Datasets:

math-eval

/

TAL-SCQ5K

like 48

[ "2014年全国迎春杯五年级竞赛初赛第12题" ]

[ [ { "aoVal": "A", "content": "$$1034$$ " } ], [ { "aoVal": "B", "content": "$$2021$$ " } ], [ { "aoVal": "C", "content": "$$2815$$ " } ], [ { "aoVal": "D", "content": "$$3036$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "要求$$\\overline{ABFG}$$最小也就是要求$$\\overline{ABCDE}$$尽量小，$$\\overline{ABCDE}$$是$$2013$$的倍数， 把$$\\overline{ABCDE}$$看成$$\\overline{AB}\\left\\textbar{} \\overline{CDE} \\right.$$，则$$\\overline{CDE}$$是$$13$$的倍数，而$$\\overline{CDEFG}$$是$$1221$$的倍数， 把$$\\overline{CDEFG}$$看成$$\\overline{CDE}\\left\\textbar\\overline{FG} \\right.$$，从$$1$$倍、$$2$$倍、\\ldots\\ldots 变化的话，前段加$$12$$，后段加$$21$$， 则每$$5$$次，$$\\overline{FG}$$向$$\\overline{CDE}$$进$$1$$， $$12\\times n+1\\times \\left[ \\frac{n}{5} \\right]$$ 是$$13$$的倍数， $$12\\times n+1\\times \\left[ \\frac{n}{5} \\right]=13n-n+\\left[ \\frac{n}{5} \\right]$$， 不难看出$$n$$最小得$$16$$， $$1221\\times16=19536$$，$$195\\div 13=15$$，$$2013\\times 15=30195$$． " ]

[ "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3题5分" ]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$26$$ " } ], [ { "aoVal": "C", "content": "$$28$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "从题干可以知道，把一根木料截成$$4$$段用$$12$$分钟； 我们要知道，锯一次可以把一根木料截成$$2$$段；锯二次可以把一根木料截成$$3$$段； 所以把一根木料截成$$4$$段需要锯三次，那么每次用时：$$12\\div3=4$$（分钟）； 照这样的速度，如果把同样的木料截成$$8$$段，即需要锯$$8-1=7$$（次）； 一共用时：$$7\\times4=28$$（分钟）；所以选择$$\\text{C}$$选项． " ]

[ "2017年第17届湖北武汉世奥赛五年级竞赛决赛第7题" ]

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

[ "拓展思维->思想->逐步调整思想" ]

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

[ "2016年第14届全国创新杯小学高年级五年级竞赛复赛第7题" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据最不利原则，刚开始可以抽到$$1$$，$$2$$，$$3$$，$$4$$；$$9$$，$$10$$，$$11$$，$$12$$；$$17$$，$$18$$，$$19$$，$$20$$；此时一共抽了$$12$$张，如果下一张不管抽到谁，必然会两个数相差$$4$$，则至少要$$13$$张． " ]

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

[ [ { "aoVal": "A", "content": "$$\\frac{1}{9}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{10}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{11}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{12}$$ " } ] ]

[ "课内体系->知识点->应用题->比例应用题", "拓展思维->拓展思维->应用题模块->浓度问题->浓度基本题型->已知溶质溶液求浓度" ]

[ "盐的质量为$$100$$千克，盐水的质量为盐与水的质量之和，为$$100$$千克$$+1$$吨$$=100$$千克$$+1000$$千克$$=1100$$（千克），所以盐与盐水的质量比为$$100:1100=1:11$$，选C。 " ]

[ "2013年IMAS小学中年级竞赛第一轮检测试题第12题4分" ]

[ [ { "aoVal": "A", "content": "$$57$$ " } ], [ { "aoVal": "B", "content": "$$58$$ " } ], [ { "aoVal": "C", "content": "$$59$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ], [ { "aoVal": "E", "content": "$$61$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "可知这四位小朋友共有$$240$$本课外读物，所以当他们的书都相等时，每人各有$$24\\div 4=6$$（本）． 而题意可知此时甲的书多了$$6-3=3$$（本）、乙的书少了$$4-3=1$$（本）、丙的书少了$$5-4=1$$（本）、丁的书少了$$6-5=1$$（本），因此甲原有$$57$$本书，而乙、丙、丁各原有$$61$$本书．故选$$\\text{A}$$． " ]

[ "2016年全国华夏杯小学低年级一年级竞赛初赛香港賽區第9题5分" ]

[ [ { "aoVal": "A", "content": "$$\\blacksquare$$ " } ], [ { "aoVal": "B", "content": "$$\\blacktriangle$$ " } ], [ { "aoVal": "C", "content": "$$\\bigstar$$ " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "知识标签->数学思想->对应思想" ]

[ "根據規律，「$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$」為一個週期 他的順序是： $$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$、$$\\blacktriangle$$、$$\\bigstar$$、$$\\blacksquare$$ 故答案為：$$\\blacksquare$$． " ]

[ "2016年IMAS小学高年级竞赛第二轮检测试题第14题20分" ]

[ [ { "aoVal": "A", "content": "请将该小题的答案写在答题卡的指定答题区域，考试结束后上传至``本讲巩固'' " } ] ]

[ "知识标签->拓展思维->组合模块->组合模块最值问题->枚举型最值问题" ]

[ "该题为解答题，请将该小题的答案写在答题卡的指定答题区域，考试结束后上传至``本讲巩固'' " ]

[ "2013年河南郑州中原网杯六年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->连线法" ]

[ "首先，``乙没有坐在$$3$$号椅子上''是错的，就可以判断出乙在$$3$$号椅子上；再看，``乙坐在丙旁边''是错的，可以判断出丙不在$$2$$，$$4$$号椅子上，所以在$$1$$号椅子；最后，``甲坐在乙、丙中间''是错的，可以判断出甲在$$4$$号，故得出丁在$$2$$号椅子上． " ]

[ "2015年湖北武汉世奥赛五年级竞赛模拟训练题（三）第8题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->鸡兔同笼问题->假设法解鸡兔同笼->基本型->变型题" ]

[ "假设按照$$2$$个苹果和$$4$$个梨进行分配，当苹果分完时，梨也正好分完，但实际梨还剩余$$23$$个，这是因为每名小朋友实际比假设少分$$43=1$$（个）梨，所以小朋友的人数为$$23\\div 1=23$$（人）． " ]

[ "2017年全国美国数学大联盟杯五年级竞赛初赛第40题" ]

[ [ { "aoVal": "A", "content": "$${t}_{1}\\textless{t}_{2}$$ " } ], [ { "aoVal": "B", "content": "$${t}_{1}={t}_{2}$$ " } ], [ { "aoVal": "C", "content": "$${t}_{1}\\textgreater{t}_{2}$$ " } ], [ { "aoVal": "D", "content": "无法判断 " } ] ]

[ "拓展思维->拓展思维->组合模块->时间问题->时间计算" ]

[ "两个塔发射每一个能量波的时间是相同的，但是他们发送能量波的间隔并不相同．塔$$A$$发射$$5$$次能量波需要正好$$5$$秒，即是$$5$$次发送能量波的时间加上$$4$$次塔$$A$$的发射时间间隔，塔$$B$$发射$$10$$次能量波需要正好$$10$$秒，即是$$10$$次发送能量波的时间加上$$9$$次塔$$B$$的发射时间间隔，而$$10$$秒的时间等于$$5$$次发送能量波的时间加上$$4$$次塔$$A$$的发射时间间隔的两倍即$$10$$次发送能量波的时间加上$$8$$次塔$$A$$的发射时间间隔，故塔$$A$$的发射时间间隔大于塔$$B$$的发射时间间隔．它们发射$$12$$次能量波的时间均为$$12$$次发射能量波的时间加上各自的$$11$$次的发射时间间隔，由于塔$$A$$的发射时间间隔大于塔$$B$$的发射时间间隔，故$${t}_{1}\\textgreater{t}_{2}$$． " ]

[ "2017年河南郑州豫才杯六年级竞赛决赛4分" ]

[ [ { "aoVal": "A", "content": "红 " } ], [ { "aoVal": "B", "content": "黄 " } ], [ { "aoVal": "C", "content": "绿 " } ], [ { "aoVal": "D", "content": "蓝 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$1+3+2+2=8$$（个）灯泡一个循环周期，$$99\\div 8=12$$（组）$$\\cdot \\cdot \\cdot 3$$（个），所以第$$99$$个小灯泡是红色． " ]

