Post Date: February 17, 2026

Welcome to the fully solved CBSE Class 10 Maths board paper 2026. Whether you are double-checking your exam performance or practicing for next year, our comprehensive guide has you covered. We have broken down every single question to provide clear, step-by-step CBSE math exam answers. Prepared by expert educators, these Class 10 board exam marking scheme solutions show you exactly how to structure your responses, apply the right formulas, and secure full marks on every problem. Section-wise solution of Class 10 Maths Exam Paper Set 1, 2 and 3 are given below:

Section A: Multiple Choice Questions (Q 1-20)
Section B: Very Short Answer Questions (Q 21-25)
Section C: Short Answer Questions (Q 26-30)
Section D: Long Answer Questions (Q 32-35)
Section E: Case Study Based Question (Q 36-38)

Class 10 Maths CBSE Exam Paper 2026 – Section A

Q. Numbers 1 to 20 are multiple choices questions of 1 marks each.
1. Devansh proved that △ABC ~ △PQR using SAS similarity criteria. If he found ∠C = ∠R, then which of the following was proved true?
(A) AC/AB = PR/PQ
(B) BC/AC = PR/QR
(C) AC/BC = PR/PQ
(D) AC/BC = PR/QR
Solution:
Since △ABC ~ △PQR (in the same order), the corresponding vertices are:
A ↔ P
B ↔ Q
C ↔ R
Given that ∠C = ∠R, side pairs corresponding to these angles are:
AC ↔ PR
BC ↔ QR
In similar triangles, ratios of corresponding sides are equal. So,
AC/BC = PR/QR
Correct Answer: (D) AC/BC = PR/QR

2. In the given figure, PQ is tangent to the circle with centre O. S is a point on the circle such that ∠SQT = 55°. The m ∠QPS is

Class 10 Maths CBSE Exam 2026 Question 2

(A) 55°
(B) 20°
(C) 35°
(D) 70°
Solution:
Since PQ is tangent at Q, the rays QP and QT form a straight line.
Given ∠SQT = 55°.
Therefore, ∠PQS = 180° − 55° = 125° (because PQ and QT are opposite rays)
By the tangent–chord theorem, the angle between tangent QT and chord QS equals the angle in the opposite arc.
So, ∠QAS = 55°.
Since AS is a diameter, ∠AQS = 90°.
Now in triangle AQS:
∠QSA = 180° − (90° + 55°)
∠QSA = 35°.
Since P, A, S are collinear, ∠QSP = 35°.
Now in triangle PQS:
∠QPS = 180° − (125° + 35°)
∠QPS = 20°.
Correct answer: (B) 20°.

3. In an A.P., if a₁₄ – a₈ = 24, then the common difference of the A.P. is
(A) 6
(B) 4
(C) ±4
(D) 3
Solution:
a₁₄ = a + 13d
a₈ = a + 7d
Given: a₁₄ − a₈ = 24
So, (a + 13d) − (a + 7d) = 24
⇒ a + 13d − a − 7d = 24
⇒ 6d = 24
⇒ d = 4
Correct answer: (B) 4

4. The value of p for which roots of the quadratic equation x² – px + 6 = 0 are rational, is
(A) 1
(B) –5
(C) 25
(D) √5
Solution:
For the quadratic equation x² − px + 6 = 0
The roots will be rational if the discriminant (D) is a perfect square.
Discriminant, D = b² − 4ac
Here, a = 1, b = −p, c = 6
So, D = (−p)² − 4(1)(6)
D = p² − 24
For rational roots, p² − 24 must be a perfect square.
Check the options:
(A) p = 1, so D = 1 − 24 = −23 (not possible)
(B) p = −5, so D = 25 − 24 = 1 (perfect square)
(C) p = 25, so D = 625 − 24 = 601 (not a perfect square)
(D) p = √5, so D = 5 − 24 = −19 (not possible)
Correct answer: (B) −5

5. In the given figure, PQ ∥ YZ such that XP : PY = 2 : 3. If PQ = 5 cm, then YZ equals

Class 10 Maths CBSE Exam 2026 Question 5

(A) 12.5 cm
(B) 10 cm
(C) 15 cm
(D) 7.5 cm
Solution:
By Basic Proportionality Theorem (BPT):
Since PQ ∥ YZ, △XPQ ~ △XYZ
XP/XY = PQ/YZ
XP/XY = XP/(XP + PY) = 2/(2+3) = 2/5
∴ 2/5 = 5/YZ
YZ = 25/2 = 12.5 cm
Option (A) 12.5 cm is correct.

