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

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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

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

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New post 18 Dec 2010, 14:29
oh apologies ...that means i had a wrong question. Thanks Bunuel
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Re: If a and b are positive [#permalink]

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New post 19 Dec 2010, 21:18
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

Not sure about the ans.


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.

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.



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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New post 28 Jul 2013, 05:16
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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New post 12 Dec 2013, 05:10
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

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

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New post 12 Dec 2013, 05:14
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

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!


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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New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 12 Dec 2013, 06:29
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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Re: If a and b are positive, is (a^(-1) + b^(-1))^(-1) less than   [#permalink] 09 Sep 2016, 11:39
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