NCERT Solutions Class 9 Maths Chapter 2 Think and Reflect — Introduction to Linear Polynomials

NCERT Class 9 Mathematics — Ganita Manjari Part I, Introduction to Linear Polynomials. All 17 Think and Reflect prompts answered.

Also here. Every exercise in this chapter is solved, one page per exercise.

Chapter 2 sets fifteen Think and Reflect boxes into its text, holding seventeen prompts between them. Each one sits directly under an example and asks you to push that example one step further — usually by running it backwards. The example works out a fare from a distance; the box asks which distance gives a particular fare.

That reversal is the chapter’s real lesson. A linear rule can be read in either direction: substitute to go forwards, solve to come back. Answering these boxes is the cheapest practice in the chapter for the skill that Exercise Set 2.5 and the End-of-Chapter questions then test properly.

Key insight. Every box here rests on the same fact: in a linear relationship the step is constant. That single property is what lets you predict the next term without computing all of them, find which term equals a given value by dividing, and recognise the pattern as a straight line when it is drawn. Where a prompt looks hard, find the step first and the rest follows.

Prompts 1 and 2 come from the first box on page 17, under Example 2, where a rectangular garden of length $l$ and width $w$ gives the total cost $200l + 160w + 50lw$.

Question 1

Can you identify the terms, variables and coefficients of this algebraic expression?

Solution. The expression is $200l + 160w + 50lw$, built by adding three separate pieces.

Terms are the pieces joined by $+$ or $-$: $\;200l$, $\;160w$, $\;50lw$.

Variables are the letters standing for quantities that can change: $l$ and $w$ — the length and width of the garden in metres.

Coefficients are the numbers multiplying each term: $200$, $160$ and $50$ respectively. Each has a meaning here — ₹$200$ per metre of length once the two long sides are counted, ₹$160$ per metre of width for the two short sides, and ₹$50$ per square metre of seed.

There is no constant term: with a garden of no size, nothing is spent.

Question 2

How is it different from the algebraic expression in Example 1?

Solution. Example 1 gave $4x + 5y + 3$ for the total number of pens and pencils. Set the two side by side:

$$4x + 5y + 3 \qquad \text{against} \qquad 200l + 160w + 50lw$$

Both use two variables and both have three terms, so the differences are in the detail:

• Example 1 has a constant term, the $3$ free pens, which is there no matter how many boxes Raju buys. Example 2 has none.
• Example 2 has a term with two variables multiplied together, $50lw$. Every term in Example 1 contains at most one variable, to the first power.

That second difference is the one that matters for this chapter. In $4x + 5y + 3$ each variable appears only to the first power, so it is linear; the term $50lw$ has total degree $2$, so the cost expression is not linear.

Prompts 3 and 4 come from the second box on page 17, under Example 3, where a $20$ cm wire bent into a rectangle of length $x$ gives the area $10x – x^2$.

Question 3

Can you identify the terms, variables and coefficients of this algebraic expression?

Solution. The expression is $10x – x^2$.

Terms: $10x$ and $-x^2$. The minus sign belongs to the second term.

Variable: just $x$, the length of the rectangle in centimetres. (The width is $10 – x$, but it has been written in terms of $x$ rather than given its own letter.)

Coefficients: $10$ for the $x$ term, and $-1$ for the $x^2$ term — an $x^2$ written with no visible number in front of it still has a coefficient, and here it is $-1$.

Question 4

Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?

Solution. Compare $4x + 5y + 3$ with $10x – x^2$.

Similarities. Both are algebraic expressions in which the parts are combined by addition and subtraction, both have numerical coefficients, and both describe a real quantity that can be evaluated once the variables are known.

Differences.

• Example 1 uses two variables, $x$ and $y$; Example 3 uses one, $x$.
• The highest power in Example 1 is $1$, so it is linear. In Example 3 it is $2$, because of the $-x^2$ term — the expression is quadratic.
• Example 1 has a constant term; Example 3 does not.

The last two matter for what follows. An expression in one variable with highest power $1$ is what the chapter will call a linear polynomial; $10x – x^2$ is one variable but the wrong degree, and $4x + 5y + 3$ is the right degree but two variables.

Prompt 5 comes from the box on page 19, under Example 4, where the perimeter of a square of side $x$ is $4x$.

