A, B, C work one after the other in that order. Each can finish a work individually in 10, 16, 12 days, respectively. After working for 3 such cycles ABC, ABC, ABC, then A and C leave the work. B alone will complete the rest of the work in:

Correct option is A

Given:

A can do work in 10 days

B can do the same work in 16 days

C can do the work in 12 days

A,B and C worked in 3 cycles

Formula Used:

Let total work be 1.

Then work done 1 day =$$\frac{1}{Total\ time\ to\ work}$$​​

Solution:

A’s one day’s work =$$\frac{1}{10}$$​​

B’s one day’s work =$$\frac{1}{16}$$​​

C’s one day’s work =$$\frac{1}{12}$$​​

Work in all three in one cycle =$$\frac{1}{10}+\frac{1}{16}+\frac{1}{12}$$​​

=$$\frac{24+15+20}{240}=\frac{59}{240}$$​​

Work done in 3 cycles $$= 3 \times \frac{59}{240} = \frac{177}{240}$$​​

Work left =$$1 - \frac{177}{240} = \frac{240-177}{240} = \frac{63}{240}$$​​

Time taken byB to complete the work $$= \frac{Work\ left}{B’s\ one\ day\ work} = \frac{\frac{63}{240}}{\frac{1}{16}} = \frac{21}{5} =4\frac{1}{5} days$$​