Datasets:

math-eval

/

TAL-SCQ5K

like 49

[ "2016年全国美国数学大联盟杯小学中年级三年级竞赛初赛" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "求积末尾$$0$$的个数，需要看$$2$$和$$5$$的因数个数．$$2$$的个数多余$$5$$的个数，因此找出前$$10$$个数里有多少个$$5$$，$$5$$本身一个，$$10=2\\times 5$$，一共两个$$5$$，因此，末尾两个$$0$$． " ]

[ "2017年第15届全国希望杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac{20122012}{20132013}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{20132013}{20142014}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{20142014}{20152015}$$ " } ], [ { "aoVal": "D", "content": "三个一样大 " } ] ]

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

[ "因为$$\\frac{20122012}{20132013}=\\frac{2012\\times 1001}{2013\\times 1001}=\\frac{2012}{2013}$$， $$\\frac{20132013}{20142014}=\\frac{2013\\times 1001}{2014\\times 1001}=\\frac{2013}{2014}$$，$$\\frac{20142014}{20152015}=\\frac{2014\\times 1001}{2015\\times 1001}=\\frac{2014}{2015}$$，$$1-\\frac{2012}{2013}=\\frac{1}{2013}$$，$$1-\\frac{2013}{2014}=\\frac{1}{2014}$$， $$1-\\frac{2014}{2015}=\\frac{1}{2015}$$． 因为$$\\frac{1}{2015}\\textless{}\\frac{1}{2014}\\textless{}\\frac{1}{2013}$$，所以$$\\frac{2014}{2015}\\textgreater\\frac{2013}{2014}\\textgreater\\frac{2012}{2013}$$．因此，三个分数中，最大的是$$\\frac{20142014}{20152015}$$． " ]

[ "2015年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{25}{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{50}{3}$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->认识单位1" ]

[ "解：设原来卖出的票数为$$1$$，则每张票售价降了： $$50-50\\times \\frac{1}{4}\\div \\frac{1}{3}$$ $$=50-\\frac{75}{2}$$ $$=\\frac{25}{2}$$（元） 答：每张票售价降了$$\\frac{25}{2}$$元。 故选：$$B$$。 " ]

[ "2016年创新杯小学高年级六年级竞赛训练题（一）第2题" ]

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

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

[ "方法一: 从甲地到乙地共有$$450$$千米，共用$$9.4$$小时，上坡车速为每小时$$45$$千米，下坡车速为每小时 50千米，可得上坡所用的时间为$$ (50 \\times 9.4-450)\\div(50-45)=4(个, $$),那么上坡路为 $$ 45 \\times 4=180(千米 $$),下坡路为$$ 450-180=270（千米） $$原路返回时，原来的上坡路变成下 坡路，原来的下坡路变成上坡路，所用时间为$$ 270 \\div 45+180+50=6+3.6=9.6（小时） $$ 方法二: 本题也可以从整体上进行考虑，由于原路返回时，原来的上坡路变成下坡路，原来的下坡路变 成了上坡路，所以往返一次相当于共走了$$450$$千米的上坡路和$$450$$千米的下坡路，那么所用的 总时间为$$ 450 \\div 45+450 \\div 50=19(小+1+) $$,其中从甲地到乙地用了$$9.4$$小时，所以原路返回 要用$$ 19-9.4=9.6(.1,1= $$ 方法三：设原来上坡用了$$x$$小时，$$45x+50\\left( 9.4-x \\right)=450$$，解得$$x=4$$，那么原来上坡是$$45\\times 4=180$$千米，下坡就是$$450-180=270$$千米，返回时下坡变为上坡，上坡变为下坡，所以返回时时间是$$\\frac{270}{45}+\\frac{180}{50}=9.6$$（小时）． " ]

[ "2017年全国小学生数学学习能力测评六年级竞赛复赛第10题3分" ]

[ [ { "aoVal": "A", "content": "$$1332$$ " } ], [ { "aoVal": "B", "content": "$$1032$$ " } ], [ { "aoVal": "C", "content": "$$1000$$ " } ], [ { "aoVal": "D", "content": "$$998$$ " } ] ]

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

[ "满足$$a\\times b\\times c=a+b+c$$的只有$$1$$，$$2$$，$$3$$，即$$1\\times 2\\times 3=1+2+3=6$$， 所以这些三位数是$$123$$，$$132$$，$$213$$，$$231$$，$$312$$，$$321$$； 和为$$123+132+213+231+312+321=1332$$． 故$$\\text{BCD}$$错误，$$\\text{A}$$正确． 故选$$\\text{A}$$． " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（五）第1题" ]

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

[ "拓展思维->思想->数形结合思想", "课内体系->思想->数形结合思想" ]

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

[ "2021年第8届鹏程杯四年级竞赛初赛第15题4分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$17$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "甲乙先过去，甲回来，用时$$2+1=3$$分钟，丙丁过去乙回来，用时$$10+2=12$$分钟，甲乙过去用时$$2$$分钟，总共用时$$3+12+2=17$$分钟， 故选$$\\text{C}$$． " ]

[ "2019年广东深圳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": "圆最大 " } ], [ { "aoVal": "B", "content": "正方形最大 " } ], [ { "aoVal": "C", "content": "长方形最大 " } ], [ { "aoVal": "D", "content": "一样大 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "假设正方形、长方形、圆的周长都是$$16$$厘米，则： （$$1$$）正方形的边长：$$16\\div 4=4$$（厘米）， 面积：$$4\\times 4=16$$（平方厘米）； （$$2$$）假设长方形的长为$$6$$厘米，宽为$$2$$厘米， 则面积：$$2\\times 6=12$$（平方厘米）； （$$3$$）圆的半径：$$16\\div 3.14\\div 2=\\frac{800}{314}$$（厘米）， 面积：$$3.14\\times \\left( \\frac{800}{314} \\right)\\times \\left( \\frac{800}{314} \\right)$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=3.14\\times \\frac{800}{314}\\times \\frac{800}{314}$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=\\frac{6400}{314}$$ $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde=20\\frac{60}{157}$$（平方厘米）； 所以，$$12$$平方厘米$$\\textless{}16$$平方厘米$$\\textless{}20\\frac{60}{157}$$平方厘米， 所以$$\\text{A}$$选项是正确的． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第7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" ]

[ "A~ 如果循环节是$$123456789$$，循环周期为$$9$$，除了已经有的$$10$$位小数以外，后面还有$$100-10=90$$（位）小数， $90\\div9=10\\cdots0$，所以最后一位是$$9$$，排除 A； B~ 如果循环节是$$23456789$$，循环周期为$$8$$，除了已经有的$$10$$位小数以外，后面还有$$100-10=90$$（位）小数， $90\\div8=11\\cdots2$，所以最后一位是$$3$$，排除 B； C 如果循环节是$$3456789$$，循环周期为$$7$$，除了已经有的$$10$$位小数以外，后面还有$$100-10=90$$（位）小数， $90\\div7=12\\cdots6$，所以最后一位是$$8$$，选择 C； D~ 如果循环节是$$456789$$，循环周期为$$6$$，除了已经有的$$10$$位小数以外，后面还有$$100-10=90$$（位）小数， $90\\div6=15\\cdots0$，所以最后一位是$$9$$，排除 D； 所以，选择 C． " ]

[ "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17题3分" ]

[ [ { "aoVal": "A", "content": "$$40$$ " } ], [ { "aoVal": "B", "content": "$$60$$ " } ], [ { "aoVal": "C", "content": "$$80$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "因为找回的是$$50$$元，则有可能是$$100-50=50$$元，$$100+10-50=60$$元，$$100+100-50=150$$元，$$100+100+10-50=160$$元，则选项中只有$$B$$符合． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛一年级竞赛初赛第8题5分" ]

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

[ "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->原型题" ]

[ "枚举法： 当有$$1$$只兔，$$3$$只鸡时，共$$1\\times4+3\\times2=10$$（条）腿． 当有$$2$$只兔，$$2$$只鸡时，共$$2\\times4+2\\times2=12$$（条）腿． 所以有$$2$$只兔． " ]

[ "2014年全国迎春杯六年级竞赛初赛第2题" ]

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

[ "知识标签->课内知识点->测量->面积->圆的面积->圆的面积公式" ]

[ "$$C$$ " ]

[ "2015年全国中环杯二年级竞赛决赛第5题" ]

