NCERT Exemplar Class 8 Maths Unit 4 Linear Equations in One Variable - Free Printable

Problem Analysis:

The task involves solving a series of mathematical problems related to age, algebraic equations, and number properties. Each problem is independent but follows a logical structure. Below, I will solve each problem step by step.

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Problem 33:

Statement:
3 years ago, the age of a boy was \( y \) years. His age 2 years ago was \( (y - 2) \) years.

Solution:
- Let the boy's current age be \( A \).
- 3 years ago, his age was \( y \). Therefore, his current age is \( A = y + 3 \).
- 2 years ago, his age was \( (y - 2) \). Therefore, his current age is also \( A = (y - 2) + 2 = y \).

This confirms the relationship between the ages. The statement is consistent with the given information.

Answer:
\[
\boxed{y}
\]

---

Problem 34:

Statement:
Shikha’s present age is \( p \) years. Reemu’s present age is 4 times the present age of Shikha. After 5 years, Reemu’s age will be \( 15p \) years.

Solution:
- Shikha’s present age = \( p \).
- Reemu’s present age = \( 4p \).
- After 5 years, Reemu’s age will be \( 4p + 5 \).
- According to the problem, after 5 years, Reemu’s age will be \( 15p \).

Set up the equation:
\[
4p + 5 = 15p
\]

Solve for \( p \):
\[
5 = 15p - 4p
\]
\[
5 = 11p
\]
\[
p = \frac{5}{11}
\]

Answer:
\[
\boxed{\frac{5}{11}}
\]

---

Problem 35:

Statement:
In a 2-digit number, the units place digit is \( x \). If the sum of the digits is 9, then the number is \( (10x - 9) \).

Solution:
- Let the tens digit be \( t \) and the units digit be \( x \).
- The number can be expressed as \( 10t + x \).
- The sum of the digits is given as 9:
\[
t + x = 9
\]
- Solve for \( t \):
\[
t = 9 - x
\]
- Substitute \( t \) back into the expression for the number:
\[
\text{Number} = 10t + x = 10(9 - x) + x = 90 - 10x + x = 90 - 9x
\]

However, the problem states the number is \( 10x - 9 \). This appears to be a misinterpretation or error in the problem statement. The correct expression for the number, based on the given conditions, is \( 90 - 9x \).

Answer:
\[
\boxed{90 - 9x}
\]

---

Problem 36:

Statement:
The sum of the ages of Anju and her mother is 65 years. If Anju’s present age is \( y \) years, then her mother’s age before 5 years was \( (60 - y) \) years.

Solution:
- Let Anju’s present age be \( y \).
- Let her mother’s present age be \( m \).
- The sum of their ages is 65:
\[
y + m = 65
\]
- Solve for \( m \):
\[
m = 65 - y
\]
- Her mother’s age before 5 years was:
\[
m - 5 = (65 - y) - 5 = 60 - y
\]

This matches the given statement.

Answer:
\[
\boxed{60 - y}
\]

---

Problem 37:

Statement:
The number of boys and girls in a class are in the ratio 5:4. If the number of boys is 9 more than the number of girls, then the number of boys is 9.

Solution:
- Let the number of boys be \( 5k \) and the number of girls be \( 4k \), where \( k \) is a positive integer.
- According to the problem, the number of boys is 9 more than the number of girls:
\[
5k = 4k + 9
\]
- Solve for \( k \):
\[
5k - 4k = 9
\]
\[
k = 9
\]
- The number of boys is:
\[
5k = 5 \times 9 = 45
\]

The problem states that the number of boys is 9, which contradicts the solution. There seems to be an error in the problem statement.

Answer:
\[
\boxed{45}
\]

---

Problem 38:

Statement:
A and B are together 90 years old. Five years ago, A was thrice as old as B was. Hence, the ages of A and B five years back would be \( (x - 5) \) years and \( (85 - x) \) years respectively.

Solution:
- Let A’s current age be \( A \) and B’s current age be \( B \).
- The sum of their ages is 90:
\[
A + B = 90
\]
- Five years ago, A’s age was \( A - 5 \) and B’s age was \( B - 5 \). According to the problem:
\[
A - 5 = 3(B - 5)
\]
- Simplify the equation:
\[
A - 5 = 3B - 15
\]
\[
A = 3B - 10
\]
- Substitute \( A = 3B - 10 \) into \( A + B = 90 \):
\[
(3B - 10) + B = 90
\]
\[
4B - 10 = 90
\]
\[
4B = 100
\]
\[
B = 25
\]
- Solve for \( A \):
\[
A = 90 - B = 90 - 25 = 65
\]
- Five years ago, A’s age was:
\[
A - 5 = 65 - 5 = 60
\]
- Five years ago, B’s age was:
\[
B - 5 = 25 - 5 = 20
\]

The problem states the ages five years back as \( (x - 5) \) and \( (85 - x) \). Setting \( x = 65 \):
\[
x - 5 = 65 - 5 = 60
\]
\[
85 - x = 85 - 65 = 20
\]

These match the calculated values.

Answer:
\[
\boxed{60 \text{ and } 20}
\]

---

Problem 39:

Statement:
Two different equations can never have the same answer.

Solution:
This statement is false. Two different equations can have the same solution if they are equivalent or if they represent the same relationship. For example:
\[
x + 1 = 3 \quad \text{and} \quad 2x + 2 = 6
\]
Both equations have the solution \( x = 2 \).

Answer:
\[
\boxed{\text{False}}
\]

---

Problem 40:

Statement:
In the equation \( 3x - 3 = 9 \), transposing \(-3\) to RHS, we get \( 3x = 9 \).

