Algebra

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by gabriel » Fri Aug 17, 2007 11:49 am
...a^2-b^2 will be greater than 6(a^2-b^2) if a^2-b^2 is negative ... so the question further reduces to is a^2-b^2<0 ..

statement 1 says b<a .. eg b =1 and a =2 .. a^2-b^2 = 4 -1 =3 > 0 ..
eg b = -3 and a = -2 in which case a^2-b^2 = 4 - 9 = -5 < 0 .. so 2 different possibilities .. so this statement alone is insufficient ..

statement 2 says a< -1 .. it says nothing about b .. so insufficient ..

combine the two statements we get b< a and a<-1 so if a = -2 , b can be -3 .. so a^2-b^2 = 4 - 9 = -5 .. which gives us a defnite answer so ans is C ..

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Re: Algebra

by bingojohn » Fri Aug 17, 2007 12:02 pm
Anonymous wrote:Is the value of a^2 - b^2 greater than the value of (3a + 3b)(2a - 2b) ?

(1) b < a

(2) a < -1


I assumed stmt 1 is enough to solve, using the followin method:
Deduced the given question as
Is a^2 - b^2 > 6(a^2 - b^2 )

Taking statement 1 - b<a, which means that the after performing the action of a^2 - b^2 the result is a positive no, which ultimately helps to answer the final question.
Stmt 2 insuff as b is not mentioned.

So, statement 1 is by itself sufficient, But the answer is C, anyone can explain why,how ???
Is a^2 - b^2 > 6(a^2 - b^2 )?

For this to be true, we need to know if a^2 - b^2 is negative or not... in other words,

Is a^2 < b^2? ............. (question)

statement(1): b < a
First consider values such that a^2 > b^2...
say a = 4 and b = -2
=> a^2 = 16 and b^2 = 4
=> a^2 - b^2 = 12
satisfies statement(1) doesn't satisfy question

Then consider values such that a^2 < b^2...
say a = 2 and b = -4
=> a^2 = 4 and b^2 = 16
=> a^2 - b^2 = -12
satisfies statement(1) and satisfies question

statement(1) Not Sufficient

Its quite obvious that statement (2) alone is Not Sufficient.

When you consider statement(1) and (2) together, if a is negative and greater than b, b has to be negative also... such that a^2 < b^2 is always true, and hence the answer is [C].