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

[ "拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找一致（同伙）" ]

[ "根据条件，牧师不可能说假话，所以牧师一定会说自己是牧师，此题只有一个说自己是牧师．所以甲就是牧师． " ]

[ "2017年河南郑州K6联赛竞赛模拟第七套第3题4分" ]

[ [ { "aoVal": "A", "content": "梨的个数$$+50=$$苹果的个数 " } ], [ { "aoVal": "B", "content": "梨的个数$$=50+$$苹果的个数 " } ], [ { "aoVal": "C", "content": "梨的个数$$+$$苹果的个数$$=50$$ " } ], [ { "aoVal": "D", "content": "梨的个数$$=50$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "梨多，所以梨的个数$$=50+$$苹果的个数． " ]

[ "2018年迎春杯五年级竞赛初赛C卷第2题8分" ]

[ [ { "aoVal": "A", "content": "$$9874$$ " } ], [ { "aoVal": "B", "content": "$$9797$$ " } ], [ { "aoVal": "C", "content": "$$9753$$ " } ], [ { "aoVal": "D", "content": "$$9371$$ " } ], [ { "aoVal": "E", "content": "$$9731$$ " } ] ]

[ "拓展思维->能力->数据处理", "海外竞赛体系->知识点->数论模块->质数与合数" ]

[ "要使这个四位数最大，高位要尽可能大，$$\\overline{AB}$$为$$100$$以内最大的质数，即$$97$$，$$\\overline{BC}$$为质数，只能是$$79$$，$$73$$，$$71$$，因为各位数字互不相同，且尽可能大，$$\\overline{BC}=73$$，$$\\overline{CD}$$为质数，只能是$$37$$，$$31$$，因为数字互不相同，$$\\overline{CD}=31$$，所以$$\\overline{ABCD}$$最大是$$9731$$． 故答案为：$$9731$$． " ]

[ "2020年新希望杯五年级竞赛第14题", "2020年希望杯五年级竞赛模拟第14题" ]

[ [ { "aoVal": "A", "content": "$$3200$$ " } ], [ { "aoVal": "B", "content": "$$3600$$ " } ], [ { "aoVal": "C", "content": "$$32000$$ " } ], [ { "aoVal": "D", "content": "$$36000$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$0.004186\\times 8812345.321\\approx 0.004\\times 9000000=36000$$． 故选$$\\text{D}$$． " ]

[ "2010年第10届全国小机灵杯三年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "$$481$$ " } ], [ { "aoVal": "B", "content": "$$471$$ " } ], [ { "aoVal": "C", "content": "$$454$$ " } ], [ { "aoVal": "D", "content": "$$544$$ " } ], [ { "aoVal": "E", "content": "$$445$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->公式类->直接求平均数", "Overseas Competition->知识点->应用题模块->平均数问题" ]

[ "组成的三位数分别为$$445$$，$$454$$，$$544$$，平均数为$$\\left( 445 + 454 + 544 \\right) \\div 3 = 481$$． " ]

[ "2009年第7届创新杯四年级竞赛初赛第7题4分" ]

[ [ { "aoVal": "A", "content": "$$87$$只与$$86$$只 " } ], [ { "aoVal": "B", "content": "$$87$$只与$$82$$只 " } ], [ { "aoVal": "C", "content": "$$80$$只与$$87$$只 " } ], [ { "aoVal": "D", "content": "$$94$$只与$$80$$只 " } ] ]

[ "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->加法拆数（应用）" ]

[ "因为$$\\left( 78+94+86+87+82+80 \\right)\\div \\left( 1+2 \\right)=169$$，又$$87+82=169$$，所以$$87$$只与$$82$$只这两筐是徒弟加工的． " ]

[ "2015年湖北武汉世奥赛五年级竞赛模拟训练题（二）第4题" ]

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

[ "拓展思维->能力->数据处理" ]

[ "根据``以多初少''可知，小乐之前考试的次数为$$\\left( 100-91 \\right)\\div \\left( 91-88 \\right)=3$$（次），这是第$$3+1=4$$（次）． " ]

[ "2015年六年级其它", "2010年第8届希望杯六年级竞赛复赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$0$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "根据题意，容易解出$$1-\\dfrac{1}{6+\\dfrac{1}{6+\\dfrac{1}{6}}}=\\dfrac{191}{228}$$， 所以$$A+\\dfrac{1}{B+\\dfrac{1}{C+\\dfrac{1}{C}}}=1+\\dfrac{37}{191}$$， 而$$B+\\dfrac{1}{C+\\dfrac{1}{C}}$$大于$$1$$， 所以$$A=1$$，同理可知，$$B=5$$，$$C=6$$，则$$\\left( A+B \\right)\\div C=1$$． " ]

[ "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11题2分" ]

[ [ { "aoVal": "A", "content": "$$35$$ " } ], [ { "aoVal": "B", "content": "$$30$$ " } ], [ { "aoVal": "C", "content": "$$25$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "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" ]

[ "两班共有$$70\\div 2=35$$（个）孩子， 假设所有人都分得$$5$$个橘子，需要$$35\\times 5=175$$（个）橘子． 则小班有$$\\left( 175-135 \\right)\\div \\left( 5-3 \\right)=20$$（个）孩子． 故选：$$\\text{D}$$． " ]

[ "2016年全国AMC六年级竞赛8第15题" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$64$$ " } ], [ { "aoVal": "E", "content": "$$128$$ " } ] ]

[ "Overseas Competition->知识点->数论模块->分解质因数", "拓展思维->拓展思维->数论模块->分解质因数->分解质因数的应用->已知乘积求因数" ]

[ "$${{13}^{4}}-{{11}^{4}}$$ $$=({{13}^{2}}-{{11}^{2}})\\times ({{13}^{2}}+{{11}^{2}})$$ $$=(13-11)\\times (13\\times 11)\\times (169+121)$$ $$=2\\times 24\\times 290$$ $$=2\\times {{2}^{3}}\\times 3\\times 2\\times 145$$ $$={{2}^{5}}\\times 3\\times 145$$ $${{2}^{5}}=32$$． " ]

[ "2017年安徽合肥庐江县小升初第19题1分", "2017年河南郑州豫才杯竞赛第14题", "2017年河南郑州小升初豫才杯第14题3分" ]

[ [ { "aoVal": "A", "content": "角的两边是两条直线 " } ], [ { "aoVal": "B", "content": "知道了物体的方向，就能确定物体的位置 " } ], [ { "aoVal": "C", "content": "平行四边形有一条对称轴 " } ], [ { "aoVal": "D", "content": "长方体、正方体、圆柱体的体积都能用``底面积✖️ 高''来计算 " } ] ]

[ "拓展思维->拓展思维->组合模块->方向与坐标->方向" ]

[ "$$A$$不对，角的两边是两条射线；$$B$$不对，缺少观察点无法确定物体的位置；$$C$$不对，平行四边形不是轴对称图形，所以没有对称轴；故选$$D$$． " ]

[ "2017年全国希望杯六年级竞赛" ]

[ [ { "aoVal": "A", "content": "甲同学花的钱最多，丙同学花的钱最少 " } ], [ { "aoVal": "B", "content": "乙同学花的钱最多，甲同学花的钱最少 " } ], [ { "aoVal": "C", "content": "乙同学花的钱最多，丙同学花的钱最少 " } ], [ { "aoVal": "D", "content": "丙同学花的钱最多，甲同学花的钱最少 " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "设商品单价为$$1$$，则甲同学花的钱为$$1+\\frac{1}{2}+\\frac{1}{3}+...+\\frac{1}{9}+\\frac{1}{10}=\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5} \\right)+\\left( \\frac{1}{6}+\\frac{1}{7}+\\frac{1}{8}+\\frac{1}{9}+\\frac{1}{10} \\right)$$；乙同学花的钱为$$\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5} \\right)\\times 2=\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5} \\right)+\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5} \\right)$$；丙同学花的钱为$$\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4} \\right)+\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6} \\right)=\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5} \\right)+\\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{6} \\right)$$． 因 为$$1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}\\textgreater1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{6}\\textgreater\\frac{1}{6}+\\frac{1}{7}+\\frac{1}{8}+\\frac{1}{9}+\\frac{1}{10}$$，所以乙同学花的钱最多，甲同学花的钱最少． " ]

