A pipe can fill a tank in x hours and another pipe can empty it in y hours. Assuming y > x so that the net effect is filling, how long will both pipes together take to fill the tank?

Introduction / Context:
This is the general formula case for one inlet and one outlet. With y > x, the emptying pipe is slower than the filler, so the net rate is positive and the tank fills eventually.

Given Data / Assumptions:

• Fill rate = 1/x tank/hour.
• Empty rate = 1/y tank/hour.
• y > x ⇒ net positive filling.

Concept / Approach:
Net rate = 1/x − 1/y = (y − x) / (xy). Time is the reciprocal of the net rate.

Step-by-Step Solution:
Net rate = 1/x − 1/y = (y − x)/(xy).Time = 1 ÷ ((y − x)/(xy)) = xy / (y − x) hours.

Verification / Alternative check:
Plug numerical values (e.g., x = 5, y = 10): net rate = 1/5 − 1/10 = 1/10 ⇒ time = 10 = xy/(y − x) = 50/5 = 10.

Why Other Options Are Wrong:
(x − y) or (y − x) hours ignore reciprocal; xy/(x − y) has the wrong sign for y > x.

Common Pitfalls:
Forgetting to invert net rate; sign confusion when forming the difference.

Final Answer:
xy / ( y - x) hours