Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2014年全国华杯赛小学中年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$83$ " } ], [ { "aoVal": "B", "content": "$99$ " } ], [ { "aoVal": "C", "content": "$96$ " } ], [ { "aoVal": "D", "content": "$98$~ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "较小数为$$a$$，较大数为$$2a$$，和为$$a+2a=3a$$． 和小于$$100$$，最大为$$99$$， $$3a=99$$ $$a=33$$． 较大数：$$2\\times 33=66$$，符合题意． 故选$$\\text{B}$$． " ]

[ "2014年迎春杯三年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "$$117$$ " } ], [ { "aoVal": "B", "content": "$$327$$ " } ], [ { "aoVal": "C", "content": "$$219$$ " } ], [ { "aoVal": "D", "content": "$$312$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->位值原理与进制->位值原理运用->位值原理的综合应用" ]

[ "解：设这个三位数是$$\\overline{abc}$$，依题意可知$$\\overline{2abc}-\\overline{abc2}=945$$或者$$\\overline{abc2}-\\overline{2abc}=945$$， 按照不完全拆分把$$abc$$看成一组， 当：$$\\overline{2abc}-\\overline{abc2}=945$$， $$2000+\\overline{abc}-10\\overline{abc}-2=945$$， $$1998-9\\overline{abc}=945$$， $$\\overline{abc}=117$$（与题中说互不相同矛盾）， 当：$$\\overline{abc2}-\\overline{2abc}=945$$， $$10\\overline{abc}+2-2000-\\overline{abc}=945$$， $$9\\overline{abc}-1998=945$$， $$\\overline{abc}=327$$（满足条件）。 故选:B。 " ]

[ "2017年IMAS小学中年级竞赛（第一轮）第10题3分" ]

[ [ { "aoVal": "A", "content": "$$43$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$46$$ " } ], [ { "aoVal": "D", "content": "$$51$$ " } ], [ { "aoVal": "E", "content": "$$61$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "由题意，十位数为$$1$$的两位数有$$12$$、$$13$$、$$14$$、$$15$$、$$16$$共$$5$$个，同理可得十位数为$$2$$、$$3$$、$$4$$的两位数也是各有$$5$$个，因此第$$21$$个两位数是$$51$$． 故选$$\\text{D}$$． " ]

[ "2018年第6届湖北长江杯六年级竞赛初赛A卷第9题3分", "六年级其它" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$66$$ " } ], [ { "aoVal": "C", "content": "$$75$$ " } ], [ { "aoVal": "D", "content": "$$88$$ " } ] ]

[ "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" ]

[ "已知服务费占总钱数的$$2 \\%$$， 服务费是$$4.5$$元，总钱数$$=4.5\\div 2 \\%=225$$（元）． $$225$$元是爸爸、妈妈和张紫亦发的总钱数， 平均每个人发$$225\\div 3=75$$（元）， 爸爸发的最多一定大于$$75$$元，选项$$\\text{D}$$正确． " ]

[ "2008年第6届创新杯六年级竞赛初赛B卷第3题5分", "2008年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$85$$个 " } ], [ { "aoVal": "B", "content": "$$67$$个 " } ], [ { "aoVal": "C", "content": "$$34$$个 " } ], [ { "aoVal": "D", "content": "$$17$$个 " } ] ]

[ "拓展思维->拓展思维->计数模块->容斥原理->二量容斥" ]

[ "在$$1\\text{、}2\\text{、}3\\text{、}\\cdots\\text{、}100$$这$$100$$个整数中，$$2$$的倍数有$$50$$个，$$3$$的倍数有$$33$$个，$$6$$的倍数有$$16$$个，所以能被$$2$$或$$3$$整除的整数有$$50+33-16=67$$（个）。 " ]

[ "2008年第6届创新杯五年级竞赛初赛B卷第3题5分" ]

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

[ "拓展思维->拓展思维->应用题模块->列方程解应用题->多元一次方程解应用题->整数系数二元一次方程组解应用题" ]

[ "设全班有$$x$$人，女生有$$y$$人， 可得$$6x=15y$$， $$x:y=15:6$$， 所以女生占全班的$$6\\div 15=\\frac{6}{15}=\\frac{2}{5}$$， 男生占$$1-\\frac{2}{5}=\\frac{3}{5}$$， 总棵树不变，男生人数$$:$$女生人数$$=3:2$$， 故只由男生完成，男生应植树$$15\\times \\frac{2}{3}=10$$（株）． 故选$$\\text{B}$$． " ]

[ "2011年北京学而思杯五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$37$$和$$18$$ " } ], [ { "aoVal": "B", "content": "$$74$$和$$3$$ " } ], [ { "aoVal": "C", "content": "$$74$$和$$3$$或$$37$$和$$18$$ " } ] ]

[ "知识标签->拓展思维->计数模块->枚举法综合->枚举法->有序枚举" ]

[ "两位数中，数字相同的两位数有$$11$$、$$22$$、$$33$$、$$44$$、$$55$$、$$66$$、$$77$$、$$88$$、$$99$$共九个，它们中的每个数都可以表示成两个整数相加的形式，例如$$33=1+32=2+31=3+30=\\cdots \\cdots =16+17$$，共有$$16$$种形式，如果把每个数都这样分解，再相乘，看哪两个数的乘积是三个数字相同的三位数，显然太繁琐了．可以从乘积入手，因为三个数字相同的三位数有$$111$$、$$222$$、$$333$$、$$444$$、$$555$$、$$666$$、$$777$$、$$888$$、$$999$$，每个数都是$$111$$的倍数，而$$111=37\\times 3$$，因此把这九个数表示成一个两位数与一个一位数或两个两位数相乘时，必有一个因数是$$37$$或$$37$$的倍数，但只能是$$37$$的$$2$$倍（想想为什么？）$$3$$倍就不是两位数了． 把九个三位数分解：$$111=37\\times 3$$、$$222=37\\times 6=74\\times 3$$、$$333=37\\times 9$$、$$444=37\\times12=74\\times 6$$、$$555=37\\times 15$$、$$666=37\\times 18=74\\times 9$$、$$777=37\\times 21$$、$$888=37\\times24=74\\times 12$$、$$999=37\\times 27$$． 把两个因数相加，只有($$74+3$$)$$=77$$和($$37+18$$)$$=55$$的两位数字相同．所以满足题意的答案是$$74$$和$$3$$，$$37$$和$$18$$． " ]

[ "2018年第6届湖北长江杯六年级竞赛初赛B卷第10题3分" ]

[ [ { "aoVal": "A", "content": "$$66$$，$$30$$ " } ], [ { "aoVal": "B", "content": "$$72$$，$$36$$ " } ], [ { "aoVal": "C", "content": "$$81$$，$$45$$ " } ], [ { "aoVal": "D", "content": "$$90$$，$$54$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "因为从第一桶倒入第二桶$$18$$千克，两桶重量相等， 所以第一桶重量$$-$$第二桶重量$$=36$$千克，选项$$\\text{ABCD}$$都满足， 且知第一桶重量$$\\times \\frac{2}{9}=$$第二桶重量$$\\times \\frac{2}{5}$$，只有$$\\text{C}$$项满足， 故答案选$$\\text{C}$$． " ]

[ "2015年第13届全国创新杯五年级竞赛复赛第8题" ]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$26$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "每次移动$$18$$张，那么移动牌的总数是$$18$$的倍数，要使红桃$$K$$又出现在最上面，那么$$18\\times $$次数的乘积应该是$$52$$的倍数，那么至少是$$26$$． " ]

[ "2017年第15届湖北武汉创新杯六年级竞赛决赛第1题" ]

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

[ "拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->双工程问题" ]

[ "工程问题．从开始到结束三人一共用了$$(400+600)\\div (25+52+48)=1000\\div 125=8$$天． " ]

[ "2014年全国迎春杯五年级竞赛复赛第5题" ]

[ [ { "aoVal": "A", "content": "$$9090909091$$ " } ], [ { "aoVal": "B", "content": "$$909090909091$$ " } ], [ { "aoVal": "C", "content": "$$10000000001$$ " } ], [ { "aoVal": "D", "content": "$$100000000001$$ " } ] ]

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "根据$$11$$乘法的特征``两边一拉，中间相加''可得到结果D 本题考察乘$$11$$的速算诀窍. " ]

[ "2016年全国华杯赛小学中年级竞赛在线模拟第3题", "2013年全国华杯赛小学中年级竞赛初赛A卷第3题" ]

[ [ { "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->真假话->找矛盾" ]

[ "由于只有一个人说对了，而小北支持小东，那么他们俩都错了，所以反对小东的小南说对了． ", "<p>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&ldquo;小东说的不对&rdquo;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．</p>\n<p>故选：$$\\text{C}$$．</p>" ]

