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by fruti_yum » Mon Sep 14, 2009 7:54 am
T is a set of y integers, where 0 < y < 7. If the average of Set T is the positive integer x, which of the following could NOT be the median of Set T?

0
x
-x
(1/3)y
(2/7)y

I dont understand how 0 and -x are even possible choices of the median considering y <0 and average is only positive number.

Can anyone explain???

Source Mgmat

OA after some discussion
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IMO

by xcusemeplz2009 » Mon Sep 14, 2009 9:00 am
IMO E = (2/7)y

value of y can be 1,2,3,4,5,6

so in set T there can be 1,2,3,4,5 or 6 no.

avg of set T is x ( a +ve int)

suppose when
y=3 set T= -1,0,1(for eg); avg=0 and hence x=0 , now median is 0 and x
keeping y=3; t can be -2,-1and 6 ; avg=1;med=-1=-x
similarly d is posible bcoz (1/3 )y= whole no. when y is 3 or 6 so an integer
but for E (2/7)y can not be a whole no. bcoz y<7, so cannot be an int.
It does not matter how many times you get knocked down , but how many times you get up
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by vkb16 » Fri Oct 02, 2009 4:46 am
MGMAT SAYS

''As for answer choice E, there is no possible way to create Set T with a median of (2/7)y. Why? We know that y is either 1, 2, 3, 4, 5, or 6. Thus, (2/7)y will yield a value that is some fraction with denominator of 7.

The possible values of (2/7)y are as follows:
2/7, 4/7, 6/7, 1 1/7, 1 3/7, 1 5/7''

My question is, how did these values pop up?

thanks
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