[ "2018年第21届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第4题5分" ]

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

[ "拓展思维->能力->推理推导->言语逻辑推理" ]

[ "在$$1\\sim 39$$里，有$$4$$个数字$$4$$，分别是$$4$$、$$14$$、$$24$$、$$34$$， 在$$40$$到$$50$$中，有十个数字$$4$$，分别是$$40$$、$$41$$、$$42$$、$$43$$、$$44$$、$$45$$、$$46$$、$$47$$、$$48$$、$$49$$， 一共出现了$$15$$次． 故选$$\\text{C}$$． " ]

[ "2019年广东广州羊排赛六年级竞赛第6题3分" ]

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

[ "拓展思维->能力->构造模型->模型思想" ]

[ "根据题意分析可知，求一个数的几分之几，用乘法所以甲乙两数之间的关系式为： 甲$$\\times \\frac{2}{3}=$$乙$$\\times \\frac{6}{7}$$， 则甲$$:$$乙$$=\\frac{6}{7}:\\frac{2}{3}$$ $$=\\frac{6}{7}\\times \\frac{3}{2}$$ $$=\\frac{9}{7}$$ $$=9:7$$． 故选$$\\text{B}$$． " ]

[ "其它改编题", "2017年全国华杯赛小学中年级竞赛初赛模拟第5题" ]

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

[ "拓展思维->思想->整体思想" ]

[ "把和为$$12$$的两个数分成一组，这样就把这$$11$$个数分成$$6$$组：$$(1,11)$$，$$\\left( 2,10 \\right)$$，$$\\left( 3,9 \\right)$$，$$(4,8)$$，$$(5,7)$$，$$(6)$$．要保证一定有两个数的和为$$12$$，就要保证至少有两个数属于同一组． 根据最不利原则，如果我们从$$6$$组中各取一个数，则取出的这$$6$$个数中，没有两个数的和是$$12$$，因此再选一个数，一定会与前面的某数在同一组，即和为$$12$$．本题的答案就是至少选出$$7$$个不同的数． " ]

[ "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": "$$4$$ " } ] ]

[ "Overseas Competition->知识点->应用题模块->列方程解应用题", "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->方程法解倍数问题" ]

[ "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 " ]

[ "2015年第27届广东广州五羊杯小学高年级竞赛第1题4分" ]

[ [ { "aoVal": "A", "content": "$$1108.8$$ " } ], [ { "aoVal": "B", "content": "$$1008.8$$ " } ], [ { "aoVal": "C", "content": "$$1018.8$$ " } ], [ { "aoVal": "D", "content": "$$1118.8$$ " } ] ]

[ "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" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\frac{1}{2}\\times \\left( 1+2015+0.1+201.5 \\right)$$ $$=1008+100.8$$ $$=1108.8$$， 本题主要考察等差数列、平均数的概念及小数的基本计算方法和简单应用． 故选$$\\text{A}$$． " ]

[ "2014年全国创新杯小学高年级五年级竞赛5分" ]

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

[ "拓展思维->拓展思维->计数模块->枚举法综合->整数分拆->整数拆分应用->乘法拆数（应用）" ]

[ "$$36=2\\times 3\\times 6$$，$$\\text{A}_{3}^{3}=6$$，有$$6$$种 $$36=3\\times 3\\times 4$$，有$$3$$种 $$36=1\\times 6\\times 6$$，有$$3$$种 共$$6+3+3=12$$种不同的得分，最多有$$12$$人． " ]

[ "2017年河南郑州豫才杯小学高年级六年级竞赛" ]

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

[ "拓展思维->能力->数感认知->数学概念理解（数）" ]

[ "再过$$\\left[ 7,9 \\right]=63$$天两人又一起值日，$$4$$月有$$30$$天，$$5$$月有$$31$$天，所以是$$6$$月$$7$$日． " ]

[ "2021年新希望杯三年级竞赛初赛第13题5分" ]

[ [ { "aoVal": "A", "content": "只有苹果 " } ], [ { "aoVal": "B", "content": "只有橘子 " } ], [ { "aoVal": "C", "content": "只有草莓 " } ], [ { "aoVal": "D", "content": "香蕉和草莓 " } ], [ { "aoVal": "E", "content": "橘子和香蕉 " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "根据题意分析可知，苹果和草莓都与橘子相邻， 所以可以确定橘子在苹果和草莓的中间， 假设苹果、橘子、草莓这样排列， 则香蕉在草莓的右边，芒果在苹果的左边， 即苹果与芒果一定相邻， 故选：$$\\text{A}$$． " ]

[ "2004年第2届创新杯六年级竞赛复赛第5题" ]

[ [ { "aoVal": "A", "content": "$$\\left( \\frac{1}{17}+\\frac{1}{19} \\right)\\times 20$$ " } ], [ { "aoVal": "B", "content": "$$\\left( \\frac{1}{24}+\\frac{1}{29} \\right)\\times30$$ " } ], [ { "aoVal": "C", "content": "$$\\left( \\frac{1}{31}+\\frac{1}{17} \\right)\\times40$$ " } ], [ { "aoVal": "D", "content": "$$\\left( \\frac{1}{41}+\\frac{1}{47} \\right)\\times50$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$\\left( \\frac{1}{17}-\\frac{1}{19} \\right)\\div 20=\\frac{2}{17\\times 19}\\div 20=\\frac{1}{17\\times 19}\\times \\frac{1}{10}$$;$$\\left( \\frac{1}{15}-\\frac{1}{21} \\right)\\div 60=\\frac{6}{15\\times 21}\\div 60=\\frac{1}{15\\times 21}\\times \\frac{1}{10}$$; $$\\left( \\frac{1}{13}-\\frac{1}{23} \\right)\\div 100=\\frac{10}{13\\times 23}\\div 100=\\frac{1}{13\\times 23}\\times \\frac{1}{10};$$$$\\left( \\frac{1}{11}-\\frac{1}{25} \\right)\\div 140=\\frac{14}{11\\times 25}\\div 140=\\frac{1}{11\\times 25}\\times \\frac{1}{10};$$ 只需比较$$\\frac{1}{17\\times 19}$$，$$\\frac{1}{15\\times 21}$$，$$\\frac{1}{13\\times 23}$$，$$\\frac{1}{11\\times 25}$$的大小，根据和一定，两数越接近乘 积越大，则$$11\\times 25 \\textless{} 13\\times 23 \\textless{} 15\\times 21 \\textless{} 17\\times 19$$，那么 $$\\frac{1}{11\\times 25}\\textgreater\\frac{1}{13\\times 23}\\textgreater\\frac{1}{15\\times 21}\\textgreater\\frac{1}{17\\times 19}$$，所以答案为$$D$$ " ]

[ "2014年全国迎春杯四年级竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "$$44$$ " } ], [ { "aoVal": "B", "content": "$$43$$ " } ], [ { "aoVal": "C", "content": "$$42$$ " } ], [ { "aoVal": "D", "content": "$$41$$ " } ] ]

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "倒推．$$44$$ 对应的是$$44-2=42$$，颠倒后是$$24$$，除以$$2$$ 为$$12$$．符合条件．其他的均不符合条件． " ]

[ "2022年第9届广东深圳鹏程杯四年级竞赛初赛第30题5分" ]

[ [ { "aoVal": "A", "content": "甲 " } ], [ { "aoVal": "B", "content": "乙 " } ], [ { "aoVal": "C", "content": "丙 " } ], [ { "aoVal": "D", "content": "甲或丙 " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->体育比赛" ]

[ "无 " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（二）第3题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{5}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{9}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{10}$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->统计与概率->概率->基本概率->可能性" ]

[ "假设分两次抽，第一次抽中小明的可能性是$$\\frac{1}{10}$$，第二次才抽中小明的可能性是$$\\frac{1}{10}$$，$$\\frac{1}{10}+\\frac{1}{10}=\\frac{1}{5}$$． " ]

[ "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->思想->对应思想" ]

