${{F}_{2}}$$分别为双曲线的左、右焦点），则该双曲线的离心率$$e$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{6}}{2}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3\\sqrt{3}}{2}$$ "}], [{"aoVal": "D", "content": "$$2\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->双曲线"], "answer_analysis": ["由题意得$$2c=3\\times \\frac{2{{a}^{2}}}{c}$$，解得$$e=\\sqrt{3}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "728", "queId": "b5fcb0ddbaea45b5a5f5cf7f8d5610e6", "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": ["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", "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": "511", "queId": "3255f1b965044787a498926489a9eaf6", "competition_source_list": ["2016年浙江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$$f\\left( x \\right)=4{{\\sin }^{3}}x-\\sin x+2{{\\left( \\sin \\frac{x}{2}-\\cos \\frac{x}{2} \\right)}^{2}}$$的最小正周期为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\mathrm{ }\\pi $$ "}], [{"aoVal": "B", "content": "$$\\frac{\\mathrm{ }\\pi }{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{3}\\mathrm{ }\\pi $$ "}], [{"aoVal": "D", "content": "$$\\mathrm{ }\\pi $$ "}]], "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->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->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->二倍角的余弦", "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"], "answer_analysis": ["$$f\\left( x \\right)=4{{\\sin }^{3}}x-\\sin x+2\\left( 1-\\sin x \\right)=-\\sin 3x+2$$.  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "142", "queId": "3cedbba643944cfbbe8dbf3a2d008949", "competition_source_list": ["2008年福建全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知函数$$f(x)$$满足对所有的实数$$x$$、$$y$$，都有$$f(x)+f(2x+y)+5xy=f(3x-y)+2{{x}^{2}}+1$$，则$$f(10)$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-49$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "25 "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->解方程（组）"], "answer_analysis": ["令$$x=10$$，$$y=5$$，得$$f(10)+f(25)+250=f(25)+200+1$$，  所以$$f(10)=-49$$．故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1102", "queId": "c617b8ec8c8f4b0db43e2d4aea893923", "competition_source_list": ["2016年湖南全国高中数学联赛竞赛初赛第1题5分", "2007年6月高考真题陕西卷理科第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设集合$$S=\\left { {{A}_{0}},{{A}_{1}},{{A}_{2}},{{A}_{3}} \\right }$$，在$$S$$上定义运算``$$\\oplus $$''为：$${{A}_{i}}\\oplus {{A}_{j}}={{A}_{k}}$$，其中$$k$$为$$i+j$$被$$4$$除的余数，$$i,j=0,1,2,3$$．则满足关系$$\\left( x\\oplus x \\right)\\oplus {{A}_{2}}={{A}_{0}}$$的$$x\\left( x\\in S \\right)$$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->集合->集合的基本关系->子集个数的计算", "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->集合中元素的性质", "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", "课内体系->素养->逻辑推理", "课内体系->素养->数学运算", "课内体系->特色题型->新定义"], "answer_analysis": ["设$$x={{A}_{i}}$$，由题意，$$2i+2$$被$$4$$除余$$0$$，$$i=1,3$$，有$$2$$个．  故选$$\\text{B}$$． ", "<p>当$$x={{A}_{0}}$$时，原式$$=\\left( {{A}_{0}}\\oplus {{A}_{0}} \\right)\\oplus {{A}_{2}}={{A}_{0}}\\oplus {{A}_{2}}={{A}_{2}}$$，不符合．</p>\n<p>当$$x={{A}_{1}}$$时，原式$$=\\left( {{A}_{1}}\\oplus {{A}_{1}} \\right)\\oplus {{A}_{2}}={{A}_{2}}\\oplus {{A}_{2}}={{A}_{0}}$$，符合．</p>\n<p>当$$x={{A}_{2}}$$时，原式$$=\\left( {{A}_{2}}\\oplus {{A}_{2}} \\right)\\oplus {{A}_{2}}={{A}_{0}}\\oplus {{A}_{2}}={{A}_{2}}$$，不符合．</p>\n<p>当$$x={{A}_{3}}$$时，原式$$=\\left( {{A}_{3}}\\oplus {{A}_{3}} \\right)\\oplus {{A}_{2}}={{A}_{2}}\\oplus {{A}_{2}}={{A}_{6}}$$，符合．</p>\n<p>综上，$$x={{A}_{1}}$$或$$x={{A}_{3}}$$成立，合计$$2$$个．</p>\n<p>故选$$\\text{B}$$．</p>"], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "15", "queId": "3c7b8420f7b04d958984a48ebb91f2d0", "competition_source_list": ["2000年第18届美国数学邀请赛（AIME）竞赛第12题"], "difficulty": "3", "qtype": "single_choice", "problem": "已知函数$$f\\left( x \\right)=f\\left( 398-x \\right)=f\\left( 2158-x \\right)=f\\left( 3214-x \\right)$$对于所有实数$$x$$均成立，则$$f\\left( 0 \\right)$$，$$f\\left( 1 \\right)$$，$$f\\left( 2 \\right)$$，．．．，$$f\\left( 999 \\right)$$中最多有个不同的值？", "answer_option_list": [[{"aoVal": "A", "content": "$$165$$ "}], [{"aoVal": "B", "content": "$$177$$ "}], [{"aoVal": "C", "content": "$$183$$ "}], [{"aoVal": "D", "content": "$$199$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "课内体系->知识点->函数的概念与性质->函数的性质->对称性"], "answer_analysis": ["由$$f\\left( 398-x \\right)=f\\left( 2158-x \\right)$$得$$f\\left( x \\right)=f\\left( x+1760 \\right)$$．  由$$f\\left( 2158-x \\right)=f\\left( 3214-x \\right)$$得，$$f\\left( x \\right)=f\\left( x+1056 \\right)$$．  故有$$f\\left( x \\right)=f\\left( x+1056 \\right)=f\\left( x+2112 \\right)=f\\left( x+352 \\right)$$．  从而$$f\\left( x \\right)$$为周期函数，且它的一个周期为$$352$$．  又由于$$f\\left( x \\right)=f\\left( 398-x \\right)$$，故$$f\\left( x \\right)$$关于$$x=199$$对称．  于是当$$f$$在$$x\\in \\left[ 23,199 \\right]$$定义后，  由$$f\\left( x \\right)$$的对称性知$$x\\in \\left[ 199,375 \\right]$$时，  $$f$$亦可定义；再由$$f\\left( x \\right)$$的周期性知对任意$$x\\in \\mathbf{R}$$，  $$f$$可定义．故在一个周期内，  每一个函数值至少对应两个不同的$$x$$，其中$$x\\ne 199+352k\\left( k\\in \\mathbf{Z} \\right)$$．  因此，$$f\\left( 0 \\right)$$，$$f\\left( 1 \\right)$$，．．．，$$f\\left( 999 \\right)$$至多有$$\\frac{1}{2}\\cdot 352+1=177$$个不同的值，  且当$$f\\left( 23 \\right)$$，$$f\\left( 24 \\right)$$，．．．，$$f\\left( 199 \\right)$$两两不等时，$$177$$可以取到．  因此，所求的最大值为$$177$$．  【点评】由已知恒等式组$$f\\left( x \\right)=f\\left( 398-x \\right)$$，$$f\\left( x \\right)=f\\left( 2158-x \\right)$$及$$f\\left( x \\right)=f\\left( 3214-x \\right)$$可得出以下恒等式组：  $$f\\left( x \\right)=f\\left( 2158-x \\right)=f\\left( 3214-\\left( 2158-x \\right) \\right)=f\\left( 1056+x \\right)$$，  $$f\\left( x \\right)=f\\left( 1056+x \\right)=f\\left( 2158-\\left( 1056+x \\right) \\right)=f\\left( 1192-x \\right)$$，  $$f\\left( x \\right)=f\\left( 1056+x \\right)=f\\left( 1102-\\left( 1056+x \\right) \\right)=f\\left( 46-x \\right)$$及  $$f\\left( x \\right)=f\\left( 46-x \\right)=f\\left( 398-\\left( 46-x \\right) \\right)=f\\left( 352+x \\right)$$．  由最后一个恒等式知$$f$$是一个周期函数，且$$f$$的一个周期为$$352$$．  故$$f\\left( 0 \\right)$$，$$f\\left( 1 \\right)$$，．．．，$$f\\left( 999 \\right)$$中每个值都可在$$f\\left( 0 \\right)$$，$$f\\left( 1 \\right)$$，．．．，$$\\left( 351 \\right)$$中找到．  恒等式$$f\\left( x \\right)=f\\left( 398-x \\right)$$表明$$f\\left( 200 \\right)$$，$$f\\left( 201 \\right)$$，．．．，$$f\\left( 351 \\right)$$可在$$f\\left( 0 \\right)$$，$$f\\left( 1 \\right)$$，．．．，$$f\\left( 199 \\right)$$中找到．  恒等式$$f\\left( x \\right)=f\\left( 46-x \\right)$$表明$$f\\left( 0 \\right)$$，$$f\\left( 1 \\right)$$，．．．，$$f\\left( 22 \\right)$$可在$$f\\left( 23 \\right)$$，$$f\\left( 24 \\right)$$，．．．，$$f\\left( 199 \\right)$$中找到．  故题中所给一列函数中最多有$$177$$个不同的值．  要说明$$f\\left( 23 \\right)$$，$$f\\left( 24 \\right)$$，．．．，$$f\\left( 199 \\right)$$可以是不同的值，  考虑函数$$f\\left( x \\right)=\\cos \\left( \\frac{360}{352}\\left( x-23 \\right) \\right)$$，$$x$$的单位为度．  用常规方法可证明恒等式$$f\\left( x \\right)=f\\left( 398-x \\right)$$，$$f\\left( x \\right)=f\\left( 2158-x \\right)$$及$$f\\left( x \\right)=f\\left( 3214-x \\right)$$，  且可看出  $$1=f\\left( 23 \\right)\\textgreater f\\left( 24 \\right)\\textgreater f\\left( 25 \\right)\\textgreater\\cdots \\textgreater f\\left( 199 \\right)=-1$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "707", "queId": "76d578ea96414e46baf905e08e2ece60", "competition_source_list": ["2010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": "圆周上有$$10$$个等分点，则以这$$10$$个等分点中的四个点为顶点的凸四边形中，梯形所占的比为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{8}{21}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4}{21}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{126}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{7}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合", "竞赛->知识点->排列组合与概率->概率初步"], "answer_analysis": ["任选$$4$$点，共有$$C_{10}^{4}=210$$个凸四边形，其中梯形的两条平行边可以从$$5$$组平行于直径的$$5$$条平行弦中选取，也可以$$5$$组不平行于直径的$$4$$条平行弦中选取，去除矩形，梯形共有$$60$$个，所以，梯形所占的比为$$\\frac{2}{7}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "196", "queId": "2658aaf8b23048ed8928d81038b5090e", "competition_source_list": ["2016年AMC10竞赛B第4题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$Zeoy$$閲讀$$15$$本書，一次閱讀$$1$$本．她花了$$1$$天讀完第一本書，又花了$$2$$天讀完第二本，再花了$$3$$天讀完第三本，其餘依此類推，即每讀完一本書要比讀完前一本書多花$$1$$天．$$Zoey$$讀完第一本書是在星期一，讀完第二本書是在星期三，若她每天都閲讀，則當她讀完第$$15$$本書時尼在星期幾？  Zoey read $$15$$ books, one at a time. The first book took her $$1$$ day to read, the second book took $$2$$ her days to read, the third book took $$3$$ her days to read, and so on, with each book taking her $$1$$ more day to read than the previous book. Zoey finished the first book on a Monday, and the second on a Wednesday. On what day of the week did she finish her $$15\\text{th}$$ book?", "answer_option_list": [[{"aoVal": "A", "content": "星期日｜$ $Sunday "}], [{"aoVal": "B", "content": "星期一｜Monday "}], [{"aoVal": "C", "content": "星期三｜Wednesday "}], [{"aoVal": "D", "content": "星期五｜Friday "}], [{"aoVal": "E", "content": "星期六｜Saturday "}]], "knowledge_point_routes": [], "answer_analysis": ["The process took $$1+2+3+\\cdot\\cdot\\cdot+13+14+15=120$$ days, so the last day was $$119$$ days after the first day. Since $$119$$ is divisible by $$7$$, both must have been the same day of the week, so the answer is $$\\left(\\text{B}\\right)$$ Monday. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "271", "queId": "4fde3c0c8c2449f091d87056540e9192", "competition_source_list": ["2015年浙江全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知一个角大于$$120{}^{}\\circ $$的三角形的三边长分别为$$m,m+1,m+2$$，则实数$$m$$的取值范围为．", "answer_option_list": [[{"aoVal": "A", "content": "$$m\\textgreater1$$ "}], [{"aoVal": "B", "content": "$$1\\textless{}m\\textless{}\\frac{3}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{2}\\textless{}m\\textless{}3$$ "}], [{"aoVal": "D", "content": "$$m\\textgreater3$$ "}]], "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": ["由题意可知：  $$\\begin{cases}m+(m+1)\\textgreater m+2    {{(m+2)}^{2}}\\textgreater{{m}^{2}}+{{(m+1)}^{2}}+m(m+1) \\end{cases}$$  解得$$1\\textless{}m\\textless{}\\frac{3}{2}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "291", "queId": "1691bd50694c473c9e62e694d052a6d8", "competition_source_list": ["2012年山东全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "$${{\\left( {{x}^{2}}-\\frac{1}{x} \\right)}^{n}}$$的展开式中，常数项为$$15$$，则$$n$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$9$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "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": ["$${{\\left( {{x}^{2}}-\\frac{1}{x} \\right)}^{n}}$$的展开式中，常数项为$$15$$，则$$\\text{C}_{n}^{k}{{\\left( {{x}^{2}} \\right)}^{k}}{{\\left( -\\frac{1}{x} \\right)}^{n-k}}=15$$，其中$$2k+k-n=0$$，即$$k=\\frac{n}{3}$$，所以$$n$$应被$$3$$整除，由此得$$\\text{C}_{n}^{\\frac{n}{3}}=15$$．当$$n=3$$时，$$\\text{C}_{3}^{1}=3\\ne 15$$，当$$n=6$$时，$$\\text{C}_{6}^{2}=15$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "394", "queId": "4bf715f9f420451eb9ba1d3354d6f2c6", "competition_source_list": ["2009年河南全国高中数学联赛高一竞赛初赛第1题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "若集合$$A$$满足：对任意$$x\\in A$$，都有$$\\frac{1}{x}\\in A$$，就称$$A$$是``和谐''集合，则在集合$$M=\\left { -1,0,\\frac{1}{3},\\frac{1}{2},1,2,3,4,5,6 \\right }$$的所有非空子集中，``和谐''集合有~\\uline{~~~~~~~~~~}~个．", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$15$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$17$$ "}]], "knowledge_point_routes": ["课内体系->特色题型->新定义", "课内体系->素养->逻辑推理", "课内体系->知识点->集合->集合的新定义问题", "课内体系->知识点->集合->集合的基本关系->子集"], "answer_analysis": ["题意为：互为倒数的两个数必须同时在一个集合或者同时不在一个集合中，所以需要将集合中的元素分组，$$\\left { 1 \\right }$$，$$\\left { -1 \\right }$$，$$\\left { 2,\\frac{1}{2} \\right }$$，$$\\left { 3,\\frac{1}{3} \\right }$$，所以和谐集有$${{2}^{4}}-1=15$$个．  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": "1099", "queId": "b88817947e264569bc77a9182d1d3c20", "competition_source_list": ["高二下学期单元测试《轨迹问题》自招第3题", "2011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": "在半径为$$1$$的$$\\odot O$$上，取一个定点$$A$$和一个动点$$B$$．设点$$P$$满足$$AP//OB$$且$$\\overrightarrow{AP}\\cdot \\overrightarrow{AB}=1$$，则$$P$$点的轨迹是．", "answer_option_list": [[{"aoVal": "A", "content": "椭圆 "}], [{"aoVal": "B", "content": "抛物线 "}], [{"aoVal": "C", "content": "双曲线 "}], [{"aoVal": "D", "content": "以上都有可能 "}]], "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->曲线与方程", "竞赛->知识点->解析几何->直线与方程"], "answer_analysis": ["不妨设$$O(0,  0),  A(1,  0),  P(x,  y)$$，由于$$AP//OB$$，可设$$B(k(x-1),  ky)$$．将这些坐标代入$$\\overrightarrow{AP}\\cdot \\overrightarrow{AB}=1$$．可得$$k=\\frac{x}{{{(x-1)}^{2}}+{{y}^{2}}}$$．最后，利用$$B$$在$$\\odot O$$上，即可得到$$(x,  y)$$满足的方程为$${{x}^{2}}={{(x-1)}^{2}}+{{y}^{2}}$$，即$${{y}^{2}}=2x-1$$，所以$$P$$的轨迹是抛物线． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "86", "queId": "0874009b64da40af95ebf024cb4db186", "competition_source_list": ["2019年AMC12竞赛A第23题"], "difficulty": "3", "qtype": "single_choice", "problem": "2019AMC12A, 23  Define binary operations $$\\diamondsuit $$ and $$\\heartsuit$$ by $$a\\diamondsuit b=a^{\\log _7(b)}$$ and $$a\\heartsuit b=a^{\\frac{1}{\\log _7(b)}}$$ for all real numbers $$a$$ and $$b$$ for which these expressions are defined. The sequence $$(a_n)$$ is defined recursively by $$a_3=3\\heartsuit 2$$ and $$a_n=(n\\heartsuit (n-1))\\diamondsuit a_{n-1}$$ for all integers $$n\\geqslant 4$$. To the nearest integer, what is $$\\log _7(a_{2019})$$?   对所有实数$$a$$, $$b$$, 定义二元运算$$\\diamondsuit $$ 和 $$\\heartsuit$$如下: $$a\\diamondsuit b=a^{\\log _7(b)}$$ , $$a\\heartsuit b=a^{\\frac{1}{\\log _7(b)}}$$. 数列 $$(a_n)$$ 满足 $$a_3=3\\heartsuit 2$$以及$$a_n=(n\\heartsuit (n-1))\\diamondsuit a_{n-1}$$ ($$n\\geq 4$$). 距离 $$\\log _7(a_{2019})$$ 最近的整数是多少? ．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "$$11$$ "}], [{"aoVal": "E", "content": "$$12$$ "}]], "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->Combination->Reasoning->Information Migration (new definition)", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Recurrence and Recursion"], "answer_analysis": ["首先化简一下题目中给的条件.  $$a_3=3\\heartsuit2=3^{\\frac{1}{\\log_7{(2)}}}=3^{\\log_2{7}}$$  $$a_n=(n\\heartsuit (n-1))\\diamondsuit a_{n-1}=(n\\heartsuit (n-1))^{\\log_7{a_{n-1}}}=(n^{\\log_{n-1}{7}})^{\\log_7{a_{n-1}}}=n^{\\log_{n-1}{a_{n-1}}}$$  $$\\Rightarrow\\log_n{a_n}=\\log_{n-1}{a_{n-1}}=\\log_3{a_3}=\\log_2{7}$$  因此$$\\log_{2019}{a_{2019}}=\\log_2{7}\\Rightarrow\\log_7{a_{2019}}=\\log_7{2019}\\log_2{7}=\\log_2{2019}$$.  这个值最接近$\\log_2{2048}=11$, 选$$\\text{D}$$. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "63", "queId": "131c4db76b4d4ee69dc796797fb0632c", "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": "528", "queId": "4ce7f529bed245c48aca2f2e387a2dc4", "competition_source_list": ["2016年AMC10竞赛第17题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "設$$N$$爲$$5$$的倍數．一個紅球與$$N$$個綠球隨意地排在一直線上．令P(N) 表示至少有$$ \\frac{3}{5} $$的綠球排在紅球同一側的概率．顯然$$P（5） =1$$，且當$$N$$非常大时P(N) 趨近於$$ \\frac{4}{5} $$．試問使得$$ P(N)\\textless{} \\frac{321}{400} $$最小的$$N$$，其各位數的數字和爲多少？  Let N be a positive multiple of 5. One red ball and N green balls are arranged in a line in randon order. Let  P(N)be the probability that at least$$ \\frac{3}{5} $$ of the green balls are on the same side of the red ball. Observe that  $$P(5)=1$$ and that P(N) approaches$$ \\frac{4}{5} $$ as N grows large. What is the sum of the digits of the least value of  $$P(N)\\textless{} \\frac{321}{400} $$", "answer_option_list": [[{"aoVal": "A", "content": "$$12$$ "}], [{"aoVal": "B", "content": "$$14$$ "}], [{"aoVal": "C", "content": "$$16$$ "}], [{"aoVal": "D", "content": "$$18$$ "}], [{"aoVal": "E", "content": "$$20$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["Let $$n=\\frac N5$$. Then, consider $$5$$ blocks of $$n$$ green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the $$N+1$$ positions between the green balls to insert the red ball.  Less than $$\\frac 35$$ of the green balls will be on the same side of the red ball if the red ball is inserted in the middle block of $$n$$ balls, and there are $$n-1$$ positions where this happens.  Thus, $$P(N)=1-\\frac {n-1}{N+1}=\\frac {4n+2}{5n+1}$$, so $$P(N)=\\frac {4n+2}{5n+1}\\textless\\frac {321}{400}$$.  Multiplying both sides of the inequality by $$400(5n+1)$$, we have $$400(4n+2)\\textless321(5n+1)$$,  and by the distributive property,  $$1600n+800\\textless1605n+321$$.  Subtracting $$1600n+321$$ on both sides of the inequality gives us $$479\\textless5n$$,  Therefore, $$N=5n\\textgreater479$$, so the least possible value of $$N$$ is $$480$$. The sum of the digits of $$480$$ is $$12$$.  Let $$N=5$$, $$P(N)=1$$(Given)  Let $$N=10$$, $$P(N)=\\frac {10}{11}$$  Let $$N=15$$, $$P(N)=\\frac {14}{16}$$  Notice that the fraction can be written as $$1-\\frac{\\frac{N}{5}-1}{N+1}$$  Now it\\textquotesingle s quite simple to write the inequality as $$1-\\frac {\\frac N5-1}{N+1}\\textless\\frac {321}{400}$$  We can subtract $$1$$ on both sides to obtain $$-\\frac {\\frac N5-1}{N+1}\\textless-\\frac {79}{400}$$  Dividing both sides by $$-1$$, we derivee $$\\frac {\\frac N5-1}{N+1}\\textgreater\\frac {79}{400}$$. (Switch the inequality sign when dividing by $$-1$$)  We then cross multiply to get $$80N-400\\textgreater79N+79$$  Finally we get $$N\\textgreater479$$  To achieve $$N=480$$  So the sum of the digits of $$N=12$$.  We are trying to find the number of places to put the red ball, such that $$\\frac 35$$ of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with $$N$$: Trying a few values, we see that the ball \"works\" in places $$1$$ to $$\\frac 25N+1$$ and spaces $$\\frac 35N+1$$ to $$N+2$$. This is a total of $$\\frac 45N+2$$ spaces, over a total possible $$N+1$$ places to put the ball. So: $$\\frac {\\frac 45N+2}{N+1}=\\frac {321}{400}\\rightarrow N=479$$. And we know that the next value is what we are looking for, so $$N+1=480$$, and the sum of it\\textquotesingle s digits is $$12$$. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "156", "queId": "1451f095ed61419b94c8a3c242dc6822", "competition_source_list": ["2008年陕西全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设等差数列$$ {{{a}_{n}} }$$的前$$n$$项和为$${{S}_{n}}$$，若$${{S}_{2}}=10$$，$${{S}_{5}}=55$$，则经过$$P(n,{{a}_{n}})$$、$$Q(n+2,{{a}_{n+2}})$$两点的直线的一个方向向量的坐标可以是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 2,\\frac{1}{2} \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( -\\frac{1}{2},  -2 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( -\\frac{1}{2},-1 \\right)$$ "}], [{"aoVal": "D", "content": "$$(-1,-1)$$ "}]], "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$$ {{{a}_{n}} }$$的公差为$$d$$，  则由$$\\begin{cases}{{S}_{2}}=2{{a}_{1}}+d=10    {{S}_{5}}=5{{a}_{1}}+10d=55 \\end{cases}$$，得$$\\begin{cases}{{a}_{1}}=3    d=4 \\end{cases}$$，  所以$${{K}_{PQ}}=\\frac{{{a}_{n+2}}-{{a}_{n}}}{(n+2)-n}=d=4$$，  故所$$B$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "26", "queId": "6f9decad8f26447bba809b1195cc4491", "competition_source_list": ["1983年全国高中数学联赛竞赛一试第1题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$p，q$$是自然数，条件甲：$${{p}^{3}}-{{q}^{3}}$$是偶数；条件乙；$$p+q$$是偶数，那么(~ ~ )．", "answer_option_list": [[{"aoVal": "A", "content": "甲是乙充分而非必要条件 "}], [{"aoVal": "B", "content": "甲是乙的必要而非充分条件 "}], [{"aoVal": "C", "content": "甲是乙的充要条件 "}], [{"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": ["若$${{p}^{3}}-{{q}^{3}}$$是偶数$$\\Leftrightarrow p，q$$有相同的奇偶性$$\\Leftrightarrow p+q$$是偶数． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "602", "queId": "7629ac945ea54c6484101cae08a04f2e", "competition_source_list": ["2010年安徽全国高中数学联赛竞赛初赛第5题8分", "2019~2020学年上海浦东新区华东师范大学第二附属中学高二下学期单元测试第17题"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$z$$是复数，则$$\\left\\textbar{} z-1 \\right\\textbar+\\left\\textbar{} z-\\text{i} \\right\\textbar+\\left\\textbar{} z+1 \\right\\textbar$$的最小值等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$1+\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$1+\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的应用"], "answer_analysis": ["法一：$$\\left\\textbar{} z-1 \\right\\textbar+\\left\\textbar{} z-\\text{i} \\right\\textbar+\\left\\textbar{} z+1 \\right\\textbar$$表示动点$$Z$$到三个定点$$A\\left( 1,0 \\right)$$，$$B\\left( 1,0 \\right)$$，$$C\\left( -1,0 \\right)$$的距离之和，而且$$\\triangle ABC$$是一个锐角三角形．所以$$Z$$是对各边张角为$$120{}^{}\\circ $$的点，即点$$Z$$落在$$y$$轴的正半轴上，且$$\\angle AZB=120{}^{}\\circ $$，从而  $${{\\left( \\left\\textbar{} z-1 \\right\\textbar+\\left\\textbar{} z-\\text{i} \\right\\textbar+\\left\\textbar{} z+1 \\right\\textbar{} \\right)}_{\\min }}=2\\frac{1}{\\sin 60{}^{}\\circ }+\\left( 1-\\frac{1}{\\tan 60{}^{}\\circ } \\right)=1+\\sqrt{3}$$．  法二：$$\\left\\textbar{} z-1 \\right\\textbar+\\left\\textbar{} z-\\text{i} \\right\\textbar+\\left\\textbar{} z+1 \\right\\textbar$$表示动点$$Z$$三个定点$$A\\left( 1,0 \\right)$$，$$B\\left( 0,1 \\right)$$，$$C\\left( -1,0 \\right)$$的距离之和，能确定的是当取得最小值时点$$Z$$落在$$y$$轴的正半轴上，故设$$\\angle ZAC=\\theta $$，且$$\\theta $$为锐角，从而$$\\left\\textbar{} z-1 \\right\\textbar+\\left\\textbar{} z-\\text{i} \\right\\textbar+\\left\\textbar{} z+1 \\right\\textbar=\\frac{2}{\\cos \\theta }+1-\\tan \\theta =1-\\frac{\\sin \\theta -2}{\\cos \\theta }{\\geqslant }1+\\sqrt{3}$$．  故答案为：$$1+\\sqrt{3}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "951", "queId": "97516743165947c6bcaf6cec978fbb86", "competition_source_list": ["2011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": "将$${{(a+b+c+d)}^{9}}$$展开之后再合并同类项，所得的多项式的项数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\text{C}_{9}^{4}$$ "}], [{"aoVal": "B", "content": "$$\\text{C}_{9}^{3}$$ "}], [{"aoVal": "C", "content": "$$\\text{C}_{12}^{4}$$ "}], [{"aoVal": "D", "content": "$$\\text{C}_{12}^{3}$$ "}]], "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": ["所得多项式中每一项都形如$$k{{a}^{{{x}_{1}}}}{{b}^{{{x}_{2}}}}{{c}^{{{x}_{3}}}}{{d}^{{{x}_{4}}}}$$，其中$$k\\textgreater0$$，  $${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=9,  {{x}_{i}}\\geqslant 0$$．  易知上式共有$$\\text{C}_{9+4-1}^{4-1}=\\text{C}_{12}^{3}$$组整数解，因此选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "140", "queId": "187912903a4d4223a6c0c1c55db13628", "competition_source_list": ["2008年辽宁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$$f(x)=\\begin{cases}(3-a)x-a,x\\textless{}1    {{\\log }_{a}}x,x\\geqslant 1 \\end{cases}$$在$$(-\\infty ,+\\infty )$$上单调增加，那么$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(0,3)$$ "}], [{"aoVal": "B", "content": "$$\\left( 1,\\frac{3}{2} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left[ \\frac{3}{2},3 \\right)$$ "}], [{"aoVal": "D", "content": "$$(1,3)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["由题知$$\\begin{cases}a\\textgreater1    3-a\\textgreater0    {{\\log }_{a}}1\\geqslant \\left( 3-a \\right)\\cdot 1-a \\end{cases}$$，解得$$\\frac{3}{2}\\leqslant a\\textless{}3$$．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "992", "queId": "a0c50a982e3e42529c3f1b5bc02b67f5", "competition_source_list": ["2009年AMC12竞赛A第24题"], "difficulty": "3", "qtype": "single_choice", "problem": "\\uline{The tower function of twos} is defined recursively as follows: $$T\\left(1\\right)=2$$ and $$T\\left(n+1\\right)=2^{T\\left(n\\right)}$$ for $$n\\geqslant 1$$. Let $$A=\\left(T\\left(2009\\right)\\right)^{T\\left(2009\\right)}$$ and $$B=\\left(T\\left(2009\\right)\\right)^{A}$$. What is the largest integer $$k$$ such that $$\\underbrace{\\log_{2}\\log_{2}\\log_{2}\\ldots\\log_{2}B}_{{k\\textasciitilde{} \\text{times}}}$$ is defined?  2的\\uline{塔函数}定义如下: $$T\\left(1\\right)=2$$ , $$T\\left(n+1\\right)=2^{T\\left(n\\right)}$$ ($$n\\geqslant 1$$). 令 $$A=\\left(T\\left(2009\\right)\\right)^{T\\left(2009\\right)}$$ , $$B=\\left(T\\left(2009\\right)\\right)^{A}$$. 使得 $$\\underbrace{\\log_{2}\\log_{2}\\log_{2}\\ldots\\log_{2}B}_{{k\\textasciitilde{} \\text{times}}}$$有意义的最大整数 $$k$$是多少?", "answer_option_list": [[{"aoVal": "A", "content": "$$2009$$ "}], [{"aoVal": "B", "content": "$$2010$$ "}], [{"aoVal": "C", "content": "$$2011$$ "}], [{"aoVal": "D", "content": "$$2012$$ "}], [{"aoVal": "E", "content": "$$2013$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Reasoning->Mapping and Correspondence", "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$\\log_2$为$\\log$.  由条件, 有$\\log{T(n+1)}=T(n)$.  我们对$B$求几次对数看看情况.  $\\log{B}=A\\log{T(2009)}=A\\cdot T(2008)=T(2009)^{T(2009)}T(2008)$  $\\log{\\log{B}}=T(2009)\\log{T(2009)}+\\log{T(2008)}=T(2009)T(2008)+T(2007)$  再往下似乎不太好求了, 不过我们可以进行一些适当的舍入.  $\\log{\\log{B}}=T(2009)T(2008)+T(2007)=T(2009)\\left(T(2008)+\\frac{T(2007)}{T(2009)}\\right)$  $\\Rightarrow\\log\\log\\log{B}=\\log{T(2009)}+\\log{\\left(T(2008)+\\frac{T(2007)}{T(2009)}\\right)}\\approx\\log{T(2009)}+\\log{T(2008)}=T(2008)+T(2007)$  考虑到$T(2007)=\\log\\log{T(2009)}\\textless\\textless T(2009)$, 这里的舍入是合理的.  故技重施,  $\\log\\log\\log\\log{B}=\\log{\\left(T(2008)\\left(1+\\frac{T(2007)}{T(2008)}\\right)\\right)}=\\log{T(2008)}+\\log{\\left(1+\\frac{T(2007)}{T(2008)}\\right)}\\approx T(2007)$  到这里就好求了, 从$T(2007)$再往下求$$2007$$层, 会得到$$1.$$ 考虑到我们在计算过程中做了一些舍入, 实际上这时的值会略大于$$1.$$ 再求一层, 得到一个略大于$$0$$的值$$.$$ 再求一层, 得到一个负值$$.$$ 之后再往下，就没有意义了.  因此, 总层数$=2+2007+4=2013$, 选E. "], "answer_value": "E"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "49", "queId": "5d084a95afb34d56a6e38926d0192121", "competition_source_list": ["2008年辽宁全国高中数学联赛竞赛初赛第7题5分", "2016~2017学年10月上海徐汇区上海市上海中学高一上学期周测第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a$$，$$b$$，$$c$$都是非负数，则$$\\frac{c}{a}+\\frac{a}{b+c}+\\frac{b}{c}$$的最小值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$0.5$$ "}]], "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", "课内体系->知识点->等式与不等式->不等式->基本不等式->多元均值不等式的应用", "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"], "answer_analysis": ["$$\\frac{c}{a}+\\frac{a}{b+c}+\\frac{b}{c}$$  $$=\\frac{c}{a}+\\frac{a}{b+c}+\\frac{b+c}{c}-1$$  $$\\geqslant 3\\sqrt[3]{\\frac{c}{a}\\cdot \\frac{a}{b+c}\\cdot \\frac{b+c}{c}}-1$$  $$=2$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "547", "queId": "639b44648ba14d239fde8603a64f86ec", "competition_source_list": ["2020~2021学年北京高三单元测试", "2020年北京海淀区北京大学自主招生（强基计划）第6题5分", "2020~2021学年北京高二单元测试", "竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$x$$，$$y$$，$$z$$均不为$$\\left( k+\\frac{1}{2} \\right)\\pi $$，其中$$k$$为整数，已知$$\\sin \\left( y+z-x \\right)$$，$$\\sin \\left( x+z-y \\right)$$，$$\\sin \\left( x+y-z \\right)$$成等差数列，则依然成等差数列的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sin x$$，$$\\sin y$$，$$\\sin z$$ "}], [{"aoVal": "B", "content": "$$\\cos x$$，$$\\cos y$$，$$\\cos z$$ "}], [{"aoVal": "C", "content": "$$\\tan x$$，$$\\tan y$$，$$\\tan z$$ "}], [{"aoVal": "D", "content": "前三个答案都不对 "}]], "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->三角恒等变换"], "answer_analysis": ["$$\\because \\cos x$$，$$\\cos y$$，$$\\cos z$$均不为$$0$$，  $$2\\sin \\left( x+z-y \\right)=\\sin \\left( x+y-z \\right)+\\sin \\left( y+z-x \\right)$$  $$=2\\sin y\\cdot \\cos \\left( z-y \\right)$$，  $$\\therefore \\sin y\\cos \\left( z-x \\right)=\\sin \\left[ \\left( x+z \\right)-y \\right]$$  $$\\Leftrightarrow \\sin y\\cos \\left( z-x \\right)=\\sin \\left( x+z \\right)\\cos y-\\cos \\left( x+z \\right)\\sin y$$  $$\\Leftrightarrow 2\\sin y\\cos x\\cos z=\\cos y\\sin x\\cos z+\\cos y\\cos x\\sin z$$  $$\\Leftrightarrow 2\\tan y=\\tan x+\\tan z$$，  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "836", "queId": "6a522fbaf51a4bcebdb0c158df03065c", "competition_source_list": ["2014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": "设复数$$x=\\frac{1+\\text{i}}{1-\\text{i}}$$（$$\\text{i}$$是虚数单位），则$$C_{2014}^{0}+C_{2014}^{1}x+C_{2014}^{2}{{x}^{2}}+\\ldots +C_{2014}^{2014}{{x}^{2014}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$${{2}^{1007}}\\text{i}$$ "}], [{"aoVal": "B", "content": "$$-{{2}^{1007}}\\text{i}$$ "}], [{"aoVal": "C", "content": "$$-{{2}^{1005}}$$ "}], [{"aoVal": "D", "content": "$$-{{2}^{1004}}$$ "}]], "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": ["因为$$x=\\frac{{{\\left( 1+\\text{i} \\right)}^{2}}}{\\left( 1-\\text{i} \\right)\\left( 1+\\text{i} \\right)}\\text{=}\\frac{2\\text{i}}{2}\\text{=i}$$，所以  $$C_{2014}^{0}+C_{2014}^{1}x+C_{2014}^{2}{{x}^{2}}+\\ldots +C_{2014}^{2014}{{x}^{2014}}={{\\left( 1+x \\right)}^{2014}}={{\\left( 1+\\text{i} \\right)}^{2014}}={{\\left( 2\\text{i} \\right)}^{1007}}=-{{2}^{1007}}\\text{i}$$ "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1117", "queId": "d850bd8d8f594313b8083c54abddd123", "competition_source_list": ["2017年AMC10竞赛A第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "定义數列 $$F_0=0$$, $$F_1=1$$, 且對於所有$$n≥2$$ 有$$F_n=F_{n−1}+F_{n−2}$$ 再除以$$3$$的餘數，序列從$$0,1,1,2,0,2$$$$\\cdots \\cdots$$ 開始，求$$F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}$$？  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 $$F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}$$?", "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": [], "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": "1176", "queId": "fdc60d8a436c4bbc99ef941064089287", "competition_source_list": ["2004年全国高中数学联赛竞赛一试第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$M= {(x,y)\\textbar{{x}^{2}}+2{{y}^{2}}=3 }$$，$$N= {(x,y)\\textbar y=mx+b }$$．若对于所有的$$m\\in R$$，均有$$M\\cap N\\ne \\varnothing $$，则$$b$$的取值范围是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ -\\frac{\\sqrt{6}}{2},\\frac{\\sqrt{6}}{2} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left( -\\frac{\\sqrt{6}}{2},\\frac{\\sqrt{6}}{2} \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( -\\frac{2\\sqrt{3}}{3},\\frac{2\\sqrt{3}}{3} \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left[ -\\frac{2\\sqrt{3}}{3},\\frac{2\\sqrt{3}}{3} \\right]$$ "}]], "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->椭圆"], "answer_analysis": ["点$$(0,b)$$在椭圆内或椭圆上$$\\Rightarrow 2{{b}^{2}}\\leqslant 3\\Rightarrow b\\in \\left[ -\\frac{\\sqrt{6}}{2},\\frac{\\sqrt{6}}{2} \\right]$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "749", "queId": "60b4e460952d45afb71b3d5694c5855f", "competition_source_list": ["2016年广西全国高中数学联赛竞赛初赛第1题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)=\\frac{1}{\\sqrt{-{{\\left( \\lg x \\right)}^{2}}+3\\lg x-2}}$$的定义域是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$(10,100)$$ "}], [{"aoVal": "B", "content": "$$(10,20)$$ "}], [{"aoVal": "C", "content": "$$(20,100)$$ "}], [{"aoVal": "D", "content": "$$(50,100)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->基本初等函数->对数函数->对数方程和对数不等式->解对数不等式", "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->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->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->函数的概念与性质->函数的概念及其表示->函数的定义域->求具体函数（包括复合函数）的定义域", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理"], "answer_analysis": ["$$-{{(\\lg x)}^{2}}+3\\lg x-2\\textgreater0$$，$$1\\textless{}\\lg x\\textless{}2$$．  故答案为： $$(10,100)$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "786", "queId": "a3e121d565f54e94aa8810e9dd4b3058", "competition_source_list": ["2013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": "函数$$f\\left( x \\right)$$对于任意实数$$x$$满足：$$f\\left( x+3 \\right)=-\\frac{1}{f(x)}$$，若$$f\\left( 0 \\right)=2$$，则$$f\\left( 2013 \\right)=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$${-}\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$2013$$ "}]], "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->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->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->函数的概念与性质->函数的性质->周期性->函数周期性判断"], "answer_analysis": ["$$f\\left( x+6 \\right)=-\\frac{1}{f(x+3)}=f\\left( x \\right)$$，所以，$$f\\left( x \\right)$$是以$$6$$为周期的函数.  所以，$$f\\left( 2013 \\right)=f\\left( 3 \\right)=-\\frac{1}{f(0)}=-\\frac{1}{2}$$. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "569", "queId": "561dac7c72c444fba8ec8f53f65e93d0", "competition_source_list": ["2008年陕西全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "把分母为$$24$$的所有最简真分数按从小到大的顺序排列，依次为$${{a}_{1}},{{a}_{2}},\\cdots ,  {{a}_{n}}$$，则$$\\sum\\limits_{i=1}^{n}{\\cos ({{a}_{i}} \\pi )}$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$0$$ "}], [{"aoVal": "D", "content": "$$-\\frac{1}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["分母为$$24$$的所有最简真分数有$$\\frac{1}{24},\\frac{5}{24},\\frac{7}{24},\\frac{11}{24},\\frac{13}{24},\\frac{17}{24},\\frac{19}{24},\\frac{23}{24}$$，  注意到$$\\frac{1}{24}+\\frac{23}{24}=\\frac{5}{24}+\\frac{19}{24}=\\frac{7}{24}+\\frac{17}{24}+\\frac{11}{24}+\\frac{13}{24}=1$$，  因此$$\\sum\\limits_{i=1}^{n}{\\cos ({{a}_{i}} \\pi )}=0$$．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "592", "queId": "b0dcdbeeae9b42c48bd620b6551530d3", "competition_source_list": ["2018年AMC10竞赛B第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$\\left {2,3,4,5,6,7,8,9\\right }$$的多少个子集至少包含一个素数？  How many subsets of $$\\left {2,3,4,5,6,7,8,9\\right }$$ contain at least one prime number?", "answer_option_list": [[{"aoVal": "A", "content": "$$128$$ "}], [{"aoVal": "B", "content": "$$192$$ "}], [{"aoVal": "C", "content": "$$224$$ "}], [{"aoVal": "D", "content": "$$240$$ "}], [{"aoVal": "E", "content": "$$256$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["Consider finding the number of subsets that do not contain any primes.There are four primes in the set：$$2$$, $$3$$, $$5$$, and $$7$$.This means that the number of subsets without any primes is the number of subsets of $$\\left {4, 6, 8,9\\right }$$,~ which is just $$2^{4}=16$$. The number of subsets with at least one prime is the number of subsets minus the number of subsets without any primes.The number of subsets is $$2^{8}=256$$. Thus, the answer is $$256-16=(\\rm D)\\textasciitilde240$$. ", "<p>Well, there are $$4$$ composite numbers, and you can list them in a $$1$$ number format, a $$2$$ number, $$3$$ number, and a $$4$$ number format. Now,&nbsp; we can use combinations.&nbsp;&nbsp;$$\\left( \\begin{array}{l} {4}\\\\{1} \\end{array}\\right)+\\left( \\begin{array}{l} {4}\\\\{2} \\end{array}\\right)+\\left( \\begin{array}{l} {4}\\\\{3} \\end{array}\\right)+\\left( \\begin{array}{l} {4}\\\\{4} \\end{array}\\right)=15$$. Using the answer choices, the only multiple of $$15$$ is $$(\\rm D)~240$$.</p>", "<p>Subsets of $$\\left\\{2,3,4,5,6,7,8,9\\right\\}$$ indude a singe digit up to all eight numbers.Therefore,we must add the combinations of all possible subsets and subtract from each of the subsets fomed by the composite numbers.</p>\n<p>Hence：</p>\n<p>$$\\left( \\begin{array}{l} {8}\\\\{1} \\end{array}\\right)-\\left( \\begin{array}{l} {4}\\\\{1} \\end{array}\\right)+\\left( \\begin{array}{l} {8}\\\\{2} \\end{array}\\right)-\\left( \\begin{array}{l} {4}\\\\{2} \\end{array}\\right)+\\left( \\begin{array}{l} {8}\\\\{3} \\end{array}\\right)-\\left( \\begin{array}{l} {4}\\\\{3} \\end{array}\\right)+\\left( \\begin{array}{l} {8}\\\\{4} \\end{array}\\right)-1+\\left( \\begin{array}{l} {8}\\\\{5} \\end{array}\\right)+\\left( \\begin{array}{l} {8}\\\\{6} \\end{array}\\right)+\\left( \\begin{array}{l} {8}\\\\{7} \\end{array}\\right)+1=(\\rm D)~240$$.</p>"], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "818", "queId": "61395a20a8dd4e00a7c4ac8c454c1ef3", "competition_source_list": ["2019~2020学年山东济南历下区山东省济南第七中学高二上学期期中第21题3分", "2019年高考真题全国卷II理科第13题5分", "2018~2019学年安徽合肥包河区合肥市第一中学高一下学期期末第16题4分", "2019~2020学年12月北京西城区北京市铁路第二中学高三上学期月考第14题5分", "2019~2020学年江苏苏州姑苏区苏州第十中学高一下学期开学考试第14题4分", "2019~2020学年8月广东广州荔湾区广东实验中学高三上学期月考理科第13题5分", "2019年高考真题全国卷II文科第14题5分", "2019~2020学年天津南开区高二上学期期末第13题5分", "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15题3分", "2019~2020学年四川成都高新区成都市第七中学（高新校区）高三上学期开学考试理科第13题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "我国高铁发展迅速，技术先进，经统计，在经停某站的高铁列车中，有$$10$$个车次的正点率为$$0.97$$，有$$20$$个车次的正点率为$$0.98$$，有$$10$$个车次的正点率为$$0.99$$，则经停该站高铁列车所有车次的平均正点率的估计值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0.97$$ "}], [{"aoVal": "B", "content": "$$0.98$$ "}], [{"aoVal": "C", "content": "$$0.99$$ "}], [{"aoVal": "D", "content": "以上皆不是 "}]], "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->知识点->统计与概率->统计->用样本估计总体->众数、中位数、平均数", "课内体系->知识点->统计与概率->统计->用样本估计总体->用样本的数字特征估计总体的数字特征问题"], "answer_analysis": ["∵经统计，在经停某站的高铁列车中，有$$10$$个车次的正点率为$$0.97$$，  有$$20$$个车次的正点率为$$0.98$$，有$$10$$个车次的正点率为$$0.99$$，  ∴经停该站高铁列车所有车次的平均正点率的估计值为：  $$\\frac{(10\\times 0.97+20\\times 0.98+10\\times 0.99)}{10+20+10}=0.98$$．  故答案为：$$0.98$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "50", "queId": "06ab4b3bc48748e39f2da343fdd846fc", "competition_source_list": ["竞赛第3题"], "difficulty": "2", "qtype": "single_choice", "problem": "定义${{f}_{M}}(x)=\\text{ } ! ! { ! !\\text{ }\\begin{array}{*{35}{l}} 1,x\\in M,     -1,x\\not\\in M,     \\end{array}M\\o\\times N=\\left { x\\text{ } ! !\\textbar ! !\\text{ }{{f}_{M}}(x)\\cdot {{f}_{N}}(x)=-1\\text{ } \\right }\\text{ }$，已知集合$A=\\left { x\\text{ } ! !\\textbar ! !\\text{ }x\\textless{} \\sqrt{2-x}\\text{ } \\right }$，集合$B=\\left { x\\text{ } ! !\\textbar ! !\\text{ }x(x+3)(x-3)\\textgreater0\\text{ } \\right }$，则不包含于$A\\o\\times B$的取值集合为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$(-\\infty ,-3]$ "}], [{"aoVal": "B", "content": "$[0,1)$ "}], [{"aoVal": "C", "content": "$[1,3)$ "}], [{"aoVal": "D", "content": "$(3,+\\infty )$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据新运算的定义可得$A\\o\\times B=\\left(A\\cap \\left(\\complement _{\\mathrm{R}}B\\right)\\right)\\cup \\left(B\\cap \\left(\\complement _{\\mathrm{R}}A\\right)\\right)$，故可求$A\\o\\times B$.\\\\ 【详解】\\\\ 根据题意，有$A=(-\\mathrm{\\infty },1),B=(-3,0)\\cup (3,+\\mathrm{\\infty })$，\\\\ 于是$A\\o\\times B=(-\\mathrm{\\infty },-3]\\cup [0,1)\\cup (3,+\\mathrm{\\infty })$，\\\\ 因此$[1,3)$是不包含于$A\\o\\times B$的取值集合，\\\\ 故选：C. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "549", "queId": "37105bc046fb47ffb61d5a0fd0a1fc3c", "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1题3分"], "difficulty": "0", "qtype": "single_choice", "problem": "对于简单随机抽样，每个个体每次被抽到的机会．", "answer_option_list": [[{"aoVal": "A", "content": "相等 "}], [{"aoVal": "B", "content": "不相等 "}], [{"aoVal": "C", "content": "无法确定 "}], [{"aoVal": "D", "content": "与抽取的次数有关 "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->知识点->统计与概率->统计->随机抽样->简单随机抽样"], "answer_analysis": ["根据简单随机抽样的定义可得，每个个体被抽到的机会都是相等的．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "674", "queId": "49d10a096f364ce599ba9162709c7c0d", "competition_source_list": ["2010年辽宁全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "数列$$\\frac{1}{2},\\frac{1}{3},\\frac{2}{3},\\frac{1}{4},\\frac{2}{4},\\frac{3}{4},\\cdots ,\\frac{1}{m+1},\\frac{2}{m+1},\\cdots ,\\frac{m}{m+1},\\cdots $$的前$$40$$项的和是．", "answer_option_list": [[{"aoVal": "A", "content": "$$23\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$19\\frac{1}{9}$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "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": ["设第$$40$$项$${{a}_{40}}$$的分母为$$m+1$$，  则$$\\frac{m(m-1)}{2}\\textless{}40\\leqslant \\frac{m(m+1)}{2}$$，  解得$$m=9$$，所以$$\\frac{m(m-1)}{2}=36$$，故$${{a}_{40}}=\\frac{4}{10}$$．  $${{S}_{40}}=\\frac{1}{2}+\\frac{1}{3}\\times (1+2)+\\frac{1}{4}\\times (1+2+3)+\\cdots +\\frac{1}{9}\\times (1+2+3+\\cdots +8)+\\frac{1}{10}\\times (1+2+3+4)$$  $$=\\frac{1}{2}+1+\\frac{3}{2}+\\cdots +\\frac{7}{2}+4+1=19$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "803", "queId": "69f3383263474033979abbc2ff3e52f5", "competition_source_list": ["第二十届全国希望杯高二竞赛初赛邀请赛第7题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)={{3}^{2-x}}+{{\\log }_{2}}\\left( 2-\\sqrt{x-1} \\right)$$的最大值是．", "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->函数->函数的图像与性质"], "answer_analysis": ["由题意可算得$$f(x)$$的定义域为$$1\\leqslant x\\textless{}5$$，而$${{3}^{2-x}}$$和$${{\\log }_{2}}(2-\\sqrt{x-1})$$在$$[1,5)$$上都是减函数，故$$f(x)$$的最大值为$$f(1)=4$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "210", "queId": "1128a0ed99fd4c4e8260c1b5f0c55b73", "competition_source_list": ["2011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": "在等差数列$$\\left { {{a}_{n}} \\right }$$中，若$$\\frac{{{a}_{11}}}{{{a}_{10}}}\\textless{}-1$$，且它的前$$n$$项和$${{S}_{n}}$$有最大值，那么当$${{S}_{n}}$$取最小正值时，$$n=$$ ．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$10$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$20$$ "}]], "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": ["设该等差数列的公差为$$d$$．显然 $$d\\textless{}0$$．由$$\\frac{{{a}_{11}}}{{{a}_{10}}}\\textless{}-1$$，知  $${{a}_{10}}\\textgreater0,{{a}_{11}}\\textless{}0,$$ 且$${{a}_{11}}+{{a}_{10}}\\textless{}0$$．因此  $${{S}_{20}}=\\frac{{{a}_{1}}+{{a}_{20}}}{2}\\times 20=10({{a}_{10}}+{{a}_{11}})\\textless{}0,$$  $${{S}_{19}}=\\frac{{{a}_{1}}+{{a}_{19}}}{2}\\times 19=19{{a}_{10}}\\textgreater0$$．  由$${{a}_{11}}+{{a}_{10}}\\textless{}0,$$ 知$$2{{a}_{1}}+19d\\textless{}0$$．从而有  $${{S}_{19}}-{{S}_{1}}=19{{a}_{1}}+\\frac{19\\times 18}{2}d-{{a}_{1}}=18{{a}_{1}}+9\\times 19d=9(2{{a}_{1}}+19d)\\textless{}0$$  所以$$n=19$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "600", "queId": "49163658d1c24596bbbee2f545e69ea6", "competition_source_list": ["2000年全国高中数学联赛竞赛一试第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "给定正数$$p$$，$$q$$，$$a$$，$$b$$，$$c$$，其中$$p\\ne q$$，若$$p$$，$$a$$，$$q$$是等比数列，$$p$$，$$b$$，$$c$$，$$q$$是等差数列，则一元二次方程$$b{{x}^{2}}-2ax+c=0$$（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "无实根 "}], [{"aoVal": "B", "content": "有两个相等实根 "}], [{"aoVal": "C", "content": "有两个同号相异实根 "}], [{"aoVal": "D", "content": "有两个异号实根 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数方程", "竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["∵$$p$$，$$a$$，$$q$$成等比，  ∴$${{a}^{2}}pq$$，  ∵$$p$$，$$b$$，$$c$$，$$q$$成等差，  ∴$$p+c=2b$$，$$b+q=2c$$，  ∴$$b=\\frac{2p+q}{3}$$，$$c=\\frac{p+2q}{3}$$，  故$$b\\cdot c=\\frac{p+p+q}{3}\\cdot \\frac{p+q+q}{3}\\textgreater\\sqrt[3]{{{p}^{2}}q}\\cdot \\sqrt[3]{p{{q}^{2}}}=pq(p\\ne q)$$，  $$\\Delta =4{{a}^{2}}-4bc=4pq-4bc=4(pq-bc) ~\\textless{} ~0$$．  因此，方程无实数根．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "538", "queId": "ff4ffd40657a468b8a6405774e85c35c", "competition_source_list": ["2002年全国全国高中数学联赛竞赛一试第5题6分"], "difficulty": "3", "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->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->函数的概念->映射", "课内体系->知识点->计数原理->排列与组合->组合"], "answer_analysis": ["本题直接考虑集合$$A$$中每一个元素在$$B$$中的象的情况非常困难．  注意到集合$$B$$中每个元素都有原象，即$$A$$中有$$50$$``组''元素分别与$$B$$中的$$50$$个元素对应；现将集合$$A$$中的$$100$$个元素按原有的顺序分成$$50$$组，每组至少一个元素；将集合$$B$$中的元素按从小到大的顺序排列为$$B=\\left { {{{{b}'}}_{1}},{{{{b}'}}_{2}},\\cdots ,{{{{b}'}}_{50}} \\right }$$；  ∵$$f\\left( {{a}_{1}} \\right)\\leqslant f\\left( {{a}_{2}} \\right)\\leqslant \\cdots \\leqslant f\\left( {{a}_{100}} \\right)$$，  ∴$$A$$中的``第$$1$$组''元素的象为$${{{b}'}_{1}}$$，``第$$2$$组''元素的象为$${{{b}'}_{2}}$$，``第$$50$$组''元素的象为$${{{b}'}_{50}}$$，此处没有排列的问题，即只要$$A$$中元素的分组确定了，映射也就随之确定了；而$$A$$中元素的分组可视为在由这$$100$$个元素所形成的$$99$$个``空''中插上$$49$$块``挡板''，所以有$$\\text{C}_{99}^{49}$$种分法，即映射共有$$\\text{C}_{99}^{49}$$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": "668", "queId": "57013e1312584ec5b51189458cbf2d8f", "competition_source_list": ["2008年贵州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$$M=\\left { x\\textbar{{x}^{2}}-2x-3\\leqslant 0\\text{ },\\text{ }x\\in \\text{R} \\right }$$，$$N=\\left { x\\textbar\\left\\textbar{} x \\right\\textbar{} \\textless{} 2\\text{ },\\text{ }x\\in \\text{R} \\right }$$，则$$M\\cap N$$等于（  ）", "answer_option_list": [[{"aoVal": "A", "content": "$$\\varnothing $$ "}], [{"aoVal": "B", "content": "$$\\left { x\\textbar-1\\leqslant x \\textless{} 2 \\right }$$ "}], [{"aoVal": "C", "content": "$$\\left { x\\textbar-2\\leqslant x \\textless{} -1 \\right }$$ "}], [{"aoVal": "D", "content": "$$\\left { x\\textbar2\\leqslant x \\textless{} 3 \\right }$$ "}]], "knowledge_point_routes": ["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", "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", "课内体系->素养->数学运算"], "answer_analysis": ["【命题立意】本题考查不等式的解法、集合的运算，难度较小. 【解题思路】化简集合后利用交集的定义求解.因为$$M=\\left { x\\textbar{{x}^{2}}-2x-3\\leqslant 0 \\right }=\\left { x\\textbar-1\\leqslant x\\leqslant 3 \\right }$$，$$N=\\left { x\\textbar\\left\\textbar{} x \\right\\textbar{} \\textless{} 2 \\right }=\\left { x\\textbar-2 \\textless{} x \\textless{} 2 \\right }$$，所以$$M\\cap N=\\left { x\\textbar-1\\leqslant x \\textless{} 2 \\right }$$，故选B. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "552", "queId": "32c5e5251351456a80092f0f9a942705", "competition_source_list": ["2012年吉林全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "若集合：$${{S}_{1}}=\\left { \\left( x,  y \\right)\\textbar\\lg \\left( 1+{{x}^{2}}+{{y}^{2}} \\right)\\leqslant 1+\\lg \\left( x+y \\right) \\right }$$，$${{S}_{2}}=\\left { \\left( x,  y \\right)\\textbar\\lg \\left( 2+{{x}^{2}}+{{y}^{2}} \\right)\\leqslant 2+\\lg \\left( x+y \\right) \\right }$$，则$${{S}_{2}}$$的面积与$${{S}_{1}}$$的面积比为", "answer_option_list": [[{"aoVal": "A", "content": "$$99:1$$ "}], [{"aoVal": "B", "content": "$$100:1$$ "}], [{"aoVal": "C", "content": "$$101:1$$ "}], [{"aoVal": "D", "content": "$$102:1$$ "}]], "knowledge_point_routes": ["课内体系->知识点->直线和圆的方程->圆与方程->圆的标准方程与一般方程->圆的标准方程问题", "课内体系->知识点->基本初等函数->对数函数->对数方程和对数不等式", "课内体系->思想->数形结合思想", "课内体系->素养->数学运算", "课内体系->素养->直观想象"], "answer_analysis": ["$${{S}_{1}}$$表示圆内部（包括圆周）：$${{\\left( x-5 \\right)}^{2}}+{{\\left( y-5 \\right)}^{2}}\\leqslant 49$$，面积为$$49 \\pi $$．$${{S}_{2}}$$表示圆内部（包括圆周）：$${{\\left( x-50 \\right)}^{2}}+{{\\left( y-50 \\right)}^{2}}\\leqslant 4998$$，面积为$$4998 \\pi $$．因此，面积比为$$102:1$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "244", "queId": "5d8969c5528440cab7fe6d7d04bc7320", "competition_source_list": ["1987年全国高中数学联赛竞赛一试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "在平面直角坐标系中，纵横坐标均为有理数的点称为有理点．若$$a$$为无理数，则过点$$(a,0)$$的所有直线中（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "有无穷多条直线，其中每条直线上至少存在两个有理点 "}], [{"aoVal": "B", "content": "恰有$$n\\left( 2\\mathsf{\\leqslant }n\\textless{}+\\infty \\right)$$条直线，其中每条直线上至少存在两个有理点 "}], [{"aoVal": "C", "content": "有且仅有一条直线至少通过两个有理点 "}], [{"aoVal": "D", "content": "每条直线至多通过一个有理点 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与方程"], "answer_analysis": ["若直线$$l$$过点$$(a,0)$$且通过两个有理点$$({{x}_{1}},{{y}_{1}})$$，$$({{x}_{2}},{{y}_{2}})$$，则必有$${{x}_{1}}\\ne {{x}_{2}}$$．  若不然，$$l$$的方程必为$$x=a$$，于是$${{x}_{1}}={{x}_{2}}=a$$，这与$$({{x}_{1}},{{y}_{1}})$$，$$({{x}_{2}},{{y}_{2}})$$，是有理点矛盾．  又直线$$l$$的斜率为$$k=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\\frac{{{y}_{1}}}{{{x}_{1}}-a}$$，  从而$${{y}_{1}}=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}({{x}_{1}}-a)$$．  因为$$a$$为无理数，所以上式成立的充要条件是$${{y}_{1}}={{y}_{2}}$$，  故过点$$(a,0)$$且过两个有理点的直线有且只有一条，它的方程为$$y=0$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "279", "queId": "1a51cd6736634351a763518401888585", "competition_source_list": ["2003年全国全国高中数学联赛竞赛一试第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x$$，$$y$$都在区间$$(-2,2)$$内，且$$xy=-1$$，则函数$$u=\\frac{4}{4-{{x}^{2}}}+\\frac{9}{9-{{y}^{2}}}$$的最小值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{8}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{24}{11}$$ "}], [{"aoVal": "C", "content": "$$\\frac{12}{7}$$ "}], [{"aoVal": "D", "content": "$$\\frac{12}{5}$$ "}]], "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->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": ["由已知得$$y=-\\frac{1}{x}$$，故$$u=1+\\frac{35}{37-\\left( 9{{x}^{2}}+\\frac{4}{{{x}^{2}}} \\right)}$$，而$$x\\in \\left( -2,\\frac{1}{2} \\right)\\cup \\left( \\frac{1}{2},2 \\right)$$，故当$$9{{x}^{2}}=\\frac{4}{{{x}^{2}}}\\Rightarrow {{x}^{2}}=\\frac{2}{3}$$时有最小值$$\\frac{12}{5}$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1125", "queId": "c1ea6dcb536640d1ac11a265dccb555b", "competition_source_list": ["竞赛第22题"], "difficulty": "3", "qtype": "single_choice", "problem": "设$S= {1,2,3,\\cdots ,2020 }$，集合\\emph{T}是\\emph{S}的\\emph{n}元子集，且其中任意两个元素互质，对任意符合要求的集合\\emph{T}，均至少包含一个质数，则\\emph{n}的最小值为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "15 "}], [{"aoVal": "B", "content": "16 "}], [{"aoVal": "C", "content": "17 "}], [{"aoVal": "D", "content": "18 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 先根据特例可得$n\\geq 16$，再证明对任意$A\\subseteq S,n=\\textbar{} A\\textbar{} =16$，\\emph{A}中任意两数互质，则\\emph{A}中必存在一个质数，从而可得正确的选项.\\\\ 【详解】\\\\ 首先，我们有$n\\geq 16$．\\\\ 事实上，取集合$A_{0}=\\left {1,2^{2},3^{2},5^{2},\\cdots ,41^{2},43^{2}\\right }$，\\\\ 其元素，除1以外，均为不超过43的素数的平方，则$A_{0}\\subseteq S,\\left\\textbar{} A_{0}\\right\\textbar{} =15,A_{0}$中任意两数互质，但其中无质数，这表明$n\\geq 16$．\\\\ 其次，我们证明：对任意$A\\subseteq S,n=\\textbar{} A\\textbar{} =16$，\\emph{A}中任意两数互质，则\\emph{A}中必存在一个质数.利用反证法，假设\\emph{A}中无质数，记$A=\\left {a_{1},a_{2},\\cdots ,a_{16}\\right }$，分两种情况讨论：\\\\ \\textbf{情形一}~~~若$1\\not\\in A$，则$a_{1},a_{2},\\cdots ,a_{16}$均为合数，\\\\ 又因为$\\left(a_{i},a_{j}\\right)=1(1\\leq i\\textless{} j\\leq 16)$，所以$a_{i}$与$a_{j}$的质因数均不相同，\\\\ 设$a_{i}$的最小质因数为$p_{i}$，不妨设$p_{1}\\textless{} p_{2}\\textless{} \\cdots \\textless{} p_{16}$，则\\\\ $a_{1}\\geq {p}_{1}^{2}\\geq 2^{2}$，\\\\ $a_{2}\\geq {p}_{2}^{2}\\geq 3^{2}$，\\\\ \\ldots，\\\\ $a_{15}\\geq {p}_{15}^{2}\\geq 47^{2}\\textgreater{} 2020$，\\\\ 矛盾．\\\\ \\textbf{情形二}~~~若$1\\in A$，则不妨设$a_{16}=1,a_{1},\\cdots ,a_{15}$均为合数，同情形一所设，同理有\\\\ $a_{1}\\geq {p}_{1}^{2}\\geq 2^{2}$，\\\\ $a_{2}\\geq {p}_{2}^{2}\\geq 3^{2}$，\\\\ \\ldots，\\\\ $a_{15}\\geq {p}_{15}^{2}\\geq 47^{2}\\textgreater{} 2020$，\\\\ 矛盾．\\\\ 综上所述，反设不成立，从而\\emph{A}中必有质数，即$n=\\textbar{} A\\textbar{} =16$时结论成立，\\\\ 因此所求\\emph{n}的最小值为16．\\\\ 故选：B. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "635", "queId": "68b53c068e4e4dccb58726d9a7dff8e7", "competition_source_list": ["2003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": "过抛物线$${{y}^{2}}=8\\left( x+2 \\right)$$的焦点$$F$$作倾斜角为$$60{}^{}\\circ $$的直线，若此直线与抛物线交于$$A$$、$$B$$两点，弦$$AB$$的中垂线与$$x$$轴交于$$P$$点，则线段$$PF$$的长等于（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{16}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{8}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{16\\sqrt{3}}{3}$$ "}], [{"aoVal": "D", "content": "$$8\\sqrt{3}$$ "}]], "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->直线和圆的方程->直线与方程->直线方程的五种形式->直线的点斜式方程", "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->抛物线->直线和抛物线的位置关系", "课内体系->知识点->圆锥曲线->抛物线->抛物线的定义、标准方程->抛物线的标准方程"], "answer_analysis": ["易知此抛物线焦点$$F$$与坐标原点重合，故直线$$AB$$的方程为$$y=\\sqrt{3}x$$，因此，$$A$$，$$B$$两点的横坐标满足方程$$3{{x}^{2}}=8x+16$$，由此求得$$AB$$中点的横坐标$${{x}_{0}}=\\frac{4}{3}$$，纵坐标$${{y}_{0}}=\\frac{4}{\\sqrt{3}}$$，进而求得其中垂线方程为$$y-\\frac{4}{\\sqrt{3}}=-\\frac{1}{\\sqrt{3}}(x-\\frac{4}{3})$$，令$$y=0$$，得$$P$$点的横坐标$$x=\\frac{16}{3}$$，即$$PF=\\frac{16}{3}$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "744", "queId": "57d2a605248b4f4598a48c2b9389a4ef", "competition_source_list": ["2009年AMC10竞赛A第19题"], "difficulty": "3", "qtype": "single_choice", "problem": "2009年$$AMC10$$竞赛$$A$$第$$19$$题  Circle $$A$$ has radius $$100$$. Circle $$B$$ has an integer radius $$r\\textless{} 100$$ and remains internally tangent to circle $$A$$ as it rolls once around the circumference of circle $$A$$. The two circles have the same points of tangency at the beginning and end of circle $$B\\textquotesingle\\rm s$$ trip. How many possible values can $$r$$ have?  圆 $$A$$ 的半径为 $$100$$。 圆 $$B$$ 有一个整数半径 $$r\\textless{} 100$$ 并且在它绕圆 $$A$$ 的圆周滚动一次时保持与圆 $$A$$ 内切。 这两个圆在圆 $$B$$ 轨迹的起点和终点具有相同的切点。 $$r$$ 可以有多少个可能的值？", "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": ["美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Circle", "课内体系->知识点->直线和圆的方程->圆与方程"], "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": "939", "queId": "f70e99a344b7498f9da83d104a2e2618", "competition_source_list": ["2014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": "\\textbf{294．}已知三棱锥$$S-ABC$$，在三棱锥内任取一点$$P$$，使得$${{V}_{p-ABC}}\\textless\\frac{1}{2}{{V}_{S-ABC}}$$的概率是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{7}{8}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{4}$$ "}]], "knowledge_point_routes": ["知识标签->素养->数学抽象", "知识标签->素养->数学运算", "知识标签->题型->概率->事件与概率->几何概型问题->与体积有关的几何概率的计算", "知识标签->知识点->概率->事件与概率->几何概型"], "answer_analysis": ["当$$P$$在三棱锥的中截面与下底面构成的三棱台内时符合要求，由几何概型知，$$P=1-\\frac{1}{8}=\\frac{7}{8}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "827", "queId": "e42ebbd7484440399c6f1f70debbfa13", "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": "$$\\frac{1}{2}$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{2}$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->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": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1047", "queId": "b352d747f9464891b0ab1fc44f6ead93", "competition_source_list": ["2007年AMC10竞赛A第22题", "2007年AMC12竞赛A第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "2007 AMC12A 第$$11$$题  A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms $$247$$, $$475$$, and $$756$$ and end with the term $$824$$. Let $$S$$ be the sum of all the terms in the sequence. What is the largest prime factor that always divides $$S$$?  三位数的有限序列具有以下性质：每项的十位和个位分别为下一项的百位和十位，最后一项的十位和个位分别为百位 和第一项的十位数字。 例如，这样的序列可能以 247、$$475$$ 和$$756$$开头，并以$$824$$结尾。 令 S 是序列中所有项的总和。 能被$$S$$整除的最大质因数是什么？", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"aoVal": "D", "content": "$$37$$ "}], [{"aoVal": "E", "content": "$$43$$ "}]], "knowledge_point_routes": ["课内体系->知识点->数列->数列的实际应用", "美国AMC10/12->Knowledge Point->Number Theory->Aliquot Theory->Prime Factorization and Factors"], "answer_analysis": ["A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let $$k$$ be the sum of the units digits in all the terms. Then $$S=111k=3\\times37k$$, so $$S$$ must be divisible by $$37\\rm(D)$$. To see that it need not be divisible by any larger prime, the sequence $$123$$, $$231$$, $$312$$ gives $$S=666=2\\cdot3^{2}\\cdot37\\Rightarrow\\boxed{\\rm(D)}$$. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "270", "queId": "3901bb261882495d9c93b6a6e4efd4c6", "competition_source_list": ["1984年全国高中数学联赛竞赛一试第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "若动点$$P\\left( x,y \\right)$$以等角速度$$\\omega $$在单位圆上逆时针运动，则点$$Q\\left( -2xy,{{y}^{2}}-{{x}^{2}} \\right)$$的运动方程是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "以角速度$$\\omega $$在单位圆上顺时针运动 "}], [{"aoVal": "B", "content": "以角速度$$\\omega $$在单位圆上逆时针运动 "}], [{"aoVal": "C", "content": "以角速度$$2\\omega $$在单位圆上顺时针运动 "}], [{"aoVal": "D", "content": "以角速度$$2\\omega $$在单位圆上逆时针运动 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->极坐标与参数方程（二试）"], "answer_analysis": ["将动点$$P\\left( x,y \\right)$$的坐标表为参数形式．  $$\\begin{cases}x=\\cos \\omega t    y=\\sin \\omega t \\end{cases}$$ $$(\\omega \\textgreater0)$$  则 $$-2xy=-\\sin 2\\omega t=\\cos \\left( -2\\omega t+\\frac{3}{2} \\pi ~\\right)$$  $${{y}^{2}}-{{x}^{2}}=-\\cos 2\\omega t=\\sin \\left( -2\\omega t+\\frac{2 \\pi }{3} \\right)$$．  则点$$Q\\left( -2xy,{{y}^{2}}-{{x}^{2}} \\right)$$运动轨迹的参数方程为  $$\\begin{cases}{x}'=\\cos \\left( -2\\omega t+\\frac{3}{2} \\pi ~\\right)    {y}'=\\sin \\left( -2\\omega t+\\frac{2 \\pi }{3} \\right) \\end{cases}$$  很显然$$\\omega t$$与$$-2\\omega t$$的旋转方向是相反的，$$P\\left( x,y \\right)$$在单位圆上逆时针运动，所以$$Q\\left( -2xy,{{y}^{2}}-{{x}^{2}} \\right)$$以角速度$$2\\omega $$在单位圆上顺时针运动． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "77", "queId": "afdbff8f14e04d94b14618bb6e8c4822", "competition_source_list": ["2011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": "设正数数列$$ {{{a}_{n}} }$$的前$$n$$项之和为$${{b}_{n}}$$，数列$$ {{{b}_{n}} }$$的前$$n$$项之积为$${{c}_{n}}$$，且$${{b}_{n}}+{{c}_{n}}=1$$，则数列$$\\left { \\frac{1}{{{a}_{n}}} \\right }$$中最接近$$2011$$的数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$1980$$ "}], [{"aoVal": "B", "content": "$$2010$$ "}], [{"aoVal": "C", "content": "$$2040$$ "}], [{"aoVal": "D", "content": "$$2070$$ "}]], "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->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["由已知$${{b}_{n}}=\\frac{{{c}_{n}}}{{{c}_{n-1}}}(n\\geqslant 2),  {{c}_{1}}={{b}_{1}}$$，所以$$\\frac{{{c}_{n}}}{{{c}_{n-1}}}+{{c}_{n}}=1,  {{c}_{1}}={{b}_{1}}=\\frac{1}{2}$$．所以$$\\frac{1}{{{c}_{n}}}-\\frac{1}{{{c}_{n-1}}}=1$$．所以$$\\left { \\frac{1}{{{c}_{n}}} \\right }$$是以$$2$$为首项，以$$1$$为公差的等差数列．  所以$$\\frac{1}{{{c}_{n}}}=2+(n-1)\\cdot 1=n+1,  {{c}_{n}}=\\frac{1}{n+1},  {{b}_{n}}=\\frac{n}{n+1}$$．  因此$${{a}_{n}}={{b}_{n}}-{{b}_{n-1}}=\\frac{n}{n+1}-\\frac{n-1}{n}=\\frac{1}{n(n+1)},\\frac{1}{{{a}_{n}}}=n(n+1)$$．  因为$$44\\times 45=1980,  45\\times 46=2070$$，所以数列$$\\left { \\frac{1}{{{a}_{n}}} \\right }$$中最接近$$2011$$的数是$$1980$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "861", "queId": "f2247ae29f7240e9bbfe8697da3592ce", "competition_source_list": ["2020年贵州全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "设$$a=\\ln 2$$，$$b=\\lg 3$$，$$c={{\\log }_{3}}2$$，则$$a$$，$$b$$，$$c$$的大小关系是．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater c\\textgreater b$$ "}], [{"aoVal": "B", "content": "$$a\\textgreater b\\textgreater c$$ "}], [{"aoVal": "C", "content": "$$b\\textgreater c\\textgreater a$$ "}], [{"aoVal": "D", "content": "$$c\\textgreater a\\textgreater b$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数", "课内体系->知识点->基本初等函数->对数函数->对数比大小", "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->对数的概念及其运算->换底公式及其变形运用->利用换底公式求值", "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->素养->数学运算"], "answer_analysis": ["因为$$c={{\\log }_{3}}2=\\frac{\\ln 2}{\\ln 3}\\textless{}\\ln 2=a$$，$$2b=\\lg 9\\textless{}1$$，$$2c={{\\log }_{3}}4\\textgreater1\\Rightarrow c\\textgreater b$$，所以$$a\\textgreater c\\textgreater b$$；  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "663", "queId": "faf0aa885f5f44578d06c352da559b88", "competition_source_list": ["2012年湖南全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "在四个函数$$y=\\sin \\left\\textbar{} x \\right\\textbar,  y=\\cos \\left\\textbar{} x \\right\\textbar,  y=\\left\\textbar{} \\tan x \\right\\textbar,  y=-\\ln \\left\\textbar{} \\sin x \\right\\textbar$$中，以$$ \\pi $$为周期、在$$\\left( 0,\\frac{ \\pi }{2} \\right)$$上单调递减且为偶函数的是", "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": "$$y=\\left\\textbar{} \\tan x \\right\\textbar$$ "}], [{"aoVal": "D", "content": "$$y=-\\ln \\left\\textbar{} \\sin x \\right\\textbar$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["由于$$y=\\sin \\left\\textbar{} x \\right\\textbar$$不是周期函数，故$$A$$错；由于$$y=\\cos \\left\\textbar{} x \\right\\textbar$$是最小正周期为$$2 \\pi $$的周期函数，故$$B$$错；由于$$y=\\left\\textbar{} \\tan x \\right\\textbar$$在$$\\left( 0,\\frac{ \\pi }{2} \\right)$$单调递增，故$$C$$错；函数$$y=-\\ln \\left\\textbar{} \\sin x \\right\\textbar$$符合题设要求． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "946", "queId": "7d53012136394e1897b0306efc0e5a80", "competition_source_list": ["竞赛第2题"], "difficulty": "2", "qtype": "single_choice", "problem": "在一次竞赛中有\\emph{A}，\\emph{B}，\\emph{C}三道题．\\\\ ①在所有参赛学生中共有30人至少解出一道题；\\\\ ②仅解出一题的学生中，解出\\emph{C}题的人数占一半；\\\\ ③解出\\emph{A}题的学生人数等于仅解出\\emph{B}题的学生人数；\\\\ ④仅解出\\emph{A}，\\emph{B}题的人数等于仅解出\\emph{B}，\\emph{C}题的人数；\\\\ ⑤仅解出\\emph{A}题的人数等于4；\\\\ ⑥仅解出\\emph{A}，\\emph{C}题的人数是仅解出\\emph{A}，\\emph{B}题的人数的一半．\\\\ 则同时解出\\emph{A}，\\emph{B}，\\emph{C}三题的学生人数为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "0 "}], [{"aoVal": "B", "content": "1 "}], [{"aoVal": "C", "content": "2 "}], [{"aoVal": "D", "content": "3 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 设只解出\\emph{A}，\\emph{B}，\\emph{C}的人数分别为\\emph{a}，\\emph{b}，\\emph{c}，仅解出$AB,BC,CA$的人数分别为\\emph{z}，\\emph{x}，\\emph{y}，同时解出\\emph{A}，\\emph{B}，\\emph{C}三题的人数为\\emph{w}，则可得关于诸参数的方程组，用$x$表示其他的量后可得其值，从而可求同时解出\\emph{A}，\\emph{B}，\\emph{C}三题的学生人数.\\\\ 【详解】\\\\ 设只解出\\emph{A}，\\emph{B}，\\emph{C}的人数分别为\\emph{a}，\\emph{b}，\\emph{c}，仅解出$AB,BC,CA$的人数分别为\\emph{z}，\\emph{x}，\\emph{y}，同时解出\\emph{A}，\\emph{B}，\\emph{C}三题的人数为\\emph{w}，则\\\\ $\\left {\\begin{array}{l} a+b+c+x+y+z+w=30,   c=a+b,   a+z+y+w=b,   z=x,   a=4,   2y=z, \\end{array}\\right.\\Rightarrow \\left {\\begin{array}{l} a=4,   b=\\frac{26-x}{3},   c=\\frac{38-x}{3},   y=\\frac{1}{2}x,   z=x,   w=\\frac{28-11x}{6}, \\end{array}\\right.$\\\\ 于是$x=2$，且$w=1$，因此同时解出\\emph{A}，\\emph{B}，\\emph{C}三题的学生人数为1．\\\\ 故选：B "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "388", "queId": "79c92517b42d4acda5297512a7640bf9", "competition_source_list": ["2013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": "设等差数列$$\\left { {{a}_{n}} \\right }$$与等比数列$$\\left { {{b}_{n}} \\right }$$满足：$$0\\textless{}{{a}_{1}}={{b}_{1}}\\textless{}{{a}_{5}}={{b}_{5}}$$，则下述四个结论：  $${}{{a}_{3}}\\textless{}{{b}_{3}}$$；~~~~~~~~~~ ②$${{a}_{3}}\\textgreater{{b}_{3}}$$；~~~~~~~~~ ③$${{a}_{6}}\\textgreater{{b}_{6}}$$；~~~~~~~~~~~~~ ④$${{a}_{6}}\\textless{}{{b}_{6}}$$  中正确的有．", "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->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["因为$${{a}_{3}}=\\frac{{{a}_{1}}+{{a}_{5}}}{2}=\\frac{{{b}_{1}}+{{b}_{5}}}{2}\\textgreater\\sqrt{{{b}_{1}}{{b}_{5}}}={{b}_{3}}$$，故②正确，从而①不正确；  令$${{a}_{n}}=1+20\\left( n-1 \\right)$$，若取$${{b}_{n}}={{3}^{n-1}}$$，可知$${{a}_{6}}\\textless{}{{b}_{6}}$$，从而③不正确；  若取$${{b}_{n}}={{\\left( -3 \\right)}^{n-1}}$$，可知$${{a}_{6}}\\textgreater{{b}_{6}}$$，从而④不正确. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "542", "queId": "44148daf0ced4331a48cc6d81894f43e", "competition_source_list": ["2008年四川全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$\\alpha \\in \\left( 0,\\frac{ \\pi }{2} \\right)$$，则$$\\frac{{{\\sin }^{3}}\\alpha }{\\cos \\alpha }+\\frac{{{\\cos }^{3}}\\alpha }{\\sin \\alpha }$$的最小值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{27}{64}$$ "}], [{"aoVal": "B", "content": "$$\\frac{3\\sqrt{2}}{5}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$\\frac{5\\sqrt{3}}{6}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["$$\\frac{{{\\sin }^{3}}\\alpha }{\\cos \\alpha }+\\frac{{{\\cos }^{3}}\\alpha }{\\sin \\alpha }=\\frac{{{\\sin }^{4}}\\alpha +{{\\cos }^{4}}\\alpha }{\\sin \\alpha \\cos \\alpha }$$  $$=\\frac{1-2{{(\\sin \\alpha \\cos \\alpha )}^{2}}}{\\sin \\alpha \\cos \\alpha }$$，  令$$t=\\sin \\alpha \\cos \\alpha =\\frac{1}{2}\\sin 2\\alpha \\in \\left( 0,\\frac{1}{2} \\right]$$，  则$$f(t)=\\frac{1-2{{t}^{2}}}{t}=\\frac{1}{t}-2t$$，  在$$t\\in \\left( 0,\\frac{1}{2} \\right]$$是单调递减的．  所以$$f{{(t)}_{\\min }}=f\\left( \\frac{1}{2} \\right)=1$$，  即$$\\frac{{{\\sin }^{3}}\\alpha }{\\cos \\alpha }+\\frac{{{\\cos }^{3}}\\alpha }{\\sin \\alpha }$$有最小值$$1$$．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1182", "queId": "f001d4615f4e4fc38c84d25ba078395e", "competition_source_list": ["2016年AMC12竞赛A第4题", "2016年AMC10竞赛A第7题"], "difficulty": "1", "qtype": "single_choice", "problem": "若 $$60$$, $$100$$, $$x$$, $$40$$, $$50$$, $$200$$, $$90$$ 這 $$7$$ 個數的平均数、中位数和众数都是$$x$$，則之値爲何？  The mean, median, and mode of the $$7$$ data values $$60$$, $$100$$, $$x$$, $$40$$, $$50$$, $$200$$, $$90$$ are all equal to $$x$$. What is the value of $$x$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$50$$ "}], [{"aoVal": "B", "content": "$$60$$ "}], [{"aoVal": "C", "content": "$$75$$ "}], [{"aoVal": "D", "content": "$$90$$ "}], [{"aoVal": "E", "content": "$$100$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["Since $$x$$ is the mean,  $$x=\\frac {60+100+x+40+50+200+90}7$$$$=\\frac {540+x}7$$.  Therefore, $$7x=540+x$$, so $$x=90$$.  Order the list: $$ {40, 50, 60, 90, 100, 200  }$$. $$x$$ must be either $$60$$ or $$90$$ because it is both the median and the mode of the set. Thus $$90$$ is correct. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1115", "queId": "f399ccdfa301442c8873e04374becbfa", "competition_source_list": ["2010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": "设$${{S}_{n}}$$是等差数列$$ {{{a}_{n}} }$$的前$$n$$项和，若$${{S}_{5}}={{S}_{9}}$$，则$${{a}_{3}}:{{a}_{5}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$9:5$$ "}], [{"aoVal": "B", "content": "$$5:9$$ "}], [{"aoVal": "C", "content": "$$3:5$$ "}], [{"aoVal": "D", "content": "$$5:3$$ "}]], "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": ["$${{S}_{5}}=\\frac{5({{a}_{1}}+{{a}_{5}})}{2}=5{{a}_{3}},  {{S}_{9}}=9{{a}_{5}}$$，由$$5{{a}_{3}}=9{{a}_{5}}$$得$${{a}_{3}}:{{a}_{5}}=9:5$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "134", "queId": "186c6d96383d4a34ae0d4be37566315f", "competition_source_list": ["2009年黑龙江全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "3", "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$$Q$$是命题$$P$$的．", "answer_option_list": [[{"aoVal": "A", "content": "充要条件 "}], [{"aoVal": "B", "content": "充分不必要条件 "}], [{"aoVal": "C", "content": "必要不充分条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->逻辑推理", "竞赛->知识点->逻辑->常用逻辑用语", "竞赛->知识点->不等式->不等式的证明", "竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["当两个不等式为$${{x}^{2}}+x+1\\textgreater0$$和$${{x}^{2}}+x+2\\textgreater0$$时，显然解集相同，但$$Q$$不成立；  反之，当$$Q$$成立，取$$\\frac{{{a}_{1}}}{{{a}_{2}}}=\\frac{{{b}_{1}}}{{{b}_{2}}}=\\frac{{{c}_{1}}}{{{c}_{2}}}=-1$$，显然两个不等式解集不同．  因此，$$Q$$既不是$$P$$的充分条件也不是必要条件． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "730", "queId": "95f05179102a49a7bf28957b0abc74c6", "competition_source_list": ["2012年山东全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$A,  B,  C,  D$$是以点$$O$$为球心的球面上四点，$$AB,  AC,  AD$$两两互相垂直，且$$AB=3\\text{cm}$$，$$AC=4\\text{cm}$$，$$AD=\\sqrt{11}\\text{cm}$$，则球的半径为．", "answer_option_list": [[{"aoVal": "A", "content": "$$3\\text{cm}$$ "}], [{"aoVal": "B", "content": "$$4\\text{cm}$$ "}], [{"aoVal": "C", "content": "$$5\\text{cm}$$ "}], [{"aoVal": "D", "content": "$$6\\text{cm}$$ "}]], "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": ["方法一 以点$$A$$为原点，分别以$$AB,  AC,  AD$$所在射线为坐标$$x$$轴、$$y$$轴和$$z$$轴，建立空间直角坐标系，则$$B,  C,  D$$的坐标分别为$$B\\left( 3,  0,  0 \\right)$$、$$C\\left( 0,  4,  0 \\right)$$，$$D\\left( 0,  0,\\sqrt{11} \\right)$$．显然球心应为下述三个平面  $$a:x=\\frac{3}{2}$$；$$\\beta :y=2$$；$$\\gamma :z=\\frac{\\sqrt{11}}{2}$$  的交点，所以球心坐标为$$O\\left( \\frac{3}{2},  2,\\frac{\\sqrt{11}}{2} \\right)$$，因此球的半径$$r=OA=3\\left( \\text{cm} \\right)$$．  方法二 以点$$A$$为原点，分别以$$AB,  AC,  AD$$所在射线为坐标$$x$$轴、$$y$$轴和$$z$$轴，建立空间直角坐标系，则$$B,  C,  D$$的坐标分别为$$B\\left( 3,  0,  0 \\right)$$、$$C\\left( 0,  4,  0 \\right)$$，$$D\\left( 0,  0,\\sqrt{11} \\right)$$．设球面方程为  $${{\\left( x-{{x}_{0}} \\right)}^{2}}+{{\\left( y-{{y}_{0}} \\right)}^{2}}+{{\\left( z-{{z}_{0}} \\right)}^{2}}={{r}^{2}}$$，  其中$$O\\left( {{x}_{0}},  {{y}_{0}},  {{z}_{0}} \\right)$$为坐标原点，$$r$$为球的半径．因为$$B$$，$$C$$，$$D$$均在球面上，得方程组  $$\\begin{cases}x_{0}^{2}+y_{0}^{2}+z_{0}^{2}={{r}^{2}},  {{\\left( {{x}_{0}}-3 \\right)}^{2}}+y_{0}^{2}+z_{0}^{2}={{r}^{2}}, \\x_{0}^{2}+{{\\left( {{y}_{0}}-4 \\right)}^{2}}+z_{0}^{2}={{r}^{2}}, \\x_{0}^{2}+y_{0}^{2}+{{\\left( {{z}_{0}}-\\sqrt{11} \\right)}^{2}}={{r}^{2}}. \\end{cases}$$解得$${{x}_{0}}=\\frac{3}{2},  {{y}_{0}}=2,  {{z}_{0}}=\\frac{\\sqrt{11}}{2}$$，$$r=3\\left( \\text{cm} \\right)$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "242", "queId": "f5a735ac28ed4253affb1ac67a294100", "competition_source_list": ["2013年AMC10竞赛A第17题"], "difficulty": "3", "qtype": "single_choice", "problem": "$$2013-AMC10A-17$$  Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next $$365-$$day period will exactly two friends visit her?  她的三个最好的朋友爱丽丝、碧翠丝和克莱尔定期拜访达芙妮。 爱丽丝每三天来一次，碧翠丝每四天来一次，克莱尔每五天来一次。 昨天三位朋友都拜访了达芙妮。 在接下来的 365天期间，正好有两个朋友来拜访她有多少天？", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$ "}], [{"aoVal": "B", "content": "$$54$$ "}], [{"aoVal": "C", "content": "$$60$$ "}], [{"aoVal": "D", "content": "$$66$$ "}], [{"aoVal": "E", "content": "$$72$$ "}]], "knowledge_point_routes": ["课内体系->知识点->集合->容斥原理", "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", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Simple Logical Reasoning"], "answer_analysis": ["The $$365-$$day time period can be split up into $$660-$$day time periods, because after $$60$$ days, all three of them visit again $$\\left({}\\right.$$ Least common multiple of $$3$$, $$4$$, and $$5$$ $$\\left.{}\\right)$$. You can find how many times each pair of visitors can meet by finding the LCM of their visiting days and dividing that number by $$60$$. Remember to subtract $$1$$, because you do not wish to count the $$60\\text{th}$$ day, when all three visit.  A and B visit $$\\dfrac{60}{3\\cdot4}-1=4$$ times. B and C visit $$\\dfrac{60}{4\\cdot5}-1=2$$ times. C and A visit $$\\dfrac{60}{3\\cdot5}-1=3$$ times.  This is a total of $$9$$ visits per $$60$$ day period. Therefore, the total number of $$2-$$person visits is $$9\\cdot6=\\left( \\text{B}\\right)54$$.  From the information above, we get that $$A=3x$$, $$B=4x$$, $$C=5x$$.  Now, we want the days in which exactly two of these people meet up The three pairs are $$\\left( A,B\\right)$$, $$\\left( B,C\\right)$$, $$\\left( A,C\\right)$$.  Notice that we are trying to find the LCM of each pair.  Hence, $$LCM\\left( A,B\\right)=12x$$, $$LCM\\left( B,C\\right)=20x$$, $$LCM\\left( A,C\\right)=15x$$.  Notice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.  Hence, $$LCM\\left(A,B,C\\right)=60x$$.  Now, we add all of the days up $$\\left({}\\right.$$ including overcount $$\\left.{}\\right)$$.  We get $$30+18+24=72$$. Now, because $$60\\left(6\\right)=360$$, we have to subtract $$6$$ days from every pair.  Hence, our answer is $$72-18=\\boxed ~{\\textasciitilde54\\textasciitilde}\\Rightarrow\\boxed ~{\\textasciitilde B\\textasciitilde}$$. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "93", "queId": "1389e6899ec2453f9518b0118e0a0af7", "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3题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知数列$$\\left { {{x}_{n}} \\right }$$满足$${{x}_{1}}=20$$，$${{x}_{n+1}}=10{{x}_{n}}-3\\times {{5}^{n+1}}$$，则$${{x}_{6}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$${{5}^{7}}$$ "}], [{"aoVal": "B", "content": "$${{6}^{7}}$$ "}], [{"aoVal": "C", "content": "$$6\\times {{5}^{7}}$$ "}], [{"aoVal": "D", "content": "$$7\\times {{5}^{5}}$$ "}]], "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->等比数列求通项问题"], "answer_analysis": ["由题得$${{x}_{n+1}}-3\\times {{5}^{n+1}}=10\\left( {{x}_{n}}-3\\times {{5}^{n}} \\right)$$，  所以数列 $$\\left { {{x}_{n}}-3\\times {{5}^{n}} \\right }$$ 是以$$5$$为首项，  以$$10$$为公比的等比数列，  所以$${{x}_{6}}-3\\times {{5}^{6}}=5\\times {{10}^{5}}$$，即$${{x}_{6}}=3\\times {{5}^{6}}+5\\times {{10}^{5}}=7\\times {{5}^{7}}$$．  故选：$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "285", "queId": "22d7096fc8f64dc4963ddf39bd6da551", "competition_source_list": ["2012年山东全国高中数学联赛竞赛初赛第9题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "某科室安排国庆节放假期间（共放假$$8$$天）甲、乙、丙、丁四人的值班表．已知甲、乙各值班四天，甲不能在第一天值班且甲、乙不在同一天值班；丙需要值班$$3$$天，且不能连续值班；丁需要值班$$5$$天；规定每天必须两人值班．问符合条件的不同的值班方案共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$400$$种 "}], [{"aoVal": "B", "content": "$$700$$种 "}], [{"aoVal": "C", "content": "$$840$$种 "}], [{"aoVal": "D", "content": "$$960$$种 "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["方法一 先考虑甲、乙．甲不在第一天值班，可在后面$$7$$天中任选四天，因此有$$\\text{C}_{7}^{4}$$种排法，余下的分派给乙为$$\\text{C}_{4}^{4}$$．  再考虑丙、丁．设丙的值班日分别安排在假期的第$${{n}_{1}}$$，第$${{n}_{2}}$$，第$${{n}_{3}}$$天，则由题意知$${{n}_{1}}$$、$${{n}_{2}}$$、$${{n}_{3}}$$互不相邻，且$$1\\leqslant {{n}_{1}}\\textless{}{{n}_{2}}\\textless{}{{n}_{3}}\\leqslant 8$$．  令$${{m}_{1}}={{n}_{1}}$$，$${{m}_{2}}={{n}_{2}}-1$$，$${{m}_{3}}={{n}_{3}}-2$$，则有$$1\\leqslant {{m}_{1}}\\textless{}{{m}_{2}}\\textless{}{{m}_{3}}\\leqslant 6$$．  显然$${{n}_{1}}$$、$${{n}_{2}}$$、$${{n}_{2}}$$互不相邻等价于$${{m}_{1}}$$、$${{m}_{2}}$$、$${{m}_{3}}$$互不相同，且$$\\left { {{n}_{1}},  {{n}_{2}},  {{n}_{3}} \\right }$$和$$\\left { {{m}_{1}},  {{m}_{2}},  {{m}_{3}} \\right }$$的对应是一个一一对应．所以丙的值班安排共有$$\\text{C}_{6}^{3}$$个不同方案．  丁的值班方案个数为$$\\text{C}_{5}^{5}$$．  由乘法原理知甲、乙、丙、丁的不同值班方案种数为$$\\text{C}_{7}^{3}\\cdot \\text{C}_{4}^{4}\\cdot \\text{C}_{6}^{3}\\cdot \\text{C}_{5}^{5}=700$$．  方法二 先考虑甲、乙．甲不在第一天值班，可在后面$$7$$天中任选四天，因此有$$\\text{C}_{7}^{4}$$种排法，余下的分派给乙为$$\\text{C}_{4}^{4}$$．  再考虑丙、丁．先排丙，如果不计丙的限制，共有$$\\text{C}_{8}^{3}$$种排法，如果丙连续三天值班，则有$$\\text{C}_{6}^{1}$$种排法，如果连续两天上班，且为前两天或者最后两天值班，则有$$\\text{C}_{2}^{1}\\text{C}_{5}^{1}$$种排法，如果是中间两天连续值班，则为$$\\text{C}_{5}^{1}\\text{C}_{4}^{1}$$种排法，因此不符合情况的排法共有$$\\text{C}_{6}^{1}+\\text{C}_{2}^{1}\\text{C}_{5}^{1}+\\text{C}_{5}^{1}\\text{C}_{4}^{1}=36$$种，因此丙丁的排法共有$$\\text{C}_{8}^{3}-\\left( \\text{C}_{6}^{1}+\\text{C}_{2}^{1}\\text{C}_{5}^{1}+\\text{C}_{5}^{1}\\text{C}_{4}^{1} \\right)=56-36=20$$种．  由乘法原理知甲、乙、丙、丁的不同值班方案种数为$$\\text{C}_{7}^{3}\\cdot \\text{C}_{4}^{4}\\cdot 20=700$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "337", "queId": "be2745a2a15647e8ae0ef6203f6370eb", "competition_source_list": ["2009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": "已知$$\\textbar\\overrightarrow{a}\\textbar$$、$$\\textbar\\overrightarrow{b}\\textbar$$都是整数，且满足$$(\\textbar\\overrightarrow{a}\\textbar+\\textbar\\overrightarrow{b}\\textbar)(\\textbar\\overrightarrow{a}\\textbar+3\\textbar\\overrightarrow{b}\\textbar)=105,(\\overrightarrow{a}+\\overrightarrow{b})(\\overrightarrow{a}+3\\overrightarrow{b})=33$$，则$$\\overrightarrow{a}$$和$$\\overrightarrow{b}$$的夹角为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$30{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$60{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$120{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$150{}^{}\\circ $$ "}]], "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"], "answer_analysis": ["令$$\\textbar\\overrightarrow{a}\\textbar+\\textbar\\overrightarrow{b}\\textbar=m$$，则  $$\\textbar m(m+2\\textbar\\overrightarrow{b}\\textbar)=105$$，  由$$\\textbar\\overrightarrow{b}\\textbar\\textless{}m\\leqslant 10$$，得  $$\\textbar\\overrightarrow{b}\\textbar=\\frac{105}{2m}-\\frac{m}{2}\\textless{}m$$，  解得$$m\\textgreater6$$．又$$105=3\\times 5\\times 7$$，所以$$m=7$$，从而有$$\\textbar\\overrightarrow{b}\\textbar=\\frac{105-49}{14}=4$$，$$\\textbar\\overrightarrow{a}\\textbar=7-4=3$$．  $$(\\overrightarrow{a}+\\overrightarrow{b})(\\overrightarrow{a}+3\\overrightarrow{b})=\\textbar\\overrightarrow{a}{{\\textbar}^{2}}+3\\textbar\\overrightarrow{b}{{\\textbar}^{2}}+4\\overrightarrow{a}\\cdot \\overrightarrow{b}$$  $$=\\textbar\\overrightarrow{a}{{\\textbar}^{2}}+3\\textbar\\overrightarrow{b}{{\\textbar}^{2}}+4\\textbar\\overrightarrow{a}\\textbar\\textbar\\overrightarrow{b}\\textbar\\cos \\alpha $$  $$=33$$  这里$$\\alpha $$为$$\\overrightarrow{a}$$与$$\\overrightarrow{b}$$的夹角，从而有  $$\\cos \\alpha =\\frac{1}{4\\textbar\\overrightarrow{a}\\textbar\\textbar\\overrightarrow{b}\\textbar}(33-\\textbar\\overrightarrow{a}{{\\textbar}^{2}}-3\\textbar\\overrightarrow{b}{{\\textbar}^{2}})=-\\frac{1}{2}$$．  又$$0{}^{}\\circ \\leqslant \\alpha \\textless{}180{}^{}\\circ $$，所以$$\\alpha =120{}^{}\\circ $$． ", "<p>令$$|\\overrightarrow{a}|+|\\overrightarrow{b}|=m,|\\overrightarrow{a}|+3|\\overrightarrow{b}|=n$$，则$$m&lt;{}n&lt;{}3m$$．</p>\n<p>$$105=1\\times 105=3\\times 35=5\\times 21=7\\times 15$$，</p>\n<p>只有$$7&lt;{}15&lt;{}3\\times 7$$．所以$$\\begin{cases}|\\overrightarrow{a}|+|\\overrightarrow{b}|=7 \\\\ |\\overrightarrow{a}|+3|\\overrightarrow{b}|=15 \\end{cases}$$，解得$$\\begin{cases}|\\overrightarrow{a}|=3 \\\\ |\\overrightarrow{b}|=4 \\end{cases}$$．</p>\n<p>下同方法$$1$$．故选$$\\text{C}$$．</p>"], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "672", "queId": "f1b315716817447493b936496a2fe33f", "competition_source_list": ["2010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$${{a}^{2}}+{{b}^{2}}=1$$，且$$c\\textless{}a+b$$恒成立，则$$c$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( -\\infty ,  -2 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( -\\infty ,  -\\sqrt{2} \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( -\\sqrt{2},\\sqrt{2} \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( -\\infty ,\\sqrt{2} \\right)$$ "}]], "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": ["令$$a=\\cos \\theta ,b=\\sin \\theta ,\\theta \\in \\mathbf{R}$$，  则$$a+b=\\cos \\theta +\\sin \\theta =\\sqrt{2}\\sin \\left( \\theta +\\frac{ \\pi }{4} \\right)\\geqslant -\\sqrt{2}$$，  所以$$c\\textless{}-\\sqrt{2}$$，故选$$\\text{B}$$. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "766", "queId": "c3dd2e9b75a44018955d772093bf6623", "competition_source_list": ["2018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": "已知$${{F}_{1}}$$、$${{F}_{2}}$$分别为椭圆$$\\frac{{{x}^{2}}}{{{a}^{2}}}+\\frac{{{y}^{2}}}{{{b}^{2}}}=1\\left( a\\textgreater b\\textgreater0 \\right)$$的左、右焦点，$$P$$为椭圆上一点，满足$$\\angle {{F}_{1}}P{{F}_{2}}=90{}^{}\\circ $$，若$$\\triangle P{{F}_{1}}{{F}_{2}}$$的面积为$$2$$，则$$b$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->椭圆"], "answer_analysis": ["设$$\\left\\textbar{} P{{F}_{1}} \\right\\textbar=m$$，$$\\left\\textbar{} P{{F}_{2}} \\right\\textbar=n$$．  则$$m+n=2a$$，  $${{m}^{2}}+{{n}^{2}}=4{{c}^{2}}=4\\left( {{a}^{2}}-{{b}^{2}} \\right)$$．  $$\\frac{1}{2}mn=2$$，  由以上三式得$${{b}^{2}}=2$$，即$$b=\\sqrt{2}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "805", "queId": "6e74ee1f48b24291821b77753718e3a5", "competition_source_list": ["2017年山东全国高中数学联赛竞赛初赛第7题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)=\\frac{\\sin x-1}{\\sqrt{3-2\\cos x-2\\sin x}}\\left( 0\\leqslant x\\leqslant 2 \\pi ~\\right)$$的值域为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$[-1,0]$$ "}], [{"aoVal": "B", "content": "$$[-1,1]$$ "}], [{"aoVal": "C", "content": "$$(-1,0]$$ "}], [{"aoVal": "D", "content": "$$(-1,0)$$ "}]], "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", "课内体系->知识点->三角函数->三角函数综合"], "answer_analysis": ["显然$$\\sqrt{3-2\\cos x-2\\sin x}=\\sqrt{3-2\\sqrt{2}\\sin \\left( x+\\frac{ \\pi }{4} \\right)}\\textgreater0$$，而$$\\sin x-1\\leqslant 0$$，  所以，$$f\\left( x \\right)\\leqslant 0$$．  又$${{\\left( \\sin x-1 \\right)}^{2}}=1+{{\\sin }^{2}}x-2\\sin x=2-{{\\cos }^{2}}x-2\\sin x\\leqslant 3-2\\cos x-2\\sin x$$，  所以，$$f\\left( x \\right)\\geqslant -1$$．  因此，$$f\\left( x \\right)$$的值域为$$\\left[ -1,0 \\right]$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "832", "queId": "d1d9d86ac8db4848868da7ead6ef0e33", "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5题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$x$、$y$为正实数，且$\\lg {{2}^{x}}+\\lg {{8}^{y}}=\\lg 4$，则$\\frac{1}{x}+\\frac{3}{y}$的最小值是（~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "4 "}], [{"aoVal": "B", "content": "8 "}], [{"aoVal": "C", "content": "12 "}], [{"aoVal": "D", "content": "16 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据对数的运算法则可得$x+3y=2$，再利用乘``1''法及基本不等式计算可得．