Can you please post the complete question ? (the options that is)

well ! this is the question -

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?
a) 0
b) x
c) -x
d) (1/3)y
e) (2/7)y

going by POE.

E for 1<y<7 median = nteger value or multiple of 0.5.
not possible since average is integer.

take y= odd| even and checking,
y=3, (3+5+7)/3 = 5 median = 5
y=4, (2+4+6+8)/4 = 5 median = 5

thanks guys!

I got confused for a bit trying to construct a SET with the given constraints. If the set is given, its easy to do the math, but coming up with construction of a set has always been a challenge thus far.

I still didn't get it. if x and y are both positive and the whole set only includes positive integers, then -x is a value that is never possible for the median. Since we are only talking positive numbers. But nehoo, something looks a bit skewed about the question..

Chufus,

I got tricked by this question too. Here's how I have understood this solution:

-- First realize that the answer choices are all different median, and the question is asking us which median is NOT POSSIBLE. My gut always tends to lean towards quickly solving a problem, but I have reminded myself that whenever I see a CANNOT type question, I will have to go through each answer choices and do "Something" with them. So, I will write down on the top right corner of the scratch pad

MEDIAN CANNOT = __ ?

-- Next realize that all the answer choices are median. So our job is to make a set with the value as Median. If we can't then, well that's our potential answer choice.

1) Start with answer choice a)0

Can we make a set of upto 6 integers with median 0? Lets just make a set and put 0 as the middle value T = { , 0 , } and then fill the remaining. Remember, the set could have 1,2,3,4,5,6,digits( for all odd numbers 1 3 5 median = integer). Since 0 is whole integer, I picked 3 integers in our set T

T={_, 0 , _}
_ could be any value as long as median does not change. IE - once you sort the set after filling in the values, 0 still remains the "middle/median" value.

T = {-1,0,1} -- Works.


2) b) x

Now x is the avg of the set. Lets ask ourselves, can we make a set where mean=median?

Lets see.
T={4,4,4} Mean= X = 4 and Median= 4 as well - Works.

3) c) -x
Can median = -(Mean).
Lets see.

T={-1,-2,9}

PS:- How I made this set was by simply thinking.. I need a -2 in the center and avg as 2. For AVG to be 2 for 3 digits set, the sum needs to be 3*2 =6 and then I just fill in other values.

4) d) 1/3y

Median = 1/3y
==> If there are 3 elements in the set ==> y=3 then

1/3*3 =1=Median. So this is certainly possible. Again, this is not possible if the number of elements in set =2 or any other even number, but our goal is to just come up with 1 value which is possible since this is a "CANNOT Problem"

5) e) 2/7y

No matter what the value of y, the median will never be an integer since y<7.

So this is it. Median can't be 2/7(y).