6. For an acute angle θ, if cos θ = 1/8, then (8 sec θ + 1)/(8 sec θ – 1) equals
(A) 64/63
(B) 0
(C) 65/63
(D) 1
Solution:
cosθ = 1/8 → secθ = 8
8 secθ = 8 × 8 = 64
So, (8 sec θ + 1)/(8 sec θ – 1) = (64 + 1)/(64 – 1) = 65/63
Option (C) 65/63 is correct.

7. A card is drawn at random from a well shuffled deck of 52 playing cards. The probability that it is either a ten or a king is
(A) 1/26
(B) 2/13
(C) 1/13
(D) 8/26
Solution:
Number of tens = 4
Number of kings = 4
Total favourable = 4 + 4 = 8
P = 8/52 = 2/13
Option (B) 2/13 is correct.

8. The line segment joining the points P(–4, –2) and Q(10, 4) is divided by y-axis in the ratio
(A) 2 : 5
(B) 1 : 2
(C) 2 : 1
(D) 5 : 2
Solution:
On y-axis, x = 0
By section formula, if ratio = m:n
x = (mx₂ + nx₁)/(m + n)
⇒ 0 = (m×10 + n×(–4))/(m+n)
⇒ 10m – 4n = 0
⇒ 10m = 4n
⇒ m/n = 4/10 = 2/5
Option (A) 2:5 is correct.

9. Simplest form of sec A / √(sec²A – 1) is
(A) sin A
(B) tan A
(C) cosec A
(D) cos A
Solution:
We know: sec²A – 1 = tan²A
= secA/√(tan²A)
= secA/tanA
= (1/cosA)/(sinA/cosA)
= (1/cosA) × (cosA/sinA)
= 1/sinA
= cosecA
Option (C) cosec A is correct.

10. A wire is attached from a point A on the ground to the top of a pole BC, making an angle of elevation as 60°. If AB = 5√3 m, then length of the wire is

Class 10 Maths CBSE Exam 2026 Question 10

(A) 10 m
(B) 10√3 m
(C) 15 m
(D) 5/2 √3 m
Solution:
AB = base = 5√3 m AC = wire (hypotenuse) ∠A = 60°
cos60° = AB/AC
1/2 = 5√3/AC
AC = 10√3 m
Option (B) 10√3 m is correct.

11. If sum and product of zeroes of a polynomial are (–3) and (–2) respectively, then a polynomial is
(A) x² – 3x – 2
(B) –x² – 3x + 2
(C) –x² + 3x – 2
(D) x² + 3x + 2
Solution:
Sum of zeroes = −3
Product of zeroes = −2
So, the quadratic equation: x² − (Sum of roots)x + Product of roots
⇒ x² + 3x – 2.
⇒ -(−x² − 3x + 2)
⇒ k(−x² − 3x + 2)
Therefore, Correct answer: (B) −x² − 3x + 2.

12. Meena calculates that the probability of her winning the first prize in a lottery is 0.08. If total 800 tickets were sold, the number of tickets bought by her, is
(A) 64
(B) 640
(C) 100
(D) 10
Solution:
Number of tickets = P × Total = 0.08 × 800 = 64
Option (A) 64 is correct.

13. A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use π = 3.14)

Class 10 Maths CBSE Exam 2026 Question 13

(A) 314√2 cm²
(B) 314 cm²
(C) 3140/3 cm²
(D) 3140√2 cm²
Solution:
For maximum volume cone inside hemisphere:
Cone radius = r = 10 cm
Cone height = h = 10 cm (= radius of hemisphere)
Slant height of cone: l = √(r² + h²) = √(100 + 100) = √200 = 10√2 cm
Curved Surface Area of cone = πrl = 3.14 × 10 × 10√2 = 314√2 cm²
Option (A) 314√2 cm² is correct.

14. While calculating mean of a grouped frequency distribution, step deviation method was used ((x – a)/h = u). It was found that x̄ = 64, h = 5 and a = 62.5. The value of ū is
(A) 0.5
(B) 1.5
(C) 0.3
(D) 7.5
Solution:
Formula: x̄ = a + ū × h
64 = 62.5 + ū × 5
1.5 = ū × 5
ū = 1.5/5 = 0.3
Option (C) 0.3 is correct.

15. The area of a sector of a circle of radius 10 cm is 55/3 cm². The value of central angle is
(A) 21°/2
(B) 42°
(C) 105°
(D) 21°
Solution:
Area of sector = (θ/360°) × πr²
55/3 = (θ/360) × (22/7) × 100
55/3 = (θ/360) × 2200/7
55/3 × 7/2200 × 360 = θ
θ = (55 × 7 × 360)/(3 × 2200)
θ = (138600)/(6600)
θ = 21°
Option (D) 21° is correct.

