A can do a piece of work in 8 days, while B can do it in 18 days. In how many days will the work be completed if they both work on alternate days starting with B ?

A can complete the work in 8 days.

B can complete the work in 18 days.

They work on alternate days, starting with B.

Work done per day = $$\frac{1}{\text{number of days to complete the work}}$$​​

Total work = 1 (since the task is to complete one unit of work)

Work done by A in 1 day = $$\frac{1}{8}$$​ of the total work.

Work done by B in 1 day = $$\frac{1}{18}$$​ of the total work. 

A and B work on alternate days, so

work done by A and B together in two days, starting with B

= $$\frac{1}{18} + \frac{1}{8} = \frac{ 4 + 9}{72} = \frac{13}{72}$$ 

In 10 days, they complete $$\frac{13 \times 5}{72} = \frac { 65}{72}$$ 

remaining work = 1 - $$\frac{65}{72} = \frac{ 72 - 65 }{72} = \frac{7}{72}$$ 

now, B's turn to  work first, 

 $$\frac{7}{72} - \frac{1}{18} = \frac{ 7 - 4}{72} = \frac{3}{72} \text{ work left}$$ 

this work is completed by A in  $$\frac{ 3 / 72}{ 1 / 8} = \frac{1}{3} days$$ 

Total time to complete the work = 10 + 1 + $$\frac{1}{3}$$ = $$11 \frac{1}{3} \text{days}$$​​

Work done in 2 days = 9 + 4 = 13 units
=> 65 units work done in 2 × 5 = 10 days.

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