\\\\ 【详解】\\\\ 解：$\\because x$、$y$为正实数，且$\\lg {{2}^{x}}+\\lg {{8}^{y}}=\\lg 4$，\\\\ $\\therefore \\lg ({{2}^{x}}\\times {{8}^{y}})=\\lg 4$，\\\\ $\\therefore {{2}^{x+3y}}={{2}^{2}}$，\\\\ $\\therefore x+3y=2$．\\\\ $\\therefore $$\\frac{1}{x}+\\frac{3}{y}=\\frac{1}{2}\\left( x+3y \\right)\\left( \\frac{1}{x}+\\frac{3}{y} \\right)=\\frac{1}{2}\\left( 10+\\frac{3y}{x}+\\frac{3x}{y} \\right)\\ge \\frac{1}{2}\\left( 10+2\\sqrt{\\frac{3y}{x}\\cdot \\frac{3x}{y}} \\right)=8$，\\\\ 当且仅当$\\frac{3y}{x}=\\frac{3x}{y}$，即$x=y=\\frac{1}{2}$时取等号．\\\\ $\\therefore $$\\frac{1}{x}+\\frac{3}{y}$的最小值是$8$．\\\\ 故选：B． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1141", "queId": "cb0c6e16be16407ead0051691dbc25a4", "competition_source_list": ["全国高中数学联赛竞赛模拟一试（十四）第7题"], "difficulty": "2", "qtype": "single_choice", "problem": "将$$4$$个$$1$$、$$4$$个$$2$$、$$4$$个$$3$$、$$4$$个$$4$$这$$16$$个数分别填入一个$$4\\times 4$$的数表中，则使得每行、每列均恰有两个奇数的填法的概率为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{144}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{143}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{169}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{121}$$ "}], [{"aoVal": "E", "content": "$$\\frac{1}{133}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理", "竞赛->知识点->组合->计数问题（二试）", "竞赛->知识点->排列组合与概率->排列与组合", "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->两个基本计数原理"], "answer_analysis": ["首先确定奇数的位置．第一行两个奇数有$$\\text{C}_{4}^{2}$$种选法再考虑这两个奇数所在的列，每列还需要再填一个奇数，设为$$a$$，$$b$$．  （$$1$$）若$$a$$，$$b$$位于同行．它们的位置有$$3$$种选择，此时，剩下的四个奇数所填的位置唯一确定；  （$$2$$）若$$a$$，$$b$$位于不同的行．它们的位置有$$6$$种选择，此时，剩下的四个奇数所填的位置有$$2$$种选择．从而，奇数的不同位置数为$$\\text{C}_{4}^{2}(3+6\\times 2)=90$$．  最后从八个奇数位置中选四个填$$1$$（其余四个填$$3$$)，从其余八个位置中选四个填$$2$$（其余四个填$$4$$)，故总填法数为$$90\\text{C}_{8}^{4}\\text{C}_{8}^{4}$$，  因此，使得每行、每列均恰有两个奇数的填法的概率为 $$\\frac{90\\text{C}_{8}^{4}\\text{C}_{8}^{4}}{\\text{C}_{16}^{2}\\text{C}_{12}^{4}\\text{C}_{8}^{4}\\text{C}_{4}^{4}}=\\frac{1}{143}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "681", "queId": "76a4a58954574b0dba626f38e017b80d", "competition_source_list": ["2018~2019学年北京西城区北京市第十五中学高三下学期开学考试文科第13题5分", "2017~2018学年1月重庆沙坪坝区重庆市第八中学高三上学期周测C卷理科第7题5分", "2013年高考真题新课标卷II理科第16题", "2013年贵州全国高中数学联赛竞赛初赛第5题8分", "2016~2017学年北京高一下学期单元测试《数列求和》第33题"], "difficulty": "2", "qtype": "single_choice", "problem": "等差数列$$\\left { {{a}_{n}} \\right }$$的前$$n$$项和为$${{S}_{n}}$$，已知$${{S}_{10}}=0$$，$${{S}_{15}}=25$$，则$$n{{S}_{n}}$$的最小值为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$-47$$ "}], [{"aoVal": "B", "content": "$$-48$$ "}], [{"aoVal": "C", "content": "$$-49$$ "}], [{"aoVal": "D", "content": "$$-50$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "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"], "answer_analysis": ["设等差数列$$\\left { {{a}_{n}} \\right }$$的首项为$${{a}_{1}}$$，公差为$$d$$，由等差数列的前$$n$$项和公式可得$$\\left {\\begin{array}{l}{10 a_{1}+\\dfrac{10 \\times 9}{2} d=0}    {15 a_{1}+\\dfrac{15 \\times 14}{2} d=25}\\end{array}\\right.$$，解得$$\\left {\\begin{array}{l}{a_{1}=-3}    {d=\\frac{2}{3}}\\end{array}\\right.$$，  ∴$$n S_{n}=n^{2} a_{1}+\\frac{n^{2}(n-1)}{2} d=-3 n^{2}+\\frac{1}{3}\\left(n^{3}-n^{2}\\right)=\\frac{1}{3} n^{3}-\\frac{10 n^{2}}{3}$$，  构造函数$$f(x)=\\frac{1}{3} x^{3}-\\frac{10}{3} x^{2}$$，则$$f^{\\prime}(x)=x^{2}-\\frac{20}{3} x$$，  令$$f^{\\prime}(x)=0$$，解得$$x=0$$或$$x=\\frac{20}{3}$$，  ∴当$$0\\textless x\\textless\\frac{20}{3}$$时，$$f(x)$$单调递减，  当$$x\\textgreater\\frac{20}{3}$$时，$$f(x)$$单调递增，  ∵$$n \\in \\mathbf{N}^{*}$$且当$$n=6$$时，$$n S_{n}=\\frac{1}{3} \\times 6^{3}-\\frac{10 \\times 6^{2}}{3}=-48$$，当$$n=7$$时，$$n S_{n}=\\frac{1}{3} \\times 7^{3}-\\frac{10 \\times 7^{2}}{3}=-49$$，  ∴当$$n=7$$时，$$n S_{n}$$取得最小值$$-49$$．  设等差数列$$\\left { {{a}_{n}} \\right }$$的首项为$${{a}_{1}}$$，公差为$$d$$，  ∵$${{S}_{10}}=10{{a}_{1}}+45d=0$$，$${{S}_{15}}=15{{a}_{1}}+105d=25$$，  ∴$${{a}_{1}}=-3$$，$$d=\\frac{2}{3}$$，  ∴$${{S}_{n}}=n{{a}_{1}}+\\frac{n\\left( n-1 \\right)}{2}d=\\frac{1}{3}{{n}^{2}}-\\frac{10}{3}n$$，  ∴$$n{{S}_{n}}=\\frac{1}{3}{{n}^{3}}-\\frac{10}{3}{{n}^{2}}$$，令$$n{{S}_{n}}=f\\left( n \\right)$$，  ∴$${{f}^{\\prime }}\\left( n \\right)={{n}^{2}}-\\frac{20}{3}n$$，  ∴当$$n=\\frac{20}{3}$$时，$$f\\left( n \\right)$$取得极值，当$$n\\textless{}\\frac{20}{3}$$ 时，$$f\\left( n \\right)$$递减，  当$$n\\textgreater\\frac{20}{3}$$时，$$f\\left( n \\right)$$递增，  因此只需比较$$f\\left( 6 \\right)$$和$$f\\left( 7 \\right)$$的大小即可，  $$f\\left( 6 \\right)=-48$$，$$f\\left( 7 \\right)=-49$$，  故$$n{{S}_{n}}$$的最小值为$$-49$$．  故答案为：$$-49$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "462", "queId": "ff2e0eeb3d2d43bb97f5d1c50a201c8e", "competition_source_list": ["2011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": "已知向量$$\\overrightarrow{m},  \\overrightarrow{n}$$的夹角为$$\\frac{ \\pi }{6}$$，且$$\\left\\textbar{} \\overrightarrow{m} \\right\\textbar=\\sqrt{3},\\left\\textbar{} \\overrightarrow{n} \\right\\textbar=2$$．在$$\\triangle ABC$$中$$\\overrightarrow{AB}=2\\overrightarrow{m}+2\\overrightarrow{n},\\overrightarrow{AC}=$$  $$2\\overrightarrow{m}-6\\overrightarrow{n}$$，$$D$$为$$BC$$边的中点，则$$\\left\\textbar{} \\overrightarrow{AD} \\right\\textbar=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "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": ["$$\\left\\textbar{} \\overrightarrow{AD} \\right\\textbar=\\left\\textbar{} \\frac{1}{2}\\left( \\overrightarrow{AB}+\\overrightarrow{AC} \\right) \\right\\textbar=\\left\\textbar{} 2\\overrightarrow{m}-2\\overrightarrow{n} \\right\\textbar$$  $$=\\sqrt{4{{\\overrightarrow{m}}^{2}}+4{{\\overrightarrow{n}}^{2}}-8\\overrightarrow{m}\\cdot \\overrightarrow{n}}=\\sqrt{4}=2$$. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "492", "queId": "abfbef1702274c4fbd444bc67af71373", "competition_source_list": ["2016年AMC10竞赛第13题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某電影院中有五個朋友坐在五個位置一起的座位上 ,這五個座位由左至右編號從$$1$$到5（此處的左和右是他們坐好時，從觀察者面對他們的觀點而言）．在電影播放时，甲去大厅买爆米花，當她回來時發現「乙」向右挪了兩個位子，「丙」向左移了一个位置;且「丁」與「戊」交換了座位，而將最旁邊的位子留给了甲，試問「甲」在去買爆米花前原來坐在幾號的位子上?  Five friends sat in a movie theater in a row containing o seats, nunbered I to 5 from left to right. (The directions \"left\" and \"right\" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had noved two seats to the right, Ceci had moved one seat to the left, and Dee and fdie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?", "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": [], "answer_analysis": ["Bash: we see that the following configuration works.  Bea-Ada-Ceci-Dee-Edie   After moving, it becomes  Ada-Ceci-Bea-Edie-Dee.   Thus, Ada was in seat $$2$$.  Process of elimination of possible configurations.  Let\\textquotesingle s say that Ada$$=A$$, Bea$$=B$$, Ceci$$=C$$, Dee$$=D$$, and Edie$$=D$$. Since $$B$$ moved more to the right than $$C$$ did left, this implies that $$B$$ was in a LEFT end seat originally: $$B$$, $$-$$, $$C\\rightarrow -$$, $$C$$, $$B$$.  This is affirmed because $$DE\\rightarrow ED$$, which there is no new seats uncovered. So $$A$$, $$B$$, $$C$$ are restricted to the same $$1$$, $$2$$, $$3$$ seats. Thus, it must be $$B$$, $$A$$, $$C\\rightarrow A$$, $$C$$, $$B$$, and more specifically: $$B,A,C,D,E\\rightarrow A,C,B,E,D$$~ So , $$A$$, Ada, was originally in seat $$2$$.  The seats are numbered $$1$$ through $$5$$, so let each letter $$(A,B,C,D,E)$$ correspond to a number. Let a move to the left be subtraction and a move to the right be addition.  We know that $$1+2+3+4+5=A+B+C+D+E=15$$. After everyone moves around, however, our equation looks like $$(A+x)+B+2+C-1+D+E=15$$ because $$D$$ and $$E$$ switched seats, $$B$$ moved two to the right, and $$C$$ moved $$1$$ to the left.  For this equation to be true, $$x$$ has to be$$-1$$, meaning $$A$$ moves $$1$$ left from her original seat. Since $$A$$ is now sitting in a corner seat, the only possible option for the original placement of $$A$$ is in seat number $$2$$. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "561", "queId": "3b7393f840644695abaf1b45ebaf9636", "competition_source_list": ["2017年湖南全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$的三个内角的余弦值分别等于$$\\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$的三个内角的正弦值，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$是锐角三角形，$$\\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$也是锐角三角形 "}], [{"aoVal": "B", "content": "$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$是钝角三角形，$$\\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$也是钝角三角形 "}], [{"aoVal": "C", "content": "$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$是锐角三角形，$$\\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$也是钝角三角形 "}], [{"aoVal": "D", "content": "$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$是钝角三角形，$$\\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$也是锐角三角形 "}]], "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$$\\cos {{A}_{1}},\\cos {{A}_{2}},\\cos {{A}_{3}}$$都大于$$0$$，所以$${{A}_{1}},{{A}_{2}},{{A}_{3}}$$都是锐角，$$\\triangle {{A}_{1}}{{B}_{1}}{{C}_{1}}$$是锐角三角形．  不妨设$$\\sin {{A}_{2}}=\\cos {{A}_{1}}$$，$$\\sin {{B}_{2}}=\\cos {{B}_{1}}$$，$$\\sin {{C}_{2}}=\\cos {{C}_{1}}$$，  则$${{A}_{2}}=90{}^{}\\circ -{{A}_{1}}$$或$${{A}_{2}}=90{}^{}\\circ +{{A}_{1}}$$，$${{B}_{2}}=90{}^{}\\circ -{{B}_{1}}$$或$${{B}_{2}}=90{}^{}\\circ +{{B}_{1}}$$，$${{C}_{2}}=90{}^{}\\circ -{{C}_{1}}$$或$${{C}_{2}}=90{}^{}\\circ +{{C}_{1}}$$．  若$${{A}_{2}}=90{}^{}\\circ -{{A}_{1}}$$，$${{B}_{2}}=90{}^{}\\circ -{{B}_{1}}$$，$${{C}_{2}}=90{}^{}\\circ -{{C}_{1}}$$，  则$${{A}_{2}}+{{B}_{2}}+{{C}_{2}}=90{}^{}\\circ $$，矛盾．  所以$$\\triangle {{A}_{2}}{{B}_{2}}{{C}_{2}}$$存在一个角为钝角． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "934", "queId": "d6dc8d20ab6440f38dd54dadf8a7ef8d", "competition_source_list": ["2008年甘肃全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$f(x)=a{{x}^{3}}+{{x}^{2}}+x+da,d\\in \\bf R$$，当$$\\textbar x\\textbar\\leqslant 1$$时，$$\\textbar f(x)\\textbar\\leqslant 1$$，则$$a$$、$$d$$一定属于．", "answer_option_list": [[{"aoVal": "A", "content": "$$[-2,0]$$ "}], [{"aoVal": "B", "content": "$$[0,2]$$ "}], [{"aoVal": "C", "content": "$$[-1,0]$$ "}], [{"aoVal": "D", "content": "$$[0,1]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["将$$x=1$$，$$-1$$分别代入$$f(x)$$可得  $$\\textbar a+1+1+d\\textbar\\leqslant 1$$和$$\\textbar-a+d\\textbar\\leqslant 1$$，  即$$-1\\leqslant a+1+1+d\\leqslant 1$$和$$-1\\leqslant -a+d\\leqslant 1$$．  则$$-3\\leqslant a+d\\leqslant -1$$①，  $$-1\\leqslant -a+d\\leqslant 1$$②，  把①和②两个不等式相加得$$-4\\leqslant 2d\\leqslant 0$$，  即$$-2\\leqslant d\\leqslant 0$$．类似可得$$-2\\leqslant a\\leqslant 0$$．故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "251", "queId": "2b43329222354ac995d6fcb1ac13a1c3", "competition_source_list": ["2008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": "三棱锥$$A-BCD$$的顶点$$A$$在底面$$BCD$$内的射影为点$$O$$，且点$$O$$到三个侧面的距离均相等，则点$$O$$一定是$$\\triangle BCD$$．", "answer_option_list": [[{"aoVal": "A", "content": "重心 "}], [{"aoVal": "B", "content": "垂心 "}], [{"aoVal": "C", "content": "内心 "}], [{"aoVal": "D", "content": "外心 "}]], "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": ["设$$OE\\bot $$面$$ABC$$于点$$E$$，$$OF\\bot $$面$$ACD$$于点$$F$$．  延长$$AE$$、$$AF$$分别交$$BC$$、$$CD$$于点$$G$$、$$H$$，  连接$$OG$$、$$OH$$．在$$\\text{Rt}\\triangle OEA$$和$$\\text{Rt}\\triangle OFA$$中，  $$OE=OF$$，$$OA=OA$$，故$$\\text{Rt}\\triangle OEA\\simeq \\text{Rt}\\triangle OFA$$，  可知$$\\angle OAE=\\angle OAF$$，于是$$\\text{Rt}\\triangle OAG\\simeq \\text{Rt}\\triangle OAH$$，$$OG=OH$$．  又因为$$OE\\bot BC$$，$$OF\\bot CD$$及$$AO\\bot $$面$$BCD$$，  因此$$BC\\bot $$面$$AOG$$，$$CD\\bot $$面$$AOH$$，  则$$BC\\bot OG$$，$$CD\\bot OH$$．  从而知$$O$$是$$\\triangle BCD$$的内心．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "510", "queId": "636a77cf07274031b31507a37f8a392b", "competition_source_list": ["2008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": "正项数列$$ {{{a}_{n}} }$$满足$$\\frac{1}{{{a}_{n}}{{a}_{n+1}}}+\\frac{1}{{{a}_{n}}{{a}_{n+2}}}+\\frac{1}{{{a}_{n+1}}{{a}_{n+2}}}=1(n\\in {{N}^{*}})$$，$${{a}_{1}}+{{a}_{3}}=6$$，$${{a}_{1}}$$、$${{a}_{2}}$$、$${{a}_{3}}$$单调递增且成等比数列，$${{S}_{n}}$$为$$ {{{a}_{n}} }$$的前$$n$$项和，则$$[{{S}_{2008}}]$$的值是($$[x]$$表示不超过实数$$x$$的最大整数)．", "answer_option_list": [[{"aoVal": "A", "content": "$$5352$$ "}], [{"aoVal": "B", "content": "$$5353$$ "}], [{"aoVal": "C", "content": "$$5357$$ "}], [{"aoVal": "D", "content": "$$5358$$ "}]], "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": ["由已知有$${{a}_{n}}+{{a}_{n+1}}+{{a}_{n+2}}={{a}_{n}}{{a}_{n+1}}{{a}_{n+2}}$$①  所以$${{a}_{1}}+{{a}_{2}}+{{a}_{3}}={{a}_{1}}{{a}_{2}}{{a}_{3}}$$，  而$${{a}_{1}}{{a}_{3}}=a_{2}^{2}$$，因此$$a_{2}^{3}={{a}_{2}}+6$$，  可知$$({{a}_{2}}-2)(a_{2}^{2}+2{{a}_{2}}+3)=0$$，  解得$${{a}_{2}}=2$$．又因为$${{a}_{1}}$$、$${{a}_{2}}$$、$${{a}_{3}}$$成等比数列，  所以$${{a}_{1}}+\\frac{4}{{{a}_{1}}}=6$$，解得$${{a}_{1}}=3\\pm \\sqrt{5}$$，  由$${{a}_{1}}$$、$${{a}_{2}}$$、$${{a}_{3}}$$单调递增知$${{a}_{1}}=3-\\sqrt{5}$$．  再由①式得$${{a}_{n+1}}+{{a}_{n+2}}+{{a}_{n+3}}={{a}_{n+1}}{{a}_{n+2}}{{a}_{n+3}}$$②  ②-①得$$({{a}_{n+3}}-{{a}_{n}})(1-{{a}_{n+1}}{{a}_{n+2}})=0$$，  因此$${{a}_{n}}={{a}_{n+3}}$$，故$$ {{{a}_{n}} }$$是以$$3$$为周期的数列可知$${{a}_{2008}}={{a}_{1}}=3-\\sqrt{5}$$，  因此$${{S}_{2008}}={{S}_{2007}}+{{a}_{2008}}$$  $$=\\frac{2007}{3}\\times (6+2)+(3-\\sqrt{5})$$  $$=5352+(3-\\sqrt{5})$$，  则$$[{{S}_{2008}}]=5352$$．故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "977", "queId": "a0a0f47a1fe24362be6d89d67f20d03d", "competition_source_list": ["2008年AMC10竞赛A第8题", "2008年AMC12竞赛A第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2008-AMC10A-8$$  Heather compares the price of a new computer at two different stores. Store $$A$$ offers $$15 \\%$$ off the sticker price followed by a ＄$$90$$ rebate, and store $$B$$ offers $$25 \\%$$ off the same sticker price with no rebate. Heather saves ＄$$15$$ by buying the computer at store $$A$$ instead of store $$B$$. What is the sticker price of the computer, in dollars?  希瑟比较了两家不同商店的新电脑的价格。 商店 $$A$$ 提供15\\%的标价折扣，然后是$$90$$美元返券，$$B$$商店提供相同标价 25\\%的折扣，没有返券。 Heather 通过在商店 $$A$$ 而不是商店 $$B$$ 购买计算机，会节省$$15$$美元。 这台电脑的标价是多少，以美元计？（", "answer_option_list": [[{"aoVal": "A", "content": "$$750$$ "}], [{"aoVal": "B", "content": "$$900$$ "}], [{"aoVal": "C", "content": "$$1000$$ "}], [{"aoVal": "D", "content": "$$1050$$ "}], [{"aoVal": "E", "content": "$$1500$$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->等式->等式的性质与方程的解集", "美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems"], "answer_analysis": ["Let the sticker price be $$x$$. The price of the computer is $$0.85x-90$$ at store $$A$$, and $$0.75x$$ at store $$B$$.  Heather saves ＄$$15$$ at store $$A$$, so $$0.85x-90+15=0.75x$$.  Solving, we find $$x=750$$, and the thus answer is $$(\\text{A})$$.  The ＄$$90$$ in store $$A$$ is ＄$$15$$ better than the additional $$10 \\%$$ off at store $$B$$.  Thus the $$10 \\%$$ off is equal to ＄$$90-$$＄$$15=$$＄$$75$$, and therefore the sticker price is ＄$$750$$. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "384", "queId": "284a5c235fef4a868df2123cd243dd67", "competition_source_list": ["2008年AMC10竞赛A第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2008-AMC10A-12$$  In a collection of red, blue, and green marbles, there are $$25 \\%$$ more red marbles than blue marbles, and there are $$60 \\%$$ more green marbles than red marbles. Suppose that there are $$r$$ red marbles. What is the total number of marbles in the collection?  在红色、蓝色和绿色弹珠的集合中，红色弹珠比蓝色弹珠多 $$25 \\%$$，绿色弹珠比红色弹珠多 $$60 \\%$$。 假设有 $$r$$ 个红色弹珠，则弹珠总数是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$2.85 r$$ "}], [{"aoVal": "B", "content": "$$3 r$$ "}], [{"aoVal": "C", "content": "$$3.4 r$$ "}], [{"aoVal": "D", "content": "$$3.85r$$ "}], [{"aoVal": "E", "content": "$$4.25r$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Application->Proportion Word Problems", "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": ["The number of blue marbles is $$\\dfrac{4}{5}r$$, the number of green marbles is $$\\dfrac{8}{5}r$$, and the number of red marbles is $$r$$. Thus, the total number of marbles is $$\\dfrac{4}{5}r+\\dfrac{8}{5}r+r=3.4r$$, and the answer is $$(\\text{C})$$. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "445", "queId": "5e6cbf552a734dcf9abe1357bb8c5ba1", "competition_source_list": ["2014年湖南全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "给定平面向量$$\\overrightarrow{a}=\\left( 1,  1 \\right)$$，那么，平面向量$$\\overrightarrow{b}=\\left( \\frac{1-\\sqrt{3}}{2},\\frac{1+\\sqrt{3}}{2} \\right)$$是向量$$\\overrightarrow{a}$$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->平面向量的概念与运算"], "answer_analysis": ["设两向量所成的角为$$\\theta $$，则  $$\\cos \\theta =\\frac{\\left( \\frac{1-\\sqrt{3}}{2},\\frac{1+\\sqrt{3}}{2} \\right)\\cdot \\left( 1,  1 \\right)}{2}=\\frac{1}{2}$$，  又$$\\theta \\in \\left[ 0{}^{}\\circ ,  180{}^{}\\circ \\right]$$，所以$$\\theta =60{}^{}\\circ $$．又  $$\\frac{1-\\sqrt{3}}{2}\\textless{}0$$，$$\\frac{1+\\sqrt{3}}{2}\\textgreater0$$，  所以$$\\text{C}$$正确． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "481", "queId": "2d89fe21674d4b368253d2827f0cd5c8", "competition_source_list": ["2008年甘肃全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "设$$A= {x\\textbar1\\leqslant x\\leqslant 9,x\\in Z }$$，$$B= {(a,b)\\textbar a,b\\in A }$$，定义$$B$$到$$Z$$的映射$$f$$：$$(a,b)\\to ab-a-b$$，则满足$$(ab)\\xrightarrow{f}11$$的有序数对共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$对 "}], [{"aoVal": "B", "content": "$$6$$对 "}], [{"aoVal": "C", "content": "$$8$$对 "}], [{"aoVal": "D", "content": "$$12$$对 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的概念"], "answer_analysis": ["因为$$f(a,b)=ab-a-b=11$$，  即$$(a-1)(b-1)=2\\times 6=3\\times 4=1\\times 12$$．  则$$\\begin{matrix}\\begin{cases}a-1=2    b-1=6 \\end{cases}\\begin{cases}a-1=6    b-1=2 \\end{cases}\\begin{cases}a-1=3    b-1=4 \\end{cases}    \\begin{cases}a-1=4    b-1=3 \\end{cases}\\begin{cases}a-1=1    b-1=12 \\end{cases}\\begin{cases}a-1=12    b-1=1. \\end{cases}   \\end{matrix}$$  解得符合题设条件的有序数对为$$\\left( 3,7 \\right)$$，$$\\left( 7,3 \\right)$$，$$\\left( 4,5 \\right)$$，$$\\left( 5,4 \\right)$$．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1018", "queId": "8f2a468128724e86b9990f77f72d5199", "competition_source_list": ["2012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": "如果$$\\triangle ABC$$中，$$A$$、$$B$$为锐角，且$${{\\sin }^{2}}A+{{\\sin }^{2}}B=\\sin C$$，则对$$\\triangle ABC$$的形状描述最准确的是．", "answer_option_list": [[{"aoVal": "A", "content": "直角三角形 "}], [{"aoVal": "B", "content": "等腰三角形 "}], [{"aoVal": "C", "content": "等腰直角三角形 "}], [{"aoVal": "D", "content": "以上均不对 "}]], "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": ["$${{\\sin }^{2}}A+{{\\sin }^{2}}B=\\sin C$$  $$\\Leftrightarrow {{\\sin }^{2}}A+{{\\sin }^{2}}B-\\sin A\\cos B-\\cos A\\sin B=0$$  $$\\Leftrightarrow \\sin A\\left( \\sin A-\\cos B \\right)+\\sin B\\left( \\sin B-\\cos A \\right)=0$$  于是当$$A\\textgreater\\frac{ \\pi }{2}-B$$或$$A\\textless{}\\frac{ \\pi }{2}-B$$时，上式均不成立．  因此$$A+B=\\frac{ \\pi }{2}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "609", "queId": "37ce037ea4c74522954ee86ca9402444", "competition_source_list": ["2004年AMC10竞赛A第12题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2004-AMC10A-12$$  Henry\\textquotesingle s Hamburger Heaven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two, or three meat patties, and any collection of condiments. How many different kinds of hamburgers can be ordered?  Henry\\textquotesingle s Hamburger Heaven 提供带有以下调味品的汉堡包：番茄酱、芥末、蛋黄酱、番茄、生菜、泡菜、奶酪和洋葱。 顾客可以选择一个、两个或三个肉饼，以及任何调味品系列。 可以订购多少种不同的汉堡包？", "answer_option_list": [[{"aoVal": "A", "content": "$$24$$ "}], [{"aoVal": "B", "content": "$$256$$ "}], [{"aoVal": "C", "content": "$$768$$ "}], [{"aoVal": "D", "content": "$$40$$,$$320$$ "}], [{"aoVal": "E", "content": "$$120$$,$$960$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Counting->Classification Counting and Step Counting", "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": ["For each condiment, a customer may either choose to order it or not. There are $$8$$ total condiments to choose from. Therefore, there are $$2^{8}=256$$ ways to order the condiments. There are also $$3$$ choices for the meat, making a total of $$256\\times3=768$$ possible hamburgers.$$\\boxed {(\\text{C}) 768}$$. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1155", "queId": "eaf91d3211ad49219a639379602b64df", "competition_source_list": ["2013年辽宁全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知实数$$x,y$$满足$$17{}({{x}^{2}}+{{y}^{2}}){}-30xy-16=0$$，则$$\\sqrt{16{{x}^{2}}+4{{y}^{2}}-16xy-12x+6y+9}$$的最大值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$${}\\sqrt{29}$$ "}], [{"aoVal": "C", "content": "$${}\\sqrt{19}$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "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->解析几何->圆与方程", "竞赛->知识点->不等式->换元技巧->三角换元"], "answer_analysis": ["$$17\\left( {{x}^{2}}+{{y}^{2}} \\right){-}30xy-16=0$$，  得$${{\\left( x+y \\right)}^{2}}+16{{\\left( x-y \\right)}^{2}}=16$$  $$16={{\\left( x+y \\right)}^{2}}+16{{\\left( x-y \\right)}^{2}}$$$\\geqslant\\dfrac{1}{25}\\left( 12\\left\\textbar{} x-y\\right\\textbar+4\\left\\textbar{} x+y\\right\\textbar\\right)^{2}\\geqslant\\dfrac{16}{25}\\left( 4x-2y\\right)^{2}$  则$-5\\leqslant4x-2y\\leqslant5$  $$f\\left( x,y \\right)=\\sqrt{16{{x}^{2}}+4{{y}^{2}}-16xy-12x+6y+9}$$  $$=\\sqrt{{{\\left( 4x-2y \\right)}^{2}}-3\\left( 4x-2y \\right)+9}$$  的最小值为7 "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "705", "queId": "64cef737172e4761ab485088e5ae5791", "competition_source_list": ["2013年浙江全国高中数学联赛竞赛初赛第9题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "设函数$$f\\left( x \\right)=x{{\\left( x-1 \\right)}^{2}}{{\\left( x-2 \\right)}^{3}}{{\\left( x-3 \\right)}^{4}}$$，则函数$$y=f\\left( x \\right)$$的极大值点为．", "answer_option_list": [[{"aoVal": "A", "content": "$$x=0$$ "}], [{"aoVal": "B", "content": "$$x=1$$ "}], [{"aoVal": "C", "content": "$$x=2$$ "}], [{"aoVal": "D", "content": "$$x=3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->导数"], "answer_analysis": ["由图象可知$$x=1$$为函数极大值点，$$x=3$$是极小值点，$$x=0{,}2$$不是极值点． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "519", "queId": "b0abaeb3143b4998a66e13b676199760", "competition_source_list": ["1991年全国高中数学联赛竞赛一试第1题"], "difficulty": "0", "qtype": "single_choice", "problem": "由一个正方体的三个顶点所能构成的正三角形的个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$8$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$24$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->两个基本计数原理"], "answer_analysis": ["在公有一个项点的三个两两互相垂直的面上，各有一对角线共同组成唯一的正三角形，这样在正三角形和顶点之间建立了一一对应，而正方体共有$$8$$个顶点，所以恰好有$$8$$个正三角形． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1085", "queId": "f7ee9c2e143043a4afacaca177e11bc8", "competition_source_list": ["竞赛"], "difficulty": "1", "qtype": "single_choice", "problem": "使得$$5x+12\\sqrt{xy}\\leqslant a\\left( x+y \\right)$$对所有正实数$$x$$，$$y$$都成立的实数$$a$$的最小值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "前三个答案都不对 "}]], "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", "竞赛->知识点->不等式->几个重要的不等式->均值"], "answer_analysis": ["$$\\because x$$，$$y$$均大于$$0$$，  $$\\therefore 5x+12\\sqrt{xy}\\leqslant a\\left( x+y \\right)\\Rightarrow 5+12\\sqrt{\\frac{y}{x}}\\leqslant a\\left( 1+\\frac{y}{x} \\right)$$，  令$$t=\\sqrt{\\frac{y}{x}}$$，$$t\\in {{\\mathbf{R}}^{*}}$$，  则$$a\\geqslant \\frac{5+12t}{1+{{t}^{2}}}$$，  要使该式对所有$$t$$都成立，则$$a\\geqslant {{\\left( \\frac{5+12t}{1+{{t}^{2}}} \\right)}_{\\max }}$$，  令$$m=5+12t$$，  则$$\\frac{5+12t}{1+{{t}^{2}}} =\\frac{144m}{{{m}^{2}}-10m+169}=\\frac{144}{m+\\frac{169}{m}-10}\\leqslant \\frac{144}{2\\times 13-10}=9$$，当且仅当$$m=13$$时，取等号，  故$${{a}_{\\min }}=9$$，故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "352", "queId": "7076d1295832466fb65a37ce3766a21c", "competition_source_list": ["1985年全国高中数学联赛竞赛一试第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$0\\textless{}a\\textless{}1$$，若$${{x}_{1}}=a,{{x}_{2}}={{a}^{{{x}_{1}}}},{{x}_{3}}={{a}^{{{x}_{2}}}},\\cdots ,{{x}_{n}}={{a}^{{{x}_{n-1}}}},\\cdots $$，则数列$$\\left { {{x}_{n}} \\right }$$．", "answer_option_list": [[{"aoVal": "A", "content": "是递增的 "}], [{"aoVal": "B", "content": "是递减的 "}], [{"aoVal": "C", "content": "奇数项是非递减，偶数项是递减的 "}], [{"aoVal": "D", "content": "偶数项是递增的，奇数项是递减的． "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的综合应用", "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": ["首先考虑数列的前三项，  $${{x}_{1}}=a{}{{x}_{2}}={{a}^{a}}{}{{x}_{3}}={{a}^{{{a}^{a}}}}$$  ∵$$0\\textless{}a\\textless{}1$$，  ∴$$a\\textless{}{{a}^{a}}\\textless{}1$$  $$a\\textless{}{{a}^{{{a}^{a}}}}\\textless{}{{a}^{a}}$$．  即 $${{x}_{1}}\\textless{}{{x}_{3}}\\textless{}{{x}_{2}}$$．  此已表明数列既非递增又非递减，且奇数项也不是递减的． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1005", "queId": "97e5c7313f9749bdb3964af8b920531a", "competition_source_list": ["2010年河北全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "从满足$${{a}_{1}}={{a}_{2}}=1$$，$${{a}_{n+2}}={{a}_{n+1}}+{{a}_{n}}(n\\geqslant 1)$$的数列$$ {{{a}_{n}} }$$中，依次抽出能被$$3$$整除的项组成数列$$ {{{b}_{n}} }$$，则$${{b}_{100}}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$${{a}_{100}}$$ "}], [{"aoVal": "B", "content": "$${{a}_{200}}$$ "}], [{"aoVal": "C", "content": "$${{a}_{300}}$$ "}], [{"aoVal": "D", "content": "$${{a}_{400}}$$ "}]], "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": ["易知$${{a}_{4k}}(k\\geqslant 1)$$能被$$3$$整除，故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "986", "queId": "97b52d7d58a240f88a9dc0980f2eed79", "competition_source_list": ["2016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": "将$$2$$，$$3$$，$$4$$，$$6$$，$$8$$，$$9$$，$$12$$，$$15$$共$$8$$个数排成一行，使得任意相邻两个数的最大公约数均大于$$1$$，则所有可能的排法共有（~ ~ ）种．", "answer_option_list": [[{"aoVal": "A", "content": "$$720$$ "}], [{"aoVal": "B", "content": "$$1014$$ "}], [{"aoVal": "C", "content": "$$576$$ "}], [{"aoVal": "D", "content": "$$1296$$ "}]], "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->两个基本计数原理->加法原理与乘法原理的综合运用", "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->计数原理->两个基本计数原理->分步乘法计数原理", "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->排列与组合->排列->特殊元素优先法"], "answer_analysis": ["这$$8$$个数的两两之间的公约数的质因子只有$$2$$和$$3$$，只含因子$$2$$的有$$3$$个数：$$2$$，$$4$$，$$8$$，只含因子$$3$$的有$$3$$个数：$$3$$，$$9$$，$$15$$，$$2$$和$$3$$因子都含的有$$2$$个数：$$6$$，$$12$$．记$$A=\\left { 2,4,8 \\right }$$，$$B=\\left { 3,9,15 \\right }$$，$$C=\\left { 6,12 \\right }$$．  $$A$$中的数和$$B$$中的数显然不能相邻，$$C$$中的数可以任意放置，于是$$A$$中的$$3$$个数必须连续排在一起或$$B$$中的$$3$$个数连续排在一起．  $$A$$或$$B$$中的数只有$$1$$个是连续排在一起时，首先$$C$$中的$$2$$个数全排，有$$2$$种方法．这$$2$$个数形成$$3$$个空隙，从$$3$$个空中挑一个放$$A$$（或$$B$$）中的$$3$$个数，有$$\\mathrm{C}_{3}^{1}\\mathrm{C}_{2}^{1}\\mathrm{A}_{3}^{3}$$种．然后剩下的两个空隙放剩下的$$3$$个数，且这$$3$$个数不连续排在一起，有$$\\mathrm{2A}_{3}^{3}$$种方法．$$2\\mathrm{C}_{3}^{1}\\mathrm{C}_{2}^{1}\\mathrm{A}_{3}^{3}\\cdot \\mathrm{2A}_{3}^{3}=864$$．  $$A$$和$$B$$中的数都连续排在一起，类似的，有$$2\\cdot \\mathrm{A}_{3}^{2}\\cdot \\mathrm{A}_{3}^{3}\\cdot \\mathrm{A}_{3}^{3}=432$$种．  综上，所有可能的排法共有$$864+432=1296$$种．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "302", "queId": "54a7e5802c1546639598e1cdda5e31fa", "competition_source_list": ["2009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": "平面上满足约束条件$$\\begin{cases}x\\geqslant 2    x+y\\leqslant 0    x-y-10\\leqslant 0 \\end{cases}$$的点$$(x,y)$$形成的区域为$$D$$，区域$$D$$关于直线$$y=2x$$对称的区域为$$E$$，则区域$$D$$和区域$$E$$中距离最近的两点的距离为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{6\\sqrt{5}}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{12\\sqrt{5}}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{8\\sqrt{3}}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{16\\sqrt{3}}{5}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与方程", "竞赛->知识点->不等式->线性规划"], "answer_analysis": ["三条直线的交点为$$A(2,-2),B(2,-8),C(5,-5)$$，区域$$D$$为$$\\triangle ABC$$，点$$A$$距直线$$y=2x$$最近，距离为$$\\frac{\\left\\textbar{} 2\\times 2-1\\times (-2) \\right\\textbar}{\\sqrt{5}}=\\frac{6\\sqrt{5}}{5}$$．所以，区域$$D$$和区域$$E$$中距离最近的两点的距离为$$\\frac{12\\sqrt{5}}{5}$$．故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "74", "queId": "98a04a0c92104c05ac138422732707f0", "competition_source_list": ["2017年上海全国高中数学联赛竞赛初赛第6题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "设集合$$A=\\left { {{a}_{1}},{{a}_{2}},{{a}_{3}} \\right }$$是集合$$\\left { 1,2,\\cdots ,16 \\right }$$的子集，满足$${{a}_{1}}+7\\leqslant {{a}_{2}}+4\\leqslant {{a}_{3}}$$，则这样的子集$$A$$共有~\\uline{~~~~~~~~~~}~个．", "answer_option_list": [[{"aoVal": "A", "content": "$$86$$ "}], [{"aoVal": "B", "content": "$$102$$ "}], [{"aoVal": "C", "content": "$$144$$ "}], [{"aoVal": "D", "content": "$$165$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合", "竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["因为$$8\\leqslant {{a}_{1}}+7\\leqslant {{a}_{2}}+4\\leqslant {{a}_{3}}\\leqslant 16$$，所以$$8\\leqslant {{a}_{1}}+7\\textless{}{{a}_{2}}+5\\textless{}{{a}_{3}}+2\\leqslant 18$$．  问题等价于从$$89\\cdots 18$$中取$$3$$个互不相同的数，有$$\\text{C}_{11}^{3}=165$$个． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "865", "queId": "ad635de4acb9434592afc40ac1e58f74", "competition_source_list": ["2010年湖南全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "关于非零向量$$a$$和$$b$$有两个命题．命题甲：$$a\\bot b$$；命题乙：函数$$f(x)=(xa+b)\\cdot (xb-a)$$为一次函数．则有．", "answer_option_list": [[{"aoVal": "A", "content": "甲是乙的充分不必要条件 "}], [{"aoVal": "B", "content": "甲是乙的必要不充分条件 "}], [{"aoVal": "C", "content": "甲是乙的充分必要条件 "}], [{"aoVal": "D", "content": "甲是乙的既不充分也不必要条件 "}]], "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->充要条件与向量结合", "课内体系->知识点->常用逻辑用语->命题->命题的概念", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的表示方法->解析法", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的表示方法->求函数的解析式->用待定系数法求解析式", "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->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->平面向量基本定理及其意义", "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->平面向量数量积运算（非坐标）", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理"], "answer_analysis": ["$$f(x)=a\\cdot b{{x}^{2}}+$$（$$b^{2}-a^{2}$$）$$x-a\\cdot b$$，$$a\\bot b\\Leftrightarrow a\\cdot b=0$$.$$f(x)$$为一次函数$$\\Rightarrow a\\cdot b=0$$，而$$a·b=0$$时$$f(x)$$有可能是常数函数，不一定为一次函数，故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "159", "queId": "105c85068fdc49f7b485883d686d2489", "competition_source_list": ["2016年天津全国高中数学联赛竞赛初赛第5题6分", "2016年高考真题天津卷"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a$$是实数，要使得对任何$$x\\in \\left( -\\infty ,1 \\right)\\cup \\left( 5,+\\infty \\right)$$，都有$${{x}^{2}}-2\\left( a-2 \\right)x+a\\textgreater0$$，所有满足上述要求的$$a$$组成的集合是（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( -\\infty ,5 \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left( 1,4 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( 1,7 \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left( 1,5 \\right]$$ "}]], "knowledge_point_routes": ["课内体系->知识点->集合->集合的概念与表示方法->集合的表示方法->区间表示法", "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", "课内体系->素养->数学运算", "课内体系->思想->分类讨论思想"], "answer_analysis": ["设$$f\\left( x \\right)={{x}^{2}}-2\\left( a-2 \\right)x+a$$，则$$\\Delta =4{{\\left( a-2 \\right)}^{2}}-4a=4\\left( {{a}^{2}}-5a+4 \\right)$$，  ①$$\\Delta \\textless{}0$$，解得$$1\\textless{}a\\textless{}4$$，满足要求；  ②$$\\Delta \\geqslant 0$$，则$$f\\left( 1 \\right)\\geqslant 0$$，$$f\\left( 5 \\right)\\geqslant 0$$，对称轴$$1\\textless{}a-2\\textless{}5$$，解得$$4\\leqslant a\\leqslant 5$$．  所以，$$a$$的取值范围是$$1\\textless{}a\\leqslant 5$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "971", "queId": "a514eb4ee33e474db26f7e721c2fbf83", "competition_source_list": ["2016年陕西全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知集合$$M=\\left { 1,2,3,\\cdots ,10 \\right }$$，$$A$$是$$M$$的子集，且$$A$$中各元素的和为$$8$$，则满足条件的子集$$A$$共有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$个 "}], [{"aoVal": "B", "content": "$$7$$个 "}], [{"aoVal": "C", "content": "$$6$$个 "}], [{"aoVal": "D", "content": "$$5$$个 "}]], "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", "课内体系->知识点->集合->集合的基本关系->子集个数的计算", "课内体系->知识点->集合->集合的基本关系->子集"], "answer_analysis": ["$$8=1+7=2+6=3+5=1+2+5=1+3+4$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "514", "queId": "325eaf3347754082a074adb2fd3341ab", "competition_source_list": ["高一上学期单元测试《二次与对勾函数》竞赛第22题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知当$$1\\leqslant x\\leqslant 5$$时，关于$$x$$的不等式$$\\textbar x{}^{2}+px+q\\textbar\\leqslant 2$$恒成立，则$$p+q=$$~\\uline{~~~~~~~~~~}~．", "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->二次函数", "竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["$f\\left( x\\right)=x^{2}+px+q$，则①$$-2\\leqslant f\\left( 1 \\right)=p+q+1\\leqslant 2$$，②$$-2\\leqslant f\\left( 5 \\right)=5p+q+25\\leqslant 2$$，进而由①+②得$$-2\\leqslant 3p+q+13\\leqslant 2$$，另一方面，有③$$-2\\leqslant f\\left( 3 \\right)=3p+q+9\\leqslant 2$$，因而必有$$3p+q+9=-2$$，且$$p+q+1=2$$，及$$5p+q+25=2$$，解出$$p=-6，q=7$$，故$$p+q=1$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "110", "queId": "0bcaa25b4edf4a8ea55f85dc51b22b3d", "competition_source_list": ["2009年第二十届全国希望杯高二竞赛复赛第3题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "若函数$$f(x)=\\sqrt{3}\\sin (x-\\theta )-\\cos (x-\\theta )$$的图像关于直线$$x=\\frac{ \\pi }{3}$$对称，则$$\\theta $$的值不能是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{4}{3} \\pi $$ "}], [{"aoVal": "B", "content": "$$-\\frac{ \\pi }{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{3} \\pi $$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{3} \\pi $$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["$$f(x)=2\\sin \\left( x-\\theta -\\frac{ \\pi }{6} \\right)$$ "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "843", "queId": "ff8080814b014992014b14c587e31706", "competition_source_list": ["2008年贵州全国高中数学联赛竞赛初赛第4题5分", "高考真题"], "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": "822", "queId": "89a44800bebb438a8c74a92a24f7c8f4", "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": "$$4$$ "}]], "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": "717", "queId": "a398c2f4338e482b86e4d33d1ff96961", "competition_source_list": ["2009年竞赛珠海市第10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "等边$$\\triangle ABC$$的边长为$$2\\sqrt{2}$$，$$AD$$是$$BC$$边上的高，将$$\\triangle ABD$$沿$$AD$$折起，使之与$$\\triangle ACD$$所在平面成$$120{}^{}\\circ $$的二面角，这时$$A$$点到$$BC$$的距离是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{26}}{2}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{13}$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$2\\sqrt{5}$$ "}]], "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": ["注意到折叠后，$$\\angle BDC$$即为二面角的平面角，从而$$D$$到$$BC$$的距离为$$\\frac{\\sqrt{2}}{2}$$，又$$AD=\\sqrt{6}$$，从而$$A$$到$$BC$$的距离是$$\\sqrt{{{\\left( \\frac{\\sqrt{2}}{2} \\right)}^{2}}+{{(\\sqrt{6})}^{2}}}=\\frac{\\sqrt{26}}{2}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "534", "queId": "329a67161a5248fb938af98a02ef5e9a", "competition_source_list": ["2016~2017学年天津高二上学期期中理科六校联考第7题", "2012年湖南全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若直线$$ax-by+1=0$$过圆$$C:{{x}^{2}}+{{y}^{2}}+2x-4y+1=0$$的圆心，则$$ab$$的取值范围是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( -\\infty ,\\frac{1}{4} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left( -\\infty ,\\frac{1}{8} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left( 0,\\frac{1}{4} \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left( 0,\\frac{1}{8} \\right]$$ "}]], "knowledge_point_routes": ["课内体系->知识点->直线和圆的方程->圆与方程->圆的标准方程与一般方程", "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", "课内体系->素养->数学运算", "课内体系->方法->均值不等式法"], "answer_analysis": ["因为直线$$ax-by+1=0$$过圆$$C:{{\\left( x+1 \\right)}^{2}}+{{\\left( y-2 \\right)}^{2}}=4$$的圆心$$\\left( -1,2 \\right)$$，所以$$-a-2b+1=0$$，即$$a+2b=1$$；当$$a\\textgreater0$$，$$b\\textgreater0$$时，$$1=a+2b\\geqslant 2\\sqrt{2ab}$$，解得$$0\\textless{}ab\\leqslant \\frac{1}{8}$$ (当且仅当$$a=2b$$时等号成立)；当$$a=1$$，$$b=0$$时，$$ab=0$$；当$$a=-1,b=1$$时，$$ab\\textless{}0$$；所以$$ab$$的取值范围是$$\\left( -\\infty ,\\frac{1}{8} \\right]$$．  选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "169", "queId": "1486890f75a6432a80179a1b0dbaa668", "competition_source_list": ["2014年天津全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "在平面直角坐标系中，方程$${{x}^{2}}+2x\\sin \\left( xy \\right)+1=0$$所表示的图形是．", "answer_option_list": [[{"aoVal": "A", "content": "直线 "}], [{"aoVal": "B", "content": "抛物线 "}], [{"aoVal": "C", "content": "一个点 "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->曲线与方程"], "answer_analysis": ["易知$$x=-1$$，$$\\sin \\left( xy \\right)=1$$，或$$x=1$$，$$\\sin \\left( xy \\right)=-1$$．从而该方程所表示的图形由点列$$\\left( -1,  2k \\pi -\\frac{ \\pi }{2} \\right)$$，$$k\\in \\mathbf{Z}$$和点列$$\\left( 1,  2k \\pi -\\frac{ \\pi }{2} \\right)$$，$$k\\in \\mathbf{Z}$$构成． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "3", "queId": "09511298acc64f4e8c1ed122e4ec5df7", "competition_source_list": ["2014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": "已知集合$${{S}_{1}}=\\left { \\left( x,  y \\right)\\left\\textbar{} {{\\log }_{2}}\\left( 1+{{x}^{2}}+{{y}^{2}} \\right)\\leqslant 1+{{\\log }_{2}}\\left( x+y \\right) \\right. \\right }$$，并且集合  $${{S}_{2}}=\\left { \\left( x,  y \\right)\\left\\textbar{} {{\\log }_{\\frac{1}{2}}}\\left( 2+{{x}^{2}}+{{y}^{2}} \\right)\\geqslant -2+{{\\log }_{\\frac{1}{2}}}\\left( x+y \\right) \\right. \\right }$$，则$${{S}_{2}}$$与$${{S}_{1}}$$的面积比为．", "answer_option_list": [[{"aoVal": "A", "content": "$$2:1$$ "}], [{"aoVal": "B", "content": "$$4:1$$ "}], [{"aoVal": "C", "content": "$$6:1$$ "}], [{"aoVal": "D", "content": "$$8:1$$ "}]], "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", "竞赛->知识点->解析几何->圆与方程"], "answer_analysis": ["$${{S}_{1}}$$表示圆内部（包括圆周）：$${{\\left( x-1 \\right)}^{2}}+{{\\left( y-1 \\right)}^{2}}\\leqslant 1$$，面积为$$ \\pi $$．$${{S}_{2}}$$表示圆内部（包括圆周）：$${{\\left( x-2 \\right)}^{2}}+{{\\left( y-1 \\right)}^{2}}\\leqslant 6$$，面积为$$6 \\pi $$．因此，$${{S}_{2}}$$与$${{S}_{1}}$$面积比为$$6:1$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "978", "queId": "97a0ee0b7d3d42cc97fab80a206bbbf1", "competition_source_list": ["2014年湖南全国高中数学联赛竞赛初赛第3题5分", "2016~2017学年北京海淀区北京石油学院附属中学高一下学期期中第10题", "2004年高考真题重庆卷理科第9题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "等差数列$$\\left { {{a}_{n}} \\right }$$中，$${{a}_{1}}\\textgreater0$$，$${{a}_{2003}}+{{a}_{2004}}\\textgreater0$$，$${{a}_{2003}}\\cdot {{a}_{2004}}\\textless{}0$$，则使前$$n$$项和$${{S}_{n}}\\textgreater0$$成立的最大自然数$$n$$为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4005$$ "}], [{"aoVal": "B", "content": "$$4006$$ "}], [{"aoVal": "C", "content": "$$4007$$ "}], [{"aoVal": "D", "content": "$$4008$$ "}]], "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->等差数列角标和性质的应用", "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": ["∵$${{a}_{2003}}+{{a}_{2004}}\\textgreater0$$，$${{a}_{2003}}\\cdot {{a}_{2004}}\\textless{}0$$，  ∴$${{a}_{2003}}$$和$${{a}_{2004}}$$两项中有一正数一负数，  ∵$${{a}_{1}}\\textgreater0$$，∴公差为负数，否则各项总为正数，  ∴$${{a}_{2003}}\\textgreater{{a}_{2004}}$$，即$${{a}_{2003}}\\textgreater0$$，$${{a}_{2004}}\\textless{}0$$．  ∴$${{S}_{4006}}=\\frac{4006({{a}_{1}}+{{a}_{4006}})}{2}=2003({{a}_{2003}}+{{a}_{2004}})\\textgreater0$$，  $${{S}_{4007}}=\\frac{4007}{2}\\cdot ({{a}_{1}}+{{a}_{4007}})=4007\\cdot {{a}_{2004}}\\textless{}0$$．  