Exercise 2.5 Class 9 Maths Solutions | Linear Equations in Real Life (Ganita Manjari)

Exercise 2.5 focuses on applying linear equations in real-life situations. In this exercise, students learn how to form equations using given data and determine unknown constants. These problems strengthen algebraic thinking and problem-solving skills.


Exercise 2.5 Solutions


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

Given: y=ax+b

From data:

When x=10, y=400

So  400=10a+b….(1)

When x=14, y=500

So  500=14a+b….(2)

Subtract (2) from  (1)

500−400=(14a+b)−(10a+b)

100=4a

⇒a=25

Substitute into (1)

400=10(25)+b

400=250+b

⇒b=150

Hence a=25,b=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 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

From data:

800=10a+b…(1)

1100=15a+b…(2)

Subtract (1) from (2)

300=5a⇒a=60

Substitute into (1):

800=600+b

⇒b=200

Hence a=60,b=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.


Solution

Given relation: °C=a(°F)+b

From data:

When °F=32, then °C=0

So  0=32a+b…(1)

When °F=212, °C=100

So 100=212a+b…(2)

Subtract (1) from (2)

100=180a

So  a=95

Substitute into (1)

0=32(95)+b

b=−9160

Hence a=95,b=−9160


Conclusion

This exercise helps students:

  • Convert real-life situations into linear equations
  • Solve equations using substitution and elimination
  • Understand practical applications like billing and temperature conversion

It is an important step toward mastering algebra and analytical thinking.


Get the solutions of other exercises of this chapter