A pipe can fill a tank in $x$ hours and another pipe can empty it in $y$ ($y > x$) hours. If both the pipes are open, in how many hours will the tank be filled?

### Concept & Net Filling Rate When one pipe fills and another empties a tank, their net work is the difference between their individual rates. Because $y > x$, the filling pipe is faster, meaning the tank will eventually fill. $$ \text{Net Work Rate} = \text{Filling Rate} - \text{Emptying Rate} $$ ### Step-by-Step Solution * **Identify Rates:** Part filled in $1$ hour by the first pipe = $\frac{1}{x}$ Part emptied in $1$ hour by the second pipe = $\frac{1}{y}$ * **Calculate Net Rate:** Net part filled in $1$ hour when both are open = $\frac{1}{x} - \frac{1}{y}$ * **Simplify the Expression:** $\frac{1}{x} - \frac{1}{y} = \frac{y - x}{xy}$ * **Find Total Time:** The total time taken to fill the tank is the reciprocal of the net hourly rate. Time taken = $\frac{1}{\frac{y - x}{xy}} = \frac{xy}{y - x}$ hours. ### Exam Strategy & Shortcut Memorize the generic algebraic formula for one inlet and one outlet pipe: $\frac{\text{Product of times}}{\text{Difference of times}}$. Since the tank is being filled, the denominator must be positive, which is $(y - x)$ because $y > x$. This instantly points to $\frac{xy}{y - x}$. ### Common Pitfall A common mistake is choosing $\frac{xy}{x - y}$. Since we are given $y > x$, $(x - y)$ would result in a negative number of hours, which is physically impossible. Always subtract the smaller time value from the larger one in the denominator for filling scenarios. ### Final Answer Therefore, the correct answer is **$\frac{xy}{y - x}$ hours**.