Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2016年全国华杯赛小学高年级竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$9981733$$ " } ], [ { "aoVal": "B", "content": "$$9884737$$ " } ], [ { "aoVal": "C", "content": "$$9978137$$ " } ], [ { "aoVal": "D", "content": "$$9871773$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "注意到由于任意三个连续排列的数字都能构成三位数，所以这个七位数的前五个数字不能是$$0$$，逐步极端分析，得$$988=13\\times 76$$，$$884=13\\times 68$$，$$847=11\\times 77$$，$$473=11\\times 43$$，$$737=11\\times 67$$ " ]

[ "2013年第11届创新杯四年级竞赛初赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$480$$ " } ], [ { "aoVal": "B", "content": "$$360$$ " } ], [ { "aoVal": "C", "content": "$$240$$ " } ], [ { "aoVal": "D", "content": "$$720$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "已知进货$$5$$元钱$$4$$瓶，每瓶是$$5$$ $$\\div4= \\frac{5}{4}$$ （元）， 售出时，$$5$$元钱$$3$$瓶，每瓶$$5 \\div 3= \\frac{5}{3}$$ 元，每瓶获和 $$\\frac{5}{3}- \\frac{5}{4}= \\frac{5}{12}$$（元），要获利$$300$$元，那么需售多少瓶，用 $$300 \\div \\frac{5}{12}=720$$（元），即可得解． $$5\\div 4=\\frac{5}{4}$$， $$5\\div 3=\\frac{5}{3}$$， $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde300\\div \\left( \\frac{5}{3}-\\frac{5}{4} \\right)$$ $$=300\\div \\left( \\frac{20}{12}-\\frac{15}{12} \\right)$$ $$=300\\div \\frac{5}{12}$$ $$=300\\times \\frac{12}{5}$$ $$=720$$（瓶）， 答：要获利$$300$$元，那么需售$$720$$瓶． 故选$$\\text{D}$$． " ]

[ "2016年全国小学生数学学习能力测评四年级竞赛复赛第9题3分" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "错把被除数$$113$$写成$$131$$，被除数增加了$$131-113=18$$， 因为余数相同，且商多$$3$$，除数增加了$$3$$倍，即$$18$$是除数的$$3$$倍， 所以除数是$$18\\div3=6$$， 那么该题的余数是$$113$$除以$$6$$所得的余数，据此解答． $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde(131-113)\\div3$$ $$=18\\div3$$ $$=6$$ $$113\\div6=18\\cdots\\cdots5$$． 答：原来的除数是$$6$$，余数是$$5$$． 故选$$\\text{A}$$． " ]

[ "2021年第8届鹏程杯五年级竞赛初赛第20题4分" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$32$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->能力->数感认知->数学概念理解（数）" ]

[ "能被$$9$$整除的八位数，则这个八位数的数字和必须是$$9$$的倍数， 又因为只能由数字$$1$$和$$2$$构成， 所以数字和只能$$9$$，则能是七个$$1$$和一个$$2$$， 所以只用给$$2$$选一个位置就可以，一共是八位数，所以有八个． 故选$$\\text{A}$$． " ]

[ "2016年第3届河南K6联赛小学高年级竞赛第7题5分" ]

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

[ "知识标签->课内知识点->数学广角->推理->解决简单逻辑推理问题" ]

[ "假设找矛盾 " ]

[ "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30题" ]

[ [ { "aoVal": "A", "content": "$$4:40 P.M.$$ " } ], [ { "aoVal": "B", "content": "$$10:00 P.M.$$ " } ], [ { "aoVal": "C", "content": "$$2:40 A.M.$$ " } ], [ { "aoVal": "D", "content": "$$1:00 A.M.$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$1000\\div60=16\\cdots40$$ 过了$$16$$个小时$$40$$分钟． " ]

[ "2011年第7届全国新希望杯五年级竞赛A卷第3题", "2016年创新杯五年级竞赛训练题（四）第1题" ]

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

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

[ "算出两种方案的差$$5\\times 15-6\\times 10=15$$；即原有$$15$$人，然后，$$(15+10)\\times 6\\div 15=10$$故选$$C$$． " ]

[ "2017年全国美国数学大联盟杯五年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$900$$ " } ], [ { "aoVal": "B", "content": "$$1000$$ " } ], [ { "aoVal": "C", "content": "$$1100$$ " } ], [ { "aoVal": "D", "content": "$$1200$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "在$$10000$$和$$100000$$之间的数都是五位数，不妨设其为$$\\overline{abcba}$$，$$a$$不为$$0$$有$$9$$种选择，$$b$$、$$c$$均有$$10$$种选择，故共有$$9\\times 10\\times 10=900$$个回文数． " ]

[ "1995年六年级竞赛创新杯", "2007年第5届创新杯六年级竞赛第1题5分", "2007年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "12 " } ], [ { "aoVal": "B", "content": "13 " } ], [ { "aoVal": "C", "content": "14 " } ], [ { "aoVal": "D", "content": "15 " } ] ]

[ "拓展思维->拓展思维->数论模块->质数与合数->特殊质数运用->常用质数" ]

[ "由于30以内奇质数的平均数为$$\\left( 3+5+7+11+13+17+19+23+29 \\right)\\div 9=14.111\\cdots $$，所以与此平均数最接近的整数是14 " ]

[ "2017年华杯赛四年级竞赛初赛", "2017年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$32$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$38$$ " } ], [ { "aoVal": "D", "content": "$$54$$ " } ] ]

[ "拓展思维->拓展思维->行程模块->直线型行程问题->多次相遇和追及" ]

[ "解：$$5-3=2$$（分钟），$$18\\times 2=36$$（分钟） 故选：B。 " ]

[ "2018年第6届湖北长江杯五年级竞赛初赛B卷第10题3分" ]

[ [ { "aoVal": "A", "content": "语文$$79$$分~ ~数学$$94$$分 " } ], [ { "aoVal": "B", "content": "语文$$79$$分~ ~道德与法治$$83$$分 " } ], [ { "aoVal": "C", "content": "数学$$100$$分~ ~英语$$79$$分 " } ], [ { "aoVal": "D", "content": "科学$$94$$分~ ~道德与法治$$89$$分 " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "五科的成绩之和是$$89\\times5=445$$（分）， 政治和数学的成绩之和是：$$91．5\\times2=183$$（分）， 语文和英语的成绩之和是：$$84\\times2=168$$（分）， 政治和英语成绩之和是：$$86\\times2=172$$（分）， 所以生物成绩是： 五科成绩之和$$-$$政治和数学成绩之和$$-$$语文和英语成绩之和， 语文成绩是（语文和英语成绩之和$$-10$$）$$\\div2$$， 用语文成绩减去$$10$$分就是英语成绩， 政治成绩$$=$$政治和英语成绩之和$$-$$英语成绩， 数学成绩$$=$$政治和数学成绩之和$$-$$政治成绩， 据此解答即可． 五科成绩之和：$$89\\times5=445$$（分）， 政治和数学成绩之和是：$$91．5\\times2=183$$（分）， 语文和英语的成绩之和是：$$84\\times2=168$$（分）， 政治和英语成绩之和是：$$86\\times2=172$$（分）， 生物成绩：$$445-183-168=94$$（分）， 语文成绩：$$(168-10)\\div2=158\\div2=79$$（分）， 英语成绩：$$79+10=89$$（分）， 政治成绩：$$172-89=83$$（分）， 数学成绩：$$183-83=100$$（分）． 答：五科成绩分别为生物$$94$$分，语文$$79$$分， 英语$$89$$分，政治$$83$$分，数学$$100$$分． 故答案为：$$\\text{B}$$． " ]

[ "2012年第10届全国创新杯小学高年级六年级竞赛第5题4分" ]

[ [ { "aoVal": "A", "content": "$$7:15$$ " } ], [ { "aoVal": "B", "content": "$$7:24$$ " } ], [ { "aoVal": "C", "content": "$$7:30$$ " } ], [ { "aoVal": "D", "content": "$$7:35$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "坏钟和标准钟的速度比为：$$165:180=11:12$$，现知道闹钟共走$$12\\times 60+17=737$$格，而标准钟可以转$$737\\times \\frac{12}{11}=804$$格，合计$$13$$个小时$$24$$分钟． " ]