故使前$$n$$项和$${{S}_{n}}\\textgreater0$$成立的最大自然数$$n$$为$$4006$$，故选$$B$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "953", "queId": "bb8aa2d421b44fe98233637b8c44d85c", "competition_source_list": ["2016年AMC10竞赛B第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "阿曼達$$4$$個表兄弟的平均年齡是$$8$$，年齡的中位數是$$5$$，則阿曼達的表兄弟中最年長者與最年輕者的年齡和爲何？  The mean age of Amanda\\textquotesingle s $$4$$ cousins is $$8$$, and their median age is $$5$$. What is the sum of the ages of Amanda\\textquotesingle s youngest and oldest cousins?", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$16$$ "}], [{"aoVal": "C", "content": "$$19$$ "}], [{"aoVal": "D", "content": "$$22$$ "}], [{"aoVal": "E", "content": "$$25$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["The sum of the ages of the cousins is $$4$$ times the mean, or $$32$$. There are an even number of cousins, so there is no single median, so $$5$$ must be the median of the two in the middle. Therefore the sum of the ages of the two in the middle is $$10$$. Subtracting $$10$$ from $$32$$ produces $$\\left(\\text{D}\\right) 22$$. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1186", "queId": "e6fce64847f34229abc14d4dcdab0442", "competition_source_list": ["2010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": "已知$$-\\frac{ \\pi }{2} ~\\textless{} ~\\alpha ~~\\textless{} ~\\frac{ \\pi }{2}$$，$$2\\tan \\beta =\\tan 2\\alpha $$．$$\\tan \\left( \\beta -\\alpha \\right)=-2\\sqrt{2}$$，则$$\\cos \\alpha =$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{3}}{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{2}}{2}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{3}}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{2}}{3}$$ "}]], "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->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->知识点->三角函数->三角恒等变换->和差角公式->两角和与差的正切"], "answer_analysis": ["令$$\\tan \\alpha =u$$．由  $$\\tan \\beta =\\frac{1}{2}\\tan 2\\alpha =\\frac{\\tan \\alpha }{1-{{\\tan }^{2}}\\alpha }=\\frac{u}{1-{{u}^{2}}}$$，得  $$\\tan \\left( \\beta -\\alpha \\right)=\\frac{\\tan \\beta -\\tan \\alpha }{1+\\tan \\alpha \\tan \\beta }$$  $$=\\frac{\\dfrac{u}{1-{{u}^{2}}}-u}{1+u\\cdot \\frac{u}{1-{{u}^{2}}}}$$  $$=\\frac{u-u\\left( 1-{{u}^{2}} \\right)}{1-{{u}^{2}}+{{u}^{2}}}={{u}^{2}}$$．  由$${{u}^{3}}=-2\\sqrt{2}$$，得$$u=\\tan \\alpha =-\\sqrt{2}$$．  已知$$-\\frac{ \\pi }{2} ~\\textless{} ~\\alpha ~~\\textless{} ~\\frac{ \\pi }{2}$$，故$$-\\frac{ \\pi }{2} ~\\textless{} ~\\alpha ~~\\textless{} ~0$$，所以  $$\\cos \\alpha =\\frac{1}{\\sqrt{1+{{\\tan }^{2}}\\alpha }}=\\frac{\\sqrt{3}}{2}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "532", "queId": "6ca5e6af61a6457ca51134bbd41d1db9", "competition_source_list": ["2003年AMC12竞赛A第10题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2003-AMC12A-10$$  Al, Bert, and Carl are the winners of a school drawing for a pile of Halloween candy, which they are to divide in a ratio of $$3:2:1$$, respectively. Due to some confusion they come at different times to claim their prizes, and each assumes he is the first to arrive. If each takes what he believes to be the correct share of candy, what fraction of the candy goes unclaimed?  Al、$$Bert$$ 和 Carl 是学校为一堆万圣节糖果绘制的获胜者，他们将分别以 3:2:1 的比例分配这些糖果。 由于一些混乱，他们在不同的时间来领奖，每个人都假设他是第一个到达的人。 如果每个人都拿走他认为正确的糖果份额，那么有多少糖果无人认领？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{18}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{2}{9}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{18}$$ "}], [{"aoVal": "E", "content": "$$\\frac{5}{12}$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Application->Proportion Word Problems", "课内体系->知识点->推理与证明->合情推理与演绎推理->合情推理->归纳推理"], "answer_analysis": ["假设有$$6$$个糖果，$$A$$来拿走一半，剩下三个，$$B$$来拿走三分之一，剩下$$2$$个，$$C$$来拿走六分之一，剩下$\\dfrac{5}{3}$个，占比$\\dfrac{5}{18}$ "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "365", "queId": "505908494b4841859cbd31cc5056c081", "competition_source_list": ["2008年山东全国高中数学联赛竞赛初赛第7题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "有$$6$$个座位连成一横排，三人就座，恰有两个空位相邻的不同排法共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$48$$种 "}], [{"aoVal": "B", "content": "$$60$$种 "}], [{"aoVal": "C", "content": "$$72$$种 "}], [{"aoVal": "D", "content": "$$96$$种 "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["方法一：将座位按顺序编号为$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$．  若相邻空位是$$1$$、$$2$$或$$5$$、$$6$$时，各有排法$$\\text{C}_{3}^{1}\\cdot \\text{A}_{3}^{3}$$种．  若相邻空位是$$2$$、$$3$$，$$3$$、$$4$$或$$4$$、$$5$$时，各有排法 $$\\text{C}_{2}^{1}\\cdot \\text{A}_{3}^{3}$$种．  所以恰有两个空位相邻的不同排法共有$$\\text{2C}_{3}^{1}\\cdot \\text{A}_{3}^{3}+3\\text{C}_{2}^{1}\\cdot \\text{A}_{3}^{3}=72$$（种）．故选$$\\text{C}$$．  方法二：先把两个空位作为一人（另一空位暂不考虑），  这样，四人就座共$$\\text{A}_{4}^{4}=24$$（种）排法．  对每一种这样的排法，再把另一空位安插到与两空位不相邻的地方，  共有$$5-2=3$$（种）插法．所以共有$$\\text{3A}_{4}^{4}=72$$（种）不同排法．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "146", "queId": "0c87fed3777d4139b74b24886394031f", "competition_source_list": ["2009年浙江全国高中数学联赛竞赛初赛第8题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "``函数$$f(x)$$在$$\\left[ 0, 1 \\right]$$上单调''是``函数$$f(x)$$在$$\\left[ 0, 1 \\right]$$上有最大值''的．", "answer_option_list": [[{"aoVal": "A", "content": "必要非充分条件 "}], [{"aoVal": "B", "content": "充分非必要条件 "}], [{"aoVal": "C", "content": "充分且必要条件 "}], [{"aoVal": "D", "content": "既非充分也非必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->逻辑推理", "竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["$$f\\left( x \\right)$$在$$\\left[ 0,1 \\right]$$上单调，则$$f\\left( x \\right)$$在$$\\left[ 0,1 \\right]$$上有最大值，充分性成立；  反之，若$$f\\left( x \\right)$$在$$\\left[ 0,1 \\right]$$上有最大值，则$$f\\left( x \\right)$$在$$\\left[ 0,1 \\right]$$上不一定单调，比如$$f\\left( x \\right)={{\\left( x-\\frac{1}{2} \\right)}^{2}}$$．  所以选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "144", "queId": "f59395ea20c94e349ecc754333888bbe", "competition_source_list": ["2008年河南全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知定义域是全体实数的函数$$y=f(x)$$满足$$f(x+2 \\pi )=f(x)$$且函数  $$g(x)=\\frac{f(x)+f(-x)}{2}$$,$$h(x)=\\frac{f(x)-f(-x)}{2}.$$  如果定义函数$$p(x)$$、$$q(x)$$为$$p(x)=\\begin{cases}\\frac{g(x)-g(x+ \\pi )}{2\\cos x},\\left( x\\ne k \\pi +\\frac{ \\pi }{2} \\right)    0,\\left( x=k \\pi +\\frac{ \\pi }{2} \\right) \\end{cases}$$，$$q(x)=\\begin{cases}\\frac{h(x)+h(x+ \\pi )}{2\\sin 2x},\\left( x\\ne \\frac{k \\pi }{2} \\right)    0,\\left( x=\\frac{k \\pi }{2} \\right) \\end{cases}$$，其中$$k\\in Z$$，那么下列关于$$p(x),q(x)$$叙述正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "都是奇函数且周期为$$ \\pi $$ "}], [{"aoVal": "B", "content": "都是偶函数且周期为$$ \\pi $$ "}], [{"aoVal": "C", "content": "均无奇偶性但都有周期性 "}], [{"aoVal": "D", "content": "均无周期性但都有奇偶性 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数综合"], "answer_analysis": ["显然，$$g(-x)=g(x)$$，$$h(-x)=-h(x)$$．  因为$$f(x+2 \\pi )=f(x)$$，  所以$$g(x+2 \\pi )=\\frac{f(x+2 \\pi )+f(-x-2 \\pi )}{2}$$  $$=\\frac{f(x)+f(-x)}{2}=g(x)$$，  $$h(x+2 \\pi )=\\frac{f(x+2 \\pi )-f(-x-2 \\pi )}{2}$$  $$=\\frac{f(x)-f(-x)}{2}=h(x)$$，  当$$x\\ne k \\pi +\\frac{ \\pi }{2}$$时，  $$p(-x)=\\frac{g(-x)-g(-x+ \\pi )}{2\\cos (-x)}=\\frac{g(x)-g(x- \\pi )}{2\\cos x}$$  $$=\\frac{g(x)-g(x+ \\pi )}{2\\cos x}=p(x)$$，  $$p(x+ \\pi )=\\frac{g(x+ \\pi )-g(x+2 \\pi )}{2\\cos (x+ \\pi )}=\\frac{g(x+ \\pi )-g(x)}{-2\\cos x}$$  $$=\\frac{g(x)-g(x+ \\pi )}{2\\cos x}=p(x)$$，  当$$x=k \\pi +\\frac{ \\pi }{2}$$时，$$p(-x)=p(x)=p(x+ \\pi )=0$$．  因此$$x\\in R$$时，$$p(-x)=p(x)p(x+ \\pi )=p(x)$$．  同理可得$$x\\in R$$时，$$q(-x)=q(x)$$及$$q(x+ \\pi )=q(x)$$．故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1039", "queId": "a14499c3b03e42ab9a6f4b0ab0c94bfb", "competition_source_list": ["2016年AMC10竞赛A第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "小美列出$$1$$到$$30$$所有的整數，小龍複製小美列出的所有整數，並將原出現數字是$$2$$的都改爲$$1$$．小美將她所列的數加起來，小龍也將他更改過的數加起來．試問小美所得的和比小龍所得的和大多少？  Ximena lists the whole numbers $$1$$ through $$30$$ once. Emilio copies Ximena\\textquotesingle s numbers, replacing each occurrence of the digit $$2$$ by the digit $$1$$. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena\\textquotesingle s sum than Emilio\\textquotesingle s?", "answer_option_list": [[{"aoVal": "A", "content": "$$13$$ "}], [{"aoVal": "B", "content": "$$26$$ "}], [{"aoVal": "C", "content": "$$102$$ "}], [{"aoVal": "D", "content": "$$103$$ "}], [{"aoVal": "E", "content": "$$110$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["For every tens digit $$2$$, we subtract $$10$$, and for every units digit $$2$$, we subtract $$1$$. Because $$2$$ appears $$10$$ times as a tens digit and $$2$$ appears $$3$$ times as a units digit, the answer is $$10\\cdot 10+1\\cdot 3=103$$. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1151", "queId": "f8b2f18f248b43d8b33c1dc8a50ab039", "competition_source_list": ["2009年第二十届全国希望杯高二竞赛复赛第10题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "若正数$$xyz$$使等式$$\\frac{6}{xyz}\\left( \\frac{1}{x}+\\frac{2}{y}+\\frac{3}{z} \\right)=1$$成立，则$$\\left( \\frac{1}{x}+\\frac{3}{z} \\right)\\left( \\frac{2}{y}+\\frac{3}{z} \\right)$$的最小值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{3}$$ "}]], "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->几个重要的不等式->均值"], "answer_analysis": ["设$$\\frac{1}{x}=a,\\frac{2}{y}=b,\\frac{3}{z}=c$$，则已知条件化为$$abc(a+b+c)=1$$．所求的式子为$$(a+c)(b+c)$$，  因为$${{\\left[ (a+c)(b+c) \\right]}^{2}}={{\\left[ ab+(ac+bc+{{c}^{2}}) \\right]}^{2}}\\geqslant 4ab(ac+bc+{{c}^{2}})=4abc(a+b+c)=4$$，所以$$(a+c)(b+c)\\geqslant 2$$，仅当$$ab=ac+bc+{{c}^{2}}$$时等好成立，故最小值为$$2$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1063", "queId": "d331d67bb43d49e8951a9cedece26f57", "competition_source_list": ["2009年山东全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "函数$$f(x)={{\\log }_{\\frac{1}{2}}}(-{{x}^{2}}-2x+3)$$的递增区间是 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-3,-1)$$ "}], [{"aoVal": "B", "content": "$$[-1, 1)$$ "}], [{"aoVal": "C", "content": "$$(-3,-1]$$ "}], [{"aoVal": "D", "content": "$$(-1,1]$$ "}]], "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->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->函数的概念与性质->函数的概念及其表示->函数的定义域->求具体函数（包括复合函数）的定义域", "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", "课内体系->知识点->函数的概念与性质->函数的性质->单调性->函数单调性的判定->求单调区间", "课内体系->知识点->函数的概念与性质->函数的性质->单调性->函数单调性的判定->判断复合函数单调性", "课内体系->知识点->导数->导数的应用->导数与单调性->利用导数求函数的单调性、单调区间", "课内体系->知识点->导数->导数的运算->复合函数的求导法则", "课内体系->知识点->导数->导数的运算->利用公式和四则运算法则求导", "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"], "answer_analysis": ["由$$-{{x}^{2}}-2x+3\\textgreater0$$，得$$-3\\textless{}x\\textless{}1$$．抛物线$$y=-{{x}^{2}}-2x+3$$的顶点坐标为$$(-1,4)$$．所以函数$$y=-{{x}^{2}}-2x+3$$在$$[-1,1)$$上递减，又底数$$\\frac{1}{2}\\textless{}1$$，从而知函数$$f(x)={{\\log }_{\\frac{1}{2}}}(-{{x}^{2}}-2x+3)$$的递增区间为$$[-1,1)$$． ", "<p>由$$-{{x}^{2}}-2x+3&gt;0$$，得$$-3&lt;{}x&lt;{}1$$．$${f}&rsquo;(x)={{\\log }_{\\frac{1}{2}}}\\text{e}\\cdot \\frac{-2(x+1)}{(x-1)(x+3)}\\geqslant 0$$得$$x+1\\geqslant 0$$，即$$x\\geqslant -1$$．所以函数$$f(x)={{\\log }_{\\frac{1}{2}}}(-{{x}^{2}}-2x+3)$$的递增区间为$$[-1,1)$$．故选$$\\text{B}$$．</p>"], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1174", "queId": "fdb933d123d04d449a42000ab6275fe9", "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~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$12$$ "}], [{"aoVal": "C", "content": "$$13$$ "}], [{"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": ["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": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "917", "queId": "85e6fd9fa6b64ea2a92fbe71f4945970", "competition_source_list": ["2014年浙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$$\\left[ x \\right]$$表示不超过$$x$$的最大整数，则方程$$3{{x}^{2}}-10\\left[ x \\right]+3=0$$的所有实数根的个数为．", "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->数论模块->取整函数->与[x]有关的方程和不等式", "竞赛->知识点->函数->二次函数"], "answer_analysis": ["由  $$3{{x}^{2}}-10\\left[ x \\right]+3=0$$  $$\\Rightarrow 10\\left( x-1 \\right)\\textless{}10\\left[ x \\right]=3{{x}^{2}}+3\\leqslant 10x$$  $$\\Rightarrow \\frac{1}{3}\\leqslant x\\leqslant 3$$．  当$$\\frac{1}{3}\\leqslant x\\textless{}1$$时，方程$$3{{x}^{2}}-10\\left[ x \\right]+3=0$$无解．  当$$1\\leqslant x\\textless{}2$$时，方程$$3{{x}^{2}}-10\\left[ x \\right]+3=0$$的解为$$x=\\sqrt{\\frac{7}{3}}$$．  当$$2\\leqslant x\\textless{}3$$时，方程$$3{{x}^{2}}-10\\left[ x \\right]+3=0$$的解为$$x=\\sqrt{\\frac{17}{3}}$$．  当$$x=3$$时，符合满足方程$$3{{x}^{2}}-10\\left[ x \\right]+3=0$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1121", "queId": "f8661ec67c9945bc90230c92567144a9", "competition_source_list": ["1993年全国高中数学联赛竞赛一试第2题", "2007年上海复旦大学自主招生千分考第17题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$f\\left( x \\right)=a\\sin x+b\\sqrt[3]{x}+4$$（$$a$$，$$b$$为实数），且$$f\\left( \\lg {{\\log }_{3}}10 \\right)=5$$，则$$f\\left( \\lg \\lg 3 \\right)=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-5$$ "}], [{"aoVal": "B", "content": "$$-3$$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "随$$a$$，$$b$$取不同值而取不同值 "}]], "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": ["∵$$f(x)-4$$是奇函数，  故$$f(-x)-4=-(f(x)-4)$$，  即$$f{}(-x){}=-f{}(x){}+8$$，而$$\\lg \\lg 3=-\\lg {{\\log }_{3}}10$$．  ∴$$f(\\lg \\lg 3)=f(-\\lg {{\\log }_{3}}10)=-f(\\lg {{\\log }_{3}}10)+8$$  $$=-5+8=3$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "794", "queId": "dae3a2e36c8b41e085b7594898b5013d", "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$$\\sum\\limits_{k=1}^{2020}{\\left( k\\cdot {{\\text{i}}^{k}} \\right)}=$$．", "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"], "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$$\\sum\\limits_{k=1}^{2020}{\\left( k\\cdot {{\\text{i}}^{k}} \\right)}=1010-1010\\text{i}$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "987", "queId": "a0ba04f70adb4930b7af2afeea266310", "competition_source_list": ["1999年全国全国高中数学联赛竞赛一试第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "平面直角坐标系中，纵、横坐标都是整数的点叫做整点，那么，满足不等式 $${{(\\left\\textbar{} x \\right\\textbar-1)}^{2}}+{{(\\left\\textbar{} y \\right\\textbar-1)}^{2}}\\textless{}2$$的整点$$(x,y)$$的个数是（  ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$17$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$25$$ "}]], "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->不等式的性质->针对不等式变形判断正误"], "answer_analysis": ["由$${{\\left( \\textbar x\\textbar-1 \\right)}^{2}}+{{\\left( \\textbar y\\textbar-1 \\right)}^{2}}\\textless{}2$$，  可得$$\\left( \\left\\textbar{} x \\right\\textbar-1,\\left\\textbar{} y \\right\\textbar-1 \\right)$$为$$\\left( 0,0 \\right)$$，$$\\left( 0,1 \\right)$$，$$\\left( 0,-1 \\right)$$，$$\\left( 1,0 \\right)$$或$$\\left( -1,0 \\right)$$．  从而，不难得到$$\\left( x,y \\right)$$共有$$16$$个． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "804", "queId": "776b929303c04a8faa9fd166c8892df2", "competition_source_list": ["2011年浙江全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "过椭圆$$\\frac{{{x}^{2}}}{2}+{{y}^{2}}=1$$的右焦点$${{F}_{2}}$$作倾斜角为$${{45}^{\\circ }}$$弦$$AB$$，则$$\\left\\textbar{} AB \\right\\textbar$$为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2\\sqrt{6}}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{4\\sqrt{6}}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{4\\sqrt{2}}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{4\\sqrt{3}}{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->椭圆"], "answer_analysis": ["椭圆的右焦点为$$（1，0）$$，则弦$$AB$$为$$y=x-1$$代入椭圆方程得  $$3{{x}^{2}}-4x=0\\Rightarrow {{x}_{1}}=0,  {{x}_{2}}=\\frac{4}{3}\\Rightarrow \\left\\textbar{} AB \\right\\textbar=\\sqrt{2{{({{x}_{1}}-{{x}_{2}})}^{2}}}=\\frac{4\\sqrt{2}}{3}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1112", "queId": "caaba1b161ab443282bbe7aa60a75ff4", "competition_source_list": ["2008年浙江全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "当$$0\\textless{}x\\textless{}1$$时，$$f(x)=\\frac{x}{\\lg x}$$，则下列大小关系正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$${{f}^{2}}(x)\\textless{}f({{x}^{2}})\\textless{}f(x)$$ "}], [{"aoVal": "B", "content": "$$f({{x}^{2}})\\textless{}{{f}^{2}}(x)\\textless{}f(x)$$ "}], [{"aoVal": "C", "content": "$$f(x)\\textless{}f({{x}^{2}})\\textless{}{{f}^{2}}(x)$$ "}], [{"aoVal": "D", "content": "$$f({{x}^{2}})\\textless{}f(x)\\textless{}{{f}^{2}}(x)$$ "}]], "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", "竞赛->知识点->函数->函数综合"], "answer_analysis": ["当$$0\\textless{}x\\textless{}1$$时，$$f(x)=\\frac{x}{\\lg x}\\textless{}0$$，  $$f({{x}^{2}})=\\frac{{{x}^{2}}}{\\lg {{x}^{2}}}\\textless{}0$$，$${{f}^{2}}(x)={{\\left( \\frac{x}{\\lg x} \\right)}^{2}}\\textgreater0$$．  又因为$$\\frac{x}{\\lg x}-\\frac{{{x}^{2}}}{\\lg {{x}^{2}}}=\\frac{2x-{{x}^{2}}}{2\\lg x}=\\frac{(2-x)x}{2\\lg x}\\textless{}0$$．  所以 $$f(x)\\textless{}f({{x}^{2}})\\textless{}{{f}^{2}}(x)$$． 故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "527", "queId": "2e49d0292f7c45e79052a0c0c901494f", "competition_source_list": ["2021年贵州全国高中数学联赛竞赛初赛第15题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$${{a}_{1}}$$，$${{a}_{2}}$$，$$\\cdots $$，$${{a}_{n}}$$均为非负实数，若有：$$\\frac{{{a}_{1}}}{a_{2}^{2}+\\cdots +a_{n}^{2}}+\\frac{{{a}_{2}}}{a_{1}^{2}+a_{3}^{2}\\cdots +a_{n}^{2}}+\\cdots +\\frac{{{a}_{n}}}{a_{1}^{2}+\\cdots +a_{n-1}^{2}}\\geqslant \\frac{k}{{{a}_{1}}+{{a}_{2}}+\\cdots +{{a}_{n}}}$$恒成立，求实数$$k$$的最大值．", "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->不等式->几个重要的不等式->柯西", "竞赛->知识点->不等式->几个重要的不等式->均值", "课内体系->知识点->等式与不等式->不等式->柯西不等式"], "answer_analysis": ["根据柯西不等式，得$$\\left( \\frac{{{a}_{1}}}{a_{2}^{2}+\\cdots +a_{n}^{2}}+\\frac{{{a}_{2}}}{a_{1}^{2}+a_{3}^{2}\\cdots +a_{n}^{2}}+\\cdots +\\frac{{{a}_{n}}}{a_{1}^{2}+\\cdots +a_{n-1}^{2}} \\right)\\left( {{a}_{1}}+{{a}_{2}}+\\cdots +{{a}_{n}} \\right)$$  $$\\geqslant {{\\left( \\sqrt{\\frac{a_{1}^{2}}{a_{2}^{2}+\\cdots +a_{n}^{2}}}+\\sqrt{\\frac{a_{2}^{2}}{a_{1}^{2}+a_{3}^{2}\\cdots +a_{n}^{2}}}+\\cdots +\\sqrt{\\frac{a_{n}^{2}}{a_{1}^{2}+\\cdots +a_{n-1}^{2}}} \\right)}^{2}}$$，  于是，只需证明  $$\\left( \\sqrt{\\frac{a_{1}^{2}}{a_{2}^{2}+\\cdots +a_{n}^{2}}}+\\sqrt{\\frac{a_{2}^{2}}{a_{1}^{2}+a_{3}^{2}\\cdots +a_{n}^{2}}}+\\cdots +\\sqrt{\\frac{a_{n}^{2}}{a_{1}^{2}+\\cdots +a_{n-1}^{2}}} \\right)\\geqslant 2$$．  根据均值不等式，有  $$\\sqrt{\\frac{a_{2}^{2}+\\cdots +a_{n}^{2}}{a_{1}^{2}}}\\leqslant \\frac{1}{2}\\left( \\frac{a_{2}^{2}+\\cdots +a_{n}^{2}}{a_{1}^{2}}+1 \\right)=\\frac{a_{1}^{2}+a_{2}^{2}+\\cdots +a_{n}^{2}}{2a_{1}^{2}}$$，  同理可得$$\\sqrt{\\frac{a_{1}^{2}+a_{3}^{2}\\cdots +a_{n}^{2}}{a_{2}^{2}}}\\leqslant \\frac{a_{1}^{2}+a_{2}^{2}+\\cdots +a_{n}^{2}}{2a_{2}^{2}}$$，$$\\cdots \\cdots $$，$$\\sqrt{\\frac{a_{1}^{2}+\\cdots +a_{n-1}^{2}}{a_{n}^{2}}}\\leqslant \\frac{a_{1}^{2}+\\cdots +a_{n-1}^{2}}{2a_{n}^{2}}$$，  所以$$\\sqrt{\\frac{a_{1}^{2}}{a_{2}^{2}+\\cdots +a_{n}^{2}}}+\\sqrt{\\frac{a_{2}^{2}}{a_{1}^{2}+a_{3}^{2}\\cdots +a_{n}^{2}}}+\\cdots +\\sqrt{\\frac{a_{n}^{2}}{a_{1}^{2}+\\cdots +a_{n-1}^{2}}}$$  $$\\geqslant \\frac{2a_{1}^{2}}{a_{1}^{2}+a_{2}^{2}\\cdots +a_{n}^{2}}+\\frac{2a_{2}^{2}}{a_{1}^{2}+a_{2}^{2}\\cdots +a_{n}^{2}}+\\cdots +\\frac{2a_{n}^{2}}{a_{1}^{2}+a_{2}^{2}\\cdots +a_{n}^{2}}=2$$，  等号成立的条件：$$n$$个数$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$$\\cdots $$，$${{a}_{n}}$$中，有两个相等，其余为$$0$$．  故值不等式得证． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "138", "queId": "0c737b48a2bc486ba5672f2f218ccdb9", "competition_source_list": ["1994年全国高中数学联赛竞赛一试第1题", "1994年全国高中数学联赛竞赛"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$a$$、$$b$$、$$c$$是实数，那么对任何实数$$x$$，不等式$$a\\sin x+b\\cos x+c\\textgreater0$$都成立的充分必要条件是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$a$$、$$b$$同时为$$0$$，且$$c\\textgreater0$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{{{a}^{2}}+{{b}^{2}}}=c$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{{{a}^{2}}+{{b}^{2}}}\\textless{}c$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{{{a}^{2}}+{{b}^{2}}}\\textgreater c$$ "}]], "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": ["当$$a$$，$$b$$不同时为零时，  $$a\\sin x+b\\cos x+c\\textgreater0$$①  $$\\Leftrightarrow \\sqrt{{{a}^{2}}+{{b}^{2}}}\\sin (x+\\varphi )+c\\textgreater0$$  $$\\Leftrightarrow \\sin (x+\\varphi )\\textgreater-\\frac{c}{\\sqrt{{{a}^{2}}+{{b}^{2}}}}$$②  而②式成立的充分必要条件是$$-\\frac{c}{\\sqrt{{{a}^{2}}+{{b}^{2}}}}\\textless{}-1$$，即$$\\sqrt{{{a}^{2}}+{{b}^{2}}}\\textless{}c$$，当$$a$$，$$b$$同时为零时，此时不等式成立的充要条件是$$c\\textgreater\\sqrt{{{a}^{2}}+{{b}^{2}}}$$，这一结论含于$$\\text{C}$$中．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "489", "queId": "6c66a3d17971472e8c292cdfac446889", "competition_source_list": ["竞赛第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "若\\emph{x}，\\emph{y}，\\emph{z}均为正整数，则下列不等式中不一定成立的是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\left(x+y+z\\right)\\ln \\frac{x+y+z}{3}\\geq x\\ln x+y\\ln y+z\\ln z$ "}], [{"aoVal": "B", "content": "$\\left(xyz\\right)^{\\frac{x+y+z}{3}}\\leq x^{x}y^{y}z^{z}$ "}], [{"aoVal": "C", "content": "$x^{2}+\\frac{1}{x^{2}}\\geq x+\\frac{1}{x}$ "}], [{"aoVal": "D", "content": "$\\textbar{} x-y\\textbar{} \\leq \\textbar{} x-z\\textbar{} +\\textbar{} y-z\\textbar$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据特例可判断A的正误，根据指数函数的性质可判断B的正误，根据因式分解可判断C的正误，根据绝对值不等式可判断D的正误.\\\\ 【详解】\\\\ 对于选项A，取$(x,y,z)=(4,1,1)$，\\\\ 则$\\left(x+y+z\\right)\\ln \\frac{x+y+z}{3}=6\\ln 2,x\\ln x+y\\ln y+z\\ln z=4\\ln 4=8\\ln 2$，\\\\ 因此选项A不一定成立．\\\\ 对于选项B，该不等式即$(xyz)^{x+y+z}\\leq x^{3x}y^{3y}z^{3z}\\Rightarrow x^{y+z}y^{z+x}z^{x+y}\\leq x^{2x}y^{2y}z^{2z}$，\\\\ 而$\\left(\\frac{x}{y}\\right)^{x-y}\\geq 1\\Rightarrow x^{x}y^{y}\\geq x^{y}y^{x}$，\\\\ 两边作轮换积即得该不等式，因此选项B一定成立.\\\\ 对于选项C，该不等式即$\\left(x+\\frac{1}{x}\\right)^{2}-\\left(x+\\frac{1}{x}\\right)-2\\geq 0\\Rightarrow \\left(x+\\frac{1}{x}+1\\right)\\left(x+\\frac{1}{x}-2\\right)\\geq 0$，\\\\ 命题成立，因此选项C一定成立，\\\\ 对于选项D，根据绝对值不等式，有$\\textbar{} x-z\\textbar{} +\\textbar{} y-z\\textbar{} \\geq \\textbar{} (x-z)-(y-z)\\textbar{} =\\textbar{} x-y\\textbar$，\\\\ 命题成立，因此选项D一定成立．\\\\ 故选：A. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "316", "queId": "e32d1402a1e442cca48d232a6d5deb5c", "competition_source_list": ["2009年第27届美国数学邀请赛（AIME）竞赛第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "有多少个小于$$1000$$的正整数$$N$$使得方程$${{x}^{\\left[ x \\right]}}=N$$有解（$$\\left[ x \\right]$$表示小于或等于$$x$$的最大整数）？", "answer_option_list": [[{"aoVal": "A", "content": "$$412$$ "}], [{"aoVal": "B", "content": "$$415$$ "}], [{"aoVal": "C", "content": "$$418$$ "}], [{"aoVal": "D", "content": "$$421$$ "}], [{"aoVal": "E", "content": "$$425$$ "}]], "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->取整函数（高斯函数）", "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": ["对于正整数$$k$$、考虑当$$\\left[ x \\right]=k$$时方程$${{x}^{\\left[ x \\right]}}=N$$有解的$$N$$的个数．  这时$$x=\\sqrt[k]{N}$$，由于$$k\\leqslant xk+1$$，故$${{k}^{k}}\\leqslant {{x}^{k}}=N\\leqslant {{\\left( k+1 \\right)}^{k}}-1$$，  因此，存在$${{\\left( k+1 \\right)}^{k}}-{{k}^{k}}$$个可能的$$N$$使得$${{x}^{k}}=N$$有解．  因$${{5}^{4}}\\textless1000$$且$${{5}^{5}}\\textgreater1000$$，  故所求的满足条件的$$N$$的个数为$$\\sum\\limits_{k=1}^{4}{\\left[ {{\\left( k+1 \\right)}^{k}}-{{k}^{k}} \\right]}=1+5+37+369=412$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "864", "queId": "b1f0f013885d4485a0619e6b473e614d", "competition_source_list": ["2018年湖北全国高中数学联赛竞赛初赛第5题10分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$G$$为$$\\triangle ABC$$的重心．若$$BG\\bot CG$$，$$BC=\\sqrt{2}$$，则$$AB+AC$$的最大值为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$2\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$2\\sqrt{5}$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->平面几何->欧拉线上的心（二试）->重心（二试）"], "answer_analysis": ["设$$BC$$的中点为$$D$$，  由$$\\triangle BCG$$为直角三角形知$$GD=\\frac{1}{2}BC=\\frac{\\sqrt{2}}{2}$$，  又因为$$G$$是$$\\triangle ABC$$的重心，  所以，$$AD=3GD=\\frac{3\\sqrt{2}}{2}$$，  由三角形中线长公式得：$$A{{D}^{2}}=\\frac{1}{4}(2A{{B}^{2}}+2A{{C}^{2}}-B{{C}^{2}})$$，  则$$A{{B}^{2}}+A{{C}^{2}}=2A{{D}^{2}}+\\frac{1}{2}B{{C}^{2}}=10$$，  故$$AB+AC\\leqslant 2\\sqrt{\\frac{A{{B}^{2}}+A{{C}^{2}}}{2}}=2\\sqrt{5}$$，  当且仅当$$AB=AC$$时，上式等号成立，  因此，$$AB+AC$$的最大值为$$2\\sqrt{5}$$．  故答案为$$2\\sqrt{5}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1190", "queId": "f969778654074707b65b100a5c8fed49", "competition_source_list": ["2005年AMC10竞赛A第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "一个边长为$n$的木质立方体的六个面都被涂成红色，然后被切割成$n^{3}$的单位立方体．单位立方体总面数中正好有四分之一是红色.$n$是多少？", "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": ["美国AMC10/12->Knowledge Point->Combination->Reasoning->Dyeing and Covering", "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": ["Since there are $$n^{2}$$ little faces on each face of the big wooden cube, there are $$6n^{2}$$ little faces painted red.  Since each unit cube has $$6$$ faces, there are $$6n^{3}$$ little faces total.  Since one-fourth of the little faces are painted red,  $$\\dfrac{6n^{2}}{6n^{3}}=\\dfrac{1}{4}$$  \\hspace{0pt}$$\\dfrac{1}{n}=\\dfrac{1}{4}$$  $$n=4\\Rightarrow\\left( \\rm B\\right)$$. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "907", "queId": "814c450ea00c4e908aa50f8cd639ee1a", "competition_source_list": ["2011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": "函数$$f(x)=\\begin{cases} 1-{{5}^{-x}} x\\geqslant 0    {{5}^{x}}-1 x\\textless{}0 \\end{cases}$$，则该函数为．", "answer_option_list": [[{"aoVal": "A", "content": "单调增加函数、奇函数 "}], [{"aoVal": "B", "content": "单调递减函数、偶函数 "}], [{"aoVal": "C", "content": "单调增加函数、偶函数 "}], [{"aoVal": "D", "content": "单调递减函数、奇函数 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["由单调性和奇偶性定义知道函数为单调增加的奇函数． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "642", "queId": "451e1095ce244026b19d86e3d828569f", "competition_source_list": ["2008年湖南全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "定义集合运算： $$A\\otimes B=\\left { z\\textbar z=xy,x\\in A,y\\in B \\right }$$．设$$A=\\left { 2,0 \\right }$$，$$B=\\left { 0,8 \\right }$$，则集合$$A\\otimes B$$的所有元素之和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$16$$ "}], [{"aoVal": "B", "content": "$$18$$ "}], [{"aoVal": "C", "content": "$$20$$ "}], [{"aoVal": "D", "content": "$$22$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["集合$$A\\otimes B$$的元素：$${{z}_{1}}=2\\times 0=0$$，$${{z}_{2}}=2\\times 8=16$$，  $${{z}_{3}}=0\\times 0=0$$，$${{z}_{4}}=0\\times 8=0$$，  故集合$$A\\otimes B$$的所有元素之和为$$16$$．故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "627", "queId": "b59a231a6dfb4ad2a648ee4c5f681c3d", "competition_source_list": ["2010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": "当$$x,  y$$满足条件$$\\left\\textbar{} x \\right\\textbar+\\left\\textbar{} y \\right\\textbar\\leqslant 1$$时，变量$$z=x-y+2$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ \\frac{\\sqrt{2}}{2},\\frac{3\\sqrt{2}}{2} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left( 1,  3 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left[ 1,  3 \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{\\sqrt{2}}{2},\\frac{3\\sqrt{2}}{2} \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["当$$\\left\\textbar{} x \\right\\textbar+\\left\\textbar{} y \\right\\textbar\\leqslant 1$$时， $$-1\\leqslant x-y\\leqslant 1$$， 所以$$z=x-y+2$$的取值范围是$$\\left[ 1,  3 \\right],$$ "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1045", "queId": "a5d4b19a530e4b9eae7820473f10a1c3", "competition_source_list": ["竞赛第11题"], "difficulty": "3", "qtype": "single_choice", "problem": "已知圆$C\\colon x^{2}+y^{2}=r^{2}$，点\\emph{P}，$P\\textquotesingle$在以\\emph{O}为起点的射线上，且满足$\\textbar{} OP\\textbar{} \\cdot \\left\\textbar{} OP\\textquotesingle\\right\\textbar{} =r^{2}$，则称点\\emph{P}，$P\\textquotesingle$关于圆周\\emph{C}对称，那么抛物线$y=x^{2}$上的点$P(x,y)$关于单位圆$x^{2}+y^{2}=1$的对称点满足的方程为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$x^{4}-y^{2}=0$ "}], [{"aoVal": "B", "content": "$\\left(x^{2}+y^{2}\\right)x^{2}=y$ "}], [{"aoVal": "C", "content": "$y^{4}-x^{2}=0$ "}], [{"aoVal": "D", "content": "$\\left(x^{2}+y^{2}\\right)y=x^{2}$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 求出抛物线的极坐标方程，利用极坐标系下曲线关于圆周的对称的曲线方程公式可求对应曲线的极坐标方程，从而可得所求的直角方程.\\\\ 【详解】\\\\ 以直角坐标系的原点为极点，$x$轴的正半轴为极轴建立极坐标系.\\\\ 于是$y=x^{2}$的极坐标方程为$\\sin \\theta =\\rho \\cos ^{2}\\theta$.\\\\ 因为$\\Gamma \\colon f(\\rho ,\\theta )=0$关于圆周\\emph{C}对称的曲线$\\textquotesingle\\colon f\\left(\\frac{r^{2}}{\\rho },\\theta \\right)=0$．\\\\ 于是方程$\\sin \\theta =\\rho \\cos ^{2}\\theta$关于单位圆对称的曲线为$\\sin \\theta =\\frac{1}{\\rho }\\cos ^{2}\\theta$，\\\\ 其对应的直角方程为$\\left(x^{2}+y^{2}\\right)y=x^{2}$．\\\\ 故选：D "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "641", "queId": "ba3e5b2f07654a5188c98dce2bdedf94", "competition_source_list": ["2017年江苏全国高中数学联赛高二竞赛复赛", "2017年江苏全国高中数学联赛竞赛复赛第8题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "从正$$1680$$边形的顶点中任取若干个，顺次相连成多边形，其中正多边形的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$3432$$ "}], [{"aoVal": "B", "content": "$$3532$$ "}], [{"aoVal": "C", "content": "$$3632$$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "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->两个基本计数原理->加法原理与乘法原理的综合运用", "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", "课内体系->知识点->计数原理->排列与组合->组合"], "answer_analysis": ["若$$a\\textbar1680$$，则存在$$\\frac{1680}{d}$$个正$$d$$边形，$$2 ~\\textless{} ~d ~\\textless{} ~1680$$，  所以，正多边形的个数为$$\\sum\\limits_{d\\textbar1680，2~ \\textless{} ~d ~\\textless{} ~1680}{d}$$，  而$$1680={{2}^{4}}\\cdot 3\\cdot 5\\cdot 7$$，正约数之和为$$\\left( 1+2+{{2}^{2}}+{{2}^{3}}+{{2}^{4}} \\right)\\left( 1+3 \\right)\\left( 1+5 \\right)\\left( 1+7 \\right)=5952$$，  正多边形个数为$$5952-1680-\\frac{1680}{2}=3432$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "781", "queId": "9f48e8540de549fd8e5589556c495af9", "competition_source_list": ["2008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": "命题$$p$$：若$$a$$，$$b\\in R$$，则$$\\textbar a+b\\textbar\\textless{}1$$是$$\\textbar a\\textbar+\\textbar b\\textbar\\textless{}1$$的充分而不必要条件；命题$$q$$：函数$$y=\\sqrt{\\textbar x+1\\textbar-2}$$的定义域是$$\\left( -\\infty ,-3 \\right]\\cup \\left[ 1,+\\infty \\right)$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "``$$p$$或$$q$$''为假命题 "}], [{"aoVal": "B", "content": "``$$p$$且$$q$$''为真命题 "}], [{"aoVal": "C", "content": "$$p$$为真命题，$$q$$为假命题 "}], [{"aoVal": "D", "content": "$$p$$为假命题，$$q$$为真命题 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->逻辑->常用逻辑用语"], "answer_analysis": ["由于$$\\textbar a+b\\textbar\\leqslant \\textbar a\\textbar+\\textbar b\\textbar$$，  因此$$\\textbar a+b\\textbar\\textless{}1$$是$$\\textbar a\\textbar+\\textbar b\\textbar\\textless{}1$$的必要而不充分条件，  命题$$p$$为假命题；  由$$\\textbar x+1\\textbar-2\\geqslant 0$$，得$$x\\leqslant -3$$或$$\\geqslant 1$$，命题$$q$$为真命题．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "359", "queId": "1f9a6318c5bf4d2781673fc5173fa15a", "competition_source_list": ["2009年湖南全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在区间$$[-1, 1]$$上随机取一个数$$x,\\cos \\frac{\\pi x}{2}$$的值介于$$0$$到$$\\frac{1}{2}$$之间的概率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{\\pi }$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->概率初步"], "answer_analysis": ["由$$\\cos \\frac{\\pi x}{2}=\\frac{1}{2}$$，得$$x=\\pm \\frac{2}{3}$$，又因为函数$$y=\\cos \\frac{\\pi x}{2}$$的周期为$$4$$，由函数$$y=\\cos \\frac{\\pi x}{2}$$的图象可知，使$$\\cos \\frac{\\pi x}{2}$$的值介于$$0$$到$$\\frac{1}{2}$$之间的点落在$$\\left[ -1, -\\frac{2}{3} \\right]$$和$$\\left[ \\frac{2}{3}, 1 \\right]$$之内．所以，所求概率$$P=\\frac{2\\times \\frac{1}{3}}{2}=\\frac{1}{3}$$．故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "679", "queId": "8c81529659c74ecca8d22ee43ec86898", "competition_source_list": ["2018年福建全国高中数学联赛竞赛初赛第9题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知整数系数多项式$$f\\left( x \\right)={{x}^{5}}+{{a}_{1}}{{x}^{4}}+{{a}_{2}}{{x}^{3}}+{{a}_{3}}{{x}^{2}}+{{a}_{4}}x+{{a}_{5}}$$，若$$f\\left( \\sqrt{3}+\\sqrt{2} \\right)=0$$，$$f\\left( 1 \\right)+f\\left( 3 \\right)=0$$，则$$f\\left( -1 \\right)=$$~\\uline{~~~~~~~~~~}~．", "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->等式与不等式->等式->代数式化简求值", "竞赛->知识点->函数->函数的概念"], "answer_analysis": ["解：设$${{x}_{0}}=\\sqrt{3}+\\sqrt{2}$$，则$${{x}_{0}}-\\sqrt{3}=\\sqrt{2}$$，$${{\\left( {{x}_{0}}-\\sqrt{3} \\right)}^{2}}=2$$，  于是$${{x}_{0}}^{2}-2\\sqrt{3}{{x}_{0}}+3=2$$，$$2\\sqrt{3}{{x}_{0}}={{x}_{0}}^{2}+1$$．  $$\\therefore {{\\left( 2\\sqrt{3}{{x}_{0}} \\right)}^{2}}={{\\left( {{x}_{0}}^{2}+1 \\right)}^{2}}$$，$${{x}_{0}}^{4}-10{{x}_{0}}^{2}+1=0$$．  $$\\therefore {{x}_{0}}=\\sqrt{3}+\\sqrt{2}$$是多项式$$g\\left( x \\right)={{x}^{4}}-10{{x}^{2}}+1$$的一个跟．  又$${{x}_{0}}=\\sqrt{3}+\\sqrt{2}$$不可能是三次整数系数多项式、二次整数系数多项式的零点．  $$\\therefore g\\left( x \\right)$$整除$$f\\left( x \\right)$$．  $$\\therefore f\\left( x \\right)=g\\left( x \\right)\\left( x-r \\right)=\\left( {{x}^{4}}-10{{x}^{2}}+1 \\right)\\left( x-r \\right)$$，$$r$$为整数．  $$\\therefore f\\left( 1 \\right)=-8\\left( 1-r \\right)=-8+8r$$，$$f\\left( 3 \\right)=-8\\left( 3-r \\right)=-24+8r$$．  由$$f\\left( 1 \\right)+f\\left( 3 \\right)=0$$，得$$\\left( -8+8r \\right)+\\left( -24+8r \\right)=0$$，$$r=2$$，  $$\\therefore f\\left( x \\right)=\\left( {{x}^{4}}-10{{x}^{2}}+1 \\right)\\left( x-2 \\right)$$，$$f\\left( -1 \\right)=24$$． "], "answer_value": "E"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "957", "queId": "a4e60b3c73b246318d2034660316d173", "competition_source_list": ["2019年吉林全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "若$$x$$、$$y$$满足$$\\left\\textbar{} y \\right\\textbar\\leqslant 2x$$，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": ["竞赛->知识点->不等式->线性规划"], "answer_analysis": ["作出可行域，知$$2x+y$$在$$x=-1$$，$$y=-3$$时，取得最小值$$-5$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1040", "queId": "bc3ee18d087446dfbf92a2dd72aefa5f", "competition_source_list": ["2016年AMC10竞赛A第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "滿足$$10^{}x\\cdot 100^{2x}=1000^{5}$$的$$x$$之値爲？  For what value $$x$$ does $$10^{}x\\cdot 100^{2x}=1000^{5}$$?", "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": [], "answer_analysis": ["We can rewrite $$10^{}x\\cdot 100^{2x}=1000^{5}$$ as $$10^{5x}=10^{15}$$.  $$10^{}x\\cdot 100^{2x}=10^{}x\\cdot (10^{2})^{2x}$$  $$10^{}x\\cdot 10^{4x}=(10^{3})^{5}$$  $$10^{5x}=10^{15}$$  Since the bases are equal, we can set the exponents equal, giving us $$5x=15$$. Solving the equation gives us $$x=3$$. ", "<p>We can rewrite this expression as $$\\log (10^x\\cdot 100^{2x})=\\log (1000^5)$$, which can be simplified to $$\\log (10^x\\cdot 10^{4x})=5\\log (1000)$$, and that can be further simplified to $$\\log (10^{5x})=5\\log (10^3)$$.This leads to $$5x=15$$. Solving this linear equation yields $$x=3$$.</p>"], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "472", "queId": "9e123702d484420886b7fa1d2ef02fae", "competition_source_list": ["2009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": "甲、乙两人之间进行一场打完$$7$$局的比赛（每局无平局），则比赛结果出现甲比乙为$$4$$ : $$3$$的概率是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{35}{128}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{16}$$ "}], [{"aoVal": "C", "content": "$$\\frac{4}{7}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{8}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合", "竞赛->知识点->排列组合与概率->概率初步"], "answer_analysis": ["符合条件的概率为$$\\frac{C_{7}^{4}}{{{2}^{7}}}=\\frac{35}{128}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "743", "queId": "b60db560b6894187b9931def2d92e713", "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、$$2$$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": "$$30$$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": "41", "queId": "412e5020cab94c018e834b27edd1d482", "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6题", "2021~2022学年重庆渝中区重庆市巴蜀中学高三上学期月考（适应性月考六）第6题", "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6题"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$x\\textgreater0$$，$$y\\textgreater0$$，设命题$$p$$：$${{2}^{x}}+{{2}^{y}}\\ge 4$$，命题$$q$$：$$xy\\ge 1$$，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": "充要条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["课内体系->知识点->常用逻辑用语->充分条件与必要条件"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 取特值，$$x=\\frac{1}{3},y=2$$，满足$${{2}^{x}}+{{2}^{y}}\\ge 4$$，不满足$$xy\\ge 1$$；运用基本不等式得$$1\\le xy\\le {{\\left( \\frac{x\\text{+}y}{2} \\right)}^{2}}$$，即$$x\\text{+}y\\ge 2$$，由指数函数的单调性得$${{2}^{x\\text{+}y}}\\ge {{2}^{2}}=4$$，运用基本不等式和充分必要条件的定义 判断可得选项.\\\\ 【详解】\\\\ 解：当$$x=\\frac{1}{3},y=2$$时，$${{2}^{\\frac{1}{3}}}+{{2}^{2}}\\ge 4$$满足$${{2}^{x}}+{{2}^{y}}\\ge 4$$，但$$xy=\\frac{1}{3}\\times 2=\\frac{2}{3}\\textless{} 1$$，不满足$$xy\\ge 1$$，所以$$p$$不是$$q$$的充分条件；\\\\ 当$$xy\\ge 1$$，$$x\\textgreater0$$，$$y\\textgreater0$$时，$$1\\le xy\\le {{\\left( \\frac{x\\text{+}y}{2} \\right)}^{2}}$$，即$$x\\text{+}y\\ge 2$$，当且仅当$$x=y$$时取等号，所以$${{2}^{x\\text{+}y}}\\ge {{2}^{2}}=4$$，即$${{2}^{x}}\\cdot {{2}^{y}}\\ge 4$$，又$$4\\le {{2}^{x}}\\cdot {{2}^{y}}\\le {{\\left( \\frac{{{2}^{x}}\\text{+}{{2}^{y}}}{2} \\right)}^{2}}$$，当且仅当$$x=y$$时取等号，\\\\ 解得$${{2}^{x}}+{{2}^{y}}\\ge 4$$，所以$$p$$是$$q$$的必要条件，\\\\ 因此，$$p$$是$$q$$的必要不充分条件.\\\\ 故选：B. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1007", "queId": "a9f2b5ebecd049f0948655967acf1177", "competition_source_list": ["2010年陕西全国高中数学联赛竞赛初赛第8题8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12题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设对任意的正整数$$n$$，都有$$\\frac{1}{n}\\left[ \\sqrt{1-{{\\left( \\frac{1}{n} \\right)}^{2}}}+\\sqrt{1-{{\\left( \\frac{2}{n} \\right)}^{2}}}+\\cdots +\\sqrt{1-{{\\left( \\frac{n-1}{n} \\right)}^{2}}} \\right]\\textless{}a$$$$\\textless{}\\frac{1}{n}\\left[ 1+\\sqrt{1-{{\\left( \\frac{1}{n} \\right)}^{2}}}+\\cdots +\\sqrt{1-{{\\left( \\frac{n-1}{n} \\right)}^{2}}} \\right]$$  则实数$$a=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{ \\pi }{2}$$ "}], [{"aoVal": "B", "content": "$$\\frac{ \\pi }{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{ \\pi }{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{ \\pi }{6}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->积分", "课内体系->知识点->导数->定积分"], "answer_analysis": ["设$$f(x)=\\sqrt{1-{{x}^{2}}}$$，  把区间$$[0,1]$$平均分成$$n$$份，那么不等式的是$$n$$个矩形的面积（底为$$\\frac{1}{n}$$，高取每个区间右侧的函数值），  右边是$$n$$个矩形的面积（底为$$\\frac{1}{n}$$高取每个区间左侧的函数值）．  当$$n\\to \\infty $$时，左边$$=$$右边$$=f\\left( x \\right)$$在$$[0,1]$$的定积分$$=\\frac{ \\pi }{4}$$（四分之一圆）．  故中间的常数只能取$$\\frac{ \\pi }{4}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "650", "queId": "5fd41c6497da49bbaebf9cac46156183", "competition_source_list": ["2013年天津全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "3", "qtype": "single_choice", "problem": "设函数$$f{}(x){}=\\sqrt{1-{{x}^{2}}}\\centerdot \\left( \\left\\textbar{} x-2 \\right\\textbar+\\frac{4}{{{2}^{x}}-1} \\right)$$．考虑命题$$p:f{(}x{)}$$是奇函数；命题$$q{:}f{(}x{)}$$是偶函数．那么，正确的命题是．", "answer_option_list": [[{"aoVal": "A", "content": "$$p{,}q$$ "}], [{"aoVal": "B", "content": "$$p{,}\\neg q$$ "}], [{"aoVal": "C", "content": "$$\\neg p{,}q$$ "}], [{"aoVal": "D", "content": "$$\\neg p{,}\\neg q$$ "}]], "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->函数的概念", "竞赛->知识点->逻辑->常用逻辑用语"], "answer_analysis": ["注意$$f\\left( x \\right)$$的定义域是$$\\left[ -1,0 \\right)\\cup \\left( 0,1 \\right]$$，所以  $$f\\left( x \\right)=\\sqrt{1-{{x}^{2}}}\\cdot \\left( 2-x+\\frac{4}{{{2}^{x}}-1} \\right)$$  $$=\\sqrt{1-{{x}^{2}}}\\cdot \\left( -x+2\\centerdot \\frac{{{2}^{x}}+1}{{{2}^{x}}-1} \\right)$$，  容易看出$$f\\left( x \\right)$$是奇函数，不是偶函数． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "800", "queId": "963940e825ae4ebdb4bb65778213be38", "competition_source_list": ["2008年湖南全国高中数学联赛竞赛初赛第11题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "多项式$${{\\left( 1+x+{{x}^{2}}+\\cdot \\cdot \\cdot +{{x}^{100}} \\right)}^{3}}$$的展开式在合并同类项后，$${{x}^{150}}$$的系数为~\\uline{~~~~~~~~~~}~．（用数字作答）", "answer_option_list": [[{"aoVal": "A", "content": "$$7431$$ "}], [{"aoVal": "B", "content": "$$7651$$ "}], [{"aoVal": "C", "content": "$$7653$$ "}], [{"aoVal": "D", "content": "$$7831$$ "}], [{"aoVal": "E", "content": "$$8315$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理", "竞赛->知识点->排列组合与概率->二项式定理及其应用"], "answer_analysis": ["由多项式乘法法则可知，  可将问题转化为求方程$$s+t+r=150$$①，  的不超过去$$100$$的自然数解的组数．  显然，方程①的自然数解的组数为$$\\text{C}_{152}^{2}$$，  下面求方程①的超过$$100$$自然数解的组数．  因其和为$$150$$，故只能有一个数超过$$100$$，  不妨设$$s\\textgreater100$$．将方程①化为$$(s-101)+t+r=49$$，  记$${s}'=s-101$$，则方程$${s}'+t+r=49$$的自然数解的组数为$$\\text{C}_{51}^{2}$$，  因此，$${{x}^{150}}$$的系数为$$\\text{C}_{152}^{2}-\\text{C}_{3}^{1}\\text{C}_{51}^{2}=7651$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "566", "queId": "5618914194a343fd9718ed78631c100a", "competition_source_list": ["2012年辽宁全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，设$$BC=a$$，$$AC=b$$，$$AB=c$$，则等式$${{\\sin }^{2}}\\frac{A}{2}+{{\\sin }^{2}}\\frac{B}{2}+{{\\sin }^{2}}\\frac{C}{2}={{\\cos }^{2}}\\frac{B}{2}$$成立的充分必要条件是．", "answer_option_list": [[{"aoVal": "A", "content": "$$c\\cdot a={{b}^{2}}$$ "}], [{"aoVal": "B", "content": "$$a+b=2c$$ "}], [{"aoVal": "C", "content": "$$b+c=2a$$ "}], [{"aoVal": "D", "content": "$$c+a=2b$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["$${{\\sin }^{2}}\\frac{A}{2}+{{\\sin }^{2}}\\frac{B}{2}+{{\\sin }^{2}}\\frac{C}{2}={{\\cos }^{2}}\\frac{B}{2}$$  $$\\Leftrightarrow \\frac{1-\\cos A}{2}+\\frac{1-\\cos B}{2}+\\frac{1-\\cos C}{2}=\\frac{1+\\cos B}{2}$$  $$\\Leftrightarrow \\cos A+\\cos C=2-2\\cos B$$  $$\\Leftrightarrow 2\\cos \\frac{A+C}{2}\\cos \\frac{A-C}{2}=4{{\\sin }^{2}}\\frac{B}{2}$$  $$\\Leftrightarrow 2\\cos \\frac{A-C}{2}=4\\sin \\frac{B}{2}$$  $$\\Leftrightarrow 2\\cos \\frac{A-C}{2}\\sin \\frac{A+C}{2}=4\\sin \\frac{B}{2}\\sin \\frac{A+C}{2}$$  $$\\Leftrightarrow \\sin A+\\sin C=2\\sin B$$  $$\\Leftrightarrow a+c=2b$$ "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1135", "queId": "d89b16b3f1364fec9d5d00ee1e26619f", "competition_source_list": ["2009年竞赛珠海市第11题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "抛两个各面上分别标有$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$的均匀的正方体玩具，``向上的两个数之和为$$3$$''的概率是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{6}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{36}$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{18}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->统计与概率->概率->事件与概率->事件与基本事件空间->基本事件与基本事件空间", "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->素养->数学运算"], "answer_analysis": ["总的基本事件的个数为$$6\\times 6=36$$种，``向上的两个数之和为$$3$$''的基本事件有$$\\left( 1,2 \\right)$$和$$\\left( 2,1 \\right)$$两种，  由古典概型知概率为$$P=\\frac{2}{36}=\\frac{1}{18}$$  故选：$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "413", "queId": "205e5727d7e447f4b468bf5aace4bf59", "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1题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知集合$A=\\left { x\\mid {{2}^{x}}\\le 18 \\right }$，则$A\\cap \\text{N}$的子集个数为（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "8 "}], [{"aoVal": "B", "content": "16 "}], [{"aoVal": "C", "content": "32 "}], [{"aoVal": "D", "content": "64 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 求出$A\\cap \\text{N= } ! ! { ! !\\text{ 0,1,2,3,4 } ! ! } ! !\\text{ }$，即可求解.\\\\ 【详解】\\\\ 由题得${{2}^{x}}\\le 18={{2}^{{{\\log }_{2}}18}}$，$\\therefore x\\le {{\\log }_{2}}18$.\\\\ 因为${{\\log }_{2}}16\\textless{} {{\\log }_{2}}18\\textless{} {{\\log }_{2}}32$，$\\therefore 4\\textless{} {{\\log }_{2}}12\\textless{} 5$.\\\\ 所以$A\\cap \\text{N= } ! ! { ! !\\text{ 0,1,2,3,4 } ! ! } ! !\\text{ }$.\\\\ 所以$A\\cap \\text{N}$的子集个数为${{2}^{5}}=32$个.\\\\ 故选：C. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "166", "queId": "0d1b565ef49d4e7db3618f049d316c16", "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": ["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->用单调性观察法求值域", "课内体系->知识点->三角函数->反三角函数", "课内体系->素养->数学运算"], "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": "478", "queId": "834b110093c34035bef97ac6dde84298", "competition_source_list": ["2020~2021学年10月上海浦东新区华东师范大学第二附属中学高一上学期月考第16题", "竞赛", "2020~2021学年11月浙江杭州滨江区浙江省杭州第二中学高一上学期周测A卷第9题", "2020~2021学年北京高二单元测试", "2020年北京海淀区北京大学自主招生（强基计划）第9题5分", "2020~2021学年北京高三单元测试"], "difficulty": "1", "qtype": "single_choice", "problem": "使得$$5x+12\\sqrt{xy}\\leqslant a\\left( x+y \\right)$$对所有正实数$$x$$，$$y$$都成立的实数$$a$$的最小值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$10$$ "}], [{"aoVal": "D", "content": "前三个答案都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->均值"], "answer_analysis": ["$$\\because x$$，$$y$$均大于$$0$$，  $$\\therefore 5x+12\\sqrt{xy}\\leqslant a\\left( x+y \\right)\\Rightarrow 5+12\\sqrt{\\frac{y}{x}}\\leqslant a\\left( 1+\\frac{y}{x} \\right)$$，  令$$t=\\sqrt{\\frac{y}{x}}$$，$$t\\in {{\\mathbf{R}}^{*}}$$，  则$$a\\geqslant \\frac{5+12t}{1+{{t}^{2}}}$$，  要使该式对所有$$t$$都成立，则$$a\\geqslant {{\\left( \\frac{5+12t}{1+{{t}^{2}}} \\right)}_{\\max }}$$，  令$$m=5+12t$$，  则$$\\frac{5+12t}{1+{{t}^{2}}} =\\frac{144m}{{{m}^{2}}-10m+169}=\\frac{144}{m+\\frac{169}{m}-10}\\leqslant \\frac{144}{2\\times 13-10}=9$$，当且仅当$$m=13$$时，取等号，  故$${{a}_{\\min }}=9$$，故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "459", "queId": "70d597d55db04b3a8fe57f42414df933", "competition_source_list": ["2011~2012学年北京东城区高一上学期期末南片第5题3分", "2016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": "函数$$f\\left( x \\right)={{\\text{e}}^{x}}+2x-3$$的零点所在的一个区间是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-1,0)$$ "}], [{"aoVal": "B", "content": "$$(0,\\frac{1}{2})$$ "}], [{"aoVal": "C", "content": "$$(\\frac{1}{2},1)$$ "}], [{"aoVal": "D", "content": "$$(1,\\frac{3}{2})$$ "}]], "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->函数的零点->求零点所在区间的问题->判定函数零点所在区间（存在性定理）", "课内体系->知识点->函数的应用->函数的零点->函数零点存在定理"], "answer_analysis": ["函数单调递增，  ∵$$f\\left( \\frac{1}{2} \\right)=\\sqrt{e}+\\frac{1}{2}-3\\textless{}0$$，  $$f\\left( 1 \\right)=e+1-3=e-2\\textgreater0$$，  ∴函数在$$(\\frac{1}{2},1)$$内存在唯一零点． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1211", "queId": "fa12e61d9e024e0cb3a73be7024e929b", "competition_source_list": ["2014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$$a$$、$$b$$为正实数，记$$P=\\sqrt{\\frac{{{a}^{2}}+{{b}^{2}}}{2}}-\\frac{a+b}{2}$$，$$Q=\\frac{a+b}{2}-\\sqrt{ab}$$，$$R=\\sqrt{ab}-\\frac{2ab}{a+b}$$，则下列判断正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$P\\geqslant Q\\geqslant R$$ "}], [{"aoVal": "B", "content": "$$Q\\geqslant P\\geqslant R$$ "}], [{"aoVal": "C", "content": "$$Q\\geqslant R\\geqslant R$$ "}], [{"aoVal": "D", "content": "$$P$$、$$Q$$、$$R$$的大小关系不能确定 "}]], "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->证明不等式的基本方法->综合法"], "answer_analysis": ["$$P\\leqslant Q\\Leftrightarrow \\sqrt{\\frac{{{a}^{2}}+{{b}^{2}}}{2}}+\\sqrt{ab}\\leqslant a+b$$  $$\\Leftrightarrow 2\\sqrt{\\frac{{{a}^{2}}+{{b}^{2}}}{2}\\cdot ab}\\leqslant \\frac{{{a}^{2}}+{{b}^{2}}}{2}+ab$$，  $$P\\geqslant R\\Leftrightarrow \\sqrt{\\frac{{{a}^{2}}+{{b}^{2}}}{2}}-\\sqrt{ab}\\geqslant \\frac{a+b}{2}-\\frac{2ab}{a+b}$$  $$\\Leftrightarrow \\frac{{{a}^{2}}+{{b}^{2}}}{2}+ab-\\sqrt{2ab\\left( {{a}^{2}}+{{b}^{2}} \\right)}\\geqslant \\frac{{{\\left( a+b \\right)}^{2}}}{4}-\\frac{2ab\\left( {{a}^{2}}+{{b}^{2}} \\right)}{{{\\left( a+b \\right)}^{2}}}$$  $$\\Leftrightarrow \\frac{{{\\left( a+b \\right)}^{2}}}{4}+\\frac{2ab\\left( {{a}^{2}}+{{b}^{2}} \\right)}{{{\\left( a+b \\right)}^{2}}}\\geqslant \\sqrt{2ab\\left( {{a}^{2}}+{{b}^{2}} \\right)}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "604", "queId": "491d646104ae442d8f8fec314647c7fc", "competition_source_list": ["2009年竞赛第20届全国希望杯数学邀请赛高二复赛第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->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", "课内体系->素养->数学运算"], "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": "463", "queId": "c2fdaaa0f77d484a901cdd3b39290822", "competition_source_list": ["2010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10月北京海淀区清华大学附属中学高一上学期月考理科第5题", "2008年高考真题辽宁卷理科第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$f(x)$$是连续的偶函数，且当$$x\\textgreater0$$时$$f(x)$$是单调函数，则满足$$f(x)=f\\left( \\frac{x+3}{x+4} \\right)$$的所有$$x$$之和为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-3$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$-8$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "knowledge_point_routes": ["知识标签->素养->逻辑推理", "知识标签->素养->数学运算", "知识标签->题型->函数->函数的性质->抽象函数的问题->抽象函数与函数性质综合", "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"], "answer_analysis": ["由$$f(x)$$的性质知$$f(x)=f\\left( \\frac{x+3}{x+4} \\right)\\Rightarrow \\left\\textbar{} x \\right\\textbar=\\left\\textbar{} \\frac{x+3}{x+4} \\right\\textbar$$，若$$x=\\frac{x+3}{x+4}$$，即$${{x}^{2}}+3x-3=0$$，有解，且两根之和为$$-3$$；  若$$-x=\\frac{x+3}{x+4}$$，即$${{x}^{2}}+5x+3=0$$，有解，且两根之和为$$-5$$，故总和为$$-8$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "227", "queId": "3d4ab39216704bb3b440b6f92a8d74ce", "competition_source_list": ["1990年全国高中数学联赛竞赛一试第4题"], "difficulty": "2", "qtype": "single_choice", "problem": "点集$$\\left { \\left( x,y \\right)\\textbar\\lg \\left( {{x}^{3}}+\\frac{1}{3}{{y}^{3}}+\\frac{1}{9} \\right)=\\lg x+\\lg y \\right }$$中元素的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "多于$$2$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的应用->函数与方程", "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->基本初等函数->对数函数->对数函数的概念", "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", "课内体系->素养->数学运算"], "answer_analysis": ["$$\\lg \\left( {{x}^{3}}+\\frac{1}{3}{{y}^{3}}+\\frac{1}{9} \\right)=\\lg x+\\lg y$$，  $$\\Leftrightarrow {{x}^{3}}+\\frac{1}{3}{{y}^{3}}+\\frac{1}{9}=xy\\mathsf{}x\\textgreater0\\mathsf{}y\\textgreater0\\mathsf{}$$，  由均值不等式，  $${{x}^{3}}+\\frac{1}{3}{{y}^{3}}+\\frac{1}{9}\\mathsf{\\geqslant }3\\sqrt[3]{({{x}^{3}})\\left( \\frac{1}{3}{{y}^{3}} \\right)\\left( \\frac{1}{9} \\right)}=xy\\mathsf{}$$，  当且仅当 $$\\begin{cases}{{x}^{3}}=\\frac{1}{9}    \\frac{1}{3}{{y}^{3}}=\\frac{1}{9}\\mathsf{}   \\end{cases}$$，  上式等号成立，解此二方程得：  $$x=\\sqrt[3]{\\frac{1}{9}}\\mathsf{}y=\\sqrt[3]{\\frac{1}{3}}\\mathsf{}$$，  故点集中有唯一点为：  $$\\left( \\sqrt[3]{\\frac{1}{9}}，\\sqrt[3]{\\frac{1}{3}} \\right)\\mathsf{}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "238", "queId": "34399ff6e56446448893d9727acfbe61", "competition_source_list": ["2008年江西全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "四面体$$ABCD$$的六条棱长分别为$$7$$、$$13$$、$$18$$、$$27$$、$$36$$、$$41$$，且知$$AB=41$$，则$$CD=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$18$$ "}], [{"aoVal": "D", "content": "$$27$$ "}]], "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": ["四面体中除$$CD$$外，其余的棱皆与$$AB$$相邻接，  若长为$$13$$的棱与$$AB$$相邻，不妨设$$BC=13$$，  根据构成三角形条件，可知$$AC\\notin \\left { 7,18,27 \\right }$$，  因此$$AC=36$$，$$BD=7$$，可得$$ {AD,CD }= {18,27 }$$，  于是$$\\triangle ABD$$中，两边之和小于第三边，矛盾．因此只有$$CD=13$$．  另一方面，使$$AB=41$$，$$CD=13$$的四面体$$ABCD$$可作出，  例如取$$BC=7$$，$$AC=36$$，$$BD=18$$，$$AD=27$$．故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1021", "queId": "ce3d9b94a6bb4de5952b86d4eb2ae3eb", "competition_source_list": ["2013年湖北全国高中数学联赛竞赛初赛第11题20分", "2013年湖北全国高中数学联赛高一竞赛初赛第11题20分", "2013年湖北全国高中数学联赛高二竞赛初赛第11题20分"], "difficulty": "1", "qtype": "single_choice", "problem": "求函数$$y={{x}^{2}}+x\\sqrt{{{x}^{2}}-1}$$的值域.", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ \\frac{1}{2},+\\infty \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left[ 1,+\\infty \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{1}{2},1 \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{1}{2},+\\infty \\right)$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->方法->构造法", "课内体系->思想->转化化归思想", "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->函数的表示方法->解析法", "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->求复合函数的值域", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域->求函数的值域->用单调性观察法求值域", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->函数的定义", "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->函数的概念与性质->函数的概念及其表示->函数的定义域->求具体函数（包括复合函数）的定义域", "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", "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": ["易求得函数的定义域为$$ {x\\textbar x{\\geqslant }1$$或$$x{\\leqslant }-1 }$$.  易知函数$$y={{x}^{2}}+x\\sqrt{{{x}^{2}}-1}$$是$$\\left[ 1,+\\infty \\right)$$上的增函数，所以，可得$$y{\\geqslant }1$$.  当$$x{\\leqslant }-1$$时，  $$y=x\\left( x+\\sqrt{{{x}^{2}}-1} \\right)=\\frac{x\\left( x+\\sqrt{{{x}^{2}}-1} \\right)\\left( x-\\sqrt{{{x}^{2}}-1} \\right)}{x-\\sqrt{{{x}^{2}}-1}}$$  $$=\\frac{x}{x-\\sqrt{{{x}^{2}}-1}}=\\frac{1}{1-\\frac{\\sqrt{{{x}^{2}}-1}}{x}}=\\frac{1}{1+\\sqrt{1-\\frac{1}{{{x}^{2}}}}}$$.  因为$$x{\\leqslant }-1$$，所以$$0{\\leqslant }1-\\frac{1}{{{x}^{2}}}\\textless{}1$$，$$1{\\leqslant }1+\\sqrt{1-\\frac{1}{{{x}^{2}}}}\\textless{}2$$，所以$$\\frac{1}{2}\\textless{}\\frac{1}{1+\\sqrt{1-\\frac{1}{{{x}^{2}}}}}{\\leqslant }1$$，即$$\\frac{1}{2}\\textless{}y{\\leqslant }1$$.  综上可知：$$y\\textgreater\\frac{1}{2}$$，故函数$$y={{x}^{2}}+x\\sqrt{{{x}^{2}}-1}$$的值域为$$\\left( \\frac{1}{2},+\\infty \\right)$$ "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "452", "queId": "3a54b80d6dde48b6ba9fc3869bcce418", "competition_source_list": ["2010年黑龙江全国高中数学联赛竞赛初赛第11题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "空间中一点$$P$$到三条两两垂直的射线$$OA,  OB,  OC$$的距离分别为$$\\sqrt{3},  2,\\sqrt{5}$$，且垂足分别为$${A}',  {B}',  {C}'$$，则三棱锥$$P-{A}'{B}'{C}'$$的体积为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\sqrt{6}}{2}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{6}}{3}$$ "}]], "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$$O{A}',  O{B}',  O{C}'$$的长度分别为$$a,  b,  c,$$则  $${{a}^{2}}+{{b}^{2}}=5,  {{b}^{2}}+{{c}^{2}}=3,  {{c}^{2}}+{{a}^{2}}=4\\Rightarrow {{a}^{2}}=3,  {{b}^{2}}=2,  {{c}^{2}}=1,$$  由体积分割可得$$V=abc,$$ $${{V}_{P-{A}'{B}'{C}'}}=abc-\\frac{4}{6}abc=\\frac{1}{3}abc=\\frac{\\sqrt{6}}{3}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "770", "queId": "91912c0ac8b54e4c9e04e9d98e27cac4", "competition_source_list": ["2017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": "已知$$\\alpha ,\\beta \\in \\left( 0 ,\\pi ~\\right)$$，$$\\tan \\alpha ,\\tan \\beta $$是方程$${{x}^{2}}+3x+1=0$$的两个根，则$$\\cos \\left( \\alpha -\\beta \\right)$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{5}}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{5}}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["由韦达定理，$$\\tan \\alpha +\\tan \\beta =-3$$，$$\\tan \\alpha \\tan \\beta =1$$，  $$\\frac{\\sin \\alpha }{\\cos \\alpha }\\cdot \\frac{\\sin \\beta }{\\cos \\beta }=1$$，得$$\\cos \\left( \\alpha +\\beta \\right)=0$$，结合$$\\tan \\alpha +\\tan \\beta =-3$$，可知$$\\alpha +\\beta =\\frac{3}{2} \\pi $$．  $$\\tan \\alpha +\\tan \\beta =\\frac{\\sin \\left( \\alpha +\\beta \\right)}{\\cos \\alpha \\cos \\beta }=-\\frac{1}{\\cos \\alpha \\cos \\beta }=-3$$，所以$$\\cos \\alpha \\cos \\beta =\\frac{1}{3}$$．  因此，$$\\cos \\left( \\alpha -\\beta \\right)=\\cos \\alpha \\cos \\beta +\\sin \\alpha \\sin \\beta =2\\cos \\alpha \\cos \\beta =\\frac{2}{3}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "24", "queId": "0237a5f592c54fbf936f86e053dcd8c4", "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6~6题"], "difficulty": "0", "qtype": "single_choice", "problem": "已知$$\\textbar\\vec{a}\\textbar=\\sqrt{2},\\textbar\\vec{b}\\textbar=1,\\vec{a}\\cdot (\\vec{a}-\\vec{b})=1$$，则$$\\vec{a}$$与$$\\vec{b}$$的夹角为(~~~~~~~)", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{2\\pi }{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\pi }{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\pi }{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{9\\pi }{4}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->平面向量->平面向量的运算->数量积"], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据向量的数量积计算运算律即可计算.\\\\ 【详解】\\\\ $$\\vec{a}\\cdot \\left( \\vec{a}-\\vec{b} \\right)=1\\Rightarrow \\textbar\\vec{a}{{\\textbar}^{2}}-\\vec{a}\\cdot \\vec{b}=1\\Rightarrow \\vec{a}\\cdot \\vec{b}=1$$，\\\\ $$\\text{cos}\\left\\langle \\vec{a},\\vec{b} \\right\\rangle =\\frac{\\vec{a}\\cdot \\vec{b}}{\\left\\textbar{} {\\vec{a}} \\right\\textbar\\left\\textbar{} {\\vec{b}} \\right\\textbar}=\\frac{1}{\\sqrt{2}\\times 1}=\\frac{\\sqrt{2}}{2}$$，\\\\ $$\\because \\left\\langle \\vec{a},\\vec{b} \\right\\rangle \\in \\left[ 0,\\pi \\right]$$，\\\\ $$\\therefore \\left\\langle \\vec{a},\\vec{b} \\right\\rangle =\\frac{\\pi }{4}$$.\\\\ 故选：C. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1107", "queId": "c6253104b54f490292a0d0c69b08b1d9", "competition_source_list": ["2019~2020学年广东广州高二上学期期末", "2019~2020学年宁夏银川兴庆区宁夏回族自治区银川一中高二下学期期中理科第12题5分", "2019~2020学年广东广州天河区华南师范大学附属中学高二上学期期末第11题3分", "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3题3分"], "difficulty": "2", "qtype": "single_choice", "problem": "从$$0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$这六个数字中任取两个奇数和两个偶数，组成没有重复数字的四位数的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$300$$ "}], [{"aoVal": "B", "content": "$$216$$ "}], [{"aoVal": "C", "content": "$$180$$ "}], [{"aoVal": "D", "content": "$$162$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->排列与组合->排除法", "课内体系->知识点->计数原理->两个基本计数原理->加法原理与乘法原理的综合运用", "课内体系->素养->逻辑推理", "课内体系->素养->数学运算"], "answer_analysis": ["由题意知，本题是一个分类计数原理，  第一类：从$$1$$，$$2$$，$$3$$，$$4$$，$$5$$中任取两个奇数和两个偶数，  组成没有重复数字的四位数的个数为$$\\text{C}_{3}^{2}\\text{A}_{4}^{4}=72$$，  第二类：取$$0$$，此时$$2$$和$$4$$只能取一个，$$0$$不能排在首位，  组成没有重复数字的四位数的个数为$$\\text{C}_{3}^{2}\\text{C}_{2}^{1}\\left( \\text{A}_{4}^{4}-\\text{A}_{3}^{3} \\right)=108$$，  $$\\therefore $$组成没有重复数字的四位数的个数为$$108+72=180$$，  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "261", "queId": "26eb2ae0826d49899d07630ace5b3b30", "competition_source_list": ["2009年AMC12竞赛A第24题"], "difficulty": "3", "qtype": "single_choice", "problem": "2009AMC12A, 24  \\uline{The tower function of twos} is defined recursively as follows: $$T\\left(1\\right)=2$$ and $$T\\left(n+1\\right)=2^{T\\left(n\\right)}$$ for $$n\\geqslant 1$$. Let $$A=\\left(T\\left(2009\\right)\\right)^{T\\left(2009\\right)}$$ and $$B=\\left(T\\left(2009\\right)\\right)^{A}$$. What is the largest integer $$k$$ such that $$\\underbrace{\\log_{2}\\log_{2}\\log_{2}\\ldots\\log_{2}B}_{{k\\textasciitilde{} \\text{times}}}$$ is defined?  2的\\uline{塔函数}定义如下: $$T\\left(1\\right)=2$$ , $$T\\left(n+1\\right)=2^{T\\left(n\\right)}$$ ($$n\\geqslant 1$$). 令 $$A=\\left(T\\left(2009\\right)\\right)^{T\\left(2009\\right)}$$ , $$B=\\left(T\\left(2009\\right)\\right)^{A}$$. 使得 $$\\underbrace{\\log_{2}\\log_{2}\\log_{2}\\ldots\\log_{2}B}_{{k\\textasciitilde{} \\text{times}}}$$有意义的最大整数 $$k$$是多少?", "answer_option_list": [[{"aoVal": "A", "content": "$$2009$$ "}], [{"aoVal": "B", "content": "$$2010$$ "}], [{"aoVal": "C", "content": "$$2011$$ "}], [{"aoVal": "D", "content": "$$2012$$ "}], [{"aoVal": "E", "content": "$$2013$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Reasoning->Mapping and Correspondence", "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$\\log_2$为$\\log$.  由条件, 有$\\log{T(n+1)}=T(n)$.  我们对$B$求几次对数看看情况.  $\\log{B}=A\\log{T(2009)}=A\\cdot T(2008)=T(2009)^{T(2009)}T(2008)$  $\\log{\\log{B}}=T(2009)\\log{T(2009)}+\\log{T(2008)}=T(2009)T(2008)+T(2007)$  再往下似乎不太好求了, 不过我们可以进行一些适当的舍入.  $\\log{\\log{B}}=T(2009)T(2008)+T(2007)=T(2009)\\left(T(2008)+\\frac{T(2007)}{T(2009)}\\right)$  $\\Rightarrow\\log\\log\\log{B}=\\log{T(2009)}+\\log{\\left(T(2008)+\\frac{T(2007)}{T(2009)}\\right)}\\approx\\log{T(2009)}+\\log{T(2008)}=T(2008)+T(2007)$  考虑到$T(2007)=\\log\\log{T(2009)}\\textless\\textless T(2009)$, 这里的舍入是合理的.  故技重施,  $\\log\\log\\log\\log{B}=\\log{\\left(T(2008)\\left(1+\\frac{T(2007)}{T(2008)}\\right)\\right)}=\\log{T(2008)}+\\log{\\left(1+\\frac{T(2007)}{T(2008)}\\right)}\\approx T(2007)$  到这里就好求了, 从$T(2007)$再往下求$$2007$$层, 会得到$$1.$$ 考虑到我们在计算过程中做了一些舍入, 实际上这时的值会略大于$$1.$$ 再求一层, 得到一个略大于$$0$$的值$$.$$ 再求一层, 得到一个负值$$.$$ 之后再往下，就没有意义了.  因此, 总层数$=2+2007+4=2013$, 选E. "], "answer_value": "E"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "127", "queId": "4ab9e8ac8b77412a95503a1784cf9113", "competition_source_list": ["全国高中数学联赛竞赛复赛一试"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$f(x)={{x}^{2}}- \\pi x$$，$$\\alpha =\\arcsin \\frac{1}{3}$$，$$\\beta =\\text{arctg}\\frac{5}{4}$$，$$\\gamma =\\arccos \\left( -\\frac{1}{3} \\right)$$，$$\\delta =\\text{arcctg}\\left( -\\frac{5}{4} \\right)$$，则（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$f(\\alpha )\\textgreater f(\\beta )\\textgreater f(\\delta )\\textgreater f(\\gamma )$$ "}], [{"aoVal": "B", "content": "$$f(\\alpha )\\textgreater f(\\delta )\\textgreater f(\\beta )\\textgreater f(\\gamma )$$ "}], [{"aoVal": "C", "content": "$$f(\\delta )\\textgreater f(\\alpha )\\textgreater f(\\beta )\\textgreater f(\\gamma )$$ "}], [{"aoVal": "D", "content": "$$f(\\delta )\\textgreater f(\\alpha )\\textgreater f(\\gamma )\\textgreater f(\\beta )$$ "}]], "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->含参二次函数的图象及性质", "课内体系->素养->逻辑推理"], "answer_analysis": ["由题意，$$f(x)$$的图像关于直线$$x=\\frac{\\mathsf{\\pi }}{2}$$对称，且在$$\\left( -\\infty ,\\frac{ \\pi }{2} \\right)$$单调减少，在$$\\left( \\frac{ \\pi }{2},+\\infty \\right)$$单调增加．所以$$\\left\\textbar{} {{x}_{1}}-\\frac{\\mathsf{\\pi }}{2} \\right\\textbar\\textgreater\\left\\textbar{} {{x}_{2}}-\\frac{\\mathsf{\\pi }}{2} \\right\\textbar$$时，有$$f({{x}_{1}})\\textgreater f({{x}_{2}})$$．又易知$$0\\textless{}\\alpha \\textless{}\\frac{\\mathsf{\\pi }}{6}$$，$$\\frac{\\mathsf{\\pi }}{4}\\textless{}\\beta \\textless{}\\frac{\\mathsf{\\pi }}{3}$$，$$\\frac{\\mathsf{\\pi }}{2}\\textless{}\\gamma \\textless{}\\frac{2\\mathsf{\\pi }}{3}$$，$$\\frac{3\\mathsf{\\pi }}{4}\\textless{}\\delta \\textless{}\\frac{5\\mathsf{\\pi }}{6}$$，  所以，$$0\\textless{}\\left\\textbar{} \\gamma -\\frac{\\mathsf{\\pi }}{2} \\right\\textbar\\textless{}\\frac{\\mathsf{\\pi }}{6}\\textless{}\\left\\textbar{} \\beta -\\frac{\\mathsf{\\pi }}{2} \\right\\textbar\\textless{}\\frac{\\mathsf{\\pi }}{4}\\textless{}\\left\\textbar{} \\delta -\\frac{\\mathsf{\\pi }}{2} \\right\\textbar\\textless{}\\frac{\\mathsf{\\pi }}{3}\\textless{}\\left\\textbar{} \\alpha -\\frac{\\mathsf{\\pi }}{2} \\right\\textbar\\textless{}\\frac{\\mathsf{\\pi }}{2}$$．故有： $$f(\\alpha )\\textgreater f(\\delta )\\textgreater f(\\beta )\\textgreater f(\\gamma )$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1079", "queId": "d7d7799a14374b52bfc9536508df744c", "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": "$$1\\textless{}a\\textless{}2$$ "}], [{"aoVal": "B", "content": "$1\\lt a\\leqslant2$ "}], [{"aoVal": "C", "content": "$a\\gt2$ "}], [{"aoVal": "D", "content": "$a\\geqslant2$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "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": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1180", "queId": "f9347a5eb63946e69ce20f7d3c761fbd", "competition_source_list": ["2013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": "若三位数$$\\overrightarrow{abc}$$被$$7$$整除，且$$a,b,c$$成公差非零的等差数列，则这样的整数共有个．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$7$$ "}], [{"aoVal": "D", "content": "$$8$$ "}]], "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": ["设三位数为$$\\overline{\\left( b-d \\right)b\\left( b+d \\right)}=111b-99d{}0{\\leqslant }b\\textless{}9{,}-9\\textless{}d\\textless{}9,d\\ne 0{}$$，由$$7\\textbar{}111b-99d{}$$ $$\\Rightarrow 7\\textbar{}b+d{}\\Rightarrow b=1,d=-1;b=2,d=-2;b=3,d=-3;b=4,d=3,-4;b=5,d=2;b=6,d=1;b=8,d=-1;$$．所以，所有的三位数为$$210$$，$$420$$，$$630$$，$$147$$，$$840$$，$$357$$，$$567$$，$$987$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1148", "queId": "d8aeded1700d4f63ad9ce132d5e60bd0", "competition_source_list": ["2007年全国全国高中数学联赛竞赛一试第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "将号码分别为$$1$$、$$2$$、\\ldots、$$9$$的九个小球放入一个袋中，这些小球仅号码不同，其余完全相同．甲从袋中摸出一个球，其号码为$$a$$，放回后，乙从此袋中再摸出一个球，其号码为$$b$$．则使不等式$$a-2b+10\\textgreater0$$成立的事件发生的概率等于（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{52}{81}$$ "}], [{"aoVal": "B", "content": "$$\\frac{59}{81}$$ "}], [{"aoVal": "C", "content": "$$\\frac{60}{81}$$ "}], [{"aoVal": "D", "content": "$$\\frac{61}{81}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->概率初步"], "answer_analysis": ["甲、乙二人每人摸出一个小球都有$$9$$种不同的结果，故基本事件总数为$${{9}^{2}}=81$$个．由不等式$$a-2b+10\\textgreater0$$得$$2b ~\\textless{} ~a+10$$，于是，当$$b=1$$、$$2$$、$$3$$、$$4$$、$$5$$时，每种情形$$a$$可取$$1$$、$$2$$、\\ldots、$$9$$中每一个值，使不等式成立，则共有$$9\\times 5=45$$种；当$$b=6$$时，$$a$$可取$$3$$、$$4$$、\\ldots、$$9$$中每一个值，有$$7$$种；当$$b=7$$时，$$a$$可取$$5$$、$$6$$、$$7$$、$$8$$、$$9$$中每一个值，有$$5$$种；当$$b=8$$时，$$a$$可取$$7$$、$$8$$、$$9$$中每一个值，有$$3$$种；当$$b=9$$时，$$a$$只能取$$9$$，有$$1$$种．于是，所求事件的概率为$$\\frac{45+7+5+3+1}{81}=\\frac{61}{81}$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "829", "queId": "df945687f70b4a6a8c512961e8e62e32", "competition_source_list": ["2009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": "若对任意的$$x\\in R$$，函数$$f(x)$$满足$$f(x+2009)=-f(x+2008)$$，且$$f(2009)=-2009$$，则$$f(-1)=$$ ．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$2009$$ "}], [{"aoVal": "D", "content": "$$-2009$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数综合"], "answer_analysis": ["$$f(x+2009)=-f(x+2008)=-f(x-1)+2009)=f(x-1+2008)=f(x+2007)$$  所以$$f(x)$$是以$$2$$为周期的周期函数，从而有  $$f(-1)=f(2009-2\\times 1005)=f(2009)=-2009$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "269", "queId": "8faabed1166544d89535e6ab1242e775", "competition_source_list": ["2010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$$5$$件产品中有$$4$$件正品、$$1$$件次品．从中任取$$2$$件，记其中含正品的个数为随机变量$$\\xi $$，则$$\\xi $$的数学期望$$E\\xi $$是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{6}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{7}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{8}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{9}{5}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["数学期望是：$$1\\times \\frac{\\text{C}_{4}^{1}}{\\text{C}_{5}^{2}}+2\\times \\frac{\\text{C}_{4}^{2}}{\\text{C}_{5}^{2}}=\\frac{8}{5}$$．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "421", "queId": "abcedcb6d8524ca2a7451922a35d9a76", "competition_source_list": ["2016~2017学年4月湖南高三下学期月考理科第11题5分", "2018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": "已知抛物线$$C$$：$${{y}^{2}}=2px$$（$$p\\textgreater0$$）与动直线$$l$$：$$y=hx+b$$（$$k$$、$$b$$为参变量，且$$k\\ne 0$$，$$b\\ne 0$$）交于$$A({{x}_{1}},{{y}_{1}})$$、$$B({{x}_{2}},{{y}_{2}})$$两点，直角坐标系原点为$$O$$，记直线$$OA$$、$$OB$$的斜率分别为$${{k}_{OA}}$$、$${{k}_{OB}}$$，若$${{k}_{OA}}{{k}_{OB}}=\\sqrt{3}$$恒成立，则当$$k$$ 变化时，直线$$l$$恒过的定点为．", "answer_option_list": [[{"aoVal": "A", "content": "$$(-\\sqrt{3}p,0)$$ "}], [{"aoVal": "B", "content": "$$(-2\\sqrt{3}p,0)$$ "}], [{"aoVal": "C", "content": "$$\\left( -\\frac{\\sqrt{3}}{3}p,0 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( -\\frac{2\\sqrt{3}}{3}p,0 \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与圆锥曲线"], "answer_analysis": ["将直线方程与抛物线方程联立，消去$$y$$后整理得$${{k}^{2}}{{x}^{2}}+(2kb-2p)x+{{b}^{2}}=0$$，  故$${{x}_{1}}+{{x}_{2}}=\\frac{-2kb+2p}{{{k}^{2}}}$$，$${{x}_{1}}{{x}_{2}}=\\frac{{{b}^{2}}}{{{k}^{2}}}$$，  因为$${{k}_{OA}}{{k}_{OB}}=\\sqrt{3}$$，  所以$${{y}_{1}}{{y}_{2}}=\\sqrt{3}{{x}_{1}}{{x}_{2}}$$，  则$${{y}_{1}}{{y}_{2}}=(k{{x}_{1}}+b)(k{{x}_{2}}+b)$$  $$={{k}^{2}}{{x}_{1}}{{x}_{2}}+kb({{x}_{1}}+{{x}_{2}})+{{b}^{2}}$$  $$=\\frac{2bp}{k}$$  $$=\\sqrt{3}{{x}_{1}}{{x}_{2}}$$  $$=\\frac{\\sqrt{3}{{b}^{2}}}{{{k}^{2}}}$$，  由于$$b\\ne 0$$，于是$$b=\\frac{2pk}{\\sqrt{3}}$$，  故$$y=kx+\\frac{2pk}{\\sqrt{3}}=k\\left( x+\\frac{2}{\\sqrt{3}}p \\right)$$，  因此，直线$$l$$恒过定点$$\\left( -\\frac{2\\sqrt{3}}{3}p,0 \\right)$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "443", "queId": "f13f7f5c8aeb4f70b7ab6de8dec7b440", "competition_source_list": ["2016年天津全国高中数学联赛竞赛初赛第4题6分", "2016年高考真题天津卷"], "difficulty": "1", "qtype": "single_choice", "problem": "掷两次骰子，用$$X$$记两次掷得点数的最大值，则下列各数中，与期望$$E\\left( X \\right)$$最接近的数是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$4.5$$ "}], [{"aoVal": "C", "content": "$$5$$ "}], [{"aoVal": "D", "content": "$$5.5$$ "}]], "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->随机变量及其分布->离散型随机变量的数字特征->离散型随机变量的数学期望", "课内体系->知识点->统计与概率->概率->随机变量及其分布->离散型随机变量的分布列"], "answer_analysis": ["$$X=1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，则：  $$P\\left( X=1 \\right)=\\frac{1}{36}$$，$$P\\left( X=2 \\right)=\\frac{3}{36}=\\frac{1}{12}$$，$$P\\left( X=3 \\right)=\\frac{5}{36}$$，  $$P\\left( X=4 \\right)=\\frac{7}{36}$$，$$P\\left( X=5 \\right)=\\frac{9}{36}$$，$$P\\left( X=6 \\right)=\\frac{11}{36}$$．  所以，$$E\\left( X \\right)=\\frac{1}{36}+\\frac{2}{12}+\\frac{15}{36}+\\frac{28}{36}+\\frac{45}{36}+\\frac{66}{36}=\\frac{161}{36}=4\\frac{17}{36}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "539", "queId": "55f625b6f0054bbba7b388490270cfd0", "competition_source_list": ["2011年江苏全国高中数学联赛竞赛复赛第2题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知函数$$f(x)=\\frac{x}{\\sqrt{{{x}^{2}}+2x+4}}$$，则$$f(x)$$的值域为（）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ \\frac{2\\sqrt{3}}{3},+\\infty~ \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left[ -\\frac{2\\sqrt{3}}{3},  1 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( 0,\\frac{2\\sqrt{3}}{3} \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left( 0,\\frac{2\\sqrt{3}}{3} \\right)$$ "}]], "knowledge_point_routes": ["知识标签->思想->分类讨论", "知识标签->方法->定义法", "知识标签->素养->数学运算", "知识标签->知识点->函数->函数及其表示->函数的值域", "知识标签->题型->函数->函数及其表示->函数的值域->求复合函数的值域"], "answer_analysis": ["$$x\\textgreater0$$时，$$f\\left( x \\right)=\\frac{1}{\\sqrt{\\frac{4}{{{x}^{2}}}+\\frac{2}{x}+1}}\\textless{}\\frac{1}{\\sqrt{1}}=1$$；  $$x=0$$时，$$f\\left( x \\right)=0$$；  $$x\\textless{}0$$时，$$f\\left( x \\right)=-\\frac{1}{\\sqrt{\\frac{4}{{{x}^{2}}}+\\frac{2}{x}+1}}=-\\frac{1}{\\sqrt{{{\\left( \\frac{2}{x}+\\frac{1}{2} \\right)}^{2}}+\\frac{3}{4}}}\\geqslant -\\frac{1}{\\sqrt{\\frac{3}{4}}}=-\\frac{2}{3}\\sqrt{3}$$．  所以，$$f\\left( x \\right)$$的值域为$$\\left[ -\\frac{2\\sqrt{3}}{3},  1 \\right)$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "558", "queId": "9523433489204afa942e73202b612f1f", "competition_source_list": ["1988年全国高中数学联赛竞赛一试第3题"], "difficulty": "1", "qtype": "single_choice", "problem": "平面上有三个点集$$M$$，$$N$$，$$P$$：（~ ）．$$M=\\left { \\left( x,y \\right)\\textbar\\textbar x\\textbar+\\textbar y\\textbar\\textless{}1 \\right }$$，$$N=\\left { \\left( x,y \\right)\\textbar\\sqrt{\\left( x-\\frac{1}{2} \\right)+{{\\left( y+\\frac{1}{2} \\right)}^{2}}},+\\sqrt{{{\\left( x+\\frac{1}{2} \\right)}^{2}}+{{\\left( y-\\frac{1}{2} \\right)}^{2}}}\\textless{}2\\sqrt{2}, \\right }$$，$$P=\\left { \\left( x,y \\right)\\textbar\\textbar x+y\\textbar\\textless{}1,\\textbar x\\textbar\\textless{}1,\\textbar y\\textbar\\textless{}1 \\right }$$则．", "answer_option_list": [[{"aoVal": "A", "content": "$$M\\subset P\\subset N$$； "}], [{"aoVal": "B", "content": "$$M\\subset N\\subset P$$； "}], [{"aoVal": "C", "content": "$$P\\subset N\\subset M$$； "}], [{"aoVal": "D", "content": "$$\\text{A,B,C}$$都不成立 "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->椭圆", "竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["$$M$$是正方形$$BCEF$$内部，$$P$$是六边形$$ABCDEF$$内部，$$N$$是从$$AD$$为长轴的椭圆内部，易知正方形顶点在椭圆内．故答Ａ． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "790", "queId": "6e5eb99439c641a880ea6553e7d1ad9c", "competition_source_list": ["2010年山东全国高中数学联赛竞赛初赛第8题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "多项式$${{(a+b+c+d)}^{8}}$$的展开式中，每一字母的指数均不为零的项共有．", "answer_option_list": [[{"aoVal": "A", "content": "$$35$$项 "}], [{"aoVal": "B", "content": "$$42$$项 "}], [{"aoVal": "C", "content": "$$45$$项 "}], [{"aoVal": "D", "content": "$$50$$项 "}]], "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": ["$${{(a+b+c+d)}^{8}}$$的展开式中的项$$p{{a}^{{{x}_{1}}}}{{b}^{{{x}_{2}}}}{{c}^{{{x}_{3}}}}{{d}^{{{x}_{4}}}}\\frac{1}{2}$$，  若其每一字母的指数不为零，则$${{x}_{i}}\\geqslant 1,i=,2,3,4,$$  且$${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=8$$．  令$${{u}_{1}}={{x}_{1}},{{u}_{2}}={{x}_{1}}+{{x}_{2}},{{u}_{3}}={{x}_{1}}+{{x}_{2}}+{{x}_{3}},{{u}_{4}}=8$$，  则有$$1\\leqslant {{u}_{1}}\\textless{}{{u}_{2}}\\textless{}{{u}_{3}}\\leqslant 7$$．  $$\\left { {{u}_{1}},{{u}_{2}},{{u}_{3}} \\right }$$是集合$$\\left { 1,2,\\cdots ,7 \\right }$$的一个$$3$$元子集．  显然，任一这样的$$3$$元子集，  只要令$${{x}_{1}}={{u}_{1}},{{x}_{i}}={{u}_{i}}-{{u}_{i-1}},i\\geqslant 2$$，  即可得一组符合要求的$$\\left { {{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}} \\right }$$．所以，所要求的项数为$$\\text{C}_{7}^{3}=35$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "119", "queId": "461296582a7545138444967466e9720d", "competition_source_list": ["2008年山西全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，$$D$$是$$BC$$边的中点，若$$\\overrightarrow{AD}\\cdot \\overrightarrow{AC}=0$$，则$$\\tan A+2\\tan C$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{\\sqrt{3}}{3}$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{3}}{3}$$ "}], [{"aoVal": "D", "content": "$$1$$ "}]], "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": ["取$$AB$$的中点$$E$$，则$$DE//AC$$，$$\\angle ADE=90{}^{}\\circ $$，$$\\angle DAE=A-90{}^{}\\circ $$，  记$$AC=b$$，$$AD=m$$，  则$$\\tan C=\\frac{m}{b}$$，$$\\tan (A-90{}^{}\\circ )=\\frac{b}{2m}$$，  即$$-\\cot A=\\frac{b}{2m}$$，  所以$$\\tan A=\\frac{-2m}{b}$$，  因此$$\\tan A+2\\tan C=0$$，  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "187", "queId": "41b6d9373aeb45ad8f195d8dac20b9aa", "competition_source_list": ["2010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": "四面体$$S-ABC$$中，三组对棱分别相等，依次为$$5,4,x$$，则$$x$$的取值范围为．", "answer_option_list": [[{"aoVal": "A", "content": "$$(2,\\sqrt{41})$$ "}], [{"aoVal": "B", "content": "$$(3,9)$$ "}], [{"aoVal": "C", "content": "$$(3,\\sqrt{41})$$ "}], [{"aoVal": "D", "content": "$$(2,9)$$ "}]], "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": ["将四面体嵌入到长方体中，设长方体的棱长分别为$$a,b,c$$，  则$$\\begin{cases}{{a}^{2}}+{{c}^{2}}={{4}^{2}}    {{b}^{2}}+{{c}^{2}}={{5}^{2}} \\end{cases}$$，从而$$0\\textless{}c\\textless{}4$$．  所以$$x=\\sqrt{{{a}^{2}}+{{b}^{2}}}=\\sqrt{{{4}^{2}}+{{5}^{2}}-2{{c}^{2}}}\\in (3,\\sqrt{41})$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "312", "queId": "2789b36ec1ab482e9fe2cdfcf46ccdcb", "competition_source_list": ["1984年全国高中数学联赛竞赛一试第3题", "2015年湖南全国高中数学联赛竞赛初赛第4题5分", "2006年上海复旦大学自主招生千分考第22题"], "difficulty": "3", "qtype": "single_choice", "problem": "对所有满足$$1\\leqslant n\\leqslant m\\leqslant 5$$的$$m$$，$$n$$，极坐标方程$$\\rho =\\frac{1}{1-\\text{C}_{m}^{n}\\cos \\theta }$$表示的不同双曲线条数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$9$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$15$$ "}]], "knowledge_point_routes": ["竞赛->知识点->极坐标系与直角坐标系"], "answer_analysis": ["$$e=\\text{C}_{m}^{n}$$，∴$$m\\textgreater n$$．注意离心率相同的情形． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "785", "queId": "60f64e54e00f413f927f4d9db5281f5c", "competition_source_list": ["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", "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年天津河北区高三一模理科第13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": "$$\\frac{1}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{2}{5}$$ "}]], "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": ["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": "661", "queId": "45562866995248aaafe34e08d0f0018d", "competition_source_list": ["2009年黑龙江全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知不等式$${{m}^{2}}+({{\\cos }^{2}}\\theta -5)m+4{{\\sin }^{2}}\\theta \\geqslant 0$$恒成立，则实数$$m$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$0\\leqslant m\\leqslant 4$$ "}], [{"aoVal": "B", "content": "$$1\\leqslant m\\leqslant 4$$ "}], [{"aoVal": "C", "content": "$$m\\geqslant 4$$或$$x\\leqslant 0$$ "}], [{"aoVal": "D", "content": "$$m\\geqslant 1$$或$$m\\leqslant 0$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["$${{m}^{2}}+({{\\cos }^{2}}\\theta -5)m+4{{\\sin }^{2}}\\theta ={{m}^{2}}+\\left( -4-{{\\sin }^{2}}\\theta \\right)m+4{{\\sin }^{2}}\\theta =\\left( m-4 \\right)\\left( m-{{\\sin }^{2}}\\theta \\right)\\geqslant 0$$，  所以$$m\\geqslant 4$$或$$m\\leqslant {{\\sin }^{2}}\\theta $$恒成立，则$$m\\geqslant 4$$或$$m\\leqslant 0$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "588", "queId": "51d3cdc633a44a809b34c3f1d248bf81", "competition_source_list": ["高二上学期单元测试《代数变形1》自招第2题", "2009年河北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$$x,y,z\\in (0,1)$$且$$x+y+z=2$$，则$$xy+yz+zx$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( 0,\\frac{4}{3} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left( 1,\\frac{4}{3} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left( 1,\\frac{4}{3} \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( \\frac{4}{3},2 \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->换元技巧->代数换元"], "answer_analysis": ["设$$S=xy+yz+zx$$，则$${{(x+y+z)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2S$$．  由于$$x,y,z\\in (0,1)$$，所以$$S\\textless{}3$$，$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}\\textless{}x+y+z=2$$，  即$$4\\textless{}2+2S$$，所以$$S\\textgreater1$$．  又$${{S}^{2}}\\leqslant {{({{x}^{2}}+{{y}^{2}}+{{z}^{2}})}^{2}}={{(4-2S)}^{2}}$$，$$(3S-4)(S-4)\\geqslant 0$$，  解得$$S\\leqslant \\frac{4}{3}$$或$$S\\geqslant 4$$，但是$$S\\textless{}3$$，所以$$S\\leqslant \\frac{4}{3}$$，  且当$$x=y=z=\\frac{2}{3}$$时，$$S=\\frac{4}{3}$$．  综上可得$$1\\textless{}S\\leqslant \\frac{4}{3}$$．故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "381", "queId": "42d39ff9fb604fb0a6a458915e6546ff", "competition_source_list": ["2009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": "函数$$y={{x}^{2}}$$的图象$${{F}_{1}}$$与它按向量$$\\overrightarrow{a}=(m,1)$$平移后的函数图象$${{F}_{2}}$$在$$x=1$$处的切线互相垂直，则实数$$m$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-\\frac{5}{4}$$ "}], [{"aoVal": "B", "content": "$$-\\frac{3}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{3}{4}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{4}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->导数"], "answer_analysis": ["因为$${y}'(1)=2$$，故$${{F}_{2}}$$的函数为$$y={{(x-m)}^{2}}-1$$，其在$$x=1$$处的切线的斜率为$${{k}_{2}}=2(1-m)$$，由$$2\\times 2(1-m)=-1$$知$$m=\\frac{5}{4}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "272", "queId": "1e5350c6ab924468afabb95972633b16", "competition_source_list": ["2014年AMC10竞赛B第11题"], "difficulty": "1", "qtype": "single_choice", "problem": "$$2014-AMC10A-11$$  For the consumer, a single discount of $$n \\%$$ is more advantageous than any of the following discounts:  ($$1$$) two successive $$15 \\%$$ discounts  ($$2$$) three successive $$10 \\%$$ discounts  ($$3$$)a $$25 \\%$$ discount followed by a $$5 \\%$$ discount  What is the smallest possible posiive integer value of $$n$$?  