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Bunuel
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Machine A can do a certain job in 8 hours. Machine B can do the same 07 Feb 2016, 10:12
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C
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65% (02:34) correct 35% (02:51) wrong based on 398 sessions

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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.
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C
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Machine A can do a certain job in 8 hours. Machine B can do the same 08 Feb 2016, 20:21
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Ans: C (1:00pm)

From 9:00-11:00c(2 hrs)

A can do 1/8 Job/hr. so in 2 hours he can do 1/4 job.
B can do 1/10 job/hr so in 2 hrs. he can do 1/5 Job.
C can do 1/12 job/hr so in 2 hours he can do 1/6 job.

Total = 1/4+1/5+1/6 = 37/60 Job

Now balance work needs to be done by B and C.
Balance Work= 23/60 Job

Combined rate of B and C = 1/10+1/12 = 11/60 job/hr.
So they will do 22/60 work in 2 hrs (approx. to the balance work of 23/60)

Hence the Job will be completed 2hrs. after 11:00 i.e. @ 1:00pm

Ans: C (1:00pm)
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Machine A can do a certain job in 8 hours. Machine B can do the same 13 Feb 2016, 12:52
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Bunuel
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.
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The numbers in this problem make it a little challenging, so it takes a bit longer to do the regular rate = work / time method:

Combined rate of machines A,B,C: (15+12+10) / 120 = 37 /120
Time for all three machines = 2 hrs
Work completed= (37/120) * 2 = 74 / 120 :?

Work left: 120 - 74 = 46 /120 :|

Combined rate of machines B and C: ( 6 + 5) / 60 = 11/60
Time to complete remaining work: (46/120) / (11/60) = 46/120 * 60/11 = 23/11 =~ 2hrs
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Machine A can do a certain job in 8 hours. Machine B can do the same 27 Nov 2017, 12:35
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Bunuel
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.
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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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Machine A can do a certain job in 8 hours. Machine B can do the same 20 Mar 2019, 16:31
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Bunuel
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.
Show more

LCM Method will help to solve this one easily .

LCM of work = 120 units.

A/hr = 15
B/hr =12
C/hr = 10.

Their 1 hour work = 37 units.

9 to 11 : 2 hrs.

37 * 2 = 74 units.

work left = 120 - 74 = 46 units.

46 units will be produced by B and C.

B + C / hr = 12 + 10 = 22 units.

46/22 = 2.00

11 + 1 + 1 = 1.00 pm.

C is the correct answer.
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Machine A can do a certain job in 8 hours. Machine B can do the same 22 Mar 2019, 11:29
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Rate Time


A: 1/8 2 hrs
B: 1/10 t(Time taken by all 3)
C: 1/12 t(Time taken by all 3)



Rate X time = Work

(1/8)*2 + (1/10)*t + (1/12)*t = 1(1 work is done i.e completing 1 task)

t ~ 4 hours

9am + 4 hours = 1pm.
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Machine A can do a certain job in 8 hours. Machine B can do the same 07 Aug 2019, 04:47
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Let total work be equal to 120 (LCM of 8 10 and 12).
Efficiency of A = 120/8 = 15 units per hour
Efficiency of B = 120/10 = 12 units per hour
Efficiency of C = 120/12 = 10 units per hour.
Combined efficiency = 37 units per hour
There are 2 hours between 9:00 and 11:00
Work completed in 2 hours = 37 x 2 = 74 units
Leftover = 120 -74 = 46 units.
Time required to complete = 46/(12+10) = 46/22= 23/11 = approximately 2 hours.

Time = 11:00+ 2 hours = 1:00pm
Therefore the right answer is C.
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Machine A can do a certain job in 8 hours. Machine B can do the same 08 Aug 2019, 08:51
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Job done in 2 hrs = (1/4 + 1/5 + 1/6)
Job left = 1 - (1/4 + 1/5 + 1/6)
After 11 am, more time required = (1 - (1/4 + 1/5 + 1/6))/(1/10 + 1/12) = Approx 2 hrs (2.11)
So Answer 1 PM
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Machine A can do a certain job in 8 hours. Machine B can do the same 25 Oct 2025, 08:58
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