Class 9 Maths Chapter 2 Exercise 2.2 Solutions | NCERT Ganita Manjari 2026

Looking for complete Class 9 Maths Chapter 2 Exercise 2.2 Solutions? Here you will get detailed step-by-step solutions of all questions from the latest NCERT Ganita Manjari 2026 textbook. These Introduction to Linear Polynomials Exercise 2.2 Solutions cover evaluation of polynomials, linear equations, age problems, ratio problems, coin problems and practical applications in a simple CBSE answer-writing format.

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Class 9 Maths Chapter 2 Exercise 2.2 Solutions – Introduction to Linear Polynomials

Welcome to the complete Class 9 Maths Chapter 2 Exercise 2.2 Solutions based on the latest NCERT Ganita Manjari (2026) textbook. In this exercise, students learn how to evaluate linear and quadratic polynomials, solve real-life word problems using linear equations, and apply algebraic concepts in practical situations.

These NCERT Class 9 Maths Exercise 2.2 Solutions are prepared in a step-by-step format following the latest CBSE examination pattern. Every question is explained with detailed calculations, important concepts, common mistakes, and exam-oriented tips.

📘 Important Formulas and Concepts Used in Exercise 2.2

Before solving the questions of Class 9 Maths Chapter 2 Exercise 2.2, let us quickly revise the important concepts and formulas used in this exercise.

Class 9 Maths Chapter 2 Exercise 2.2 Solutions

Below are the complete step-by-step NCERT Ganita Manjari 2026 Exercise 2.2 Solutions prepared according to the latest CBSE pattern.

Question 1

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

• (i) x = 0
• (ii) x = −1
• (iii) x = 2

Given:

p(x)=5x−3

To Find:

The value of the polynomial for the given values of x.

Solution:

To find the value of a polynomial, substitute the given value of the variable and simplify.

(i) When x = 0

p(0)=5(0)−3

=0−3

=−3

(ii) When x = −1

p(−1)=5(−1)−3

=−5−3

=−8

(iii) When x = 2

p(2)=5(2)−3

=10−3

=7

Key Concept Used: To evaluate a polynomial, substitute the given value of the variable and simplify the expression.

Common Mistake: Students often forget to use brackets while substituting negative values, which leads to sign errors.

Exam Tip: Always substitute negative values inside brackets first and then perform the arithmetic operations step-by-step.

Question 2

Find the value of the quadratic polynomial

7s² − 4s + 6

if:

• (i) s = 0
• (ii) s = −3
• (iii) s = 4

Given:

p(s)=7s²−4s+6

To Find:

The value of the polynomial for the given values of s.

Solution:

To find the value of a polynomial, substitute the given value of the variable and simplify the expression using the order of operations.

(i) When s = 0

p(0)=7(0)²−4(0)+6

=7(0)−0+6

=6

(ii) When s = −3

p(−3)=7(−3)²−4(−3)+6

=7(9)+12+6

=63+12+6

=81

(iii) When s = 4

p(4)=7(4)²−4(4)+6

=7(16)−16+6

=112−16+6

=102

Key Concept Used: To evaluate a polynomial, substitute the given value of the variable and simplify according to the BODMAS rule.

Common Mistake: Students often make mistakes while squaring negative numbers, such as writing (−3)² = −9 instead of 9.

Exam Tip: Always write negative values inside brackets before squaring. This helps avoid sign errors and ensures accurate calculations.

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.

Given:

• Salil’s mother’s present age is three times Salil’s present age.
• After 5 years, the sum of their ages will be 70 years.

To Find:

The present ages of Salil and his mother.

Solution:

Let Salil’s present age be

x years

Then, Salil’s mother’s present age will be

3x years

After 5 years,

• Salil’s age = (x + 5) years
• Mother’s age = (3x + 5) years

According to the question,

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

x + 5 + 3x + 5 = 70

4x + 10 = 70

4x = 70 − 10

4x = 60

x = 15

Therefore,

Salil’s present age = 15 years

Mother’s present age = 3 × 15 = 45 years

Verification:

After 5 years,

Salil’s age = 15 + 5 = 20 years

Mother’s age = 45 + 5 = 50 years

20 + 50 = 70 ✓

Key Concept Used: Represent the unknown age by a variable and form a linear equation using the given conditions.

Common Mistake: Students often write the mother’s age after 5 years as 3(x + 5), whereas the correct expression is 3x + 5.

Exam Tip: In age-related problems, always assume the present age first and then carefully write the ages after the given number of years.

Question 4

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

Given:

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

To Find:

The two positive integers.

Solution:

Since the ratio of the two integers is

2 : 5,

let the two integers be

2x and 5x.