对于消费者来说，单次折扣$n \\%$$比以下任何一种折扣都更具优势：  ($$1$$) 连续两次 $$15 \\%$$ 折扣  ($$2$$) 连续三次 $$10 \\%$$ 折扣  (3) 25\\%折扣，然后是 5\\%折扣  $$n$$ 的最小可能正整数值是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$27$$ "}], [{"aoVal": "B", "content": "$$28$$ "}], [{"aoVal": "C", "content": "$$29$$ "}], [{"aoVal": "D", "content": "$$31$$ "}], [{"aoVal": "E", "content": "$$33$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Algebra->Application->Equation Word Problems", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->分段函数"], "answer_analysis": ["Let the orginal price be $$x$$. Then, for option $$1$$, the discounted price is $$(1-0.15)(1-0.15)x=0.7225x$$. For option $$2$$, the discounted price is $$(1- 0.1)(1-0.1)(1-0.1)x=0.729 x$$. Finally, for option $$3$$, the discounted price is $$(1-0.25)(1-0.05)=0.7125x$$. Therefore, $$n$$ must be greater than max $$(x-0.7225x,x-0.729x,x-0.7125x)$$. It follows $$n$$ must be greater than $$0.2875$$. We multiply this by $$100$$ to get the percent value, and then round up because $$n$$ is the smallest integer that provides a greater discount than $$28.75$$, leaving us with the answer of $$\\boxed{(\\text{C})29}$$. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "323", "queId": "235b9ee687304251915ef4d9f479dcda", "competition_source_list": ["2011年浙江全国高中数学联赛竞赛初赛第10题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$a\\in [-1,  1]$$，则$${{x}^{2}}+(a-4)x+4-2a\\textgreater0$$的解为．", "answer_option_list": [[{"aoVal": "A", "content": "$$x\\textgreater3$$或$$x\\textless{}2$$ "}], [{"aoVal": "B", "content": "$$x\\textgreater2$$或$$x\\textless{}1$$ "}], [{"aoVal": "C", "content": "$$x\\textgreater3$$或$$x\\textless{}1$$ "}], [{"aoVal": "D", "content": "$$1\\textless{}x\\textless{}3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["不等式的左端看成$$a$$的一次函数，$$f(a)=(x-2)a+({{x}^{2}}-4x+4)$$  由$$f(-1)={{x}^{2}}-5x+6\\textgreater0,  f(1)={{x}^{2}}-3x+2\\textgreater0\\Rightarrow x\\textless{}1$$或$$x\\textgreater3$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "91", "queId": "0b45753c6ef941b1a68cd50a952413e9", "competition_source_list": ["2012年天津全国高中数学联赛竞赛初赛第10题9分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$y=\\left\\textbar{} x-1 \\right\\textbar+\\left\\textbar{} x-2 \\right\\textbar+\\cdots +\\left\\textbar{} x-10 \\right\\textbar$$的最小值是~\\uline{~~~~~~~~~~}~.", "answer_option_list": [[{"aoVal": "A", "content": "$$65$$ "}], [{"aoVal": "B", "content": "$$24$$ "}], [{"aoVal": "C", "content": "$$25$$ "}], [{"aoVal": "D", "content": "$$18$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数基础->函数的概念", "竞赛->知识点->函数基础->函数综合"], "answer_analysis": ["当$$x\\in \\left[ 5,6 \\right]$$时取得最小值$$25$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "983", "queId": "edf9491bb6424ef4956a7d718bafa670", "competition_source_list": ["2008年甘肃全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "六个家庭依次编号为$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$．每家三人，大家一起聚会做游戏，游戏按每组三人依次进行，那么同一组的成员来自不同家庭的概率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{5}{68}$$ "}], [{"aoVal": "B", "content": "$$\\frac{15}{68}$$ "}], [{"aoVal": "C", "content": "$$\\frac{45}{68}$$ "}], [{"aoVal": "D", "content": "$$\\frac{5}{204}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->概率初步", "竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["从$$18$$个人中选出$$3$$人，不同的选法共有$$\\text{C}_{18}^{3}=\\frac{18\\times 17\\times 16}{3\\times 2\\times 1}=816$$．  因为一个家庭的三个人编号是相同的，为使所选的组员编号不同，  应先从$$6$$个编号中选取$$3$$个号，有$$\\text{C}_{6}^{3}$$种选法，对每一个编号再选人，  又有$$3$$种选法．从而，选出的小组成员来自不同家庭的选法为  $$\\text{C}_{6}^{3}\\cdot {{3}^{3}}=\\frac{6\\times 5\\times 4}{3\\times 2\\times 1}\\times {{3}^{3}}=5\\times 4\\times {{3}^{3}}=540$$．  故一个小组的三个成员来自不同家庭的概率为  $$\\frac{540}{816}=\\frac{5\\times 4\\times {{3}^{3}}}{3\\times 17\\times 16}\\times \\frac{5\\times {{3}^{2}}}{17\\times 4}=\\frac{45}{68}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "698", "queId": "88f4bcfa24464ea78df570cb21b210eb", "competition_source_list": ["2008年高考真题福建卷理科第15题4分", "2008年高考真题福建卷文科第15题4分", "2009年河南全国高中数学联赛高一竞赛初赛第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "若三棱锥的三个侧面两两垂直，且侧棱长均为$$\\sqrt{3}$$，则其外接球的表面积是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$3 \\pi $$ "}], [{"aoVal": "B", "content": "$$6 \\pi $$ "}], [{"aoVal": "C", "content": "$$9 \\pi $$ "}], [{"aoVal": "D", "content": "以上都不对 "}]], "knowledge_point_routes": ["课内体系->知识点->立体几何初步->基本立体图形->空间几何体的体积、表面积->空间几何体的体积", "竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["三棱锥的三个侧面两两垂直，说明三棱锥的三条侧棱两两垂直，  设其外接球的半径为$$R$$，则有$${{(2R)}^{2}}={{(\\sqrt{3})}^{2}}+{{(\\sqrt{3})}^{2}}+{{(\\sqrt{3})}^{2}}=9$$，  所以外接球的表面积为$$S=4\\pi {{R}^{2}}=9\\pi $$．  在棱长为$$\\sqrt{3}$$的正方体$$ABCD-{{A}_{1}}{{B}_{1}}{{C}_{1}}{{D}_{1}}$$中，以截面$$BD{{A}_{1}}$$为底面，$$A$$为顶点的三棱锥的三个侧面两两垂直，且侧棱长均为$$\\sqrt{3}$$的三棱锥，而正方体一定有一个外接球，该外接球也是所述三棱锥的外接球，正方体的体对角线是其外接球的直径，于是，所求三棱锥外接球的表面积是$$4 \\pi \\times {{\\left( \\frac{1}{2}\\times \\sqrt{3}\\times \\sqrt{3} \\right)}^{2}}=9 \\pi $$．  故答案为：$$9 \\pi $$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1124", "queId": "c663cc4b62ec4bfc8e73dca03820947a", "competition_source_list": ["2008年辽宁全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "有一个正四棱锥，它的底面边长与侧棱长均为$$a$$，现用一张正方形包装纸将其完全包住（不能裁剪纸，但可以折叠），那么包装纸的最小边长应就为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{2}+\\sqrt{6}}{2}a$$ "}], [{"aoVal": "B", "content": "$$(\\sqrt{2}+\\sqrt{6})a$$ "}], [{"aoVal": "C", "content": "$$\\frac{1+\\sqrt{3}}{2}a$$ "}], [{"aoVal": "D", "content": "$$(1+\\sqrt{3})a$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["将四棱锥的侧面向外展开到底面，则$$4$$个侧面三角形的顶点所构成的正方形即为最小正方形，边长为$$\\frac{\\sqrt{2}+\\sqrt{6}}{2}a$$，故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "778", "queId": "8d08398705e244478159ff48bfb8dbf1", "competition_source_list": ["2004年高考真题北京卷理科第12题5分", "2008年江苏全国高中数学联赛竞赛初赛第1题6分", "2004年高考真题安徽卷理科第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f(x)={{\\cos }^{4}}x+{{\\sin }^{2}}x(x\\in R)$$的最小正周期是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{ \\pi }{4}$$ "}], [{"aoVal": "B", "content": "$$\\frac{ \\pi }{2}$$ "}], [{"aoVal": "C", "content": "$$ \\pi $$ "}], [{"aoVal": "D", "content": "$$2 \\pi $$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["方法一：$$f\\left( x+\\frac{ \\pi }{2} \\right)={{\\sin }^{4}}x+{{\\cos }^{2}}x$$  $$={{\\sin }^{4}}x+{{\\cos }^{4}}x+{{\\cos }^{2}}x{{\\sin }^{2}}x $$  $$={{\\cos }^{4}}x+{{\\sin }^{2}}x=f(x)$$，  又因为$$f(0)=1$$，$$f\\left( \\frac{ \\pi }{4} \\right)=\\frac{1}{4}+\\frac{1}{2}\\ne f(0)$$，故选$$\\text{B}$$．  方法二：$$f(x)={{\\cos }^{4}}x+1-{{\\cos }^{2}}x$$  $$={{\\cos }^{2}}x({{\\cos }^{2}}x-1)+1=1-{{\\cos }^{2}}x{{\\sin }^{2}}x$$  $$=1-\\frac{1}{4}{{\\sin }^{2}}2x=\\frac{1}{8}\\cos 4x+\\frac{7}{8}$$，  可知$$f(x)$$的最小正周期为$$\\frac{2 \\pi }{4}=\\frac{ \\pi }{2}$$．故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "393", "queId": "39ed3c23dc3c4c95aabcc50bac6a8b34", "competition_source_list": ["2011年全国高中数学联赛竞赛复赛一试第7题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "直线$$x-2y-1=0$$与抛物线$${{y}^{2}}=4x$$交于$$A$$，$$B$$两点，$$C$$为抛物线上的一点，$$\\angle ACB=90{}^{}\\circ $$，则$$C$$点的坐标为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$(1,-2)$$或$$(9,-6)$$ "}], [{"aoVal": "B", "content": "$$(1,2)$$或$$(9,6)$$ "}], [{"aoVal": "C", "content": "$$(1,2)$$或$$(9,-6)$$ "}], [{"aoVal": "D", "content": "$$(1,-2)$$或$$(9,6)$$ "}]], "knowledge_point_routes": ["课内体系->知识点->平面向量->平面向量的运算->数量积->数量积的坐标表达式", "课内体系->知识点->平面向量->平面向量基本定理及其坐标表示->坐标表示平面向量的垂直", "课内体系->知识点->直线和圆的方程->直线与方程->直线方程的五种形式->直线的一般式方程", "课内体系->知识点->圆锥曲线->椭圆->直线和椭圆的位置关系", "课内体系->知识点->圆锥曲线->直线与圆锥曲线问题->定点问题", "课内体系->知识点->圆锥曲线->抛物线->直线和抛物线的位置关系", "课内体系->知识点->圆锥曲线->抛物线->抛物线的定义、标准方程->抛物线的标准方程", "课内体系->素养->数学运算", "课内体系->素养->数学抽象"], "answer_analysis": ["设$$A({{x}_{1}},{{y}_{1}})$$，$$B({{x}_{2}},{{y}_{2}})$$，$$C({{t}^{2}},2t)$$，由$$\\begin{cases}x-2y-1=0    {{y}^{2}}=4x         \\end{cases}$$  得$${{y}^{2}}-8y-4=0$$，则$${{y}_{1}}+{{y}_{2}}=8$$，$${{y}_{1}}\\cdot {{y}_{2}}=-4$$．  又$${{x}_{1}}=2{{y}_{1}}+1$$，$${{x}_{2}}=2{{y}_{2}}+1$$，所以  $${{x}_{1}}+{{x}_{2}}=2({{y}_{1}}+{{y}_{2}})+2=18$$，$${{x}_{1}}\\cdot {{x}_{2}}=4{{y}_{1}}\\cdot {{y}_{2}}+2({{y}_{1}}+{{y}_{2}})+1=1$$．  因为$$\\angle ACB=90{}^{}\\circ $$，所以$$\\overrightarrow{CA}\\cdot \\overrightarrow{CB}=0$$，即有  $$({{t}^{2}}-{{x}_{1}})({{t}^{2}}-{{x}_{2}})+(2t-{{y}_{1}})(2t-{{y}_{2}})=0$$，  即$${{t}^{4}}-14{{t}^{2}}-16t-3=0$$，  即$$({{t}^{2}}+4t+3)({{t}^{2}}-4t-1)=0$$．  显然$${{t}^{2}}-4t-1\\ne 0$$，否则$${{t}^{2}}-2\\cdot 2t-1=0$$，则点$$C$$在直线$$x-2y-1=0$$上，从而点$$C$$与点$$A$$或点$$B$$重合．  所以$${{t}^{2}}+4t+3=0$$，解得$${{t}_{1}}=-1$$，$${{t}_{2}}=-3$$．  故所求点$$C$$的坐标为$$(1,-2)$$或$$(9,-6)$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "222", "queId": "6b5b7cf916bb460ab4596354cda21f03", "competition_source_list": ["2017年湖南全国高中数学联赛竞赛初赛第4题5分", "2016年四川全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "对任意正整数$$n$$与$$k$$（$$k\\leqslant n$$），$$f(n,k)$$表示不超过$$\\left[ \\frac{n}{k} \\right]$$，且与$$n$$互质的正整数的个数，则$$f\\left( 100,3 \\right)=$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$11$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$14$$ "}], [{"aoVal": "D", "content": "$$19$$ "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->函数的定义", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的概念->判定是否为函数", "课内体系->特色题型->新定义"], "answer_analysis": ["不超过$$33$$且与$$100$$互质的正整数，即$$1$$，$$3$$，$$5$$，$$7$$，$$\\cdots $$，$$33$$中去掉$$5$$的倍数$$5$$，$$15$$，$$25$$，有$$14$$个．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "906", "queId": "7cd033fe59094e70b4e618644404c436", "competition_source_list": ["2006年全国高中数学联赛竞赛一试第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "数码$${{a}_{1}},{{a}_{2}},{{a}_{3}},\\cdots ,{{a}_{2006}}$$中有奇数个$$9$$的$$2007$$位十进制数$$\\overline{2{{a}_{1}}{{a}_{2}}\\cdots {{a}_{2006}}}$$的个数为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{1}{2}({{10}^{2006}}+{{8}^{2006}})$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{2}({{10}^{2006}}-{{8}^{2006}})$$ "}], [{"aoVal": "C", "content": "$${{10}^{2006}}+{{8}^{2006}}$$ "}], [{"aoVal": "D", "content": "$${{10}^{2006}}-{{8}^{2006}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["出现奇数个$$9$$的十进制数个数有$$A=\\mathrm{C}_{2006}^{1}{{9}^{2005}}+\\mathrm{C}_{2006}^{3}{{9}^{2003}}+\\cdots +\\mathrm{C}_{2006}^{2005}9$$．  又由于$${{(9+1)}^{2006}}=\\sum\\limits_{k=0}^{2006}{\\mathrm{C}_{2006}^{k}{{9}^{2006-k}}}$$以及$${{(9-1)}^{2006}}=\\sum\\limits_{k=0}^{2006}{\\mathrm{C}_{2006}^{k}{{(-1)}^{k}}{{9}^{2006-k}}}$$，从而得  $$A=\\mathrm{C}_{2006}^{1}{{9}^{2005}}+\\mathrm{C}_{2006}^{3}{{9}^{2003}}+\\cdots +\\mathrm{C}_{2006}^{2005}9=\\frac{1}{2}\\left( {{10}^{2006}}-{{8}^{2006}} \\right)$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "216", "queId": "f0f710efcaf74eba885903be4e567505", "competition_source_list": ["2015年浙江全国高中数学联赛竞赛初赛第7题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若过点$$P\\left( 1,0 \\right)$$，$$Q\\left( 2,0 \\right)$$，$$R\\left( 4,0 \\right)$$，$$S\\left( 8,0 \\right)$$作四条直线构成一个正方形，则该正方形的面积不可能等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{16}{17}$$ "}], [{"aoVal": "B", "content": "$$\\frac{36}{5}$$ "}], [{"aoVal": "C", "content": "$$\\frac{26}{5}$$ "}], [{"aoVal": "D", "content": "$$\\frac{196}{53}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与方程"], "answer_analysis": ["不妨设四条直线交成的正方形在第一象限，且边长为$$a$$，面积为$$S$$，过$$P$$的直线的倾斜角为$$\\theta (0\\textless{}\\theta \\textless{}\\frac{ \\pi }{2})$$．  当过点$$P,Q$$的直线为正方形的对边所在的直线时，  $$\\left\\textbar{} PQ \\right\\textbar\\sin \\theta =a=\\left\\textbar{} RS \\right\\textbar\\cos \\theta \\Leftrightarrow \\sin \\theta =4\\cos \\theta \\Leftrightarrow \\sin \\theta =\\frac{4}{\\sqrt{17}}$$，  此时正方形的面积$$S={{(\\left\\textbar{} PQ \\right\\textbar\\sin \\theta )}^{2}}=\\frac{16}{17}$$．  同理，当过点$$P,R$$的直线为正方形的对边所在的直线时，$$S=\\frac{36}{5}$$；  当过点$$P,S$$的直线为正方形的对边所在的直线时，$$S=\\frac{196}{53}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "436", "queId": "b5128ccc87504a488c18e33f79d7af05", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "体育课下课后，老师要求体育委员把$$3$$个相同的篮球、$$2$$个相同的排球、$$2$$个相同的橄榄球排成一排放好，则不同的放法有．", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$种 "}], [{"aoVal": "B", "content": "$$60$$种 "}], [{"aoVal": "C", "content": "$$120$$种 "}], [{"aoVal": "D", "content": "$$720$$种 "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->两个基本计数原理->分步乘法计数原理", "竞赛->知识点->排列组合与概率->两个基本计数原理", "竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["总共$$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$$种. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1032", "queId": "b32f8c4f14cc479c95955cca128ac53f", "competition_source_list": ["2001年全国高中数学联赛竞赛一试第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "命题$$1$$：长方体中，必存在到各顶点距高相等的点．  命题$$2$$：长方体中，必存在到各条棱距离相等的点．  命题$$3$$：长方体中，必存在到各个面距离相等的点．  以上三个命题中正确的有（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$个 "}], [{"aoVal": "B", "content": "$$1$$个 "}], [{"aoVal": "C", "content": "$$2$$个 "}], [{"aoVal": "D", "content": "$$3$$个 "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间中的角与距离"], "answer_analysis": ["由于长方体的中心到各顶点的距离相等，所以命题$$1$$正确．对于命题$$2$$和命题$$3$$，一般的长方体（除正方体外）中不存在到各条棱距离相等的点，也不存在到各个面距离相等的点．因此，本题只有命题$$1$$正确．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "516", "queId": "67fba1f30e054224aba0d2adcde1f009", "competition_source_list": ["2013年辽宁全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "内直径为$$\\frac{4\\sqrt{3}}{3}+2$$，高为$$20$$的圆柱形容器中最多可以放入直径为$$2$$的小球的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$30$$ "}], [{"aoVal": "B", "content": "$$33$$ "}], [{"aoVal": "C", "content": "$$36$$ "}], [{"aoVal": "D", "content": "$$39$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["圆柱的底部可以放$$3$$个两两相切的小球，上一层也放$$3$$个两两相切的小球，使每个小球与下层$$2$$个小球相切，按此方法依次向上放置小球，设共放$$k$$层，则总高度  $$H=2+\\left( k-1 \\right)\\frac{2\\sqrt{6}}{3}$$  当$$k=12$$时，$$H=\\left( 2+\\frac{22\\sqrt{6}}{3} \\right)\\approx 19{}96\\textless{}20$$；  当$$k=13$$时，$$H=\\left( 2+\\frac{24\\sqrt{6}}{3} \\right)\\approx 21{}60\\textgreater20$$．  故最多可放人$$12$$层共$$36$$个， "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "453", "queId": "cc37ac95f8f849fe885377714c31a4a0", "competition_source_list": ["2019年吉林全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "在正方形$$ABCD$$中，$$M$$为边$$BC$$的中点．若$$\\overrightarrow{AC}=\\lambda \\overrightarrow{AM}+\\mu \\overrightarrow{BD}$$，则$$\\lambda +\\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$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算"], "answer_analysis": ["注意到，  $$\\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 +\\mu =\\dfrac{5}{3}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "471", "queId": "4c71b2c3fdec4652866e132fff7ffd5e", "competition_source_list": ["2009年山东全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知两个一元二次方程：$$a{{x}^{2}}+bx+c=0,u{{x}^{2}}+vx+w=0$$都有实根，这里$$a\\ne u$$．若交换这两个方程的二次项系数，则$$wc\\textgreater0$$是系数交换之后所得两个二次方程中至少有一个有实根的 ．", "answer_option_list": [[{"aoVal": "A", "content": "充分但不必要条件 "}], [{"aoVal": "B", "content": "必要但不充分条件 "}], [{"aoVal": "C", "content": "充分且必要条件 "}], [{"aoVal": "D", "content": "既不充分又不必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数方程"], "answer_analysis": ["原来两个二次方程的判别式分别为$${{\\Delta }_{1}}={{b}^{2}}-4ac$$；$${{\\Delta }_{2}}={{v}^{2}}-4uw$$．  二次项系数交换后两个方程的判别式分别为$${{{\\Delta }'}_{1}}={{b}^{2}}-4uc$$；$${{{\\Delta }'}_{2}}={{v}^{2}}-4aw$$．  $$({{\\Delta }_{1}}-{{{\\Delta }'}_{1}})({{\\Delta }_{2}}-{{{\\Delta }'}_{2}})=16wc(u-a)(a-u)=-16wc{{(u-a)}^{2}}$$．  若$$wc\\textgreater0$$，则$$({{\\Delta }_{1}}-{{{\\Delta }'}_{1}})({{\\Delta }_{2}}-{{{\\Delta }'}_{2}})\\textless{}0$$，所以，或有$${{{\\Delta }'}_{1}}\\textgreater{{\\Delta }_{1}}\\geqslant 0$$，或有$${{{\\Delta }'}_{2}}\\textgreater{{\\Delta }_{2}}\\geqslant 0$$，即二次项系数交换后两个方程中至少有一个方程有实根，所以条件``$$wc\\textgreater0$$''是充分的．  条件不必要，事实上，若$$wc=0$$，二次项系数交换后两个方程中至少有一个有$$x=0$$的实根．故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1082", "queId": "c151666557554a779c7d9143d733541c", "competition_source_list": ["2011年四川全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$y=\\sqrt{x-5}+\\sqrt{24-3x}$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$2\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$3\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "课内体系->知识点->函数的概念与性质"], "answer_analysis": ["法一：$$f(x)$$的定义域为$$5\\leqslant x\\leqslant 8$$，  由$${f}'(x)=\\frac{1}{2\\sqrt{x-5}}+\\frac{-3}{2\\sqrt{24-3x}}=\\frac{\\sqrt{24-3x}-3\\sqrt{x-5}}{2\\sqrt{x-5}\\cdot \\sqrt{24-3x}}=0$$，解得$$x=\\frac{23}{4}$$．  因为$$f(5)=3$$，$$f(\\frac{23}{4})=2\\sqrt{3}$$，$$f(8)=\\sqrt{3}$$，于是$$f{{(x)}_{\\max }}=f(\\frac{23}{4})=2\\sqrt{3}$$．  法二：$$f(x)$$的定义域为$$5\\leqslant x\\leqslant 8$$，  $${{f}^{2}}(x)={{(1\\cdot \\sqrt{x-5}+\\sqrt{3}\\cdot \\sqrt{8-x})}^{2}}\\leqslant (1+3)(x-5+8-x)=12$$  当且仅当$$\\frac{\\sqrt{x-5}}{1}=\\frac{\\sqrt{8-x}}{\\sqrt{3}}$$，即$$x=\\frac{23}{4}$$时，$$f(x)$$取到最大值$$2\\sqrt{3}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "425", "queId": "479969aeeb8f48efb6bc2ce177e67608", "competition_source_list": ["2010年辽宁全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若正方体的棱长为$$a$$，则与正方体对角线垂直的最大截面的面积为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3\\sqrt{2}}{4}{{a}^{2}}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{2}{{a}^{2}}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{3}}{2}{{a}^{2}}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3\\sqrt{3}}{4}{{a}^{2}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["最大截面过正方体的中心，是边长为$$\\frac{\\sqrt{2}}{2}a$$的正六边形，面积为  $$\\frac{\\sqrt{3}}{4}\\times \\frac{1}{2}{{a}^{2}}\\times 6=\\frac{3\\sqrt{3}}{4}{{a}^{2}}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "748", "queId": "89409f4b938b4c688c52e761a59e9a4c", "competition_source_list": ["全国全国高中数学联赛竞赛一试"], "difficulty": "2", "qtype": "single_choice", "problem": "已知三个平面$$\\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$$相交于一点．  则．", "answer_option_list": [[{"aoVal": "A", "content": "甲是乙的充分非必要条件 "}], [{"aoVal": "B", "content": "甲是乙的必要非充分条件 "}], [{"aoVal": "C", "content": "甲是乙的充分必要条件 "}], [{"aoVal": "D", "content": "甲是乙的非充分非必要条件 "}]], "knowledge_point_routes": ["课内体系->知识点->常用逻辑用语->充分条件与必要条件->充要条件与立体几何结合", "课内体系->知识点->立体几何初步->基本图形位置关系->空间中的基本事实与定理", "课内体系->素养->逻辑推理"], "answer_analysis": ["设$$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 $$） "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "204", "queId": "e3075926ef9b45d79f24dad1c6d42758", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "下列四个函数中，最小正周期为$$ \\pi $$，且图象关于直线$$x=\\frac{5}{12} \\pi $$对称的函数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$y=\\sin \\left( \\frac{x}{2}+\\frac{ \\pi }{3} \\right)$$ "}], [{"aoVal": "B", "content": "$$y=\\sin \\left( \\frac{x}{2}-\\frac{ \\pi }{3} \\right)$$ "}], [{"aoVal": "C", "content": "$$y=\\sin \\left( 2x-\\frac{ \\pi }{3} \\right)$$ "}], [{"aoVal": "D", "content": "$$y=\\sin \\left( 2x+\\frac{ \\pi }{3} \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["对于函数$$y=\\sin (\\omega x+\\varphi )$$，由最小正周期为：$$T=\\frac{2 \\pi }{w}= \\pi $$，求得$$w=2$$，  ∴$$y=\\sin (2x+\\varphi )$$，  又∵函数的图象关于直线$$x=\\frac{ 5\\pi }{12}$$对称，  ∴$$2\\times \\frac{ 5\\pi }{12}+\\varphi =\\frac{ \\pi }{2}+k \\pi $$，$$k\\in \\mathbf{Z}$$，  ∴$$\\varphi =-\\frac{ \\pi }{3}+k \\pi $$，$$k\\in \\mathbf{Z}$$，  结合选项，$$\\varphi $$可取$$-\\frac{ \\pi }{3}$$，此时$$y=\\sin \\left( 2x-\\frac{ \\pi }{3} \\right)$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "295", "queId": "7e22d3c9b5a44d98bd3ecc0b8f6eb1ba", "competition_source_list": ["2009年辽宁全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$M$$是正方体各条棱的中点的集合，则过且仅过$$M$$中$$3$$个点的平面的个数是．", "answer_option_list": [[{"aoVal": "A", "content": "$$56$$ "}], [{"aoVal": "B", "content": "$$81$$ "}], [{"aoVal": "C", "content": "$$136$$ "}], [{"aoVal": "D", "content": "$$145$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合"], "answer_analysis": ["通过$$M$$中$$3$$个点的平面一共有$$\\text{C}_{12}^{3}=220$$（个）（包含重复的），共中过且仅过$$M$$中$$4$$个点的平面共$$\\text{C}_{4}^{3}\\times (9+12)=84$$（个）．  过$$M$$中$$6$$个点的平面共$$\\text{C}_{6}^{3}\\times 4=80$$（个）；所以过且仅过$$M$$中$$3$$个点的平面共有$$220-84-80=56$$（个）．故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "587", "queId": "5f4dc031c8574dfd8b966b71f20020dd", "competition_source_list": ["2016年全国高中数学联赛竞赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$${{a}_{4}}$$是$$1$$，$$2$$，$$\\cdots $$，$$100$$中的$$4$$个互不相同的数，满足$$\\left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \\right)\\left( a_{2}^{2}+a_{3}^{2}+a_{4}^{2} \\right)={{\\left( {{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+{{a}_{3}}{{a}_{4}} \\right)}^{2}}$$，则这样的有序数组$$\\left( {{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}} \\right)$$的个数为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$38$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$42$$ "}], [{"aoVal": "D", "content": "$$44$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->二项式定理->整除问题", "课内体系->知识点->等式与不等式->不等式->柯西不等式", "竞赛->知识点->数论模块->取整函数->取整函数的定义与性质", "竞赛->知识点->数列与数学归纳法->等差数列与等比数列", "竞赛->知识点->不等式->几个重要的不等式->柯西"], "answer_analysis": ["由柯西不等式知，$$\\left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \\right)\\left( a_{2}^{2}+a_{3}^{2}+a_{4}^{2} \\right)\\geqslant {{\\left( {{a}_{1}}{{a}_{2}}+{{a}_{2}}{{a}_{3}}+{{a}_{3}}{{a}_{4}} \\right)}^{2}}$$，等号成立的充分必要条件是$$\\frac{{{a}_{1}}}{{{a}_{2}}}=\\frac{{{a}_{2}}}{{{a}_{3}}}=\\frac{{{a}_{3}}}{{{a}_{4}}}$$，即$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$${{a}_{4}}$$成等比数列．于是问题等价于计算满足$$\\left { {{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}} \\right }\\subseteq \\left { 1,2,3,\\cdots ,100 \\right }$$的等比数列$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$${{a}_{4}}$$的个数．设等比数列的公比$$q\\ne 1$$，且$$q$$为有理数．记$$q=\\frac{n}{m}$$，其中$$m$$，$$n$$为互素的正整数，且$$m\\ne n$$．  先考虑$$n\\textgreater m$$的情况．  此时$${{a}_{4}}={{a}_{1}}\\cdot {{\\left( \\frac{n}{m} \\right)}^{3}}=\\frac{{{a}_{1}}{{n}^{3}}}{{{m}^{3}}}$$，注意到$${{m}^{3}}$$，$${{n}^{3}}$$互素，故$$l=\\frac{{{a}_{1}}}{{{m}^{3}}}$$为正整数．相应地，$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$${{a}_{4}}$$分别等于$${{m}^{3}}l$$，$${{m}^{2}}nl$$，$$m{{n}^{2}}l$$，$${{n}^{3}}l$$，它们均为正整数．这表明，对任意给定的$$q=\\frac{n}{m}\\textgreater1$$，满足条件并以$$q$$为公比的等比数列$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$${{a}_{4}}$$的个数，即为满足不等式$${{n}^{3}}l\\leqslant 100$$的正整数$$l$$的个数，即$$\\left[ \\frac{100}{{{n}^{3}}} \\right]$$．  由于$${{5}^{3}}\\textgreater100$$，故仅需考虑$$q=2$$，$$3$$，$$\\frac{3}{2}$$，$$4$$，$$\\frac{4}{3}$$这些情况，相应的等比数列的个数为$$\\left[ \\frac{100}{8} \\right]+\\left[ \\frac{100}{27} \\right]+\\left[ \\frac{100}{27} \\right]+\\left[ \\frac{100}{64} \\right]+\\left[ \\frac{100}{64} \\right]=12+3+3+1+1=20$$．  当$$n\\textless{}m$$时，由对称性可知，亦有$$20$$个满足件的等比数列$${{a}_{1}}$$，$${{a}_{2}}$$，$${{a}_{3}}$$，$${{a}_{4}}$$．  综上可知，共有$$40$$个满足条件的有序数组$$\\left( {{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}} \\right)$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "856", "queId": "f6bf9791128844d4aacbd18ec503be2f", "competition_source_list": ["2017年四川全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$\\triangle ABC$$中，$$\\overrightarrow{AB}\\cdot \\overrightarrow{BC}=3\\overrightarrow{CA}\\cdot \\overrightarrow{AB}$$，则$$\\frac{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar+\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}{\\left\\textbar{} \\overrightarrow{BC} \\right\\textbar}$$的最大值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{5}}{2}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$\\sqrt{5}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->平面向量的概念与运算"], "answer_analysis": ["因为$$\\overrightarrow{CB}\\cdot \\overrightarrow{BC}=\\overrightarrow{CA}\\cdot \\overrightarrow{BC}+\\overrightarrow{AB}\\cdot \\overrightarrow{BC}=\\overrightarrow{CA}\\cdot \\overrightarrow{BC}+3\\overrightarrow{CA}\\cdot \\overrightarrow{AB}=\\overrightarrow{CA}\\cdot \\overrightarrow{AC}+2\\overrightarrow{CA}\\cdot \\overrightarrow{AB}$$，  所以$${{\\left\\textbar{} \\overrightarrow{BC} \\right\\textbar}^{2}}={{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}^{2}}+2\\overrightarrow{AC}\\cdot \\overrightarrow{AB}$$．  因为$$\\overrightarrow{AB}\\cdot \\overrightarrow{BA}=\\overrightarrow{AB}\\cdot \\overrightarrow{BC}+\\overrightarrow{AB}\\cdot \\overrightarrow{CA}=4\\overrightarrow{CA}\\cdot \\overrightarrow{AB}$$，  所以$${{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}^{2}}=4\\overrightarrow{AB}\\cdot \\overrightarrow{AC}$$，则$${{\\left\\textbar{} \\overrightarrow{BC} \\right\\textbar}^{2}}={{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}^{2}}+\\frac{1}{2}{{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}^{2}}$$．  于是$$\\frac{{{\\left( \\left\\textbar{} \\overrightarrow{AC} \\right\\textbar+\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar{} \\right)}^{2}}}{{{\\left\\textbar{} \\overrightarrow{BC} \\right\\textbar}^{2}}}=\\frac{{{\\left( \\left\\textbar{} \\overrightarrow{AC} \\right\\textbar+\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar{} \\right)}^{2}}}{{{\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar}^{2}}+\\frac{1}{2}{{\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar}^{2}}}\\leqslant 3$$，当$$2\\left\\textbar{} \\overrightarrow{AC} \\right\\textbar=\\frac{1}{2}\\left\\textbar{} \\overrightarrow{AB} \\right\\textbar$$时取等号． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "943", "queId": "db79de255df54f62b0bad1a80143b651", "competition_source_list": ["2009年浙江全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知集合$$M= {1, 2 }, N= {2a-1\\textbar a\\in M }$$，则$$M\\cap N=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$ {1 }$$ "}], [{"aoVal": "B", "content": "$$ {1,2 }$$ "}], [{"aoVal": "C", "content": "$$ {1,2,3 }$$ "}], [{"aoVal": "D", "content": "空集 "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["由于$$N= {2a-1\\textbar a\\in M }= {1,3 }$$，所以$$M\\cap N= {1 }$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "608", "queId": "e3a84b300c1c4d82bd9d1b6bc66b4c0a", "competition_source_list": ["2018~2019学年11月广东深圳福田区深圳市实验学校高二上学期周测B卷第10题5分", "2015年四川全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "记函数$$f(x)=\\sqrt{2-x}+\\sqrt{3x+12}$$的最大值为$$M$$，最小值为$$m$$，则$$\\frac{M}{m}$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\sqrt{6}}{2}$$ "}], [{"aoVal": "B", "content": "$$\\sqrt{2}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的概念"], "answer_analysis": ["$$f\\left( x \\right)=\\sqrt{2-x}+\\sqrt{3}\\sqrt{x+4}$$，则$${{\\left[ f\\left( x \\right) \\right]}^{2}}\\leqslant \\left( 1+3 \\right)\\left( 2-x+x+4 \\right)=24$$，$$f\\left( x \\right)\\leqslant 2\\sqrt{6}$$，所以$$M=2\\sqrt{6}$$．  又$$-4\\leqslant x\\leqslant 2$$，所以$$f\\left( x \\right)\\geqslant \\sqrt{2-x}\\geqslant \\sqrt{2-\\left( -4 \\right)}=\\sqrt{6}$$，$$m=\\sqrt{6}$$．  因此$$\\frac{M}{m}=2$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "147", "queId": "c2804a7693ce4ff5b7c3a59ed140c1f6", "competition_source_list": ["2021年吉林全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设实数$$x$$，$$y$$满足不等式组$$\\begin{cases} x-y\\leqslant 0    4x-y\\geqslant 0    x+y\\leqslant 3 \\end{cases}$$，则$$z=x+2y-\\frac{1}{x}$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{77}{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{56}{15}$$ "}], [{"aoVal": "C", "content": "$$\\frac{23}{6}$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->几个重要的不等式->均值", "课内体系->素养->数学运算", "课内体系->知识点->等式与不等式->不等式->不等式的性质->利用不等式性质求代数式范围或最值", "课内体系->知识点->等式与不等式->不等式->基本不等式->利用基本不等式求最值"], "answer_analysis": ["$$z=x+2y-\\frac{1}{x}=2(x+y)-\\left( x+\\frac{1}{x} \\right)\\leqslant 2\\times 3-\\left( x+\\frac{1}{x} \\right)\\leqslant 2\\times 3-2=4$$当且仅当$$\\begin{cases} x=1    y=2 \\end{cases}$$时，等号成立．  选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "619", "queId": "f1921ea19c5c46ca90d2a31d0c5d84f4", "competition_source_list": ["2013年吉林全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知函数$$f\\left( x \\right)=ax+b\\left( x\\in \\left[ 0,1 \\right] \\right)$$，则``$$a+2b\\textgreater0$$''是``$$f\\left( x \\right)\\textgreater0$$恒成立''的．", "answer_option_list": [[{"aoVal": "A", "content": "充分非必要条件 "}], [{"aoVal": "B", "content": "必要非充分条件 "}], [{"aoVal": "C", "content": "充要条件 "}], [{"aoVal": "D", "content": "既不是充分条件，也不是必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->常用逻辑用语"], "answer_analysis": ["由题意可知，选$$\\text{B}$$， "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "121", "queId": "825b1b13e34b42d2807bef96cc886e69", "competition_source_list": ["2022~2023学年湖南邵阳隆回县湖南省隆回县第二中学高一上学期月考（竞赛）第1题"], "difficulty": "1", "qtype": "single_choice", "problem": "命题``$\\exists x\\in\\text R,x^{3}-x^{2}+1\\textgreater0$''的否定是", "answer_option_list": [[{"aoVal": "A", "content": "$\\exists x\\in\\mathrm R,x^{3}-x^{2}+1\\textless{} 0$ "}], [{"aoVal": "B", "content": "$\\forall x\\in\\mathrm R,x^{3}-x^{2}+1\\leq0$ "}], [{"aoVal": "C", "content": "$\\exists x\\in\\mathrm R,x^{3}-x^{2}+1\\leq0$ "}], [{"aoVal": "D", "content": "不存在$x\\in\\mathrm R$，${{x}^{3}}-{{x}^{2}}+1\\textgreater0$ "}]], "knowledge_point_routes": ["课内体系->知识点->常用逻辑用语->命题->全称量词命题与存在量词命题的否定"], "answer_analysis": ["根据存在性命题的否定知，命题$\\exists x\\in R,{{x}^{3}}-{{x}^{2}}+1\\textgreater0$的否定是$\\forall x\\in R,{{x}^{3}}-{{x}^{2}}+1\\leqslant 0$.故选：$$\\text{B}$$ "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "314", "queId": "46fcd11f83304a37b0cf62245aaffa88", "competition_source_list": ["2015年湖南全国高中数学联赛竞赛初赛第11题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "记集合$$T= {0,1,2,3,4,5,6 },M= {\\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 }$$，将$$M$$中的元素按从大到小顺序排列，则第$$2015$$个数是~\\uline{~~~~~~~~~~}~", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{385}{2401}$$ "}], [{"aoVal": "B", "content": "$$\\frac{386}{2401}$$ "}], [{"aoVal": "C", "content": "$$\\frac{387}{2401}$$ "}], [{"aoVal": "D", "content": "$$\\frac{388}{2401}$$ "}], [{"aoVal": "E", "content": "$$\\frac{389}{2401}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->立体几何初步", "竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["$$M$$中总共$${{7}^{4}}=2401$$个数，所求的数从小到大是第$$387$$个数，所以为$$\\frac{386}{2401}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "557", "queId": "e38b456cf393474d9f7c61fe586e104d", "competition_source_list": ["2013年四川全国高中数学联赛竞赛初赛第12题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知函数$$f\\left( x \\right)=\\frac{a}{x}-x$$，对任意$$x\\in \\left( 0,1 \\right)$$，有$$f\\left( x \\right)\\cdot f\\left( 1-x \\right){\\geqslant }1$$恒成立，则实数$$a$$的取值范围是~\\uline{~~~~~~~~~~}~.", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[-\\frac{1}{4},1\\right]$$ "}], [{"aoVal": "B", "content": "$$\\left[-1,\\frac{1}{4}\\right]$$ "}], [{"aoVal": "C", "content": "$$\\left(-\\infty,-\\frac{1}{4}\\right]\\cup\\left[1,+\\infty\\right)$$ "}], [{"aoVal": "D", "content": "$$\\left(-\\infty,-1\\right]\\cup\\left[\\frac{1}{4},+\\infty\\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法", "竞赛->知识点->不等式->换元技巧->代数换元"], "answer_analysis": ["记$$y=1-x\\in \\left( 0,1 \\right)$$，则$$f\\left( x \\right)\\cdot f\\left( 1-x \\right)=f\\left( x \\right)f\\left( y \\right)=\\left( \\frac{a}{x}-x \\right)\\left( \\frac{a}{y}-y \\right)$$  $$=\\frac{{{x}^{2}}{{y}^{2}}-a\\left( {{x}^{2}}+{{y}^{2}} \\right)+{{a}^{2}}}{xy}$$  $$=\\frac{{{\\left( xy \\right)}^{2}}-a\\left[ {{\\left( x+y \\right)}^{2}}-2xy \\right]+{{a}^{2}}}{xy}$$.  令$$xy=t$$，则$$t\\in \\left( 0{}\\frac{1}{4} \\right]$$，令  $$g\\left( t \\right)=f\\left( x \\right)\\cdot f\\left( 1-x \\right)=\\frac{{{t}^{2}}-a\\left( 1-2t \\right)+{{a}^{2}}}{t}{\\geqslant }1$$，即$${{t}^{2}}+\\left( 2a-1 \\right)t+{{a}^{2}}-a{\\geqslant }0$$.  ①若$$\\frac{1}{2}-a{\\leqslant }0$$，即$$\\frac{1}{2}{\\leqslant }a$$，只需$$g\\left( 0 \\right){\\geqslant }0$$，即$${{a}^{2}}-a{\\geqslant }0$$，解得：$$a{\\geqslant }1$$；  ②若$$0\\textless{}\\frac{1}{2}-a{\\leqslant }\\frac{1}{4}$$，即$$\\frac{1}{4}{\\leqslant }a\\textless{}\\frac{1}{2}$$，只须$$g\\left( \\frac{1}{2}-a \\right){\\geqslant }0$$，无解；  ③若$$\\frac{1}{2}-a\\textgreater\\frac{1}{4}$$，即$$a\\textless{}\\frac{1}{4}$$，只需$$g\\left( \\frac{1}{4} \\right){\\geqslant }0$$，即$${{\\left( a-\\frac{1}{4} \\right)}^{2}}{\\geqslant }\\frac{1}{4}$$，解得，$$a{\\leqslant }-\\frac{1}{4}$$.  所以，实数$$a$$的取值范围为$$ {a\\textbar a{\\geqslant }1$$或$$a{\\leqslant }-\\frac{1}{4} }$$. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "876", "queId": "c4438c4b834f4351aab3654a6f36c53d", "competition_source_list": ["2008年山东全国高中数学联赛竞赛初赛第9题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "向量$$\\overrightarrow{a}=(1,-1)$$，$$\\overrightarrow{b}=(-1,2)$$，则$$( 2\\overrightarrow{a}+\\overrightarrow{b})\\cdot \\overrightarrow{a}=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["知识标签->题型->平面向量->平面向量的运算->平面向量的数量积->线性运算和数量积综合问题", "知识标签->题型->平面向量->平面向量的运算->平面向量的数量积->几何、代数、坐标进行的数量积运算", "知识标签->素养->数学运算", "知识标签->知识点->平面向量->平面向量的运算->平面向量的线性运算->平面向量的加法运算及运算规则", "知识标签->知识点->平面向量->平面向量的运算->数量积->向量的数量积的定义", "知识标签->知识点->平面向量->平面向量的运算->数量积->数量积的坐标表达式"], "answer_analysis": ["$$\\overrightarrow{a}=\\left( 1,-1 \\right)$$，$$\\overrightarrow{b}=\\left( -1,2 \\right)$$，$$\\therefore ( 2\\overrightarrow{a}+\\overrightarrow{b})\\cdot \\overrightarrow{a}=\\left( 1,0 \\right)\\cdot \\left( 1,-1 \\right)=1$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "274", "queId": "46be171ee60c4673ba2d4f983e4425a9", "competition_source_list": ["2017年AMC12竞赛B第18题"], "difficulty": "2", "qtype": "single_choice", "problem": "半径为$$2$$的圆的直径$$AB$$被延伸到圆外的一点$$D$$，使$$BD=3$$.选择$$E$$点，使$$ED=5$$，直线$$ED$$与直线$$AD$$垂直．线段$$AE$$与圆相交于$$A$$和$$E$$之间的点$$C$$.请问三角形$$ABC$$的面积是多少？", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{120}{37}$$ "}], [{"aoVal": "B", "content": "$$\\frac{140}{39}$$ "}], [{"aoVal": "C", "content": "$$\\frac{145}{39}$$ "}], [{"aoVal": "D", "content": "$$\\frac{140}{37}$$ "}], [{"aoVal": "E", "content": "$$\\frac{120}{31}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->直线和圆的方程", "美国AMC10/12->Knowledge Point->Geometry->Calculation of Plane Geometry->Circle"], "answer_analysis": ["Solution 1  Notice that $$ADE$$ and $$ABC$$ are right triangles.  Then $$AE=\\sqrt{7^{2}+5^{2}}=\\sqrt{74}$$,$$\\sin DAE=\\sin ABC=\\dfrac{5}{\\sqrt{74}}=\\sin BAE=\\dfrac{BC}{4}$$,so $$BC=\\dfrac{20}{\\sqrt{74}}$$. We also find that $$AC=\\dfrac{28}{\\sqrt{74}}$$, and thus the area of $$ABC$$ is $$\\dfrac{\\dfrac{20}{\\sqrt{74}}\\cdot\\dfrac{28}{\\sqrt{74}}}{2}=\\dfrac{\\dfrac{560}{74}}{2}=\\dfrac{140}{37}$$.  Solution 2  We note that $$\\triangle ACB\\sim\\triangle ADE$$ by $$AA$$ similarity. Also, since the area of $$\\triangle ADE=\\dfrac{7\\cdot5}{2}=\\dfrac{35}{2}$$ and $$AE=\\sqrt {74}$$,$$\\dfrac{\\left[ ABC\\right]}{\\left[ ADE\\right]}=\\dfrac{\\left[ ABC\\right]}{\\dfrac{35}{2}}=\\left( \\dfrac{4}{\\sqrt{74}}\\right)^{2}$$ , so the area of $$\\triangle ABC=\\frac {140}{37}$$.  Solution 3  As stated before, note that $$\\triangle ACB$$,$$\\triangle ADE$$. By similarity, we note that $$\\dfrac{\\overline{AC}}{\\overline{BC}}$$ is equivalent to $$\\frac 75$$. We set $$\\overline{AC}$$ to $$7x$$~ and $$\\overline{BC}$$ to $$5x$$. By the Pythagorean Theorem, $$\\left( 7x\\right)^{2}+\\left( 5x\\right)^{2}=4^{2}$$. Combining, $$49x^{2}+25x^{2}=16$$. We can add and divide to get $$x^{2}=\\frac 8{37}$$. We square root and rearrange to get $$x=\\dfrac{2\\sqrt{74}}{37}$$. We know that the legs of the triangle are $$7x$$ and $$5x$$. Mulitplying $$x$$ by $$7$$ and $$5$$ eventually gives us $$\\dfrac{14\\sqrt{74}}{37}\\dfrac{10\\sqrt{74}}{37}$$. We divide this by $$2$$, since $$\\frac 12bh$$ is the formula for a triangle. This gives us $$\\frac {140}{37}$$.  Solution 4  Let\\textquotesingle s call the center of the circle that segment $$AB$$ is the diameter of,$$O$$. Note that $$\\triangle ODE$$ is an isosceles right triangle. Solving for side $$OE$$, using the Pythagorean theorem, we find it to be $$5\\sqrt 2$$. Calling the point where segment $$OE$$ intersects circle $$O$$, the point $$I$$, segment $$IE$$ would be $$5\\sqrt 2-2$$. Also, noting that $$\\triangle ADE$$ is a right triangle, we solve for side $$AE$$, using the Pythagorean Theorem, and get $$\\sqrt {74}$$. Using Power of Point on point $$E$$, we can solve for $$CE$$. We can subtract $$CE$$ from $$AE$$ to find $$AC$$ and then solve for $$CB$$ using Pythagorean theorem once more $$(AE)(CE)$$. $$=$$ (Diameter of circle $$O+IE$$ ),$$ \\left( IE\\right)\\rightarrow\\sqrt{74}\\left( CE\\right)=\\left( 5\\sqrt{2}+2\\right)\\left( 5\\sqrt{2}-2\\right)\\Rightarrow CE=\\dfrac{23\\sqrt{74}}{37}$$,$$AC=AE-CE\\rightarrow AC=\\sqrt{74}-\\dfrac{23\\sqrt{74}}{37}\\Rightarrow AC=\\dfrac{14\\sqrt{74}}{37}$$,Now to solve for $$CB$$:$$AB^{2}-AC^{2}=CB^{2}\\rightarrow4^{2}+\\dfrac{14\\sqrt{74}^{2}}{37}=CB^{2}\\Rightarrow CB=\\dfrac{10\\sqrt{74}}{37}$$ ,  Note that $$\\triangle ABC$$ is a right triangle because the hypotenuse is the diameter of the circle. Solving for area using the bases $$AC$$ and $$BC$$, we get the area of triangle $$ABC$$ to be $$\\frac {140}{37}$$. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "850", "queId": "e8ddccdee5c64604ac2931542d2af173", "competition_source_list": ["2009年竞赛珠海市第5题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知数列$$\\sqrt{3}$$，$$\\sqrt{7}$$，$$\\sqrt{11}$$，$$\\sqrt{15}$$，\\ldots，则$$3\\sqrt{11}$$是它的．", "answer_option_list": [[{"aoVal": "A", "content": "第$$23$$项 "}], [{"aoVal": "B", "content": "第$$24$$项 "}], [{"aoVal": "C", "content": "第$$19$$项 "}], [{"aoVal": "D", "content": "第$$25$$项 "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["$$3\\sqrt{11}=\\sqrt{99}$$，$$\\frac{99-3}{4}+1=25$$ "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "35", "queId": "035b17fbcbfe4a3ba0af2cec515a551a", "competition_source_list": ["2008年河南全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "函数$$f(x)=\\log _{2}^{(x\\textbar x\\textbar)}$$的单调递增区间是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(0,+\\infty )$$ "}], [{"aoVal": "B", "content": "$$(-\\infty ,0)$$ "}], [{"aoVal": "C", "content": "$$(-\\infty ,0)\\cup (0,+\\infty )$$ "}], [{"aoVal": "D", "content": "$$(-\\infty ,+\\infty )$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["$$f(x)={{\\log }_{2}}^{(x\\textbar x\\textbar)}=2{{\\log }_{2}}x$$，故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1022", "queId": "f76ef37db8a444c19a4ebf8284921eb2", "competition_source_list": ["2018年吉林全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "三棱锥$$P-ABC$$的底面$$\\triangle ABC$$是边长为$$4$$的正三角形，$$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{39}$$ "}], [{"aoVal": "C", "content": "$$\\sqrt{11}$$ "}], [{"aoVal": "D", "content": "$$2\\sqrt{3}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->立体几何初步->基本立体图形->空间几何体的体积、表面积->三棱锥体积问题->等体积法", "竞赛->知识点->立体几何与空间向量"], "answer_analysis": ["解：$${{V}_{P-ABC}}={{V}_{A-PBC}}=\\frac{1}{3}\\times \\left( \\frac{1}{2}\\times 3\\times 4 \\right)\\times \\sqrt{{{4}^{2}}-{{\\left( \\frac{5}{2} \\right)}^{2}}}=\\sqrt{39}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1119", "queId": "ea8c4824b153452a8af487074f0dc2b7", "competition_source_list": ["2008年四川全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "某学校的数学课外小组有$$8$$个女生和$$6$$个男生，要从他们中挑选$$4$$人组成代表队去参加比赛，则代表队包含男女生各$$2$$人的概率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{10}{143}$$ "}], [{"aoVal": "B", "content": "$$\\frac{30}{143}$$ "}], [{"aoVal": "C", "content": "$$\\frac{60}{143}$$ "}], [{"aoVal": "D", "content": "$$\\frac{70}{143}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理->排列与组合->组合", "课内体系->知识点->计数原理->两个基本计数原理->分步乘法计数原理", "课内体系->知识点->统计与概率->概率->事件与概率->古典概型", "课内体系->知识点->统计与概率->概率->事件与概率->概率的概念->概率的基本性质", "课内体系->知识点->统计与概率->概率->事件与概率->概率的概念->频率与概率问题", "课内体系->知识点->统计与概率->概率->事件与概率->事件与基本事件空间->基本事件与基本事件空间", "课内体系->素养->数学运算", "课内体系->素养->数据分析"], "answer_analysis": ["所求的概率为$$\\frac{\\text{C}_{8}^{2}\\text{C}_{6}^{2}}{\\text{C}_{14}^{4}}=\\frac{60}{143}$$．