A can do $$\frac{1}{5}$$​th of some work in 12 days, B can do 20% of the same work in 10 days, C can do $$\frac{1}{6}$$​th of the work in 8 days and D can do $$\frac{1}{5}$$​th of the work in 12 days. Who will complete the work first if all four started to work at the same time?

Correct option is D

Given:

A can do $$\frac{1}{5}$$​ of the work in 12 days.

B can do 20% of the work in 10 days.

C can do $$\frac{1}{6}$$​ of the work in 8 days.

D can do $$\frac{1}{5}$$​ of the work in 12 days.

Concept Used:

Time to complete the work $$= \frac{1}{\text{work rate}}$$​​

Solution:
A does $$\frac{1}{5}$$​​ of the work in 12 days.
Work rate of A $$= \frac{\frac{1}{5}}{12} = \frac{1}{60}​$$​ (work per day)
Time to complete the work =$$\frac{1}{\frac{1}{60}}$$​= 60 days
B does 20% (or $$\frac{1}{5}$$​) of the work in 10 days.
Work rate of B =$$\frac{\frac{1}{5}}{10} = \frac{1}{50}​$$​(work per day)
Time to complete the work =$$\frac{1}{\frac{1}{50}} =$$​50 days
C does $$\frac{1}{6}​$$ of the work in 8 days.
Work rate of C =$$\frac{\frac{1}{6}}{8} = \frac{1}{48}​$$​ (work per day)
Time to complete the work $$=\frac{1}{\frac{1}{48}} = 48$$​ days
D does $$\frac{1}{5}​$$​ of the work in 12 days.
Work rate of D =$$\frac{\frac{1}{5}}{12} = \frac{1}{60}​$$​ (work per day)
Time to complete the work = $$\frac{1}{\frac{1}{60}}$$​= 60 days

Thus, C will complete the work first as they will finish in 48 days.