VeritasPrepKarishma wrote:

nades09 wrote:

a is an odd integer and a <>-b

Is \(\frac{a^2}{|b+a|}> a-b ?\)

(1) ab=0

(2)\(a^3>a\)

PLease explain the shortest way to solve this

Thanks

NAD

Alternate approach:

I am assuming a <>-b means 'a is not equal to -b' since a + b is in the denominator so it should not be 0.

First thing that strikes me about the question stem is that \(\frac{a^2}{|b+a|}\) is always positive or 0 (if a is 0) while a - b can be +ve, 0 or -ve. I do not know if this observation will help me here but it does give me some level of confidence.

Stmnt 1: ab = 0 means either a = 0 or b = 0 but not both since a should not be equal to -b.

If a = 0, question becomes is 0 > - b. We do not know. If b is positive, 0 will be greater than -b. If b is negative 0 will not be greater than -b. So not sufficient.

Note: I do not need to consider 'if b = 0' since already I have both possibilities, a YES and a NO.

Thats a wrong assumption. Question tells us a is an odd integer.

Stmnt 2: \(a^3>a\). This only happens when a > 1 or when -1 < a < 0. Since a is odd and positive integer, a > 1. This alone is again not sufficient since the answer YES or NO depends on the value of b.

Using both together, b = 0 and a is odd positive integer. Question stem becomes is |a| > a? Answer is definite NO. (They are both equal.) Hence sufficient.

Answer (C).

In equation

\(\frac{a^2}{|b+a|} + b-a > 0\),

the \(b+a\) part can be either positive or negetive, so we must consider both to remove the MOD. The two conditions we get, after applying a little algebra, are:

\(\frac{b^2}{b+a} > 0\) --------------------------------E1

and

\(\frac{b^2 - 2a^2}{b+a} > 0\)-------------------------E2

any statement must satisfy both the equations to be considered good.

S1 gives us b=0 (since a is an odd integer), putting that in E1 and E2 above we get:

E1: LHS = 0, RHS = 0. Answers the question.

E2: \(-2a>0 ?\). Does not answer the question.

S1 Not sufficientS2 gives us a>0, putting that in E1 and E2 above does not answer the question.

S2 Not sufficientPutting b=0 and a>0 in E1 and E2 above:

E1: LHS = 0, RHS = 0. Answers the question.

E2: \(-2a>0 ?\)(where a is positive int). Answers the question.

Both together are sufficient. C

_________________

Respect,

Vaibhav

PS: Correct me if I am wrong.