Ex 13.4 Class 9 Maths Ganita Manjari Solutions | Sid Classes
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 13 Exercise 13.4 Framing Pairs of Linear Equations in Two Variables from Word Problems and Everyday Situations. Designed to reflect how top-scoring students write their solutions and reasoning in school examinations, this walkthrough covers translating real-world scenarios involving movie tickets, snack boxes, taxi fares, and sports meet ticket sales into algebraic linear equations entirely clear.
🔑 Key Concepts & Strategies:
- Variable Assignment: Clearly defining unknown quantities using variables (such as \(x\) and \(y\)) based on the problem statement.
- Equation Framing: Translating statements of total cost, fixed charges, variable rates, and combined quantities into standard algebraic linear equations in two variables.
NCERT Class 9 Maths Ganita Manjari Chapter 13 Exercise 13.4 Solutions
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.
Solution:
Given:
* Cost of one movie ticket = ₹\(x\)
* Cost of one snack box = ₹\(y\)
(i) First day situation:
Buying 2 movie tickets and 3 snack boxes for a total of ₹850 gives the linear equation:
\(\mathbf{2x + 3y = 850}\)
(ii) Second day situation:
Buying 4 movie tickets and 1 snack box for a total of ₹1100 gives the linear equation:
\(\mathbf{4x + y = 1100}\)
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.
Solution:
Given:
* Fixed charge for a trip = ₹\(x\)
* Additional charge per km = ₹\(y\)
(i) Sahil's travel:
Traveling 6 km means paying the fixed charge plus 6 times the per-kilometre charge for a total of ₹122:
\(\mathbf{x + 6y = 122}\)
(ii) Meena's travel:
Traveling 8 km means paying the fixed charge plus 8 times the per-kilometre charge for a total of ₹160:
\(\mathbf{x + 8y = 160}\)
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.
Solution:
Given:
* Number of adult tickets sold = \(x\) (costing ₹150 each)
* Number of children's tickets sold = \(y\) (costing ₹100 each)
1. Total people equation:
The total number of people who attended is 200:
\(x + y = 200\)
2. Total collection equation:
The total amount collected from selling \(x\) adult tickets at ₹150 and \(y\) children's tickets at ₹100 is ₹25,000:
\(150x + 100y = 25000\)
Thus, the required pair of linear equations is:
\(\mathbf{x + y = 200}\)
\(\mathbf{150x + 100y = 25000}\) (which can also be simplified as \(\mathbf{3x + 2y = 500}\))
Frequently Asked Questions (FAQs)
Q1: How do you form linear equations from word problems involving fixed and variable costs?
A: Assign variables to the fixed base amount and the per-unit variable rate, then set up equations representing total amounts as their sum.
Q2: Can a pair of linear equations represent different conditions of the same real-world scenario?
A: Yes, different instances or days provide independent equations that share the same underlying variable values.
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