Pump A can fill a water reservoir in 6 hours, while a drain pipe can empty the full reserv… — JAMB UTME

Pump A can fill a water reservoir in $6$ hours, while a drain pipe can empty the full reservoir in $15$ hours. Pump A is switched on to fill an empty reservoir while the drain pipe is accidentally left open. After $3$ hours, an identical pump, Pump B, is also switched on to assist Pump A while the drain pipe remains open. How many total hours will it take for the reservoir to become completely full?

Answer: 5.625 hours

Answer

The total time required to fill the reservoir completely is

5.625

hours.

To solve multi-stage work and rate problems involving opposing forces (filling vs. draining), calculate the net rate of change per unit of time for each stage. In stage one, Pump A adds

\frac{1}{6}

while the drain removes

\frac{1}{15}

, giving a net rate of

\frac{1}{10}

per hour. In 3 hours,

\frac{3}{10}

of the reservoir is filled, leaving

\frac{7}{10}

. In stage two, adding identical Pump B increases the filling rate to

2 \times \frac{1}{6} = \frac{1}{3}

. Subtracting the drain rate

\frac{1}{15}

yields a net rate of

\frac{4}{15}

per hour. Dividing the remaining

\frac{7}{10}

\frac{4}{15}

gives

2.625

hours. Adding the initial 3 hours yields a total of

5.625

hours.

Step-by-Step Solution

Determine individual hourly rates

Pump A rate =

+\frac{1}{6}

reservoir/hr, Drain rate =

-\frac{1}{15}

reservoir/hr

Rate is the reciprocal of the time required to complete the full job.

Calculate net rate and progress for the first 3 hours

Net rate =

\frac{1}{6} - \frac{1}{15} = \frac{1}{10}

reservoir/hr. Progress in 3 hours =

3 \times \frac{1}{10} = \frac{3}{10}

of the reservoir.

Only Pump A and the drain pipe are active during the initial 3-hour period.

Calculate remaining fraction of reservoir to be filled

Remaining portion =

1 - \frac{3}{10} = \frac{7}{10}

The total capacity of the reservoir is represented by

1

whole unit.

Calculate the combined rate after Pump B is added

New net rate =

\frac{1}{6} + \frac{1}{6} - \frac{1}{15} = \frac{1}{3} - \frac{1}{15} = \frac{4}{15}

reservoir/hr

Pump B is identical to Pump A, so its rate is also

\frac{1}{6}

reservoir/hr.

Find additional time needed and total time elapsed

Additional time =

\frac{7/10}{4/15} = \frac{7}{10} \times \frac{15}{4} = \frac{21}{8} = 2.625

hours. Total time =

3 + 2.625 = 5.625

hours.

Time equals remaining work divided by combined net rate, then added to elapsed time.

Key Concept

Work-Rate and Simultaneous Operations (Combined Filling and Emptying Rates)

Estimated Time:2m 30s

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