Exercise Set 2.4 Class 9 · Ganita Manjari
📘 Chapter 2 — Polynomials
📌 Linear Growth & Decay
📐
Linear growth occurs when a quantity increases by a constant amount each period:
y = a + bt (where b > 0).
Linear decay occurs when a quantity decreases by a constant amount each period: y = a − bt (where b > 0).
Linear decay occurs when a quantity decreases by a constant amount each period: y = a − bt (where b > 0).
1
Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.
(i) Find the height after 7 months.
(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
(i) Find the height after 7 months.
(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
Given:
- Initial height = 1.75 feet
- Growth per month = 0.5 feet
(i)
Height after 7 months
Height growth in each month = 0.5 feet
h = 1.75 + (0.5 × 7)
= 1.75 + 3.5
= 5.25 feet
= 1.75 + 3.5
= 5.25 feet
(ii)
Table of values
| t (months) | h (feet) |
|---|---|
| 0 | 1.75 |
| 1 | 2.25 |
| 2 | 2.75 |
| 3 | 3.25 |
| 4 | 3.75 |
| 5 | 4.25 |
| 6 | 4.75 |
| 7 | 5.25 |
| 8 | 5.75 |
| 9 | 6.25 |
| 10 | 6.75 |
(iii)
Expression relating h and t
Let height after t months = h
h = 1.75 + 0.5t
✅
This represents linear growth because height increases by a constant amount (0.5 feet) every month.
2
A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.
(i) Find the value of the phone after 3 years.
(ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
(i) Find the value of the phone after 3 years.
(ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
Given:
- Initial value = ₹10,000
- Decrease per year = ₹800
(i)
Value after 3 years
Decrease in value after 1 year = ₹800
v = 10000 − (800 × 3)
= 10000 − 2400
= ₹7600
= 10000 − 2400
= ₹7600
(ii)
Table of values
| t (years) | v (₹) |
|---|---|
| 0 | 10000 |
| 1 | 9200 |
| 2 | 8400 |
| 3 | 7600 |
| 4 | 6800 |
| 5 | 6000 |
| 6 | 5200 |
| 7 | 4400 |
| 8 | 3600 |
(iii)
Expression relating v and t
Let value after t years = v
v = 10000 − 800t
✅
This represents linear decay because the value decreases by a constant amount (₹800) every year.
3
The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.
(i) Find the population of the village after 6 years.
(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
(iii) Find an expression that relates P and t, and explain why it represents linear growth.
(i) Find the population of the village after 6 years.
(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
(iii) Find an expression that relates P and t, and explain why it represents linear growth.
(i)
Population after 6 years
P = 750 + (50 × 6)
= 750 + 300
= 1050
= 750 + 300
= 1050
(ii)
Table of values
| t (years) | P (population) |
|---|---|
| 0 | 750 |
| 1 | 800 |
| 2 | 850 |
| 3 | 900 |
| 4 | 950 |
| 5 | 1000 |
| 6 | 1050 |
| 7 | 1100 |
| 8 | 1150 |
| 9 | 1200 |
| 10 | 1250 |
(iii)
Expression relating P and t
Let population after t years = P
P = 750 + 50t
✅
This represents linear growth because the population increases by a constant number (50 people) every year.
4
A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after recharge.
(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x) reduces with time.
(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x) reduces with time.
(i)
Equation for remaining balance
Let remaining balance after x days = b(x)
b(x) = 600 − 15x
✅
This represents linear decay because the balance decreases by a constant amount (₹15) every day.
(ii)
When will the balance run out?
When balance runs out:
b(x) = 0
600 − 15x = 0
15x = 600
x = 40
600 − 15x = 0
15x = 600
x = 40
📅
So, the balance will run out after 40 days.
(iii)
Table of values
| x (days) | b(x) (₹) |
|---|---|
| 1 | 585 |
| 2 | 570 |
| 3 | 555 |
| 4 | 540 |
| 5 | 525 |
| 6 | 510 |
| 7 | 495 |
| 8 | 480 |
| 9 | 465 |
| 10 | 450 |