Mathematics Basic (241) – Solutions CBSE Board Exam 2026-27

MATHEMATICS BASIC (241) SAMPLE QUESTION PAPER

Class X (2026 – 27) – Detailed Solutions

Subject Code: 241 Maximum Marks: 80 Time Allowed: 3 Hours

SECTION A (MCQs & Assertion-Reason) 20 Marks

Q1. Real Numbers [1 Mark]

If HCF (2520, 6600) = 40 and LCM (2520, 6600) = 252 × k, then the value of k is:

(A) 165

(B) 1600

(D) 1650

Correct Option: (D) 1650

Formula Used:

HCF(a, b) × LCM(a, b) = a × b

Step-by-Step Substitution & Calculation:

Given: a = 2520, b = 6600, HCF = 40, LCM = 252 × k
40 × (252 × k) = 2520 × 6600
10080 × k = 16,632,000
k = 2520 × 660040 × 252
k = 10 × 660040 = 6600040 = 1650

Conceptual Rationale: The product of two numbers is equal to the product of their HCF and LCM. Substituting the given parameters yields k = 1650.

Q2. Polynomials [1 Mark]

If α and β are the zeroes of the polynomial 2x² − 4x − 5, the value of (α − β)² is:

(A) 4

(B) 6

(D) 56

Correct Option: (C) 14

Formula Used:

α + β = −ba, αβ = ca, and (α − β)² = (α + β)² − 4αβ

Step-by-Step Substitution & Calculation:

For polynomial 2x² − 4x − 5: a = 2, b = −4, c = −5
α + β = −(−4)2 = 2
αβ = −52
(α − β)² = (2)² − 4×(−52) = 4 + 10 = 14

Conceptual Rationale: Expressing (α − β)² in terms of the sum and product of roots gives (2)² + 10 = 14.

Q3. Pair of Linear Equations in Two Variables [1 Mark]

For what value of p does the pair of linear equations 4x + py + 8 = 0 and 2x + 2y + 2 = 0 have a unique solution?

(A) p = 4 only

(B) p = 2 only

(D) p ≠ 2

Correct Option: (C) p ≠ 4

Formula Used:

Condition for unique solution: a₁a₂ ≠ b₁b₂

Step-by-Step Substitution & Calculation:

Here a₁ = 4, b₁ = p; a₂ = 2, b₂ = 2
42 ≠ p2
2 ≠ p2 ⇒ p ≠ 4

Conceptual Rationale: For a unique solution, the ratio of x-coefficients must not equal the ratio of y-coefficients, giving p ≠ 4.

Q4. Pair of Linear Equations in Two Variables [1 Mark]

The sum of the numerator and denominator of a fraction is 11. If the denominator is increased by 1, the fraction becomes 12, then the fraction is:

(A) 29

(B) 38

(D) 56

Correct Option: (C) 47

Formula Used & System Formulation:

Let fraction be xy. Given: x + y = 11 and xy + 1 = 12

Step-by-Step Substitution:

2x = y + 1 ⇒ 2x − y = 1
Adding (x + y = 11) and (2x − y = 1): 3x = 12 ⇒ x = 4
y = 11 − 4 = 7
Required Fraction = 47

Conceptual Rationale: Solving the linear system yields numerator 4 and denominator 7, making the original fraction 4/7.

Q5. Quadratic Equations [1 Mark]

If one root of the quadratic equation ax² + bx + c = 0 is the reciprocal of the other, then:

(A) b = c

(B) a = b

(D) a = c

Correct Option: (D) a = c

Formula Used:

Product of roots of ax² + bx + c = 0 is ca

Step-by-Step Substitution:

Let roots be α and 1α
Product of roots = α × 1α = 1
ca = 1 ⇒ a = c

Conceptual Rationale: Since reciprocal roots multiply to 1, setting the product of roots c/a = 1 proves a = c.

