The first pump empties half the pond in 3 hours, so in 1/6 that time (1/2 hour), it empties (1/6)·(1/2) = 1/12 of the pond.
The second pump empties the other 5/12 of the pond in that half hour, so has a pumping rate of (1/2 h)/(5/12 pond) = (6/5 h)/pond.
The second pump could do the entire job alone in 1 hour and 12 minutes.
Answer:
The second pump to drain the pond if it had to do the same job alone is [tex]\frac{3}{5}[/tex] hour or 0.6 hour.
Step-by-step explanation:
Given : A pond is being drained by a pump. After 3 hours, the pond is half empty. A second pump is put into operation, and together the two pumps finish emptying the pond in half an hour.
To find : How long would it take the second pump to drain the pond if it had to do the same job alone?
Solution :
According to question,
Pump A drained in 3 hours.
So, Work done by pump in 1 hour is [tex]\frac{1}{3}[/tex]
Together the two pumps finish emptying the pond in half an hour.
i.e. Pump A + Pump B drained in [tex]\frac{1}{2}[/tex] hour
Work done by pump in 1 hour is [tex]\frac{1}{\frac{1}{2}}=2[/tex] hour
The work done by pump B is [tex]2-\frac{1}{3}[/tex] hour.
The work done by pump B is [tex]\frac{5}{3}[/tex] hour.
So, Pump B drained in [tex]\frac{3}{5}[/tex] hour.
Therefore, The second pump to drain the pond if it had to do the same job alone is [tex]\frac{3}{5}[/tex] hour or 0.6 hour.
