Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2017年全国华杯赛小学高年级竞赛公开题" ]

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

[ "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" ]

[ "设乘坐公交车$$x$$分钟，乘地铁$$\\left(40-6-x\\right)$$， $$\\frac{1}{50}x+\\frac{1}{30}\\left(40-6-x\\right)=1$$ 解得$$x=10$$． " ]

[ "2017年安徽合肥庐江县小升初第16题1分", "2019年湖南郴州六年级下学期小升初模拟（6）第11题", "2017年河南郑州小升初豫才杯第11题3分", "2019~2020学年北京四年级上学期期末模拟（北京版）（四）第9题", "2017年河南郑州豫才杯竞赛第11题" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$400$$ " } ], [ { "aoVal": "C", "content": "$$4000$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数乘除->整数除法运算->带余除法" ]

[ "$$3400\\div 600=5\\cdot \\cdot \\cdot \\cdot \\cdot \\cdot 400$$，故选$$B$$． " ]

[ "2010年六年级竞赛创新杯", "2010年六年级竞赛创新杯" ]

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

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

[ "$$6$$个人，每两个人都要赛一场，一共有$$6\\times 5\\div 2=15$$（场），总分$$30$$分，第一名最多胜$$5$$场得$$10$$分。若第一名不为$$10$$分，则由条件$$6$$人得分最多分别为$$8,6,6,4,2,2$$分，和为$$28\\textless30$$，矛盾，故第一名得$$10$$分。 " ]

[ "2017年IMAS小学中年级竞赛（第一轮）第17题4分" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "个位与十位数码之和为$$9$$的二位数有$$18$$、$$27$$、$$36$$、$$45$$、$$54$$、$$63$$、$$72$$、$$81$$、$$90$$共$$9$$个数，将这些二位数乘以$$5$$，得到的乘积分别为$$90$$、$$135$$、$$180$$、$$225$$、$$270$$、$$315$$、$$360$$、$$405$$、$$450$$，它们的数码之和仍为$$9$$，所有数都符合题目条件﹐因此总共有$$9$$个． 故逐$$\\text{E}$$． " ]

[ "2017年全国华杯赛竞赛初赛模拟3第2题" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "因为$$1729=12\\times12\\times12+1$$，所以，至少$$2$$人出生在同一个月、同一个时辰，且有相同的生肖． " ]

[ "2014年全国学而思杯一年级竞赛第5题" ]

[ [ { "aoVal": "A", "content": "单 " } ], [ { "aoVal": "B", "content": "双 " } ] ]

[ "拓展思维->能力->逻辑分析", "海外竞赛体系->知识点->应用题模块->分百应用题->认识单位1" ]

[ "四个数中三个单数一个双数，三个单数和是单数，再加一个双数和是单数． " ]

[ "2017年河南郑州模拟考试k6考试第6题", "2017年河南郑州K6联赛竞赛模拟第6题" ]

[ [ { "aoVal": "A", "content": "每段长$$\\frac{3}{5}$$米 " } ], [ { "aoVal": "B", "content": "每段长度是全长的$$\\frac{3}{5}$$ " } ], [ { "aoVal": "C", "content": "每段长度是全长的$$\\frac{1}{5}$$ " } ], [ { "aoVal": "D", "content": "每段长度是$$1$$米的$$\\frac{3}{5}$$ " } ] ]

[ "课内体系->知识点->应用题->分数百分数应用题->已知单位一和部分量，求分率", "拓展思维->拓展思维->计算模块->分数->分数基础" ]

[ "每段长$$3\\div 5=\\frac{3}{5}（\\text{m}）$$，每段长度是全长的$$1\\div 5=\\frac{1}{5}$$，每段长度是$$1$$米的$$\\frac{3}{5}\\div 1=\\frac{3}{5}$$，所以选项$$\\text{B}$$不正确． " ]

[ "2018年第6届湖北长江杯五年级竞赛初赛B卷第5题3分" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$60$$ " } ], [ { "aoVal": "C", "content": "$$90$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公倍数与最小公倍数->两数的最小公倍数" ]

[ "$$10=2\\times 5$$，$$15=3\\times 5$$，$$20=2\\times 2\\times 5$$， 所以$$10$$，$$15$$，$$20$$的最小公倍数为：$$2\\times 2\\times 3\\times 5=60$$， 所以至少经过$$60$$分钟这三种路线再次同时发车， 故选$$\\text{B}$$． " ]

[ "2012年IMAS小学高年级竞赛第一轮检测试题第1题3分" ]

[ [ { "aoVal": "A", "content": "$$90909$$ " } ], [ { "aoVal": "B", "content": "$$10101$$ " } ], [ { "aoVal": "C", "content": "$$100000$$ " } ], [ { "aoVal": "D", "content": "$$900000$$ " } ], [ { "aoVal": "E", "content": "$$10000$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数加减->整数减法运算->减法横式" ]

[ "$$101010-10101=90909$$． " ]

[ "2016年IMAS小学高年级竞赛第一轮检测试题第2题3分" ]

[ [ { "aoVal": "A", "content": "$$1.2\\times 3.4=12\\times 3.4$$ " } ], [ { "aoVal": "B", "content": "$$0.98\\times 0.99\\textgreater0.99$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{2}-\\frac{1}{3}\\textless{}\\frac{1}{3}-\\frac{1}{4}$$ " } ], [ { "aoVal": "D", "content": "$$10.4\\times 0.1\\textless{}1.04$$ " } ], [ { "aoVal": "E", "content": "$$1.1\\times 1.1\\textgreater1.1$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->两数相减法" ]

[ "$$\\text{A}$$：$$1.2\\times 3.4\\textless{}12\\times 3.4$$~~~~~~~~~~~~ $$\\text{B}$$：$$0.98\\times 0.99\\textless{}0.99$$ $$\\text{C}$$：$$\\frac{1}{2}-\\frac{1}{3}=\\frac{1}{6}$$，$$\\frac{1}{3}-\\frac{1}{4}=\\frac{1}{12}$$，$$\\frac{1}{2}-\\frac{1}{3}\\textgreater\\frac{1}{3}-\\frac{1}{4}$$ $$\\text{D}$$：$$10.4\\times 0.1=1.04$$ " ]

[ "2004年五年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->计算模块->比较与估算->比较大小综合" ]

[ "$$\\frac{2}{3}=0.\\dot{6}$$，那么这10个小数都小于$$\\frac{2}{3}$$，又$$\\frac{2}{3}\\times 7=\\frac{14}{3}$$小于5，所以至少需要7个以上的小数它们的和才有可能大于5，当取8个小数时，其和$$0.6+0.66+0.666+\\ldots +0.66666666\\textgreater0.66\\times 7+0.6=5.22$$.故选C. " ]

[ "2010年四年级竞赛明心奥数挑战赛" ]

[ [ { "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->计算模块->定义新运算->直接运算型->普通型" ]

[ "$$5\\times 2=10=11-1$$ " ]

[ "2008年全国迎春杯三年级竞赛初赛第10题" ]

[ [ { "aoVal": "A", "content": "$$110$$ " } ], [ { "aoVal": "B", "content": "$$112$$ " } ], [ { "aoVal": "C", "content": "$$114$$ " } ], [ { "aoVal": "D", "content": "$$116$$ " } ], [ { "aoVal": "E", "content": "$$118$$ " } ] ]

[ "海外竞赛体系->知识点->数论模块->质数与合数->特殊质数运用->特殊质数2", "拓展思维->能力->空间想象->立体几何加工" ]

[ "显然大正方体表面不能全为蓝色，$$8$$个顶点上的正方体木块表面积是$$3$$平方厘米，棱上的正方体木块表面积是$$2$$平方厘米，面上的正方体木块表面积是$$1$$平方厘米，所以要先在顶点和棱上放蓝色的正方体木块，剩下的放在面上，不放在内部．所以让正方体顶点和棱均为蓝色方块时蓝色表面积最大．此时蓝色的面积为：$$8\\times3+2\\times（5-2）\\times12+（62-8-36）=114$$（平方厘米）. " ]

[ "2014年全国迎春杯六年级竞赛初赛第1题" ]

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

[ "拓展思维->能力->运算求解" ]

