Chapter 2
Introduction to Linear Polynomials
Important Questions Bank — Short Answer & Long Answer | Based on Ganita Manjari Grade 9
10 Long Answer Qs
Theory + Application
Step-by-Step Solutions
Exam Ready
What is a polynomial? Define the degree of a polynomial and give one example each of a linear, quadratic, and cubic polynomial.
TheoryEasy
An algebraic expression involving one variable and its whole-number powers is called a polynomial. The degree of a polynomial is the highest power of the variable in it.
- Linear polynomial (degree 1): 3z + 7
- Quadratic polynomial (degree 2): x² + 5x + 1
- Cubic polynomial (degree 3): 5y³ + y² + 2y − 1
Find the degrees of the following polynomials: (i) 2x² − 5x + 3 (ii) y³ + 2y − 1 (iii) −9 (iv) 4z − 3
NumericalEasy
- 2x² − 5x + 3 → highest power of x is 2 → Degree = 2 (quadratic)
- y³ + 2y − 1 → highest power of y is 3 → Degree = 3 (cubic)
- −9 → constant, written as −9x⁰ → Degree = 0 (constant polynomial)
- 4z − 3 → highest power of z is 1 → Degree = 1 (linear)
In the polynomial x⁴ − 3x³ + 6x² − 2x + 7, find the coefficients of x² and x³. Also state the constant term.
NumericalEasy
Writing the polynomial term by term: x⁴ − 3x³ + 6x² − 2x + 7
- Coefficient of x³ = −3
- Coefficient of x² = 6
- Constant term = 7 (the term with no variable)
The coefficient of a term is the number in front of the variable. For a negative term like −3x³, the coefficient is −3, not just 3.
Find the value of the linear polynomial 5x − 3 for (i) x = 0 (ii) x = −1 (iii) x = 2
NumericalEasy
Substituting each value into 5x − 3:
- When x = 0: 5(0) − 3 = −3
- When x = −1: 5(−1) − 3 = −5 − 3 = −8
- When x = 2: 5(2) − 3 = 10 − 3 = 7
The perimeter of a square of side x cm is 4x. Is this a linear polynomial? What is the perimeter when the side is 3.5 cm?
TheoryEasy
Yes, 4x is a linear polynomial in the variable x because the highest power of x is 1 (degree = 1).
When x = 3.5 cm: Perimeter = 4 × 3.5 = 14 cm.
This is an example of a linear relationship — as the side increases by a fixed amount, the perimeter also increases by a fixed amount (4 times as much).
The sum of two numbers is 64. One number is 10 more than the other. Set up a linear equation and find both numbers.
ApplicationEasy
Let the smaller number be x. Then the larger number is x + 10.
A chess club charges a joining fee of ₹200 plus ₹50 for every match played. Write a linear polynomial for the total amount paid if a player plays m matches. How much will a player pay if they play 11 matches?
ApplicationEasy
Total amount paid = ₹(200 + 50m) → This is a linear polynomial in m.
For 11 matches: 200 + 50 × 11 = 200 + 550 = ₹750
What is a linear pattern? How is it different from other patterns? Give one example from daily life.
TheoryMedium
A linear pattern is a sequence of numbers where the difference between two consecutive terms is constant.
For example, in the sequence 1, 3, 5, 7, 9, … the difference between each consecutive pair is always 2. This is a linear pattern.
Daily life example: Bela has ₹100 and spends ₹5 every day. The amounts remaining — 100, 95, 90, 85, … — form a linear pattern with a constant difference of −5.
Linear patterns are different from other patterns (like quadratic) where the differences between consecutive terms are not constant.
Define linear growth and linear decay. Give one real-life example of each.
TheoryEasy
A quantity increases by a fixed (constant) amount over equal intervals of time or steps.
Example: The cost of a journey C(d) = 100 + 60d increases by ₹60 for every additional km.
A quantity decreases by a fixed (constant) amount over equal intervals.
Example: The height of water h(t) = 3 − 0.5t decreases by 0.5 m each month.
Linear growth → straight line with positive slope; linear decay → straight line with negative slope.
Bela has ₹100 for pocket money and spends ₹5 every day. Write a linear expression for the amount left after n days. After how many days will she have ₹40 left?
ApplicationEasy
Amount left on the nth day = ₹(100 − 5n)
Setting 100 − 5n = 40:
An auto-rickshaw fare is ₹25 for the first 2 km, then increases by ₹15 per km. What is the fare for 10 km? Find the linear expression for the fare for n km (where n ≥ 2).