16. A camping tent in hemispherical shape of radius 1.4 m, has a door opening of area 0.50 m². Outer surface area of the tent is
(A) 11.78 m²
(B) 12.32 m²
(C) 11.82 m²
(D) 12.86 m²
Solution:
Curved Surface Area of hemisphere = 2πr²
= 2 × 22/7 × 1.4 × 1.4
= 2 × 22/7 × 1.96
= 2 × 6.16
= 12.32 m²
Outer surface area = CSA – door area = 12.32 – 0.50 = 11.82 m²
Option (C) 11.82 m² is correct.

17. Which of the following can not be the probability of an event?
(A) 39/100
(B) 0.001/20
(C) 10/0.2
(D) 10%
Solution:
Probability must be between 0 and 1.
(A) 39/100 = 0.39 (Valid)
(B) 0.001/20 = 0.00005 (Valid)
(C) 10/0.2 = 50 (Greater than 1, NOT valid)
(D) 10% = 0.10 (Valid)
Option (C) 10/0.2 is correct.

18. The value of k for which the equation kx² – 6x – 4 = 0 has real and equal roots, is
(A) 9/4
(B) –4
(C) –9/4
(D) –2
Solution:
For real and equal roots: D = 0
⇒ b² – 4ac = 0
⇒ (–6)² – 4(k)(–4) = 0
⇒ 36 + 16k = 0
⇒ 16k = –36
⇒ k = –36/16 = –9/4
Option (C) –9/4 is correct.

(Assertion and Reason Based Questions)

Directions: Questions number 19 and 20 are Assertion and Reason based questions. Two statements are given, one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below :
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

19. Assertion (A) : The system of linear equations 3x – 5y + 7 = 0 and –6x + 10y + 14 = 0 is inconsistent.
Reason (R) : When two linear equations don’t have unique solution, they always represent parallel lines.
Solution:
Assertion (A): 3x – 5y + 7 = 0 and –6x + 10y + 14 = 0
Check: a₁/a₂ = 3/–6 = –1/2
b₁/b₂ = –5/10 = –1/2
c₁/c₂ = 7/14 = 1/2
Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂
→ Lines are parallel → inconsistent, So A is TRUE.
Reason (R): When two equations don’t have unique solution, they always represent parallel lines.
This is FALSE — they could also be coincident lines (infinite solutions), which is also not unique but not parallel.
Option (C) Assertion (A) is true, but Reason (R) is false is correct.

20. Assertion (A) : H.C.F. (36 m², 18 m) = 18 m, where m is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.
Solution: Assertion (A): HCF(36m², 18m) = 18m, where m is prime
36m² = 2² × 3² × m²
18m = 2 × 3² × m
HCF = 2 × 3² × m = 18m ✓ A is TRUE
Reason (R): HCF of two numbers is always less than or equal to the smaller number.
Smaller number = 18m
HCF = 18m ≤ 18m ✓ R is TRUE
And R correctly explains why HCF = 18m (it equals the smaller number here).
Option (A) Both A and R are true and R is the correct explanation of A is correct.

Class 10 Maths CBSE Exam Paper 2026 – Section B

Q. Numbers 21 to 25 are very short answer questions of 2 marks each.
21. In the given figure, AB ∥ DE and AC ∥ DF. Show that △ABC ~ △DEF. If BC = 10 cm, EB = CF = 5 cm and AB = 7 cm, then find the length DE.

Class 10 Maths CBSE Exam 2026 Question 21

Solution:
Given: AB ∥ DE, AC ∥ DF, BC = 10 cm, EB = CF = 5 cm, AB = 7 cm
BE = CF = 5 cm, BC = 10 cm
∴ BE/BC = CF/BC → BE = CF (given)
Also, BE + BC = EC → EC = 15 cm
Similarly, BF = BC + CF = 15 cm
In △ABC and △DEF:
AB ∥ DE → ∠ABE = ∠DEB (alternate angles)
AC ∥ DF → ∠ACF = ∠DFC (alternate angles)
Since AB ∥ DE and AC ∥ DF:
∠BAC = ∠EDF (corresponding angles)
Also: AB/DE = BC/EF = AC/DF
Now, BE/BC = 5/10 = 1/2
∴ DE/AB = EC/BC = 15/10 = 3/2
∴ DE = AB × 3/2 = 7 × 3/2 = DE = 10.5 cm

