dataset_name
stringclasses 4
values | dataset_version
unknown | qid
stringlengths 1
5
| queId
stringlengths 32
32
| competition_source_list
sequence | difficulty
stringclasses 5
values | qtype
stringclasses 1
value | problem
stringlengths 6
1.51k
| answer_option_list
list | knowledge_point_routes
sequence | answer_analysis
sequence | answer_value
stringclasses 7
values |
---|---|---|---|---|---|---|---|---|---|---|---|
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2991 | c9a5b99bd27f4bccb633cfd24ce74123 | [
"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"
] | 0 | single_choice | 1. 小数乘法:$$0.025\times 0.04$$的结果的小数位数有位. | [
[
{
"aoVal": "A",
"content": "$$2$$ "
}
],
[
{
"aoVal": "B",
"content": "$$3$$ "
}
],
[
{
"aoVal": "C",
"content": "$$4$$ "
}
],
[
{
"aoVal": "D",
"content": "$$5$$ "
}
]
] | [
"知识标签->拓展思维->计算模块->小数->小数乘除->小数乘法运算"
] | [
"$$0.025\\times 0.04=0.001$$ "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 539 | ebf0b6df87354cb6b9335fcbfcc7c741 | [
"2020年希望杯五年级竞赛模拟第33题",
"2020年新希望杯五年级竞赛第33题"
] | 0 | single_choice | 【$$2020$$五年级卷第$$33$$题】斯普林特老师在$$3$$个小箱中各放一个有颜色的球,让四只忍者神龟猜箱子中球的颜色. 李奥纳多说:``$$1$$号箱中放红球,$$2$$号箱中放黑球,$$3$$号箱中放黄球.'' 拉斐尔说:``$$1$$号箱中放橙球,$$2$$号箱中放黑球,$$3$$号箱中放绿球.'' 米开朗琪罗说:``$$1$$号箱中放蓝球,$$2$$号箱中放橙球,$$3$$号箱中放紫球.'' 多纳泰罗说:``$$1$$号箱中放橙球,$$2$$号箱中放绿球,$$3$$号箱中放蓝球.'' 斯普林特老师说:``你们中有一个人恰好猜对了两个,其余三人都只猜对一个.'' 那么$$3$$号箱中放的是球. | [
[
{
"aoVal": "A",
"content": "黄 "
}
],
[
{
"aoVal": "B",
"content": "黑 "
}
],
[
{
"aoVal": "C",
"content": "红 "
}
],
[
{
"aoVal": "D",
"content": "橙 "
}
],
[
{
"aoVal": "E",
"content": "蓝 "
}
],
[
{
"aoVal": "F",
"content": "紫 "
}
],
[
{
"aoVal": "G",
"content": "绿 "
}
]
] | [
"拓展思维->思想->对应思想"
] | [
"共猜对$$5$$次,由题意知:$$1$$号箱被猜$$2$$次橙,$$2$$号箱被猜$$2$$次黑,则$$1$$号箱为橙或$$2$$号箱为黑. 假设$$1$$号为橙色,则拉斐尔和多纳泰罗一个猜中$$1$$次,一人猜中两次. 假设拉斐尔猜中$$1$$次,则$$2$$号不为黑,$$3$$号不为绿,则李奥纳多猜中$$3$$号为黄,米开朗琪罗全猜错. 假设有问题,则拉斐尔猜中$$2$$次,$$1$$号为橙,$$2$$号不为黑,$$3$$号为绿,则李奥纳多全猜错,则$$1$$号为橙,$$2$$号为黑,$$3$$号不为绿,拉斐尔猜中$$2$$次,李奥纳多猜中$$2$$号为黑,多纳泰罗猜中$$1$$号为橙,米开朗琪罗猜中$$3$$号为紫. "
] | F |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 123 | 1a49b835c49a4037b98bd3a778b616da | [
"2021年新希望杯二年级竞赛初赛第30题5分"
] | 2 | single_choice | 谎言岛有一半的人只在星期三、星期五、星期六说谎,另一半的人只在星期二、星期四、星期日说谎.某一天,岛上的所有人都说:``我明天说真话.''那么,这一天是.(2021年新希望杯二年级竞赛初赛数学试卷) | [
[
{
"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 "
}
]
] | [
"拓展思维->思想->逐步调整思想"
] | [
"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$$\\text{C}$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1268 | 4712f1af1dd9498d93c46e5013627dfb | [
"2009年第7届创新杯六年级竞赛初赛第5题4分"
] | 2 | single_choice | $$4$$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,葡萄还剩吨. | [
[
{
"aoVal": "A",
"content": "$$1$$ "
}
],
[
{
"aoVal": "B",
"content": "$$2$$ "
}
],
[
{
"aoVal": "C",
"content": "$$3$$ "
}
],
[
{
"aoVal": "D",
"content": "$$4$$ "
}
]
] | [
"拓展思维->能力->构造模型->模型思想"
] | [
"$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde4\\times (1-99 \\%)\\div (1-98 \\%)$$ $$=4\\times 1 \\%\\div 2 \\%$$ $$=0.04\\div 2 \\%$$ $$=2$$(吨), 答:运抵武昌后,葡萄还剩$$2$$吨. 故选$$\\text{B}$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1130 | 6fe3ce8a6b2a4a39a187c73dd16a2e1e | [
"2011年第9届全国创新杯小学高年级六年级竞赛第1题"
] | 1 | single_choice | 某人年初买了一种股票,该股票当年下跌了$$20 \%$$,第二年应上涨$$ \%$$才能恢复原值. | [
[
{
"aoVal": "A",
"content": "$$20$$ "
}
],
[
{
"aoVal": "B",
"content": "$$25$$ "
}
],
[
{
"aoVal": "C",
"content": "$$30$$ "
}
],
[
{
"aoVal": "D",
"content": "$$35$$ "
}
]
] | [
"拓展思维->拓展思维->应用题模块->分百应用题->求分率"
] | [
"$$1\\div \\left( 1-20 \\% \\right)-1=25 \\%$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1028 | 03e2d52cb208439485f4d0dfa13286e1 | [
"2017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"
] | 1 | single_choice | 小玲在计算一道除法题时,把被除数$$1250$$写成$$1205$$,结果得到的商是$$48$$还余$$5$$,正确的商是. | [
[
{
"aoVal": "A",
"content": "$$50$$ "
}
],
[
{
"aoVal": "B",
"content": "$$52$$ "
}
],
[
{
"aoVal": "C",
"content": "$$55$$ "
}
]
] | [
"拓展思维->能力->数据处理"
] | [
"先用错误的被除数$$1205$$减去余数$$5$$,再除以商$$48$$求出除数;再用原题的被除数$$1250$$除以除数求出商. $$(1205-5)\\div 48$$, $$=1200\\div 48$$, $$=25$$, $$1250\\div 25=50$$. 答:正确的商是$$50$$. 故选$$\\text{A}$$. "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 665 | 27a40db3f4184c2ebad37a63f02b653e | [
"2009年亚太杯竞赛初赛第19题5分"
] | 1 | single_choice | 在阿简所拥有的枝条中,有九根为$$1$$厘米长,六根为$$2$$厘米长及三根为$$4$$厘米长.已知阿简必须用尽所有的枝条,问:她能造多少个长方形? | [
[
{
"aoVal": "A",
"content": "无法造 "
}
],
[
{
"aoVal": "B",
"content": "一个 "
}
],
[
{
"aoVal": "C",
"content": "两个 "
}
],
[
{
"aoVal": "D",
"content": "三个 "
}
],
[
{
"aoVal": "E",
"content": "以上各项皆否 "
}
]
] | [
"拓展思维->思想->对应思想"
] | [
"长方形的长为$$a$$,宽为$$b$$,那么长方形的周长为$$2a+2b$$,一定是$$2$$的倍数,但是该题中要求用尽所有的枝条,所有枝条的长度和为$$9\\times 1+6\\times 2+3\\times 4=33$$(厘米),是奇数,所以矛盾,无法造. "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3455 | d4203e6194354e3eb8cadafb72e65791 | [
"2016年第12届全国新希望杯小学高年级六年级竞赛复赛第3题4分"
] | 1 | single_choice | 口袋里有大小相同的$$6$$个球,其中$$3$$个红球,$$3$$个白球,从中任意摸出$$2$$个球,都摸到红球的可能性是(~ ). | [
[
{
"aoVal": "A",
"content": "$$\\frac{1}{9}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{1}{5}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{1}{6}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{1}{3}$$ "
}
]
] | [
"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"
] | [
"$$6$$个球中选$$2$$个,一共有$$C_{6}^{2}=15$$种情况,$$3$$个红球中选$$2$$个,一共有$$C_{3}^{2}=3$$种.则都是红球的概率为$$P=\\frac{C_{3}^{2}}{C_{6}^{2}}=\\frac{3}{15}=\\frac{1}{5}$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 492 | dca0ec8702d34b34b31bdbe35fa2616a | [
"2015年华杯赛四年级竞赛初赛"
] | 2 | single_choice | 一次考试共有$$6$$道选择题,评分规则如下:每人先给$$6$$分,答对一题加$$4$$分,答错一题减$$1$$分,不答得$$0$$分。现有$$51$$名同学参加考试,那么,至少有( )人得分相同。 | [
[
{
"aoVal": "A",
"content": "$$3$$ "
}
],
[
{
"aoVal": "B",
"content": "$$4$$ "
}
],
[
{
"aoVal": "C",
"content": "$$5$$ "
}
],
[
{
"aoVal": "D",
"content": "$$6$$ "
}
]
] | [
"拓展思维->拓展思维->组合模块->抽屉原理->构造型抽屉原理"
] | [
"因为每人先给$$6$$分,所以得分情况有以下$$25$$种: $$0$$、$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$、$$10$$、$$11$$、$$12$$、$$13$$、$$14$$、$$15$$、$$16$$、$$17$$、$$18$$、$$20$$、$$21$$、$$22$$、$$25$$、$$26$$、$$30$$,因此根据抽屉原理,$$51\\div 25=2\\cdots \\cdots 1$$,所以至少有$$3$$人得分相同。 "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3174 | 542373ca5cba4836b6a1c9b2316be641 | [
"2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第7题5分"
] | 1 | single_choice | \textbf{(2019 Youth Mathematics Olympics, Primary 5, Question \#7)} $$990$$有很多因数,这些因数的平均数是($\textasciitilde\textasciitilde\textasciitilde\textasciitilde\textasciitilde\textasciitilde$). | [
[
{
"aoVal": "A",
"content": "$$110$$ "
}
],
[
{
"aoVal": "B",
"content": "$$115$$ "
}
],
[
{
"aoVal": "C",
"content": "$$117$$ "
}
],
[
{
"aoVal": "D",
"content": "$$120$$ "
}
]
] | [
"拓展思维->思想->对应思想",
"Overseas Competition->知识点->数论模块->因数与倍数->因数个数定理"
] | [
"$$990$$的因数有:$$1$$;$$990$$;$$2$$;$$495$$;$$3$$;$$330$$;$$5$$;$$198$$;$$6$$;$$165$$;$$9$$;$$110$$;$$10$$;$$99$$;$$11$$;$$90$$;$$15$$;$$66$$;$$18$$;$$55$$;$$22$$;$$45$$;$$30$$;$$33$$,一共$$24$$个, 平均数为:$$(1+990+2+495+3+330+5+198+6+165+9+110+10+99+11+90+15+66+18+55+22+45+30+33)\\div 24$$ $$=2808\\div 24$$ $$=117$$, 故选答案$$\\text{C}$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2519 | 16c1a6a222f04d31966a004c27ad6491 | [
"2014年全国迎春杯六年级竞赛复赛第2题"
] | 2 | single_choice | 对于任何自然数,定义$$n!=1\times 2\times 3\times \cdots \times n$$.那么请问算式$$2022!-3!$$的计算结果的个位数字是. | [
[
{
"aoVal": "A",
"content": "$$2$$ "
}
],
[
{
"aoVal": "B",
"content": "$$4$$ "
}
],
[
{
"aoVal": "C",
"content": "$$6$$ "
}
],
[
{
"aoVal": "D",
"content": "$$8$$ "
}
]
] | [
"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"
] | [
"由新定义:$$n!=1\\times 2\\times 3\\times \\ldots\\times n$$得: $$2014!=1\\times 2\\times 3\\times 4\\times 5\\times \\ldots\\times 2021\\times 2022$$ $$=1\\times 3\\times 4\\times 6\\times 7\\times 8\\times \\ldots\\times 2021\\times 2022\\times 10$$ 所以$$1\\times 3\\times 4\\times 6\\times 7\\times 8\\times \\ldots\\times 2021\\times 2022\\times 10$$是$$10$$的倍数, 所以$$2022!$$的个位数为$$0$$; $$3!=1\\times 2\\times 3=6$$ 所以$$2022!-3!$$的个位数也就为:$$10-6=4$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 420 | a4db558e7d4f4f6ca32ee5cc399821c1 | [
"2013年华杯赛四年级竞赛初赛",
"2013年华杯赛四年级竞赛初赛"
] | 2 | single_choice | 有一个人捡到了一条红领巾,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。 | [
[
{
"aoVal": "A",
"content": "小东 "
}
],
[
{
"aoVal": "B",
"content": "小西 "
}
],
[
{
"aoVal": "C",
"content": "小南 "
}
],
[
{
"aoVal": "D",
"content": "小北 "
}
]
] | [
"拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理"
] | [
"解:根据题干分析可得,小南与小北说的话是相互矛盾的,所以两人中一定有一个人说的是正确的。 假设小北说的是正确的,则小南说``小东说的不对''是错,可得:小东说的对,这样与已知只有一个人说对了相矛盾,所以此假设不成立,故小南说的是正确的。 故选:$$\\text{C}$$。 "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2744 | ff8080814518d5240145201b5b8e0a7b | [
"2014年全国迎春杯三年级竞赛复赛第2题"