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

[ "2012年IMAS小学高年级竞赛第一轮检测试题第12题4分" ]

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

[ "拓展思维->思想->整体思想" ]

[ "由题意可知$$6$$根香蕉与$$8$$个橘子的重量相同，从而$$5$$个苹果与$$8$$个橘子的重量相同．所以$$16$$个橘子与$$10$$个苹果的重量相同． " ]

[ "2016年创新杯六年级竞赛训练题（四）第2题" ]

[ [ { "aoVal": "A", "content": "$$77$$ " } ], [ { "aoVal": "B", "content": "$$154$$ " } ], [ { "aoVal": "C", "content": "$$231$$ " } ], [ { "aoVal": "D", "content": "$$286$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->数字谜->横式数字谜->与数论的结合" ]

[ "将原式中的数分解质因数，数出每个质数的个数，发现$$3$$，$$7$$，$$11$$为奇数个，故提出来． $$\\quad 12\\times 11\\times 10\\times 9\\times 8\\times 7\\times 6\\times 5\\times 4\\times 3\\times 2\\times 1=2\\times 2\\times 3\\times 11\\times 2\\times 5\\times 3\\times 3\\times 2\\times 2\\times 2\\times 7\\times 2\\times 3\\times 5\\times 2\\times 2\\times 3\\times 2\\times 1$$ $$=\\left( 2\\times 2\\times 2\\times 2\\times 2\\times 3\\times 3\\times 5 \\right)\\div \\left( 2\\times 2\\times 2\\times 2\\times 2\\times 3\\times 3\\times 5 \\right)\\times 3\\times 7\\times 11$$ $$=\\left( 12\\times 10\\times 9\\times 4 \\right)\\div \\left( 8\\times 6\\times 5\\times 3\\times 2 \\right)\\times 3\\times 7\\times 11$$ $$=3\\times 7\\times 21$$ $$=231$$． " ]

[ "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->拓展思维->组合模块->时间问题->认识钟表" ]

[ "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． " ]

[ "2011年北京小学高年级五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$660$$ " } ], [ { "aoVal": "B", "content": "$$580$$ " } ], [ { "aoVal": "C", "content": "$$750$$ " } ], [ { "aoVal": "D", "content": "$$800$$ " } ] ]

[ "拓展思维->思想->逐步调整思想", "课内体系->能力->运算求解" ]

[ "两人原有钱数之比为$$6$$:$$5$$，如果甲得到$$180$$元，乙得到$$150$$元，那么两人的钱数之比仍为$$6$$:$$5$$，现在甲得到$$180$$元，乙只得到$$30$$元，相当于少得到了$$120$$元，现在两人钱数之比为$$18$$:$$11$$，可以理解为：两人的钱数分别增加$$180$$元和$$150$$元之后，钱数之比为$$18$$：$$15$$，然后乙的钱数减少$$120$$元，两人的钱数之比变为$$18$$:$$11$$，所以$$120$$元相当于$$4$$份，$$1$$份为$$30$$元，后来两人的钱数之和为$$30 \\times (18 + 15) = 990$$元，所以原来两人的总钱数之和为$$990 - 180 - 150 = 660$$元． 设甲、乙原来各有$$6x$$、$$5x$$元，那么有$$(6x+180):(5x+30)=18:11$$，解比例方程得$$x=60$$，原来共有$$60\\times (6+5)=660$$（元）． 由于甲、乙得到的钱数不同，所以差并非不变．为了让得到的钱数相同，可以考虑将乙原有的钱数和得到的钱数都翻$$6$$倍，那么甲、乙原有钱数之比变为$$1:5$$，现有钱数之比变为$$3:11$$，统一差，变为$$2:10$$以及$$3:11$$，一份就是$$180$$元，所以原有钱数之和为$$(2+10\\div 6)\\times 180=660$$． 故答案为：$$660$$元． " ]

[ "2014年IMAS小学中年级竞赛第一轮检测试题第2题3分" ]

[ [ { "aoVal": "A", "content": "$$298$$ " } ], [ { "aoVal": "B", "content": "$$312$$ " } ], [ { "aoVal": "C", "content": "$$231$$ " } ], [ { "aoVal": "D", "content": "$$357$$ " } ], [ { "aoVal": "E", "content": "$$101$$ " } ] ]

[ "拓展思维->能力->数感认知->整数数字加工" ]

[ "因为$$101\\textless{}231\\textless{}298\\textless{}312\\textless{}357$$，所以最小的数为$$101$$．故选$$\\text{E}$$． " ]

[ "2017年第13届湖北武汉新希望杯小学高年级六年级竞赛决赛第1题" ]

[ [ { "aoVal": "A", "content": "$$1:23$$ " } ], [ { "aoVal": "B", "content": "$$1:24$$ " } ], [ { "aoVal": "C", "content": "$$1:25$$ " } ], [ { "aoVal": "D", "content": "$$1:26$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "不及格率$$1-96 \\%=4 \\%$$，不及格人数与总人数的比是$$4 \\%:100 \\%=1:25$$． " ]

[ "2015年湖北武汉世奥赛五年级竞赛模拟训练题（一）第3题" ]

[ [ { "aoVal": "A", "content": "$$1.65$$ " } ], [ { "aoVal": "B", "content": "$$16.5$$ " } ], [ { "aoVal": "C", "content": "$$25.06$$ " } ], [ { "aoVal": "D", "content": "$$28.45$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "把这个一位小数的末尾与$$8.56$$对齐，相当于把这个一位小数缩小到原数的$$\\frac{1}{10}$$，又因为$$10.21-8.56=1.65$$，所以原来的一位小数是$$16.5$$．正确的结果应是$$16.5+8.56=25.06$$ " ]

[ "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模拟11第9题3分", "2015年湖北武汉世奥赛五年级竞赛模拟训练题（四）第4题" ]

[ [ { "aoVal": "A", "content": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$55$$ " } ], [ { "aoVal": "D", "content": "$$66$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "从最不利的情况考虑，第$$1$$把锁需要试$$9$$次；第$$2$$把锁需要试$$8$$次$$\\ldots \\ldots $$；第$$9$$把锁需要试$$1$$次．一共需要试$$1+2+3+\\cdots \\cdots +9=45$$（次）． " ]

[ "2018年湖北武汉新希望杯小学高年级五年级竞赛训练题（三）第2题" ]

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

[ "拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理正应用->总个数" ]

[ "$$42=2\\times 3\\times 7$$，$$(1+1)\\times (1+1)\\times (1+1)=8$$（个）． " ]

[ "2019年第24届YMO三年级竞赛决赛第2题3分" ]

[ [ { "aoVal": "A", "content": "星期日 " } ], [ { "aoVal": "B", "content": "星期一 " } ], [ { "aoVal": "C", "content": "星期二 " } ], [ { "aoVal": "D", "content": "不确定 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "根据题意分析可知，某年的$$2$$月份有$$5$$个星期一， 因为$$2$$月有$$28$$天则只有$$4$$个周一， 所以这一年的$$2$$月有$$29$$天，且$$2$$月$$1$$日为星期一， 则$$2$$月$$22$$日为星期一． 故选$$\\text{B}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛决赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$0.21$$ " } ], [ { "aoVal": "C", "content": "$$0.32$$ " } ], [ { "aoVal": "D", "content": "$$0.43$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "原式$$=\\left( 1+0.21+0.32 \\right)\\times \\left( 0.21+0.32 \\right)+\\left( 1+0.21+0.32 \\right)\\times 0.43-\\left( 1+0.21+0.32 \\right)\\times \\left( 0.21+0.32 \\right)-0.43\\times \\left( 0.21+0.32 \\right)$$ $$=\\left[ \\left( 1+0.21+0.32 \\right)\\times \\left( 0.21+0.32 \\right)-\\left( 1+0.21+0.32 \\right)\\times \\left( 0.21+0.32 \\right) \\right]+\\left[ \\left( 1+0.21+0.32 \\right)\\times 0.43-\\left( 0.21+0.32 \\right)\\times 0.43 \\right]$$ $$=0+0.43\\times \\left[ \\left( 1+0.21+0.32 \\right)-\\left( 0.21+0.32 \\right) \\right]$$ $$=0.43\\times 1$$ $$=0.43$$． 故答案选：$$\\text{D}$$． " ]

