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stonecold
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GRE 1: Q169 V154
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mbaprep2016
answer will be E
x = integer
x = 2p+1 = integer here p can be any n/2 and n can be even or odd , matters alot 2 get cancelled out
if n were even x would odd
if n odd x would be even
Similarly other

Hey mbaprep2016,

as per your statement, "here p can be any n/2 and n can be even or odd , matters alot 2 get cancelled out ", if n is even, 2 at the denominator would cancel the 2 at the numerator, which in turn would make p an integer. but it is given in the question that p is NOT an integer.

hence, E is not the right answer.
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Statement 1 => If X is an integer and X=2p+1 where p is not an integer.
From 1, X is an integer, p is not an integer. Therefore, for X to be integer, p denominator should be 2 and since p is not an integer, Nr of the number should be only odd. Therefore o+o= even

Statement 2 => If Y is an integer and Y=2q where q is not an integer
From 2, q is also an odd no. using above logic.

Method 2 but same logic:

p= X-1/2 therefore X-1 is not divisible by 2, X-1 is odd therefore X is even.

Similarly, q=Y/2. In this case Y is not divisible by 2. Hence, Odd
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