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标题: 费费(二)18\19,(三)11 [打印本页]

作者: 虚无飘渺    时间: 2006-5-9 06:56     标题: 费费(二)18\19,(三)11

(二)18、300 个病人有3种症状A、B、C(每人至少各有一症状),A为35%, B为45%, C为40%, 有且仅有两种症状的为10%, 问有且仅有一种症状的为多少人?

(思路) 假设同时有三种症状的百分比为X,而有且仅有一种症状的百分比为Y,
则35%+45%+40%-10%-3X+X=1--->X=5%

>为什么要还要减个3X?

19、问根号n是否大于100?
(1)根号(n+1)>100
(2)最大的4个数都大于8

>请问(2)的题意怎么理解??

(三)11、已知整数K前所有整数之和为K(K+1)/2,问M和N之间,包含M和N,整数的和为多少(M<N)?

(答案) N(N+1)/2-M(M-1)/2

>我一眼看不出题目隐藏着1-K的和为K(K+1)/2呀……换我做,我的答案会是N(N+1)/2-M(M+1)/2,为什么应该减M(M-1)/2呢?


作者: knight9514    时间: 2006-5-9 13:24

For the 1st question, my explantion is as follows: since symptoms A,B, and C each contain AnBnC(denoted by x), so 3x must be substracted.

Note: AuBuC=A+B+C-AnB-AuC-BnC +AnBnC

Master of Mathemitics.


作者: sunjueliang    时间: 2006-5-10 19:51

For the Q19, (2) is wrong, which should be radical (n-1) is less than 99.

For Q11 in the Part3, I give you hint as follow:

1+2+3+...+K=K(K+1)/2


作者: 虚无飘渺    时间: 2006-5-13 07:37

谢谢!!!!




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