[ "2013年华杯赛六年级竞赛决赛", "2013年华杯赛五年级竞赛决赛" ]

[ [ { "aoVal": "A", "content": "$$1:1.6\\pi$$ " } ], [ { "aoVal": "B", "content": "$$\\pi :10$$ " } ], [ { "aoVal": "C", "content": "$$3:4$$ " } ], [ { "aoVal": "D", "content": "$$3\\pi :40$$ " } ] ]

[ "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" ]

[ "骑车人$$A$$：$$10$$分钟共走$$3\\pi $$千米，骑车人$$B$$：$$5$$分钟走$$20$$千米，即$$10$$分钟走$$40$$千米，速度比等于路程比是$$3\\pi :40$$。 " ]

[ "2017年第15届全国希望杯小学高年级六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac{p}{q}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{q-p}{q}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{p+q}{p}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{p+q}{q}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->方程基础->等量代换->代数式" ]

[ "由$$p$$，$$q$$是非$$0$$的自然数，并且$$p\\textless{}q$$，可知$$0\\textless{}\\frac{p}{q}\\textless{}1$$，于是有$$\\frac{q-p}{q}=1-\\frac{p}{q}\\textless{}1$$，$$\\frac{p+q}{p}=1+\\frac{q}{p}\\textgreater1+1=2$$，$$1\\textless{}\\frac{p+q}{q}=1+\\frac{p}{q}\\textless{}2$$，故四个式子：$$\\frac{p}{q}$$，$$\\frac{q-p}{q}$$，$$\\frac{p+q}{p}$$，$$\\frac{p+q}{q}$$中，值在$$1$$和 $$2$$之间的是$$\\frac{p+q}{q}$$． " ]

[ "2018年美国数学大联盟杯四年级竞赛初赛第29题5分" ]

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

[ "拓展思维->思想->对应思想", "Overseas Competition->知识点->应用题模块->工程问题" ]

[ "根据内容分析可知：工作时间$$=$$工作总量$$\\div $$工作效率． 假设工作总量为$$1$$，用工作总量减去共同完成的工作量，再除以汤姆的工作效率即可得到剩下的工作汤姆单独完成需要多长时间．故列式计算如下： $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left[ 1-\\left( \\frac{1}{2}+\\frac{1}{4} \\right)\\times 1 \\right]\\div \\frac{1}{4}$$ $$=\\left[ 1-\\left( \\frac{2}{4}+\\frac{1}{4} \\right)\\times 1 \\right]\\div \\frac{1}{4}$$ $$=\\left( 1-\\frac{3}{4}\\times 1 \\right)\\div \\frac{1}{4}$$ $$=\\frac{1}{4}\\times 4$$ $$=1$$（小时）． 故选答案$$\\text{A}$$． " ]

[ "2020年长江杯五年级竞赛复赛A卷", "2020年长江杯五年级竞赛复赛A卷第5题5分" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$31$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->智巧趣题->数学趣题->砝码问题" ]

[ "（$$1$$）$$1$$个砝码可以称出的重量有$$5$$种：$$1$$克，$$2$$克，$$4$$克，$$8$$克，$$16$$克； （$$2$$）$$2$$个砝码可以称出的重量有$$10$$种： $$1+2=3$$（克），$$1+4=5$$（克），$$1+8=9$$（克）， $$1+16=17$$（克）； $$2+4=6$$（克），$$2+8=10$$（克），$$2+16=18$$（克）； $$4+8=12$$（克），$$4+16=20$$（克）， $$8+16=24$$（克）； （$$3$$）$$3$$个砝码可以称出的重量有$$10$$种： $$1+2+4=7$$（克），$$1+2+8=11$$（克）， $$1+2+16=19$$（克）， $$1+4+8=13$$（克），$$1+4+16=21$$（克）， $$1+8+16=25$$（克）， $$2+4+8=14$$（克），$$2+4+16=22$$（克）， $$2+8+16=26$$（克）， $$4+8+16=28$$（克）； （$$4$$）$$4$$个砝码可以称出的重量有$$5$$种： $$1+2+4+8=15$$（克），$$1+2+4+16=23$$（克）， $$1+2+8+16=27$$（克）， $$1+4+8+16=29$$（克），$$2+4+8+16=30$$（克）； （$$5$$）$$5$$个砝码可以称出的重量有$$1$$种： $$1+2+4+8+16=31$$（克）． 因为$$5+10+10+5+1=31$$（种）， 所以最多可以称出$$31$$种不同的重量，它们是$$1$$克$$-31$$克． 答：最多能称出$$31$$种不同重量的物体． 故选$$\\text{D}$$． " ]

[ "2011年北京学而思杯五年级竞赛" ]

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

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

[ "因为必须是调换相邻的两卷，将第$$5$$卷调至原来第$$1$$卷的位置最少需$$4$$次，得到的顺序为$$51234$$； 现在将第$$4$$卷调至此时第$$1$$卷的位置最少需$$3$$次，得到的顺序为$$54123$$； 现在将第$$3$$卷调至此时第$$1$$卷的位置最少需$$2$$次，得到的顺序为$$54312$$； 最后将第$$1$$卷和第$$2$$卷对调即可． 所以，共需调换$$4+3+2+1=10$$（次）． " ]

[ "2019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": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$3$$个球一组，可分为$$3$$组．分别标号第①组，第②组，第③组． 第$$1$$步：将第①组与第②组称$$1$$次．若平衡，则次品在第③组，第$$2$$步：在第③组的$$3$$个球中任选两个称第$$2$$次，即可判断．若第$$1$$次不平衡，哪组轻，就在哪一组进行第$$2$$次称重．故至少称$$2$$次，即可保证一定将这个次品找出来． " ]

[ "2013年第11届全国创新杯五年级竞赛第5题5分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$13$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "首位为$$1$$，有 $$12312$$，$$12331$$，$$12333$$这三种；首位为$$2$$，有$$23123$$，$$23312$$，$$23331$$，$$23333$$这四种；首位为$$3$$，有$$31231$$，$$31233$$，$$33123$$，$$33312$$，$$33331$$，$$33333$$这六种， 所以共$$(4+100)\\div 2=52$$． " ]

[ "2008年第6届创新杯四年级竞赛初赛B卷第7题5分" ]

[ [ { "aoVal": "A", "content": "$$1957$$ " } ], [ { "aoVal": "B", "content": "$$1959$$ " } ], [ { "aoVal": "C", "content": "$$2008$$ " } ], [ { "aoVal": "D", "content": "$$2009$$ " } ] ]

[ "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" ]

[ "所得的$$101$$个新数之和为：$$2008-1+2-3+4-\\cdots -99+100-101=2008+\\underbrace{1+1+1+\\cdots +1}_{50个1}-101=1957$$． " ]

[ "2017年IMAS小学中年级竞赛（第一轮）第14题4分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "可知这次考试比所欲达到的平均分数多的$$98-90=8$$分要分配到前几次的考试才能使平均分数增加$$90-88=2$$分，因此小明之前共考了$$8\\div 2=4$$次数学考试，所以连同这一次他总共考了$$4+1=5$$次数学考试．故选$$\\text{C}$$． ", "<p>可知最后一次考试要比所欲达到的平均分数多$$98-88=10$$分才能使平均分数从$$88$$分增加到$$90$$分，即须增加$$2$$分，因此连同这一次小明总共考了$$10\\div 2=5$$次数学考试．故选$$\\text{C}$$．</p>" ]

[ "2016年创新杯五年级竞赛训练题（一）第6题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ] ]

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "一个数除以$$9$$的余数等于这个数各个数字之和除以$$9$$的余数，每次操作将数的和变为数字和，不改变除以$$9$$的余数，$$1+2+3+\\cdots +2013=2014\\times 2013\\div 2=1007\\times 2013$$，$$1007$$除以$$9$$的余数相同，余数为$$3$$，那么最后剩下的四个数字的和除以$$9$$也是余$$3$$的；将$$27$$拆分成四个数的乘积；$$27=3\\times 3\\times 3\\times 1=3\\times 9\\times 1\\times 1=27\\times 1\\times 1\\times 1$$，和分别为$$10$$，$$14$$，$$30$$，只有$$30$$除以$$9$$余数为$$3$$，所以四个数的和为$$30$$． " ]

