NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.5 (2026-27)

Question 1.
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:
The monthly bill is given by the linear relation:

y = ax + b

Here, x is the number of modules accessed, and y is the monthly bill.

When x = 10, y = 400:

400 = 10a + b …(i)

When x = 14, y = 500:

500 = 14a + b …(ii)

Subtract equation (i) from equation (ii):

500 - 400 = (14a + b) - (10a + b)
100 = 4a
a = 25

Now, substitute a = 25 into equation (i):

400 = 10(25) + b
400 = 250 + b
b = 150

Therefore, a = 25 and b = 150.

So, the monthly bill can be written as:

y = 25x + 150

Question 2.
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 depends on the hours of use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.

Solution:
The monthly bill is given by the linear relation:

y = ax + b

Here, x is the number of hours the badminton court is used, and y is the monthly bill.

When x = 10, y = 800:

800 = 10a + b …(i)

When x = 15, y = 1100:

1100 = 15a + b …(ii)

Subtract equation (i) from equation (ii):

1100 - 800 = (15a + b) - (10a + b)
300 = 5a
a = 60

Now, substitute a = 60 into equation (i):

800 = 10(60) + b
800 = 600 + b
b = 200

Therefore, a = 60 and b = 200.

So, the gym’s monthly bill can be written as:

y = 60x + 200

Question 3.
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:
The relation between Celsius and Fahrenheit is given as:

°C = a°F + b

When °C = 0 and °F = 32:

0 = 32a + b …(i)

When °C = 100 and °F = 212:

100 = 212a + b …(ii)

From equation (i):

b = -32a

Substitute b = -32a into equation (ii):

100 = 212a - 32a
100 = 180a
a = 100/180
a = 5/9

Now, substitute a = 5/9 in b = -32a:

b = -32 × 5/9
b = -160/9

Therefore, a = 5/9 and b = -160/9.

So, the linear relationship between Celsius and Fahrenheit is:

°C = (5/9)°F - 160/9

Think and Reflect (NCERT Textbook Page No. 27)

Can you guess what the numbers 20 and 150 in the equation y = 20x + 150 represent?

Solution:
In the equation y = 20x + 150, y represents the monthly bill, and x represents the internet data used in GB.

Here, 20 represents the extra charge for each GB of data used, and 150 represents the fixed monthly charge.

Key Points of Vedantu’s Class 9 Maths Chapter 2 Exercise 2.5

• Understand how a linear relation can be written as y = ax + b.

• Find the values of a and b using two given values of x and y.

• Apply linear equations to practical situations such as monthly bills and usage-based charges.

• Understand the meaning of the slope-like coefficient and fixed value in a linear relation.

• Use given Celsius and Fahrenheit values to form a linear relationship.

• Connect algebraic expressions with real-life quantities such as fees, hours, data usage, and temperature.

Access Exercise Wise NCERT Solutions for Chapter 2 Maths Class 9

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