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a^2 - b^2

Expert replies
by arora007 » Wed Jun 23, 2010 11:25 am
Is the value of a^2 - b^2 greater than the value of (3a+3b)(2a-2b) ?
1) b<a
2) a<-1

OA is C, dunno know why i got this as A probably had been going too fast

The above is from peterson master the GMAT 2010 , test 2
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Source: — Data Sufficiency |

by hemanthquartz1 » Wed Jun 23, 2010 1:48 pm
In stm 1

lets take a= 1 b = -2
Here b<a
then a^2 - b^2 > 6 (a^2 - b^2)

take a = 6 b=5
Again b<a
In this case a^2 - b^2 < 6 (a^2 - b^2)

Insuff

Stm 2 nothing is given about b so insuff

Combining 1 and 2

Lets a = -2 b = -3 which satisfies both statements
so in this care a^2 - b^2 is always greater than than 6 (a^2 - b^2)
Suff
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by bharatishiv » Thu Jun 24, 2010 7:15 pm
hemanthquartz1 wrote:In stm 1

lets take a= 1 b = -2
Here b<a
then a^2 - b^2 > 6 (a^2 - b^2)

take a = 6 b=5
Again b<a
In this case a^2 - b^2 < 6 (a^2 - b^2)

Insuff

Stm 2 nothing is given about b so insuff

Combining 1 and 2

Lets a = -2 b = -3 which satisfies both statements
so in this care a^2 - b^2 is always greater than than 6 (a^2 - b^2)
Suff


Further simplification of the q would lead to the final q as : Is b^2 > a^2

As per stmt 1: b < a

If b = 3, a = 5, then b^2 < a^2

but, if b = -4 and a = -3, then b^12 > a^2

so, insuff.

er stmt

As per stmt 2: a < -1
nothing is given about b, so insufficient.

combining, we see that b^2 > a^2 and the ans. to q is YES.
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by hemanthquartz1 » Thu Jun 24, 2010 8:56 pm
Good explanation. Thanks
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by Haaress » Fri Jun 25, 2010 9:54 am
First off, IMO this is a level 700 question.

Is the value of a^2 - b^2 greater than the value of (3a+3b)(2a-2b) ?
1) b<a
2) a<-1

First, its a good idea to rephrase such a question.

So the rephrase of a^2 - b^2 > 6( a^2 - b^2 ) ? is b^2 > a^2?

Stmt 1. b < a . b^2 < a^2 if both are positives , however, b^2 > a2 is both are negatives. So Insuff.

Stmt 2. If a < -1 , so a^2 > 1 , thus the question would be Is b^2 > 1. Yes if it is an integer ( both +ve and -ve integers) but No if it is a fraction. Therefore, Insuff.

Combining the stmts.
a is a negative integer (-1) is stmt 2 and when a is negative in stmt 1, then b^2 > a2 so suff.
In other words, it only works when both are negative integers.
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