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14 Aug 2017, 22:19
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B
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:
71% (02:48) correct
29%
(02:48)
wrong
based on 348
sessions
History
Date
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Result
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A and B together complete a piece of job in 10 days, B and C together complete the same piece of job in 15 days, whereas A and C together complete it in 8 days. How many days will it take to complete the job if all of them work together?
A. \(3 \frac{3}{7}\)
B. \(6 \frac{6}{7}\)
C. \(7\)
D. \(7 \frac{23}{35}\)
E. \(7 \frac{6}{7}\)
A. \(3 \frac{3}{7}\)
B. \(6 \frac{6}{7}\)
C. \(7\)
D. \(7 \frac{23}{35}\)
E. \(7 \frac{6}{7}\)
Bunuel
(A+B)'s 1 day job = \(\frac{1}{10}\)
(B+C)'s 1 day job = \(\frac{1}{15}\)
(A+C)'s 1 day job = \(\frac{1}{8}\)
Adding the three equations we get -
\(2\)(A+B+C)'s 1 day job = \(\frac{1}{10}+\frac{1}{15}+\frac{1}{8} = \frac{35}{120}\)
or (A+B+C)'s 1 day job = \(\frac{35}{120}*\frac{1}{2}\) \(=\frac{7}{48}\)
Hence time taken by All to complete the job = \(\frac{48}{7}\) = \(6 \frac{6}{7}\)
Option B
General Discussion
14 Aug 2017, 22:58
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If A and B can together do the piece of work in \(10(2*5)\) days, B and C do the work in \(15(3*5)\) days
, and A and C do the work in \(8(2^3)\) days
Let us assume the work to be \(120(2^3 * 3 * 5)\) units.
Therefore A & B's combined rate is 12 units, B & C's combined rate is 8 units, and A & C's combined rate is 15 units.
Together they would do half of 35(12+8+15) units in a day.
So their combined rate is \(\frac{35}{2}\) units per day.
In order to finish 120 units at that rate, it would require \(120 * (\frac{2}{35}) = \frac{120*2}{35} = \frac{24*2}{7} = \frac{48}{7} = 6\frac{6}{7}\) days(Option B)
, and A and C do the work in \(8(2^3)\) days
Let us assume the work to be \(120(2^3 * 3 * 5)\) units.
Therefore A & B's combined rate is 12 units, B & C's combined rate is 8 units, and A & C's combined rate is 15 units.
Together they would do half of 35(12+8+15) units in a day.
So their combined rate is \(\frac{35}{2}\) units per day.
In order to finish 120 units at that rate, it would require \(120 * (\frac{2}{35}) = \frac{120*2}{35} = \frac{24*2}{7} = \frac{48}{7} = 6\frac{6}{7}\) days(Option B)
