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PLs help : Kaplan Q !

Expert replies
Source: — Data Sufficiency |

by sahilchaudhary » Mon Dec 09, 2013 1:19 am
We have to found whether m^3>1. If m^3>1 then we have a yes, if m^3<=1, then we have a no.

1. Put p=0 in the equation m= -p+(-2)^p, you get m=1, so m^3=1, so the answer is no.
Put p=2 in the equation m= -p+(-2)^p, you get m=2, so m^3=8, so the answer is yes.
Since we do not have a definite answer, statement 1 is not sufficient.

2. p^3<=-1 can be written as:
a. p^3=-1, so p=-1. Put p=-1 in the equation m= -p+(-2)^p, you get m=3/2, so m^3>1, so the answer is yes.
b. p^3<-1, so p<-1. Put p=-2 (since p could only be an integer) in the equation m= -p+(-2)^p, you get m=9/4, so m^3>1, so the answer is yes.
Since we have a definite yes, statement 2 is sufficient.

So, the answer is B .
Sahil Chaudhary
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by himu » Mon Dec 09, 2013 5:06 am
Hi Sahil,

Thanks for your ans & that is the correct ans.

but my Q is zero is neither even nor odd

so how can we put p=0 ?


Please suggest

thanks & Regards
~Himu.
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by Brent@GMATPrepNow » Mon Dec 09, 2013 6:39 am
himu wrote:Hi Sahil,

Thanks for your ans & that is the correct ans.

but my Q is zero is neither even nor odd

so how can we put p=0 ?


Please suggest

thanks & Regards
~Himu.
Hi Himu,

Allow me to jump in and say that zero is, indeed, even.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by sahilchaudhary » Mon Dec 09, 2013 6:59 am
himu wrote:Hi Sahil,

Thanks for your ans & that is the correct ans.

but my Q is zero is neither even nor odd

so how can we put p=0 ?


Please suggest

thanks & Regards
~Himu.
Hi Himu,

Any number divisible by 2 is even. So, 0 is even.
Any number not divisible by 2 is odd.

I would recommend everyone to read page 108 of OG 13 at least once before going for the actual GMAT exam.
Sahil Chaudhary
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