Prepare for the CBSE board exams using the Class 10 Maths Standard (Code 041) Previous Year Question Paper 2025. The paper is set according to the new CBSE exam pattern and contains significant questions in Algebra, Geometry, Trigonometry, Mensuration, and Statistics. Practising this previous year paper enables the students to understand the level of preparation, the pattern of questions, and enhance time management skills in exams. It’s a worth reading tool for revision, self-check, and last minute practice prior to the board exam. Download PDF now and increase your chances of achieving high marks in Class 10 Maths Standard.
Series GE1FH
Question Paper Code 30/1/1 Set 1
MATHEMATICS (STANDARD)
(Session 2024-25)
Time allowed : 3 hours
Maximum Marks : 80
General Instructions: Read the following instructions very carefully and strictly follow them:
(i) This question paper contains 38 questions. All questions are compulsory.
(ii) This question paper is divided into five Sections A, B, C, D and E.
(iii) In Section A, Questions no. 1 to 18 are multiple choice questions (MCQs) and questions number 19 and 20 are Assertion-Reason based questions of 1 mark each.
(iv) In Section B, Questions no. 21 to 25 are very short answer (VSA) type questions, carrying 2 marks each.
(v) In Section C, Questions no. 26 to 31 are short answer (SA) type questions, carrying 3 marks each.
(vi) In Section D, Questions no. 32 to 35 are long answer (LA) type questions carrying 5 marks each.
(vii) In Section E, Questions no. 36 to 38 are case study based questions carrying 4 marks each. Internal choice is provided in 2 marks questions in each case study.
(viii) There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 2 questions in Section C, 2 questions in Section D and 3 questions in Section E.
(ix) Draw neat diagrams wherever required. Take = wherever required, if not stated.
(x) Use of calculator is not allowed.
SECTION A
This section has 20 Multiple Choice Questions (MCQs) carrying 1 mark each. 20X1=20
1. If α and β are the zeroes of polynomial 3x2 + 6x + k such that α + β + αβ = -2/3, then the value of k is :
a) -8
b) 8
c) -4
d) 4
2. If x=1 and y=2 is a solution of the pair of linear equations 2x – 3y + a = 0 and 2x + 3y – b = 0, then:
a) a = 2b
b) 2a = b
c) a + 2b = 0
d) 2a + b = 0
3. The mid-point of the line segment joining the points P(-4, 5) and Q(4, 6) lies on :
a) x-axis
b) y-axis
c) origin
d) neither x-axis nor y-axis
4. If θ is an acute angle and 7+4 sin θ=9, then the value of θ is :
a) 90°
b) 30°
c) 45°
d) 60°
5. The value of tan2θ – (1 / cosθ x secθ) is:
a) 1
b) 0
c) -1
d) 2
6. If HCF(98, 28) = m and LCM(98, 28) = n, then the value of n – 7m is:
a) 0
b) 28
c) 98
d) 198
7. The tangents drawn at the extremities of the diameter of a circle are always:
a) parallel
b) perpendicular
c) equal
d) intersecting
8. In triangles ABC and DEF, ∠B = ∠E, ∠F = ∠C and AB = 3 DE. Then, the two triangles are :
a) congruent but not similar
b) congruent as well as similar
c) neither congruent nor similar
d) similar but not congruent
9. If (-1)n + (-1)8 = 0, then n is:
a) any positive integer
b) any negative integer
c) any odd number
d) any even number
10. Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is:

a) 3
b) 5
c) 2
d) 4
11. If the sum of first m terms of an AP is 2m2 + 3m, then its second term is:
a) 10
b) 9
c) 12
d) 4
12. Mode and Mean of a data are 15x and 18x, respectively. Then the median of the data is:
a) x
b) 11x
c) 17x
d) 34x
13. A card is selected at random from a deck of 52 playing cards. The probability of it being a red face card is:
a) 3/13
b) 2/13
c) 1/2
d) 3/26
14. Which of the following is a rational number between √3 and √5?
a) 1·4142387954012 . . . .
b) 2.326¯
c) π
d) 1·857142
15. If a sector of a circle has an area of 40 sq. units and a central angle of 72°, the radius of the circle is:
a) 200 units
b) 100 units
c) 20 units
d) 10√2 units
16. In the given figure, PA is a tangent from an external point P to a circle with centre O. If ∠POB = 115° , then ∠APO is equal to:

