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DS INTEGER

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Source: — Data Sufficiency |
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by gmatclubmember » Thu Sep 22, 2011 8:24 am
[email protected] wrote:Q: IF M AND N ARE INTEGERS, AND X= 3^N AND Y= 3^M .IS X >2Y ?

A: N= M+1
B: N= 2M
Consider A:
x>2y=> 3^n > 2.3^(n-1) => 3^n > 2/3 * 3^n ---- sufficient.
Consider B:
x>2y => 3^2m > 2.3^m => insufficient.

So answer is A.

Cheers
Ami/-
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by cans » Thu Sep 22, 2011 8:29 am
X= 3^N AND Y= 3^M . IS X >2Y ??
3^n > 2*3^m ??
A) n=m+1. thus lhs = 3*3^m. as 3>2 sufficient
B) n=2M.
3^(2m) > 2* 3^m ?? or 3^m * 3^m > 2*3^m or 3^m>2???
if m=0, 1>2 false
if m=1, 3>2 true
Insufficient
IMO A
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by [email protected] » Thu Sep 22, 2011 11:33 pm
cans wrote:X= 3^N AND Y= 3^M . IS X >2Y ??
3^n > 2*3^m ??
A) n=m+1. thus lhs = 3*3^m. as 3>2 sufficient
B) n=2M.
3^(2m) > 2* 3^m ?? or 3^m * 3^m > 2*3^m or 3^m>2???
if m=0, 1>2 false
if m=1, 3>2 true
Insufficient
IMO A
THANKS FOR YOUR POST.
BUT WE DON'T KNOW THAT BOTH M AND N ARE POSITIVE OR NEGATIVE INTEGER. IF BOTH ARE NEGATIVE THAN I THINK THE ANSWER WOULD BE NOT SAME AS WHEN BOTH ARE POSITIVE
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by ikaplan » Thu Sep 22, 2011 11:47 pm
If you plug-in negative values for M in the equality 3^m*3>2/3 *3^m you will realize that the equality works out both for negative numbers as well
"Commitment is more than just wishing for the right conditions. Commitment is working with what you have."
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