Reoti, Ballia (U.P.) · Affiliated to CBSE · www.gjms.edu.in
SAMPLE QUESTION PAPER · Session 2026-27 MATHEMATICS (Standard) — Class IX
Time Allowed: 3 HoursMaximum Marks: 80
General Instructions:
This question paper contains 39 questions. All questions are compulsory.
The paper is divided into five Sections — A, B, C, D and E.
Section A has 20 questions of 1 mark each (Q1–Q20), comprising Multiple-Choice and Assertion–Reason type questions.
Section B has 5 Very Short Answer (VSA) questions of 2 marks each (Q21–Q25).
Section C has 6 Short Answer (SA) questions of 3 marks each (Q26–Q31).
Section D has 4 Long Answer (LA) questions of 5 marks each (Q32–Q35).
Section E has 3 case-study / competency-based questions of 4 marks each (Q36–Q38, with sub-parts and an internal choice in the last part).
There is no overall choice. However, an internal choice (OR) has been provided in 2 questions of Section B, 2 questions of Section C, 2 questions of Section D and in the 4-mark part of each case study in Section E. Attempt only one of the choices in such questions.
Multiple-Choice Questions (Q1–Q18) and Assertion–Reason (Q19–Q20). Each carries 1 mark.
1.1 Which of the following is an irrational number?
0.3¯ (0.333…)
√36
√7
22/7
2.1 The value of √5 × √20 is
10
√25
100
2√5
3.1 After rationalising the denominator, 1 / (3 − √2) equals
(3 + √2)/7
(3 − √2)/7
(3 + √2)/11
3 + √2
4.1 The zero of the polynomial p(x) = 2x + 5 is
5/2
−5/2
2/5
−2/5
5.1 If x + 1 is a factor of p(x) = x³ + 3x² + 3x + k, then the value of k is
0
1
−1
3
6.1 For the arithmetic progression 7, 11, 15, 19, …, the 10th term is
39
43
47
51
7.1 In the geometric progression 3, 6, 12, 24, …, the common ratio is
2
3
1/2
6
8.1 Using an identity, the value of 98 × 102 is
9998
9996
10004
10000
9.1 The linear equation 2x + 3y = 6 in two variables has
a unique solution
only two solutions
no solution
infinitely many solutions
10.1 The point (−4, 3) lies in the
I quadrant
II quadrant
III quadrant
IV quadrant
11.1 The distance between the points A(0, 0) and B(6, 8) is
10 units
14 units
√14 units
2 units
12.1 Euclid's statement "A straight line may be drawn from any one point to any other point" is a
theorem
definition
postulate
proof
13.1 Two supplementary angles are in the ratio 2 : 3. The larger angle is
72°
108°
60°
120°
14.1 In ▵ABC and ▵PQR, AB = PQ, ∠A = ∠P and ∠B = ∠Q. The congruence criterion that applies is
SSS
SAS
ASA
RHS
15.1 The diagonals of a rhombus
are equal
bisect each other at right angles
do not bisect each other
are parallel
16.1 The angle subtended by a semicircle at the centre is
90°
180°
60°
360°
17.1 The total surface area of a sphere of radius 7 cm is (take π = 22/7)
154 cm²
616 cm²
308 cm²
1232 cm²
18.1 A die is thrown once. The empirical/theoretical probability of getting a number greater than 4 is
1/6
1/2
1/3
2/3
Assertion–Reason (Q19–Q20). In each, choose:
(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.
19.1Assertion (A): The sum of the two interior angles of a triangle equals its exterior angle at the third vertex. Reason (R): The sum of the three interior angles of a triangle is 180°.
20.1Assertion (A): √2 + √3 is an irrational number. Reason (R): The sum of two irrational numbers is always irrational.
SECTION B — Very Short Answer 5 × 2 = 10 marks
21.2 Represent √9.3 on the number line, describing the construction steps briefly.
22.2 Find the value of the polynomial p(x) = 5x² − 3x + 7 at x = 2, and hence state whether x = 2 is a zero of p(x).
OR
Factorise: 6x² + 17x + 5.
23.2 Which term of the arithmetic progression 21, 18, 15, … is the first negative term?
24.2 Find the coordinates of the midpoint of the segment joining P(−3, 5) and Q(7, −1). Also find the distance PQ.
OR
The angles of a triangle are (x + 10)°, (2x − 30)° and (x + 40)°. Find each angle.
25.2 Two coins are tossed simultaneously 500 times. Two heads appeared 130 times, one head 260 times and no head 110 times. Find the empirical probability of getting at most one head.
SECTION C — Short Answer 6 × 3 = 18 marks
26.3 If a = (5 + √3)/(5 − √3) and b = (5 − √3)/(5 + √3), find the value of a + b.
27.3 Factorise the cubic polynomial x³ − 23x² + 142x − 120, given that x = 1 is one of its zeroes.
OR
Without actually calculating the cubes, find the value of 30³ + (−18)³ + (−12)³.
28.3 Draw the graph of the linear equation x + 2y = 6. From the graph, read off the value of y when x = 2, and the value of x when y = 0.