a) 25°
b) 65°
c) 90°
d) 35°
17. A kite is flying at a height of 150 m from the ground. It is attached to a string inclined at an angle of 30° to the horizontal. The length of the string is:
a) 100√3 m
b) 300 m
c) 150√2 m
d) 150√3 m
18. A piece of wire 20 cm long is bent into the form of an arc of a circle of radius 60/π cm. The angle subtended by the arc at the centre of the circle is:
a) 30°
b) 60°
c) 90°
d) 50°
Questions number 19 and 20 are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (a) 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 probability of selecting a number at random from the numbers 1 to 20 is 1.
Reason (R): For any event E, if P(E) = 1, then E is called a sure event.
Ans: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
20. Assertion (A): If we join two hemispheres of same radius along their bases, then we get a sphere.
Reason (R): Total Surface Area of a sphere of radius r is 3πr2.
Ans: (c) Assertion (A) is true, but Reason (R) is false.
SECTION B
This section has 5 Very Short Answer (VSA) type questions carrying 2 marks each. 5X2=10
21. (a) If x cos 60° + y cos 0° + sin 30° – cot 45° = 5, then find the value of x + 2y.

OR
(b) Evaluate: (tan260°)/(sin260°+cos230°)

22. Find the zeroes of the polynomial p(x) = x2 + (4/3)x – (4/3).

23. The coordinates of the centre of a circle are (2a, a-7). Find the value(s) of ‘a’ if the circle passes through the point (11, -9) and has diameter 10√2 units.

24. (a) If △ABC ~ △PQR in which AB = 6 cm, BC = 4 cm, AC = 8 cm and PR = 6 cm, then find the length of (PQ+QR).

OR
(b) In the given figure, QR/QS = QT/PR and ∠1 = ∠2, show that △PQS ~ △TQR.


25. A person is standing at P outside a circular ground at a distance of 26 m from the centre of the ground. He found that his distances from the points A and B on the ground are 10 m (PA and PB are tangents to the circle). Find the radius of the circular ground.


SECTION C
This section has 6 Short Answer (SA) type questions carrying 3 marks each. 6X3=18
26. (a) In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that ∠BAC + ∠ACD = 90°.

Solution: In △OAC,
OA = OC
∠OCA = ∠OAC
Now, ∠OCD = 90°
⇒∠OCA + ∠ACD = 90°
⇒∠OAC + ∠ACD = 90°
or ∠BAC + ∠ACD = 90°
OR
(b) Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:

△OAP ≅ △OAS
∴ ∠1 = ∠2
Similarly, ∠3 = ∠4 , ∠5 = ∠6 , ∠7 = ∠8
Also, ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
⇒ 2 (∠1 + ∠4 + ∠5 + ∠8) = 360°
⇒ ∠AOB + ∠COD = 180°
Similarly, ∠BOC + ∠AOD = 180°


28. Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also find the point of intersection.

29. Prove that 1/√5 is an irrational number.

30. A room is in the form of a cylinder surmounted by a hemispherical dome. The base radius of the hemisphere is half of the height of the cylindrical part. If the room contains 1408/21 m3 of air, find the height of the cylindrical part. (Use π=22/7).

31. Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 2.
Solution: Total outcomes = 36
Number of Outcomes with difference of the numbers on the two dice is 2 = 8
(1,3) (3,1) (4,2) (2,4) (5,3) (3,5) (4,6) (6,4)
P (difference of the numbers on the two dice is 2) = 8/36 = 2/9
SECTION D
This section has 4 Long Answer (LA) type questions carrying 5 marks each. 4X5=20
32. Vijay invested certain amounts of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. He received Rs. 1,860 as the total annual interest. However, had he interchanged the amounts of investments in the two schemes, he would have received Rs. 20 more as annual interest. How much money did he invest in each scheme?

33. (a) The diagonal BD of a parallelogram ABCD intersects the line segment AE at the point F, where E is any point on the side BC. Prove that DF x EF = FB x FA.

OR
(b) In △ABC, if AD⊥BC and AD2 = BD x DC, then prove that ∠BAC = 90°.

