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?

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

answer: 5100 just subtract

step-by-step explanation: