Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
| Degree | Name | Example |
|---|---|---|
| 3 | Cubic | 5y³ + y² + 2y − 1 |
| 2 | Quadratic | x² + 5x + 1 |
| 1 | Linear | 3z + 7 |
| 0 | Constant | 8 (= 8x⁰) |
This chapter is about linear polynomials: degree 1, such as 2x + 3 or 5 − 4y.
| Matches m | 1 | 2 | 3 | 4 | 5 | m |
|---|---|---|---|---|---|---|
| Amount (₹) | 250 | 300 | 350 | 400 | 450 | 200 + 50m |
Each stage adds 2 tiles. Stage n has 2n − 1 tiles (one less than twice the stage number). This is a linear polynomial and the pattern is a linear relationship.
| d (km) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| C (₹) | 100 | 160 | 220 | 280 | 340 | 400 |
| t (months) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| h (m) | 3 | 2.5 | 2 | 1.5 | 1 |
Here 20 is the cost per GB and 150 is the fixed monthly fee.
| Idea | Rule |
|---|---|
| Degree | Highest power of the variable |
| Linear polynomial | ax + b, a ≠ 0 |
| Linear pattern | Constant difference; nth term an + b |
| Linear growth / decay | Positive / negative slope |
| Line y = ax + b | Slope a, y-intercept b, cuts y-axis at (0, b) |
| Parallel lines | Same a, different b |
Try each problem first, then open the solution to check your working.
For 7x³ − 4x² + x − 9, write the terms, the coefficient of each power of x, the constant term and the degree.
Terms: 7x³, −4x², x, −9.
Coefficients: x³: 7; x²: −4; x: 1.
Constant term: −9. The highest power is 3, so the degree is 3 (a cubic).
Answer: degree 3; coefficients 7, −4, 1; constant −9
State the degree and the kind of polynomial: (a) 6 − 2z (b) 5 (c) y³ − 2y (d) 4x² + x.
(a) degree 1, linear. (b) 5 = 5x⁰, degree 0, constant. (c) degree 3, cubic. (d) degree 2, quadratic.
Answer: linear, constant, cubic, quadratic
Find p(t) = 3t² − 5t + 2 at t = −2 and at t = ⅓.
Use: Substitute the value of t
p(−2) = 3(4) − 5(−2) + 2 = 12 + 10 + 2 = 24.
p(⅓) = 3(1/9) − 5/3 + 2 = 1/3 − 5/3 + 2 = −4/3 + 2 = 2/3.
Answer: 24 and 2/3
A notebook costs ₹x and a pen costs ₹y. Write the cost of 5 notebooks and 8 pens with ₹20 delivery, and find it for x = 30, y = 10.
Cost = 5x + 8y + 20.
For x = 30, y = 10: 150 + 80 + 20 = 250.
Answer: 5x + 8y + 20; ₹250
The sum of two numbers is 90 and one is 14 more than the other. Find them.
Smaller x, larger x + 14. So x + (x + 14) = 90, 2x = 76 and x = 38.
Larger = 52.
Answer: 38 and 52
The ratio of two numbers is 3 : 7 and their difference is 36. Find the numbers.
Numbers 3k and 7k. Then 7k − 3k = 4k = 36, so k = 9.
Numbers: 27 and 63.
Answer: 27 and 63
Meena has twice as many ₹10 notes as ₹20 notes, in total ₹400. How many of each?
Let ₹20 notes be n; ₹10 notes are 2n.
20n + 10(2n) = 40n = 400, so n = 10.
Answer: 10 notes of ₹20 and 20 notes of ₹10
The length of a rectangle is 5 cm more than its width and its perimeter is 50 cm. Find the sides.
Use: Perimeter = 2(l + w)
Width w, length w + 5. 2(2w + 5) = 50, so 2w + 5 = 25 and w = 10.
Answer: width 10 cm, length 15 cm
For the pattern 5, 8, 11, 14, … find the nth term, the 20th term and which term is 101.
Difference is 3, so the nth term is 3n + 2 (check n = 1: 5).
20th term: 3(20) + 2 = 62.
3n + 2 = 101 gives n = 33.
Answer: 3n + 2; 62; the 33rd term
A row of squares is made with matchsticks: 1 square needs 4 sticks and each extra square adds 3 sticks. Find the sticks for n squares, for 25 squares, and how many squares 100 sticks make.
Sticks = 4 + 3(n − 1) = 3n + 1.
n = 25: 76 sticks.
3n + 1 = 100 gives n = 33.
Answer: 3n + 1; 76; 33 squares
A tank has 500 L and loses 20 L every minute. Write the volume after t minutes, the volume after 12 minutes and when it is empty.
