Chapter 2 of Ganita Manjari, Class 9 Mathematics, introduces students to algebraic expressions in one variable and builds up to the idea of a linear polynomial. Starting from everyday situations — buying pens and pencils in sealed boxes, fencing a rectangular garden, chess club fees, auto-rickshaw fares and mobile phone depreciation — the chapter shows how real-life patterns can be written as algebraic expressions, and then narrows the focus to polynomials of degree 1. Students learn to find the degree of a polynomial, evaluate polynomials at given values, recognise linear growth and linear decay, form linear relationships of the form y = ax + b, and finally plot such relationships as straight lines on a coordinate plane, understanding the roles of slope and y-intercept. This page provides complete NCERT solutions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials, with detailed answers to every Exercise Set question from the Ganita Manjari textbook.
Last Updated: September 23, 2026
2.1 Introduction — Exercise Set 2.1 (Page 18-19)
Q1. Find the degrees of the following polynomials: (i) 2x² – 5x + 3 (ii) y³ + 2y – 1 (iii) – 9 (iv) 4z – 3 — Answer: The degree of a polynomial is the highest power of the variable that appears in…
Answer: The degree of a polynomial is the highest power of the variable that appears in it.
(i) 2x² – 5x + 3 has the highest power of x equal to 2, so its degree is 2.
(ii) y³ + 2y – 1 has the highest power of y equal to 3, so its degree is 3.
(iii) – 9 is a constant, which can be written as –9x⁰, so its degree is 0.
(iv) 4z – 3 has the highest power of z equal to 1, so its degree is 1 (it is a linear polynomial).
Q2. Write polynomials of degrees 1, 2 and 3 — Answer: Many correct answers are possible since we simply need the highest power of the…
Answer: Many correct answers are possible since we simply need the highest power of the variable to match the required degree. For example:
Degree 1 (linear polynomial): 3x + 2
Degree 2 (quadratic polynomial): x² + 2x + 1
Degree 3 (cubic polynomial): x³ – 2x² + x – 5
Q3. What are the coefficients of x² and x³ in the polynomial x⁴ – 3x³ + 6x² – 2x + 7? — Answer: In the polynomial x⁴ – 3x³ + 6x² – 2x + 7, the term in x³ is –3x³,…
Answer: In the polynomial x⁴ – 3x³ + 6x² – 2x + 7, the term in x³ is –3x³, so the coefficient of x³ is –3. The term in x² is 6x², so the coefficient of x² is 6.
Q4. What is the coefficient of z in the polynomial 4z³ + 5z² – 11? — Answer: The polynomial 4z³ + 5z² – 11 has no term containing z to the first power…
Answer: The polynomial 4z³ + 5z² – 11 has no term containing z to the first power (only z³, z² and a constant term appear). This means the term in z is effectively 0z, so the coefficient of z is 0.
Q5. What is the constant term of the polynomial 9x³ + 5x² – 8x – 10? — Answer: The constant term is the term that does not contain the variable x, which is the…
Answer: The constant term is the term that does not contain the variable x, which is the number that stands alone. In 9x³ + 5x² – 8x – 10, the constant term is –10.
2.2 Linear Polynomials — Exercise Set 2.2 (Page 21)
Q1. Find the value of the linear polynomial 5x – 3 if: (i) x = 0 (ii) x = –1 (iii) x = 2 — Answer: (i) At x = 0: 5(0) – 3 = 0 – 3 = –3 (ii) At x = –1: 5(–1) – 3 = –5…
Answer:
(i) At x = 0: 5(0) – 3 = 0 – 3 = –3
(ii) At x = –1: 5(–1) – 3 = –5 – 3 = –8
(iii) At x = 2: 5(2) – 3 = 10 – 3 = 7
Q2. Find the value of the quadratic polynomial 7s² – 4s + 6 if: (i) s = 0 (ii) s = –3 (iii) s = 4 — Answer: (i) At s = 0: 7(0)² – 4(0) + 6 = 0 – 0 + 6 = 6 (ii) At s = –3: 7(–3)²…
Answer:
(i) At s = 0: 7(0)² – 4(0) + 6 = 0 – 0 + 6 = 6
(ii) At s = –3: 7(–3)² – 4(–3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81
(iii) At s = 4: 7(4)² – 4(4) + 6 = 7(16) – 16 + 6 = 112 – 16 + 6 = 102
Q3. The present age of Salil's mother is three times Salil's present age. After 5 years, their ages will add up to 70 years. Find their present ages — Answer: Let Salil's present age be x years. Then his mother's present age is 3x years.…
Answer: Let Salil’s present age be x years. Then his mother’s present age is 3x years.