[ "1990年全国华杯赛竞赛复赛", "1990年第1届全国华杯赛小学高年级竞赛复赛第10题" ]

[ [ { "aoVal": "A", "content": "$$\\dfrac{2}{9}$$ " } ], [ { "aoVal": "B", "content": "$$\\dfrac{4}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\dfrac{5}{9}$$ " } ], [ { "aoVal": "D", "content": "$$\\dfrac{1}{3}$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->认识单位1" ]

[ "不妨认为第二堆全是黑子，第一堆全是白子，（即将第一堆黑子与第二堆白子互换） 第二堆黑子是全部棋子的$$\\dfrac{1}{3}$$， 同时，又是黑子的$$1-\\dfrac{2}{5}$$， 所以黑子占全部棋子的：$$\\dfrac{1}{3}\\div~ (1-\\dfrac{2}{5})=\\dfrac{5}{9}$$， 白子占全部棋子的：$$1-\\dfrac{5}{9}=\\dfrac{4}{9}$$． " ]

[ "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6题2分" ]

[ [ { "aoVal": "A", "content": "$$18:15$$ " } ], [ { "aoVal": "B", "content": "$$19:15$$ " } ], [ { "aoVal": "C", "content": "$$22:05$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->时间问题->时间计算" ]

[ "根据题意分析可知，$$60$$分钟$$=1$$小时，$$85$$分钟$$=1$$小时$$25$$分钟， 用结束的时间减去放映的时长即可得到开始放映的时间， 故列式为：$$20:40-1:25=19:15$$． 故选$$\\text{B}$$． " ]

[ "2017年全国美国数学大联盟杯小学中年级三年级竞赛初赛第38题" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$26$$ " } ], [ { "aoVal": "C", "content": "$$27$$ " } ], [ { "aoVal": "D", "content": "$$28$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$40-(40\\times 4-95)\\div (4+1)=27$$． " ]

[ "2009年全国迎春杯小学中年级四年级竞赛初赛第5题" ]

[ [ { "aoVal": "A", "content": "$$223$$ " } ], [ { "aoVal": "B", "content": "$$251$$ " } ], [ { "aoVal": "C", "content": "$$252$$ " } ], [ { "aoVal": "D", "content": "$$446$$ " } ] ]

[ "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" ]

[ "假设原来有红球$$x$$个，那么第一次操作放入了$$2（x-1）$$个黄球，球数增加到$$3x-2$$个，第二次操作放入了$$2（3x-2-1）=6x-6$$个篮球，球数增加到$$3x-2+6x-6=9x-8$$个．所以$$9x-8=2008$$，解得：$$x=224$$,所以黄球有$$2（x-1）=446$$． " ]

[ "2012年全国希望杯六年级竞赛初赛第17题" ]

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

[ "拓展思维->思想->转化与化归的思想" ]

[ "有$$3$$倍关系的数放入一组：（$$1$$，$$3$$，$$9$$）、（$$2$$，$$6$$）、（$$4$$，$$12$$）、（$$5$$，$$15$$）， 其余$$7$$个数每个数放入一组，第一组最多取$$2$$个（$$1$$和$$9$$），其余每组最多只能取$$1$$个， 因此最多取$$12$$个保证没有$$3$$倍关系，再多取一个就可以保证有一个数是另一个数的$$3$$倍． " ]

[ "2008年第6届创新杯六年级竞赛初赛A卷第8题5分", "2008年六年级竞赛创新杯" ]

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

[ "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" ]

[ "显然，三个最小的连续的奇质数是$$3$$、$$5$$、$$7$$，如果取出$$1$$张，$$3$$、$$5$$、$$7$$都为质数； 如果取出两张，可以组成的两位数有$$35$$、$$53$$、$$37$$、$$73$$、$$57$$、$$75$$， $$35=5\\times 7$$，$$57=3\\times 19$$，$$75=3\\times 5\\times 5$$，所以$$3$$、$$5$$、$$7$$组成的两位数质数有$$53$$、$$37$$、$$73$$； 如果取出三张，可以组成三位数，由于三个数字的和是$$3+5+7=15$$是$$3$$的倍数，所以组成的任何一个三位数都不是质数． 综上，所有的质数有：$$3$$、$$5$$、$$7$$、$$53$$、$$37$$、$$73$$，一共有$$6$$个． " ]

[ "2017年北京学而思杯六年级竞赛年度教学质量监测第2题5分" ]

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

[ "拓展思维->拓展思维->计数模块->统计与概率->概率->计数求概率" ]

[ "狼人牌有$$3$$张，一共有$$10$$张卡牌．所以抽到狼人牌的概率是$$\\frac{3}{10}=30 \\% $$． " ]

[ "2017年四川成都六年级竞赛“全能明星”选拔赛第8题2分" ]

[ [ { "aoVal": "A", "content": "$$5{}^{}\\circ $$ " } ], [ { "aoVal": "B", "content": "$$10{}^{}\\circ $$ " } ], [ { "aoVal": "C", "content": "$$15{}^{}\\circ $$ " } ], [ { "aoVal": "D", "content": "$$20{}^{}\\circ $$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "在下午$$2$$点$$10$$分，分针从数字$$12$$开始转了$$10\\times 6{}^{}\\circ =60{}^{}\\circ $$，时针从数字$$2$$开始转了$$10\\times 0.5{}^{}\\circ =5{}^{}\\circ $$，这时时针与分针所成的锐角为$$60{}^{}\\circ +10\\times 0.5{}^{}\\circ -60{}^{}\\circ =5{}^{}\\circ $$． " ]

[ "2016年北京学而思杯小学高年级五年级竞赛冲刺讲义" ]

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

[ "知识标签->拓展思维->应用题模块->经济问题" ]

[ "甲：$$30\\times 0.75\\times 100=2250$$（元）； 乙：$$100\\div 7=14\\cdots\\cdots 2$$，$$(14\\times 5+2)\\times 30=2160$$（元）； 丙：$$40\\times 30+30\\times 30\\times 0.7+30\\times 30\\times 0.5=76\\times 30=2280$$（元）． 丙\\textgreater 甲\\textgreater 乙． 丙花费最多． " ]

[ "2018年第23届华杯赛小学高年级竞赛初赛第5题", "2019年广东深圳福田区深圳市高级中学小升初第5题", "2018年华杯赛小学高年级竞赛初赛第5题10分" ]

[ [ { "aoVal": "A", "content": "$$19$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$22$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "将$$1\\sim 20$$分成如下$$10$$组： $$\\left( 1,20 \\right)$$，$$\\left( 2,19 \\right)$$，$$\\left( 3,18 \\right)$$，$$\\cdots \\cdots $$，$$\\left( 10,11 \\right)$$，每组和均为$$21$$，根据抽屉原理，$$10$$组取$$11$$个数，至少有$$2$$个在同一组和为$$21$$，所以选 $$\\text{C}$$． " ]

[ "2016年第14届全国创新杯五年级竞赛复赛第5题" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "假设免费携带$$x$$千克 根据两次付费可得比例关系$$\\frac{100-2x}{100-x}=\\frac{10}{15}$$，解得$$x=25$$． " ]

[ "2016年创新杯小学高年级五年级竞赛训练题（四）第4题", "2012年全国创新杯五年级竞赛第1题5分" ]

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

[ "拓展思维->拓展思维->计数模块->容斥原理->三量容斥" ]

[ "让三项都会的人最少，即让所有人都只会两项，则一共会$$48\\times 2=96$$项，二十几会$$27+33+40=100$$项，即至少有$$100-96=4$$人会三项． 构造如下：$$8$$人会游泳$$+$$自行车；$$15$$人会游泳$$+$$乒乓球；$$21$$人会自行车$$+$$乒乓球；$$4$$人全会． " ]

[ "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->思想->转化与化归的思想" ]

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

[ "2014年全国迎春杯三年级竞赛初赛第12题" ]

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

[ "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" ]

