The problem should read as follows:
If | x - 9/2 | = 5/2, and if y is the median of a set of p consecutive integers, where p is odd, which of the following must be true?
I. xyp is odd
II. xy(p² + p) is even
III. x²y²p² is even
II only
III only
I and III
II and III
I, II, and III
Case 1: No signs changed
x - 9/2 = 5/2
x = 14/2
x = 7.
Case 2: Signs changed on one side
x - 9/2 = -5/2
x = 4/2
x = 2.
Try to prove that the statements DON'T have to be true.
Let p=1, implying that there is only one integer in the set.
Let the set = {1}.
Since the median of the set is 1, y=1.
Statement I: xyp is odd
If x=2, y=1 and p=1, then xyp = 2*1*1 = 2, which is NOT odd.
Since statement I does not have to be true, eliminate C and E.
Statement II: xy(p² + p) is even
If x=2, y=1 and p=1, then xy(p² + p) =(2)(1)(1² + 1) = 4, which is even.
If x=7, y=1 and p=1, then xy(p² + p) =(7)(1)(1² + 1) = 14, which is even.
Hold onto statement II.
Statement III: x²y²p² is even
If x=2, y=1 and p=1, then x²y²p²= 2²1²1² = 4, which is even.
If x=7, y=1 and p=1, then x²y²p²= 7²1²1² = 49, which is NOT even.
Since statement III does not have to be true, eliminate B and D.
The correct answer is
A.
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