it seems to be easy

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it seems to be easy

by ricaototti » Thu Sep 04, 2008 8:29 pm
If a and b are integers and a is not equal to b, is ab > 0?

(1) a^b > 0

(2) a^b is a non-zero integer

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Re: it seems to be easy

by aspire750 » Thu Sep 04, 2008 10:53 pm
ricaototti wrote:If a and b are integers and a is not equal to b, is ab > 0?

(1) a^b > 0

(2) a^b is a non-zero integer
Is the ans C?

Lets plug values

Stm1. assume a=2, b=3, ab>0 but when a=2, b=-3 in this case ab<0 so, insufficient

Stm2. a^b is non zero, means it could be negative or positive similarly it is insufficient.

Combining 1+2, it is sufficient, when a=2, b=-3,

Hence, C

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Re: it seems to be easy

by Stuart@KaplanGMAT » Fri Sep 05, 2008 12:13 am
ricaototti wrote:If a and b are integers and a is not equal to b, is ab > 0?

(1) a^b > 0

(2) a^b is a non-zero integer
Starting with the question:

is ab > 0?

In other words, do a and b have the same sign?

Let's examine (1):

a^b > 0

a and b could both be positive, which gives us a "yes" answer.

However, a could be negative and b could be an even positive number, giving us a "no" answer.

We can get both a "yes" and a "no": insufficient.

(If you prefer to pick numbers, we could have picked:

a=2 and b=3, and 2*3 IS greater than 0; or

a=-2 and b=2, and -2*2 is NOT greater than 0.)

(2) we can pick the exact same numbers to show that (2) gives us both a "yes" and a "no": insufficient.

Since the same set of numbers satisfy both (1) and (2), the correct answer is (E), not enough information.
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