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a is a positive integer, k and m are integers. If k>m, is

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a is a positive integer, k and m are integers. If k>m, is [#permalink]

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New post 01 Aug 2008, 04:40
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

a is a positive integer, k and m are integers. If k>m, is a^k>a^m?
(1) a^k<1
(2) a^m<1
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.

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New post 01 Aug 2008, 04:56
I think E, because a can be equal to 1.
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New post 01 Aug 2008, 05:04
but, if a=1....., then how does it help ??a^m,and a^k is what important... hmm i feel.
we want to know a^m and a^k relation. so even if a=1. since from given conditions, m and k are both -ve... that is what we should consider right? :?:

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New post 01 Aug 2008, 05:17
"a is a positive integer, k and m are integers. If k>m, is a^k>a^m?"

If a = 1, then a^k = a^m.
If a > 1, then a^k > a^m.

So, unless we have more info on a, we cannot anwer the question.
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Re: exponents.... [#permalink]

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New post 01 Aug 2008, 05:28
id pick D for this one

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New post 01 Aug 2008, 05:56
there could be 3 options

1. m>0
then k>0
and a^k > a^m allways

2. m<0
which goes into 2 options
2.1. k>0. a^k > a^m allways
2.2 k<0 a^k < a^m allways

now lets see what else we know

1) a^k<1
means k<0 for all a>0
thus a^k < a^m
sufficient

2)a^m<1
means m<0 for all a>0
it stays unclear for k; k >0 or k e <0
insufficient

1) and 2)
k<0
m<0
k>m
then
a^k < a^m allways

my bet is C

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Re: exponents.... [#permalink]

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New post 01 Aug 2008, 08:32
Nerdboy wrote:
I think E, because a can be equal to 1.



a is a positive integer, k and m are integers. If k>m, is a^k>a^m?
(1) a^k<1
(2) a^m<1

State 1:
a^k<1 -- a can't be equal to 1..
if a=1 for any value of K( -ve or +ve) a^k<1 == 1<1 (so "a" not equal to 1" ) a>1
a must be >1 from statement (1)
a^k<1 --> a>1 and k --> -ve integer
k>m must be -ve integer
( k=-2 m=-3)
a^k>a^m --> 1/a^2 > 1/a^3

sufficient

State 2:
also sufficient use the above logic.


D for me.

what is OA
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New post 01 Aug 2008, 09:05
hmmm..... the ans is D.
so i think we can uncork a champagne bottle huh!!! :lol:
or if someone feels otherwise , they are most welcome to stop the party...
cos suggestions and rectifications are part of the learning process..
so for guys who said D hmmmmm welll
i'd got it as D too
:bouncer :drunk :band :cool :2gunfire: :banana :punk :woohoo :plasma :wave :shot

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Re: exponents.... [#permalink]

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New post 01 Aug 2008, 09:14
BUT SURESH, IF a=1 and k=-2 and m=-3,
a^k<1 and also a^m<1.
so why cant a be 1?

This isn't true.

If you have a negative exponent, such as \(2^{-2}\) that is the same as \(\frac{1}{2^2}\) or \(\frac{1}{4}\).

If a = 1 and k = -2, then you have \(\frac{1}{1^k} = \frac{1}{1}\) so this cannot be < 1. same applies for a^m.
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New post 01 Aug 2008, 09:17
hmmm... allen ... thanks..
sorry suresh.. guess u were right abt that one.. :lol:
cheers!!!

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Re: exponents.... [#permalink]

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New post 01 Aug 2008, 11:19
arjtryarjtry wrote:
a is a positive integer, k and m are integers. If k>m, is a^k>a^m?

to answer this we wanna make sure if:

a=0
a=1
k=m=0 (not possible since k>m)


(1) a^k<1 and a is a positive integer

either k is -ve or a =0
if K is -ve and k>m then Is a^k>a^m True
if a =0 then Is a^k>a^m False

INSUFF


(2) a^m<1 and a is a positive integer

either m is -ve or a=0 INSUFF
if M is -ve and k>m then Is a^k>a^m True
if a =0 then Is a^k>a^m False

INSUFF

Together

either a=0 or M and K are -ve

if K and M are -ve and k>m then Is a^k>a^m True
if a =0 then Is a^k>a^m False

INSUFF



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.


IMO E

BTW what is OA?

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Re: exponents.... [#permalink]

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New post 02 Aug 2008, 23:44
arjtryarjtry wrote:
a is a positive integer, k and m are integers. If k>m, is a^k>a^m?
(1) a^k<1
(2) a^m<1
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.


Please let me know what are the sources of these questions which u post ?

a^m < a^k if m<k if a>1
if 0<a<1 then a^m > a^k if m<k

(1) does not indicate whether a<1 or not since either k <0 or a <1
if k<0 then a^m < a^k
if k>0 and a<1 then a^m > a^k => insufficient

(2) does not indicate whether a<1 or not since either m <0 or a <1
if m<0 then a^m < a^k
if m>0 and a<1 then a^m > a^k => insufficient

(1) and (2) taken together :
says that a^k<1 and a^m<1

this is possible when a <1 and k,m are positive or a> 1 and k,m both are < 0 in both the cases => opposite results
hence A,B,C,D eliminated
IMO E
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Re: exponents....   [#permalink] 02 Aug 2008, 23:44
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