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解析1995小学数学奥林匹克试题决赛

解析1995小学数学奥林匹克试题决赛
解析1995小学数学奥林匹克试题决赛

1995小学数学奥林匹克试题决赛

1.计算:

2.下面是一个残缺的乘法竖式,那么乘积是__________.

3.如图所示,每一条线段的端点上的两数之和算作线段的长度,那么图上六条线段的长度之和是__________.

4. 甲、乙两数的最小公倍数是90,乙、丙两数的最小公倍数是105,甲、丙两数的最小公倍数是126,那么甲数是__________.

5.某次考试,张、王、李、陈四人的成绩统计如下:

张、王、李平均分91分

王、李、陈平均分89分

张、陈平均分95分

那么张得了__________分.

6.甲数数字和是29,乙数数字和是18,当甲、乙两数用竖式相加时,有三位进位,那么这两数和的数字和是__________.

7.有一串数如下:

1 2 4 7 11 16 …

它的规律是:由1开始,加1,加2,加3,…,依次逐个产生这串数,直到产生第50个数为止,那么在这50个数中,被3除余1的数有__________个.

8.如图,ABCG是4×7的长方形,DEFG是2×5的长方形,那么三角形BCM的面积与三角形DEM的面积之差是__________.

9.某次会议,昨天参加会议的男代表比女代表多700人,今天男代表减少10%,女代表增加了

5%,今天共1995人出席会议,那么昨天参加会议的有__________人.

10.有甲、乙两项工作,张单独完成甲工作要10天,单独完成乙工作要15天;李单独完成甲工作要8天,单独完成乙工作要20天,如果每项工作都可以由两人合作,那么这两项工作都完成最少需要__________天.

11.小明买红、蓝两种笔各1支,共用17元,两种笔的单价都是整元,并且红笔比蓝笔贵,小强打算用35元买这两种笔(也允许只买其中一种),可是他无论怎么买,都不能把35元恰好用完,那么红笔的单价是__________元.

12.甲、乙两人同时从山脚开始爬山,到达山顶后就由原路立即下山,他们两人的下山速度都是各自上山速度的1.5倍,而且甲比乙速度快,开始后1小时。甲与乙在离山顶600米处相遇,当乙到达山顶时,甲恰好下到半山腰,那么甲回到出发点共用__________小时.

1. 【解】原式====9

2. 【解】乘数的个位数字与被乘数相乘得22,所以乘数的个位数字是2,被乘数是11,由于被乘数与乘数的十位数字相乘,积的个位数字是9(否则这积与2相加不会发生进位)。因此乘数是92,乘积是1O12

3. 【解】所求的和=3×(++0.6+0.875)=1+++=

4. 【解】90=2×3×5,105=3×5×7,126=2×3×7。所以乙不被2整除,甲被2整除。甲不被5整除。丙不被3整除,甲被3整除,从而甲数是2×3=18

5. 【解】(91×3+95×2-89×3)÷2=196÷2=98(分)

答:张得98分.

6. 【解】29+18-9×3=20

答:两数和的数字和为20。

7. 【解】1,2,4,7,11,16,22,29,37,…被3除,所得余数是1,2,1,1,2,1,1,2,1,1,……

又50=3×16+2

于是,被3除余1的数有2×16十1=33(个)

8. 【解】延长BC交EF于N.

三角形BCM与三角形DEM的面积差=三角形BNE与长方形EDCN的面积差

=×(10-7)×(4+2)-(10-7)×2=3(平方单位)

9. 【解】1995-700×(1-10%)是昨天女代表人数的(1+5%)+(1-10%),因此昨天参加会议的有

[1995-700×(1-10%)]÷[(1-10%)+(1+5%)]×2+700

=1365÷195%×2+700

=700×2+700

=2100(人)

10. 【解】合理安排应是李先单独完成甲工作,同时,张单独先做(8天)乙工作,然后张李合作完成乙工作的剩余量,共用

(1-×8)÷(+)+8=÷+8=12(天).

因此,这两项工作都完成至少需要12天.

11. 【解】蓝笔单价不能是35的约数.而且应当小于红笔单价,所以蓝笔单价只能是2、3、4、6、8(元),相应的红笔单价是15,14,13,11,9(元),但

35=15×1+2×10=14×1+3×7=11×1十4×6=9×3+8×1

所以红笔的单价应当是13元.

