It is currently 23 Nov 2017, 21:02

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Current Student
Joined: 11 May 2008
Posts: 554

Kudos [?]: 226 [0], given: 0

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

### Show Tags

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.

Kudos [?]: 226 [0], given: 0

Senior Manager
Joined: 16 Jul 2008
Posts: 289

Kudos [?]: 16 [0], given: 4

### Show Tags

01 Aug 2008, 04:56
I think E, because a can be equal to 1.
_________________

http://applicant.wordpress.com/

Kudos [?]: 16 [0], given: 4

Current Student
Joined: 11 May 2008
Posts: 554

Kudos [?]: 226 [0], given: 0

### Show Tags

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?

Kudos [?]: 226 [0], given: 0

Senior Manager
Joined: 16 Jul 2008
Posts: 289

Kudos [?]: 16 [0], given: 4

### Show Tags

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.
_________________

http://applicant.wordpress.com/

Kudos [?]: 16 [0], given: 4

SVP
Joined: 28 Dec 2005
Posts: 1543

Kudos [?]: 186 [0], given: 2

### Show Tags

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

Kudos [?]: 186 [0], given: 2

Intern
Joined: 24 Jul 2008
Posts: 3

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

### Show Tags

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

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

SVP
Joined: 07 Nov 2007
Posts: 1790

Kudos [?]: 1091 [0], given: 5

Location: New York

### Show Tags

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
_________________

Smiling wins more friends than frowning

Kudos [?]: 1091 [0], given: 5

Current Student
Joined: 11 May 2008
Posts: 554

Kudos [?]: 226 [0], given: 0

### Show Tags

01 Aug 2008, 09:05
hmmm..... the ans is D.
so i think we can uncork a champagne bottle huh!!!
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

Kudos [?]: 226 [0], given: 0

SVP
Joined: 30 Apr 2008
Posts: 1863

Kudos [?]: 624 [0], given: 32

Location: Oklahoma City
Schools: Hard Knocks

### Show Tags

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.
_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a.

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 624 [0], given: 32

Current Student
Joined: 11 May 2008
Posts: 554

Kudos [?]: 226 [0], given: 0

### Show Tags

01 Aug 2008, 09:17
hmmm... allen ... thanks..
sorry suresh.. guess u were right abt that one..
cheers!!!

Kudos [?]: 226 [0], given: 0

Director
Joined: 20 Sep 2006
Posts: 653

Kudos [?]: 135 [0], given: 7

### Show Tags

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?

Kudos [?]: 135 [0], given: 7

VP
Joined: 17 Jun 2008
Posts: 1374

Kudos [?]: 426 [0], given: 0

### Show Tags

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
_________________

cheers
Its Now Or Never

Kudos [?]: 426 [0], given: 0

Re: exponents....   [#permalink] 02 Aug 2008, 23:44
Display posts from previous: Sort by

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

Moderator: chetan2u

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.