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

Answer:

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

Step-by-step 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