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If x>y and y is a positive integer, is y even?

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If x>y and y is a positive integer, is y even? [#permalink]

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New post 14 Nov 2017, 07:46
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If \(x>y\) and \(y\) is a positive integer, is \(y\) even?

(1) \(xy\) is odd.
(2) \(x=y+2\)


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[Reveal] Spoiler: OA

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Re: If x>y and y is a positive integer, is y even? [#permalink]

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New post 14 Nov 2017, 08:34
We are not given whether x is an integer or not. But we know that y is a positive integer.

Statement 1: xy is odd. Now, if x is an integer, then y cannot be even, it has to be odd. Because product of two integers is odd ONLY if both are odd.
But if x is not an integer, then we cannot say. Eg - if x is 2.5 and y is 2 then xy = 2.5*2 = 5 (odd). Here y is even still xy is odd.
So Insufficient.

Statement 2: x = y+2. This tells us that x is also an integer, and also either both x/y are odd, or both x/y are even. Insufficient.

Combining the two, we know that both x/y are integers and the product is odd. So BOTH x/y have to be odd. Thus y cannot be even. Sufficient.

Hence C answer
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Re: If x>y and y is a positive integer, is y even? [#permalink]

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New post 14 Nov 2017, 08:37
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chetan2u wrote:
If \(x>y\) and \(y\) is a positive integer, is \(y\) even?

(1) \(xy\) is odd.
(2) \(x=y+2\)


self made


y = even ?

1) xy = od
x = 3, y = 1 => xy = odd NO, y is odd
x = \(\frac{5}{2}\), y = 2 => xy = odd YES, but y is even

2 different answers. Insufficient.

2) x = y + 2
Insufficient. Y could be odd or even

1+2)
xy = odd
x = y + 2 => x is also a positive integer
=> y has to be odd.

Sufficient. C is the answer.
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Re: If x>y and y is a positive integer, is y even?   [#permalink] 14 Nov 2017, 08:37
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