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2015上海第13届走美杯三年级初赛试题答案

2015上海第13届走美杯三年级初赛试题答案
2015上海第13届走美杯三年级初赛试题答案

小学三年级试卷

注意事项:

1.考生要按要求在密封线内填好考生的有关信息.2.不允许使用计算器.

3.为方便决赛通知,务必填写联系电话.电话:一、填空题(每小题8分,共40分)

1.135797992014++++++-= .

【分析】486

考点:等差数列计算;原式250201425002014486=-=-=.

2.在右图的每个方框中填入一个数字,使得乘法竖式成立.那么,这个算式的乘积是

1

3

7?

【分析】407或777考点:乘法数字谜;

由乘积个位是7可知乘数的个位与被乘数的乘积是37,进而得到被乘数即为37,如图所示:

3

713

77

?

由于乘数的十位与37相乘所得结果为两位数,因此该位置可能是1或2;①如果乘数的十位填入1,结果如下图所示:

②如果乘数的十位填入2,结果如下图所示:

3

71137

374

7?

372137

7477

7?

因此这个算式的乘积是407或777.

3.有一堆红球与白球,球的总数不超过50.已知红球个数是白球个数的3倍,那么,红球最多有个.

【分析】36个考点:和差倍问题;

由于红球个数是白球个数的3倍,因此球的总数应为白球个数的4倍,可得球的总数一定是4的倍数;红球最多的情况即对应了球的总数最多的情况,而不超过50的最大的4的倍数为48;

因此球的总数最多有48个,此时红球最多有484336÷

?=个.4.一袋奶糖分给几位小朋友,如果每人得8颗,还剩4颗;如果每人得11颗,就有一位小朋友拿不到.一共有位小朋友.

【分析】5位

考点:盈亏问题;

如果每人得11颗,就有一位小朋友拿不到,意味着此时奶糖少了11颗,因此此题为“盈亏”型;小朋友人数:(

)()4111185+÷-=位.5.数一数,图中共有个三角形.

【分析】12个

考点:图形计数;

如果首先去掉三角形右侧内部的斜线,得到如下图形:

此时应有(

)21228+?+=个三角形;之后加上被去掉的线,此时会增加4个三角形,如下图所示:

因此原图中一共有8412+=个三角形.

二、填空题(每小题10分,共50分)

6.某小学三年级的部分学生排成一个实心正方形方阵,最外面3层有学生72人,这个方阵共有学生

人.

【分析】81人

考点:间隔与方阵;

次外层的人数:72324÷

=人;最外层的人数:24832+=人;最外层每边的人数:32419÷+=人;方阵总人数:9981?

=人.7.把48粒棋子放入9个盒子中,每个盒子至少放1粒,每盒棋子数都不一样,棋子最多的盒子里最多可

以放粒棋子.【分析】12粒

考点:最值问题;

当棋子总数一定时,要使棋子最多的盒子里棋子尽可能的多,另外8个盒子的棋子总数就要尽可能的少;

而由于每盒棋子数都不一样,这8个盒子的棋子总数最少为:1234567836+++++++=粒;因此棋子最多的盒子里最多可以放483612-=粒棋子.

8.,A B 两地相距1000米,甲从A 地出发,1小时后到达B 地.乙在甲出发后20分钟从B 地出发,40分

钟到达A 地.甲、乙二人相遇点距A 地米.【分析】600米

考点:行程问题——相遇;

由乙40分钟可走1000米,得到乙的速度为10004025÷=米/分钟;

甲60分钟可走1000米,而乙60分钟可走25601500?

=米;由1000与1500的关系不难看出,相同时间内若甲走2份路程,则乙可走3份;

现在甲比乙早出发20分钟,即为乙比甲晚出发20分钟;

可构造一种情形:乙先向后退20分钟甲再出发,即为乙后退2520500?=米;此时甲、乙二人的实际距离为10005001500+=米;甲、乙二人相遇点与A 地的距离即为相遇时甲所走的路程;

在二人的路程和1500米当中,甲所走的路程为()1500232600÷

+?=米;所以甲、乙二人相遇点距A 地600米.

9.小明说:“我妈妈比我大24岁,两年前妈妈的年龄是我的4倍.”小明今年岁.

