--- title: "NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4" url: https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-4 dateModified: 2026-10-07T15:41:27+00:00 --- # NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4 Chapter 2: Introduction to Linear Polynomials. Every question from Exercise 2.4, with full working and the final answer. Free PDF (24 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-9/swavid-ncert-solutions-class-9-maths-chapter-2-introduction-to-linear-polynomials-1ce112dddf.pdf ## Exercise Set 2.4 ### Question 1 *3 marks · Short answer* 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. **Part (i)** 1. Initial height = $1.75$ feet and growth rate = $0.5$ feet per month. 2. Height after $7$ months $= 1.75 + 7 \times 0.5 = 1.75 + 3.5 = 5.25$ feet. Answer (i): $5.25$ feet **Part (ii)** 1. For $t = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, the height $h$ is $1.75, 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75$ feet respectively. Answer (ii): Table of values showing $h$ increasing by $0.5$ each month **Part (iii)** 1. The expression relating $h$ and $t$ is $h = 1.75 + 0.5t$. 2. This represents linear growth because the height increases by a constant amount ($0.5$ feet) over equal intervals of time ($1$ month). Answer (iii): $h = 1.75 + 0.5t$ **Answer:** The height after 7 months is 5.25 feet, the expression relating $h$ and $t$ is $h = 1.75 + 0.5t$, representing linear growth. > Common mistake: Adding the initial height multiple times or multiplying incorrectly. ### Question 2 *3 marks · Short answer* 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. **Part (i)** 1. Initial value = ₹10,000 and annual depreciation = ₹800. 2. Value after $3$ years $= 10000 - 3 \times 800 = 10000 - 2400 = ₹7600$. Answer (i): ₹7600 **Part (ii)** 1. For $t = 0, 1, 2, 3, 4, 5, 6, 7, 8$, the value $v$ is ₹$10000, 9200, 8400, 7600, 6800, 6000, 5200, 4400, 3600$ respectively. Answer (ii): Table of values showing $v$ decreasing by ₹800 each year **Part (iii)** 1. The expression relating $v$ and $t$ is $v = 10000 - 800t$. 2. This represents linear decay because the value decreases by a constant amount (₹800) over equal intervals of time ($1$ year). Answer (iii): $v = 10000 - 800t$ **Answer:** The value of the phone after 3 years is ₹7600, the expression relating $v$ and $t$ is $v = 10000 - 800t$, representing linear decay. > Common mistake: Adding the depreciation instead of subtracting. ### Question 3 *3 marks · Short answer* 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. **Part (i)** 1. Initial population = $750$ and annual increase = $50$ people. 2. Population after $6$ years $= 750 + 6 \times 50 = 750 + 300 = 1050$. Answer (i): $1050$ **Part (ii)** 1. For $t = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, the population $P$ is $750, 800, 850, 900, 950, 1000, 1050, 1100, 1150, 1200, 1250$ respectively. Answer (ii): Table of values showing $P$ increasing by $50$ each year **Part (iii)** 1. The expression relating $P$ and $t$ is $P = 750 + 50t$. 2. This represents linear growth because the population increases by a constant amount ($50$) over equal intervals of time ($1$ year). Answer (iii): $P = 750 + 50t$ **Answer:** The population after 6 years is 1050, the expression relating $P$ and $t$ is $P = 750 + 50t$, representing linear growth. > Common mistake: Multiplying the initial population by the number of years. ### Question 4 *3 marks · Short answer* 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. **Part (i)** 1. Initial balance = ₹600 and daily reduction = ₹15. 2. The remaining balance after $x$ days is given by the linear equation $b(x) = 600 - 15x$. 3. This represents linear decay because the balance decreases by a constant amount (₹15) over equal intervals of time ($1$ day). Answer (i): $b(x) = 600 - 15x$ **Part (ii)** 1. Set the remaining balance to $0$: $600 - 15x = 0$. 2. $15x = 600$, which gives $x = \frac{600}{15} = 40$ days. Answer (ii): $40$ days **Part (iii)** 1. For $x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, the balance $b(x)$ is $585, 570, 555, 540, 525, 510, 495, 480, 465, 450$ respectively. Answer (iii): Table of values showing $b(x)$ reducing by ₹15 each day **Answer:** The balance is modelled by $b(x) = 600 - 15x$, it runs out after 40 days, and decreases by ₹15 daily. > Common mistake: Forgetting to equate the balance to zero when finding when it runs out. ## Related pages - [All Chapter 2 NCERT solutions](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions) - [Exercise 2.1](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-1) - [Exercise 2.2](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-2) - [Exercise 2.3](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-3) - [Exercise 2.5](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-5) - [Exercise 2.6](https://www.swavid.com/maths/class/9/chapter/introduction-to-linear-polynomials/ncert-solutions/exercise-2-6) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. 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