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If a and b are positive, is (a^(1) + b^(1))^(1) less than (a^(1)*b
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Updated on: 08 Oct 2017, 08:46
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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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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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Re: If a and b are positive, is (a^(1) + b^(1))^(1) less than (a^(1)*b
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18 Dec 2010, 14:16
ajit257 wrote: If a and b are positive, is (a1 + b1)1 less than (a1b1)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, is (a^(1) + b^(1))^(1) less than (a^(1)*b
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12 Dec 2013, 05:10
Bunuel wrote: ajit257 wrote: If a and b are positive, is (a1 + b1)1 less than (a1b1)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 (a^(1)*b
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12 Dec 2013, 05:14
unceldolan wrote: Bunuel wrote: ajit257 wrote: If a and b are positive, is (a1 + b1)1 less than (a1b1)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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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 (a^(1)*b
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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 (a^(1)*b
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21 Feb 2019, 22:31
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Re: If a and b are positive, is (a^(1) + b^(1))^(1) less than (a^(1)*b
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21 Feb 2019, 22:31






