Cedagmat wrote:74. Is k^(m + n) > 0?
(1) k < 0.
(2) k^(m - n) < 0.

OA is E.
I first simplified the question to: K^m(K^n)>0, which means that either K^m and K^n are both negative, or K^m and K^n are both positive.

S1: k<0, insufficient since it does not tell us about m or n.
S2: Simplify into K^m/K^n<0. This means that K^m or K^n are less than 0. I took this to mean that since K^m or K^n can be negative, that K^m(K^n) will always give a negative answer and thus not be less than zero.

What is the problem with my reasoning here?

THere are two possibilities for k^(m+n) to be > 0.
First, If k >0 m+n can have any value i.e. m+n>0 or m+n<0, the result k^(m+n) will always be > 0
check - 2^(2+1) >0 ; 2^(-2-1) = 1/8>0
Second, If k<0 then for k^(m+n) to be >0 m+n should be even only. m+n cannot be odd as (-ve)^even power will always be >0.
Now check if the statements provide any such information.
1) k <0 , refer to the second scenario, if k < 0 we need additional info about m+n i.e. whether m+n is even or odd.
Not sufficient.
2) k^(m-n) < 0 this statement tells us that k < 0; Any integer/fraction raised to -ve or +ve integer will only be -ve if the base is -ve. This stmt provides same info as in stmt 1. Not sufficient.
Combining 1 & 2 too provides no additional info.
Hence E.

I think you shouldn't have split the question stem as the base is same, it involved too much thinking.

beat_gmat_09 wrote:

Cedagmat wrote:74. Is k^(m + n) > 0?
(1) k < 0.
(2) k^(m - n) < 0.

OA is E.
I first simplified the question to: K^m(K^n)>0, which means that either K^m and K^n are both negative, or K^m and K^n are both positive.

S1: k<0, insufficient since it does not tell us about m or n.
S2: Simplify into K^m/K^n<0. This means that K^m or K^n are less than 0. I took this to mean that since K^m or K^n can be negative, that K^m(K^n) will always give a negative answer and thus not be less than zero.

What is the problem with my reasoning here?

THere are two possibilities for k^(m+n) to be > 0.
First, If k >0 m+n can have any value i.e. m+n>0 or m+n<0, the result k^(m+n) will always be > 0
check - 2^(2+1) >0 ; 2^(-2-1) = 1/8>0
Second, If k<0 then for k^(m+n) to be >0 m+n should be even only. m+n cannot be odd as (-ve)^even power will always be >0.
Now check if the statements provide any such information.
1) k <0 , refer to the second scenario, if k < 0 we need additional info about m+n i.e. whether m+n is even or odd.
Not sufficient.
2) k^(m-n) < 0 this statement tells us that k < 0; Any integer/fraction raised to -ve or +ve integer will only be -ve if the base is -ve. This stmt provides same info as in stmt 1. Not sufficient.
Combining 1 & 2 too provides no additional info.
Hence E.

I think you shouldn't have split the question stem as the base is same, it involved too much thinking.

Stat. (2) does tell you something that stat. (1) didn't - that m-n is not even - otherwise a -ve base when raised to an even power becomes positive.
If m and n were integers, cedagmat's analysis would've been correct, because stat. (2) would then have indicated that m-n (which is an integer) must be odd, and so m+n will also have been odd, and k^(odd power) would've definitely been negative.
The problem is that m and n do not have to be integers. Think of an m and n that:
when added give you an even power (turning the question stem into a positive power), but
when subtracted do not give you an even power (to satisfy stat. (2): Take m=1.5, n=0.5, k=-2.

Stat. (2): (-2)^(1.5-0.5) = (-2)^1 = -2, which is indeed <0 so this satisfies stat. (2), BUT
question stem: (-2)^(1.5+0.5) = (-2)^2 = 4 - which IS >0, and thus provides the counter example needed to show stat.(2) (and the combination) insufficient.

Good question - tough one.

Dani@MasterGMAT wrote:

beat_gmat_09 wrote:

Cedagmat wrote:74. Is k^(m + n) > 0?
(1) k < 0.
(2) k^(m - n) < 0.

OA is E.
I first simplified the question to: K^m(K^n)>0, which means that either K^m and K^n are both negative, or K^m and K^n are both positive.

S1: k<0, insufficient since it does not tell us about m or n.
S2: Simplify into K^m/K^n<0. This means that K^m or K^n are less than 0. I took this to mean that since K^m or K^n can be negative, that K^m(K^n) will always give a negative answer and thus not be less than zero.

What is the problem with my reasoning here?

THere are two possibilities for k^(m+n) to be > 0.
First, If k >0 m+n can have any value i.e. m+n>0 or m+n<0, the result k^(m+n) will always be > 0
check - 2^(2+1) >0 ; 2^(-2-1) = 1/8>0
Second, If k<0 then for k^(m+n) to be >0 m+n should be even only. m+n cannot be odd as (-ve)^even power will always be >0.
Now check if the statements provide any such information.
1) k <0 , refer to the second scenario, if k < 0 we need additional info about m+n i.e. whether m+n is even or odd.
Not sufficient.
2) k^(m-n) < 0 this statement tells us that k < 0; Any integer/fraction raised to -ve or +ve integer will only be -ve if the base is -ve. This stmt provides same info as in stmt 1. Not sufficient.
Combining 1 & 2 too provides no additional info.
Hence E.

I think you shouldn't have split the question stem as the base is same, it involved too much thinking.

Stat. (2) does tell you something that stat. (1) didn't - that m-n is not even - otherwise a -ve base when raised to an even power becomes positive.
If m and n were integers, cedagmat's analysis would've been correct, because stat. (2) would then have indicated that m-n (which is an integer) must be odd, and so m+n will also have been odd, and k^(odd power) would've definitely been negative.
The problem is that m and n do not have to be integers. Think of an m and n that:
when added give you an even power (turning the question stem into a positive power), but
when subtracted do not give you an even power (to satisfy stat. (2): Take m=1.5, n=0.5, k=-2.

Stat. (2): (-2)^(1.5-0.5) = (-2)^1 = -2, which is indeed <0 so this satisfies stat. (2), BUT
question stem: (-2)^(1.5+0.5) = (-2)^2 = 4 - which IS >0, and thus provides the counter example needed to show stat.(2) (and the combination) insufficient.

Good question - tough one.

Thanks Dani.
Didn't think of non-integers.
Very helpful.