Question

Working alone at its constant rate, pump X pumped out \small \frac{1}{3} of the water in a pool in 4 hours. Then pump Y started working and the two pumps, working simultaneously at their respective constant rates, pumped out the rest of the water in 6 hours. How many hours would it have taken pump Y, working alone at its constant rate, to pump out all of the water that was pumped out of the pool?

Answer 1

P(y) take 36 h to do the job alone

Explanation:

P(x) quantity of water pump by Pump X and

P(y) quantity of water pump by Pump Y

Then if P(x) pumped 1/3 of the water in a pool in 4 hours

Then in 1 hour P(x) will pump

1/3 ⇒ 4 h

? x ⇒ 1 h x = 1/3/4 ⇒ x = 1/12

Then in 1 hour P(x) will pump 1/12 of the water of the pool

Now both pumps P(x) and P(y) finished 2/3 of the water in the pool (left after the P(x) worked alone ) in 6 hours. Then

P(x) + P(y) in 6 h ⇒ 2/3

in 1 h ⇒ x ?? x = (2/3)/6 x = 2/18 x = 1/9

Then P(x) + P(y) pump 1/9 of the water of the pool in 1 h. We find out how long will take the two pumps to empty the pool

water in a pool is 9/9 ( the unit) then

1 h ⇒ 1/9

x ?? ⇒ 9/9 x = ( 9/9)/( 1/9) ⇒ x = 9 h

The two pumps would take 9 hours working together from the beggining

And in 1 hour of work, both pump 1/9 of the water, and P(x) pump 1/12 in 1 hour

Then in 1 hour P(y)

P(y) = 1/9 - 1/12 ⇒ P(y) = 3/108 P(y) = 1/36

And to pump all the water (36/36) P(y) will take

1 h 1/36

x ?? 36/36 x = (36/36)/1/36

x = 36 h

P(y) take 36 h to do the job alone