The average age of A, B and C is $20$ years. Two years ago the sum of the ages of A and B was $6$ more than that of C. Find the present age of C.

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**Solution:**

1. Given the average age of A, B, and C is $20$ years.

2. Calculate the sum of their present ages:
   - Sum = Average × Number of persons = $20 	imes 3 = 60$ years.

3. Calculate the sum of their ages 2 years ago:
   - Each person was 2 years younger, so total reduction = $2 	imes 3 = 6$ years.
   - Sum of ages 2 years ago = $60 - 6 = 54$ years.

4. According to the problem, 2 years ago, the sum of ages of A and B was 6 more than that of C:
   - Let age of C 2 years ago be $x$.
   - Then, sum of ages of A and B 2 years ago = $x + 6$.

5. Total sum 2 years ago = sum of A and B + age of C = $(x + 6) + x = 2x + 6$.

6. Equate this to the total sum 2 years ago:
   $$2x + 6 = 54$$

7. Solve for $x$:
   $$2x = 54 - 6 = 48$$
   $$x = 24$$

8. Present age of C = age 2 years ago + 2 = $24 + 2 = 26$ years.

**Answer:** $26$ years

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