[ "2012年第8届全国新希望杯小学高年级六年级竞赛复赛第6题4分" ]

[ [ { "aoVal": "A", "content": "$$\\overline{**45}$$ " } ], [ { "aoVal": "B", "content": "$$\\overline{19*8}$$ " } ], [ { "aoVal": "C", "content": "$$\\overline{23*1}$$ " } ], [ { "aoVal": "D", "content": "$$\\overline{3*49}$$ " } ] ]

[ "拓展思维->知识点->数论模块->完全平方数->平方数的尾数特征" ]

[ "完全平方数的个位只能为$$0$$、$$1$$、$$4$$、$$5$$、$$6$$、$$9$$，因此排除$$\\text{B}$$选项．如果一个完全平方数是 $$5$$的倍数，那么它至少是$$25$$的倍数，因此排除$$\\text{A}$$选项．估算$${{48}^{2}}=2304$$，$${{49}^{2}}=2401$$，排除 $$\\text{C}$$选项．经检验$${{57}^{2}}=3249$$． " ]

[ "2018年湖北武汉新希望杯六年级竞赛训练题（五）第1题" ]

[ [ { "aoVal": "A", "content": "$$2000$$元 " } ], [ { "aoVal": "B", "content": "$$2100$$元 " } ], [ { "aoVal": "C", "content": "$$2200$$元 " } ], [ { "aoVal": "D", "content": "$$2300$$元 " } ] ]

[ "拓展思维->拓展思维->计算模块->比和比例->比例->化连比" ]

[ "甲：乙$$=3:2$$，乙：丙$$=3:4$$，甲：乙：丙$$=9:6:8$$，$$100\\div (9-8)\\times (9+6+8)=2300$$（元）． " ]

[ "2014年华杯赛四年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->应用题模块->和差倍问题->和倍问题->二量和倍问题->两量和倍" ]

[ "解：根据$$3$$的倍数特征，不难判断$$83$$和$$98$$都不是$$3$$的倍数，$$99$$和$$96$$都是，但$$99\\textgreater96$$，所以这两个数的最大值是$$99$$。 故选：B。 " ]

[ "2020年第23届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第8题3分" ]

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

[ "拓展思维->思想->方程思想" ]

[ "设$$10$$年前儿子的年龄是$$x$$岁，那么父亲的年龄是$$7x$$； 根据题意可得： $$7x+10+15=\\left( x+10+15 \\right)\\times 2$$， $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde7x+25=2x+50$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde5x=25$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde x=5$$， 现在儿子的年龄是：$$x+10=5+10=15$$（岁）， 父亲的年龄是：$$7x+10=7\\times 5+10=45$$（岁）． $$45\\div 15=3$$（倍）． 故选$$\\text C$$． " ]

[ "2014年全国迎春杯三年级竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "红珠 " } ], [ { "aoVal": "B", "content": "黄珠 " } ], [ { "aoVal": "C", "content": "绿珠 " } ], [ { "aoVal": "D", "content": "白珠 " } ] ]

[ "知识标签->拓展思维->应用题模块->周期问题->基本排列的周期问题" ]

[ "$$252\\div (4+3+2+1)=25······2$$．红色 " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛初赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->列方程解应用题->一元一次方程解应用题->方程法解其他问题" ]

[ "题目中有$$2$$个时间线，一是现在的年龄和$$35$$岁的现在时，二是当哥哥像弟弟现在这样大时的过去时． 在过去时中：弟弟的年龄是$$1$$份，则哥哥为$$2$$份，所以哥哥比弟弟多$$1$$份． 到现在时的时候：弟弟的年龄$$2$$份，则哥哥为$$3$$份．差仍然是$$1$$份． 所以现在：$$35\\div$$ （2+3）=7（岁/份） 弟弟：$$2\\times$$ 7=14（岁） 哥哥：$$3\\times$$ 7=21（岁） " ]

[ "2019年美国数学大联盟杯五年级竞赛初赛第34题5分" ]

[ [ { "aoVal": "A", "content": "$$392$$ " } ], [ { "aoVal": "B", "content": "$$448$$ " } ], [ { "aoVal": "C", "content": "$$512$$ " } ], [ { "aoVal": "D", "content": "$$620$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->加乘原理->加乘原理综合", "Overseas Competition->知识点->计数模块->加乘原理" ]

[ "本题题意为``大于$$99$$且小于$$1000$$的整数既不包含数字$$1$$也不包含数字$$2$$有多少个？''本题相当于问三位数中即不包括数字$$1$$也不包括数字$$2$$的有多少个，本题考查排列组合，先考虑个位上可能出现的数字有$$0$$、$$3$$、$$4$$、$$5$$、$$6$$、$$7$$、$$8$$、$$9$$总共$$8$$个，同理十位也是$$8$$个，百位上去掉不能为$$0$$的情况有$$7$$个数字，因此一共有$$7\\times 8\\times 8=448$$（个）． 故选$$\\text{B}$$． " ]

[ "2013年全国走美杯五年级竞赛初赛B卷第11题", "2013年全国走美杯六年级竞赛初赛第11题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ] ]

[ "知识标签->学习能力->七大能力->数据处理" ]

[ "本题可以算出答案，但用递推推出答案会使过程更简单，由定义式可推得：$$1 \\And3=3\\times 5-2$$，$$1 \\And3 \\And5=(1\\times3\\times 5-2+2)\\times 7-2=1\\times 3\\times 5\\times 7-2$$$$\\cdots$$以此类推，$$1 \\And3 \\And5 \\And 7 \\And 9 \\And 11=1\\times 3\\times 5\\times 7\\times 9\\times11\\times 13-2$$，故原式答案为$$2$$． " ]

[ "2017年第16届湖北武汉创新杯小学高年级六年级竞赛邀请赛训练题（二）" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$42$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "一班原来的人数占两班总人数的$$\\frac{8}{15}$$，$$4+5=9$$，如果将一班的$$8$$名同学调到二班去，那么一班的人数占两班总人数的$$\\frac{4}{9}$$．原来两班总人数是:$$8\\div \\left( 8\\div154\\div9 \\right)=90$$，因此，原来一班有：$$90\\times \\frac{8}{15}=48$$人，二班有：$$90\\times \\frac{7}{15}=42$$人． " ]

[ "2018年第12届北京学而思杯四年级竞赛初赛线上第13题8分" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$50$$ " } ] ]

[ "知识标签->课内知识点->式与方程->数量关系->路程=速度×时间" ]

[ "火车的路程为$$240+360=600$$（米），所以需要$$\\left( 240+360 \\right)\\div 20=30$$（秒）． " ]

[ "2019年全国小学生数学学习能力测评五年级竞赛复赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "无需将锁打开，只需匹配成功，故最多：$$3+2+1=6$$（次）． 故选$$\\text{B}$$． " ]

[ "2019年第24届YMO一年级竞赛决赛第3题3分" ]

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

[ "拓展思维->思想->对应思想" ]

[ "五堆共有$$20$$个苹果，要使数量最多的那堆最多， 则其它$$4$$堆尽量小， 而其它$$4$$堆最小为$$1$$，$$2$$，$$3$$，$$4$$个苹果， 则数量最多那堆最多， $$20-1-2-3-4=10$$（个）． 故选$$\\text{C}$$． " ]

[ "2017年第15届全国希望杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac{11}{10}$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{10}{11}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{9}{10}$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "原式$$=\\frac{\\frac{234-2}{990}+\\frac{84}{495}}{\\frac{568-56}{900}-\\frac{56}{450}}$$ $$=\\frac{\\frac{232}{990}+\\frac{168}{990}}{\\frac{512}{900}-\\frac{112}{900}}$$ $$=\\frac{\\frac{400}{990}}{\\frac{400}{900}}=\\frac{10}{11}$$ " ]