[ "2016年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$1152$$ " } ], [ { "aoVal": "B", "content": "$$864$$ " } ], [ { "aoVal": "C", "content": "$$576$$ " } ], [ { "aoVal": "D", "content": "$$288$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->加乘原理->乘法原理" ]

[ "首先求出$$1,2,3,4,5,6,7$$的和是$$28$$，判断出$$8$$的两边各数之和都是$$14$$，然后分为$$4$$种情况： （$$1$$）$$8$$的一边是$$1,6,7$$，另一边是$$2,3,4,5$$时； （$$2$$）$$8$$的一边是$$2,5,7$$，另一边是$$1,3,4,6$$时； （$$3$$）$$8$$的一边是$$3,4,7$$，另一边是$$1,2,5,6$$时； （$$4$$）$$8$$的一边是$$1,2,4,7$$，另一边是$$3,5,6$$时。 求出每种情况下各有多少种不同的排法，即可求出共有多少种不同的排法。 解：$$1+2+3+4+5+6+7=28$$ $$8$$的两边各数之和是：$$28\\div 2=14$$ （$$1$$）$$8$$的一边是$$1,6,7$$，另一边是$$2,3,4,5$$时， 不同的排法一共有： $$\\left( 3\\times 2\\times 1 \\right)\\times \\left( 4\\times 3\\times 2\\times 1 \\right)\\times 2$$ $$=6\\times 24\\times 2$$ $$=288$$（种） （$$2$$）$$8$$的一边是$$2,5,7$$，另一边是$$1,3,4,6$$时， 不同的排法一共有$$288$$种。 （$$3$$）$$8$$的一边是$$3,4,7,$$另一边是$$1,2,5,6$$时， 不同的排法一共有$$288$$种。 （$$4$$）$$8$$的一边是$$1,2,4,7,$$另一边是$$3,5,6$$时， 不同的排法一共有$$288$$种。 因为$$288\\times 4=1152$$（种）， 所以共有$$1152$$种不同的排法。 答：共有$$1152$$种不同的排法。 故选：A " ]

[ "2021年新希望杯五年级竞赛模拟（考前培训100题）第61题" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$19$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$471$$ " } ] ]

[ "拓展思维->能力->公式记忆->符号化数学原理" ]

[ "$$M$$取最小时乘积应为$$4$$位数，根据$$13$$的整除特征，三位截断，积的末$$3$$位$$123$$与最高位数字的差必为$$13$$的倍数，而$$123 \\div13$$$\\cdots\\cdots$$$6$$， 故取最高位数字为$$6$$，则$$M=6123\\div13=471$$． " ]

[ "2008年第6届创新杯五年级竞赛初赛B卷第9题5分", "2008年五年级竞赛创新杯" ]

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

[ "拓展思维->拓展思维->行程模块->发车问题" ]

[ "设人，电车的速度分别为$${{v}_{1}}，{{v}_{2}}$$，两辆相邻电车之间的距离为$$s$$，则$$\\frac{s}{{{v}_{2}}-{{v}_{1}}}=12$$，$$\\frac{s}{{{v}_{1}}+{{v}_{2}}}=4$$，从而$$\\frac{{{v}_{2}}-{{v}_{1}}}{s}=\\frac{1}{12}$$，$$\\frac{{{v}_{2}}+{{v}_{1}}}{s}=\\frac{1}{4}$$，两等式相加得$$\\frac{2{{v}_{2}}}{s}=\\frac{1}{12}+\\frac{1}{4}=\\frac{1}{3}$$，所以$$\\frac{s}{{v}_{2}}=6$$，即电车每隔$$6$$分钟发出一辆． " ]

[ "2018年IMAS小学高年级竞赛（第一轮）第6题3分", "2018年IMAS小学中年级竞赛（第一轮）第17题4分" ]

[ [ { "aoVal": "A", "content": "$$5$$小时$$20$$分钟 " } ], [ { "aoVal": "B", "content": "$$10$$小时$$20$$分钟 " } ], [ { "aoVal": "C", "content": "$$15$$小时$$20$$分钟 " } ], [ { "aoVal": "D", "content": "$$16$$小时$$20$$分钟 " } ], [ { "aoVal": "E", "content": "$$17$$小时$$20$$分钟 " } ] ]

[ "拓展思维->能力->实践应用->度量单位认知" ]

[ "可知列车在出发当日共用了$$24$$小时减去$$8$$小时$$30$$分钟，即$$15$$小时$$30$$分钟；在第二日用了$$1$$小时$$50$$分钟，故全程所用的时间是$$16$$小时又$$80$$分钟，即$$17$$小时$$20$$分钟． 故选$$\\text{E}$$． " ]

[ "2013年第25届广东广州五羊杯六年级竞赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$280$$ " } ], [ { "aoVal": "B", "content": "$$281$$ " } ], [ { "aoVal": "C", "content": "$$282$$ " } ], [ { "aoVal": "D", "content": "$$283$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "这本书有$$741$$个数字，故页码在$$100\\sim 999$$之间， $$\\frac{741-9-90 \\times 2}{3}=184$$，$$9+90+184=283$$（页）． 故选$$\\text{D}$$． " ]

[ "2005年六年级竞赛创新杯", "2005年第3届创新杯六年级竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "12 " } ], [ { "aoVal": "B", "content": "6 " } ], [ { "aoVal": "C", "content": "4 " } ], [ { "aoVal": "D", "content": "3 " } ] ]

[ "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" ]

[ "设这个长方体的长宽高分别为$$a，b，c$$，依题意有$$abc=12，ab=3，ac=12$$，那么$$b=1，a=3，c=4$$，所以另一对侧面的面积为$$bc=4$$，选$$C$$. " ]

[ "2020年北京迎春杯六年级竞赛模拟初赛第2题3分", "2017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（六）第2题3分" ]

[ [ { "aoVal": "A", "content": "$\\dfrac{1}{5}$ " } ], [ { "aoVal": "B", "content": "$\\dfrac{1}{6}$ " } ], [ { "aoVal": "C", "content": "$\\dfrac{1}{7}$ " } ], [ { "aoVal": "D", "content": "$\\dfrac{2}{7}$ " } ] ]

[ "课内体系->能力->逻辑分析", "拓展思维->拓展思维->计算模块->分数->分数运算->分小四则混合运算" ]

[ "$\\dfrac{1}{7}$ " ]

[ "2013年全国华杯赛小学高年级竞赛初赛C卷第4题" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$31$$ " } ] ]

[ "拓展思维->七大能力->逻辑分析" ]

[ "根据``$$A$$的二分之一是完全平方数''可以知道，$$\\alpha -1$$、$$\\beta $$、$$\\gamma $$都是$$2$$的倍数． 根据``$$A$$的三分之一是完全立方数''可以知道，$$\\alpha $$、$$\\beta -1$$、$$\\gamma $$都是$$3$$的倍数． 根据``$$A$$的五分之一是某个自然数的五次方''可以知道，$$\\alpha $$、$$\\beta $$、$$\\gamma -1$$都是$$5$$的倍数． 同时满足三个条件的$$\\alpha $$的最小值恰好是$$\\left[ 3,5 \\right]=15$$；$$\\beta $$的最小值恰好是$$\\left[2,5 \\right]=10$$；$$\\gamma $$的最小值恰好是$$\\left[ 2,3 \\right]=6$$． 所以，$$\\alpha +\\beta +\\gamma $$的最小值是$$15+10+6=31$$． " ]

[ "2013年河南郑州中原网杯六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$41$$ " } ], [ { "aoVal": "B", "content": "$$43$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$47$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "第二次平均分比第一次平均分多$$84.7-84.1=0.6$$，其中小红第二次分数比第一次多$$96-69=27$$，显然，平均分的差$$\\times $$班级人数$$=$$小红少加的分数．故有$$27\\div 0.6=45$$人． " ]

[ "2009年希望杯四年级竞赛复赛", "2009年希望杯三年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "$$115$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$125$$ " } ], [ { "aoVal": "D", "content": "$$130$$ " } ] ]

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

[ "被除数$$\\div 13=9\\cdots\\cdots 8$$，则被除数$$=13\\times 9+8=125$$。 " ]

[ "2012年世界少年奥林匹克数学竞赛三年级竞赛初赛A卷第6题5分" ]

