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ricaototti
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 Posted: Thu Sep 04, 2008 8:29 pm Post subject: it seems to be easy |
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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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aspire750
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| Posted: Thu Sep 04, 2008 10:53 pm Post subject: Re: it seems to be easy |
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| 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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Stuart Kovinsky
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| Posted: Fri Sep 05, 2008 12:13 am Post subject: Re: it seems to be easy |
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| 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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