[ "2021年河南郑州二七区京广实验学校小升初第5题2分", "2017年江苏镇江句容市句容市崇明小学小升初第16题1分", "2017年河南郑州联合杯竞赛决赛第5题2分" ]

[ [ { "aoVal": "A", "content": "直角 " } ], [ { "aoVal": "B", "content": "锐角 " } ], [ { "aoVal": "C", "content": "钝角 " } ], [ { "aoVal": "D", "content": "平角 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "当$$9:30$$时，时针和分针所组成的角是钝角． " ]

[ "2020年希望杯二年级竞赛模拟第15题" ]

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

[ "拓展思维->拓展思维->应用题模块->应用题模块排队问题->单主角求总数", "Overseas Competition->知识点->应用题模块->应用题模块排队问题" ]

[ "根据题意分析可知，用黄小鸭前面的数量加上后面的数量再加上自己所以一共有： $$2+3+1=6$$（只）小鸭，再加上鸭妈妈 故答案为：$$\\text{D}$$． " ]

[ "2018年IMAS小学高年级竞赛（第一轮）第2题3分" ]

[ [ { "aoVal": "A", "content": "$$202$$ " } ], [ { "aoVal": "B", "content": "$$212$$ " } ], [ { "aoVal": "C", "content": "$$218$$ " } ], [ { "aoVal": "D", "content": "$$224$$ " } ] ]

[ "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" ]

[ "$$\\square ~~\\textless{} ~\\frac{2018}{9}=224\\frac{2}{9}$$，最大的整数为$$224$$． 故选$$\\text{D}$$． " ]

[ "2014年第10届全国新希望杯小学高年级六年级竞赛复赛第4题4分" ]

[ [ { "aoVal": "A", "content": "$$3200$$ " } ], [ { "aoVal": "B", "content": "$$3100$$ " } ], [ { "aoVal": "C", "content": "$$2012.5$$ " } ], [ { "aoVal": "D", "content": "$$3500$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->比例应用题->和不变" ]

[ "和不变：$$1:7=4:28$$和$$8$$份$$\\times 4$$ $$9:23$$和$$32$$份 故$$1$$份为$$500\\div \\left( 9-4 \\right)=100$$（米） 全长为$$32\\times 100=3200$$（米） 选$$A$$． " ]

[ "2016年第14届全国创新杯五年级竞赛初赛第8题" ]

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

[ "拓展思维->思想->整体思想" ]

[ "斐波那契数列从第三项开始，等于前两项的和，即$$1$$、$$2$$、$$3$$、$$5$$、$$8$$、$$13$$、$$21$$． 即第二长的为$$13$$． " ]

[ "2019年第7届湖北长江杯六年级竞赛复赛A卷第10题3分" ]

[ [ { "aoVal": "A", "content": "$$476$$ " } ], [ { "aoVal": "B", "content": "$$400$$ " } ], [ { "aoVal": "C", "content": "$$470$$ " } ], [ { "aoVal": "D", "content": "$$486$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "设个位是$$a$$，十位是$$a+1$$，百位$$17-a-a-1=16-2a$$． 根据题意列出方程：$$100a+10(a+1)+16-2a-100(16-2a)-(10a+1)-a=198$$， 解这个方程，求出个位数字，然后再求十位与百位数字，解决问题． 设原数个位为$$a$$，则十位为$$a+1$$，百位为$$16-2a$$， 根据题意列方程 $$100a+10(a+1)+16-2a-100(16-2a)-(10a+1)$$， 解得$$a=6$$，则$$a+1=7$$，$$16-2a=4$$； 答：原数为$$476$$． " ]

[ "2018年第8届北京学而思综合能力诊断一年级竞赛年度教学质量第12题" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$20$$~ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "一排的人数：$$5+3-1=7$$（人） 两排的人数：$$7+7=14$$（人） 同学的人数：$$14-1=13$$（人）． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛决赛第9题5分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->思想->逆向思想" ]

[ "一个数的一半的一半是$$4$$，则这个数的一半为$$4+4=8$$，则这个数为$$8+8=16$$，故选$$\\text{D}$$． " ]

[ "1989年华杯赛六年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "如果在最不利的情况下能完成目标，则保证在任何情况下都能完成。最不利的情况是：抽的前$$12$$张是$$4$$种花色各$$3$$张，再抽两张王牌，这时抽第$$15$$张，无论是哪种花色，都能保证凑成$$4$$张牌同一花色。所以至少要抽$$15$$张牌，才能保证有四张牌是同一花色的。 " ]

[ "2015年湖北武汉世奥赛六年级竞赛模拟训练题（四）第1题" ]

[ [ { "aoVal": "A", "content": "$$28$$ " } ], [ { "aoVal": "B", "content": "$$29$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$31$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "经推算此月星期天的日期分别为$$7$$号、$$14$$号、$$21$$号、$$28$$号，又星期三的天数比星期四的天数多，那么最后一天应是星期三，$$31$$号． " ]

[ "2014年世界少年奥林匹克数学竞赛三年级竞赛初赛B卷第4题5分" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$46$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->年龄问题->年龄问题基本关系->年龄和" ]

[ "今年年龄和：$$32+4=36$$（岁），两人共长：$$50-36=14$$（岁），经过的年头：$$14\\div 2=7$$（年）． 答：当父子年龄和是$$50$$时，应该是$$7$$年以后的事． " ]

[ "2016年环亚太杯四年级竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "$$51.2$$ " } ], [ { "aoVal": "B", "content": "$$60$$ " } ], [ { "aoVal": "C", "content": "$$70$$ " } ], [ { "aoVal": "D", "content": "$$80$$ " } ], [ { "aoVal": "E", "content": "$$90$$ " } ], [ { "aoVal": "F", "content": "$$100$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "依题意得，这是一道有关路程的题，列式如下： $$64\\times (21-1)\\div (21-4-1)$$ $$=64\\times 20\\div 16$$ $$=80$$． 故答案为：$$80$$．选$$\\text{D}$$. " ]

[ "2017年河南郑州豫才杯六年级竞赛决赛4分", "2013年四川成都小升初某师大一中第18题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->简单抽屉原理" ]

[ "把三种颜色看做三个抽屉，从极端考虑：先摸出的是红色球、黄色球和蓝色球各$$9$$个，共$$27$$个球，再摸第$$28$$个球则一定有一种球是同色的，因此至少要摸出$$28$$个球． " ]

[ "2020年长江杯五年级竞赛复赛B卷第4题5分" ]

[ [ { "aoVal": "A", "content": "$$73$$ " } ], [ { "aoVal": "B", "content": "$$77$$ " } ], [ { "aoVal": "C", "content": "$$90$$ " } ], [ { "aoVal": "D", "content": "$$78$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "要使年龄最大，那么要使其它的年龄最小，最小的$$17$$岁，只有两个人年龄相同， 那么其它$$5$$个人年龄是$$17$$、$$18$$、$$19$$、$$20$$、$$21$$； 最大的一个人是$$190-17-17-18-19-20-21=78$$（岁）． 故选$$\\text{D}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第3题5分" ]

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

[ "拓展思维->拓展思维->组合模块->组合模块最值问题->和一定型最值问题->构造和一定最值原理" ]

[ "欲求最大的数至少为多少，则这$$4$$个不同的自然数越接近越好．则为$$8$$，$$9$$，$$11$$，$$12$$．故最大的数至少为$$12$$，所以选择$$\\text{C}$$． " ]

[ "2015年第27届广东广州五羊杯小学高年级竞赛第9题4分" ]

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

[ "拓展思维->思想->转化与化归的思想" ]

[ "显然其中的奇数都可以，这可以利用平方差公式与和差问题说明；偶数必须是$$4$$的倍数，所以有$$2000$$，$$2004$$，$$2008$$，$$2012$$可以做得到．因而共有$$2002$$，$$2006$$，$$2010$$，$$2014$$四个不能．答案为$$\\text{C}$$． " ]

[ "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->思想->数形结合思想" ]

[ "后一个数等于前一个数$$\\times$$ 2 " ]

[ "2019年第24届YMO一年级竞赛决赛第1题3分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "Overseas Competition->知识点->计算模块->方程基础->等量代换", "拓展思维->能力->逻辑分析" ]

