A, B and C can complete a piece of work in 8 days, 16 days and 24 days, respectively. If they work together, how many days will it take to complete a task that is eleven times the original work?

Correct option is C

Given:

A can finish work in 8 days

B can finish work in 16 days

C can finish work in 24 days

Total work required = 11 times the original work

Need: Time if A, B, and C work together

Concept Used:

Work rate = $$\frac{1}{\text{time}}$$​​

Time for work =$$\frac{\text{Total work}}{\text{Combined rate}}$$​​

Solution:

A’s 1-day work =$$\frac{1}{8}$$​​

B’s 1-day work =$$\frac{1}{16}$$​​

C’s 1-day work =$$\frac{1}{24}$$​​

Combined 1-day work $$=\frac{1}{8} + \frac{1}{16} + \frac{1}{24}$$​

​$$= \frac{6}{48} + \frac{3}{48} + \frac{2}{48} = \frac{11}{48}$$​

So, together they finish $$\frac{11}{48}$$​ of work per day.

Time to complete 1 work =$$\frac{48}{11}$$​ days

Time to complete 11 works =$$11 \times \frac{48}{11} =$$ 48 days

Alternate Method :

Take LCM of 8, 16, 24 = 48 units (assume total work = 48 units).

Work done per day:

A = 48 ÷ 8 = 6 units/day

B = 48 ÷ 16 = 3 units/day

C = 48 ÷ 24 = 2 units/day

Together per day = 6 + 3 + 2 = 11 units/day

Time for 1 work (48 units) = $$\frac{48}{11} \ \text{days}$$​

Time for 11 works (11 × 48 = 528 units):

​$$\frac{528}{11} = 48 \ \text{days}$$​