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CBSE Class 10 Mathematics Standard (041) Sample Paper 5 - 2026-27

CBSE Class 10 Mathematics Standard (041) Sample Paper 5 - 2026-27
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CBSE Sample Practice Paper — 5   Session 2026–27)
Class: X (10th)
Subject: Mathematics (Standard) — Code: 041
Time: 3 Hours
Max. Marks: 80
Student Name: ______________________ Roll No.: __________ Section: __________

General Instructions:

1. This Question Paper has 5 Sections A, B, C, D and E.

2. Section A comprises 20 MCQs (including 2 Assertion-Reason questions) of 1 mark each.

3. Section B comprises 5 Very Short Answer (VSA) questions of 2 marks each.

4. Section C comprises 6 Short Answer (SA) questions of 3 marks each.

5. Section D comprises 4 Long Answer (LA) questions of 5 marks each.

6. Section E comprises 3 Case-Study based questions of 4 marks each, with sub-parts of 1, 1 and 2 marks.

7. All questions are compulsory. Internal choices have been provided in 2 questions of Section B, 3 questions of Section C, 2 questions of Section D, and in one sub-part of each Case-Study question of Section E.

8. This paper is a High-Level / HOTS practice edition — questions emphasise multi-step reasoning and application. Draw neat figures wherever required. Take π = 22/7 wherever not stated otherwise.

