Is a^3-b^20?

Question

a^3 - b^2 > 0? 
1. b > 0
2. a > |b|

chetan2u · Answer

Hi,

I do not think the answer should be B, it has to be E and this is WHY...
 a^3-b^2>0 ...
Here b^2 will always be POSITIVE, so -b^2 will be NEGATIVE..
a^3 will depend on a...

Let's see the statements..
1) b>0..
We have b^2 which is always POSITIVE, so b >0 really doesn't help much..
Insufficient 
2) a>|b|
This tells us that a is positive and also GREATER than numeric value of b..
If a >1, a^3 will be greater than b^2 and so ans will always be YES..
But if a is between 0 and 1, and will vary..
Say a=0.2 and b=0.1..... 
Hence a>|b| but a^3=0.2^3=0.008 and b^2=0.1^2=0.01.... 0.01>0.008 or b^2>a^3
Now if a=0.9 and b=0.1 ....... a>|b| and a^3>b^2
So different answers 
Insufficient..

Combined nothing new ..
Insufficient

E

mcmoorthy · Answer

IMHO B
Statement 1: b>0 doesn't tell anything about a. So insufficient.

Statement 2: a>|b| tells us a is positive and its value greater than absolute value of b.

Therefore a^2 >b^2 and since a is positive a^3>b^2
a^3-b^2>0
Sufficient 
Ans B

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mcmoorthy · Answer

Clearly missed the a<1 scenario.
E is right

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rajarshee · Answer

Sorry folks, as you pointed out the answer provided was incorrect. Picked it up from a compiled set of questions in a pdf and so not a very reliable source for an answer.

exc4libur · Answer

(1) insufic no info about a (2) insufic a≥0 and a>|b| a,b=(3,2): 27-4>0 a,b=(1/4,1/5): 1/64-1/25<0 (1/2) insufic a≥0 and a>|b| a,b=(3,2): 27-4>0 a,b=(1/4,1/5): 1/64-1/25<0 ans (E)

Is a^3-b^2>0?

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