Q6. Arithmetic Progressions [1 Mark]

The first term of an AP is p and the common difference is q, then its 10th term is:

(A) q + 10p

(B) p − 9q

(D) p + 10q

Correct Option: (C) p + 9q

Formula Used:

n-th term of an AP: a_n = a + (n − 1)d

Step-by-Step Substitution:

Given: a = p, d = q, n = 10
a₁₀ = p + (10 − 1)q = p + 9q

Conceptual Rationale: Substituting first term p and common difference q into the standard n-th term formula gives p + 9q.

Q7. Arithmetic Progressions [1 Mark]

Which term of the AP: 21, 42, 63, 84 … is 210?

(A) 9th

(B) 10th

(D) 12th

Correct Option: (B) 10th

Formula Used:

a_n = a + (n − 1)d

Step-by-Step Substitution:

a = 21, d = 42 − 21 = 21, a_n = 210
210 = 21 + (n − 1)21
189 = (n − 1)21 ⇒ n − 1 = 9 ⇒ n = 10

Conceptual Rationale: Solving 21 + (n − 1)×21 = 210 yields n = 10, making 210 the 10th term.

Q8. Coordinate Geometry [1 Mark]

If the point P(5, 2) divides the line segment joining A(8, 5) and B(4, y) in the ratio 3 : 1, then the value of y is:

(A) 4

(B) 3

(D) 1

Correct Option: (D) 1

Formula Used:

Section formula for y-coordinate: y_P = m₁y₂ + m₂y₁m₁ + m₂

Step-by-Step Substitution:

m₁ : m₂ = 3 : 1, y₁ = 5, y₂ = y, y_P = 2
2 = 3(y) + 1(5)3 + 1
2 = 3y + 54 ⇒ 8 = 3y + 5 ⇒ 3y = 3 ⇒ y = 1

Conceptual Rationale: Applying the section formula to the ordinate of point P yields y = 1.

Q9. Probability [1 Mark]

A letter from the word INDEPENDENCE is selected at random. What is the probability that the letter selected is a vowel which occurs the maximum number of times in the given word?

(A) 112

(B) 14

(D) 512

Correct Option: (C) 13

Formula Used:

P(E) = Favourable OutcomesTotal Outcomes

Step-by-Step Analysis:

Total letters in INDEPENDENCE = 12
Frequency of vowels: ‘I’ = 1, ‘E’ = 4
Maximum occurring vowel = ‘E’ (Count = 4)
P(E) = 412 = 13

Conceptual Rationale: ‘E’ is the most frequent vowel (4 occurrences out of 12 letters), giving probability 4/12 = 1/3.

Q10. Statistics [1 Mark]

For the following distribution, the lower limit of the median class is:

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

(A) 5

(B) 10

(D) 20

Correct Option: (B) 10

Cumulative Frequency Table:

0-5: f = 10, cf = 10
5-10: f = 15, cf = 25
10-15: f = 12, cf = 37
15-20: f = 20, cf = 57
20-25: f = 9, cf = 66

Step-by-Step Determination:

N = 66 ⇒ N2 = 33
Cumulative frequency just greater than 33 is 37 (Class 10-15).
Lower limit of median class 10-15 = 10

Conceptual Rationale: Since N/2 = 33 falls into the cumulative frequency bin of 37 (class 10-15), its lower limit is 10.

Q11. Circles [1 Mark]

The length of a tangent drawn from a point at a distance of 10 cm from the centre of the circle is 8 cm. The radius of the circle is:

(A) 4 cm

(B) 5 cm

(D) 7 cm

Correct Option: (C) 6 cm

Formula Used:

Pythagoras Theorem in right-angled triangle OTP: OP² = OT² + PT²

Step-by-Step Substitution:

Distance from centre (OP) = 10 cm, Tangent length (PT) = 8 cm
10² = r² + 8²
100 = r² + 64 ⇒ r² = 36 ⇒ r = 6 cm

Conceptual Rationale: Tangent is perpendicular to the radius at point of contact. r = √(10² − 8²) = 6 cm.

