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

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.

Good explanation. Thanks

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.