[ "分析可知：$$\\square =\\bigcirc +\\bigcirc =2\\bigcirc $$， 把$$\\square =2\\bigcirc $$代入$$\\bigcirc +\\square +\\square =30$$中得： $$\\bigcirc +2\\bigcirc +2\\bigcirc =5\\bigcirc $$，$$5\\bigcirc =30$$， 所以$$\\bigcirc =30\\div 5=6$$， 那么$$\\square =2\\bigcirc =2\\times 6=12$$， 由此可知，$$\\square $$与$$\\bigcirc $$的差为$$12-6=6$$． 故选$$\\text{C}$$． " ]

[ "2017年全国希望杯六年级竞赛" ]

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

[ "拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->平均速度->设数法" ]

[ "设全程的路程为$$\\text{s}$$，时间为$$\\text{t}$$，则学生$$A$$走全程的平均速度为$$\\text{v=}\\frac{\\text{s}}{\\text{t}}$$，学生$$A$$下坡的时间为 $$\\frac{\\frac{\\text{s}}{2}}{2\\text{v}}=\\frac{\\text{s}}{4\\text{v}}=\\frac{\\text{s}}{4\\times \\frac{\\text{s}}{\\text{t}}}=\\frac{\\text{t}}{4}$$，所以，学生$$A$$上坡的时间为$$\\text{t-}\\frac{\\text{t}}{4}=\\frac{3}{4}\\text{t}$$，故学生$$A$$上坡的速度是$$\\frac{\\frac{\\text{s}}{2}}{\\frac{3\\text{t}}{4}}=\\frac{2\\text{s}}{3\\text{t}}=\\frac{2}{3}\\text{v}$$， 即上坡的速度是他走全程平均速度$$\\frac{2}{3}$$倍． " ]

[ "2013年华杯赛四年级竞赛初赛", "2013年华杯赛四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "小东 " } ], [ { "aoVal": "B", "content": "小西 " } ], [ { "aoVal": "C", "content": "小南 " } ], [ { "aoVal": "D", "content": "小北 " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理" ]

[ "解：根据题干分析可得，小南与小北说的话是相互矛盾的，所以两人中一定有一个人说的是正确的。 假设小北说的是正确的，则小南说``小东说的不对''是错，可得：小东说的对，这样与已知只有一个人说对了相矛盾，所以此假设不成立，故小南说的是正确的。 故选：$$\\text{C}$$。 " ]

[ "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第17题3分" ]

[ [ { "aoVal": "A", "content": "甲 " } ], [ { "aoVal": "B", "content": "乙 " } ], [ { "aoVal": "C", "content": "丙 " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "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" ]

[ "假设甲会英语， 那么甲和乙说的是真话， 所以和已知矛盾， 所以甲不会英语； 假设乙会英语， 那么甲和乙说的是假话， 丙说的是真话，符合题意； 假设丙会英语， 那么乙和丙说的是真话，和题意矛盾， 所以丙不会英语， 所以会英语的是乙． 选$$\\text{B}$$． " ]

[ "2004年第2届创新杯六年级竞赛初赛第4题", "2004年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$220$$人 " } ], [ { "aoVal": "B", "content": "$$280$$人 " } ], [ { "aoVal": "C", "content": "$$360$$人 " } ], [ { "aoVal": "D", "content": "$$420$$人 " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->认识单位1" ]

[ "由题意可知，$$480$$人加上$$60$$人正好是全校学生人数的$$\\frac{3}{5}$$，因此全校学生人数为：$$\\left( 480+60 \\right)\\div \\frac{3}{5}$$，然后减去男生人数，就是女生人数．据此解答． $$\\left( 480+60 \\right)\\div \\frac{3}{5}-480=540\\times \\frac{5}{3}-480=900-480=420$$（人）． 答：这个学校女生的人数是$$420$$人． 故选$$\\text{D}$$． " ]

[ "2013年第11届全国小机灵杯三年级竞赛决赛第6题" ]

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

[ "知识标签->课内知识点->数学广角->简单的周期->简单的周期计算" ]

[ "$$3$$月有$$31$$天，从中任取连续$$28$$天，正好是四周，恰好有$$4$$个周三和$$4$$个周六， 因此，剩下的$$3$$天中不能有周三和周六 不妨取$$3$$月$$4$$日到$$3$$月$$31$$日，这$$28$$天中恰有$$4$$个周三和$$4$$个周六 剩下的$$1$$日到$$3$$日是连续的三天，其中有不能有周三和周六，发现这三天只能是周日、周一、周二 因此，$$3$$月$$1$$日是周日． " ]

[ "2016年全国小学生数学学习能力测评五年级竞赛初赛第8题3分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{8}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{8}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{5}{8}$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "全部情况有：$$2\\times 2=4$$（种）， 最终得到一个正面一个反面可有以下$$2$$种情况： 正反、反正， 故所求概率为$$\\frac{2}{4}$$． 故答案为：$$\\text{B}$$． " ]

[ "2012年IMAS小学中年级竞赛第一轮检测试题第19题4分" ]

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

[ "拓展思维->思想->对应思想" ]

[ "因为$$A\\textgreater E\\textgreater C\\textgreater D\\textgreater B$$，所以第三大的数为$$\\text{C}$$． " ]

[ "小学高年级其它华杯教程", "2019年华杯赛小学高年级竞赛第9题10分" ]

[ [ { "aoVal": "A", "content": "$$41$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$38$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "用$$13$$，$$26$$，$$39$$，$$52$$，$$65$$，$$78$$，$$91$$和$$17$$，$$34$$，$$51$$，$$68$$，$$85$$拼出一个八位整数$$39178526$$．除此，无其他解答． " ]

[ "2020年长江杯六年级竞赛复赛A卷第6题5分" ]

[ [ { "aoVal": "A", "content": "$$650$$ " } ], [ { "aoVal": "B", "content": "$$550$$ " } ], [ { "aoVal": "C", "content": "$$450$$ " } ], [ { "aoVal": "D", "content": "$$400$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "百位数字和个位数字相等的数有∶ $$1\\times 10\\times 10\\times 1=100$$（个）， 因为个位数字比百位数字大的和比百位数字小的各占一半， 所以满足条件的数有$$\\left( 1000-100 \\right)\\div 2=450$$（个）． 故选：$$\\text{C}$$． " ]

[ "2017年全国小升初八中入学备考课程", "2014年全国迎春杯三年级竞赛复赛第8题" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$26$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->还原问题->多量还原问题" ]

[ "还原法，初二没喝之前有$$(2+1)\\times 2=6$$（瓶），初一没喝之前有$$6\\times 2=12$$（瓶），开始有$$(12-1)\\times 2=22$$（瓶）． " ]

[ "2019年第22届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第3题3分" ]

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

[ "拓展思维->能力->公式记忆->符号化数学原理" ]

[ "从题干可以知道，用$$26$$厘米长的铁丝围成长方形，那么长$$+$$宽$$=26\\div 2=13$$（厘米）， 由于长和宽都是整厘米数， $$13=1+12=2+11=3+10=4+9=5+8=6+7$$， 所以不同的围法有$$6$$种． " ]

[ "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第13题3分" ]

[ [ { "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->平均数问题->综合题->综合题普通型" ]

[ "从题干可以知道，五个数的平均数是$$12$$， $$12\\times5=60$$， 那么五个数之和是$$60$$； 如果把其中一个数改为$$1$$， 这时五个数的平均数是$$11$$， $$11\\times5=55$$， 此时五个数之和是$$55$$； $$60-55+1=6$$； 那么这个被改动的数原来是$$6$$． " ]

[ "2014年迎春杯四年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "$$53$$ " } ], [ { "aoVal": "B", "content": "$$56$$ " } ], [ { "aoVal": "C", "content": "$$103$$ " } ], [ { "aoVal": "D", "content": "$$106$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->四则混合运算" ]

[ "解：$$2014\\div (2\\times 2+2\\times 3+3\\times 3)$$ $$=2014\\div (4+6+9)$$ $$=2014\\div 19$$ $$=106$$ 故选：D。 " ]