[ "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年春蕾杯六年级竞赛第12题2分", "2021年春蕾杯六年级竞赛第7题2分" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "$$B$$队两场平局，得$$2$$分，因得分为奇数，所以还有一场应胜，得$$3$$分，共得分$$5$$分． 那么$$A$$队虽是第一，也不可能全胜，则$$A$$队得$$7$$分， 其中平局只能是与$$B$$队了，则$$A$$队胜了$$C$$队和$$D$$队． 因此，$$B$$队胜的一场一定是与$$D$$队，与$$A$$和$$C$$队打平． 因为$$C$$队也必须是奇数，所以$$C$$队一定负于$$D$$队，积$$1$$分，$$D$$队得分$$3$$分． 故选：$$\\text{B}$$． " ]

[ "2018年第21届世界少年奥林匹克数学竞赛四年级竞赛复赛（中国区）第8题5分" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "因为通过题干可知，把这根长$$20$$米的彩带剪成两段，设较长的一段长$$x$$米，则较短的一段长$$(20-x)$$米，较长的比较短的长$$4$$米，则可列方程为：$$x-(20-x)=4$$，解得：$$x=12$$，所以较长的一段是$$12$$米，较短的一段是：$$20-12=8$$（米）． 故选：$$\\text{A}$$． " ]

[ "2007年华杯赛六年级竞赛初赛", "2007年华杯赛五年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$12$$分钟 " } ], [ { "aoVal": "B", "content": "$$15$$分钟 " } ], [ { "aoVal": "C", "content": "$$18$$分钟 " } ], [ { "aoVal": "D", "content": "$$20$$分钟 " } ] ]

[ "拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->已知工时反推->多人合作" ]

[ "解：$$\\frac{1}{\\frac{1}{30}+\\frac{1}{45}}=\\frac{1}{\\frac{5}{6\\times 15}}=18$$（分钟） " ]

[ "2015年湖北武汉世奥赛五年级竞赛模拟训练题（四）第5题" ]

[ [ { "aoVal": "A", "content": "一 " } ], [ { "aoVal": "B", "content": "二 " } ], [ { "aoVal": "C", "content": "三 " } ], [ { "aoVal": "D", "content": "四 " } ] ]

[ "拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期" ]

[ "因为这年的二月份有$$5$$个星期二，说明这是一闰年，且$$2$$月$$1$$号是星期二，$$17\\div 7=2$$（周）$$\\ldots \\ldots 3$$（天），所以$$2$$月$$17$$号也是星期四． " ]

[ "2012年全国创新杯五年级竞赛第4题5分" ]

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

[ "拓展思维->拓展思维->计数模块->容斥原理->多量容斥的最值问题" ]

[ "每个数字用$$10$$次，故所有数字和为$$(16\\times 10+18\\times 10)\\div 10=34$$，而出现了$$10$$ 次$$16$$和$$10$$次$$18$$，故除了最小的，其他$$5$$个都相同，所以只能为$$4$$、$$6$$、$$6$$、$$6$$、$$6$$、$$6$$． " ]

[ "2009年全国迎春杯小学中年级四年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$48$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$54$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "同倍数变化问题．把$$1$$个田格本和$$5$$个练习本捆绑成一组，那么每发$$3$$个横格本，就发一组田格本和练习本，田格本和练习本的总量是横线本的$$2$$倍，每次发的数量也是$$2$$倍，所以剩下的也是$$2$$倍，即$$48$$本． " ]

[ "2015年全国华杯赛小学高年级竞赛复赛A卷第9题", "2015年北京华杯赛小学高年级竞赛复赛A卷第9题" ]

[ [ { "aoVal": "A", "content": "$$552$$，$$115$$ " } ], [ { "aoVal": "B", "content": "$$232$$，$$435$$ " } ], [ { "aoVal": "C", "content": "$$552$$，$$115$$或$$232$$，$$435$$ " } ] ]

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "$$120={{2}^{3}}\\times 3\\times 5$$，$$667=23\\times 29$$．（*） 设这两个自然数是$$da$$和$$db$$，其中$$a$$和$$b$$是互质，$$d$$是这两个自然数的最大公约数，$$d$$是$$667$$的因子，且大于$$1$$，小于$$667$$．无妨设$$a\\textgreater b$$，则有： （1）当$$d=23$$时，由（*），$$a+b=29$$，$$ab={{2}^{3}}\\times 3\\times 5$$，此时易得：$$a=24$$，$$b=5$$，这两个自然数是$$435$$，$$232$$． （2）当$$d=29$$时，由（*），$$a+b=23$$，$$ab={{2}^{3}}\\times 3\\times 5$$，此时易得：$$a=15$$，$$b=8$$，这两个自然数是$$552$$，$$115$$． " ]

[ "2003年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$2002$$ " } ], [ { "aoVal": "B", "content": "$$2003$$ " } ], [ { "aoVal": "C", "content": "$$2004$$ " } ], [ { "aoVal": "D", "content": "$$4005$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数加减->整数加减巧算之分组法" ]

[ "注意到$$2003+2002-2001-2000=4$$，$$1999+1998-1997-1996=4$$，$$7+6-5-4=4$$，如果把$$2003$$到$$4$$的$$2000$$个数依次$$4$$个分为一组共有$$(2000\\div 4)$$组，且每组的计算结果都为$$4$$，那么这$$2000\\div 4$$组的和是$$2000\\div 4\\times 4=2000$$，再加上最后的$$3$$个数，结果是：$$2000+3+2-1=2004$$。 " ]

[ "2021年陕西西安六年级下学期小升初模拟分类专题五十七 （年龄问题 平均数问题）第26题", "2021年陕西西安雁塔区西安高新第一中学小升初（五）第6题3分", "2016年北京走美杯四年级竞赛冲刺讲义", "2019年重庆九龙坡区重庆市育才中学小升初（八）第1题3分", "2018年重庆九龙坡区重庆市育才中学小升初（三）第3题3分", "2014年陕西西安小升初高新一中入学真卷（二）第6题3分" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ], [ { "aoVal": "E", "content": "$$10$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "$$(7\\times6+11-8\\times4)\\div3=7$$． " ]

[ "2014年全国迎春杯四年级竞赛复赛第1题" ]

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

[ "拓展思维->能力->运算求解" ]

[ "$$2014\\div 19=106$$． " ]

[ "2016年第12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": "$$125$$ " } ], [ { "aoVal": "B", "content": "$$130$$ " } ], [ { "aoVal": "C", "content": "$$135$$ " } ], [ { "aoVal": "D", "content": "$$140$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "去的过程中一共花了$$4\\times 60+20=260$$分钟，$$260\\div 50=5\\ldots \\ldots 10$$说明走路时间一共是$$5\\times 40+10=210$$分．回来过程中速度为原来的$$2$$倍，走路的时间则为原来的$$\\frac{1}{2}$$，即$$210\\times \\frac{1}{2}=105$$分钟$$105\\div 35=3$$，即一共走了$$3$$次，第$$3$$次走完之后不用休息，则一共花了$$\\left( 35+15 \\right)\\times 3-15=135$$分钟． " ]

[ "2018年第22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6分" ]

[ [ { "aoVal": "A", "content": "$$198$$ " } ], [ { "aoVal": "B", "content": "$$199$$ " } ], [ { "aoVal": "C", "content": "$$200$$ " } ] ]

[ "拓展思维->能力->公式记忆->符号化数学原理", "Overseas Competition->知识点->计算模块->数列与数表->等差数列" ]

[ "根据题意：公差为$$5$$， 所以项数：$$\\left( 1003-13 \\right)\\div 5+1=199$$。 故选：$$\\text{B}$$． " ]

[ "2015年湖北武汉世奥赛小学高年级六年级竞赛模拟训练题（三）第2题" ]

[ [ { "aoVal": "A", "content": "$$8300$$ " } ], [ { "aoVal": "B", "content": "$$9300$$ " } ], [ { "aoVal": "C", "content": "$$10080$$ " } ], [ { "aoVal": "D", "content": "$$12080$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->折扣问题" ]

[ "$$\\left( 1276-268 \\right)\\div \\left( 95 \\%- 85 \\% \\right)=10080$$（元）． " ]

[ "2018年第21届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第8题3分" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ] ]

