Linear polynomials: Questions and Answers

73 linear polynomials questions solved step by step. Open a question for the full working, the marks for each step and exam practice.

• Raju went to a shop where there were sealed boxes of different colours on sale. The shop owner told him that the red boxes have 4 pens each and the blue boxes have 5 pencils each. Now, if Raju bought x red boxes and y blue boxes, how can he quickly figure out the total quantity of pens and pencils? Also, if he got 3 extra pens free, how many pens and pencils did he get altogether?Answer: Pens and pencils in the boxes = 4x + 5y. With 3 free pens, the total is 4x + 5y + 3.
• A rectangular garden of length l metres and width w metres has to be fenced and decorated. A wire fence is to be laid along the length costing ₹100 per metre and a wooden fence is to be built along the width costing ₹80 per metre. Special seeds have to be sown throughout the garden which will cost ₹50 per square metre. What will be the total cost incurred?Answer: Total cost = ₹(200l + 160w + 50lw).
• Thus, 200l + 160w + 50lw is the algebraic expression for the total cost.
  1. Can you identify the terms, variables and coefficients of this algebraic expression?
  2. How is it different from the algebraic expression 4x + 5y + 3 for Raju’s pens and pencils?Answer: 1. Terms: 200l, 160w, 50lw; variables: l and w; coefficients: 200 (of l), 160 (of w), 50 (of lw). 2. It has a term 50lw in which two variables are multiplied, and it has no constant term; 4x + 5y + 3 has no product term and has a constant 3.
• A wire of length 20 cm is bent in different ways to form rectangles. For example, we can have a rectangle with length 7 cm and width 3 cm. We can also have one of length 5.5 cm and width 4.5 cm. (Think of a few more ways of forming such rectangles.) Can you write an expression for the area of such rectangles?Answer: If the length is x cm, the width is (10 − x) cm, so the area is x(10 − x) = 10x − x² cm². Other rectangles: 6 cm × 4 cm, 8 cm × 2 cm, 9 cm × 1 cm, 5 cm × 5 cm.
• The expression for the area of these rectangles is x(10 − x) or 10x − x².
  1. Can you identify the terms, variables and coefficients of this algebraic expression?
  2. Can you point out any similarity or difference between the algebraic expressions obtained for Raju’s pens and pencils (4x + 5y + 3) and for these rectangles?Answer: 1. Terms: 10x and −x²; variable: x; coefficients: 10 (of x) and −1 (of x²). 2. Both are algebraic expressions with terms and coefficients, but 4x + 5y + 3 uses two variables, each only to power 1, and has a constant; 10x − x² uses one variable, has an x² term and no constant.
• Find the degrees of the following polynomials:Answer: (i) 2 (ii) 3 (iii) 0 (iv) 1
• Write polynomials of degrees 1, 2 and 3.Answer: For example: degree 1: 3x + 5; degree 2: x² − 4x + 1; degree 3: 2x³ + x − 7. (Many answers are correct.)
• What are the coefficients of x² and x³ in the polynomial x⁴ − 3x³ + 6x² − 2x + 7?Answer: Coefficient of x² = 6; coefficient of x³ = −3.
• What is the coefficient of z in the polynomial 4z³ + 5z² − 11?Answer: The coefficient of z is 0, because there is no z term.
• What is the constant term of the polynomial 9x³ + 5x² − 8x −10?Answer: The constant term is −10.
• The perimeter of a square of side x is 4x, which is a linear polynomial in the variable x.Answer: Perimeter = x + x + x + x = 4x. The highest power of x is 1, so 4x is a linear polynomial.
• Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?Answer: Perimeters: 4 cm, 6 cm, 8 cm, 10 cm, 12 cm. Each time the side increases by 0.5 cm, the perimeter increases by 2 cm.
• A chess club charges a joining fee of ₹200 plus ₹50 for every match played. The following table shows the amount a player will have to pay as the number of matches varies.Answer: For m matches the total cost is ₹(200 + 50m), a linear polynomial in m. The amount rises by a constant ₹50 per extra match.
• A chess club charges a joining fee of ₹200 plus ₹50 for every match played. If a player paid ₹750, how many matches did he play?Answer: 200 + 50m = 750 gives m = 11. He played 11 matches.
• The sum of two numbers is 64. One of the numbers is 10 more than the other. What are the two numbers?Answer: Let the smaller number be x: x + (x + 10) = 64 gives x = 27. The numbers are 27 and 37.
