BTGmoderatorDC wrote:A tank has two water pumps Alpha and Beta and one drain Gamma. Pump Alpha alone can fill the whole tank in x hours, and pump Beta alone can fill the whole tank in y hours. The drain can empty the whole tank in z hours such that z>x. When the tank was empty, pumps Alpha and Beta started pumping water in the tank and the drain Gamma simultaneously was draining water from the tank. When finally the tank is full, which of the following represents the amount of water in terms of the fraction of the tank which pump Alpha pumped into the tank?

(A) yz/(x+y+z)
(B) yz/(yz + xz - xy)
(C) yz/(yz + xz + xy)
(D) xyz/(yz + xz - xy)
(E) (yz + xz - xy)/yz

Let x = 2 hours, y = 3 hours and z = 4 hours.
Let the tank = 12 gallons.

Since Alpha takes 2 hours to fill the 12-gallon tank, Alpha's rate = w/t = 12/2 = 6 gallons per hour.
Since Beta takes 3 hours to fill the 12-gallon tank, Beta's rate = w/t = 12/3 = 4 gallons per hour.
Since Gamma takes 4 hours to empty the 12-gallon tank, Gamma's rate = w/t = 12/4 = 3 gallons per hour.

When all 3 pumps work together -- Alpha and Beta increasing the volume by 6 gallons per hour and 4 gallons per hour, Gamma reducing the volume by 3 gallons per hour -- the net gain per hour = 6+4-3 = 7 gallons.
Since the net gain for all 3 pumps = 7 gallons per hour, and Alpha's rate alone = 6 gallons per hour, the fraction attributed to Alpha = 6/7.

The correct answer must yield a value of 6/7 when x=2, y=3 and z=4.
Only B works:
yz/(yz + xz - xy) = (3*4)/(3*4 + 2*4 - 2*3) = 12/14 = 6/7.

The correct answer is B.

Work done by Alpha in 1 hour = 1/x
Work done by Beta in 1 hour = 1/y
Work done by Gamma in 1 hour = 1/z
Total work done by 3 pumps in 1 hour = 1/x + 1/y + 1/z = (zy+xz-xy)/(xyz)

Alpha's contribution to this work will be

(1/x)/((zy+xz-xy)/(xyz))= (yz)/(zy+xz-xy).

Hence, the correct answer is B