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "$$90$$除以$$1+2$$的和，结果是$$30$$，从后向前推算，剩下的是$$1$$份，那么苹果就是$$2$$份， 其他水果也是$$2$$份，所以可以得到是$$30$$个苹果．故选：$$\\text{C}$$． " ]

[ "2016年创新杯五年级竞赛训练题（二）第5题" ]

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

[ "拓展思维->思想->转化与化归的思想" ]

[ "设一台抽水机$$1$$分钟抽$$1$$份水，则每分钟的抽水量为$$\\left( 5\\times 45-10\\times 20 \\right)\\div \\left( 45-20 \\right)=1$$（份），船内原有水量为$$5\\times 45-1\\times 45=180$$（份），所以$$30$$分钟抽完需要$$180\\div 30+1=7$$（台）． " ]

[ "2020年新希望杯六年级竞赛（2月）第13题" ]

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

[ "拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->接力施工问题" ]

[ "哥哥工效：$$\\frac{1}{6}$$，弟弟工效：$$\\frac{1}{9}$$， 则$$1\\div \\left( \\frac{1}{6}+\\frac{1}{9} \\right)=\\frac{18}{5}=3$$（周期）$$\\cdots \\cdots \\frac{3}{5}$$， $$1-3\\times \\left( \\frac{1}{6}+\\frac{1}{9} \\right)=\\frac{1}{6}$$， $$\\frac{1}{6}\\div\\frac{1}{6}=1$$（小时）， 共$$3\\times 2+1=7$$（小时）． " ]

[ "2015年湖北武汉世奥赛六年级竞赛模拟训练题（二）第2题" ]

[ [ { "aoVal": "A", "content": "红 " } ], [ { "aoVal": "B", "content": "黑 " } ], [ { "aoVal": "C", "content": "蓝 " } ], [ { "aoVal": "D", "content": "无法确定 " } ] ]

[ "拓展思维->拓展思维->组合模块->逻辑推理->条件型逻辑推理->表格法" ]

[ "戴黑帽的同学穿的不是红鞋，只能穿蓝鞋，红衣服，所以戴蓝帽的同学不可能穿红衣服，只能穿黑衣服． " ]

[ "2018年第6届湖北长江杯六年级竞赛初赛B卷第4题3分" ]

[ [ { "aoVal": "A", "content": "亏了 " } ], [ { "aoVal": "B", "content": "赚了 " } ], [ { "aoVal": "C", "content": "不赚不亏 " } ], [ { "aoVal": "D", "content": "无法比较 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "从题目中我们知道，两件衣服都按$$120$$元出售， 也就是一共入账：$$120\\times2=240$$ （元）； 其中一件赚了$$20 \\%$$，那么成本是：$$120\\div (1+20 \\%)=100$$（元）， 另一件亏了$$20 \\%$$，那么成本是：$$120\\div (1-20 \\%)=150$$（元）； 总成本是：$$100+150=250$$（元），也就是实际上还是亏了． 故选：$$\\text{A}$$． " ]

[ "2017年河南郑州豫才杯小学高年级五年级竞赛初赛第15题" ]

[ [ { "aoVal": "A", "content": "$$4$$小时 " } ], [ { "aoVal": "B", "content": "$$4.5$$小时 " } ], [ { "aoVal": "C", "content": "$$5$$小时 " } ], [ { "aoVal": "D", "content": "$$8$$小时 " } ] ]

[ "知识标签->学习能力->七大能力->实践应用" ]

[ "设最多在停车场停了$$x$$小时，$$3+\\left( x-1 \\right)\\times 3=13.5$$，解得$$x=4.5$$，所以他们的车在停车场最多停$$4.5$$小时． " ]

[ "2015年第11届全国新希望杯小学高年级六年级竞赛复赛第4题" ]

[ [ { "aoVal": "A", "content": "$$18$$分钟 " } ], [ { "aoVal": "B", "content": "$$12$$分钟 " } ], [ { "aoVal": "C", "content": "$$9$$分钟 " } ], [ { "aoVal": "D", "content": "$$6$$分钟 " } ] ]

[ "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" ]

[ "设三人的时间分别为$$2x$$，$$3x$$，$$4x$$，那么有$$2\\div \\left( \\frac{1}{2x}+\\frac{1}{3x}+\\frac{1}{4x} \\right)=72$$，解得$$x=39$$，那么甲的效率是$$\\frac{1}{78}$$，丙的效率是$$\\frac{1}{156}$$，那么丙帮$$A$$的时间是$$\\left( 1-\\frac{72}{78} \\right)\\div \\frac{1}{156}=12$$分钟． " ]

[ "2006年第4届创新杯六年级竞赛初赛A卷第2题" ]

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

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

[ "$$(1)1\\frac{3}{17}-1\\frac{1}{19}，(2)1\\frac{1}{4}+1\\frac{1}{29}，(3)1\\frac{9}{31}+1\\frac{3}{37}，(4)1\\frac{9}{41}+1\\frac{3}{47}$$，显然($$3$$)最大． " ]

[ "2003年第1届创新杯五年级竞赛复赛第1题", "2003年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$0.413$$ " } ], [ { "aoVal": "B", "content": "$$4.13$$ " } ], [ { "aoVal": "C", "content": "$$41.3$$ " } ], [ { "aoVal": "D", "content": "$$413$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->小数基础->小数点的移动规律" ]

[ "小数点向右移动三位再向左移动两位，实际上是向右移动了一位，向右移动一位后得到的数是$$4.13$$，那么把小数点往左移动一位即得原数，故原数是$$0.413$$。 " ]

[ "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1题3分" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "把一根木料锯成$$3$$段，需要锯$$3-1=2$$（次），要用$$6$$分钟，每次需要$$6\\div 2=3$$（分钟），用同样的速度把这根木料锯成$$6$$段，需要锯$$6-1=5$$（次），需要$$5\\times 3=15$$（分钟）． 故选$$\\text{C}$$． " ]

[ "2014年中环杯五年级竞赛初赛" ]

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

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "$$60\\div 15=4$$（种） $$4+1=5$$（块） 答：一次至少取出$$5$$块才能保证其中至少有$$2$$块木块颜色相同。 " ]

[ "2013年全国希望杯五年级竞赛初赛第9题" ]

[ [ { "aoVal": "A", "content": "$$392$$ " } ], [ { "aoVal": "B", "content": "$$416$$ " } ], [ { "aoVal": "C", "content": "$$420$$ " } ], [ { "aoVal": "D", "content": "$$445$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列截断求和" ]

[ "方法一：求前$$13$$个数中间数：$$247\\div13=19$$，则后$$13$$个数的第一个数为$$19+6+1=26$$，最后一个数为$$26+12=38$$，所以后$$13$$个数的和是$$(26+38)\\times13\\div2=416$$． 方法二：后$$13$$个数比前$$13$$个数按顺序一一对应，后$$13$$个数的每个数比前$$13$$个数的每个数都多$$13$$． 所以后$$13$$个数的和是$$247+13\\times13=416$$． " ]

[ "2019年广东广州学而思综合能力诊断五年级竞赛第11题12分" ]

[ [ { "aoVal": "A", "content": "$$360$$ " } ], [ { "aoVal": "B", "content": "$$380$$ " } ], [ { "aoVal": "C", "content": "$$390$$ " } ], [ { "aoVal": "D", "content": "$$420$$ " } ], [ { "aoVal": "E", "content": "$$430$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理逆应用", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1" ]

[ "本题是因数个数定理的逆应用，先从$$A$$、$$B$$、$$C$$三个的因数个数开始讨论：$$C$$有$$11$$个因数，$$11$$是质数，所以$$C$$一定是一个质数的$$10$$次方，又因为三个数的和为$$2021$$，小于$$3$$的$$10$$次方，所以$$C$$只能是$$2$$的$$10$$次方，等于$$1024$$，则$$A+B=2021-1024=997$$，$$A$$有$$8$$个因数，则$$A$$的分解质因数形式可以是$$a$$的$$7$$次方、$$a$$的$$1$$次方乘$$b$$的$$3$$次方、$$a\\times b\\times c$$这$$3$$种可能，$$B$$的分解质因数形式可以是$$c$$的$$9$$次方、$$c$$的$$1$$次方乘$$d$$的$$4$$次方，因为$$B$$和$$C$$互质，所以$$B$$不含质因数$$2$$，根据和的大小可以确定$$B$$不等于$$c$$的$$9$$次方，又$$5$$的$$4$$次方等于$$625$$，$$625$$乘$$3$$等于$$1275$$大于$$997$$，所以$$d$$只能等于$$3$$，$$3$$的$$4$$次方等于$$81$$，$$81$$乘$$5$$等于$$405$$，$$997-405=592$$不符合$$A$$的分解质因数条件，$$81$$乘$$7$$等于$$567$$，$$997-567=430$$符合条件，则$$A$$等于$$430$$． " ]

