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If x and y are positive real numbers satisfying x + y = 102,

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sarthakaggarwal
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If x and y are positive real numbers satisfying x + y = 102, [#permalink] New post   Updated on: 08 Sep 2021, 09:51
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63% (02:30) correct 37% (02:12) wrong based on 51 sessions
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If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601(1 + 1/x)(1 + 1/y) is
A 2704
B 2800
C 2700
D 2698
E 2500
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A

Originally posted by sarthakaggarwal on 08 Sep 2021, 09:27.
Last edited by sarthakaggarwal on 08 Sep 2021, 09:51, edited 1 time in total.
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GMATWhizTeam
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Re: If x and y are positive real numbers satisfying x + y = 102, [#permalink] New post   08 Sep 2021, 09:41
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sarthakaggarwal wrote:
If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601(1 + 1x)(1 + 1y) is
A 2704
B 2800
C 2700
D 2698
E 2500


Solution:

Looking at the answer I guess the question is wrongly stated. It should be "find the minimum value of \(2601(1 + \frac{1}{x})(1 + \frac{1}{y})\)"

So, we have \(2601(1 + \frac{1}{x})(1 + \frac{1}{y})\)

\(⇒ 2601(\frac{(1+x)}{x})(\frac{(1+y)}{y})\)

\(⇒ \frac{2601(1+x)(1+y)}{xy}\)

We know that for a fraction to be minimum, its denominator \(xy\) has to be maximum.

We know \(x+y=102\) and to maximize \(xy\), \(x\) has to be equal to \(y\). So, if \(x=y\), then \(x=y=\frac{102}{2}=51\)

\(⇒ (\frac{2601(1+x)(1+y)}{xy})_{min}=(\frac{2601(1+51)(1+51)}{51\times 51})\)

\(⇒ 2704\)

Hence the right answer is Option A.
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sarthakaggarwal
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Re: If x and y are positive real numbers satisfying x + y = 102, [#permalink] New post   08 Sep 2021, 09:53
GMATWhizTeam wrote:
sarthakaggarwal wrote:
If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601(1 + 1x)(1 + 1y) is
A 2704
B 2800
C 2700
D 2698
E 2500


Solution:

Looking at the answer I guess the question is wrongly stated. It should be "find the minimum value of \(2601(1 + \frac{1}{x})(1 + \frac{1}{y})\)"

So, we have \(2601(1 + \frac{1}{x})(1 + \frac{1}{y})\)

\(⇒ 2601(\frac{(1+x)}{x})(\frac{(1+y)}{y})\)

\(⇒ \frac{2601(1+x)(1+y)}{xy}\)

We know that for a fraction to be minimum, its denominator \(xy\) has to be maximum.

We know \(x+y=102\) and to maximize \(xy\), \(x\) has to be equal to \(y\). So, if \(x=y\), then \(x=y=\frac{102}{2}=51\)

\(⇒ (\frac{2601(1+x)(1+y)}{xy})_{min}=(\frac{2601(1+51)(1+51)}{51\times 51})\)

\(⇒ 2704\)

Hence the right answer is Option A.


I didn't noticed the mistake, will correct it
Thank you!
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Crytiocanalyst
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Re: If x and y are positive real numbers satisfying x + y = 102, [#permalink] New post   08 Sep 2021, 11:39
sarthakaggarwal wrote:
If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601(1 + 1/x)(1 + 1/y) is
A 2704
B 2800
C 2700
D 2698
E 2500


2601(1+(x+y) / xy )

=>2601(1+ 102/ xy)

=> 2601 + 2601 * 102 /xy (option E is eleminated )
=>only when x=y=51 is when the value is minimised
=> 2601 +102

Therefore IMO A
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Re: If x and y are positive real numbers satisfying x + y = 102, [#permalink]
08 Sep 2021, 11:39
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