A can do a piece of work in 8 days and B in 24 days. They begin together, but 4 days before the completion of the work, A leaves. Find the number of days taken by A and B to complete the entire work.

Correct option is C

​Given:
A completes its work in 8 days and B completes it work in 24 days but before A leaves 4 days before completion of work.

Solution:

Total work = LCM(8, 24) = 24 units

Efficiencies of A =$$\frac{24}8 = 3$$​ units/day

Efficiencies of B =$$\frac{24}{24}= 1$$ unit/day

If A does not leave the work, then 4 days of A’s work more will be completed.

Then, total work = 24 + 4 × (3) = 36 units

Required time =$$\frac {36}{3 + 1}$$​= 9 days

Alternate Solution:

We can calculate the number taken to complete the work using the formula

Days =  $$\frac{y(x + n)}{(x + y)}$$​​

Where x and y are the number of days taken by A and B. n is the number of days before A left. 

Here we have x = 8, y = 24 and n = 4

​$$\text{Hence, days required} = 24 \times \frac{8 + 4}{8 + 24} \\= 24 \times \frac{12}{32} \\= 9 \, \text{days}$$​​