NCERT Solutions Class 9 Maths Chapter 2 End-of-Chapter Exercises — Introduction to Linear Polynomials

NCERT Class 9 Mathematics — Ganita Manjari Part I, Introduction to Linear Polynomials, End-of-Chapter Exercises. All 14 questions solved.

These fourteen questions gather the whole chapter. Four skills cover all of them:

• Reading a polynomial — its degree, its coefficients, its value at a given number.
• Turning a sentence into an equation, solving it, and reading the answer back into the words.
• Building a linear polynomial from conditions — two points, or a point and a direction, determine $y = ax + b$ completely.
• Reading a graph — $a$ is the slope, $b$ the $y$-intercept, and equal slopes mean parallel lines.

Questions $6$ to $14$ carry an asterisk in the book, marking them as enrichment that is not assessed. They are also the most interesting questions here, and all are solved in full.

Key insight. A linear polynomial has exactly two unknowns, $a$ and $b$, so two independent facts always pin it down — two points it passes through, or one point plus a slope, or a sum and a difference. Questions 10, 11 and 13 look quite different from one another but are the same exercise in counting facts: gather two, write two equations, solve.

Question 1

Write a polynomial of degree $3$ in the variable $x$, in which the coefficient of the $x^2$ term is $-7$.

Solution. Two conditions must hold. Degree $3$ means an $x^3$ term must be present with a non-zero coefficient and no higher power may appear. The coefficient of $x^2$ must be exactly $-7$.

Everything else is free — the coefficient of $x$ and the constant term may be anything at all, including zero.

$$x^3 – 7x^2 + 4x + 9$$

Check: the highest power is $3$, so the degree is $3$ ✓, and the $x^2$ term is $-7x^2$ ✓.

Question 2

Find the values of the following polynomials at the indicated values of the variables.

Solution. Substitute and evaluate. The second part substitutes a letter rather than a number, which changes nothing about the method.

(i) $5x^2 – 3x + 7$ if $x = 1$

$$5(1)^2 – 3(1) + 7 = 5 – 3 + 7 = 9$$

(ii) $4t^3 – t^2 + 6$ if $t = a$

Replace every $t$ by $a$:

$$4a^3 – a^2 + 6$$

There is nothing further to simplify. Substituting a letter gives an expression, not a number — and that is a complete answer.

Question 3

If we multiply a number by $\dfrac{5}{2}$ and add $\dfrac{2}{3}$ to the product, we get $-\dfrac{7}{12}$. Find the number.

Solution. Let the number be $x$. Translating the sentence in the order it is written:

$$\frac{5}{2}x + \frac{2}{3} = -\frac{7}{12}$$

Move the $\frac{2}{3}$ across. Both fractions go over $12$:

$$\frac{5}{2}x = -\frac{7}{12} – \frac{2}{3} = -\frac{7}{12} – \frac{8}{12} = -\frac{15}{12} = -\frac{5}{4}$$

Now divide by $\frac{5}{2}$, which means multiplying by $\frac{2}{5}$:

$$x = -\frac{5}{4} \times \frac{2}{5} = -\frac{10}{20} = -\frac{1}{2}$$

Check: $\frac{5}{2} \times \left(-\frac{1}{2}\right) + \frac{2}{3} = -\frac{5}{4} + \frac{2}{3} = -\frac{15}{12} + \frac{8}{12} = -\frac{7}{12}$ ✓

Question 4

A positive number is $5$ times another number. If $21$ is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Solution. Let the smaller number be $x$; then the other is $5x$. After adding $21$ to each, the new numbers are $x + 21$ and $5x + 21$.

“One of the new numbers becomes twice the other” does not say which, so both readings must be tested.

Reading A — the larger is twice the smaller:

$$5x + 21 = 2(x + 21)$$ $$5x + 21 = 2x + 42$$ $$3x = 21 \quad\Longrightarrow\quad x = 7$$

giving the numbers $7$ and $35$. Check: adding $21$ gives $28$ and $56$, and $56 = 2 \times 28$ ✓.

