Four years ago, the ratio of the ages of A and B was $3 : 4$. The average of the present ages of A, B, and C is 26 years. C is 11 years younger than B. What is the present age of B?

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**Solution:**

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

2. Given that four years ago, the ratio of ages of A and B was $3 : 4$, so:
   $$\frac{x - 4}{y - 4} = \frac{3}{4}$$
   Cross-multiplying:
   $$4(x - 4) = 3(y - 4)$$
   $$4x - 16 = 3y - 12$$
   $$4x - 3y = 4 \quad \ldots (i)$$

3. The average of the present ages of A, B, and C is 26 years, so:
   $$\frac{x + y + z}{3} = 26 \implies x + y + z = 78 \quad \ldots (ii)$$

4. C is 11 years younger than B, so:
   $$z = y - 11 \quad \ldots (iii)$$

5. Substitute $z$ from (iii) into (ii):
   $$x + y + (y - 11) = 78$$
   $$x + 2y = 89 \quad \ldots (iv)$$

6. From (i), express $x$ in terms of $y$:
   $$4x - 3y = 4 \implies 4x = 3y + 4 \implies x = \frac{3y + 4}{4}$$

7. Substitute $x$ into (iv):
   $$\frac{3y + 4}{4} + 2y = 89$$
   Multiply both sides by 4:
   $$3y + 4 + 8y = 356$$
   $$11y + 4 = 356$$
   $$11y = 352$$
   $$y = 32$$

8. Find $x$:
   $$x = \frac{3(32) + 4}{4} = \frac{96 + 4}{4} = \frac{100}{4} = 25$$

9. Find $z$:
   $$z = y - 11 = 32 - 11 = 21$$

**Answer:** The present age of B is $32$ years.

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