Question : A, B and C can do a piece of work in 11 days, 20 days and 55 days respectively. How soon can the work be done if A works with B on the first day, A works with C on the second day and so on?
Option 1: $8 \frac{1}{2}$ days
Option 2: $9\frac{1}{2}$ days
Option 3: 8 days
Option 4: 9 days
Correct Answer: 8 days
Solution :
Total work = LCM (11, 20 and 55) = 220 units
Efficiency of A = $\frac{220}{11}$ = 20 units/day
Efficiency of B = $\frac{220}{20}$ = 11 units/day
Efficiency of C = $\frac{220}{55}$ = 4 units/day
Combined work of A and B = 20 + 11 = 31 units
Now A and C work together = 20 + 4 = 24 units
2 days' work = 31 + 24 = 55 units
We know, total work = 220 units
$\therefore$ Time taken = $2\times\frac{220}{55}=2\times4= 8$ days
Hence, the correct answer is 8 days.
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