NCERT Solutions Class 9 Maths Exercise 2.4 — Introduction to Linear Polynomials

NCERT Class 9 Mathematics — Ganita Manjari Part I, Introduction to Linear Polynomials, Exercise Set 2.4. All 4 questions solved.

This exercise gives names to the two patterns built in Exercise Set 2.3:

• Linear growth — a quantity increases by the same fixed amount over equal intervals, so the model is $\text{start} + (\text{step}) \times t$ with the step positive.
• Linear decay — it decreases by the same fixed amount, and the step is negative.

What makes either of them linear, and not merely “increasing” or “decreasing”, is that the amount added or removed never changes. A savings account paying interest grows too, but by a larger amount every year, and that is not linear.

Each question asks for the same three things: a value at one moment, a table, and the expression with a reason. The tables are the point — they are what makes the constant step visible.

Key insight. In every model here, the number multiplying $t$ is the step per interval and the constant term is the value at $t = 0$. So you can read a model straight off the words without any algebra: “₹$10{,}000$, falling ₹$800$ a year” is $10000 – 800t$ and nothing else. The table then confirms it, because the difference between consecutive rows must come out as that same step every time.

Question 1

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

Solution. The starting height is $1.75$ feet and each month adds $0.5$ feet, so with $t$ in months and $h$ in feet the model is $h = 1.75 + 0.5t$.

(i) Find the height after $7$ months

$$h = 1.75 + 0.5 \times 7 = 1.75 + 3.5 = 5.25 \text{ feet}$$

(ii) Make a table of values for $t$ varying from $0$ to $10$ months and show how the height, $h$, increases every month

Reading across, every entry is $0.5$ more than the one before it — never more, never less. The height at $t = 7$ agrees with part (i).

(iii) Find an expression that relates $h$ and $t$, and explain why it represents linear growth

$$h = 1.75 + 0.5t$$

It is linear growth because the coefficient of $t$ is a constant, $0.5$, and it is positive. Constant means the plant gains the same half-foot every month, whatever month it is; positive means the gain is an increase. Those two facts together are exactly what “linear growth” means, and in the table they show up as a first difference that never changes.

Question 2

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

Solution. The value starts at ₹$10{,}000$ and falls by ₹$800$ a year, so with $t$ in years and $v$ in rupees the model is $v = 10000 – 800t$.

(i) Find the value of the phone after $3$ years

$$v = 10000 – 800 \times 3 = 10000 – 2400 = ₹7600$$

(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

Each year the value drops by exactly ₹$800$. The entry at $t = 3$ matches part (i).

(iii) Find an expression that relates $v$ and $t$, and explain why it represents linear decay

$$v = 10000 – 800t$$

It is linear decay because the coefficient of $t$ is constant, $-800$, and negative: the same fixed amount is lost every year regardless of how old the phone is.

It is worth noticing that this is a model, and a crude one. Real phones lose value fastest in the first year and more slowly later, which is not linear at all. This model would also reach zero at $t = 12.5$ years and go negative afterwards, which no second-hand phone does.

Question 3

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

Solution. With $t$ in years and $P$ the population, the model is $P = 750 + 50t$.

(i) Find the population of the village after $6$ years

$$P = 750 + 50 \times 6 = 750 + 300 = 1050$$

(ii) Make a table of values for $t$ varying from $0$ to $10$ years and show how the population, $P$, increases every year

The population rises by exactly $50$ each year; the value at $t = 6$ agrees with part (i).

(iii) Find an expression that relates $P$ and $t$, and explain why it represents linear growth

$$P = 750 + 50t$$

Linear growth, because the coefficient of $t$ is the constant $+50$: the same number of people arrive each year.

This is a modelling assumption, and a strong one — it counts only the $50$ arrivals and ignores births, deaths and anyone leaving. A population that grew by a fixed percentage each year instead would not be linear, since the yearly increase would itself get bigger.

Question 4

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

Solution. The balance starts at ₹$600$ and falls by ₹$15$ a day.

(i) Write an equation that models the remaining balance $b(x)$ after using the scheme for $x$ days. Explain why it represents linear decay

$$b(x) = 600 – 15x$$

Linear decay: the coefficient of $x$ is the constant $-15$, so the same ₹$15$ is taken off each day however much is left.

(ii) After how many days will the balance run out?

The balance runs out when $b(x) = 0$:

$$600 – 15x = 0$$ $$15x = 600$$ $$x = 40$$

Here the division is exact, so the balance lands precisely on zero on the $40^{\text{th}}$ day — unlike the rally in Exercise Set 2.3, where the corresponding calculation gave $13\tfrac{1}{3}$ and the model simply stopped making sense partway through an hour.

(iii) Make a table of values for $x$ varying from $1$ to $10$ days and show how the balance $b(x)$ reduces with time

The balance falls by ₹$15$ each day. Extending the same pattern to $x = 40$ gives $600 – 600 = ₹0$, confirming part (ii).

Common mistakes

• Starting the table at the wrong row. Questions 1 to 3 ask for $t$ from $0$, and $t=0$ means before the first month or year, so the first entry is the starting value itself — $1.75$, $10000$, $750$. Question 4(iii), by contrast, asks for $x$ from $1$, so its table begins at $585$ and ₹$600$ never appears in it. Read the range each question actually asks for.
• Question 1, writing $h = 0.5 + 1.75t$. The numbers are swapped. The constant term is where the plant starts; the coefficient of $t$ is how much it gains each month. This model would have a seedling of half a foot growing nearly two feet a month.
• Questions 2 and 4, writing $800t – 10000$ or $15x – 600$. These have the subtraction reversed and give negative values from the first year. The quantity that shrinks starts from its full value, so the constant comes first.
• Calling any decrease “linear decay”. Linear decay is a fixed amount per interval. A phone losing $20\%$ of its value each year loses ₹$2000$ in the first year and ₹$1600$ in the second — decreasing, but not linearly.
• Question 4(ii), answering $40$ without checking it is a whole number. It is here, which is why the balance reaches exactly zero. When the division is not exact the model runs out partway through a day, and saying which case you are in is part of the answer.
• Forgetting that these are models. All four expressions eventually give negative values — a plant growing for ever, a phone worth less than nothing. The algebra does not know about the situation, so the sensible range has to be stated separately.

Practise next

• Exercise Set 2.5 — the inverse problem. Here the starting value and the step were given and you built the model; there, two readings are given and you must find the starting value and the step from them.
• Exercise Set 2.3 — worth revisiting for the contrast in Question 4(ii): its rally question is the decay model that does not reach zero at a whole number of steps.