Bunuel wrote:
Machine A can do a certain job in 8 hours. Machine B can do the same job in 10 hours. Machine C can do the same job in 12 hours. All three machines start the job at 9:00 a.m. Machine A breaks down at 11:00 a.m., and the other two machines finish the job. Approximately what time will the job be finished?
A. Noon
B. 12:30 p.m.
C. 1:00 p.m.
D. 1:30 p.m.
E. 2:00 p.m.
We can let the rate of machines A, B, and C be 1/8, 1/10, and 1/12, respectively.
When they all work for two hours they complete the following portion of the job:
(1/8) x 2 + (1/10) x 2 + (1/12) x 2 = 1/4 + 1/5 + 1/6
15/60 + 12/60 + 10/60 = 37/60
So 23/60 is left to be done after 11:00 am.
The combined rate of machines B and C is 1/10 + 1/12 = 6/60 + 5/60 = 11/60.
Thus, the time for machines B and C to finish the job is:
(23/60)/(11/60) = 23/60 x 60/11 = 23/11 = 2 1/11 ≈ 2 hours
Thus, the job was completed at about 11:00 am + 2 hours = 1:00 pm.
Answer: C
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