Exercise Set 2.4 Class 9

Class 9 Mathematics Ganita Manjari Part 1 NCERT Solutions
Chapter-2: Introduction to Linear Polynomials

Exercise Set 2.4

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 $\bm{t}$ varying from 0 to 10 months and show how the height, $\bm{h}$ , increases every month.
(iii) Find an expression that relates $\bm{h}$ and $\bm{t}$ , and explain why it represents linear growth.

Solution:
(i) Initial height of the plant $=1.75\text{ feet}$
Plant grows every month $=0.5\text{ feet}$

So, Height after 7 months $=1.75+0.5\times7$
$=1.75+3.5=5.25\text{ feet}$

(ii) Table of values

Time, $t$ (months)	Height, $h$ (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

The height increases by

0.5

feet every month.

(iii) Let the height after $t$ months $=h$
∴ $\boxed{h=1.75+0.5t}$
It represents linear growth because the height increases by a constant amount of $0.5\text{ feet}$ 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 $\bm{t}$ varying from 0 to 8 years and show how the value of the phone, $\bm{v}$ , depreciates with time.
(iii) Find an expression that relates $\bm{v}$ and $\bm{t}$ , and explain why it represents linear decay.

Solution:

(i) The initial value of phone $=₹10,000$
Value decreases every year $=₹800$
The value after 3 years $=10,000+800\times3$
$=10,000-2,400=₹7,600$

(ii) Table of values

Time, $t$ (years)	Value, $v$ (₹)
0	10,000
1	9,200
2	8,400
3	7,600
4	6,800
5	6,000
6	5,200
7	4,400
8	3,600

The value decreases by

₹800

every year.

(iii) Let the value after $t$ years $=v$
∴ $\boxed{v=10,000-800t}$
It represents linear decay because the value decreases by a constant amount of $₹800$ 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 $\bm{t}$ varying from 0 to 10 years and show how the population, $\bm{P}$ , increases every year.
(iii) Find an expression that relates $\bm{P}$ and $\bm{t}$ , and explain why it represents linear growth.

Solution:

(i) Initial population of village $=750$
Population increases per year $=50$
Population after 6 years $=750+50\times6$
$=750+300=1050$

(ii) Table of values

Time, $t$ (years)	Population, $P$
0	750
1	800
2	850
3	900
4	950
5	1000
6	1050
7	1100
8	1150
9	1200
10	1250

The population increases by

50

people every year.

(iii) Let the population after $t$ years $=P$
∴ $\boxed{P=750+50t}$

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 $\bm{b(x)}$ after using the scheme for $\bm{x}$ days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table of values for $\bm{x}$ varying from 1 to 10 days and show how the balance $\bm{b(x)}$ reduces with time.

Solution:

(i) The initial prepaid balance $=₹600$
The balance reduced per day $=₹15$

Let the remaining balance after $x$ days $b(x)$
∴ $\boxed{b(x)=600-15x}$
It represents linear decay because the balance decreases by a constant amount of $₹15$ every day.

(ii) When the balance becomes zero:
$b(x)=0$
∴ $600-15x=0$
$15x=600$
$x=\frac{600}{15}$
$x=40$
Hence, the balance will run out after $\boxed{40\text{ days}}$

(iii) Table of values

Time, $x$ (days)	Balance, $b(x)$ (₹)
0	600
1	585
2	570
3	555
4	540
5	525
6	510
7	495
8	480
9	465
10	450

The balance decreases by

₹15

every day.