 # A can do a piece of work in 24 hours while B alone can do it in 16 hours.

Question:

A can do a piece of work in 24 hours while B alone can do it in 16 hours. If A, B and C working together can finish it in 8 hours, in how many hours can C alone finish the work?

Solution:

Time taken by A to complete the piece of work $=24 \mathrm{~h}$

Work done per hour by $\mathrm{A}=\frac{1}{24}$

Time taken by $\mathrm{B}$ to complete the work $=16 \mathrm{~h}$

Work done per hour by $\mathrm{B}=\frac{1}{16}$

Total time taken when $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ work together $=8 \mathrm{~h}$

Work done per hour by $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}=\frac{1}{8}$

Work done per hour by $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}=($ work done per hour by $\mathrm{A})+($ work done per hour by $\mathrm{B})+$ (work done per hour by $\mathrm{C}$ )

(Work done per hour by $\mathrm{C}$ ) $=($ work done per hour by A, B and C $)-($ work done per hour by A $)-$ (work done per hour by B)

$=\frac{1}{8}-\frac{1}{24}-\frac{1}{16}=\frac{6-2-3}{48}=\frac{1}{48}$

Thus, $\mathrm{C}$ alone will take $48 \mathrm{~h}$ to complete the work.