Question : A, B, and C can do a piece of work in 10, 20, and 30 days, respectively. In how many days can A do the work if he is assisted by B and C on every third day?
Option 1: 9 days
Option 2: $8 \frac{5}{11}$ days
Option 3: $8 \frac{2}{11}$ days
Option 4: $6 \frac{5}{11}$ days
Correct Answer: $8 \frac{2}{11}$ days
Solution :
Total Work = Least Common Multiple (10, 20, 30) = 60
The rate of A
⇒ $\frac{60}{10}$ = 6 units
The rate of B
⇒ $\frac{60}{20}$ = 3 units
The rate of C
⇒ $\frac{60}{30}$ = 2 units
In 3 days, Total work done = 6 × 2 + (6 + 3 + 2) = 23 units
⇒ total work done in 6 days = 23 × 2 = 46 units
Next 2 days, work done = 6 + 6 = 12 units
⇒ Remaining Work = 60 - 46 - 12 = 2
Time taken for the remaining work = $\frac{2}{6+3+2}$ = $\frac{2}{11}$
⇒ Total time = 8 + $\frac{2}{11}$ days
Hence, the correct answer is 8$\frac{2}{11}$ days
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