If a is a positive integer (number?) and k>m, whether a^k>a^m?
1) a^k<1
2) a^m<1
IMO : A please elaborate ur answers.
integer
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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..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:
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I might be wrong here, but I'm getting D.
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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-
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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
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
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How did you conclude that a^k>a^m only if k,m>1 ?????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...
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.
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- cubicle_bound_misfit
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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?
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?
Cubicle Bound Misfit