问一道JJ里的题目 没有思路

XIAOTINGTING

44.set X has 5 numbers, and the average is greater than medium; set Y has 7 numbers, and the average is greater than medium. set Z combined x and y, is the average greater than medium?
(1) medium of set X is greater than medium of set Y (2)average of set X is greater than average of set Y

mymengming

顶，这题偶也没有思路

yijijuechen

This is my analysis:

A: the average of X.
B:the average of Y.

Z: the average of Z.

a: the medium of X.

b: the medium of Y.

z: the medium of Z.

according to (1): we get a>z>b

according to (2): 5A+7B=12Z,we get A>Z>B

But we do not know if B>a, or if A>b.

so we can not get a solution. the correct selection is (E)

But if (1) is changed to a<b, and (2) remain unchanged, we get:

a<z<b,   A>Z>B, and B>b, so the outcome is Z>z.

If (2) is changed to A<B, and (1) remain unchanged. We get:

a>z>b, A<Z<B, and A>a, so the outcome is Z>z

paulfrank11

顶一下，我刚才用了最笨的方法代数法，可还是想听听高人的详解

诗琳

why we can get  a>z>b  from (1)?  thank you for your explaination

mymengming

以下是引用yijijuechen在2006-12-9 3:46:00的发言：

This is my analysis:

A: the average of X.
B:the average of Y.

Z: the average of Z.

a: the medium of X.

b: the medium of Y.

z: the medium of Z.

according to (1): we get a>z>b

according to (2): 5A+7B=12Z,we get A>Z>B

But we do not know if B>a, or if A>b.

so we can not get a solution. the correct selection is (E)

But if (1) is changed to a<b, and (2) remain unchanged, we get:

a<z<b,   A>Z>B, and B>b, so the outcome is Z>z.

If (2) is changed to A<B, and (1) remain unchanged. We get:

a>z>b, A<Z<B, and A>a, so the outcome is Z>z

同意这种解法

minititan

以下是引用诗琳在2006-12-9 17:34:00的发言：
why we can get  a>z>b  from (1)?  thank you for your explaination

a>b，对于集合Z来说，那么比a大的最多5个，所以z<a；同样，比b小的最多5个，所以z>b。

mccoist

以下是引用诗琳在2006-12-9 17:34:00的发言：
why we can get  a>z>b  from (1)?  thank you for your explaination

a>b，对于集合Z来说，那么比a大的最多5个，所以z<a；同样，比b小的最多5个，所以z>b。

高深啊～～

请问是否可以由此推论：如果两个集合相加，新的medium在原先两个medium之间，新的average在原先两个average之间？

XIAOTINGTING

有点晕

需要仔细琢磨一下

不过高人就是高人