NCERT Solutions Class 9 Maths Chapter 2: Linear Polynomials

The perimeters are 4, 6, 8, 10 and 12 cm. Every time the side grows by 0.5 cm, the perimeter grows by 2 cm.

Perimeter of a square = 4 × side.

The perimeter goes up by the same amount (2 cm) each time, so it follows a linear pattern: P = 4s.

15th stage: 29 tiles. 26th stage: 51 tiles. 21 tiles are in stage 11 and 47 tiles are in stage 24.

1. Tiles in a given stagen = 15: 2(15) − 1 = 29n = 26: 2(26) − 1 = 51
2. Stage for a given number of tilesPut 2n − 1 equal to the number of tiles.2n − 1 = 21 ⇒ 2n = 22 ⇒ n = 112n − 1 = 47 ⇒ 2n = 48 ⇒ n = 24

y = 3x + 1 rises from left to right (slope 3). y = −3x + 1 falls from left to right (slope −3). Both cut the y-axis at (0, 1).

Both lines pass through the same point (0, 1), but one slopes up and the other slopes down.

Yes. When a is fixed and b varies, the lines are parallel. Only their position changes.

• The slope a is the same, so every line has the same inclination (tilt).
• The y-intercept b changes, so each line shifts up or down and cuts the y-axis at a different point.

The coefficient of z is 0.

The polynomial has no term with z¹ (that is, z). So the coefficient of z is 0.

The constant term is −10.

The constant term is the term without any variable, which is −10.

Coefficient of x² = 6. Coefficient of x³ = −3.

For example: 2x + 5 (degree 1), x² + 3x + 1 (degree 2) and x³ − 2x² + x + 4 (degree 3).

Answers may differ: any polynomial whose highest power is 1, 2 or 3 is correct.

(i) 2 (ii) 3 (iii) 0 (iv) 1

The degree of a polynomial is the highest power of its variable.

The values are −3, −8 and 7.

1. x = 05(0) − 3 = −3
2. x = −15(−1) − 3 = −5 − 3 = −8
3. x = 25(2) − 3 = 10 − 3 = 7

The values are 6, 81 and 102.

1. s = 07(0)² − 4(0) + 6 = 6
2. s = −37(−3)² − 4(−3) + 6 = 63 + 12 + 6 = 81
3. s = 47(4)² − 4(4) + 6 = 112 − 16 + 6 = 102

Salil is 15 years old and his mother is 45 years old.

1. Choose the unknownLet Salil’s age = x years. Then his mother’s age = 3x years.
2. After 5 yearsSalil: x + 5. Mother: 3x + 5.
3. Make the equation(x + 5) + (3x + 5) = 70 ⇒ 4x + 10 = 70 ⇒ 4x = 60 ⇒ x = 15
4. AnswerSalil = 15 years, mother = 3 × 15 = 45 years.

Check: after 5 years, 20 + 50 = 70.

The integers are 42 and 105.

1. Let the integers be 2x and 5x5x − 2x = 63 ⇒ 3x = 63 ⇒ x = 21
2. The integers2x = 42 and 5x = 105

Check: 105 − 42 = 63 and 42 : 105 = 2 : 5.

Ruby has 8 five-rupee coins and 24 two-rupee coins.

1. Let the number of five-rupee coins = xThen the number of two-rupee coins = 3x.
2. Total value5x + 2(3x) = 88 ⇒ 11x = 88 ⇒ x = 8

Five-rupee coins: 8. Two-rupee coins: 3 × 8 = 24. Check: 40 + 48 = 88.

The shorter piece is 60 feet and the longer piece is 240 feet.

1. Let the shorter piece = x feetThe longer piece = 4x feet.
2. Total lengthx + 4x = 300 ⇒ 5x = 300 ⇒ x = 60

Longer piece = 4 × 60 = 240 feet.

The width is 3 cm and the length is 9 cm.

1. Let the width = x cmThen the length = 2x + 3 cm.
2. Perimeter = 2(length + width)2[(2x + 3) + x] = 24 ⇒ 6x + 6 = 24 ⇒ 6x = 18 ⇒ x = 3

Length = 2(3) + 3 = 9 cm. Check: 2(9 + 3) = 24.

