This is not a EASY one arora bhai!!

st 1:
ab=0

a is an odd integer (Given); b = 0

LHS = a^2 / |( a+0)|

= a^2 / a -- ( This will always be a positive no:)

LHS = a

RHS : a-b

: a- 0
: a

Since we dont know teh sign of "a" we cant decide whether LHS is > RHS; for example if a = -1 then LHS is > RHS.
If a= 1, then LHS = RHS
Insufficient

St 2:
a^3 - a >0
a( a^2 -1)>0

which leads to a >0 or a^2 >1-- Here "a" can be postitive or negative.

No idea about b. So insufficient

Combining st1 & st2:

still , we are NOT sure of sign of "a".

So PicK E

yeah machoman... its not easy... and the OA is C

arora007 wrote:yeah machoman... its not easy... and the OA is C

dude,,.. Iam going NUTS now!!

all the DS of GMATfix tripped me!!

arora007 wrote:yeah machoman... its not easy... and the OA is C

dude, i am not doubting patrick..but can u plz post OE for this one..

I guess the last 2-3 steps would have confused...have a closer look...

I am not sure my approach is right.Please correct me I I am wrong.

stmt1,

ab=0

Since we know a is an odd integer ,b=0

LHS=a^2/|a|=a^2/a which is +ve

RHS=a

We dont know sign of a

Insuff

stmt2,

a^3>a

This implies a must be +ve.But we dont know value and sign of b.

Insuff

Combining 1 and 2,

ab=0 and a^3>a

we know a is +ve odd integer and b=0

LHS=a^2/|a|=a^2/a which is +ve

RHS=a,which is +ve.

( a^2/ |b+a| ) = a-b

Suff

Pick C

read the FOIL-> "Is a^2 > a^2-b^2"
where b is zero
the answer is no,
thus it gives the answer to
Is ( a^2/ |b+a| ) > a-b , answer is No. pick C.

e-GMAT wrote:gmatmachoman, your analysis is all correct and you would get the answer as C as well if you plug the condition that a is an "odd integer" in a^3 > a. This means that a is positive and > 1.

Hope that helps.

-Payal

Yes!! Payal...Somehow i didn't see this reply..now i understood y i am missing my target !!

I am too careless!!!