Shobhita can complete a painting work in 20 hours while Saroj can complete the same work in 15 hours. In how much time (in hours) can the entire work of painting be completed, if both Shobhita and Saroj work simultaneously?

Correct option is B

Given:
Time taken by Shobhita to finish a job working alone is 20 hours.
Time taken by Saroj to finish the job is 15 hours.
Formula Used:
Let total work be 1.
Then work done in each hour will be $$\frac{\text{Total Work}}{\text{Total Time}}$$​​
Solution:
Work done by Shobhita in each hour will be $$\frac{\text{Total Work}}{\text{Total Time}}$$​= $$\frac 1{20}$$​​
Work done by Saroj in each hour will be $$\frac 1{15}$$​
Work done by Shobhita and Saroj in each hour will be $$= \frac{1}{20} + \frac{1}{15} = \frac{3 + 4}{60} = \frac{7}{60}$$​​

Time taken by Shobhita and Saroj to finish the job will be $$\frac{60}7 hours = 8 \frac47 hours$$​​
Alternate Method:
Total Work = Efficiency $$\times$$​ Time
Total Work = LCM of 20 and 15 = 60
Efficiency of Shobhita = $$\frac{60}{20}$$​ = 3 units per hour.
Efficiency of Saroj = $$\frac{60}{15}$$​= 4 units per hour.
Then efficiency of both Shobhita and Saroj will be 3 +4 =7 unit per hour.
Hence Time taken by Shobhita and Saroj will be = $$\frac{60}7$$​ = 8 $$\frac47$$​hours.