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Guys, you have to write the question stem correctly. Use brackets!

[(4^A)*((1/3)^B)]<1 now all is clear

(1) A=2B; thus, (16^B)*(1/3^B)=(16/3)^B if B=1, then it is >1, if B=-3000, then it is <1
(2) B=4 we have nothing on A, so it is not enough. Consider A=1000 and A=-1000.

Combine: (16/3)^4>1, which is OK. Thus, it is C.
The problem seems so simple...

Re: GMAT-DS: a and b [#permalink]
21 Feb 2008, 13:32

in the question, notwithstanding the correction, there is an explicit statement of A and B being positive integers. if so, then (16/3)^b is always greater than 1. Hence (i) is sufficient. however, for (ii), you end up with 2^{2a}/81; where for a < 7, the given statement is true. while for a > 6, this isn't. so (ii) is insufficient.

Re: GMAT-DS: a and b [#permalink]
23 Feb 2008, 19:33

kanyshkae wrote:

in the question, notwithstanding the correction, there is an explicit statement of A and B being positive integers. if so, then (16/3)^b is always greater than 1. Hence (i) is sufficient. however, for (ii), you end up with 2^{2a}/81; where for a < 7, the given statement is true. while for a > 6, this isn't. so (ii) is insufficient.

Hence A.

yep. has to be A. _________________

-Underline your question. It takes only a few seconds! -Search before you post.

gmatclubot

Re: GMAT-DS: a and b
[#permalink]
23 Feb 2008, 19:33