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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.** A and B together can finish a job in 6 days. B alone finishes the same job in 15 days. How long does A take alone?
A$8$ days
B$9$ days
C$10$ days
D$12$ days
Answer & Solution
Correct answer: C. $10$ days
A's one-day work = (combined) − (B's). $\tfrac{1}{A} = \tfrac{1}{6} - \tfrac{1}{15} = \tfrac{5 - 2}{30} = \tfrac{1}{10}$. So A alone takes $10$ days. Option B (9) and D (12) come from algebra slips; C is the consistent answer.
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