If a certain number of workmen can do a piece of work in 25 hours, in how many hours will another set of an equal number of men, do a piece of work, twice as great, supposing that 2 men of the first set can do as much work in an hour, as 3 men of the second set do in an hour?

### Concept & Efficiency-Adjusted Work Rule When work amounts and group efficiencies vary between two identically sized groups, apply the modified MDH formula integrating the efficiency multipliers ($E$) and work units ($W$). $$ \frac{M_1 \times H_1 \times E_1}{W_1} = \frac{M_2 \times H_2 \times E_2}{W_2} $$ ### Step-by-Step Solution 1. **Determine the efficiency ratio:** - 2 men of first set ($E_1$) = 3 men of second set ($E_2$). - $2E_1 = 3E_2 \implies \frac{E_1}{E_2} = \frac{3}{2}$. Let $E_1 = 3$ and $E_2 = 2$. 2. **Identify variables:** - Set 1: $M_1 = x$, $H_1 = 25$, $W_1 = 1$ - Set 2: $M_2 = x$, $H_2 = ?$, $W_2 = 2$ (since the work is twice as great) 3. **Substitute into the formula:** - $\frac{x \times 25 \times 3}{1} = \frac{x \times H_2 \times 2}{2}$ 4. **Solve for $H_2$:** - The variable $x$ cancels out on both sides. - $75 = \frac{H_2 \times 2}{2}$ - $H_2 = 75$ ### Exam Strategy & Shortcut Look at the pure multipliers conceptually. The work doubles, so hours should intuitively double ($25 \times 2 = 50$). However, the new workers are less efficient (efficiency ratio is $2/3$ compared to the old group). Therefore, they will take $3/2$ times longer to complete the same effort. $50 \times \frac{3}{2} = 75$ hours. ### Common Pitfall Assuming that because "2 of the first set equal 3 of the second set", the first set is slower. It actually means the first set is faster/more efficient, hence properly assigning $E_1 = 3$ and $E_2 = 2$ is critical. ### Final Answer Therefore, the correct answer is **75**.