Reading B — the smaller is twice the larger:

$$x + 21 = 2(5x + 21)$$ $$x + 21 = 10x + 42$$ $$-21 = 9x \quad\Longrightarrow\quad x = -\frac{7}{3}$$

This must be rejected: the question says the numbers are positive, and $-\frac{7}{3}$ is neither positive nor a whole number. Reading B also fails common sense — the smaller number cannot overtake the larger when both are increased by the same amount.

Question 5

If you have ₹$800$ and you save ₹$250$ every month, find the amount you have after (i) $6$ months, (ii) $2$ years. Express this as a linear pattern.

Solution. Starting at ₹$800$ and adding ₹$250$ each month, the amount after $m$ months is

$$A(m) = 800 + 250m$$

(i) After $6$ months:

$$A(6) = 800 + 250 \times 6 = 800 + 1500 = ₹2300$$

(ii) Two years is $24$ months — the units must be converted before substituting:

$$A(24) = 800 + 250 \times 24 = 800 + 6000 = ₹6800$$

Question 6

*The digits of a two-digit number differ by $3$. If the digits are interchanged, and the resulting number is added to the original number, we get $143$. Find both the numbers.

Solution. Let the tens digit be $a$ and the units digit be $b$. Then the number is $10a + b$, and reversing the digits gives $10b + a$.

Their sum is

$$(10a + b) + (10b + a) = 11a + 11b = 11(a + b)$$

This is the useful observation: the sum of a two-digit number and its reverse is always $11$ times the sum of the digits. So

$$11(a + b) = 143 \quad\Longrightarrow\quad a + b = 13$$

The digits also differ by $3$:

$$\lvert a – b \rvert = 3$$

Two digits adding to $13$ and differing by $3$ must be $8$ and $5$. (Adding the two conditions: if $a – b = 3$ then $2a = 16$, so $a = 8$ and $b = 5$.)

So the two numbers are $85$ and $58$.

Check: $85 + 58 = 143$ ✓, and the digits $8$ and $5$ differ by $3$ ✓.

Question 7

*Draw the graph of the following equations, and identify their slopes and $y$-intercepts. Also, find the coordinates of the points where these lines cut the $y$-axis. Are any of the lines parallel?

Solution. Each equation must first be rearranged into the form $y = ax + b$, because only then can the slope and intercept be read off directly. Dividing through by the coefficient of $y$ is all that is needed.

(i) $y = -3x + 4$ is already in that form: slope $-3$, $y$-intercept $4$.

(ii) $2y = 4x + 7$. Divide by $2$:

$$y = 2x + \frac{7}{2}$$

slope $2$, $y$-intercept $\frac{7}{2}$.

(iii) $5y = 6x – 10$. Divide by $5$:

$$y = \frac{6}{5}x – 2$$

slope $\frac{6}{5}$, $y$-intercept $-2$.

(iv) $3y = 6x – 11$. Divide by $3$:

$$y = 2x – \frac{11}{3}$$

slope $2$, $y$-intercept $-\frac{11}{3}$.

A line cuts the $y$-axis where $x = 0$, and there $y = b$. So each line crosses at $(0, b)$ with $b$ the intercept already found.

Which are parallel? Parallel lines are exactly those with equal slopes. Comparing $-3$, $2$, $\frac{6}{5}$ and $2$: lines (ii) and (iv) both have slope $2$, and their intercepts differ ($\frac{7}{2}$ against $-\frac{11}{3}$), so they are parallel and distinct. No other pair matches.

It is worth noticing why (iii) is not parallel to them even though $6$ and $2$ both appear in it: the slope is $\frac{6}{5}$, not $6$, because the equation had to be divided by $5$ first.

Question 8

*If the temperature of a liquid can be measured in Kelvin units as $x$ K and in Fahrenheit units as $y$ °F, the relation between the two systems of measurement of temperature is given by the linear equation $y = \dfrac{9}{5}(x – 273) + 32$.

Solution. The relation converts Kelvin to Fahrenheit. Going the other way means solving for $x$ instead of substituting.