After 5 years: Salil’s age = (x + 5) years, mother’s age = (3x + 5) years.
Given: (x + 5) + (3x + 5) = 70
4x + 10 = 70
4x = 60
x = 15
So, Salil’s present age is 15 years and his mother’s present age is 3 × 15 = 45 years.
Check: after 5 years, ages are 20 and 50, which add up to 70. ✓
Q4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers — Answer: Let the two integers be 2k and 5k (since their ratio is 2:5). Given: 5k – 2k =…
Answer: Let the two integers be 2k and 5k (since their ratio is 2:5).
Given: 5k – 2k = 63
3k = 63
k = 21
So, the two integers are 2(21) = 42 and 5(21) = 105.
Check: 105 – 42 = 63 and 42:105 = 2:5. ✓
Q5. Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total ₹88, how many coins does she have of each type? — Answer: Let the number of five-rupee coins be x. Then the number of two-rupee coins is…
Answer: Let the number of five-rupee coins be x. Then the number of two-rupee coins is 3x.
Total value: 5(x) + 2(3x) = 88
5x + 6x = 88
11x = 88
x = 8
So, Ruby has 8 five-rupee coins and 3 × 8 = 24 two-rupee coins (32 coins in total).
Check: 5(8) + 2(24) = 40 + 48 = 88. ✓
Q6. A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces? — Answer: Let the shorter piece be x feet. Then the longer piece is 4x feet. x + 4x = 300…
Answer: Let the shorter piece be x feet. Then the longer piece is 4x feet.
x + 4x = 300
5x = 300
x = 60
So, the shorter piece is 60 feet and the longer piece is 4 × 60 = 240 feet.
Check: 60 + 240 = 300. ✓
Q7. If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle? — Answer: Let the width be w cm. Then the length is (2w + 3) cm. Perimeter = 2(length +…
Answer: Let the width be w cm. Then the length is (2w + 3) cm.
Perimeter = 2(length + width) = 24
length + width = 12
(2w + 3) + w = 12
3w + 3 = 12
3w = 9
w = 3
So, the width is 3 cm and the length is 2(3) + 3 = 9 cm.
Check: Perimeter = 2(9 + 3) = 2(12) = 24 cm. ✓
2.3 Exploring Linear Patterns — Exercise Set 2.3 (Page 24-25)
Q1. A student has ₹500 in her savings bank account. She gets ₹150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the nth month — Answer: Starting amount = ₹500, and ₹150 is added every month. Amount at end of…
Answer: Starting amount = ₹500, and ₹150 is added every month.
Amount at end of month 1 = 500 + 150 = ₹650
Amount at end of month 2 = 500 + 2(150) = ₹800
Amount at end of month 3 = 500 + 3(150) = ₹950
Amount at end of month 4 = 500 + 4(150) = ₹1100
In general, the amount she will have at the end of the nth month is given by the linear expression A(n) = 500 + 150n.
Q2. A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3, … hours? Find a linear expression to represent the number of members at the end of the nth hour — Answer: Starting members = 120, decreasing by 9 every hour. After 1 hour: 120 – 9 =…
Answer: Starting members = 120, decreasing by 9 every hour.
After 1 hour: 120 – 9 = 111 members
After 2 hours: 120 – 2(9) = 102 members
After 3 hours: 120 – 3(9) = 93 members
In general, the number of members remaining at the end of the nth hour is given by the linear expression M(n) = 120 – 9n.