ApplicationMedium
For 10 km: first 2 km costs ₹25. Remaining = 10 − 2 = 8 km at ₹15 per km.
General formula for n km (where n ≥ 2):
In the equation y = ax + b, what do a and b represent geometrically? Where does this line cut the y-axis?
TheoryMedium
In the equation y = ax + b:
- a = slope of the line. It determines how steep the line is. If a > 1, the line is steeper than y = x; if a < 1, it is less steep.
- b = y-intercept. It is the value of y when x = 0, meaning the line cuts the y-axis at the point (0, b).
Example: For y = 3x − 2, the line cuts the y-axis at (0, −2), meaning 2 units below the origin.
When b = 0, the line y = ax passes through the origin (0, 0).
A growing pattern of square tiles has 1 tile at Stage 1, 3 tiles at Stage 2, 5 tiles at Stage 3. Write the linear polynomial for the number of tiles at Stage n. How many tiles are at Stage 15?
NumericalMedium
The pattern is 1, 3, 5, 7, 9, … — each stage adds 2 tiles. At each stage n, the number of tiles is one less than twice n.
Verify: Stage 2 → 2(2) − 1 = 3 ✓ | Stage 5 → 2(5) − 1 = 9 ✓
At Stage 15: 2(15) − 1 = 30 − 1 = 29
What are parallel lines in the context of linear equations? How can you tell from the equations y = 2x − 1, y = 2x + 1, and y = 2x + 5 that they are parallel?
TheoryMedium
Lines of the form y = ax + b are parallel when they have the same slope (a) but different y-intercepts (b). Parallel lines never meet.
In y = 2x − 1, y = 2x + 1, and y = 2x + 5:
- All three have slope a = 2 → they are equally steep.
- Their y-intercepts are −1, 1, and 5 respectively → different positions on the y-axis.
When a (slope) is fixed but b changes, the lines shift up or down but remain parallel. They will never intersect.
The cost function of a journey is C(d) = 100 + 60d. What is the cost for 15 km? For how many km will the cost be ₹700? Does this represent linear growth or decay? Justify.
ApplicationHard
Cost for 15 km: C(15) = 100 + 60 × 15 = 100 + 900 = ₹1000
When C(d) = 700:
Linear growth or decay? As d increases by 1, C increases by a fixed ₹60. This is linear growth — represented by a straight line with a positive slope (60).
Define univariate polynomial, its degree, and explain the four types: constant, linear, quadratic, and cubic. Give one real-life example that leads to a linear polynomial and one that leads to a quadratic polynomial.
TheoryMedium
- Univariate Polynomial: An algebraic expression in one variable involving only whole-number powers of that variable. Example: x² + 5x + 3.
- Degree: The highest power of the variable in the polynomial. For 5y³ + y² + 2y − 1, the degree is 3.
- Constant polynomial (degree 0): 8 or 8x⁰. The value never changes.
- Linear polynomial (degree 1): 3z + 7. The variable appears only to the first power.
- Quadratic polynomial (degree 2): x² + 5x + 1. The variable appears squared.
- Cubic polynomial (degree 3): 5y³ + y² + 2y − 1. Highest power is 3.
- Real-life linear example: Perimeter of a square of side x = 4x (degree 1). As x increases, perimeter increases at a constant rate.
- Real-life quadratic example: A wire of 20 cm is bent into a rectangle. If one side is x, the other is (10 − x), and the area = x(10 − x) = 10x − x² (degree 2).
The degree of a polynomial tells you its type and how “curved” its graph is. Linear = straight line; quadratic = parabola; cubic = S-curve.
Solve the following word problems using linear equations: (i) The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70. Find their present ages. (ii) Ruby has 3 times as many ₹2 coins as she has ₹5 coins. She has a total of ₹88. How many coins of each type does she have?
ApplicationMedium
Part (i) — Ages:
- Let Salil’s present age = x years. Mother’s age = 3x years.
- After 5 years: Salil = (x + 5), Mother = (3x + 5).
- Sum of ages after 5 years = 70: (x + 5) + (3x + 5) = 70
- Simplify: 4x + 10 = 70 → 4x = 60 → x = 15
Part (ii) — Ruby’s coins:
- Let number of ₹5 coins = y. Then ₹2 coins = 3y.
