$A$ 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?

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Solution

We know that,

Number of hours $A$ required to do a piece of work $= 24$ hours

Number of $B$ required to do a piece of work $= 16$ hours

Let us consider number of hours $C$ required to do a piece of work $= x$ hours

Number of hours $A, B$ and $C$ together required to do piece of work $= 8$ hours

$∴$ We can calculate, work done by $A$ in $1$ hour $= \frac{1}{24}$

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

Work done by $C$ in $1$ hour $= \frac{1}{x}$

Work done by $A, B$ and $C$ together in a $1$ hour $= \frac{1}{8}$

Now, work done by $A, B$ and $C$ together in $1$ hour is also equal to the sum of work done by $A, B$ and $C$ in $1$ hour
$= \frac{1}{24} + \frac{1}{16} + \frac{1}{x} = \frac{5}{48} + \frac{1}{x}$

Therefore,

$\frac{5}{48} + \frac{1}{x} = \frac{1}{8}$

$\frac{(5 x + 48)}{48 x} = \frac{1}{8}$

$40 x + 384 = 48 x$

$8 x = 384$

$x = \frac{384}{8} = 48$

$∴ C$ can do the work in $48$ hours.

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