Which of the following CANNOT be the median of the 3 . . .

M7MBA wrote:Which of the following CANNOT be the median of the 3 positive integers x, y, and z?

A. x
B. z
C. x+z
D. (x+z)/2
E. (x+z)/3

When the 3 positive integers are arranged in ascending order, the median will be the middle value.
Answer choice C suggests that the median will be greater than x AND greater than z
Since the median of 3 numbers cannot be greater than 2 of the numbers, the correct answer is C

Cheers,
Brent

M7MBA wrote:Which of the following CANNOT be the median of the 3 positive integers x, y, and z?

A. x
B. z
C. x+z
D. (x+z)/2
E. (x+z)/3

The OA is C.

Experts how can I show that C can not be the median? Is there a formula?

The median of three positive integers must be one of the integers. We see that the quantities in choices A, B, and D can be the median of the integers; since choices A and B are one of the integers, and the quantity in choice D (which is always between x and z) could equal the third integer (which is y). Thus they can’t be the correct answer and it narrows down to either choice C or E. Let’s analyze them.

If x + z is the median of the three integers, then neither x nor z can be the median since x + z is greater than either of them. This leaves us that the only possible candidate for the median is y. Therefore, y = x + z. However, since y = x + z, y is greater than both of x and z. Thus y is greatest integer and it can’t be the median. Therefore, we see that it’s impossible for x + z to be the median (and we will leave the readers as an exercise to show that (x + z)/3 can be also the median).

Answer: C