29.3(IKS — Baudhāyana / Śulba-sūtra.) The Baudhāyana Śulba-sūtra (c. 800 BCE) states a rule equivalent to the Pythagoras theorem for laying out sacrificial altars. A rectangular altar is planned with sides 12 units and 5 units. Using the Baudhāyana rule, find the length of its diagonal, and verify that a right triangle of legs 5 and 12 is a valid "sacred" triangle.
OR
In ▵ABC, AB = AC and the bisector of ∠A meets BC at D. Prove that ▵ABD ≅ ▵ACD, and hence that BD = DC.
30.3 The perimeter of a triangular field is 144 m and its sides are in the ratio 3 : 4 : 5. Using Heron's formula, find its area.
31.3 The following table shows the daily wages (in ₹) of 40 workers of a factory:
Daily wage (₹)
200
250
300
350
No. of workers
8
12
15
5
Find the weighted mean daily wage of the workers.
SECTION D — Long Answer 4 × 5 = 20 marks
32.5(Computational thinking — Tower of Hanoi.) In the Tower of Hanoi puzzle, the minimum number of moves needed to shift a tower of n discs follows the pattern M(n) = 2·M(n−1) + 1, with M(1) = 1.
Write the number of moves for n = 1, 2, 3, 4, 5 discs.
These numbers 1, 3, 7, 15, 31 form a sequence. Show that M(n) = 2ⁿ − 1 satisfies the rule, by checking n = 4 and n = 5.
How many moves are needed for 10 discs?
OR
The sum of the first three terms of an arithmetic progression is 27 and the sum of their squares is 293.
Taking the terms as (a − d), a, (a + d), find a.
Find the common difference d.
Write the three terms.
33.5 Prove that the angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle. Using this result, if an arc subtends an angle of 130° at the centre, find the angle it subtends at a point on the major arc.
34.5 Prove that the diagonals of a parallelogram bisect each other. Conversely, using this idea, show that if the diagonals of a quadrilateral bisect each other then it is a parallelogram.
OR
ABCD is a parallelogram. E and F are the mid-points of sides AB and DC respectively. Prove that the segments AF and CE trisect the diagonal BD.
35.5 A solid is in the shape of a cone mounted on a hemisphere, both of radius 7 cm. The total height of the solid is 21 cm.
Find the height of the cone.
Find the volume of the solid (take π = 22/7).
Find the curved surface area of the hemispherical part.
SECTION E — Case-Study / Competency-Based 3 × 4 = 12 marks
36.4
Case Study I — Designing a rangoli grid. For a school festival, students draw a rangoli on a square coordinate grid. Four decorative diyas are placed at the points A(1, 1), B(5, 1), C(5, 5) and D(1, 5). A central lamp is to be kept exactly at the centre of the design.
Write the coordinates of the midpoint of diagonal AC. 1
Show that the midpoint of diagonal BD is the same point, and state where the central lamp should be placed. 1
Find the length of one side AB and hence the perimeter of the square ABCD.
OR
Find the length of the diagonal AC. 2
37.4
Case Study II — Water tank on the terrace. A cylindrical water tank of radius 1.4 m and height 3 m is installed on a building terrace. Water is supplied to fill it (take π = 22/7).
Find the area of the circular base of the tank. 1
Find the volume of water (in m³) the tank can hold when completely full. 1
The tank is painted on its curved outer surface only. Find the curved surface area to be painted.
OR
If 1 m³ = 1000 litres, find the capacity of the tank in litres. 2
38.4
Case Study III — Class survey on transport. A student surveys how the 45 pupils of Class IX travel to school. The stacked data collected are shown below.
Mode
Walk
Cycle
Bus
Car
No. of students
9
15
18
3
If one student is chosen at random, find the probability that the student comes by bus. 1
Find the probability that a randomly chosen student does not come by car. 1
To show this data as a stacked bar, the 45 students are drawn as one bar of length 9 cm. Find the length of the "Bus" portion of the bar.
OR
Two students are surveyed one after another (with replacement). Using a tree diagram idea, find the probability that both walk to school. 2
— END OF PAPER —
MARKING SCHEMEfor teachers
Mathematics (Standard), Class IX · Session 2026-27 · Max Marks 80. Award full marks for any correct alternative method. Deduct only where a step is genuinely missing.
