3 pumps, working 8 hours a day, can empty a tank in 2 days. How many hours a day must 4 pumps work to empty the tank in 1 day?

### Concept & Work Equivalence Formula The total pump-hours required to complete the task of emptying the tank remain constant: $$ P_1 \times H_1 \times D_1 = P_2 \times H_2 \times D_2 $$ where $P$ is the number of pumps, $H$ is hours per day, and $D$ is days. ### Step-by-Step Solution * **Given**: - $P_1 = 3$, $H_1 = 8$, $D_1 = 2$ - $P_2 = 4$, $D_2 = 1$, $H_2 = x$ * **Calculation**: $$ 3 \times 8 \times 2 = 4 \times x \times 1 $$ $$ 48 = 4x $$ $$ x = \frac{48}{4} = 12 $$ ### Exam Strategy & Shortcut Total work units = $3 \times 8 \times 2 = 48$ pump-hours. To finish these $48$ pump-hours using $4$ pumps in $1$ day, divide by $4$: $\frac{48}{4} = 12$ hours per day. ### Common Pitfall Treating the number of pumps as inversely related to hours in a non-standard way, or forgetting to multiply by the number of days. ### Final Answer Therefore, the correct answer is **12**.