[ "方法一：星期六有：$$21\\to 28\\to 4(35)\\to 11\\to 18\\to 25$$，所以 $$31$$日是星期五． 方法二：$$10+31=41$$（天），$$41\\div7=5\\cdots 6$$ ，差一天是星期六，所以$$31$$日是星期五． " ]

[ "2013年IMAS小学高年级竞赛第一轮检测试题第17题4分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{13}{24}$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{1}{4}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->定义新运算->反解未知数型" ]

[ "可观察出新运算即计算两数之平均值，首先把括弧看成一个整体，则括弧里的数与$$\\frac{3}{4}$$之平均等于$$\\frac{1}{2}$$，即$$\\frac{1}{6}\\Theta \\square =1-\\frac{3}{4}=\\frac{1}{4}$$，从而$$\\square $$与$$\\frac{1}{6}$$之平均等于$$\\frac{1}{4}$$，所以可得$$\\square =\\frac{1}{4}\\times 2-\\frac{1}{6}=\\frac{1}{3}$$． 故选$$\\text{B}$$． " ]

[ "2006年五年级竞赛创新杯", "2006年第4届创新杯五年级竞赛复赛第1题" ]

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

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

[ "不妨假设这段路程为$$6a$$，则$$6a\\times 2\\div \\left( 6a\\div a+6a\\div 2a \\right)=\\frac{4}{3}a$$ " ]

[ "2015年第13届全国创新杯五年级竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "$$50$$ " } ], [ { "aoVal": "B", "content": "$$51$$ " } ], [ { "aoVal": "C", "content": "$$120$$ " } ], [ { "aoVal": "D", "content": "$$131$$ " } ] ]

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

[ "等差数列中，$${{a}_{2}}+{{a}_{93}}=2{{a}_{50}}$$，则$${{a}_{50}}=262\\div 2=131$$．（$${{a}_{n}}$$表示这列数中第$$n$$个数）． " ]

[ "2008年四年级竞赛创新杯", "2008年第6届创新杯四年级竞赛复赛第7题4分" ]

[ [ { "aoVal": "A", "content": "$$5.25 \\textless{} a \\leqslant 5.35$$ " } ], [ { "aoVal": "B", "content": "$$5.25 \\textless{} a \\textless{} 5.35$$ " } ], [ { "aoVal": "C", "content": "$$5.25\\leqslant a\\leqslant 5.35$$ " } ], [ { "aoVal": "D", "content": "$$5.25\\leqslant a \\textless{} 5.35$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->小数基础->取近似值" ]

[ "当$$a=5.35$$时，$$a$$四舍五入得到的近似值为$$5.4$$，不合题意，排除$$\\text{A}$$、$$\\text{C}$$，当$$a=5.25$$时，$$a$$四舍五入得到的近似值为$$5.3$$，符合题意，排除$$\\text{B}$$，故选$$\\text{D}$$． " ]

[ "2009年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$0.25$$ " } ], [ { "aoVal": "B", "content": "$$0.125$$ " } ], [ { "aoVal": "C", "content": "$$0.50$$ " } ], [ { "aoVal": "D", "content": "以上都不对 " } ] ]

[ "拓展思维->拓展思维->计数模块->统计与概率->概率->计数求概率" ]

[ "小红、小军星期天到四个地方（东湖、黄鹤楼、中山公园或古琴台）去玩，共有$$4\\times 4=16$$种情况，他们碰巧到同一地方玩，只有4种情况，因此，小红与小军星期天到同一地方玩的机会有$$4\\div 16=0.25$$。 " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$39$$ " } ], [ { "aoVal": "B", "content": "$$40$$ " } ], [ { "aoVal": "C", "content": "$$41$$ " } ], [ { "aoVal": "D", "content": "$$43$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "将从$$0$$开始的连续自然数倒序书写，则相邻的奇数与偶数之差为$$1$$，共计$$20$$组． 所以会得到$$20$$个奇数，$$20$$个偶数．共计$$40$$个数，所以从$$0\\sim 39$$会有$$40$$个数． 故选择$$\\text{A}$$． " ]

[ "2019年第7届湖北长江杯五年级竞赛复赛A卷第2题3分" ]

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

[ "拓展思维->拓展思维->应用题模块->分百应用题->量率对应求单位1" ]

[ "如果甲、乙都错的题，都是$$3$$道， 则题目总数为：$$3\\div \\frac{1}{6}=18$$（道）， 则甲错题数不是整数，故不可以； 甲、乙都错的题是一道， 则题目总数为$$6$$道， 则甲错题数也不是整数，故不可以； 甲、乙都错的题目是$$2$$道， 则题目总数为：$$2\\div \\frac{1}{6}=12$$（道）， 则甲错题数为：$$12\\times \\frac{1}{4}=3$$（道）， 所以甲、乙答错的题目数量为：$$2+1+1=4$$（道）， 所以剩下的就是甲乙都答对的题目数：$$12-4=8$$（道）． 故选$$\\text{B}$$． " ]

[ "2013年第11届全国创新杯五年级竞赛第4题5分" ]

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

[ "拓展思维->七大能力->逻辑分析" ]

[ "依题意可知比赛总场次为$$24$$场 现在所有队伍获得的总分值为：$$22+19+14+12=67$$（分），$$24$$场比赛，最高有$24\\times3=72$分，所以平局有$$72-67=5$$场． " ]

[ "2017年第16届湖北武汉创新杯小学高年级六年级竞赛邀请赛训练题（四）" ]

[ [ { "aoVal": "A", "content": "$$8$$天 " } ], [ { "aoVal": "B", "content": "$$10$$天 " } ], [ { "aoVal": "C", "content": "$$12$$天 " } ], [ { "aoVal": "D", "content": "$$15$$天 " } ] ]

[ "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" ]

[ "求三队工效和：$$\\left( \\frac{1}{12}+\\frac{1}{15}+\\frac{1}{20} \\right)\\div 2=\\frac{1}{10}$$，$$1\\div \\frac{1}{10}=10$$．选$$\\text{B}$$． " ]

[ "2012年第10届全国创新杯小学高年级六年级竞赛第6题4分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->分数->繁分数->繁分数方程" ]

[ "$$A+\\frac{1}{B+\\frac{1}{C+1}}=4+\\frac{4}{5}=4+\\frac{1}{1+\\frac{1}{4}}=4+\\frac{1}{1+\\frac{1}{3+1}}$$，那么$$A=4$$，$$B=1$$，$$C=3$$，则$$A+2B+3C=4+2+9=15$$． " ]

[ "2015年湖北武汉世奥赛五年级竞赛模拟训练题（二）第1题", "五年级其它" ]

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

[ "拓展思维->思想->逆向思想" ]

[ "每人至少订一种，共有$$3+3+1=7$$种不同的订法，又因为$$47\\div 7=6\\cdots \\cdots 5$$，所以至少有$$6+1=7$$名小朋友订的报刊相同． " ]

[ "2004年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$56$$ " } ], [ { "aoVal": "B", "content": "$$78$$ " } ], [ { "aoVal": "C", "content": "$$84$$ " } ], [ { "aoVal": "D", "content": "$$96$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->公因数与最大公因数->两数的最大公因数" ]

[ "因为两个两位数的最大公因数为$$8$$，可设这两个数分别为$$8a，8b$$($$a，b$$互质)，那么$$8\\times a\\times b=96$$，即$$a\\times b=12$$，则$$a，b$$为$$\\left( 3,4 \\right),\\left( 1,12 \\right)$$，又因为这两个数皆为两位数，所以这两个数只能为$$24$$和$$32$$，$$24+32=56$$，选 A. " ]

[ "2019年第24届YMO二年级竞赛决赛第10题3分" ]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "根据题意分析可知，假设$$20$$道题全部做对， 则得：$$20\\times5=100$$（分），现在比满分少了$$100-76=24$$（分）， 又因为答错或不答一题比答对一题少$$1+5=6$$（分）， 也就是做错了：$$24\\div6=4$$（题）， 所以做对了：$$20-4=16$$（题）， 故选答案$$\\text{D}$$． " ]