[ "2007年第5届创新杯五年级竞赛第1题5分" ]

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

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "有奇数个因数的数必是完全平方数，如果不是完全平方数，它们的余数都是成对出现的，因此原题等价于求$$1000$$到$$2007$$之间的完全平方数，$${{32}^{2}}=1024$$，$${{33}^{2}}=1089$$，$${{34}^{2}}=1156$$，$$\\cdots \\cdots $$，$${{42}^{2}}=1764$$，$${{43}^{2}}=1849$$，$${{44}^{2}}=1936$$，$${{45}^{2}}=2025\\textgreater2007$$，所以一共有：$$44-32+1=13$$（个）． 故选$$\\text{D}$$． " ]

[ "2014年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$56$$ " } ], [ { "aoVal": "B", "content": "$$48$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$321$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "底边上的高为$$8$$，高将等腰三角形分成了两个相等的直角三角形，周长为$$32\\div 2+8=24$$，其中一条直角边（高）为$$8$$另一条（底边的一半）记作$$a$$、则斜边（等腰三角形的腰）应为$$16-a$$，根据勾股定理$${{(16-a)}^{2}}={{8}^{2}}+{{a}^{2}}$$，求得$$a=6$$所以三角形的面积为$$48$$． " ]

[ "2011年第7届全国新希望杯小学高年级六年级竞赛第6题4分" ]

[ [ { "aoVal": "A", "content": "$$108$$ " } ], [ { "aoVal": "B", "content": "$$96$$ " } ], [ { "aoVal": "C", "content": "$$88$$ " } ], [ { "aoVal": "D", "content": "$$81$$ " } ], [ { "aoVal": "E", "content": "$$75$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "先算卡片的总值． 最小值$$100\\div 10=10$$次，共减$$10\\times 27$$； 最多$$14$$次，共减去$$14\\times 27$$， 因此差为$$(14-10)\\times 27=108$$． " ]

[ "2020年第23届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第3题3分" ]

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

[ "拓展思维->思想->对应思想" ]

[ "如果每只猴子分$$6$$个，剩$$57$$个桃子；如果每只猴子分$$9$$个，就有$$5$$只猴子一个也分不到，还有一只猴子只分到$$3$$个，证明少了$$5\\times9+6=51$$（个），猴子共有：$$(57+51)\\div (9-6)=36$$（只）；桃子共有：$$36\\times6+57=273$$（个） 故选：$$\\text{C}$$． " ]

[ "2017年IMAS小学高年级竞赛（第一轮）第14题4分" ]

[ [ { "aoVal": "A", "content": "$$7:30$$ " } ], [ { "aoVal": "B", "content": "$$8:20$$ " } ], [ { "aoVal": "C", "content": "$$9:40$$ " } ], [ { "aoVal": "D", "content": "$$10:00$$ " } ], [ { "aoVal": "E", "content": "$$11:00$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "由题意可知，每隔$$\\left[ 8,10 \\right]=40$$分钟两条线路同时发出一辆车． 而$$7:30$$与$$6:00$$相差$$90$$分钟、$$8:20$$与$$6:00$$相差$$140$$分钟、$$9:40$$与$$6:00$$相差$$220$$分钟、$$10:00$$与$$6:00$$相差$$240$$分钟、$$11:00$$与$$6:00$$相差$$300$$分钟，其中仅$$10:00$$与$$6:00$$相差的分钟数是$$40$$的倍数． 故选：$$\\text{D}$$． " ]

[ "2017年第17届世奥赛六年级竞赛决赛第2题" ]

[ [ { "aoVal": "A", "content": "$$100n\\left( n+1 \\right)+25$$ " } ], [ { "aoVal": "B", "content": "$$100\\left( n+1 \\right)+25$$ " } ], [ { "aoVal": "C", "content": "$$100{{n}^{2}}+25$$ " } ], [ { "aoVal": "D", "content": "$$100{{n}^{2}}+100n+5$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->完全平方数->平方数的简单应用" ]

[ "$${{\\left( 10n+5 \\right)}^{2}}=100n\\left( n+1 \\right)+25=100{{n}^{2}}+100n+25$$． " ]

[ "2017年北京迎春杯小学高年级五年级竞赛模拟" ]

[ [ { "aoVal": "A", "content": "$$\\frac{3}{8}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{11}{16}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{16}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{7}{16}$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$\\rm B$$ " ]

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

[ "拓展思维->拓展思维->组合模块->逻辑推理->体育比赛->单循环赛" ]

[ "每周比赛要分出胜负分差必须在$$2$$分或以上，题中又给出每局比赛分差不超过$$3$$分，故每局比赛的分差只有两种可能：差$$2$$分或$$3$$分．且分差为$$3$$分的那局彭老师得分为$$11$$分，总分差为$$4839=9$$分，故必有$$3$$场分差为$$2$$分，另一场分差为$$3$$分；即有一场的比分为$$118$$，另两场的总比分为$$3731$$，有以下四种情况：①$$11:9$$，$$11:9$$，$$15:13$$②$$11:9$$，$$12:10$$，$$14:12$$③$$11:9$$，$$13:11$$，$$13:11$$④$$12:10$$，$$12:10$$，$$13:11$$．故一共有$$4$$种情况． " ]

[ "其它改编自2015年全国希望杯六年级竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{199}{302}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{201}{299}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{201}{302}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{199}{299}$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "分子是公差为$$2$$的等差数列，第$$100$$项是：$$1+\\left( 100-1 \\right)\\times 2=199$$， 分母是公差为$$3$$的等差数列，第$$100$$项是$$2+\\left( 100-1 \\right)\\times 3=299$$， 所以第$$100$$个数是$$\\frac{199}{299}$$． " ]

[ "2017年河南郑州联合杯竞赛附加赛第一场第1题2分" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$51$$ " } ], [ { "aoVal": "C", "content": "$$75$$ " } ], [ { "aoVal": "D", "content": "$$100$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "因为张数相同，所以先求出$$1$$张$$10$$元和$$5$$元共有$$15$$元，再看一下选项中哪个是$$15$$的倍数即可．容易看出$$C$$正确． " ]

[ "2020年新希望杯二年级竞赛初赛（团战）第24题" ]

[ [ { "aoVal": "A", "content": "甲、丁、乙、丙 " } ], [ { "aoVal": "B", "content": "丙、甲、乙、丁 " } ], [ { "aoVal": "C", "content": "丙、乙、丁、甲 " } ], [ { "aoVal": "D", "content": "乙、丙、丁、甲 " } ], [ { "aoVal": "E", "content": "丙、甲、丁、乙 " } ] ]

[ "拓展思维->能力->推理推导->言语逻辑推理" ]

[ "暂无 " ]

[ "2017年河南郑州联合杯竞赛第7题4分" ]

[ [ { "aoVal": "A", "content": "$$153$$ " } ], [ { "aoVal": "B", "content": "$$203$$ " } ], [ { "aoVal": "C", "content": "$$206$$ " } ], [ { "aoVal": "D", "content": "$$211$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "页码问题；$$1\\sim 99$$中数字``$$1$$''出现了$$20$$次，$$100\\sim 199$$中数字``$$1$$''出现了$$120$$个，剩下的$$200\\sim 500$$有$$3\\times 20=60$$（个），$$501\\sim 513$$中数字``$$1$$''出现了$$6$$次，所以一共有$$20+120+60+6=206$$（次）． " ]

[ "2014年第2届广东广州羊排赛六年级竞赛第3题1分" ]

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

[ "拓展思维->能力->实践应用" ]

[ "人数为$$48$$，则比的总份数必须是$$48$$的因数，只有$$7+5=12$$符合． " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（五）第1题" ]

[ [ { "aoVal": "A", "content": "$$32$$ " } ], [ { "aoVal": "B", "content": "$$34$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$38$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->方阵问题->实心方阵->实心方阵的增减" ]

[ "$$9\\times 9-7\\times 7=32$$（人）． " ]

[ "2018年环亚太杯三年级竞赛初赛第12题" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "吃掉的苹果加上篮子里剩下的苹果就是原来一共的苹果数量．白雪公主吃了一个苹果，王子吃了一个苹果，再加上七个小矮人各吃了一个苹果，所以一共吃了$$1+1+7=9$$个苹果，再加上篮子里还剩下的$$4$$个苹果，那么原来没吃苹果之前就有$$9+4=13$$个苹果． " ]

[ "2015年IMAS小学高年级竞赛第一轮检测试题第1题4分" ]

[ [ { "aoVal": "A", "content": "$$88075$$ " } ], [ { "aoVal": "B", "content": "$$88800$$ " } ], [ { "aoVal": "C", "content": "$$88200$$ " } ], [ { "aoVal": "D", "content": "$$74000$$ " } ], [ { "aoVal": "E", "content": "$$80800$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$32\\times 37\\times 75$$ $$=4\\times 8\\times 37\\times 3\\times 25$$ $$=(4\\times 25)\\times (37\\times 3)\\times 8$$ $$=100\\times 111\\times 8$$ $$=88800$$． " ]