According to the question,

5x − 2x = 63

3x = 63

x = 21

Therefore,

First integer = 2 × 21 = 42

Second integer = 5 × 21 = 105

Verification:

Difference of the two integers:

105 − 42 = 63 ✓

Ratio of the two integers:

42 : 105

= 42 105 = 2 5 ✓

Key Concept Used: When the ratio of two quantities is given, represent them as multiples of a common variable and then form a linear equation.

Common Mistake: Students often assume the numbers as x and x + 63 instead of using the given ratio.

Exam Tip: Whenever a ratio is given, first express the quantities in terms of a common variable before forming the equation.

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?

Given:

• The number of two-rupee coins is three times the number of five-rupee coins.
• The total amount is ₹88.

To Find:

The number of two-rupee coins and five-rupee coins.

Solution:

Let the number of five-rupee coins be

x

Then, the number of two-rupee coins will be

3x

Step 1: Form the equation using the total amount

According to the question,

5x + 6x = 88

11x = 88

x = 8

Step 2: Find the number of each type of coin

Number of five-rupee coins = x = 8

Number of two-rupee coins = 3x

= 3 × 8

= 24

Verification:

Value of five-rupee coins = 8 × 5 = ₹40

Value of two-rupee coins = 24 × 2 = ₹48

₹40 + ₹48 = ₹88 ✓

Key Concept Used: Represent the unknown quantity by a variable and form a linear equation using the total value of the coins.

Common Mistake: Students often write the value of the two-rupee coins as 3x instead of 2(3x)=6x.

Exam Tip: In coin problems, first prepare a table showing the number of coins, value of each coin, and total value before forming the equation.

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?

Given:

• The total length of the fence is 300 feet.
• The longer piece is four times the shorter piece.

To Find:

The lengths of the two pieces.

Solution:

Let the length of the shorter piece be

x feet

Then, the length of the longer piece will be

4x feet

Step 1: Form the equation

According to the question,

x + 4x = 300

5x = 300

x = 60

Step 2: Find the lengths of the two pieces

Length of the shorter piece = 60 feet

Length of the longer piece = 4 × 60

= 240 feet

Verification:

60 + 240 = 300 ✓

240 = 4 × 60 ✓

Key Concept Used: Represent the unknown quantity by a variable and form a linear equation using the given relationship.

Common Mistake: Students often assume the longer piece as x and the shorter piece as 4x, which leads to an incorrect equation.

Exam Tip: In ratio or comparison problems, assign the smaller quantity as the variable whenever possible to simplify calculations.

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?

Given:

• The length of the rectangle is three more than twice its width.
• The perimeter of the rectangle is 24 cm.

To Find:

The length and width of the rectangle.

Solution:

Let the width of the rectangle be

x cm

Then, the length of the rectangle will be

(2x + 3) cm

Step 1: Use the formula for perimeter

We know that,

Perimeter = 2(Length + Width)

Substituting the given values,

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

2(3x + 3) = 24

6x + 6 = 24

6x = 24 − 6

6x = 18

x = 3

Step 2: Find the dimensions of the rectangle

Width = x = 3 cm

Length = 2x + 3

= 2(3) + 3

= 9 cm

Verification:

Perimeter = 2(9 + 3)

= 2(12)

= 24 cm ✓

Key Concept Used: Represent the unknown dimension by a variable and use the perimeter formula of a rectangle.

Common Mistake: Students often forget to multiply by 2 while applying the perimeter formula of a rectangle.

Exam Tip: In geometry word problems, always write the formula first and then substitute the given information carefully.

What We Learned in Class 9 Maths Chapter 2 Exercise 2.2

In this exercise, we learned how to evaluate linear and quadratic polynomials, solve real-life problems using linear equations, and represent practical situations using algebraic expressions. These concepts form the foundation for higher algebra in Class 9 and Class 10 Mathematics.

What We Learned in Class 9 Maths Chapter 2 Exercise 2.2 Solutions

In these Class 9 Maths Chapter 2 Exercise 2.2 Solutions, we learned how to evaluate linear and quadratic polynomials, solve real-life problems using linear equations, and represent practical situations using algebraic expressions. These NCERT Class 9 Maths Chapter 2 Solutions Exercise 2.2 Solutions are based on the latest Ganita Manjari 2026 textbook and help students build a strong foundation in Introduction to Linear Polynomials. The concepts covered in Exercise 2.2 Solutions are important for higher algebra topics in Class 9 and Class 10 Mathematics.

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About the Author

This solution has been prepared by Rakesh Kumar Singh, founder of Newton Study Point, with more than 18 years of Mathematics teaching experience. The solutions are prepared according to the latest NCERT Ganita Manjari 2026 and CBSE examination pattern.