] | 1 | single_choice | 下式中,$$\square $$和△分别代表(~~~~~ ). $$\square $$+$$\square $$+$$\square $$+△+△+△$$=27$$ △+△+$$\square $$$$=12$$ | [
[
{
"aoVal": "A",
"content": "$$3$$和$$4$$ "
}
],
[
{
"aoVal": "B",
"content": "$$3$$和$$6$$ "
}
],
[
{
"aoVal": "C",
"content": "$$4$$和$$6$$ "
}
],
[
{
"aoVal": "D",
"content": "$$6$$和$$3$$ "
}
]
] | [
"拓展思维->思想->对应思想"
] | [
"由``$$\\square $$+$$\\square $$+$$\\square $$$$+$$△$$+$$△$$+$$△$$=27$$''可得``$$\\square $$+△$$=9$$'',又由于``$$△$$+$$△$$+$$\\square $$$$=12$$''可得△$$=3$$,$$\\square $$=$$6$$.题目问的是$$\\square $$和△分别代表什么,所以选$$\\text{D}$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1930 | e09083f6d6a24c49930dca942c84d66a | [
"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"
] | 1 | single_choice | 甲乙两人在春节一共得$$200$$元压岁钱,甲给乙$$40$$元,还比乙多$$10$$元,那么,原来甲得了~\uline{~~~~~~~~~~}~元压岁钱. | [
[
{
"aoVal": "A",
"content": "$$145$$ "
}
],
[
{
"aoVal": "B",
"content": "$$140$$ "
}
],
[
{
"aoVal": "C",
"content": "$$125$$ "
}
],
[
{
"aoVal": "D",
"content": "$$120$$ "
}
]
] | [
"拓展思维->知识点->应用题模块->和差倍问题->和差问题->二量和差问题->明和暗差"
] | [
"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) "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 155 | 3514d857ffb64b5d8463f4d021994705 | [
"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9题3分"
] | 1 | single_choice | 从$$1$$至$$9$$这$$9$$个整数中,至少取个数,才能保证其中有两个数的和等于$$10$$. | [
[
{
"aoVal": "A",
"content": "$$3$$ "
}
],
[
{
"aoVal": "B",
"content": "$$4$$ "
}
],
[
{
"aoVal": "C",
"content": "$$5$$ "
}
],
[
{
"aoVal": "D",
"content": "$$6$$ "
}
]
] | [
"拓展思维->拓展思维->组合模块->抽屉原理->最不利原则"
] | [
"从$$1$$至$$9$$这$$9$$个整数中, $$1+9=2+8=3+7=4+6$$,都等于$$10$$; 根据最不利原则,取的数都是$$(1,9)$$、$$(2,8)$$、$$(3,7)$$、$$(4,6)$$组合中的$$1$$个, 再取一个$$5$$,即取第$$6$$个数时,能保证其中有两个数的和等于$$10$$. 故选$$\\text{D}$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 145 | 4707e9d5d2d74e44aac6e679ac412bbb | [
"2010年迎春杯六年级竞赛复赛",
"六年级其它",
"2010年迎春杯五年级竞赛复赛"
] | 4 | single_choice | 黑板上一共写了$$10100$$个数字,包括$$2018$$个$$1$$,$$2019$$个$$2$$,$$2020$$个$$3$$,$$2021$$个$$4$$,$$2022$$个$$5$$.每次操作都擦去其中$$4$$个不同的数字并写上一个第$$5$$种数字(例如擦去$$1$$、$$2$$、$$3$$、$$4$$各$$1$$个,写上$$1$$个$$5$$;或者擦去$$2$$、$$3$$、$$4$$、$$5$$各一个,写上一个$$1$$;\ldots);如果经过有限次操作后,黑板上恰好剩下了两个数字,那么这两个数字的乘积是~\uline{~~~~~~~~~~}~. | [
[
{
"aoVal": "A",
"content": "$$2$$ "
}
],
[
{
"aoVal": "B",
"content": "$$6$$ "
}
],
[
{
"aoVal": "C",
"content": "$$8$$ "
}
],
[
{
"aoVal": "D",
"content": "$$12$$ "
}
],
[
{
"aoVal": "E",
"content": "以上都不正确 "
}
]
] | [
"Overseas Competition->知识点->组合模块->构造与论证",
"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"
] | [
"每次操作,每个数字个数的奇偶性都会变化.$$1$$、$$3$$、$$5$$原来都是偶数个,它们的个数奇偶性永远保持一致;$$2$$、$$4$$原来都是奇数个,它们的个数奇偶性也永远保持一致,而且和$$1$$、$$3$$、$$5$$的个数奇偶性不同$$.$$ 最后剩下$$2$$个数字,只能是$$2$$和$$4$$,乘积为$$8$$,答案选$$\\text{C}$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 450 | 8aac50a7508d5d410150d532383276fe | [
"2015年北京华杯赛小学高年级竞赛初赛A卷第1题"
] | 2 | single_choice | 现在从甲、乙、丙、丁四个人中选出两个人参加一项活动.规定:如果甲去,那么乙也去;如果丙不去,那么乙也不去;如果丙去,那么丁不去.最后去参加活动的两个人是(~ ~ ). | [
[
{
"aoVal": "A",
"content": "甲、乙 "
}
],
[
{
"aoVal": "B",
"content": "乙、丙 "
}
],
[
{
"aoVal": "C",
"content": "甲、丙 "
}
],
[
{
"aoVal": "D",
"content": "乙、丁 "
}
]
] | [
"拓展思维->能力->逻辑分析"
] | [
"题目要求有两个人去,可以使用假设法,若甲去,则乙去,乙去则丙也去.三个人去,矛盾,所以甲不去.若丙不去则乙不去,那么只有丁去,矛盾,所以丙去.丙去则丁不去,由两个人去得到结论,乙要去. 所以答案是$$\\text{B}$$,丙和乙去. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1009 | 029a024cca5e4ee590c9bf8dbd2a2f95 | [
"2020年新希望杯五年级竞赛决赛(8月)第17题"
] | 1 | single_choice | $$2019$$年国庆节是星期二,则$$2020$$年国庆节是. | [
[
{
"aoVal": "A",
"content": "星期一 "
}
],
[
{
"aoVal": "B",
"content": "星期二 "
}
],
[
{
"aoVal": "C",
"content": "星期三 "
}
],
[
{
"aoVal": "D",
"content": "星期四 "
}
]
] | [
"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"
] | [
"$$2020\\div 4=505$$,$$2020$$年是闰年. 从$$2019$$年国庆节经过$$366$$天是$$2020$$年国庆节. $$366\\div 7=52$$(周)$$\\cdots \\cdots 2$$(天), 故$$2020$$年的国庆节是星期四. 故选$$\\text{D}$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1073 | 1c845b4a765b411f8e1f9138b6f9b113 | [
"2008年陈省身杯小学高年级六年级竞赛"
] | 0 | single_choice | 一堆煤$$2$$吨,每天用去它的$$\frac{1}{25}$$,$$3$$天后还剩(~ )吨. | [
[
{
"aoVal": "A",
"content": "$$\\frac{6}{25}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{22}{25}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{44}{25}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$1\\frac{22}{25}$$ "
}
]
] | [
"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"
] | [
"根据题意得,一堆煤$$2$$吨,每天用去它的$$\\frac{1}{25}$$,$$3$$天后还剩( )吨,可列式$$2-3\\times \\frac{1}{25}\\times 2=\\frac{44}{25}$$,故答案为$$\\text{C}$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2318 | e15edc75c4f94b98ab03d44d40ac6c96 | [
"2017年全国华杯赛小学高年级竞赛"
] | 2 | single_choice | 小张从甲地去乙地,然后再返回甲地.去时他的速度为$$a$$千米/小时,回来时他的速度为$$b$$千米/小时,那么他往返的平均速度是(~ )千米/小时. | [
[
{
"aoVal": "A",
"content": "$$\\frac{ab}{a+b}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{a+b}{2}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{2ab}{a+b}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{2a+2b}{ab}$$ "
}
]
] | [
"拓展思维->思想->转化与化归的思想"
] | [
"设甲、乙两地之间的距离为$$s$$千米.去时所用的时间为$$\\frac{s}{a}$$,返回时所用的时间为$$\\frac{s}{b}$$,往返的平均速度为 $$\\frac{2s}{\\frac{s}{a}+\\frac{s}{b}}=\\frac{2s}{\\frac{bs+as}{ab}}=\\frac{2s\\times ab}{as+bs}=\\frac{2ab}{a+b}$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2801 | dab968be167c44bcae1128fe64a12c43 | [
"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35题"
] | 1 | single_choice | $$1+2+3+\cdots +15$$的个位数是多少? | [
[
{
"aoVal": "A",
"content": "$$0$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1$$ "
}
],
[
{
"aoVal": "C",
"content": "$$2$$ "
}
],
[
{
"aoVal": "D",
"content": "$$3$$ "
}
]
] | [
"拓展思维->能力->运算求解"
] | [
"原式$$=120$$,各位是$$0$$. "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 129 | 1e967de8608a4dac83f21a606e1df6a6 | [
"2020年广东广州羊排赛四年级竞赛第9题4分"
] | 2 | single_choice | 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~\uline{~~~~~~~~~~}~分. | [
[
{
"aoVal": "A",
"content": "$$20$$ "
}
],
[
{
"aoVal": "B",
"content": "$$50$$ "
}
],
[
{
"aoVal": "C",
"content": "$$40$$ "
}
],
[
{
"aoVal": "D",
"content": "$$30$$ "
}
],
[
{
"aoVal": "E",
"content": "$$34$$ "
}
]
] | [
"拓展思维->能力->逻辑分析->代数逻辑推理"
] | [
"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:$$\\text{D}$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2603 | 28e9c890287e4ce8919f041e0cafc7dc | [
"2018年湖北武汉新希望杯六年级竞赛训练题(五)第14题"
] | 2 | single_choice | $$A=1\frac{11}{111}+2\frac{11}{112}+3\frac{11}{113}+\cdots +11\frac{11}{121}$$, find the integral part of $$A$$. 已知$$A=1\frac{11}{111}+2\frac{11}{112}+3\frac{11}{113}+\cdots +11\frac{11}{121}$$,那么$$A$$的整数部分是~\uline{~~~~~~~~~~}~. | [
[
{
"aoVal": "A",
"content": "$$66$$ "
}
],
[
{
"aoVal": "B",
"content": "$$67$$ "
}
],
[
{
"aoVal": "C",
"content": "$$68$$ "
}
],
[
{
"aoVal": "D",
"content": "$$69$$ "
}
],
[
{
"aoVal": "E",
"content": "$$70$$ "
}
]
] | [
"海外竞赛体系->知识点->计算模块->比较与估算",
"拓展思维->拓展思维->计算模块->比较与估算->放缩法->首尾放缩法"
] | [
"$$A=(1+2+3+\\cdots +11)+\\left( \\frac{11}{111}+\\frac{11}{112}+\\frac{11}{113}+\\cdots +\\frac{11}{121} \\right)\\textgreater66+\\frac{11}{121}\\times 11=67$$,$$A\\textless{}66+\\frac{11}{111}\\times 11\\textless{}68$$,所以$$67\\textless{}A\\textless{}68$$,$$A$$的整数部分是$$67$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 921 | ee4afca412024e8597a8987053c078f0 | [
"2013年希望杯三年级竞赛复赛",
"2013年希望杯二年级竞赛复赛",
"2013年希望杯四年级竞赛复赛",
"2013年希望杯五年级竞赛复赛",
"2013年希望杯六年级竞赛复赛",
"2013年希望杯一年级竞赛复赛"
] | 2 | single_choice | $$42$$的因数共有( )个。 | [
[
{
"aoVal": "A",
"content": "$$7$$ "
}
],
[
{
"aoVal": "B",
"content": "$$8$$ "
}
],
[
{
"aoVal": "C",
"content": "$$9$$ "
}
],
[
{
"aoVal": "D",
"content": "$$10$$ "
}
]
] | [
"拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理"
] | [
"$$42=2\\times 3\\times 7$$,因数个数等于指数加$$1$$再连乘,$$\\left( 1+1 \\right)\\times \\left( 1+1 \\right)\\times \\left( 1+1 \\right)=8$$(个)。所以选B。 "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3355 | a3f78a6d920649c78a55ec9d267e4e84 | [
"2019年第24届YMO二年级竞赛决赛第4题3分"
] | 1 | single_choice | 把$$15$$个玻璃球分成数量不同的$$2$$堆,共有种不同的分法. | [
[
{
"aoVal": "A",
"content": "$$4$$ "
}
],
[
{
"aoVal": "B",
"content": "$$5$$ "
}
],
[
{
"aoVal": "C",
"content": "$$6$$ "
}
],
[
{
"aoVal": "D",
"content": "$$7$$ "
}
]
] | [
"拓展思维->能力->逻辑分析"
] | [
"根据题意可得: $$15=1+2+3+9$$;$$15=1+2+4+8$$; $$15=1+2+5+7$$;$$15=1+3+4+7$$; $$15=1+3+5+6$$;$$15=2+3+4+6$$; 一共有$$6$$种. 答:共有$$6$$种不同的分法. 故选:$$\\text{C}$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3042 | d7dc77f9582a4ae5a815649af7ff8d66 | [
"2018年IMAS小学中年级竞赛(第二轮)第1题4分"
] | 1 | single_choice | 请问$$100-97+94-91+88-85+\cdots +4-1$$之值等于多少? | [
[
{
"aoVal": "A",
"content": "$$45$$ "
}
],
[
{
"aoVal": "B",
"content": "$$48$$ "
}
],
[
{
"aoVal": "C",
"content": "$$51$$ "
}
],
[
{
"aoVal": "D",
"content": "$$54$$ "
}
],
[
{
"aoVal": "E",
"content": "$$57$$ "
}
]