[ "2020年广东深圳龙岗区亚迪学校迎春杯五年级竞赛模拟第20题2分", "2019~2020学年陕西宝鸡渭滨区五年级上学期期末第22题1分", "2019~2020学年山东济南高新区五年级下学期期末第10题1分" ]

[ [ { "aoVal": "A", "content": "$$31$$，$$71$$，$$91$$ " } ], [ { "aoVal": "B", "content": "$$13$$，$$21$$，$$37$$ " } ], [ { "aoVal": "C", "content": "$$17$$，$$37$$，$$85$$ " } ], [ { "aoVal": "D", "content": "$$43$$，$$53$$，$$73$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$\\text{A}$$中$$91$$不是质数，故$$\\text{A}$$错误； $$\\text{B}$$中$$$21$$不是质数，故$$\\text{B}$$错误． $$\\text{C}$$中$$85$$不是质数，故$$\\text{C}$$错误． 故选$$\\text{D}$$． " ]

[ "2016年河南郑州K6联赛六年级竞赛第14题2分" ]

[ [ { "aoVal": "A", "content": "$$1:6$$和$$1:5$$ " } ], [ { "aoVal": "B", "content": "$$1:5$$和$$1:6$$ " } ], [ { "aoVal": "C", "content": "$$1:4$$和$$1:5$$ " } ], [ { "aoVal": "D", "content": "$$1:4$$和$$1:6$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "盐与水的比为$$20:100=1:5$$，盐与盐水的比是$$20:\\left( 20+100 \\right)=1:6$$． " ]

[ "2004年第2届创新杯五年级竞赛复赛第6题" ]

[ [ { "aoVal": "A", "content": "$$17$$束 " } ], [ { "aoVal": "B", "content": "$$18$$束 " } ], [ { "aoVal": "C", "content": "$$19$$束 " } ], [ { "aoVal": "D", "content": "$$20$$束 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "先求出有月季花、马蹄莲、白兰花的一共有多少束，然后减去有其中$$2$$种花的束数，再加上三种花都有束数，就是三种花至少有一种的有多少束，再用总数$$50$$束，减去至少有一种的束数，就是三种都没有的束数． $$3$$种花至少有一种的有： $$16+15+21-7-8-10+5$$ $$=52-\\left( 7+8+10 \\right)+5$$ $$=52-25+5$$ $$=32$$ （束） $$50-32=18$$ （束） 答：月季花、马蹄莲、白兰花三种花都没有的花束共有$$18$$束． 故选$$\\text{B}$$． " ]

[ "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": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$77$$ " } ], [ { "aoVal": "C", "content": "$$160$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "由整除关系容易得到$$N$$的最后一位是$$7$$，显然$$N=7$$不成立，即$$K$$大于等于$$2$$，于是当$$K$$大于等于$$2$$时可以设$$N\\text{=X7}$$（表示$$X$$与$$7$$连在一起，$$X$$是一个数，不是$$X*7$$，例如$$N=857$$，则$$\\text{X=85}$$），于是$$\\text{M=1X71}$$（同上，时表示连在一起的数），把$$M$$、$$N$$表示如下：$$M={{10}^{\\wedge }}\\left( K+1 \\right)+X*100+71$$，$$N=10*X+7$$，由条件$$M=23*N$$，展开$${{10}^{\\wedge }}\\left( K+1 \\right)+X*100+71=\\left( 10*7 \\right)*23=230*X+161$$，化简得$${{10}^{\\wedge }}K=13*X+9$$，容易看出，当$$K=2$$时，$$X=7$$符合条件，所以$$K$$最小为$$2$$时，$$N=7$$，$$\\text{M=1771}$$． " ]

[ "2013年华杯赛六年级竞赛初赛" ]

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

[ "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" ]

[ "（$$1$$）根据互质数的意义，公因数只有$$1$$的两个数叫做互质数，在这$$11$$个中，质数有$$2$$、$$3$$、$$5$$、$$7$$、$$11$$，任何两个质数一定是互质数，又因为在这$$11$$个数中偶数有$$2$$、$$4$$、$$6$$、$$8$$、$$10$$五个，奇数有六个，所以任意取出$$6$$个数，其中必有两个数互质。 （$$2$$）比如取$$5$$个偶数一个奇数，在这$$5$$个偶数中$$4$$、$$6$$、$$8$$、$$10$$都是$$2$$的倍数，如果取$$1$$、$$3$$、$$5$$、$$7$$、$$9$$、$$11$$，其中$$9$$是$$3$$的倍数；如果取$$6$$、$$7$$、$$8$$、$$9$$、$$10$$、$$11$$，就没有倍数关系； （$$3$$）比如$$1$$的$$2$$倍是$$2$$，$$2$$是$$2$$的倍数，$$2$$的$$2$$倍是$$4$$，$$4$$是$$4$$的倍数.据此解答。 " ]

[ "2014年全国创新杯五年级竞赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->完全平方数->平方数的简单应用" ]

[ "质数有$$4$$个，偶数有$$5$$个，完全平方数有$$4$$个（$$0$$也是）$$4+5+4=13$$． " ]

[ "2010年第8届创新杯六年级竞赛初赛第1题4分", "2010年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{6}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{7}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{8}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{9}$$ " } ] ]

[ "拓展思维->拓展思维->行程模块->比例解行程问题->行程中的比例" ]

[ "速度比为$${{v}_{1}}:{{v}_{2}}=1:\\left( 1-\\frac{1}{7} \\right)=7:6$$，路程一定，时间比为$${{t}_{1}}:{{t}_{2}}=6:7$$，则在路上的时间增加了$$\\frac{1}{6}$$。 " ]

[ "2014年IMAS小学高年级竞赛第一轮检测试题第8题3分" ]

[ [ { "aoVal": "A", "content": "$$38$$ " } ], [ { "aoVal": "B", "content": "$$43$$ " } ], [ { "aoVal": "C", "content": "$$17$$ " } ], [ { "aoVal": "D", "content": "$$96$$ " } ], [ { "aoVal": "E", "content": "$$87$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数" ]

[ "重新排列后，除了最前面与最后面的数码外，每个数码都要出现偶数次，现只有数码$$1$$与$$6$$各出现一次，故$$17$$必须是第一个数，接下来的排列方式为：$$17$$、$$74$$、$$43$$、$$38$$、$$87$$、$$79$$、$$96$$，故第四个数是$$38$$．故选$$\\text{A}$$． " ]

[ "2014年全国迎春杯六年级竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "$$4027$$ " } ], [ { "aoVal": "B", "content": "$$4029$$ " } ], [ { "aoVal": "C", "content": "$$2013$$ " } ], [ { "aoVal": "D", "content": "$$2015$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->分数->分数运算->分数四则混合运算" ]

[ "$$2013\\times \\frac{2015}{2014}\\textgreater2013$$，$$2014\\times\\frac{2016}{2015}\\textgreater2014$$结果大于$$4027$$．结果为B． " ]

[ "2013年华杯赛四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$34$$ " } ], [ { "aoVal": "B", "content": "$$37$$ " } ], [ { "aoVal": "C", "content": "$$43$$ " } ], [ { "aoVal": "D", "content": "$$68$$ " } ] ]

[ "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" ]

[ "$$C=111-31\\times 2=49$$ $$B=111-37\\times 2=37$$ $$\\left( 49+37 \\right)\\div 2$$ $$=86\\div 2$$ $$=43$$ 答：$$B$$、$$C$$的平均数是$$43$$． 故选：$$\\text{C}$$． " ]

[ "2009年六年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->计算模块->方程基础->一元一次方程->分数、小数系数方程" ]

[ "不妨称不含水的葡萄为``干葡萄''。设运抵武昌后葡萄还剩$$x$$吨，由于运输途中``干葡萄''的重量不变，所以$$x\\times \\left( 1-40 \\% \\right)=4\\times \\left( 1-55 \\% \\right)$$，解得$$x=3$$，所以运抵武昌后葡萄剩下$$3$$吨。 " ]

[ "2019年第7届湖北长江杯六年级竞赛复赛A卷第2题3分" ]

[ [ { "aoVal": "A", "content": "$$20:1$$ " } ], [ { "aoVal": "B", "content": "$$1:10$$ " } ], [ { "aoVal": "C", "content": "$$10:9$$ " } ], [ { "aoVal": "D", "content": "$$1:9$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "原有的盐有$$100$$克$$\\times 5 \\%=5$$克，所以原有水$$95$$克． 加入$$10$$克盐和$$40$$克水后，有盐$$5$$克$$+10$$克$$=15$$克，有水$$95$$克$$+40$$克$$=135$$克． 所以盐$$:$$水$$=15$$克$$:135$$克$$=1:9$$． 故选$$\\text{D}$$． " ]