SECTION A — Answer Key 20 × 1 = 20
MCQ / A-R key (1 mark each):
1 (c) √7 · 2 (a) 10 · 3 (a) (3+√2)/7 · 4 (b) −5/2 · 5 (b) 1 · 6 (b) 43 · 7 (a) 2 · 8 (b) 9996 · 9 (d) infinitely many solutions · 10 (b) II quadrant · 11 (a) 10 units · 12 (c) postulate · 13 (b) 108° · 14 (c) ASA · 15 (b) bisect each other at right angles · 16 (b) 180° · 17 (b) 616 cm² · 18 (c) 1/3 · 19 (a) · 20 (c). Working notes: Q2: √5·√20 = √100 = 10. Q3: multiply by (3+√2)/(3+√2); denom 9−2=7. Q5: p(−1)=−1+3−3+k=0 ⟹ k=1. Q6: a₁₀=7+9·4=43. Q8: 98·102=(100−2)(100+2)=10000−4=9996. Q11: √(6²+8²)=√100=10. Q13: 2x+3x=180 ⟹ x=36, larger=3·36=108°. Q17: 4πr²=4·(22/7)·49=616. Q18: {5,6} of 6 ⟹ 2/6=1/3. Q19: exterior-angle theorem is a consequence of the 180° angle-sum — R correctly explains A ⟹ (a). Q20: A true, but R false (e.g. √2+(−√2)=0 is rational) ⟹ (c).
SECTION B — Marking 5 × 2 = 10
21. Draw AB = 9.3 units on the line, extend by BC = 1 unit (so AC = 10.3). Mark midpoint O of AC; draw a semicircle with radius OA = 5.15. Erect a perpendicular to AC at B meeting the semicircle at D; then BD = √9.3. (Construction 1 mark + correct justification/labelling 1 mark.)
22. p(2)=5·4−3·2+7 = 20−6+7 = 21 (1½). Since p(2)=21 ≠ 0, x = 2 is not a zero (½). OR 6x²+17x+5: split 17x = 15x+2x ⟹ 6x²+15x+2x+5 = 3x(2x+5)+1(2x+5) = (2x+5)(3x+1). (split 1, factor 1.)
23. a = 21, d = −3, aₙ = 21+(n−1)(−3) = 24−3n (1). First negative ⟹ 24−3n < 0 ⟹ n > 8 ⟹ n = 9. Check a₉ = 24−27 = −3 < 0, a₈ = 0. So the 9th term (= −3) is the first negative term (1).
27. p(1)=1−23+142−120 = 0 ✓, so (x−1) is a factor (1). Dividing: x³−23x²+142x−120 = (x−1)(x²−22x+120) (1). x²−22x+120 = (x−10)(x−12). So p(x) = (x−1)(x−10)(x−12) (1). OR Here a+b+c = 30−18−12 = 0, so a³+b³+c³ = 3abc (1). = 3·30·(−18)·(−12) (1) = 3·6480 = 19440 (1).
28. x+2y=6 ⟹ table e.g. (0,3), (6,0), (2,2) (1). Correct straight-line graph through these points (1). From graph: when x = 2, y = 2; when y = 0, x = 6 (1).
29. By the Baudhāyana rule (Pythagoras), diagonal = √(12²+5²) = √(144+25) = √169 = 13 units (2). Since 5²+12² = 13², (5,12,13) is a Pythagorean triple, so the triangle is right-angled — a valid "sacred" (right) triangle (1). OR In ▵ABD and ▵ACD: AB = AC (given), ∠BAD = ∠CAD (AD bisects ∠A), AD = AD (common) ⟹ ▵ABD ≅ ▵ACD by SAS (2). Hence BD = DC by CPCT (1).
30. Sides = 3k,4k,5k; perimeter 12k = 144 ⟹ k = 12 ⟹ sides 36, 48, 60 m (1). s = 144/2 = 72. Area = √[72(72−36)(72−48)(72−60)] = √[72·36·24·12] (1) = √746496 = 864 m² (1). (Note 3:4:5 is right-angled, ½·36·48 = 864 also accepted.)
33. Given circle with centre O, arc subtending ∠AOB at centre and ∠ACB at point C on the remaining circle. Join CO and produce to D. In ▵OAC, OA = OC (radii) ⟹ ∠OAC = ∠OCA; exterior ∠AOD = ∠OAC+∠OCA = 2∠OCA. Similarly ∠BOD = 2∠OCB. Adding, ∠AOB = 2∠ACB (correct figure + proof, 3½). Application: reflex ∠AOB isn't needed here — angle at centre = 130°, so angle at a point on the major arc = 130°/2 = 65° (1½).
34. Parallelogram ABCD, diagonals AC, BD meet at O. In ▵AOB and ▵COD: AB = CD (opp. sides), ∠OAB = ∠OCD and ∠OBA = ∠ODC (alternate angles, AB∥DC) ⟹ ▵AOB ≅ ▵COD (ASA) ⟹ OA = OC, OB = OD, i.e. diagonals bisect each other (3). Converse: if OA = OC and OB = OD, then ▵AOB ≅ ▵COD (SAS, vertically opposite angles) ⟹ AB = CD and AB∥CD ⟹ ABCD is a parallelogram (2). OR AE = ½AB = ½DC = FC and AE∥FC ⟹ AECF is a parallelogram ⟹ AF∥EC. In ▵DQC (Q on BD from AF), F is midpoint of DC and FP∥CQ ⟹ P is midpoint of DQ ⟹ DP = PQ (2½). Similarly in ▵ABP, E midpoint of AB and EQ∥AP ⟹ Q is midpoint of PB ⟹ PQ = QB (2½). Hence DP = PQ = QB, i.e. AF and CE trisect BD.