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "529", "queId": "ff4cd33e2439478aab58e7acc40da070", "competition_source_list": ["2017年天津全国高中数学联赛竞赛初赛第5题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "实数$$a,b$$满足$$\\left\\textbar{} a \\right\\textbar\\leqslant 1$$，$$\\left\\textbar{} a+b \\right\\textbar\\leqslant 1$$，则$$\\left( a+1 \\right)\\left( b+1 \\right)$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ 0,\\frac{9}{4} \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left[ -2,\\frac{9}{4} \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left[ 0,2 \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left[ -2,2 \\right]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["$$\\left( a+1 \\right)\\left( b+1 \\right)\\leqslant \\frac{{{\\left[ \\left( a+1 \\right)+\\left( b+1 \\right) \\right]}^{2}}}{4}\\leqslant \\frac{{{3}^{2}}}{4}=\\frac{9}{4}$$，当$$a=b=\\frac{1}{2}$$时取等号．  $$\\left( a+1 \\right)\\left( b+1 \\right)=ab+\\left( a+b \\right)+1\\geqslant ab\\geqslant -\\left\\textbar{} b \\right\\textbar\\geqslant -2$$，当$$a=1b=-2$$时取到等号． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "75", "queId": "d4fff5d764174754b8bd99af5e98b875", "competition_source_list": ["2000年全国高中数学联赛竞赛一试第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设全集是实数，若$$A=\\left { x\\textbar\\sqrt{x-2}\\leqslant 0 \\right }$$，$$B=\\left { x\\textbar{{10}^{{{x}^{2}}-2}}={{10}^{x}} \\right }$$，则$$A\\cap \\overline{B}$$是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$ {2 }$$ "}], [{"aoVal": "B", "content": "$$ {-1 }$$ "}], [{"aoVal": "C", "content": "$$\\left { x\\textbar x\\leqslant 2 \\right }$$ "}], [{"aoVal": "D", "content": "$$\\varnothing $$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["由$$\\sqrt{x-2}\\leqslant 0$$，得$$x=2$$，故$$A=\\left { 2 \\right }$$；由$${{10}^{{{x}^{2}}-2}}={{10}^{x}}$$，得$${{x}^{2}}-x-2=0$$，故$$B= {-1,2 }$$．所以$$A\\cap \\overline{B}=\\varnothing $$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "614", "queId": "99f2e1f1b0fd4abba6ec1cdfbb51d6f2", "competition_source_list": ["2018年天津全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$f(x)=\\cos \\omega x$$的最小正周期为$$6$$．则$$\\sum_{i=1}^{2018} f(i)$$的值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$1$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{2}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{3}}{2}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->函数的概念与性质->函数的性质->周期性->函数周期性判断", "课内体系->知识点->三角函数->三角函数的图象与性质->余弦函数的图象和性质", "课内体系->知识点->三角函数->三角函数的图象与性质->余弦型三角函数的图象与性质", "课内体系->素养->数学运算", "课内体系->素养->逻辑推理"], "answer_analysis": ["由最小正周期为$$6\\Rightarrow 6 \\omega=2 \\pi \\Rightarrow \\omega=\\frac{\\pi}{3}$$，  于是，当$$k$$为整数时，  $$\\sum_{i=1}^{6} f(6 k+i)=0$$，  即每个完整周期内的六个函数值之和为零，  注意到，$$2018=6 \\times 336+2$$，  故原式$$=f(1)+f(2)=\\cos \\frac{\\pi}{3}+\\cos \\frac{2 \\pi}{3}=0$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "466", "queId": "7ea61bd7847e47deb70051d7604c0861", "competition_source_list": ["2011年山东全国高中数学联赛竞赛初赛第9题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知函数$$f(x)=(\\frac{1}{{{a}^{x}}-1}+\\frac{1}{2}){{x}^{2}}+bx+6$$ （$$a,  b$$为常数，$$a\\textgreater1$$），且$$f(\\lg {{\\log }_{8}}1000)=8$$，则$$f(\\lg \\lg 2)$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$-4$$ "}], [{"aoVal": "D", "content": "$$-8$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->基本初等函数"], "answer_analysis": ["由已知可得  $$f(\\lg {{\\log }_{8}}1000)=f(\\lg \\frac{3}{3\\lg 2})=f(-\\lg \\lg 2)=8.$$  又$$\\frac{1}{{{a}^{-x}}-1}+\\frac{1}{2}=\\frac{{{a}^{x}}}{1-{{a}^{x}}}+\\frac{1}{2}=-1+\\frac{1}{1-{{a}^{x}}}+\\frac{1}{2}=-\\frac{1}{{{a}^{x}}-1}-\\frac{1}{2}$$  令$$F(x)=f(x)-6$$，则有$$F(-x)=-F(x)$$ 从而有  $$f(-\\lg \\lg 2)=F(\\lg \\lg 2)+6=F(\\lg \\lg 2)+6=8$$  即知$$F(\\lg \\lg 2)=-2,  f(\\lg \\lg 2)=F(\\lg \\lg 2)+6=4$$ "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "470", "queId": "90394426126a4335bc10f22ddd84b417", "competition_source_list": ["2019年北京高一竞赛初赛（中学生数学竞赛）第3题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "对于无理数$$\\text{e}=2.71828182845\\cdots $$，定义函数$$f(n)=k$$，定义域为正整数，其中，$$k$$为$$\\text{e}$$的小数点后第$$n$$位的数字，规定$$f(0)=2$$，则$$f(f(f(n)))$$的值域为．", "answer_option_list": [[{"aoVal": "A", "content": "$$ {3,4,6,9 }$$ "}], [{"aoVal": "B", "content": "$$ {2,0,1,9 }$$ "}], [{"aoVal": "C", "content": "$$ {0,3,5,9 }$$ "}], [{"aoVal": "D", "content": "$$ {1,2,7,8 }$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的概念"], "answer_analysis": ["易知，$$f(n)$$的值域为$$ {0,1,\\cdots ,9 }$$，  $$f(f(n))$$的值域为$$ {1,2,7,8 }$$．  则$$f(f(f(n)))$$的值域为$$ {1,2,7,8 }$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "493", "queId": "c7b151d8598147abb9b1fc39a20a314a", "competition_source_list": ["2007年AMC12竞赛B第21题"], "difficulty": "3", "qtype": "single_choice", "problem": "2007AMC12B, 21  The first $$2007$$ positive integers are each written in base $$3$$. How many of these base- $$3$$ representations are palindromes? (A palindrome is a number that reads the same forward and backward.)．  把前$$2007$$个正整数用$$3$$进制来表示。这些$3$进制表示中有多少是回文数？（回文数，指正着或倒着读都一样的数。）．", "answer_option_list": [[{"aoVal": "A", "content": "$$100$$ "}], [{"aoVal": "B", "content": "$$101$$ "}], [{"aoVal": "C", "content": "$$102$$ "}], [{"aoVal": "D", "content": "$$103$$ "}], [{"aoVal": "E", "content": "$$104$$ "}]], "knowledge_point_routes": ["课内体系->知识点->算法与框图->算法案例->进位制", "美国AMC10/12->Knowledge Point->Combination->Reasoning->Information Migration (new definition)"], "answer_analysis": ["首先把$2007_{10}$转化成$$3$$进制, 得到$2202100_3$.  于是我们就要数出有多少不超过$2202100_3$的$$3$$进制正回文数$$.$$ 按位数分类计数, 然后求和, 是一个可行的思路.  1位的$$3$$进制回文数有$$2$$个, 1和$$2$$（注意题目要求正整数）.  2位的有$$2$$个, $11$和$22$.  3位的有$2\\times 3=6$个, $1(\\cdot)1$和$2(\\cdot)2$.  4位的有$2\\times 3=6$个, $1(\\cdot)(\\cdot)1$和$2(\\cdot)(\\cdot)2$.  算到这里, 我们发现$2k-1$位的回文数和$2k$位的回文数的个数理应是相同的, 且都等于$2k-2$位的回文数的个数$\\times 3$.  于是$$5$$位和$$6$$位的回文数各有$6\\times 3=18$个.  7位的回文数有$18\\times 3=54$个.  注意我们要求的是不超过$2202100_3$的回文数, 因此再之后的就就不用求了, 而且我们还需要剔除几个太大的数.  7位且大于$2202100_3$的回文数只能是$22(x)(y)(x)22$的形式, 且$x\\neq0$.因此需要剔除的数共有$2\\times3=6$个.  于是我们的结果是$2\\times(2+6+18)+54-6=100$个, 选$$\\text{A}$$.  拓展: 如果把这前$$2007$$个正整数用$$10$$进制表示, 其中的回文数又有多少个? 用其它进制表示呢? 回文数的数量与进制存在关系吗? "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "114", "queId": "0bd99b0620374f55afbb796a9998ff6c", "competition_source_list": ["2019年贵州全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$a=\\pi -3$$，$$b=\\ln \\pi -\\ln 3$$，$$c=\\text{e}^{\\pi}-\\text{e}^{3}$$，其中，$$\\pi $$是圆周率，$$\\text{e}$$是自然对数的底数．则$$a$$、$$b$$、$$c$$的关系为 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textless b\\textless c$$ "}], [{"aoVal": "B", "content": "$$b\\textless c\\textless a$$ "}], [{"aoVal": "C", "content": "$$c\\textless b\\textless a$$ "}], [{"aoVal": "D", "content": "$$b\\textless a\\textless c$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["设$$f(x)=x-\\ln x$$．  则$$f^{\\prime}(x)= \\frac{x-1}{x}$$ ．  于是，$$f(x)$$在区间$$(1,+\\infty)$$上单调递增．  故$$f(\\pi )\\textgreater f(3)$$，  $$\\Rightarrow \\pi -\\ln \\pi \\textgreater3-\\ln 3$$，  $$\\Leftrightarrow \\pi -3\\textgreater\\ln \\pi -\\ln 3$$，  $$\\Rightarrow a\\textgreater b$$．  设$$g(x)=\\text{e}^{}x-x$$．  则$$g^{}\\prime (x)=\\text{e}^{}x-1$$．  于是，$$g(x)$$在区间$$(0,+\\infty )$$上单调递增．  故$$g(\\pi )\\textgreater g(3)$$，  $$\\Rightarrow \\text{e}^{x}--\\pi \\textgreater\\text{e}^{3}-3$$  $$\\Leftrightarrow \\text{e}\\pi -\\text{e}^{3}\\textgreater\\pi -3$$  $$\\Rightarrow c\\textgreater a$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "191", "queId": "66a80a0f7f714f9e9fcc6b852873c2c1", "competition_source_list": ["2005年AMC12竞赛A第21题"], "difficulty": "3", "qtype": "single_choice", "problem": "2005AMC12A, 21  How many ordered triples of integers $$(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$$ "}]], "knowledge_point_routes": ["课内体系->知识点->基本初等函数->对数的概念及其运算->指对化简求值", "美国AMC10/12->Knowledge Point->Algebra->Calculation->Solve Quadratic Equations"], "answer_analysis": ["由$\\log_a{b}=c^{2005}$可知$b=a^{c^{2005}}$.  又由$a+b+c=2005$知$a\\textless2005$, 故$c=0$或$c=1$.  $c=0$时, $b=a^{0}=1$, $a=2005-b-c=2004$.  $c=1$时, $b=a^{1}=a$, 再由$a+b+c=2005$知$(a,b,c)=(1002, 1002, 1)$.  共两种, 选$$\\text{C}$$. "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "690", "queId": "4e603777290f47d1b1a08cb422ed0df1", "competition_source_list": ["竞赛第8题"], "difficulty": "2", "qtype": "single_choice", "problem": "设实数\\emph{a}，\\emph{b}，\\emph{c}满足$a,b,c\\geq 1$且$ab\\sqrt{c-1}+ac\\sqrt{b-1}+bc\\sqrt{a-1}=\\frac{3}{2}abc$，则\\emph{a}，\\emph{b}，\\emph{c}之间的大小关系是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$a\\textgreater{} b\\textgreater{} c$ "}], [{"aoVal": "B", "content": "$a=b=c$ "}], [{"aoVal": "C", "content": "$a\\textless{} b\\textless{} c$ "}], [{"aoVal": "D", "content": "不能比较大小 "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 利用二次函数的性质或基本不等式可求$a=b=c=2$，故可得正确的选项.\\\\ 【详解】\\\\ \\textbf{解法一}~~~题中等式可以变形为$\\sqrt{\\frac{c-1}{c^{2}}}+\\sqrt{\\frac{b-1}{b^{2}}}+\\sqrt{\\frac{a-1}{a^{2}}}=\\frac{3}{2}$，\\\\ 而$\\sqrt{\\frac{c-1}{c^{2}}}=\\sqrt{\\frac{1}{4}-\\left(\\frac{1}{c}-\\frac{1}{2}\\right)^{2}}\\leq \\frac{1}{2}$，所以只能有$\\sqrt{\\frac{c-1}{c^{2}}}=\\sqrt{\\frac{b-1}{b^{2}}}=\\sqrt{\\frac{a-1}{a^{2}}}=\\frac{1}{2}$，\\\\ 解得$a=b=c=2$．\\\\ \\textbf{解法二}~~~令$x=\\sqrt{a-1},y=\\sqrt{b-1},z=\\sqrt{c-1}$，\\\\ 则有$x,y,z\\geq 0$且题中条件变为：\\\\ $2z\\left(x^{2}+1\\right)\\left(y^{2}+1\\right)+2y\\left(x^{2}+1\\right)\\left(z^{2}+1\\right)+2x\\left(z^{2}+1\\right)\\left(y^{2}+1\\right)=$\\\\ $3\\left(x^{2}+1\\right)\\left(y^{2}+1\\right)\\left(z^{2}+1\\right)$，\\\\ 而$2z\\left(x^{2}+1\\right)\\left(y^{2}+1\\right)\\leq \\left(z^{2}+1\\right)\\left(x^{2}+1\\right)\\left(y^{2}+1\\right)$，\\\\ $2y\\left(x^{2}+1\\right)\\left(z^{2}+1\\right)\\leq \\left(y^{2}+1\\right)\\left(x^{2}+1\\right)\\left(z^{2}+1\\right)$，\\\\ $2x\\left(y^{2}+1\\right)\\left(z^{2}+1\\right)\\leq \\left(x^{2}+1\\right)\\left(y^{2}+1\\right)\\left(z^{2}+1\\right)$，\\\\ 故$2z\\left(x^{2}+1\\right)\\left(y^{2}+1\\right)+2y\\left(x^{2}+1\\right)\\left(z^{2}+1\\right)+2x\\left(z^{2}+1\\right)\\left(y^{2}+1\\right)\\leq$\\\\ $3\\left(x^{2}+1\\right)\\left(y^{2}+1\\right)\\left(z^{2}+1\\right)$，当且仅当$x=y=z=1$时等号成立，\\\\ 故$a=b=c=2$.\\\\ 故选：B "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "524", "queId": "3b1e328c748945848eaccc9e51845ab9", "competition_source_list": ["2015年广西全国高中数学联赛竞赛初赛第7题8分"], "difficulty": "2", "qtype": "single_choice", "problem": "$$\\triangle ABC$$内有$$2015$$个点，加上三个顶点，共有$$2018$$个点，把这些点连线，形成互不重叠的小三角形，则一共可以形成小三角形的个数是~\\uline{~~~~~~~~~~}~", "answer_option_list": [[{"aoVal": "A", "content": "$$4031$$ "}], [{"aoVal": "B", "content": "$$4032$$ "}], [{"aoVal": "C", "content": "$$4033$$ "}], [{"aoVal": "D", "content": "$$4034$$ "}], [{"aoVal": "E", "content": "$$4036$$ "}]], "knowledge_point_routes": ["竞赛->知识点->逻辑->逻辑推理", "课内体系->知识点->计数原理"], "answer_analysis": ["所有小三角形的内角和为$$2015\\cdot 2 \\pi + \\pi =4031 \\pi $$，所以小三角形的个数有$$4031$$个． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "622", "queId": "521a8d9203ab4e23bc8b63e806138fee", "competition_source_list": ["2018年陕西全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知$$a={{\\log }_{8}}5$$，$$b={{\\log }_{4}}3$$，$$c=\\frac{2}{3}$$，则$$a$$、$$b$$、$$c$$的大小关系是(  )", "answer_option_list": [[{"aoVal": "A", "content": "$$a\\textgreater b\\textgreater c$$ "}], [{"aoVal": "B", "content": "$$b\\textgreater c\\textgreater a$$ "}], [{"aoVal": "C", "content": "$$b\\textgreater a\\textgreater c$$ "}], [{"aoVal": "D", "content": "$$c\\textgreater b\\textgreater a$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式"], "answer_analysis": ["解：$$a={{\\log }_{2}}\\sqrt[3]{5}\\textgreater0$$，$$b={{\\log }_{2}}\\sqrt{3}\\textgreater1$$，$$c=\\frac{2}{3} \\textless{} 1$$，  $$\\because {{5}^{2}} \\textless{} {{3}^{3}}$$，$$\\therefore \\sqrt[3]{5} \\textless{} \\sqrt{3}$$，$$\\therefore c \\textless{} a \\textless{} b$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "108", "queId": "13c0cced41834a0cb1c79a4d7072857f", "competition_source_list": ["2018年辽宁全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$A=\\left[ -2,4 \\right)$$，$$B=\\left { x\\left\\textbar{} {{x}^{2}}-ax-4\\leqslant 0 \\right. \\right }$$，若$$B\\subseteq A$$，则实数$$a$$取值范围为(  )", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left[ -3,0 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left[ -2,0 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left[ 0,2 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left[ 0,3 \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合"], "answer_analysis": ["由$$f\\left( x \\right)={{x}^{2}}-ax-4$$开口向上，  且$$\\left\\textbar{} x \\right\\textbar{{x}^{2}}-ax-4\\leqslant 0\\left\\textbar{} \\subseteq \\left[ -2,4 \\right) \\right.$$，  知$$f\\left( -2 \\right)\\geqslant 0$$，$$f\\left( 4 \\right)\\textgreater0\\Rightarrow a\\in \\left[ 0,3 \\right)$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "541", "queId": "a2e7ecec0d874e26b1b09a93ec084d16", "competition_source_list": ["2008年江西全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "若对所有实数$$x$$，均有$${{\\sin }^{k}}x\\cdot \\sin kx+{{\\cos }^{k}}x\\cdot \\cos kx={{\\cos }^{k}}2x$$，则$$k$$为．", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$5$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["记$$f\\left( x \\right)={{\\sin }^{k}}x\\cdot \\sin kx+{{\\cos }^{k}}x\\cdot \\cos kx-{{\\cos }^{k}}2x$$ ，  则由题意知对于任意的$$x$$，$$f\\left( x \\right)$$恒为$$0$$．  若取$$x=\\frac{ \\pi }{2}$$，可得$$\\sin \\frac{k \\pi }{2}={{\\left( -1 \\right)}^{k}}$$，  故选项中只有$$k=3$$满足题意．故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "839", "queId": "617f15f1926f4751839423f47d7be09a", "competition_source_list": ["2019年贵州全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "一个不透明箱子内装有$$20$$个大小，形状均相同的小球，分别标有号码$$1$$，$$2$$，$$\\cdots $$，$$20$$，从中任意取出两个球．则这两个球的号码之和能被$$3$$整除的概率为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{38}$$ "}], [{"aoVal": "B", "content": "$$\\frac{32}{95}$$ "}], [{"aoVal": "C", "content": "$$\\frac{11}{95}$$ "}], [{"aoVal": "D", "content": "$$\\frac{37}{190}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合", "竞赛->知识点->排列组合与概率->概率初步"], "answer_analysis": ["易知，从$$1\\sim 20$$被$$3$$整除的数有$$6$$个，被$$3$$除余$$1$$的数有$$7$$个，被$$3$$除余$$2$$的数有$$7$$个．故两数之和能被$$3$$整除的概率为  $$P=\\frac{\\text{C}_{6}^{2}+\\text{C}_{7}^{1}\\text{C}_{7}^{1}}{\\text{C}_{20}^{2}}=\\frac{64}{190}=\\frac{32}{95}$$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "491", "queId": "43a0e263e33148caa0aa9568b58f73c9", "competition_source_list": ["2012年辽宁全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$${{A}_{k}}=\\left { x\\textbar x=kt+\\frac{1}{kt},\\frac{1}{{{k}^{2}}}\\leqslant t\\leqslant 1 \\right }\\left( k=2,  3,\\cdots ,  2012 \\right)$$，则所有$${{A}_{k}}$$的交集为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\varnothing $$ "}], [{"aoVal": "B", "content": "$$\\left { 2 \\right }$$ "}], [{"aoVal": "C", "content": "$$\\left[ 2,\\frac{5}{2} \\right]$$ "}], [{"aoVal": "D", "content": "$$\\left[ 2,\\frac{{{2012}^{2}}+1}{2012} \\right]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["由``对号''函数性质$${{A}_{k}}=\\left[ 2,  k+\\frac{1}{k} \\right]$$，$$k+\\frac{1}{k}$$在$$k\\in \\left[ 2,  +\\infty \\right)$$递增，因此所有$${{A}_{k}}$$的交集为$${{A}_{2}}=\\left[ 2,\\frac{5}{2} \\right]$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "868", "queId": "80f7d6d1b9ab45bfa908abd9f64220d6", "competition_source_list": ["1982年全国高中数学联赛竞赛第5题"], "difficulty": "1", "qtype": "single_choice", "problem": "对任何$$\\varphi \\in (0，\\frac{\\pi }{2})$$，都有(~ ~ )．~", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sin \\sin \\varphi ~~\\textless{} ~\\cos \\varphi ~~\\textless{} ~\\cos \\cos \\varphi $$ "}], [{"aoVal": "B", "content": "$$\\sin \\sin \\varphi \\textgreater\\cos \\varphi \\textgreater\\cos \\cos \\varphi $$ "}], [{"aoVal": "C", "content": "$$\\sin \\cos \\varphi \\textgreater\\cos \\varphi \\textgreater\\cos \\sin \\varphi $$ "}], [{"aoVal": "D", "content": "$$\\sin \\cos \\varphi ~~\\textless{} ~\\cos \\varphi ~~\\textless{} ~\\cos \\sin \\varphi $$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角函数的图像与性质"], "answer_analysis": ["由于$$(0\\mathsf{，}\\frac{ \\pi }{2})$$$$$$中，$$\\sin x ~\\textless{} ~x$$  易知$$0 ~\\textless{} ~\\cos x ~\\textless{} ~1 ~\\textless{} ~\\frac{ \\pi }{2}$$  所以$$\\sin \\mathsf{(}\\cos \\varphi \\mathsf{)} ~\\textless{} ~\\cos \\varphi $$  由在$$(0\\mathsf{，}\\frac{ \\pi }{2})$$中，$$\\varphi \\textgreater\\sin \\varphi \\mathsf{，}\\cos x$$单调递减得  $$\\cos \\varphi ~~\\textless{} ~\\cos \\mathsf{(}\\sin \\varphi \\mathsf{)}$$．  $$\\therefore $$ $$\\sin \\cos \\varphi ~~\\textless{} ~\\cos \\varphi ~~\\textless{} ~\\cos \\sin \\varphi $$成立． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "115", "queId": "2a60f9ff5ad44ae383964c61ee98d25a", "competition_source_list": ["2008年天津全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "3", "qtype": "single_choice", "problem": "考虑集合$$S=\\left { 1  ,2  ,\\cdots ,  10 \\right }$$的所有非空子集，若一个非空子集中的偶数的数目不少于奇数的数目，称这个子集是``好子集''，则``好子集''的数目有个．", "answer_option_list": [[{"aoVal": "A", "content": "$$631$$ "}], [{"aoVal": "B", "content": "$$633$$ "}], [{"aoVal": "C", "content": "$$635$$ "}], [{"aoVal": "D", "content": "$$637$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["方法一：设一个``好子集''中有$$i\\left( i=1,2,3,4,5 \\right)$$个偶数，则奇数的数目可以有$$j=0,1,\\cdots ,i$$个，因此``好子集''的数目为  $$\\sum\\limits_{i=1}^{5}{\\left( \\text{C}_{5}^{i}\\sum\\limits_{j=0}^{i}{\\text{C}_{5}^{j}} \\right)}=\\text{C}_{5}^{1}\\left( \\text{C}_{5}^{0}+\\text{C}_{5}^{1} \\right)+\\text{C}_{5}^{2}\\left( \\text{C}_{5}^{0}+\\text{C}_{5}^{1}+\\text{C}_{5}^{2} \\right)$$  $$+\\text{C}_{5}^{3}\\left( \\text{C}_{5}^{0}+\\text{C}_{5}^{1}+\\text{C}_{5}^{2}+\\text{C}_{5}^{3} \\right)$$  $$+\\text{C}_{5}^{4}\\left( \\text{C}_{5}^{0}+\\text{C}_{5}^{1}+\\text{C}_{5}^{2}+\\text{C}_{5}^{3}+\\text{C}_{5}^{4} \\right)$$  $$+\\text{C}_{5}^{5}\\left( \\text{C}_{5}^{0}+\\text{C}_{5}^{1}+\\text{C}_{5}^{2}+\\text{C}_{5}^{3}+\\text{C}_{5}^{4}+\\text{C}_{5}^{5} \\right)=637$$.  故选$$\\text{D}$$．  方法二：$$S$$的非空子集共有$${{2}^{10}}-1=1023$$（个），根据子集中偶数与奇数个数的多少可为三类：（1）偶数多于奇数；（2）奇数多于偶数；（3）奇数与偶数个数相等．由于$$S$$中的$$10$$个元素偶数与奇数的个数相等，所以（1）、（2）的子集数相等．现考虑第三类，分别考虑含有$$2$$、$$4$$、$$6$$、$$8$$、$$10$$个元素子集的数目，则共有子集数为$$\\text{C}_{5}^{1}\\cdot \\text{C}_{5}^{1}+\\text{C}_{5}^{2}\\cdot \\text{C}_{5}^{2}+\\text{C}_{5}^{3}\\cdot \\text{C}_{5}^{3}+\\text{C}_{5}^{4}\\cdot \\text{C}_{5}^{4}+1=251$$．  所以，第一类子集数为$$\\frac{1}{2}(1023-251)=386$$．  因此，``好子集''的数目为$$386+251=637$$（个）．故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "607", "queId": "5f78b77382ae461ba171072d2b9d1fbc", "competition_source_list": ["2013年辽宁全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "已知实数$$x,y$$满足$$17{}{{x}^{2}}+{{y}^{2}}{}-30xy-16=0$$，则$$\\sqrt{16{{x}^{2}}+4{{y}^{2}}-16xy-12x+6y+9}$$的最大值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$7$$ "}], [{"aoVal": "B", "content": "$${}\\sqrt{29}$$ "}], [{"aoVal": "C", "content": "$${}\\sqrt{19}$$ "}], [{"aoVal": "D", "content": "$$3$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->圆与方程", "竞赛->知识点->不等式->换元技巧->三角换元"], "answer_analysis": ["$$17\\left( {{x}^{2}}+{{y}^{2}} \\right){-}30xy-16=0$$，  得$${{\\left( x+y \\right)}^{2}}+16{{\\left( x-y \\right)}^{2}}=16$$，$${{\\left( \\frac{x+y}{4} \\right)}^{2}}+{{\\left( x-y \\right)}^{2}}=1$$．  令$$\\begin{cases}x+y=4\\cos \\theta    x-y=\\sin \\theta \\end{cases}{}\\theta \\in \\mathbf{R}{}$$，得  $$f\\left( x,y \\right)=\\sqrt{16{{x}^{2}}+4{{y}^{2}}-16xy-12x+6y+9}$$  $$=\\sqrt{{{\\left( 4x-2y \\right)}^{2}}-3\\left( 4x-2y \\right)+9}$$  $$=\\sqrt{[3\\left( x-y \\right)+\\left( x+y \\right)\\left] ^{2}-3 \\right[3\\left( x-y \\right)+\\left( x+y \\right)]+9}$$  $$=\\sqrt{{{\\left( 3\\sin \\theta +4\\cos \\theta \\right)}^{2}}-3\\left( 3\\sin \\theta +4\\cos \\theta \\right)+9}$$  $$=\\sqrt{[5\\sin \\left( \\theta +\\alpha \\right)\\left] ^{2}-3 \\right[5\\sin \\left( \\theta +\\alpha \\right)]+9}$$，  其中$$\\alpha =\\text{arc}\\sin \\frac{4}{5}$$;当$$\\theta =\\frac{3 \\pi }{2}-\\alpha $$时，$$\\sin \\left( \\theta +\\alpha \\right)=-1,f{(}x,y{)}$$取得最大值$$7$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "189", "queId": "8277e5467f0f42459a7e5ddf10148c76", "competition_source_list": ["2014年天津全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$\\triangle ABC$$中，$$\\tan A$$、$$\\tan B$$、$$\\tan C$$都是整数，且$$A\\textgreater B\\textgreater C$$，则以下说法中错误的是．", "answer_option_list": [[{"aoVal": "A", "content": "$$A\\textless{}80{}^{}\\circ $$ "}], [{"aoVal": "B", "content": "$$B\\textless{}60{}^{}\\circ $$ "}], [{"aoVal": "C", "content": "$$C\\textless{}50{}^{}\\circ $$ "}], [{"aoVal": "D", "content": "$$A\\textgreater65{}^{}\\circ $$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["由于$$A\\textgreater B\\textgreater C$$，所以$$B$$、$$C$$都是锐角，$$\\tan B$$、$$\\tan C$$都是正整数，这样  $$\\tan A=-\\tan \\left( B+C \\right)=\\frac{\\tan B+\\tan C}{\\tan B\\tan C-1}\\textgreater0$$，  可见$$A$$也是锐角．这时，$$\\tan C\\geqslant 1$$，$$\\tan B\\geqslant 2$$，$$\\tan A\\geqslant 3$$．我们有  $$\\frac{\\tan A+\\tan B}{\\tan A\\tan B-1}=\\tan C\\geqslant 1$$，  即$$\\left( \\tan A-1 \\right)\\left( \\tan B-1 \\right)\\leqslant 2$$．但是$$\\tan A-1\\geqslant 2$$，$$\\tan B-1\\geqslant 1$$，比较可知只可能$$\\tan A=3$$，$$\\tan B=2$$，$$\\tan C=1$$．因此，$$C=45{}^{}\\circ $$，选项$$\\text{C}$$正确．由$$\\tan B\\textgreater\\sqrt{3}$$可知$$B\\textgreater60{}^{}\\circ $$，选项$$B$$是错误的．至于选项$$A$$和$$D$$，由$$\\tan 75{}^{}\\circ =2+\\sqrt{3}\\textgreater\\tan A$$可知$$A\\textgreater75{}^{}\\circ $$，选项$$\\text{A}$$正确；由$$A+B=135{}^{}\\circ $$和$$A\\textgreater B$$可知$$A\\textgreater65{}^{}\\circ $$，选项$$\\text{D}$$正确． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "417", "queId": "552d7709943243099f5ca29413443c9f", "competition_source_list": ["2009年湖南全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "(★★)函数$$f(x)$$的定义域为$$R$$，若$$f(x+1)$$与$$f(x-1)$$都是奇函数，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$f(x)$$是偶函数 "}], [{"aoVal": "B", "content": "$$f(x)$$是奇函数 "}], [{"aoVal": "C", "content": "$$f(x+3)$$是奇函数 "}], [{"aoVal": "D", "content": "$$f(x+3)$$是偶函数 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "课内体系->知识点->函数的概念与性质->函数的性质->奇偶性->函数奇偶性的判定->利用定义判断函数奇偶性"], "answer_analysis": ["因为$$f(x+1)$$是奇函数，所以  $$f(x+1)=-f(-x+1), f(x)=-f(-x+2)$$．①  同理，由$$f(x-1)$$是奇函数，可得  $$f(x)=-f(-x-2)$$．②  由①、②可得，$$f(-x+2)=f(-x-2)$$，所以$$f(x+2)=f(x-2), f(x+4)=f(x)$$，即$$f(x)$$是周期为$$4$$的函数．  又因为$$f(x-1)$$是奇函数，所以$$f(x+3)$$是奇函数，故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "612", "queId": "37d281007f4e4b6a80cf74a83186d7ae", "competition_source_list": ["1990年全国高中数学联赛竞赛一试第5题"], "difficulty": "2", "qtype": "single_choice", "problem": "设非零复数$$x$$，$$y$$满足$${{x}^{2}}+xy+{{y}^{2}}=0$$，则代数$${{\\left( \\frac{x}{x+y} \\right)}^{1990}}+{{\\left( \\frac{y}{x+y} \\right)}^{1990}}$$的值是（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{2}^{-1989}}$$ "}], [{"aoVal": "B", "content": "$$－1$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "以上答案都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算", "竞赛->知识点->复数与平面向量->模、辐角与单位根"], "answer_analysis": ["令$$y=\\omega x\\mathsf{}\\omega \\ne 1\\mathsf{}$$代入已知条件得  $$1+\\omega +{{\\omega }^{2}}=0\\Rightarrow (1-\\omega )(1+\\omega +{{\\omega }^{2}})=0$$，  进而  $${{\\omega }^{3}}=1\\mathsf{}$$，  故 原式=$$\\frac{1}{{{(1+\\omega )}^{1900}}}+\\frac{{{\\omega }^{1990}}}{{{(1+\\omega )}^{1900}}}$$，  $$=\\frac{1+{{\\omega }^{1990}}}{{{(1-{{\\omega }^{2}})}^{1990}}}=\\frac{1+\\omega }{{{\\omega }^{2}}}=-1$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "646", "queId": "76701479256948889e95c62ab79b03dd", "competition_source_list": ["2009年湖南全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "设$$z$$是复数，$$a(z)$$表示满足$${{z}^{n}}=1$$的最小正整数$$n$$，则对虚数单位$$\\text{i},\\alpha (\\text{i})=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的应用"], "answer_analysis": ["因为$${{\\text{i}}^{1}}=1, {{\\text{i}}^{2}}=-1, {{\\text{i}}^{3}}=-1, {{\\text{i}}^{4}}=1$$，所以满足$${{z}^{n}}=1$$的最小正整数$$n=4$$．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "522", "queId": "da3c11f50d3643fd981df405bcbf419f", "competition_source_list": ["2015年天津全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "在正方体的$$12$$条面对角线和$$4$$条体对角线中随机的选取两条对角线，则这两条对角线所在的直线为异面直线的概率等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{7}{30}$$ "}], [{"aoVal": "B", "content": "$$\\frac{9}{20}$$ "}], [{"aoVal": "C", "content": "$$\\frac{7}{15}$$ "}], [{"aoVal": "D", "content": "以上结果都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->排列与组合", "竞赛->知识点->排列组合与概率->概率初步"], "answer_analysis": ["与任一条面对角线异面的对角线有$$7$$条，与任一条体对角线异面的对角线有$$6$$条，所以异面直线对的数目为$$\\frac{1}{2}\\left( 7\\times 12+6\\times 4 \\right)=54$$，所求概率$$\\frac{54}{\\text{C}_{16}^{2}}=\\frac{9}{20}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "292", "queId": "94612cd4292943b6bfeba5be7fcc7f1c", "competition_source_list": ["高二上学期单元测试《代数变形（3）》自招第10题", "2008年河北全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "把$$2008$$表示成两个整数的平方差的形式，则不同的表示方法有种．", "answer_option_list": [[{"aoVal": "A", "content": "$$4$$ "}], [{"aoVal": "B", "content": "$$6$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数论模块->不定方程->因式分解与恒等变形"], "answer_analysis": ["设$${{x}^{2}}-{{y}^{2}}=2008$$，即$$(x+y)(x-y)=2008$$．$$2008$$有$$8$$个正因数，分别为$$1$$、$$2$$、$$4$$、$$8$$、$$251$$、$$502$$、$$1004$$、$$2008$$．而且$$(x+y)(x-y)$$只能同为偶数，因此对应的方程组为  $$\\begin{cases}x+y=-2    x-y=-1 004 \\end{cases},\\begin{cases}x+y=-4    x-y=-502 \\end{cases}$$  $$\\begin{cases}x+y=-502    x-y=-4 \\end{cases},\\begin{cases}x+y=-1 004    x-y=-2 \\end{cases}$$  $$\\begin{cases}x+y=2    x-y=1 004 \\end{cases}, \\begin{cases}x+y=4    x-y=502 \\end{cases}$$  $$\\begin{cases}x+y=502    x-y=4 \\end{cases},\\begin{cases}x+y=1 004    x-y=2 \\end{cases}$$  故$$\\left( x,y \\right)$$共有$$8$$组不同的值：$$\\left( 503,501 \\right)$$，$$\\left( -503,-501 \\right)$$，$$\\left( -503,501 \\right)$$，$$\\left( 503,-501 \\right)$$，$$\\left( 253,249 \\right)$$，$$\\left( -253,-249 \\right)$$，$$\\left( -253,249 \\right)$$，$$\\left( 253,-249 \\right)$$．故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "881", "queId": "858a0b7718e54a0c8176e419d6d4f5e8", "competition_source_list": ["2003年AMC12竞赛B第8题", "2003年AMC10竞赛B第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "让$$♣(x)$$表示正整数$x$的所有数字之和．例如, $$♣(8)=8$$ ， $$♣(123)=1+2+3=6$$. 有多少个两位数 $$x$$ 满足 $$♣(♣(x))=3$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$9$$ "}], [{"aoVal": "E", "content": "$$10$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Reasoning->Information Migration (new definition)", "课内体系->知识点->等式与不等式->等式->方程组的解集"], "answer_analysis": ["Let $$a$$ and $$b$$ be the digits of $$x$$.  ♣$$($$♣$$(x))=a+b=3$$.  Clearly ♣$$(x)$$ can only be $$3$$, $$12$$, $$21$$, or $$30$$ and only $$3$ and $$12$$ are possible to have two digits sum to.  If ♣$$(x)$$ sums to $$3$$, there are $$3$$ different solutions: $$12$$, $$21$$, or $$30$$.  If ♣$$(x)$$ sums to $$12$$, there are $$7$$ different solutions: $$39$$, $$48$$, $$57$$, $$66$$, $$75$$, $$84$$, or $$93$$.  The total number of solutions is $$3+7=10\\Rightarrow \\rm E$$. "], "answer_value": "E"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "2", "queId": "00494c945cef42da8ed38ed17c82d7a1", "competition_source_list": ["2018年四川南充高三零模理科第11题5分", "2016年陕西全国高中数学联赛竞赛初赛第7题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "设函数$$f(x)={{x}^{3}}+a{{x}^{2}}+bx+c$$（$$a$$，$$b$$，$$c$$均为非零整数）．若$$f(a)={{a}^{3}}$$，$$f(b)={{b}^{3}}$$，则$$c$$的值是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$-16$$ "}], [{"aoVal": "B", "content": "$$-4$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$16$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->方法->构造法", "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->基本初等函数->实数指数幂运算", "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": ["由题设，$$f\\left( x \\right)-{{x}^{3}}=a{{x}^{2}}+bx+c$$有$$2$$个根$$ab$$，所以  $$a+b=-\\frac{b}{a}$$，$$ab=\\frac{c}{a}$$  因为$$abc$$都是非零整数，所以$$a\\textbar b$$，$$a\\textbar c$$，设$$b=ka$$，$$k\\in \\mathbf{Z}$$，则  $$a+ka=-k$$，即$$\\left( a+1 \\right)\\left( k+1 \\right)=1$$  只能$$a+1=k+1=-1$$，$$a=k=-2$$，$$b=ka=4$$，  因此，$$c={{a}^{2}}b=16$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "758", "queId": "5385c378c2ce41debf2b14c8776a3886", "competition_source_list": ["2013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": "如果不等式$${{x}^{2}}\\textless{}\\left\\textbar{} x-1 \\right\\textbar+a$$的解集是区间$$\\left( -3,3 \\right)$$的子集，则实数$$a$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( -\\infty ,7 \\right)$$ "}], [{"aoVal": "B", "content": "$$\\left( -\\infty ,7 \\right]$$ "}], [{"aoVal": "C", "content": "$$\\left( -\\infty ,5 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( -\\infty ,5 \\right]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的划分与覆盖"], "answer_analysis": ["当$$x{\\geqslant }1$$时，原不等式成为$${{x}^{2}}-x+1-a\\textless{}{0}$$，其解  集中不含任何$$\\geqslant 3$$的数，故当$$x{\\geqslant }3$$时总有$${{x}^{2}}-x+1{-}a{\\geqslant 0}$$成立，由此得到$$a{\\leqslant }7$$．  当$$x\\textless{}1$$时，原不等式成为$${{x}^{2}}+x-1-a\\textless{}0$$，其解集中不含任何小于等于$$-3$$的数，故当$$x{\\leqslant -}3$$时总有$${{x}^{2}}+x-1-a{\\geqslant }0$$成立，由此得到$$a{\\leqslant }5$$．  综上，$$a\\in \\left( -\\infty ,5 \\right]$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "735", "queId": "650b378bb14248459006c0f584c5432f", "competition_source_list": ["2016年AMC10竞赛B第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "若$$n❤m=n^{3}m^{2}$$, 则$$\\dfrac{2❤4}{4❤2}$$為何值？  If $$n❤m=n^{3}m^{2}$$, what is $$\\dfrac{2❤4}{4❤2}$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$\\dfrac{1}{4}$$ "}], [{"aoVal": "B", "content": "$$\\dfrac{1}{2}$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}], [{"aoVal": "E", "content": "$$4$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["$$\\dfrac{2^{3}\\left( 2^{2}\\right)^{2}}{\\left( 2^{2}\\right)^{3}2^{2}}=\\dfrac{2^{7}}{2^{8}}=\\dfrac{1}{2}$$ which is $$\\left(\\text{B}\\right)$$. ", "<p>We can replace $$2$$ and $$4$$ with a and b respectively. Then substituting with n and m we can get $$\\dfrac{a^{3}b^{2}}{b^{3}a^{2}}=\\dfrac{a}{b}$$ and substitute to get $$\\dfrac{2}{4}=\\boxed &nbsp;{~~\\dfrac{1}{2}~~}$$ which is $$\\left(\\text{B}\\right)$$.</p>"], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "396", "queId": "5e360032ce934d4e855a6f284e8760a7", "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7题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$f\\left( x \\right)=\\frac{{{3}^{x}}}{{{3}^{x}}+1}-\\frac{1}{3}$，若$\\left[ x \\right]$表示不超过$x$的最大整数，则函数$y=\\left[ f\\left( x \\right) \\right]$的值域是（~~~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\left { 0,-1 \\right }$ "}], [{"aoVal": "B", "content": "$\\left { 0,1 \\right }$ "}], [{"aoVal": "C", "content": "$\\left { -1,1 \\right }$ "}], [{"aoVal": "D", "content": "$\\left { -1,0,1 \\right }$ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 先化简分析$f\\left( x \\right)=\\frac{{{3}^{x}}}{{{3}^{x}}+1}-\\frac{1}{3}$的值域,再分析$y=\\left[ f\\left( x \\right) \\right]$的值域即可.\\\\ 【详解】\\\\ $f\\left( x \\right)=\\frac{{{3}^{x}}}{{{3}^{x}}+1}-\\frac{1}{3}=\\frac{1}{1+\\frac{1}{{{3}^{x}}}}-\\frac{1}{3}$,因为${{3}^{x}}\\in \\left( 0,+\\infty \\right)$,故$1+\\frac{1}{{{3}^{x}}}\\in \\left( 1,+\\infty \\right)$,故$\\frac{1}{1+\\frac{1}{{{3}^{x}}}}\\in \\left( 0,1 \\right)$\\\\ 故$\\frac{1}{1+\\frac{1}{{{3}^{x}}}}-\\frac{1}{3}\\in \\left( -\\frac{1}{3},\\frac{2}{3} \\right)$,由$\\left[ x \\right]$表示不超过$x$的最大整数知$\\left[ f\\left( x \\right) \\right]=-1$或0\\\\ 故选A\\\\ 【点睛】\\\\ 本题主要考查函数的值域求解以及取整函数的值域等.属于基础题型. "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1073", "queId": "d347d7b785a9446280815be02dd82dae", "competition_source_list": ["2010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": "已知关于$$x$$的不等式$$\\sqrt{x}+\\sqrt{2-x}\\geqslant k$$有实数解，则实数$$k$$的取值范围是．", "answer_option_list": [[{"aoVal": "A", "content": "$$(0,2]$$ "}], [{"aoVal": "B", "content": "$$(-\\infty ,0]$$ "}], [{"aoVal": "C", "content": "$$(-\\infty ,0)$$ "}], [{"aoVal": "D", "content": "$$(-\\infty ,2]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->二次函数"], "answer_analysis": ["令$$y=\\sqrt{x}+\\sqrt{2-x}$$，$$0\\leqslant x\\leqslant 2$$  则$${{y}^{2}}=x+(2-x)+2\\sqrt{x(2-x)}\\leqslant 4$$．  故$$0\\textless{}y\\leqslant 2$$，且$$x=2-x$$，即$$x=1$$时等号成立，  所以实数$$k$$的取值范围是$$(-\\infty ,2]$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "88", "queId": "087b52e84e5c450098b909fe283dd3d0", "competition_source_list": ["2016年AMC10竞赛B第13题"], "difficulty": "1", "qtype": "single_choice", "problem": "波力士醫院在一年中多胞胎嬰兒出生數的統計如下：雙胞胎、三胞胎及四胞胎共出生$$1000$$個婴儿，其中三胞胎之组數是四胞胎組數的$$4$$倍，雙胞胎組数是三胞胎组数的$$3$$倍，則在此$$1000$$個嬰兒中四胞胎的婴儿人數總共多少？  At Megapolis Hospital one year, multiple$$-$$birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for $$1000$$ of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these $$1000$$ babies were in sets of quadruplets?", "answer_option_list": [[{"aoVal": "A", "content": "$$25$$ "}], [{"aoVal": "B", "content": "$$40$$ "}], [{"aoVal": "C", "content": "$$64$$ "}], [{"aoVal": "D", "content": "$$100$$ "}], [{"aoVal": "E", "content": "$$160$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["We can set up a system of equations where $$a$$ is the sets of twins, $$b$$ is the sets of triplets, and $$c$$ is the sets of quadruplets. $$\\begin{array}{l} {2a+3b+4c=1000}  {b=4c}  {a=3b} \\end{array}$$.  Solving for $$c$$ and $$a$$ in the second and third equations and substituting into the first equation yields $$\\begin{array}{l} {2\\left( 3b\\right)+3b+4\\left( 0.25b\\right)=1000}  {6b+3b+b=1000}  {b=100} \\end{array}$$.  Since we are trying to find the number of babies and NOT the number of sets of quadruplets, the solution is not $$c$$, but rather $$4c$$. Therefore, we strategically use the second initial equation to realize that $$b=4c$$, leaving us with the number of babies born as quadruplets equal to$$\\left(\\text{D}\\right) 100$$. ", "<p>Say there are $$12x$$ sets of twins, $$4x$$ sets of triplets, and $$x$$ sets of quadruplets.</p>\n<p>That&#39;s $$12x\\cdot2=24x$$ twins, $$4x\\cdot3=12x$$ triplets, and $$x\\cdot4=4x$$ quadruplets. A tenth of the babies are quadruplets and that&#39;s $$\\dfrac{1}{10}\\left( 1000\\right)=\\left( \\text{D}\\right)100$$.</p>"], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1108", "queId": "caa7e5f8fd1d470ca80b1dc327a354ae", "competition_source_list": ["1981年全国高中数学联赛竞赛一试第3题"], "difficulty": "0", "qtype": "single_choice", "problem": "设$$\\alpha \\ne \\frac{k \\pi }{2}(k=0,\\pm 1,\\pm 2,\\cdots )$$，$$T=\\frac{\\sin \\alpha +\\tan\\alpha }{\\cos \\alpha +\\cot\\alpha }$$，则．", "answer_option_list": [[{"aoVal": "A", "content": "$$T$$取负值 "}], [{"aoVal": "B", "content": "$$T$$取非负值 "}], [{"aoVal": "C", "content": "$$T$$取正值 "}], [{"aoVal": "D", "content": "$$T$$取值可正可负 "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["作恒等变形，易得  $$T=\\frac{\\sin \\alpha+\\tan\\alpha}{\\cos \\alpha+\\cot\\alpha}=\\frac{\\tan\\alpha(\\cos \\alpha+1)}{\\cot\\alpha(\\sin \\alpha+1)}=\\frac{{\\cot^{2}}\\alpha(\\cos \\alpha+1)}{\\sin \\alpha+1}\\textgreater0$$  （其中$$\\alpha\\ne \\frac{k \\pi }{2}$$，$$k=0$$，$$\\pm 1$$，$$\\pm 2$$，\\ldots）． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "872", "queId": "a8e16c52af644c8f969690dd2e265c83", "competition_source_list": ["2012年山东全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知集合$$A=\\left { 1,  b,  a+b \\right },  B=\\left { a-b,  ab \\right }$$，且$$A\\cap B=\\left { -1,  0 \\right }$$，则$$a,  b$$的值分别为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$，$$0$$ "}], [{"aoVal": "B", "content": "$$0,  -1$$ "}], [{"aoVal": "C", "content": "$$-1,  1$$ "}], [{"aoVal": "D", "content": "$$1,  -1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["提示：因为$$1\\notin A\\cap B$$，所以$$b$$和$$a+b$$中必有一个为$$0$$，一个为$$-1$$，但是$$a\\ne 0$$，否则集合$$A$$中有两个元素相同，因此必有$$a=-1,  b=0$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "799", "queId": "583554207c8c4038b9628214af16c7f2", "competition_source_list": ["2016年陕西全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "在棱长为$$1$$的正四面体$$ABCD$$中，$$G$$为$$\\triangle BCD$$的重心，$$M$$是线段$$AG$$的中点，则三棱锥$$M-BCD$$的外接球的表面积为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$ \\pi $$ "}], [{"aoVal": "B", "content": "$$\\frac{3}{2} \\pi $$ "}], [{"aoVal": "C", "content": "$$\\frac{\\sqrt{6}}{4} \\pi $$ "}], [{"aoVal": "D", "content": "$$\\frac{\\sqrt{6}}{8} \\pi $$ "}]], "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->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->基本立体图形->空间几何体的概念->空间几何体的内切球、外接球问题", "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"], "answer_analysis": ["易算出$$AG=\\frac{\\sqrt{6}}{3}$$，所以$$MG=\\frac{\\sqrt{6}}{6}$$．因为$$GD=\\frac{\\sqrt{3}}{3}\\textgreater MG$$，所以外心$$O$$在$$MG$$的延长线上．  设$$MG$$的延长线交外接球于另一点$$N$$，由射影定理，$$D{{M}^{2}}=MG\\cdot MN$$，  即$$D{{G}^{2}}+M{{G}^{2}}=MG\\cdot 2R$$（$$R$$是外接球半径），解得$$R=\\frac{\\sqrt{6}}{4}$$．  所以外接球表面积$$S=4 \\pi {{R}^{2}}=\\frac{3}{2} \\pi $$．  故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1041", "queId": "bc3ffd734fe84b6f95d5addc73189729", "competition_source_list": ["2008年湖南全国高中数学联赛竞赛初赛第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "$$5$$名志愿者随进入$$3$$个不同的奥运场馆参加接待工作，则每个场馆至少有一名志愿者的概率为（~ ）", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{3}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{1}{15}$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{8}$$ "}], [{"aoVal": "D", "content": "$$\\frac{50}{81}$$ "}]], "knowledge_point_routes": ["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"], "answer_analysis": ["$$5$$名志愿者随进入$$3$$个不同的奥运场馆的方法数为$${{3}^{5}}=243$$种．每个场馆至少有一名志愿者的情形可分两类考虑：第$$1$$类 ，一个场馆去$$3$$人，剩下两场馆各去$$1$$人，此类的方法数为$$\\text{C}_{3}^{1}\\cdot \\text{C}_{5}^{3}\\cdot \\text{A}_{2}^{2}=60$$种；第$$2$$类，一场馆去$$1$$人，剩下两场馆各$$2$$人，此类的方法数为$$\\text{C}_{3}^{1}\\cdot \\text{C}_{5}^{1}\\cdot \\text{C}_{4}^{2}=90$$种．故每个场馆至少有一名志愿者的概率为$$P=\\frac{60+90}{243}=\\frac{50}{81}$$．  故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1146", "queId": "f3f97edd30fd4dc6adc9ba0e5c280a85", "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1题"], "difficulty": "1", "qtype": "single_choice", "problem": "已知集合$M= {-1,1,3,5\\text{ } },N= {-2,1,2,3,5\\text{ } }$，则$M\\cap N=$（~~~~~）", "answer_option_list": [[{"aoVal": "A", "content": "$\\left { -1,1,3 , \\right }$ "}], [{"aoVal": "B", "content": "$\\left { 1,2,5 , \\right }$ "}], [{"aoVal": "C", "content": "$\\left { 1,3,5 , \\right }$ "}], [{"aoVal": "D", "content": "$\\phi $ "}]], "knowledge_point_routes": [], "answer_analysis": ["\\hfill\\break 【分析】\\\\ 根据交集的定义，直接运算求解即可.