[ "2017年全国美国数学大联盟杯小学中年级四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$33$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$3080=2^{3}\\times5\\times7\\times11$$, $$5+7+8+11=31$$． " ]

[ "2022年陕西西安六年级下学期小升初模拟《推理问题》第9题", "2016年陕西西安小升初工大附中入学真题4第10题", "2021年陕西西安碑林区西安铁一中龙岗中学小升初（三）第10题3分", "2016年陕西西安小升初工大附中", "2010年天津陈省身杯三年级竞赛第13题" ]

[ [ { "aoVal": "A", "content": "甲 " } ], [ { "aoVal": "B", "content": "乙 " } ], [ { "aoVal": "C", "content": "丙 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "这是逻辑推理题，理清思路即可．假设甲说的是真话，那么是乙干的，这时丙说的话是真话，与只有一人说真话产生矛盾，因此甲说的是假话，即不是乙干的，所以乙说的是真话，从而丙说的是假话，故是丙干的． " ]

[ "2016年环亚太杯三年级竞赛初赛第9题" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$2.5$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$3.5$$ " } ], [ { "aoVal": "F", "content": "$$4.5$$ " } ] ]

[ "拓展思维->能力->逻辑分析", "课内体系->知识点->数学广角->优化->解决问题策略" ]

[ "$$\\frac{7}{2}=3.5$$公升$$\\textgreater3$$公升，所以只能将$$3$$公升的水桶装满，且$$7$$公升水桶中仍有$$3$$公升，则$$7-3=4$$公升，即给$$4$$公升水给$$3$$公升水桶，则$$1$$公升溢出，两个容器水一样多． 故答案为：$$4$$．选$$\\text{D}$$. " ]

[ "2008年六年级竞赛创新杯" ]

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

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "对每个选项进行检验，如（$$C$$）：$$7$$瓶，第一天喝了$$3.5+0.5=4$$（瓶），剩下$$3$$瓶；第二天喝了$$1.5+0.5=2$$（瓶），剩下$$1$$瓶；第三天喝了$$0.5+0.5=1$$（瓶），正好把妈妈买的$$7$$瓶饮料喝光。 " ]

[ "2019年美国数学大联盟杯五年级竞赛初赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$180+240$$ " } ], [ { "aoVal": "B", "content": "$$180+540$$ " } ], [ { "aoVal": "C", "content": "$$1800+2400$$ " } ], [ { "aoVal": "D", "content": "$$1800+5400$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$60\\times 120=7200$$， $$\\text{A}$$选项中，$$180+240=420$$； $$\\text{B}$$选项中，$$180+540=720$$； $$\\text{C}$$选项中，$$1800+2400=4200$$； $$\\text{D}$$选项中，$$1800+5400=7200$$． 故选：$$\\text{D}$$． " ]

[ "2006年第4届创新杯六年级竞赛初赛B卷第1题" ]

[ [ { "aoVal": "A", "content": "$$2:6$$ " } ], [ { "aoVal": "B", "content": "$$5:3$$ " } ], [ { "aoVal": "C", "content": "$$3:5$$ " } ], [ { "aoVal": "D", "content": "$$6:2$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "把工作总量看作``$$1$$''，根据工作总量$$\\div $$工作时间$$=$$工作效率，分别求出甲、乙的工作效率，再写出对应的比，根据比的基本性质化成最简整数比． $$\\left( 1\\div \\frac{1}{5} \\right):\\left( 1\\div \\frac{1}{3} \\right)=5:3$$． 答：甲与乙的工作效率比是$$5:3$$． 故选$$\\text{B}$$． " ]

[ "2017年第13届湖北武汉新希望杯六年级竞赛决赛第5题" ]

[ [ { "aoVal": "A", "content": "$$131$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$133$$ " } ], [ { "aoVal": "D", "content": "$$134$$ " } ] ]

[ "拓展思维->知识点->应用题模块->页码问题->数码综合问题", "课内体系->七大能力->逻辑分析" ]

[ "百位上是$$1$$：$$100$ $193$$共$$94$$个; 十位上是$$1$$：$$10$ $19$$，$$110$ $119$$共$$20$$个; 个位上是$$1$$：$$1,11,21,31,\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 191$$共$$20$$个; 总共$$94+20+20=134$$（个）． " ]

[ "2014年迎春杯三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$5\\times 99+1$$ " } ], [ { "aoVal": "B", "content": "$$100+25\\times 4$$ " } ], [ { "aoVal": "C", "content": "$$88\\times 4\\text{ }+37\\times 4$$ " } ], [ { "aoVal": "D", "content": "$$100\\times 0\\times 5$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数乘法巧算之提取公因数（普通型）" ]

[ "$$88\\times 4+37\\times 4=\\left( 88+37 \\right)\\times 4=125\\times 4=500$$。 " ]

[ "2017年第22届湖北武汉华杯赛三年级竞赛复赛（华罗庚金杯）第1题10分" ]

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

[ "拓展思维->思想->转化与化归的思想" ]

[ "前后括号同时除以$$111$$，即$$(8+7)\\div (6+5+4)=1$$． 故选$$\\text{A}$$． " ]

[ "2008年第6届创新杯四年级竞赛初赛A卷第4题5分", "2008年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "6991 " } ], [ { "aoVal": "B", "content": "6992 " } ], [ { "aoVal": "C", "content": "6993 " } ], [ { "aoVal": "D", "content": "6994 " } ] ]

[ "拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举" ]

[ "在这些``创新数对''中，被减数最大的是9999，此时减数是7991，而减数最小的为1000，所以``创新数对''共有$$7991-1000+1=6992$$个 " ]

[ "2018年第22届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都没有->锯木头类型问题" ]

[ "锯成$$4$$段，则需要锯$$3$$次，则所花时间为：$$3+3+3=9$$（分）． 故选$$\\text{C}$$． " ]

[ "2017年陕西西安碑林区西北工业大学附属中学小升初（二）第3题3分", "2017年陕西西安碑林区西北工业大学附属中学小升初(十)第3题3分", "2016年创新杯六年级竞赛训练题（四）第4题", "2017年陕西西安小升初某工大附中", "2018年湖南长沙雨花区中雅培粹中学小升初第3题3分", "2018年陕西西安小升初分类卷15第5题" ]

[ [ { "aoVal": "A", "content": "$$2018$$ " } ], [ { "aoVal": "B", "content": "$$2017$$ " } ], [ { "aoVal": "C", "content": "$$2016$$ " } ], [ { "aoVal": "D", "content": "$$2015$$ " } ] ]

[ "拓展思维->能力->抽象概括" ]

[ "每取出$$1$$个，盒中就增加$$6$$个，有$$7+6n=2017$$，∴$$n=335$$． 而$$\\rm A$$中，$$7+6n=2018$$，$$\\rm C$$中$$7+6n=2016$$，$$\\text{D}$$中$$7+6n=2015$$，$$n$$都不为整数． " ]

[ "2014年全国华杯赛小学高年级竞赛初赛B卷第4题" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$17$$ " } ] ]

[ "知识标签->数学思想->转化与化归的思想" ]

[ "小华所带的``快表''每小时快了$$4$$分钟，说明准确时间走$$60$$分钟的时候，``快表''已经走了$$64$$分钟了，这样我们就可以得到$$\\frac{快表}{准表}=\\frac{64}{60}=\\frac{16}{15}$$；现在快表走了$$4\\times60=240$$，那么标准表走了$$240\\times 15\\div 16=225$$；所以实际上早到了$$240-225=15$$，选$$B$$． " ]

[ "2020年希望杯一年级竞赛模拟第29题" ]

[ [ { "aoVal": "A", "content": "$$\\bigcirc +\\bigcirc $$ " } ], [ { "aoVal": "B", "content": "$$\\bigcirc +\\bigcirc+\\bigcirc $$ " } ], [ { "aoVal": "C", "content": "$$\\bigcirc +\\bigcirc+\\bigcirc +\\bigcirc $$ " } ], [ { "aoVal": "D", "content": "$$\\bigcirc +\\bigcirc+\\bigcirc +\\bigcirc +\\bigcirc $$ " } ], [ { "aoVal": "E", "content": "$$\\bigcirc +\\bigcirc+\\bigcirc +\\bigcirc +\\bigcirc +\\bigcirc $$ " } ] ]

[ "拓展思维->拓展思维->计算模块->方程基础->等量代换->可直接计算的算式代换" ]