22. (a) For acute angles A and B, if sec (2A – B) = √2 and cosec (A + B) = 2, then find the values of A and B.
Solution:
Given: sec(2A – B) = √2 and cosec(A + B) = 2
sec(2A – B) = √2 = sec 45° ∴ 2A – B = 45° …(i)
cosec(A + B) = 2 = cosec 30° ∴ A + B = 30° …(ii)
Adding (i) and (ii): 3A = 75° → A = 25°
From (ii): B = 30° – 25° = B = 5°

22. (b) Evaluate : (2 cos 30° – cot³ 60°) / tan 30°
Solution:
Known values:
cos 30° = √3/2
cot 60° = 1/√3
tan 30° = 1/√3
Numerator: = 2 × (√3/2) – (1/√3)³ = √3 – 1/(3√3) = √3 – 1/(3√3) = (3√3 × √3 – 1) / (3√3) = (9 – 1) / (3√3) = 8/(3√3)
Denominator: 1/√3
= [8/(3√3)] ÷ [1/√3] = 8/(3√3) × √3 = 8/3
∴ Answer = 8/3

23. Prove that 4 – 2√5 is an irrational number given that √5 is irrational.
Solution:
Proof (by contradiction):
Assume 4 – 2√5 is rational.
Then 4 – 2√5 = p/q, where p, q are integers, q ≠ 0
⇒ 2√5 = 4 – p/q
⇒ √5 = (4q – p) / 2q
Since p and q are integers, (4q – p)/2q is rational
∴ √5 is rational — Contradiction!
Since √5 is given to be irrational, our assumption is wrong.
∴ 4 – 2√5 is irrational.

24. A bag contains 25 balls. Some of them are yellow and others are green. One ball is drawn at random. If probability of getting a green ball is 3/5, then find the number of yellow balls.
Solution:
Given: Total balls = 25, P(green) = 3/5
Number of green balls = 3/5 × 25 = 15
Number of yellow balls = 25 – 15 = 10

25. (a) Vertices of a right triangle ABC with ∠B = 90° are A(3, 4), B(1, 1) and C(–8, 7). Find the value of tan A.
Solution:
Given: ∠B = 90°, A(3,4), B(1,1), C(–8,7)
BC = √[(–8–1)² + (7–1)²] = √[81 + 36] = √117 = 3√13
AB = √[(3–1)² + (4–1)²] = √[4 + 9] = √13
AC = √[(3–(–8))² + (4–7)²] = √[121 + 9] = √130
In right △ABC, ∠B = 90°
tan A = BC/AB = 3√13/√13 = 3
∴ tan A = 3

25. (b) Using distance formula, prove that the points A(2, 3), B(–7, 0) and C(–1, 2) are collinear.
Solution:
Prove A(2,3), B(–7,0), C(–1,2) are collinear
AB = √[(–7–2)² + (0–3)²] = √[81 + 9] = √90 = 3√10
BC = √[(–1–(–7))² + (2–0)²] = √[36 + 4] = √40 = 2√10
AC = √[(–1–2)² + (2–3)²] = √[9 + 1] = √10
Check: BC + AC = 2√10 + √10 = 3√10 = AB
Since AB = BC + AC, points A, B, C are collinear.

Class 10 Maths CBSE Exam Paper 2026 – Section C

Q. Numbers 26 to 31 are short answer questions of 3 marks each:
26. (a) A circle centered at (2, 1) passes through the points A(5, 6) and B(–3, K). Find the value(s) of K. Hence find length of chord AB.
Solution:
Centre = (2, 1), passes through A(5, 6) and B(–3, K)
Radius using A: r² = (5–2)² + (6–1)² = 9 + 25 = 34
Using B: r² = (–3–2)² + (K–1)² = 25 + (K–1)²
∴ 25 + (K–1)² = 34 (K–1)² = 9 K – 1 = ±3 K = 4 or K = –2
Length of chord AB (using K = 4): B = (–3, 4) AB = √[(5–(–3))² + (6–4)²] = √[64 + 4] = √68 = 2√17 cm

(b) Prove that the point P dividing the line segment joining the points A(–1, 7) and B(4, –3) in the ratio 3 : 2, lies on the line x – 3y = –1. Also find length of PA and PB.
Solution:
P divides A(–1,7) and B(4,–3) in ratio 3:2
By section formula: P = [(3×4 + 2×(–1))/(3+2), (3×(–3) + 2×7)/(3+2)] P = [(12–2)/5, (–9+14)/5] P = [10/5, 5/5] P = (2, 1)
Check on x – 3y = –1: 2 – 3(1) = 2 – 3 = –1 ✓ P lies on the line.
PA: = √[(2–(–1))² + (1–7)²] = √[9 + 36] = √45 = 3√5 units
PB: = √[(2–4)² + (1–(–3))²] = √[4 + 16] = √20 = 2√5 units

