Problem 1350

1. We start with the given equations:
$$A = \frac{B + C}{4}$$
$$B = \frac{C + A}{6}$$

2. Rearrange the first equation to express $C$ in terms of $A$ and $B$:
$$4A = B + C \implies C = 4A - B$$

3. Rearrange the second equation to express $C$ in terms of $A$ and $B$:
$$6B = C + A \implies C = 6B - A$$

4. Now we have two expressions for $C$:
$$C = 4A - B$$
$$C = 6B - A$$

5. Set the two expressions for $C$ equal to each other:
$$4A - B = 6B - A$$

6. Solve for $A$ in terms of $B$:
$$4A + A = 6B + B \implies 5A = 7B \implies A = \frac{7}{5}B$$

7. Substitute $A = \frac{7}{5}B$ back into one of the original equations to find $C$ in terms of $B$:
$$C = 4A - B = 4\left(\frac{7}{5}B\right) - B = \frac{28}{5}B - B = \frac{28}{5}B - \frac{5}{5}B = \frac{23}{5}B$$

8. We need to find the ratio of $C$ to $A + B$:
$$A + B = \frac{7}{5}B + B = \frac{7}{5}B + \frac{5}{5}B = \frac{12}{5}B$$

9. Therefore, the ratio of $C$ to $A + B$ is:
$$\frac{C}{A + B} = \frac{\frac{23}{5}B}{\frac{12}{5}B} = \frac{23}{12}$$

The final answer is $\boxed{\frac{23}{12}}$.