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# Which one of (a^-1+b^-1)^-1 and (a^-1*b^-1)^-1 is greater?

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Which one of (a^-1+b^-1)^-1 and (a^-1*b^-1)^-1 is greater? [#permalink]  02 Feb 2005, 09:37
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Which one of (a^-1+b^-1)^-1 and (a^-1*b^-1)^-1 is greater?
1) a=2b
2) a+b>1
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Re: DS [#permalink]  02 Feb 2005, 21:44
pb_india wrote:
Which one of (a^-1+b^-1)^-1 and (a^-1*b^-1)^-1 is greater?
1) a=2b
2) a+b>1

(a^-1+b^-1)^-1 =>ab/(a+b)
(a^-1*b^-1)^-1 => ab

1) a=2b => not sufficient.

2) a+b>1 => not sufficient.

togather also insufff...........
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1/a+1/b=(a+b)/ab
S=(1/a+1/b)-1/ab=1/ab*(a+b-1)

(I) a=2b
S=1/2b^2*(3b-1) depending on b's value. insufficient
(II) a+b>1
a+b-1>0
depending on the signs of a and b
Combine
2b^2>0
a+b-1>0
S>0
Sufficient

(C)
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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

Last edited by banerjeea_98 on 03 Feb 2005, 09:27, edited 1 time in total.
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ab could be a negative number, baner.
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HongHu wrote:
ab could be a negative number, baner.

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

explain pls......... i believe from i and ii, ab cannot be a -ve.
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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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Ans is C [#permalink]  04 Feb 2005, 07:50
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.

Thanks,

Anirban
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Folaa3 wrote:
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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