[ "看图可知$$3$$个正方形相加$$=2$$个三角形相加，$$1$$个三角形$$=3$$个圆形相加，所以$$2$$个三角形$$=3+3=6$$（个）圆形相加，所以$$3$$个正方形相加$$=6$$个圆形相加，可以得到$$1$$个正方形$$=2$$个圆形相加，故选项$$\\text{A}$$正确． " ]

[ "2012年全国创新杯五年级竞赛第6题5分" ]

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

[ "拓展思维->能力->实践应用" ]

[ "$$123\\cdot \\cdot \\cdot 9$$为$$9$$的倍数；$$1011\\cdot \\cdot \\cdot 27$$为$$9$$的倍数；$$282930$$的数字和为$$24$$，除以$$9$$ 余$$6$$． " ]

[ "2018年湖北武汉新希望杯六年级竞赛训练题（一）第4题", "2017年新希望杯六年级竞赛训练题（一）第4题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{5}{14}\\textless{}\\frac{10}{27}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{12}{31}\\textless{}\\frac{20}{53}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ " } ] ]

[ "拓展思维->能力->数感认知->分数数字加工" ]

[ "通分子$$\\frac{5}{14}=\\frac{60}{168}$$， $$\\frac{10}{27}=\\frac{60}{162}$$， $$\\frac{12}{31}=\\frac{60}{155}$$， $$\\frac{20}{53}=\\frac{60}{159}$$． " ]

[ "2013年河南郑州中原网杯小学高年级六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$1000$$ " } ], [ { "aoVal": "B", "content": "$$1250$$ " } ], [ { "aoVal": "C", "content": "$$1729$$ " } ], [ { "aoVal": "D", "content": "$$2009$$ " } ] ]

[ "拓展思维->能力->图形认知" ]

[ "求至少有一面的块数我们可以先求一面都没有的块数，从第三个正方体开始我们可以去掉外层的正方体得到中间的$$1$$个正方体，以此类推可得到一组新的正方体，$$1$$，$$2$$，$$3$$，$$\\cdot \\cdot \\cdot $$，$$8$$，那我们就可以按照块数得到一组新的数列$${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+\\cdot \\cdot \\cdot +{{8}^{3}}=1296$$，而总块数是$${{1}^{3}}+{{2}^{3}}+{{3}^{3}}+\\cdot \\cdot \\cdot +{{10}^{3}}=3025$$，则至少有一面的块数为$$3025-1296=1729$$个．数列立方和公式：$$1+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+\\cdot \\cdot \\cdot +{{n}^{3}}={{\\left[ \\frac{n\\left( n+1 \\right)}{2} \\right]}^{2}}$$． " ]

[ "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" ]

[ [ { "aoVal": "A", "content": "小$$Y$$ " } ], [ { "aoVal": "B", "content": "小$$M$$ " } ], [ { "aoVal": "C", "content": "小$$O$$ " } ], [ { "aoVal": "D", "content": "小$$T$$ " } ] ]

[ "Overseas Competition->知识点->组合模块->逻辑推理", "拓展思维->能力->创新思维" ]

[ "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$$\\text{B}$$． " ]

[ "2016年第3届广东深圳鹏程杯小学高年级六年级竞赛集训材料第十章从算术到代数第6题" ]

[ [ { "aoVal": "A", "content": "$$y=7\\frac{1}{2}$$ " } ], [ { "aoVal": "B", "content": "$$y=8\\frac{1}{2}$$ " } ], [ { "aoVal": "C", "content": "$$y=9\\frac{1}{2}$$ " } ], [ { "aoVal": "D", "content": "$$y=10\\frac{1}{2}$$ " } ] ]

[ "知识标签->课内知识点->式与方程->简易方程（一元一次）->解简易方程->整数简易方程" ]

[ "解：把$$a+2$$代入方程$$2(x+1)=3(x-1)$$ 得：$$2(a+2+1)=3(a+2-1)$$ $$2(a+3)=3(a+1)$$ $$2a+6=3a+3$$ $$3=3a-2a$$ $$a=3$$ 把$$a=3$$代入方程$$2[2(y+3)-3(y-a)]=3a$$ 得：$$2[2(y+3)-3(y-3)]=3\\times 3$$ $$2(2y+6-3y+9)=9$$ $$2(15-y)=9$$ $$30-2y=9$$ $$30-9=2y$$ $$21=2y$$ $$y=10\\frac{1}{2}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛决赛第10题5分" ]

[ [ { "aoVal": "A", "content": "$$1.5$$ " } ], [ { "aoVal": "B", "content": "$$1\\frac{3}{4}$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3\\frac{1}{2}$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "本题主要考查分数的应用， 首先计算出三个人搬仓库用时，再分别计算甲、乙完成了多少， 然后可知丙完成的为$$1$$减去甲的完成量加上$$1$$减去乙的完成量， 再计算丙帮甲、乙分别用时即可， 三个人都搬了同样长的时间， 甲每小时搬$$\\frac{1}{6}$$，乙每小时搬$$\\frac{1}{7}$$，丙每小时搬$$\\frac{1}{14}$$， 三人每小时共可般$$\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{14}=\\frac{8}{21}$$， 则搬完两个仓库共要$$2\\div \\frac{8}{21}=\\frac{21}{4}$$时， 即三人都同样工作了$$\\frac{21}{4}$$小时， 而$$\\frac{21}{4}$$小时内甲完成$$\\frac{21}{4}\\times \\frac{1}{6}=\\frac{7}{8}$$， 乙完成$$\\frac{21}{4}\\times \\frac{1}{7}=\\frac{3}{4}$$， 即甲有$$\\frac{1}{8}$$是丙完成的，乙有$$\\frac{1}{4}$$是丙帮忙完成的， 丙帮甲、乙分别用时$$\\frac{1}{8}\\div \\frac{1}{14}=\\frac{7}{4}$$时，$$\\frac{1}{4}\\div \\frac{1}{14}=\\frac{7}{2}$$小时． 答：丙帮甲、乙分别用时$$\\frac{7}{4}$$时、$$\\frac{7}{2}小时$$， $$\\frac{7}{4}=1\\frac{3}{4}$$． 故选$$\\text{B}$$． " ]

[ "2016年新希望杯六年级竞赛训练题（二）第3题" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$27$$ " } ], [ { "aoVal": "C", "content": "$$35$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "丁今年$$22+3=25$$（岁），丙今年$$25+2=27$$（岁），甲、乙年龄和是$$127-25-27=75$$（岁），甲比乙大$$5$$岁，则乙的年龄是$$\\left( 75-5 \\right)\\div 2=35$$（岁）． " ]

[ "2011年全国美国数学大联盟杯小学高年级五年级竞赛初赛第32题" ]

[ [ { "aoVal": "A", "content": "$$120$$ " } ], [ { "aoVal": "B", "content": "$$70$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "是$$3+4=7$$的倍数． " ]

[ "2021年第8届鹏程杯四年级竞赛初赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2020$$ " } ], [ { "aoVal": "D", "content": "$$2021$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->能力->运算求解->程序性计算" ]

[ "原式$$=2020\\times 2021\\times 10001-2021\\times 2020\\times 10001=0$$． 故选$$\\text{A}$$． " ]

[ "2020年希望杯六年级竞赛模拟第25题" ]

[ [ { "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$$\\text{B}$$． " ]

[ "2006年五年级竞赛创新杯", "2006年第4届创新杯五年级竞赛初赛A卷第3题" ]

[ [ { "aoVal": "A", "content": "45 " } ], [ { "aoVal": "B", "content": "50 " } ], [ { "aoVal": "C", "content": "55 " } ], [ { "aoVal": "D", "content": "60 " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "要使某选手得分最低，其他选手得分必须尽可能的高，又知每人得分互不相同，所以前四名选手得分依次为90，89，88，87，因此，得出得分最低的选手至少得$$404-\\left( 90+89+88+87 \\right)=50$$分 " ]

[ "2019年第7届湖北长江杯六年级竞赛复赛B卷第2题3分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "天平是用来称量物体质量的工具，此题并不是称量物体的质量，而是使用天平来比较物体质量的大小，所以，在调好的天平两盘中分别放上物体，当哪边的托盘上升，则说明这边托盘中的物体质量偏小． 将$$16$$个零件分成$$6$$、$$6$$、$$4$$三组，先称量$$6$$，$$6$$的一组如果重量相等，则将$$4$$的那一组分为$$2$$，$$2$$的两组，接着称量，找出较重的一组，分为$$1$$，$$1$$的两组，进行第三次称量，找出次品；如果$$6$$，$$6$$的重量不相等，找出较重的一组分为$$3$$，$$3$$的两组，进行第二次称量，找出较重的一组，分为$$1$$，$$1$$，$$1$$的三组，任意挑选其中的两个，如果重量一样，那么另一个就是次品，再或者可以直接称量出来．这样只需$$3$$次即可找出次品． 故选$$\\text{B}$$． " ]