Q3. Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area of the rectangle — Answer: Area of rectangle = length × breadth = 13 × breadth. (i) Breadth 12 cm: Area =…
Answer: Area of rectangle = length × breadth = 13 × breadth.
(i) Breadth 12 cm: Area = 13 × 12 = 156 cm²
(ii) Breadth 10 cm: Area = 13 × 10 = 130 cm²
(iii) Breadth 8 cm: Area = 13 × 8 = 104 cm²
If b represents the breadth, the linear pattern representing the area is Area = 13b, which is a linear expression in b since the length (13) is fixed.
Q4. Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern representing the volume of the rectangular box — Answer: Volume of the box = length × breadth × height = 7 × 11 × height = 77 ×…
Answer: Volume of the box = length × breadth × height = 7 × 11 × height = 77 × height.
(i) Height 5 cm: Volume = 77 × 5 = 385 cm³
(ii) Height 9 cm: Volume = 77 × 9 = 693 cm³
(iii) Height 13 cm: Volume = 77 × 13 = 1001 cm³
If h represents the height, the linear pattern representing the volume is Volume = 77h.
Q5. Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern — Answer: Pages read in 15 days = 20 × 15 = 300 pages. Pages left = 500 – 300 = 200…
Answer: Pages read in 15 days = 20 × 15 = 300 pages.
Pages left = 500 – 300 = 200 pages.
If n represents the number of days, the linear pattern for the pages left is P(n) = 500 – 20n.
2.4 Linear Growth and Linear Decay — Exercise Set 2.4 (Page 26-27)
Q1. Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month. (i) Find the height after 7 months. (ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month. (iii) Find an expression that relates h and t, and explain why it represents linear growth — Answer: (i) Height after 7 months = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25 feet (ii) Table of…
Answer:
(i) Height after 7 months = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25 feet
(ii) Table of values (h = 1.75 + 0.5t):
t = 0: h = 1.75 ft; t = 1: h = 2.25 ft; t = 2: h = 2.75 ft; t = 3: h = 3.25 ft; t = 4: h = 3.75 ft; t = 5: h = 4.25 ft; t = 6: h = 4.75 ft; t = 7: h = 5.25 ft; t = 8: h = 5.75 ft; t = 9: h = 6.25 ft; t = 10: h = 6.75 ft
(iii) The expression relating h and t is h = 1.75 + 0.5t. This represents linear growth because for every increase of 1 month in t, the height h increases by a fixed amount (0.5 feet), which is the defining feature of linear growth.
Q2. A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year. (i) Find the value of the phone after 3 years. (ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time. (iii) Find an expression that relates v and t, and explain why it represents linear decay — Answer: (i) Value after 3 years = 10000 – 800(3) = 10000 – 2400 = ₹7600 (ii) Table…
Answer:
(i) Value after 3 years = 10000 – 800(3) = 10000 – 2400 = ₹7600
(ii) Table of values (v = 10000 – 800t):
t = 0: v = ₹10000; t = 1: v = ₹9200; t = 2: v = ₹8400; t = 3: v = ₹7600; t = 4: v = ₹6800; t = 5: v = ₹6000; t = 6: v = ₹5200; t = 7: v = ₹4400; t = 8: v = ₹3600
(iii) The expression relating v and t is v = 10000 – 800t. This represents linear decay because for every increase of 1 year in t, the value v decreases by a fixed amount (₹800), which is the defining feature of linear decay.
Q3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village. (i) Find the population of the village after 6 years. (ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year. (iii) Find an expression that relates P and t, and explain why it represents linear growth — Answer: (i) Population after 6 years = 750 + 50(6) = 750 + 300 = 1050 (ii) Table of…
Answer:
(i) Population after 6 years = 750 + 50(6) = 750 + 300 = 1050
(ii) Table of values (P = 750 + 50t):
t = 0: P = 750; t = 1: P = 800; t = 2: P = 850; t = 3: P = 900; t = 4: P = 950; t = 5: P = 1000; t = 6: P = 1050; t = 7: P = 1100; t = 8: P = 1150; t = 9: P = 1200; t = 10: P = 1250
(iii) The expression relating P and t is P = 750 + 50t. This represents linear growth because the population increases by a fixed amount (50 people) for every additional year.
