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[小学] 也来问奥数题了

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发表于 22-3-2012 14:35:04|来自:新加坡 | 显示全部楼层 |阅读模式
A language school has 100pupils in which 69%of the pupils study French ,79%study German, 89% study Japanese and 99% study English.Give that at least x%of the students study all four languages,find the value of x.

谢谢。
发表于 22-3-2012 14:41:35|来自:新加坡 | 显示全部楼层
小狮租房
Peter and Jane are to turns to subtract perfect squares from a given whole number and the one who reaches zero first is the winner.If the whole number is 29,and Peter is the first player,what number must he subtract in order for him to definitely win.[Note:4,9and16 are examples of perfect squares]
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发表于 22-3-2012 15:39:41|来自:新加坡 | 显示全部楼层
第一题:

不学法语的100-69=31
不学德语的100-79=21
不学日语的100-89=11
不学英语的100-99=1

至少不学一门语言的人最大可能是31+21+11+1=64

所以答案是100-64=36

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谢谢。  详情 回复 发表于 22-3-2012 20:51
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发表于 22-3-2012 15:42:01|来自:新加坡 | 显示全部楼层
本帖最后由 haloha23 于 22-3-2012 15:42 编辑

哈哈,很久没有动脑筋了(自己的孩子还没有学到这么深),不知现在还有有没有小学生的水平.:L
试一下第一题,答错不要笑咱啊!

Let's take only english and Japanese first. 99 students take english, 89 students take Japanese. so 1 student does not take English and 11 don't take japanese. the 1 student that does not take English has a possibility of taking only Japanese or does not take both jap and eng. So if this student does not take both, the 1 can be deducted from the 89 student who take Japanese. Until now, at least 88 students take both English and Japanese.

Using the same logic, we add in German. Now 88 students take both English and Japanese. 12 students may take only 1 subject or take none at all. 79 students take German. the remaining 21 do not. So we deduct 21 and 12 from the total number of 100 students, leaving a balance of 67 students, who take all 3 subjects.

Now we add in French. 69 take and 31 do not take. Add 31 to 21 and 12 = 64. This is the number of students that study 3 languages or less. 100-64=36%.

At least 36% of students take all 4 subjects. Answer x= 36%

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谢谢你的详细解释。  详情 回复 发表于 22-3-2012 20:52
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发表于 22-3-2012 15:59:11|来自:新加坡 | 显示全部楼层
再来试一下第二题

The answer is 16.

given 29 to start with, if Peter deducts 16, Jane is only left with 13. She can either deduct 4 or 9, and then Peter would be able to deduct the other, leaving a balance of 0. Peter will win at the third try.

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2 5 7 10 12 15 17 20 22 25 are winning positions, keep yourself always at these positions, so from 29 you can either move to 25 or 20, for sure move to 20 is more efficient  详情 回复 发表于 24-3-2012 07:27
1 is also perfect square. who occupy 13-1 =12 will win  详情 回复 发表于 23-3-2012 10:40
我儿子他们做出来的答案也是16。参考答案是9。  详情 回复 发表于 22-3-2012 21:00
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发表于 22-3-2012 21:04:45|来自:新加坡 | 显示全部楼层
Esther picks n composite numbers ,all less than 2007,such that the HCF (Highest Common Factor) of any two of them is 1.Find the largest possible value of n.
(a)10
(b)12
(c)13
(d)14
(e)None of the above
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发表于 23-3-2012 09:10:43|来自:新加坡 | 显示全部楼层
这是几年级的奥数题啊?

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六年级的。  详情 回复 发表于 23-3-2012 10:23
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发表于 23-3-2012 10:25:11|来自:新加坡 | 显示全部楼层
本帖最后由 davidbin 于 23-3-2012 16:43 编辑
yyhclt001212 发表于 22-3-2012 21:04
Esther picks n composite numbers ,all less than 2007,such that the HCF (Highest Common Factor) of an ...


2 3 5 7 11 13 17 19 23 29 31 37 41 || 43 47 53 59 61 67 71 73 79  83  89  97 101
41X43= 1763 <2007
43X47>2007
故分割线画在43前,左右1——1配对,(2X101),....(41X43)共有13组

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another answer: 2^2, 3^2, 5^2 7^2 11^2 13^2 17^2 19^2 23^2 29^2 31^2 37^2 41^2 43^2  详情 回复 发表于 23-3-2012 18:17
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发表于 23-3-2012 10:35:39|来自:新加坡 | 显示全部楼层
男人如树 发表于 22-3-2012 15:39
第一题:

不学法语的100-69=31

反向解法相当漂亮,正向也不难
A=69 B=79,C=89,D=99
least A & B = 69+79 -100 = 48
least  C & D = 89 + 99 -100 = 88
least A&B&C&D = 48 + 88 -100 = 36
Valid formula for 正/反向 is same
A+B+C+D - 300


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谢谢。  详情 回复 发表于 23-3-2012 10:48
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发表于 23-3-2012 10:46:03|来自:新加坡 | 显示全部楼层
老同志过了下数学瘾,该下课去了

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热心人!欢迎继续来上课。  详情 回复 发表于 23-3-2012 11:43
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