[ [ { "aoVal": "A", "content": "爸爸 " } ], [ { "aoVal": "B", "content": "妈妈 " } ], [ { "aoVal": "C", "content": "哥哥 " } ], [ { "aoVal": "D", "content": "姐姐 " } ], [ { "aoVal": "E", "content": "我 " } ] ]

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

[ "由题干可推拍照顺序可能为爸爸~ 我~ 妈妈~ 姐姐~ 哥哥； 或者 哥哥~ 姐姐~ 妈妈~ 我~ 爸爸，因此妈妈在中间． " ]

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

[ [ { "aoVal": "A", "content": "$$145$$ " } ], [ { "aoVal": "B", "content": "$$140$$ " } ], [ { "aoVal": "C", "content": "$$125$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

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

[ "因为甲给乙$$40$$元，还比乙多$$10$$元，所以甲比乙多$$40\\times 2+10=90$$元， 所以甲：$$(200+90)\\div 2=145$$元． " ]

[ "2014年全国迎春杯三年级竞赛初赛第5题", "2017年全国小升初八中入学备考课程" ]

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

[ "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" ]

[ "鸡兔同笼，假设$$12$$枚都是$$5$$分硬币，则1角硬币共$$(90-12\\times 5)\\div (10-5)=6$$枚，每种硬币各$$6$$枚． ", "<p>$$5$$分的数量：</p>\n<p>$$(12\\times 1-9)\\div \\left( 1-0.5 \\right)$$</p>\n<p>$$=3\\div 0.5$$</p>\n<p>$$=6$$（枚）；</p>\n<p>$$1$$角的硬币数量为：$$12-6=6$$（枚）．</p>\n<p>答：每种硬币各$$6$$枚．</p>\n<p>故选：$$\\text{C}$$．</p>" ]

[ "2015年第14届春蕾杯一年级竞赛初赛第5题1分" ]

[ [ { "aoVal": "A", "content": "小王 " } ], [ { "aoVal": "B", "content": "小胖 " } ], [ { "aoVal": "C", "content": "小朱 " } ], [ { "aoVal": "D", "content": "园园 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "根据题意分析可知，小胖比小王高，所以小胖的身高高于小王，小胖说比园园矮，所以园园的身高高于小胖，小朱没有小王高，所以小王的身高高于小朱，由此可知，四个人的身高从高到低位：园园$$\\textgreater$$小胖$$\\textgreater$$小王$$\\textgreater$$小朱． 故选$$\\text{D}$$． " ]

[ "2012年IMAS小学中年级竞赛第一轮检测试题第1题3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->思想->整体思想" ]

[ "木棍被锯成了$$20\\div 4=5$$（根），因此只需要锯$$5-1=4$$（次）． " ]

[ "2018年IMAS小学高年级竞赛（第二轮）第1题4分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left( 2019-2018 \\right)\\times \\left( 2019-2017 \\right)\\times \\cdots \\times \\left( 2019-2012 \\right)\\times \\left( 2019-2011 \\right)$$ $$=1\\times 2\\times 3\\times 4\\times 5\\times 6\\times 7\\times 8$$ $$=2\\times 3\\times \\left( 2\\times 2 \\right)\\times 5\\times \\left( 2\\times 3 \\right)\\times 7\\times \\left( 2\\times 2\\times 2\\times 2 \\right)$$． 包含不同的质因数有$$2$$、$$3$$、$$5$$、$$7$$，共有$$4$$个． 故选$$\\text{C}$$． " ]

[ "2017年全国美国数学大联盟杯小学中年级三年级竞赛初赛第40题" ]

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

[ "拓展思维->思想->整体思想" ]

[ "$$(40+20)\\div 2=30$$． " ]

[ "2015年北京华杯赛小学高年级竞赛初赛A卷第1题" ]

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

[ "拓展思维->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾", "Overseas Competition->知识点->组合模块->逻辑推理" ]

[ "题目要求有两个人去，可以使用假设法，若甲去，则乙去，乙去则丙也去．三个人去，矛盾，所以甲不去．若丙不去则乙不去，那么只有丁去，矛盾，所以丙去．丙去则丁不去，由两个人去得到结论，乙要去． 所以答案是$$\\text{B}$$，丙和乙去． " ]

[ "2017年全国小升初八中入学备考课程", "2014年全国华杯赛小学高年级竞赛初赛B卷第5题" ]

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

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

[ "甲乙的年龄差是$$22-16=6$$（岁）；当甲$$19$$岁时， $$13$$岁；至少一年前甲$$22$$岁，所以当甲$$19$$岁的时候，此时至少是$$4$$年前的年龄，那么甲今年至少是$$23$$岁；甲$$19$$岁时，丙的年龄是丁的$$3$$倍，假设丁为$$1$$岁，丙为$$3$$岁，此时四人的年龄和至少是$$19+13+1+3=36$$（岁）；且甲今年的年龄至多为$$19+\\left(72-36 \\right)\\div 4=28$$（岁）；所以甲今年的年龄可能是$$23$$，$$24$$，$$25$$，$$26$$，$$27$$，$$28$$；共$$6$$种，所以选$$\\text{B}$$． " ]

[ "2017年IMAS小学中年级竞赛（第一轮）第13题4分" ]

[ [ { "aoVal": "A", "content": "$$128$$ " } ], [ { "aoVal": "B", "content": "$$208$$ " } ], [ { "aoVal": "C", "content": "$$2008$$ " } ], [ { "aoVal": "D", "content": "$$2012$$ " } ], [ { "aoVal": "E", "content": "$$2020$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "由题意与范例可判断出运算结果为先写出两个数的乘积后紧接着写下两个数的和．因$$10\\times2=20$$、$$10+2=12$$，所以$$10*2=2012$$．故选$$\\text{D}$$． 答案：$$\\text{D}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛四年级竞赛初赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$666$$ " } ], [ { "aoVal": "B", "content": "$$22200$$ " } ], [ { "aoVal": "C", "content": "$$6660$$ " } ], [ { "aoVal": "D", "content": "$$66660$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "每一个数字在个位、十位、百位、千位均会出现$$3\\times2\\times1=6$$次． 故总和：$$(1+2+3+4)\\times6666=66660$$． 故选择$$\\text{D}$$． " ]

[ "2012年IMAS小学中年级竞赛第一轮检测试题第2题3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->计算模块->整数->整数加减->整数减法运算->减法横式" ]

[ "13-2=11；11-5=6 " ]

[ "2013年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$31$$ " } ] ]

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

[ "根据``$$A$$的二分之一是完全平方数''可以知道，$$\\left( \\alpha -1 \\right)$$、$$\\beta $$、$$\\gamma $$都是$$2$$的倍数。根据``$$A$$的三分之一是完全立方数''可以知道，$$\\alpha $$、$$\\left( \\beta -1 \\right)$$、$$\\gamma $$都是$$3$$的倍数。根据``$$A$$的五分之一是某个自然数的五次方''可以知道，$$\\alpha $$、$$\\beta $$、$$\\left( \\gamma -1 \\right)$$都是$$5$$的倍数。同时满足三个条件的$$\\alpha $$的最小值恰好是$$\\left[ 3 \\right.$$，$$\\left. 5 \\right]=15$$；$$\\beta $$的最小值恰好是$$\\left[ 2 \\right.$$，$$\\left. 5 \\right]=10$$；$$\\gamma $$的最小值恰好是$$\\left[ 2 \\right.$$，$$\\left. 3 \\right]=6$$．所以，$$\\alpha +\\beta +\\gamma $$的最小值是$$15+10+6=31$$。 " ]

[ "2005年五年级竞赛创新杯", "2005年第3届创新杯五年级竞赛复赛第4题" ]

[ [ { "aoVal": "A", "content": "一 " } ], [ { "aoVal": "B", "content": "二 " } ], [ { "aoVal": "C", "content": "四 " } ], [ { "aoVal": "D", "content": "五 " } ] ]

[ "拓展思维->拓展思维->应用题模块->周期问题" ]

[ "从星期三开始，每7天会播出5集，$$83\\div 5=16(周)\\ldots 3（集）$$那么16周后再播出三集就全部播放完毕，那么16周后从星期三开始，播放这最后3集中的第一集，星期四为第二集，星期五为最后一集，选D. " ]

[ "2016年全国华杯赛小学高年级竞赛在线模拟第6题" ]

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

[ "拓展思维->七大能力->运算求解" ]

