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Inequality DS

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Source: — Data Sufficiency |
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by Geva@EconomistGMAT » Tue Aug 23, 2011 3:44 am
knight247 wrote:Is A^x>B^x?
(1)A>B, x>0
(2)1/a<1/b and x<|x|

OA is E
There's a problem with the question: From stat. (1) x is positive, but from stat. (2) x<|x| - which could only happen if x is negative. The statements contradict each other, yet neither of then is sufficient on their own to answer to the question, so you are forced to consider them combined - which is illogical.
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by GmatKiss » Tue Aug 23, 2011 5:21 am
knight247 wrote:Is A^x>B^x?
(1)A>B, x>0
(2)1/a<1/b and x<|x|

OA is E
1 is sufficient!
2 has some problem, its ambiguous

IMO: D if (2) is 1/a<1/b
Please correct if wrong!
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by Geva@EconomistGMAT » Tue Aug 23, 2011 5:24 am
GmatKiss wrote:
knight247 wrote:Is A^x>B^x?
(1)A>B, x>0
(2)1/a<1/b and x<|x|

OA is E
1 is sufficient!
2 has some problem, its ambiguous

IMO: D if (2) is 1/a<1/b
Please correct if wrong!
Plug in x=2, and then two pairs of A and B where A>B:

A=3, B=2: 3^2 is greater than 2^2, so the answer to the question "Is A^x>B^x?" is "yes".
A=-2, B=-3: (-2)^2 is not greater than (-3)^2, so the answer to the question "Is A^x>B^x" is "No".

Thus, stat. (1) is insufficient.
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by knight247 » Tue Aug 23, 2011 8:15 am
Thanks Geva...And there definitely is a problem with this question...But just for the sake of improving my math knowledge, can u please provide an explanation for statement 2. Thanks
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by Geva@EconomistGMAT » Tue Aug 23, 2011 8:53 am
knight247 wrote:Thanks Geva...And there definitely is problem with this question...But just for the sake of improving my math knowledge, can u please provide an explanation for statement 2. Thanks
Begin with an easy plug in for A and B. 1/3 is smaller than 1/2, right? So let A=3 and B=2, so that 1/A is indeed smaller than 1/B. What does that do to the question stem, given a negative x (for example x=-2?)

3^-2 = 1/3^2 = 1/9

2^-2 = 1/2^2 = 1/4

Since 1/9 is smaller than 1/4, this means that A^x is NOT greater than B^x, so the answer is NO.

Fine, now can we fine a Yes, where A^x IS greater than B^x?
a negative A and positive B will assure that 1/a is smaller than 1/b, since negative is smaller than positive. But if we keep x as -2, the even power will return A^x back to the positive side.

B=2
A=-1


1/A is smaller than 1/B, since 1/-1 < 1/2

Go to the question stem, keeping X=-2:

A^x = (-1)^-2 = 1
B^x = 2^-2 = 1/2^2 = 1/4.

1 is greater than 1/4, so here A^x is greater than B^x and the answer is "Yes".

Since stat. (2) allows both a yes and a no answer, it is insufficient.
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