Q4. A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge. (i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay. (ii) After how many days will the balance run out? (iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time — Answer: (i) The remaining balance is modelled by b(x) = 600 – 15x. This represents…
Answer:
(i) The remaining balance is modelled by b(x) = 600 – 15x. This represents linear decay because the balance decreases by a fixed amount (₹15) for every additional day.
(ii) The balance runs out when b(x) = 0:
600 – 15x = 0
15x = 600
x = 40
So the balance will run out after 40 days.
(iii) Table of values (b(x) = 600 – 15x):
x = 1: b = ₹585; x = 2: b = ₹570; x = 3: b = ₹555; x = 4: b = ₹540; x = 5: b = ₹525; x = 6: b = ₹510; x = 7: b = ₹495; x = 8: b = ₹480; x = 9: b = ₹465; x = 10: b = ₹450
2.5 Linear Relationships — Exercise Set 2.5 (Page 28)
Q1. A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b — Answer: Substituting the two given conditions into y = ax + b: 400 = 10a + b … (1) 500…
Answer: Substituting the two given conditions into y = ax + b:
400 = 10a + b … (1)
500 = 14a + b … (2)
Subtracting (1) from (2): 100 = 4a, so a = 25.
Substituting back in (1): 400 = 10(25) + b = 250 + b, so b = 150.
The relationship is y = 25x + 150 (fixed monthly fee ₹150, plus ₹25 per module).
Q2. A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b — Answer: Substituting the two given conditions into y = ax + b: 800 = 10a + b … (1)…
Answer: Substituting the two given conditions into y = ax + b:
800 = 10a + b … (1)
1100 = 15a + b … (2)
Subtracting (1) from (2): 300 = 5a, so a = 60.
Substituting back in (1): 800 = 10(60) + b = 600 + b, so b = 200.
The relationship is y = 60x + 200 (fixed monthly fee ₹200, plus ₹60 per hour).
Q3. Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a °F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit — Answer: Substituting the two given conditions into °C = a °F + b: 0 = 32a + b … (1)…
Answer: Substituting the two given conditions into °C = a °F + b:
0 = 32a + b … (1)
100 = 212a + b … (2)
Subtracting (1) from (2): 100 = 180a, so a = 100/180 = 5/9.
Substituting back in (1): 0 = 32(5/9) + b = 160/9 + b, so b = –160/9.
The relationship is °C = (5/9)°F – 160/9, which is the same as the familiar formula °C = (5/9)(°F – 32).
2.6 Visualising Linear Relationships — Exercise Set 2.6 (Page 37)
Q1. Draw the graphs of the following sets of lines. In each case, reflect on the role of 'a' and 'b'. (i) y = 4x, y = 2x, y = x (ii) y = –6x, y = –3x, y = –x (iii) y = 5x, y = –5x (iv) y = 3x – 1, y = 3x, y = 3x + 1 (v) y = –2x – 3, y = –2x, y = 2x + 3 — Answer: Each equation is of the form y = ax + b, where a is the slope and b is the…
Answer: Each equation is of the form y = ax + b, where a is the slope and b is the y-intercept (the point where the line cuts the y-axis, at (0, b)). Two convenient points are enough to plot each line.
(i) y = 4x, y = 2x, y = x: All three lines pass through the origin (0, 0), since b = 0 in each case. Using a second point for each: y = 4x passes through (1, 4); y = 2x passes through (1, 2); y = x passes through (1, 1). As the value of a increases (4 > 2 > 1), the line becomes steeper — a larger positive slope means a steeper upward line.
(ii) y = –6x, y = –3x, y = –x: All three lines again pass through the origin. Second points: y = –6x passes through (1, –6); y = –3x passes through (1, –3); y = –x passes through (1, –1). As the magnitude of the negative slope increases (–6 is “steeper” than –3, which is steeper than –1), the line falls more sharply from left to right.