27. The dimensions of a window are 156 cm × 216 cm. Arjun wants to put grill on the window creating complete squares of maximum size. Determine the side length of the square and hence find the number of squares formed.
Solution:
Given: Window = 156 cm × 216 cm
Side of largest square = HCF(156, 216)
Finding HCF by Euclid’s division: 216 = 1 × 156 + 60 156 = 2 × 60 + 36 60 = 1 × 36 + 24 36 = 1 × 24 + 12 24 = 2 × 12 + 0
HCF = 12 cm (side of each square)
Number of squares = (156/12) × (216/12) = 13 × 18 = 234 squares

28. Use graphical method to solve the system of linear equations : y = –3 and x + 2y = 4.
Solution:
Equation 1: y = –3 (horizontal line)
x: 0 |2 | 4
y: –3|–3|–3
Equation 2: x + 2y = 4 → x = 4 – 2y
x: 4|0|–2
y: 0|2| 3
Substituting y = –3 in x + 2y = 4: x + 2(–3) = 4 x – 6 = 4 x = 10
∴ Answer: x = 10, y = –3 → Point (10, –3)

29. (a) In an A.P., 15th term exceeds the 8th term by 21. If sum of first 10 terms is 55, then form the A.P.
Solution:
Given: a₁₅ – a₈ = 21, S₁₀ = 55
a₁₅ – a₈ = (a + 14d) – (a + 7d) = 7d = 21 ∴ d = 3
S₁₀ = 10/2 × [2a + 9d] = 55 5[2a + 27] = 55 2a + 27 = 11 2a = –16 a = –8
∴ A.P. = –8, –5, –2, 1, 4, 7, …

(b) The sum of first n terms of an A.P. is 2n² + 13n. Find its nᵗʰ term and hence 10th term.
Solution:
Given: Sₙ = 2n² + 13n
nᵗʰ term: aₙ = Sₙ – Sₙ₋₁ = [2n² + 13n] – [2(n–1)² + 13(n–1)] = 2n² + 13n – [2n² – 4n + 2 + 13n – 13] = 2n² + 13n – 2n² + 4n – 2 – 13n + 13 = aₙ = 4n + 11
10th term: a₁₀ = 4(10) + 11 = 40 + 11 = 51.

30. A circle of diameter 20 cm is equally divided into five sectors. Find the area and perimeter of one of the sectors.
Solution:
Given: Diameter = 20 cm → r = 10 cm, 5 equal sectors
Each sector angle = 360°/5 = 72°
Area of one sector: = (θ/360°) × πr² = (72/360) × π × 100 = (1/5) × 100π = 20π = 62.8 cm²
Perimeter of one sector: = 2r + arc length = 2(10) + (72/360) × 2π × 10 = 20 + (1/5) × 20π = 20 + 4π = 20 + 12.56 = 32.56 cm

31. Prove that :
(1 + cot θ – cosec θ) (1 + tan θ + sec θ) = 2
Solution:
LHS: = (1 + cosθ/sinθ – 1/sinθ)(1 + sinθ/cosθ + 1/cosθ)
= [(sinθ + cosθ – 1)/sinθ] × [(cosθ + sinθ + 1)/cosθ]
= [(sinθ + cosθ – 1)(sinθ + cosθ + 1)] / (sinθ · cosθ)
= [(sinθ + cosθ)² – 1²] / (sinθ · cosθ)
= [sin²θ + 2sinθcosθ + cos²θ – 1] / (sinθ · cosθ)
= [1 + 2sinθcosθ – 1] / (sinθ · cosθ)
= [2sinθcosθ] / (sinθ · cosθ)
= 2 = RHS

Class 10 Maths CBSE Exam Paper 2026 – Section D

Q. numbers 32 to 35 are long answer questions of 5 marks each.
32. By selling an article for ₹48, a trader loses as much percent as half of the cost price of the article. Calculate the cost price and loss amount of the article.
Solution:
Let cost price = ₹x
Loss% = x/2 % (half of cost price as percentage)
Loss amount = (x/2)/100 × x = x²/200
SP = CP – Loss
⇒ 48 = x – x²/200
⇒ x² – 200x + 9600 = 0
⇒ x² – 120x – 80x + 9600 = 0
⇒ x(x – 120) – 80(x – 120) = 0
⇒ (x – 80)(x – 120) = 0
⇒ x = 80 or x = 120
If x = 80: Loss% = 40%
⇒ Loss = 32 and SP = 48
If x = 120: Loss% = 60%
⇒ Loss = 72 and SP = 48
Therefore, Cost Price = ₹80 and Loss = ₹32
OR
Cost Price = ₹120, Loss = ₹72.

33. PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

Class 10 Maths CBSE Exam 2026 Question 33

Solution:
Given: Circle with centre O, Radius (r) = OQ = OR = OC = 5 cm
OP = 13 cm
PQ and PR are tangents from external point P to the circle.
AB is a tangent to the circle at point C, where C lies on OP
Since PQ is a tangent and OQ is radius:
∠OQP = 90° (Radius ⊥ Tangent)
In right triangle OQP:
PQ² = OP² − OQ²
⇒ PQ² = 13² − 5²
⇒ PQ² = 169 − 25
⇒ PQ² = 144
⇒ PQ = √144 = 12 cm
Since C lies on OP and OC = radius = 5 cm:
⇒ PC = OP − OC
⇒ PC = 13 − 5
⇒ PC = 8 cm
By the Equal Tangents Theorem (tangents from same external point are equal):
From point A: AQ = AC …(1)
From point B: BR = BC …(2)
From point P: PQ = PR = 12 cm …(3)
Now:
PA + AC = PQ [Since AC = AQ]
⇒ PA + AC = 12 …(i)
Similarly PB + BC = PR
⇒ PB + BC = 12 …(ii)
By symmetry of the figure: PA = PB and AC = BC
So, AB = AC + BC = 2 × AC
Using Pythagoras in right triangle PCB (OC ⊥ AB, so ∠PCB = 90°):
PB² = PC² + BC²
⇒ (12 − BC)² = 8² + BC² [Since PB = 12 − BC from eq.(ii)]
⇒ 144 − 24·BC + BC² = 64 + BC²
⇒ 144 − 24·BC = 64
⇒ 24·BC = 144 − 64 = 80
⇒ BC = 80 ÷ 24 = 10/3 cm
So, AC = BC = 10/3 cm (by symmetry)
AB = AC + BC = 10/3 + 10/3 = 20/3 cm
PA = PQ − AQ = PQ − AC = 12 − 10/3 = 36/3 − 10/3 = 26/3 cm
AB = 20/3 cm ≈ 6.67 cm PA = 26/3 cm ≈ 8.67 cm.

34. (a) D is the mid-point of side BC of △ABC. CE and BF intersect at O, a point on AD. AD is produced to G such that OD = DG. Prove that

Class 10 Maths CBSE Exam 2026 Question 34-Part a

(i) OBGC is a parallelogram.
Solution:
Given: D is midpoint of BC, OD = DG
(i) D is midpoint of BC and also mid point of OG.
So, OD = DG and BD = DC.
⇒ Diagonals of OBGC bisect each other at D.
∴ OBGC is a parallelogram.

(ii) EF ∥ BC
Solution:
Since OBGC is a parallelogram [From part (i)]:
GC || OB (which means GC || BF)
BG || OC (which means BG || CE)
Now, consider Triangle ABG:
Since OC || BG (and E lies on OC), we have OE || BG.
By Basic Proportionality Theorem (BPT) in Triangle ABG:
AE / EB = AO / OG — (1)
Now, consider Triangle ACG:
Since OB || GC (and F lies on OB), we have OF || GC.
By Basic Proportionality Theorem (BPT) in Triangle ACG:
AF / FC = AO / OG — (2)
From Equation 1 and Equation 2, we get: AE / EB = AF / FC
In Triangle ABC, since the line EF divides sides AB and AC in the same ratio:
By Converse of BPT: EF || BC
(Hence Proved)

(iii) △AEF ~ △ABC
Solution:
In Triangle AEF and Triangle ABC:
1. Angle A = Angle A (Common angle)
2. Angle AEF = Angle ABC (Corresponding angles, since EF || BC)
3. Angle AFE = Angle ACB (Corresponding angles, since EF || BC)
By AA (Angle-Angle) Similarity Criterion: △AEF ~ △ABC.