12. 【解】因为乙到达山顶时,甲恰好下到半山腰,所以如果甲全程都用下山速度,那么所

行路程是山脚到山顶的+1×1.5=2倍,即甲下山的速度正好是乙上山的速度的2倍.乙上到与甲相会处用1小时,甲从这里到山脚应当用1÷2=(小时),因此甲回到出发点共用1十=1.5(小时)

89届小学数学奥林匹克竞赛初赛

1.计算:

= 。

2.1到1989这些自然数中的所有数字之和是。

3.把若干个自然数,2,3,……乘到一起,如果已知这个乘积的最末13位恰好都是零,那么最后出现的自然数最小应该是。

4.在1,,,,,…,,中选出若干个数,使它们的和大于3,至少要选个数。

5.在右边的减法算式中,每一个字母代表一个数字,不同的字母代表不同的数字,那么D+G= 。

6.如图,ABFD和CDEF都是矩形,AB的长是4厘米,BC的长是3厘米,那么图中阴影部分的面积是平方厘米。

7.甲乙两包糖的重量比是4:1,如果从甲包取出10克放入乙包后,甲乙两包糖的重量比变为7:5,那么两包糖重量的总和是克。

8.设1,3,9,27,81,243是六个给定的数,从这六个数中每次或者取一个,或者取几个不同的数求和(每个数只能取一次),可以得到一个新数,这样共得到63个新数。如果把它们从小到大依次排列起来是1,3,4,9,12……那么第60个数是。

9.有甲、乙、丙三辆汽车各以一定的速度从A地开往B地,乙比丙晚出发10分钟,出发后40分钟追上丙。甲比乙又晚出发20分钟,出发后1小时40分追上丙,那么甲出发后需用分钟才能追上乙。

10.有一个俱乐部,里面的成员可以分成两类,第一类是老实人,永远说真话;第二类是骗子,永远说假话。某天俱乐部全体成员围着一张圆桌坐下,每个老实人的两旁都是骗子,每个骗子的两旁都是老实人。记者问俱乐部成员张三:俱乐部共有多少成员?张三回答:有45人。李四说:张三是老实人。那么张三是老实人还是骗子?张三是。

11.某工程如果由第一、二、三小队合干需要12天才能完成;如果由第一、三、五小队合干需要7天完成;如果由第二、四、五小队合干4天完成;如果由第一、三、四小队合干需要42天才能完成。那么这五个小队一起合干需要天才能完成这项工程。

12把一个两位数的个位数字与其十位数字交换后得到一个新数,它与原来的数加起来恰好是某个自然数的平方,这个和数是。

13.把自然数1,2,3,……,998,999分成三组,如果每一组数的平均数恰好相等地,那么这三个平均数的和是。

14.某种商品的价格是:每一个1分钱,每五个4分钱,每九个7分钱。小赵的钱至多能买50个,小李的钱至多能买500个。小李的钱比小赵的钱多分钱。

15.一个自行车选手在相距950千米的甲、乙两地之间训练,从甲地出发,去时每90千米休息一次;到达乙地并休息一天后再沿原路返回,每100千米休息一次。他发现恰好有一个休息的地点与去时的一个休息地点相同,那么这个休息地点距甲地有千米。

16.现有四个自然数,它们的和是1111,如果要求这四个数的公约数尽可能地大,那么这四个数的公约数最大可能是。

17.桌面上有一条长度为100厘米的红色直线,另外有直径分别是2、3、7、15厘米的圆形纸片若干个,现在用这些圆形纸片将桌上的红线盖住,如果要使所用纸片的圆周长总和最短,那么这个周长总和是。

18.右图是一个边长为2厘米的正方体,在正方体的上面的正中向下挖一个边长为1厘米的正方体小洞;接着在小

洞的底面正中再向下挖一个边长为厘米的小洞;第三个小洞的挖法与前两个相同,边长为

厘米,那么最后得到的立体图形的表面积是平方厘米。

19.小明在左衣袋和右衣袋中分别装有6枚和8枚硬币,并且两衣袋中硬币的总钱数相等,当任意从左边衣袋取出两个硬币和右边衣袋的任意两个硬币交换时,左边衣袋的总钱数要么比原来的钱数多二分,要么比原来钱数少二分。那么两个衣袋中共有钱。

20.从1,3,5,7,…97,99中最多可以选出个数,使它们当中的每一个数都不是另一个数的倍数。

1.【解】将1~1989中的每个数看成“四位数”,位数不够的前面补“0”,从0000~1999,所有数的数字之和是(0十1+2十…+9)×300×2+1×1000

=45×600+1OOO

=28000

而从1990~1999中的所有数的数字之和为

1×10+9×2×10十(0+1+ (9)

