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

Book page 36–39 Updated on2026-09-08

Q1.

Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is –7.

Answer

For example: x³ − 7x² + 2x + 5

Other correct answers: 4x³ − 7x², −x³ − 7x² + 1, 2x³ − 7x² − 3x.

The two conditions that must be met:
• Degree 3 means the coefficient of x³ must be non-zero. Writing −7x² + 2x + 5 would satisfy the second condition but has degree 2, not 3.
• The x² term must be exactly −7x², sign included.
Everything else is free: the x term and the constant may be any numbers at all, including 0. This is why the answer is not unique — the conditions fix two of the four coefficients and leave two to choose.

Check it yourself: in x³ − 7x² + 2x + 5 the coefficients in order are 1, −7, 2, 5. The first is not zero (degree 3 ✓) and the second is −7 ✓.

Q2.

Find the values of the following polynomials at the indicated values of the variables. (i) 5x² – 3x + 7 if x = 1 (ii) 4t³ – t² + 6 if t = a

Answer

(i) Substitute x = 1.

5(1)² − 3(1) + 7
= 5 − 3 + 7
= 9

(ii) Substitute t = a. The input is a letter, so the output is an expression.

4a³ − a² + 6
= 4a³ − a² + 6

Why (ii) is not a trick question: evaluating a polynomial means replacing the variable by whatever is given — a number, another letter, even a whole expression. Nothing says the input must be numerical. Here every t becomes an a and no arithmetic can be done, so the answer stays in symbols. Note also that in (i) the value at x = 1 equals the sum of the coefficients, 5 + (−3) + 7 = 9, because every power of 1 is 1 — a quick check worth remembering.

Try This: evaluate the same polynomial in (ii) at t = 2a. You should get 4(2a)³ − (2a)² + 6 = 32a³ − 4a² + 6 — brackets are essential.

Q3.

If we multiply a number by 5/2 and add 2/3 to the product, we get –7/12. Find the number.

Answer

Let the number be x. The description translates directly into a linear equation.

(5/2)x + 2/3 = −7/12
(5/2)x = −7/12 − 2/3
= −7/12 − 8/12
= −15/12 = −5/4
x = (−5/4) × (2/5)
x = −10/20 = −1/2

Check: (5/2)(−1/2) = −5/4; and −5/4 + 2/3 = −15/12 + 8/12 = −7/12. ✓

Why the LCM 12 is the right common denominator: to subtract 2/3 from −7/12 both must be written over the same denominator, and 12 is the LCM of 12 and 3. Then dividing by 5/2 is the same as multiplying by its reciprocal 2/5 — that is what “undoing a multiplication” means for fractions. The polynomial (5/2)x + 2/3 is linear, so this equation has exactly one solution.

Tip: an alternative is to clear all fractions at the start: multiply the whole equation by 12 to get 30x + 8 = −7, so 30x = −15 and x = −½. Same answer, no fraction arithmetic in the middle.

Q4.

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?

Answer

Let the smaller number be x. The other is 5x. After adding 21 they become x + 21 and 5x + 21.

Since 5x + 21 is the larger of the two, it must be the one that is twice the other.

5x + 21 = 2(x + 21)
5x + 21 = 2x + 42
3x = 21
x = 7

The numbers are 7 and 35.

Check: adding 21 gives 28 and 56, and 56 = 2 × 28. ✓

Why the other case is rejected: the phrase “one becomes twice the other” leaves two possibilities, and both should be tested. Taking x + 21 = 2(5x + 21) gives x + 21 = 10x + 42, so 9x = −21 and x = −7/3. That is not a positive number, and the question says the numbers are positive, so this branch is discarded. Checking both branches and then ruling one out is the honest way to answer — the answer is unique, but only after the positivity condition is used.

Check it yourself: adding the same 21 to both numbers shrinks the ratio from 5 : 1 down to 2 : 1, which is why a single value of x lands it exactly on 2 : 1.

Q5.

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.

Answer

Let n be the number of months. The amount is A = 800 + 250n.

(i) n = 6: 800 + 250(6) = 800 + 1500 = ₹2300
(ii) 2 years = 24 months, so n = 24:
800 + 250(24) = 800 + 6000 = ₹6800

Linear pattern: A = ₹(800 + 250n)

Month n	0	1	2	3	6	12	24
Amount (₹)	800	1050	1300	1550	2300	3800	6800

Why it is a linear pattern: consecutive amounts differ by the constant ₹250, so A(n + 1) − A(n) = 250 for every n. The ₹800 is the value at n = 0 — the constant term, or y-intercept — and 250 is the slope. Because the step never changes, we can leap straight to n = 24 instead of adding ₹250 twenty-four times.