[ "2020年新希望杯二年级竞赛初赛（个人战）第4题", "2020年新希望杯二年级竞赛决赛（8月）第4题" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "根据$$1$$斤菠萝可以做$$2$$个菠萝蛋糕，那么要做$$10$$个菠萝蛋糕，需要$$10\\div2=5$$（斤）菠萝．故选择$$\\text{B}$$． " ]

[ "2014年全国迎春杯六年级竞赛复赛第13题" ]

[ [ { "aoVal": "A", "content": "$$188$$ " } ], [ { "aoVal": "B", "content": "$$178$$ " } ], [ { "aoVal": "C", "content": "$$168$$ " } ], [ { "aoVal": "D", "content": "$$158$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "设第$$1$$段有$$n$$个，则第$$2$$段有$$n+1$$个，第一个擦的奇数是$$2n+1$$，第二个擦的奇数是$$4n+5$$，和为$$6n+6$$，是$$6$$的倍数．只有$$168$$符合． " ]

[ "2009年华杯赛三年级竞赛初赛", "2009年华杯赛四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "6 " } ], [ { "aoVal": "B", "content": "9 " } ], [ { "aoVal": "C", "content": "12 " } ], [ { "aoVal": "D", "content": "15 " } ] ]

[ "拓展思维->拓展思维->应用题模块->和差倍问题->差倍->两量差倍->暗差型二量差倍问题" ]

[ "由于开学前6天时小强比小明多做了60道题，而开学时两人做的题一样多，所以这6天中小明比小强多做了60道题，而这6天中小明做的题的数目是小强的3倍，所以这6天小明做了$$60\\div \\left( 3-1 \\right)\\times 3=90$$(道) 题，他平均每天做$$90\\div 6=15$$题.正确答案为 D. " ]

[ "2013年第12届全国小机灵杯小学中年级三年级竞赛初赛第18题15分" ]

[ [ { "aoVal": "A", "content": "$$990$$ " } ], [ { "aoVal": "B", "content": "$$996$$ " } ], [ { "aoVal": "C", "content": "$$998$$ " } ], [ { "aoVal": "D", "content": "$$999$$ " } ], [ { "aoVal": "E", "content": "$$1009$$ " } ], [ { "aoVal": "F", "content": "$$1010$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->周期问题->数列操作周期问题->数的周期" ]

[ "第一个数为$$6$$，第二个数为$$3$$， 根据``从第二个数起，每个数都比它前面的那个数与后面的那个数的和小$$5$$''，可以得到第三个数为$$3+5-6=2$$； 同理，后面的数依次得到为:$$6$$、$$3$$、$$2$$、$$4$$、$$7$$、$$8$$、$$6$$、$$3$$、$$2$$、$$4$$\\ldots\\ldots． 易知，每$$6$$个数为一个周期循环．$$200\\div 6=33$$（组）$$\\cdots \\cdots 2$$（个）． 所以前$$200$$个数的和为$$33\\times \\left( 6+3+2+4+7+8 \\right)+6+3=999$$． " ]

[ "2017年河南郑州豫才杯六年级竞赛决赛4分" ]

[ [ { "aoVal": "A", "content": "$$196$$ " } ], [ { "aoVal": "B", "content": "$$285$$ " } ], [ { "aoVal": "C", "content": "$$273$$ " } ], [ { "aoVal": "D", "content": "$$308$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "每辆$$A$$种汽车$$1$$小时可运$$144\\div 3\\div 4=12$$吨，则$$5$$辆$$A$$种汽车$$2$$小时可运$$12\\times 5\\times 2=120$$吨，每辆$$B$$种汽车$$1$$小时可运$$\\left( 296-120 \\right)\\div 8\\div 2=11$$吨，所以$$7$$辆$$B$$种汽车$$4$$小时可运$$11\\times 7\\times 4=308$$吨． " ]

[ "2015年第13届全国创新杯六年级竞赛第3题", "小学高年级六年级其它2015年数学思维能力等级测试初试第3题4分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{4}{33}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{33}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{8}{33}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{11}$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "原式$$=\\frac{1}{3}-\\frac{1}{11}=\\frac{8}{33}$$． " ]

[ "2013年走美杯五年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac{8}{9}+\\frac{1}{2}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{8}{9}-\\frac{1}{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{8}{9}\\times \\frac{1}{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{8}{9}\\div \\frac{1}{2}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数基准数法" ]

[ "解：$$\\frac{8}{9}+\\frac{1}{2}\\textgreater\\frac{8}{9}$$， $$\\frac{8}{9}-\\frac{1}{2} \\textless{} \\frac{8}{9}$$， $$\\frac{8}{9}\\times \\frac{1}{2} \\textless{} \\frac{8}{9}$$ $$\\frac{8}{9}\\div \\frac{1}{2}\\textgreater\\frac{8}{9}$$， 又因为，$$\\frac{8}{9}+\\frac{1}{2} \\textless{} \\frac{8}{9}\\div \\frac{1}{2}$$， 所以，结果最大的是$$\\frac{8}{9}\\div \\frac{1}{2}$$。 故选：D。 " ]

[ "2016年创新杯五年级竞赛训练题（四）第6题", "2015年第11届全国新希望杯五年级竞赛复赛第6题" ]

[ [ { "aoVal": "A", "content": "$$0.7$$千米/小时 " } ], [ { "aoVal": "B", "content": "$$1.4$$千米/小时 " } ], [ { "aoVal": "C", "content": "$$0.9$$千米/小时 " } ], [ { "aoVal": "D", "content": "$$1.8$$千米/小时 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "原来的速度和是船速和$$60\\div 2=30$$，那么船速是$$30\\div 2=15$$千米/时，第二次时间为$$60\\div 40=1.5$$小时，那么水速是$$0.45\\div (2-1.5)=0.9$$千米/时． " ]

[ "2014年IMAS小学高年级竞赛第一轮检测试题第2题3分" ]

[ [ { "aoVal": "A", "content": "$$6$$、$$17$$、$$59$$ " } ], [ { "aoVal": "B", "content": "$$4$$、$$17$$、$$53$$ " } ], [ { "aoVal": "C", "content": "$$2$$、$$13$$、$$59$$ " } ], [ { "aoVal": "D", "content": "$$2$$、$$19$$、$$53$$ " } ], [ { "aoVal": "E", "content": "$$2$$、$$23$$、$$29$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "选项$$\\text{A}$$乘积的末位数为$$8$$，故不合； 选项$$\\text{B}$$的乘积$$4\\times 17\\times 53\\textgreater60\\times 50=3000$$，故不合； 选项$$\\text{C}$$的乘积$$2\\times 13\\times 59\\textless{}30\\times 60=1800$$，故不合； 选项$$\\text{E}$$的乘积$$2\\times 23\\times 29\\textless{}50\\times 30=1500$$，故不合； 只有选项$$\\text{D}$$的乘积$$2\\times 19\\times 53=2014$$，符合所求． 故选$$\\text{D}$$． " ]

[ "2016年第16届世奥赛六年级竞赛决赛第8题", "2016年全国世奥赛竞赛A卷第8题" ]

[ [ { "aoVal": "A", "content": "$$1262$$ " } ], [ { "aoVal": "B", "content": "$$1222$$ " } ], [ { "aoVal": "C", "content": "$$1300$$ " } ], [ { "aoVal": "D", "content": "$$2496$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "找到石油月初与月末单价的等量关系列出方程，设月初每盎司黄金价格是$$x$$美元，$$x\\div \\frac{47}{5}-70=\\left( x-70 \\right)\\div \\frac{96}{5}$$，解得$$x=1222$$． " ]

[ "2013年全国华杯赛小学中年级竞赛初赛A卷第6题" ]

