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$A$ can do a piece of work in $24$ days, $A\& B$ can do it in 16 days and $A,B\& C$ in $10\dfrac{2}{3}$ days. In how many days can $A\& C$ do it?

Last updated date: 18th Jun 2024
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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.