NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 13 Two Variables One Line Exercise 13.4

Exercise Set 13.4

Question 1. On two different days a family buys movie tickets and snack boxes. Let the cost of one movie ticket be ₹x and the cost of one snack box be ₹y. Frame a pair of linear equations in x and y to represent the following situations:
(i) On the first day, the family buys 2 movie tickets and 3 snack boxes for ₹850.
(ii) On the second day, the family buys 4 movie tickets and 1 snack box for ₹1100.
Answer:
(i) 2 tickets cost 2x and 3 snack boxes cost 3y: 2x + 3y = 850.
(ii) 4 tickets cost 4x and 1 snack box costs y: 4x + y = 1100.
Required pair: 2x + 3y = 850 and 4x + y = 1100.
(Solving: y = 1100 − 4x, so 2x + 3300 − 12x = 850, giving x = 245 and y = 120. A ticket costs ₹245 and a snack box ₹120.)

Question 2. Two friends, Sahil and Meena, travelled by taxi. Let the fixed charge for one taxi trip be ₹x and the additional charge per kilometre be ₹y. Frame a pair of linear equations in x and y to represent the following situation:
(i) Sahil travelled 6 km and paid a total fare of ₹122.
(ii) Meena travelled 8 km and paid a total fare of ₹160.
Answer:
Fare = fixed charge + (charge per km × distance).
(i) Sahil: x + 6y = 122.
(ii) Meena: x + 8y = 160.
Required pair: x + 6y = 122 and x + 8y = 160.
(Solving: subtracting gives 2y = 38, so y = 19 and x = 122 − 114 = 8. Fixed charge ₹8, charge per km ₹19.)

Teacher's Note:
1. The fixed charge is paid once per trip, so it has coefficient 1 in both equations.
2. Subtracting the equations removes the fixed charge immediately.

Question 3. In a sports meet, tickets for adults cost ₹150 each and tickets for children cost ₹100 each. A total of 200 people attended the meet, and the total amount collected from ticket sales was ₹25,000. (Let the number of adult tickets sold be x and the number of children's tickets sold be y.) Frame a pair of linear equations in x and y to represent this situation.
Answer:
Number of people: x + y = 200.
Money collected: 150x + 100y = 25000 (which simplifies, on dividing by 50, to 3x + 2y = 500).
Required pair: x + y = 200 and 150x + 100y = 25000.
(Solving: 3x + 2y = 500 and 2x + 2y = 400 give x = 100 and y = 100, so 100 adult and 100 children's tickets were sold.)