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ajit257
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If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than (a^(-1)*b Updated on: 08 Oct 2017, 08:46
Originally posted by ajit257 on 18 Dec 2010, 14:01.
Last edited by Bunuel on 08 Oct 2017, 08:46, edited 1 time in total.
Renamed the topic and edited the question.
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63% (01:52) correct 37% (02:11) wrong based on 356 sessions

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If a and b are positive, is is \((a^{-1}+b^{-1})^{-1}\) less than \((a^{-1}*b^{-1})^{-1}\)?

(1) a = 2b
(2) a + b > 1
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If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than (a^(-1)*b 18 Dec 2010, 14:16
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ajit257
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

Not sure about the ans.
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Question: is \((a^{-1}+b^{-1})^{-1}<(a^{-1}*b^{-1})^{-1}\)?

Is \((\frac{1}{a}+\frac{1}{b})^{-1}<(\frac{1}{ab})^{-1}\) ?

Is \(\frac{ab}{a+b}<ab\) ? Since \(a\) 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.

Answer: B.

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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If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than (a^(-1)*b 12 Dec 2013, 05:10
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Bunuel
ajit257
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

Not sure about the ans.

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.

Answer: B.

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

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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If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than (a^(-1)*b 12 Dec 2013, 05:14
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Bunuel
ajit257
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

Not sure about the ans.

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.

Answer: B.

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!
Show more

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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If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than (a^(-1)*b 12 Dec 2013, 06:29
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Bunuel


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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Yeah, now I see it. Guess my head was just overloaded with math --> it's really clear now! Thanks!
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If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than (a^(-1)*b 04 Apr 2022, 12:47
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brunel How did the last stage of ab/a+b<ab < become > in a+b>1 ? Thanks
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If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than (a^(-1)*b 05 Apr 2022, 01:58
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brunel How did the last stage of ab/a+b<ab < become > in a+b>1 ? Thanks
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\(\frac{ab}{a+b}<ab\). Since \(a\) and \(b\) are positive, we can reduce by \(ab\):

\(\frac{1}{a+b}<1\). Since \(a\) and \(b\) are positive, we can multiply by \(a+b\):

\(1 < a+b\).

Hope it's clear.
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If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than (a^(-1)*b 13 Apr 2022, 04:18
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Great thanks brunel makes sense now.
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If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than (a^(-1)*b 14 Jun 2025, 01:05
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