] | [
"拓展思维->拓展思维->计算模块->数列与数表->数列规律"
] | [
"$$100-97+94-91+88-85+\\cdots +4-1$$ $$=\\left( 100-97 \\right)+\\left( 94-91 \\right)+\\left( 88-85 \\right)+\\cdots +\\left( 4-1 \\right)$$ $$=\\underbrace{3+3+3+\\cdots +3}_{17项}$$ $$=3\\times 17$$ $$=51$$. 故选$$\\text{C}$$. ",
"<p>$$100-97+94-91+88-85+\\cdots +4-1$$</p>\n<p>$$=\\left( 100+94+88+\\cdots +4 \\right)-\\left( 97+91+85+\\cdots +1 \\right)$$</p>\n<p>$$=\\frac{\\left( 100+4 \\right)\\times 17}{2}-\\frac{\\left( 97+1 \\right)\\times 17}{2}$$</p>\n<p>$$=\\left( 104-98 \\right)\\times \\frac{17}{2}$$</p>\n<p>$$=6\\times \\frac{17}{2}$$</p>\n<p>$$=51$$.</p>\n<p>故选$$\\text{C}$$.</p>"
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3291 | 8c0064d3ec4546fa9d8d8cf858123b7e | [
"2005年第3届创新杯六年级竞赛初赛第10题",
"2005年六年级竞赛创新杯"
] | 1 | single_choice | 有五个砝码,其中重$$1$$克的砝码$$1$$个,重$$3$$克的砝码$$2$$个,重$$5$$克的砝码$$2$$个。如果规定砝码只能放在天平的同一边,那么在$$1\sim17$$克所有整数克的重量中,不能称出的两个重量是( )。 | [
[
{
"aoVal": "A",
"content": "$$2$$克和$$9$$克 "
}
],
[
{
"aoVal": "B",
"content": "$$2$$克和$$11$$克 "
}
],
[
{
"aoVal": "C",
"content": "$$2$$克和$$13$$克 "
}
],
[
{
"aoVal": "D",
"content": "$$2$$克和$$15$$克 "
}
]
] | [
"拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举"
] | [
"我们只用分析所给四个选项中的重量能否称出,而$$9=5+3+1\\text{,}11=5+5+1\\text{,}13=5+5+3$$都可称出,选$$D$$。 "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2518 | 22fa6501a31e41319a1941678ef43fdb | [
"2020年希望杯二年级竞赛模拟第4题"
] | 1 | single_choice | 请你根据数串的规律再横线上填上正确的答案:3,6,9,12~\uline{~~~~~~~~~~}~,18. | [
[
{
"aoVal": "A",
"content": "$$10$$ "
}
],
[
{
"aoVal": "B",
"content": "$$12$$ "
}
],
[
{
"aoVal": "C",
"content": "$$14$$ "
}
],
[
{
"aoVal": "D",
"content": "$$15$$ "
}
]
] | [
"拓展思维->思想->数形结合思想"
] | [
"后一个数等于前一个数$$+$$ 3 "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1920 | a12889b5e75941d2873393dfc5a0177d | [
"2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第5题5分"
] | 1 | single_choice | 今年哥哥$$16$$岁,弟弟$$8$$岁,问年前,哥哥的年龄是弟弟的$$3$$倍. | [
[
{
"aoVal": "A",
"content": "$$2$$ "
}
],
[
{
"aoVal": "B",
"content": "$$4$$ "
}
],
[
{
"aoVal": "C",
"content": "$$6$$ "
}
],
[
{
"aoVal": "D",
"content": "$$8$$ "
}
]
] | [
"拓展思维->能力->运算求解"
] | [
"从题干中可以知道,今年哥哥$$16$$岁,弟弟$$8$$岁,二者相差,$$16-8=8$$(岁), 那么题目问几年前,哥哥的年龄是弟弟的$$3$$倍, 也就是二者年龄差应该是弟弟的$$2$$倍,$$8\\div 2=4$$(岁),$$8-4=4$$(年), 即$$4$$年前,哥哥的年龄是弟弟的$$3$$倍. 我们可以验证一下,四年前,哥哥的年龄是:$$16-4=12$$(岁),弟弟的年龄是:$$8-4=4$$(岁). $$12$$岁正好是$$4$$岁的$$3$$倍. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1016 | 03257410e3be488f80e86d90b23ed285 | [
"2019年浙江杭州滨江区杭州江南实验学校五年级竞赛模拟(江南杯)第12题1分"
] | 0 | single_choice | 学校举行``六一欢乐大卖场''活动,$$505$$班同学批发了一些小玩具参加活动,共卖出玩具$$22$$个,最便宜的售价$$10.5$$元$$/$$个,最贵的售价$$15.5$$元$$/$$个,这些玩具大约卖出元. | [
[
{
"aoVal": "A",
"content": "$$340-400$$ "
}
],
[
{
"aoVal": "B",
"content": "$$200-230$$ "
}
],
[
{
"aoVal": "C",
"content": "$$230-350$$ "
}
]
] | [
"拓展思维->能力->运算求解"
] | [
"假设都按照最低价$$10.5$$一个,$$22$$个总价是$$231$$元,假设都按照最高价$$15.5$$元一个,总价是$$341$$元,所以这些玩具的总价不少于$$230$$元,也不高于$$340$$元.据此解答. $$10.5\\times22=231$$(元), $$15.5\\times22=341$$(元), 所以这些玩具的总价不少于$$230$$元,也不高于$$340$$元. 故答案为∶$$\\text{C}$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2057 | d8d03079467946c7b57a0116cdd60e61 | [
"2015年全国世奥赛竞赛A卷第6题"
] | 2 | single_choice | 在编号为$$1$$,$$2$$,$$3$$的三个相同的杯子里,分别盛着半杯液体.$$1$$好杯中溶有$$100$$克糖,$$2$$号杯中是水,$$3$$号杯中溶有$$100$$克盐.先将$$1$$号杯中液体的一半及$$3$$号杯中液体的$$\frac{1}{4}$$倒入$$2$$号杯,然后搅匀.再从$$2$$号杯倒出所盛液体的$$\frac{2}{7}$$到$$1$$号杯.这时$$1$$号杯含盐量与含糖量之比是~\uline{~~~~~~~~~~}~.(写成最简整数比) | [
[
{
"aoVal": "A",
"content": "$$2:9$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1:9$$ "
}
],
[
{
"aoVal": "C",
"content": "$$1:8$$ "
}
],
[
{
"aoVal": "D",
"content": "$$1:7$$ "
}
]
] | [
"知识标签->拓展思维->应用题模块->比例应用题->复杂的比例问题"
] | [
"第一次倒完后,$$1$$号杯中剩下$$50$$克糖,$$2$$号杯中有$$50$$克糖和$$25$$克盐.此时把$$2$$号杯的$$\\frac{2}{7}$$倒入$$1$$号杯,那么$$2$$号杯的糖和盐都倒了$$\\frac{2}{7}$$,即糖倒了$$50\\times \\frac{2}{7}=\\frac{100}{7}$$(克),盐倒了$$25\\times \\frac{2}{7}=\\frac{50}{7}$$(克).那么$$1$$号杯的糖就变成了$$50+\\frac{100}{7}=\\frac{450}{7}$$(克),盐$$\\frac{50}{7}$$克. 所以含盐量和含糖量之比为$$\\frac{50}{7}:\\frac{450}{7}=1:9$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2738 | ff8080814518d5240145191ee20103f9 | [
"2014年全国迎春杯四年级竞赛复赛第5题"
] | 1 | single_choice | 对于任何自然数,定义$$n!=1\times 2\times 3\times \cdots \times n$$,如$$8!=1\times2\times 3\times \cdots \times 8$$;那么,算式:$$2014!+2013!-2012!+2011!+\cdots -4!+3!-2!+1!$$,计算结果的个位数字是(~~~~~~ ). | [
[
{
"aoVal": "A",
"content": "$$0$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1$$ "
}
],
[
{
"aoVal": "C",
"content": "$$3$$ "
}
],
[
{
"aoVal": "D",
"content": "$$9$$ "
}
]
] | [
"拓展思维->拓展思维->数论模块->尾数特征->末一位数"
] | [
"因为从$$5!=1\\times 2\\times 3\\times 4\\times 5$$ 开始,都含有$$2$$和$$5$$,那么个位必然是$$0$$,那么我们只要计算$$-4!+3!-2!+1!$$的个位数字即可,得出个位数字为$$1$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1283 | 1b6f00c74c024851b5f48f5e6a24377c | [
"2019年全国小学生数学学习能力测评四年级竞赛复赛第8题3分"
] | 0 | single_choice | 游乐场游玩,每人第一个小时付款$$15$$元,以后每小时付款$$8$$元,军军和弟弟游玩后共付了$$62$$元,他们在游乐场玩了个小时. | [
[
{
"aoVal": "A",
"content": "$$6$$ "
}
],
[
{
"aoVal": "B",
"content": "$$4$$ "
}
],
[
{
"aoVal": "C",
"content": "$$3$$ "
}
]
] | [
"拓展思维->能力->运算求解"
] | [
"除去第$$1$$小时以外的费用:$$62-15\\times2=32$$(元); 每人除了第一小时以外所花的时间:$$32\\div 2\\div 8=2$$(小时); 共时长:$$2+1=3$$(小时). 故选$$\\text{C}$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 28 | 031197a17061487db7badbb2648908d1 | [
"2021年新希望杯二年级竞赛初赛第30题5分"
] | 2 | single_choice | 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. | [
[
{
"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 "
}
]
] | [
"拓展思维->思想->分类讨论思想"
] | [
"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$$\\text{C}$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 752 | 88856aef562c4200ac34e26b6c173c0d | [
"2014年第15届上海中环杯小学低年级二年级竞赛初赛第16题"
] | 2 | single_choice | 有一种靶子,上面有$$1$$环、$$3$$环、$$5$$环、$$7$$环和$$9$$环表示射中的环数. 甲说:``我打了$$5$$枪,每枪都中靶,共中了$$35$$环.'' 乙说:``我打了$$6$$枪,每枪都中靶,共中了$$36$$环.'' 丙说:``我打了$$3$$枪,每枪都中靶,共中了$$24$$环.'' 丁说:``我打了$$4$$枪,只有$$1$$枪没中靶,共中了$$21$$环.'' 已知他们四人中,只有一人说了假话.那么说假话的是. | [
[
{
"aoVal": "A",
"content": "甲 "
}
],
[
{
"aoVal": "B",
"content": "乙 "
}
],
[
{
"aoVal": "C",
"content": "丙 "
}
],
[
{
"aoVal": "D",
"content": "丁 "
}
]
] | [
"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"
] | [
"单双数加减的性质是:单数个单数和为单,双数个单数和为双. 甲:$$9+9+9+1+7=35$$,正确. 乙:$$9+9+9+1+1+7=36$$,正确. 丙:三个单数的和为单数,结果$$24$$为双数,错误. 丁:$$7+7+7+0=21$$,正确. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3277 | 5a6baa9d60b74ffebe8bbdef38d05c48 | [
"2007年第5届创新杯六年级竞赛第6题5分"
] | 2 | single_choice | 有$$2007$$盏亮着的灯,各有一个拉线开关控制着,拉一下拉线开关灯会由亮变灭,再拉一下又由灭变亮,现按其顺序将灯编号为$$1$$,$$2$$,$$\cdots$$,$$2007$$,然后将编号为$$2$$的倍数的灯线都拉一下,再将编号为$$3$$的倍数的灯线都拉一下,最后将编号为$$5$$的倍数的灯线都拉一下,三次拉完后亮着的灯有盏. | [
[
{
"aoVal": "A",
"content": "$$1004$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1002$$ "
}
],
[
{
"aoVal": "C",
"content": "$$1000$$ "
}
],
[
{
"aoVal": "D",
"content": "$$998$$ "
}
]
] | [
"拓展思维->思想->整体思想"
] | [
"∵有$$2007$$盏亮着的电灯,现按其顺序编号为$$1$$,$$2$$,$$\\cdots$$,$$2007$$, ∴编号为$$2$$的倍数的灯有$$(2007-1)\\div2=1003$$(只), 编号为$$3$$的倍数的灯有$$2007\\div3=669$$(只), 编号为$$5$$的倍数的灯的有$$(2007-2)\\div5=401$$(只), 其中既是$$3$$的倍数也是$$5$$的倍数有$$(2007-12)\\div15=133$$(只), 既是$$2$$的倍数也是$$3$$的倍数有$$(2007-3)\\div6=334$$(只), 既是$$2$$的倍数也是$$5$$的倍数有$$(2007-7)\\div10=200$$(只), 既是$$2$$的倍数也是$$5$$的倍数,还是$$3$$的倍数有$$(2007-27)\\div30=66$$(只), 只拉$$1$$次的:$$1003-334-200+66=535$$, $$669-334-133+66=268$$, $$401-200-133+66=134$$, 拉$$3$$次的$$66$$, 所以亮的就是$$2007-535-268-134-66=1004$$(只). 故选$$\\text{A}$$. "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1934 | aa560020cf464d41bfadb4edc70c3a78 | [
"2017年全国华杯赛小学高年级竞赛"
] | 1 | single_choice | 甲瓶盐水浓度为$$5 \%$$,乙瓶盐水浓度为$$10 \%$$,两瓶混合后盐水的浓度(~ ). | [
[
{
"aoVal": "A",
"content": "是$$5 \\%$$ "
}
],
[
{
"aoVal": "B",
"content": "是$$10 \\%$$ "
}
],
[
{
"aoVal": "C",
"content": "是$$7.5 \\%$$ "
}
],
[
{
"aoVal": "D",
"content": "不能确定 "
}
]
] | [
"拓展思维->思想->对应思想"
] | [
"如果甲瓶盐水为$$a$$升,则含盐$$5a \\%$$,如果乙瓶盐水为$$b$$升,则含盐$$10 \\%b$$,依照浓度的计算公式,混合后盐水的浓度是:$$\\frac{5 \\%a+10 \\%b}{a+b}$$,因为不知$$a$$和$$b$$的数值,所以两瓶混合后盐水的浓度不能确定. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1557 | ff8080814502fa2401450775bb5d0aa1 | [
"2014年全国迎春杯四年级竞赛复赛第8题"
] | 2 | single_choice | 小元和小芳合作进行一项$$10000$$字的打字作业,但他们都非常马虎,小元每打$$10$$个字,就会打错$$1$$个;小芳每打字$$10$$个,就会打错$$2$$个,最后,当两人完成工作时,小元打正确的字数恰好是小芳打正确的字数的$$2$$倍,那么,两人打正确的字共有(~~~~~~~ )个. | [
[
{
"aoVal": "A",
"content": "$$5000$$ "
}
],
[
{
"aoVal": "B",
"content": "$$7320$$ "
}
],
[
{
"aoVal": "C",
"content": "$$8000$$ "
}
],
[
{
"aoVal": "D",
"content": "$$8640$$ "
}
]
] | [
"拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->已知工时反推->两人合作"
] | [