V = 500 − 20t.
t = 12: 500 − 240 = 260 L.
V = 0 when t = 25.
Answer: V = 500 − 20t; 260 L; 25 minutes
Rohan has ₹1,200 and saves ₹150 every week. When will he have ₹3,000?
S = 1200 + 150w.
1200 + 150w = 3000 gives 150w = 1800, so w = 12.
Answer: after 12 weeks
A taxi charges ₹40 plus ₹12 per km. Find the fare for 15 km and the distance for ₹148.
F(d) = 40 + 12d.
F(15) = 40 + 180 = 220.
40 + 12d = 148 gives d = 9.
Answer: ₹220; 9 km
Say whether each is linear growth or linear decay: (a) tea temperature T = 80 − 4t (b) plant height h = 2 + 0.3t (c) petrol in a tank P = 40 − 6t.
Use: Look at the sign of the slope
(a) slope −4: decay. (b) slope +0.3: growth. (c) slope −6: decay.
Answer: (a) decay (b) growth (c) decay
y = ax + b gives y = 11 when x = 2 and y = 27 when x = 6. Find a and b.
11 = 2a + b and 27 = 6a + b. Subtract: 16 = 4a, so a = 4.
b = 11 − 8 = 3. The relation is y = 4x + 3.
Answer: a = 4, b = 3
°F = a°C + b. Water freezes at 0 °C = 32 °F and boils at 100 °C = 212 °F. Find the relation and the Fahrenheit value of 37 °C.
32 = b. 212 = 100a + 32, so a = 180/100 = 9/5.
F = (9/5)C + 32. At 37 °C: 66.6 + 32 = 98.6.
Answer: F = (9/5)C + 32; 98.6 °F
Find the slope and y-intercept of 2y = 6x − 8, and where it cuts both axes.
y = 3x − 4: slope 3, y-intercept −4.
Cuts the y-axis at (0, −4). For the x-axis put y = 0: 3x = 4, x = 4/3, so (4/3, 0).
Answer: slope 3; intercept −4; (0, −4) and (4/3, 0)
Does (3, 11) lie on y = 4x − 1? Does (−2, −7)?
x = 3: 4(3) − 1 = 11. Yes.
x = −2: 4(−2) − 1 = −9, not −7. No.
Answer: (3, 11) is on the line; (−2, −7) is not
Which of these are parallel to y = 3x + 5: (a) 6x − 2y = 7 (b) y = −3x + 5 (c) y = 3x − 8?
Use: Parallel lines have equal slopes
(a) y = 3x − 7/2: slope 3, parallel. (b) slope −3, not parallel. (c) slope 3, parallel.
Answer: (a) and (c)
A line passes through (2, 5) and (6, 13). Find its equation.
a = (13 − 5)/(6 − 2) = 2.
5 = 2(2) + b gives b = 1.
Check (6, 13): 2(6) + 1 = 13.
Answer: y = 2x + 1
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every question in the chapter, including Think and Reflect. Try each one first, then tap Show answer.
(i) 2 (ii) 3 (iii) 0 (a non-zero constant, −9 = −9x⁰) (iv) 1.
Answers may differ. Degree 1: 3x + 2. Degree 2: x² − 4x + 1. Degree 3: 2x³ − x + 5.
Coefficient of x²: 6. Coefficient of x³: −3.
There is no z term, so the coefficient of z is 0.
The constant term is −10.
Terms: 200l, 160w, 50lw. Variables: l and w. Coefficients: 200, 160, 50. Unlike 4x + 5y + 3, it has no constant term and it has a product term 50lw (degree 2 in the two variables).
Terms: 10x and −x². One variable, x. Coefficients: 10 and −1. Like Example 1 it has no product term; unlike Examples 1 and 2 it has only one variable, so it is a polynomial (of degree 2).
Perimeters 4, 6, 8, 10, 12 cm. Each 0.5 cm increase in the side adds 2 cm to the perimeter.
200 + 50m = 750, so 50m = 550 and m = 11.
Input x (the length), output the area. For x = 6: 10(6) − 6² = 60 − 36 = 24 cm².
The sequence is 1, 3, 5, 7, 9, 11, 13. The next three stages (5, 6, 7) have 9, 11, 13 squares.
15th: 2(15) − 1 = 29. 26th: 2(26) − 1 = 51. 2n − 1 = 21 gives n = 11. 2n − 1 = 47 gives n = 24.
100 − 5(15) = ₹25. 100 − 5n = 0 gives n = 20 days.
15n − 5 = 130 gives 15n = 135 and n = 9 km.
