The present age of A is 12 years more than that of C, while C is 3 years younger than B. The ratio of the present age of A to B is $6 : 5$. What will be the sum of the ages of B and C after 10 years?

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**Solution:**

1. Let the present age of C be $x$ years.

2. Then, the present age of A is $x + 12$ years.

3. Since C is 3 years younger than B, the present age of B is $x + 3$ years.

4. According to the problem, the ratio of A's age to B's age is:
   $$\frac{x + 12}{x + 3} = \frac{6}{5}$$

5. Cross-multiplying, we get:
   $$5(x + 12) = 6(x + 3)$$
   $$5x + 60 = 6x + 18$$

6. Rearranging terms:
   $$60 - 18 = 6x - 5x$$
   $$42 = x$$

7. So, the present age of C is $42$ years.

8. Present age of B is $42 + 3 = 45$ years.

9. After 10 years, the age of C will be $42 + 10 = 52$ years, and the age of B will be $45 + 10 = 55$ years.

10. The sum of their ages after 10 years is:
    $$52 + 55 = 107$$ years.

**Answer:** $107$

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