[ "任何三个连续排列的数字都构成一个三位数，说明没有数字$$0$$，$$\\left. 11 \\right\\textbar990$$，$$\\left. 13 \\right\\textbar988$$则首三位为$$988$$，选择题这是已经得到答案了！接着往下分析$$88$$，$$13\\left\\textbar{} 884 \\right.$$，$$11\\left\\textbar{} 880 \\right.$$，所以第四位最大为$$4$$，依次类推，得到答案，$$9884737$$． " ]

[ "2017年全国美国数学大联盟杯小学中年级四年级竞赛初赛" ]

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

[ "Overseas Competition->知识点->数论模块->分解质因数", "拓展思维->思想->转化与化归的思想" ]

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

[ "2016年创新杯六年级竞赛训练题（一）第3题" ]

[ [ { "aoVal": "A", "content": "$$4 \\%$$ " } ], [ { "aoVal": "B", "content": "$$6 \\%$$ " } ], [ { "aoVal": "C", "content": "$$10 \\%$$ " } ], [ { "aoVal": "D", "content": "$$30 \\%$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "$$\\left( 1-20 \\% \\right)\\times \\left( 1+30 \\% \\right)-1=0.04=4 \\%$$． " ]

[ "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12题2分" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$34$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->操作与策略->统筹规划->简单时间统筹问题->时间总和问题" ]

[ "要使它们等候时间（等候时间包括接水时间）的总和最少，应该让接水用时少的先接水，即接水顺序是：小鹿、小马、小熊、小牛． 等候时间总和最少是： $$2\\times 4+3\\times 3+5\\times 2+7=8+9+10+7=34$$（分钟）， 故选：$$\\text{C}$$． " ]

[ "其它改编自2015年全国希望杯六年级竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1371}{30}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{322}{15}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{211}{60}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{109}{60}$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$\\left { \\frac{2015}{3} \\right }+\\left { \\frac{315}{4} \\right }+\\left { \\frac{412}{5} \\right }=\\frac{2}{3}+\\frac{3}{4}+\\frac{2}{5}=\\frac{109}{60}$$． " ]

[ "走美杯三年级竞赛", "走美杯四年级竞赛" ]

[ [ { "aoVal": "A", "content": "因为$$92\\times 8\\approx$$ 720（元），$$92\\times$$ 8\\textgreater720，所以$$92\\times$$ 8\\textgreater720，够 " } ], [ { "aoVal": "B", "content": "因为$$92\\times 8\\approx$$ 800（元），$$92\\times$$ 8\\textless800，所以$$92\\times$$ 8\\textless720，够 " } ], [ { "aoVal": "C", "content": "因为$$92\\times 8\\approx$$ 720（元），$$92\\times$$ 8\\textgreater720，所以$$92\\times$$ 8\\textgreater720，不够 " } ], [ { "aoVal": "D", "content": "因为$$92\\times 8\\approx$$ 800（元），$$92\\times$$ 8\\textgreater800，所以$$92\\times$$ 8\\textgreater720，不够 " } ] ]

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

[ "92\\textgreater90，总价大于$$720$$元，所以不够 " ]

[ "2006年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->归一归总问题->归总问题" ]

[ "实际每天加工零件的个数为$$80\\times 5\\div 4=100$$（个）；实际每天比原计划每天多加工的个数是$$100-80=20$$（个），故选$$\\text{C}$$．错选$$\\text{A}$$选项是不能根据条件先算出零件的总数量，进而解决问题；错选$$\\text{B}$$、$$\\text{D}$$选项是计算出错． " ]

[ "2009年第8届全国新希望杯三年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$2012$$ " } ], [ { "aoVal": "B", "content": "$$2013$$ " } ], [ { "aoVal": "C", "content": "$$2014$$ " } ], [ { "aoVal": "D", "content": "$$2022$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "为了简化计算，我们常把一些数拆开来使它与其他数字之和是整十、整百、整千$$ \\cdots ~\\cdots $$这种方法叫作借数凑整法，我们发现如果将$$1258$$和$$159$$分别拆成$$1250 + 8$$和$$150 + 9$$时，便可以凑出整百，$$365$$可以拆成$$360 + 5$$，也可以拆成$$370 - 5$$，但是由于后面要加上$$230$$，所以拆成$$370 - 5$$时会与$$230$$凑成整百，所以总的花费 $$ = \\left( 1250 + 150 \\right) + \\left( 370 + 230 \\right) + \\left( 8 + 9 - 5 \\right) = 1400 + 600 + 12 = 2012$$（元）． " ]

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

[ [ { "aoVal": "A", "content": "$$78$$ " } ], [ { "aoVal": "B", "content": "$$62$$ " } ], [ { "aoVal": "C", "content": "$$84$$ " } ], [ { "aoVal": "D", "content": "$$124$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "根据题意，要求这个五位数比$$20000$$大，则首位必须是$$2$$、$$3$$、$$4$$、$$5$$这四个数字，当首位是$$3$$时，百位数不会是数字$$3$$，共有$$A44=24$$种情况，当首位是$$2$$、$$4$$、$$5$$时，由于百位数不能是数字$$3$$，有$$3\\left( A44A33 \\right)=54$$种情况，有分步计数原理，共有$$54+24=78$$个数字． " ]

[ "2019年第7届湖北长江杯五年级竞赛复赛B卷第10题3分" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "因为$$120$$的因数有：$$1$$、$$2$$、$$3$$、$$4$$、$$5$$、$$6$$、$$8$$、$$10$$、$$12$$、$$15$$、$$20$$、$$24$$、$$30$$、$$40$$、$$60$$、$$120$$，所以一共有$$16$$个因数． 故选$$\\text{D}$$． " ]

[ "2012年IMAS小学中年级竞赛第一轮检测试题第8题3分" ]

[ [ { "aoVal": "A", "content": "$$\\bigcirc $$填入$$+$$，$$\\square $$填入$$\\times $$ " } ], [ { "aoVal": "B", "content": "$$\\bigcirc $$填入$$\\times $$，$$\\square $$填入$$-$$ " } ], [ { "aoVal": "C", "content": "$$\\bigcirc $$填入$$+$$~ ，$$\\square $$填入$$\\div $$ " } ], [ { "aoVal": "D", "content": "$$\\bigcirc $$填入$$\\times $$，$$\\square $$填入$$\\div $$ " } ] ]

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "等式左边等于$$13$$，当算式右边依次填入$$\\times $$、$$-$$时，2\\times 8-3=13，等式成立．经验算，其余选项均不正确. " ]

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

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

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

[ "利润问题．假设成本为$$100$$元，则利润为$$20$$元，现在成本是$$100\\times (1+25 \\%)=125$$元，$$20\\div 125\\times 100 \\%=16 \\%$$． 以进价为单位``1''，利润为20\\%，第二次进价为1+25\\%=125\\%，则第二次利润率$$20 \\%\\div 125 \\%\\times 100 \\%=16 \\%$$． " ]

[ "2017年湖北武汉中环杯四年级竞赛初赛第14题" ]

[ [ { "aoVal": "A", "content": "$$37$$ " } ], [ { "aoVal": "B", "content": "$$38$$ " } ], [ { "aoVal": "C", "content": "$$39$$ " } ], [ { "aoVal": "D", "content": "$$41$$ " } ] ]

[ "知识标签->拓展思维->组合模块->逻辑推理->假设型逻辑推理->真假话->找矛盾" ]

[ "逻辑推理．由第一句话可知，小明超过$$40$$岁； 由第二句话可知，小红不是$$38$$岁，且两人年龄相差不超过$$4$$岁，即小红的年龄大于$$36$$岁； 由第三句话可知，小红的年龄小于$$39$$岁． 则小红的年龄为$$37$$岁． " ]

[ "2019年第24届YMO一年级竞赛决赛第5题3分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "路程一样，则用时越少，跑的越快．根据时间排序，从少到多为$$10$$秒，$$11$$秒，$$12$$秒，$$14$$秒，故第二名为$$11$$秒的，即小$$C$$． 故选$$\\text{C}$$． " ]

[ "2015年上海走美杯三年级竞赛初赛", "2015年走美杯三年级竞赛初赛" ]

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

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

[ "$$24\\div \\left( 4-1 \\right)+2=24\\div 3+2=8+2=10$$ （岁） " ]

[ "2016年IMAS小学高年级竞赛第一轮检测试题第15题4分" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->余数问题->余数问题带余除法->除法中四量关系" ]