[ "2006年第4届创新杯六年级竞赛初赛B卷第10题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{2}$$，$$\\frac{1}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{2}$$，$$\\frac{1}{6}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{3}$$，$$\\frac{1}{4}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{3}$$，$$\\frac{1}{5}$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}=\\frac{5}{6}+\\frac{9}{20}+\\frac{1}{6}=1\\frac{9}{20}=1.45$$， $$\\frac{6}{7}\\approx 0.857$$， $$1.45-0.857=0.593$$， 所以题目即需要从五个分数中选出两个，使他们的和最接近$$0.593$$，比较后可得应选$$\\frac{1}{3}$$和$$\\frac{1}{4}$$，故$$\\text{C}$$正确． 故选$$\\text{C}$$． " ]

[ "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": "$$10$$；$$100$$ " } ], [ { "aoVal": "B", "content": "$$20$$；$$100$$ " } ], [ { "aoVal": "C", "content": "$$40$$；$$100$$ " } ], [ { "aoVal": "D", "content": "$$80$$；$$200$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "若通过$$860$$米隧道时速度不变则需要$$24\\times 2=48$$（秒），火车过桥速度：$$\\left( 860-320 \\right)\\div \\left( 48-21 \\right)=20$$（米/秒）：火车车身长：$$21\\times 20-320=100$$（米）． " ]

[ "2007年四年级竞赛创新杯", "2007年第5届创新杯四年级竞赛第4题5分" ]

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

[ "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->剪绳子" ]

[ "将绳折成3段再对折，相当于折成6段，一刀与这6段共有6个交点，所以将绳剪成7段 " ]

[ "2006年第4届创新杯四年级竞赛初赛A卷第3题", "2006年四年级竞赛创新杯", "2006年三年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$68$$ " } ], [ { "aoVal": "B", "content": "$$64$$ " } ], [ { "aoVal": "C", "content": "$$54$$ " } ], [ { "aoVal": "D", "content": "$$52$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->鸡兔同笼问题" ]

[ "$$4$$只正常猫一共有$$18\\times 4=72$$（个）爪，$$4$$只伤残猫失去的腿都不同，一共失去$$5\\times 2+4\\times 2=18$$（个）爪，现在$$4$$只伤残猫一共有$$72-18=54$$（个）爪。 " ]

[ "2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第10题2分" ]

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

[ "拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期" ]

[ "①$$2009$$年是平年，平年$$2$$月只有$$28$$天． ②$$21$$时用普通计时法表示是晚上$$9$$点． ③东西相对南北相对，正确． ④小数$$1.6$$和$$1.8$$之间有无数个数． " ]

[ "2014年全国迎春杯五年级竞赛复赛第3题" ]

[ [ { "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" ]

[ "根据题意可知，$$1$$份的啤酒可以变成$$3$$份的泡沫．球球倒的啤酒一半是泡沫，那么我们可以把球球倒的每杯酒分成$$6$$份，那么每倒一杯酒只有$$4$$份．而一瓶啤酒可以倒$$4$$杯共有$$4\\times6＝24$$份．球球倒的每杯酒为$$4$$份，她共可以倒的杯数为：$$24\\div4＝6$$． " ]

[ "2004年希望杯三年级竞赛复赛", "2004年第2届希望杯五年级竞赛复赛第3题", "2004年希望杯四年级竞赛复赛", "2004年希望杯六年级竞赛复赛", "2004年希望杯二年级竞赛复赛", "2004年希望杯一年级竞赛复赛", "2004年希望杯五年级竞赛复赛" ]

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

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

[ "$$2$$、$$2$$、$$4$$、$$8$$、$$2$$、$$6$$、$$2$$、$$2$$、$$4$$、$$8$$\\ldots 发现$$6$$个一循环，$$2004\\div 6=334$$（组），所以第$$2004$$个数和第$$6$$个相同为$$6$$． " ]

[ "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6题5分" ]

[ [ { "aoVal": "A", "content": "$$\\bigcirc $$ " } ], [ { "aoVal": "B", "content": "$$\\triangle $$ " } ], [ { "aoVal": "C", "content": "$$\\square $$ " } ], [ { "aoVal": "D", "content": "☆ " } ] ]

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

[ "从图中可知五个一循环，$$30\\div 5=6\\cdots \\cdots 0$$，所以第三十个与第五个相同． 故选：$$\\text{D}$$． " ]

[ "2021年第8届鹏程杯五年级竞赛初赛第24题4分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->拓展思维->计算模块->比较与估算->放缩法->首尾放缩法" ]

[ "$$A\\textgreater1+\\frac{1}{2}+\\frac{1}{4}\\times 2+\\frac{1}{8}\\times 4+\\frac{1}{16}\\times 8=3$$， $$A\\textless{}1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}\\times 4+\\frac{1}{8}\\times 8+\\frac{1}{16}\\textless{}4$$， ∴$$\\left[ A \\right]=3$$． 答：$$\\text{B}$$． " ]

[ "2011年全国学而思杯五年级竞赛第10题" ]

[ [ { "aoVal": "A", "content": "前锋、前卫 " } ], [ { "aoVal": "B", "content": "前卫、后腰 " } ], [ { "aoVal": "C", "content": "守门员、前锋 " } ], [ { "aoVal": "D", "content": "后腰、守门员 " } ] ]

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

[ "甲不是前腰和后卫，前腰和后卫是两个人，乙不是后卫，那么丙是后卫，根据$$6$$可知道甲是后腰，那么根据$$2$$可知道，乙是前腰，根据$$1$$，则乙不是前锋，根据$$4$$，甲也不能是前锋，所以丙是前锋，根据$$3$$，前卫只能是乙，那么剩下的守门员只能是甲，所以甲的位置是后腰和守门员． " ]

[ "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": "$$\\frac{10}{27}\\textless{}\\frac{5}{14}\\textless{}\\frac{20}{53}\\textless{}\\frac{12}{31}$$ " } ] ]

[ "知识标签->数学思想->转化与化归的思想" ]

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

[ "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$148$$ " } ], [ { "aoVal": "B", "content": "$$248$$ " } ], [ { "aoVal": "C", "content": "$$348$$ " } ], [ { "aoVal": "D", "content": "$$648$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "首先还原为十进制： $${{(\\overline{abc})}_{7}}=a\\times{{7}^{2}}+b\\times {{7}^{1}}+c\\times {{7}^{0}}=49a+7b+c$$；$${{(\\overline{cba})}_{9}}=c\\times{{9}^{2}}+b\\times {{9}^{1}}+a\\times {{9}^{0}}=81c+9b+a$$． 于是$$49a+7b+c=81c+9b+a$$；得到$$48a=80c+2b$$，即$$24a=40c+b$$． 因为$$24a$$是$$8$$的倍数，$$40c$$也是$$8$$的倍数，所以$$b$$也应该是$$8$$的倍数，于是$$b=0$$或$$8$$． 但是在$$7$$进制下，不可能有$$8$$这个数字．于是$$b=0$$，$$24a=40c$$，则$$3a=5c$$． 所以$$a$$为$$5$$的倍数，$$c$$为$$3$$的倍数． 所以，$$a=0$$或$$5$$，但是，首位不可以是$$0$$，于是$$a=5$$，$$c=3$$； 所以$${{(\\overline{abc})}_{7}}={{(503)}_{7}}=5\\times 49+3=248$$ ． 于是，这个三位数在十进制中为$$248$$． " ]

[ "2017年第16届湖北武汉创新杯小学高年级五年级竞赛邀请赛训练题（一）" ]

[ [ { "aoVal": "A", "content": "$$6$$个 " } ], [ { "aoVal": "B", "content": "$$30$$个 " } ], [ { "aoVal": "C", "content": "$$36$$个 " } ], [ { "aoVal": "D", "content": "$$42$$个 " } ] ]