[ "2022年袋鼠数学竞赛(Math Kangaroo)小学高年级竞赛第19题4分" ]

[ [ { "aoVal": "A", "content": "$$2kg$$ " } ], [ { "aoVal": "B", "content": "$$3kg$$ " } ], [ { "aoVal": "C", "content": "$$4kg$$ " } ], [ { "aoVal": "D", "content": "$$5kg$$ " } ], [ { "aoVal": "E", "content": "$$6kg$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->枚举法综合->整数分拆" ]

[ "According to the given information, the 2nd heaviest dog is 28kg, so the sum of the remaining 3 dogs is: 60 - 28 = 32kg. So the heaviest one has at least 29kg, and the sum of the weight of the 2 remaining dogs is 3kg. Since none of them weigh the same, the weight of the 2 remaining dogs is 2kg and 1kg respectively. Choose A. 本题中，第二重的狗体重为$$28$$千克，剩余$$3$$只狗的体重和为：$$60-28=32$$（千克）．此时最重的狗体重至少为$$29$$千克，那么剩下的两只狗体重之和至多只有$$3$$千克，则它们的体重只能是$$1$$千克和$$2$$千克．因此第三重的狗体重为$$2$$千克．选$$\\text{A}$$． " ]

[ "2010年北京学而思综合能力诊断三年级竞赛第3题" ]

[ [ { "aoVal": "A", "content": "$$116$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$140$$ " } ], [ { "aoVal": "D", "content": "$$156$$ " } ], [ { "aoVal": "E", "content": "$$180$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数" ]

[ "每两个数差值为$$3$$，$$6$$，$$10$$，$$15$$，$$\\cdots$$ 它们的差值为$$3$$，$$4$$，$$5$$，$$\\cdots$$ 所以第$$8$$个数是$$120$$． $$1+3=4$$，$$4+6=10$$，$$10+10=20$$，$$20+15=35$$，$$35+21=56$$，$$56+28=84$$，$$84+36=120$$，每次加上的数$$3$$，$$6$$，$$10$$，$$15$$，$$21$$，$$28$$，$$36$$又有规律，$$3+3=6$$，$$6+4=10$$，$$10+5=15$$，$$15+6=21$$，$$21+7=28$$，$$28+8=36$$，所以第$$8$$个数是$$120$$． " ]

[ "2008年五年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->应用题模块->列方程解应用题->设而不求" ]

[ "设共有[6,15]=90棵数，所以共有90÷6=15位同学，其中女生有90÷15=6，那么男生有15-6=9位，平均植树90÷9=10棵，故选B " ]

[ "2017年全国华杯赛竞赛初赛模拟试卷3第1题" ]

[ [ { "aoVal": "A", "content": "$$480$$ " } ], [ { "aoVal": "B", "content": "$$510$$ " } ], [ { "aoVal": "C", "content": "$$450$$ " } ], [ { "aoVal": "D", "content": "$$620$$ " } ] ]

[ "拓展思维->拓展思维->行程模块->环形跑道->环形跑道中的相遇->环形相遇同时同地出发" ]

[ "设环形路长为$$M$$米，则甲、乙相遇$$1$$次所需时间是： $$\\frac{M}{5+7}=\\frac{M}{12}$$（秒），所以$$14\\times60\\div\\frac{M}{12}=21$$，因此，$$M=480$$． " ]

[ "2019年浙江杭州滨江区杭州江南实验学校五年级竞赛模拟（江南杯）第11题1分" ]

[ [ { "aoVal": "A", "content": "$$9.2729$$ " } ], [ { "aoVal": "B", "content": "$$9.276$$ " } ], [ { "aoVal": "C", "content": "$$9.286$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "要考虑$$9.28$$是一个三位小数的近似数，有两种情况：``四舍''得到的$$9.28$$最大是$$9.284$$，``五入''得到的$$9.28$$最小是$$9.275$$，即该数的取值是大于或等于$$9.275$$并且小于或等于$$9.284$$；由此进行选择即可． ``四舍''得到的$$9.28$$最大是$$9.284$$，``五入''得到的$$9.28$$最小是$$9.275$$，即该数的取值是大于或等于$$9.275$$并且小于或等于$$9.284$$，结合选项可知：准确值可能是$$9.276$$，故选$$\\text{B}$$． 故答案为：$$\\text{B}$$． " ]

[ "2015年第13届全国创新杯小学高年级五年级竞赛复赛第5题", "2016年创新杯小学高年级五年级竞赛训练题（三）第5题" ]

[ [ { "aoVal": "A", "content": "$$203$$ " } ], [ { "aoVal": "B", "content": "$$205$$ " } ], [ { "aoVal": "C", "content": "$$207$$ " } ], [ { "aoVal": "D", "content": "$$209$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "末尾是$$3$$，三个奇数的末尾只能是$$7$$、$$9$$、$$1$$，考虑$$6$$位数，三个奇数是两位数，且首位在$$6 \\tilde{ } 7$$之间，那么有$$67\\times 69\\times 71=328233$$满足，和是$$207$$． " ]

[ "2017年第13届湖北武汉新希望杯六年级竞赛决赛第2题" ]

[ [ { "aoVal": "A", "content": "$$148$$~ " } ], [ { "aoVal": "B", "content": "$$138$$ " } ], [ { "aoVal": "C", "content": "$$150$$~ " } ], [ { "aoVal": "D", "content": "$$151$$ " } ] ]

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "$$11:06至22:56$$共$$11$$小时$$50$$分，其中停车$$50$$分钟，运行$$11$$小时． 平均速度为$$1518\\div11=138\\rm km/h$$. " ]

[ "小学中年级三年级上学期其它", "2017年全国华杯赛小学中年级竞赛初赛模拟第1题" ]

[ [ { "aoVal": "A", "content": "$$150$$ " } ], [ { "aoVal": "B", "content": "$$140$$ " } ], [ { "aoVal": "C", "content": "$$125$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->和差倍问题->和差问题->二量和差问题->明和暗差" ]

[ "因为甲给乙$$40$$元，还比乙多$$10$$元，所以甲比乙多$$40\\times 2+10=90$$元， 所以甲：$$(210+90)\\div 2=150$$ " ]

[ "2009年第7届创新杯六年级竞赛初赛第10题4分", "2009年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "49 " } ], [ { "aoVal": "B", "content": "81 " } ], [ { "aoVal": "C", "content": "01 " } ], [ { "aoVal": "D", "content": "69 " } ] ]

[ "拓展思维->拓展思维->数论模块->余数问题->余数的性质->余数的可乘性" ]

[ "显然$${{1949}^{2009}}$$的末两位数字与$${{49}^{2009}}$$的末两位数字相同.由于$${{49}^{1}}$$的末两位数字为49，$${{49}^{2}}$$的末两位数字为01，$${{49}^{3}}$$的末两位数字为$$49$$，$${{49}^{4}}$$的末两位数字为01，\\ldots，可见$${{49}^{n}}\\left( n=1，2，3，4，... \\right)$$的末两位数字是以2为周期循环出现49和01的，所以$${{49}^{2009}}$$的末两位数字与$${{49}^{1}}$$的末两位数字相同，都是49. " ]

[ "2018年第6届湖北长江杯五年级竞赛初赛A卷第2题3分" ]

[ [ { "aoVal": "A", "content": "$$a\\textless{}b$$ " } ], [ { "aoVal": "B", "content": "$$a\\textgreater b$$ " } ], [ { "aoVal": "C", "content": "$$a=b$$ " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->算式比较大小" ]

[ "因为$$0.988\\times a$$的积和$$0.999\\times b$$的积相等， 而$$0.988\\textless{}0.999$$， 即一个因数变大，要使积不变，另一个因数需要变小， 所以$$a\\textgreater b$$． 故选$$\\text{B}$$． " ]

[ "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$a\\textgreater b$$ " } ], [ { "aoVal": "B", "content": "$$a\\textless{b}$$ " } ], [ { "aoVal": "C", "content": "$$a=b$$ " } ], [ { "aoVal": "D", "content": "与$$a$$和$$b$$的大小无关 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "可根据题意算出买进和卖出鱼所花钱的总数，买进一共花了$$(3a+2b)$$元，一共卖了$$\\frac{a+b}{2}$$$$\\times5$$元，前者大于后者，即可推导出$$a\\textgreater b$$． " ]

[ "2013年第9届全国新希望杯小学高年级六年级竞赛复赛第1题4分" ]

[ [ { "aoVal": "A", "content": "$$10$$岁 " } ], [ { "aoVal": "B", "content": "$$10.5$$岁 " } ], [ { "aoVal": "C", "content": "$$11$$岁 " } ], [ { "aoVal": "D", "content": "$$11.5$$岁 " } ] ]

