A and B can do a piece of work in 12 days, B and C in 8 days and C and A in 6 days. How long would B take to do the same work alone ?

Answer: Option D

$$\begin{array}{rl}& \left(\text{A}+\text{B}\right)\text{’s 1 day’s work}=\frac{1}{12}\\ & \left(\text{B}+\text{C}\right)\text{’s 1 day’s work}=\frac{1}{8}\\ & \left(\text{A}+\text{C}\right)\text{’s 1 day’s work}=\frac{1}{62}\end{array}$$
[ (A + B)'s 1 day's work + (B + C)'s 1 day's work ] - (A + C)'s 1 day's work
$$\begin{array}{rl}& =\frac{1}{12}+\frac{1}{8}-\frac{1}{6}\\ & \Rightarrow 2\left(\text{B’s 1 day’s work}\right)=\frac{1}{24}\\ & \Rightarrow \text{B’s 1 day’s work}=\frac{1}{48}\end{array}$$
Hence, B alone can do the work in 48 days.