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CBSE Class 9 · Maths · 2026-27

CBSE Class 9 Maths Sample Paper 2026-27

This is a full Class 9 Maths sample paper built to the CBSE board blueprint — 38 questions in Sections A to E, 80 marks, with MCQs, Assertion-Reason, short and long answers, and case-based questions. A complete answer key with solutions is included below.

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Annual Examination 2026-27

Mathematics Sample Question Paper

Class 9 · Mathematics

Time: 3 HoursMax Marks: 80

General Instructions

  • This question paper contains 38 questions. All questions are compulsory.
  • Section Section A: Questions 1–20 — Multiple Choice Questions & Assertion–Reason Questions (20 marks).
  • Section Section B: Questions 21–25 — Short Answer Questions (10 marks).
  • Section Section C: Questions 26–31 — Short Answer Questions (18 marks).
  • Section Section D: Questions 32–35 — Long Answer Questions (20 marks).
  • Section Section E: Questions 36–38 — Case Study Based Questions (12 marks).
  • Marks for each question are indicated against it.

Section A — Multiple Choice Questions & Assertion–Reason Questions

(20 marks)

Choose the correct option. Each question has an Assertion (A) and a Reason (R). Choose the correct option.

  1. Q1. What are the coordinates of a point on the positive x-axis that is 5 units away from the origin?

    [1]
    (A) (0, 5)(B) (5, 0)(C) (0, 0)(D) (5, 5)

  2. Q2. A rectangle has vertices at (1, 2), (1, 6), (5, 6), and (5, 2). What is the area of the rectangle?

    [1]
    (A) 8 square units(B) 12 square units(C) 16 square units(D) 20 square units

  3. Q3. Find the coordinates of the point on the x-axis that is equidistant from the points (2, 4) and (6, -4).

    [1]
    (A) (0, 0)(B) (4, 0)(C) (2, 0)(D) (6, 0)

  4. Q4. What is the degree of the polynomial 8 - 3x?

    [1]
    (A) 0(B) 1(C) 2(D) 3

  5. Q5. A taxi charges a fixed fare of ₹30 plus ₹12 per kilometre. If a passenger travels x km, which linear polynomial represents the total fare in rupees?

    [1]
    (A) 12x + 30(B) 30x + 12(C) 42x(D) 12x - 30

  6. Q6. A book costs ₹10 more than twice the cost of a pen. Two such books and three pens together cost ₹230. If the cost of a pen is ₹x, which linear equation correctly models this situation?

    [1]
    (A) 2(2x + 10) + 3x = 230(B) 2(x + 10) + 3x = 230(C) (2x + 10) + 3x = 230(D) 2(2x + 10) + 3 = 230

  7. Q7. Which of the following is equivalent to the fraction -1/2?

    [1]
    (A) -2/4(B) 1/-2(C) Both -2/4 and 1/-2(D) None of these

  8. Q8. A spice trader takes a loan of ₹850, makes a profit of ₹1,200 the next day, and then incurs a loss of ₹450 the following week. What is his final financial standing?

    [1]
    (A) ₹100 profit(B) ₹100 debt(C) ₹2,500 profit(D) ₹2,500 debt

  9. Q9. If a, b, and c are integers, which of the following expressions always results in a negative integer?

    [1]
    (A) a + b + c(B) a - b - c(C) a × b × c when all are negative(D) a + b - c

  10. Q10. If (x+5)^2 = x^2 + kx + 25, what is the value of k?

    [1]
    (A) 5(B) 10(C) 25(D) 20

  11. Q11. Factor 50a^2 + 60ab + 18b^2 completely.

    [1]
    (A) 2(5a + 3b)^2(B) (5a + 3b)^2(C) 2(25a + 9b)^2(D) 2(5a - 3b)^2

  12. Q12. Without directly squaring the numbers, evaluate 106^2 - 94^2 using a suitable identity.

    [1]
    (A) 1200(B) 2400(C) 3600(D) 4800

  13. Q13. A chord of a circle that passes through its centre is called the:

    [1]
    (A) Radius(B) Diameter(C) Tangent(D) Secant

  14. Q14. Two chords AB and CD of a circle are equal in length. Which of the following is true?

    [1]
    (A) They subtend different angles at the centre(B) They subtend equal angles at the centre(C) One chord is longer than the other(D) They are parallel

  15. Q15. In a circle, a chord of length 16 cm is at a distance of 6 cm from the centre. Another chord is at a distance of 8 cm from the centre. What is the length of the other chord?

