i am getting D

If state 1 is true then why isnt statement 2 also true. for both case a becomes a fractions and K>m so a^k<a^m

m i missing something? :roll:

PAB2706 wrote:i am getting D

If state 1 is true then why isnt statement 2 also true. for both case a becomes a fractions and K>m so a^k<a^m

m i missing something? :roll:

IMO yes, take the positive and negative possibilities of k and m, u'll get 4 combinations. by statement 1 only one combination is satisfying it and with statement 2, 2 combinations are satisfying. hence IMO A..

I might be wrong here, but I'm getting D.

-------------------------------------------------

since a is a positive integer, a^k and a^m will be positive regardless of what k and m are. so, the question stem can be manupulated without worrying about the inequality sign.

is a^k > a^m ?
is (a^k)/(a^m) > 1?
is a^(k-m) > 1?

since a is a positive integer, we can have the following outcome:

(i) if a = 1, then regardless of k and m, a^(k-m) = 1. Answer to question stem is NO.

(ii) if a > 1, then the question stem reduces to: is k-m > 0?. we already know from the question that k>m, so this will give the answer YES.

so we really only need to know whether a=1 or a>1 from the statements.

statement 1: a^k < 1

so a must not be 1, because 1 to the power of anything will always give you 1. therefore a must be greater than 1. sufficient.

statement 2: a^m < 1

similar reasoning to statement 1, a must be greater than 1. sufficient.

choose D.

-BM-

a is positive, so no need to worry about negative numbers
k>m so a^k > a^m as long as k,m > 1 otherwise false... but we can say k will deffinitely be >1 if m>1

conversly if k<1 then m must be <1

so if we can answer is m>1 then the answer is deffinitively yes
or if we can answer is k<1 then the answer is deffinitively no
both however are sufficient


so in a)
a^k < 1 therefore k<1 which means m<1 so suff

in b)
a^m < 1 therefore m<1 which doesn't meet either conditions above so insuff

ans (IMO) should be A

m&m wrote:a is positive, so no need to worry about negative numbers
k>m so a^k > a^m as long as k,m > 1 otherwise false...

How did you conclude that a^k>a^m only if k,m>1 ?????

Lets assume a=2
If k=-1 and m=-2 ,

2^-1 = 1/2
2^-2 = 1/4

since 1/2>1/4 even if k,m<1 the equation holds true!!!

If k=1 and m=-1

2^1 = 2
2^-1 = 1/2

since 2>1/2 the original equation holds true!!!

I go with D.

m&M ,

The problem is that

in b,


a^m < 1 hence m<0 as a is +ve int

we already know ,

k>m

so whaever the value of k (positive or negative) a^k is always > than a^m

hence choose D.

what is OA?

thanks for ur replies..
OA is D. I got it when i solved it twice....
thanks