Question 5

Find the perimeter of squares with sides $1$ cm, $1.5$ cm, $2$ cm, $2.5$ cm and $3$ cm. What will happen to the perimeters if the sides increase by $0.5$ cm?

Solution. The perimeter is $4x$, so multiply each side length by $4$:

Side $x$ (cm)	$1$	$1.5$	$2$	$2.5$	$3$
Perimeter $4x$ (cm)	$4$	$6$	$8$	$10$	$12$

Each time the side goes up by $0.5$ cm the perimeter goes up by exactly $2$ cm — and it is the same $2$ cm every time, not a growing amount.

The reason is visible in the formula: adding $0.5$ to $x$ adds $4 \times 0.5 = 2$ to $4x$. In general, increasing the input of a linear expression by a fixed amount increases the output by a fixed amount, and that is precisely what makes it linear.

Prompt 6 comes from the second box on page 19, under Example 5, where a chess club charges a joining fee of ₹$200$ plus ₹$50$ a match, so the amount paid for $m$ matches is $200 + 50m$.

Question 6

If a player paid ₹$750$, how many matches did he play?

Solution. This runs the example backwards: the amount is known and the number of matches is not.

$$200 + 50m = 750$$ $$50m = 550$$ $$m = 11$$

Check: $200 + 50 \times 11 = 200 + 550 = ₹750$ ✓

It is worth subtracting the joining fee first, as the working does. The ₹$200$ buys no matches at all; only the remaining ₹$550$ is spent on play, and at ₹$50$ each that is $11$ matches.

Prompt 7 comes from the box on page 20, returning to Example 3 and the area $10x – x^2$.

Question 7

We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, $10x – x^2$, is a function of $x$. Can you interpret this as an input-output process? What value does the expression take when $x = 6$ cm?

Solution. Yes — and thinking of it that way is what the word function means.

The expression is a machine. Feed in a length $x$; it works out the matching width $10 – x$, multiplies the two, and returns the area. One input, one output, every time:

$$x = 3 \;\rightarrow\; 3 \times 7 = 21 \qquad x = 4 \;\rightarrow\; 4 \times 6 = 24 \qquad x = 5 \;\rightarrow\; 5 \times 5 = 25$$

At $x = 6$:

$$10(6) – 6^2 = 60 – 36 = 24 \text{ cm}^2$$

The rectangle is $6$ cm by $4$ cm, and $6 \times 4 = 24$ ✓.

Something worth noticing: $x = 4$ and $x = 6$ both give $24 \text{ cm}^2$. A $4$ cm by $6$ cm rectangle and a $6$ cm by $4$ cm rectangle are the same shape turned round. So this machine can give the same output for two different inputs — which a linear rule never does, and is a first hint at why quadratics behave differently.

Prompts 8 and 9 come from the two boxes on page 22, under the growing pattern of square tiles in Fig. 2.4, whose stages hold $1$, $3$, $5$ and $7$ tiles — the rule $2n – 1$.

Question 8

Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage $7$ of the pattern.

Solution. Stages $1$ to $4$ hold $1$, $3$, $5$ and $7$ tiles, going up by $2$ each time. Continuing the same step:

$$\text{Stage } 5: 7 + 2 = 9 \qquad \text{Stage } 6: 9 + 2 = 11 \qquad \text{Stage } 7: 11 + 2 = 13$$

Question 9

Using the expression $2n – 1$, can you find out how many tiles will be there in the $15^{th}$ stage and the $26^{th}$ stage of the pattern? Also, which stage will contain $21$ tiles and $47$ tiles?

Solution. The first half substitutes; the second half solves.

Forwards:

$$n = 15: \; 2(15) – 1 = 29 \qquad n = 26: \; 2(26) – 1 = 51$$

Backwards:

$$2n – 1 = 21 \;\Longrightarrow\; 2n = 22 \;\Longrightarrow\; n = 11$$ $$2n – 1 = 47 \;\Longrightarrow\; 2n = 48 \;\Longrightarrow\; n = 24$$

Both come out whole, as they must for a real stage to exist. If a tile count gave a fractional $n$ — an even number would, since $2n – 1$ is always odd — there would be no such stage at all.

