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# Is x > k? (1) 2^x 2^k = 4 (2) 9^x 3^k = 81 A. Statement

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Director
Joined: 01 Apr 2008
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Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Is x > k? (1) 2^x 2^k = 4 (2) 9^x 3^k = 81 A. Statement [#permalink]

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22 Mar 2009, 23:28
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Question Stats:

100% (00:01) correct 0% (00:00) wrong based on 7 sessions

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Is x > k?
(1) 2^x • 2^k = 4
(2) 9^x • 3^k = 81
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Kudos [?]: 843 [0], given: 18

Director
Joined: 14 Aug 2007
Posts: 726

Kudos [?]: 212 [0], given: 0

Re: DS: Is x > k ? [#permalink]

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23 Mar 2009, 00:03
Economist wrote:
(1) 2^x • 2^k = 4

common base 2 => 2^(x + k ) = 4 = 2^2
i.e x+k=2
x and k can have different values (0,2)(2,0)(1,1)
insuff

Economist wrote:
(2) 9^x • 3^k = 81

3^2x * 3^k = 3^4
i.e 2x + k = 4
again x and k can have different values : (0,4) (1,2) (2,0)

together

2x+2k=4
2x + k = 4
k = 0, x = 2
Suff

C

Kudos [?]: 212 [0], given: 0

Re: DS: Is x > k ?   [#permalink] 23 Mar 2009, 00:03
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