"我们可以理解为小元每打$$10$$份的字数就会打错$$1$$份,小芳每打$$10$$份的字数就会打错$$2$$份,即小芳打$$5$$份的字数只能正确$$4$$份. 假设小元打的总字数为$$80$$份,那么他正确的为$$8\\times 9=72$$份,根据题意可知小芳正确的为$$72\\div 2=36$$份, 那么小芳打字的总分数为$$36\\div 4\\times 5=45$$份.小元和小芳的总字数份为$$80+45=125$$份,共$$10000$$字,即每份字数为$$10000\\div125=80$$. 小元和小芳共正确的字数为$$(72+36)\\times 80=8640$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1282 | 629f89421d444e8f8b7a124bf4538e77 | [
"2016年陈省身杯六年级竞赛国际青少年数学解题能力展示活动第11题"
] | 3 | single_choice | 文具店以每支$$8$$元的价格购进一批钢笔,并标价每支$$20$$元出售.当售出这批钢笔的$$\frac{3}{4}$$时发现,不仅已经收回了全部成本,且获利$$224$$元.那么这批钢笔共有~\uline{~~~~~~~~~~}~支. | [
[
{
"aoVal": "A",
"content": "$$16$$ "
}
],
[
{
"aoVal": "B",
"content": "$$32$$ "
}
],
[
{
"aoVal": "C",
"content": "$$48$$ "
}
],
[
{
"aoVal": "D",
"content": "$$64$$ "
}
]
] | [
"拓展思维->思想->对应思想",
"课内体系->能力->运算求解"
] | [
"成本$$:$$利润$$:$$售价$$=8:12:20=2:3:5$$,卖出$$\\frac{3}{4}$$后,售价为$$\\frac{3}{4}\\times 5=\\frac{15}{4}$$(份),此时获利份数为$$\\frac{15}{4}-2=\\frac{7}{4}$$(份),全成本为$$224\\div \\frac{7}{4}\\times 2=256$$(元),钢笔共有$$256\\div 8=32$$(支). "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 734 | 3748cf45c9de4b53a0a03c675453b1b8 | [
"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"
] | 1 | single_choice | 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(~ ). | [
[
{
"aoVal": "A",
"content": "$$67$$ "
}
],
[
{
"aoVal": "B",
"content": "$$73$$ "
}
],
[
{
"aoVal": "C",
"content": "$$158$$ "
}
],
[
{
"aoVal": "D",
"content": "$$22$$ "
}
]
] | [
"拓展思维->能力->运算求解"
] | [
"设这个数为$$a$$,则$$\\left { \\begin{matrix}a\\div 3\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 2 a\\div 5\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 3 a\\div 7\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 4 \\end{matrix} \\right.$$,整理得$$\\left { \\begin{matrix}a\\div 3\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 11 a\\div 5\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 3 a\\div 7\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 11 \\end{matrix} \\right.$$,所以$$a=21k+11$$ 所以当$$k=7$$时,$$a=158$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 993 | 0548a2551c534d1daef0cc71b97a7308 | [
"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"
] | 1 | single_choice | 有$$26$$个小朋友排队做操,排在涛涛前面的有$$10$$人,排在涛涛后面的有人. | [
[
{
"aoVal": "A",
"content": "$$10$$ "
}
],
[
{
"aoVal": "B",
"content": "$$15$$ "
}
],
[
{
"aoVal": "C",
"content": "$$16$$ "
}
],
[
{
"aoVal": "D",
"content": "$$20$$ "
}
]
] | [
"拓展思维->能力->运算求解"
] | [
"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:$$\\text{B}$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1538 | 4087c73c729549e4aa282363fe0f76f0 | [
"2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第8题5分"
] | 1 | single_choice | 小明今年$$14$$岁,小强今年$$8$$岁,$$2$$年后,小明比小强大岁. | [
[
{
"aoVal": "A",
"content": "$$5$$ "
}
],
[
{
"aoVal": "B",
"content": "$$6$$ "
}
],
[
{
"aoVal": "C",
"content": "$$7$$ "
}
],
[
{
"aoVal": "D",
"content": "$$8$$ "
}
]
] | [
"拓展思维->拓展思维->应用题模块->年龄问题->年龄问题基本关系->年龄差"
] | [
"从题干中可以知道,小明今年$$14$$岁,小强今年$$8$$岁, 那么今年二者的年龄差是$$14-8=6$$(岁); 无论经过几年,小明和小强的年龄差还是不变的; 所以$$2$$年后,小明比小强大$$6$$岁; 可以换另一种方法检验一下:$$2$$年后小明$$16$$岁,小强$$10$$岁;$$16-10=6$$(岁). 所以选择$$\\text{B}$$选项. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 491 | e1263b323754425ebb457db711514639 | [
"2013年第11届全国创新杯五年级竞赛第4题5分"
] | 2 | single_choice | (3分)在垒球比赛中,若赢$$1$$场得$$3$$分,平$$1$$场得$$1$$分,输$$1$$场不得分.每个队都与其他队交锋$$4$$场,这时四个参赛队的总积分为:$$A$$队$$22$$分,$$B$$队$$19$$分,$$C$$队$$14$$分,$$D$$队$$12$$分.那么有场比赛为平局. | [
[
{
"aoVal": "A",
"content": "$$3$$ "
}
],
[
{
"aoVal": "B",
"content": "$$4$$ "
}
],
[
{
"aoVal": "C",
"content": "$$5$$ "
}
],
[
{
"aoVal": "D",
"content": "$$7$$ "
}
]
] | [
"拓展思维->七大能力->逻辑分析"
] | [
"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. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 445 | e05104d6577e498b8bdd505b82ac5abc | [
"2008年二年级竞赛学而思杯"
] | 1 | single_choice | 古时候某国有两座城,一座``真城'',一座``假城'',真城的人都说真话,假城的人都说假话.一天,一个国外游客来到其中的一座城,他向遇到的一位该国国民提了一个问题,就明白了自己到的是真城还是假城.以下( )最可能是游客提的问题. | [
[
{
"aoVal": "A",
"content": "你是真城的人吗? "
}
],
[
{
"aoVal": "B",
"content": "你是假城的人吗? "
}
],
[
{
"aoVal": "C",
"content": "你是说真话的人吗? "
}
],
[
{
"aoVal": "D",
"content": "你是说假话的人吗? "
}
],
[
{
"aoVal": "E",
"content": "你是这座城的人吗? "
}
]
] | [
"拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理"
] | [
"对于A,B,C,D,不论问哪个城里的人,很明显得到的是相同的答案,无法判断真假,只有问E时,真城里的人会回答``是'',而假城里的人的答案是``不是''. "
] | E |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1740 | 6145fde6be8d42abbb324a70fe516de8 | [
"2012年第10届创新杯四年级竞赛初赛第1题6分"
] | 1 | single_choice | 两位篮球运动员的体重分别为$$75$$千克和$$87.5$$千克,第$$3$$位运动员的体重介于这两者之间,下列哪一个不可能是这$$3$$位运动员的平均体重? | [
[
{
"aoVal": "A",
"content": "$$80.5$$ "
}
],
[
{
"aoVal": "B",
"content": "$$81.5$$ "
}
],
[
{
"aoVal": "C",
"content": "$$82.5$$ "
}
],
[
{
"aoVal": "D",
"content": "$$83.5$$ "
}
]
] | [
"拓展思维->能力->实践应用"
] | [
"排除法: $$80.5\\times 3-(75+87.5)=241.5-162.5=79$$; $$81.5\\times 3-(75+87.5)=244.5-162.5=82$$; $$82.5\\times 3-(75+87.5)=247.5-162.5=85$$; $$83.5\\times 3-(75+87.5)=250.5-162.5=88\\textgreater87.5$$. 故选$$\\text{D}$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 465 | c9dda81a33f44f8bb8c34025b7ae45dd | [
"2010年五年级竞赛创新杯",
"2010年六年级竞赛创新杯"
] | 3 | single_choice | 要想用天平称出$$1\sim40$$克所有整数克的质量,如果允许两边放砝码,至少用( )个砝码。 | [
[
{
"aoVal": "A",
"content": "$$4$$ "
}
],
[
{
"aoVal": "B",
"content": "$$5$$ "
}
],
[
{
"aoVal": "C",
"content": "$$6$$ "
}
],
[
{
"aoVal": "D",
"content": "$$7$$ "
}
]
] | [
"拓展思维->思想->枚举思想"
] | [
"若只有一个砝码,那么这个砝码有放左、放右和不放$$3$$种不同选择,其中左、右都不放的情况不计,剩下的情况会出现左右重复,只能称出$$\\frac{3-1}{2}=1$$(种)不同的质量;取$$1$$克,可以称出$$1$$克; 若有两个砝码,共有$$3\\times 3=9$$(种)放法,最多称出$$\\frac{{{3}^{2}}-1}{2}=4$$(种)不同的质量;取$$1$$克和$$3$$克,即可称出$$1$ $4$$克; 若有三个砝码,共有$${{3}^{3}}=27$$(种)放法,最多称出$$\\frac{{{3}^{3}}-1}{2}=13$$(种)不同的质量;取$$1$$克、$$3$$克和$$9$$克,即可称出$$1$ $13$$克; 若有四个砝码,共有$${{3}^{4}}=81$$(种)放法,最多称出$$\\frac{{{3}^{4}}-1}{2}=40$$(种)不同的质量.取$$1$$克、$$3$$克、$$9$$克和$$27$$克,即可称出$$1$ $40$$克. 由此我们只需$$1$$克、$$3$$克、$$9$$克、$$27$$克这样的$$4$$个砝码,即可称出$$1$ $40$$克所有整数克的质量. "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1641 | 914492a0de9d4509aa9dd207c900bc41 | [
"2015年华杯赛六年级竞赛初赛"
] | 2 | single_choice | 桌上有编号为$$1\sim 20$$的$$20$$张卡片,小明每次取出$$2$$张卡片,要求一张卡片的编号是另一张卡片的$$2$$倍多$$2$$,则小明最多取出( )张卡片。 | [
[
{
"aoVal": "A",
"content": "$$12$$ "
}
],
[
{
"aoVal": "B",
"content": "$$14$$ "
}
],
[
{
"aoVal": "C",
"content": "$$16$$ "
}
],
[
{
"aoVal": "D",
"content": "$$18$$ "
}
]
] | [
"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"
] | [
"解:设另一张卡号是$$x$$。则: $$2x+2\\leqslant 20$$ $$2x+2-2\\leqslant 20-2$$ $$2x\\leqslant 18$$ $$2x\\div 2\\leqslant 18\\div 2$$ $$x\\leqslant 9$$ 又因为$$x\\geqslant 1$$ $$9\\times 2=18$$(张) $$\\because 4$$、$$6$$、$$8$$即可以做为一倍量的数,也可以作为$$2$$倍量多$$2$$的数, $$\\therefore $$即总共可以取出:$$18-3\\times 2=12$$(张)。 答:小明最多可以取出$$12$$张卡片。 故选:$$A$$。 "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1573 | ff8080814518d5240145190941300312 | [
"2014年全国迎春杯三年级竞赛初赛第6题"
] | 1 | single_choice | 奶奶折一个纸鹤用$$3$$分钟,每折好一个需要休息$$1$$分钟,奶奶从$$2$$时$$30$$分开始折,她折好第$$5$$个纸鹤时已经到了(~~~~~~~ ). | [
[
{
"aoVal": "A",
"content": "$$2$$时$$45$$分 "
}
],
[
{
"aoVal": "B",
"content": "$$2$$时$$49$$分 "
}
],
[
{
"aoVal": "C",
"content": "$$2$$时$$50$$分 "
}
],
[
{
"aoVal": "D",
"content": "$$2$$时$$53$$分 "
}
]
] | [
"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"
] | [
"$$3\\times 5+1\\times 4=19$$,折好第$$5$$个时已经到了$$2$$时$$49$$分. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2810 | 6078a63ffcd241b68ad97f9f5a7907bc | [
"2019年华夏杯一年级竞赛复赛(华南赛区)第16题6分"
] | 0 | single_choice | 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? | [
[
{
"aoVal": "A",
"content": "小奥 "
}
],
[
{
"aoVal": "B",
"content": "小林 "
}
],
[
{
"aoVal": "C",
"content": "小匹 "
}
],
[
{
"aoVal": "D",
"content": "小克 "
}
]
] | [
"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"
] | [
"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. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1220 | 3da244afc15143e182441a9c76d6fe2a | [
"2014年全国迎春杯五年级竞赛初赛第3题"
] | 0 | single_choice | (5分)一辆大卡车一次最多可以装煤$$2.5$$吨,现在要一次运走$$48$$吨煤,那么至少需要辆这样的大卡车. | [
[
{
"aoVal": "A",
"content": "$$18$$ "
}
],
[
{
"aoVal": "B",
"content": "$$19$$ "
}
],
[
{
"aoVal": "C",
"content": "$$20$$ "
}
],
[
{
"aoVal": "D",
"content": "$$21$$ "
}
]
] | [
"知识标签->学习能力->七大能力->逻辑分析"
] | [
"$$48\\div 2.5=19\\cdots 0.5$$,$$19+1=20$$(辆). "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 209 | 4c7692da34b74807ac24febc1229a0b8 | [
"2011年全国希望杯四年级竞赛初赛第16题"
] | 2 | single_choice | 王强步行去公园,回来时坐车,往、返用了一个半小时,如果他来回都步行,则需要$$2$$个半小时,那么,他来回都坐车,则需~\uline{~~~~~~~~~~}~分钟. | [
[
{
"aoVal": "A",
"content": "$$60$$ "
}
],
[
{
"aoVal": "B",
"content": "$$40$$ "
}
],
[
{
"aoVal": "C",
"content": "$$30$$ "
}
],
[
{
"aoVal": "D",
"content": "$$20$$ "
}
]
] | [
"知识标签->数学思想->对应思想"
] | [
"一个半小时是$$90$$分钟,$$2$$个半小时是$$2\\times 60+30=150$$分钟.易知王强步行单程需要$$150\\div 2=75$$分钟,则坐车单程要$$90-75=15$$分钟,所以来回都坐车要$$15\\times 2=30$$分钟. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3421 | d7733d6409ad4b6ba55131ce03af17e4 | [