- Total value: 5y + 2(3y) = 88
- Simplify: 5y + 6y = 88 → 11y = 88 → y = 8
- Number of ₹2 coins = 3 × 8 = 24. Verify: 5(8) + 2(24) = 40 + 48 = 88 ✓
A growing pattern of square tiles has 1 tile at Stage 1, adding 2 tiles at each stage. (i) Write the number of tiles at Stages 1 to 7 in a table. (ii) Find the linear polynomial for Stage n. (iii) How many tiles are at Stage 15 and Stage 26? (iv) Which stage has 21 tiles and which has 47 tiles?
NumericalHard
Part (i) — Table:
| Stage | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Tiles | 1 | 3 | 5 | 7 | 9 | 11 | 13 |
- Linear polynomial: Each stage n has one less than twice the stage number. So: Tiles at Stage n = 2n − 1
- Stage 15: 2(15) − 1 = 30 − 1 = 29 tiles
- Stage 26: 2(26) − 1 = 52 − 1 = 51 tiles
- Stage with 21 tiles: 2n − 1 = 21 → 2n = 22 → n = 11 → Stage 11
- Stage with 47 tiles: 2n − 1 = 47 → 2n = 48 → n = 24 → Stage 24
The constant difference (2) between consecutive terms is the slope of the linear relationship. This is a key property of all linear patterns.
A telecom company charges a fixed monthly fee and an additional cost per GB of internet used. A student’s bill was ₹350 for 10 GB and ₹550 for 20 GB. (i) Find the linear relationship y = ax + b between the bill y and data used x. (ii) What do the values of a and b represent in real life? (iii) What will the bill be for 15 GB?
ApplicationHard
- We are given: when x = 10, y = 350, and when x = 20, y = 550.
- Substituting into y = ax + b:Equation 1: 350 = 10a + b
Equation 2: 550 = 20a + b
- Subtract Equation 1 from Equation 2:550 − 350 = (20a + b) − (10a + b) → 200 = 10a → a = 20
- Substitute a = 20 into Equation 1:350 = 10(20) + b → 350 = 200 + b → b = 150
- Linear relationship: y = 20x + 150
- Meaning: a = 20 means the cost per GB of data is ₹20. b = 150 is the fixed monthly fee of ₹150.
- Bill for 15 GB: y = 20(15) + 150 = 300 + 150 = ₹450
Explain the concept of slope and y-intercept using the graphs of y = 2x − 1, y = 2x + 1, and y = 2x + 5. What do you conclude when: (i) the slope is fixed but y-intercept changes? (ii) the y-intercept is fixed but slope changes?
TheoryHard
- Slope (a): In y = ax + b, ‘a’ is the slope. It measures the steepness of the line — how much y changes for each unit increase in x.
- y-intercept (b): It is the point where the line crosses the y-axis, at coordinates (0, b).
- For y = 2x − 1: slope = 2, y-intercept = −1. Line cuts y-axis at (0, −1).
- For y = 2x + 1: slope = 2, y-intercept = 1. Line cuts y-axis at (0, 1).
- For y = 2x + 5: slope = 2, y-intercept = 5. Line cuts y-axis at (0, 5).
- Conclusion (i) — Fixed slope, changing b: All three lines are parallel to each other (same steepness, same slope = 2), but they cut the y-axis at different heights. Changing b shifts the line up or down without changing its direction.
- Conclusion (ii) — Fixed b, changing slope: Lines have the same y-intercept (they all start from the same point on the y-axis), but different slopes make them more or less steep. A larger slope makes the line steeper; a negative slope makes it fall to the right.
Lines with equal slopes are parallel. Lines with the same y-intercept pass through the same point on the y-axis but fan out at different angles.
A mobile phone is bought for ₹10,000 and its value decreases by ₹800 every year. (i) Make a table of values for t = 0 to 5 years. (ii) Write a linear expression for its value after t years. (iii) Find its value after 3 years. (iv) After how many years will its value become ₹3,600? (v) Explain why this is linear decay.
ApplicationHard
- Table of values:
Years (t) 0 1 2 3 4 5 Value (₹) 10000 9200 8400 7600 6800 6000 - Linear expression: Starting value ₹10,000, decreasing by ₹800 per year.v(t) = 10000 − 800t
- Value after 3 years: v(3) = 10000 − 800(3) = 10000 − 2400 = ₹7600
- When v(t) = ₹3600: 10000 − 800t = 3600 → 800t = 6400 → t = 8 years
- Why linear decay: The value decreases by a fixed amount (₹800) over equal intervals (1 year). This is the definition of linear decay. On a graph, this is a straight line with negative slope (−800).
