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Introduction to Linear Polynomials Class 9 Mathematics Ganita Manjari [LATEST] Solutions Exercise Set 2.5 in English - CBSE Study

Introduction to Linear Polynomials Mathematics Ganita Manjari Class 9 exercise - [LATEST] Solutions Exercise Set 2.5 cbse board school study materials like cbse notes in English medium, all chapters and exercises are covered the ncert latest syllabus 2026 - 27.

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Introduction to Linear Polynomials Class 9 Mathematics Ganita Manjari [LATEST] Solutions Exercise Set 2.5 in English - CBSE Study

NCERT Solutions for Class 9 Mathematics Ganita Manjari are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important Introduction to Linear Polynomials with detailed explanations and step-by-step answers for better exam preparation. Each Exercise Set 2.5 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 9 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics Ganita Manjari.

Class 9 English Medium Mathematics Ganita Manjari All Chapters:

Introduction to Linear Polynomials

5. Exercise Set 2.5

Exercise Set 2.5

Q1. 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 relation:

y = ax + b

When x = 10, y = 400

So,

400 = 10a + b      ......(1)

When x = 14, y = 500

So,

500 = 14a + b      ......(2)

Subtract equation (1) from equation (2):

500 − 400 = 14a − 10a

100 = 4a

a = 100/4

a = 25

Now substitute a = 25 in equation (1):

400 = 10(25) + b

400 = 250 + b

b = 400 − 250

b = 150

Therefore:

a = 25

b = 150

Q2. 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 relation:

y = ax + b

When x = 10, y = 800

So,

800 = 10a + b      ......(1)

When x = 15, y = 1100

So,

1100 = 15a + b      ......(2)

Subtract equation (1) from equation (2):

1100 − 800 = 15a − 10a

300 = 5a

a = 300/5

a = 60

Now substitute a = 60 in equation (1):

800 = 10(60) + b

800 = 600 + b

b = 800 − 600

b = 200

Therefore:

a = 60

b = 200

Q3. 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.

Solution:

Given relation:

°C = a°F + b

When °C = 0 and °F = 32

So,

0 = 32a + b      ......(1)

When °C = 100 and °F = 212

So,

100 = 212a + b      ......(2)

Subtract equation (1) from equation (2):

100 − 0 = 212a − 32a

100 = 180a

a = 100/180

a = 5/9

Now substitute a = 5/9 in equation (1):

0 = 32(5/9) + b

0 = 160/9 + b

b = −160/9

Therefore:

a = 5/9

b = −160/9

Hence, the linear relationship between °C and °F is:

°C = (5/9)°F − 160/9

 

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