A, B and C can do a piece of work in 30, 45, and 90 days respectively. In how…

Answer: Option B

Explanation:

Let us first assume the total work as the LCM of (30, 45, 90)

So, total work = LCM of (30, 45, 90) = 90 units

If the total work is 90 units then,

The efficiency of A = 90/30 = 3 u/d;

The efficiency of B = 90/45 = 2 u/d;

The efficiency of C = 90/90 = 1 u/d

Given that A alone starts the work and he is assisted by B and C on every 4th day

So, the work completed at the end of first 4 days = Work completed by A in the first 3 days + Work completed by A, B and C together in the fourth day

Work completed at the end of first 4 days = (3 × 3) + ((3 + 2 + 1) × 1) = (9 + 6) = 15 units

This cycle of A working alone for 3 days and being accompanied by B and C on the fourth day is repeated.

So, it is very clear that a total of 15 units of work is completed in every 4 days.

In this pattern the number of cycles required to complete the total work of 90 units = 90/15 = 6 cycles where each cycle is comprised of 4 days

Thus, the total number of days required = (6 × 4) = 24 days