NCERT Solutions Class 9 Maths Exercise 2.5 — Introduction to Linear Polynomials

NCERT Class 9 Mathematics — Ganita Manjari Part I, Introduction to Linear Polynomials, Exercise Set 2.5. All 3 questions solved.

Every question here is the same problem in different clothes. A relationship is known to have the form

$$y = ax + b$$

and two readings are given. Two unknowns, two facts — so both $a$ and $b$ can be pinned down.

The quickest route, and the one worth learning, is subtraction. Write the two readings as equations and take one from the other: the $b$ terms are identical, so they cancel and leave $a$ on its own. Then put $a$ back into either equation to get $b$.

$$ \begin{aligned} y_2 &= a x_2 + b \\ y_1 &= a x_1 + b \\[2pt] \hline y_2 – y_1 &= a(x_2 – x_1) \end{aligned} $$

which says $a = \dfrac{y_2 – y_1}{x_2 – x_1}$ — the change in $y$ per unit change in $x$.

Key insight. $a$ and $b$ are never just numbers in these problems; each one means something. $a$ is the rate — the cost of one more module, one more hour, one more degree — and $b$ is the fixed part that is there even when $x = 0$. Once the two readings give you $a$, the value of $b$ follows from asking “what would the bill be with nothing used at all?”

Question 1

A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed $10$ modules, her bill was ₹$400$. When she accessed $14$ modules, her bill was ₹$500$. If the monthly bill $y$ depends on the number of modules accessed, $x$, according to the relation $y = ax + b$, find the values of $a$ and $b$.

Solution. Substitute the two readings into $y = ax + b$:

$$10a + b = 400 \qquad \text{(i)}$$ $$14a + b = 500 \qquad \text{(ii)}$$

Subtract (i) from (ii). The $b$ cancels:

$$4a = 100 \quad\Longrightarrow\quad a = 25$$

Now substitute $a = 25$ into (i):

$$10(25) + b = 400 \quad\Longrightarrow\quad 250 + b = 400 \quad\Longrightarrow\quad b = 150$$

So the relation is $y = 25x + 150$.

Check both readings: $25(10) + 150 = 400$ ✓ and $25(14) + 150 = 500$ ✓.

Reading the answer in words: each module costs ₹$25$, and there is a fixed monthly fee of ₹$150$ payable even if no module is accessed. The four extra modules cost $500 – 400 = ₹100$ between them, which is ₹$25$ each — the same number, arrived at directly.

Question 2

A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for $10$ hours, her bill was ₹$800$. When she used it for $15$ hours, her bill was ₹$1100$. If the monthly bill $y$ depends on the hours of use of the badminton court, $x$, according to the relation $y = ax + b$, find the values of $a$ and $b$.

Solution. The two readings give

$$10a + b = 800 \qquad \text{(i)}$$ $$15a + b = 1100 \qquad \text{(ii)}$$

Subtracting (i) from (ii):

$$5a = 300 \quad\Longrightarrow\quad a = 60$$

Substituting into (i):

$$10(60) + b = 800 \quad\Longrightarrow\quad 600 + b = 800 \quad\Longrightarrow\quad b = 200$$

So $y = 60x + 200$.

Check: $60(10) + 200 = 800$ ✓ and $60(15) + 200 = 1100$ ✓.

In words: the court costs ₹$60$ an hour and the fixed monthly membership is ₹$200$.

Question 3

Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by $°\mathrm{C} = a \, °\mathrm{F} + b$. Find $a$ and $b$, given that ice melts at $0$ degrees Celsius and $32$ degrees Fahrenheit, and water boils at $100$ degrees Celsius and $212$ degrees Fahrenheit.

Solution. Read the form of the relation carefully before substituting. It is written $°\mathrm{C} = a \, °\mathrm{F} + b$, so Fahrenheit is the input and Celsius the output — $°\mathrm{F}$ plays the part of $x$ and $°\mathrm{C}$ the part of $y$.

The two facts give:

• ice melting: $°\mathrm{F} = 32$ gives $°\mathrm{C} = 0$, so $32a + b = 0$   (i)
• water boiling: $°\mathrm{F} = 212$ gives $°\mathrm{C} = 100$, so $212a + b = 100$   (ii)

Subtract (i) from (ii):

$$180a = 100 \quad\Longrightarrow\quad a = \frac{100}{180} = \frac{5}{9}$$

Substitute into (i):

$$32 \times \frac{5}{9} + b = 0 \quad\Longrightarrow\quad \frac{160}{9} + b = 0 \quad\Longrightarrow\quad b = -\frac{160}{9}$$

So the relationship is

$$°\mathrm{C} = \frac{5}{9} \, °\mathrm{F} – \frac{160}{9}$$

which factorises into the form usually quoted:

$$°\mathrm{C} = \frac{5}{9}\left( °\mathrm{F} – 32 \right)$$

Check: at $°\mathrm{F} = 32$ this gives $0$ ✓, and at $°\mathrm{F} = 212$ it gives $\frac{5}{9}(180) = 100$ ✓.

The meaning of $a = \frac{5}{9}$ is worth stating: a rise of one Fahrenheit degree is a rise of only $\frac{5}{9}$ of a Celsius degree, because the $180$ Fahrenheit degrees between freezing and boiling cover the same range as $100$ Celsius degrees.

Common mistakes

• Questions 1 and 2, dividing the bill by the usage. Reading Question 1 as “₹$400$ for $10$ modules, so ₹$40$ each” ignores the fixed fee entirely, and it contradicts the second reading ($500 \div 14 \approx ₹35.7$). Two different “rates” from the same tariff is the signal that there is a fixed part to separate out.
• Questions 1 and 2, adding the equations instead of subtracting. Adding gives $24a + 2b = 900$ — one equation with two unknowns still in it. Subtracting is what removes $b$, because its coefficient is the same in both.
• Question 3, answering $a = \frac{9}{5}$, $b = 32$. That is the relation the other way round, $°\mathrm{F} = \frac{9}{5}\,°\mathrm{C} + 32$, which is the more familiar form and the one most students reach for. The question defines $°\mathrm{C} = a\,°\mathrm{F} + b$, so the roles are reversed. Substituting $°\mathrm{F}=32$ into the wrong version gives $°\mathrm{C} = 89.6$, not $0$ — always test your answer against both given readings.
• Question 3, rounding $\frac{5}{9}$ to $0.56$ too early. Then $b$ comes out as about $-17.92$ instead of $-\frac{160}{9} = -17.\overline{7}$, and the boiling point check fails by more than a degree. Keep the fraction until the end.
• Losing the minus sign on $b$. From $32a + b = 0$ it follows that $b = -32a$. Writing $b = +\frac{160}{9}$ would put the melting point of ice at about $35\,°\mathrm{C}$.
• Treating $b$ as “the bill for one unit”. $b$ is the value when $x = 0$, not when $x = 1$. In Question 2 the bill for one hour is $60 + 200 = ₹260$, while $b$ is ₹$200$.

Practise next

• Exercise Set 2.6 — takes the same $y = ax + b$ and asks what $a$ and $b$ look like once the relation is drawn, which is the natural companion to computing them.
• The End-of-Chapter Exercises, Questions 10, 11 and 13 — all three find a linear polynomial from conditions, with Question 13 adding the idea that parallel graphs share the same $a$.