Rohan can alone finish a work in 12 days and Hari can alone in 16 days. They both work for 3 days and then Rohan left. After that only Hari works and finishes the work. The total work will be completed in:

Correct option is A

Given:
Time taken by Rohan to finish a job working alone is 12 days.
Time taken by Hari to finish the job is 16 days.
Formula Used:
Let total work be 1.
Then work done in each day= $$\frac{\text{Total Work}}{\text{Total Time}}$$​
Solution:
Work done by Rohan in each day = $$\frac1{12}$$​​
Work done by Hari in each day =$$\frac{1}{16}$$​
Work done by Rohan and Hari in each day= $$\frac{1}{12} + \frac{1}{16} = \frac{4 + 3}{48} = \frac{7}{48}$$​​
Work done by Rohan and Hari in 3 days will be = $$\frac7{48} \times 3 =\frac7 {16}$$​​
Work left = $$1 - \frac{7}{16} = \frac{9}{16}$$​​
Time taken by Hari to finish the job $$\frac{\frac{9}{16}}{\frac{1}{16}} = 9 \, \text{days}$$​​
Total time taken = 9 +3 =12days
Alternate Method:
Total Work = Efficiency $$\times$$​ Time
Total Work = LCM of 12 and 16 = 48
Efficiency of Rohan = $$\frac{48}{12}$$​ = 4 units per day.
Efficiency of Hari =$$\frac{ 48}{16}$$​ = 3 units per day.
Then efficiency of both Rohan and Hari will be 3+4=7 unit per day.
Work done by Rohan and Hari in 3 days will be = 3 $$\times$$​7 = 21units
Work left =48-21= 27units
Time taken by Hari to finish the job $$\frac{27}{3}$$​ =9 days
Total time taken = 9 +3 =12days