Tricky Problem

IMO C

1) not sufficient
A = -1/2 ---> -1/8 > -1/2
A = -2 ----> -8 < -2
2) not sufficient

A = 2 ---> 8 > 2
A = 1/2 ---> 1/8 < 1/2

both together -1 < A < 0 A = -1/2 ---> -1/8 > -1/2 sufficient

mikeCoolBoy wrote:IMO C

1) not sufficient
A = -1/2 ---> -1/8 > -1/2
A = -2 ----> -8 < -2
2) not sufficient

A = 2 ---> 8 > 2
A = 1/2 ---> 1/8 < 1/2

both together -1 < A < 0 A = -1/2 ---> -1/8 > -1/2 sufficient

Not able to understand A = -2 ----> -8 < -2.
If A= -2 cube root of A will be greater than -2. Even take more -ve number the cube root will always be greater than A. So (A) should be the answer.

bjp2008 wrote:

mikeCoolBoy wrote:IMO C

1) not sufficient
A = -1/2 ---> -1/8 > -1/2
A = -2 ----> -8 < -2
2) not sufficient

A = 2 ---> 8 > 2
A = 1/2 ---> 1/8 < 1/2

both together -1 < A < 0 A = -1/2 ---> -1/8 > -1/2 sufficient

Not able to understand A = -2 ----> -8 < -2.
If A= -2 cube root of A will be greater than -2. Even take more -ve number the cube root will always be greater than A. So (A) should be the answer.

As mikeCoolBoy showed, if you have -1/2, -1/2^3=-1/8, which is greater than -1/2,so a greater cube of negative numbers is relevant for NEGATIVE FRACTIONS ONLY
And -2^3=-8, so -8<-2, so a cube will be less than the number for EVERYTHING LESS THAN ONE
And you can also notice than -1^3=-1, so here is another condition that proves that the first statement is INSUFF