Three water pumps, , , and , operate at constant individual rates to fill a large reservoir. The ratio of the pumping rate of to that of is , and the ratio of the pumping rate of to that of is .
At 8:00 AM, all three pumps begin filling an empty reservoir together. At 10:00 AM, pump shuts down, while and continue operating at their original rates. At 11:00 AM, the operating rate of is decreased by , and the operating rate of is increased by . The two remaining pumps continue at these adjusted rates until the reservoir is completely full at 1:00 PM.
If pump were to fill the empty reservoir working alone at its original constant rate, how many hours would it take?
Answer: 9.1 hours
Answer
It would take pump exactly hours (or when entered numerically) to fill the empty reservoir alone at its original constant rate.
By unifying the given ratios and , we obtain the relative rates , , and . Summing the work done across the three intervals (8–10 AM at rate , 10–11 AM at rate , and 11 AM–1 PM at rate ) yields a total capacity of . Dividing total work by 's original rate of gives hours.
Step-by-Step Solution
Key Concept
Multi-stage work rates, compound ratio unification, and percentage rate adjustments
Estimated Time:3m 0s