[ "被除数$$=$$商$$\\times $$除数$$+$$余数，用$$304-24=280$$，$$280$$一定是该两位数的倍数， 将$$280$$分解因数：$$280=2\\times 2\\times 2\\times 5\\times 7$$， 则符合条件的两位数有$$70$$、$$56$$、$$40$$、$$35$$、$$28$$，共$$5$$种． " ]

[ "2014年迎春杯三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

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

[ "解：$$3\\times 4\\div 2$$ $$=12\\div 2$$ $$=6$$（千克） 答：每盒苹果重$$6$$千克。 故选:B。 " ]

[ "2011年北京五年级竞赛" ]

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

[ "知识标签->数学思想->对应思想" ]

[ "设乙瓶盐水的浓度是$$x \\% $$，甲瓶盐水的浓度是$$3x \\% $$，有$$100 \\times 3x \\%~ + 300\\times x \\%~ = (100 + 300) \\times 15 \\% $$，解得$$x = 10$$，即甲瓶盐水的浓度是$$30 \\% $$． " ]

[ "2021年春蕾杯六年级竞赛第2题2分", "2020年春蕾杯六年级竞赛第7题2分" ]

[ [ { "aoVal": "A", "content": "$$54$$ " } ], [ { "aoVal": "B", "content": "$$64$$ " } ], [ { "aoVal": "C", "content": "$$52$$ " } ], [ { "aoVal": "D", "content": "$$56$$ " } ], [ { "aoVal": "E", "content": "$$60$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

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

[ "2016年新希望杯六年级竞赛训练题（六）第2题" ]

[ [ { "aoVal": "A", "content": "$$53$$ " } ], [ { "aoVal": "B", "content": "$$57$$ " } ], [ { "aoVal": "C", "content": "$$63$$ " } ], [ { "aoVal": "D", "content": "$$67$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "这个分数的分母比分子的$$\\frac{9}{7}$$倍多$$7$$，比分子的$$\\frac{4}{3}$$倍少$$3$$． 设分子为$$x$$，$$\\frac{9}{7}x+7=\\frac{4}{3}x-3$$，解得$$x=210$$，分母$$\\frac{4}{3}x-3=277$$． 分母比分子多$$277-210=67$$． " ]

[ "2011年其它", "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$14\\frac{1}{5}$$ " } ], [ { "aoVal": "B", "content": "$$14\\frac{1}{3}$$ " } ], [ { "aoVal": "C", "content": "$$7\\frac{1}{5}$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ] ]

[ "知识标签->数学思想->整体思想" ]

[ "① 若甲、乙两人合作共需多少小时？ ~~ $$1 \\div \\left( {\\frac{1}{{12}} + \\frac{1}{{18}}} \\right) = 1 \\div \\frac{5}{{36}} = 7\\frac{1}{5}$$（小时）． ~②甲、乙两人各单独做$$7$$小时后，还剩多少？ ~~ $$1 - 7 \\times \\left( {\\frac{1}{{12}} + \\frac{1}{{18}}} \\right) = 1 - \\frac{{35}}{{36}} = \\frac{1}{{36}}$$． ~③余下的$$\\frac{1}{{36}}$$由甲独做需要多少小时？ ~~ $$\\frac{1}{{36}} \\div \\frac{1}{{12}} = \\frac{1}{3}$$（小时）． ~④共用了多少小时？ ~~ $$7 \\times 2 + \\frac{1}{3} = 14\\frac{1}{3}$$（小时）． 在工程问题中，转换条件是常用手法． 本题中，甲做$$1$$小时，乙做$$1$$小时，相当于他们合作$$1$$小时，也就是每$$2$$小时，相当于两人合做$$1$$小时． 这样先算一下一共进行了多少个这样的$$2$$小时，余下部分问题就好解决了． " ]

[ "2017年第16届湖北武汉创新杯六年级竞赛邀请赛训练题（三）" ]

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

[ "知识标签->拓展思维->组合模块->逻辑推理->体育比赛->2-1-0 积分制" ]

[ "无论每场比赛的结果如何，每场比赛两队的总得分都是$$2$$分，一共有$$4$$个队，会有$$6$$场比赛，共$$2\\times 6=12$$分．每个人打$$3$$场比赛，第一名甲最多全胜得$$6$$分，乙$$+$$丙$$+$$丁$$=6$$分，所以乙和丙大于$$6\\div 3=2$$分．如果乙$$=$$丙$$=4$$分，那么甲$$4$$分，此时总分数大于$$12$$分，不可能，所以乙丙只能得$$3$$分． " ]

[ "2011年五年级竞赛明心奥数挑战赛" ]

[ [ { "aoVal": "A", "content": "39 " } ], [ { "aoVal": "B", "content": "40 " } ], [ { "aoVal": "C", "content": "47 " } ], [ { "aoVal": "D", "content": "49 " } ], [ { "aoVal": "E", "content": "53 " } ] ]

[ "拓展思维->拓展思维->数论模块->余数问题->中国剩余定理->逐级满足法" ]

[ "这个数除以2余1，除以3余2，除以5余2，是$$17+30k\\left( k=0，1，2\\cdots \\right)$$ 类型的数，选项中47符合，所以选C. " ]

[ "2020年第1届广东深圳超常思维竞赛五年级竞赛初赛第21题4分" ]

[ [ { "aoVal": "A", "content": "$$\\text{Torino}$$（$$1522$$年$$2$$月） " } ], [ { "aoVal": "B", "content": "$$\\text{Cremona}$$（$$1712$$年$$3$$月） " } ], [ { "aoVal": "C", "content": "$$\\text{Pavia}$$（$$1512$$年$$2$$月） " } ], [ { "aoVal": "D", "content": "$$\\text{Marengo}$$（$$1800$$年$$6$$月） " } ], [ { "aoVal": "E", "content": "$$\\text{Castiglione}$$（$$1796$$年$$8$$月） " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "设$$h=$$载的长度； $$n=$$该战役的月份中的天数； $$p=$$死期与发现日期之间的年数； 且$$a=$$该上尉死时的年龄． 则 $$h\\times n\\times \\frac{p}{2}\\times \\frac{a}{2}=451066$$，或$$h\\times n\\times p\\times a={{2}^{3}}\\times 7\\times 11\\times 29\\times 101$$，对$$n$$仅有的可能性是$$n=28$$或$$n=29$$，因此该月必是二月，首先设$$n=28$$，$$n={{2}^{2}}\\times 7$$． 我们注意到$$p$$不可能是$$2$$，$$29$$，$$11$$，$$22$$，$$58$$也不能是$$29\\times 101$$，仅有的其他可能性是： （$$1$$）$$p=2\\times 101$$，这种情况下$$a\\times h=29\\times 11$$， 但$$a\\ne 11$$（太年轻）且$$h\\ne 11$$（太长）． （$$2$$）$$p=29\\times 11$$，这种情况下$$a=2$$或$$101$$，这两者均不可能， 所以$$n\\ne 28$$． 因此$$n$$必须是$$29$$，且这年是闰年．故选$$\\text{C}$$． 注：尽管上述理由已足以回答此问题，以下的理由也说明该年是$$1512$$年．猜测$$p$$必须接近于$$400$$，唯一的可能性是$$p={{2}^{2}}\\times 101=404$$， 所以该上尉死于$$1914-404$$和$$1918-404$$之间，即$$1510$$年和$$1514$$年之间，仅有的闰年是$$1512$$年．注意，这也给出$$a\\times h=2\\times 7\\times 11$$，得出$$a=22$$，$$h=7$$． " ]

[ "2017年第13届湖北武汉新希望杯六年级竞赛决赛第5题" ]

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

[ "拓展思维->思想->分类讨论思想" ]

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

[ "2020年新希望杯五年级竞赛第11题", "2020年希望杯五年级竞赛模拟第11题" ]

[ [ { "aoVal": "A", "content": "$$27$$ " } ], [ { "aoVal": "B", "content": "$$17.28$$ " } ], [ { "aoVal": "C", "content": "$$1.92$$ " } ], [ { "aoVal": "D", "content": "$$19.2$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "因为由题干可知，车百公里油耗为$$8$$升，每天上下班共行驶$$30$$公里，则上下班共耗油$$30\\div 100\\times 8=2.4$$（升），油价按每升$$7.2$$元计算，所以每天上下班需要支付的油费为$$7.2\\times 2.4=17.28$$（元）． 故选$$\\text{B}$$． " ]

