Problem Solving

How I and II is true??

For example x consecutive even numbers are 2,4,6,8 then median is 5 that is y is not even. Y is odd.

Again , X consecutive even number can be 2,4,6,8,10 then median is 6 that is y is not odd. Y is even.

So y can be even and also can be odd. This means Y is neither always even nor always odd.

If there are x consecutive even numbers that have a median of y, which of the following could be true?

I: y is even
II: x is even
III: y>x

A) I only
B) II only
C) II and III
D) I and III
E) I, II and III

The question stem asks which of the three statements COULD be true.
If we can find even ONE case in which a statement is true, then that statement COULD be true and must be included in the correct answer choice.

If x=2, then there are two consecutive even integers.
Let the two integers be 2 and 4.
y = median = (2+4)/2 = 3.
In this case, x is even and y>x.
Thus, the correct answer choice must include II and III.
Eliminate A, B, and D.

If x=3, then there are three consecutive even integers.
Let the three integers be 2, 4 and 6.
y = median = 4.
In this case, y is even.
Thus, the correct answer choice must include I.
Eliminate C.

The correct answer is E.