A municipal water treatment facility uses three pumps—Pump , Pump , and Pump —to fill a main reservoir. Working alone at their respective constant rates, Pump can fill the reservoir in 20 hours, Pump in 30 hours, and Pump in 40 hours.
All three pumps begin filling the empty reservoir simultaneously. After 4 hours, Pump shuts off. Pumps and continue operating together for another 8 hours before Pump is also shut off. Pump continues to run alone until the reservoir is completely filled.
How many total hours does it take to fill the reservoir from the start of the process until it is completely filled?
Cevap: 16 hours
Cevap
The total time required to fill the reservoir from start to finish is 16 hours.
Converting the individual completion times into work rates per hour (1/20, 1/30, and 1/40 of the reservoir per hour) allows us to determine the combined output per phase. In the first 4 hours, all three pumps fill 52/120 of the reservoir. In the next 8 hours, Pumps Q and R fill 56/120 of the reservoir. This leaves 12/120 (or 1/10) of the job remaining. Pump R, working at a rate of 1/40 per hour, takes 4 hours to complete the final 1/10. Adding all three durations (4 + 8 + 4) gives a total of 16 hours.
Adım Adım Çözüm
Anahtar Kavram
Work Rate and Combined Work in Multi-Stage Processes
Alternatif Yöntem
Assume a convenient total reservoir volume equal to the least common multiple of the times: 120 units. Pump P produces 6 units/hr, Pump Q produces 4 units/hr, and Pump R produces 3 units/hr. Stage 1 (4 hrs): 4 × (6 + 4 + 3) = 52 units. Stage 2 (8 hrs): 8 × (4 + 3) = 56 units. Total filled = 108 units. Remaining = 12 units. Stage 3 (Pump R alone): 12 / 3 = 4 hrs. Total time = 4 + 8 + 4 = 16 hours.
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