[ "知识标签->拓展思维->应用题模块->平均数问题->公式类->加权平均数" ]

[ "$$\\left( 9\\times 11+11\\times 2+13\\times 3 \\right)\\div 16=10$$． " ]

[ "2006年五年级竞赛创新杯", "2006年第4届创新杯五年级竞赛初赛A卷第8题", "2006年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "2 " } ], [ { "aoVal": "B", "content": "3 " } ], [ { "aoVal": "C", "content": "4 " } ], [ { "aoVal": "D", "content": "5 " } ] ]

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

[ "解法一:设这个相同的数为$$8m+n$$或$$9\\left( m-1 \\right)+a$$，即$$8m+n=9\\left( m-1 \\right)+a$$，又$$m+a=13$$，所以$$n=m+a-9=13-9=4$$，故选C. 解法二:设这个相同的数为a，则$$\\begin{cases}a=8m+n\\left( 0\\leqslant n \\textless{} 8 \\right) a=9{{m}^{'}}+{{n}^{'}} m+{{n}^{'}}=13 \\end{cases}$$① 所以，$$a=9{{m}^{'}}+13-m$$② ①$$-$$②得$$9m-9{{m}^{'}}+n-13=0$$，所以$$9\\left( m-{m}' \\right)=13-n$$，又$$5 \\textless{} 13-n\\leqslant 13$$，所以$$5 \\textless{} 9\\left( m-{m}' \\right)\\leqslant 13$$，因此，$$m-{{m}^{'}}=1$$，从而$$9=13-n$$，故$$n=4$$. " ]

[ "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": "$$3$$ " } ], [ { "aoVal": "E", "content": "$$4$$ " } ] ]

[ "拓展思维->能力->运算求解", "Overseas Competition->知识点->计数模块->加乘原理->排队问题" ]

[ "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：$$\\text{A}$$． " ]

[ "2018年全国小学生数学学习能力测评六年级竞赛复赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "第一次弹起：$$75\\times \\frac{2}{5}=30$$（米）， 第二次弹起：$$30\\times \\frac{2}{5}=12$$（米）． 故选$$\\text{D}$$． " ]

[ "2008年第6届创新杯四年级竞赛初赛B卷第4题5分", "2008年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->枚举法综合->枚举法->有序枚举" ]

[ "和可以是$$3$$，$$5$$，$$7$$，$$9$$，$$11$$，$$13$$，$$15$$和$$17$$，所以和的个数是$$8$$。 " ]

[ "2021年新希望杯四年级竞赛初赛第15题" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$300$$ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数除法巧算之提取公除数（普通型）" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde2788\\div 4\\div 27+565\\div (27\\times 5)$$ $$=697\\div 27+565\\div 5\\div 27$$ $$=697\\div 27+113\\div 27$$ $$=(697+113)\\div 27$$ $$=810\\div 27$$ $$=30$$． 故答案为：$$30$$． " ]

[ "2007年四年级竞赛创新杯", "2007年第5届创新杯四年级竞赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$2000$$ " } ], [ { "aoVal": "B", "content": "$$2005$$ " } ], [ { "aoVal": "C", "content": "$$2007$$ " } ], [ { "aoVal": "D", "content": "$$2009$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "为使$$\\left( m+n \\right)\\div \\left( m-n \\right)$$最大，应使$$m+n$$尽可能大，而$$m-n$$尽可能小，所以$$m=1004$$，$$n=1003$$，$$\\left( m+n \\right)\\div \\left( m-n \\right)=2007$$为最大值。 " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第7题5分" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->加减法应用->同增同减应用题->连筐（桶、瓶等）带物（考虑筐桶瓶等重量）" ]

[ "倒了$$44-14-10=20$$千克，倒了相同多，小桶倒了$$20\\div 2=10$$千克，小桶原来$$10+10=20$$千克。 故选：$$\\text{C}$$． " ]

[ "2017年河南郑州豫才杯六年级竞赛初赛第5题" ]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$55$$ " } ] ]

[ "拓展思维->拓展思维->行程模块->比例解行程问题->行程中的比例" ]

[ "比例解行程；相等的时间内，甲跑了$$200$$米时，乙跑了$$160$$米，甲乙的速度比是$$200:160=5:4$$，要使两人同时到达终点，乙跑了$$200$$米的时间内甲要跑$$200\\div 4\\times 5=250$$米，所以同学甲的起跑线应后退$$250-200=50$$米． " ]

[ "2020年希望杯二年级竞赛模拟第4题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数" ]

[ "后一个数等于前一个数$$\\times$$ 2 " ]

[ "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 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "小明后面有$$5$$人，前面有$$4$$人，则他此时排在第$$5$$，之后，没有人超过他，小明又超过了$$3$$人后，他此时就是第二名． " ]

[ "2016年创新杯小学高年级六年级竞赛训练题（四）第7题" ]

[ [ { "aoVal": "A", "content": "$$883$$ " } ], [ { "aoVal": "B", "content": "$$882$$ " } ], [ { "aoVal": "C", "content": "$$888$$ " } ], [ { "aoVal": "D", "content": "$$1509$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "假设相邻两数为$$a$$、$$b$$，且$$a$$大$$b$$小，其实我们知道，$$a-b$$的差小于$$1$$时，$$[a]=[b]$$或者$$[a]-[b]=1$$；当$$a-b$$的差大于等于$$1$$时，$$[a]$$和$$[b]$$一定不同． 从小到大看，取整符号中的第$$n$$项是$$\\frac{{{\\left( 2n-1 \\right)}^{2}}}{2017}$$，第$$\\left( n+1 \\right)$$项是$$\\frac{{{\\left( 2n+1 \\right)}^{2}}}{2017}$$， 相减并利用平方差公式为$$\\frac{{{\\left( 2n+1 \\right)}^{2}}}{2017}-\\frac{{{\\left( 2n-1 \\right)}^{2}}}{2017}=\\frac{8n}{2017}$$，要想大于等于$$1$$，则$$n$$至少取$$253$$．所以取整符号中的第$$253$$项之后的数之差均大于$$1$$，即各不相同． 第$$253$$项为$$\\frac{{{505}^{2}}}{2017}=126$$，$$1-253$$项中，取整后的数每个整数都取过$$0-126$$有$$127$$个不一样的数．从第$$254-1009$$项中，每个数都不相同， 有$$1009-253=756$$个数．$$127+756=883$$（个）． " ]

[ "2003年第1届创新杯六年级竞赛复赛第5题" ]

[ [ { "aoVal": "A", "content": "$$5$$个 " } ], [ { "aoVal": "B", "content": "$$9$$个 " } ], [ { "aoVal": "C", "content": "$$11$$个 " } ], [ { "aoVal": "D", "content": "$$12$$个 " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "如果$$\\overline{2ab3}$$能被$$9$$整除，那么$$(2+a+b+3)$$一定是$$9$$的倍数，即$$(a+b+5)$$是$$9$$的倍数．所以$$(a+b+5)$$只能等于$$9$$或$$18$$． 经枚举得，当$$a+b+5=9$$时，这个四位数有：$$2043$$、$$2133$$、$$2223$$、$$2313$$、$$2403$$．共$$5$$种情况． 当$$a+b+5=18$$时，这个四位数有：$$2943$$、$$2853$$、$$2763$$、$$2673$$、$$2583$$、$$2493$$．共$$6$$种情况． 综上，这样的四位数的个数有$$5+6=11$$（个）． ", "<p>若$$9\\left| \\overline{2ab3} \\right.$$，则$$9\\left| \\left( 2+a+b+3 \\right) \\right.$$，则$$\\left( a+b \\right)$$除以$$9$$余$$4$$， 所以$$\\overline{ab}$$除以$$9$$余$$4$$，</p>\n<p>所以$$\\overline{ab}=4,4+9,4+9\\times 2,4+9\\times 3,\\ldots ,4+9\\times 10$$， 所以有$$11$$个．</p>" ]

[ "2014年全国迎春杯四年级竞赛初赛第4题" ]

[ [ { "aoVal": "A", "content": "$$10$$，$$10$$，$$10$$，$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$，$$8$$，$$20$$，$$5$$ " } ], [ { "aoVal": "C", "content": "$$9$$，$$11$$，$$5$$，$$20$$ " } ], [ { "aoVal": "D", "content": "$$9$$，$$11$$，$$12$$，$$13$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "设相同的结果为$$x$$，根据题意有：$$x-１+x+１+2x+0.5x=45$$，$$\\Rightarrow x=10$$ 易知原来的$$4$$ 个数依次是$$9$$，$$11$$，$$5$$，$$20$$． " ]