She has ₹800, ₹950, ₹1100, … at the end of months 2, 3, 4, … The linear expression is A_n = 500 + 150n.

The amount rises by ₹150 every month, so the pattern is linear: A_n = 150n + 500.

111, 102, 93, … members remain after 1, 2, 3, … hours. The expression is M_n = 120 − 9n.

The number falls by 9 every hour, so the pattern is linear: M_n = 120 − 9n.

The areas are 156 cm², 130 cm² and 104 cm². The linear pattern is A = 13x.

If the breadth is x cm, then A = 13x. Each 1 cm of breadth adds 13 cm² of area.

The volumes are 385 cm³, 693 cm³ and 1001 cm³. The linear pattern is V = 77h.

Volume = length × breadth × height = 7 × 11 × h = 77h.

200 pages will be left after 15 days. The linear pattern is P_n = 500 − 20n.

1. Pages read in 15 days20 × 15 = 300
2. Pages left500 − 300 = 200
3. After n daysP_n = 500 − 20n

(i) 5.25 feet. (iii) h = 1.75 + 0.5t, which is linear growth.

1. (i) Height after 7 monthsh = 1.75 + 0.5 × 7 = 1.75 + 3.5 = 5.25 feet
2. (ii) Table of values

   t (months)	0	1	2	3	4	5	6	7	8	9	10
   h (feet)	1.75	2.25	2.75	3.25	3.75	4.25	4.75	5.25	5.75	6.25	6.75

3. (iii) Expressionh = 1.75 + 0.5t. The height goes up by the same amount (0.5 feet) every month, so this is linear growth.

(i) ₹7600. (iii) v = 10000 − 800t, which is linear decay.

1. (i) Value after 3 yearsv = 10000 − 800 × 3 = 10000 − 2400 = ₹7600
2. (ii) Table of values

   t (years)	0	1	2	3	4	5	6	7	8
   v (₹)	10000	9200	8400	7600	6800	6000	5200	4400	3600

3. (iii) Expressionv = 10000 − 800t. The value falls by the same amount (₹800) every year, so this is linear decay.

(i) 1050 people. (iii) P = 750 + 50t, which is linear growth.

1. (i) Population after 6 yearsP = 750 + 50 × 6 = 750 + 300 = 1050
2. (ii) Table of values

   t (years)	0	1	2	3	4	5	6	7	8	9	10
   P	750	800	850	900	950	1000	1050	1100	1150	1200	1250

3. (iii) ExpressionP = 750 + 50t. The population goes up by the same number (50) every year, so this is linear growth.

(i) b(x) = 600 − 15x. (ii) The balance runs out after 40 days.

1. (i) Equationb(x) = 600 − 15x. The balance falls by the same amount (₹15) every day, so this is linear decay.
2. (ii) When does the balance run out?Put b(x) = 0.600 − 15x = 0 ⇒ 15x = 600 ⇒ x = 40
3. (iii) Table of values

   x (days)	1	2	3	4	5	6	7	8	9	10
   b(x) (₹)	585	570	555	540	525	510	495	480	465	450

a = 25 and b = 150.

1. Use the two bills10a + b = 400 …(1)14a + b = 500 …(2)
2. Subtract (1) from (2)4a = 100 ⇒ a = 25
3. Find b from (1)10(25) + b = 400 ⇒ b = 150

So the bill is y = 25x + 150: a fixed fee of ₹150 and ₹25 per module. Check: 25(14) + 150 = 500.

a = 60 and b = 200.

1. Use the two bills10a + b = 800 …(1)15a + b = 1100 …(2)
2. Subtract (1) from (2)5a = 300 ⇒ a = 60
3. Find b from (1)10(60) + b = 800 ⇒ b = 200

So y = 60x + 200. Check: 60(15) + 200 = 1100.

a = 59 and b = −1609. So °C = 59(°F − 32).

1. Use the two points (°F, °C) = (32, 0) and (212, 100)0 = 32a + b …(1)100 = 212a + b …(2)
2. Subtract (1) from (2)100 = 180a ⇒ a = 100180 = 59
3. Find b from (1)0 = 32 × 59 + b ⇒ b = −1609

So °C = 59°F − 1609, which can also be written as °C = 59(°F − 32).

