Datasets:

math-eval

/

TAL-SCQ5K

like 45

"2023-07-07T00:00:00"

[ "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" ]

[ [ { "aoVal": "A", "content": "$$216$$ " } ], [ { "aoVal": "B", "content": "$$324$$ " } ], [ { "aoVal": "C", "content": "$$273$$ " } ], [ { "aoVal": "D", "content": "$$301$$ " } ] ]

[ "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" ]

[ "每只猴子多分了$$3$$个，分$$5\\times 9+\\left( 9-3 \\right)+57=108$$（个），那么共$$108\\div 3=36$$（只）猴子，共$$36\\times 6+57=273$$（个）桃子。 " ]

"2023-07-07T00:00:00"

[ "2006年四年级竞赛创新杯", "2006年第4届创新杯四年级竞赛复赛第4题" ]

[ [ { "aoVal": "A", "content": "红桃A " } ], [ { "aoVal": "B", "content": "红桃K " } ], [ { "aoVal": "C", "content": "黑桃A " } ], [ { "aoVal": "D", "content": "黑桃K " } ] ]

[ "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" ]

[ "从左至右，顺次有三张牌，编号为①②③由第一、二个条件得①为A，②为K，③为K.因为，黑桃左边有两张可得③为黑桃，即可得出③为黑桃K，且①、②中至少有一张是红桃.因为红桃右边有两张，可得①为红桃，即得出①为红桃A，且②、③中至少有一张红桃.因为，③为黑桃K，所以②为红桃.因此得出②为红桃K，即中间的一张为红桃K.故选B " ]

"2023-07-07T00:00:00"

[ "2011年四年级其它", "2016年第3届广东深圳鹏程杯六年级竞赛集训材料第十五章综合训练题（五）第4题" ]

[ [ { "aoVal": "A", "content": "$$23$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$25$$ " } ], [ { "aoVal": "D", "content": "$$26$$ " } ] ]

[ "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" ]

[ "每千克$$25$$元和每千克$$30$$元的糖果共收入了$$1970$$元，则每千克$$20$$元的收入：$$2570-1970=600$$元， 所以卖出：$$600\\div 20=30$$千克， 所以卖出每千克$$25$$元和每千克$$30$$克的糖果共$$100-30=70$$千克， 相当于将题目转换成： 卖出每千克$$25$$元和每千克$$30$$克的糖果共$$70$$千克，收入$$1970$$元，问：每千克$$25$$元的糖果售出了多少千克？ 转换成了最基本的鸡兔同笼问题． 假设全是每千克$$25$$元的，$$\\left( 1970-25\\times 70 \\right)\\div \\left( 30-25\\right)=44$$（千克）， 所以$$30$$元的是$$44$$千克，所以$$25$$元的有：$$70-44=26$$（千克）． 关键：将三种以及更多的动物/东西，转化为两种最基本模型，即：抓住转化后的``头''与``脚''． " ]

"2023-07-07T00:00:00"

[ "1993年四年级竞赛创新杯", "2006年第4届创新杯四年级竞赛初赛B卷第7题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "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" ]

[ "无论赵的哪半句对了，钱的后半句一定错了，那前半句一定对了，丙是$$4$$号。 " ]

"2023-07-07T00:00:00"

[ "2020年广东广州羊排赛六年级竞赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{5}{7}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{13}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{4}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{10}{17}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数化小数法" ]

[ "$$\\frac{5}{7}=0.714285714285\\cdots \\cdots $$，$$\\frac{2}{13}=0.153846153846\\cdots \\cdots $$， $$\\frac{3}{4}=0.75$$，$$\\frac{10}{17}=0.588235294\\cdots \\cdots $$， $$0.75\\textgreater0.714285714285\\cdots \\cdots \\textgreater0.588235294\\cdots \\cdots \\textgreater$$ $$0.153846153846\\cdots \\cdots $$，所以$$\\frac{3}{4}\\textgreater\\frac{5}{7}\\textgreater\\frac{10}{17}\\textgreater\\frac{2}{13}$$， 则从大到小排列排在第二位的是$$\\frac{5}{7}$$． 故选$$\\text{A}$$． " ]

"2023-07-07T00:00:00"

[ "2021年鹏程杯六年级竞赛初赛第5题" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$100$$ " } ], [ { "aoVal": "C", "content": "$$400$$ " } ], [ { "aoVal": "D", "content": "$$2020$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$2$$，$$4$$，$$6$$，$$8$$这$$4$$个数字，每个在个位出现$$10$$次，在十位出现$$10$$次， 所以$$E(1)+E(2)+···+E(100)=(2+4+6+8)\\times (10+10)=400$$． 故选$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2011年全国美国数学大联盟杯小学高年级五年级竞赛初赛第34题" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$27$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->因数与倍数基础" ]

[ "最大完全平方数是$$36$$，最小完全平方数是$$1$$． " ]

"2023-07-07T00:00:00"

[ "2021年新希望杯六年级竞赛初赛第19题5分" ]

[ [ { "aoVal": "A", "content": "$$44$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$28$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "假设这一批产品每件成本为$$1$$元，一共有$$100$$件，总成本为$$1\\times 100=100$$（元）， 原价：$$1\\times (1+100 \\%)=2$$（元）， 打折后：$$2\\times 0.6=1.2$$（元）， 原价售出$$30 \\%$$：$$2\\times 100\\times 30 \\%=60$$（元）， 打折后全部售出：$$1.2\\times 100\\times(1-30 \\%)=84$$（元）， 总售价：$$60+84=144$$（元）， 利润率：$$\\frac{144-100}{100}=44 \\%$$． 答：实际利润率为$$44 \\%$$． " ]

"2023-07-07T00:00:00"

[ "2013年第25届广东广州五羊杯六年级竞赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$43$$ " } ], [ { "aoVal": "B", "content": "$$46$$ " } ], [ { "aoVal": "C", "content": "$$49$$ " } ], [ { "aoVal": "D", "content": "$$52$$ " } ] ]

[ "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" ]

[ "设小红今年$$x$$岁．则能列出方程：$$3(x+3)=x+26+3$$， 解出$$x=10$$，即小红今年$$10$$岁，父亲今年$$36$$岁，所以年龄和为$$46$$． " ]

"2023-07-07T00:00:00"

[ "2012年全国学而思杯三年级竞赛第15题" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ], [ { "aoVal": "E", "content": "$$12$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->比较型逻辑推理", "Overseas Competition->知识点->组合模块->逻辑推理" ]

[ "不超过表示等于或小于均可．现在相安无事，说明猫比狗多，鼠比猫多，猫与狗之和比鼠多，从再进来任意$$1$$只，都会打破平衡可知，猫比狗多$$1$$只，鼠比猫多$$1$$只，猫与狗之和比鼠多$$1$$只，经尝试知有$$4$$只鼠，$$3$$只猫，$$2$$只狗． 再进$$1$$只猫，会打破平衡，所以鼠比猫多$$1$$只； 再进$$1$$只狗，会打破平衡，所以猫比狗多$$1$$只； 再进$$1$$只鼠，会打破平衡，所以猫与狗的和比鼠多$$1$$只； 比较第一个条件和第三个条件可知，猫有$$3$$只，进而求得鼠有$$4$$只，狗有$$2$$只． 共有$$4+3+2=9$$（只）小动物． " ]

"2023-07-07T00:00:00"

[ "2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$30$$辆 " } ], [ { "aoVal": "B", "content": "$$60$$辆 " } ], [ { "aoVal": "C", "content": "$$61$$辆 " } ], [ { "aoVal": "D", "content": "$$120$$辆 " } ] ]

[ "拓展思维->七大能力->实践应用" ]

[ "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． " ]

"2023-07-07T00:00:00"

[ "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" ]

[ [ { "aoVal": "A", "content": "$$54$$ " } ], [ { "aoVal": "B", "content": "$$64$$ " } ], [ { "aoVal": "C", "content": "$$52$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$\\left( 79-4 \\right)\\div 3=75\\div 3=25$$（岁）， $$79-25=54$$（岁）， 答：父亲现在的年龄是$$54$$岁． 故选：$$\\text{A}$$． " ]