[ "2013年走美杯六年级竞赛初赛", "2013年走美杯六年级竞赛", "2013年浙江杭州走美杯六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$217017$$ " } ], [ { "aoVal": "B", "content": "$$207217$$ " } ], [ { "aoVal": "C", "content": "$$207216$$ " } ], [ { "aoVal": "D", "content": "$$217016$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之凑整法->整数基准数法" ]

[ "解：$$183\\times279\\times361-182\\times278\\times360$$ $$=\\left(182+1\\right)\\times\\left(278+1\\right)\\times\\left(360+1\\right)-182\\times278\\times360$$ $$=182\\times278\\times360+182\\times278+182\\times360+278\\times360+182+278+360+1-182\\times278\\times360$$ $$=182\\times278+182\\times360+278\\times360+182+278+360+1$$ $$=50596+65520+100080+182+278+360+1$$ $$=217017$$ 故答案是$$217017$$。 " ]

[ "2009年全国迎春杯小学中年级竞赛复赛第7题10分" ]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$90$$ " } ], [ { "aoVal": "C", "content": "$$120$$ " } ], [ { "aoVal": "D", "content": "$$150$$ " } ], [ { "aoVal": "E", "content": "$$180$$ " } ] ]

[ "拓展思维->思想->对应思想", "Overseas Competition->知识点->计数模块->加乘原理" ]

[ "假如给小强的是智力拼图，则有$$2\\times 5\\times 4\\times 3=120$$种方法． 假如给小强的是遥控汽车，则有$$1\\times 5\\times 4\\times 3=60$$种方法． 总共有$$120+60=180$$种方法． 若将遥控汽车给小强，则学习机要给小玉，此时另外$$3$$个孩子在剩余$$5$$件礼物中任选$$3$$件，有$$5\\times 4\\times 3=60$$种方法；若将遥控汽车给小玉，则智力拼图要给小强，此时也有$$60$$种方法；若遥控汽车既不给小强、也不给小玉，则智力拼图要给小强，学习机要给小玉，此时仍然有$$60$$种方法. 所以共有$$60+60+60=180$$种方法. " ]

[ "2014年迎春杯三年级竞赛复赛" ]

[ [ { "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->拓展思维->应用题模块->列方程解应用题->列方程解应用题等量代换" ]

[ "解：$$\\square +\\square +\\square +\\triangle +\\triangle +\\triangle =27$$， 得$$\\square +\\triangle =9$$①， $$\\triangle +\\triangle +\\square =12$$得$$2\\triangle +\\square =12$$②， ②$$-$$①得：$$\\triangle =3$$， 把$$\\triangle =3$$代入①得： $$\\square =6$$ 故选：D。 " ]

[ "2013年IMAS小学高年级竞赛第一轮检测试题第12题4分" ]

[ [ { "aoVal": "A", "content": "$$120$$ " } ], [ { "aoVal": "B", "content": "$$136$$ " } ], [ { "aoVal": "C", "content": "$$160$$ " } ], [ { "aoVal": "D", "content": "$$172$$ " } ], [ { "aoVal": "E", "content": "$$190$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->分解质因数->分解质因数的应用" ]

[ "解法$$1$$：将$$24$$、$$28$$、$$32$$、$$36$$分别表示成两个数码之积的行驶，即可得 $$24=3\\times 8=4\\times 6$$ $$28=4\\times 7$$ $$32=4\\times 8$$ $$36=4\\times 9=6\\times 6$$ 因这四个两位数为四个连续的两位数，故此时可判断出这四个连续的正整数为$$46$$、$$47$$、$$48$$、$$49$$，其和为$$190$$．故选$$\\text{E}$$． 解法$$2$$：由题意可知，这些所得到的乘积不包含数码$$0$$，所以这四个两位数的十位数必须相同．设它们的十位数码为$$a$$．由于两个相邻的个位数是互质的，所以$$24$$、$$28$$、$$32$$、$$36$$的最大公因数为$$a$$． 因此，$$a=4$$，且这四个数的个位数分别为$$6$$、$$7$$、$$8$$、$$9$$．即这四个整数之和是$$46+47+48+49=190$$．故选$$\\text{E}$$． " ]

[ "2017年第1届重庆华杯赛竞赛" ]

[ [ { "aoVal": "A", "content": "$$1234$$ " } ], [ { "aoVal": "B", "content": "$$1243$$ " } ], [ { "aoVal": "C", "content": "$$1343$$ " } ], [ { "aoVal": "D", "content": "$$1244$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "~ ~ $$\\frac{2013\\times 2013}{2014\\times 2014+2012}=\\frac{2013\\times 2013}{\\left( 2013+1 \\right)\\times \\left( 2013+1 \\right)+2013-1}$$ $$=\\frac{2013\\times 2013}{2013\\times 2013+2013\\times 2+1+2013-1}$$ $$=\\frac{2013\\times 2013}{2013\\times 2013+3\\times 2013} $$ $$=\\frac{671}{672}$$ $$n+m=671+672=1343$$ " ]

[ "2016年创新杯六年级竞赛训练题（二）第4题" ]

[ [ { "aoVal": "A", "content": "$$216$$ " } ], [ { "aoVal": "B", "content": "$$288$$ " } ], [ { "aoVal": "C", "content": "$$324$$ " } ], [ { "aoVal": "D", "content": "$$396$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "设$$1$$份草地原有草量为$$x$$，每天生长量为$$y$$，第三块草地需要$$n$$天吃完． 根据牛每天吃的草量相等可得： $$\\frac{x+12y}{12}=\\frac{2x+2\\left( 12y+48y \\right)}{48}$$ 解得$$x=36y$$，这群牛的每天吃草量为$$4y$$． 故第三块草地的等量关系为$$\\frac{3x+3\\left( 12y+48y+ny \\right)}{4y}=n$$，解得$$n=288$$． " ]

[ "2018年第8届北京学而思综合能力诊断六年级竞赛年度教学质量监测第12题" ]

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

[ "拓展思维->能力->逻辑分析", "课内体系->能力->逻辑分析" ]

[ "每场平局两队共得$$2$$分，如果分出胜负则两队共得$$3$$分．$$6$$支球队共要比$$6\\times 5\\div 2=15$$场比赛， 其中有$$4$$场平局，所以有$$15-4=11$$场分出了胜负，那么$$6$$支球队总得分为$$2\\times 4+3\\times 11=41$$分， 其中$$5$$支球队共得了$$31$$分，所以第$$6$$支球队得了$$41-31=10$$分． " ]

[ "2021年新希望杯六年级竞赛初赛第11题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{199}{201}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{199}{201}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{99}{100}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{100}{101}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater\\frac{99}{100}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{99}{100}\\textgreater{} \\frac{199}{201}$$ " } ] ]

[ "拓展思维->能力->数感认知->分数数字加工" ]

[ "$$\\frac{99}{100}=1- \\frac{1}{100}=1- \\frac{2}{200}$$， $$\\frac{100}{101}=1- \\frac{1}{101}=1- \\frac{2}{202}$$， $$\\frac{199}{201}=1- \\frac{2}{201}$$， 因为$$\\frac{2}{202}\\textless{} \\frac{2}{201}\\textless{} \\frac{2}{200}$$， 所以$$1- \\frac{2}{202}\\textgreater1- \\frac{2}{201}\\textgreater1- \\frac{2}{200}$$， 即$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{99}{100}$$． 故选$$\\text{D}$$． " ]

[ "2018年IMAS小学中年级竞赛（第二轮）第1题4分" ]

[ [ { "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" ]

[ "方法一：等差数列的通项公式 第$$n$$个数$$=$$第$$1$$个数+（n-1）\\times 公差 $$100-97+94-91+88-85+\\cdots +4-1$$ $$=\\left( 100-97 \\right)+\\left( 94-91 \\right)+\\left( 88-85 \\right)+\\cdots +\\left( 4-1 \\right)$$ $$=\\underbrace{3+3+3+\\cdots +3}_{17项}$$ $$=3\\times 17$$ $$=51$$． ①先分组再等差数列求有多少组：以$$4$$为首项，公差为6 ②先等差数列求有多少个数再除以$$2$$求有多少组：以$$1$$为首项，公差为3 故选$$\\text{C}$$． 方法二：等差数列求和公式 等差数列的和=（第$$1$$个数$$+$$第$$n$$个数）\\times 项数$$\\div$$ 2 $$100-97+94-91+88-85+\\cdots +4-1$$ $$=\\left( 100+94+88+\\cdots +4 \\right)-\\left( 97+91+85+\\cdots +1 \\right)$$ $$=\\frac{\\left( 100+4 \\right)\\times 17}{2}-\\frac{\\left( 97+1 \\right)\\times 17}{2}$$ $$=\\left( 104-98 \\right)\\times \\frac{17}{2}$$ $$=6\\times \\frac{17}{2}$$ $$=51$$． 故选$$\\text{C}$$ " ]