12 men and 18 boys, working $7 \frac{1}{2}$ hours a day, can do a piece of work in 60 days. If a man works equal to 2 boys, then how many boys will be required to help 21 men to do twice the work in 50 days, working 9 hours a day?

### Concept & Work Equivalence Formula When total work scales (like doing "twice the work"), we use the generalized time and work equation: $$ \frac{M_1 \times D_1 \times H_1}{W_1} = \frac{M_2 \times D_2 \times H_2}{W_2} $$ We must also convert all workers into a single baseline unit using the given equivalency ($1M = 2B$). ### Step-by-Step Solution * **Convert Units:** We are given $1$ Man ($M$) = $2$ Boys ($B$). Let's convert everyone to Boys. * **Group 1:** $12M + 18B = 12(2B) + 18B = 24B + 18B = 42B$. * They work $D_1 = 60$ days, $H_1 = 7.5$ hours/day. Let their work be $W_1 = 1$. * **Group 2:** Let $x$ be the number of required boys. The group is $21M + xB = 21(2B) + xB = 42B + xB = (42 + x)B$. * They work $D_2 = 50$ days, $H_2 = 9$ hours/day. Their work is twice as much, so $W_2 = 2$. * **Apply Formula:** * $\frac{42 \times 60 \times 7.5}{1} = \frac{(42 + x) \times 50 \times 9}{2}$ * $42 \times 450 = \frac{(42 + x) \times 450}{2}$ * **Solve for $x$:** Cancel $450$ from both sides. * $42 = \frac{42 + x}{2}$ * $84 = 42 + x$ * $x = 42$ boys. ### Exam Strategy & Shortcut Recognize that $60 \times 7.5 = 450$ hours and $50 \times 9 = 450$ hours. Since the total hours are identical, and the work is doubled, the total "manpower" (in terms of boys) must simply be doubled! Initial manpower = $42$ boys. Required manpower = $42 \times 2 = 84$ boys. Since $21$ men = $42$ boys, we need exactly $84 - 42 = 42$ more boys. ### Common Pitfall Forgetting to account for the "twice the work" condition ($W_2 = 2$) and incorrectly equating the two sides directly, which would result in a negative or nonsensical number of boys. ### Final Answer Therefore, the correct answer is **42**.