"2023-07-07T00:00:00"

[ "2014年迎春杯三年级竞赛初赛", "2014年迎春杯四年级竞赛初赛", "2014年迎春杯三年级竞赛初赛", "2014年迎春杯四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$25526250$$ " } ], [ { "aoVal": "B", "content": "$$26650350$$ " } ], [ { "aoVal": "C", "content": "$$27775250$$ " } ], [ { "aoVal": "D", "content": "$$28870350$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法" ]

[ "解：（$$1$$）四个选项都是$$8$$位数； （$$2$$）四选项都是$$25$$的倍数，C的数字和是$$35$$不是$$3$$的倍数，排除C； （$$3$$）都满足条件； （$$4$$）都满足条件； （$$5$$）A百位数字和万位数字相等，D百位数字比万位数字少$$4$$，不满足条件； （$$6$$）B满足条件。 故选：B。 " ]

"2023-07-07T00:00:00"

[ "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" ]

[ [ { "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 " } ] ]

[ "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" ]

[ "选$$\\text{B}$$ 假设甲、乙原来长度都是$$a$$米， 甲剩下($$\\frac{2}{3}a-\\frac{1}{3}$$)米； 乙剩下$$\\frac{2}{3}（a-\\frac{1}{3})$$=$$\\frac{2}{3}a-\\frac{2}{9}$$米 所以乙剩下的部分长. " ]

"2023-07-07T00:00:00"

[ "2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（三）" ]

[ [ { "aoVal": "A", "content": "$$2$$；$$100$$ " } ], [ { "aoVal": "B", "content": "$$4$$；$$100$$ " } ], [ { "aoVal": "C", "content": "$$8$$；$$100$$ " } ], [ { "aoVal": "D", "content": "$$8$$；$$200$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "若通过$$860$$米隧道时速度不变则需要$$120\\times 2=240$$（秒），火车过桥速度：$$\\left( 860-320 \\right)\\div \\left( 240-105 \\right)=4$$（米/秒）：火车车身长：$$105\\times 4-320=100$$（米）． " ]

"2023-07-07T00:00:00"

[ "2014年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之和倍型" ]

[ "已知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；当甲是$$19$$岁时，则乙$$13$$岁，丙的年龄是丁的$$3$$倍（此时丁至少$$1$$岁），因为此时是至少$$4$$年前，所以$$4$$个人的年龄和不超过$$72-4\\times 4=56$$（岁）。 设此时丁的年龄是$$x$$岁，丙的年龄是$$3x$$岁，则$$19+13+x+3x\\leqslant 56$$，解得$$x\\leqslant 6$$，又知道$$x\\geqslant 1$$，所以丁的年龄有$$6$$种情况。 如果甲、乙、丙、丁四个人的年龄互不相同，所以今年甲的年龄可以有$$6$$种情况。 " ]

"2023-07-07T00:00:00"

[ "2016年第3届广东深圳鹏程杯六年级竞赛集训材料第十九章综合训练题（九）第2题" ]

[ [ { "aoVal": "A", "content": "$$147$$ " } ], [ { "aoVal": "B", "content": "$$148$$ " } ], [ { "aoVal": "C", "content": "$$149$$ " } ], [ { "aoVal": "D", "content": "$$150$$ " } ] ]

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "解：阿里读的页数是$$67-54+1=14$$， 同理苏明读的页数是$$135-95+1=41$$， 梅莉阅读的页数是$$273-180+1=94$$， 他们共读$$14+41+94=149$$页． " ]

"2023-07-07T00:00:00"

[ "2010年第11届上海中环杯小学中年级三年级竞赛第11题12分" ]

[ [ { "aoVal": "A", "content": "$$35$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$27$$ " } ], [ { "aoVal": "E", "content": "$$25$$ " } ] ]

[ "拓展思维->能力->逻辑分析", "Overseas Competition->知识点->应用题模块->年龄问题" ]

[ "$$3$$年前，四人年龄之和是$$68-3\\times 4=56$$（岁），但实际上是$$57$$岁，这说明$$3$$ 年前小儿子还没有出生，是$$2$$年前出生的，所以小儿子今年是$$2$$岁．则其余$$3$$人年龄之和是$$66$$岁．$$5$$年前，$$3$$人的年龄和应该是$$66-5\\times 3=51$$（岁），但实际上是$$52$$岁，这说明$$5$$年前大儿子还没有出生，是$$4$$年前出生的，所以大儿子今年是$$4$$岁．则爸爸妈妈的年龄和是$$62$$岁，而爸爸比妈妈大两岁，所以爸爸今年$$32$$岁，妈妈今年$$30$$岁． $3$ years ago from now, their sum of ages should be $$68-3\\times 4=56$$ but actually it was $57$, which means the little brother was not born $3$ years ago. He was born $2$ years ago. Thus, the little brother this year is $2$ years old and the sum of the other three members this year is $68-2=66$ years old. $5$ years ago from now, the sum of the three members ages should be $66-5\\times 3=51$ but actually it was $52$, whcih means the elder brother was not born $5$ years ago. He was born $4$ years ago. Thus, the elder brother is $4$ years old this year. Thus, the sum of ages of mom and dad this year is $66-4=62$. Dad is $2$ years older than mom, which means dad is $32$ years old and mom is $30$ years old. " ]

"2023-07-07T00:00:00"

[ "2016年创新杯六年级竞赛训练题（三）第5题" ]

[ [ { "aoVal": "A", "content": "$$4\\frac{4}{7}$$小时 " } ], [ { "aoVal": "B", "content": "$$6\\frac{4}{7}$$小时 " } ], [ { "aoVal": "C", "content": "$$7\\frac{4}{7}$$小时 " } ], [ { "aoVal": "D", "content": "$$8\\frac{4}{7}$$小时 " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "略 " ]

"2023-07-07T00:00:00"

[ "2019年美国数学大联盟杯竞赛" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "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" ]

[ "$$8$$千克的香蕉是$$12$$美元．$$12$$千克的香蕉是多少钱？ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde12\\div 8\\times 12$$ $$=12\\times 12\\div 8$$ $$=144\\div 8$$ $$=18$$（美元）． 故选$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2017年第17届湖北武汉世奥赛五年级竞赛决赛第8题" ]

[ [ { "aoVal": "A", "content": "像山一样高 " } ], [ { "aoVal": "B", "content": "像平房(一层楼)一样高 " } ], [ { "aoVal": "C", "content": "像成人一样高 " } ], [ { "aoVal": "D", "content": "不足$$1$$米 " } ] ]

[ "拓展思维->拓展思维->计算模块->乘方->乘方的认识" ]

[ "叠纸高$$0.1\\times {{2}^{14}}=1638.4$$（毫米）．像成人一样高． " ]

"2023-07-07T00:00:00"

[ "2019年广东深圳全国小学生数学学习能力测评四年级竞赛初赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "最后读作：九百万零五，所以$$9$$在百万位，$$5$$在个位．故中间需要添$$5$$个$$0$$． 故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2007年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "10 " } ], [ { "aoVal": "B", "content": "11 " } ], [ { "aoVal": "C", "content": "12 " } ], [ { "aoVal": "D", "content": "13 " } ] ]

[ "拓展思维->拓展思维->数论模块->完全平方数->偶指奇因->奇因" ]

[ "若为$$n\\textgreater1$$的整数，则$$n=p_{1}^{{{\\alpha }_{1}}}p_{2}^{{{\\alpha }_{2}}}\\cdots p_{r}^{{{\\alpha }_{r}}}$$为质因数分解式，其中$${{p}_{1}} \\textless{} {{p}_{2}} \\textless{} \\cdots \\textless{} {{p}_{r}}$$为质数，所以$$n$$的因数个数为$$d\\left( n \\right)=\\left( {{\\alpha }_{1}}+1 \\right)\\cdots \\left( {{\\alpha }_{r}}+1 \\right)$$，因此$$d\\left( n \\right)$$为奇数时，$${{\\alpha }_{1}}\\text{，}\\alpha {}_{2}\\text{，}\\cdots \\text{，}{{\\alpha }_{r}}\\text{，}$$均为偶数，$$n$$为完全平方数，反过来也成立. $$1000$$到$$2007$$中完全平方数有$$13$$个，即$${{32}^{2}}=1024\\text{，}{{33}^{2}}\\text{，}{{34}^{2}}\\text{，}\\cdots \\text{，}{{44}^{2}}=1936\\left( {{45}^{2}}=2025\\textgreater2007 \\right)$$，从而$$1000$$到$$2007$$中有奇数个因数的整数有$$13$$个. " ]

