49 pumps can empty a reservoir in $6 \frac{1}{2}$ days, working 8 hours a day. If 196 pumps are used for 5 hours each day, then the same work will be completed in

### Concept & The Chain Rule This problem involves three variables affecting total work: Number of workers/machines (M), Days (D), and Hours per day (H). Because the total amount of work (emptying one reservoir) is constant, the product of these variables must remain constant across different scenarios. $$ M_1 D_1 H_1 = M_2 D_2 H_2 $$ ### Step-by-Step Solution * Identify the variables for the first scenario: $M_1 = 49 \text{ pumps}$, $D_1 = 6 \frac{1}{2} = \frac{13}{2} \text{ days}$, $H_1 = 8 \text{ hours}$. * Identify the variables for the second scenario: $M_2 = 196 \text{ pumps}$, $H_2 = 5 \text{ hours}$, $D_2 = x \text{ days}$. * Substitute the values into the formula: $49 \times \frac{13}{2} \times 8 = 196 \times x \times 5$. * Simplify the left side: The $2$ in the denominator cancels with $8$ to give $4$. The equation becomes $49 \times 13 \times 4 = 196 \times 5 \times x$. * Notice that $49 \times 4 = 196$. Substitute this into the equation: $196 \times 13 = 196 \times 5 \times x$. * Cancel $196$ from both sides: $13 = 5x$. * Solve for $x$: $x = \frac{13}{5} = 2 \frac{3}{5} \text{ days}$. ### Exam Strategy & Shortcut Look for multiples before multiplying large numbers. Seeing $49$ and $196$ should immediately trigger the realization that $196 = 49 \times 4$. Keeping the equation as factors ($49 \times 13 \times 4$) rather than computing the product ($2548$) allows for instant cancellation, saving valuable time. ### Common Pitfall A common mistake is incorrectly converting the mixed fraction $6 \frac{1}{2}$ into an improper fraction, or forgetting to multiply by the hours per day when setting up the initial total work equivalence. ### Final Answer Therefore, the correct answer is **$2 \frac{3}{5}$ days**.