【分析】10岁

考点:年龄问题;

由于2个人年龄差不变,两年前妈妈也比小明大24岁;因此两年前小明的年龄是:()24418÷

-=岁;所以小明今年的年龄是:8210+=岁.

10.将数字1~9放入图中的小方格中,每格一个数,可得到四条线上三个数的和都相等,请问*应该是.

【分析】8

考点:数阵图;

由于在图中只有1,4,2这三个数字位于其中的两条线上,各被重复计算过一次;因此图中四条线的总和是:12345678914252+++++++++++=;得到每条线上三个数的和应为:52413÷=;

由*所在的线可得:*1314

8=

--=.三、填空题(每小题12分,共60分)

11.右图是可以一笔画出的,一共有种不同的一笔画法(起点、终点或顺序只要有一样不同,就

算不同的画法).

【分析】12种

考点:一笔画;

首先将图中各点命名如下:

由于,A B 两点均为奇点,因此画法必定是从A 开始到B 结束,或是从B 开始到A 结束,且不难想到这两种画法的种类数相同;

下面以从A 开始到B 结束为例:

如果先从A 画到B ,则接下来剩余的正方形只有顺时针和逆时针2种画法,即ABCADB 和ABDACB ;如果先从A 画到C ,那么接下来必定画到B ,之后会有2种选择:一是先直接画到A ,再从D 画到B ,即ACBADB ;二是经过D 画到A ,再从A 画到B ,即ACBDAB ;

如果先从A 画到D ,根据图形的对称性其种类数应与先从A 画到C 相同,也是2种;综上所述,从A 开始到B 结束的画法一共有2226++=种,类似的从B 开始到A 结束的画法也有6种;

因此该图形一共有6612+

=种不同的一笔画法.

12.有五个互不相等的非零自然数,最小的一个数是7.如果其中一个减少20,另外四个数都加5,那么得到的仍然是这五个数.这五个数的和是.【分析】85

考点:等差数列;

由于7不可能是减少20的数,因此这五个数当中一定有7512+=;

同理这五个数当中一定还有12517+

=和17522+=;如果减少20的数是22,那么这五个数当中一定有22202-=,但27<不满足条件;因此这五个数当中一定还有22527+=,此时27205-=满足条件;

即这五个数是7,12,17,22,27,它们的和是71217222785+

+++=.13.一个正方体的6个面分别标着,,,,,A B C D E F 六个字母,从3个不同角度看正方体如图所示,字母C 的

对面是字母.

【分析】D

考点:图形规律;

由图1和图2可得字母D 与字母,,,A B E F 均为邻面,因此其对面为字母C ;另:类似可得字母A 的对面是字母E ,字母B 的对面是字母F .

14.24点游戏:用加、减、乘、除、括号等运算符号把4,4,10,10这四个数连起来,使结果等于24,

.【分析】(

)10104424?-÷=考点:24点计算;过程略.

的方格表内有四个筹码,这些筹码一面为白色另一面为黑色.每一次操作可以任选一个筹码跳15.在15

过一个、二个或三个筹码到空位上,但不可以用走动的.被跳过的筹码都必须翻面,但跳的筹码不翻面.现欲经过六次的操作,将下左图的情况变成下右图的情况.如果依次将跳动的筹码跳动前所在位置的号码记录下来,就可以得到一个六位数.请给出可能完成任务的一个六位数.(填出一个即可).

【分析】251425或152415

考点:操作性问题;

251425操作如下:

152415操作如下:

2015年上海市中考英语试卷-答案

上海市2015年初中毕业统一学业考试 英语答案解析 Part 1 Listening Ⅰ.Listening comprehension 1.【答案】E 2.【答案】F 3.【答案】B 4.【答案】C 5.【答案】D 6.【答案】G 7.【答案】C 8.【答案】A 9.【答案】D 10.【答案】A 11.【答案】D 12.【答案】B 13.【答案】B 14.【答案】C 15.【答案】F 16.【答案】T 17.【答案】T 18.【答案】F 19.【答案】T 20.【答案】F 21.【答案】have fun 22.【答案】work hard 23.【答案】in class 24.【答案】11 o'clock