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

[ "数字之和为偶数的数有$$123$$、$$125$$、$$134$$、$$145$$、$$235$$、$$345$$，而每种选法又可以有五种不同三位数，所以$$6\\times 6=36$$个不同的三位数． " ]

[ "2004年六年级竞赛创新杯", "2004年第2届创新杯六年级竞赛初赛第9题" ]

[ [ { "aoVal": "A", "content": "$$7:1$$ " } ], [ { "aoVal": "B", "content": "$$7:2$$ " } ], [ { "aoVal": "C", "content": "$$24:7$$ " } ], [ { "aoVal": "D", "content": "$$31:9$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->比例应用题->按比分配" ]

[ "两个相同的瓶子装满酒精溶液，设一个瓶子装溶液的体积是单位``$$1$$''，那么第一个瓶子中酒精体积是$$3\\div (1+3)=\\frac{3}{4}$$，第二个瓶子中酒精体积是$$4\\div (1+4)=\\frac{4}{5}$$，把两瓶酒精溶液混合，两个瓶子的酒精体积和是$$\\frac{3}{4}+\\frac{4}{5}=\\frac{31}{20}$$， 水体积和$$=1+1-\\frac{31}{20}=\\frac{9}{20}$$， 那么酒精与水的体积比是$$\\frac{31}{20}:\\frac{9}{20}=31:9$$． 故选$$\\text{D}$$． " ]

[ "2020年广东广州羊排赛四年级竞赛第16题5分" ]

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

[ "课内体系->思想->方程思想", "拓展思维->拓展思维->计算模块->方程基础->一元一次方程" ]

[ "$$\\begin{eqnarray}10+2x+3\\&=\\&19 2x\\&=\\&19-3-10 2x\\&=\\&6 x\\&=\\&3．\\end{eqnarray}$$ " ]

[ "2022年第九届鹏程杯四年级竞赛初赛第20题5分" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$55$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "\\[70\\] " } ], [ { "aoVal": "E", "content": "\\[75\\] " } ] ]

[ "拓展思维->拓展思维->计算模块->定义新运算" ]

[ "无 " ]

[ "2014年迎春杯三年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$8$$厘米 " } ], [ { "aoVal": "B", "content": "$$16$$厘米 " } ], [ { "aoVal": "C", "content": "$$24$$厘米 " } ] ]

[ "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" ]

[ "暗和型： 周长是$$48$$厘米，那么长$$+$$宽的和：$$48\\div$$ 2=24（厘米） 宽：$$24\\div$$ （2+1）=8（厘米） 长：$$8\\times$$ 2=16（厘米） " ]

[ "2019年广东深圳全国小学生数学学习能力测评四年级竞赛初赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$9:05$$ " } ], [ { "aoVal": "B", "content": "$$9:35$$ " } ], [ { "aoVal": "C", "content": "$$9:55$$ " } ] ]

[ "拓展思维->能力->实践应用->度量单位认知" ]

[ "根据选项$$\\text{C}$$，现在是$$9:55$$，那么$$5$$分钟前分针的位置是指向$$10$$，$$5$$分钟后时针也是指向$$10$$．所以$$\\text{C}$$选项答案是满足题目要求的，所以选择$$\\text{C}$$． " ]

[ "2018年全国小学生数学学习能力测评五年级竞赛初赛第10题3分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->时间问题->认识钟表" ]

[ "先算出从第一次记录到第十二次记录中间经过的时间是多少，第$$1$$次到第$$12$$次之间有个$$11$$间隔：$$5\\times11=55$$（小时）． 时针每过$$12$$小时就会转一圈回到原来的状态，$$55\\div12=4$$（圈）$$\\cdots \\cdots 7$$（小时）． 即时针转了$$4$$圈以后，又经过了$$7$$个小时，时针指向$$9$$， 所以原来第一次时针指向：$$9-7=2$$． 因此，选$$\\text{A}$$． " ]

[ "2018年湖北武汉新希望杯六年级竞赛训练题（一）第6题", "2017年新希望杯六年级竞赛训练题（一）第6题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "将这个$$500$$位数从左到右依次编号第$$1$$位、第$$2$$位、第$$3$$位、$$\\cdots $$、第$$500$$位，且编号不变． 第$$1$$次删数后剩下$$2$$的倍数，第$$2$$次删数后剩下$$22$$的倍数，第$$3$$次删数后剩下$$23$$的倍数，$$\\cdots \\cdots $$，因为$$256\\textless{}500\\textless{}512$$，所以最后剩下的数字在第$$256$$位，$$256\\div 5=51$$（组）$$\\cdots \\cdots 1$$（个），第$$1$$位上的数字是$$1$$，所以最后剩下的数字是$$1$$． " ]

[ "2017年第22届全国华杯赛小学中年级竞赛初赛第2题10分" ]

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

[ "知识标签->数学思想->枚举思想" ]

[ "将和为$$10$$的两个数作为一组，有$$\\left( 1,9 \\right)$$，$$\\left( 2,8 \\right)$$，$$\\left( 3,7 \\right)$$，$$\\left( 4,6 \\right)$$，另外$$\\left( 5\\right)$$，$$\\left(10 \\right)$$单独一组，考虑极端情况，若在$$6$$组中每组各取一个数字，也没有两数和为$$10$$，然后在剩下的数中取任何一个，都会有两数能凑成$$10$$．所以至少需要取$$7$$个数． " ]

[ "2012年第10届创新杯四年级竞赛初赛第5题6分" ]

[ [ { "aoVal": "A", "content": "$$71$$ " } ], [ { "aoVal": "B", "content": "$$255$$ " } ], [ { "aoVal": "C", "content": "$$316$$ " } ], [ { "aoVal": "D", "content": "$$377$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "因为分钟的十位最大为$$5$$，所以下一次数字相同的时刻为$$11:11$$，$$11:11$$距$$5:55$$有$$5$$小时$$16$$分，即$$316$$分钟．所以选$$\\text{C}$$． " ]

[ "2006年第4届创新杯四年级竞赛初赛B卷第2题" ]

[ [ { "aoVal": "A", "content": "$$37$$ " } ], [ { "aoVal": "B", "content": "$$59$$ " } ], [ { "aoVal": "C", "content": "$$73$$ " } ], [ { "aoVal": "D", "content": "$$86$$ " } ] ]

[ "拓展思维->思想->逆向思想" ]

[ "从最后一步推起，``除以$$8$$，其结果等于$$80$$''可以求出被除数：$$80\\times 8=640$$；再看倒数第$$2$$步，``减去$$8$$''得$$640$$，可以求出被减数：$$640+8=648$$；然后看倒数第$$3$$步，``乘$$8$$''得$$648$$，可以求出另一个因数：$$648\\div 8=81$$；最后看第$$1$$步，``某数加上$$8$$''得$$81$$，某数为$$81-8=73$$，由此即可解决问题． $$(80\\times 8+8)\\div 8-8=648\\div 8-8=81-8=73$$． 答：这个数是$$73$$． 故选$$\\text{C}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛决赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$66$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "第一天$$5$$个大字， 第二天$$5+2=7$$个大字， 第三天$$7+2=9$$个大字， 第四天$$9+2=11$$个大字， 第五天$$11+2=13$$个大字， 第六天$$13+2=15$$个大字， 六天共写了$$5+7+9+11+13+15=60$$个大字． 故选：$$\\text{C}$$． " ]

[ "2018年IMAS小学中年级竞赛（第一轮）第2题3分" ]

[ [ { "aoVal": "A", "content": "$$503$$ " } ], [ { "aoVal": "B", "content": "$$504$$ " } ], [ { "aoVal": "C", "content": "$$505$$ " } ], [ { "aoVal": "D", "content": "$$506$$ " } ] ]

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

[ "$$\\left( \\Delta \\times 2-1 \\right)\\times 2=2018$$， $$\\Delta \\times 2-1=2018\\div 2=1009$$， $$\\Delta \\times 2=1009+1=1010$$， $$\\Delta =1010\\div 2=505$$． 故选$$\\text{C}$$． " ]

[ "2015年湖北武汉世奥赛六年级竞赛模拟训练题（二）第3题" ]

