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# A can do a work in $6$ hours, while B can do it in $8$ hours. With the help of C, they completed the work in $2\dfrac{1}{2}$ hours. In what time can C alone complete the work?

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Hint: In this problem we use a unitary method to solve time and work problems.
It is given that A can do a work in $6$ hours and B can do it in $8$ hours
All three together finish work in $2\dfrac{1}{2}$ hours.
Then we find C takes complete the work alone.

Complete step-by-step answer:
If a person can complete a work in days, we will take work done by a person in $1$ day is $\dfrac{1}{n}$
Now, we take
Work done by A in $1$ hour$= \dfrac{1}{6}$
Also, Work done by B in $1$ hour$= \dfrac{1}{8}$
Let C can do same work in $c$ hours
Work done by C in $1$ hour$= \dfrac{1}{c}$
All three together finish works in $2\dfrac{1}{2}$ hours
Now, work done by all three together$= \dfrac{2}{5}$
Work done by A+B+C = work done by all together,
We get,
$\dfrac{1}{6} + \dfrac{1}{8} + \dfrac{1}{c} = \dfrac{2}{5}$
Taking $\dfrac{1}{c}$on LHS, and remaining in subtracting to the RHS,
$\dfrac{1}{c} = \dfrac{2}{5} - \dfrac{1}{6} - \dfrac{1}{8}$
Now, finding the LCM of 5, 6, 8 and evaluate it
$\dfrac{1}{c} = \dfrac{{48 - 20 - 15}}{{120}}$
This implies that,
$\dfrac{1}{c} = \dfrac{{13}}{{120}}$
Taking the reciprocal on both side,
$c = \dfrac{{120}}{{13}}$
Now to divide the above terms and convert into the missed fraction.
$= 9\dfrac{3}{{13}}$ Hours.
Therefore, C takes $9\dfrac{3}{{13}}$ hours to finish the work alone.

Note: Unitary-method is all about finding value to a single unit.
If A can complete a work in days, work done by A in $1$ day is $\dfrac{1}{n}$. And if A can complete $\dfrac{1}{n}$ part of the work in $1$ day, then A will complete the work in days.