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Class 9 Ganita ManjariClass 9 Maths Solutions | 2026–27
Chapter 2: Introduction to Linear PolynomialsPage 27 – 28All 3 Questions on This Page

Class 9 Maths Chapter 2 Exercise Set 2.5 Solutions

Complete step-by-step solutions for Class 9 Mathematics Ganita Manjari Chapter 2 Exercise Set 2.5 (Questions 1–3). Covers finding the constants a and b in linear relationships of the form y = ax + b from two known value pairs, interpreting fixed and variable components, and establishing temperature conversion between Celsius and Fahrenheit.

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Key Concepts for Exercise Set 2.5

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1

Linear Relationship in Two Variables: y = ax + b

A relationship where one variable depends linearly on another has the standard algebraic form y = ax + b. Here, a represents the uniform rate of change, and b represents the fixed initial or baseline value.

y = ax + b (Linear Relationship)
2

Determining Parameters from Two Known Pairs

Given two known pairs of values (x₁, y₁) and (x₂, y₂), substituting them into y = ax + b yields two simultaneous linear equations in a and b. Subtracting the equations eliminates b to find a, which is then substituted back to determine b.

y₂ − y₁ = a(x₂ − x₁) ⇒ Eliminate b to find a and b
3

Physical Interpretation of Constants a and b

In billing and cost models, the constant term b represents the fixed charge (such as a monthly subscription or base fee), while the coefficient a represents the variable cost per unit of usage (such as per module or per hour).

Total Bill y = (Variable Rate a × Usage x) + Fixed Fee b
4

Linear Temperature Conversion

Physical temperature scales with fixed reference points (such as melting and boiling points of water) are related by linear equations. Setting ice melting (32°F, 0°C) and water boiling (212°F, 100°C) uniquely determines the linear conversion constants.

°C = (5/9)(°F) − 160/9 or 9(°C) = 5(°F) − 160
Question 1
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A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.

Solution
Given,
y = ax + b

When x = 10, the bill is ₹400:

400 = 10a + b...(1)

When x = 14, the bill is ₹500:

500 = 14a + b...(2)
Subtracting (1) from (2),
500 − 400 = 14a − 10a
100 = 4a
a = 25
Substituting a = 25 in (1),
400 = 10(25) + b
400 = 250 + b
b = 150
Therefore,
a = 25
b = 150

Here, a = ₹25 represents the additional cost per module, while b = ₹150 represents the fixed monthly fee.

The linear relationship is:

y = 25x + 150
Answer:a = 25 | b = 150 | Linear relationship: y = 25x + 150
Question 2
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A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.

Solution
Given,
y = ax + b

When x = 10, the bill is ₹800:

800 = 10a + b...(1)

When x = 15, the bill is ₹1100:

1100 = 15a + b...(2)
Subtracting (1) from (2),
1100 − 800 = 15a − 10a
300 = 5a
a = 60
Substituting a = 60 in (1),
800 = 10(60) + b
800 = 600 + b
b = 200
Therefore,
a = 60
b = 200

Here, a = ₹60 represents the additional cost per hour, while b = ₹200 represents the fixed monthly fee.

The linear relationship is:

y = 60x + 200
Answer:a = 60 | b = 200 | Linear relationship: y = 60x + 200
Question 3
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Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a °F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit. (Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)

Solution

Given the linear relationship,

°C = a(°F) + b

When °C = 0 and °F = 32,

0 = 32a + b...(1)

When °C = 100 and °F = 212,

100 = 212a + b...(2)
Subtracting (1) from (2),
100 = 180a
a = 100180
= 59
Substituting a = 5/9 in (1),
0 = 32(59) + b
0 = 1609 + b
b = −1609
Therefore,
a = 59
b = −1609
Therefore, the linear relationship between Celsius and Fahrenheit is:
°C = (59)(°F) − 1609

Equivalently:

9(°C) = 5(°F) − 160
Verification

For 32°F:

°C = (59)(32) − 1609
= 1609 − 1609
= 0°C

For 212°F:

°C = (59)(212) − 1609
= 10609 − 1609
= 9009
= 100°C
Thus, the relationship satisfies both given temperature values.
Answer:a = 59 | b = −1609 | Linear relationship: °C = (59)(°F) − 1609

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