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Ganita Manjari · Class 9 · Part II · Chapter 13

Ganita Manjari Class 9 Maths Chapter 13 Two Variables, One Line Exercise 13.4 Solutions

Three questions on framing pairs of linear equations: movie tickets and snacks, a taxi fare with a fixed charge, and tickets at a sports meet.

Last updated 6 October 2026

  • 3Questions
  • 0Diagrams
  • Ch 13Two Variables, One Line

How to answer every question

  1. Name the unknowns Say clearly what x and y stand for.
  2. Write one fact per equation Each sentence of the problem gives one equation.
  3. Use units consistently All terms of an equation must be in the same unit, such as rupees.
  4. Check with numbers Test your equation with a simple example.

Pair of linear equations: two equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.

Fixed charge: a part of the cost that does not depend on the quantity.

Rate: the cost per unit, such as ₹ per km.

Framing equations

Chapter 13 Exercise 13.4 Question 1 Solution One movie ticket costs ₹x and one snack box costs ₹y. (i) On the first day the family buys 2 tickets and 3 snack boxes for ₹850. (ii) On the second day they buy 4 tickets and 1 snack box for ₹1100. Frame the equations.

Answer: 2x + 3y = 850; 4x + y = 1100
Cost = number × price, added
DayTicketsSnack boxesEquation
First232x + 3y = 850
Second414x + y = 1100

2x + 3y = 850 and 4x + y = 1100.

Chapter 13 Exercise 13.4 Question 2 Solution A taxi trip has a fixed charge ₹x and ₹y per km. (i) Sahil travelled 6 km and paid ₹122. (ii) Meena travelled 8 km and paid ₹160. Frame the equations.

Answer: x + 6y = 122; x + 8y = 160
Fare = fixed charge + distance × rate
PersonDistanceFareEquation
Sahil6 km₹122x + 6y = 122
Meena8 km₹160x + 8y = 160

x + 6y = 122 and x + 8y = 160.

Chapter 13 Exercise 13.4 Question 3 Solution Adult tickets cost ₹150 and children's tickets ₹100. 200 people attended and ₹25,000 was collected. With x adult tickets and y children's tickets, frame the equations.

Answer: x + y = 200; 150x + 100y = 25000
Two facts
  • Number of people: x + y = 200.
  • Money collected: 150x + 100y = 25000 (dividing by 50 gives 3x + 2y = 500).

x + y = 200 and 150x + 100y = 25000.

Answers at a glance

Each sentence of the problem gives one equation.

QuestionWhat is askedKey ideaAnswer
Q1 Tickets and snacksOne fact per equation2x + 3y = 850, 4x + y = 1100
Q2 Taxi fareFixed + rate × kmx + 6y = 122, x + 8y = 160
Q3 Sports meetPeople and moneyx + y = 200, 150x + 100y = 25000

Other chapters of Ganita Manjari (Class 9)

Source of the questions: NCERT, Ganita Manjari, Grade 9, Part II, Chapter 13, Exercise Set 13.4. The solutions, explanations and diagrams on this page are our own working.

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