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**Hint**: These type of problems are solved by following some rule which are-

Suppose if a person can finish a work in $n$ days. Then work done by that person in $1$ day \[ = \] $1/{n^{th}}$part of work.

Suppose that work done by a person in $1$ day is $\dfrac{1}{n}$ then time taken by that person to finish the whole work$ = n$ days.

**:**

__Complete step-by-step answer__In the question it is given that

Time taken by $A$ to finish the work $ = 24$ days

Work done by $A$ in 1 day$ = \dfrac{1}{{24}}$

Time taken by $(A + B)$ to finish the work$ = 16$ days

Work done by $(A + B)$in $1$ day$ = \dfrac{1}{{16}}$

Time taken by $\left( {A + B + C} \right)$ to finish the work$ = 10\dfrac{2}{3}$days

$ = \dfrac{{32}}{3}$ days

Work done by $\left( {A + B + C} \right)$ in 1 day$ = \dfrac{3}{{32}}$

Now we can calculate the work done by the $C$ which is

Work done by $\left( {A + B + C} \right)$ in 1 day$ - $Work done by $(A + B)$ in 1 day \[ = \dfrac{3}{{32}} - \dfrac{1}{{16}}\]

$ = \dfrac{{3 - 2}}{{32}} = \dfrac{1}{{32}}$

Now work done by the $C$ in 1 day$ = \dfrac{1}{{32}}$

Further the work done by the $\left( {A + C} \right)$ in 1 day$ = $Work done by $A$ in 1 day$ + $work done by the $C$ in 1 day

$ = \dfrac{1}{{24}} + \dfrac{1}{{32}}$

$ = \dfrac{{4 + 3}}{{96}} = \dfrac{7}{{96}}$

Then the time taken by$\left( {A + C} \right)$to finish work$ = \dfrac{1}{{\left( {\dfrac{7}{{96}}} \right)}}$days

$ = \dfrac{1}{7} \times 96 = \dfrac{{96}}{7}$

$ = 13\dfrac{5}{7}$ days

**So, the correct answer is “Option C”.**

**Note**: In these types of questions we can add or subtract quantities of only work done. We cannot add or subtract time which is taken to finish work. We need to avoid any type of calculation mistake.

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