34. (a) The perimeter of a right triangle is 60 cm and its hypotenuse is 25 cm. Find the lengths of other two sides of the triangle.
Solution: Let the other two sides be x cm and y cm
ATQ
x + y + 25 = 60
y = 35 – x
Now,
x2 + y2 = (25)2
x2 + (35 – x)2 = 625
x2 – 35x + 300 = 0
(x – 20) (x – 15) = 0
⇒ x = 20, 15
x = 20 ⇒ y = 15
x = 15 ⇒ y = 20
Hence sides are 15 cm and 20 cm.
OR
(b) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.

35. Find the missing frequency ‘f’ in the following table, if the mean of the given data is 18. Hence find the mode.
Daily Allowance | Number of Children |
11-13 | 7 |
13-15 | 6 |
15-17 | 9 |
17-19 | 13 |
19-21 | f |
21-23 | 5 |
23-25 | 4 |
Daily Allowance | xi | fi | fixi |
11-13 | 12 | 7 | 84 |
13-15 | 14 | 6 | 84 |
15-17 | 16 | 9 | 144 |
17-19 | 18 | 13 | 234 |
19-21 | 20 | f | 20f |
21-23 | 22 | 5 | 110 |
23-25 | 24 | 4 | 96 |
Total | 44+f | 752+20f |
SECTION E
This section has 3 case study based questions carrying 4 marks each. 3X4=12
Case Study 1
36. A school is organizing a charity run to raise funds for a local hospital. The run is planned as a series of rounds around a track, with each round being 300 metres. To make the event more challenging and engaging, the organizers decide to increase the distance of each subsequent round by 50 metres. For example, the second round will be 350 metres, the third round will be 400 metres and so on. The total number of rounds planned is 10.

Based on the information given above, answer the following questions:
(i) Write the fourth, fifth and sixth term of the Arithmetic Progression so formed.
A.P. formed is 300, 350, 400……
Solution: a4=450
a5=500
a6=550
(ii) Determine the distance of the 8th round.
Solution: a8 = 300 +7 × 50 = 650 m
(iii) (a) Find the total distance run after completing all 10 rounds.
Ans: S10 = (10/2) × (2 × 300 + 9 × 50) = 5250 m
OR
(iii) (b) If a runner completes only the first 6 rounds, what is the total distance run by the runner?
Solution: S6 = (6/2) × (2 × 300 + 5 × 50) = 2250 m
Case Study 2
37. A brooch is a decorative piece often worn on clothing like jackets, blouses or dresses to add elegance. Made from precious metals and decorated with gemstones, brooches come in many shapes and designs.

One such brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the figure.

Based on the above given information, answer the following questions:
(i) Find the central angle of each sector.
Solution: central angle = 360°/10 = 36°
(ii) Find the length of the arc ACB.
Solution: length of arc ACB = (1/10) × 2 × (22/7) × (35/2) = 11mm
(iii) (a) Find the area of each sector of the brooch.
Solution: Area of each sector of the brooch = (1/10) × (22/7) × (35/2) × (35/2)
= 385/4 mm2 or 96.25 mm2
OR
(iii) (b) Find the total length of the silver wire used.
Solution: length of silver wire used = 2 × (22/7) × (35/2) + 5 × 35 = 285 mm
Case Study 3
38. Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be 60°. Then, she climbed a nearby observation deck, 40 metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be 45°.

Based on the above given information, answer the following questions:
(i) If CD is h metres, find the distance BD in terms of ‘h’.
Solution: h/BD = tan 45° = 1
⇒ BD = h m
(ii) Find distance BC in terms of ‘h’.
Solution: h/BC = sin 45° = 1/√2
⇒ BC = √2h m
(iii) (a) Find the height CE of the lighthouse [Use √3=1·73]
Solution: tan 60° = EC/AE
⇒ √3 = (h+40)/h
⇒ h=20 (√3+1) = 20 x 2.73 = 54.6 m
∴ CE = 54.6 + 40 = 94.6 m
OR
(iii) (b) Find distance AE, if AC = 100 m.
Solution: cos 60° = AE/AC
⇒ 1/2 = AE/100
∴ AE = 50 m