Ten years ago, the average age of A and B was $30$ years. Eight years from now, the ratio of the age of A to that of B will be $5 : 7$. If the age of C five years ago was $\frac{7}{9}$ of the age of B three years ago, what is the present age of C (in years)?

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**Solution:**

1. Given that ten years ago, the average age of A and B was 30 years, so the total age of A and B at that time was:
   
   $$2 	imes 30 = 60 	ext{ years}$$

2. Therefore, the sum of their present ages is:
   
   $$60 + 10 + 10 = 80 	ext{ years}$$
   
   (Since 10 years have passed for both A and B, add $10 + 10 = 20$ years to the total.)

3. Let the present ages of A and B be $5x - 8$ and $7x - 8$ respectively, based on the ratio given for 8 years hence:
   
   - Eight years from now, ages will be $5x$ and $7x$ respectively.
   - So, present ages are $5x - 8$ and $7x - 8$.

4. According to the sum of present ages:
   
   $$ (5x - 8) + (7x - 8) = 80 $$
   $$ 12x - 16 = 80 $$
   $$ 12x = 96 $$
   $$ x = 8 $$

5. Present age of B is:
   
   $$7x - 8 = 7 	imes 8 - 8 = 56 - 8 = 48 	ext{ years}$$

6. Age of B three years ago:
   
   $$48 - 3 = 45 	ext{ years}$$

7. Age of C five years ago was $\frac{7}{9}$ of B's age three years ago:
   
   $$	ext{Age of C five years ago} = \frac{7}{9} 	imes 45 = 35 	ext{ years}$$

8. Therefore, present age of C is:
   
   $$35 + 5 = 40 	ext{ years}$$

**Answer:** $40$

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