"2023-07-07T00:00:00"

[ "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" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ] ]

[ "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" ]

[ "$$(40+20)\\div 2=30$$． " ]

"2023-07-07T00:00:00"

[ "2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$500$$ " } ], [ { "aoVal": "B", "content": "$$800$$ " } ], [ { "aoVal": "C", "content": "$$1000$$ " } ], [ { "aoVal": "D", "content": "$$1400$$ " } ] ]

[ "知识标签->数学思想->逆向思想" ]

[ "$$500+600-300=800$$个． " ]

"2023-07-07T00:00:00"

[ "2017年全国华杯赛小学高年级竞赛决赛" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "不超过$$100$$的所有质数包含$$2$$、$$5$$， 所以乘积个位一定为$$0$$． 不超过$$100$$的所有个位数字为$$3$$或$$7$$的质数： $$3$$、$$7$$、$$13$$、$$17$$、$$23$$、$$37$$、$$43$$、 $$47$$、$$53$$、$$67$$、$$73$$、$$83$$、$$97$$，$$3\\times7$$的个位为$$1$$， 那么不超过$$100$$的所有个位数字为$$3$$或$$7$$的质数的乘积个位是$$3$$， 差的个位是$$7$$． " ]

"2023-07-07T00:00:00"

[ "2017年第17届湖北武汉世奥赛五年级竞赛决赛第7题" ]

[ [ { "aoVal": "A", "content": "$$123$$ " } ], [ { "aoVal": "B", "content": "$$495$$ " } ], [ { "aoVal": "C", "content": "$$594$$ " } ], [ { "aoVal": "D", "content": "$$954$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$123$$、$$198$$、$$792$$、$$693$$、$$594$$、$$495$$、$$495$$、$$495$$，可以发现选项的$$4$$个数字都会陷入$$495$$的数字黑洞，所以数字黑洞是$$495$$．故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2006年第4届创新杯四年级竞赛初赛B卷第4题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "因为$${{3}^{n}}$$的个位数字是$$3$$，$$9$$，$$7$$，$$1$$四个一循环，$$2006\\div 4=501$$（组）$$\\cdots \\cdots 2$$（个）， 所以$${{3}^{2006}}$$个位数字和$${{3}^{2}}$$的个位数字是相同的，即为$$9$$； 因为$${{7}^{n}}$$的个位数字是$$7$$，$$9$$，$$3$$，$$1$$四个一循环，$$100\\div 4=25$$， 所以$${{7}^{100}}$$个位数字和$${{7}^{4}}$$的个位数字是相同的，即为$$1$$； 所以$$\\underbrace{3\\times 3\\times \\cdots \\times 3}_{2006个3}$$减去$$\\underbrace{7\\times 7\\times \\cdots \\times 7}_{100个7}$$，得数的个位数字是$$8$$． 故选$$\\text{D}$$． " ]

"2023-07-07T00:00:00"

[ "2017年希望杯四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->和差倍问题->差倍->两量差倍->整数差倍" ]

[ "解：父子年龄差是：$$31-5=26$$（岁）， 爸爸的年龄是小军的$$3$$倍时， 小军的年龄是：$$26\\div \\left( 3-1 \\right)$$ $$=26\\div 2$$ $$=13$$（岁）， $$13-5=8$$（年）， 则再过$$8$$年，爸爸的年龄是小军的$$3$$倍。 故答案为：$$8$$。 " ]

"2023-07-07T00:00:00"

[ "2004年五年级竞赛创新杯", "2004年第2届创新杯五年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "扩大$$10$$倍 " } ], [ { "aoVal": "B", "content": "缩小$$10$$倍 " } ], [ { "aoVal": "C", "content": "扩大$$1000$$倍 " } ], [ { "aoVal": "D", "content": "缩小$$1000$$倍 " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->小数加减->小数加减巧算之位值原理" ]

[ "把除数为两位小数的小数点去掉，那么除数扩大了$$100$$倍，商就会缩小$$100$$倍.又把被除数缩小$$10$$倍，那么商又缩小了$$10$$倍，一共缩小了$$1000$$倍，选D. " ]

"2023-07-07T00:00:00"

[ "2013年中环杯三年级竞赛决赛" ]

[ [ { "aoVal": "A", "content": "叠被子$$-$$洗碗 " } ], [ { "aoVal": "B", "content": "用洗衣机洗衣服$$-$$晾衣服 " } ], [ { "aoVal": "C", "content": "拖地板$$-$$削土豆皮 " } ], [ { "aoVal": "D", "content": "洗碗$$-$$晾衣服 " } ] ]

[ "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" ]

[ "解： 先用洗衣机洗衣服，然后晾衣服。 " ]

"2023-07-07T00:00:00"

[ "2018年IMAS小学高年级竞赛（第一轮）第7题3分" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ], [ { "aoVal": "E", "content": "$$100$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->归一归总问题->双归一问题" ]

[ "因为$$3$$个纸箱所装的球鞋数量与$$1$$个木箱所装的球鞋数量一样．所以$$9$$个纸箱所装的球鞋数量与$$9\\div 3=3$$个木箱所装的球鞋数量一样．因此$$300$$双球鞋相当于装在$$2+3=5$$个木箱中．故每个木箱装$$300\\div5=60$$双球鞋． 故选$$\\text{D}$$． " ]

"2023-07-07T00:00:00"

[ "2012年IMAS小学高年级竞赛第一轮检测试题第5题3分" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "$$13$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "因为小明以$$2$$分钟从第$$1$$根电线杆开始跑到第$$5$$根电线杆共跑了$$4$$个间隔，所以$$4$$分钟可以跑到第$$9$$根电线杆，$$6$$分钟可以跑到第$$13$$根电线杆． 故$$n=13$$． " ]

"2023-07-07T00:00:00"

[ "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第6题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第6题3分", "2020年第24届YMO四年级竞赛决赛第6题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第6题3分", "2019年第24届YMO四年级竞赛决赛第6题3分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "拓展思维->思想->分类讨论思想" ]

[ "小于$$2000$$的四位数千位数字是$$1$$，要它数字和为$$24$$，只需其余三位数字和是$$23$$，因为十位、个位数字和最多为$$9+9=18$$，因此，百位数字至少是$$5$$，于是： 百位为$$5$$时，只有$$1599$$一个； 百位为$$6$$时，只有$$1689$$、$$1698$$两个； 百位为$$7$$时，只有$$1779$$、$$1788$$、$$1797$$三个； 百位为$$8$$时，只有$$1869$$、$$1878$$、$$1887$$、$$1896$$四个； 百位为$$9$$时，只有$$1959$$、$$1968$$、$$1977$$、$$1986$$、$$1995$$五个； 总计共$$1+2+3+4+5=15$$ （个）． 故选$$\\text{A}$$． " ]

"2023-07-07T00:00:00"

[ "2014年全国迎春杯四年级竞赛复赛第4题" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ] ]

[ "拓展思维->能力->抽象概括" ]

[ "根据题意得$$({{a}_{1}}+{{a}_{8}})\\times 8\\div 2=({{a}_{9}}+{{a}_{12}})\\times4\\div 2$$，因为$${{a}_{8}}={{a}_{1}}+14$$，$${{a}_{9}}={{a}_{1}}+16$$，$${{a}_{12}}={{a}_{1}}+22$$， 所以$$({{a}_{1}}+{{a}_{1}}+14)\\times 8\\div2=({{a}_{1}}+16+{{a}_{1}}+22)\\times 4\\div 2$$，解得$${{a}_{1}}=5$$，因此$${{a}_{2}}=5+2=7$$． " ]

