Question : A, B, and C can do a piece of work in 11 days, 20 days, and 55 days respectively. If B works daily and is supported by A and C on alternate days beginning with A, then in how many days will the work be finished?
Option 1: $10$
Option 2: $9 \frac{1}{3}$
Option 3: $12$
Option 4: $8$
Correct Answer: $9 \frac{1}{3}$
Solution :
A, B, and C can do a piece of work in 11 days, 20 days, and 55 days respectively.
Total work = LCM(11, 20 and 55) = 220
Efficiency of A = $\frac{220}{11}$ = 20
Efficiency of B = $\frac{220}{20}$ = 11
Efficiency of C = $\frac{220}{55}$ = 4
1st day work = work done by A and B = (20 + 11) = 31
2nd day work = work done by B and C = (11 + 4) = 15
So, 2 days' work = 31 + 15 = 46
Total pairs of days required to work
= $\frac{220}{46}$
= $4 + \frac{36}{46}$
So, 4 × 2 = 8 days and remaining work = 36
On the next day, 31 of 36 will be done and the rest 5 will be done in $\frac{5}{15} = \frac{1}{3}$ days.
So total days = $9 + \frac{1}{3}$
∴ Total days required = $9\frac{1}{3}$.
Hence, the correct answer is $9\frac{1}{3}$.
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