C(15) = 100 + 60(15) = ₹1000. 100 + 60d = 700 gives d = 10 km.
h(5) = 3 − 0.5(5) = 0.5 m.
(i) −3 (ii) 5(−1) − 3 = −8 (iii) 10 − 3 = 7.
(i) 6 (ii) 7(9) + 12 + 6 = 81 (iii) 7(16) − 16 + 6 = 102.
Salil x, mother 3x. (x + 5) + (3x + 5) = 70, so 4x = 60 and x = 15.
Salil 15 years, mother 45 years.
Numbers 2k and 5k. 5k − 2k = 3k = 63, so k = 21.
42 and 105.
Let ₹5 coins be n; ₹2 coins are 3n. 5n + 2(3n) = 11n = 88, so n = 8.
8 coins of ₹5 and 24 coins of ₹2 (32 coins in all).
Shorter x, longer 4x. 5x = 300, so x = 60.
60 ft and 240 ft.
Width w, length 2w + 3. 2(3w + 3) = 24, so 3w + 3 = 12 and w = 3.
Width 3 cm, length 9 cm.
Taking ₹500 as the balance at the start and ₹150 added at the end of each month, the balance after n months is 500 + 150n. After month 2: ₹800; month 3: ₹950; month 4: ₹1100, and so on.
After 1, 2, 3 hours: 111, 102, 93. After n hours: 120 − 9n.
Areas: 13 × 12 = 156, 13 × 10 = 130, 13 × 8 = 104. For breadth b the area is 13b. The areas fall by 26 each time the breadth falls by 2, a linear pattern.
Base area = 77 cm². Volumes: 77 × 5 = 385, 77 × 9 = 693, 77 × 13 = 1001 cm³. Volume = 77h, a linear pattern in h (it rises by 308 for every 4 cm of height).
Pages left after n days: 500 − 20n. After 15 days: 500 − 300 = 200 pages.
(i) 1.75 + 0.5(7) = 5.25 ft.
(ii)
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| h | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |
(iii) h = 1.75 + 0.5t. The height rises by a fixed 0.5 ft each month, so this is linear growth.
(i) 10000 − 2400 = ₹7600.
(ii)
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| v (₹) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
(iii) v = 10000 − 800t. It falls by a fixed ₹800 a year: linear decay.
(i) 750 + 300 = 1050.
(ii)
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| P | 750 | 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 |
(iii) P = 750 + 50t: it rises by 50 every year, so linear growth.
(i) b(x) = 600 − 15x; it falls by a fixed ₹15 a day, so linear decay.
(ii) 600 − 15x = 0 gives x = 40 days.
(iii)
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| b(x) | 585 | 570 | 555 | 540 | 525 | 510 | 495 | 480 | 465 | 450 |
400 = 10a + b and 500 = 14a + b. Subtract: 100 = 4a, so a = 25. Then b = 400 − 250 = 150.
a = 25, b = 150 (y = 25x + 150).
800 = 10a + b and 1100 = 15a + b. Subtract: 300 = 5a, so a = 60. Then b = 800 − 600 = 200.
a = 60, b = 200 (y = 60x + 200).
0 = 32a + b and 100 = 212a + b. Subtract: 100 = 180a, so a = 5/9. Then b = −32a = −160/9.
a = 5/9, b = −160/9, so °C = (5/9)°F − 160/9 = (5/9)(°F − 32).
Points: (0, 0) and (1, 4); (0, 0) and (1, 2); (0, 0) and (1, 1). All pass through the origin (b = 0). The bigger a is, the steeper the line.
Points (0, 0) and (1, −6); (0, 0) and (1, −3); (0, 0) and (1, −1). All fall to the right through the origin; the bigger |a| is, the steeper the fall.
Points (0, 0), (1, 5) and (0, 0), (1, −5). Same steepness, one rising and one falling; each is the mirror image of the other in the y-axis (and in the x-axis).
Points: (0, −1), (1, 2); (0, 0), (1, 3); (0, 1), (1, 4). Same slope 3, so the lines are parallel; b = −1, 0, 1 are the y-intercepts.
Points: (0, −3), (1, −5); (0, 0), (1, −2); (0, 3), (1, 1). Same slope −2 (falling), so they are parallel; the y-intercepts are −3, 0, 3.
In each point y is three times x, so y = 3x.
