Mathematics solution NCERT
Class 9 - Chapter 2: Introduction to Linear Polynomials
Q1: 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 that the monthly bill is represented by
y = ax + b
where:
- x = number of modules accessed
- y = total monthly bill
- a = cost per module
- b = fixed monthly fee
From the question:
| Modules (x) | Bill (y) |
|---|---|
| 10 | ₹400 |
| 14 | ₹500 |
Substituting the first pair of values:
400 = 10a + b ..........(1)
Substituting the second pair of values:
500 = 14a + b ..........(2)
Subtracting equation (1) from equation (2):
500 − 400 = 14a − 10a
100 = 4a
a = 25
Substituting a = 25 in equation (1):
400 = 10(25) + b
400 = 250 + b
b = 150
Therefore,
a = 25 and b = 150
Required Linear Equation:
y = 25x + 150
Answer: Cost per module = ₹25 and fixed monthly fee = ₹150.
Q2: 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 use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.
Solution:
The monthly bill follows the relation
y = ax + b
where:
- x = hours of badminton court usage
- y = total bill
- a = charge per hour
- b = fixed monthly fee
| Hours (x) | Bill (y) |
|---|---|
| 10 | ₹800 |
| 15 | ₹1100 |
Substituting the first pair:
800 = 10a + b ..........(1)
Substituting the second pair:
1100 = 15a + b ..........(2)
Subtract equation (1) from equation (2):
1100 − 800 = 15a − 10a
300 = 5a
a = 60
Substitute a = 60 into equation (1):
800 = 10(60) + b
800 = 600 + b
b = 200
Therefore,
a = 60 and b = 200
Required Linear Equation:
y = 60x + 200
Answer: Court charge = ₹60 per hour and fixed monthly fee = ₹200.
Q3: 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°C and 32°F, and water boils at 100°C and 212°F.
Solution:
The relation is given by
°C = a°F + b
We are given two temperature points:
| °F | °C |
|---|---|
| 32 | 0 |
| 212 | 100 |
Using the first point (32, 0):
0 = 32a + b ..........(1)
Using the second point (212, 100):
100 = 212a + b ..........(2)
Subtract equation (1) from equation (2):
100 − 0 = 212a − 32a
100 = 180a
a = 100/180
a = 5/9
Substituting a = 5/9 into equation (1):
0 = 32(5/9) + b
0 = 160/9 + b
b = −160/9
Therefore,
a = 5/9
b = −160/9
The required relationship becomes:
°C = (5/9)°F − 160/9
Taking 5/9 common:
°C = (5/9)(°F − 32)
This is the standard formula used to convert Fahrenheit temperature into Celsius temperature.
Answer:
a = 5/9
b = −160/9
Linear Relationship:
°C = (5/9)(°F − 32)