=10十180+45

=235

从而,所求所有数字之和为28000—235=27765

2.【解】l×2×…×50中有10+2=12(个)因数5(在25、50中,因数5各出现2次,在5的其它倍数

中各出现一次)

于是,l×2×…×55的末尾有13个0,且55为最小的这样的数,

即最后出现的自然数最小为55

3.【解】首先A=1,B=0,E=9。再由十位的运算可知F=8,从而C=7,并且10+D-G =8即G-D=2,G可能为6,5,4,相应地,D为4、3、2。于是D+G=10、8、6

4.【解】阴影部分的面积和

=100×3—144-2×42

=72(平方厘米)

5.【解】两包糖重量的总和是

10÷()

=10÷

=(克)

6.【解】根据题意,丙行50分钟的路程乙只需40分钟,所以∶=

4∶5;丙行130分钟的路程。甲只需100分钟,∶=10∶13

从而∶=26∶25

因为乙早出发加分钟,所以甲出发后追上乙所花的时间为

25×20÷(26-25)=500(分钟).

7.【解】张三是骗子因为骗子与老实人是相间地围着圆桌坐的,所以两者人数相等,俱乐部的人数必定是偶数,张三讲的是假话,他是骗子.

8.【解】设原来的两位数为,则交换十位数字与个位数字后的两位数为,两个数的和为

+=10x+y+10y+x=11×(x+y)

是11的倍数,因为它是平方数,所以也是11×11=121的倍数.但这个和<100+100=200<121×2,所以这个和数为121。

9.【解】小赵的钱至多能买50个,而50=9×5十5×1

因此,小赵有7×5+4×1=39(分)

小李的钱至多能买500个。而

500=9×55+5×1

因此,小李有7×55+4×1=389(分)

于是小李比小赵多389-39=350(分)

10.【解】设这个休息地距甲地有a公里,显然a为90的倍数.且a-50

为100的倍数,此时a就只能为450.

从而这个休息地距甲地有450公里.

11.【解】这些圆纸片的直径的和≥100.所以它们的周长的和≥lOOπ≈314(厘米)

另一方面,这些圆可以恰好将长为lOO厘米的红线盖住(例如用10个7厘米,2个15厘米的圆,或50个2厘米的圆).

因此,圆周长总和最短时,这个周长总和是314厘米.

12.【解】2×2×6+l×l×4+××4+××4

=29.25(平方厘米)

13.【解】原式=

=1-(1-)-()-()-…-()

=1-()

=1-(1-)

14.【解】

=2+<2+×4=3

=2+++(+)+(+)<2++

++=3

=2++++

=2++(+)+(+)+(++)>2++++++=3+

(+)->3

所以至少要选11个数

15.【解】最大的(即第63个数)是

1+3+9+27+81+243=364

第60个数(倒数第4个数)是

364-1-3=360.

16.【解】1÷[(++×2+)÷3]=4(天).’

即5个小队合干需要4天.

【注】第二、四、五3个小队合干也只需要4天,所以在本题中第一、三这2个小队实际上没有人干活,这是不符合实际的。命题者考虑不够周到.

17.【解】若设每一组的平均数均为a

别总和为999a=

a=500

500×3=1500

从而这三组平均教的和为1500.

18.【解】这4个数的公约数必为1111的约数,

而1111=11×101

又11=1十2+3+5

所以,101,2×101,3×101,15×101的和为1111,且最大公约数为101

因此,这四个数的公约数最大是101

19.【解】设右边衣袋的硬币“、比左边的、多2分.右边的、比左边的、

少2分,于是这8枚硬币的钱数正好相等.

由于两边钱数相等,所以左边剩下的、比右边的、多2分,比,右边的、

也多2分,从而、的钱数是2分,、的钱数也是2分,而、的钱数是4分.

由于左边的两枚硬币可以任意选取,而且不可能比2分钱少2分,所以左边每两枚的钱数是4分,左边6枚共12分,两个衣袋共有24分钱.

20.【解】35,37,…,99这33个数中,每一个数都不是另一个数的倍数(因为35×3>99).

另一方面,将1,3,5,…,99这50个数,每一个都写成·t的形式.其中α是0或自然数,t是不能被3整除的自然数,由于1,3,…,99中有17个数是3的倍数,剩下50-17=33

不是3的倍数,所以t的值只有33种.于是从1,3,5,…,99中任取34个数,其中必有两个数的t相同,从而一个数是另一个数的倍数.

因此答案是33.

2014年英语真题含答案

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2010考研英语二真题以及答案

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