\\\\ 【详解】\\\\ $\\because $$M= {-1,1,3,5\\text{ } },N= {-2,1,2,3,5\\text{ } }$，\\\\ $\\therefore $$M\\cap N=$$\\left { 1,3,5 , \\right }$\\\\ 故选：C "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "947", "queId": "863f783b2bb9453ca22bd04ff54aceb8", "competition_source_list": ["2009年竞赛珠海市第3题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "由甲城市到乙城市$$t$$分钟的电话费由函数$$g(t)=1.06\\times (0.75[t]+1)$$给出，其中$$t\\textgreater0$$，$$[t]$$表示大于或等于$$t$$的最小整数，则从甲城市到乙城市$$5.5$$分钟的电话费为．", "answer_option_list": [[{"aoVal": "A", "content": "$$5.83$$元 "}], [{"aoVal": "B", "content": "$$5.25$$元 "}], [{"aoVal": "C", "content": "$$5.56$$元 "}], [{"aoVal": "D", "content": "$$5.04$$元 "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->数论模块->取整函数->取整函数的定义与性质"], "answer_analysis": ["$$g(5.5)=1.06\\times (0.75\\times 6+1)=5.83$$ "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1137", "queId": "ef51dd1068cc48c0bcde52c75a71c74b", "competition_source_list": ["1989年全国高中数学联赛竞赛一试第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f\\left( x \\right)=\\text{arctg}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]$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的值域->求函数的值域->用单调性观察法求值域", "课内体系->知识点->三角函数->反三角函数"], "answer_analysis": ["$$f(x)$$的定义域是$$[-1,1]$$，此时  $$-\\frac{ \\pi }{4}\\mathsf{\\leqslant }\\text{arctg}x\\mathsf{\\leqslant }\\frac{ \\pi }{4}$$，$$-\\frac{ \\pi }{4}\\mathsf{\\leqslant }\\frac{1}{2}\\arcsin x\\mathsf{\\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}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "149", "queId": "143bd3e81e154baca461790f02a86d3b", "competition_source_list": ["竞赛第12题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "隨意從$$1$$、$$2$$、$$3$$、$$4$$、$$5$$五數中選出$$2$$個不同的數相乘，試問乘積爲偶數的機率是多少？  Two different numbers are selected at random from ($$1$$,$$ 2$$, $$3$$,$$4$$,$$5$$) and multiplied together. What is the probability  that the product is even?", "answer_option_list": [[{"aoVal": "A", "content": "$$0.2$$ "}], [{"aoVal": "B", "content": "$$0.4$$ "}], [{"aoVal": "C", "content": "$$0.5$$ "}], [{"aoVal": "D", "content": "$$0.7$$ "}], [{"aoVal": "E", "content": "$$0.8$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["The product will be even if at least one selected number is even, and odd if none are. Using complementary counting, the chance that both numbers are odd is $$\\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$$. "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "315", "queId": "5020b871c1bb43f5a6de66ba53711dcd", "competition_source_list": ["2018年黑龙江全国高中数学联赛竞赛初赛第7题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "小方，小明，小马，小红四人参加完某项比赛，当问到四人谁得第一时，回答如下：小方：``我得第一名''；小明：``小红没得第一名''；小马：``小明没得第一名''；小红：``我得第一名''．已知他们四个人中只有一人说真话，且只有一人得第一．根据以上信息可以判断出得第一名的人是．", "answer_option_list": [[{"aoVal": "A", "content": "小明 "}], [{"aoVal": "B", "content": "小马 "}], [{"aoVal": "C", "content": "小红 "}], [{"aoVal": "D", "content": "小方 "}]], "knowledge_point_routes": ["课内体系->素养->逻辑推理", "课内体系->知识点->推理与证明->合情推理与演绎推理->演绎推理"], "answer_analysis": ["由小明与小红所说，知两人一真一假．  若小明所说为真，则其余三人所说均为假，由此推出小明得第一名，满足条件．  故选$$\\text{A}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "732", "queId": "b5ff2f57d0eb4fa2b28d00e9eb4fcd5b", "competition_source_list": ["2016~2017学年北京海淀区中国人民大学附属中学高一下学期期中第20题6分", "2016年广东全国高中数学联赛高三竞赛初赛第2题8分", "2015~2016学年上海闵行区上海市七宝中学高一上学期期中第10题4分", "2016~2017学年江苏连云港赣榆县赣榆县海头中学高二上学期期中联考第7题5分", "2018~2019学年天津南开区天津中学高二上学期期中第15题", "2016~2017学年江苏连云港海州区江苏省海州高级中学高二上学期期中联考第7题5分", "2018~2019学年天津和平区天津市第一中学高二上学期期中第14题"], "difficulty": "2", "qtype": "single_choice", "problem": "若正数$$x$$，$$y$$满足$$x+3y=5xy$$，则$$3x+4y$$的最小值为~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$5$$ "}], [{"aoVal": "B", "content": "$\\frac{14}{3}$ "}], [{"aoVal": "C", "content": "$3$ "}], [{"aoVal": "D", "content": "$\\frac{\\sqrt{15}}{2}$ "}]], "knowledge_point_routes": ["课内体系->知识点->等式与不等式->不等式->基本不等式->基本不等式的概念", "课内体系->知识点->等式与不等式->不等式->基本不等式->利用基本不等式求最值", "课内体系->素养->数学运算"], "answer_analysis": ["由$$x+3y=5xy$$，$$x\\textgreater0$$，$$y\\textgreater0$$，可得$$\\frac{1}{5y}+\\frac{3}{5x}=1$$，  所以$$3x+4y=\\left( 3x+4y \\right)\\left( \\frac{1}{5y}+\\frac{3}{5x} \\right)=\\frac{9}{5}+\\frac{4}{5}+\\frac{3x}{5y}+\\frac{12y}{5x}\\geqslant \\frac{13}{5}+2\\sqrt{\\frac{3x}{5y}\\cdot \\frac{12y}{5x}}=5$$．  当且仅当$$\\frac{3x}{5y}=\\frac{12y}{5x}$$，即$$x=1$$，$$y=\\frac{1}{2}$$时，等号成立，  此时$$3x+4y$$取得最小值为$$5$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "725", "queId": "848de2b3929b4caaa19c2e4e2995074a", "competition_source_list": ["2003年AMC12竞赛A第24题"], "difficulty": "3", "qtype": "single_choice", "problem": "2003AMC12A, 24  If $$a\\geqslant b\\textgreater1$$, what is the largest possible value of $$\\log_a\\left(\\frac{a}{b}\\right)+\\log_b\\left(\\frac{b}{a}\\right)$$?  若$$a\\geqslant b\\textgreater1$$, $$\\log_a\\left(\\frac{a}{b}\\right)+\\log_b\\left(\\frac{b}{a}\\right)$$的最大值为? ．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$3$$ "}], [{"aoVal": "E", "content": "$$4$$ "}]], "knowledge_point_routes": ["美国AMC10/12->Knowledge Point->Combination->Reasoning->Simple Logical Reasoning", "课内体系->知识点->基本初等函数->对数的概念及其运算->对数的运算"], "answer_analysis": ["$$\\log_a\\left(\\frac{a}{b}\\right)+\\log_b\\left(\\frac{b}{a}\\right)=1-\\log_a{b}+1-\\log_b{a}=2-(\\log_a{b}+\\log_b{a})$$  $$=2-\\left(\\log_a{b}+\\frac{1}{\\log_a{b}}\\right)\\leq 2-2=0$$, 选$$\\text{B}$$. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "764", "queId": "6e324ada07c144cc8b1087b9e29c6505", "competition_source_list": ["1982年全国高中数学联赛竞赛第2题"], "difficulty": "1", "qtype": "single_choice", "problem": "极坐标方程$$\\rho =\\frac{1}{1-\\cos \\theta +\\sin \\theta }$$所确定的曲线是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "圆 "}], [{"aoVal": "B", "content": "椭圆 "}], [{"aoVal": "C", "content": "双曲线 "}], [{"aoVal": "D", "content": "抛物线 "}]], "knowledge_point_routes": ["竞赛->知识点->极坐标系与直角坐标系"], "answer_analysis": ["原方程化简为 $$\\rho =\\frac{1}{1-\\sqrt{2}\\cos \\left( \\theta +\\frac{ \\pi }{4} \\right)}$$．  由离心率$$e=\\sqrt{2}\\textgreater1$$，知所确定的曲线是双曲线．  另解 原方程变形为．$$\\rho -\\rho \\cos \\theta +\\rho \\sin \\theta =1$$，化为直角坐标方程为  $$\\sqrt{{{x}^{2}}+{{y}^{2}}}=x-y+1$$．  化简得 $$2xy-2x+2y=1$$  即 $$\\left( x+1 \\right)\\left( y-1 \\right)=\\frac{1}{2}$$可知，它所确定的曲线是双曲线． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "167", "queId": "aff4f0540be2466382108a14c26be076", "competition_source_list": ["2015年黑龙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "设$$k=\\int_{0}^{ \\pi }{(\\sin x-\\cos x)\\text{d}x}$$，若$${{(1-kx)}^{8}}={{a}_{0}}+{{a}_{1}}x+\\cdots +{{a}_{8}}{{x}^{8}}$$，则$${{a}_{1}}+{{a}_{2}}+\\cdots +{{a}_{8}}$$等于", "answer_option_list": [[{"aoVal": "A", "content": "$$-1$$ "}], [{"aoVal": "B", "content": "$$0$$ "}], [{"aoVal": "C", "content": "$$1$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->积分", "竞赛->知识点->排列组合与概率->二项式定理及其应用"], "answer_analysis": ["$$k=2$$ "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1140", "queId": "fd275963b69745e18706d5381f2c2220", "competition_source_list": ["2009年吉林全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "一个等比数列的前三项的积为$$2$$，最后三项的积为$$4$$，且所有项的积为$$64$$，则该数列有．", "answer_option_list": [[{"aoVal": "A", "content": "$$10$$项 "}], [{"aoVal": "B", "content": "$$11$$项 "}], [{"aoVal": "C", "content": "$$12$$项 "}], [{"aoVal": "D", "content": "$$13$$项 "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["设等比数列为$${{a}_{1}},{{a}_{2}},\\cdots ,{{a}_{n}}$$，由题设，$${{a}_{1}}{{a}_{2}}{{a}_{3}}=2$$，$${{a}_{n-2}}{{a}_{n-1}}{{a}_{n}}=4$$，$${{a}_{1}}{{a}_{2}}\\cdots {{a}_{n}}=64$$．  由等比数列的性质得，$$a_{2}^{3}=2$$，$$a_{n-1}^{3}=4$$，$${{\\left( {{a}_{2}}{{a}_{n-1}} \\right)}^{\\frac{n}{2}}}=64$$．  因此，$${{2}^{\\frac{n}{2}}}=64$$，解得$$n=12$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1051", "queId": "e09eb2b74f3c4382a6cd085c0e1854f6", "competition_source_list": ["第二十届全国希望杯高二竞赛初赛邀请赛第3题4分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知点$$P\\left( \\cos \\alpha ,\\sin \\alpha \\right)$$在直线$$l$$：$$\\frac{x}{a}+\\frac{y}{b}=1$$上，且$$l\\bot OP$$（$$O$$为坐标原点），则．", "answer_option_list": [[{"aoVal": "A", "content": "$$a+b=1$$ "}], [{"aoVal": "B", "content": "$${{a}^{2}}+{{b}^{2}}=1$$ "}], [{"aoVal": "C", "content": "$$\\frac{1}{a}+\\frac{1}{b}=1$$ "}], [{"aoVal": "D", "content": "$$\\frac{1}{{{a}^{2}}}+\\frac{1}{{{b}^{2}}}=1$$ "}]], "knowledge_point_routes": ["竞赛->知识点->解析几何->直线与方程"], "answer_analysis": ["由题设$$\\frac{\\cos \\alpha }{a}+\\frac{\\sin \\alpha }{b}=1,\\frac{\\sin \\alpha }{\\cos \\alpha }=\\frac{a}{b}$$，不难推出$$\\frac{1}{{{a}^{2}}}+\\frac{1}{{{b}^{2}}}=1$$；当然也可以数形结合，由$$\\frac{1}{2}\\sqrt{{{a}^{2}}+{{b}^{2}}}\\cdot 1=\\frac{1}{2}ab$$，可得$$\\frac{1}{{{a}^{2}}}+\\frac{1}{{{b}^{2}}}=1$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "67", "queId": "c71396c17182418585a62ee644193345", "competition_source_list": ["2011年四川全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "函数$$f(x)=\\sqrt{x-5}+\\sqrt{24-3x}$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$3$$ "}], [{"aoVal": "C", "content": "$$2\\sqrt{3}$$ "}], [{"aoVal": "D", "content": "$$3\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["法一：$$f(x)$$的定义域为$$5\\leqslant x\\leqslant 8$$，  由$${f}'(x)=\\frac{1}{2\\sqrt{x-5}}+\\frac{-3}{2\\sqrt{24-3x}}=\\frac{\\sqrt{24-3x}-3\\sqrt{x-5}}{2\\sqrt{x-5}\\cdot \\sqrt{24-3x}}=0$$，解得$$x=\\frac{23}{4}$$．  因为$$f(5)=3$$，$$f(\\frac{23}{4})=2\\sqrt{3}$$，$$f(8)=\\sqrt{3}$$，于是$$f{{(x)}_{\\max }}=f(\\frac{23}{4})=2\\sqrt{3}$$．  法二：$$f(x)$$的定义域为$$5\\leqslant x\\leqslant 8$$，  $${{f}^{2}}(x)={{(1\\cdot \\sqrt{x-5}+\\sqrt{3}\\cdot \\sqrt{8-x})}^{2}}\\leqslant (1+3)(x-5+8-x)=12$$  当且仅当$$\\frac{\\sqrt{x-5}}{1}=\\frac{\\sqrt{8-x}}{\\sqrt{3}}$$，即$$x=\\frac{23}{4}$$时，$$f(x)$$取到最大值$$2\\sqrt{3}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "774", "queId": "ed456ab20bc64d80963d89b3ad8fec68", "competition_source_list": ["2016年高考真题天津卷", "2016年天津全国高中数学联赛竞赛初赛第7题9分"], "difficulty": "1", "qtype": "single_choice", "problem": "椭圆$${{x}^{2}}+k{{y}^{2}}=1$$与双曲线$$\\frac{{{x}^{2}}}{4}-\\frac{{{y}^{2}}}{5}=1$$有相同的准线，则$$k$$等于 ．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{14}{7}$$ "}], [{"aoVal": "B", "content": "$$\\frac{15}{7}$$ "}], [{"aoVal": "C", "content": "$$\\frac{16}{7}$$ "}], [{"aoVal": "D", "content": "$$\\frac{17}{7}$$ "}]], "knowledge_point_routes": ["知识标签->素养->数学运算", "知识标签->知识点->平面解析几何->双曲线->双曲线的定义、标准方程->双曲线的标准方程", "知识标签->知识点->平面解析几何->双曲线->双曲线的定义、标准方程->双曲线的定义", "知识标签->知识点->平面解析几何->椭圆->椭圆的定义、标准方程->椭圆的标准方程", "知识标签->知识点->平面解析几何->椭圆->椭圆的定义、标准方程->椭圆的定义", "知识标签->题型->平面解析几何->圆锥曲线中确定基本量的问题->椭圆、双曲线、抛物线的基本量求解", "知识标签->题型->平面解析几何->圆锥曲线的第二定义问题"], "answer_analysis": ["双曲线的准线方程是  $$x=\\pm \\frac{4}{\\sqrt{4+5}}=\\pm \\frac{4}{3}$$，  椭圆的准线方程是$$x=\\pm \\frac{1}{\\sqrt{1-\\frac{1}{k}}}$$，因此由方程$$\\frac{1}{\\sqrt{1-\\frac{1}{k}}}=\\frac{4}{3}$$得$$1-\\frac{1}{k}=\\frac{9}{16}$$，解得$$k=\\frac{16}{7}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "993", "queId": "8aac50a74ed448af014efe4f3e401f9a", "competition_source_list": ["2014年吉林全国高中数学联赛竞赛初赛第2题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，$$\\angle A={{120}^{\\circ }}$$，$$AB=5$$，$$BC=7$$，则$$\\frac{\\sin B}{\\sin C}$$的值为（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{8}{5}$$ "}], [{"aoVal": "B", "content": "$$\\frac{5}{8}$$ "}], [{"aoVal": "C", "content": "$$\\frac{5}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{3}{5}$$ "}]], "knowledge_point_routes": ["课内体系->知识点->解三角形->余弦定理", "课内体系->知识点->解三角形->正弦定理", "课内体系->素养->数学运算"], "answer_analysis": ["由余弦定理，得$$B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-2AB\\cdot AC\\cdot \\cos A$$，  即$$49=25+A{{C}^{2}}-2\\times 5\\cdot AC\\cdot \\cos {{120}^{\\circ }}$$，解得$$AC=3(AC=-8$$舍去$$)$$，  由正弦定理，得$$\\frac{\\sin B}{\\sin C}=\\frac{AC}{AB}=\\frac{3}{5}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "568", "queId": "cc78bffb19d346a7a24d3958741b4427", "competition_source_list": ["2019~2020学年辽宁高一上学期期中（辽南协作体）第11题5分", "2008年浙江全国高中数学联赛竞赛初赛第3题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "【2019辽南期中11】  设$$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)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质", "竞赛->知识点->函数->函数的概念", "课内体系->知识点->等式与不等式->不等式->不等式的性质"], "answer_analysis": ["函数$$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}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "645", "queId": "40d3edd7a0844f799a05ae5ac5025bb6", "competition_source_list": ["2008年吉林全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "对于集合$$M$$、$$N$$，定义$$M-N= {x\\textbar x\\in M$$，且$$x\\notin N }$$，$$M\\oplus N=(M-N)\\cup (N-M)$$，设$$A= {y\\textbar y={{x}^{2}}-3x,x\\in R }$$，$$B= {y\\textbar y=-{{2}^{x}},x\\in R }$$，则$$A\\oplus B=$$．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\left( -\\frac{9}{4},0 \\right]$$ "}], [{"aoVal": "B", "content": "$$\\left[ -\\frac{9}{4},0 \\right)$$ "}], [{"aoVal": "C", "content": "$$\\left( -\\infty ,-\\frac{9}{4} \\right)\\cup \\left[ 0,+\\infty \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( -\\infty ,-\\frac{9}{4} \\right)\\cup \\left( 0,+\\infty \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->集合->集合的概念与运算"], "answer_analysis": ["由题意可得$$A=\\left[ -\\frac{9}{4},+\\infty \\right)$$，$$B=\\left( -\\infty ,0 \\right)$$，  所以$$A\\oplus B=\\left( -\\infty ,-\\frac{9}{4} \\right)\\cup \\left[ 0,+\\infty \\right)$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "753", "queId": "803854d3f70f4c24a3057aea24d73fb9", "competition_source_list": ["2016年湖北全国高中数学联赛高二竞赛初赛第6题9分"], "difficulty": "1", "qtype": "single_choice", "problem": "以正十三边形的顶点为顶点的形状不同的三角形共有~\\uline{~~~~~~~~~~}~个．（全等的三角形视为形状相同）", "answer_option_list": [[{"aoVal": "A", "content": "$$14$$ "}], [{"aoVal": "B", "content": "$$13$$ "}], [{"aoVal": "C", "content": "$$12$$ "}], [{"aoVal": "D", "content": "$$11$$ "}], [{"aoVal": "E", "content": "$$10$$ "}]], "knowledge_point_routes": ["课内体系->知识点->计数原理", "竞赛->知识点->组合->图论（二试）"], "answer_analysis": ["问题等价于将$$13$$分成$$3$$个正整数的和的方式的种数，  最小数是$$1$$时，$$13=1+1+11=1+2+10=1+3+9=1+4+8=1+5+7=1+6+6$$，有$$6$$种；  最小数是$$2$$时，$$13=2+2+9=2+3+8=2+4+7=2+5+6$$，有$$4$$种；  最小数是$$3$$时，$$13=3+3+7=3+4+6=3+5+5$$，有$$3$$种；  最小数是$$4$$时，$$13=4+4+5$$，有$$1$$种．  总共$$6+4+3+1=14$$种．  故答案为：$$14$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "247", "queId": "4fc5d1877e414591af64dcab072a310f", "competition_source_list": ["2012年浙江全国高中数学联赛竞赛初赛第2题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "下列函数中，既是奇函数，又是区间$$\\left( -\\infty ,  +\\infty \\right)$$上单调递增的函数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$y={{x}^{2}}+x$$ "}], [{"aoVal": "B", "content": "$$y=x+2\\sin x$$ "}], [{"aoVal": "C", "content": "$$y={{x}^{3}}+x$$ "}], [{"aoVal": "D", "content": "$$y=\\tan x$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数->函数的图像与性质"], "answer_analysis": ["逐一验证． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "259", "queId": "420ff9ba2d9648f2a6c2db17fa37df45", "competition_source_list": ["2010年吉林全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "一个正六面体的各个面和一个正八面体的各个面都是边长为$$a$$的正三角形，这样的两个  多面体的内切球的半径之比是一个最简分数$$\\frac{m}{n}$$,那么积$$m\\cdot n$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$$3$$ "}], [{"aoVal": "B", "content": "$$4$$ "}], [{"aoVal": "C", "content": "$$6$$ "}], [{"aoVal": "D", "content": "$$12$$ "}]], "knowledge_point_routes": ["竞赛->知识点->立体几何与空间向量->空间几何体中的求值问题"], "answer_analysis": ["利用等体积法，可以求出$$\\frac{m}{n}=\\frac{2}{3}$$，所以$$m\\cdot n$$等于$$6$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1075", "queId": "bcb1cad47b064e408b6601b6a9fe5421", "competition_source_list": ["2008年吉林全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "0", "qtype": "single_choice", "problem": "已知数列$$ {{{a}_{n}} }$$的通项$${{a}_{n}}={{\\left( \\frac{2}{3} \\right)}^{n-1}}\\left[ {{\\left( \\frac{2}{3} \\right)}^{n-1}}-1 \\right]$$，则下列表述正确的是．", "answer_option_list": [[{"aoVal": "A", "content": "最大项为$${{a}_{1}}$$，最小项为$${{a}_{4}}$$ "}], [{"aoVal": "B", "content": "最大项为$${{a}_{1}}$$，最小项不存在 "}], [{"aoVal": "C", "content": "最大项不存在，最小项为$${{a}_{3}}$$ "}], [{"aoVal": "D", "content": "最大项为$${{a}_{1}}$$，最小项为$${{a}_{3}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的通项与求和"], "answer_analysis": ["由题意可得$${{a}_{1}}=0$$，$${{a}_{2}}=-\\frac{2}{9}$$，$${{a}_{3}}=-\\frac{20}{81}$$．  当$$n\\geqslant 3$$时，$$t={{\\left( \\frac{2}{3} \\right)}^{n-1}}$$为减函数，且$$0\\textless{}{{\\left( \\frac{2}{3} \\right)}^{n-1}}\\leqslant {{\\left( \\frac{2}{3} \\right)}^{2}}\\textless{}\\frac{1}{2}$$．  又因为$${{a}_{n}}=t(t-1)$$为$$\\left( 0,\\frac{1}{2} \\right)$$上的减函数，  所以$${{a}_{n}}$$单调递增，即有$$0\\textgreater{{a}_{n}}\\textgreater{{a}_{n-1}}\\textgreater\\cdots \\textgreater{{a}_{4}}\\textgreater{{a}_{3}}$$．  因此$${{a}_{1}}$$最大，$${{a}_{3}}$$最小．故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "182", "queId": "0d6e3efd801f46cdb340116f274e30e1", "competition_source_list": ["2016年重庆沙坪坝区重庆市第一中学高三零模理科第5题5分", "2001年全国高中数学联赛竞赛一试第4题6分", "2017~2018学年10月甘肃兰州城关区甘肃省兰州第一中学高二上学期月考第9题5分", "2019~2020学年四川成都郫都区郫都区成都外国语学校高一下学期期末理科第10题5分", "2019~2020学年四川成都郫都区郫都区成都外国语学校高一下学期期末文科第10题5分", "2019~2020学年3月四川成都锦江区四川师范大学附属中学高一下学期月考第8题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "如果满足$$\\angle ABC=60{}^{}\\circ $$，$$AC=12$$，$$BC=k$$的$$\\triangle ABC$$恰有一个，那么$$k$$的取值范围是（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$$k=8\\sqrt{3}$$ "}], [{"aoVal": "B", "content": "$$0\\textless{}k\\leqslant 12$$ "}], [{"aoVal": "C", "content": "$$k\\geqslant 12$$ "}], [{"aoVal": "D", "content": "$$0\\textless{}k\\leqslant 12$$或$$k=8\\sqrt{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角形中的问题->解三角形"], "answer_analysis": ["这是``已知三角形的两边及其一边的对角，解三角形''这类问题的一个逆向问题，由课本结论知，应选结论$$\\text{D}$$．  故选$$\\text{D}$$． ", "<p>（$$1$$）当$$AC&lt;{}BC\\sin \\angle ABC$$，即$$12&lt;{}k\\sin 60{}^\\circ $$，即$$k&gt;8\\sqrt{3}$$，三角形无解；</p>\n<p>（$$2$$）当$$AC=BC\\sin \\angle ABC$$，即$$12=k\\sin 60{}^\\circ $$，即$$k=8\\sqrt{3}$$时，三角形有一解；</p>\n<p>（$$3$$）当$$AC&lt;{}BC\\sin \\angle ABC&lt;{}BC$$，即$$k\\sin 60{}^\\circ &lt;{}12&lt;{}k$$，</p>\n<p>即$$12&lt;{}k&lt;{}8\\sqrt{3}$$，三角形有二个解；</p>\n<p>（$$4$$）当$$0&lt;{}BC\\leqslant AC$$，即$$0&lt;{}k\\leqslant 12$$时，三角形有一个解．</p>\n<p>综上所述：当$$0&lt;{}k\\leqslant 12$$或$$k=8\\sqrt{3}$$时，三角形恰有一个解．</p>\n<p>故选$$\\text{D}$$．</p>"], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "762", "queId": "bf469c752e664448a9124dff830eebfa", "competition_source_list": ["2014年吉林全国高中数学联赛竞赛初赛第4题6分", "2016年北京海淀区中国人民大学附属中学高三零模理科第6题5分"], "difficulty": "2", "qtype": "single_choice", "problem": "某地举行一次民歌大奖赛，六个省中每个省各有两名歌手参加决赛，现要选出$$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}$$ "}]], "knowledge_point_routes": ["知识标签->题型->概率->事件与概率->古典概型问题->古典概型的概率计算公式", "知识标签->知识点->计数原理->排列与组合->组合", "知识标签->知识点->概率->事件与概率->概率的概念->概率的基本性质", "知识标签->知识点->概率->事件与概率->古典概型", "知识标签->素养->数学运算"], "answer_analysis": ["由题意知本题是一个古典概型，  试验发生的总事件是从$$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}$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "954", "queId": "8e56c60a82be48b4b2ca865727b1cd12", "competition_source_list": ["2018年黑龙江全国高中数学联赛竞赛初赛第1题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "复数$$z={{\\left( 1+\\text{i} \\right)}^{2}}\\left( 2+\\text{i} \\right)$$的虚部为．", "answer_option_list": [[{"aoVal": "A", "content": "$$-2\\text{i}$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$4\\text{i}$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["课内体系->知识点->复数->复数的概念及几何意义->复数的基本概念", "课内体系->知识点->复数->复数的运算->复数的乘法和除法", "课内体系->知识点->复数->复数的运算->复数的加法和减法", "课内体系->知识点->复数->复数的运算->复数的四则运算综合", "课内体系->素养->数学运算"], "answer_analysis": ["$$z={{(1+\\text{i})}^{2}}(2+\\text{i})=2\\text{i}(2+\\text{i})=4\\text{i}-2$$，虚部为$$4$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "168", "queId": "e7a3a8ff9fad472e9c36de59a1d04818", "competition_source_list": ["2009年竞赛珠海市第6题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知等差数列$$\\left { {{a}_{n}} \\right }$$的公差不为零，$$\\left { {{a}_{n}} \\right }$$中的部分项$${{a}_{{{k}_{1}}}}$$，$${{a}_{{{k}_{2}}}}$$，$${{a}_{{{k}_{3}}}}$$，\\ldots，$${{a}_{{{k}_{n}}}}$$，\\ldots 构成等比数列，其中$${{k}_{1}}=1$$，$${{k}_{2}}=5$$，$${{k}_{3}}=17$$，则$${{k}_{1}}+{{k}_{2}}+{{k}_{3}}+\\cdots +{{k}_{n}}$$等于．", "answer_option_list": [[{"aoVal": "A", "content": "$${{3}^{n}}-n-1$$ "}], [{"aoVal": "B", "content": "$${{3}^{n}}+n-1$$ "}], [{"aoVal": "C", "content": "$${{3}^{n}}-n+1$$ "}], [{"aoVal": "D", "content": "都不对 "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->等差数列与等比数列"], "answer_analysis": ["设公差为$$d$$，由题设，$${{a}_{5}}^{2}={{a}_{1}}{{a}_{17}}$$，即$${{({{a}_{1}}+4d)}^{2}}={{a}_{1}}({{a}_{1}}+16d)\\Rightarrow {{a}_{1}}=2d$$，所以$${{a}_{n}}=(n+1)d$$，于是$${{a}_{5}}=6d，{{a}_{17}}=18d$$，公比为$$3$$，$${{a}_{{{k}_{n}}}}=2d\\times {{3}^{n-1}}$$，因而$${{k}_{n}}=2\\times {{3}^{n-1}}-1$$，  因此$${{k}_{1}}+{{k}_{2}}+\\cdots +{{k}_{n}}=2\\times (1+3+\\cdots +{{3}^{n-1}})-n={{3}^{n}}-n-1$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "739", "queId": "7294c61641ef41a3bc5152f0068bd63f", "competition_source_list": ["2015年山东全国高中数学联赛竞赛初赛第9题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，角$$A,B,C$$满足$$A\\textless{}B\\textless{}C$$，且$$\\frac{\\sin A+\\sin B+\\sin C}{\\cos A+\\cos B+\\cos C}=\\sqrt{3}$$，则$$\\angle B=$$~\\uline{~~~~~~~~~~}~", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{\\pi}{6}$$ "}], [{"aoVal": "B", "content": "$$\\frac{\\pi}{4}$$ "}], [{"aoVal": "C", "content": "$$\\frac{\\pi}{3}$$ "}], [{"aoVal": "D", "content": "$$\\frac{\\pi}{2}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["通分后利用辅助角公式，得$$\\sin \\left( A-60{}^{}\\circ \\right)+\\sin \\left( B-60{}^{}\\circ \\right)+\\sin \\left( C-60{}^{}\\circ \\right)=0$$．  由和差化积公式，$$2\\sin \\frac{A+C-120{}^{}\\circ }{2}\\cos \\frac{A-C}{2}+\\sin \\left( B-60{}^{}\\circ \\right)=0$$，  即$$2\\sin \\frac{60{}^{}\\circ -B}{2}\\cos \\frac{A-C}{2}+2\\sin \\frac{B-60{}^{}\\circ }{2}\\cos \\frac{B-60{}^{}\\circ }{2}=0$$，  $$\\sin \\frac{B-60{}^{}\\circ }{2}\\left( -\\cos \\frac{A-C}{2}+\\cos \\frac{B-60{}^{}\\circ }{2} \\right)=0$$，  因为$$90{}^{}\\circ \\textgreater\\frac{A-C}{2}\\textgreater\\left\\textbar{} \\frac{B-60{}^{}\\circ }{2} \\right\\textbar$$，所以$$-\\cos \\frac{A-C}{2}+\\cos \\frac{B-60{}^{}\\circ }{2}\\ne 0$$，  因此$$\\sin \\frac{B-60{}^{}\\circ }{2}=0$$，$$B=60{}^{}\\circ $$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "392", "queId": "993972681f514d9296a36658b69f6c8b", "competition_source_list": ["2011年山东全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$x,  y$$均为正实数，则$$\\frac{x}{2x+y}+\\frac{y}{x+2y}$$的最大值为．", "answer_option_list": [[{"aoVal": "A", "content": "$$2$$ "}], [{"aoVal": "B", "content": "$$\\frac{2}{3}$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$\\frac{4}{3}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->导数", "竞赛->知识点->不等式->几个重要的不等式->均值"], "answer_analysis": ["解法一 令$$s=2x+y,,t=x+2y$$，则  $$x=\\frac{1}{3}(2s-t),  y=\\frac{1}{3}(2t-s)$$  所以$$\\frac{x}{2x+y}+\\frac{y}{x+2y}=\\frac{4}{3}-\\frac{1}{3}(\\frac{t}{s}+\\frac{s}{t})\\leqslant \\frac{2}{3}$$．  解法二 令$$t=\\frac{y}{x}$$， 则$$t\\in (0,+\\infty )$$， 此时  $$\\frac{x}{2x+y}+\\frac{y}{x+2y}=\\frac{1}{t+2}+\\frac{t}{2t+1}=f(t)$$，  即有$$f^{}\\prime (t)=-\\frac{3({{t}^{2}}-1)}{{{(t+2)}^{2}}{{(2t+1)}^{2}}}$$．  显然当$$t\\textless{}1$$时，$$f^{}\\prime (t)\\textgreater0$$；当$$t\\textgreater1$$时，$$f^{}\\prime (t)\\textless{}0$$，所以函数$$f(t)$$在$$t=1$$, 即$$x=y$$时取得最大值$$f(1)=\\frac{2}{3}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "90", "queId": "0f372bb423fd45f49793dfd2b29045a6", "competition_source_list": ["2010年四川全国高中数学联赛竞赛初赛第4题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知函数$$f(x)=\\frac{{{x}^{4}}+({{k}^{2}}+2k-4){{x}^{2}}+4}{{{x}^{4}}+2{{x}^{2}}+4}$$的最小值是$$0$$，则非零实数$$k$$的值是．", "answer_option_list": [[{"aoVal": "A", "content": "$$-4$$ "}], [{"aoVal": "B", "content": "$$-2$$ "}], [{"aoVal": "C", "content": "$$2$$ "}], [{"aoVal": "D", "content": "$$4$$ "}]], "knowledge_point_routes": ["竞赛->知识点->不等式->不等式的解法"], "answer_analysis": ["$$f(x)=1+({{k}^{2}}+2k-6)\\frac{{{x}^{2}}}{{{x}^{4}}+2{{x}^{2}}+4}$$，  因为$${{x}^{4}}+4\\geqslant 4{{x}^{2}}$$，故  $$0\\leqslant \\frac{{{x}^{2}}}{{{x}^{4}}+2{{x}^{2}}+4}\\leqslant \\frac{1}{6}$$．  当$${{k}^{2}}+2k-6\\geqslant 0$$时，$${{f}_{\\min }}=1$$，不合题意；  当$${{k}^{2}}+2k-6\\textless{}0$$时，$${{f}_{\\max }}=1,  {{f}_{\\min }}=1+\\frac{1}{6}({{k}^{2}}+2k-6)$$，  由条件知  $$1+\\frac{1}{6}({{k}^{2}}+2k-6)=0$$，  解得$$k=-2$$或0（舍去）．故选$$\\text{B}$$. "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "378", "queId": "d558143a9a6d4206a30d731a6dfbf007", "competition_source_list": ["2001年全国高中数学联赛竞赛一试第5题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "若$${{\\left( 1+x+{{x}^{2}} \\right)}^{1000}}$$的展开式为$${{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+\\cdots +{{a}_{2000}}{{x}^{2000}}$$，则$${{a}_{0}}+{{a}_{3}}+{{a}_{6}}+{{a}_{9}}+\\cdots +{{a}_{1998}}$$的值为（~ ~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "$${{3}^{333}}$$ "}], [{"aoVal": "B", "content": "$${{3}^{666}}$$ "}], [{"aoVal": "C", "content": "$${{3}^{999}}$$ "}], [{"aoVal": "D", "content": "$${{3}^{2001}}$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->二项式定理及其应用"], "answer_analysis": ["由于要求的是展开式中每间隔两项系数的和，所以联想到$$1$$的单位根，用特殊值法．  取$$\\omega =-\\frac{1}{2}+\\frac{\\sqrt{3}}{2}\\mathrm{i}$$，则$${{\\omega }^{3}}=1$$，$${{\\omega }^{2}}+\\omega +1=0$$．  令$$x=1$$，得$${{3}^{1000}}={{a}_{0}}+{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+\\cdots +{{a}_{2000}}$$；  令$$x=\\omega $$，得  $$0={{a}_{0}}+{{a}_{1}}\\omega +{{a}_{2}}{{\\omega }^{2}}+\\cdots +{{a}_{2000}}{{\\omega }^{2000}}$$；  令$$x={{\\omega }^{2}}$$，得  $$0={{a}_{0}}+{{a}_{1}}{{\\omega }^{2}}+{{a}_{2}}{{\\omega }^{4}}+{{a}_{3}}{{\\omega }^{6}}+\\cdots +{{a}_{2000}}{{\\omega }^{4000}}$$．  三个式子相加得  $${{3}^{1000}}=3\\left( {{a}_{0}}+{{a}_{3}}+{{a}_{6}}+\\cdots +{{a}_{1998}} \\right)$$．  ∴$${{a}_{0}}+{{a}_{3}}+{{a}_{6}}+\\cdots +{{a}_{1998}}={{3}^{999}}$$．  故选$$\\text{C}$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "476", "queId": "a752a03819ef42b482094c26a94fd74a", "competition_source_list": ["2018年浙江竞赛第5题8分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知虚数$$z$$满足$${{z}^{3}}+1=0$$，$$z\\ne -1$$．则$${{\\left( \\frac{z}{z-1} \\right)}^{2018}}+{{\\left( \\frac{1}{z-1} \\right)}^{2018}}=$$~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$$1$$ "}], [{"aoVal": "B", "content": "$$-1$$ "}], [{"aoVal": "C", "content": "$$z$$ "}], [{"aoVal": "D", "content": "$$-z$$ "}]], "knowledge_point_routes": ["竞赛->知识点->复数与平面向量->复数的概念与运算", "课内体系->知识点->复数->复数的运算"], "answer_analysis": ["由$${{z}^{3}}+1=0$$  $$\\Rightarrow \\left( z+1 \\right)\\left( {{z}^{2}}-z+1 \\right)=0$$  $$\\Rightarrow {{z}^{2}}-z+1=0$$，  则$${{\\left( \\frac{z}{z-1} \\right)}^{2018}}+{{\\left( \\frac{1}{z-1} \\right)}^{2018}}$$  $$=\\frac{{{z}^{2018}}+1}{{{\\left( {{z}^{2}} \\right)}^{2018}}}=\\frac{{{z}^{2}}+1}{{{z}^{4}}}=\\frac{{{z}^{2}}+1}{-z}=-1$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1169", "queId": "e6a6322cfa7d4ae6917ba1d05cd2866e", "competition_source_list": ["2008年山西全国高中数学联赛竞赛初赛第5题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "使得$$\\frac{x+1}{{{x}^{2}}-3x+3}$$的值为整数的无理数$$x$$的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$0$$ "}], [{"aoVal": "B", "content": "$$2$$ "}], [{"aoVal": "C", "content": "$$4$$ "}], [{"aoVal": "D", "content": "$$6$$ "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->解方程（组）"], "answer_analysis": ["设$$\\frac{x+1}{{{x}^{2}}-3x+3}=k$$，则$$k{{x}^{2}}-(3k+1)x+3k-1=0$$，  若$$k=0$$，则$$x=-1$$；  若$$k\\ne 0$$，由$$\\Delta \\geqslant 0$$，得$$\\frac{5-2\\sqrt{7}}{3}\\leqslant k\\leqslant \\frac{5+2\\sqrt{7}}{3}$$，  则$$k$$只能取$$0$$、$$1$$、$$2$$、$$3$$；  当$$k=1$$时，$$x=2\\pm \\sqrt{2}$$；  当$$k=2$$时，$$x=1$$，$$\\frac{5}{2}$$；  当$$k=3$$时，$$x=2$$，$$\\frac{4}{3}$$，  因此这种无理数只有$$x=2\\pm \\sqrt{2}$$．故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "317", "queId": "1f1d69dde11f4a5c9556882b6d16856f", "competition_source_list": ["2009年山东全国高中数学联赛竞赛初赛第4题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "在$$\\triangle ABC$$中，$$(\\sqrt{3}\\sin B-\\cos B)(\\sqrt{3}\\sin C-\\cos C)=4\\cos B\\cos C$$，且$$AB+AC=4$$，则$$BC$$的取值范围为．", "answer_option_list": [[{"aoVal": "A", "content": "$$(2,4)$$ "}], [{"aoVal": "B", "content": "$$(2,4]$$ "}], [{"aoVal": "C", "content": "$$[2,4)$$ "}], [{"aoVal": "D", "content": "$$[2,4]$$ "}]], "knowledge_point_routes": ["竞赛->知识点->三角函数->三角形中的问题->解三角形", "竞赛->知识点->三角函数->三角恒等变换"], "answer_analysis": ["由已知  $$ 3\\sin B\\sin C-\\sqrt{3}\\sin B\\cos C-\\sqrt{3}\\cos B\\sin C+\\cos B\\cos C=4\\cos B\\cos C$$  $$\\sqrt{3}(\\sin B\\cos C+\\cos B\\sin C)=-3(\\cos B\\cos C-\\sin A\\sin C)$$，  $$\\sqrt{3}\\sin (B+C)=-3\\cos (B+C)$$，  $$\\tan (B+C)=-\\sqrt{3}$$．  因为$$0\\textless{}B+C\\textless{}\\pi $$，所以$$B+C=\\frac{2\\pi }{3}$$，$$A=\\pi -(B+C)=\\frac{\\pi }{3}$$．  由$$AB+AC=4$$，得$$AB=4-AC$$．  $$B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}-2AB \\cdot ~AC\\cos A$$  $$={{(4-AC)}^{2}}+A{{C}^{2}}-(4-AC)\\cdot AC$$  $$=3A{{C}^{2}}-12AC+16$$  $$=3{{(AC-2)}^{2}}+4$$  $$\\geqslant 4$$  当且仅当$$AC=2$$时，上式取等号，所以$$BC\\geqslant 2$$．又$$BC\\textless{}AB+AC=4$$．所以$$BC$$的取值范围为$$[2,4)$$．故选$$\\text{C}$$． ", "<p>由已知</p>\n<p>$$(\\sqrt{3}\\tan B-1)(\\sqrt{3}\\tan C-1)=4$$，</p>\n<p>$$3\\tan B\\tan C-\\sqrt{3}\\tan B-\\sqrt{3}\\tan C+1=4$$，</p>\n<p>$$\\sqrt{3}(\\tan B+\\tan C)=3(\\tan B\\tan C-1)$$，</p>\n<p>$$\\tan (B+C)=\\frac{\\tan B+\\tan C}{1-\\tan B\\tan C}=-\\sqrt{3}$$．</p>\n<p>以下同方法$$1$$．</p>\n"], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "488", "queId": "cc4e81b79f2e4fee82be1e1ebbcdd404", "competition_source_list": ["2016年高考真题山东卷文科", "2016年山东全国高中数学联赛竞赛初赛第3题8分"], "difficulty": "1", "qtype": "single_choice", "problem": "若实数$$x$$满足$$\\arcsin x\\textgreater\\arccos x$$，则关系式$$f\\left( x \\right)=\\sqrt{2{{x}^{2}}-x+3}+{{2}^{\\sqrt{{{x}^{2}}-x}}}$$的取值范围是~\\uline{~~~~~~~~~~}~．", "answer_option_list": [[{"aoVal": "A", "content": "$[3,+\\infty)$ "}], [{"aoVal": "B", "content": "$[2,+\\infty)$ "}], [{"aoVal": "C", "content": "$$3$$ "}], [{"aoVal": "D", "content": "$$2$$ "}]], "knowledge_point_routes": ["课内体系->素养->数学运算", "课内体系->知识点->三角函数->反三角函数", "课内体系->知识点->函数的概念与性质->函数的概念及其表示->函数的定义域->求具体函数（包括复合函数）的定义域"], "answer_analysis": ["若$$-1\\leqslant x\\leqslant 0$$，则$$\\arcsin x\\in \\left[ -\\frac{\\pi }{2},0 \\right]$$，$$\\arccos x\\in \\left[ \\frac{\\pi }{2},\\pi \\right]$$，  此时$$\\arcsin x \\textless{} \\arccos x$$，不合条件，故$$0 \\textless{} x\\leqslant 1$$．  此时$$\\arcsin x\\in \\left( 0,\\frac{\\pi }{2} \\right]$$，$$\\arccos x\\in \\left[ 0,\\frac{\\pi }{2} \\right)$$．  而$$y=\\sin x$$在区间$$\\left[ 0,\\frac{\\pi }{2} \\right]$$上为增函数，所以由条件得$$\\sin \\left( \\arcsin x \\right)\\textgreater\\sin \\left( \\arccos x \\right)$$，  即$$x\\textgreater\\sqrt{1-{{x}^{2}}}$$，解得$$\\left\\textbar{} x \\right\\textbar\\textgreater\\frac{\\sqrt{2}}{2}$$，又$$0 \\textless{} x\\leqslant 1$$，所以$$\\frac{\\sqrt{2}}{2} \\textless{} x\\leqslant 1$$．  而由$$\\begin{cases}2{{x}^{2}}-x+3\\geqslant 0    {{x}^{2}}-x\\geqslant 0   \\end{cases}$$，知$$x\\in \\left( -\\infty ,0 \\right]\\cup \\left[ 1,+\\infty \\right)$$．  因此$$x\\in \\left { 1 \\right }$$，故关系式$$f\\left( x \\right)$$的取值为$$\\left { 3 \\right }$$． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1159", "queId": "e1e61e219e444f009b975071003ecea2", "competition_source_list": ["2008年安徽全国高中数学联赛竞赛初赛第6题6分"], "difficulty": "2", "qtype": "single_choice", "problem": "设多项式$${{(1+x)}^{2008}}={{a}_{0}}+{{a}_{1}}x+\\ldots +{{a}_{2 008}}{{x}^{2008}}$$，则$${{a}_{0}}{{a}_{1}}\\ldots {{a}_{2008}}$$中偶数的个数为．", "answer_option_list": [[{"aoVal": "A", "content": "$$127$$ "}], [{"aoVal": "B", "content": "$$1003$$ "}], [{"aoVal": "C", "content": "$$1005$$ "}], [{"aoVal": "D", "content": "$$1881$$ "}]], "knowledge_point_routes": ["竞赛->知识点->排列组合与概率->二项式定理及其应用"], "answer_analysis": ["熟知$$n!$$中含素数$$p$$的次数为$${{v}_{p}}\\left( n! \\right)=\\frac{n-{{\\tau }_{p}}\\left( n \\right)}{p-1}$$，其中$${{\\tau }_{p}}\\left( n \\right)$$表示$$n$$在$$p$$进制下的数字和$$\\text{C}_{2008}^{k}=\\frac{2008!}{k!\\left( 2008-k \\right)!}$$中含$$2$$的次数为  $${{v}_{2}}\\left( \\text{C}_{2008}^{k} \\right)=\\left( 2008-{{\\tau }_{2}}\\left( 2008 \\right) \\right)-\\left( k-{{\\tau }_{2}}\\left( k \\right) \\right)-\\left( 2008-k-{{\\tau }_{2}}\\left( 2008-k \\right) \\right)$$  $$={{\\tau }_{2}}\\left( k \\right)+{{\\tau }_{2}}\\left( 2008-k \\right)-{{\\tau }_{2}}\\left( 2008 \\right)$$  $${{\\tau }_{2}}\\left( \\text{C}_{2008}^{k} \\right)=0$$当且仅当在二进制下$$k$$与$$2008-k$$相加时不产生进位．  因为$$2008={{\\left( 11111011000 \\right)}_{2}}$$，所以$$k$$的选择有$${{2}^{7}}=128$$种，因此$${{\\tau }_{2}}\\left( \\text{C}_{2008}^{k} \\right)\\textgreater0$$的情况共有$$2009-{{2}^{7}}=1881$$，即$${{a}_{0}},{{a}_{1}},\\cdots ,{{a}_{2018}}$$中共有$$1881$$个偶数． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "217", "queId": "38a50050220c4128a26632d7c1586b36", "competition_source_list": ["2010年四川全国高中数学联赛竞赛初赛第6题5分"], "difficulty": "0", "qtype": "single_choice", "problem": "对任意实数$$m$$，过函数$$f(x)={{x}^{2}}+mx+1$$图象上的点$$(2,  f(2))$$的切线恒过一定点$$P$$，则点$$P$$的坐标为．", "answer_option_list": [[{"aoVal": "A", "content": "$$(0,  3)$$ "}], [{"aoVal": "B", "content": "$$(0,  -3)$$ "}], [{"aoVal": "C", "content": "$$\\left( \\frac{3}{2},  0 \\right)$$ "}], [{"aoVal": "D", "content": "$$\\left( -\\frac{3}{2},  0 \\right)$$ "}]], "knowledge_point_routes": ["竞赛->知识点->导数模块->导数"], "answer_analysis": ["因为$${f}'(x)=2x+m$$，故$${f}'(2)=4+m$$．于是过$$(2,  f(2))$$的切线方程是：$$y-(5+2m)=(4+m)(x-2)$$，即$$y=(m+4)x-3$$，因此切线方程恒过$$(0,  -3)$$．故选$$\\text{B}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "912", "queId": "e475813fcb5b4666b1a900ed43771102", "competition_source_list": ["1985年全国高中数学联赛竞赛一试第6题"], "difficulty": "2", "qtype": "single_choice", "problem": "设$$0\\textless{}a\\textless{}1$$，若$${{x}_{1}}=a,{{x}_{2}}={{a}^{{{x}_{1}}}},{{x}_{3}}={{a}^{{{x}_{2}}}},\\cdots ,{{x}_{n}}={{a}^{{{x}_{n-1}}}},\\cdots $$，则数列$$\\left { {{x}_{n}} \\right }$$（~ ）．", "answer_option_list": [[{"aoVal": "A", "content": "是递增的 "}], [{"aoVal": "B", "content": "是递减的 "}], [{"aoVal": "C", "content": "奇数项是递丧失，偶数项是递减的 "}], [{"aoVal": "D", "content": "偶数项是递增的，奇数项是递减的． "}]], "knowledge_point_routes": ["竞赛->知识点->数列与数学归纳法->数列的综合应用"], "answer_analysis": ["首先考虑数列的前三项，  $${{x}_{1}}=a\\mathsf{}{{x}_{2}}={{a}^{a}}\\mathsf{}{{x}_{3}}={{a}^{{{a}^{a}}}}$$  ∵$$0\\textless{}a\\textless{}1$$，  ∴$$a\\textless{}{{a}^{a}}\\textless{}1$$  $$a\\textless{}{{a}^{{{a}^{a}}}}\\textless{}{{a}^{a}}$$．  即 $${{x}_{1}}\\textless{}{{x}_{3}}\\textless{}{{x}_{2}}$$．  此已表明数列既非递增又非递减，且奇数项也不是递减的． "], "answer_value": "C"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "499", "queId": "4ca306b606ab4608a6e1a02040766f00", "competition_source_list": ["2008年福建全国高中数学联赛竞赛初赛第1题6分"], "difficulty": "1", "qtype": "single_choice", "problem": "已知$$0\\textless{}x\\textless{}\\frac{ \\pi }{2}$$，$$\\sin x-\\cos x=\\frac{ \\pi }{4}$$，若$$\\tan x+\\frac{1}{\\tan x}$$可以表示成$$\\frac{a}{b-{{ \\pi }^{c}}}$$的形式，其中$$a$$、$$b$$、$$c$$是正整数，则$$a+b+c$$= ．", "answer_option_list": [[{"aoVal": "A", "content": "$$8$$ "}], [{"aoVal": "B", "content": "$$32$$ "}], [{"aoVal": "C", "content": "$$48$$ "}], [{"aoVal": "D", "content": "$$50$$ "}]], "knowledge_point_routes": ["课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式的化简和求值", "课内体系->知识点->三角函数->三角函数的概念->任意角的三角函数->同角三角函数的基本关系式", "课内体系->素养->数学运算"], "answer_analysis": ["由条件知，$${{(\\sin x-\\cos x)}^{2}}={{\\left( \\frac{ \\pi }{4} \\right)}^{2}}$$，因此$$\\sin x\\cos x=\\frac{16-{{ \\pi }^{2}}}{32}$$，  所以$$\\tan x+\\frac{1}{\\tan x}=\\frac{\\sin x}{\\cos x}+\\frac{\\cos x}{\\sin x}=\\frac{1}{\\sin x\\cos x}=\\frac{32}{16-{{ \\pi }^{2}}}$$，  因此$$a+b+c=32+16+2=50$$．故选$$\\text{D}$$． "], "answer_value": "D"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "615", "queId": "5aff5a51f6094cc29ab0b94129b6e95a", "competition_source_list": ["1991年全国高中数学联赛竞赛一试第4题"], "difficulty": "0", "qtype": "single_choice", "problem": "设函数$$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$$ "}]], "knowledge_point_routes": ["竞赛->知识点->函数基础->函数的图像与性质"], "answer_analysis": ["若$$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{\\pm }{{\\alpha }_{1}}$$，$$3{\\pm }{{\\alpha }_{2}}$$，$$3{\\pm }{{\\alpha }_{3}}$$．  于是它们的和等于$$18$$． "], "answer_value": "A"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "835", "queId": "9b038e0fc6da4b68abc4b4b44da0ad71", "competition_source_list": ["2011年黑龙江全国高中数学联赛竞赛初赛第11题5分"], "difficulty": "3", "qtype": "single_choice", "problem": "已知球的半径为$$2$$，球面上的三个大圆所在的平面两两垂直，则以三个大圆的交点为顶点的八面体的体积为．", "answer_option_list": [[{"aoVal": "A", "content": "$$\\frac{16}{3}$$ "}], [{"aoVal": "B", "content": "$$\\frac{32}{3}$$ "}], [{"aoVal": "C", "content": "$$\\frac{64}{3}$$ "}], [{"aoVal": "D", "content": "$$32$$ "}]], "knowledge_point_routes": ["课内体系->思想->数形结合思想", "课内体系->知识点->立体几何初步->基本立体图形->空间几何体的体积、表面积->组合体求体积、表面积问题", "课内体系->知识点->立体几何初步->基本立体图形->空间几何体的体积、表面积->空间几何体的体积", "课内体系->知识点->立体几何初步->基本立体图形->空间几何体的概念->圆柱、圆锥、圆台和球的结构特征", "课内体系->知识点->立体几何初步->基本立体图形->空间几何体的概念->空间几何体的内切球、外接球问题", "课内体系->素养->直观想象", "课内体系->素养->数学运算"], "answer_analysis": ["$$\\frac{1}{3}\\cdot \\left( 4\\cdot 4\\cdot \\frac{1}{2} \\right)\\cdot 2\\times 2=\\frac{32}{3}$$． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "1164", "queId": "cfe7cbf256af44bbba83a57668f74ffc", "competition_source_list": ["2009年竞赛珠海市第1题5分"], "difficulty": "1", "qtype": "single_choice", "problem": "如果$$a$$，$$b$$，$$c$$都是实数，那么$$p$$：$$ac\\textless{}0$$，是$$q$$：关于$$x$$的方程$$a{{x}^{2}}+bx+c=0$$有一个正根和一个负根的．", "answer_option_list": [[{"aoVal": "A", "content": "必要而不充分条件 "}], [{"aoVal": "B", "content": "充要条件 "}], [{"aoVal": "C", "content": "充分而不必要条件 "}], [{"aoVal": "D", "content": "既不充分也不必要条件 "}]], "knowledge_point_routes": ["竞赛->知识点->多项式与方程->解方程（组）"], "answer_analysis": ["若$$ac\\textless{}0$$，则$$a\\ne 0$$，且$$\\Delta ={{b}^{2}}-4ac\\textgreater0$$，两根之积$$\\frac{c}{a}\\textless{}0$$，充分性成立；  若$$a{{x}^{2}}+bx+c=0$$有一个正根和一个负根，则$$\\frac{c}{a}\\textless{}0$$，从而$$ac\\textless{}0$$，必要性也成立． "], "answer_value": "B"}
{"dataset_name": "high_math_competition_ch_single_choice_1.2K_dev", "dataset_version": "2023-07-07", "qid": "40", "queId": "0627fe65425b4f8c82b2144bde3d64d2", "competition_source_list": ["2016年AMC12竞赛A第6题"], "difficulty": "1", "qtype": "single_choice", "problem": "一個用$$2016$$個銅板所排成的三角形陣列， 第一列排放$$1$$個銅板，第二列排放$$2$$個銅板，第三列排放$$3$$個銅板， 依此繼續排放，直到第$$N$$列排放$$N$$個銅板恰好排完．試問$$N$$各個位數的數字和爲多少？  A triangular array of $$2016$$ coins has $$1$$ coin in the first row, $$2$$ coins in the second row, $$3$$ coins in the third row, and so on up to $$N$$ coins in the Nth row. What is the sum of the digits of $$N$$?", "answer_option_list": [[{"aoVal": "A", "content": "$$6$$ "}], [{"aoVal": "B", "content": "$$7$$ "}], [{"aoVal": "C", "content": "$$8$$ "}], [{"aoVal": "D", "content": "$$9$$ "}], [{"aoVal": "E", "content": "$$10$$ "}]], "knowledge_point_routes": [], "answer_analysis": ["We are trying to find the value of $$N$$ such that $$1+2+3\\cdots +(N-1)+N=\\frac {N(N+1)}2=2016$$.  Noticing that $$\\frac {63\\cdot 64}2=2016$$, we have $$N=63$$, so our answer is $$9$$.  Notice that we were attempting to solve $$\\frac {N(N+1)}2=2016\\Rightarrow N(N+1)=2016\\cdot 2=4032$$. Approximating $$N(N+1)\\approx N^{2}$$, we were looking for a square that is close to, but less than $$4032$$. Since $$64^{2}=4096$$, we see that $$N=63$$ is a likely candidate. Multiplying $$63\\cdot 64$$ confirms that our assumption is correct. "], "answer_value": "D"}