Solution:
- Start with the equation:
\[
3x - 3 = 9
\]
- Transpose \(-3\) to the right-hand side:
\[
3x = 9 + 3
\]
\[
3x = 12
\]

The statement is false because transposing \(-3\) results in \( 3x = 12 \), not \( 3x = 9 \).

Answer:
\[
\boxed{\text{False}}
\]

---

Problem 41:

Statement:
In the equation \( 2x = 4 - x \), transposing \(-x\) to LHS, we get \( x = 4 \).

Solution:
- Start with the equation:
\[
2x = 4 - x
\]
- Transpose \(-x\) to the left-hand side:
\[
2x + x = 4
\]
\[
3x = 4
\]
- Solve for \( x \):
\[
x = \frac{4}{3}
\]

The statement is false because transposing \(-x\) results in \( x = \frac{4}{3} \), not \( x = 4 \).

Answer:
\[
\boxed{\text{False}}
\]

---

Problem 42:

Statement:
If \( \frac{15}{8} - 7x = 9 \), then \( -7x = 9 + \frac{15}{8} \).

Solution:
- Start with the equation:
\[
\frac{15}{8} - 7x = 9
\]
- Transpose \( \frac{15}{8} \) to the right-hand side:
\[
-7x = 9 - \frac{15}{8}
\]

The statement is false because transposing \( \frac{15}{8} \) results in \( -7x = 9 - \frac{15}{8} \), not \( -7x = 9 + \frac{15}{8} \).

Answer:
\[
\boxed{\text{False}}
\]

---

Problem 43:

Statement:
If \( \frac{x}{3} + 1 = \frac{7}{15} \), then \( \frac{x}{3} = \frac{6}{15} \).

Solution:
- Start with the equation:
\[
\frac{x}{3} + 1 = \frac{7}{15}
\]
- Transpose 1 to the right-hand side:
\[
\frac{x}{3} = \frac{7}{15} - 1
\]
- Convert 1 to a fraction with a denominator of 15:
\[
\frac{x}{3} = \frac{7}{15} - \frac{15}{15}
\]
\[
\frac{x}{3} = \frac{7 - 15}{15}
\]
\[
\frac{x}{3} = \frac{-8}{15}
\]

The statement is false because \( \frac{x}{3} = \frac{-8}{15} \), not \( \frac{x}{3} = \frac{6}{15} \).

Answer:
\[
\boxed{\text{False}}
\]

---

Problem 44:

Statement:
If \( 6x = 18 \), then \( 18x = 54 \).

Solution:
- Start with the equation:
\[
6x = 18
\]
- Solve for \( x \):
\[
x = \frac{18}{6} = 3
\]
- Substitute \( x = 3 \) into \( 18x \):
\[
18x = 18 \times 3 = 54
\]

The statement is true.

Answer:
\[
\boxed{\text{True}}
\]

---

Problem 45:

Statement:
If \( \frac{x}{11} = 15 \), then \( x = \frac{11}{15} \).

Solution:
- Start with the equation:
\[
\frac{x}{11} = 15
\]
- Solve for \( x \):
\[
x = 15 \times 11 = 165
\]

The statement is false because \( x = 165 \), not \( x = \frac{11}{15} \).

Answer:
\[
\boxed{\text{False}}
\]

---

Problem 46:

Statement:
If \( x \) is an even number, then the next even number is \( 2(x + 1) \).

Solution:
- If \( x \) is an even number, the next even number is \( x + 2 \).
- The statement suggests the next even number is \( 2(x + 1) \), which simplifies to:
\[
2(x + 1) = 2x + 2
\]

This is true because \( x + 2 \) is indeed the next even number.

Answer:
\[
\boxed{\text{True}}
\]

---

Problem 47:

Statement:
If the sum of two consecutive numbers is 93 and one of them is \( x \), then the other number is \( 93 - x \).

Solution:
- Let the two consecutive numbers be \( x \) and \( x + 1 \).
- Their sum is given as 93:
\[
x + (x + 1) = 93
\]
\[
2x + 1 = 93
\]
\[
2x = 92
\]
\[
x = 46
\]
- The other number is:
\[
x + 1 = 47
\]

The statement is true because the other number is indeed \( 93 - x \).

Answer:
\[
\boxed{\text{True}}
\]

---

Problem 48:

Statement:
Two numbers differ by 40. When each number is increased by 8, the bigger becomes thrice the lesser number. If one number is \( x \), then the other number is \( 40 - x \).

Solution:
- Let the two numbers be \( x \) and \( x + 40 \) (since they differ by 40).
- When each number is increased by 8:
- The first number becomes \( x + 8 \).
- The second number becomes \( x + 40 + 8 = x + 48 \).
- According to the problem, the bigger number (after increase) is thrice the lesser number:
\[
x + 48 = 3(x + 8)
\]
- Solve for \( x \):
\[
x + 48 = 3x + 24
\]
\[
48 - 24 = 3x - x
\]
\[
24 = 2x
\]
\[
x = 12
\]
- The other number is:
\[
x + 40 = 12 + 40 = 52
\]

The statement is false because the other number is \( x + 40 \), not \( 40 - x \).

Answer:
\[
\boxed{\text{False}}
\]

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Final Answers:

\[
\boxed{y, \frac{5}{11}, 90 - 9x, 60 - y, 45, 60 \text{ and } 20, \text{False}, \text{False}, \text{False}, \text{False}, \text{True}, \text{False}, \text{True}, \text{True}, \text{False}}
\]