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huongguyen
Joined: 23 May 2020
Last visit: 26 Nov 2022
Posts: 158
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Concentration: Statistics, Finance
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Updated on: 12 Sep 2021, 06:29
Originally posted by huongguyen on 11 Sep 2021, 05:06.
Last edited by Bunuel on 12 Sep 2021, 06:29, edited 2 times in total.
Last edited by Bunuel on 12 Sep 2021, 06:29, edited 2 times in total.
Edited the question.
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
54% (01:38) correct
46%
(01:44)
wrong
based on 117
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If a and b are positive and ab = 1, what is the value of a + b?
(1) \(\frac{2}{\frac{1}{a} + \frac{1}{b}} =1\)
(2) \(a^b = b^a\)
(1) \(\frac{2}{\frac{1}{a} + \frac{1}{b}} =1\)
(2) \(a^b = b^a\)
11 Sep 2021, 22:39
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If a and b are positive and ab = 1, what is the value of a + b?
(1) \(\frac{2}{\frac{1}{a} + \frac{1}{b}} =1\)
\(\frac{2}{\frac{a+b}{ab}} =1\)
\(\frac{2}{\frac{a+b}{1}} =1\)
\(a+b=2\)
Sufficient
(2) \(a^b = b^a\)
Now we know ab=1.
So one solution is \(1^1=1^1\) or a+b=1+1=2
But can there be another solution.
If we decrease any of the two, the other will increase.
So say a<1, b becomes greater than 1.
Can then \((<1)^{>1}=(>1)^{<1}\)? No.
Because \(a^b=x\), where a<1 and b>1 will result in a value lesser than a, that is 0<x<a<1.
But \(b^a=y\), where a<1 and b>1 will result in a value greater than 1, that is y>1.
Can x=y then?. NO
Hence only possibility is a=b=1
Sufficient
D
(1) \(\frac{2}{\frac{1}{a} + \frac{1}{b}} =1\)
\(\frac{2}{\frac{a+b}{ab}} =1\)
\(\frac{2}{\frac{a+b}{1}} =1\)
\(a+b=2\)
Sufficient
(2) \(a^b = b^a\)
Now we know ab=1.
So one solution is \(1^1=1^1\) or a+b=1+1=2
But can there be another solution.
If we decrease any of the two, the other will increase.
So say a<1, b becomes greater than 1.
Can then \((<1)^{>1}=(>1)^{<1}\)? No.
Because \(a^b=x\), where a<1 and b>1 will result in a value lesser than a, that is 0<x<a<1.
But \(b^a=y\), where a<1 and b>1 will result in a value greater than 1, that is y>1.
Can x=y then?. NO
Hence only possibility is a=b=1
Sufficient
D
11 Sep 2021, 22:54
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huongguyen
(1) \(\frac{2}{\frac{1}{a} + \frac{1}{b}} =1\)
2 =a+b can be achieved through simple simplification
Clearly sufficient
(2) \(a^b = b^a\)
This can be satisfied for a=b=1 or a=b=-1 since only positive values can be satisfied therfore
Only possibility that is probable =a=b=1
Therfore IMO D