"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"
] | 1 | single_choice | 五一到了,花花想要出去旅游,她有$$2$$顶帽子、$$3$$条项链、$$5$$种头花、$$2$$双鞋、$$3$$条裙子,每类物品中各选一类进行搭配.那么,一共有种不同的搭配方法. | [
[
{
"aoVal": "A",
"content": "$$200$$ "
}
],
[
{
"aoVal": "B",
"content": "$$160$$ "
}
],
[
{
"aoVal": "C",
"content": "$$180$$ "
}
],
[
{
"aoVal": "D",
"content": "$$170$$ "
}
]
] | [
"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"
] | [
"$$2\\times 2\\times 3=12$$(种). "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2105 | ebc55cce51d540668553066fffa571d0 | [
"2017年河南郑州联合杯竞赛第2题4分"
] | 1 | single_choice | 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. | [
[
{
"aoVal": "A",
"content": "$$3$$次 "
}
],
[
{
"aoVal": "B",
"content": "$$4$$次 "
}
],
[
{
"aoVal": "C",
"content": "$$5$$次 "
}
],
[
{
"aoVal": "D",
"content": "$$6$$次 "
}
]
] | [
"拓展思维->拓展思维->应用题模块->周期问题->蜗牛爬井问题->蜗牛爬井求天数(爬出去类)"
] | [
"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. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2478 | 1599856f795f4ab2963e1d566c6a2ccd | [
"2020年希望杯六年级竞赛模拟第25题"
] | 1 | single_choice | 比较大小: $$1+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\cdots +\frac{1}{{{2020}^{2}}}$$~\uline{~~~~~~~~~~}~$$2$$. | [
[
{
"aoVal": "A",
"content": "$$\\textgreater$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\textless{}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$=$$ "
}
]
] | [
"拓展思维->能力->运算求解"
] | [
"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$$\\text{B}$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 597 | 6b24da2c7391400081fb0ff2a5b3aa57 | [
"2020年广东广州羊排赛五年级竞赛第7题3分"
] | 1 | single_choice | 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. | [
[
{
"aoVal": "A",
"content": "$$5$$ "
}
],
[
{
"aoVal": "B",
"content": "$$6$$ "
}
],
[
{
"aoVal": "C",
"content": "$$7$$ "
}
],
[
{
"aoVal": "D",
"content": "$$8$$ "
}
],
[
{
"aoVal": "E",
"content": "$$2$$ "
}
]
] | [
"拓展思维->拓展思维->数论模块->因数与倍数->因数与倍数基础",
"Overseas Competition->知识点->数论模块->质数与合数->质数与合数判定"
] | [
"最小质数是$$2$$,所以千位是$$2$$; 最小合数是$$4$$,所以百位是$$4$$;最小的自然数是$$0$$,所以十位是$$0$$,这个四位数是$$3$$的倍数,那么这个四位数各个位数之和是$$3$$的倍数,$$2+4+0=6$$. $$\\text{A}$$选项:$$6+5=11$$,不是$$3$$的整数倍,故$$\\text{A}$$错误; $$\\text{B}$$选项:$$6+6=12$$,是$$3$$的整数倍,故$$\\text{B}$$正确; $$\\text{C}$$选项:$$6+7=13$$,不是$$3$$的整数倍,故$$\\text{C}$$错误; $$\\text{D}$$选项:$$6+8=14$$,不是$$3$$的整数倍,故$$\\text{D}$$错误; $$\\text{E}$$选项:$$6+2=8$$,不是$$3$$的整数倍,故$$\\text{E}$$错误. 故选$$\\text{B}$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1733 | bf6caaf45c044ca4accdeecc12f9c345 | [
"2020年新希望杯二年级竞赛决赛(8月)第4题",
"2020年新希望杯二年级竞赛初赛(个人战)第4题"
] | 1 | single_choice | 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. | [
[
{
"aoVal": "A",
"content": "$$2$$ "
}
],
[
{
"aoVal": "B",
"content": "$$5$$ "
}
],
[
{
"aoVal": "C",
"content": "$$10$$ "
}
],
[
{
"aoVal": "D",
"content": "$$8$$ "
}
]
] | [
"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"
] | [
"根据$$1$$斤菠萝可以做$$2$$个菠萝蛋糕,那么要做$$10$$个菠萝蛋糕,需要菠萝$$10\\div2=5$$(斤).故选择$$\\text{B}$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 381 | 6f26597099f24f42acb20ca89c9a2818 | [
"2017年IMAS小学高年级竞赛(第一轮)第15题4分"
] | 1 | single_choice | 在算式$$\overline{ab}+ \overline{cd}= \overline{ef}$$ 中, $$\overline{ab}、 \overline{cd}、 \overline{ef}$$ 各代表一个二位数.且$$a$$、$$b$$、$$c$$、$$d$$、$$e$$、$$f$$六个数码两两不同.请问$$\overline{ef}$$的最小可能值是. | [
[
{
"aoVal": "A",
"content": "$$30$$ "
}
],
[
{
"aoVal": "B",
"content": "$$34$$ "
}
],
[
{
"aoVal": "C",
"content": "$$36$$ "
}
],
[
{
"aoVal": "D",
"content": "$$39$$ "
}
],
[
{
"aoVal": "E",
"content": "$$41$$ "
}
]
] | [
"拓展思维->能力->逻辑分析"
] | [
"因$$a+c\\geqslant 1+2=3$$.故$$e\\geqslant 3$$,因此取$$e=3$$,此时可判断出$$a$$、$$c$$分别为$$1$$与$$2$$,且知$$b$$、$$d$$均不能为$$0$$,否则数码$$f$$会与$$b$$或$$d$$相等,故$$b+d\\geqslant 4+5=9$$,即$$f=9$$,例如$$14+25=39$$、$$15+24=39$$. 故选$$\\text{D}$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2555 | 62a6616b10194309912fc66792abcd8d | [
"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"
] | 1 | single_choice | $$9\div 11$$的商的小数部分第$$50$$位上的数字是. | [
[
{
"aoVal": "A",
"content": "$$8$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1$$ "
}
],
[
{
"aoVal": "C",
"content": "$$9$$ "
}
]
] | [
"拓展思维->思想->对应思想",
"课内体系->能力->数据处理"
] | [
"列竖式 故$$9\\div 11=0.\\dot{8}\\dot{1}$$. 周期为$$2$$故$$50\\div 2=25$$, 故第$$50$$位为$$1$$. 故答案为:$$\\text{B}$$选项. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1347 | 478d3e6c4aea43fa835715a1f7e96e68 | [
"2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第2题5分"
] | 1 | single_choice | 幼儿园老师给小朋友分糖果,每个小朋友分$$5$$颗糖果,就多出$$24$$颗;每个小朋友分$$7$$颗糖果,就少$$16$$颗糖果.一共有糖果颗. | [
[
{
"aoVal": "A",
"content": "$$120$$ "
}
],
[
{
"aoVal": "B",
"content": "$$124$$ "
}
],
[
{
"aoVal": "C",
"content": "$$126$$ "
}
],
[
{
"aoVal": "D",
"content": "$$130$$ "
}
]
] | [
"拓展思维->能力->运算求解"
] | [
"两次分糖剩余之差:$$24+16=40$$(颗), 每次分糖数之差:$$7-5=2$$(颗). 小朋友人数:$$40\\div2=20$$(人), 糖果总数:$$20\\times5+24=124$$(颗). 所以选择$$\\text{B}$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2932 | a057a12b8efe42f690a0f0c1a50969a0 | [
"2017年全国美国数学大联盟杯小学高年级五年级竞赛初赛第36题"
] | 2 | single_choice | 小罗星期一工作了$$2$$个小时.在那之后的每一天,他的工作时间是前一天的两倍.请问他从星期一到星期四总共工作了多少小时? | [
[
{
"aoVal": "A",
"content": "$$14$$ "
}
],
[
{
"aoVal": "B",
"content": "$$16$$ "
}
],
[
{
"aoVal": "C",
"content": "$$28$$ "
}
],
[
{
"aoVal": "D",
"content": "$$30$$ "
}
]
] | [
"拓展思维->能力->运算求解"
] | [
"这是一个等比数列问题,$${{a}_{n}}={{a}_{1}}{{q}^{n-1}}={{2}^{n}}$$,$${{S}_{n}}=\\frac{{{a}_{1}}\\left( 1-{{q}^{n}} \\right)}{1-q}=\\frac{2\\left( 1-{{2}^{n}} \\right)}{1-2}=2\\left( {{2}^{n}}-1 \\right)=30$$,所以选$$D$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2882 | e8f3d290583b471c830124ea9c437e15 | [
"2017年河南郑州豫才杯四年级竞赛初赛第6题"
] | 1 | single_choice | 聪聪在数学课上总能专心听讲,每次完成练习之后,总要进行自检自查,发现错误能够及时更正过来.这次他在检查一道除法笔算时,发现自己把除数$$65$$写成了$$56$$,得到的商是$$15$$,余数是$$11$$.请你根据以上信息选择正确的结果. | [
[
{
"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$$2$$ "
}
]
] | [
"拓展思维->能力->运算求解"
] | [
"$$56\\times 15+11=851$$,$$851\\div 65=13\\cdots \\cdots 6$$,所以$$\\text{B}$$选项正确. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1655 | ff8080814767499401477b3b55cc15bb | [
"2013年全国华杯赛小学高年级竞赛初赛B卷第4题"
] | 2 | single_choice | 某日,甲学校买了$$56$$千克水果糖,每千克$$8.06$$元.过了几日,乙学校也需要买同样的$$56$$千克水果糖,不过正好赶上促销活动,每千克水果糖降价$$0.56$$元,而且只要买水果糖都会额外赠送$$5 \%$$同样的水果糖.那么乙学校将比甲学校少花(~ ~ ~ ~)元. | [
[
{
"aoVal": "A",
"content": "$$20$$ "
}
],
[
{
"aoVal": "B",
"content": "$$51.36$$ "
}
],
[
{
"aoVal": "C",
"content": "$$31.36$$ "
}
],
[
{
"aoVal": "D",
"content": "$$10.36$$ "
}
]
] | [
"拓展思维->能力->实践应用"
] | [
"甲学校花钱$$56\\times 8.06=451.36$$元;乙学校要买糖$$56\\div (1+5 \\%)=\\frac{160}{3}$$kg,单价$$8.06-0.56=7.5$$元,甲学校花钱$$\\frac{160}{3}\\times7.5=400$$元;乙学校将比甲学校少花$$51.36$$元. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2682 | 5ab31d10e4764e8c8560806baf7ec117 | [
"2022年第九届鹏程杯四年级竞赛初赛第3题5分"
] | 2 | single_choice | 计算:(1+2+3+···+29+30)$\times6-6\times128=\left( \right)$ | [
[
{
"aoVal": "A",
"content": "$$2022$$ "
}
],
[
{
"aoVal": "B",
"content": "$$2021$$ "
}
],
[
{
"aoVal": "C",
"content": "$$2020$$ "
}
],
[
{
"aoVal": "D",
"content": "$$2019$$ "
}
],
[
{
"aoVal": "E",
"content": "$$2018$$ "
}
]
] | [
"拓展思维->拓展思维->计算模块->整数->整数乘除->整数乘除法巧算"
] | [
"无 "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 893 | d6d49a392b294375a8ca78fcf994262f | [
"2013年第25届广东广州五羊杯六年级竞赛第8题5分"
] | 1 | single_choice | 已知一个素数的三倍与另一个素数的五倍的和是$$133$$,则这两个素数和为. | [
[
{
"aoVal": "A",
"content": "$$38$$ "
}
],
[
{
"aoVal": "B",
"content": "$$43$$ "
}
],
[
{
"aoVal": "C",
"content": "$$48$$ "
}
],
[
{
"aoVal": "D",
"content": "$$53$$ "
}
]
] | [
"拓展思维->思想->对应思想"
] | [
"显然两个素数中至少有一个是偶数,而素数中只有$$2$$是偶数. 将$$2$$代入,可得另一素数为$$41$$,故和为$$43$$. 故选$$\\text{B}$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3304 | 3c3c0d6ae897454ab801254e069dc3e1 | [
"2016年河南郑州联合杯小学高年级六年级竞赛复赛第17题2分"
] | 1 | single_choice | 五一大假期间,甲商场以打五折的方式优惠,乙商场以每满$$200$$送$$100$$元优惠券的措施优惠,妈妈打算花$$1000$$元购物,去(~ )商场比较合算. | [
[
{
"aoVal": "A",
"content": "甲 "
}
],
[
{
"aoVal": "B",
"content": "乙 "
}
],
[
{
"aoVal": "C",
"content": "甲、乙均可 "
}
],
[
{
"aoVal": "D",
"content": "不知道 "
}
]
] | [
"拓展思维->能力->实践应用"
] | [
"经济问题;甲:$$1000$$元可以买到$$1000\\div 50 \\% =2000$$元,乙:$$1000\\div 200=5$$,$$100\\times 5=500$$元,$$1000$$元可以买到:$$1000+500=1500$$元,$$2000\\textgreater1500$$,所以去甲商场合算. "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 337 | 6e3bd81aa1404e508056ac65434053ce | [
"2019年第24届YMO六年级竞赛决赛第7题3分"
] | 3 | single_choice | 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. | [
[
{
"aoVal": "A",
"content": "$$149$$ "
}
],
[
{
"aoVal": "B",
"content": "$$244$$ "
}
],
[
{
"aoVal": "C",
"content": "$$264$$ "
}
],
[
{
"aoVal": "D",
"content": "$$274$$ "
}
]
] | [
"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"
