This page from a Class 8 math textbook features practice questions on algebra and ratios, accompanied by an answer key for earlier exercises.

Problem: Solve the given math problems step by step.

#### Problem 34:
If a father is twice as old as his son and also 29 years older than his son, what is the age of the father?

Solution:
Let the age of the son be \( x \) years.
According to the problem:
1. The father's age is twice the son's age: \( 2x \).
2. The father's age is also 29 years more than the son's age: \( x + 29 \).

We can set up the equation:
\[
2x = x + 29
\]

Solve for \( x \):
\[
2x - x = 29
\]
\[
x = 29
\]

So, the son's age is \( 29 \) years.
The father's age is:
\[
2x = 2 \times 29 = 58 \text{ years}
\]

Answer:
\[
\boxed{58 \text{ years}}
\]

---

#### Problem 35:
Solve for \( x \): \( 9x + 36 = 4x + 91 \).

Solution:
Start with the given equation:
\[
9x + 36 = 4x + 91
\]

Rearrange the terms to isolate \( x \):
\[
9x - 4x = 91 - 36
\]
\[
5x = 55
\]

Solve for \( x \):
\[
x = \frac{55}{5} = 11
\]

Answer:
\[
\boxed{11}
\]

---

#### Problem 36:
Solve for \( x \): \( \frac{-2 + x}{3 + x} = 6 \).

Solution:
Start with the given equation:
\[
\frac{-2 + x}{3 + x} = 6
\]

Cross-multiply to eliminate the fraction:
\[
-2 + x = 6(3 + x)
\]

Expand the right-hand side:
\[
-2 + x = 18 + 6x
\]

Rearrange the terms to isolate \( x \):
\[
x - 6x = 18 + 2
\]
\[
-5x = 20
\]

Solve for \( x \):
\[
x = \frac{20}{-5} = -4
\]

Answer:
\[
\boxed{-4}
\]

---

#### Problem 37:
Simplify: \( (x + 7)^2 - (x - 7)^2 \).

Solution:
Use the difference of squares formula:
\[
(a + b)^2 - (a - b)^2 = 4ab
\]

Here, \( a = x \) and \( b = 7 \). Apply the formula:
\[
(x + 7)^2 - (x - 7)^2 = 4(x)(7) = 28x
\]

Answer:
\[
\boxed{28x}
\]

---

#### Problem 38:
Two numbers are in the ratio 4:7. If the sum of the numbers is 143, find the numbers.

Solution:
Let the two numbers be \( 4x \) and \( 7x \), where \( x \) is a common multiplier.
According to the problem, their sum is 143:
\[
4x + 7x = 143
\]

Combine like terms:
\[
11x = 143
\]

Solve for \( x \):
\[
x = \frac{143}{11} = 13
\]

Now, find the two numbers:
\[
4x = 4 \times 13 = 52
\]
\[
7x = 7 \times 13 = 91
\]

Answer:
\[
\boxed{52 \text{ and } 91}
\]

---

#### Problem 39:
The sides of a rectangle are in the ratio 14:3. If the perimeter of the rectangle is 170 cm, find the length and breadth.

Solution:
Let the length be \( 14x \) cm and the breadth be \( 3x \) cm, where \( x \) is a common multiplier.
The perimeter of a rectangle is given by:
\[
2(\text{length} + \text{breadth}) = 170
\]

Substitute the expressions for length and breadth:
\[
2(14x + 3x) = 170
\]

Simplify inside the parentheses:
\[
2(17x) = 170
\]
\[
34x = 170
\]

Solve for \( x \):
\[
x = \frac{170}{34} = 5
\]

Now, find the length and breadth:
\[
\text{Length} = 14x = 14 \times 5 = 70 \text{ cm}
\]
\[
\text{Breadth} = 3x = 3 \times 5 = 15 \text{ cm}
\]

Answer:
\[
\boxed{70 \text{ cm and } 15 \text{ cm}}
\]

---

#### Problem 40:
Find three consecutive odd numbers whose sum is 147.

Solution:
Let the three consecutive odd numbers be \( x \), \( x + 2 \), and \( x + 4 \).
According to the problem, their sum is 147:
\[
x + (x + 2) + (x + 4) = 147
\]

Combine like terms:
\[
3x + 6 = 147
\]

Solve for \( x \):
\[
3x = 147 - 6
\]
\[
3x = 141
\]
\[
x = \frac{141}{3} = 47
\]

Now, find the three consecutive odd numbers:
\[
x = 47, \quad x + 2 = 49, \quad x + 4 = 51
\]

Answer:
\[
\boxed{47, 49, 51}
\]

---

Final Answers:

1. Problem 34: \(\boxed{58 \text{ years}}\)
2. Problem 35: \(\boxed{11}\)
3. Problem 36: \(\boxed{-4}\)
4. Problem 37: \(\boxed{28x}\)
5. Problem 38: \(\boxed{52 \text{ and } 91}\)
6. Problem 39: \(\boxed{70 \text{ cm and } 15 \text{ cm}}\)
7. Problem 40: \(\boxed{47, 49, 51}\)