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22 Dec 2020, 09:28
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D
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Difficulty:
35%
(medium)
Question Stats:
76% (02:20) correct
24%
(02:41)
wrong
based on 284
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Working together machines A and B can complete a certain job in 6 hours. If it takes machine A 40% longer than it takes A and B together to complete the same job, then how long would it take machine B to complete the job working alone?
(A) 9
(B) 12
(C) 18
(D) 21
(E) 24
(A) 9
(B) 12
(C) 18
(D) 21
(E) 24
22 Dec 2020, 19:43
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sjuniv32
Machine A takes 40% more time than A and B together. So time taken by A alone = 1.4 * 6 = 8.4 Hours.
Work done by A in 1 hour = \(\frac{1}{8.4} = \frac{10}{84} = \frac{5}{42}\)
Work done by B in 1 hour = \(\frac{1}{B}\)
Work done together in 1 hour = \(\frac{1}{6}\)
Therefore \(\frac{5}{42}+ \frac{1}{B} = \frac{1}{6}\)
\(\frac{1}{B} = \frac{1}{6} - \frac{5}{42} = \frac{7 - 5}{42} = \frac{2}{42}= \frac{1}{21}\)
Therefore B = 21 hours
Option D
Arun Kumar
22 Dec 2020, 21:06
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sjuniv32
It takes A: \(6*1.4 = 8.4 = 8+\frac{2}{5} = \frac{42}{5}\) hours to complete the job. A and B together take 6 hours to complete the job. The rate of A+B is \(\frac{1}{6}\) while the rate of A is \(\frac{1}{8.4} = \frac{5}{42}\).
Then B's rate alone would be \(\frac{1}{6} - \frac{5}{42} = \frac{7 - 5}{42} = \frac{2}{42} = \frac{1}{21}\). Thus it takes B 21 days alone.
Ans: D