OR
(b) Through the mid-point Q of side CD of a parallelogram ABCD, the line AR is drawn which intersects BD at P and produced BC at R. Prove that
(i) AQ = QR
(ii) AP = 2PQ
(iii) PR = 2AP

Class 10 Maths CBSE Exam 2026 Question 34-Part b

Solution:
Given: ABCD parallelogram, Q is midpoint of CD, AR intersects BD at P, BC produced at R.
(i) AQ = QR:
In △DQA and △CQR:
DQ = QC (Q is midpoint of DC)
∠DQA = ∠CQR (vertically opposite angles)
∠QDA = ∠QCR (alternate interior angles, AD ∥ BC)
By AAS: △DQA ≅ △CQR
∴ AQ = QR

(ii) AP = 2PQ:
From (i): △DQA ≅ △CQR
⇒ DA = CR
In parallelogram ABCD, AD = BC.
Therefore, BR = BC + CR = AD + AD = 2AD.
In triangle APD and triangle RPB:
1. Angle PAD = Angle PRB (Alternate interior angles, AD || BR)
2. Angle ADP = Angle RBP (Alternate interior angles, AD || BR)
By AA Similarity:
Triangle APD is similar to Triangle RPB.
So, the ratio of sides is:
AP/PR = AD/BR
⇒ AP/PR = AD/2AD
⇒ AP/PR = 1/2
⇒ PR = 2AP
Now, we know: AQ = AP + PQ
Since AQ = QR (from part i), then AR = 2AQ.
Also, AR = AP + PR.
Substitute PR = 2AP: AP + 2AP = 2(AP + PQ)
⇒ 3AP = 2AP + 2PQ
⇒ AP = 2PQ (Hence Proved)

(iii) As derived in the similarity step in part (ii):
Since Triangle APD is similar to Triangle RPB,
AP/PR = AD/BR
Substitute BR = 2AD: AP/PR = AD/2AD
⇒ AP/PR = 1/2
⇒ PR = 2AP (Hence Proved)

35. (a) Find the mean and mode of the following frequency distribution:

Class Interval	Frequency
400 – 450	15
450 – 500	18
500 – 550	20
550 – 600	23
600 – 650	22
650 – 700	12

Solution:

Class 10 Maths CBSE Exam 2026 Question 35-Part a

Mean = a + (Σfd/Σf) × h = 575 + (-2750/110) = 575 – 25 = 550
Mode: Maximum frequency = 23, Modal class = 550–600
Mode = l + [(f₁ – f₀)/(2f₁ – f₀ – f₂)] × h
= 550 + [(23–20)/(46–20–22)] × 50
= 550 + [3/4] × 50
= 550 + 37.5
= 587.5

OR
(b) If the median of the following frequency distribution is 32.5 and sum of all frequencies is 40, then find the values of f₁ and f₂ :

Class Interval	Frequency
0 – 10	3
10 – 20	f₁
20 – 30	9
30 – 40	12
40 – 50	6
50 – 60	f₂
60 – 70	2

Solution:
Given: Median = 32.5, Σf = 40
Total: 32 + f₁ + f₂ = 40 → f₁ + f₂ = 8 …(i)
Median = 32.5, so median class = 30–40
Median = l + [(N/2 – cf)/f] × h
32.5 = 30 + [(20 – (12+f₁))/12] × 10
2.5 = [(8 – f₁)/12] × 10
2.5 × 12 = 10(8 – f₁)
30 = 80 – 10f₁
10f₁ = 50
f₁ = 5
From (i): f₂ = 8 – 5 = f₂ = 3

Class 10 Maths CBSE Exam Paper 2026 – Section E

This section (Q. 36 to 38) has 3 case study based questions of 4 marks each.
36. Elevated water storage tanks are built to store and supply water to nearby colonies. In the diagram given above, AB is an elevated water tank and CD is a nearby multistorey building. The building is 54 metres away from the water tank.
From a window (W) of the building, the angle of elevation of top of the tank is 45° and angle of depression of its foot is 30°.

Class 10 Maths CBSE Exam 2026 Question 36

(i) Write a relation between d (the height of window) and y.
(ii) Determine the value of h.
(iii) (a) Determine height of the water tank.
Solution:
AC = 54 m, angle of elevation of top = 45°
Angle of depression to foot = 30°
Here, d = height of window W from ground (XA = WC = d)
h = height of top of tank above window level
(i) Relation between d and y:
In triangle AXW, sin 30° = d/y
1/2 = d/y
y = 2d

(ii) Value of h:
In triangle BXW, tan 45° = h/XW
⇒ 1 = h/54
⇒ h = 54 m

(iii)(a) Height of water tank:
In triangle AXW, tan 30° = d/XW
⇒ 1/√3 = d/54
⇒ d = 54/√3 = 18√3
Height of water tank = h + d = 54 + 18√3
= 54 + 18 × 1.732
= 54 + 31.17
= 85.17 m ≈ 54 + 18√3 m

(iii) (b) Find the value of x and height of the window above ground level.
Solution:
Value of x and height of window:
In triangle BXW, cos 45° = XW/x
⇒ 1/√2 = 54/x
⇒ x = 54√2

37. An arch of a railway bridge, built on Chenab riverbed, is shown in the above diagram. It is a parabolic arch connecting two hills at P and Q. If the parabolic curve is represented by the polynomial p(x) = –0.0025x² – 0.025x + 136.