    [1]
    (A) 12 cm(B) 14 cm(C) 16 cm(D) 20 cm

  16. Q16. If the radius of a circle is doubled, how does its circumference change?

    [1]
    (A) It remains the same(B) It becomes two times(C) It becomes three times(D) It becomes four times

  17. Q17. Assertion (A): The point (0, -3) lies on the y-axis. Reason (R): Every point with x-coordinate zero lies on the y-axis.

    [1]
    (A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.

  18. Q18. Assertion (A): The point (0, 4) lies in Quadrant I. Reason (R): Quadrant I is the region where both coordinates are positive.

    [1]
    (A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.

  19. Q19. Assertion (A): A point with positive abscissa and negative ordinate lies in the fourth quadrant. Reason (R): Abscissa refers to the x-coordinate and ordinate refers to the y-coordinate.

    [1]
    (A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.

  20. Q20. Assertion (A): The polynomial 9 is a constant polynomial. Reason (R): A constant polynomial has degree 0.

    [1]
    (A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.

Section B — Short Answer Questions

(10 marks)

Answer the following questions briefly.

  1. Q21. A point has coordinates (0, −5). On which axis does it lie? Justify your answer.

    [2]

  2. Q22. A point lies in Quadrant III. Its distance from the x-axis is 5 units and from the y-axis is 2 units. Determine its coordinates.

    [2]

  3. Q23. Points A and B are on the coordinate plane such that A lies on the x-axis and B lies on the y-axis. If the distance AB is 13 units and both coordinates of A and B are positive integers, find one possible set of coordinates for A and B.

    [2]

  4. Q24. What is the coefficient of y in the linear polynomial 9y – 15?

    [2]

  5. Q25. The length of a rectangle is 5 cm more than twice its width. If the width is w cm, express the length in terms of w. If the perimeter is 70 cm, form an equation and solve for w.

    [2]

Section C — Short Answer Questions

(18 marks)

Answer the following questions.

  1. Q26. State the quadrant or axis on which the following points lie: (i) (-3, 5) (ii) (4, -2) (iii) (0, -6). Also, find the distance of the point (4, -2) from the x-axis.

    [3]

  2. Q27. The coordinates of three vertices of a rectangle are A(2,3), B(2, -1), and C(-4, -1). Find the coordinates of the fourth vertex D. Also, find the lengths of the sides AB and BC.

    [3]

  3. Q28. A treasure is hidden at a point in a park that is exactly 5 units away from the fountain at (0,0) and also 5 units away from the statue at (6,0). Find all possible coordinates for the treasure.

    [3]

  4. Q29. Find the value of the linear polynomial \(4x - 7\) when \(x = -3\).

    [3]

  5. Q30. In a pattern of matchsticks, the number of matchsticks at stage \(n\) is given by \(5n + 2\). How many matchsticks are needed for stage 10? At what stage will 47 matchsticks be used?

    [3]

  6. Q31. A mobile plan charges a fixed monthly fee plus a per-minute call charge. For a usage of 300 minutes, the bill is ₹450. For 500 minutes, the bill is ₹650. Let the fixed fee be ₹\(F\) and per-minute charge be ₹\(R\). Express the total bill for \(x\) minutes as a linear polynomial in \(x\). Then find the bill for 400 minutes.

    [3]

Section D — Long Answer Questions

(20 marks)

Answer the following questions in detail.

  1. Q32. Plot the points O(0, 0), P(4, 0), Q(4, 3), and R(0, 3) on a coordinate plane. Join them in order O → P → Q → R → O. What type of quadrilateral do you get? Find the lengths of its sides, its perimeter, and its area.

    [5]

  2. Q33. On a city map, a hospital is at H(2, -5) and a fire station is at F(-6, 7). Distances are in kilometres. Find the coordinates of the police station P which is exactly midway between H and F. Also calculate the straight-line distance between the hospital and the fire station.