Plot the line y = 2x + 1 by finding at least four points. Identify the slope and y-intercept. Verify whether (7, 15) lies on this line. Also compare: does the point (5, 12) lie on this line?
NumericalMedium
- Find points on y = 2x + 1:
x 0 1 2 3 5 7 y = 2x+1 1 3 5 7 11 15 - Plot A(0, 1), B(1, 3), C(2, 5), D(3, 7). Join them — all lie on a straight line.
- Slope: a = 2. For every 1 unit increase in x, y increases by 2.
- y-intercept: b = 1. The line cuts the y-axis at (0, 1).
- Check (7, 15): Substitute x = 7: y = 2(7) + 1 = 15 ✓ → (7, 15) lies on the line.
- Check (5, 12): Substitute x = 5: y = 2(5) + 1 = 11 ≠ 12 ✗ → (5, 12) does NOT lie on the line.
A point (x, y) lies on a line if and only if its coordinates satisfy the equation of the line.
A farmer cuts a 300 feet fence into two pieces. The longer piece is 4 times as long as the shorter piece. Also, separately: If the length of a rectangle is 3 more than twice its width and its perimeter is 24 cm, find the dimensions.
ApplicationMedium
Part 1 — Fence:
- Let the shorter piece = x ft. Then the longer piece = 4x ft.
- Total: x + 4x = 300 → 5x = 300 → x = 60 ft
Part 2 — Rectangle:
- Let width = w cm. Then length = 2w + 3 cm.
- Perimeter = 2(l + w) = 24 → l + w = 12
- Substitute: (2w + 3) + w = 12 → 3w + 3 = 12 → 3w = 9 → w = 3 cm
- Length = 2(3) + 3 = 9 cm. Verify: 2(9 + 3) = 24 ✓
You have ₹800 and save ₹250 every month. (i) Find the amount after 6 months and after 2 years. (ii) Write this as a linear pattern and identify the linear polynomial. (iii) After how many months will you have ₹3,300?
ApplicationHard
- Linear polynomial: Amount after n months = 800 + 250n
- After 6 months: 800 + 250(6) = 800 + 1500 = ₹2300
- After 2 years = 24 months: 800 + 250(24) = 800 + 6000 = ₹6800
- Linear pattern: 800, 1050, 1300, 1550, … with constant difference 250. This is linear growth.
- When amount = ₹3300:800 + 250n = 3300 → 250n = 2500 → n = 10 months
The constant addition of ₹250 per month makes this linear growth. The slope of the corresponding line y = 250n + 800 is 250.
Draw the graphs of y = 3x and y = −2x by selecting suitable points. State: (i) the slope of each. (ii) Do they pass through the origin? (iii) Which line is steeper? (iv) Which represents linear growth and which represents linear decay?
NumericalHard
- Points on y = 3x: (0,0), (1,3), (2,6), (−1,−3). Plot and join to get a straight line rising steeply to the right.
- Points on y = −2x: (0,0), (1,−2), (2,−4), (−1,2). Plot and join to get a straight line falling to the right.
- Slopes: Slope of y = 3x is a = 3. Slope of y = −2x is a = −2.
- Do they pass through origin? Yes — both equations have b = 0, so both lines pass through (0, 0). All lines of form y = ax pass through the origin.
- Steeper line: y = 3x is steeper because its slope (3) has a larger absolute value than y = −2x (|−2| = 2).
- Linear growth vs decay: y = 3x has positive slope (+3) → linear growth (y increases as x increases). y = −2x has negative slope (−2) → linear decay (y decreases as x increases).
Students think a more negative slope means “slower decay.” A slope of −3 is actually steeper (faster decay) than a slope of −1. Always compare absolute values for steepness.
To check if a point lies on a line, substitute its x-coordinate into the equation and check if the result equals the y-coordinate.
In word problems, always define your variable clearly (e.g., “let x = number of matches played”) before writing the equation.
For linear growth, the slope is positive; for linear decay, the slope is negative. The absolute value of the slope tells you how fast the change is.
Confusing the degree of a polynomial with the number of terms. 3x² + 2x + 1 has degree 2 (quadratic), not degree 3.
Forgetting that parallel lines have the SAME slope, not the same y-intercept. y = 2x + 3 and y = 2x − 5 are parallel (same slope = 2).
5-mark long answers: 1 mark for setting up the equation/expression, 2 marks for working, 1 mark for correct answer, 1 mark for proper conclusion/verification.
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