"2023-07-07T00:00:00"

[ "2011年第7届全国新希望杯小学高年级五年级竞赛A卷第6题" ]

[ [ { "aoVal": "A", "content": "$$105$$ " } ], [ { "aoVal": "B", "content": "$$205$$ " } ], [ { "aoVal": "C", "content": "$$305$$ " } ], [ { "aoVal": "D", "content": "$$405$$ " } ] ]

[ "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" ]

[ "只用计算末三位数字和为$$1+12+123+234+345+456+567+678+789=3205$$，故答案为$$205$$，选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2013年IMAS小学中年级竞赛第二轮检测试题第3题4分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "依次枚举，$$30+20=50$$（元），$$30+50=80$$（元），$$30+80=110$$（元），$$60+80=140$$（元），$$90+80=170$$（元），共$$5$$种． " ]

"2023-07-07T00:00:00"

[ "2012年第10届全国创新杯小学高年级六年级竞赛第4题4分" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$31$$ " } ], [ { "aoVal": "D", "content": "$$35$$ " } ] ]

[ "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" ]

[ "错误的题共有$$260-181=79$$题，假设得$$2$$分的和得$$3$$分的为$$a$$人，得$$4$$分的为$$b$$人，可以得到 $$79=28+5a+b$$ 有$$5a+b=51$$，只有$$31$$满足． " ]

"2023-07-07T00:00:00"

[ "2015年第20届全国华杯赛小学中年级四年级竞赛初赛A卷第5题10分" ]

[ [ { "aoVal": "A", "content": "快$$12$$分 " } ], [ { "aoVal": "B", "content": "快$$6$$分 " } ], [ { "aoVal": "C", "content": "慢$$6$$分 " } ], [ { "aoVal": "D", "content": "慢$$12$$分 " } ] ]

[ "拓展思维->拓展思维->行程模块->时钟问题->坏钟问题" ]

[ "时针速度为每分钟$$0.5$$度，分针速度为每分钟$$6$$度．分钟每比时针多跑一圈，即多跑$$360$$度，时针分针重合一次．经过$$\\frac{360}{6-0.5}=\\frac{720}{11}$$分钟，旧钟时针分针重合一次，需要经过标准时间$$66$$分钟；则旧钟的$$24$$小时，相当于标准时间的$$\\frac{24\\times 60}{\\frac{720}{11}}\\times 66=1452$$分钟，所以比标准时间$$24$$小时对应的$$24\\times 60=1440$$分钟多了$$1452-1440=12$$分钟，即慢了$$12$$分钟． " ]

"2023-07-07T00:00:00"

[ "2021年第8届鹏程杯四年级竞赛初赛第23题4分" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$66$$ " } ], [ { "aoVal": "C", "content": "$$132$$ " } ], [ { "aoVal": "D", "content": "$$144$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->能力->公式记忆->符号化数学原理" ]

[ "日期$$\\leqslant 12$$的都会产生误会，$$12\\times12=144$$天，但日期和月份相同时不会产生误会，这样的时间共有$$12$$天，所以共有$$144-12=132$$天． 故选：$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2019年第24届YMO二年级竞赛决赛第1题3分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "根据题意分析可知，小$$Y$$后面有$$42-22=20$$（人），$$22-20-1=1$$（人）．那么小$$M$$应该在小$$Y$$的正前方，所以小$$Y$$和小$$M$$之间有$$0$$个人． 故选：$$\\text{A}$$． " ]

"2023-07-07T00:00:00"

[ "1993年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "15 " } ], [ { "aoVal": "B", "content": "18 " } ], [ { "aoVal": "C", "content": "20 " } ], [ { "aoVal": "D", "content": "24 " } ] ]

[ "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" ]

[ "插板法，$$\\text{C}_{6}^{3}=20$$ " ]

"2023-07-07T00:00:00"

[ "2020年希望杯六年级竞赛（个人赛）第4题", "2020年新希望杯六年级竞赛初赛（个人战）第4题" ]

[ [ { "aoVal": "A", "content": "$$11:8$$ " } ], [ { "aoVal": "B", "content": "$$33:55$$ " } ], [ { "aoVal": "C", "content": "$$32:55$$ " } ], [ { "aoVal": "D", "content": "$$41:32$$ " } ], [ { "aoVal": "E", "content": "$$55:32$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "已知$$x:y=4:7$$，$$z:x=3:5$$，那么$$y:x=7:4$$，$$x:z=5:3$$， $$y:x=\\left( 7\\times 5 \\right):\\left( 4\\times 5 \\right)=35:20$$，$$x:z=\\left( 5\\times 4 \\right):\\left( 3\\times 4 \\right)=20:12$$， $$y:x:z=35:20:12$$． 假设$$y=35a$$，$$x=20a$$，$$z=12a$$，其中$$a\\ne 0$$． $$x+y=20a+35a=55a$$，$$z+x=12a+20a=32a$$， 那么$$\\left( x+y \\right):\\left( z+x \\right)=55a:32a=55:32$$． 故选$$\\text{E}$$． " ]

"2023-07-07T00:00:00"

[ "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" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "$$14$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "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． " ]

"2023-07-07T00:00:00"

[ "2005年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "135 " } ], [ { "aoVal": "B", "content": "165 " } ], [ { "aoVal": "C", "content": "180 " } ], [ { "aoVal": "D", "content": "195 " } ] ]

[ "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" ]

[ "这两个整数的最大公因数为15，可设这两个数分别为15a，15b($$a \\textless{} b$$且$$a\\text{，}b$$互质)，.那么$$15\\times a\\times b=360$$，即$$a\\times b=24$$，又$$a\\text{，}b$$互质，则$$\\left( a\\text{，}b \\right)=\\left( 1\\text{，}24 \\right)$$或$$\\left( 3\\text{，}8 \\right)$$.当$$\\left( a\\text{，}b \\right)=\\left( 1\\text{，}24 \\right)$$时，这两个整数的和为$$15\\times 1+15\\times 24=15\\times 25=375$$;当$$\\left( a\\text{，}b \\right)=\\left( 3\\text{，}8 \\right)$$时，这两个整数的和为$$15\\times 3+15\\times 8=15\\times 11=165$$.四个选项中只有B符合结论，故选B " ]

"2023-07-07T00:00:00"

[ "2021年新希望杯一年级竞赛初赛第14题5分" ]

[ [ { "aoVal": "A", "content": "鼠和鸡 " } ], [ { "aoVal": "B", "content": "鸡和兔 " } ], [ { "aoVal": "C", "content": "兔和狗 " } ], [ { "aoVal": "D", "content": "兔和鼠 " } ], [ { "aoVal": "E", "content": "鼠和狗 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据题意分析可知，狗和猫都与鸡相邻，即鸡的左右两边是猫和狗，又因为猫与老鼠都与兔相邻，所以猫的旁边是兔子和老鼠，兔子和老鼠在猫的同一侧，由此可知，与猫相邻的是鸡和兔． 故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2019年全国小学生数学学习能力测评六年级竞赛复赛第9题3分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{4}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{5}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{41}$$ " } ], [ { "aoVal": "D", "content": "无选项 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "盐水：$$0.5+20=20.5\\left( \\text{kg} \\right)$$， 盐占盐水：$$0.5\\div 20.5=\\frac{1}{41}$$． 故选：$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "六年级其它", "2017年河南郑州豫才杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$\\left( 2\\oplus 3 \\right)=2+1=3$$，$$\\left( 5\\oplus 4 \\right)=2\\times 4-1=7$$，$$\\left( 2\\oplus 3 \\right)\\oplus \\left( 5\\oplus 4 \\right)=3\\oplus 7=3+1=4$$． " ]

"2023-07-07T00:00:00"

[ "2016年第12届全国新希望杯小学高年级六年级竞赛复赛第4题4分" ]

[ [ { "aoVal": "A", "content": "$$49$$ " } ], [ { "aoVal": "B", "content": "$$48$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$51$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->认识单位1" ]