Q12. Statistics [1 Mark]

Mean and median of certain data are 32 and 30 respectively. Using empirical formula, the value of mode is:

(A) 36

(B) 26

(D) 20

Correct Option: (B) 26

Formula Used:

Empirical Relationship: Mode = 3 × Median − 2 × Mean

Step-by-Step Substitution:

Given: Median = 30, Mean = 32
Mode = 3(30) − 2(32) = 90 − 64 = 26

Conceptual Rationale: Direct substitution into the empirical formula yields Mode = 90 − 64 = 26.

Q13. Triangles / Heights & Distances [1 Mark]

A ladder 15 m long reaches a window 12 m above the ground. The distance of the foot of the ladder from the base of the wall is:

(A) 8 m

(B) 9 m

(D) 13 m

Correct Option: (B) 9 m

Formula Used:

Pythagoras Theorem: Hypotenuse² = Perpendicular² + Base²

Step-by-Step Substitution:

Ladder = 15 m, Height = 12 m
15² = 12² + Base² ⇒ Base² = 225 − 144 = 81 ⇒ Base = 9 m

Conceptual Rationale: In the right-angled triangle, base distance = √(225 − 144) = 9 m.

Q14. Introduction to Trigonometry [1 Mark]

The value of sin² 60° − 2 tan² 45° − cos² 30° is:

(A) −2

(B) −1

(D) 2

Correct Option: (A) −2

Trigonometric Values Used:

sin 60° = √32, tan 45° = 1, cos 30° = √32

Step-by-Step Substitution:

Expression = 34 − 2(1) − 34 = −2

Conceptual Rationale: Since sin 60° = cos 30°, their squares cancel, leaving −2.

Q15. Triangles [1 Mark]

The perimeter of two similar triangles is 28 cm and 35 cm respectively. If one side of the first triangle is 8 cm, then the corresponding side of the second triangle is:

(A) 10 cm

(B) 12 cm

(D) 16 cm

Correct Option: (A) 10 cm

Theorem Used:

Ratio of perimeters of two similar triangles = Ratio of their corresponding sides.

Step-by-Step Substitution:

2835 = 8x ⇒ 45 = 8x ⇒ 4x = 40 ⇒ x = 10 cm

Conceptual Rationale: Corresponding linear dimensions of similar triangles maintain identical ratios, giving 10 cm.

Q16. Triangles [1 Mark]

If in ΔABC and ΔPQR, ∠B = ∠Q, ∠R = ∠C and AB = 2PQ, then the two triangles are:

(A) Congruent but not similar

(B) Similar but not congruent

(D) Congruent as well as similar

Correct Option: (B) Similar but not congruent

Criteria Analysis:

By AA Similarity Criterion: Since ∠B = ∠Q and ∠C = ∠R, ΔABC ~ ΔPQR.
Side ratio = 2 ≠ 1, so sides are unequal. Hence, they are similar but not congruent.

Conceptual Rationale: Equal corresponding angles satisfy AA similarity, but the 2:1 side ratio violates congruence conditions.

Q17. Circles [1 Mark]

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠AOB is equal to:

(A) 60°

(B) 70°

(D) 100°

Correct Option: (D) 100°

Property Used:

∠AOB and ∠APB are supplementary (∠AOB + ∠APB = 180°).

Step-by-Step Substitution:

∠AOB + 80° = 180° ⇒ ∠AOB = 100°

Conceptual Rationale: In quadrilateral OAPB, opposite angles at centre and external point are supplementary (180° − 80° = 100°).

Q18. Surface Areas and Volumes [1 Mark]

Two cubes each of volume 64 cm³ are joined end to end to form a cuboid. The total surface area of the resulting cuboid is:

(A) 128 cm²

(B) 160 cm²

(D) 192 cm²

Correct Option: (B) 160 cm²

Step-by-Step Substitution:

Edge of cube a = (64)^1/3 = 4 cm
Cuboid Dimensions: l = 8 cm, b = 4 cm, h = 4 cm
TSA = 2(8×4 + 4×4 + 4×8) = 2(32 + 16 + 32) = 160 cm²

Conceptual Rationale: Combining two 4 cm cubes length-wise forms an 8×4×4 cm cuboid with surface area 160 cm².