Prompt 10 comes from the box on page 23, under the example where ₹$100$ is spent at ₹$5$ a day, leaving ₹$(100 – 5n)$ after $n$ days.

Question 10

What amount will be left on the $15^{th}$ day? How many days will it take for the entire amount to be spent?

Solution. Substitute for the first part, solve for the second.

$$n = 15: \quad 100 – 5(15) = 100 – 75 = ₹25$$

The money runs out when nothing is left:

$$100 – 5n = 0 \;\Longrightarrow\; 5n = 100 \;\Longrightarrow\; n = 20$$

The division is exact, so the last rupee goes on day $20$ and the balance lands precisely on zero rather than skipping past it.

Prompt 11 comes from the second box on page 23, under Example 8, where an auto-rickshaw fare is ₹$25$ for the first $2$ km and ₹$15$ for each kilometre after that, giving $15n – 5$ for $n \geq 2$.

Question 11

For how many km will the fare be ₹$130$?

Solution. Set the fare formula equal to $130$:

$$15n – 5 = 130$$ $$15n = 135$$ $$n = 9$$

Check against the original description rather than just the formula: the first $2$ km cost ₹$25$, and the remaining $7$ km cost $7 \times 15 = ₹105$, giving $25 + 105 = ₹130$ ✓

The check matters here because the formula $15n – 5$ is only valid for $n \geq 2$; below that the fare is a flat ₹$25$. An answer of $9$ km is comfortably inside the valid range.

Prompt 12 comes from the box on page 24, under Example 9, where the cost of a journey of $d$ km is $C(d) = 100 + 60d$.

Question 12

What is the cost for travelling $15$ km? For how many kilometres will the cost of the journey be ₹$700$?

Solution. Substitute, then solve.

$$C(15) = 100 + 60(15) = 100 + 900 = ₹1000$$

$$100 + 60d = 700 \;\Longrightarrow\; 60d = 600 \;\Longrightarrow\; d = 10 \text{ km}$$

Notice that the ₹$100$ must be removed before dividing. Dividing ₹$700$ by $60$ straight away gives about $11.7$ km, which is wrong because it charges the fixed ₹$100$ as though it bought distance.

Prompt 13 comes from the box on page 25, under Example 10, where the height of water in a tank after $t$ months is $h(t) = 3 – 0.5t$ metres.

Question 13

What will be the height of the water at the end of $5$ months?

Solution.

$$h(5) = 3 – 0.5 \times 5 = 3 – 2.5 = 0.5 \text{ m}$$

Half a metre of water is left. It is worth going one step further than the question: at $t = 6$ the formula gives $0$, so the tank empties in exactly six months, and beyond that the model has nothing left to describe.

Prompt 14 comes from the box on page 27, under Example 11, where a telecom bill for $x$ GB of data comes to $y = 20x + 150$.

Question 14

Can you guess what the numbers $20$ and $150$ in the equation $y = 20x + 150$ represent?

Solution. Read each number by asking what it does to the bill.

$150$ is the fixed monthly fee. Putting $x = 0$ gives $y = 150$: the bill with no data used at all. It is the part you pay for having the connection.

$20$ is the cost of one GB. Each extra GB adds ₹$20$ to the bill, since replacing $x$ by $x + 1$ raises $20x + 150$ by $20$.

Check against the original readings: $10$ GB gives $200 + 150 = ₹350$ and $20$ GB gives $400 + 150 = ₹550$, both as stated in the example. The extra $10$ GB cost ₹$200$ — ₹$20$ each ✓

In the language of the next section, $20$ is the slope and $150$ the $y$-intercept.

Prompt 15 comes from the box on page 28, under Fig. 2.5, the straight line $y = 2x + 1$.

Question 15

Identify other points on the line by completing the following table.

Solution. A point lies on the line exactly when its coordinates satisfy $y = 2x + 1$, so double each $x$ and add one.

$x$	$1$	$2$	$5$	$7$	$9$	$12$	$20$
$y$	$3$	$5$	$11$	$15$	$19$	$25$	$41$

The two values already printed in the book, $y = 3$ at $x = 1$ and $y = 15$ at $x = 7$, both agree ✓

So $(1,3)$, $(2,5)$, $(5,11)$, $(7,15)$, $(9,19)$, $(12,25)$ and $(20,41)$ all lie on the line — and there are infinitely many more, one for every real number $x$.