[ [ { "aoVal": "A", "content": "$$24 \\% $$ " } ], [ { "aoVal": "B", "content": "$$36 \\% $$ " } ], [ { "aoVal": "C", "content": "$$42 \\% $$ " } ], [ { "aoVal": "D", "content": "$$44 \\% $$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "设$$A$$瓶盐水的浓度是$$x \\%$$，$$B$$瓶盐水的浓度是$$y \\%$$． $$1\\centerdot x \\%+1\\centerdot y \\%=2\\centerdot 30 \\%$$ ① $$2\\centerdot 30 \\% +1\\centerdot x \\%=3\\centerdot 32 \\%$$ ② 解得$$x=36$$，$$y=24$$． " ]

[ "2019年第24届YMO三年级竞赛决赛第5题3分", "2020年第24届YMO三年级竞赛决赛第5题3分" ]

[ [ { "aoVal": "A", "content": "$$56$$ " } ], [ { "aoVal": "B", "content": "$$58$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$62$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->间隔问题->封闭型->封闭型植树问题->封闭植树类型问题(段数大于10)" ]

[ "根据题意分析可知，四个角都插上一面小红旗，每边都插$$26$$面， 所以每一边有小红旗：$$26-1=25$$（面）， 因为是正方形，每边都有$$25$$面，那么一共插了：$$25\\times4=100$$（面）红旗， 故答案为：$$\\text{C}$$． " ]

[ "2016年陕西西安小升初工大附中入学真题2第3题", "2017年全国小学生数学学习能力测评五年级竞赛初赛第10题3分", "2016年陕西西安小升初某工大附中", "2018年陕西西安小升初分类卷13第14题" ]

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

[ "拓展思维->思想->逐步调整思想" ]

[ "对折一次剪一刀是三段$$(2+1)$$， 再把它折成相等的三折，是$$(2\\times 3+1)$$， 接着再对折剪一刀是$$2\\times 3\\times 2+1=13$$（段）． " ]

[ "2020年四川绵阳涪城区绵阳东辰国际学校超越杯四年级竞赛第2题2分" ]

[ [ { "aoVal": "A", "content": "$$32$$ " } ], [ { "aoVal": "B", "content": "$$33$$ " } ], [ { "aoVal": "C", "content": "$$34$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "根据题意分析可知，利用关系式：数量$$=$$总价$$\\div $$单价，用爸爸带的钱除以每个笔记本的单价即可得到最多能买多少个笔记本，列式为：$$269\\div 8=33$$（个）$$\\cdots \\cdots 5$$（元），剩下的$$5$$元不足以再买一个笔记本， 所以最多能买$$33$$个， 故选$$\\text{B}$$． " ]

[ "2021年鹏程杯六年级竞赛初赛第5题" ]

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

[ "拓展思维->能力->符号代换->代数运算" ]

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

[ "2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第14题2分" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$80$$ " } ], [ { "aoVal": "D", "content": "$$100$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "暂无 " ]

[ "2016年第6届广东广州羊排赛六年级竞赛第9题1分" ]

[ [ { "aoVal": "A", "content": "$$164$$ " } ], [ { "aoVal": "B", "content": "$$168$$ " } ], [ { "aoVal": "C", "content": "$$172$$ " } ], [ { "aoVal": "D", "content": "$$176$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->综合题->容斥求平均数" ]

[ "琦琦$$3$$次共跳$$156\\times 3=468$$（下）， 要想跳$$4$$次后达到平均每次跳$$160$$下，$$4$$次一共跳$$4\\times 160=640$$（下）， 他第$$4$$次要跳$$640-468=172$$（下）． 故选$$\\text{C}$$． " ]

[ "2003年第1届创新杯六年级竞赛初赛第5题", "六年级竞赛创新杯", "2003年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{2}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{6}$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->统计与概率->概率->基本概率" ]

[ "投掷这个骰子时，6个面都有可能朝上，每个面朝上的可能性是$$\\frac{1}{6}$$，有两个面上写着数字2，数字2朝上的可能性为$$2\\times \\frac{1}{6}=\\frac{1}{3}$$. " ]

[ "2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（一）" ]

[ [ { "aoVal": "A", "content": "亏欠了 $$3$$元 " } ], [ { "aoVal": "B", "content": "赠了$$3$$元 " } ], [ { "aoVal": "C", "content": "亏了$$1$$元 " } ], [ { "aoVal": "D", "content": "赚了$$1$$元 " } ] ]

[ "拓展思维->拓展思维->应用题模块->经济问题->基本经济概念->折扣问题" ]

[ "甲第一次卖给乙价格为：$$1000\\times \\left( 1+10 \\% ~\\right)=1100$$元，甲盈利了$$100$$元； 乙转卖给甲价格：$$1100\\times \\left( 110 \\% ~\\right)=99$$元 综上，甲在此次交易过程中盈利了$$1$$元． " ]

[ "2008年第6届创新杯四年级竞赛初赛B卷第2题5分" ]

[ [ { "aoVal": "A", "content": "$$68$$米 " } ], [ { "aoVal": "B", "content": "$$70$$米 " } ], [ { "aoVal": "C", "content": "$$71$$米 " } ], [ { "aoVal": "D", "content": "$$72$$米 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "在两根电线杆中间等距加$$22$$根电线杆， 有$$22+1=23$$（个）间隔， 那么$$115\\div 23=5$$（米）一个间隔， 从第$$2$$根到第$$16$$根共有$$16-2=14$$（个）间隔， $$14\\times 5=70$$（米）． 故选$$\\text{B}$$． " ]

[ "2020年第24届YMO四年级竞赛决赛第4题3分", "2019年第24届YMO四年级竞赛决赛第4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4题3分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$100$$ " } ], [ { "aoVal": "C", "content": "$$120$$ " } ], [ { "aoVal": "D", "content": "$$216$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "依次填涂， $$\\text{Y}$$有$$6$$中颜色可选， 则$$\\text{M}$$有$$6-1=5$$（种）颜色可选， 则$$\\text{O}$$有$$6-1-1=4$$（种）颜色可选， 则有$$6\\times 5\\times 4=120$$（种）涂法． 故选$$\\text{C}$$． " ]

[ "2018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": "$$17$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$13$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

[ "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" ]

[ "四个数之和为$$13.75\\times 4=55$$，三个大数之和为$$15\\times 3=45$$，最小数为$$55-45=10$$，三个小数之和为$$12\\times 3=36$$，最大数为$$55-36=19$$．中间两数之和为$$45-19=26$$，中间两数平均数为$$26\\div 2=13$$．第二大数是奇数且比$$13$$大比$$19$$小，只有$$15$$和$$17$$两种可能．若第二大数为$$17$$，则第三大数为$$26-17=9$$，比最小数$$10$$小，不成立；所以第二大数为$$15$$，此时第三大数为$$26-15=11$$，成立． 故选$$\\text{B}$$． " ]

[ "2019年第7届湖北长江杯五年级竞赛复赛A卷第5题3分" ]

[ [ { "aoVal": "A", "content": "$$77$$ " } ], [ { "aoVal": "B", "content": "$$80$$ " } ], [ { "aoVal": "C", "content": "$$83$$ " } ], [ { "aoVal": "D", "content": "$$79$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "设爷爷今年$$x$$岁， $$\\left[ \\left( x-15 \\right)\\times \\frac{1}{4}-6 \\right]\\times 10=100$$ $$\\left( x-15 \\right)\\times \\frac{1}{4}-6=100\\div 10$$ $$\\left( x-15 \\right)\\times \\frac{1}{4}-6=10$$ $$\\frac{1}{4}\\left( x-15 \\right)=10+6$$ $$\\frac{1}{4}\\left( x-15 \\right)=16$$ $$x-15=16\\div \\frac{1}{4}$$ $$x-15=64$$ $$x=64+15$$ $$x=79$$． 所以小明的爷爷今年$$79$$岁，故选：$$\\text{D}$$． " ]

[ "2010年第8届创新杯五年级竞赛初赛第8题4分" ]

