Arithmetic

Time & Work

Hard

A, B and C can do a work in 10, 12 and 15 days respectively. In how many days will the work be completed if B is assisted by both A and C on every third day ?

$7\frac{1}{5}$

$7\frac{2}{3}$

✅ Correct Option: 3

B's 1-day work = $\frac{1}{12}$ . Combined A+B+C = $\frac{1}{10}+\frac{1}{12}+\frac{1}{15}=\frac{6+5+4}{60}=\frac{15}{60}=\frac{1}{4}$ . Every 2 days B alone: $2\times\frac{1}{12}=\frac{1}{6}$ ; 3rd day all three: $\frac{1}{4}$ . Per 3-day cycle: $\frac{1}{6}+\frac{1}{4}=\frac{5}{12}$ . In 6 days: $\frac{10}{12}$ . Remaining $\frac{2}{12}=\frac{1}{6}$ needs B 2 more days. Total = $6+2=8$ days.

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Solution

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