Q19. Assertion-Reason (Probability) [1 Mark]

ASSERTION (A): The probability of getting number 8 on rolling a die is zero (0).

REASON (R): The probability of an impossible event is zero (0).

(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).

(D) (A) is false but (R) is true.

Correct Option: (A)

Conceptual Rationale: Rolling 8 on a single die is an impossible event, whose probability is 0 as stated in Reason (R).

Q20. Assertion-Reason (Triangles) [1 Mark]

ASSERTION (A): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

REASON (R): Line drawn from midpoint of one side of triangle parallel to another will bisect the third side.

(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).

(D) (A) is false but (R) is true.

Correct Option: (B)

Conceptual Rationale: Both statements are true geometric theorems, but Reason (R) is the Mid-point theorem, not the explanation for Converse of BPT in Assertion (A).

SECTION B (Very Short Answers) 10 Marks

Q21. Real Numbers [2 Marks]

Q21 (A): Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

Final Answer: 4663

Step-by-Step Calculation:

Prime factorization of 520 = 2³ × 5 × 13
Prime factorization of 468 = 2² × 3² × 13
LCM(520, 468) = 2³ × 3² × 5 × 13 = 4680
Required number = 4680 − 17 = 4663

CBSE Marking Scheme: Finding LCM (4680) [1M] | Subtracting 17 to get 4663 [1M]

Q21 (B): Show that (25)ⁿ will never end with digit zero for any natural number n.

Conclusion: (25)ⁿ never ends with digit 0

Proof:

(25)ⁿ = (5²)ⁿ = 5²ⁿ
By Fundamental Theorem of Arithmetic, prime factorization is unique. Since factor 2 is missing, it cannot end with 0.

Q22. Introduction to Trigonometry [2 Marks]

If 3 sin θ = 4 cos θ, find the value of sin θ + cos² θ.

Final Answer: 2925 (1.16)

Step-by-Step Evaluation:

tan θ = 4/3 ⇒ sin θ = 4/5, cos θ = 3/5
sin θ + cos² θ = 45 + (35)² = 45 + 925 = 2925

Q23. Polynomials [2 Marks]

If one zero of the quadratic polynomial 2x² − 3x + p is 3, find the value of p. Also, find the other zero.

Final Answer: p = −9 and Other Zero = −32

Calculation:

p(3) = 0 ⇒ 2(3)² − 3(3) + p = 0 ⇒ 18 − 9 + p = 0 ⇒ p = −9
Sum of zeroes = 3 + β = 3/2 ⇒ β = −3/2

Q24. Circles [2 Marks]

In the given figure, PA is a common tangent and QB and PC are tangents from Q and P. If QB = 5 cm and PC = 9 cm, evaluate length of PQ.

Final Answer: PQ = 4 cm (VI Variant: 8 cm)

QA = QB = 5 cm, and PA = PC = 9 cm (tangents from external point)
PQ = PA − QA = 9 − 5 = 4 cm

Q25. Coordinate Geometry [2 Marks]

Q25 (A): Points A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find values of a and b.

Final Answer: a = 6, b = 3

Midpoint AC = Midpoint BD ⇒ (3+a2, 1+b2) = (92, 2)
3 + a = 9 ⇒ a = 6 | 1 + b = 4 ⇒ b = 3

Q25 (B): Find a linear relation between x and y such that P(x, y) is equidistant from A(1, 4) and B(−1, 2).

Final Answer: x + y − 3 = 0

PA² = PB² ⇒ (x − 1)² + (y − 4)² = (x + 1)² + (y − 2)²
Simplifying gives 4x + 4y − 12 = 0 ⇒ x + y − 3 = 0

SECTION C (Short Answers) 18 Marks

Q26. Real Numbers [3 Marks]

Given that √5 is irrational, prove that 2 + 3√5 is irrational.

