A municipal water treatment facility uses three pumps—Pump P, Pump Q, and Pump R—to fill a… — GMAT

A municipal water treatment facility uses three pumps—Pump $P$ , Pump $Q$ , and Pump $R$ —to fill a main reservoir. Working alone at their respective constant rates, Pump $P$ can fill the reservoir in 20 hours, Pump $Q$ in 30 hours, and Pump $R$ in 40 hours.

All three pumps begin filling the empty reservoir simultaneously. After 4 hours, Pump $P$ shuts off. Pumps $Q$ and $R$ continue operating together for another 8 hours before Pump $Q$ is also shut off. Pump $R$ 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?

Answer: 16 hours

Answer

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.

Step-by-Step Solution

Determine the individual hourly rates of work for each pump.

Pump P completes

\frac{1}{20}

of the job per hour, Pump Q completes

\frac{1}{30}

of the job per hour, and Pump R completes

\frac{1}{40}

of the job per hour.

Work rate is the reciprocal of the total time required to complete one whole task.

Calculate the work completed during Stage 1 (4 hours with all 3 pumps working).

Combined rate =

\frac{1}{20} + \frac{1}{30} + \frac{1}{40} = \frac{6 + 4 + 3}{120} = \frac{13}{120}

. Work done =

4 \times \frac{13}{120} = \frac{52}{120}

Work done equals combined rate multiplied by time spent working together.

Calculate the work completed during Stage 2 (8 hours with Pumps Q and R working).

Combined rate of Q and R =

\frac{1}{30} + \frac{1}{40} = \frac{4 + 3}{120} = \frac{7}{120}

. Work done =

8 \times \frac{7}{120} = \frac{56}{120}

Only Pumps Q and R are active during this second period.

Calculate the remaining fraction of work after Stage 1 and Stage 2.

Total work done so far =

\frac{52}{120} + \frac{56}{120} = \frac{108}{120} = \frac{9}{10}

. Remaining work =

1 - \frac{9}{10} = \frac{1}{10}

Subtracting completed work from the whole (1) leaves the remaining work to be completed.

Calculate the time required for Pump R to finish the remaining work alone.

Time =

\frac{1/10}{1/40} = \frac{40}{10} = 4

hours.

Time equals remaining work divided by the individual rate of Pump R.

Sum the time spent across all stages.

Total time =

4 \text{ hours (Stage 1)} + 8 \text{ hours (Stage 2)} + 4 \text{ hours (Stage 3)} = 16

hours.

The total elapsed time is the sum of the durations of each distinct phase.

Key Concept

Work Rate and Combined Work in Multi-Stage Processes

Alternative Method

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.

Estimated Time:2m 0s

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