• We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider the example where the wire is bent to form a rectangle. Here, the area of the rectangle, 10x − x², is a function of x. Can you interpret this as an input-output process? What value does the expression take when x = 6 cm?Answer: Yes: the input is the length x, the machine works out 10x − x², and the output is the area. For x = 6 cm the area is 10(6) − 6² = 24 cm².
• Find the value of the linear polynomial 5x − 3 if:Answer: (i) −3 (ii) −8 (iii) 7
• Find the value of the quadratic polynomial 7s² − 4s + 6 if:Answer: (i) 6 (ii) 81 (iii) 102
• 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: Salil is 15 years old and his mother is 45 years old.
• The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.Answer: The integers are 42 and 105.
• 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: Ruby has 24 two-rupee coins and 8 five-rupee coins.
• 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: Shorter piece 60 feet, longer piece 240 feet.
• 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: Width = 3 cm, length = 9 cm.
• Observe the following growing pattern of square tiles.
  Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage 7 of the pattern.Answer: Stages 5, 6 and 7 have 9, 11 and 13 squares. The sequence up to Stage 7 is 1, 3, 5, 7, 9, 11, 13.
• This leads us to conclude that the number of squares at Stage n is given by 2n − 1.
  Using the expression 2n − 1, can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles?Answer: 15th stage: 29 tiles; 26th stage: 51 tiles. 21 tiles: Stage 11; 47 tiles: Stage 24.
• Bela has ₹100 for pocket money. She spends ₹5 every day. After how many days will she be left with ₹40?Answer: Amount left after n days = ₹(100 − 5n). 100 − 5n = 40 gives n = 12: she has ₹40 left after 12 days.
• Bela has ₹100 for pocket money. She spends ₹5 every day. Observe that the amount left on the n^th day will be ₹(100 − 5n).
  What amount will be left on the 15th day? How many days will it take for the entire amount to be spent?Answer: On the 15th day ₹(100 − 75) = ₹25 is left. The money runs out when 100 − 5n = 0, that is, after 20 days.
• An auto-rikshaw fare starts at ₹25 and remains the same for the initial 2 km. Then it increases by ₹15 per km. What will be the fare for a travel of 10 km?Answer: Fare = 25 + 15 × (10 − 2) = 25 + 120 = ₹145. In general, for n km (n ≥ 2) the fare is ₹(15n − 5).
• Observe that the total fare for a travel of n km will be ₹25 + 15 × (n − 2) = 15n − 5, when n ≥ 2.
  For how many km will the fare be ₹130?Answer: 15n − 5 = 130 gives n = 9. The fare is ₹130 for 9 km.
• 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 n^th month.Answer: End of month 1: ₹650; then ₹800, ₹950, ₹1100, … (₹150 more each month). Amount at the end of the nth month = ₹(500 + 150n).
• 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 n^th hour.Answer: After 1, 2, 3, 4, … hours: 111, 102, 93, 84, … members. At the end of the nth hour: 120 − 9n members.
• 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: (i) 156 cm² (ii) 130 cm² (iii) 104 cm². Linear pattern: Area = 13b cm² for breadth b cm; every 2 cm less breadth gives 26 cm² less area.
• 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: (i) 385 cm³ (ii) 693 cm³ (iii) 1001 cm³. Linear pattern: V = 77h cm³; each 4 cm more height adds 308 cm³.
• 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 left after n days = 500 − 20n. After 15 days: 500 − 300 = 200 pages.
• The cost of a journey is given by the linear function C(d) = 100 + 60d, where C indicates total cost in rupees and d the distance travelled in km. Let us make a table of values for d varying from 0 to 10 km and show how the cost increases for every km.Answer: Costs for d = 0 to 10 km: ₹100, 160, 220, 280, 340, 400, 460, 520, 580, 640, 700. The cost rises by a fixed ₹60 for every km: this is linear growth.
• The cost of a journey is given by the linear function C(d) = 100 + 60d, where C indicates total cost in rupees and d the distance travelled in km.
  What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be ₹700?Answer: C(15) = 100 + 900 = ₹1000. 100 + 60d = 700 gives d = 10 km.
• The height of water in a cylindrical tank is 3 m at the start of summer. The height h m at the end of t months is given by the linear function h(t) = 3 − 0.5t.Answer: h(0) = 3, h(1) = 2.5, h(2) = 2, h(3) = 1.5, h(4) = 1 m. The height falls by a fixed 0.5 m every month: linear decay.
• The height of water in a cylindrical tank is 3 m at the start of summer. The height h m at the end of t months is given by the linear function h(t) = 3 − 0.5t.
  What will be the height of the water at the end of 5 months?Answer: h(5) = 3 − 0.5 × 5 = 0.5 m.
• Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.Answer: (i) 5.25 feet (ii) 1.75, 2.25, 2.75, …, 6.75 feet for t = 0 to 10 (up 0.5 ft a month) (iii) h = 1.75 + 0.5t; linear growth because h increases by the same 0.5 ft every month.
• A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.Answer: (i) ₹7,600 (ii) ₹10,000, 9,200, 8,400, …, 3,600 for t = 0 to 8 (iii) v = 10000 − 800t; linear decay because the value falls by the same ₹800 every year.
• The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.Answer: (i) 1050 (ii) 750, 800, 850, …, 1250 for t = 0 to 10 (iii) P = 750 + 50t; linear growth because the population rises by the same 50 every year.
• A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.Answer: (i) b(x) = 600 − 15x, linear decay (falls by ₹15 each day) (ii) 40 days (iii) ₹585, 570, 555, …, 450 for x = 1 to 10.
• A telecom company charges a fixed monthly fee and an additional cost per GB of the internet data used. A student observes that when she used 10 GB, her bill was ₹350. When she used 20 GB, her bill was ₹550. If the monthly bill y depends on the amount of data used, x (in GB), according to the relation y = ax + b, find the values of a and b.Answer: 350 = 10a + b and 550 = 20a + b give a = 20, b = 150, so y = 20x + 150.
• Thus, y = 20x + 150 represents the linear relationship between y, the bill amount in Rupees, and x, the number of GB of the internet data used.
  Can you guess what the numbers 20 and 150 in the equation y = 20x + 150 represent?Answer: 20 is the cost per GB of data (₹20 for each GB). 150 is the fixed monthly fee (₹150), paid even if no data is used.
• 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: a = 25, b = 150, so y = 25x + 150 (₹25 per module, ₹150 fixed fee).
• 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: a = 60, b = 200, so y = 60x + 200 (₹60 per hour, ₹200 fixed fee).
• 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.
  (Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)Answer: a = 59 and b = −1609, so °C = 59 °F − 1609, that is, °C = 59(°F − 32).
• Identify other points on the line by completing the following table.
  x: 1, 2, 5, 7, 9, 12, 20
  y: 3, __, __, 15, __, __, __Answer: For y = 2x + 1: x = 1, 2, 5, 7, 9, 12, 20 give y = 3, 5, 11, 15, 19, 25, 41. So (2, 5), (5, 11), (9, 19), (12, 25) and (20, 41) are also on the line.
• Let us plot the points (−1, −3), (0, 0), (1, 3), (3, 9), (4, 12) in the coordinate plane on a graph paper as shown in the figure. Join the points (−1, −3) and (4, 12) using a ruler. Doing so, observe that all five points lie on a straight line. Can you guess the equation of this line by looking at the relationship between the x and y coordinates of each point?Answer: In every point the y-coordinate is three times the x-coordinate, so the line is y = 3x.
• Let us plot the points (− 3, 6), (− 2, 4), (0, 0), (1, − 2), (2, − 4), (3, − 6) in the coordinate plane on a graph paper as shown in the figure. Join the points (− 3, 6) and (3, − 6) using a ruler. Doing so, observe that all five points lie on a straight line. Can you guess the equation of this line by looking at the relationship between the x and y coordinates of each point?Answer: In every point the y-coordinate is −2 times the x-coordinate, so the line is y = −2x.
• Draw the graphs of y = 12x, y = x, y = 2x by selecting suitable points on these lines.
  (Hint: In order to graph y = 12x, we could take the points (0, 0) and (4, 2). Can you verify that these lie on the line?)Answer: Use (0, 0) and (4, 2) for y = ½x, (0, 0) and (2, 2) for y = x, (0, 0) and (1, 2) for y = 2x. All three lines pass through the origin; y = 2x is the steepest and y = ½x the least steep. (0, 0) and (4, 2) do lie on y = ½x: 0 = ½ × 0 and 2 = ½ × 4.
• The figure shows all the three graphs on the same axes. Does this help you to conclude anything about the linear equation y = ax, a > 0 as a varies? What happens when a > 1 and when a < 1?
  (Hint: You may also plot the equations y = 3x and y = 13x on the same axes.)Answer: Every line y = ax passes through the origin (0, 0), and rises from left to right when a > 0. The bigger a is, the steeper the line: for a > 1 it is steeper than y = x, and for a < 1 it is less steep than y = x. The number a is called the slope.
• Now let us draw the graphs of y = −13x, y = −x, y = −3x by selecting suitable points on these lines.Answer: Use (0, 0) and (3, −1) for y = −⅓x, (0, 0) and (2, −2) for y = −x, (0, 0) and (1, −3) for y = −3x. All three lines pass through the origin and go down from left to right; y = −3x is the steepest and y = −⅓x the least steep.