Proof Completed

Assume 2 + 3√5 = a/b (where a, b are co-prime integers, b ≠ 0).
3√5 = a − 2bb ⇒ √5 = a − 2b3b
Since a, b are integers, RHS is rational, which means √5 is rational.
This contradicts the fact that √5 is irrational. Hence, 2 + 3√5 is irrational.

Q27. Circles [3 Marks]

Q27 (A): Prove that AP = 12(AB + BC + CA).

Proof Completed

Tangents: AP = AR, BP = BQ, CQ = CR
Perimeter ΔABC = AB + BC + CA = AB + (BQ + QC) + CA = (AB + BP) + (CR + CA) = AP + AR = 2AP
Hence, AP = 12(AB + BC + CA).

Q27 (B): Prove that ∠APB = 2∠OAB.

Proof Completed

In ΔPAB, PA = PB ⇒ ∠PAB = 90° − θ/2 (where θ = ∠APB)
∠OAP = 90° ⇒ ∠OAB = 90° − (90° − θ/2) = θ/2 ⇒ ∠APB = 2∠OAB.

Q28. Coordinate Geometry [3 Marks]

Determine ratio in which (−6, y) divides segment joining A(−3, −1) and B(−8, 9). Also find y.

Final Answer: Ratio = 3 : 2 and y = 5

−6 = k(−8) − 3k + 1 ⇒ −6k − 6 = −8k − 3 ⇒ 2k = 3 ⇒ k = 3/2
y = (3/2)(9) + 1(−1)(3/2) + 1 = 25/25/2 = 5

Q29. Introduction to Trigonometry [3 Marks]

In right ΔACB, AB = 29, BC = 21, ∠ABC = θ. Find: (i) 1 + tan² θ, (ii) cos² θ − sin² θ.

Final Answers: (i) 841441 | (ii) 41841

AC = √(29² − 21²) = √400 = 20
(i) 1 + tan² θ = 1 + (20/21)² = 841/441
(ii) cos² θ − sin² θ = (21/29)² − (20/29)² = 41/841

Q30. Statistics [3 Marks]

Q30 (A): Mean of distribution is 48, total frequency is 50. Find missing frequencies x and y.

Final Answer: x = 12, y = 13

Equation 1: 25 + x + y = 50 ⇒ x + y = 25
Equation 2: Σf_ix_i / 50 = 48 ⇒ 9x + 13y = 277
Solving gives x = 12, y = 13

Q30 (B): Find the Modal weight of 50 students.

Final Answer: 60.56 kg

Modal class = 55 − 65 (f₁ = 20, f₀ = 10, f₂ = 12, h = 10, l = 55)
Mode = 55 + [20 − 1040 − 22] × 10 = 55 + 5.56 = 60.56 kg

Q31. Pair of Linear Equations [3 Marks]

The sum of digits of a two-digit number is 9. Nine times this number is twice the reversed number. Find the number.

Final Answer: 18

x + y = 9 and 9(10x + y) = 2(10y + x) ⇒ 8x − y = 0
Adding equations: 9x = 9 ⇒ x = 1, y = 8. Number = 18.

SECTION D (Long Answers – 5 Marks) 20 Marks

Q32. Triangles (Thales Theorem / BPT) [5 Marks]

Prove that if a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the sides in the same ratio.

Proof Completed

Proof Outline:

Area(ΔADE)/Area(ΔBDE) = (1/2)×AD×EN(1/2)×DB×EN = AD/DB
Area(ΔADE)/Area(ΔDEC) = (1/2)×AE×DM(1/2)×EC×DM = AE/EC
Since ΔBDE and ΔDEC lie on same base DE and between same parallels DE || BC, Area(ΔBDE) = Area(ΔDEC).
Therefore, AD/DB = AE/EC. Hence Proved!

Q33. Quadratic Equations [5 Marks]

Q33 (A): A train travels 360 km at uniform speed. If speed had been 5 km/h more, it would take 1 hour less. Find speed of train.

Final Answer: 40 km/h

360/x − 360/(x + 5) = 1 ⇒ x² + 5x − 1800 = 0
(x − 40)(x + 45) = 0 ⇒ Speed = 40 km/h (rejecting negative speed).