9. Use of calculators is not permitted.

SECTION A  (1 Mark each) — Q1 to Q20
Q1. If a = 2³×3×5² and b = 2²×3³×5, then HCF(a, b) is: 1
(A) 60(B) 120(C) 30(D) 90
Q2. Which of the following is an irrational number? 1
(A) √16(B) 22/7(C) √12(D) 0.3333...
Q3. The decimal expansion of 13/3125 terminates after how many decimal places? 1
(A) 3(B) 4(C) 5(D) 6
Q4. If the sum of the zeroes of kx² + 2x + 3k equals their product, then k equals: 1
(A) −2/3(B) 2/3(C) −3/2(D) 3/2
Q5. The value of k for which x + 2y = 3 and 5x + ky + 7 = 0 have no solution is: 1
(A) 10(B) −10(C) 5(D) −5
Q6. If the equation x² − 2(k+1)x + k² = 0 has real and equal roots, then k equals: 1
(A) −1/2(B) 1/2(C) −2(D) 2
Q7. In an AP, if the mth term is n and the nth term is m (m ≠ n), then the (m+n)th term is: 1
(A) 0(B) m+n(C) mn(D) 1
Q8. The HCF of two numbers is 18 and their LCM is 720. If one number is 144, the other number is: 1
(A) 80(B) 90(C) 100(D) 110
Q9. If AD is a median of ΔABC, then by Apollonius' theorem, AB² + AC² equals: 1
(A) 2AD² + BC²/2(B) AD² + BC²(C) 2AD² − BC²/2(D) AD² + BC²/2
Q10. Two poles of height 6 m and 11 m stand on a plane ground, with their feet 12 m apart. The distance between their tops is: 1
(A) 12 m(B) 13 m(C) 14 m(D) 15 m
Q11. From an external point 13 cm from the centre of a circle of radius 5 cm, the length of a tangent drawn to the circle is: 1
(A) 10 cm(B) 12 cm(C) 14 cm(D) 8 cm
Q12. If 3cotA = 4, then (5sinA − 2cosA)/(5sinA + 2cosA) equals: 1
(A) 7/23(B) 23/7(C) 1/2(D) 2
Q13. sin⁶θ + cos⁶θ + 3sin²θcos²θ equals: 1
(A) 0(B) 1(C) 2(D) −1
Q14. A sphere of radius 6 cm is melted and recast into small spheres of radius 1 cm each. The number of small spheres formed is: 1
(A) 108(B) 216(C) 36(D) 72
Q15. Two dice are thrown together. The probability of getting a doublet of odd numbers is: 1
(A) 1/12(B) 1/6(C) 1/4(D) 1/3
Q16. If the mode of a distribution exceeds its mean by 12, then the mode exceeds the median by: 1
(A) 4(B) 6(C) 8(D) 10
Q17. If α, β are the zeroes of 2x² − 5x + 7, then 1/α + 1/β equals: 1
(A) 5/7(B) 7/5(C) 2/5(D) 5/2
Q18. If the ratio of the sum of the first m and n terms of an AP is m² : n², the ratio of its mth and nth terms is: 1
(A) (2m−1):(2n−1)(B) m:n(C) (m−1):(n−1)(D) (2m+1):(2n+1)
Q19. Assertion (A): If a pair of linear equations is consistent, the lines represented by them are either intersecting or coincident.
Reason (R): If a₁/a₂ ≠ b₁/b₂, the pair of linear equations is consistent. 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.
Q20. Assertion (A): Two concentric circles have radii 5 cm and 3 cm; the length of a chord of the larger circle that touches the smaller circle is 8 cm.
Reason (R): The perpendicular drawn from the centre of a circle to a chord bisects the chord. 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.
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SECTION B  (2 Marks each) — Q21 to Q25
Q21. Find the HCF and LCM of 96 and 404 by the prime factorisation method, and verify that HCF × LCM = product of the two numbers. 2
Q22. Find the ratio in which the point P(−1, y) divides the line segment joining A(−3, 10) and B(6, −8), and hence find the value of y. 2
— OR —
The point C(−1, 2) divides the line segment joining A(2, 5) and B(x, y) in the ratio 3 : 4. Find the coordinates of B. 2
Q23. Prove that: (cosecθ − sinθ)(secθ − cosθ) = 1/(tanθ + cotθ). 2
Q24. If tanA = n·tanB and sinA = m·sinB, prove that cos²A = (m² − 1)/(n² − 1). 2
— OR —
Evaluate: (5cos²60° + 4sec²30° − tan²45°) / (sin²30° + cos²30°). 2
Q25. A bag contains only red and white balls. The probability of drawing a red ball is 3/8. If the bag contains 25 white balls, find the total number of balls and the number of red balls. 2
SECTION C  (3 Marks each) — Q26 to Q31
Q26. Prove that √3 is an irrational number, and hence show that 5 + 2√3 is irrational. 3
Q27. Solve the pair of linear equations: 8x + 5y = 9 and 3x + 2y = 4. 3
— OR —
The sum of the digits of a two-digit number is 8. If 18 is added to the number, the digits get reversed. Find the number. 3
Q28. Find the sum of the first 40 positive integers divisible by 6. 3