] | [
"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:$$\\text{D}$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 221 | 634d324416b945a19d65585ef1ec0736 | [
"2014年第25届亚太杯四年级竞赛决赛第18题5分"
] | 2 | single_choice | 小董并非既懂英文又懂法语.如果上述判断为真,那么下述判定为真的是. | [
[
{
"aoVal": "A",
"content": "小董懂英文但不懂法语. "
}
],
[
{
"aoVal": "B",
"content": "小董懂法语但不懂英文. "
}
],
[
{
"aoVal": "C",
"content": "小董既不懂英文也不懂法语. "
}
],
[
{
"aoVal": "D",
"content": "如果小董懂英文,小董一定不懂法语. "
}
],
[
{
"aoVal": "E",
"content": "如果小董不懂法语,那么他一定懂英文. "
}
]
] | [
"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"
] | [
"逻辑推理. 小董并非既懂英文又懂法语. 则有以下$$3$$种情况: $$(1)$$小董懂英文但不懂法语. $$(2)$$小董懂法语但不懂英文. $$(3)$$小董既不懂英文也不懂法语. 故选$$\\text{D}$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3100 | f45eb434a57f416fbf21dc522fee4c23 | [
"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"
] | 2 | single_choice | 对自然数$$n$$进行如下操作:如果$$n$$是偶数,就把它除以$$2$$,如果$$n$$是奇数,就把它加上$$7$$.现在对$$154$$进行有限次操作,得到的结果不可能是(~ ). | [
[
{
"aoVal": "A",
"content": "$$11$$ "
}
],
[
{
"aoVal": "B",
"content": "$$14$$ "
}
],
[
{
"aoVal": "C",
"content": "$$21$$ "
}
],
[
{
"aoVal": "D",
"content": "$$28$$ "
}
]
] | [
"知识标签->学习能力->七大能力->运算求解"
] | [
"154按照操作依次为$$77$$,$$84$$,$$42$$,$$21$$,$$28$$,$$14$$,$$7$$,$$14$$,7\\ldots\\ldots,不可能得到$$11$$. ",
"<p>154和$$7$$都是$$7$$点倍数,每次操作后最终结果都是$$7$$的倍数,不可能得到11.</p>"
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1050 | d4fd9d727c534fdd9a5a1d7fb6023fd2 | [
"2019年第7届湖北长江杯六年级竞赛复赛B卷第5题3分"
] | 1 | single_choice | 学生问老师的年龄,老师说:``当我是你这么大的时候,你刚$$3$$岁,当你是我这么大的时候,我已经$$39$$岁了,''这位老师今年岁. | [
[
{
"aoVal": "A",
"content": "$$15$$ "
}
],
[
{
"aoVal": "B",
"content": "$$26$$ "
}
],
[
{
"aoVal": "C",
"content": "$$27$$ "
}
],
[
{
"aoVal": "D",
"content": "$$28$$ "
}
]
] | [
"拓展思维->思想->方程思想"
] | [
"根据年龄差不会变这一特征,从年龄差入手,年龄差$$+$$老师现在的年龄$$=39$$, 所以老师$$+$$学生$$=42$$,设老师今年岁数为$$x$$, 则学生的岁数是$$42-x$$岁,再根据年龄差$$+$$老师现在的年龄$$=39$$, 列出方程解决问题, $$x-(42---x)+x=39$$, $$3x---42=39$$, $$3x=39+42$$, $$3x=81$$, $$x=27$$, 也就是老师今年$$27$$岁, 所以选择$$\\text{C}$$选项. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1331 | 42ef1153cc1e49d2a405ff5ab85d019a | [
"2017年河南郑州豫才杯五年级竞赛初赛第2题"
] | 1 | single_choice | 欢欢在今年的国庆节$$8$$天假期里,计划先读完老师推荐的青少年版的《红楼梦》,字数总计约为$$20$$万字,她的阅读速度可以达到$$500$$字/分钟,她需要平均每天坚持阅读(~ )分钟左右,才能完成阅读计划. | [
[
{
"aoVal": "A",
"content": "$$5$$ "
}
],
[
{
"aoVal": "B",
"content": "$$40$$ "
}
],
[
{
"aoVal": "C",
"content": "$$50$$ "
}
],
[
{
"aoVal": "D",
"content": "$$500$$ "
}
]
] | [
"拓展思维->能力->逻辑分析"
] | [
"$$200000\\div 500\\div 8=50$$(分钟),所以她需要平均每天坚持阅读$$50$$分钟才能完成阅读计划. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 385 | 96bda4b0141b43bc8e12fe59c6d60073 | [
"2019年第24届YMO五年级竞赛决赛第6题3分",
"2020年第24届YMO五年级竞赛决赛第6题3分"
] | 1 | single_choice | 小$$Y$$、小$$M$$、小$$O$$、小$$T$$四人中只有$$1$$人会开车.小$$Y$$说:``我会开''.小$$M$$说:``我不会开''.小$$O$$说:``小$$Y$$不会开''.小$$T$$什么也没说.已知小$$Y$$、小$$M$$、小$$O$$三人只有一人说了真话.会开车的是. | [
[
{
"aoVal": "A",
"content": "小$$Y$$ "
}
],
[
{
"aoVal": "B",
"content": "小$$M$$ "
}
],
[
{
"aoVal": "C",
"content": "小$$O$$ "
}
],
[
{
"aoVal": "D",
"content": "小$$T$$ "
}
]
] | [
"拓展思维->思想->对应思想"
] | [
"对比发现,小$$O$$与小$$Y$$说的矛盾,相互对立, 则小$$O$$与小$$Y$$必一对一错, 则小$$M$$说假话,则小$$M$$会开车,选$$\\text{B}$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1074 | 0f55a722f9a544ddafcf00b266282d95 | [
"2008年四年级竞赛创新杯"
] | 1 | single_choice | 两根电线杆之间相隔115米,在它们之间等距离增加22根电线杆后,第2根和第16根电线杆之间相隔 ( )米. | [
[
{
"aoVal": "A",
"content": "68 "
}
],
[
{
"aoVal": "B",
"content": "70 "
}
],
[
{
"aoVal": "C",
"content": "71 "
}
],
[
{
"aoVal": "D",
"content": "72 "
}
]
] | [
"拓展思维->拓展思维->应用题模块->间隔问题"
] | [
"24根电线杆之间有23个间隔,每个间隔的长是$$115\\div 23=5$$(米),又第2根和第16根电线杆之间有14个间隔,所以,它们相隔$$5\\times 14=70$$(米) "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 484 | b850000f15514a08992ffaf46c7d12b4 | [
"2020年第24届YMO三年级竞赛决赛第8题3分",
"2017年华杯赛四年级竞赛初赛"
] | 2 | single_choice | 从$$1$$至$$10$$这$$10$$个整数中,至少取个数,才能保证其中有两个数的和等于$$10$$。(把和为$$10$$的两个数看成是一双袜子,利用抽屉原理来做) | [
[
{
"aoVal": "A",
"content": "$$4$$ "
}
],
[
{
"aoVal": "B",
"content": "$$5$$ "
}
],
[
{
"aoVal": "C",
"content": "$$6$$ "
}
],
[
{
"aoVal": "D",
"content": "$$7$$ "
}
]
] | [
"拓展思维->拓展思维->组合模块->抽屉原理->简单抽屉原理"
] | [
"先找到和为$$10$$的组合:$$(1$$,$$9)$$、$$(2$$,$$8)$$、$$(3$$,$$7)$$、$$(4$$,$$6)$$,剩下$$5$$、$$10$$,若前面四组数中各取一个,后面$$5$$、$$10$$全取,这样还不能满足题目要求的两数和为$$10$$,如果再取剩下四个中的任意一个,必定会凑出和为$$10$$的组合,故最少要取$$6+1=7$$(个)。 "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3430 | fc407edef8184f58ae240303ce42114c | [
"2012年第8届全国新希望杯五年级竞赛复赛第2题"
] | 1 | single_choice | 某十字路口的交通信号灯,黄灯亮$$3$$秒,绿灯亮$$9$$秒,红灯亮$$24$$秒,那么某一时刻亮绿灯的可能性为(~ ). | [
[
{
"aoVal": "A",
"content": "$$\\frac{1}{12}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{1}{4}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{2}{3}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$9$$ "
}
]
] | [
"拓展思维->拓展思维->计数模块->统计与概率->概率->计数求概率"
] | [
"亮绿灯的可能性为:$$9\\div (3+9+24)=\\frac{1}{4}$$,故选$$\\text{B}$$. ~ "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3439 | ceef9867d23e4dd9ace97ece28971669 | [
"2012年第10届创新杯四年级竞赛初赛第2题6分"
] | 1 | single_choice | 整数$$2012$$具有如下的性质:它是$$4$$的倍数,它的各位数字和为$$5$$.在具有这两个性质的整数中,按由小到大顺序排列,$$2012$$是第个. | [
[
{
"aoVal": "A",
"content": "$$9$$ "
}
],
[
{
"aoVal": "B",
"content": "$$10$$ "
}
],
[
{
"aoVal": "C",
"content": "$$11$$ "
}
],
[
{
"aoVal": "D",
"content": "$$12$$ "
}
]
] | [
"拓展思维->能力->公式记忆->符号化数学原理"
] | [
"计数------分类枚举. 一位数:$$0$$个, 两位数:$$32$$,共$$1$$个, 三位数:$$104$$、$$140$$、$$212$$、$$320$$、$$500$$,共$$5$$个, 四位数:$$1004$$、$$1040$$、$$1112$$、$$1220$$、$$1400$$、$$2012\\cdots $$ 故$$2012$$为第$$12$$个. 故选$$\\text{D}$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2077 | dd94a0a3269843598a48014e32466c04 | [
"2018年湖北武汉新希望杯五年级竞赛训练题(五)第2题"
] | 2 | single_choice | 现有甲、乙两个仓库,甲仓库的货物是乙仓库的$$2.5$$倍.如果甲仓库减少$$100$$吨,乙仓库增加$$80$$吨,则两个仓库的货物一样重,两个仓库的货物一共有吨. | [
[
{
"aoVal": "A",
"content": "$$400$$ "
}
],
[
{
"aoVal": "B",
"content": "$$420$$ "
}
],
[
{
"aoVal": "C",
"content": "$$440$$ "
}
],
[
{
"aoVal": "D",
"content": "$$460$$ "
}
]
] | [
"拓展思维->能力->实践应用"
] | [
"甲仓库比乙仓库多$$100+80=180$$(吨),乙仓库:$$180\\div (2.5-1)=120$$(吨),甲仓库:$$120+180=300$$(吨),$$120+300=420$$(吨). "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2973 | a9b4639778b54679972496c3c1d0f969 | [
"2014年第10届全国新希望杯五年级竞赛复赛第4题"
] | 2 | single_choice | 欣欣、希希、望望和贝贝四人共给希望小学捐赠了$$49$$本课外书,其中欣欣捐赠的课外书本数是希希和望望捐赠的课外书本数之和,希希捐赠的课外书本数是望望和贝贝捐赠的课件书本数之和.如果贝贝捐赠了$$3$$本课外书,那么欣欣捐赠了(~~~ )本课外书. | [
[
{
"aoVal": "A",
"content": "$$10$$ "
}
],
[
{
"aoVal": "B",
"content": "$$13$$ "
}
],
[
{
"aoVal": "C",
"content": "$$20$$ "
}
],
[
{
"aoVal": "D",
"content": "$$23$$ "
}
]
] | [
"拓展思维->思想->方程思想"
] | [
"欣+希+望+贝=$$49$$ 欣=希+望=$$2$$望+$$3$$ 希=望+贝=望+$$3$$ 则$$4$$望+$$9=49$$,所以望=$$10$$, 那么欣欣捐了$$2\\times 10+3=23$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1670 | 9a84c66d5f004d55a4ba4ca886da2658 | [
"2020年新希望杯六年级竞赛(2月)第17题"
] | 1 | single_choice | 在一个奇怪的动物村庄里只住着猫和狗,有$$20 \% $$的狗认为它们自己是猫,有$$20 \% $$的猫认为它们自己是狗,其余的猫和狗都是正常的,所有的猫和狗中,有$$36 \% $$认为自己是猫,如果这个奇怪的动物村庄里狗比猫多$$140$$只,那么狗有只. | [
[
{
"aoVal": "A",
"content": "$$40$$ "
}
],
[
{
"aoVal": "B",
"content": "$$180$$ "
}
],
[
{
"aoVal": "C",
"content": "$$80$$ "
}
],
[
{
"aoVal": "D",
"content": "$$220$$ "
}
]
] | [
"拓展思维->能力->实践应用"
] | [
"$$80 \\%$$的猫认为自己是猫,$$20 \\%$$的狗认为自己是猫,猫狗混合后有$$36 \\%$$的认为自己是猫,所以狗:猫$$=\\left(80 \\%-36 \\%\\right):\\left(36 \\%-20 \\%\\right)=11:4$$, 狗:$$140\\div \\left( 11-4 \\right)\\times 11=220$$(只). "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 765 | 68c7685cb8aa4d418419879aae54f2c9 | [
"2021年第8届鹏程杯五年级竞赛初赛第22题4分"
] | 2 | single_choice | $$\underbrace{99\cdots99}_{2021个}\times\underbrace{99\cdots99}_{2021个}+1\underbrace{99\cdots99}_{2021个}$$,能被$$10^{}n$$整除,不能被$$10^{n+1}$$整除,则$$n=$$. | [
[
{
"aoVal": "A",
"content": "$$2020$$ "
}
],
[
{
"aoVal": "B",
"content": "$$2021$$ "
}
],
[
{
"aoVal": "C",
"content": "$$4040$$ "
}
],
[
{
"aoVal": "D",
"content": "$$4042$$ "
}
],
[
{
"aoVal": "E",
"content": "以上都不对 "
}
]
] | [
"拓展思维->拓展思维->数论模块->整除->整除特征->整除特征综合"
] | [
"$$\\underbrace{99\\cdots99}_{2021个}\\times\\underbrace{99\\cdots99}_{2021个}+1\\underbrace{99\\cdots99}_{2021个}$$ $$=\\underbrace{99\\cdots99}_{2021个}\\times\\underbrace{99\\cdots99}_{2021个}+1\\underbrace{00\\cdots00}_{2021个}+\\underbrace{99\\cdots99}_{2021个}$$ $$=\\underbrace{99\\cdots99}_{2021个}\\times(\\underbrace{99\\cdots99}_{2021个}+1)+1\\underbrace{00\\cdots00}_{2021个}$$ $$=\\underbrace{99\\cdots99}_{2021个}\\times 1\\underbrace{00\\cdots00}_{2021个}+1\\underbrace{00\\cdots00}_{2021个}$$ $$=(\\underbrace{99\\cdots99}_{2021个}+1)\\times 1\\underbrace{00\\cdots00}_{2021个}$$ $$=1\\underbrace{00\\cdots00}_{2021个}+\\times 1\\underbrace{00\\cdots00}_{2021个}$$ $$=1\\underbrace{00\\cdots00}_{4042个}$$ $$n$$最大取$$4042$$. 故选:$$\\text{D}$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1698 | 9611ffc45cfa43d1b9383b16546a1b10 | [
