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?

9$\frac{1}{3}$

A = 11D, B = 20D, C = 55D,

Since B is working daily and A & C assist him on alternate days,starting with A, the efficiencies will be as follows:

1st day (B + A) = 11 + 20 = 31 units

2nd day (B + C) = 11 + 4 = 15 units

Every 2 days work =            46 units

⇒Hence, in 8 days, 184 units of work will be completed

and in 9 days, 215 units of work will be completed,

⇒The remaining work i.e; 5 units will be completed by (B + C) = \(\frac{5}{15}\) = \(\frac{1}{3}\) days.

⇒Therefore, The Total time taken to complete work = 9 + \(\frac{1}{3}\) = \( {9 }_{3 }^{1 } \) days.