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S99-10

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S99-10 [#permalink]

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New post 16 Sep 2014, 01:53
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

72% (01:09) correct 28% (01:07) wrong based on 50 sessions

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New post 16 Sep 2014, 01:53
Official Solution:


For the sum of \(x\) and \(y\) to be even, the two variables must both be even or both be odd.

Statement (1): INSUFFICIENT. We know that at least one of the variables is even, but we do not know whether they are both even.

Statement (2): INSUFFICIENT. We know that \(x = y \times \text{(some even integer)}\). Since an even integer multiplied by any other integer is also even, we know that \(x\) is even. However, we do not know whether \(y\) is even.

Statements (1) & (2): INSUFFICIENT. Using both statements, we only know that \(x\) is even. Meanwhile, y could be even, but it does not have to be. As a result, we cannot determine whether the sum of \(x\) and \(y\) is even or odd. All of these results can be confirmed by picking numbers.


Answer: E
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Re: S99-10 [#permalink]

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New post 15 Aug 2017, 01:11
Shouldn't answer be B, because if we divide x + y with y, we get x/y + 1, we know x/y is even, so even + 1 will be odd and therefore, B should be sufficient to answer the Q.
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New post 15 Aug 2017, 01:16
ravjai81 wrote:
Shouldn't answer be B, because if we divide x + y with y, we get x/y + 1, we know x/y is even, so even + 1 will be odd and therefore, B should be sufficient to answer the Q.



We should find whether x+y is even, not (x + y)/y is even.
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S99-10 [#permalink]

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New post 16 Aug 2017, 01:23
If \(x\) and \(y\) are positive integers, is \(x + y\) even?


(1) \(xy\) is even

Let x = 12 & y =3....Answer to question is N0

Let x = 12 & y =2....Answer to question is Yes

Insufficient

(2) \(\frac{x}{y}\) is even

Let x = 12 & y =3....Answer to question is N0

Let x = 12 & y =2....Answer to question is Yes

Insufficient

Combining 1 &

Use same examples in both statements. We still can reach a solution

Answer: E
S99-10   [#permalink] 16 Aug 2017, 01:23
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