[ "2011年全国学而思杯五年级竞赛第5题", "2011年北京学而思综合能力诊断六年级竞赛第3题", "2011年全国学而思杯四年级竞赛第5题" ]

[ [ { "aoVal": "A", "content": "$$30\\times 3$$ " } ], [ { "aoVal": "B", "content": "$$30\\times 2$$ " } ], [ { "aoVal": "C", "content": "$$30\\times 30$$ " } ], [ { "aoVal": "D", "content": "$$30\\times 30\\times 30$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "$$8$$级和$$5$$级差了$$3$$级，所以$$8$$级是$$5$$级的$$30\\times 30\\times 30$$倍. " ]

[ "2014年IMAS小学高年级竞赛第二轮检测试题第3题4分" ]

[ [ { "aoVal": "A", "content": "$$100$$ " } ], [ { "aoVal": "B", "content": "$$110$$ " } ], [ { "aoVal": "C", "content": "$$117$$ " } ], [ { "aoVal": "D", "content": "$$181$$ " } ], [ { "aoVal": "E", "content": "$$271$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "如果一个两位数是完全平方数，那么这个两位数的所有可能值是： $$16$$，$$25$$，$$36$$，$$49$$，$$64$$，$$81$$， 而$$3+6={{3}^{2}}$$，$$8+1={{3}^{2}}$$， 所以符合条件的两位数为$$36$$，$$81$$， $$36+81=117$$． " ]

[ "2016年全国华杯赛小学中年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$$350$$ " } ], [ { "aoVal": "B", "content": "$$360$$ " } ], [ { "aoVal": "C", "content": "$$37$$ " } ], [ { "aoVal": "D", "content": "$$380$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "原式$$=\\left( 124+106 \\right)+\\left( 129+141 \\right)+\\left( 237+113 \\right)-500$$ ~~~~~~~~$$=230+270+350-500$$ ~ ~ ~ ~ $$=500-500+350$$ ~ ~ ~ ~ $$=350$$ " ]

[ "2020年新希望杯二年级竞赛决赛（8月）第6题", "2020年新希望杯二年级竞赛初赛（个人战）第6题" ]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$25$$ " } ] ]

[ "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->分百应用题->认识单位1" ]

[ "加上李叔叔自己，一共：$$11+1+12=24$$（人）， 故选择：$$\\text{D}$$． " ]

[ "2017年第13届湖北武汉新希望杯六年级竞赛决赛第6题" ]

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

[ "知识标签->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" ]

[ "甲、乙$$2$$小时一个周期：$$\\frac{1}{4}+\\frac{1}{6}=\\frac{5}{12}$$，两个周期后还剩下：$$1-2\\times \\frac{5}{12}=\\frac{1}{6}$$，剩下的甲做：$$\\frac{1}{6}\\div \\frac{1}{6}=1$$小时，共需$$2\\times 2+1=5$$小时． " ]

[ "2019年第24届YMO二年级竞赛决赛第1题3分" ]

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

[ "拓展思维->能力->运算求解", "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，小$$Y$$后面有$$42-22=20$$（人），$$22-20-1=1$$（人）．那么小$$M$$应该在小$$Y$$的正前方，所以小$$Y$$和小$$M$$之间有$$0$$个人． 故选：$$\\text{A}$$． " ]

[ "2021年鹏程杯六年级竞赛初赛第22题" ]

[ [ { "aoVal": "A", "content": "$$210$$ " } ], [ { "aoVal": "B", "content": "$$220$$ " } ], [ { "aoVal": "C", "content": "$$230$$ " } ], [ { "aoVal": "D", "content": "$$240$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "设原来的车速为$$v$$千米/小时，预定时间为$$t$$小时， 由第一个条件可知$$2v+\\left( 1+\\frac{1}{5} \\right)v\\times \\left( t-2-\\frac{30}{60} \\right)=vt$$， 两边同时除以$$v$$，$$2+\\left( 1+\\frac{1}{5} \\right)\\left( t-2-\\frac{30}{60} \\right)=t$$， 解得$$t=5$$， 由第二个条件可得$$80+\\left( 1+\\frac{1}{4} \\right)v\\times \\left( t-\\frac{80}{v}-\\frac{40}{60} \\right)=vt$$， 把$$t=5$$代入，解得$$v=48$$， 所以总路程$$=48\\times 5=240$$（千米）． 故选$$\\text{D}$$． " ]

[ "2021年世界少年奥林匹克数学竞赛五年级竞赛复赛第9题12分" ]

[ [ { "aoVal": "A", "content": "$$660$$ " } ], [ { "aoVal": "B", "content": "$$590$$ " } ], [ { "aoVal": "C", "content": "$$340$$ " } ], [ { "aoVal": "D", "content": "$$560$$ " } ], [ { "aoVal": "E", "content": "$$230$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "根据题意则相距：$$250\\times 3-160=750-160=590$$米． 答：相距$$590$$米． " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（二）第1题" ]

[ [ { "aoVal": "A", "content": "两个小数相乘，积一定是小数 " } ], [ { "aoVal": "B", "content": "一个小数四舍五入后是$$2.34$$，这个小数最大可能是$$2.344$$ " } ], [ { "aoVal": "C", "content": "小数都比整数小 " } ], [ { "aoVal": "D", "content": "循环小数都是无限小数 " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->小数基础->取近似值" ]

[ "$$A$$项，$$2.5\\times 0.4=1$$，不是小数； $$B$$项，这个小数没有最大值，例如：$$2.3499\\cdots $$ $$C$$项，不一定，例如：$$3.6\\textgreater2$$． " ]

[ "2016年新希望杯六年级竞赛训练题（三）第15题" ]

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

[ "拓展思维->思想->方程思想" ]

[ "设答错$$x$$道，则答对$$\\frac{95+x}{3}$$，要使$$x$$最大，则不答的题应当最少． 当不答的题为$$0$$道时，$$x+\\frac{95+x}{3}=50$$，$$x=\\frac{55}{4}$$（舍），当不答的题为$$1$$道时，$$x+\\frac{95+x}{3}+1=50$$，$$x=13$$． " ]

[ "2015年湖北武汉世奥赛六年级竞赛模拟训练题（三）第5题" ]

[ [ { "aoVal": "A", "content": "$$425$$ " } ], [ { "aoVal": "B", "content": "$$426$$ " } ], [ { "aoVal": "C", "content": "$$427$$ " } ], [ { "aoVal": "D", "content": "$$428$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$\\left( 0+2+4+6+8 \\right)\\times 10+\\left( 1+2+\\ldots \\ldots +9 \\right)\\times 5+1=426$$． " ]

[ "2017年全国小学生数学学习能力测评六年级竞赛初赛第8题3分" ]

[ [ { "aoVal": "A", "content": "现价高 " } ], [ { "aoVal": "B", "content": "原价高 " } ], [ { "aoVal": "C", "content": "相等 " } ], [ { "aoVal": "D", "content": "无法比较 " } ] ]

[ "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" ]

[ "设商品的原价为$$x$$， ∴现价为：$$x\\cdot \\left( 1-\\frac{1}{8} \\right)\\cdot \\left( 1+\\frac{1}{8} \\right)=\\frac{63}{64}x$$， ∴$$x\\textgreater\\frac{63}{64}x$$ ， ∴原价$$\\textgreater$$现价． 故选$$\\text{B}$$． " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（五）第4题" ]

[ [ { "aoVal": "A", "content": "$$1343$$ " } ], [ { "aoVal": "B", "content": "$$1344$$ " } ], [ { "aoVal": "C", "content": "$$1345$$ " } ], [ { "aoVal": "D", "content": "$$1346$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->奇数与偶数->奇数与偶数的认识" ]

[ "这串数的奇偶性：奇，奇，偶，奇，奇，偶，奇，奇，偶，$$\\cdots \\cdots $$周期为$$3$$，$$2017\\div 3=762\\cdots \\cdots 1$$，奇数共有$$2\\times 672+1=1345$$（个）． " ]

[ "2017年第15届湖北武汉创新杯小学高年级五年级竞赛决赛第1题" ]