Class 10 Maths CBSE Exam 2026 Question 37

Observe the diagram and based on above information, answer the following questions:
(i) Write the co-ordinates of point A.
(ii) Find the span of the arch.
(iii) (a) Write the zeroes of the polynomial using diagram and verify the relationship between sum of zeroes and polynomials.

(iii) (b) Find the values of p(x) at x = 100 and x = –100. Are they same ?
Solution:
(i) Co-ordinates of point A:
Point A is the highest point (vertex), at x = 0 (from diagram, A is on Y-axis):
p(0) = –0.0025(0) – 0.025(0) + 136 = 136
∴ A = (0, 136)

(ii) Span of the arch:
Span = distance from Q to P = 238.5 + 228.5 = 467 m
(From diagram: Q = (–238.5, 0) and P = (228.5, 0))

(iii)(a) Zeroes and verification:
Zeroes from diagram: x = –238.5 and x = 228.5
Sum of zeroes = –238.5 + 228.5 = –10
From polynomial: –b/a = –(–0.025)/(–0.0025) = –10
Product of zeroes = –238.5 × 228.5 = –54,507.25
From polynomial: c/a = 136/(–0.0025) = –54,400 ≈ –54,507.25

(iii)(b) Find p(100) and p(–100):
p(100) = –0.0025(10000) – 0.025(100) + 136 = –25 – 2.5 + 136 = 108.5
p(–100) = –0.0025(10000) – 0.025(–100) + 136 = –25 + 2.5 + 136 = 113.5
No, they are not the same. p(100) = 108.5 ≠ p(–100) = 113.5

38. A wall mounted lamp, made of fabric, is shown below. Lamp has cuboidal shape, open from top and bottom. A spherical bulb of diameter 7 cm is latched with a very thin rod. (Ignore the rod while making calculations.)

Class 10 Maths CBSE Exam 2026 Question 38

Dimensions of the cuboid are 24 cm × 12 cm × 17 cm.
(i) Find the surface area of the bulb.
(ii) What could be the maximum diameter of the bulb if at least 1 cm space is left from each side?
(iii) (a) Find the area of the fabric used if there is a fold of 2 cm on top and bottom edges.
OR
(iii) (b) Find the space available inside the lamp.
Solution:
(i) Surface area of spherical bulb:
SA = 4πr² = 4 × 22/7 × (3.5)² = 4 × 22/7 × 12.25 = 4 × 22 × 1.75 = 154 cm²

(ii) Maximum diameter of bulb:
Smallest dimension of cuboid base = 12 cm
Leaving 1 cm from each side: 12 – 2 = 10 cm
Maximum diameter = 10 cm

(iii)(a) Area of fabric used:
Cuboid is open from top and bottom, so there is only 4 lateral faces.
With fold of 2 cm on top and bottom edges, effective height
= 17 + 2×2 = 21 cm
Area of fabric = 2(24 + 12) × 21 = 1512 cm²

(iii)(b) Space available inside lamp:
Volume of cuboid – Volume of bulb
Volume of cuboid = 24 × 12 × 17 = 4896 cm³
Volume of bulb = (4/3)πr³
= (4/3) × 22/7 × (3.5)³
= (4/3) × 22/7 × 42.875
= 179.67 cm³
Space available = 4896 – 179.67 = 4716.33 cm³

Finished reviewing the 2026 CBSE Maths paper? Keep your preparation strong by practicing all textbook chapters. Explore our comprehensive, step-by-step Class 10 Maths Solutions to master every concept for your board exams.

Class 10 Maths CBSE Exam Paper 2026 Solutions

Class 10 Maths CBSE Exam Paper 2026 – Section A

Class 10 Maths CBSE Exam Paper 2026 – Section B

Class 10 Maths CBSE Exam Paper 2026 – Section C

Class 10 Maths CBSE Exam Paper 2026 – Section D

Class 10 Maths CBSE Exam Paper 2026 – Section E