A tap can fill a tank in $48$ minutes whereas another tap can empty it in $2$ hours. If both the taps are opened at $11 \text{ : } 40 \text{ A.M.}$, then the tank will be filled at

### Concept & Unit Conversion with Net Rates Before calculating the combined rate of filling and emptying pipes, all time values must be converted to the same unit (e.g., minutes). The net work is then found by subtracting the emptying rate from the filling rate. ### Step-by-Step Solution * **Unit Conversion:** Filling pipe time = $48$ minutes. Emptying pipe time = $2$ hours = $2 \times 60 = 120$ minutes. * **Calculate Individual Rates:** Part filled in $1$ minute = $\frac{1}{48}$ Part emptied in $1$ minute = $\frac{1}{120}$ * **Calculate Net Rate:** Net part filled in $1$ minute = $\frac{1}{48} - \frac{1}{120}$ LCM of $48$ and $120$ is $240$. Net part filled = $\frac{5}{240} - \frac{2}{240} = \frac{3}{240} = \frac{1}{80}$ * **Calculate Total Time & Final Clock Time:** The tank will be full in $80$ minutes, which is equal to $1$ hour and $20$ minutes. Start time: $11:40$ A.M. Add $1$ hour $20$ minutes: $11:40$ A.M. $+ 1$ hour $= 12:40$ P.M. $12:40$ P.M. $+ 20$ minutes $= 1:00$ P.M. ### Exam Strategy & Shortcut Using the $\frac{ab}{b - a}$ shortcut: $\frac{48 \times 120}{120 - 48} = \frac{48 \times 120}{72}$. Cancel $24$ from $48$ and $72$ to get $\frac{2 \times 120}{3} = 2 \times 40 = 80$ minutes. Adding $80$ mins to $11:40$ gives $1:00$ P.M. quickly. ### Common Pitfall A frequent error is forgetting to convert $2$ hours into $120$ minutes, leading to an incorrect setup like $\frac{1}{48} - \frac{1}{2}$, which produces negative (impossible) results. ### Final Answer Therefore, the correct answer is **$1 \text{ : } 00 \text{ P.M.}$**.