[ "2016年创新杯五年级竞赛训练题（一）第3题" ]

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

[ "拓展思维->思想->整体思想" ]

[ "我们首先考虑在$$2012$$之前有多少个数，在$$2012$$之前的数考虑其各个数位可以选择的数字，我们把$$2012$$前面的所有数当做是四位数，没有的数位拿$$0$$占位置，此时千位数字：$$0$$或$$1$$；百位、十位、个位数字为$$0$$，$$1$$，$$2$$；共$$2\\times 3\\times 3\\times 3=54$$（个）．当千位数字为$$2$$时：即自第$$55$$个开始：$$2000$$，$$2001$$，$$2002$$，$$2010$$，$$2011$$，$$2012$$为第$$60$$个数． " ]

[ "2021年世界少年奥林匹克数学竞赛六年级竞赛初赛第15题" ]

[ [ { "aoVal": "A", "content": "$$10$$\\% " } ], [ { "aoVal": "B", "content": "$$30$$\\% " } ], [ { "aoVal": "C", "content": "$$32$$\\% " } ], [ { "aoVal": "D", "content": "$$40$$\\% " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "第一天剩下纯果汁：$$1-\\frac{1}{5}=\\frac{4}{5}$$， 第二天剩下纯果汁：$$:\\frac{4}{5}\\times \\left( 1-\\frac{1}{5} \\right)=\\frac{16}{25}$$， 第三天剩下纯果汁：$$\\frac{16}{25}\\times \\left( 1-\\frac{1}{2} \\right)=\\frac{8}{25}$$， 这时果汁的浓度是：$$\\frac{8}{25}\\div 1\\times 100 \\%=32 \\%$$． 答：这时果汁的浓度是$$32 \\%$$． " ]

[ "2015年湖北武汉世奥赛六年级竞赛模拟训练题（一）第3题" ]

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

[ "拓展思维->能力->运算求解" ]

[ "设男学生有$$a$$人，女学生有$$b$$人，则老师人数为$$\\frac{a+b}{3}$$． 可得不定方程$$13a+10b+15\\times \\frac{a+b}{3}=186$$，化简得到$$6a+5b=62$$，考虑到$$a+b$$为$$3$$的倍数，只有一组解$$a=2$$，$$b=10$$．那么老师有$$\\left( 2+10 \\right)\\div 3=4$$人． " ]

[ "2017年第20届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第1题3分" ]

[ [ { "aoVal": "A", "content": "$$296$$ " } ], [ { "aoVal": "B", "content": "$$301$$ " } ], [ { "aoVal": "C", "content": "$$306$$ " } ], [ { "aoVal": "D", "content": "$$311$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列求通项" ]

[ "根据题意，这个等差数列的首项是$$1$$，公差是$$6-1=5$$，项数是$$61$$， 根据末项$$=$$首项$$+$$公差$$\\times $$（项数$$-1$$）来计算：$$1+5\\times (61-1)=301$$， 所以第$$61$$项是$$301$$． 故选$$\\text{B}$$． " ]

[ "2018年第6届湖北长江杯六年级竞赛初赛A卷第3题3分" ]

[ [ { "aoVal": "A", "content": "赚了 " } ], [ { "aoVal": "B", "content": "亏了 " } ], [ { "aoVal": "C", "content": "不亏不赚 " } ], [ { "aoVal": "D", "content": "不能确定 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "售价为$$200$$元，$$A$$赚了$$10 \\%$$， $$A$$的进价为： $$200\\div (1+10 \\%)=\\frac{2000}{11}$$（元）， $$B$$亏了$$10 \\%$$，$$B$$的进价为： $$200\\div (1-10 \\%)=\\frac{2000}{9}$$（元）， $$\\frac{2000}{11}+\\frac{2000}{9}=\\frac{40000}{99}$$（元）$$\\textgreater400$$元． 进价$$\\textgreater$$售价，亏了．选$$\\text{B}$$． " ]

[ "2017年第22届全国华杯赛小学高年级竞赛初赛第1题10分" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$19$$ " } ] ]

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

[ "设这两个数为$$a$$、$$b$$，则有$$7\\times 10\\textless{}ab\\textless{}8\\times 11$$，所以$$70\\textless{}ab \\textless{}88$$，其中的整数部分有$$70$$，$$71$$，$$72$$，$$\\ldots $$，$$87$$，共有$$18$$种． " ]

[ "2017年第15届湖北武汉创新杯六年级竞赛决赛第1题" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$25$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

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

[ "工程问题．从开始到结束三人一共用了$$(1000+1250)\\div (28+32+30)=2250\\div 90=25$$天． " ]

[ "2019年新加坡高级学府数学竞赛（SASMO）五年级竞赛第14题" ]

[ [ { "aoVal": "A", "content": "$$Kieran$$ " } ], [ { "aoVal": "B", "content": "$$Franklin$$ " } ], [ { "aoVal": "C", "content": "$$Mabel$$ " } ], [ { "aoVal": "D", "content": "$$Jeff$$ " } ] ]

[ "Overseas Competition->知识点->组合模块->逻辑推理->条件型逻辑推理->表格法", "拓展思维->拓展思维->组合模块->逻辑推理" ]

[ "如果$$Kieran$$说谎，那么他最矮，则和$$Mabel$$矛盾，同理$$Mabel$$也不能说谎，$$Franklin$$如果说谎的话则和$$Jeff$$或者$$Kieran$$矛盾，所以只能$$Jeff$$说谎，那么$$Jeff$$和$$Franklin$$ 不是最高，$$Mabel$$最矮，所以$$Kieran$$最高. " ]

[ "2014年迎春杯三年级竞赛复赛" ]

[ [ { "aoVal": "A", "content": "$$18$$、$$6$$ " } ], [ { "aoVal": "B", "content": "$$22$$、$$6$$ " } ], [ { "aoVal": "C", "content": "$$25$$、$$6$$ " } ] ]

[ "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" ]

[ "解：注意到： $$4$$是$$2$$的平方， $$9$$是$$3$$的平方， $$16$$是$$4$$的平方， $$25$$是$$5$$的平方， $$36$$是$$6$$的平方， $$\\cdots $$ 根据这个规律，可知中间两个括号分别应填$$25$$和$$6$$。 故选:C。 " ]

[ "2014年全国迎春杯三年级竞赛初赛第8题" ]

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

[ "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" ]

[ "$$2021\\div (4+3+2+1)=201（组）\\ldots\\ldots1（个）$$． " ]

[ "2018年河南郑州K6联赛竞赛初赛第23题3分" ]

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

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

[ "可能性；要想红色朝上次数最少，则$$1$$面涂红色，$$2$$面染黄色． " ]

[ "2019年第7届湖北长江杯五年级竞赛复赛B卷第1题3分" ]

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

[ "拓展思维->能力->运算求解" ]

[ "$$\\frac{1}{29}\\times 10\\textless{}\\frac{1}{20}+\\frac{1}{21}+\\cdots +\\frac{1}{29}\\textless{}\\frac{1}{20}\\times 10$$， $$\\frac{10}{29}\\textless{}\\frac{1}{20}+\\frac{1}{21}+\\cdots +\\frac{1}{29}\\textless{}\\frac{1}{2}$$， $$\\frac{2}{1}\\textless{}\\frac{1}{\\dfrac{1}{20}+\\dfrac{1}{21}+\\cdots +\\dfrac{1}{29}}\\textless{}\\frac{29}{10}$$， 所以它的整数部分应是$$2$$，故选$$\\text{D}$$． " ]

