A and B can do a work in 30 days and 10 days, respectively. If they work on alternate days starting with A, then in how many days will the work be completed?

Correct option is C

Given:

Time taken by A to complete the work = 30 days

Time taken by B to complete the work  = 10 days

Formula Used:

Let total work to fill the tank be 1.

Then work done by in one hour = $$\frac{1}{Total\ time\ to\ complete}$$​​
Solution:

Total Work = Efficiency× Time

Total Work = LCM of 30,10 =30 unit

Efficiency of A$$=\frac{30}{30}$$= 1 units per hour

Efficiency of B =$$\frac{30}{10}$$​= 3 units per hour

Total efficiency of A and B = 1+3 = 4 units

In 14 days, 28 unit of work will be completed by them
Next  on 15th Day, A will completed one unit of work 
Next on 16th Day, 1 unit work left, completed by B in $$\frac{1}{3}$$ days

Total Time = 15 + $$\frac{1}{3}$$ = $$15\frac{1}{3}$$ days​