All three lines pass through the origin (b = 0). The larger the value of a, the steeper the line.

• All lines have the form y = ax with b = 0, so they pass through the origin (0, 0).
• The value of a (the slope) decides the steepness: 4 > 2 > 1, so y = 4x is the steepest.

All three lines pass through the origin. They slope downward because a is negative, and the larger the size of a, the steeper the line.

• All lines have the form y = ax with b = 0, so they pass through the origin.
• A negative a means the line slopes downward from left to right.
• A larger size of a (6 > 3 > 1) means a steeper downward slope.

Both lines pass through the origin. y = 5x slopes upward and y = −5x slopes downward. They have the same steepness but opposite directions.

The two lines are mirror images of each other in the axes. Only the sign of a changes the direction; its size (5) gives the same steepness.

All three lines have the same slope a = 3, so they are parallel. A different b only shifts the line up or down.

• b = −1: the line is below the origin, cutting the y-axis at (0, −1).
• b = 0: the line passes through the origin.
• b = 1: the line is above the origin, cutting the y-axis at (0, 1).

y = −2x − 3 and y = −2x have the same slope (−2), so they are parallel. y = 2x + 3 has a positive slope, so it slopes the other way.

Conclusion for Q29 to Q33:

• a (the coefficient of x) controls the slope: steepness and direction.
• b (the constant term) controls the vertical shift: the y-intercept.

One such polynomial is x³ − 7x² + 2x + 1.

A polynomial of degree 3 has the form ax³ + bx² + cx + d. Taking b = −7 gives x³ − 7x² + 2x + 1.

Answers may differ: any degree 3 polynomial with −7 as the coefficient of x² is correct.

(i) 9. (ii) 4a³ − a² + 6.

1. (i) Put x = 15(1)² − 3(1) + 7 = 5 − 3 + 7 = 9
2. (ii) Put t = a4a³ − a² + 6

The number is −12.

1. Let the number be x52x + 23 = −712
2. Subtract 23 from both sides52x = −712 − 812 = −1512 = −54
3. Multiply both sides by 25x = −54 × 25 = −12

Check: 52 × (−12) + 23 = −54 + 23 = −712.

The numbers are 7 and 35.

1. Let the smaller number be xThe larger number is 5x. After adding 21: x + 21 and 5x + 21.
2. Make the equation5x + 21 = 2(x + 21) ⇒ 5x + 21 = 2x + 42 ⇒ 3x = 21 ⇒ x = 7

Numbers: 7 and 5 × 7 = 35. Check: 35 + 21 = 56 = 2 × 28 = 2(7 + 21).

After 6 months: ₹2300. After 2 years: ₹6800. The linear pattern is A = 800 + 250n.

1. PatternAfter n months the amount is A = 800 + 250n.
2. (i) 6 months800 + 250(6) = 800 + 1500 = ₹2300
3. (ii) 2 years = 24 months800 + 250(24) = 800 + 6000 = ₹6800

The two numbers are 85 and 58.

1. Let the tens digit be x and the units digit be yOriginal number = 10x + y. Interchanged number = 10y + x.
2. The sum is 143(10x + y) + (10y + x) = 143 ⇒ 11(x + y) = 143 ⇒ x + y = 13 …(1)
3. The digits differ by 3x − y = 3 …(2)
4. SolveAdd (1) and (2): 2x = 16 ⇒ x = 88 + y = 13 ⇒ y = 5

The original number is 85 and the interchanged number is 58. Check: 85 + 58 = 143 and 8 − 5 = 3.

If the tens digit is the smaller one (x − y = −3), we get the original number 58 and the interchanged number 85, which is the same pair.

Lines (ii) and (iv) are parallel (both have slope 2). The slopes and y-intercepts are in the table.

Write each equation in the form y = ax + b. Then a is the slope and b is the y-intercept.

Parallel lines: two lines are parallel if they have the same slope. Lines (ii) and (iv) both have slope 2, so they are parallel.

(i) 104 °F. (ii) 343 K.

1. (i) Put x = 313y = 95(313 − 273) + 32 = 95(40) + 32 = 72 + 32 = 104 °F
2. (ii) Put y = 158158 = 95(x − 273) + 32 ⇒ 126 = 95(x − 273)x − 273 = 126 × 59 = 70 ⇒ x = 343 K

The equation is w = 3d. When d = 2, the work done is w = 6 units.