(iii) y = 5x, y = –5x: Both pass through the origin. y = 5x passes through (1, 5) and rises steeply left to right; y = –5x passes through (1, –5) and falls steeply left to right. The two lines are mirror images of each other in the x-axis (and in the y-axis), since their slopes are equal in magnitude but opposite in sign.
(iv) y = 3x – 1, y = 3x, y = 3x + 1: All three lines have the same slope a = 3, so they are all equally steep and are parallel to one another. Their y-intercepts differ: –1, 0 and 1 respectively, so the lines are shifted up or down but never meet.
(v) y = –2x – 3, y = –2x, y = 2x + 3: The first two lines, y = –2x – 3 and y = –2x, both have slope –2, so they are parallel to each other, with y-intercepts –3 and 0. The third line, y = 2x + 3, has a different slope (2, positive) and y-intercept 3, so it is not parallel to the other two — it rises instead of falling, and crosses both of them.
General conclusion (as given in the chapter): In y = ax + b, a is the slope (it controls the steepness and direction of the line) and b is the y-intercept (it controls where the line crosses the y-axis). Changing a while keeping b fixed rotates the line about the point (0, b); changing b while keeping a fixed shifts the line up or down without changing its steepness, producing a family of parallel lines.
End-of-Chapter Exercises (Page 37-39)
Q1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is –7 — Answer: Many correct answers are possible; we only need a cubic term (so the polynomial…
Answer: Many correct answers are possible; we only need a cubic term (so the polynomial has degree 3) together with an x² term whose coefficient is –7. For example: p(x) = 2x³ – 7x² + 4x – 1.
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) At x = 1: 5(1)² – 3(1) + 7 = 5 – 3 + 7 = 9 (ii) At t = a: 4t³ – t²…
Answer:
(i) At x = 1: 5(1)² – 3(1) + 7 = 5 – 3 + 7 = 9
(ii) At t = a: 4t³ – t² + 6 becomes 4a³ – a² + 6 (we simply replace every t with a).
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. According to the condition: (5/2)x + 2/3 = –7/12 (5/2)x =…
Answer: Let the number be x. According to the condition:
(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) = –10/20 = –1/2
Check: (5/2)(–1/2) + 2/3 = –5/4 + 2/3 = –15/12 + 8/12 = –7/12. ✓
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, so the larger number is 5x. After adding 21 to…
Answer: Let the smaller number be x, so the larger number is 5x.
After adding 21 to both: new numbers are (x + 21) and (5x + 21).
Since x is positive, 5x + 21 > x + 21, so the larger new number must be twice the smaller new number:
5x + 21 = 2(x + 21)
5x + 21 = 2x + 42
3x = 21
x = 7
So the two original numbers are 7 and 35 (since 5 × 7 = 35).
Check: New numbers are 7 + 21 = 28 and 35 + 21 = 56, and 56 = 2 × 28. ✓
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: Linear pattern: A(n) = 800 + 250n, where n is the number of months. (i) After 6…
Answer: Linear pattern: A(n) = 800 + 250n, where n is the number of months.
(i) After 6 months: A(6) = 800 + 250(6) = 800 + 1500 = ₹2300
(ii) After 2 years (24 months): A(24) = 800 + 250(24) = 800 + 6000 = ₹6800
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 x and the units digit be y, so the original number is (10x…
Answer: Let the tens digit be x and the units digit be y, so the original number is (10x + y). The number with digits interchanged is (10y + x).
Given: (10x + y) + (10y + x) = 143
11x + 11y = 143
11(x + y) = 143
x + y = 13
Also, the digits differ by 3, so x – y = 3 (taking x as the larger digit).
Adding the two equations: 2x = 16, so x = 8, and y = 13 – 8 = 5.
So the original two-digit number is 85, and with digits interchanged it is 58.
Check: 85 + 58 = 143, and 8 – 5 = 3. ✓ (Either 85 or 58 may be taken as the “original” number — both satisfy the condition since the equation is symmetric.)
