Q:

Working at a constant rate, pump x pumped out half of the water in a flooded basement in 4 hours. then pump y was started and the two pumps, working independently at their respective constant rates, pumped out the rest of the water in 3 hours. how many hours would it have taken pump y, operating alone at its own constant rate, to pump out all of the water that was pumped out of the basement

Accepted Solution

A:
Answer:24 hours.Step-by-step explanation:This is one of those problems where the easiest way to do is just to reason it out without trying to devise a formula. Call the water in the basement = wTherefore w/2 = 4*x where x is the rate of the pump. The statement means that 1/2 the water in the basement is pumped out by x in 4 hours.Now x and y work together. They (together ) pump out the rest of the water which is also w/2w/2 = 3x + 3y              Now you can equate the two values for w/24x = 3x + 3y                Subtract 3x from both sides.4x-3x = 3x-3x + 3yx = 3yNow you have to be very careful in interpreting what you found. It takes pump y three times as long to do anything as it does x.So here's the reasoning part. x by itself will empty the basement in 8 hours. It takes x four hours to get 1/2 the water out.It will take pump y 3 times as long to pump out that basement. 3*8 = 24 hours