Each y is −2 times x, so y = −2x.
y = ½x: points (0, 0) and (4, 2). y = x: (0, 0) and (1, 1). y = 2x: (0, 0) and (1, 2). All three pass through the origin; 2x is steepest and ½x flattest.
| Value | Result |
|---|---|
| x = 1 | y = 3 |
| x = 2 | y = 5 |
| x = 5 | y = 11 |
| x = 7 | y = 15 |
| x = 9 | y = 19 |
| x = 12 | y = 25 |
| x = 20 | y = 41 |
The lines pass through the origin. When a > 1 the line is steeper than y = x; when a < 1 it is flatter. So a measures steepness (the slope). y = 3x is steeper than y = 2x; y = ⅓x is flatter than y = ½x.
The lines fall from left to right. A larger a gives a steeper fall. y = 3x + 1 rises (slope 3) and y = −3x + 1 falls (slope −3), but both cut the y-axis at (0, 1).
With a = 2 and b = −1, 1, 5 the lines are parallel: changing b shifts the line up or down but does not change its slope.
20 is the cost per GB (the slope: the bill rises ₹20 per GB). 150 is the fixed monthly fee (the y-intercept: the bill when x = 0).
Answers may differ. For example x³ − 7x² + 2x + 5.
(i) 5 − 3 + 7 = 9.
(ii) 4a³ − a² + 6.
(5/2)x + 2/3 = −7/12, so (5/2)x = −7/12 − 8/12 = −15/12 = −5/4.
x = (−5/4) × (2/5) = −1/2. Check: −5/4 + 2/3 = −15/12 + 8/12 = −7/12 ✓.
Numbers x and 5x. After adding 21: x + 21 and 5x + 21. Take 5x + 21 = 2(x + 21): 3x = 21 and x = 7. (The other way, x + 21 = 2(5x + 21), gives a negative x.)
7 and 35. Check: 28 and 56 ✓.
Amount after n months = 800 + 250n. (i) 800 + 1500 = ₹2300. (ii) 2 years = 24 months: 800 + 6000 = ₹6800.
Digits a, b: (10a + b) + (10b + a) = 11(a + b) = 143, so a + b = 13. With |a − b| = 3: a, b = 8, 5.
The numbers are 85 and 58 (either could be the original; 85 + 58 = 143).
(i) slope −3, intercept 4, cuts at (0, 4).
(ii) y = 2x + 7/2: slope 2, intercept 7/2, cuts at (0, 7/2).
(iii) y = (6/5)x − 2: slope 6/5, intercept −2, cuts at (0, −2).
(iv) y = 2x − 11/3: slope 2, intercept −11/3, cuts at (0, −11/3).
Lines (ii) and (iv) are parallel (both have slope 2).
(i) (9/5)(40) + 32 = 72 + 32 = 104 °F.
(ii) 158 − 32 = 126 = (9/5)(x − 273), so x − 273 = 70 and x = 343 K.
w = 3d. The graph is a straight line through the origin, for example through (0, 0) and (1, 3).
For d = 2: w = 6; the point (2, 6) lies on the graph, which verifies it.
(i) a = (11 − 5)/(3 − 1) = 3 and b = 5 − 3 = 2, so p(x) = 3x + 2.
(ii) y-axis: (0, 2). x-axis: 3x + 2 = 0 gives x = −2/3, so (−2/3, 0).
(iii) Plot (1, 5), (3, 11), (0, 2) and (−2/3, 0): all lie on one line.
(i) b = 5. (iii) a + c = 6 and b + d = 4, so d = −1.
(ii) p − q = (a − c)x + (b − d) = (a − c)x + 6 is zero at x = 3: 3(a − c) + 6 = 0, so a − c = −2.
a + c = 6 and a − c = −2 give a = 2 and c = 4.
p(x) = 2x + 5, q(x) = 4x − 1. Check: p − q = −2x + 6 (zero at 3) and p + q = 6x + 4 ✓.
Stage 1 has 6 sticks and each new hexagon adds 5 (it shares one side).
(i) Stage 4: 21 sticks; Stage 5: 26 sticks.
(ii)
| Stage n | 1 | 2 | 3 | 4 | 5 | n |
|---|---|---|---|---|---|---|
| Matchsticks | 6 | 11 | 16 | 21 | 26 | 5n + 1 |
(iii) 5n + 1. (iv) 5(15) + 1 = 76.
(v) 5n + 1 = 200 gives n = 39.8, not a whole number, so no.
p: a = (11 − 3)/4 = 2, b = 3 − 4 = −1, so p(x) = 2x − 1.
q is parallel, so c = 2. −1 = 2(4) + d gives d = −9, so q(x) = 2x − 9.
Meeting the x-axis: p at (1/2, 0); q at (9/2, 0).
f(−1) = −a + a = 0, so every such line passes through the point (−1, 0) (it cuts the x-axis at x = −1), while the slope a varies.