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, rewrite each equation in the form y = ax + b: (i) y = –3x + 4 → slope…
Answer: First, rewrite each equation in the form y = ax + b:
(i) y = –3x + 4 → slope = –3, y-intercept = 4, cuts the y-axis at (0, 4)
(ii) 2y = 4x + 7 → y = 2x + 3.5 → slope = 2, y-intercept = 3.5, cuts the y-axis at (0, 3.5)
(iii) 5y = 6x – 10 → y = 1.2x – 2 → slope = 1.2, y-intercept = –2, cuts the y-axis at (0, –2)
(iv) 3y = 6x – 11 → y = 2x – 11/3 → slope = 2, y-intercept = –11/3 (≈ –3.67), cuts the y-axis at (0, –11/3)
Comparing the slopes, lines (ii) and (iv) both have slope 2, so 2y = 4x + 7 and 3y = 6x – 11 are parallel to each other. The other two lines have different slopes (–3 and 1.2) and are not parallel to any other line in the set. To plot each line, use the y-intercept as one point and any second point obtained by substituting a convenient value of x.
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) At x = 313 K: y = (9/5)(313 – 273) + 32 = (9/5)(40) + 32 = 72 + 32 = 104…
Answer:
(i) At x = 313 K:
y = (9/5)(313 – 273) + 32 = (9/5)(40) + 32 = 72 + 32 = 104 °F
(ii) At y = 158 °F:
158 = (9/5)(x – 273) + 32
126 = (9/5)(x – 273)
x – 273 = 126 × 5/9 = 70
x = 273 + 70 = 343 K
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: Since Work = Force × Distance, and the constant force is 3 units, the linear…
Answer: Since Work = Force × Distance, and the constant force is 3 units, the linear equation is w = 3d. This is a straight line through the origin with slope 3; it passes through points such as (0, 0), (1, 3) and (2, 6).
Work done when the distance travelled is 2 units: w = 3(2) = 6 units. Plotting d = 2 on the horizontal axis and reading off the graph confirms the corresponding point on the line is (2, 6), verifying the answer.
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) Let p(x) = ax + b. Substituting the two given points: 5 = a(1) + b → a + b…
Answer:
(i) Let p(x) = ax + b. Substituting the two given points:
5 = a(1) + b → a + b = 5 … (1)
11 = a(3) + b → 3a + b = 11 … (2)
Subtracting (1) from (2): 2a = 6, so a = 3. Then b = 5 – 3 = 2.
So, p(x) = 3x + 2.
(ii) The graph cuts the y-axis where x = 0: p(0) = 2, giving the point (0, 2).
The graph cuts the x-axis where p(x) = 0: 3x + 2 = 0 → x = –2/3, giving the point (–2/3, 0).
(iii) Plotting the points (1, 5) and (3, 11) and drawing a straight line through them, and extending the line, confirms that it crosses the y-axis at (0, 2) and the x-axis at (–2/3, 0).
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: From (i): p(0) = a(0) + b = b = 5, so b = 5. From (iii): p(x) + q(x) = (ax + b)…
Answer:
From (i): p(0) = a(0) + b = b = 5, so b = 5.
From (iii): p(x) + q(x) = (ax + b) + (cx + d) = (a + c)x + (b + d) = 6x + 4 for all x, so a + c = 6 and b + d = 4. Since b = 5, we get d = 4 – 5 = –1.
From (ii): p(x) – q(x) = (ax + b) – (cx + d) = (a – c)x + (b – d). This expression equals 0 when x = 3:
(a – c)(3) + (b – d) = 0
3(a – c) + (5 – (–1)) = 0
3(a – c) + 6 = 0
a – c = –2
Now solve a + c = 6 and a – c = –2 together: adding gives 2a = 4, so a = 2, and c = 6 – 2 = 4.
So, p(x) = 2x + 5 and q(x) = 4x – 1.
Check: p(0) = 5 ✓; p(x) + q(x) = 6x + 4 ✓; p(x) – q(x) = –2x + 6, which is 0 at x = 3 ✓.
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 table for Stage Number 1 to n. (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: A single hexagon needs 6 matchsticks. Each new hexagon added shares one side…
Answer: A single hexagon needs 6 matchsticks. Each new hexagon added shares one side with the previous hexagon, so it only needs 6 – 1 = 5 new matchsticks.
Stage 1: 6 matchsticks
Stage 2: 6 + 5 = 11 matchsticks
Stage 3: 11 + 5 = 16 matchsticks
(i) Stage 4: 16 + 5 = 21 matchsticks; Stage 5: 21 + 5 = 26 matchsticks
(ii) Table: Stage 1 → 6; Stage 2 → 11; Stage 3 → 16; Stage 4 → 21; Stage 5 → 26; Stage n → 5n + 1
(iii) The rule is: number of matchsticks at Stage n = 5n + 1 (starting stick count of 6, plus 5 more for every additional hexagon after the first).
(iv) At Stage 15: 5(15) + 1 = 75 + 1 = 76 matchsticks
(v) We check whether 5n + 1 = 200 has a whole-number solution: 5n = 199, so n = 39.8, which is not a whole number. So, 200 matchsticks cannot form any stage of this pattern, since the number of matchsticks used is always of the form 5n + 1 (one more than a multiple of 5), and 200 is a multiple of 5, not one more than a multiple of 5.
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: From (i): substituting the two points into p(x) = ax + b: 3 = 2a + b … (1) 11…
Answer:
From (i): substituting the two points into p(x) = ax + b:
3 = 2a + b … (1)
11 = 6a + b … (2)
Subtracting (1) from (2): 8 = 4a, so a = 2. Then b = 3 – 2(2) = –1.
So, p(x) = 2x – 1.
Since q(x) is parallel to p(x), it has the same slope: c = a = 2.
From (ii): q(4) = –1, so 2(4) + d = –1 → 8 + d = –1 → d = –9.
So, q(x) = 2x – 9.
Points where the lines meet the x-axis (set each polynomial to 0):
p(x) = 0: 2x – 1 = 0 → x = 1/2, so p(x) meets the x-axis at (1/2, 0).
q(x) = 0: 2x – 9 = 0 → x = 9/2, so q(x) meets the x-axis at (9/2, 0).
Q14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common? — Answer: We can factor f(x) = ax + a as f(x) = a(x + 1). No matter what positive value a…
Answer: We can factor f(x) = ax + a as f(x) = a(x + 1). No matter what positive value a takes, when x = –1: f(–1) = a(–1) + a = –a + a = 0.
So, every linear function of the form f(x) = ax + a (a > 0) passes through the same point, (–1, 0), on the x-axis, even though each has a different (positive) slope a and a different y-intercept (also equal to a). In other words, this whole family of lines pivots around the fixed point (–1, 0).
Practice more: Extra Questions for Class 9 Mathematics Chapter 2
Quick revision: Revision Notes for Class 9 Mathematics Chapter 2
- Chapter 1: Orienting Yourself: The Use of Coordinates – Free PDF Download
- Chapter 3: The World of Numbers – Free PDF Download
- Chapter 4: Exploring Algebraic Identities – Free PDF Download
- Chapter 5: I'm Up and Down, and Round and Round – Free PDF Download
- Chapter 6: Measuring Space: Perimeter and Area – Free PDF Download
- Chapter 7: The Mathematics of Maybe: Introduction to Probability – Free PDF Download
- Chapter 8: Predicting What Comes Next?: Exploring Sequences and Progressions – Free PDF Download
Frequently Asked Questions
What is a polynomial, and what makes an expression NOT a polynomial?
A polynomial is an algebraic expression with only non-negative integer powers of the variable (e.g. 3x^2+2x+1); an expression is NOT a polynomial if the variable appears with a negative or fractional exponent, or under a root, or in a denominator.
What is the difference between a linear polynomial and its zero?
A linear polynomial has the form ax+b (degree 1); its zero is the specific value of the variable that makes the polynomial equal to zero, found by solving ax+b=0, which gives exactly one zero for any linear polynomial.
Chapter Quiz — Test Your Understanding
Class 9 Mathematics Chapter 2: Introduction to Linear Polynomials – Notes and Extra Questions
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