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M24-06

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M24-06  [#permalink]

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New post 16 Sep 2014, 01:20
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

61% (00:53) correct 39% (00:56) wrong based on 143 sessions

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Re M24-06  [#permalink]

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New post 16 Sep 2014, 01:20
1
2
Official Solution:


(1) \(ab\) is even. If \(a=b=-2\), then \(a^b=\frac{1}{4} \gt 0\) but if \(a=-2\) and \(b=-1\), then \(a^b=-\frac{1}{2} \lt 0\). Not sufficient.

(2) \(a+b\) is even. If \(a=b=-2\), then \(a^b=\frac{1}{4} \gt 0\) but if \(a=-1\) and \(b=-1\), then \(a^b=-1 \lt 0\). Not sufficient.

(1)+(2) From \(a+b=even\) it follows that \(a\) and \(b\) are either both odd or both even. Now, if both are odd then \(ab=even\) won't hold true, so both \(a\) and \(b\) must be even. Hence \(a^b=negative^{even}=positive\). Sufficient.


Answer: C
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Re: M24-06  [#permalink]

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New post 22 Nov 2014, 19:55
We know that both a and b are negative integers. Therefore, a is being raised to the power of a negative number. This means that for a^b to be positive, b must be even (because a is neg. ex: a=(-2), (-2)^2 =4, while (-2)^3=(-8).

stmt 1: all we know is that at least one of a or b is even, because you need a min. of one even number in a set, for the product of a set of numbers to be even. a, or b, or both could be even. Insuff.

stmt 2: all we know is that a and b are either both odd or both even, because o+o=e and e+e=e but o+e=o.

stmts 1&2: stmt 1 tells us that at least a or b is even. combined with stmt 2, we know that both terms are even, therefore, b is even.
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Re: M24-06  [#permalink]

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New post 13 Apr 2018, 04:16
+1 for option C.

The question requires us to find out if b is even or not.

St 1 : ie a is even , b is even or both are even. We can't be sure of b is even or not. NS
St 2 : i.e both are even or both are odd. NS

Both together : Yes , sufficient ! We now know that both are even. Hence the answer is C
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Re: M24-06 &nbs [#permalink] 13 Apr 2018, 04:16
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