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17 Dec 2012, 06:29
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C
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Difficulty:
5%
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Question Stats:
84% (00:55) correct
16%
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It would take one machine 4 hours to complete a large production order and another machine 3 hours to complete the same order. How many hours would it take both machines, working simultaneously at their respective constant rates, to complete the order?
(A) \(\frac{7}{12}\)
(B) \(1 \frac{1}{2}\)
(C) \(1 \frac{5}{7}\)
(D) \(3 \frac{1}{2}\)
(E) 7
(A) \(\frac{7}{12}\)
(B) \(1 \frac{1}{2}\)
(C) \(1 \frac{5}{7}\)
(D) \(3 \frac{1}{2}\)
(E) 7
16 Jun 2016, 06:06
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Walkabout
We can classify this problem as a “combined worker” problem. To solve this type of problem we should use the formula:
Work (done by worker 1) + Work (done by worker 2) = Total Work Completed
It takes machine one 4 hours to complete a job, so the rate of machine one is ¼. It takes machine two 3 hours to complete a job, so the rate of machine two is 1/3. Since we know they are both working together to complete the job, we can label this unknown time as “t” for each machine during the time that both machines are working together. Since rate x time = work, we can multiply to get the work completed for each machine.
Finally, we can plug our two work values into the combined work formula and determine t. Since the job is completed, the total work completed is 1.
Work (done by worker 1) + Work (done by worker 2) = Total Work Completed
(1/4)t + (1/3)t = 1
Multiplying the entire equation by 12 gives us:
3t + 4t = 12
7t = 12
t = 12/7 = 1 5/7
Answer is C.
17 Dec 2012, 06:33
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Walkabout
The rate of the first machine is 1/4 job per hour;
The rate of the second machine is 1/3 job per hour;
Thus, the combined rate of the machines is \(\frac{1}{4}+\frac{1}{3}=\frac{7}{12}\) job per hour, which means that it takes \(\frac{1}{(\frac{7}{12})}=\frac{12}{7}=1 \frac{5}{7}\) hours both machines to do the job.
Answer: C.
can you help to understand the logic behind this \(\frac{1}{(\frac{7}{12})}=\frac{12}{7}=1 \\
can you help to understand the logic behind this \(\frac{1}{(\frac{7}{12})}=\frac{12}{7}=1 \\