[ [ { "aoVal": "A", "content": "周日 " } ], [ { "aoVal": "B", "content": "周六 " } ], [ { "aoVal": "C", "content": "周二 " } ], [ { "aoVal": "D", "content": "周一 " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "一般这种题目问这个月的第几天是周几时，我们要弄清楚这个月的周几被重复出现$$5$$次．计算$$5\\times 60\\div 20=15$$（天）说明这个月周一、周六、周日都是$$5$$次，说明这个月$$31$$天，且第一天是周六，所以第$$10$$天是$$10\\div 7=1\\cdots3$$，周六、周日、周一这个月的第$$10$$天是周一． " ]

[ "2016年全国世奥赛五年级竞赛A卷第11题5分" ]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$1\\sim 30$$个数中有两组这样相邻自然数：$$5$$和$$6$$；$$24$$和$$25$$．$$\\left[ 5,6 \\right]=30$$，同时考虑$$30$$个数为周期． $$5$$和$$6$$一组，下组分别为$$35$$和$$36$$；$$65$$和$$66$$；$$95$$和$$96$$；$$125$$和$$126$$；$$155$$和$$156$$；$$185$$和$$186$$，共$$7$$组． $$24$$和$$25$$为一组，下组分别为$$54$$和$$55$$；$$84$$和$$85$$；$$114$$和$$115$$；$$144$$和$$145$$；$$174$$和$$175$$，共$$6$$组． 故满足条件的数组共有$$7+6=13$$（组）． " ]

[ "2004年六年级竞赛创新杯", "2004年第2届创新杯六年级竞赛复赛第2题" ]

[ [ { "aoVal": "A", "content": "$$1232$$毫米 " } ], [ { "aoVal": "B", "content": "$$616$$毫米 " } ], [ { "aoVal": "C", "content": "$$750$$毫米 " } ], [ { "aoVal": "D", "content": "$$480$$毫米 " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->综合题" ]

[ "从$$1993$$年初到$$2002$$年底这$$10$$年的平均降雨量为$$631$$毫米，那么这$$10$$年的总降雨量为$$631\\times 10=6310$$毫米，一年后，再统计近$$10$$年的平均降雨量为$$601$$毫米，为$$1994$$年到$$2003$$年的平均降雨量，那么这$$10$$年的总降雨量为$$601\\times 10=6010$$毫米，又$$2003$$年的降雨量为$$450$$毫米，那么$$1994$$到$$2002$$年这$$9$$年的总降雨量为$$6010-450=5560$$毫米，所以$$1993$$年的降雨量为$$6310-5560=750$$毫米，选C。 " ]

[ "2016年第16届世奥赛六年级竞赛决赛第15题", "2016年全国世奥赛竞赛A卷第15题" ]

[ [ { "aoVal": "A", "content": "$$132$$ " } ], [ { "aoVal": "B", "content": "$$158$$ " } ], [ { "aoVal": "C", "content": "$$224$$ " } ], [ { "aoVal": "D", "content": "$$264$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据题目要求，我们先补充完这个作品．观察图例，第$$1$$条线为横线，第$$2$$条线为横线，第$$3$$条线为竖线，第$$4$$条线为横线，第$$5$$条线为横线，第$$6$$条线为竖线，$$\\cdot \\cdot \\cdot $$．可以推出规律，每话$$3$$条线，是先画两条横线，再画一条竖线，所以可以画到$$45$$条线时，横线有$$30$$条，竖线有$$15$$条．画到$$47$$条，则需要再画两条横线，也就是说$$47$$条线中，共有$$32$$条横线，$$15$$条竖线．纸上面共有$$\\left( 32+1 \\right)\\times \\left( 15+1 \\right)=528$$个四边形．相邻的两个四边形颜色不能相同的，可以参考国际象棋的棋盘，所以红色和黄色的四边形应该各占一半，也就是说黄色四边形的个数为$$528\\div 2=264$$个． " ]

[ "2012年美国数学大联盟杯六年级竞赛初赛第12题5分(每题5分)" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{16}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{9}{16}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{3}{4}$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "（$$\\frac{3}{4}$$的倒数）$$\\div $$（$$\\frac{4}{3}$$的倒数）$$=$$乘积是$$1$$的两个数互为倒数，所以$$\\frac{3}{4}$$的倒数是$$1\\div \\frac{3}{4}=\\frac{4}{3}$$，$$\\frac{4}{3}$$的倒数是$$1\\div \\frac{4}{3}=\\frac{3}{4}$$，$$\\frac{4}{3}\\div \\frac{3}{4}=\\frac{16}{9}$$． 故选$$\\text{B}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第9题5分" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$26$$ " } ] ]

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

[ "草地上三种兔子一共有： $$(16+17+15)\\div 2$$ $$=48\\div 2$$ $$=24$$（只）． 故选：$$\\text{C}$$． " ]

[ "2017年第15届全国希望杯五年级竞赛第1试试题第18题", "其它改编题" ]

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

[ "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" ]

[ "$$A$$和$$E$$说的话对立，$$C$$和$$D$$说的话对立，必有两对两错，故$$B$$说的是错的，则是$$E$$射中的． " ]

[ "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 " } ] ]

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

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

[ "2020年新希望杯三年级竞赛初赛（团战）第60题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$5$$ " } ] ]

[ "拓展思维->能力->推理推导->言语逻辑推理" ]

[ "暂无 " ]

[ "2018年美国数学大联盟杯三年级竞赛初赛第31题5分" ]

[ [ { "aoVal": "A", "content": "even $$\\times$$ odd " } ], [ { "aoVal": "B", "content": "odd $$\\times$$ even " } ], [ { "aoVal": "C", "content": "odd $$+$$ even " } ], [ { "aoVal": "D", "content": "odd $$+$$ odd " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "本题题意为``下面结果哪个是奇数''， $$\\text{A}$$选项是偶数$$\\times $$奇数$$=$$偶数， $$\\text{B}$$选项是奇数$$\\times $$偶数$$=$$偶数， $$\\text{C}$$选项是奇数$$+$$偶数$$=$$奇数， $$\\text{D}$$选项是奇数$$+$$奇数$$=$$偶数， 综上结果为奇数的只有$$\\text{C}$$选项． 故选$$\\text{C}$$． " ]

[ "2016年第7届广东广州羊排赛六年级竞赛第6题1分" ]

[ [ { "aoVal": "A", "content": "$$2002.0$$ " } ], [ { "aoVal": "B", "content": "$$201602016$$ " } ], [ { "aoVal": "C", "content": "$$20160201$$ " } ], [ { "aoVal": "D", "content": "$$1001.0001$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->小数基础->小数的认识" ]

[ "选项$$\\text{A}$$读法为二千零二点零，选项$$\\text{B}$$读法为二亿零一百六十万二千零一十六，选项$$\\text{C}$$读法为二千零一十六万零二百零一，选项$$\\text{D}$$读法为一千零一点零零零一． " ]

[ "2016年新希望杯六年级竞赛训练题（四）第5题" ]

[ [ { "aoVal": "A", "content": "$$90$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$130$$ " } ], [ { "aoVal": "D", "content": "$$180$$ " } ] ]

[ "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" ]

[ "设小强跳了$$x$$个，则小亮跳了$$\\frac{3}{2}x$$个，小明跳了$$\\frac{3}{4}x$$个．$$x+\\frac{3}{2}x+\\frac{3}{4}x=130\\times 3$$，解得：$$x=120$$，$$\\frac{3}{2}x=180$$，小亮跳了$$180$$个． " ]

[ "2011年世界少年奥林匹克数学竞赛六年级竞赛初赛第3题5分", "2021年陕西西安六年级下学期小升初模拟分类专题五十六（和差倍问题 盈亏问题）第6题", "2018年海南海口琼山区华中师范大学海南附属中学小升初华师杯第10题3分", "2018年浙江金华六年级下学期小升初模拟第8题2分", "2019年陕西西安灞桥区铁一中滨河中学小升初入学真卷2第12题3分", "2016年陕西西安小升初某西北大附中" ]

[ [ { "aoVal": "A", "content": "$$23$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$26$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "都不对 " } ] ]

[ "拓展思维->能力->构造模型->模型思想", "Overseas Competition->知识点->应用题模块->和差倍问题->和差问题" ]