Prompt 16 comes from the box on page 33, after Fig. 2.11 has compared lines of the form $y = -ax$.

Question 16

Differentiate between the graphs of the equations $y = 3x + 1$ and $y = -3x + 1$.

Solution. Compare them term by term.

What is the same. Both have $b = 1$, so both cut the $y$-axis at $(0, 1)$ — they pass through the same point. Both have slopes of the same size, $3$, so they are equally steep.

What differs. The signs of the slopes are opposite. $y = 3x + 1$ rises from left to right; $y = -3x + 1$ falls. Where they meet the $x$-axis differs too:

$$3x + 1 = 0 \;\Longrightarrow\; x = -\tfrac{1}{3} \qquad\qquad -3x + 1 = 0 \;\Longrightarrow\; x = \tfrac{1}{3}$$

so one crosses at $\left(-\frac{1}{3}, 0\right)$ and the other at $\left(\frac{1}{3}, 0\right)$.

Put together: the two lines form a symmetrical “$\vee$” shape opening upwards from their crossing point $(0,1)$, each the mirror image of the other in the $y$-axis.

Prompt 17 comes from the box on page 35, after Fig. 2.13 has drawn $y = 2x – 1$, $y = 2x + 1$ and $y = 2x + 5$ on the same axes.

Question 17

Does this help you to conclude anything about the linear equation $y = ax + b$ when $a$ is fixed but $b$ varies?

Solution. Yes, and it is one of the two central facts about straight-line graphs.

All three lines have the same $a = 2$, so all three have the same direction and the same steepness. What changes is only where each one sits: they cut the $y$-axis at $-1$, $1$ and $5$ — exactly their values of $b$.

So with $a$ fixed and $b$ varying:

• the lines are all parallel to one another;
• they never meet, however far they are extended, since at any given $x$ they differ by the same constant gap that separates their values of $b$;
• changing $b$ slides the line vertically without tilting it — increasing $b$ lifts it, decreasing $b$ lowers it.

The companion fact, from Fig. 2.11, is the other way round: fixing $b$ and varying $a$ pivots the lines about their shared point $(0, b)$.

Common mistakes

• Question 2, stopping at “both have three terms”. The comparison the chapter is building towards is about degree: $50lw$ has two variables multiplied, which is what disqualifies the garden cost from being linear.
• Question 3, giving the coefficient of $x^2$ as $1$. The term is $-x^2$, so the coefficient is $-1$. A term written with no number in front of it still has one.
• Question 6, dividing ₹$750$ by ₹$50$. That gives $15$ matches and ignores the joining fee. Subtract the fixed part first, then divide. The same slip appears in Question 12, where dividing ₹$700$ by $60$ gives $11.7$ km instead of $10$.
• Question 7, expecting one input per output. $x = 4$ and $x = 6$ both give $24 \text{ cm}^2$, because the two rectangles are the same shape turned round. A linear rule never does this; a quadratic one can.
• Question 9, forgetting to check that $n$ is a whole number. $2n – 1$ is always odd, so no stage ever holds an even number of tiles. A tile count of $20$ would have no stage at all.
• Question 11, using $15n – 5$ outside its range. The formula only holds for $n \geq 2$ km; for a shorter trip the fare is a flat ₹$25$. Checking the answer against the words rather than the formula catches this.
• Question 16, calling the two lines parallel. They have the same steepness but opposite signs, so they are not parallel — they cross at $(0, 1)$. Parallel needs the slopes to be equal, not merely equal in size.
• Question 17, thinking $b$ shifts the line sideways. $b$ moves the line vertically. The line $y = 2x + 5$ is $y = 2x$ lifted five units, and it crosses the $y$-axis at $5$.

Practise next

• Exercise Set 2.5 — the formal version of what Questions 6, 11 and 12 do informally: two readings given, the fixed part and the rate to be found.
• Exercise Set 2.6 — draws the very lines Questions 16 and 17 describe, so that “same slope means parallel” becomes something you can see.
• The End-of-Chapter Exercises, Question 12 — the matchstick pattern, which is Questions 8 and 9 here with a harder rule to spot.