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一道数学题~~求分析~~~谢谢

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楼主
发表于 2010-8-18 23:46:45 | 只看该作者 回帖奖励 |倒序浏览 |阅读模式
Set S consists of five consecutive integers, and set T consists of seven consecutive integers. Is the median of the numbers
in set S equal to the median of the numbers in set T ?
(1) The median of the numbers in set S is 0.
(2) The sum of the numbers in set S is equal to the sum of the numbers in set T.
答案是C
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沙发
发表于 2010-8-19 00:33:48 | 只看该作者
S有5个连续整数, T有7个连续整数
(1) 中数是0,S{-2.-1.0.1.2}
(2)S的和是0,T的和也是0的话,T又有七个连续整数,易得T是-3到3

问两个中数是否相等,当然要两个条件啦 哈哈
板凳
发表于 2010-8-19 00:40:00 | 只看该作者
Another way of solving this::==>

Let S consist of x,x+1,x+2,x+3 & x+4 integers
& T consist of y,y+1,y+2,y+3, y+4, y+5 & y+6 integers.

Now median for S = x+2
& for T = y+3.

Ques asked is :=> are they equal i.e is x+2 = y+3 ?
in other words is x-y = 1 ??

1.
Now St 1 tells us:
x+2 = 0 or x = -2.
So, nos are -2,-1,0,1 & 2.

But it tells nothing about y. So cant say if x-y = 1.
Hence Insufficient.

2.
Now St 2 says that
5x + 10 = 7y + 21
=> 5x-7y = 11
Hence cant say if x-y = 1.

3.
Now when both the statements are combined we get:
5x + 10 = 7y + 21
=> 5(x+2) = 7(y+3)
but x+2 =0
=> 0 = 7(y+3)
=> y=-3.
So now x-y = -2-(-3) = 1

hence option C is the right one!!
地板
发表于 2010-8-19 00:41:21 | 只看该作者
(1)only肯定insufficient的
(2)的话设S的median为x,T的median为y,因为S、T数列为连续整数数列,所以S的sum可以表示为5x,T的sum可以表示为7y,由(2)可知,5x=7y,那么xy有无穷多组解,例如x=7 y=5, x=0 y=0
(1)+(2),由(1)可知x=0,那么由5x=7y可知,y=0
所以x=y sufficient
5#
 楼主| 发表于 2010-8-19 09:55:49 | 只看该作者
非常感谢~~~
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