(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is $313$ K

$$y = \frac{9}{5}(313 – 273) + 32 = \frac{9}{5}(40) + 32 = 72 + 32 = 104$$

(ii) If the temperature is $158$ °F, then find the temperature in Kelvin

Substitute $y = 158$ and unwind the operations in reverse order — subtract the $32$ first, then undo the $\frac{9}{5}$, then undo the $-273$:

$$158 = \frac{9}{5}(x – 273) + 32$$ $$126 = \frac{9}{5}(x – 273)$$ $$x – 273 = 126 \times \frac{5}{9} = 70$$ $$x = 343$$

Check: $\frac{9}{5}(343 – 273) + 32 = \frac{9}{5}(70) + 32 = 126 + 32 = 158$ ✓

Question 9

*The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work $w$ and distance $d$), and draw its graph by taking the constant force as $3$ units. What is the work done when the distance travelled is $2$ units? Verify it by plotting it on the graph.

Solution. In words, work $=$ force $\times$ distance. Writing $F$ for the constant force:

$$w = F d$$

Taking $F = 3$ units gives the linear equation

$$w = 3d$$

This is $y = ax + b$ with $a = 3$ and $b = 0$, so the graph is a straight line through the origin of slope $3$. Two points are enough to draw it:

$d$	$0$	$1$	$2$	$3$
$w$	$0$	$3$	$6$	$9$

At $d = 2$:

$$w = 3 \times 2 = 6 \text{ units}$$

On the graph, going across to $d = 2$ and up to the line lands at the point $(2, 6)$, which is the same answer read off the picture.

The constant term is zero for a good physical reason: no distance travelled means no work done, so the line has to pass through the origin.

Question 10

*The graph of a linear polynomial $p(x)$ passes through the points $(1, 5)$ and $(3, 11)$.

Solution. Write $p(x) = ax + b$. Two points give two equations.

(i) Find the polynomial $p(x)$

$$a + b = 5 \qquad \text{(from } (1,5))$$ $$3a + b = 11 \qquad \text{(from } (3,11))$$

Subtracting the first from the second removes $b$:

$$2a = 6 \quad\Longrightarrow\quad a = 3$$

Then $3 + b = 5$, so $b = 2$.

$$p(x) = 3x + 2$$

Check: $p(1) = 5$ ✓ and $p(3) = 11$ ✓.

(ii) Find the coordinates where the graph of $p(x)$ cuts the axes

The $y$-axis is where $x = 0$:

$$p(0) = 2 \quad\Longrightarrow\quad (0, 2)$$

The $x$-axis is where $y = 0$:

$$3x + 2 = 0 \quad\Longrightarrow\quad x = -\frac{2}{3} \quad\Longrightarrow\quad \left(-\frac{2}{3},\, 0\right)$$

(iii) Draw the graph of $p(x)$ and verify your answers

Plot $(1, 5)$ and $(3, 11)$, join them and extend the line both ways. It should pass through $(0, 2)$ on the $y$-axis and cross the $x$-axis a little left of $-\frac{2}{3}$ — about two thirds of a unit before the origin. A third point such as $(2, 8)$ confirms the line.

Question 11

*Let $p(x) = ax + b$ and $q(x) = cx + d$ be two linear polynomials such that: (i) $p(0) = 5$; (ii) the polynomial $p(x) – q(x)$ cuts the $x$-axis at $(3, 0)$; (iii) the sum $p(x) + q(x)$ is equal to $6x + 4$ for all real $x$. Find the polynomials $p(x)$ and $q(x)$.

Solution. There are four unknowns — $a$, $b$, $c$, $d$ — so four facts are needed. The three conditions supply them, because (iii) is really two facts.

From (i): $p(0) = a(0) + b = b$, so

$$b = 5$$

From (iii): $p(x) + q(x) = (a + c)x + (b + d)$, and this must equal $6x + 4$ for every $x$. Two linear polynomials agree everywhere only if their coefficients match term by term:

$$a + c = 6 \qquad \text{and} \qquad b + d = 4$$

Since $b = 5$, the second gives $d = 4 – 5 = -1$.

