Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.2 Solutions: Introduction to Linear Polynomials

Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.2 Solutions cover evaluation of linear and quadratic polynomials, along with word problems based on linear equations. This exercise from Introduction to Linear Polynomials helps students substitute values in algebraic expressions and form equations from real-life situations.

Chapter 2 introduces students to linear polynomial Class 9 concepts through examples such as match fees, ages, coins, fencing and rectangle dimensions. In Ganita Manjari Class 9 Chapter 2 Exercise 2.2, students solve questions on the value of polynomial Class 9, linear equations and basic algebraic modelling. These Class 9 Maths Chapter 2 Exercise 2.2 Solutions explain each answer step by step in a clear, exam-ready format.

Key Takeaways

Polynomial Value: Substitute the given variable value and simplify.
Linear Equation: Word problems can be written as equations in one variable.
Ratio Method: Ratio-based numbers can be written as 2x and 5x.
Perimeter Application: Rectangle dimensions can be found using algebraic expressions.

Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.2 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.2 Solutions

Exercise Set 2.2 is based on finding values of polynomials and solving word problems using linear equations Class 9. Students first evaluate a linear polynomial and a quadratic polynomial by substituting given values. Then they solve application-based questions involving ages, ratios, coins, fence lengths and rectangle dimensions.

These Class 9 Maths linear polynomial exercise answers also revise earlier ideas from algebraic expressions Class 9, including variables, coefficients, constants and substitution.

Exercise 2.2 Question 1

Find the value of the linear polynomial 5x − 3 if:

(i) x = 0

Solution:

Given polynomial:

5x − 3

Substitute x = 0.

5x − 3 = 5(0) − 3

5x − 3 = 0 − 3

5x − 3 = −3

Answer: The value of the polynomial is −3.

(ii) x = −1

Solution:

Substitute x = −1 in 5x − 3.

5x − 3 = 5(−1) − 3

5x − 3 = −5 − 3

5x − 3 = −8

Answer: The value of the polynomial is −8.

(iii) x = 2

Solution:

Substitute x = 2 in 5x − 3.

5x − 3 = 5(2) − 3

5x − 3 = 10 − 3

5x − 3 = 7

Answer: The value of the polynomial is 7.

Exercise 2.2 Question 2

Find the value of the quadratic polynomial 7s² − 4s + 6 if:

(i) s = 0

Solution:

Given polynomial:

7s² − 4s + 6

Substitute s = 0.

7s² − 4s + 6 = 7(0)² − 4(0) + 6

7s² − 4s + 6 = 0 − 0 + 6

7s² − 4s + 6 = 6

Answer: The value of the polynomial is 6.

(ii) s = −3

Solution:

Substitute s = −3.

7s² − 4s + 6 = 7(−3)² − 4(−3) + 6

7s² − 4s + 6 = 7(9) + 12 + 6

7s² − 4s + 6 = 63 + 12 + 6

7s² − 4s + 6 = 81

Answer: The value of the polynomial is 81.

(iii) s = 4

Solution:

Substitute s = 4.

7s² − 4s + 6 = 7(4)² − 4(4) + 6

7s² − 4s + 6 = 7(16) − 16 + 6

7s² − 4s + 6 = 112 − 16 + 6

7s² − 4s + 6 = 102

Answer: The value of the polynomial is 102.

Exercise 2.2 Question 3

The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70 years. Find their present ages.

Solution:

Let Salil’s present age be x years.

Then his mother’s present age is 3x years.

After 5 years:

Salil’s age = x + 5

Mother’s age = 3x + 5

Their ages will add up to 70 years.

So,

(x + 5) + (3x + 5) = 70

4x + 10 = 70

4x = 70 − 10

4x = 60

x = 15

So, Salil’s present age is 15 years.

Mother’s present age:

3x = 3(15) = 45

Answer: Salil is 15 years old and his mother is 45 years old.

Exercise 2.2 Question 4

The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.

Solution:

The ratio of the two integers is 2:5.

Let the two integers be:

2x and 5x

Their difference is 63.

So,

5x − 2x = 63

3x = 63

x = 21

Now,

Smaller integer:

2x = 2(21) = 42

Larger integer:

5x = 5(21) = 105

Answer: The two integers are 42 and 105.

Exercise 2.2 Question 5

Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total ₹88, how many coins does she have of each type?

Solution:

Let the number of five-rupee coins be x.

Then the number of two-rupee coins is 3x.

Value of five-rupee coins:

5x

Value of two-rupee coins:

2 × 3x = 6x

Total value is ₹88.

So,

5x + 6x = 88

11x = 88

x = 8

Therefore:

Number of five-rupee coins = 8

Number of two-rupee coins:

3x = 3(8) = 24

Answer: Ruby has 8 five-rupee coins and 24 two-rupee coins.

Exercise 2.2 Question 6

A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?

Solution:

Let the shorter piece be x feet.

Then the longer piece is 4x feet.

Total length of fence = 300 feet.

So,

x + 4x = 300

5x = 300

x = 60

Shorter piece = 60 feet

Longer piece:

4x = 4(60) = 240 feet

Answer: The two pieces are 60 feet and 240 feet long.

Exercise 2.2 Question 7

If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?

Solution:

Let the width of the rectangle be x cm.

Length is three more than twice the width.

So,

Length = 2x + 3

Perimeter of rectangle:

2(length + width) = 24

Substitute the values.

2[(2x + 3) + x] = 24

2(3x + 3) = 24

6x + 6 = 24

6x = 24 − 6

6x = 18

x = 3

So, width = 3 cm

Length:

2x + 3 = 2(3) + 3

2x + 3 = 6 + 3

2x + 3 = 9

Answer: The rectangle has width 3 cm and length 9 cm.

Final Answers for Exercise 2.2

Concept Used in Introduction to Linear Polynomials Exercise 2.2

Introduction to Linear Polynomials Exercise 2.2 uses two main ideas: evaluating polynomials and solving linear equations.

To find the value of a polynomial, students substitute the given value of the variable in the expression. For example, in the linear polynomial 5x − 3, substituting x = 2 gives 7. In the quadratic polynomial Class 9 example 7s² − 4s + 6, substituting s = 4 gives 102.

For word problems, students form a linear equation from the given condition and solve it step by step. These ideas are central to Class 9 Ganita Manjari linear polynomials solutions and help students understand how algebra can model real-life situations.

About Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.2

Ganita Manjari Class 9 Chapter 2 Exercise 2.2 is part of Introduction to Linear Polynomials. It comes after students learn terms, variables, coefficients, constant terms and degree of polynomials.

These Class 9 Maths Chapter 2 Exercise 2.2 Solutions prepare students for later parts of the chapter, such as:

• linear polynomials,
• polynomial evaluation,
• linear equations,
• linear patterns,
• linear growth and decay,
• linear relationships,
• graphing equations of the form y = ax + b.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 2