Q33 (B): John and Jivanti have 45 marbles. Both lost 5 marbles each, product of remaining is 124. Find initial marbles.

Final Answer: 36 and 9 marbles

(x − 5)(40 − x) = 124 ⇒ x² − 45x + 324 = 0
(x − 36)(x − 9) = 0 ⇒ Initial counts were 36 and 9.

Q34. Surface Areas and Volumes [5 Marks]

Hemisphere drilled out of wooden cube of side 21 cm. Find (i) volume of remaining wood, (ii) total surface area of solid.

Final Answers: (i) 6835.5 cm³ | (ii) 2992.5 cm²

Radius r = 21/2 = 10.5 cm.
(i) Volume = a³ − (2/3)πr³ = 9261 − 2425.5 = 6835.5 cm³
(ii) TSA = 6a² − πr² + 2πr² = 2646 + 346.5 = 2992.5 cm²

Q35. Heights & Distances [5 Marks]

Q35 (A): Drone observes ambulance with depression angles 30° and 60° (12 minutes later). Find total time taken.

Final Answer: 18 minutes

AP₁ = h√3, AP₂ = h/√3 ⇒ Distance P₁P₂ = 2h/√3 in 12 min.
Remaining distance AP₂ takes 6 min ⇒ Total Time = 12 + 6 = 18 minutes.

Q35 (B): Statue 1.6 m tall on pedestal. Elevation to top is 60°, to bottom is 45°. Find height of pedestal.

Final Answer: 0.8(√3 + 1) m (approx. 2.19 m)

tan 45° = h/x ⇒ x = h; tan 60° = (h + 1.6)/h ⇒ h(√3 − 1) = 1.6
h = 1.6/(√3 − 1) = 0.8(√3 + 1) m (≈ 2.19 m).

SECTION E (Case Study Competency Questions) 12 Marks

Q36. Case Study 1: Areas Related to Circles [4 Marks]

Hand fan sector boundary radius R = 10 cm, inner radius r = 4 cm, θ = 60°.

(i) Area of fan: (60/360)×(22/7)×100 = 1100/21 cm² (52.38 cm²) [1M]
(ii) Area of black sheet: (60/360)×(22/7)×(100 − 16) = 44 cm² [1M]
(iii)(A) Tape length: Outer arc + Inner arc + 2(R − r) = 44/3 + 12 = 80/3 cm (26.67 cm) [2M]
OR (iii)(B) Central angle for 29.6 cm tape: θ = 72° [2M]

Q37. Case Study 2: Arithmetic Progressions [4 Marks]

Tree plantation: Trees per section = 2n + 3 (2 sections per class). T_n = 4n + 6.

(i) Common difference: T₁ = 10, T₂ = 14 ⇒ d = 4 [1M]
(ii) Trees by Class VI: 4(6) + 6 = 30 trees [1M]
(iii)(A) Total trees (I to XII): S₁₂ = (12/2)(10 + 54) = 384 trees [2M]
OR (iii)(B) Class that planted 34 trees: 4n + 6 = 34 ⇒ Class VII [2M]

Q38. Case Study 3: Probability [4 Marks]

400 T-shirts: 312 Good, 54 Minor defects, 34 Major defects.

(i) P(Harish buys): 312/400 = 39/50 (0.78) [1M]
(ii) P(Trader buys): 366/400 = 183/200 (0.915) [1M]
(iii)(A) P(Not good): 88/400 = 11/50 (0.22) [2M]
OR (iii)(B) P(Neither buys): 34/400 = 17/200 (0.085) [2M]

Math Sample Paper Class 10 2027 With Solutions PDF Download.

MATHEMATICS BASIC (241) SAMPLE QUESTION PAPER

Class 10 Solutions Hub

Leave a Comment Cancel reply

Class 10 Solutions Hub

Leave a Comment Cancel reply

BestFunda

UP Board Exam 2027

Entrance Exams

Tools & Support