[ "设去年参加比赛的为单位``$$1$$''，女生占$$\\dfrac{1}{5}$$，即女生人数为$$\\dfrac{1}{5}\\times 1=\\dfrac{1}{5}$$人，今年参加比赛的人数增加$$20 \\%$$，则今年参加比赛人数为$$1\\times \\left( 1+20 \\% \\right)=1.2$$，女生占$$\\dfrac{1}{4}$$，则女生人数为$$1.2\\times \\dfrac{1}{4}=0.3$$人．今年比去年增加了$$\\dfrac{\\left( 0.3-\\dfrac{1}{5} \\right)}{\\dfrac{1}{5}}=50 \\%$$． " ]

"2023-07-07T00:00:00"

[ "2016年全国华杯赛小学高年级竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "$$109$$ " } ], [ { "aoVal": "B", "content": "$$110$$ " } ], [ { "aoVal": "C", "content": "$$111$$ " } ], [ { "aoVal": "D", "content": "$$112$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "$$1$$到$$2016$$中，数字和最大$$28$$， 数字和$$28$$的数只有$$1999$$， 最坏情况：取数字和$$1$$到$$27$$的数各$$4$$个，以及$$1999$$，共$$109$$个数， 再多取一个数就保证有$$5$$个数字和相等，$$n=110$$． " ]

"2023-07-07T00:00:00"

[ "2016年新希望杯六年级竞赛训练题（一）第1题" ]

[ [ { "aoVal": "A", "content": "$$10250$$ " } ], [ { "aoVal": "B", "content": "$$9975$$ " } ], [ { "aoVal": "C", "content": "$$10000$$ " } ], [ { "aoVal": "D", "content": "$$9750$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->已知成本利润求售价" ]

[ "$$10\\times 1000\\times \\left( 1-5 \\% \\right)\\times \\left( 1+5 \\% \\right)=9975$$（元）． " ]

"2023-07-07T00:00:00"

[ "2014年迎春杯三年级竞赛初赛", "2014年迎春杯五年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$19$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "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：$$48\\div 2.5=19.2\\approx 20$$（辆），由实际情况可知需要$$20$$辆。 答：至少需要$$20$$辆这样的大卡车。 故选：C。 " ]

"2023-07-07T00:00:00"

[ "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）" ]

[ [ { "aoVal": "A", "content": "$$2017$$ " } ], [ { "aoVal": "B", "content": "$$6012$$ " } ], [ { "aoVal": "C", "content": "$$7012$$ " } ], [ { "aoVal": "D", "content": "$$8120$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->加乘原理->组数问题->有特殊要求的组数问题" ]

[ "组成的四位数一共有$$8\\times 8\\times 7\\times 6=2688$$（个），其中千位为$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$的各有$$336$$个．$$2017336\\times 6=1$$（个），即千位为$$7$$的第$$1$$个数为$$7012$$，选$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2017年迎春杯五年级竞赛决赛二试第2题10分", "2017年迎春杯六年级竞赛决赛二试第3题10分" ]

[ [ { "aoVal": "A", "content": "$$34$$ " } ], [ { "aoVal": "B", "content": "$$43$$ " } ], [ { "aoVal": "C", "content": "$$44$$ " } ], [ { "aoVal": "D", "content": "$$49$$ " } ], [ { "aoVal": "E", "content": "$$55$$ " } ] ]

[ "拓展思维->能力->逻辑分析", "Overseas Competition->知识点->数论模块" ]

[ "$$A$$、$$A+1$$、$$B$$、$$B+1$$均不为质数；也不能是质数的$$n$$次方．所以，$$B$$只能是$$14$$．（$$B$$为$$6$$、$$10$$时，$$B+1$$都是质数），此时$$B+1$$为$$15$$，$$B(B+1)$$含有质因数$$2$$、$$3$$、$$5$$、$$7$$；最小符合条件的$$A$$为$$20$$，所以，$$A+B$$最小值为$$34$$． " ]

"2023-07-07T00:00:00"

[ "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" ]

[ [ { "aoVal": "A", "content": "二进制计数法 " } ], [ { "aoVal": "B", "content": "五进制计数法 " } ], [ { "aoVal": "C", "content": "十进制计数法 " } ] ]

[ "拓展思维->七大能力->实践应用", "课内体系->七大能力->实践应用" ]

[ "古时候的原始人捕猎运用现在的数学中的十进制计数法． " ]

"2023-07-07T00:00:00"

[ "2017年第23届浙江杭州华杯赛小学高年级竞赛卓学堂高端备考活动第7题", "2014年全国中环杯五年级竞赛决赛第7题" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

[ "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" ]

[ "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）． " ]

"2023-07-07T00:00:00"

[ "2021年新希望杯一年级竞赛初赛第14题5分" ]

[ [ { "aoVal": "A", "content": "鼠和鸡 " } ], [ { "aoVal": "B", "content": "鸡和兔 " } ], [ { "aoVal": "C", "content": "兔和狗 " } ], [ { "aoVal": "D", "content": "兔和鼠 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据题意分析可知，狗和猫都与鸡相邻，即鸡的左右两边是猫和狗，又因为猫与老鼠都与兔相邻，所以猫的旁边是兔子和老鼠，兔子和老鼠在猫的同一侧，由此可知，与猫相邻的是鸡和兔． 故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2012年IMAS小学高年级竞赛第二轮测试真题第3题" ]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$51$$ " } ], [ { "aoVal": "E", "content": "$$60$$ " } ] ]

[ "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" ]

[ "用排除法，$$12$$个人共握手$$C_{12}^{2}=12\\times 11\\div 2=66$$（次） 男宾与妻子握手$$6$$次， 女宾与女宾握手$$\\text{C}_{6}^{2}=6\\times 5\\div 2=15$$（次） ∴$$66-6-15=45$$（次）． ", "<p>用排除法，$$12$$个人共握手$$12\\times 11\\div 2=66$$（次）</p>\n<p>男宾与妻子握手$$6$$次，</p>\n<p>女宾与女宾握手$$6\\times 5\\div 2=15$$（次）</p>\n<p>&there4;$$66-6-15=45$$（次）．</p>" ]

"2023-07-07T00:00:00"

[ "2016年第14届全国创新杯五年级竞赛初赛第9题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{45}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{75}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{7}{225}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{8}{225}$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "三位数一共有$$900$$个，满足条件的魔力数有$$\\text{C}_{7+2-1}^{2}=28$$个，则$$\\frac{28}{900}=\\frac{7}{225}$$． " ]

"2023-07-07T00:00:00"

[ "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" ]

[ [ { "aoVal": "A", "content": "$$S$$ " } ], [ { "aoVal": "B", "content": "$$Q$$ " } ], [ { "aoVal": "C", "content": "$$O$$ " } ], [ { "aoVal": "D", "content": "$$T$$ " } ] ]

[ "知识标签->拓展思维->组合模块->字母规律->数字与字母结合" ]

[ "$$2015\\div 6\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 5$$，则$$2015$$个数应为$$S$$． " ]

"2023-07-07T00:00:00"

[ "2016年新希望杯六年级竞赛训练题（五）第3题" ]

[ [ { "aoVal": "A", "content": "$$1600$$ " } ], [ { "aoVal": "B", "content": "$$1800$$ " } ], [ { "aoVal": "C", "content": "$$3200$$ " } ], [ { "aoVal": "D", "content": "$$3600$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->工程问题->进水与排水问题" ]

[ "每小时水池的水减少$$2\\times 80-3\\times 50=10$$立方米，出完水需要的时间是$$200\\div 10=20$$小时，共出水$$20\\times 2\\times 80=3200$$立方米． " ]

"2023-07-07T00:00:00"

[ "2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第4题5分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "云云比芳芳多：$$18-8=10$$（朵）， 所以要使两人一样多，应将多余的$$10$$朵花一半分， 即每人$$5$$朵，所以云云给芳芳$$5$$朵． 故选择$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2014年全国迎春杯三年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$$5\\times99+1$$ " } ], [ { "aoVal": "B", "content": "$$100+25\\times4$$ " } ], [ { "aoVal": "C", "content": "$$88\\times4+37\\times4$$ " } ], [ { "aoVal": "D", "content": "$$100\\times0\\times5$$ " } ] ]

