Welcome to Sid Classes. In this detailed guide, we provide accurate, step-by-step solutions for NCERT Class 9 Maths Ganita Manjari Chapter 2 Exercise 2.5. This exercise focuses on finding the constants \(a\) and \(b\) in a linear relation \(y = ax + b\) given specific pairs of values, which introduces students to simultaneous linear conditions.
Our solutions utilize clean standard mathematical notation (\(\text{LaTeX}\)) and breakdown logic to ensure you score 100% marks in your school tests and board examinations.
🔑 Key Concepts to Remember:
- Linear Equation Constants: In the general equation \(y = ax + b\), \(a\) represents the rate of change (slope) and \(b\) represents the fixed base value (\(y\)-intercept).
- Method of Substitution: Form two separate linear equations using the given data pairs, then subtract or substitute them to isolate variables and solve for \(a\) and \(b\).
NCERT Class 9 Maths Ganita Manjari Chapter 2 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 the linear relation: $$y = ax + b$$
Step 1: Form the first equation using the first case.
When \(x = 10\) (modules), \(y = 400\) (bill amount):
$$400 = 10a + b \quad \text{--- (Equation 1)}$$
Step 2: Form the second equation using the second case.
When \(x = 14\) (modules), \(y = 500\) (bill amount):
$$500 = 14a + b \quad \text{--- (Equation 2)}$$
Step 3: Solve for \(a\) by subtracting Equation 1 from Equation 2.
$$(500 - 400) = (14a + b) - (10a + b)$$
$$100 = 4a$$
$$a = \frac{100}{4} = \mathbf{25}$$
Step 4: Solve for \(b\) by substituting the value of \(a\) back into Equation 1.
$$400 = 10(25) + b$$
$$400 = 250 + b$$
$$b = 400 - 250 = \mathbf{150}$$
Final Answer: The value of \(a\) is 25 (additional cost per module) and the value of \(b\) is 150 (fixed monthly fee).
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 the linear relation: $$y = ax + b$$
Step 1: Form the first equation.
When court hours \(x = 10\), total bill \(y = 800\):
$$800 = 10a + b \quad \text{--- (Equation 1)}$$
Step 2: Form the second equation.
When court hours \(x = 15\), total bill \(y = 1100\):
$$1100 = 15a + b \quad \text{--- (Equation 2)}$$
Step 3: Subtract Equation 1 from Equation 2 to find \(a\).
$$(1100 - 800) = (15a + b) - (10a + b)$$
$$300 = 5a$$
$$a = \frac{300}{5} = \mathbf{60}$$
Step 4: Substitute \(a = 60\) into Equation 1 to find \(b\).
$$800 = 10(60) + b$$
$$800 = 600 + b$$
$$b = 800 - 600 = \mathbf{200}$$
Final Answer: The value of \(a\) is 60 (cost per court hour) and the value of \(b\) is 200 (fixed monthly gym fee).
Question 3
Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by \(\text{°C} = a\text{°F} + b\). 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 the temperature relation: $$\text{°C} = a(\text{°F}) + b$$
Step 1: Set up Equation 1 using the melting point of ice.
When \(\text{°C} = 0\), \(\text{°F} = 32\):
$$0 = a(32) + b \implies 32a + b = 0 \quad \text{--- (Equation 1)}$$
Step 2: Set up Equation 2 using the boiling point of water.
When \(\text{°C} = 100\), \(\text{°F} = 212\):
$$100 = a(212) + b \implies 212a + b = 100 \quad \text{--- (Equation 2)}$$
Step 3: Subtract Equation 1 from Equation 2 to eliminate \(b\).
$$(212a + b) - (32a + b) = 100 - 0$$
$$180a = 100$$
$$a = \frac{100}{180} = \mathbf{\frac{5}{9}}$$
Step 4: Find \(b\) by substituting \(a = \frac{5}{9}\) into Equation 1.
$$32\left(\frac{5}{9}\right) + b = 0$$
$$\frac{160}{9} + b = 0$$
$$b = \mathbf{-\frac{160}{9}}$$
Final Answer: The value of \(a\) is \(\mathbf{\frac{5}{9}}\) and the value of \(b\) is \(\mathbf{-\frac{160}{9}}\).
(Note: This results in the standard conversion formula: \(\text{°C} = \frac{5}{9}(\text{°F} - 32)\)).
Frequently Asked Questions (FAQs)
Q1: What do the constants 'a' and 'b' signify in these linear expressions?
A: The constant \(a\) is the coefficient of variable \(x\) representing the variable rate of change. The constant \(b\) is an independent term representing the initial static or fixed baseline cost.
Q2: How do you solve simultaneous linear equations formed from real-world contexts?
A: The easiest method is subtraction elimination. Because both equations contain a single \(+ b\) term with a coefficient of 1, subtracting one equation from the other removes \(b\) instantly, letting you isolate \(a\).
Q3: Can the value of constants 'a' or 'b' be negative numbers?
A: Yes. As shown in the temperature conversion question (Question 3), constants can be negative fractions or integers depending on the intersection points of the physical variables.
Q4: What NCERT book includes Exercise 2.5 patterns?
A: This chapter corresponds directly to the latest NCERT Grade 9 Ganita Manjari Part I Mathematics syllabus tracker guidelines.
Q5: Is this system distinct from standard linear expressions studied in prior modules?
A: Yes, earlier exercises gave you the expressions outright. Exercise 2.5 requires you to find the inner mathematical system constraints from baseline numeric data observations.
Q6: Will these dynamic responsive frames work cleanly inside darker layout themes?
A: Yes, the container boundaries use alpha-channel color codes (`rgba`) that cleanly tint background styling elements across both theme variants automatically.
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