• The figure shows all the three graphs on the same axes. Does this help you to conclude anything about the linear equation y = − ax, a > 0, as a varies? What will happen when a > 1 and when a < 1?Answer: Every line y = −ax (a > 0) passes through the origin and goes down from left to right (negative slope). The bigger a is, the steeper the line: for a > 1 it is steeper than y = −x, and for a < 1 it is less steep than y = −x.
• Differentiate between the graphs of the equations y = 3x + 1, and y = −3x + 1.Answer: Both lines cut the y-axis at the same point (0, 1) and are equally steep. But y = 3x + 1 has slope 3 and rises from left to right, while y = −3x + 1 has slope −3 and falls. They are mirror images in the y-axis.
• Let us now draw the graphs of y = 2x − 1, y = 2x + 1, y = 2x + 5, first individually (as shown in the figure) and then on the same axes (as shown in the figure).Answer: Points: y = 2x − 1 through (0, −1) and (2, 3); y = 2x + 1 through (0, 1) and (2, 5); y = 2x + 5 through (0, 5) and (−2, 1). On the same axes the three lines are parallel (all have slope 2) and cut the y-axis at −1, 1 and 5.
• Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?
  (Hint: In these equations a = 2, and b takes the values −1, 1 and 5, respectively.)Answer: When a is fixed and b changes, the line keeps the same slope but shifts up or down: the lines are parallel, and each cuts the y-axis at (0, b).
• Now let us draw the graphs of the equations y = x + 3, y = 2x + 5 and y = 3x − 2. See the figure and observe where these lines cut the y-axis.Answer: y = 2x + 5 cuts the y-axis at A (0, 5), y = x + 3 at B (0, 3) and y = 3x − 2 at C (0, −2). In general y = ax + b cuts the y-axis at (0, b); b is the y-intercept.
• Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’.Answer: (i) All through the origin (b = 0); larger a = steeper. (ii) All through the origin, falling; larger size of a = steeper. (iii) Equally steep, one rising and one falling, mirror images in the y-axis. (iv) Parallel (same a = 3), cutting the y-axis at −1, 0, 1. (v) y = −2x − 3 and y = −2x are parallel; y = 2x + 3 rises and meets y = −2x − 3 on the x-axis at (−1.5, 0).
• 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 + 1 (any polynomial whose highest power is x³ and whose x² term is −7x²).
• Find the values of the following polynomials at the indicated values of the variables.Answer: (i) 9 (ii) 4a³ − a² + 6
• If we multiply a number by 52 and add 23 to the product, we get −712. Find the number.Answer: 52x + 23 = −712 gives x = −12.
• 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: The numbers are 7 and 35.
• 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: Amount after n months = ₹(800 + 250n). (i) After 6 months: ₹2300. (ii) After 2 years (24 months): ₹6800.
• 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: The numbers are 85 and 58.
• 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.Answer: (i) slope −3, y-intercept 4, (0, 4) (ii) slope 2, y-intercept 72, (0, 72) (iii) slope 65, y-intercept −2, (0, −2) (iv) slope 2, y-intercept −113, (0, −113). Lines (ii) and (iv) are parallel (both slope 2).
• 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 = 95(x − 273) + 32.Answer: (i) 313 K = 104 °F (ii) 158 °F = 343 K.
• 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: w = Fd; with force 3 units, w = 3d. When d = 2, w = 6 units, and the point (2, 6) lies on the graph.
• The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).Answer: (i) p(x) = 3x + 2 (ii) y-axis at (0, 2), x-axis at (−23, 0) (iii) the line through (1, 5) and (3, 11) passes through both these points.
• 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: p(x) = 2x + 5 and q(x) = 4x − 1.
• 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.Answer: (i) Stage 4: 21, Stage 5: 26 matchsticks (ii) 6, 11, 16, 21, 26, …, 5n + 1 (iii) 5n + 1 (iv) 76 (v) No: 5n + 1 = 200 gives n = 39.8, not a whole number.
• 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: p(x) = 2x − 1 and q(x) = 2x − 9. They meet the x-axis at (12, 0) and (92, 0) respectively.
• What do all linear functions of the form f(x) = ax + a, a > 0, have in common?Answer: f(x) = a(x + 1), so every such graph passes through (−1, 0). Also, in each one the slope and the y-intercept are the same number a, and since a > 0 every line rises from left to right (linear growth) and cuts the y-axis above the origin.

All maths questions