A, B and C working together can do a piece of work in 8 hours. A alone can do it in 20 hours and B alone can do it in 24 hours. In how many hours will C alone do the same work?
Time taken by A to do the work $=20$ hours
Time taken by $\mathrm{B}$ to do the work $=24$ hours
Time taken by $(\mathrm{A}+\mathrm{B}+\mathrm{C})$ to do the work $=8$ hours
Now,
Work done by $\mathrm{A}=\frac{1}{20}$
Work done by $\mathrm{B}=\frac{1}{24}$
Work done by $(\mathrm{A}+\mathrm{B}+\mathrm{C})=\frac{1}{8}$
$\therefore$ Work done by $\mathrm{C}=\frac{1}{8}-\left(\frac{1}{20}+\frac{1}{24}\right)$
$=\frac{1}{8}-\left(\frac{6}{120}+\frac{5}{120}\right)=\frac{1}{8}-\left(\frac{11}{120}\right)$
$=\frac{15-11}{120}=\frac{4}{120}=\frac{1}{30}$
Thus, $C$ can do the work in 30 hours.
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