NCERT Solutions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials Exercise 2.2 | 2026-27

Think and Reflect

1. Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?

Solution

Since the perimeter of a square of side x cm = 4x cm

Perimeter of a square of side 1 cm = 4 × 1=4 cm

Perimeter of a square of side 1.5 cm = 4 × 1.5 = 6 cm

Perimeter of a square of side 2 cm =4 × 2 = 8 cm

Perimeter of a square of side 2.5 cm = 4 × 2.5 = 10 cm

Perimeter of a square of side 3cm= 4 × 3 = 12 cm

Observation

The perimeter increases by exactly 2 cm whenever the side length increases by 0.5 cm. This fixed rate of change represents a key characteristic of linear polynomials.

Think and Reflect

1. If a player paid ₹ 750, how many matches did he play? 

Solution

Given

Joining fee = ₹200

Cost for each match = ₹50

Total amount paid = ₹(200 + 50m), where m is the number of matches played.

A player paid ₹750.

200 + 50m = 750

Subtract 200 from both sides:

50m = 750 - 200         50m = 550

Divide both sides by 50:

m = 550/50

m = 11

Therefore, the player played 11 matches.

Think and Reflect

1. We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, 10x − x², is a function of x. Can you interpret this as an input-output process? What value does the expression take when x = 6 cm?

Solution

Given expression:

10x − x²

Here, x is the input and the value of 10x − x² is the output.

For x = 6 cm,

10x − x²

Substituting x = 6,

         = 10(6) − 6²

                              = 60 − 36

                              = 24

Therefore, when x = 6 cm, the value of the expression 10x − x² is 24.

Exercise Set 2.2 

1. Find the value of the linear polynomial 5x – 3 if: 

(i) x = 0               (ii) x = –1                (iii) x = 2 

(i) Solution

When x = 0; the linear polynomial

5x – 3 = 5 × 0 – 3 = -3

(ii) Solution

When x = -1; the linear polynomial

5x – 3 = 5 × (-1) – 3 = -5 – 3 = -8

(iii)Solution

When x = 2; the linear polynomial

5x – 3 = 5 × 2 – 3 = 10 – 3 = 7

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

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

Solution: 

Given polynomial: 7s² − 4s + 6 

(i) 7s² − 4s + 6 = 7(0)² − 4(0) + 6 

      = 0 − 0 + 6 

       = 6 

Therefore, the value of the polynomial when s = 0 is 6. 

(ii) 7s² − 4s + 6 = 7(−3)² − 4(−3) + 6 

        = 7(9) + 12 + 6 

        = 63 + 12 + 6 

        = 81 

Therefore, the value of the polynomial when s = −3 is 81. 

(iii) 7s² − 4s + 6 = 7(4)² − 4(4) + 6 

        = 7(16) − 16 + 6 

        = 112 − 16 + 6 

                  = 102 

Therefore, the value of the polynomial when s = 4 is 102. 

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.

Salil’s mother’s present age = 3x years

After 5 years:

Salil’s age = x + 5 years

Salil’s mother’s age = 3x + 5 years

According to the question,

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

4x + 10 = 70

4x = 70 − 10

4x = 60

x = 60 ÷ 4

x = 15

Therefore,

Salil’s present age = 15 years

Salil’s mother’s present age = 3 × 15 = 45 years

Hence, Salil’s present age is 15 years, and his mother’s present age is 45 years.

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

Solution:

Let the two positive integers be 2x and 5x.

5x − 2x = 63

3x = 63

x = 63 ÷ 3

x = 21

Therefore,

First integer = 2x = 2 × 21 = 42

Second integer = 5x = 5 × 21 = 105

Hence, the two integers are 42 and 105. 

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.

Number of two-rupee coins = 3x

Value of five-rupee coins = 5x

Value of two-rupee coins = 2 × 3x = 6x

According to the question,

5x + 6x = 88

11x = 88

x = 88 ÷ 11

x = 8

Therefore,

Number of five-rupee coins = 8

Number of two-rupee coins = 3 × 8 = 24

Hence, Ruby has 8 five-rupee coins and 24 two-rupee coins.

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 long. 

Length of the longer piece = 4x feet 

x + 4x = 300 

5x = 300 

x = 300 ÷ 5 

x = 60 

Therefore, 

Length of the shorter piece = 60 feet 

Length of the longer piece = 4 × 60 = 240 feet 

Hence, the two pieces of the fence are 60 feet and 240 feet long. 

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 of the rectangle = 2x + 3 cm 

The perimeter of a rectangle is: 

2(length + width) = 24 

Substituting the values, 

2((2x + 3) + x) = 24 

2(3x + 3) = 24 

6x + 6 = 24 

6x = 24 − 6 

6x = 18 

x = 18 ÷ 6 

x = 3 

Therefore, the width of the rectangle = 3 cm 

Length of the rectangle = 2 × 3 + 3 = 9 cm 

Hence, the dimensions of the rectangle are: 

Length = 9 cm and Width = 3 cm. 

Key Benefits of Vedantu’s NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2

• Helps students understand the concepts covered in Class 9 Maths Chapter 2 Exercise 2.2 clearly.

• Provides accurate and step-by-step solutions for all NCERT textbook questions.

• Makes it easier to solve Class 9 Ex 2.2 problems with proper methods.

• Improves calculation accuracy and logical thinking skills.

• Helps students revise important concepts quickly before exams.

• Supports effective exam preparation with easy-to-understand explanations.

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