Tip: The commonest error is to use n = 2 for “2 years”. The saving is monthly, so the unit of n must be months: 2 years = 24 months.

Q6.

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.

Answer

Let the tens digit be t and the units digit be u.

Original number = 10t + u
Reversed number = 10u + t
Sum = 11t + 11u = 11(t + u) = 143
t + u = 143 ÷ 11 = 13

Also the digits differ by 3, so the two digits add to 13 and differ by 3:

digits = (13 + 3)/2 = 8 and (13 − 3)/2 = 5

The two numbers are 85 and 58.

Check: 85 + 58 = 143 ✓ and 8 − 5 = 3. ✓

Why the sum is always a multiple of 11: adding a two-digit number to its reverse gives (10t + u) + (10u + t) = 11(t + u). Each digit is counted once as a ten and once as a unit, so it contributes 11 times itself. This is why 143 — which is 11 × 13 — works at all: had the total been, say, 145, no two-digit number could produce it. Notice that the sum depends only on t + u, so it cannot by itself tell 85 from 58; the “differ by 3” condition fixes the pair of digits, and both orders are valid answers.

Check it yourself: digits adding to 13 are (4, 9), (5, 8), (6, 7). Only (5, 8) differs by 3, so the pair 85 / 58 is the only one that works.

Q7.

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. (i) y = –3x + 4 (ii) 2y = 4x + 7 (iii) 5y = 6x – 10 (iv) 3y = 6x – 11. Are any of the lines parallel?

Answer

First put every equation in the form y = ax + b by dividing through by the coefficient of y. Then a is the slope and b the y-intercept, and the line cuts the y-axis at (0, b).

Equation	In the form y = ax + b	Slope a	y-intercept b	Cuts y-axis at
(i) y = −3x + 4	y = −3x + 4	−3	4	(0, 4)
(ii) 2y = 4x + 7	y = 2x + 7/2	2	7/2 = 3.5	(0, 3.5)
(iii) 5y = 6x − 10	y = (6/5)x − 2	6/5 = 1.2	−2	(0, −2)
(iv) 3y = 6x − 11	y = 2x − 11/3	2	−11/3	(0, −11/3)

The four lines. (ii) and (iv) never meet — they have the same slope 2.

Two convenient points for each line:

Line	Point A	Point B
(i)	(0, 4)	(2, −2)
(ii)	(0, 3.5)	(2, 7.5)
(iii)	(0, −2)	(5, 4)
(iv)	(1, −5/3)	(4, 13/3)

Parallel lines: yes — (ii) and (iv).

Why (ii) and (iv) are parallel and the others are not: both reduce to slope 2, and their y-intercepts 7/2 and −11/3 are different, so they can never meet: setting 2x + 7/2 = 2x − 11/3 cancels the 2x and leaves 7/2 = −11/3, which is false. Slopes −3, 2, 6/5, 2 show that no other pair matches. It is essential to divide first: (ii) and (iv) look quite unlike each other as printed, and only the reduced form reveals that they have the same slope.

Tip: the y-intercept can always be found without rearranging — put x = 0. In (iv) that gives 3y = −11, so y = −11/3 straight away.

Q8.

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 = (9/5)(x – 273) + 32. (i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K. (ii) If the temperature is 158 °F, then find the temperature in Kelvin.

Answer

(i) Put x = 313.

y = (9/5)(313 − 273) + 32
= (9/5)(40) + 32
= 72 + 32
= 104 °F

(ii) Put y = 158 and work backwards.

158 = (9/5)(x − 273) + 32
158 − 32 = (9/5)(x − 273)
126 = (9/5)(x − 273)
x − 273 = 126 × (5/9) = 70
x = 70 + 273 = 343 K

Check: (9/5)(343 − 273) + 32 = (9/5)(70) + 32 = 126 + 32 = 158 °F. ✓

Why the equation is built this way: (x − 273) converts Kelvin to Celsius, because the two scales have the same size of degree but Kelvin starts 273 lower. Multiplying by 9/5 converts a Celsius interval to a Fahrenheit interval — the reciprocal of the 5/9 found in Exercise Set 2.5 Q3 — and adding 32 shifts the zero to where Fahrenheit puts freezing. Undoing it in part (ii) means reversing those three steps in the opposite order: subtract 32, multiply by 5/9, add 273.

