The ratio of ages of A and B 4 years ago was $5 : 3$. The sum of present ages of A, B and C is $80$ years. If the present age of C is equal to the sum of present ages of A and B, find the present age of A.

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**Solution:**

1. Let the present ages of A and B be $x$ and $y$ years respectively.

2. Given the ratio of their ages 4 years ago:
   $$\frac{x - 4}{y - 4} = \frac{5}{3}$$

3. Cross-multiply:
   $$3(x - 4) = 5(y - 4)$$
   $$3x - 12 = 5y - 20$$
   $$3x = 5y - 8 \quad \ldots (i)$$

4. Let the present age of C be $z$ years.

5. Given the sum of present ages:
   $$x + y + z = 80$$

6. Also, present age of C equals sum of present ages of A and B:
   $$z = x + y$$

7. Substitute $z = x + y$ into the sum equation:
   $$x + y + (x + y) = 80$$
   $$2(x + y) = 80$$
   $$x + y = 40 \quad \ldots (ii)$$

8. From equations (i) and (ii), solve for $x$ and $y$:

From (ii):
   $$y = 40 - x$$

Substitute into (i):
   $$3x = 5(40 - x) - 8$$
   $$3x = 200 - 5x - 8$$
   $$3x + 5x = 192$$
   $$8x = 192$$
   $$x = 24$$

9. Therefore, the present age of A is $24$ years.

**Answer:** 24 years

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