Exercise Set 2.5 Class 9 · Ganita Manjari
📘 Chapter 2 — Polynomials
📌 Finding a and b in y = ax + b
📐
When a linear relationship is given by y = ax + b, we can find the constants
a and b by using two known pairs of values
(x, y) and solving the resulting system of equations.
1
A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed.
A student observed 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.
Given:
- When x = 10, y = 400
- When x = 14, y = 500
Using y = ax + b:
For x = 10: 400 = 10a + b …(1)
For x = 14: 500 = 14a + b …(2)
For x = 14: 500 = 14a + b …(2)
Subtracting (1) from (2), we get:
500 − 400 = 14a − 10a
⇒ 100 = 4a
⇒ a = 25
⇒ 100 = 4a
⇒ a = 25
Substituting a = 25 in (1), we get:
400 = 10(25) + b
⇒ 400 = 250 + b
⇒ b = 150
⇒ 400 = 250 + b
⇒ b = 150
✅
Therefore, a = 25 and b = 150.
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.
Given:
- When x = 10, y = 800
- When x = 15, y = 1100
Using y = ax + b:
For x = 10: 800 = 10a + b …(1)
For x = 15: 1100 = 15a + b …(2)
For x = 15: 1100 = 15a + b …(2)
Subtracting (1) from (2), we have:
1100 − 800 = 15a − 10a
⇒ 300 = 5a
⇒ a = 60
⇒ 300 = 5a
⇒ a = 60
Substitute a = 60 in (1), we get:
800 = 10(60) + b
⇒ 800 = 600 + b
⇒ b = 200
⇒ 800 = 600 + b
⇒ b = 200
✅
Therefore, a = 60 and b = 200.
3
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 degrees Celsius
and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.
(Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)
(Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)
Given relation: °C = a°F + b
Using the point (°F, °C) = (32, 0), we have:
0 = 32a + b …(1)
Using the point (°F, °C) = (212, 100), we have:
100 = 212a + b …(2)
Subtracting (1) from (2), we get:
100 − 0 = 212a − 32a
⇒ 100 = 180a
⇒ a = 100/180
⇒ a = 5/9
⇒ 100 = 180a
⇒ a = 100/180
⇒ a = 5/9
Substitute a = 5/9 in (1), we get:
0 = 32(5/9) + b
⇒ 0 = 160/9 + b
⇒ b = −160/9
⇒ 0 = 160/9 + b
⇒ b = −160/9
✅
Therefore, a = 5/9 and b = −160/9.
Hence, the linear relationship is:
°C = (5/9)°F − 160/9
It can also be written as:
°C = (5/9)(°F − 32)