Kaplan:Is m^3>1?

[shubhamkumar]

If p is an integer and m = -p +(-2)^p, is m^3>1?
1) P is even
2) P^3 <= -1

OA C

[diebeatsthegmat]

If p is an integer and m = -p +(-2)^p, is m^3>1?
1) P is even
2) P^3 <= -1

OA C

with this kind of problem, i prefer to add number and count...
statement1: p is even so p could be 0,2,4,6...
if P=0 so m=1
if P=2 so m=2 >>>insufficient
statement 2 P must be odd, P could be -1,-2,-3
if P=-1 m=-3/2
if P=-2 m=9/4
>>>>insuficient
(1+2) P must be even and #0

[sanju09]

If p is an integer and m = -p +(-2)^p, is m^3>1?
1) P is even
2) P^3 <= -1

OA C

m^3 > 1 only if m > 1, so the question reduces to

If p is an integer, is -p + (-2) ^p > 1?

I. If p is even, let's say p = 0, then -p + (-2) ^p = 1 (NOT > 1); let's say p = -2, then -p + (-2) ^p = 2¼ (> 1); Insufficient

II. If p^3 ≤ -1, then p ≤ -1. In this case, if p = -1, then -p + (-2) ^p is less than 1, and if p = -2, -3, ...then -p + (-2) ^p is always greater than 1. Insufficient

Combining I & II

Since p is an even negative integer, hence p cannot be -1, and [spoiler]hence -p + (-2) ^p > 1. Sufficient

Take C[/spoiler]