[ "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" ]

[ "A等于$$5\\times(100-1)+1=500-5+1=496$$，$$B$$等于$$100+100=200$$ ，$$C$$等于$$(88+37)\\times4=125\\times 4=500$$ ，$$D$$等于$$0$$． " ]

"2023-07-07T00:00:00"

[ "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" ]

[ [ { "aoVal": "A", "content": "$$10070$$ " } ], [ { "aoVal": "B", "content": "$$16112$$ " } ], [ { "aoVal": "C", "content": "$$22154$$ " } ], [ { "aoVal": "D", "content": "$$24168$$ " } ] ]

[ "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" ]

[ "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$$24168$$． " ]

"2023-07-07T00:00:00"

[ "2014年IMAS小学中年级竞赛第一轮检测试题第6题3分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$21$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "因为$$11$$到$$19$$号同学原地不动，所以还有$$19-11+1=9$$位同学原地不动．故选$$\\text{A}$$． " ]

"2023-07-07T00:00:00"

[ "2016年创新杯小学高年级六年级竞赛训练题（一）第2题" ]

[ [ { "aoVal": "A", "content": "$$9.4$$ " } ], [ { "aoVal": "B", "content": "$$9.5$$ " } ], [ { "aoVal": "C", "content": "$$9.6$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "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" ]

[ "设原来上坡用了$$x$$小时，$$45x+50\\left( 9.4-x \\right)=450$$，解得$$x=4$$，那么原来上坡是$$45\\times 4=180$$千米，下坡就是$$450-180=270$$千米，返回时下坡变为上坡，上坡变为下坡，所以返回时时间是$$\\frac{270}{45}+\\frac{180}{50}=9.6$$（小时）． " ]

"2023-07-07T00:00:00"

[ "2020年广东广州羊排赛四年级竞赛第18题10分" ]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题->方程法解鸡兔同笼" ]

[ "根据题目设她答对了$$x$$道题， 因为共$$12$$道题，所以答错了$$(12-x)$$道题，答对一题得$$5$$分，不答或答错一题倒扣$$2$$分． 薇儿最终分数是$$39$$分，可以列方程为：$$5x-(12-x)\\times 2=39$$， 解得$$x=9$$，所以她答对了$$9$$题． " ]

"2023-07-07T00:00:00"

[ "2015年第2届广东深圳鹏程杯四年级竞赛第6题8分" ]

[ [ { "aoVal": "A", "content": "在答题卡上提交 " } ] ]

[ "知识标签->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题->方程法解倍数问题" ]

[ "设小明今年是$$x$$岁，则爸爸今年是$$(3x+4)$$岁，再过$$6$$年爸爸就是$$[(3x+4)+6]$$岁， 而六年后，小明长大了$$6$$岁，爸爸也长大了$$6$$岁， 因此，六年后，爸爸年龄是$$[2(x+6)+10]$$岁． 据题意得到$$(3x+4)+6=2(x+6)+10$$，解得$$x=12$$． " ]

"2023-07-07T00:00:00"

[ "2014年第10届全国新希望杯小学高年级五年级竞赛复赛第1题" ]

[ [ { "aoVal": "A", "content": "$$1.9$$ " } ], [ { "aoVal": "B", "content": "$$1.6$$ " } ], [ { "aoVal": "C", "content": "$$1.7$$ " } ], [ { "aoVal": "D", "content": "$$2.1$$ " } ] ]

[ "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" ]

[ "添加一个数后这列数有奇数个，中位数是最中间的数（由小到大排），则添加的数为$$1.9$$． " ]

"2023-07-07T00:00:00"

[ "竞赛" ]

[ [ { "aoVal": "A", "content": "$$80$$ " } ], [ { "aoVal": "B", "content": "$$85$$ " } ], [ { "aoVal": "C", "content": "$$82$$ " } ], [ { "aoVal": "D", "content": "$$81$$ " } ] ]

[ "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" ]

[ "按照$$A$$、$$B$$、$$C$$的顺序理发，最短时间为$$7\\times 3+10\\times 2+40=81$$（分）． " ]

"2023-07-07T00:00:00"

[ "2018年迎春杯四年级竞赛决赛一试第3题10分" ]

[ [ { "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） " } ] ]

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "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． " ]

"2023-07-07T00:00:00"

[ "2012年河南郑州二七区郑州实验外国语中学小升初第25题5分", "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$42$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "将整个工程的工作量看作``$$1$$''个单位，那么甲每天完成总量的$$\\frac{1}{30}$$，甲、乙合作每天完成总量的$$\\frac{1}{12}$$，乙单独做每天能完成总量的$$\\frac{1}{12}-\\frac{1}{30}=\\frac{1}{20}$$，所以乙单独做$$1\\div\\frac{1}{20}=20$$天能完成． " ]

"2023-07-07T00:00:00"

[ "2019年第7届湖北长江杯六年级竞赛复赛A卷第6题3分", "2020年湖北宜昌小升初第15题1分" ]

[ [ { "aoVal": "A", "content": "$$69$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "因为圆的周长增加$$30 \\%$$，所以圆的半径也增加$$30 \\%$$．设圆的半径为$$r$$，则增加后的半径为$$(1+30 \\%)r=1.3r$$，所以原来的圆的面积为$$\\pi {{r}^{2}}$$，半径增加后的圆的面积为$$\\pi {{(1.3r)}^{2}}=1.69\\pi {{r}^{2}}$$，故面积增加$$(1.69\\pi {{r}^{2}}-\\pi {{r}^{2}})\\div \\pi {{r}^{2}}=69 \\%$$． " ]

"2023-07-07T00:00:00"

[ "2018年湖北武汉新希望杯小学高年级六年级竞赛训练题（四）第4题" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$165-21=144$$，$$220-4=216$$，$$398-38=360$$，（$$144$$，$$216$$，$$260$$）$$=72$$，所以小朋友的人数应该是$$72$$的因数，又$$38\\textgreater36$$，所以人数只能是$$72$$人，$$144\\div 72+216\\div 72+360\\div 72=10$$（个）． " ]

"2023-07-07T00:00:00"

[ "2013年第11届全国创新杯五年级竞赛第7题5分" ]

[ [ { "aoVal": "A", "content": "$$23\\leqslant M\\leqslant 67$$ " } ], [ { "aoVal": "B", "content": "$$19\\leqslant M\\leqslant 67$$ " } ], [ { "aoVal": "C", "content": "$$21\\leqslant M\\leqslant 69$$ " } ], [ { "aoVal": "D", "content": "$$17\\leqslant M\\leqslant 69$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->等差数列->中项定理应用" ]

[ "我们尝试考虑最大数$$M$$的最极端情况，当数$$M$$要去的最大值时，其他四个数必须最小，那么分别取$$1$$，$$3$$，$$5$$，$$7$$，此时$$M$$最大值为$$69$$，排除$$A$$、$$B$$选项，为了满足$$M$$是五个数最大的数且最小，那么最好极端的情况应该是这五个数恰好是一个等差的奇数列，根据中项定理易得$$M$$的最小值应为：$$85\\div 5+2+2=21$$． " ]

"2023-07-07T00:00:00"

[ "2021年第8届鹏程杯四年级竞赛初赛第19题4分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "速度和$$=360\\div 12=30$$，$$270\\div 30=9$$（秒）． 故选：$$\\text{A}$$． " ]

"2023-07-07T00:00:00"

[ "2013年IMAS小学中年级竞赛第二轮检测试题第1题4分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ], [ { "aoVal": "E", "content": "$$80$$ " } ] ]

[ "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" ]

[ "$$86\\div 10=8\\cdots \\cdots 6$$，$$\\text{A}$$符合； $$86\\div 20=4\\cdots \\cdots 6$$，$$\\text{B}$$符合； $$86\\div 30=2\\cdots \\cdots 26$$，$$\\text{C}$$不符合； $$86\\div 80=1\\cdots \\cdots 6$$，$$\\text{D}$$符合． " ]

