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

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

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18 Dec 2010, 13:01
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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
[Reveal] Spoiler: OA

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Ajit

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Re: If a and b are positive [#permalink]

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18 Dec 2010, 13:16
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ajit257 wrote:
If a and b are positive, is (a-1 + b-1)-1 less than (a-1b-1)-1?
(1) a = 2b
(2) a + b > 1

Question: is $$(a^{-1}+b^{-1})^{-1}<(a^{-1}*b^{-1})^{-1}$$? --> $$(\frac{1}{a}+\frac{1}{b})^{-1}<(\frac{1}{ab})^{-1}$$ --> $$\frac{ab}{a+b}<ab$$, as $$a$$ and $$b$$ are positive we can reduce by $$ab$$ and finally question becomes: is $$a+b>1$$?

(1) a = 2b --> is $$3b>1$$ --> is $$b>\frac{1}{3}$$, we don't know that, hence this statement is not sufficient.
(2) a + b > 1, directly gives an answer. Sufficient.

P.S. ajit257 you should type the question so that it's clear which is an exponent, which is subtraction, and so on.
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Re: If a and b are positive [#permalink]

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18 Dec 2010, 13:29
oh apologies ...that means i had a wrong question. Thanks Bunuel
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Ajit

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Re: If a and b are positive [#permalink]

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19 Dec 2010, 20:18
Bunuel wrote:
ajit257 wrote:
If a and b are positive, is (a-1 + b-1)-1 less than (a-1b-1)-1?
(1) a = 2b
(2) a + b > 1

Question: is $$(a^{-1}+b^{-1})^{-1}<a^{-1}*b^{-1}$$? --> $$(\frac{1}{a}+\frac{1}{b})^{-1}<(\frac{1}{ab})^{-1}$$ --> $$\frac{ab}{a+b}<ab$$, as $$a$$ and $$b$$ are positive we can reduce by $$ab$$ and finally question becomes: is $$a+b>1$$?

(1) a = 2b --> is $$3b>1$$ --> is $$b>\frac{1}{3}$$, we don't know that, hence this statement is not sufficient.
(2) a + b > 1, directly gives an answer. Sufficient.

P.S. ajit257 you should type the question so that it's clear which is an exponent, which is subtraction, and so on.

Very short cut way to the solution.. Really Nice...
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Re: If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than [#permalink]

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28 Jul 2013, 04:16
Didn't read the stem properly was simpler than I thought....
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Re: If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than [#permalink]

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12 Dec 2013, 04:10
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Bunuel wrote:
ajit257 wrote:
If a and b are positive, is (a-1 + b-1)-1 less than (a-1b-1)-1?
(1) a = 2b
(2) a + b > 1

Question: is $$(a^{-1}+b^{-1})^{-1}<(a^{-1}*b^{-1})^{-1}$$? --> $$(\frac{1}{a}+\frac{1}{b})^{-1}<(\frac{1}{ab})^{-1}$$ --> $$\frac{ab}{a+b}<ab$$, as $$a$$ and $$b$$ are positive we can reduce by $$ab$$ and finally question becomes: is $$a+b>1$$?

(1) a = 2b --> is $$3b>1$$ --> is $$b>\frac{1}{3}$$, we don't know that, hence this statement is not sufficient.
(2) a + b > 1, directly gives an answer. Sufficient.

P.S. ajit257 you should type the question so that it's clear which is an exponent, which is subtraction, and so on.

Hey Bunuel,

once again, this is a little bit fast for me.

I follow your first and third step to reduce the question, but I don't get the second.

I'd explained myself that $$(\frac{1}{ab})^{-1}$$ = $$1*(\frac{ab}{1})$$ so we have ab on the right side.

But I don't follow what you did do reduce the left side. Could you explain in detail?

Thank you!
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Re: If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than [#permalink]

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12 Dec 2013, 04:14
unceldolan wrote:
Bunuel wrote:
ajit257 wrote:
If a and b are positive, is (a-1 + b-1)-1 less than (a-1b-1)-1?
(1) a = 2b
(2) a + b > 1

Question: is $$(a^{-1}+b^{-1})^{-1}<(a^{-1}*b^{-1})^{-1}$$? --> $$(\frac{1}{a}+\frac{1}{b})^{-1}<(\frac{1}{ab})^{-1}$$ --> $$\frac{ab}{a+b}<ab$$, as $$a$$ and $$b$$ are positive we can reduce by $$ab$$ and finally question becomes: is $$a+b>1$$?

(1) a = 2b --> is $$3b>1$$ --> is $$b>\frac{1}{3}$$, we don't know that, hence this statement is not sufficient.
(2) a + b > 1, directly gives an answer. Sufficient.

P.S. ajit257 you should type the question so that it's clear which is an exponent, which is subtraction, and so on.

Hey Bunuel,

once again, this is a little bit fast for me.

I follow your first and third step to reduce the question, but I don't get the second.

I'd explained myself that $$(\frac{1}{ab})^{-1}$$ = $$1*(\frac{ab}{1})$$ so we have ab on the right side.

But I don't follow what you did do reduce the left side. Could you explain in detail?

Thank you!

Sure.

$$(\frac{1}{a}+\frac{1}{b})^{-1}$$;

$$(\frac{b+a}{ab})^{-1}$$;

$$\frac{ab}{b+a}$$.

Does this make sense?
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Re: If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than [#permalink]

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12 Dec 2013, 05:29
Bunuel wrote:

Sure.

$$(\frac{1}{a}+\frac{1}{b})^{-1}$$;

$$(\frac{b+a}{ab})^{-1}$$;

$$\frac{ab}{b+a}$$.

Does this make sense?

Yeah, now I see it. Guess my head was just overloaded with math --> it's really clear now! Thanks!
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Re: If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than [#permalink]

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09 Sep 2016, 10:39
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Re: If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than   [#permalink] 09 Sep 2016, 10:39
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