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
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Respect,
Vaibhav
PS: Correct me if I am wrong.