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Difficulty: Very hardWork Rate and Combined Work

Three industrial machines, AA, BB, and CC, operating independently at their respective constant rates, can complete a production order of 1,2001,200 units in 1212 hours, 1515 hours, and 2020 hours, respectively. All three machines start working simultaneously on the order at 8:00 AM. At 10:00 AM, Machine AA experiences a mechanical failure and stops working permanently. Machine BB and Machine CC continue working together until 11:00 AM, at which point Machine BB's operating efficiency drops by 50%50\% due to overheating, while Machine CC continues at its original rate. At what time will the production order of 1,2001,200 units be fully completed?

  1. 4:48 PMAnswer
  2. B
    3:09 PM
  3. C
    5:12 PM
  4. D
    4:18 PM
  5. E
    6:30 PM

Answer

The production order will be completed at 4:48 PM.
The correct answer of 4:48 PM is derived by dividing the job into three distinct stages. From 8:00 AM to 10:00 AM (2 hours), Machines A, B, and C produce 240 units/hour, yielding 480 units. From 10:00 AM to 11:00 AM (1 hour), Machines B and C produce 140 units/hour, yielding 140 units, leaving 580 units remaining. After 11:00 AM, Machine B operates at 40 units/hour and Machine C at 60 units/hour, producing a combined rate of 100 units/hour. Dividing 580 remaining units by 100 units/hour gives 5.8 hours (5 hours 48 minutes), placing completion at exactly 4:48 PM.

Step-by-Step Solution

1
Determine individual production rates for each machine.
Machine AA produces 1,20012=100\frac{1,200}{12} = 100 units/hour. Machine BB produces 1,20015=80\frac{1,200}{15} = 80 units/hour. Machine CC produces 1,20020=60\frac{1,200}{20} = 60 units/hour.
Establishing hourly rates is necessary to calculate work completed in each time interval.
2
Calculate work completed during Stage 1 (8:00 AM to 10:00 AM).
Combined rate of A+B+C=100+80+60=240A + B + C = 100 + 80 + 60 = 240 units/hour. In 22 hours, work completed =240×2=480= 240 \times 2 = 480 units. Remaining units =1,200−480=720= 1,200 - 480 = 720 units.
All three machines work together for 2 full hours before Machine A breaks down.
3
Calculate work completed during Stage 2 (10:00 AM to 11:00 AM).
Combined rate of B+C=80+60=140B + C = 80 + 60 = 140 units/hour. In 11 hour, work completed =140×1=140= 140 \times 1 = 140 units. Remaining units =720−140=580= 720 - 140 = 580 units.
Machine A is inactive, leaving only Machines B and C operating at full capacity for 1 hour.
4
Calculate the time required for Stage 3 (from 11:00 AM until completion).
Machine BB's reduced rate =80×0.50=40= 80 \times 0.50 = 40 units/hour. New combined rate of B+C=40+60=100B + C = 40 + 60 = 100 units/hour. Time needed =580100=5.8= \frac{580}{100} = 5.8 hours =5= 5 hours and 4848 minutes.
Machine B operates at half speed while Machine C continues at full speed to finish the remaining 580 units.
5
Determine the final completion clock time.
11:00 AM +5+ 5 hours and 4848 minutes =4:48= 4:48 PM.
Adding the duration of the final stage to the starting time of 11:00 AM gives the exact completion time.

Key Concept

Multi-stage combined work problems require calculating individual rates, tracking partial work completed in each phase, and adjusting combined rates whenever active entities or their individual rates change.
Estimated Time:3m 0s
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Topics

Work Rate and Combined Work

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