Question

A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its large and small drains are opened together, the tank can be emptied in 2 hours. By itself, it takes the small drain 3 hours longer to empty the tank than it takes the large drain to empty the tank on its own. How much time would it take for each drain to empty the pool on its own?

Answer 1

Let the large and the small drains take x hours and y hours respectively to empty the aquarium pool on their own, respectively. Since it takes the small drain 3 hours longer to empty the tank than it takes the large drain to empty the tank on its own, hence y = x +3…(1).

If the volume of the tank is V units, then the large and the small drains empty V/x and V/y units respectively in 1 hour. In 2 hours, the large and the small drains together empty 2V/x +2V/y units so that 2V/x +2V/y = V or, 2/x+2/y = 1 or 2x+2y = xy …(2).

Now, on substituting y = x+3 in the 2^nd equation, we get 2x+2(x+3) = x(x+3) or, 4x+6 = x²+3x or, x² –x -6 = 0 or, (x+2)(x-3) = 0 so that either x = -2 or, x = 3. Since x cannot be negative, we must have x = 3. Then y = x+3 = 6. Thus, the large and the small drains will take 3 hours and 6 hours respectively to empty the aquarium pool on their own, respectively.