"2023-07-07T00:00:00"

[ "2012年全国美国数学大联盟杯小学高年级竞赛初赛第33题", "2013年美国数学大联盟杯小学高年级竞赛初赛第33题5分" ]

[ [ { "aoVal": "A", "content": "$$17$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$26$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$6$$个连续的整数的和$$(25+30)\\times6 \\div2=165$$， $$10$$个连续整数最中间$$165\\div5=33=16+17$$， $$17+4=21$$． " ]

"2023-07-07T00:00:00"

[ "2018年全国小学生数学学习能力测评五年级竞赛初赛第7题3分" ]

[ [ { "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" ]

[ "根据``每次摆放都是下面一层比上面一层多放$$1$$本书''得知，每层摆放书的数目成等差数列，要求``至多''摆放几层，需要第一层是$$1$$本，第二层是$$2$$本，第三层是$$3$$本$$\\cdots \\cdots $$， 假设摆放$$n$$层，则：$$1+2+3+\\cdots \\cdots +n=\\frac{n(1+n)}{2}$$， $$66=\\frac{n(1+n)}{2}$$． 得$$n=11$$，所以选$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2018年美国数学大联盟杯五年级竞赛初赛第13题5分" ]

[ [ { "aoVal": "A", "content": "$$111111$$ " } ], [ { "aoVal": "B", "content": "$$222222$$ " } ], [ { "aoVal": "C", "content": "$$444444$$ " } ], [ { "aoVal": "D", "content": "$$888888$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "在两个连续整数中，其中一个是双数，一个是单数．一个双数和一个单数的和一定是一个单数，所以两个连续整数的和也定是单数． 在$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$，$$4$$个选项中，只有$$\\text{A}$$是单数，其余$$3$$个都是双数． 故答案选：$$\\text{A}$$ " ]

"2023-07-07T00:00:00"

[ "2020年广东广州羊排赛三年级竞赛第14题8分" ]

[ [ { "aoVal": "A", "content": "一 " } ], [ { "aoVal": "B", "content": "四 " } ], [ { "aoVal": "C", "content": "五 " } ], [ { "aoVal": "D", "content": "六 " } ], [ { "aoVal": "E", "content": "日 " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "$$87$$集的电视连续剧在某个星期日开播，每周日、周一、周四、周五、周六都要播一集，五天一周期，$$87\\div 5=17$$（周）$$\\cdots \\cdots 2$$（天），所以最后一集在周一播出，选项$$\\text{A}$$正确． " ]

"2023-07-07T00:00:00"

[ "2016年全国世奥赛竞赛A卷第3题", "2016年第16届世奥赛六年级竞赛决赛第3题" ]

[ [ { "aoVal": "A", "content": "杨辉，$$22$$，$$335$$ " } ], [ { "aoVal": "B", "content": "祖冲之，$$22$$，$$355$$ " } ], [ { "aoVal": "C", "content": "韩信，$$11$$，$$355$$ " } ], [ { "aoVal": "D", "content": "赵爽，$$12$$，$$335$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->位值原理与进制->进制的性质与应用->进制的认识" ]

[ "第一空考察的是数学常识．根据题干中的信息可以知道约率和密率都是接近$$3.14$$的，所以$$3.14\\times 7=21.98\\approx 22$$，$$3.14\\times 113=354.82\\approx 355$$． " ]

"2023-07-07T00:00:00"

[ "2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第13题2分" ]

[ [ { "aoVal": "A", "content": "$$47.2$$ " } ], [ { "aoVal": "B", "content": "$$37.5$$ " } ], [ { "aoVal": "C", "content": "$$24.5$$ " } ], [ { "aoVal": "D", "content": "$$10.5$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "本题考查相遇问题； 客车比货车迟发$$2$$小时，所以客车是$$8$$时出发，到$$12$$时，经过$$4$$小时到达相遇点，而货车到达相遇点共用了$$4+2=6$$（小时），那么甲、乙两地之间的距离就可求出，用货车 速度乘相遇时货车总共用的时间加上客车速度乘相遇时客车所用时间，即$$45\\times \\left( 4+2 \\right)+60\\times 4=510$$（千米）；客车行完全程所需的时间： $$510\\div \\left( 45+15 \\right)=8.5$$（小时）；货车行完全程所需的时间：$$8.5+2=10.5$$（小时）；客车到达甲地时，货车距乙地的距离，用全程减去货车已经行的路程，即$$510-45\\times 10.5=37.5$$（千米）． 客车速度： $$45+15=60$$（千米$$/$$时） 两地距离： $$45\\times \\left( 12-6 \\right)+60\\times \\left( 12-6-2 \\right)=510$$（千米） 客车行完全程所需时间： $$510\\div 60=8.5$$（小时） 客车到达甲地时，货车距乙地的距离： $$510-45\\times \\left( 8.5+2 \\right)=37.5$$（千米） 答：客车到达甲地时，货车离乙地还有$$37.5$$千米． 故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2018年美国数学大联盟杯五年级竞赛初赛第14题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{9}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{9}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{4}{9}$$ " } ] ]

[ "拓展思维->能力->实践应用", "Overseas Competition->知识点->计算模块->分数" ]

[ "本题题意为``三幕戏剧的第二幕占整个戏剧长度的$$\\frac{1}{3}$$，如果第一幕是第三幕的两倍，那么第三幕占戏剧的几分？'' 根据题意可得，第二幕占整个戏剧长度的$$\\frac{1}{3}$$， 因此第一幕和第三幕占整个长度的$$\\frac{2}{3}$$，第一幕是第三幕的两倍， 因此假设第三幕时长为$$N$$，第一幕的长度为$$2N$$， 因此$$N+2N=\\frac{2}{3}$$， 因此$$N=\\frac{2}{9}$$， 故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2017年江苏南京小升初工程问题第27题", "2016年北京西城区小升初四中八中分班", "2012年世界少年奥林匹克数学竞赛六年级竞赛初赛A卷第16题10分", "2016年第3届广东深圳鹏程杯六年级竞赛集训材料第一章工程问题第9题" ]

[ [ { "aoVal": "A", "content": "$$240$$ " } ], [ { "aoVal": "B", "content": "$$280$$ " } ], [ { "aoVal": "C", "content": "$$300$$ " } ], [ { "aoVal": "D", "content": "$$360$$ " } ] ]

[ "课内体系->能力->运算求解", "拓展思维->拓展思维->计算模块->分数->分数基础->分数的认识" ]

[ "解题关键在于①甲先做$$16$$天然后乙再做$$12$$天，还剩这批零件的$$\\frac{2}{5}$$，想成甲乙合干$$12$$天，甲独干$$4$$天完成全工程的$$1-\\frac{2}{5}=\\frac{3}{5}$$，从而分别求出甲、乙工作效率．②找出甲比乙每天多加工$$3$$个零件对应的甲乙工效差，即可使问题得解． 解：甲乙合作$$12$$天完成总工程几分之几． $$\\frac{1}{24}\\times 12=\\frac{1}{2}$$ 甲工作效率 $$\\left( 1-\\frac{2}{5}-\\frac{1}{2} \\right)\\div (16-12)=\\frac{1}{40}$$ 乙工作效率 $$\\frac{1}{24}-\\frac{1}{40}=\\frac{1}{60}$$ 这批零件共多少个？ $$3\\div \\left( \\frac{1}{40}-\\frac{1}{60} \\right)=360$$（个） 答：这批零件共$$360$$个． " ]

"2023-07-07T00:00:00"

[ "2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（二）" ]

[ [ { "aoVal": "A", "content": "$$218$$ " } ], [ { "aoVal": "B", "content": "$$208$$ " } ], [ { "aoVal": "C", "content": "$$198$$ " } ], [ { "aoVal": "D", "content": "$$200$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "由题意知：$$a*b=13\\times a-b\\div 8$$， 则$$17*24=13\\times 17-24\\div 8=221-3=218$$． " ]

"2023-07-07T00:00:00"

[ "2012年IMAS小学中年级竞赛第一轮检测试题第10题3分" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "拓展思维->七大能力->逻辑分析" ]

[ "姐姐与妹妹的年龄差是$$4$$岁，当两人年龄和为$$50$$岁时，妹妹的年龄等于$$(50-4)\\div 2=23$$（岁）． " ]

"2023-07-07T00:00:00"

[ "2014年全国华杯赛小学中年级竞赛决赛第8题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "设$$A$$，$$B$$，$$C$$，$$D$$分到的铅笔数分别是$$A$$，$$B$$，$$C$$，$$D$$，由$$B+C=2D$$，知$$C$$、$$D$$、$$B$$依次成等差数列，设公差为$$K$$；由$$A+B=2C$$，知$$A$$、$$C$$、$$B$$依次成等差数列，则公差为$$2K$$；由$$4$$人铅笔数相差不会超过$$4$$，所以$$K=0$$或$$1$$；若$$K=0$$，则$$4\\times B=53$$，但$$53$$不是$$4$$的整数倍； 若$$K=1$$，$$A\\textgreater C\\textgreater D\\textgreater B$$，则$$4\\times C-1=53$$，但$$54$$不是$$4$$的整数倍． 综上所述，$$B$$分到$$15$$支铅笔． " ]

"2023-07-07T00:00:00"

[ "2004年第2届创新杯六年级竞赛初赛第6题", "2004年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$120$$千米 " } ], [ { "aoVal": "B", "content": "$$180$$千米 " } ], [ { "aoVal": "C", "content": "$$315$$千米 " } ], [ { "aoVal": "D", "content": "$$360$$千米 " } ] ]

[ "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" ]

[ "当客车到达甲城时，货车离乙城的距离为$$45$$千米，这说明在相同的时间里，客车比货车多行了$$45$$千米。又已知货车的速度是客车的$$\\frac{3}{4}$$，那么在相同的时间里，货车所走的路程也是客车所走路程的$$\\frac{3}{4}$$，所以客车所走的路程为$$45\\div (1-\\frac{3}{4})=180$$千米，那么甲、乙两城之间的距离为$$180\\times 2=360$$千米，选D。 " ]

"2023-07-07T00:00:00"

[ "2020年新希望杯二年级竞赛初赛（团战）第12题" ]

[ [ { "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 " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "刚开始小明在的名次是第$$5$$名，再超过三个人，因此$$5-3=2$$，小明变成了第$$2$$名. " ]

"2023-07-07T00:00:00"

[ "2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$\\frac{3}{21}=\\frac{1}{7}=0.142857$$，则$$2019\\div6\\cdots\\cdots3$$， 所以小数点后第$$2019$$个数字是$$2$$． 故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2019年四川绵阳涪城区绵阳东辰国际学校小升初第2题3分", "2018年第6届湖北长江杯六年级竞赛初赛B卷第5题3分" ]

[ [ { "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 " } ] ]

[ "拓展思维->拓展思维->几何模块->直线型->图形认知->图形认知角->三角形内外角度" ]

[ "$$2+3=5$$，则为直角三角形． 两个较小角之和大于最大角时为锐角三角形，等于时为直角三角形，小于时为钝角三角形，也可求出最大角，再判断． " ]

"2023-07-07T00:00:00"

[ "2017年第15届湖北武汉创新杯六年级竞赛决赛第7题" ]

[ [ { "aoVal": "A", "content": "$$2016$$ " } ], [ { "aoVal": "B", "content": "$$2017$$ " } ], [ { "aoVal": "C", "content": "$$2018$$ " } ], [ { "aoVal": "D", "content": "$$2019$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "位值原理．假设四位数是$$\\overline{ABCD}$$，那么$$\\overline{ABCD}-(A+B+C+D)=999A+99B+9C=200\\square $$，$$200\\square $$一定为$$9$$的倍数即$$2007$$，$$111A+11B+C=223$$，$$A=2$$，$$B=0$$，$$C=1$$，$$D$$从$$9$$开始从大到小排列第三个是$$7$$，即第三个数是$$2017$$． " ]

"2023-07-07T00:00:00"

[ "2016年创新杯六年级竞赛训练题（三）第4题" ]

[ [ { "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 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "略 " ]

"2023-07-07T00:00:00"

[ "2020年第25届YMO五年级竞赛决赛第18题" ]

[ [ { "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->牛吃草问题->牛吃草转化型->草衰减牛吃草" ]

[ "设每头牛每天吃$$1$$份，则草在$$6-5=1$$天内减少$$20\\times 5-15\\times 6=10$$份，原草量为$$\\left( 20+10 \\right)\\times 5=150$$份．若吃$$10$$天，可有$$150\\div 10-10=5$$头牛． " ]

"2023-07-07T00:00:00"

[ "2019年第24届YMO一年级竞赛决赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "最后四棵树上的鸟一样多， 则每棵树有$$40\\div 4=10$$（只）鸟． 则第三棵树上的鸟飞到第四棵树之前有$$10+5=15$$（只）鸟， 则第三棵树一开始有$$15-4=11$$（只）鸟． 故选$$\\text{C}$$． " ]

"2023-07-07T00:00:00"

[ "2021年第4届山东青岛市南区京山杯六年级竞赛决赛A卷第5题4分" ]

[ [ { "aoVal": "A", "content": "$$2.4$$小时 " } ], [ { "aoVal": "B", "content": "$$3.2$$小时 " } ], [ { "aoVal": "C", "content": "$$5$$小时 " } ], [ { "aoVal": "D", "content": "$$10$$小时 " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "设完成浇水任务需要$$x$$小时，依题意有： $$\\left( \\frac{1}{4}+\\frac{1}{6} \\right)x=1$$， 解得$$x=2.4$$． 故完成浇水任务需要$$2.4$$小时． 故选$$\\text{A}$$． " ]

"2023-07-07T00:00:00"

[ "2014年全国迎春杯六年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$81$$ " } ], [ { "aoVal": "D", "content": "$$144$$ " } ] ]

[ "知识标签->拓展思维->应用题模块->分百应用题->量率对应已知单位1" ]

[ "由条件，面积变为原来的$${{\\left( 1+200 \\% \\right)}^{2}}$$，所以可供$$9\\times {{\\left( 1+200 \\% \\right)}^{2}}=81$$（个）人吃饱． " ]

"2023-07-07T00:00:00"

[ "2020年新希望杯六年级竞赛（2月）第25题" ]

[ [ { "aoVal": "A", "content": "$$\\textgreater$$ " } ], [ { "aoVal": "B", "content": "$$\\textless{}$$ " } ], [ { "aoVal": "C", "content": "$$=$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}} ~\\textless{} ~\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{1999\\times 2020}$$ $$=1-\\frac{1}{2020}=\\frac{2019}{2020}$$． ∴$$1+\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}} ~\\textless{} ~1+\\frac{2019}{2020} ~\\textless{} ~2$$． 故选$$\\text{B}$$． " ]

"2023-07-07T00:00:00"

[ "2009年第7届创新杯四年级竞赛初赛第6题4分", "2009年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "1.25 " } ], [ { "aoVal": "B", "content": "2.0 " } ], [ { "aoVal": "C", "content": "2.5 " } ], [ { "aoVal": "D", "content": "2.25 " } ] ]

[ "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" ]

[ "上山用了$$60\\times 3+50=230$$（分），由于$$230\\div \\left( 30+10 \\right)=5\\cdots \\cdots 30$$，所以上山途中休息了$$5$$次，因此走了$$230-10\\times 5=180$$（分）。由于下山速度是上山速度的$$1.5$$倍，所以下山共走了$$180\\div 1.5=120$$（分），又由于下山每走$$30$$分钟休息$$5$$分钟，而在走完最后一个$$30$$分钟后已到达山下，不需要休息，所以下山所用总时间为$$\\left( 30+5 \\right)\\times 3+30=135$$（分），即$$2.25$$小时，选$$D$$。 " ]