[ "由题意的第一个桶比第二个桶多$$12$$斤，所以利用和差关系： 第一桶水：$$(40+12)\\div 2=26$$（斤）． " ]

[ "2021年新希望杯五年级竞赛初赛第39题5分" ]

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

[ "拓展思维->拓展思维->数论模块->整除->整除特征->和系整除特征应用" ]

[ "$$21=3\\times 7$$，故能被$$21$$整除必能被$$3$$和$$7$$整除，满足被$$3$$整除为各位数字之和为$$3$$的倍数， 故$$2+0+2+1=5$$故余下$$2$$位数字和最大为$$9+9=18$$， 故可取$$1$$，$$4$$，$$7$$，$$10$$，$$13$$，$$16$$， $$1$$：$$1+0$$，$$4$$：$$4+0$$，$$3+1$$，$$2+2$$，$$7$$：$$7+0$$，$$6+1$$，$$5+2$$，$$4+3$$， $$10$$：$$9+1$$，$$8+2$$，$$7+3$$，$$6+4$$，$$5+5$$， $$13$$：$$9+4$$，$$8+5$$，$$7+6$$， $$16$$：$$9+7$$，$$8+8$$， 上述除$$0$$不能在首位外共有：$$1$$：$$1$$种；$$4$$：$$1+2+1=4$$种，$$7$$：$$1+2+2+2=7$$种；$$10$$：$$2+2+2+2+1=9$$种；$$13$$：$$2+2+2=6$$种；$$16$$：$$2+1=3$$种， 能被$$7$$整除特点为（除去个位数字后的数$$-2\\times $$个位）能被$$7$$整除， 故一共有：$$420210$$，$$520212$$，$$620214$$，$$720216$$，$$820218$$共$$5$$个． 故答案为：$$5$$． " ]

[ "2017年全国美国数学大联盟杯小学高年级六年级竞赛初赛第20题5分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$38$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ] ]

[ "拓展思维->能力->逻辑分析", "Overseas Competition->知识点->数论模块->因数与倍数->公因数与公倍数" ]

[ "$$8640$$有多少个因数？ whole-number ~整数；$$factor$$ ~因数． 它的因数有$$\\left( 6+1 \\right)\\times \\left( 3+1 \\right)\\times \\left( 1+1 \\right)=56$$． " ]

[ "2018年第22届广东世界少年奥林匹克数学竞赛六年级竞赛决赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$75$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "骰子的点数为$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$． 其中质数有$$2$$，$$3$$，$$5$$， 故$$1$$个骰子掷出质数的概率为$$\\frac{1}{2}$$， 则两次均为质数的概率为$$\\frac{1}{2}\\times\\frac{1}{2}=\\frac{1}{4}=25 \\%$$， 故选$$\\text{B}$$． " ]

[ "2021年第8届鹏程杯五年级竞赛初赛第15题4分" ]

[ [ { "aoVal": "A", "content": "$$75$$ " } ], [ { "aoVal": "B", "content": "$$76$$ " } ], [ { "aoVal": "C", "content": "$$77$$ " } ], [ { "aoVal": "D", "content": "$$78$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "总分最高是$$10+10\\times3=40$$分，最低是$$10-10\\times1=0$$，$$0\\sim 40$$共$$41$$种得分都有可能取到，从两端极值考虑，最小端，$$1=10-9\\times1-0\\times1$$，$$2=10-8\\times1-0\\times2$$，$$3=10-7\\times1-0\\times3$$，则后面的分数都能取到，再考虑最大端，若做对$$9$$题，还有一题可以错，也可以不做，分别得到$$36$$分和$$37$$分，则$$38$$、$$39$$取不到，若做对$$8$$题，剩余两题可能全错，全不做，错一题不做一题，得分分别是$$32$$、$$34$$、$$33$$，则$$35$$分取不到，依次类推可以得到$$0\\sim 40$$中$$35$$、$$38$$、$$39$$三种得分取不到，所以得分情况只有$$41-3=38$$种，要想保证有$$3$$人分一样，则每种得分至少有$$2$$人，再多一人就可以满足至少有$$3$$人得分一样，所以是$$38\\times2+1=77$$人． 故选：$$\\text{C}$$． " ]

[ "2017年第15届湖北武汉创新杯六年级竞赛决赛第4题", "2018年浙江杭州西湖区小学高年级六年级上学期单元测试《第三单元》第8题3分", "2018年四川成都锦江区四川师范大学附属第一实验中学小升初（四）第4题3分" ]

[ [ { "aoVal": "A", "content": "$$25 \\%$$ " } ], [ { "aoVal": "B", "content": "$$20 \\% $$ " } ], [ { "aoVal": "C", "content": "$$16 \\% $$ " } ], [ { "aoVal": "D", "content": "$$12.5 \\%$$ " } ] ]

[ "课内体系->知识点->数的运算->估算->百分数的简单实际问题->折扣、成数、税率、利率", "拓展思维->知识点->应用题模块->经济问题->基本经济概念->利润基本公式->已知利润成本求利润率" ]

[ "方法一：利润问题．假设成本为$$100$$元，则利润为$$20$$元，现在成本是$$100\\times (1+25 \\%)=125$$元，$$20\\div 125\\times 100 \\%=16 \\%$$． 方法二：$$20\\%\\div （1+25\\%）=16\\%$$ " ]

[ "2015年湖北武汉世奥赛小学高年级六年级竞赛模拟训练题（三）第4题" ]

[ [ { "aoVal": "A", "content": "$$0$$和$$6$$ " } ], [ { "aoVal": "B", "content": "$$1$$和$$6$$ " } ], [ { "aoVal": "C", "content": "$$0$$和$$7$$ " } ], [ { "aoVal": "D", "content": "$$1$$和$$5$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "$$E$$拿的两张卡片的和为$$\\left( 0+1+\\ldots \\ldots +9 \\right)\\left( 5+12+10+12 \\right)=6$$，则$$E$$拿的卡片有三种情况：①$$2+4$$；②$$1+5$$； ③$$0+6$$．而$$5=0+5=1+4=2+3$$，$$12=3+9=4+8=5+7$$，依次假设①②③种情况，进行推理得出矛盾，最后满足条件的是③． " ]

[ "2019年第24届YMO四年级竞赛决赛第1题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第1题3分", "2020年第24届YMO四年级竞赛决赛第1题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第1题3分", "2021年第24届世界少年奥林匹克数学竞赛四年级竞赛决赛第1题3分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "桌子的价钱：$$640\\div 5=128$$（元）， 椅子的价钱：$$(128\\times 3+96)\\div 5=480\\div 5=96$$（元）， 所以乙有椅子的数量为$$640\\div (128-96)=640\\div 32=20$$（把）． 故选$$\\text{C}$$． " ]

[ "2008年第6届创新杯五年级竞赛初赛A卷第4题5分", "2008年五年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->应用题模块->鸡兔同笼问题" ]

[ "当全打中时得$$20\\times 5=100$$（分），没打中$$\\left(100-51\\right) \\div \\left(5+2\\right) =7$$（枪），故打中$$20-7=13$$（枪）。 " ]

[ "2014年迎春杯六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$34$$ " } ], [ { "aoVal": "B", "content": "$$68$$ " } ], [ { "aoVal": "C", "content": "$$144$$ " } ], [ { "aoVal": "D", "content": "$$72$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->分数->分数巧算->乘法分配律" ]

[ "原式$$=2014\\times \\frac{1}{19}-2014\\times \\frac{1}{53}=106-38=68$$ " ]

[ "2017年IMAS小学中年级竞赛（第一轮）第1题3分" ]

[ [ { "aoVal": "A", "content": "$$2750$$ " } ], [ { "aoVal": "B", "content": "$$2850$$ " } ], [ { "aoVal": "C", "content": "$$2900$$ " } ], [ { "aoVal": "D", "content": "$$2950$$ " } ], [ { "aoVal": "E", "content": "$$3000$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$25\\times 99+55\\times 5=25\\times (99+11)=25\\times 110=2750$$． 故选$$\\text{A}$$． " ]

[ "2018~2019学年浙江杭州西湖区杭州市行知小学五年级上学期期中期中竞赛第17题1分" ]

