A can do as much work as B and C together can do. A and B can together do a piece of work in 9 hours 36 minutes and C can do it in 48 hours. The time in hours that B needs to do the work alone, is ?

Answer: Option B

9 hours 36 minutes
$$\begin{array}{rl}& =9+\frac{36}{60}\text{hours}\\ & =9\frac{3}{5}=\frac{48}{5}\text{hours}\end{array}$$
(A + B)’s 1 hour’s work = $$\frac{5}{48}$$
C’s 1 hour’s work = $$\frac{1}{48}$$
(A + B + C)’s 1 hour’s work = $$\frac{5}{48}+\frac{1}{48}$$   = $$\frac{1}{8}$$ . . . . . . .(i)
A’s 1 hour’s work = (B + C)’s 1 hour’s work . . . . . . . . (ii)
From equation (i) and (ii),
2 × (A’s 1 hour’s work) = $$\frac{1}{8}$$
A’s 1 hour’s work = $$\frac{1}{16}$$
∴ B’s 1 hour’s work
$$\begin{array}{rl}& =\frac{5}{48}-\frac{1}{16}\\ & =\frac{5-3}{48}\\ & =\frac{1}{24}\end{array}$$
∴ B alone will finish the work in 24 hours.