Exercise 2.4 focuses on linear growth and linear decay in real-life situations. Students learn how to form equations, create tables, and understand patterns where quantities increase or decrease at a constant rate. Below are clear and step-by-step solutions.
Exercise 2.4 Solutions
Question 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.
Solution
(i) Height after 7 months: h=1.75+0.5×7=1.75+3.5=5.25 feet
Answer: 5.25 feet
(ii) Table of values (t = 0 to 10)
| 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) Linear expression: h=1.75+0.5t
It is linear because height increases by a constant amount (0.5).
Question 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.
Solution
(i) Value after 3 years:
v=10000−800×3=7600
Answer: ₹7600
(ii) Table (t = 0 to 8)
| t (years) | Value (₹) |
|---|---|
| 0 | 10000 |
| 1 | 9200 |
| 2 | 8400 |
| 3 | 7600 |
| 4 | 6800 |
| 5 | 6000 |
| 6 | 5200 |
| 7 | 4400 |
| 8 | 3600 |
(iii) Linear expression: v=10000−800t
Linear decay because value decreases uniformly.
Question 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.
Solution
(i) Population after 6 years:
P=750+50×6=1050
Answer: 1050 people
(ii) Table (t = 0 to 10)
| t (years) | Population |
|---|---|
| 0 | 750 |
| 1 | 800 |
| 2 | 850 |
| 3 | 900 |
| 4 | 950 |
| 5 | 1000 |
| 6 | 1050 |
| 7 | 1100 |
| 8 | 1150 |
| 9 | 1200 |
| 10 | 1250 |
(iii) Linear expression:
P=750+50t
Linear growth because increase is constant.
Question 4
A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the 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.
Solution
(i) Equation: b(x)=600−15x
It is Linear decay because there is constant decrease in balance.
(ii) When balance becomes zero:
600−15x=0
15x=600
⇒x=40
So balance ends after 40 days
(iii) Table (x = 1 to 10)
| Days | Balance (₹) |
|---|---|
| 1 | 585 |
| 2 | 570 |
| 3 | 555 |
| 4 | 540 |
| 5 | 525 |
| 6 | 510 |
| 7 | 495 |
| 8 | 480 |
| 9 | 465 |
| 10 | 450 |
Conclusion
This exercise helps students understand:
- Linear growth (increase at constant rate)
- Linear decay (decrease at constant rate)
- Real-life applications of algebra