"2009年全国迎春杯小学中年级四年级竞赛初赛第2题"
] | 2 | single_choice | 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. | [
[
{
"aoVal": "A",
"content": "$$48$$ "
}
],
[
{
"aoVal": "B",
"content": "$$50$$ "
}
],
[
{
"aoVal": "C",
"content": "$$54$$ "
}
],
[
{
"aoVal": "D",
"content": "$$56$$ "
}
]
] | [
"拓展思维->能力->实践应用"
] | [
"同倍数变化问题.把$$1$$个田格本和$$5$$个练习本捆绑成一组,那么每发$$3$$个横格本,就发一组田格本和练习本,田格本和练习本的总量是横线本的$$2$$倍,每次发的数量也是$$2$$倍,所以剩下的也是$$2$$倍,即$$48$$本. "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1624 | 52ca58de9a024ff3b74c267e240c7916 | [
"2013年第9届全国新希望杯小学高年级五年级竞赛复赛第1题"
] | 1 | single_choice | 在期末考试中,方方五科的总分是$$445$$分,除数学外的四科平均分是$$87.25$$分,方方的数学是(~ ~ ~ ~). | [
[
{
"aoVal": "A",
"content": "$$95$$分 "
}
],
[
{
"aoVal": "B",
"content": "$$96$$分 "
}
],
[
{
"aoVal": "C",
"content": "$$97$$分 "
}
],
[
{
"aoVal": "D",
"content": "$$98$$分 "
}
]
] | [
"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"
] | [
"$$445-87.25\\times 4=96$$.~ "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 967 | f5048b4ef467401b925e9f27b852ece2 | [
"2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛决赛第7题5分"
] | 1 | single_choice | 在整数$$1$$、$$2$$、$$3$$、$$\cdots \cdots 999$$、$$1000$$中,所有偶数之和比所有奇数之和多. | [
[
{
"aoVal": "A",
"content": "$$50$$ "
}
],
[
{
"aoVal": "B",
"content": "$$100$$ "
}
],
[
{
"aoVal": "C",
"content": "$$500$$ "
}
],
[
{
"aoVal": "D",
"content": "$$1000$$ "
}
]
] | [
"拓展思维->能力->逻辑分析"
] | [
"$$1$$到$$1000$$中,一共有$$500$$个奇数,$$500$$个偶数, 奇数之和$$=1+3+5+\\cdots \\cdots +997+999$$, 偶数之和$$=2+4+6+\\cdots \\cdots +998+1000$$, 可以用比每个奇数大$$1$$的偶数去减该奇数,正好所有的奇数都可以找到唯一比它大$$1$$的偶数, 所有偶数之和比所有奇数之和多$$500$$, 故选$$\\text{C}$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3143 | 09ef30e2f87641fa950e0e8d329ac6e7 | [
"2013年小机灵杯三年级竞赛初赛"
] | 2 | single_choice | 罗马数字是由罗马人发明的,它一共由( )个数字组成。 | [
[
{
"aoVal": "A",
"content": "$$5$$ "
}
],
[
{
"aoVal": "B",
"content": "$$6$$ "
}
],
[
{
"aoVal": "C",
"content": "$$7$$ "
}
]
] | [
"拓展思维->拓展思维->计数模块->加乘原理->加法原理"
] | [
"解:罗马数字是由罗马人发明的,它一共由$$7$$个数字组成; 故选:C。 "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1232 | 4ff40ea4baa74c22af81246b414424fc | [
"六年级竞赛创新杯"
] | 1 | single_choice | 有两根同样长的钢管,第一根用去$$\frac{3}{10}$$米,第二根用去$$\frac{3}{10}$$,比较这两根钢管剩下部分的长度,结果是( ) | [
[
{
"aoVal": "A",
"content": "第一根钢管剩下的部分长些 "
}
],
[
{
"aoVal": "B",
"content": "第二根钢管剩下的部分长些 "
}
],
[
{
"aoVal": "C",
"content": "两根钢管剩下的部分同样长 "
}
],
[
{
"aoVal": "D",
"content": "以上三种结果都有可能 "
}
]
] | [
"拓展思维->拓展思维->应用题模块->分百应用题->认识单位1"
] | [
"当两根钢管都是$$1$$米时,则剩下的钢管相等;当两根钢管都大于$$1$$米时,则用去$$\\frac{3}{10}$$米的那一根剩下的多;当两根钢管都小于$$1$$米时,用去$$\\frac{3}{10}$$的那一根剩下的多。 "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3293 | defb04cc83bc444ab4855ff5c37e78f5 | [
"2017年全国华杯赛小学高年级竞赛初赛模拟"
] | 1 | single_choice | 六个小朋友排成一排照相,其中有四个男生和两个女生,两个女生必须站在一起而且不能站在边上,则一共有~\uline{~~~~~~~~~~}~种不同的排列方式. | [
[
{
"aoVal": "A",
"content": "$$96$$ "
}
],
[
{
"aoVal": "B",
"content": "$$120$$ "
}
],
[
{
"aoVal": "C",
"content": "$$144$$ "
}
],
[
{
"aoVal": "D",
"content": "$$240$$ "
}
]
] | [
"拓展思维->能力->抽象概括"
] | [
"$$\\text{P}_{4}^{4}\\times \\text{P}_{3}^{2}=4\\times 3\\times 2\\times 3\\times 2=144$$. 另一种方法:先捆绑,再插空,$$2!\\times 4!\\times C_{4}^{3}=144$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3208 | 3037baf8c9944deb9eb8974698a96b9b | [
"2017年全国美国数学大联盟杯小学中年级四年级竞赛初赛"
] | 1 | single_choice | 一共有四种硬币,penny(一美分),nickel(五美分),dime(十美分),quarter(二十五美分),每种硬币你可以取任意多个(也可以是$$0$$个)放在你的袋子里(在放这些硬币之前,你的袋子是空的),规则是你袋子例的硬币无法凑成一美元(一美元$$=100$$美分).你的袋子里最多可以有多少钱? | [
[
{
"aoVal": "A",
"content": "$$$0.90$$ "
}
],
[
{
"aoVal": "B",
"content": "$$$0.99$$ "
}
],
[
{
"aoVal": "C",
"content": "$$$1.19$$ "
}
],
[
{
"aoVal": "D",
"content": "$$$1.29$$ "
}
]
] | [
"拓展思维->能力->实践应用"
] | [
"$$3$$个quarter(二十五美分)$$+4$$个dime(十美分)$$+4$$个penny(一美分). "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3033 | c12412b0521b420e8b2803884a2ea2db | [
"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"
] | 1 | single_choice | 如果甲数$$\times 1.2=$$乙数$$\div 1.2$$(甲数、乙数$$\ne 0$$),那么. | [
[
{
"aoVal": "A",
"content": "甲数$$=$$乙数 "
}
],
[
{
"aoVal": "B",
"content": "甲数$$\\textgreater$$乙数 "
}
],
[
{
"aoVal": "C",
"content": "甲数$$\\textless{}$$乙数 "
}
]
] | [
"知识标签->拓展思维->计算模块->小数->小数乘除->小数除法运算"
] | [
"$$1.2=\\frac{6}{5}$$,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. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 345 | 84d92520b8804f53ab2be91393d55dd4 | [
"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"
] | 2 | single_choice | 有$$20$$间房间,有的开着灯,有的关着灯,在这些房间里的人都希望与大多数房间保持一致。现在,从第一间房间的人开始,如果其余$$19$$间房间的灯开着的多,就把灯打开,否则就把灯关上,如果最开始开灯与关灯的房间各$$10$$间,并且第一间的灯开着。那么, 这$$20$$间房间里的人轮完一遍后,关着灯的房间有( )间。 | [
[
{
"aoVal": "A",
"content": "$$20$$ "
}
],
[
{
"aoVal": "B",
"content": "$$10$$ "
}
],
[
{
"aoVal": "C",
"content": "$$11$$ "
}
],
[
{
"aoVal": "D",
"content": "$$0$$ "
}
]
] | [
"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"
] | [
"解:因为最开始开灯和关灯的房间各是$$10$$间,由于第一间的灯是开着的, 所以,第一间人看到的,开灯的$$9$$间,关灯的$$10$$间, 之后,他就关灯, 以后无论开灯的出来看,还是关灯的出来看,始终关灯的多, 即:一轮结束,灯全部会关闭。 故选:A。 "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2835 | a86f33a95fc24cfebfd1f16c2b4b7378 | [
"2018年湖北武汉新希望杯五年级竞赛训练题(三)第1题"
] | 0 | single_choice | 下列说法正确的是. | [
[
{
"aoVal": "A",
"content": "一个分数的分母越小,它的分数单位就越小 "
}
],
[
{
"aoVal": "B",
"content": "分数都比整数小 "
}
],
[
{
"aoVal": "C",
"content": "假分数都大于$$1$$ "
}
],
[
{
"aoVal": "D",
"content": "$$2$$米的$$\\frac{1}{3}$$和$$1$$米的$$\\frac{2}{3}$$一样长 "
}
]
] | [
"拓展思维->能力->逻辑分析"
] | [
"$$2$$米的$$\\frac{1}{3}$$和$$1$$米的$$\\frac{2}{3}$$一样长. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2967 | c4d0b3b5732949fca460651b6e45a280 | [
"2016年全国学而思杯小学高年级六年级竞赛"
] | 2 | single_choice | 找规律:有一组式子:$$a^{2}$$,$$-\dfrac{{{a}^{3}}}{2}$$,$$\dfrac{{{a}^{4}}}{3}$$,$$-\dfrac{{{a}^{5}}}{4}$$,$$\cdots$$,从左到右数的第$$10$$个式子是下面算式的第( ~ ~ ~ ~)个. | [
[
{
"aoVal": "A",
"content": "$$\\dfrac{{{a}^{11}}}{10}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$-\\dfrac{{{a}^{11}}}{10}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$-\\dfrac{{{a}^{10}}}{11}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$-\\dfrac{{{a}^{11}}}{11}$$ "
}
]
] | [
"拓展思维->思想->逐步调整思想"
] | [
"分别观察分子、分母的规律,以及符号的规律.从左到右数的第$$10$$个式子是$$-\\dfrac{{{a}^{11}}}{10}$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 802 | 5bc30212ca964e3bbd97831632eea516 | [
"2018年第6届湖北长江杯五年级竞赛初赛A卷第7题3分"
] | 2 | single_choice | 一块砖的长、宽、高分别为$$18$$厘米、$$12$$厘米、$$6$$厘米,要堆成一个正方体,至少需要这样的砖块. | [
[
{
"aoVal": "A",
"content": "$$21$$ "
}
],
[
{
"aoVal": "B",
"content": "$$28$$ "
}
],
[
{
"aoVal": "C",
"content": "$$14$$ "
}
],
[
{
"aoVal": "D",
"content": "$$36$$ "
}
]
] | [
"拓展思维->拓展思维->几何模块->立体图形->长方体与正方体->长方体与正方体基本概念运用->正方体的基本概念"
] | [
"正方体的棱长是$$18$$、$$12$$、$$6$$的倍数, 其中$$36$$是它们的最小公倍数, $$36\\div 18=2$$(层), $$36\\div 12=3$$(层), $$36\\div 6=6$$(层), 需要砖块为:$$2\\times 3\\times 6=36$$(块). 故答案选:$$\\text{D}$$. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2465 | 11699cb51a6c4589b2bf092fe2ae694d | [
"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"
] | 1 | single_choice | $$9\div 11$$的商的小数部分第$$50$$位上的数字是. | [
[
{
"aoVal": "A",
"content": "$$8$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1$$ "
}
],
[
{
"aoVal": "C",
"content": "$$9$$ "
}
]
] | [
"课内体系->知识点->数的认识->认、读、写数->小数->循环小数",
"拓展思维->思想->对应思想"
] | [
"列竖式 故$$9\\div 11=0.\\dot{8}\\dot{1}$$. 周期为$$2$$故$$50\\div 2=25$$, 故第$$50$$位为$$1$$. 故答案为:$$\\text{B}$$选项. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1129 | 98beb05ae828404c9bffb7dbea15e240 | [
"2017年河南郑州豫才杯四年级竞赛初赛第14题"
] | 1 | single_choice | 在班级图书角里,聪聪特别喜欢《哈利波特与魔法石》,字数总计达$$200$$千字,他打算在国庆的$$7$$天假期里看完,他的阅读速度可以达到每分钟$$300$$字.按照这个计划,他平均每天要读(~ )才能保证读完这本书. | [
[
{
"aoVal": "A",
"content": "$$10$$分钟~~~ "
}
],
[
{
"aoVal": "B",
"content": "$$50$$分钟~~~ "
}
],
[
{
"aoVal": "C",
"content": "$$1$$小时$$30$$分钟 "
}
],
[
{
"aoVal": "D",
"content": "$$100$$分钟 "
}
]
] | [
"拓展思维->拓展思维->应用题模块->归一归总问题->双归一问题"
] | [
"$$200000\\div 300\\div 7\\approx 100$$分钟. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 768 | ff8080814518d5240145192480f704c5 | [
"2014年全国迎春杯五年级竞赛初赛第8题",
"2014年全国迎春杯六年级竞赛初赛第9题"
] | 2 | single_choice | 已知$$4$$个质数的积是它们和的$$11$$倍,则它们的和为(~~~~~~~ ). | [
[
{
"aoVal": "A",
"content": "$$46$$ "
}
],
[
{
"aoVal": "B",
"content": "$$47$$ "
}
],
[
{
"aoVal": "C",
"content": "$$48$$ "
}
],
[
{
"aoVal": "D",
"content": "没有符合条件的数 "
}
]
] | [
"拓展思维->思想->逐步调整思想"
] | [
"由已知条件,$$4$$个质数中一定有$$11$$,那么则满足$$a\\times b\\times c=a+b+c+11$$,其中$$a$$、$$b$$、$$c$$都是质数.若$$a$$、$$b$$、$$c$$都是奇数,那么等式左边是奇数,右边为偶数,矛盾.若$$a$$、$$b$$、$$c$$中有$$1 $$个偶数,那么一定是$$2$$.即$$a\\times b\\times 2=a+b+2+11$$,此时,根据奇偶性,$$a$$、$$b$$中也必有一个偶数为$$2$$,解得$$a$$、$$b$$、$$c$$、$$d$$为$$2$$、$$2$$、$$5$$、$$11$$和为$$20$$.选项中ABC均不符合条件,故选D. "
