Exercise 2.4 Practice
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Overview
This page provides comprehensive Ch 2: Polynomials - Exercise 2.4 Practice. Practice modeling linear growth and linear decay processes, creating tables of values, and determining relationships in real-world scenarios. Free step-by-step NCERT solutions and explanations.
Applications of Linear Growth and Decay
Q1: Plant Growth
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.
(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.
(i) Height after 7 months:
Initial height = $1.75\text{ ft}$.
Growth rate = $0.5\text{ ft/month}$.
Height after 7 months: $$h = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25\text{ feet}$$
Initial height = $1.75\text{ ft}$.
Growth rate = $0.5\text{ ft/month}$.
Height after 7 months: $$h = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25\text{ feet}$$
(ii) Table of values for $t$ from 0 to 10 months:
| 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 |
(iii) Linear Expression and Explanation:
The expression relating $h$ and $t$ is: $$h(t) = 1.75 + 0.5t$$ This represents linear growth because the rate of change is constant. For every unit increase in time $t$ (1 month), the height $h$ increases by exactly the same amount ($0.5$ feet). The graph is a straight line with a constant positive slope of $0.5$.
The expression relating $h$ and $t$ is: $$h(t) = 1.75 + 0.5t$$ This represents linear growth because the rate of change is constant. For every unit increase in time $t$ (1 month), the height $h$ increases by exactly the same amount ($0.5$ feet). The graph is a straight line with a constant positive slope of $0.5$.
Height after 7 months = 5.25 ft; Expression: $h(t) = 1.75 + 0.5t$
Q2: Mobile Phone Depreciation
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.
(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.
(i) Value after 3 years:
Initial value = ₹10,000.
Depreciation rate = ₹800/year.
Value after 3 years: $$v = 10000 - 800(3) = 10000 - 2400 = \text{₹}7,600$$
Initial value = ₹10,000.
Depreciation rate = ₹800/year.
Value after 3 years: $$v = 10000 - 800(3) = 10000 - 2400 = \text{₹}7,600$$
(ii) Table of values for $t$ from 0 to 8 years:
| 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 |
(iii) Linear Expression and Explanation:
The expression relating $v$ and $t$ is: $$v(t) = 10000 - 800t$$ This represents linear decay because the value reduces by a constant amount (₹800) for every unit increase in time $t$ (1 year). The rate of change is constant and negative, meaning its graph is a straight line with a constant negative slope of $-800$.
The expression relating $v$ and $t$ is: $$v(t) = 10000 - 800t$$ This represents linear decay because the value reduces by a constant amount (₹800) for every unit increase in time $t$ (1 year). The rate of change is constant and negative, meaning its graph is a straight line with a constant negative slope of $-800$.
Value after 3 years = ₹7,600; Expression: $v(t) = 10000 - 800t$
Q3: Village Population
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.
(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.
(i) Population after 6 years:
Initial population = $750$.
Increase rate = $50\text{ people/year}$.
Population after 6 years: $$P = 750 + 50(6) = 750 + 300 = 1050$$
Initial population = $750$.
Increase rate = $50\text{ people/year}$.
Population after 6 years: $$P = 750 + 50(6) = 750 + 300 = 1050$$
(ii) Table of values for $t$ from 0 to 10 years:
| 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 |
(iii) Linear Expression and Explanation:
The expression relating $P$ and $t$ is: $$P(t) = 750 + 50t$$ This represents linear growth because the population increases by a fixed, constant amount of $50$ people per year. The relationship represents a straight line with a constant positive slope of $50$.
The expression relating $P$ and $t$ is: $$P(t) = 750 + 50t$$ This represents linear growth because the population increases by a fixed, constant amount of $50$ people per year. The relationship represents a straight line with a constant positive slope of $50$.
Population after 6 years = 1050; Expression: $P(t) = 750 + 50t$
Q4: Prepaid Balance Decay
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.
(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.
(i) Model Equation & Explanation:
Initial balance = ₹600.
Reduction per day = ₹15.
The equation representing the remaining balance $b(x)$ after $x$ days is: $$b(x) = 600 - 15x$$ This represents linear decay because the balance decreases by a constant rate of ₹15 every single day. The equation is of the form $y = mx + c$ with a negative slope $m = -15$.
Initial balance = ₹600.
Reduction per day = ₹15.
The equation representing the remaining balance $b(x)$ after $x$ days is: $$b(x) = 600 - 15x$$ This represents linear decay because the balance decreases by a constant rate of ₹15 every single day. The equation is of the form $y = mx + c$ with a negative slope $m = -15$.
(ii) Time to run out:
The balance runs out when $b(x) = 0$: $$600 - 15x = 0$$ $$15x = 600$$ $$x = 40\text{ days}$$
The balance runs out when $b(x) = 0$: $$600 - 15x = 0$$ $$15x = 600$$ $$x = 40\text{ days}$$
(iii) Table of values for $x$ from 1 to 10 days:
| Days $x$ | Remaining Balance $b(x)$ (₹) |
|---|---|
| 1 | 585 |
| 2 | 570 |
| 3 | 555 |
| 4 | 540 |
| 5 | 525 |
| 6 | 510 |
| 7 | 495 |
| 8 | 480 |
| 9 | 465 |
| 10 | 450 |
Equation: $b(x) = 600 - 15x$; Balance runs out in 40 days