From (ii): $p(x) – q(x) = (a – c)x + (b – d)$, and it is zero at $x = 3$:

$$3(a – c) + (b – d) = 0$$

With $b = 5$ and $d = -1$, we have $b – d = 6$, so

$$3(a – c) = -6 \quad\Longrightarrow\quad a – c = -2$$

Solve for $a$ and $c$. Adding $a + c = 6$ and $a – c = -2$:

$$2a = 4 \quad\Longrightarrow\quad a = 2, \qquad c = 4$$

$$p(x) = 2x + 5, \qquad q(x) = 4x – 1$$

Check all three conditions: $p(0) = 5$ ✓; $p(x) – q(x) = -2x + 6$, which is zero at $x = 3$ ✓; $p(x) + q(x) = 6x + 4$ ✓.

Question 12

*Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.

Solution. The counting rule comes from the phrase “shares a side”. The first hexagon needs $6$ matchsticks. Every hexagon after it borrows one side from its neighbour, so it needs only $5$ new matchsticks, not $6$.

$$\text{Stage } 1: 6 \qquad \text{Stage } 2: 6 + 5 = 11 \qquad \text{Stage } 3: 11 + 5 = 16$$

Stage 3 in the picture does indeed use $16$ matchsticks, which confirms the rule.

(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?

Stage 4 is four hexagons in a row, each sharing a side with the previous one; Stage 5 is five.

$$\text{Stage } 4: 16 + 5 = 21 \qquad \text{Stage } 5: 21 + 5 = 26$$

(ii) Complete the following table

Stage number	$1$	$2$	$3$	$4$	$5$	…	$n$
Number of matchsticks	$6$	$11$	$16$	$21$	$26$	…	$5n + 1$

(iii) Find a rule to determine the number of matchsticks required for the $n^{th}$ stage

The count goes up by $5$ at every stage, so the rule is linear in $n$. Starting from $6$ and adding $5$ a further $(n-1)$ times:

$$M(n) = 6 + 5(n – 1) = 5n + 1$$

Check: $M(1) = 6$ ✓, $M(3) = 16$ ✓, $M(5) = 26$ ✓.

(iv) How many matchsticks will be required for the $15^{th}$ stage of the pattern?

$$M(15) = 5 \times 15 + 1 = 76$$

(v) Can $200$ matchsticks form a stage in this pattern? Justify your answer.

Set the rule equal to $200$ and see whether $n$ comes out a whole number:

$$5n + 1 = 200 \quad\Longrightarrow\quad 5n = 199 \quad\Longrightarrow\quad n = \frac{199}{5} = 39.8$$

There is no $39.8^{\text{th}}$ stage, so the answer is no.

A neater way to say the same thing: every term $5n + 1$ leaves a remainder of $1$ when divided by $5$, so the counts always end in $1$ or $6$ — $6, 11, 16, 21, 26, \ldots$ — and $200$ ends in $0$. Stage $39$ uses $196$ matchsticks and stage $40$ uses $201$, so $200$ is skipped over.

Question 13

*Let $p(x) = ax + b$ and $q(x) = cx + d$ be two linear polynomials such that: (i) the graph of $p(x)$ passes through the points $(2, 3)$ and $(6, 11)$; (ii) the graph of $q(x)$ passes through the point $(4, -1)$; (iii) the graph of $q(x)$ is parallel to the graph of $p(x)$. Find the polynomials $p(x)$ and $q(x)$. Also, find the coordinates of the point where these lines meet the $x$-axis.

Solution. Take the conditions in order; each one supplies exactly what is needed next.

Find $p(x)$ from its two points:

$$2a + b = 3, \qquad 6a + b = 11$$

Subtracting: $4a = 8$, so $a = 2$; then $4 + b = 3$ gives $b = -1$.

$$p(x) = 2x – 1$$

Use the parallel condition. Parallel graphs have equal slopes, so $c = a = 2$. This is the fact that makes the single point for $q$ enough — without it, one point could not determine a line.

Find $d$ from the point $(4, -1)$:

$$2(4) + d = -1 \quad\Longrightarrow\quad 8 + d = -1 \quad\Longrightarrow\quad d = -9$$

$$q(x) = 2x – 9$$

Where each meets the $x$-axis — set each polynomial to zero:

$$2x – 1 = 0 \;\Longrightarrow\; x = \tfrac{1}{2} \qquad\qquad 2x – 9 = 0 \;\Longrightarrow\; x = \tfrac{9}{2}$$

Question 14

*What do all linear functions of the form $f(x) = ax + a$, $a > 0$, have in common?