] | D |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2120 | ec11d2d42e05470a851a48ef537807ab | [
"2015年第13届全国创新杯六年级竞赛第6题",
"小学高年级六年级其它2015年数学思维能力等级测试初试第6题4分"
] | 1 | single_choice | 六年级(一)班三名同学,结伴骑自行车从学校出发到东湖磨山春游,已经走了全程的$$\frac{2}{7}$$,如果再行$$9$$千米,已行路程和剩下路程之比为$$5:2$$,那学校到东湖磨山的路程是(~ )千米. | [
[
{
"aoVal": "A",
"content": "$$21$$ "
}
],
[
{
"aoVal": "B",
"content": "$$22$$ "
}
],
[
{
"aoVal": "C",
"content": "$$24$$ "
}
],
[
{
"aoVal": "D",
"content": "$$28$$ "
}
]
] | [
"拓展思维->能力->运算求解"
] | [
"第一次走了$$\\frac{2}{7}$$,第二次一共走了$$\\frac{5}{7}$$,则全程是$$9\\div \\left( \\frac{5}{7}-\\frac{2}{7} \\right)=21$$千米. "
] | A |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1738 | 7312aff182924b72a2092744292a6f02 | [
"2018年IMAS小学中年级竞赛(第一轮)第12题4分",
"2018年IMAS小学高年级竞赛(第一轮)第3题3分"
] | 1 | single_choice | \uline{小杨}在期末考试中﹐语文与数学两科的平均分数为$$97$$分,而英语考了$$94$$分,请问他三科的平均分数为多少分?(~ ) | [
[
{
"aoVal": "A",
"content": "$$94$$ "
}
],
[
{
"aoVal": "B",
"content": "$$94.5$$ "
}
],
[
{
"aoVal": "C",
"content": "$$95$$ "
}
],
[
{
"aoVal": "D",
"content": "$$95.5$$ "
}
],
[
{
"aoVal": "E",
"content": "$$96$$ "
}
]
] | [
"拓展思维->思想->对应思想"
] | [
"可知\\uline{小杨}在期末考试中,语文与数学两科的总分为$$97\\times 2=194$$分, 所以三科的总分为$$194+94=288$$分, 即三科的平均分为$$288\\div 3=96$$分. 故选$$\\text{E}$$. ",
"<p>可知<u>小杨</u>在期末考试中,语文与数学两科的平均分数为$$97$$分,而英语考了$$94$$分,</p>\n<p>所以语文与数学两科总共比英语多了$$\\left( 97-94 \\right)\\times 2=6$$分,</p>\n<p>所以三科的总平均分数比英语多了$$6\\div 3=2$$分,</p>\n<p>即三科的平均分为$$94+2=96$$分.</p>\n<p>故选$$\\text{E}$$.</p>"
] | E |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 1246 | f5b9913961ec42b888bac23828b8b3f7 | [
"2013年第25届广东广州五羊杯六年级竞赛第10题5分"
] | 2 | single_choice | 莉莉在期末考试中语数英的平均分是$$85$$分,已知她三科的成绩均不相同,且分数都是整数.最高分的科目语文比最低分的科目英语高$$4$$分,则莉莉的英语考了分. | [
[
{
"aoVal": "A",
"content": "$$82$$ "
}
],
[
{
"aoVal": "B",
"content": "$$83$$ "
}
],
[
{
"aoVal": "C",
"content": "$$84$$ "
}
],
[
{
"aoVal": "D",
"content": "$$85$$ "
}
]
] | [
"拓展思维->能力->逻辑分析"
] | [
"三科总分为$$85\\times 3=255$$(分). 设最低分的科目英语为$$x$$分, 则有$$x+4\\textgreater255-2x-4\\textgreater x$$, 解得$$\\begin{cases}x\\textless{}84 x\\textgreater82 \\end{cases}$$, 所以$$x=83$$(分). "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2993 | 8f2a4f3724184d30a2ce768c0824fa02 | [
"2017年第15届全国希望杯六年级竞赛"
] | 3 | single_choice | 计算$$\frac{1}{1\times 2\times 3}+\frac{1}{2\times 3\times 4}+\frac{1}{3\times 4\times 5}+...+\frac{1}{8\times 9\times 10}$$=~\uline{~~~~~~~~~~}~. | [
[
{
"aoVal": "A",
"content": "$$\\frac{9}{10}$$ "
}
],
[
{
"aoVal": "B",
"content": "$$\\frac{9}{20}$$ "
}
],
[
{
"aoVal": "C",
"content": "$$\\frac{11}{45}$$ "
}
],
[
{
"aoVal": "D",
"content": "$$\\frac{22}{45}$$ "
}
]
] | [
"拓展思维->能力->归纳总结->归纳推理"
] | [
"因为$$\\frac{1}{n\\left( n+1 \\right)\\left( n+2 \\right)}=\\frac{1}{2}\\times \\frac{\\left( n+2 \\right)-n}{n\\left( n+1 \\right)\\left( n+2 \\right)}=\\frac{1}{2}\\left[ \\frac{1}{n\\left( n+1 \\right)}-\\frac{1}{\\left( n+1 \\right)\\left( n+2 \\right)} \\right]$$ 所以原式$$=\\frac{1}{2}\\times \\left( \\frac{1}{1\\times 2}-\\frac{1}{2\\times 3}+\\frac{1}{2\\times 3}-\\frac{1}{3\\times 4}+\\frac{1}{3\\times 4}-\\frac{1}{4\\times 5}...+\\frac{1}{8\\times 9}-\\frac{1}{9\\times 10} \\right)$$ $$=\\frac{1}{2}\\times \\left[ \\frac{1}{2}-\\frac{1}{90} \\right]=\\frac{11}{45}$$. "
] | C |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 3229 | 3556a2d16ab448259b1b8ebf87fd2740 | [
"2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第5题5分"
] | 1 | single_choice | $$600$$的因数有很多,这些因数中有个是$$6$$的倍数. | [
[
{
"aoVal": "A",
"content": "$$6$$ "
}
],
[
{
"aoVal": "B",
"content": "$$9$$ "
}
],
[
{
"aoVal": "C",
"content": "$$12$$ "
}
],
[
{
"aoVal": "D",
"content": "$$15$$ "
}
]
] | [
"拓展思维->拓展思维->数论模块->因数与倍数->因数与倍数基础"
] | [
"$$600$$的因数有$$1$$,$$2$$,$$3$$,$$4$$,$$5$$,$$6$$,$$8$$,$$10$$,$$12$$,$$15$$,$$20$$,$$24$$,$$25$$,$$30$$,$$40$$,$$50$$,$$60$$,$$75$$,$$100$$,$$120$$,$$150$$,$$200$$,$$300$$,$$600$$,其中是$$6$$的倍数的有$$6$$,$$12$$,$$24$$,$$30$$,$$60$$,$$120$$,$$150$$,$$300$$,$$600$$这$$9$$个. 故选$$\\text{B}$$. "
] | B |
prime_math_competition_ch_single_choice_3.2K_dev | "2023-07-07T00:00:00" | 2582 | 2cb5f63a7a8641b09383b4c0857b61a8 | [
"2019年广东深圳全国小学生数学学习能力测评六年级竞赛初赛第7题3分"
] | 1 | single_choice | 糖水质量是$$900$$克,糖与水的质量比是$$1:9$$,若再加入$$100$$克糖,则糖与水的质量比是. | [
[
{
"aoVal": "A",
"content": "$$9:11$$ "
}
],
[
{
"aoVal": "B",
"content": "$$1:10$$ "
}
],
[
{
"aoVal": "C",
"content": "$$19:81$$ "
}
],
[
{
"aoVal": "D",
"content": "$$19:100$$ "
}
]
] | [
"拓展思维->能力->逻辑分析"
] | [
"依题意得,糖水质量是$$900$$克,糖与水的质量比是$$1:9$$,那么糖的质量是$$90$$克,水的质量是$$900-90=810$$(克),又加入了$$100$$克糖,现在糖与水的质量比是:$$(90+100):810=190:810=19:81$$. 故选$$\\text{C}$$. "
] | C |
TAL-SCQ5K
Dataset Description
Dataset Summary
TAL-SCQ5K-EN/TAL-SCQ5K-CN are high quality mathematical competition datasets in English and Chinese language created by TAL Education Group, each consisting of 5K questions(3K training and 2K testing). The questions are in the form of multiple-choice and cover mathematical topics at the primary,junior high and high school levels. In addition, detailed solution steps are provided to facilitate CoT training and all the mathematical expressions in the questions have been presented as standard text-mode Latex.
Supported Tasks and Leaderboards
[More Information Needed]
Languages
The text in TAL-SCQ5K-EN is in English and TAL-SCQ5K-CN is in Chinese.
Dataset Structure
Data Instances
{
"dataset_name": "prime_math_competition_en_single_choice_8K_dev",
"dataset_version": "2023-07-07",
"qid": "244",
"queId": "8afc802a8c304199b1040f11ffa2e92a",
"competition_source_list": [],
"difficulty": "2",
"qtype": "single_choice",
"problem": "A $14$-digit. number $666666 XY 444444$ is a multiple of $26$. If $X$ and $Y$ are both positive, what is the smallest vaue of $X+ Y$? ",
"answer_option_list": [
[{
"aoVal": "A",
"content": "$$3$$ "
}],
[{
"aoVal": "B",
"content": "$$4$$ "
}],
[{
"aoVal": "C",
"content": "$$9$$ "
}],
[{
"aoVal": "D",
"content": "$$14$$ "
}],
[{
"aoVal": "E",
"content": "None of the above "
}]
],
"knowledge_point_routes": ["Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules"],
"answer_analysis": ["Since $1001$ is a multiple of $13$, $111111 = 111 \\times 1001$ is also a multiple of $13$. It follows that both $666666$ and $444444$ are both multiples of $26$. $666666XY 444444 = 66666600000000 + XY 000000 + 444444$ $\\Rightarrow XY$ must be divisible by $13$. Smallest $X+Y=1+3=4$. "],
"answer_value": "B"
}
Data Fields
- "dataset_name": identification of the source dataset name from which TAL-SCQ5K-EN/TAL-SCQ5K-CN has been created, use only for inner of TAL education group, please ignore.
- "dataset_version": identification of the source dataset version from which TAL-SCQ5K-EN/TAL-SCQ5K-CN has been created, use only for inner of TAL education group, please ignore.
- "qid": identification of local id of the question in the source dataset from which TAL-SCQ5K-EN/TAL-SCQ5K-CN has been created, use only for inner of TAL education group, please ignore.
- "queId": identification of global id of the question, use only for inner of TAL education group, please ignore.
- "competition_source_list": identification of math competitions in which the questions appeared, if have been logged.
- "difficulty": difficulty level of the questions, value ranged from 0 to 4
- "qtype": question type, valued as "single_choice" for all the questions in this dataset indicates that all the questions are multiple-choice questions with unique ground-truth answer.
- "problem": the question string to a math competition question.
- "answer_option_list": answer choices to be selected
- "knowledge_point_routes": knowledge point route from coarse-grained to fine-grained.
- "answer_analysis": step-by-step answer analysis of the questions, which helps CoT training
- "answer_value": value of the ground-truth answer choice
Data Splits
name | train | test |
---|---|---|
TAL-SCQ5K-EN | 3K | 2K |
TAL-SCQ5K-CN | 3K | 2K |
Usage
Each of the above datasets is located in a separate sub-directory. To load an individual subset, use the data_dir argument of the load_dataset() function as follows:
from datasets import load_dataset
# Load all subsets (share the same schema)
dataset = load_dataset("math-eval/TAL-SCQ5K")
# Load TAL-SCQ5K-EN
dataset = load_dataset("math-eval/TAL-SCQ5K", data_dir="TAL-SCQ5K-EN")
# Load TAL-SCQ5K-CN
dataset = load_dataset("math-eval/TAL-SCQ5K", data_dir="TAL-SCQ5K-CN")
Additional Information
Dataset Curators
[More Information Needed]
Licensing Information
The TAL-SCQ5K dataset is licensed under the MIT License
Citation Information
[More Information Needed]
Contact
The original authors host this dataset on GitHub here: https://github.com/math-eval/TAL-SCQ5K You can submit inquiries to: matheval.ai@gmail.com
- Downloads last month
- 352