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pb_india
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MA
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EDITED

"C".....altho I am assuming from the eqn that a and b <> 0...as it will make 1/b or 1/a...infinity,
now....we have ab/a+b and ab as 2 eqn

I....if a = 2b....then.....first one becomes 2b/3 and second is 2b^2.....if b < 0....then 2nd one is > ....if b
is >=1....then 2nd is >....if 0<b<1....say b = 1/4.....then first is >....so insuff

II...if a+b > 1.....first ab/a+b is always < than ab, if ab is +ve and > ab is ab is -ve...so insuff...


Combine
a = 2b and a+b > 1
b > 1/3....so b +ve......and a is +ve.....so from II.....we get ab/a+b < ab...suff
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ab could be a negative number, baner.
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banerjeea_98
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HongHu
ab could be a negative number, baner.

Yep...I realized it as soon as I had hit the submit, I edited my post.... :oops:
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MA
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HongHu
ab could be a negative number, baner.



explain pls......... i believe from i and ii, ab cannot be a -ve.
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HongHu
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Sorry, was responding to baner's post before he edited it. ab could be negative for (II).
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Can someone please explain how statement II is sufficient? I have tried all possible means and only statement I seems suffiicient
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Question is ab/a+b > or < ab

(I) a=2b ==> ab is positive as a and b have the same sign. Now a+b could be either >1 or less than 1 so can't say. Eliminate A and D

(II) a+b>1. Don't say about the signs of a and b so ab could be either +ve or -ve. Can't compare. Eliminate B

Both (I) and (II) ==> ab>0 and a+b>0 so ofc ab/a+b < ab.

Answer is C

Thanks,

Anirban
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Folaa3
Can someone please explain how statement II is sufficient? I have tried all possible means and only statement I seems suffiicient


Both I and II are insufficient. But combining them is sufficient.



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