Introduction to Linear Polynomials Class 9 Mathematics Ganita Manjari [LATEST] Solutions Exercise Set 2.4 in English - CBSE Study
NCERT Solutions for Class 9 Mathematics Ganita Manjari are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important Introduction to Linear Polynomials with detailed explanations and step-by-step answers for better exam preparation. Each Exercise Set 2.4 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 9 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics Ganita Manjari.
Class 9 English Medium Mathematics Ganita Manjari All Chapters:
Introduction to Linear Polynomials
4. 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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