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Class 9 Ganita ManjariClass 9 Maths Solutions | 2026–27
Chapter 2: Introduction to Linear PolynomialsPage 20 – 22All 7 Questions on This Page

Class 9 Maths Chapter 2 Exercise Set 2.2 Solutions

Complete step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 2 Exercise Set 2.2 (Questions 1–7). Covers evaluating linear and quadratic polynomials by substitution, solving linear age word problems, integer ratio problems, currency denomination problems, fence division, and rectangle dimensions.

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Key Concepts for Exercise Set 2.2

Jump to Question 1
1

Evaluating a Polynomial by Substitution

To find the value of a polynomial for a given value of the variable, substitute that numerical value in place of the variable wherever it appears in the expression and simplify using standard arithmetic order of operations.

For p(x) at x = a, calculate p(a)
2

Order of Operations with Powers and Signs

When evaluating terms with powers and negative coefficients, evaluate the exponent first before multiplying, and apply sign rules carefully: (−a)² = a² (positive) and −c(−a) = +ca.

(−3)² = 9 and −4(−3) = +12
3

Forming Linear Equations from Age Problems

Assign a variable (e.g. x) to the unknown base age. Express related ages in terms of x using the given relationships. After n years, add n to each person's age before setting up the equation according to the problem statement.

Present: x and 3x ⇒ After 5 years: (x + 5) + (3x + 5) = 70
4

Translating Ratios into Algebraic Quantities

When two quantities are in the ratio a:b, represent them as ax and bx using a common multiplier x > 0. Then set up a linear equation using their given sum, difference, or other condition.

Ratio 2:5 ⇒ Numbers are 2x and 5x ⇒ 5x − 2x = 63
5

Coin and Value Word Problems

When dealing with different denominations of currency, let x represent the unknown number of one denomination and express the other count in terms of x. Multiply each count by its respective coin value to set up the total value equation.

Total Value = (₹5 × x) + [₹2 × (3x)] = ₹88
6

Linear Equations for Geometric Dimensions

For problems involving a rectangle's perimeter, express one dimension in terms of the other (e.g. width = x, length = 2x + 3), and substitute into the standard perimeter formula 2(length + width) = P to solve for x.

Perimeter = 2(length + width) ⇒ 2[(2x + 3) + x] = 24

Find the value of the linear polynomial 5x − 3 if: (i) x = 0 (ii) x = −1 (iii) x = 2

Solution
Given linear polynomial:
5x − 3
(i) x = 0
Substituting x = 0,
5(0) − 3
= 0 − 3
= −3
Therefore:
−3
(ii) x = −1
Substituting x = −1,
5(−1) − 3
= −5 − 3
= −8
Therefore:
−8
(iii) x = 2
Substituting x = 2,
5(2) − 3
= 10 − 3
= 7
Therefore:
7
Answer:(i) −3 | (ii) −8 | (iii) 7

Find the value of the quadratic polynomial 7s² − 4s + 6 if: (i) s = 0 (ii) s = −3 (iii) s = 4

Solution
Given quadratic polynomial:
7s² − 4s + 6
(i) s = 0
Substituting s = 0,
7(0)² − 4(0) + 6
= 6
Therefore:
6
(ii) s = −3
Substituting s = −3,
7(−3)² − 4(−3) + 6
= 7(9) + 12 + 6
= 63 + 12 + 6
= 81
Therefore:
81
(iii) s = 4
Substituting s = 4,
7(4)² − 4(4) + 6
= 7(16) − 16 + 6
= 112 − 16 + 6
= 102
Therefore:
102
Answer:(i) 6 | (ii) 81 | (iii) 102

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 then add up to 70 years.

Therefore:

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

4x + 10 = 70
4x = 60
x = 15
Therefore, Salil’s present age is:

15 years

His mother’s present age is:

3(15) = 45 years
Hence:
Salil = 15 years
Mother = 45 years
Verification
After 5 years:
Salil = 20 years
Mother = 50 years
20 + 50 = 70
So the answer satisfies the given condition.
Answer:Salil is 15 years old and his mother is 45 years old.

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

Since the difference between them is 63:

5x − 2x = 63
3x = 63
x = 21
Therefore, the two integers are:
2(21) = 42

and

5(21) = 105
Hence:

42 and 105

Verification
Difference:
105 − 42 = 63
Ratio:
42:105 = 2:5
Thus both given conditions are satisfied.
Answer:The two positive integers are 42 and 105.

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 ₹5 coins be x.

Then the number of ₹2 coins is:

3x

The total value of the coins is ₹88.

Therefore:
5x + 2(3x) = 88
5x + 6x = 88
11x = 88
x = 8
So, the number of ₹5 coins is:
8

and the number of ₹2 coins is:

3(8) = 24
Therefore:

24 two-rupee coins and 8 five-rupee coins

Verification
Value of 24 two-rupee coins:
24 × ₹2 = ₹48
Value of 8 five-rupee coins:
8 × ₹5 = ₹40
Total:
₹48 + ₹40 = ₹88
Hence:

24 two-rupee coins and 8 five-rupee coins

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

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

Since the total length is 300 feet:

x + 4x = 300
5x = 300
x = 60
Therefore, the shorter piece is:

60 feet

and the longer piece is:

4(60) = 240 feet
Hence:

60 feet and 240 feet

Verification
60 + 240 = 300

and

240 = 4 × 60

Both given conditions are satisfied.

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

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.

Then the length is:

2x + 3 cm

The perimeter of a rectangle is:

2(length + width)

Given perimeter = 24 cm:

2[(2x + 3) + x] = 24
2(3x + 3) = 24
6x + 6 = 24
6x = 18
x = 3
Therefore, the width is:

3 cm

and the length is:

2(3) + 3 = 9 cm
Hence:
Length = 9 cm
Width = 3 cm
Verification
Length condition:
2(3) + 3 = 9
Perimeter:
2(9 + 3) = 24 cm

Both conditions are satisfied.

Answer:The dimensions of the rectangle are Length = 9 cm and Width = 3 cm.

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