[ "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "一样大 " } ], [ { "aoVal": "B", "content": "$$A$$ " } ], [ { "aoVal": "C", "content": "$$D$$ " } ], [ { "aoVal": "D", "content": "$$E$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "抽中的概率为$$\\frac{1}{6}$$，没抽到的概率为$$\\frac{5}{6}$$，如果$$A$$没抽中，那么$$B$$有$$\\frac{1}{5}$$的概率抽中，如果$$A$$抽中，那么$$B$$抽中的概率为$$0$$，所以$$B$$抽中的概率为$$\\frac{5}{6}\\times \\frac{1}{5}=\\frac{1}{6}$$． 同理，$$C$$抽中的概率为$$\\frac{5}{6}\\times \\frac{4}{5}\\times \\frac{1}{4}=\\frac{1}{6}$$，$$D$$抽中的概率为$$\\frac{5}{6}\\times\\frac{4}{5}\\times \\frac{3}{4}\\times \\frac{1}{3}=\\frac{1}{6}$$， $$E$$抽中的概率为$$\\frac{5}{6}\\times\\frac{4}{5}\\times \\frac{3}{4}\\times \\frac{2}{3}\\times \\frac{1}{2}=\\frac{1}{6}$$，$$F$$抽中的概率为$$\\frac{5}{6}\\times\\frac{4}{5}\\times \\frac{3}{4}\\times \\frac{2}{3}\\times \\frac{1}{2}\\times1=\\frac{1}{6}$$． 由此可见六人抽中的概率相等，与抽签的先后顺序无关． " ]

[ "2018年湖北武汉新希望杯小学高年级五年级竞赛训练题（四）第6题" ]

[ [ { "aoVal": "A", "content": "$$512$$ " } ], [ { "aoVal": "B", "content": "$$729$$ " } ], [ { "aoVal": "C", "content": "$$768$$ " } ], [ { "aoVal": "D", "content": "$$972$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$3\\times 3=9$$，$$2\\times 2\\times 2=8$$，$$9\\textgreater8$$，所以分成$$2$$个$$3$$比分成$$3$$个$$2$$乘积大； $$3\\times 1=3$$，$$2\\times 2=4$$，$$3\\textless{}4$$，所以分成两个$$2$$比分成$$1$$个$$3$$和$$1$$个$$1$$乘积大，$$19=3+3+3+3+3+2+2$$，乘积为$$3\\times 3\\times 3\\times 3\\times 3\\times 2\\times 2=972$$． " ]

[ "2013年IMAS小学中年级竞赛第一轮检测试题第6题3分" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ], [ { "aoVal": "E", "content": "$$33$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "从家到学校，若小王选择第一种方式需要$$8+15=23$$分钟，选择第二种方式则需要$$10+10=20$$分钟，所以小王从家到学校至少需要$$20$$分钟．故选$$\\text{B}$$． " ]

[ "2017年第15届湖北武汉创新杯六年级竞赛初赛第1题" ]

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

[ "拓展思维->思想->转化与化归的思想" ]

[ "原式$$=1\\div 3\\times 5\\div 5\\times 7\\div 7\\times 9\\div 9\\times 13\\div 13\\times 15$$$$=1\\div 3\\times 15$$$$=5$$． " ]

[ "2021年新希望杯一年级竞赛初赛第14题5分" ]

[ [ { "aoVal": "A", "content": "Rat and Rooster鼠和鸡 " } ], [ { "aoVal": "B", "content": "Chicken and Rabbit鸡和兔 " } ], [ { "aoVal": "C", "content": "Rabbit and dog兔和狗 " } ], [ { "aoVal": "D", "content": "Rabbit and Rat兔和鼠 " } ], [ { "aoVal": "E", "content": "Rat and dog鼠和狗 " } ] ]

[ "拓展思维->能力->实践应用", "Overseas Competition->知识点->应用题模块->分百应用题->认识单位1" ]

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

[ "2008年第6届创新杯四年级竞赛初赛A卷第8题5分" ]

[ [ { "aoVal": "A", "content": "$$251$$棵 " } ], [ { "aoVal": "B", "content": "$$252$$棵 " } ], [ { "aoVal": "C", "content": "$$502$$棵 " } ], [ { "aoVal": "D", "content": "$$504$$棵 " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde(2008\\div 8+1)\\times 2$$ $$=(251+1)\\times 2$$ $$=252\\times 2$$ $$=504$$（棵） 故选$$\\text{D}$$． " ]

[ "2015年第13届全国创新杯六年级竞赛第1题", "小学高年级六年级其它2015年数学思维能力等级测试初试第1题4分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{15}{4}$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{20}{3}$$ " } ], [ { "aoVal": "D", "content": "以上都不对 " } ] ]

[ "拓展思维->拓展思维->计算模块->分数->分数运算->分数乘法运算" ]

[ "$$\\frac{3}{4}\\times 4=3$$分钟． " ]

[ "2013年IMAS小学高年级竞赛第一轮检测试题第15题4分" ]

[ [ { "aoVal": "A", "content": "$$70$$ " } ], [ { "aoVal": "B", "content": "$$80$$ " } ], [ { "aoVal": "C", "content": "$$90$$ " } ], [ { "aoVal": "D", "content": "$$100$$ " } ], [ { "aoVal": "E", "content": "$$180$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "由题意可知甲桶的容量的$$\\frac{1}{2}-\\frac{1}{6}=\\frac{1}{3}$$为$$60$$升，因此甲桶的容量为$$60\\div \\frac{1}{3}=180$$升，故乙桶的容量为$$180\\times \\frac{1}{2}=90$$升． 故选$$\\text{C}$$． " ]

[ "2012年IMAS小学中年级竞赛第一轮检测试题第3题3分" ]

[ [ { "aoVal": "A", "content": "$$201$$万 " } ], [ { "aoVal": "B", "content": "$$2100000$$ " } ], [ { "aoVal": "C", "content": "$$102$$万 " } ], [ { "aoVal": "D", "content": "$$20100000$$ " } ], [ { "aoVal": "E", "content": "$$210$$万 " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$\\text{A}$$等于数$$2010000$$、$$\\text{B}$$大于数$$2010000$$、$$\\text{C}$$小于数$$2010000$$、$$\\text{D}$$大于数$$2010000$$、$$\\text{E}$$大于数$$2010000$$． " ]

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

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

[ "拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数通分法->通分母", "课内体系->能力->运算求解" ]

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

[ "2019年第23届广东世界少年奥林匹克数学竞赛六年级竞赛初赛第9题5分" ]

[ [ { "aoVal": "A", "content": "$$61$$ " } ], [ { "aoVal": "B", "content": "$$85$$ " } ], [ { "aoVal": "C", "content": "$$99$$ " } ], [ { "aoVal": "D", "content": "$$61$$或$$99$$ " } ] ]

[ "拓展思维->思想->逆向思想" ]

[ "质数是指在大于$$1$$的自然数中，除了$$1$$和它本身以外不再有其他因数的自然数． 在所有质数中，只有$$2$$这一个偶数；其他质数都是奇数； 但是奇数的$$3$$倍或者$$5$$倍都是奇数；两个奇数之和不可能是奇数； 所以这两个质数必然有$$2$$； 若第一个质数是$$2$$，$$2\\times 3+$$质数$$\\times 5=301$$，第二个质数$$=\\left( 301-6 \\right)\\div 5=59$$； 若第二个质数是$$2$$，质数$$\\times 3+2\\times 5=301$$，第一个质数$$=\\left( 301-10 \\right)\\div 3=97$$； 所以这两个质数的和$$61$$或$$99$$． " ]

[ "2017年第20届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第10题3分" ]

[ [ { "aoVal": "A", "content": "$$70$$秒 " } ], [ { "aoVal": "B", "content": "$$80$$秒 " } ], [ { "aoVal": "C", "content": "$$105$$秒 " } ], [ { "aoVal": "D", "content": "$$120$$秒 " } ] ]

[ "拓展思维->拓展思维->应用题模块->间隔问题->直线型两端都有->爬楼梯问题" ]

[ "从$$1$$层走到$$3$$层走的楼梯层数是：$$3-1=2$$个，走一个楼层用时为：$$30\\div2=15$$秒，那么从$$3$$层走到$$10$$层走的楼梯层数是：$$10-3=7$$个，要用时为：$$15\\times7=105$$秒，据此解答． $$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde30\\div (3-1)\\times (10-3)$$ $$=15\\times7$$ $$=105$$（秒）， 答：他从$$3$$层走到$$10$$层需要$$105$$秒． 故答案为：$$105$$． 故选$$\\text{C}$$． " ]

[ "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": "$$200$$ " } ], [ { "aoVal": "D", "content": "$$201$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

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

[ "2019年第23届广东世界少年奥林匹克数学竞赛五年级竞赛初赛第2题5分" ]

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

[ "拓展思维->能力->逻辑分析" ]

[ "欲求两个质数中大数减小数的差的最小值即这两个质数越接近越好． 则这两个质数分别为$$47$$和$$53$$． 则差：$$53-47=6$$． 故选$$\\text{D}$$． " ]