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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.

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.

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.