[ [ { "aoVal": "A", "content": "$$2.3\\times 3\\times 7$$ " } ], [ { "aoVal": "B", "content": "$$13.8\\times 7\\div 2$$ " } ], [ { "aoVal": "C", "content": "$$2\\times 3\\times 7+0.9\\times 7$$ " } ], [ { "aoVal": "D", "content": "$$7\\times 7-7$$ " } ] ]

[ "知识标签->拓展思维->计算模块->小数->小数乘除->小数乘法运算" ]

[ "$$7\\times 7-7=7\\times \\left( 7-1 \\right)=7\\times 6=42$$，提取公因数后发现$$\\text{D}$$式变为$$7\\times 6$$，而非原式的$$6.9\\times 7$$，所以错误． 故选$$\\text{D}$$． " ]

[ "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$35$$千克 " } ], [ { "aoVal": "B", "content": "$$45$$千克 " } ], [ { "aoVal": "C", "content": "$$55$$千克 " } ] ]

[ "知识标签->数学思想->对应思想" ]

[ "相当于把石灰水分成了$$101$$份，石灰占$$1$$份．所以石灰有$$4545\\times \\frac{1}{{1{ + }100}}{ = }45$$（千克）． " ]

[ "2014年迎春杯三年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "取$$1$$根 " } ], [ { "aoVal": "B", "content": "取$$3$$根 " } ], [ { "aoVal": "C", "content": "取$$4$$根 " } ], [ { "aoVal": "D", "content": "无论怎么取都无法获胜 " } ] ]

[ "拓展思维->拓展思维->组合模块->智巧趣题->数学趣题->火柴棒问题" ]

[ "解：无论甲怎么取，乙只要让最后火柴棍剩两根，甲这时只能取$$1$$根，乙胜； 在这之前只要保证火柴剩下$$5$$根，甲取$$1$$根，则乙取$$2$$根，剩$$2$$根，乙胜； 或者甲取$$3$$根，乙取$$2$$根，乙胜；或者甲取$$4$$根，乙取$$1$$根，乙胜。 所以甲无论怎么取都无法获胜。 故选:D。 " ]

[ "2014年全国迎春杯五年级竞赛初赛第4题", "2017年全国小升初八中入学备考课程" ]

[ [ { "aoVal": "A", "content": "大于$$12.345$$ " } ], [ { "aoVal": "B", "content": "小于$$12.345$$ " } ], [ { "aoVal": "C", "content": "等于$$12.345$$ " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "排除法，A、B、C三个选项均可找到反例，故无法确定． " ]

[ "2020年春蕾杯六年级竞赛第10题2分", "2021年春蕾杯六年级竞赛第5题2分" ]

[ [ { "aoVal": "A", "content": "$$381$$ " } ], [ { "aoVal": "B", "content": "$$382$$ " } ], [ { "aoVal": "C", "content": "$$383$$ " } ], [ { "aoVal": "D", "content": "$$384$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->多数的最小公倍数" ]

[ "$$2$$、$$3$$、$$8$$的最小公倍数是$$24$$，所以这些书号都是$$24$$的倍数， $$24\\times 16=384$$（本）． 故选$$\\text{D}$$． " ]

[ "2013年第12届上海小机灵杯小学中年级四年级竞赛初赛第4题1分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "蜗牛爬井问题． 若最后一天青蛙向上爬$$3$$米到达井口，就不会向下掉了，则之前至少爬$$20-3=17$$（米）； 每天向上爬$$3$$米，晚上休息时会掉下$$2$$米，实际每天向上$$1$$米； $$17\\div 1+1=18$$（天）； 青蛙第$$18$$天才能爬出这口井； 选$$\\text{C}$$． " ]

[ "2014年第10届全国新希望杯小学高年级六年级竞赛复赛第6题4分" ]

[ [ { "aoVal": "A", "content": "$$22$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->体育比赛->3-1-0 积分制" ]

[ "设胜了$$x$$场，平$$y$$场，负$$\\left( 30-x-y \\right)$$ 有$$3x+y=77$$ $$30-\\left( x+y \\right)\\textless{}y$$，则$$x=24$$ 选$$\\text{B}$$． " ]

[ "2015年第20届全国华杯赛小学中年级四年级竞赛初赛A卷", "2016年第20届四川成都华杯赛小学中年级竞赛B卷第1~6题60分" ]

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

[ "拓展思维->思想->分类讨论思想" ]

[ "设答对了$$x$$题，答错$$y$$题，$$x+y$$≤$$6$$； 当$$x=6$$时，得分$$30$$分； 当$$x=5$$时，$$y=0$$，$$1$$，对应得分$$26$$，$$25$$； 当$$x=4$$时，$$y=0$$，$$1$$，$$2$$，对就得分$$22$$，$$21$$，$$20$$； 当$$x=3$$时，$$y=0$$，$$1$$，$$2$$，$$3$$，对应得分$$18$$，$$17$$，$$16$$，$$15$$； 当$$x=2$$时，$$y=0$$，$$1$$，$$2$$，$$3$$，$$4$$，对应得$$14$$，$$13$$，$$12$$，$$11$$，$$10$$； 当$$x=1$$时，$$y=0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，对应得$$10$$，$$9$$，$$8$$，$$7$$，$$6$$，$$5$$； 当$$x=0$$时，$$y=0$$，$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，对应得分$$6$$，$$5$$，$$4$$，$$3$$，$$2$$，$$1$$，$$0$$； 共计$$25$$种得分，$$51\\div 25=2\\cdots 1$$，则至少$$2+1=3$$人得分相同. " ]

[ "2005年第3届创新杯六年级竞赛初赛第9题", "2005年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "2003个 " } ], [ { "aoVal": "B", "content": "2004个 " } ], [ { "aoVal": "C", "content": "2005个 " } ], [ { "aoVal": "D", "content": "2006个 " } ] ]

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

[ "把1个球变成7个球，那么球的数量就增加了6个，则无论这个魔术师怎么变，球增加的数量为6的倍数，又原有7个球，那么球的总数被6除的余数总是1，而$$A，B，C，D$$这4个选项中，只有$$C$$选项的2005除以6的余数为1，所以选$$C$$. " ]

[ "2013年第25届广东广州五羊杯六年级竞赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$90$$ " } ], [ { "aoVal": "B", "content": "$$105$$ " } ], [ { "aoVal": "C", "content": "$$150$$ " } ], [ { "aoVal": "D", "content": "$$180$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "设全校有男生有$$x$$人，则$$2x+300-x=510$$， 可解得$$x=210$$， 所以女生有$$90$$人． 故选：$$\\text{A}$$． " ]

[ "2021年鹏程杯六年级竞赛初赛第9题" ]

[ [ { "aoVal": "A", "content": "$$25\\frac{1003}{1008}$$ " } ], [ { "aoVal": "B", "content": "$$25\\frac{611}{1014}$$ " } ], [ { "aoVal": "C", "content": "$$25\\frac{609}{1026}$$ " } ], [ { "aoVal": "D", "content": "$$25\\frac{597}{1040}$$ " } ], [ { "aoVal": "E", "content": "$$25\\frac{620}{1001}$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "要使算式的值尽可能大，则$$a$$，$$d$$，$$g$$应尽可能大，可取$$a=9$$，$$d=8$$，$$g=7$$．而$$b$$，$$e$$，$$h$$则应尽可能小，所以可取$$b=1$$，$$e=2$$，$$A=3$$， 由于$$\\frac{1}{m+\\dfrac{1}{n+1}}\\textgreater\\frac{1}{(m+1)+\\dfrac{1}{n}}$$，故取$$c=6$$，而$$f$$，$$i$$只剩下$$4$$和$$5$$可供选择，故可取$$f=5$$，$$i=4$$，故最大值为$$9+\\frac{1}{1+\\dfrac{1}{6}}+8+\\frac{1}{2+\\dfrac{1}{5}}+7+\\frac{1}{3+\\dfrac{1}{4}}=25\\frac{620}{1001}$$． 故选$$\\text{E}$$． " ]