10. (10 points) In 3 pencil cases, there are a total of 15 pencils and 14 pens. Each pencil case must contain at least 4 pencils and 2 pens. If the number of pencils in each pencil case is not less than the number of pens, then the pencil case with the most pens and pencils combined can have at most ( ) pens and pencils.
A. 12
B. 14
C. 15
D. 16
【Answer】: Solution: Pencils are only 1 more than pens, so in each bag, either pencils $=$ pens, or pencils $=$ pens +1. The bag with the most should have the most,

other bags should be as few as possible, the least can only be 4 pencils and 3 pens, the remaining: $15-4-4=7$ (pencils), $14-3-3=8$ (pens), which does not meet the requirement, so we take away one pen, the number of pens and pencils in the bags is as shown in the table below:
\begin{tabular}{|l|l|l|l|}
\hline & 4 & 4 & 7 \\
\hline Pencils & & & \\
\hline Pens & 3 & 4 & 7 \\
\hline
\end{tabular}

Obviously, the bag with the most pens and pencils has: $7+7=14$ pens and pencils.
Therefore, the answer is: $B$.