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26 Jan 2022, 05:45
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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:
35%
(medium)
Question Stats:
79% (02:39) correct
21%
(03:26)
wrong
based on 150
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If 6 machines of type A can complete a certain job in 15 days, working alone at their respective constant rates, and 9 machines of type B can do the same job in 12 days, then how long would it take 2 machines of type A and 3 machines of type B, working together, to complete the same job?
(A) 36
(B) 30
(C) 24
(D) 20
(E) 18
(A) 36
(B) 30
(C) 24
(D) 20
(E) 18
26 Jan 2022, 09:16
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Finding rate of a single machine A:
\(\frac{6}{x}=\frac{1}{15}\)
\(x = 90\) days for one machine to finish the job.
Finding rate of a single machine B:
\(\frac{9}{x}=\frac{1}{12}\)
\(x = 108\) days for one machine to finish the job.
How long would it take 2 machines of type A and 3 machines of type B, working together, to complete the same job?:
\(\frac{2}{90}+ \frac{3}{108} = \frac{1}{c}\)
\(\frac{12+15}{540} = \frac{1}{c}\)
\(\frac{27}{540} = \frac{1}{c}\)
\(27c = 540\)
\(c = 20\)
Answer D
\(\frac{6}{x}=\frac{1}{15}\)
\(x = 90\) days for one machine to finish the job.
Finding rate of a single machine B:
\(\frac{9}{x}=\frac{1}{12}\)
\(x = 108\) days for one machine to finish the job.
How long would it take 2 machines of type A and 3 machines of type B, working together, to complete the same job?:
\(\frac{2}{90}+ \frac{3}{108} = \frac{1}{c}\)
\(\frac{12+15}{540} = \frac{1}{c}\)
\(\frac{27}{540} = \frac{1}{c}\)
\(27c = 540\)
\(c = 20\)
Answer D
31 Jan 2022, 20:55
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Assume N units of work. Make N as lcm of 6,15,9,12 for ease of calculation. LCM(6,15,9,12) = 180.
So, assume total work to be 180 units.
6 machines of A can do it in 15 day => 1 A can do 2 units/day.
9 machines of B can do it in 12 days => 1 B can do 5/3 units/day
Combined rate of 2 As and 3 Bs: 2A + 3B = 2x2 + 3(5/3) = 9 units/day
There are 180 units, so 180/9 = 20 days for 2A & 3B together to do the work!
So, assume total work to be 180 units.
6 machines of A can do it in 15 day => 1 A can do 2 units/day.
9 machines of B can do it in 12 days => 1 B can do 5/3 units/day
Combined rate of 2 As and 3 Bs: 2A + 3B = 2x2 + 3(5/3) = 9 units/day
There are 180 units, so 180/9 = 20 days for 2A & 3B together to do the work!