    [5]

  3. Q34. A point P lies on the x-axis and is equidistant from the points A(3, -4) and B(-5, 8). Find the coordinates of P.

    [5]

  4. Q35. A fruit vendor sells apples at ₹x per kg and oranges at ₹(x+20) per kg. A customer buys 3 kg apples and 2 kg oranges. (a) Write a linear polynomial in x that represents the total cost. (b) If the total cost is ₹310, find the price per kg of apples and oranges. (c) What is the coefficient of x in the polynomial?

    [5]

Section E — Case Study Based Questions

(12 marks)

Read the passage and answer the questions that follow.

  1. Q36. A classroom floor is represented on a coordinate plane with the origin O(0,0) at the teacher's desk. The x-axis runs along the front wall, and the y-axis runs along the left wall. Students A and B are seated at A(3,2) and B(5,2). A bookshelf stands at (8,0).

    [4]

  2. Q37. A triangular flower bed in a park has vertices at A(1,1), B(5,1), and C(5,5) on a coordinate plane. A gardener places a sprinkler exactly at the midpoint of the hypotenuse of the triangle.

    [4]

  3. Q38. A rectangle ABCD is drawn on a coordinate plane with A(0,0), B(8,0), C(8,6), D(0,6). A point P lies inside the rectangle such that its distance from side AB is 3 units and its distance from side AD is 4 units.

    [4]

Answer Keyincluded with every paper

  1. Q1. (5, 0)

    Any point on the x-axis has y-coordinate 0. Since it is on the positive x-axis at a distance of 5 units from the origin, its x-coordinate is 5. Hence, coordinates are (5, 0).

  2. Q2. 16 square units

    Length = difference in y-coordinates = 6 - 2 = 4 units. Width = difference in x-coordinates = 5 - 1 = 4 units. Area = length × width = 4 × 4 = 16 square units.

  3. Q3. (4, 0)

    Let the point on the x-axis be (x, 0). Distance to (2, 4) = √[(x-2)² + 16]. Distance to (6, -4) = √[(x-6)² + 16]. Equating: (x-2)² + 16 = (x-6)² + 16 => (x-2)² = (x-6)² => x² - 4x + 4 = x² - 12x + 36 => 8x = 32 => x = 4. Thus, the point is (4, 0).

  4. Q4. B) 1

    The polynomial can be written as -3x¹ + 8. The highest power of the variable x is 1, so the degree is 1.

  5. Q5. A) 12x + 30

    Fixed fare is ₹30 (constant term) and per km charge ₹12 (coefficient of x). Total = 12x + 30.

  6. Q6. A) 2(2x + 10) + 3x = 230

    Cost of pen = x. Cost of book = 2x + 10. Two books and three pens: 2(2x+10) + 3x = 230.

  7. Q7. Both -2/4 and 1/-2

    Equivalent fractions are obtained by multiplying numerator and denominator by the same non-zero integer, and the negative sign can be in numerator, denominator, or in front. So both -2/4 and 1/-2 equal -1/2.

  8. Q8. ₹100 debt

    Initial debt = -850. Add profit: -850 + 1200 = 350. Subtract loss: 350 - 450 = -100. So he has a debt of ₹100.

  9. Q9. a × b × c when all are negative

    The product of three negative integers is negative, because (-1)^3 = -1. The other options depend on the specific values of a, b, c.

  10. Q10. 10

    Expanding (x+5)^2 gives x^2 + 2*x*5 + 5^2 = x^2 + 10x + 25. Hence, k = 10.

  11. Q11. 2(5a + 3b)^2

    First take out the common factor 2: 50a^2 + 60ab + 18b^2 = 2(25a^2 + 30ab + 9b^2). Then factor the inner perfect square: 25a^2 + 30ab + 9b^2 = (5a)^2 + 2*(5a)*(3b) + (3b)^2 = (5a + 3b)^2. So the complete factorization is 2(5a + 3b)^2.

  12. Q12. 2400

    Use the difference of squares identity: a^2 - b^2 = (a+b)(a-b). Here a = 106, b = 94. Then (106+94)(106-94) = 200 * 12 = 2400.

  13. Q13. Diameter

    A chord passing through the centre of a circle is called a diameter. It is the longest chord.

