P and Q can complete a piece of work in 28 days and 35 days, respectively. P begins to do the work and they work alternatively one at a time for one day each. In how many days will the whole work be completed?

Correct option is A

Given:

P can complete the work in 28 days, so A’s work per day =$$\frac{1}{28}$$​​

Q can complete the work in 35 days, so B’s work per day =$$\frac{1}{35}$$​​

P starts the work and they work on alternate days.

Solution:

Work done in 2 days =$$\left( \frac{1}{28} + \frac{1}{35} \right)$$= ​$$\frac{5}{140} + \frac{4}{140} = \frac{9}{140}$$​

In 2 days, work done =$$\frac{9}{140}$$​​

how many full cycles are needed

15 ×$$\frac{9}{140} = \frac{135}{140}$$​

So, after 15 full cycles (30 days)

Remaining work:  1 -$$\frac{135}{140} = \frac{5}{140} = \frac 1{28}$$​​

P works on the 31th day

A’s work per day$$\frac{1}{28}$$, Hence, Remaining work is also same. 
So, Total 31 days to complete the work

Alternate Method: 

Total Work = 140 Unit 

Work Done in 2 Days = 4 + 5 = 9 

15 $$\times$$ 9 = 135 Unit  

So, after 15 full cycles (30 days) 

Remaining work = 140 - 135 = 5 Unit  

P works on the 31th day, Remaining 5 unit of work, should be completed by P on that day 

Total time = 30+1 = 31 days