Introduction to Linear Polynomials Mathematics Ganita Manjari class 9 in English Medium ncert book solutions Exercise Set 2.4

Exercise Set 2.4
Q1. 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.
Q2. 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.
Q3. 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.
Q4. 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.

Solutions:

Q1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.

Solution:

Initial height of plant = 1.75 feet

Growth every month = 0.5 feet

(i) Height after 7 months

h = 1.75 + (0.5 × 7)

h = 1.75 + 3.5

h = 5.25 feet

So, the height after 7 months is 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

h = 1.75 + 0.5t

This represents linear growth because the height increases by the same amount every month.

Q2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.

Solution:

Initial value of phone = ₹10,000

Decrease every year = ₹800

(i) Value after 3 years

v = 10000 − (800 × 3)

v = 10000 − 2400

v = ₹7600

So, the value after 3 years is ₹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

v = 10000 − 800t

This represents linear decay because the value decreases by the same amount every year.

Q3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.

Solution:

Initial population = 750

Increase every year = 50 people

(i) Population after 6 years

P = 750 + (50 × 6)

P = 750 + 300

P = 1050

So, the population after 6 years is 1050.

(ii) Table of values

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

(iii) Expression relating P and t

P = 750 + 50t

This represents linear growth because the population increases by the same amount every year.

Q4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.

Solution:

Initial balance = ₹600

Reduction every day = ₹15

(i) Equation representing remaining balance

b(x) = 600 − 15x

This represents linear decay because the balance decreases by the same amount every day.

(ii) Number of days after which balance becomes zero

600 − 15x = 0

15x = 600

x = 600/15

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

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