[ [ { "aoVal": "A", "content": "$$30$$，$$31$$ " } ], [ { "aoVal": "B", "content": "$$6$$，$$155$$ " } ], [ { "aoVal": "C", "content": "$$5$$，$$186$$ " } ], [ { "aoVal": "D", "content": "$$1$$，$$930$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->分解质因数->分解质因数（式）" ]

[ "$$930=30\\times 31$$． " ]

[ "2017年第15届湖北武汉创新杯五年级竞赛初赛第7题" ]

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

[ "拓展思维->思想->整体思想" ]

[ "最不利原则．要保证拿到一种颜色至少有$$3$$个，则根据最不利原则，可先取每种颜色$$2$$个，最后不论取哪种颜色，都一定可以满足条件，即需要$$2\\times 3+1=7$$个． " ]

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

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

[ "拓展思维->能力->逻辑分析" ]

[ "方法一：枚举法（不推荐）$$600$$的因数有$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$6$$，$$8$$，$$10$$，$$12$$，$$15$$，$$20$$，$$24$$，$$25$$，$$30$$，$$40$$，$$50$$，$$60$$，$$75$$，$$100$$，$$120$$，$$150$$，$$200$$，$$300$$，$$600$$，其中是$$6$$的倍数的有$$6$$，$$12$$，$$24$$，$$30$$，$$60$$，$$120$$，$$150$$，$$300$$，$$600$$这$$9$$个． 故选$$\\text{B}$$． 方法二：因数个数定理。 600=2\\textsuperscript{3}$$\\times $$3$$\\times $$5\\textsuperscript{2}。 思路1：根据加乘原理，$$2$$与$$3$$必须至少选一个，故$$2$$的选法有$$3$$种，$$3$$的选法有$$1$$种，$$5$$的选法有$$3$$种（不选$$5$$、选$$1$$个$$5$$、选$$2$$个5），故$$3\\times 1\\times$$ （2+1）=9（个）； 思路2: $$600\\div 6=100$$， 100=2\\textsuperscript{2}$$\\times $$5\\textsuperscript{2~}$$，再根据因数个数定理：$$ （2+1）$$\\times （2+1）=9$$（个） " ]

[ "2017年第17届全国中环杯三年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$$33700$$ " } ], [ { "aoVal": "B", "content": "$$65000$$ " } ], [ { "aoVal": "C", "content": "$$32500$$ " } ], [ { "aoVal": "D", "content": "$$325000$$ " } ] ]

[ "拓展思维->七大能力->运算求解", "课内体系->七大能力->运算求解" ]

[ "原式$$=325\\times 337+325\\times 2\\times 330+325\\times 3$$ $$=325\\times (337+2\\times 330+3)$$ $$=325\\times 1000$$ $$=325000$$． " ]

[ "2017年全国美国数学大联盟杯小学高年级六年级竞赛第17题5分" ]

[ [ { "aoVal": "A", "content": "$$66$$ " } ], [ { "aoVal": "B", "content": "$$110$$ " } ], [ { "aoVal": "C", "content": "$$129$$ " } ], [ { "aoVal": "D", "content": "$$176$$ " } ] ]

[ "知识标签->拓展思维->应用题模块->比例应用题->按比分配" ]

[ "停车场只有黑白两种车，白车与黑车的比率是$$3:8$$，黑白车的总数是$$242$$辆．请问黑车比白车多多少辆？ ratio比率； a total of 总数； more．．than 比．．多． 白车数量：$$\\frac{3}{11}\\times 242=66$$辆，黑车数量：$$\\frac{8}{11}\\times 242=176$$辆．黑车比白车多$$176-66=110$$辆． " ]

[ "2003年第1届创新杯六年级竞赛复赛第9题" ]

[ [ { "aoVal": "A", "content": "$$2274$$ " } ], [ { "aoVal": "B", "content": "$$2003$$ " } ], [ { "aoVal": "C", "content": "$$2300$$ " } ], [ { "aoVal": "D", "content": "$$2320$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "由于$$9+10+11+\\cdots 68=2310$$，由此可知$$2320$$是错误的．由于$$2274$$、$$2003$$、$$2300$$都小于小于$$2310$$，所以减的数较多，由于减一个数，总和里面就要少这个数的$$2$$倍，如减$$2$$，则是$$2310-2\\times 2=2306$$，所以只要是小于$$2310$$．据此分析即 $$9+10+11+\\cdots 68=2310$$，$$2320\\textgreater2310$$，故$$\\text{D}$$错误； $$\\left( 2310-2274 \\right)\\div 2=18$$，$$18\\div 2=9$$，所以在$$9$$前是减号即可，符合题意． $$\\left( 2310-2003 \\right)=307\\textgreater68$$，错误 $$\\left( 2310-2000 \\right)\\div 2=155\\textgreater68$$，错误． 故选$$\\text{A}$$． " ]

[ "2017年湖北武汉中环杯五年级竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$61$$ " } ], [ { "aoVal": "C", "content": "$$62$$ " } ], [ { "aoVal": "D", "content": "$$63$$ " } ] ]

[ "拓展思维->思想->对应思想", "Overseas Competition->知识点->计算模块->数列与数表->等差数列->等差数列求公差" ]

[ "先求公差：$$(32-5)\\div (10-1)=3$$，于是第$$20$$项是$$3\\times (20-1)+5=62$$． " ]

[ "2017年IMAS小学高年级竞赛（第一轮）第4题3分" ]

[ [ { "aoVal": "A", "content": "$$2.7\\overline{18}\\textless2.\\overline{718}\\textless2.71828\\textless2.71\\overline{8}$$ " } ], [ { "aoVal": "B", "content": "$$2.71828\\textless2.7 \\overline{18}\\textless2.\\overline{718}\\textless2.71 \\overline{8}$$ " } ], [ { "aoVal": "C", "content": "$$2.7 \\overline{18}\\textless2.71828\\textless2.\\overline{718}\\textless2.71 \\overline{8}$$ " } ], [ { "aoVal": "D", "content": "$$2.71828\\textless2.71\\overline{8}\\textless2.7\\overline{18}\\textless2.\\overline{718}$$ " } ], [ { "aoVal": "E", "content": "$$2.7 \\overline{18}\\textless2.\\overline{718}\\textless2.71\\overline{8}\\textless2.71828$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$2.\\overline{718}=2.718718\\cdots $$、$$2.7\\overline{18}=2.71818\\cdots $$、$$2.71\\overline{8}=2.71888\\cdots $$与$$2.71828$$，比较它们小数点后第四位可得$$2.7 \\overline{18}\\textless2.71828\\textless2.\\overline{718}\\textless2.71 \\overline{8}$$． 故选$$\\text{C}$$． " ]

[ "2014年全国迎春杯五年级竞赛复赛第9题" ]

[ [ { "aoVal": "A", "content": "$$1512$$ " } ], [ { "aoVal": "B", "content": "$$3024$$ " } ], [ { "aoVal": "C", "content": "$$1510$$ " } ], [ { "aoVal": "D", "content": "$$3020$$ " } ] ]

[ "拓展思维->思想->对应思想", "课内体系->思想->对应思想" ]

[ "考察是计数问题中的排列组合． $$0～9$$这$$10$$个数中任意挑选$$5$$个都可以组成``神马数''，$$\\text{C}_{10}^{5}=\\frac{10\\times9\\times 8\\times 7\\times 6}{5\\times 4\\times 3\\times 2\\times 1}=252$$种；在被挑选的$$5$$个数中，最小的放中间，剩下的$$4$$个数进行组合，从中任意挑选$$2$$个可以放在左边或者右边，$$\\text{C}_{4}^{2}=6$$种； 在此一定要注意：$$4$$个数中任选$$2$$个放在左边然后再放到右边数的顺序改变了． 所以共有``神马数''$$252\\times 6=1512$$个． " ]

[ "2014年全国迎春杯六年级竞赛复赛第11题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$25$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "假设甲走$$60$$千米时，乙走了$$a$$千米，甲到达$$B$$地时，乙车应走$$\\frac{a}{60}\\times a=\\frac{{{a}^{2}}}{60}$$千米，此时甲、乙相差最远为$$a-\\frac{a^{2}}{60}=\\frac{1}{60}\\times(60-a)\\times a$$，和一定，差小积大，$$60-a=a$$，$$a=30$$．甲、乙最远相差$$30-\\frac{900}{60}=15$$（千米）． " ]

