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):
\(100 - 0 = 212a - 32a\)
\(100 = 180a\)
\(a = \frac{100}{180}\)
\(a = \frac{5}{9}\)
\(Substituting a = \frac{5}{9} into equation (1):\)
\(0 = 32(\frac{5}{9}) + b\)
\(0 = \frac{160}{9} + b\)
\(b = -\frac{160}{9}\)
Therefore,
\(a = \frac{5}{9}\)
\(b = -\frac{160}{9}\)
The required relationship becomes:
\(°C = (\frac{5}{9})°F - \frac{160}{9}\)
\(Taking \frac{5}{9} common:\)
\(°C = (\frac{5}{9})(°F - 32)\)
This is the standard formula used to convert Fahrenheit temperature into Celsius temperature.
Answer:
\(a = \frac{5}{9}\)
\(b = -\frac{160}{9}\)
Linear Relationship:
\(°C = (\frac{5}{9})(°F - 32)\)