jmc

Let $a$ denote the length of time, in minutes, that Anna holds her breath, $b$ denote the time that Bertram holds his breath, $c$ denote the time that Carli holds her breath, and $d$ denote the time that David holds his breath. Using the information in the problem, we can form the following system of linear equations (note that $\frac{2}{5}$ of an hour is the same as $24$ minutes): $$\begin{array}{rl}3a& =b+c+d\\ 4b& =a+c+d\\ 2c& =a+b+d\\ 8a+10b+6c& =24\end{array}$$ Subtracting the third equation from the first gives $3a-2c=c-a$, which simplifies to $4a=3c$. Subtracting the third equation from the second gives $4b-2c=c-b$, so $5b=3c$. We thus have that $4a=5b=3c$. Call this value $x$. Substituting $x$ for $4a$, $5b$, and $3c$ in the fourth equation gives $6x=24$, so $x=4$. Therefore, $a=\frac{4}{4}=1$, $b=\frac{4}{5}$, and $c=\frac{4}{3}$. Substituting these values into the first equation yields $3=\frac{4}{5}+\frac{4}{3}+d$, so $d=\frac{13}{15}$. Finally, the problem asks for the sum of the numerator and the denominator, so our answer is $\boxed{28}$.