Check it yourself: 313 K is 40 °C and 343 K is 70 °C. Since 30 Celsius degrees separate them, the Fahrenheit readings should differ by (9/5)(30) = 54 — and 158 − 104 = 54. ✓

Q9.

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.

Answer

In words: work = force × distance. With the force fixed at 3 units,

w = F × d
w = 3d

This is a linear equation in the two variables w and d, with slope 3 and constant term 0.

Distance d (units)	0	1	2	3	4
Work w = 3d (units)	0	3	6	9	12

At d = 2: w = 3 × 2 = 6 units

The graph of w = 3d. Going up from d = 2 on the horizontal axis meets the line at height 6, confirming the calculation.

Why the line must pass through the origin: the constant term is 0, because a body that has not moved has had no work done on it — d = 0 forces w = 0. The slope 3 is the force itself: each extra unit of distance adds 3 units of work, always the same amount, which is exactly the assumption that the force is constant. A force that changed with distance would not give a straight line.

Tip: Reading a value off a graph should confirm a calculation, never replace it. Here the arithmetic gives 6 exactly; the graph shows that 6 is where the line really is.

Q10.

The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11). (i) Find the polynomial p(x). (ii) Find the coordinates where the graph of p(x) cuts the axes. (iii) Draw the graph of p(x) and verify your answers.

Answer

(i) A linear polynomial has the form p(x) = ax + b. The two points give two equations.

a(1) + b = 5 …(1)
a(3) + b = 11 …(2)
(2) − (1): 2a = 6, so a = 3
From (1): b = 5 − 3 = 2

p(x) = 3x + 2

(ii) The graph cuts the y-axis where x = 0 and the x-axis where p(x) = 0.

x = 0: p(0) = 2 → (0, 2) on the y-axis
p(x) = 0: 3x + 2 = 0 → x = −2/3 → (−2/3, 0) on the x-axis

(iii)

p(x) = 3x + 2, cutting the y-axis at (0, 2) and the x-axis at (−2/3, 0). The given points (1, 5) and (3, 11) lie on it.

Verification: p(1) = 3 + 2 = 5 ✓ and p(3) = 9 + 2 = 11. ✓

Why the slope is found by subtracting: between the two given points, x rises by 3 − 1 = 2 while y rises by 11 − 5 = 6. The rise per unit of run is 6 ÷ 2 = 3, and that is what a means. Subtracting equation (1) from (2) does the same arithmetic algebraically, and has the advantage of eliminating b at the same time.
Why there is exactly one x-intercept: 3x + 2 = 0 is a linear equation with a non-zero coefficient of x, so it has a single solution. Every non-constant linear polynomial has precisely one zero, which is why its graph crosses the x-axis exactly once.

Check it yourself: −2/3 ≈ −0.67, so the crossing sits just left of the origin — consistent with a line that is already at height 2 when x = 0 and climbing at 3 per unit.

Q11.

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).

Answer

Turn each condition into an equation in a, b, c, d.

(i) p(0) = b = 5

(iii) p(x) + q(x) = (a + c)x + (b + d) = 6x + 4
    so a + c = 6 and b + d = 4
    with b = 5 this gives d = −1

(ii) p(x) − q(x) = (a − c)x + (b − d) is 0 at x = 3
    3(a − c) + (5 − (−1)) = 0
    3(a − c) + 6 = 0, so a − c = −2

Now solve a + c = 6 and a − c = −2.

Adding: 2a = 4, so a = 2
Then c = 6 − 2 = 4

p(x) = 2x + 5 and q(x) = 4x − 1

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

Why condition (iii) gives two equations from one statement: “equal for all real x” is much stronger than equal at one value. Two linear polynomials agree everywhere only if they are the same polynomial, which forces the coefficients of x to match and the constants to match separately. That single condition therefore supplies two of the four equations we need — and with (i) and (ii) making four in all, the four unknowns are pinned down uniquely.

Tip: Order the work as above — (i), then (iii), then (ii). Getting b and d first makes the constant term of p − q a known number, so (ii) becomes a single equation in a − c.

Q12.

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. (i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages? (ii) Complete the following table. (iii) Find a rule to determine the number of matchsticks required for the nth stage. (iv) How many matchsticks will be required for the 15th stage of the pattern? (v) Can 200 matchsticks form a stage in this pattern? Justify your answer.

Answer

(i) Stage 4 has four hexagons in the chain and Stage 5 has five.

