In two days A, B and C together can finish 1/2 of a work and in another 2 days B and C together can finish 310 part of the work. Then A alone can complete the whole work in ?

Answer: Option B

$$\begin{array}{rl}& \frac{3}{10}\left(\text{B}+\text{C }\right)=2\text{ days}\\ & \left(\text{B}+\text{C }\right)=2\times \frac{10}{3}\\ & =\frac{20}{3}\text{ days}\\ & \frac{1}{2}\left(\text{A}+\text{B}+\text{C }\right)=2\text{ days}\\ & \text{A}+\text{B}+\text{C}=\text{4 days}\end{array}$$
L.C.M of total work = 20
One day work of A + B + C = $$\frac{20}{4}$$ = 5 unit/day
One day work of B + C = $$\frac{20}{\frac{20}{3}}$$ = 3 unit/day
A = 5 - 3 = 2
A alone will complete the work
$$\begin{array}{rl}& =\frac{20}{2}\text{days}\\ & =\text{10 days}\end{array}$$