Solution. The point of the question is that the same letter appears as both the slope and the constant term. Factorising makes the consequence visible:

$$f(x) = ax + a = a(x + 1)$$

Now ask where the function is zero. Since $a > 0$, the factor $a$ is never zero, so

$$f(x) = 0 \quad\Longleftrightarrow\quad x + 1 = 0 \quad\Longleftrightarrow\quad x = -1$$

Every one of these lines passes through $(-1, 0)$, whatever $a$ is. They all cut the $x$-axis at the same place, and form a fan of lines pivoting about that single point.

Two further things they share, both following from $a > 0$:

• every one of them rises from left to right, since the slope $a$ is positive;
• every one cuts the $y$-axis at $(0, a)$, a positive height — so the $y$-intercept equals the slope, and both are above zero.

Try $a = 1, 2, 3$: the lines $x + 1$, $2x + 2$ and $3x + 3$ all vanish at $x = -1$ and cross the $y$-axis at $1$, $2$ and $3$.

Common mistakes

• Question 1, writing a polynomial of degree $2$. $-7x^2 + 4x + 9$ satisfies the coefficient condition but has degree $2$. The $x^3$ term must be there, with a non-zero coefficient.
• Question 2(ii), trying to get a number. Substituting $t = a$ gives the expression $4a^3 – a^2 + 6$. There is nothing to evaluate, and an expression is the answer.
• Question 3, dividing by $\frac{5}{2}$ as though it were multiplying. After reaching $\frac{5}{2}x = -\frac{5}{4}$, the next step multiplies by $\frac{2}{5}$, not by $\frac{5}{2}$; the latter gives $-\frac{25}{8}$.
• Question 4, testing only one reading. “One of the new numbers becomes twice the other” is genuinely ambiguous, and a complete answer checks both and says why the second is rejected — the question specifies positive numbers.
• Question 5(ii), substituting $m = 2$ for two years. The rate is per month, so two years is $m = 24$. Using $m = 2$ gives ₹$1300$ instead of ₹$6800$.
• Question 6, answering with only one number. The question asks for both numbers, and $85$ and $58$ are a pair — either could be called the original.
• Question 7(iii), reading the slope as $6$. The equation $5y = 6x – 10$ must be divided by $5$ first; the slope is $\frac{6}{5}$. Skipping that step also makes (iii) look parallel to (ii) and (iv), which it is not.
• Question 7, calling lines with the same $y$-intercept parallel. Parallel is about the slope. Two lines sharing a $y$-intercept actually cross each other, at that very point.
• Question 8(ii), undoing the operations in the wrong order. Dividing by $\frac{9}{5}$ before subtracting the $32$ gives $x – 273 = 87.8$ and a wrong answer. Unwind the outermost operation first.
• Question 11, matching only the $x$-coefficients in condition (iii). “Equal for all real $x$” forces both the $x$-coefficients and the constant terms to match, and it is the constant match that yields $d = -1$.
• Question 12, counting $6$ matchsticks for every hexagon. That gives $6n$ — $18$ at stage 3, where the picture plainly shows $16$. Each added hexagon shares a side, so it costs $5$.
• Question 12(v), answering “yes, about stage 40”. The question asks whether $200$ matchsticks can form a stage, and $n$ must be a whole number. $39.8$ means no.
• Question 13, trying to find $q(x)$ from one point alone. One point does not determine a line. The parallel condition supplies the missing slope, and without using it the problem cannot be solved.
• Question 14, answering “they are all straight lines”. True of every linear function, so it is not what these have in common. The shared feature is the specific point $(-1, 0)$, and it comes from factorising $ax + a$ as $a(x+1)$.

Practise next

• Chapter 3, The World of Numbers — the next chapter, where the number system itself is put on a firmer footing before it is used any further.
• Exercise Set 2.5 — worth returning to after Questions 10, 11 and 13 here, since all four are the same technique: gather two independent facts about a line and solve for $a$ and $b$.
• Exercise Set 2.6 — the companion to Question 7. Drawing the sets of lines there makes the equal-slopes-means-parallel rule something you can see rather than something to recall.