Q29. In ΔABC, a line segment PQ intersects AB at P and AC at Q such that PQ ∥ BC and divides ΔABC into two parts of equal area. Find the ratio BP : AB. 3
Q30. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. 3
— OR —
PA and PB are tangents drawn from an external point P to a circle with centre O. If ∠OPA = 30°, find ∠AOB and ∠APB. 3
Q31. Find the mode of the following frequency distribution:
Class0–2020–4040–6060–8080–100
Frequency15618105
3
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SECTION D  (5 Marks each) — Q32 to Q35
Q32. A pole has to be erected at a point on the boundary of a circular park of diameter 20 m so that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 4 m. Find the distances of the pole from the two gates. 5
Q33. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Using this result: ΔABC ~ ΔDEF such that ar(ABC) = 64 cm² and ar(DEF) = 121 cm². If EF = 15.4 cm, find BC. 5
Q34. Two ships are sailing in the sea on either side of a lighthouse. The angles of depression of the two ships, as observed from the top of the lighthouse, are 60° and 45° respectively. If the distance between the ships is 200√3 m, find the height of the lighthouse. 5
— OR —
A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole, as observed from a point A on the ground, is 60°, and the angle of elevation of the top of the tower from A is 45°. Find the height of the tower. 5
Q35. A container, open at the top, is in the form of a frustum of a cone of height 24 cm, with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at ₹50 per litre, and the cost of the metal sheet used to make the container, if it costs ₹10 per 100 cm². (Use π = 3.14) 5
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SECTION E — Case Study Based Questions  (4 Marks each) — Q36 to Q38
Q36. Case Study — Coordinate Geometry. A town-planning map shows a quadrilateral plot with corners A(−5, 7), B(−4, −5), C(−1, −6) and D(4, 5), where 1 unit = 1 m.4
(i) Find the length of AB. (1)
(ii) Find the length of CD. (1)
(iii) Find the midpoint of diagonal BD. (2)
[OR for (iii): Find the area of quadrilateral ABCD.]
Q37. Case Study — Mensuration. A metallic sphere of radius 10.5 cm is melted and recast into small cones, each of radius 3.5 cm and height 3 cm.4
(i) Find the volume of the sphere, in cm³. (1)
(ii) Find the volume of one cone, in cm³. (1)
(iii) Find the number of cones formed. (2)
[OR for (iii): Find the total surface area of one cone, including its base.]
Q38. Case Study — Probability. In a game show, a wheel is numbered from 1 to 50. A player wins a prize if the spinner stops on a number that is a multiple of 4 or a multiple of 6.4
(i) Find the probability that the number is a multiple of 4. (1)
(ii) Find the probability that the number is a multiple of 6. (1)
(iii) Find the probability that the number is a multiple of 4 or a multiple of 6. (2)
[OR for (iii): Find the probability that the number is neither a multiple of 4 nor a multiple of 6.]
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Design of Question Paper — Blueprint & Analysis
1. Blueprint — Section-wise Design
SectionQuestion TypeNo. of QuestionsMarks per QuestionTotal Marks
AMCQ (incl. 2 Assertion–Reason)20120
BVery Short Answer (VSA)5210
CShort Answer (SA)6318
DLong Answer (LA)4520
ECase Study Based (1+1+2)3412
Total80
2. Chapter-wise / Unit-wise Marks Distribution
Unit / Chapter GroupSec ASec BSec CSec DSec ETotal
Number Systems323008
Algebra (Polynomials, Linear Eq., Quadratic Eq., AP)7060013
Coordinate Geometry020046
Geometry (Triangles, Circles)30610019
Trigonometry2405011
Mensuration1005410
Statistics & Probability4230413
Total201018201280
3. Difficulty Level Analysis (Bloom's Taxonomy) — High-Level Edition
Cognitive LevelDescriptionMarksPercentage
Remembering & UnderstandingRecall of facts, definitions, direct formula-based questions2430%
ApplyingApplication of concepts to solve standard problems2734%
Analysing, Evaluating & Creating (HOTS)Multi-step reasoning, proof-based, case-study and real-life application questions with combined concepts2936%
Total80100%