  14. Q14. They subtend equal angles at the centre

    By the theorem, equal chords of a circle subtend equal angles at the centre.

  15. Q15. 12 cm

    For the first chord: half-length = 8 cm, distance = 6 cm. Radius r = √(8² + 6²) = 10 cm. For the second chord: distance = 8 cm, half-length = √(10² − 8²) = 6 cm, so full length = 12 cm.

  16. Q16. It becomes two times

    Circumference = 2πr. If radius becomes 2r, new circumference = 2π(2r) = 4πr = 2 × (2πr), so it doubles.

  17. Q17. Both A and R are true and R is the correct explanation of A.

    A is true because the point (0,-3) has x-coordinate 0, so it lies on the y-axis. R is also true: any point with x-coordinate 0 lies on the y-axis. R correctly explains A.

  18. Q18. A is false but R is true.

    A is false because (0,4) lies on the y-axis, not in any quadrant. R correctly states that Quadrant I requires both coordinates positive (x>0, y>0), so R is true.

  19. Q19. Both A and R are true but R is not the correct explanation of A.

    A is true because positive x and negative y indeed places a point in Quadrant IV. R is true, but it only defines terms; it does not explain the quadrant location. The correct explanation would be that in Quadrant IV, x>0 and y<0.

  20. Q20. Both A and R are true and R is the correct explanation of A.

    9 can be written as 9x^0, so its degree is 0, making it a constant polynomial. R states the defining property of constant polynomials. Thus, both are true and R explains A.

  21. Q21. It lies on the y-axis because its x-coordinate is 0.

    Any point with x-coordinate 0 lies on the y-axis. Here, (0, −5) has x=0, so it is on the y-axis, 5 units below the origin.

  22. Q22. (−2, −5)

    In Quadrant III, both coordinates are negative. Distance from x-axis (perpendicular) is the absolute y-coordinate: |y| = 5 ⇒ y = −5. Distance from y-axis is the absolute x-coordinate: |x| = 2 ⇒ x = −2. Hence, the point is (−2, −5).

  23. Q23. A(5, 0) and B(0, 12) (or A(12,0) and B(0,5))

    Let A = (p, 0) and B = (0, q). Then AB = √(p² + q²) = 13 ⇒ p² + q² = 169. Since p, q are positive integers, one possible integer pair is (5, 12) because 5² + 12² = 25 + 144 = 169. So A(5,0), B(0,12) or swapped.

  24. Q24. 9

    The coefficient of y is the numerical factor multiplying y, which is 9.

  25. Q25. length = 2w + 5; w = 10 cm

    Length L = 2w + 5. Perimeter = 2(L + w) = 2(2w+5 + w) = 2(3w+5) = 6w + 10. Set 6w + 10 = 70 → 6w = 60 → w = 10 cm.

  26. Q26. (i) Quadrant II; (ii) Quadrant IV; (iii) negative y-axis; Distance = 2 units.

    (i) x negative, y positive => Quadrant II. (ii) x positive, y negative => Quadrant IV. (iii) x=0, y negative => lies on the negative y-axis. Distance from x-axis = |y| = |-2| = 2 units.

  27. Q27. D = (-4, 3); AB = 4 units; BC = 6 units.

    Since opposite sides of a rectangle are parallel, AB is vertical (same x-coordinate) and BC is horizontal (same y-coordinate). Thus D must have same x as C and same y as A, so D=(-4,3). Length AB = |3 - (-1)| = 4 units. Length BC = |2 - (-4)| = 6 units.

  28. Q28. (3, 4) and (3, -4).

    Let the treasure point be (x,y). From condition 1: √(x²+y²)=5 => x²+y²=25. From condition 2: √((x-6)²+y²)=5 => (x-6)²+y²=25. Subtract second equation from first: x² - (x-6)² = 0 => x² - (x² -12x +36)= 12x -36=0 => x=3. Substitute x=3 into first: 9+y²=25 => y²=16 => y=±4. So points are (3,4) and (3,-4).

  29. Q29. \(-19\)

    Substitute \(x = -3\) into \(4x - 7\): \(4(-3) - 7 = -12 - 7 = -19\).