[ "2011年北京学而思综合能力诊断六年级竞赛第3题", "2011年全国学而思杯四年级竞赛第5题", "2011年全国学而思杯五年级竞赛第5题" ]

[ [ { "aoVal": "A", "content": "$$30\\times 3$$ " } ], [ { "aoVal": "B", "content": "$$30\\times 2$$ " } ], [ { "aoVal": "C", "content": "$$30\\times 30$$ " } ], [ { "aoVal": "D", "content": "$$30\\times 30\\times 30$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "$$7$$级和$$4$$级差了$$3$$级，所以$$7$$级是$$4$$级的$$30\\times 30\\times 30$$倍. " ]

[ "2018年湖北武汉创新杯小学高年级五年级竞赛初赛数学思维能力等级测试第8题4分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{5}{14}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{5}{24}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{7}{24}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2}{7}$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "设雨中步行了$$7$$份，还剩下$$5$$份，全程$$\\left( 7+5 \\right)\\times 2=24$$份，小明在雨中步行的路程是全程的$$\\frac{7}{24}$$． " ]

[ "2017年北京学而思杯小学中年级三年级竞赛年度教学质量测评第19题3分" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$19$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$23$$ " } ] ]

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "斐波那契数串，从第$$3$$个数开始，每个数等于前两个数的和． " ]

[ "2019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（八）第6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（五）第5~0题3分", "2014年全国迎春杯四年级竞赛初赛第5题" ]

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

[ "拓展思维->思想->转化与化归的思想" ]

[ "每只猴子多分了$$3$$个，分了$$5\\times 9+(9-3)+57=108$$ （个），那么共$$108\\div 3=36$$（只）猴子．共$$36\\times6+57=273$$（个）桃子． " ]

[ "2007年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$240$$ $$2$$ " } ], [ { "aoVal": "B", "content": "$$120$$ $$2$$ " } ], [ { "aoVal": "C", "content": "$$240$$ $$1$$ " } ], [ { "aoVal": "D", "content": "$$120$$ $$1$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->牛吃草问题->牛吃草基本型->求头数" ]

[ "$$5$$头牛$$30$$天共吃$$30\\times5=150$$（份）草，$$4$$头牛$$40$$天共吃$$4\\times40=160$$（份）草，总草量相差部分是因为生长天数不同，故草的生长速度为$$(160-150)\\div(40-30)=1$$（份/天），草场原有$$150-1\\times30=120$$（份）草。 " ]

[ "2017年第15届湖北武汉创新杯五年级竞赛初赛第6题" ]

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

[ "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" ]

[ "$$(72-54)\\div (5-3)=18\\div 2=9$$（年）． " ]

[ "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": "会 " } ], [ { "aoVal": "B", "content": "不会 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "根据规律，这串数是$$2，0，1，6，9，6，2，3，0，1，6，0，7，4，7，8，6，5，6，5，... $$呈现的规律是：偶偶奇偶奇，而$$2，0，1，7$$是：偶偶奇奇 按照上述的规律两个奇数不可能相邻，所以不可能出现$$2，0，1，7$$这$$4$$个数． " ]

[ "2017年第15届全国希望杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$11.24$$ " } ], [ { "aoVal": "B", "content": "$$11.28$$ " } ], [ { "aoVal": "C", "content": "$$11.29$$ " } ], [ { "aoVal": "D", "content": "$$11.24$$ 或$$11.29$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "因为$$11.195\\times 17=190.315$$，$$11.295\\times 17=192.015$$，所以$$17$$个自然数的和是$$191$$或$$192$$，$$191\\div 17\\approx 11.24$$，$$192\\div 17\\approx 11.29$$，故正确答案是$$11.24$$ 或$$11.29$$ ． " ]

[ "2014年IMAS小学中年级竞赛第一轮检测试题第19题4分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ], [ { "aoVal": "E", "content": "$$31$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->加乘原理->加乘原理综合" ]

[ "由题意可知，绿色盒子有$$4\\times 6=24$$个，蓝色盒子有$$6$$个，红色盒子有$$1$$个，所以小明总共 有$$24+6+1=31$$个盒子，故选$$\\text{E}$$． " ]

[ "2019年第24届YMO五年级竞赛决赛第10题3分", "2020年第24届YMO五年级竞赛决赛第10题3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->计数模块->加乘原理->组数问题->有特殊要求的组数问题" ]

[ "①$$6$$个数字互不相同， 有$$6\\times 5\\times 4\\times 3\\times 2\\times 1=720$$个不同六位数码舍去； ②只有$$2$$个数字相同， 有$$6\\times 5\\times 4\\times 3=360$$个不同六位数码，舍去； ③只有$$3$$个数字相同， 有$$6\\times 5\\times 4=120$$个不同六位数码，舍去； ④有$$2$$对数字分别相同， 有$$6\\times 5\\times \\left( 3+2+1 \\right)=180$$个不同六位数码，符合， 且$$6$$个数字里没有$$3$$，则最小为$$0$$，$$0$$，$$1$$，$$1$$，$$2$$，$$4$$， 故和最小为$$0+0+1+1+2+4=8$$； ⑤其他情况均小于$$180$$个六位数码； 综上，故选$$\\text{B}$$． " ]

[ "2014年全国华杯赛小学高年级竞赛初赛B卷第4题" ]

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

[ "拓展思维->思想->对应思想" ]

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

[ "其它改编自2015年全国希望杯六年级竞赛初赛第17题" ]

[ [ { "aoVal": "A", "content": "$$8$$；$$30$$ " } ], [ { "aoVal": "B", "content": "$$8$$；$$36$$ " } ], [ { "aoVal": "C", "content": "$$9$$；$$30$$ " } ], [ { "aoVal": "D", "content": "$$9$$；$$36$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "甲效率$$\\frac{1}{6}$$，乙效率$$\\frac{1}{8}$$，丙效率$$\\frac{1}{10}$$， 乙，丙都工作$$12-8=4$$（小时）， 甲工作：$$\\left( 1-\\frac{1}{8}\\times 4-\\frac{1}{10}\\times 4 \\right)\\div \\frac{1}{6}=\\frac{3}{5}$$小时$$=36$$（分）， 甲离开时间为：$$8:36$$． " ]

[ "四年级其它小学奥数优秀生培养教程7级", "2005年第6届中环杯四年级竞赛复赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "按照$$*$$运算法则为前数$$\\times 2-$$后数$$\\times 3$$，按照$$\\triangle $$运算法则为$$2\\times $$两数之积． 在解方程时，由于括号内含有未知数，所以要由外向内化简方程，一步一步展开． 也可以将$$2*x$$看作一个整体，先求结果，再求$$x$$． 有括号时先算括号里的，$$2*x=2\\times 2+3x=4+3x$$，$$\\left( 2*x \\right)\\triangle 2=\\left( 4+3x \\right)\\triangle 2=2\\times \\left( 4+3x \\right)\\times 2=16+12x$$，得一元一次方程$$16+12x=64$$，解得$$x=4$$． 设$$2*x=A$$，$$A\\Delta 2=2\\times 2\\times A=64,A=16.$$所以，$$2*x=2\\times 2+3x=16$$，解得$$x=4$$． " ]

[ "2017年全国华杯赛竞赛初赛模拟试卷2第2题" ]

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

[ "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" ]

[ "这列数为$$2016，$$21$$，$$12$$，$$6$$，$$9$$，$$15$$，$$15$$，$$12$$，$$9$$，$$12$$，$$12$$，$$6$$，$$9$$，$$15$$，$$15$$，$$12$$，$$9$$，\\cdots，$$从$$a_{4}$$开始每$$8$$个数一循环．$$2016=3+8\\times251+5$$，所以，$$a_{2016}=a_{8}=12$$． " ]

[ "2014年全国迎春杯六年级竞赛复赛第3题" ]

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

[ "知识标签->课内题型->综合与实践->应用题->还原问题->抄错数得出结果，求正确的" ]

[ "除数$$=(472-427)\\div 5=9$$，$$472\\div 9=52\\cdots \\cdots 4$$，所以余数是$$4$$． " ]