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**Time and Work — formulas used here.** - Work done in unit time. If a worker finishes a job in $m$ days, the fraction of work done in one day is $\tfrac{1}{m}$. - Combined work (two workers). A finishes in $X$ days, B in $Y$ days; together they finish in $\tfrac{XY}{X+Y}$ days. - Combined work (three workers). A, B, C take $X$, $Y$, $Z$ days; together they take $\tfrac{XYZ}{XY+YZ+ZX}$ days. - Efficiency–time inverse. If A is $k$ times as efficient as B, A takes $\tfrac{1}{k}$ as long. - Workforce equivalence. $M_1 D_1 T_1 = M_2 D_2 T_2$ for the same work, where $M$ = workers, $D$ = days, $T$ = hours/day. **Question.** $8$ men complete a piece of work in $10$ days, working $9$ hours a day. In how many days will $6$ men complete the same work if they work $8$ hours a day?
A$12$
B$15$
C$18$
D$20$
Answer & Solution
Correct answer: B. $15$
The workforce-equivalence identity $M_1 D_1 T_1 = M_2 D_2 T_2$ holds when the work to be done is the same.
Plugging in: $8 \times 10 \times 9 = 6 \times D_2 \times 8 \;\Rightarrow\; 720 = 48 D_2 \;\Rightarrow\; D_2 = 15$ days.
Distractor check: option A (12) drops one factor; option D (20) inverts the ratio. The clean way to spot the right answer: fewer men working fewer hours should give *more* days, so anything less than 10 is impossible.
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