[ [ { "aoVal": "A", "content": "$$28$$个 " } ], [ { "aoVal": "B", "content": "$$36$$个 " } ], [ { "aoVal": "C", "content": "$$45$$个 " } ], [ { "aoVal": "D", "content": "$$55$$个 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "设三位数为$$\\overline{abc}$$，则$$b=a+c$$， $$a=1$$时，$$c=0$$，$$1$$，$$2$$，$$\\ldots 8$$，共$$9$$个； $$a=2$$时，$$c=0$$，$$1$$，$$2$$，$$\\ldots 7$$，共$$8$$个； $$a=3$$时，$$c=0$$，$$1$$，$$2$$，$$\\ldots 6$$，共$$7$$个； $$a=4$$时，$$c=0$$，$$1$$，$$2$$，$$\\ldots 5$$，共$$6$$个； $$a=5$$时，$$c=0$$，$$1$$，$$2$$，$$3$$，$$4$$，共$$5$$个； $$a=6$$时，$$c=0$$，$$1$$，$$2$$，$$3$$，共$$4$$个； $$a=7$$时，$$c=0$$，$$1$$，$$2$$，共$$3$$个； $$a=8$$时，$$c=0$$，$$1$$，共$$2$$个； $$a=9$$时，$$c=0$$，共$$1$$个； 一共有$$1+ 2+ \\ldots 9= \\frac{9 \\times(1+9)}{2}=45$$个， 故选：$$\\text{C}$$． " ]

[ "2018年美国数学大联盟杯五年级竞赛初赛第34题5分" ]

[ [ { "aoVal": "A", "content": "$$2.5 \\%$$ " } ], [ { "aoVal": "B", "content": "$$2.6 \\%$$ " } ], [ { "aoVal": "C", "content": "$$5 \\%$$ " } ], [ { "aoVal": "D", "content": "$$6 \\%$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$2$$升$$2 \\%$$浓度的脂肪乳$$+3$$升$$3 \\%$$浓度的脂肪乳$$=5$$升浓度的脂肪乳． $$\\frac{总含量}{总体积}\\times 100 \\%$$， 即$$\\frac{2\\times 2 \\%+3\\times 3 \\%}{2+3}\\times 100 \\%=2.6 \\%$$． 故选$$\\text{B}$$． " ]

[ "2018年湖北武汉新希望杯六年级竞赛训练题（五）第4题" ]

[ [ { "aoVal": "A", "content": "$$11101$$ " } ], [ { "aoVal": "B", "content": "$$1011$$ " } ], [ { "aoVal": "C", "content": "$$10101$$ " } ], [ { "aoVal": "D", "content": "$$11001$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "25$\\div$2=12\\ldots\\ldots1 12$\\div$2=6\\ldots\\ldots0 6$\\div$2=3\\ldots\\ldots0 3$\\div$2=1\\ldots\\ldots1 1$\\div$2=0\\ldots\\ldots1 倒取余数，结果为（11001）\\textsubscript{2} 知识来源------四年级春季第二讲《机器人的聊天记录》 " ]

[ "2017年全国希望杯小学高年级五年级竞赛初赛考前100题" ]

[ [ { "aoVal": "A", "content": "$$19$$ " } ], [ { "aoVal": "B", "content": "$$21$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$26$$ " } ] ]

[ "知识标签->数学思想->整体思想" ]

[ "因为奇数的平方还是奇数，所以这三个数中有一个是$$2$$．因为$$390-{{2}^{2}}=386$$， 所以，通过质数平方的个位特点，判断得还有一个质数应该是$$5$$． 又$$386-{{5}^{2}}=361$$，且$$361={{19}^{2}}$$． 所以这三个质数分别是$$2$$，$$5$$，$$19$$，和是$$26$$． " ]

[ "2020年第24届YMO三年级竞赛决赛第3题3分", "2019年第24届YMO三年级竞赛决赛第3题3分" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$45$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "根据题意，$$10$$个相同的小球分给$$3$$个人，每人至少$$1$$个，就是将$$10$$个球分成$$3$$组， 一个人最多分：$$10-1-1=8$$（支）， $$8+7+6+5+4+3+2+1=36$$（种）， 故选$$\\text{C}$$． " ]

[ "2007年第5届创新杯五年级竞赛第2题5分", "2007年第5届创新杯五年级竞赛第2题5分", "2007年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$26$$ " } ] ]

[ "拓展思维->能力->公式记忆->符号化数学原理" ]

[ "不超过$${{13}^{2}}=169$$的$$6$$的倍数是$$1\\times 6$$，$$2\\times 6$$，$$\\cdots $$，$$28\\times 6$$有$$28$$个，其中不能写成$$a\\times b\\left( 1\\leqslant a\\leqslant b\\leqslant 13 \\right)$$形式的数是：$$17\\times 6$$，$$19\\times 6$$，$$21\\times 6$$，$$23\\times 6$$，$$25\\times 6$$，$$27\\times 6$$，$$28\\times 6$$，共有$$7$$个。因此，符合要求的整数共有$$28-7=21$$（个）。 " ]

[ "2019年第7届湖北长江杯五年级竞赛复赛A卷第4题3分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$55$$ " } ] ]

[ "拓展思维->能力->运算求解->程序性计算" ]

[ "$$1\\div \\left( 2\\div 3 \\right)\\div \\left( 3\\div 4 \\right)\\div \\left( 4\\div 5 \\right)\\div \\left( 5\\div 6 \\right)$$ $$=1\\div 2\\times 3\\div 3\\times 4\\div 4\\times 5\\div 5\\times 6$$ $$=1\\div 2\\times 6$$ $$=3$$． 故选$$\\text{A}$$． " ]

[ "2004年四年级竞赛创新杯", "2004年第2届创新杯六年级竞赛初赛第8题", "2004年五年级竞赛创新杯", "2004年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$28$$个 " } ], [ { "aoVal": "B", "content": "$$40$$个 " } ], [ { "aoVal": "C", "content": "$$68$$个 " } ], [ { "aoVal": "D", "content": "无穷多个 " } ] ]

[ "拓展思维->拓展思维->数论模块->余数问题->同余->同余定理" ]

[ "设符合要求的自然数为$$n$$，则$$n\\textgreater2004$$，且$$n=69q+q$$，$$0\\leqslant q\\leqslant 68$$。故而$$n=70q\\textgreater2004$$得$$q\\geqslant 29$$，因此$$29\\leqslant q\\leqslant 68$$，即$$q$$有$$68-29+1=40$$个，符合要求的自然数有$$40$$个，选B。 " ]

[ "2006年第11届全国华杯赛竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "$$48$$ " } ], [ { "aoVal": "B", "content": "$$72$$ " } ], [ { "aoVal": "C", "content": "$$96$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->思想->分类讨论思想" ]

[ "贝贝在左、妮妮在右相邻的排法有$$4\\times 3\\times 2\\times 1=24$$（种），贝贝在右、妮妮在左相邻的排法也有$$4\\times 3\\times 2\\times 1=24$$（种），总的排法$$5\\times 4\\times 3\\times 2\\times 1=120$$（种）．所以贝贝和妮妮不相邻的排法是$$120-2\\times 24=72$$（种）． " ]

[ "2015年华杯赛四年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$42$$ " } ], [ { "aoVal": "C", "content": "$$46$$ " } ], [ { "aoVal": "D", "content": "$$52$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->归一归总问题->乘除法应用->除法应用" ]

[ "$$\\left( 100-10 \\right)\\div \\left( 2+2+1 \\right)$$ $$=90\\div 5$$ $$=18$$（人） $$18\\times 2+10$$ $$=36+10$$ $$=46$$（人） 答：舞蹈队招收$$46$$人。 故选：$$C$$。 " ]

[ "2015年湖北武汉世奥赛六年级竞赛模拟训练题（三）第6题" ]

[ [ { "aoVal": "A", "content": "$$105$$分钟 " } ], [ { "aoVal": "B", "content": "$$130$$分钟 " } ], [ { "aoVal": "C", "content": "$$135$$分钟 " } ], [ { "aoVal": "D", "content": "$$150$$分钟 " } ] ]

[ "拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->单人简单行程问题" ]

[ "$$60\\times 4+20=50\\times 5+10$$，去时共走了$$40\\times 5+10=210$$（分钟），则返回时共需走$$210\\div 2=105$$（分钟），$$105\\div 35=3$$，加上休息的时间共需要$$105+15\\times \\left( 31 \\right)=135$$（分钟）． " ]