NCERT Solutions for Class 9 Maths Chapter 2 — Exercise Set 2.4#
A relation $y = a t + b$ is called linear because $y$ changes by a constant amount $a$ for each unit change in $t$. If $a > 0$ we call it linear growth; if $a < 0$ we call it linear decay.
1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month. #
The plant starts at 1.75 feet and grows 0.5 feet each month, so after $t$ months its height in feet is:
$$h = 1.75 + 0.5t$$(i) Find the height after 7 months.
We find $h$ when $t = 7$:
$$= 1.75 + 0.5(7)$$$$= 1.75 + 3.5$$
$$= 5.25$$
∴ After 7 months the plant is 5.25 feet tall.
(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
| $t$ (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 |
Each month the height increases by the same amount, 0.5 feet.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
This is a linear expression in $t$. The height increases by the same fixed amount, 0.5 feet, for every month that passes, so it represents linear growth.
2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year. #
The phone is worth ₹10,000 at first and loses ₹800 every year, so after $t$ years its value in rupees is:
$$v = 10000 - 800t$$(i) Find the value of the phone after 3 years.
We find $v$ when $t = 3$:
$$= 10000 - 800(3)$$$$= 10000 - 2400$$
$$= 7600$$
∴ After 3 years the phone is worth ₹7600.
(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.
| $t$ (years) | $v$ (₹) |
|---|---|
| 0 | 10000 |
| 1 | 9200 |
| 2 | 8400 |
| 3 | 7600 |
| 4 | 6800 |
| 5 | 6000 |
| 6 | 5200 |
| 7 | 4400 |
| 8 | 3600 |
Each year the value falls by the same amount, ₹800.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
This is a linear expression in $t$. The value decreases by the same fixed amount, ₹800, every year, so it represents linear decay.
3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village. #
The village starts with 750 people and gains 50 every year, so after $t$ years its population is:
$$P = 750 + 50t$$(i) Find the population of the village after 6 years.
We find $P$ when $t = 6$:
$$= 750 + 50(6)$$$$= 750 + 300$$
$$= 1050$$
∴ After 6 years the population is 1050.
(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
| $t$ (years) | $P$ |
|---|---|
| 0 | 750 |
| 1 | 800 |
| 2 | 850 |
| 3 | 900 |
| 4 | 950 |
| 5 | 1000 |
| 6 | 1050 |
| 7 | 1100 |
| 8 | 1150 |
| 9 | 1200 |
| 10 | 1250 |
Each year the population rises by the same amount, 50 people.
(iii) Find an expression that relates P and t, and explain why it represents linear growth.
This is a linear expression in $t$. The population increases by the same fixed amount, 50 people, every year, so it represents linear growth.
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.
The balance starts at ₹600 and falls by ₹15 each day, so after $x$ days:
$$b(x) = 600 - 15x$$The balance decreases by the same fixed amount, ₹15, every day, so it represents linear decay.
(ii) After how many days will the balance run out?
The balance runs out when $b(x)$ becomes 0, so:
$$600 - 15x = 0$$$$\Rightarrow 15x = 600$$
$$\Rightarrow x = 40$$
∴ The balance runs out after 40 days.
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.
| $x$ (days) | $b(x)$ (₹) |
|---|---|
| 1 | 585 |
| 2 | 570 |
| 3 | 555 |
| 4 | 540 |
| 5 | 525 |
| 6 | 510 |
| 7 | 495 |
| 8 | 480 |
| 9 | 465 |
| 10 | 450 |
Each day the balance falls by the same amount, ₹15.