Master Chef

Relevant wiki: Harmonic Mean

The average cooking time in the problem is, in fact, a harmonic mean within the group, and we know that the mean of the initial three chefs equals to that of the five chefs.

Now let $x_{1}$ be the number of work-time in the initial group, and $x_{2}$ be that of the group after new hiring.

$F = \dfrac{3}{\sum \dfrac{1}{x_1}} = \dfrac{5}{\sum \dfrac{1}{x_2}}$

Thus, $\dfrac{3}{F} = \sum \dfrac{1}{x_1}$ and $\dfrac{5}{F} = \sum \dfrac{1}{x_2}$ .

$\dfrac{5-3}{F} = \sum \dfrac{1}{x_2} - \sum \dfrac{1}{x_1}$

The difference of the sums of the groups will belong to the new chef fractions:

$\dfrac{2}{F} = \dfrac{1}{6} + \dfrac{1}{9}$

$\dfrac{2}{\dfrac{1}{6} + \dfrac{1}{9}} = F$

As a result, the mean cooking time between the two new chefs will equal to the initial group:

$F = \dfrac{2}{\dfrac{1}{6} + \dfrac{1}{9}} = \dfrac{3}{\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{x}}$

$\dfrac{3}{8} + \dfrac{1}{x} = (\dfrac{3}{2})(\dfrac{2+3}{18})$

$\dfrac{1}{x} = \dfrac{5}{12} - \dfrac{3}{8} = \dfrac{10 - 9}{24} = \dfrac{1}{24}$

Finally, chef $C$ can cook one dish in $\boxed{24}$ minutes.

5(1/4 + 1/8 + 1/x) = 3(1/4 + 1/8 + 1/x + 1/6 + 1/9) so x=24