25.【答案】angry with Part 2 Phonetics, Grammar and Vocabulary Ⅱ.Choose the best answer 26.【答案】A 【解析】A.shape的发音为/?e?p/,是重读开音节,字母a发本身的音/e?/;B.sheep的发音为/?i?p/;C.shop 的发音为/??p/;D.ship的发音为/??p/。 【考点】单词的读音。 27.【答案】D 【解析】根据句意可知,本题考查“by oneself”的用法。 【考点】代词辨析。 28.【答案】B 【解析】in在…里面;of…的;on在…上面;at在,后跟小地点或具体的时间;a picture of sb.某人的照片,of表所属关系。 【考点】介词辨析。 29.【答案】C 【解析】wait for为固定搭配,表目的。waiting for the opening ceremony to start做crowd的定语。 【考点】介词辨析。 30.【答案】C 【解析】根据句意可知,此处应为序数词,排除A、B,又sixteenths常用在分数表达中。 【考点】数词辨析。 31.【答案】D 【解析】根据句意,这是一个有关比较级的试题。much作为程度副词,应该放于比较级之前。 【考点】副词辨析。 32.【答案】A 【解析】由题意可知,选A句意通顺。 【考点】连词辨析。 33.【答案】C 【解析】A.If意思为如果;B.Since意思为自…起;C.Although意思为尽管;D.Because意思为因为。【考点】连词辨析。 34.【答案】D

走美杯三年级试卷

第六届“走进美妙的数学花园’’中国青少年数学论坛 趣味数学解题技能展示大赛初赛 小学三年级试卷(B卷) 一、填空题I(每题8分,共40分) +-?÷=( 777 ) 1. 777777777777777 【分析】原式=777+777-777=777 2. 在方框中添加适当运算符号(不能添加括号),使等式成立. 【分析】9+3+4+19-8-5+4=26 3.两个整数,差为l6,一个是另一个的5倍.这两个数分别是( 4 )和( 20 ) 【分析】本题属于和差问题。小数:16÷(5-1)=4;大数:4×5=20或4+16=20。 4.用若干个1分、2分、5分的硬币组成一角钱(不要求每种硬币都有),共有( 10 )种不同的方法. 【分析】此题采用枚举法,具体如下:

5. 100位同学都面向主席台,排成l0行10列的方阵.小明在方阵中,他的正左方有2位同学,正前方有4位同学.若整个方阵的同学向右转,则小明的正左方有( 4 )位同学,正前方有( 7 )位同学. 【分析】小明的正左方有2位同学,正前方有4位同学.那么他的正右方有7个同学,正后方有5个同学。现在小明向右转,转动之后他现在的正左方是原来的正 前方有4个同学,现在的正前方是原来的正右方有7个同学。 二、填空题Ⅱ(每题l0分,共50分) 6.某小学三年级的学生排成一个实心的正方形方阵,最外面一层有学生40 人.这个方阵共有学生( 121 )人. 【分析】最外面一层有学生40 人,那么这个方阵每边有40÷4+1=11(人),最后算出这个方阵共有学生11×11=121人。 7.将一个两位数的数字相乘,称为一次“操作”.如果积仍是二个两位数,重复 →?=→?=(停止)以上操作,直到得到一个一位数.例如:292918188 共经历两次操作.一个两位数经过3次如上操作,最终得到一位数.这个两位数最小是( 39 ). →?=→?==?=。 【分析】这个两位数最小是39,3939272714144

2015年上海中考英语真题试卷含答案

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2015上海中考英语首字母系列1

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2015年上海市中考英语模拟试卷2

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2015上海中考英语试卷听力完整文本精校版

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第十三届“走进美妙的数学花园”青少年展示交流活动 趣味数学解题技能展示大赛初赛(上海决赛) 小学五年级试卷(B 卷)
2015 年 3 月 8 日 满分 150 分 上午 10:45——12:15
一、填空题(每小题 8 分,共 40 分) 【第 1 题】计算: 20150308 = 101× (100000 + 24877 ×
)
【第 2 题】将
2 5 15 10 , , , 按照从小到大顺序排列 3 8 23 17

【第 3 题】 像 2,3,5,7 这样只能被 1 和自身整除的大于 1 的自然数叫做质数或素数。将 2015 分拆成 100 个质数之和,要求其中最大的质数尽可能小,那么这个最大质数是 。
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? ?
4? a? ? ? × 7 = 24 ,我们将满足 ? a ? ? × b = 24 的 7? b? ?

牌组 {a,a,b,b}称为“王亮牌组”,请再写出一组不同的“王亮牌组”
第 1 页 共 4 页

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