Mathematics solution NCERT
Class 9 - Chapter 2: Introduction to Linear Polynomials
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.
The plant is initially 1.75 feet tall.
It grows by 0.5 feet every month.
\(Growth in 7 months = 7 \times 0.5 = 3.5 feet\)
\(Total height after 7 months = 1.75 + 3.5\)
\(= 5.25 feet\)
Answer: The height of the plant after 7 months is 5.25 feet.
(ii) Table showing the height of the plant from 0 to 10 months
| Month (t) | 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) Expression relating h and t
\(Initial height = 1.75 feet\)
\(Monthly growth = 0.5 feet\)
Therefore,
\(h = 1.75 + 0.5t\)
This represents linear growth because the height increases by the same amount (0.5 feet) every month.
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.
\(Original value = ₹10,000\)
\(Depreciation per year = ₹800\)
\(Depreciation in 3 years = 3 \times 800 = ₹2,400\)
\(Value after 3 years = 10,000 - 2,400\)
\(= ₹7,600\)
Answer: Value after 3 years = ₹7,600.
(ii) Table showing depreciation from 0 to 8 years
| Years (t) | Value 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
The value decreases by ₹800 every year.
\(v = 10000 - 800t\)
This represents linear decay because the value decreases by a fixed amount every year.
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 after 6 years.
\(Initial population = 750\)
\(Increase every year = 50 people\)
\(Increase in 6 years = 6 \times 50 = 300\)
\(Population after 6 years = 750 + 300\)
\(= 1050\)
Answer: Population after 6 years = 1050 people.
(ii) Table showing population from 0 to 10 years
| Years (t) | 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) Expression relating P and t
\(Initial population = 750\)
\(Increase every year = 50\)
\(P = 750 + 50t\)
This is a linear growth model 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.
(i) Write an equation for the remaining balance b(x) after x days.
\(Initial balance = ₹600\)
\(Daily reduction = ₹15\)
Therefore,
\(b(x) = 600 - 15x\)
This represents linear decay because the balance decreases by a fixed amount every day.
(ii) After how many days will the balance run out?
When the balance becomes zero,
\(600 - 15x = 0\)
\(15x = 600\)
\(x = 40\)
Answer: The balance will run out after 40 days.
(iii) Table showing the remaining balance from Day 1 to Day 10
| 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 |