Note: This edition intentionally weights more marks toward the HOTS band than a standard-difficulty paper, to give stronger practice for board-level application questions.

4. Learning Outcome (LO) Mapping
UnitKey Learning Outcomes Assessed
Number SystemsApplies prime factorisation to verify the HCF–LCM–product relation; distinguishes rational from irrational numbers; analyses denominators for termination of decimal expansions.
AlgebraUses sum/product conditions to find unknown coefficients; solves simultaneous equations and digit-based word problems; derives AP terms from symmetric term conditions; sums an arithmetic sequence of multiples.
Coordinate GeometryApplies the section formula in both directions (finding ratio and finding an endpoint); computes lengths, midpoints and areas of quadrilaterals from coordinates.
GeometryApplies Apollonius' theorem and the Pythagoras theorem to real-life distance problems; proves and applies area-ratio and tangent-angle theorems for similar triangles and circles.
TrigonometryManipulates trigonometric identities and ratio-based expressions; evaluates compound trigonometric-ratio expressions; solves two-observer height-and-distance problems.
MensurationComputes volume and surface area of a frustum and applies unit-cost calculations; recasts a sphere into multiple smaller solids and finds the resulting count.
Statistics & ProbabilityFinds the mode of grouped data using the modal-class formula; relates mean, median and mode using the empirical relation; computes probability using the addition rule for combined events.
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Complete Answer Key & Step-wise Marking Scheme
Section A — Answer Key (1 mark each)
QAnsQAnsQAnsQAns
1(A) 606(A) −1/211(B) 12 cm16(C) 8
2(C) √127(A) 012(A) 7/2317(A) 5/7
3(C) 58(B) 9013(B) 118(A) (2m−1):(2n−1)
4(A) −2/39(A) 2AD²+BC²/214(B) 21619(B)
5(A) 1010(B) 13 m15(A) 1/1220(A)
Section B — Key Answers (2 marks each)
Q21. HCF = 4, LCM = 9696; verification: 4 × 9696 = 38,784 = 96 × 404 ✓.
Q22. Ratio = 2 : 7; y = 6. OR: B = (−5, −2).
Q23. Both sides simplify to sinθcosθ (proved).
Q24. Standard algebraic proof using tanA=n·tanB and sinA=m·sinB (proved). OR: Value = 67/12.
Q25. Total balls = 40; number of red balls = 15.
Section C — Key Answers (3 marks each)
Q26. Standard proof by contradiction for √3; hence 5+2√3 is irrational (else √3 would be forced rational).
Q27. x = −2, y = 5. OR: The number is 35.
Q28. First term 6, d = 6, n = 40 ⇒ Sum = 20 × 246 = 4920.
Q29. Since ar(APQ) = ½·ar(ABC), (AP/AB)² = 1/2 ⇒ AP/AB = 1/√2 ⇒ BP/AB = 1 − 1/√2 = (2−√2)/2.
Q30. Standard proof using the cyclic quadrilateral OAPB (angles sum to 360°, two right angles at A and B). OR: ∠AOB = 120°, ∠APB = 60°.
Q31. Modal class = 40–60; Mode = 40 + [(18−6)/(36−6−10)]×20 = 40+12 = 52.
Section D — Key Answers (5 marks each)
Q32. Using AP²+BP²=AB² (angle in semicircle=90°) and AP−BP=4, solving gives distances = 16 m and 12 m.
Q33. Standard area-ratio proof using equal heights of similar triangles. Numerical part: BC/EF = 8/11 ⇒ BC = 15.4×8/11 = 11.2 cm.
Q34. Height of lighthouse = 300(√3−1) m ≈ 219.6 m. OR: Height of tower ≈ 6.83 m (solving with tan45° and tan60°).
Q35. Volume ≈ 15,674.9 cm³ (≈15.67 litres); cost of milk ≈ ₹783.74. Slant height = √720 ≈ 26.83 cm; total metal area ≈ 2559.9 cm²; cost of sheet ≈ ₹256.
Section E — Key Answers (Case Studies, 4 marks each)
Q36. (i) AB = √145 m; (ii) CD = √146 m; (iii) Midpoint of BD = (0, 0). OR: Area of ABCD = 72 sq. m.
Q37. (i) Volume of sphere ≈ 4851 cm³; (ii) Volume of one cone = 38.5 cm³; (iii) Number of cones = 4851/38.5 = 126. OR: Total surface area of one cone ≈ 89.2 cm².
Q38. (i) P(multiple of 4) = 12/50 = 6/25; (ii) P(multiple of 6) = 8/50 = 4/25; (iii) P(multiple of 4 or 6) = (12+8−4)/50 = 16/50 = 8/25. OR: P(neither) = 1 − 8/25 = 17/25.
Step-wise Marking Scheme — General Guidelines
• Section A (MCQ/AR): 1 mark for the correct option only; no partial credit.
• Section B/C/D (VSA/SA/LA): Marks are awarded step-wise — correct method/formula (partial marks even if the final numeric answer is wrong due to a minor computational slip), correct substitution of values, and correct final answer with unit. Alternative correct methods must be given full credit.
• Proof-based questions (e.g. Q26, Q30, Q33): Marks are distributed across statement of the theorem/given-to-prove, construction (if any), logical steps of the proof, and the concluding statement.
• Section E (Case Study): Each sub-part is marked independently as per its allotted marks (1, 1, 2); no negative marking for an incorrect sub-part affecting others.
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