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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.