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Pepe
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Is x^2 - y^2 even? (1) x + y is odd (2) x - y is odd 21 Nov 2010, 10:27
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

75% (hard)

Question Stats:

41% (01:18) correct 59% (01:26) wrong based on 105 sessions

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Is x^2 - y^2 even?

(1) x + y is odd

(2) x - y is odd

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My answer would be D. But the OA is not D.

If x+y is odd I couldn't think for a possibility to get x² - y² even.
The same for x-y is odd. Again, I can't find a possibility to get x² - y² even.
So the answer has to be D, as for both statements alone we can say "NO", x² - y² is not even.

Did I miss something? :

Source: Veritas Prep GMAT Simulator­
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C
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Bunuel
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Is x^2 - y^2 even? (1) x + y is odd (2) x - y is odd 21 Nov 2010, 10:38
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Pepe
Is x² - y² even?

1) x + y is odd

2) x - y is odd

My answer would be D. But the OA is not D.

If x+y is odd I couldn't think for a possibility to get x² - y² even.
The same for x-y is odd. Again, I can't find a possibility to get x² - y² even.
So the answer has to be D, as for both statements alone we can say "NO", x² - y² is not even.

Did I miss something? :?:

Source: Veritas Prep GMAT Simulator
Show more

Note that we are not told that \(x\) and \(y\) are integers.

Is x^2-y^2 even?

(1) x + y is odd --> now, if \(x=0.5\) and \(y=0.5\) then \(x^2-y^2=(x-y)(x+y)=0=even\) but \(x=2\) and \(y=1\) then \(x^2-y^2=(x-y)(x+y)=3=odd\). Two different answers, not sufficient.

(2) x - y is odd --> now, if \(x=1.5\) and \(y=0.5\) then \(x^2-y^2=(x-y)(x+y)=2=even\) but \(x=2\) and \(y=1\) then \(x^2-y^2=(x-y)(x+y)=3=odd\). Two different answers, not sufficient.

(1)+(2) \(x+y=odd\) and \(x-y=odd\) --> \(x^2-y^2=(x-y)(x+y)=odd*odd=odd\). Sufficient.

Answer: C.

P.S. There is one more thing: \(x+y=odd\) doesn't mean that \(x^2-y^2=integer\), for example if \(x=0.7\) and \(y=0.3\) --> \(x^2-y^2=(x-y)(x+y)=0.4\neq{integer}\);
Similarly \(x-y=odd\) doesn't mean that \(x^2-y^2=integer\), for example if \(x=1.3\) and \(y=0.3\) --> \(x^2-y^2=(x-y)(x+y)=1.6\neq{integer}\).

Hope it's clear.
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Is x^2 - y^2 even? (1) x + y is odd (2) x - y is odd 21 Nov 2010, 12:29
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Bunuel, thanks a lot.

I was aware that I have also to proof non-integers, but unfortunately failed to figure out the right ones. :(
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Is x^2 - y^2 even? (1) x + y is odd (2) x - y is odd 21 Nov 2010, 12:42
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Bunuel
Pepe
Is x² - y² even?

1) x + y is odd

2) x - y is odd

My answer would be D. But the OA is not D.

If x+y is odd I couldn't think for a possibility to get x² - y² even.
The same for x-y is odd. Again, I can't find a possibility to get x² - y² even.
So the answer has to be D, as for both statements alone we can say "NO", x² - y² is not even.

Did I miss something? :?:

Source: Veritas Prep GMAT Simulator

Note that we are not told that \(x\) and \(y\) are integers.

Is x^2-y^2 even?

(1) x + y is odd --> now, if \(x=0.5\) and \(y=0.5\) then \(x^2-y^2=(x-y)(x+y)=0=even\) but \(x=2\) and \(y=1\) then \(x^2-y^2=(x-y)(x+y)=3=odd\). Two different answers, not sufficient.

(2) x - y is odd --> now, if \(x=1.5\) and \(y=0.5\) then \(x^2-y^2=(x-y)(x+y)=2=even\) but \(x=2\) and \(y=1\) then \(x^2-y^2=(x-y)(x+y)=3=odd\). Two different answers, not sufficient.

(1)+(2) \(x+y=odd\) and \(x-y=odd\) --> \(x^2-y^2=(x-y)(x+y)=odd*odd=odd\). Sufficient.

Answer: C.

P.S. There is one more thing: \(x+y=odd\) doesn't mean that \(x^2-y^2=integer\), for example if \(x=0.7\) and \(y=0.3\) --> \(x^2-y^2=(x-y)(x+y)=0.4\neq{integer}\);
Similarly \(x-y=odd\) doesn't mean that \(x^2-y^2=integer\), for example if \(x=1.3\) and \(y=0.3\) --> \(x^2-y^2=(x-y)(x+y)=1.6\neq{integer}\).

Hope it's clear.
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x^2 - y^2 = (x + y) * (x - y)
product of 2 nos (nos - because there is no possible combination for both x+y & x-y to be odd) will always be even
hence C...
i don't think we need to go to non-integers in this case...
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Is x^2 - y^2 even? (1) x + y is odd (2) x - y is odd 21 Nov 2010, 12:49
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vaibhavw
x^2 - y^2 = (x + y) * (x - y)
product of 2 nos (nos - because there is no possible combination for both x+y & x-y to be odd) will always be even
hence C...
i don't think we need to go to non-integers in this case...
Show more

Actually when \(x+y=odd\) and \(x-y=odd\) then \(x^2-y^2=(x-y)(x+y)=odd*odd=odd\) and not even as you stated. Also it's not clear what you mean by the red part.

We need to consider non-integer values for x and y in (1) and (2) because if we were told that x and y are integers then knowing that x+y=odd in (1) or x-y=odd in (2) would be sufficeint to say that x^2-y^2=odd and the answer would be D, not C.
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Is x^2 - y^2 even? (1) x + y is odd (2) x - y is odd 19 May 2024, 13:10
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