  30. Q30. 52 matchsticks; stage 9

    For \(n = 10\): \(5(10) + 2 = 50 + 2 = 52\). To find \(n\) when the expression equals 47: set \(5n + 2 = 47\). Subtract 2: \(5n = 45\). Divide by 5: \(n = 9\). So, stage 9 uses 47 matchsticks.

  31. Q31. Bill = \(150 + x\); ₹550

    The two bills give the equations: \(F + 300R = 450\) and \(F + 500R = 650\). Subtract the first from the second: \(200R = 200\) → \(R = 1\). Substitute \(R = 1\) into the first equation: \(F + 300 = 450\) → \(F = 150\). Thus, the linear polynomial for bill is \(150 + 1x = 150 + x\). For \(x = 400\), bill = \(150 + 400 = 550\).

  32. Q32. The quadrilateral is a rectangle. Side lengths are OP = 4 units, PQ = 3 units, QR = 4 units, RO = 3 units. Perimeter = 14 units. Area = 12 square units.

    Plot the points on graph paper. OP is horizontal from (0,0) to (4,0) so its length is 4. PQ is vertical from (4,0) to (4,3) so length is 3. QR is horizontal from (4,3) to (0,3) so length is 4. RO is vertical from (0,3) to (0,0) so length is 3. All angles are 90° because sides are parallel to axes. Thus it is a rectangle. Perimeter = 2 × (4 + 3) = 14 units. Area = 4 × 3 = 12 square units.

  33. Q33. Midpoint P = (-2, 1). Distance HF = √208 = 4√13 km.

    Midpoint of H(2,-5) and F(-6,7) is ((2 + (-6))/2, (-5 + 7)/2) = (-4/2, 2/2) = (-2, 1). So P = (-2, 1). Distance HF = √[(-6 - 2)² + (7 - (-5))²] = √[(-8)² + (12)²] = √(64 + 144) = √208 = √(16 × 13) = 4√13 km.

  34. Q34. P = (-4, 0).

    Let P = (x, 0) because it is on the x-axis. Then PA = PB → PA² = PB². (x - 3)² + (0 - (-4))² = (x - (-5))² + (0 - 8)² → (x - 3)² + 16 = (x + 5)² + 64. Expand: x² - 6x + 9 + 16 = x² + 10x + 25 + 64 → -6x + 25 = 10x + 89 → -6x - 10x = 89 - 25 → -16x = 64 → x = -4. So P = (-4, 0).

  35. Q35. (a) Total cost = 5x + 40; (b) Apple price = ₹54 per kg, Orange price = ₹74 per kg; (c) Coefficient of x = 5

    (a) Cost of apples for 3 kg = 3x; cost of oranges for 2 kg = 2(x+20) = 2x+40. Total cost = 3x + 2x + 40 = 5x + 40. (b) Given total cost = 310, so 5x + 40 = 310 ⟹ 5x = 270 ⟹ x = 54. Thus, apple price = ₹54, orange price = 54+20 = ₹74. (c) In 5x+40, coefficient of x is 5.

  36. Q36. (i) (3,2) (ii) Both A and B, because their y-coordinates are 2. (iii) 2 units

    (i) The coordinates of A are given as (3,2). (ii) Distance from x-axis equals absolute y-coordinate; both have y=2. (iii) A and B share the same y-coordinate, so distance = |5 - 3| = 2 units.

  37. Q37. (i) Angle at B, because AB is horizontal (same y) and BC is vertical (same x). (ii) (3,3) (iii) 2√2 units

    (i) AB from A(1,1) to B(5,1) is horizontal, BC from B(5,1) to C(5,5) is vertical → angle at B is 90°. (ii) Hypotenuse is AC; midpoint = ((1+5)/2, (1+5)/2) = (3,3). (iii) Distance A(1,1) to (3,3) = √[(3-1)²+(3-1)²] = √(4+4) = √8 = 2√2 ≈2.83.

  38. Q38. P(4, 3); PC = 5 units

    Distance from AB (x-axis) gives the y-coordinate of P, which is 3. Distance from AD (y-axis) gives the x-coordinate, which is 4. So P(4,3). Distance PC = √((8-4)² + (6-3)²) = √(4² + 3²) = √(16+9) = √25 = 5 units.

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