Zaur wrote:
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
(A) 0: {-1,0,2}; y=3; median =0; x>0; Possible
(B) x: {-4x,3x,4x}; y=3; median = x; mean = x; Possible for any x>0; Possible
(C) -x: {-x,-x,-x,7x}; y=4; median = -x, mean = x; possible for x>0; Possible
One quicker way to solve between D & E(It took me 2+ minutes to think about this hack - but that is a learning)Let us think about (D) and (E)
(E) can never be an integer, since y<7. Although that is fine, since median of a set of integers can be a non integer, we need to think about the origin of '7' in the denominator.
Median of a set of integers with even terms can only have 2 in the denominator; So how did 7 end up there?
Let us analyse (D) first:
(D) (1/3)y has the same problem: The origin of 3 in the denominator
A possible case is: y=3 and median = 1;
So, median = 1 = (1/3)*3 = (1/3)y = (D) Therefore, (D) is possible
Thus, the answer is (E)
But why?
In (E), let try to apply the same logic, that we used in (D)
median = 2; and y = some multiple of 7 = 7k
Thus, median = 2 = (2/7)*7k = (2/7)y.
But, according to the question, 0<y<7; Thus, y<7(1)
So, 7 can never be in denominator.
Thus, the answer is (E)
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