Stages 4 and 5. Each new hexagon is attached along one existing side, so it costs only 5 fresh matchsticks.

Stage 4: 6 + 5 + 5 + 5 = 21 matchsticks
Stage 5: 21 + 5 = 26 matchsticks

(ii)

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

(iii)

Number of matchsticks at Stage n = 5n + 1

(iv)

n = 15: 5(15) + 1 = 75 + 1 = 76 matchsticks

(v) Set the rule equal to 200.

5n + 1 = 200
5n = 199
n = 199/5 = 39.8

Since n must be a whole number, no — 200 matchsticks cannot form a stage of this pattern. Stage 39 uses 196 and Stage 40 uses 201.

Why the rule is 5n + 1 and not 6n: the first hexagon costs all 6 of its sides. Every hexagon after it is glued onto the previous one along a side that is already there, so it needs only 5 new matchsticks. Stage n therefore costs 6 + 5(n − 1) = 5n + 1. The constant difference of 5 between consecutive stages is what makes this a linear pattern, and 5 is the slope of the corresponding line.
A second way to see the “+1”: think of each hexagon as costing 5 sticks, with one extra stick needed at the very start to close off the first hexagon.
Why (v) can be settled without listing: 5n + 1 always leaves remainder 1 on division by 5, so every possible total is one of 6, 11, 16, 21, … — all ending in 1 or 6. 200 ends in 0, so it is not in the list.

Check it yourself: the differences 11 − 6, 16 − 11, 21 − 16, 26 − 21 are all 5 ✓, and 5(1) + 1 = 6 matches Stage 1.

Q13.

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.

Answer

Finding p(x): the two points give two equations.

2a + b = 3 …(1)
6a + b = 11 …(2)
(2) − (1): 4a = 8, so a = 2
From (1): b = 3 − 4 = −1

p(x) = 2x − 1

Finding q(x): parallel means the same slope, so c = 2. Then use the point (4, −1).

2(4) + d = −1
8 + d = −1
d = −9

q(x) = 2x − 9

Where they meet the x-axis (set each polynomial to 0):

2x − 1 = 0 → x = 1/2 → (1/2, 0)
2x − 9 = 0 → x = 9/2 → (9/2, 0)

p(x) = 2x − 1 and q(x) = 2x − 9 are parallel, meeting the x-axis at (1/2, 0) and (9/2, 0).

Check: p(2) = 3 ✓, p(6) = 11 ✓, q(4) = 8 − 9 = −1. ✓

Why “parallel” is enough to determine c: two lines are parallel exactly when their slopes are equal, so c = a = 2 with no further work. Only the intercept of q is then unknown, and one point is enough to fix one unknown — which is why the question supplies just one point for q but two for p.
Notice the horizontal gap between the two x-intercepts: 9/2 − 1/2 = 4. That agrees with the constant vertical gap of 8 between the lines, since a line of slope 2 needs a run of 8 ÷ 2 = 4 to climb 8.

Tip: the zero of ax + b is always x = −b/a. Here that is −(−1)/2 = 1/2 and −(−9)/2 = 9/2 — a one-line check on both answers.

Q14.

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

Answer

Factorise, and the answer falls out.

f(x) = ax + a = a(x + 1)
f(−1) = a(−1 + 1) = a × 0 = 0 for every a

Every such line passes through the point (−1, 0) — they all cut the x-axis at the same place.

Three further things they share, all consequences of a > 0:

The slope equals the y-intercept, since both are a: the line meets the y-axis at (0, a).
All have positive slope, so all rise from left to right — every one of them is a picture of linear growth.
Each is positive for x > −1 and negative for x < −1, because the sign of a(x + 1) is decided by the bracket alone.

y = x + 1, y = 2x + 2 and y = 3x + 3 — different slopes, but all three pass through (−1, 0).

Why the common point had to exist: a family of lines whose two coefficients are tied together has only one free parameter, so it is a family of lines rotating about a single point rather than a general collection. Writing f(x) = a(x + 1) shows this at once: a multiplies everything, so wherever the bracket is zero the whole expression is zero no matter what a is. That fixed zero, x = −1, is the pivot. Since a > 0, the family sweeps out every positive slope through that pivot, but never a horizontal or falling line.

Check it yourself: take a = 1, 2 and 10. The lines y = x + 1, y = 2x + 2 and y = 10x + 10 give y = 0 at x = −1 in all three cases, while their y-intercepts 1, 2 and 10 are all different.

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