1. The equationWork = force × distance, so w = 3d.
2. Table of points

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

3. Work done when d = 2w = 3 × 2 = 6 units

Verification: taking d on the x-axis and w on the y-axis, the point (2, 6) lies on the line, so the work done is 6 units.

(i) p(x) = 3x + 2. (ii) It cuts the y-axis at (0, 2) and the x-axis at (−23, 0).

1. (i) Let p(x) = ax + ba + b = 5 …(1)3a + b = 11 …(2)Subtract (1) from (2): 2a = 6 ⇒ a = 33 + b = 5 ⇒ b = 2
2. (ii) Cuts the y-axisPut x = 0: p(0) = 2, so the point is (0, 2).
3. Cuts the x-axisPut y = 0:3x + 2 = 0 ⇒ x = −23The point is (−23, 0).
4. (iii) GraphThe graph below passes through (1, 5) and (3, 11) and cuts the axes at (0, 2) and (−23, 0), which verifies the answers.

p(x) = 2x + 5 and q(x) = 4x − 1.

1. Use p(0) = 5a(0) + b = 5 ⇒ b = 5, so p(x) = ax + 5
2. Use p(x) + q(x) = 6x + 4(ax + 5) + (cx + d) = 6x + 4 ⇒ (a + c)x + (5 + d) = 6x + 4a + c = 6 …(1)   and   5 + d = 4 ⇒ d = −1, so q(x) = cx − 1
3. Use p(3) − q(3) = 0(3a + 5) − (3c − 1) = 0 ⇒ 3a − 3c + 6 = 0 ⇒ a − c = −2 …(2)
4. Solve (1) and (2)Add: 2a = 4 ⇒ a = 22 + c = 6 ⇒ c = 4

Check: p(x) + q(x) = 6x + 4 and p(x) − q(x) = −2x + 6, which is zero at x = 3.

(i) Stage 4: 21 matchsticks, stage 5: 26 matchsticks. (iii) The rule is M_n = 5n + 1. (iv) 76 matchsticks. (v) No, 200 matchsticks cannot form a stage.

1. How the pattern growsOne hexagon needs 6 matchsticks. Each new hexagon shares one side with the previous one, so it adds only 5 new matchsticks.Stage 1 = 6, stage 2 = 6 + 5 = 11, stage 3 = 11 + 5 = 16
2. (i) Next two stagesStage 4 = 16 + 5 = 21Stage 5 = 21 + 5 = 26
3. (ii) Table

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

4. (iii) RuleFirst term 6, common difference 5.6 + (n − 1) × 5 = 5n + 1So M_n = 5n + 1.
5. (iv) 15th stageM₁₅ = 5(15) + 1 = 76
6. (v) Can 200 matchsticks form a stage?5n + 1 = 200 ⇒ 5n = 199 ⇒ n = 39.8Since n is not a whole number, 200 matchsticks cannot form any stage.

p(x) = 2x − 1 and q(x) = 2x − 9. They meet the x-axis at (12, 0) and (92, 0).

1. Find p(x)Slope = 11 − 36 − 2 = 2, so p(x) = 2x + b.3 = 2(2) + b ⇒ b = −1, so p(x) = 2x − 1
2. Find q(x)q is parallel to p, so its slope is also 2: q(x) = 2x + d.−1 = 2(4) + d ⇒ d = −9, so q(x) = 2x − 9
3. Meeting the x-axis (put y = 0)p: 2x − 1 = 0 ⇒ x = 12q: 2x − 9 = 0 ⇒ x = 92The points are (12, 0) and (92, 0).

All have a positive slope, cut the y-axis above the origin, and pass through the fixed point (−1, 0).

1. Rewritef(x) = ax + a = a(x + 1).
2. SlopeThe slope is a and a > 0, so every line rises from left to right.
3. y-interceptf(0) = a > 0, so every line cuts the y-axis at (0, a), above the origin.
4. x-intercepta(x + 1) = 0 ⇒ x = −1 (since a ≠ 0)So every such line passes through (−1, 0).