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If 72.42=k(24+n/100)
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20 Jul 2021, 11:51
20 Jul 2021, 11:51
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20% (03:27) wrong based on 5 sessions
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If \(72.42=k(24+\frac{n}{100})\), where \(k\) and \(n\) are positive integers and \( n<100\), then \(k+n=\)
(A) 17
(B) 16
(C) 15
(D) 14
(E) 13
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
(A) 17
(B) 16
(C) 15
(D) 14
(E) 13
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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If 72.42=k(24+n/100)
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22 Jul 2021, 00:45
22 Jul 2021, 00:45
2
Carcass wrote:
If \(72.42=k(24+\frac{n}{100})\), where \(k\) and \(n\) are positive integers and \( n<100\), then \(k+n=\)
(A) 17
(B) 16
(C) 15
(D) 14
(E) 13
(A) 17
(B) 16
(C) 15
(D) 14
(E) 13
\(72.42 = k(24 + \frac{n}{100})\)
\(72 + 0.42 = k(24 + \frac{n}{100})\)
\(72 + 0.42 = 24k + \frac{nk}{100}\)
On comparison, \(k = 3\) and \(nk = 42\) i.e. \(n = 14\)
\(k + n = 3 + 14 = 17\)
Hence, option A
gmatclubot
If 72.42=k(24+n/100) [#permalink]
22 Jul 2021, 00:45
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