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If a and b are positive, is (a^-1+b^-1)^-1 less than (a^-1

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Senior Manager
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If a and b are positive, is (a^-1+b^-1)^-1 less than (a^-1 [#permalink]

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New post 12 Jun 2008, 00:05
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If a and b are positive, is (a^-1+b^-1)^-1 less than (a^-1 *b^-1)^-1?
(1) a = 2b
(2) a + b > 1
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The world is continuous, but the mind is discrete

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Director
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Re: DS question [#permalink]

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New post 12 Jun 2008, 02:14
vdhawan1 wrote:
If a and b are positive, is (a^-1+b^-1)^-1 less than (a^-1 *b^-1)^-1?
(1) a = 2b
(2) a + b > 1



the question reads
is \(\frac{1}{\frac{1}{a}+\frac{1}{b}} < ab\)

which can be simplified to is b+a > 1

as seen 1) is insufficient and 2) is sufficient

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Re: DS question [#permalink]

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New post 12 Jun 2008, 02:28
Hi
\(\frac{1}{\frac{1}{a}+\frac{1}{b}} < ab\)

simplifies to
\(\frac{ab}{a+b} < ab\)
since \(a\) and \(b\) are positive we can safely multiply both sides by \(ab\) and \(a+b\)

Hence it can be simplified to

\(a+b>1\)

B is sufficient.

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Manager
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Re: DS question [#permalink]

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New post 12 Jun 2008, 02:36
question to our experts: can anyone tell me why A is not sufficient?

cheers

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Director
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Re: DS question [#permalink]

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New post 12 Jun 2008, 02:42
domleon wrote:
question to our experts: can anyone tell me why A is not sufficient?

cheers


after you simplify the question, it simplifies to is a+b > 1 ?

the stem says a and b are both positive, but it doesnt say that they are integers

so lets say if a=0.4 and b = 0.2 a+b < 1
but if a=6 and b = 3 a+b > 1

so 1) isn't sufficient

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Re: DS question [#permalink]

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New post 12 Jun 2008, 04:52
i said E because although stem says positive numbers, it doesnt mention that they have to be positive INTEGERS ...

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Re: DS question   [#permalink] 12 Jun 2008, 04:52
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If a and b are positive, is (a^-1+b^-1)^-1 less than (a^-1

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