Item 7.7.3.3.
Use the geometric distribution five times with changing values for p. For the first book p = 1 means you are certain to get a new title. For the second book title the probability of success is p=4/5; for the third book title the probability of success is p=3/5; for the fourth the probability is p=2/5; and for the last the probability is p = 1/5. Using the mean as 1/p in each case and accumulating these gives the total expected number of meals to purchase as
\begin{align*}
& 1 + \frac{1}{\frac{4}{5}} + \frac{1}{\frac{3}{5}} + \frac{1}{\frac{2}{5}} + \frac{1}{\frac{1}{5}} \\
& = 1 + \frac{5}{4} + \frac{5}{3} + \frac{5}{2} + \frac{5}{1} \\
& = \frac{12 + 15 + 25 + 30 + 60}{12} \\
& = \frac{142}{12} = 11.833
\end{align*}
and so you would need 12 kids meals. If this were to happen, please be certain to donate the "extra" books to an organization that works with kids or directly to some kids that you might know.