CBSE Class 10 · Maths · board exam 2027
CBSE Class 10 Maths Question Paper 2025
The board’s 2025 question paper, set 30/1/1 — all 38 questions as printed; 20 objective answers from the board’s own key; full model answers for 3 written questions, how the answer begins for 10 more. Each question names its chapter, where that chapter’s most-asked questions are.
Full marks 80 · Time 3 h
Section A · Objective questions (20)
If alpha and beta are the zeroes of polynomial 3x^2 + 6x + k such that alpha + beta + alpha*beta = - 2/3, then the value of k is :
- (A) - 8
- (B) 8
- (C) - 4
- (D) 4 ✓
Answer (the board’s key): (D)
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
Answer (the board’s key): (B)
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
Answer (the board’s key): (B)
If theta is an acute angle and 7 + 4 sin theta = 9, then the value of theta is :
- (A) 90 degree
- (B) 30 degree ✓
- (C) 45 degree
- (D) 60 degree
Answer (the board’s key): (B)
The value of tan^2 theta - (1/cos theta * sec theta) is :
- (A) 1
- (B) 0
- (C) - 1 ✓
- (D) 2
Answer (the board’s key): (C)
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
Answer (the board’s key): (C)
The tangents drawn at the extremities of the diameter of a circle are always :
- (A) parallel ✓
- (B) perpendicular
- (C) equal
- (D) intersecting
Answer (the board’s key): (A)
In triangles ABC and DEF, angle B = angle E, angle F = angle 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 ✓
Answer (the board’s key): (D)
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
Answer (the board’s key): (C)
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
Answer (the board’s key): (C)
If the sum of first m terms of an AP is 2m^2 + 3m, then its second term is :
- (A) 10
- (B) 9 ✓
- (C) 12
- (D) 4
Answer (the board’s key): (B)
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
Answer (the board’s key): (C)
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 ✓
Answer (the board’s key): (D)
Which of the following is a rational number between sqrt(3) and sqrt(5) ?
- (A) 1.4142387954012 ....
- (B) 2.326 (bar over 26)
- (C) pi
- (D) 1.857142 ✓
Answer (the board’s key): (D)
If the area of a sector of a circle is 40pi square units and the angle at the centre is 72 degrees, then the radius of the circle is :
- (A) 200 units
- (B) 100 units
- (C) 20 units
- (D) 10 sqrt(2) units ✓
Answer (the board’s key): (D)
In the given figure, PA is a tangent drawn from an external point P to a circle with centre O. If angle POB = 115 degrees, then angle APO is equal to :
- (A) 25 degrees ✓
- (B) 65 degrees
- (C) 90 degrees
- (D) 35 degrees
Answer (the board’s key): (A)
A kite is flying at a height of 150 m from the ground. The string attached to it makes an angle of 30 degrees with the horizontal direction of the ground. The length of the string is :
- (A) 100 sqrt(3) m
- (B) 300 m ✓
- (C) 150 sqrt(2) m
- (D) 150 sqrt(3) m
Answer (the board’s key): (B)
A piece of wire 20 cm long is bent into the form of an arc of a circle of radius 60/pi cm. The angle subtended by the arc at the centre of the circle is :
- (A) 30°
- (B) 60° ✓
- (C) 90°
- (D) 50°
Answer (the board’s key): (B)
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.
- (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.
Answer (the board’s key): (A)
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 3pi r^2.
- (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.
Answer (the board’s key): (C)
Section B · Written questions (18)
(a) If x cos 60° + y cos 0° + sin 30° - cot 45° = 5, then find the value of x + 2y. OR (b) Evaluate : (tan^2 60°)/(sin^2 60° + cos^2 30°)
Find the zeroes of the polynomial p(x) = x^2 + 4/3 x - 4/3.
Medha Ma’am
AskMedha’s AI teacher
Model answer · zeroes of quadratic polynomial
- 1.Given: the quadratic polynomial p(x) = x^2 + (4/3)x − 4/3. Taking 1/3 common: p(x) = 1/3(3x^2 + 4x − 4)0.5 marks
- 2.By the method of splitting the middle term, 4x is written as 6x − 2x: p(x) = 1/3(3x^2 + 6x − 2x − 4) = 1/3[3x(x + 2) − 2(x + 2)]0.5 marks
- 3.Taking (x + 2) common: p(x) = 1/3(3x − 2)(x + 2)0.5 marks
- 4.So the value of p(x) is zero when 3x − 2 = 0 or x + 2 = 0, i.e., when x = 2/3 or x = −2. Therefore, the zeroes of p(x) are 2/3 and −2.0.5 marks
- First take 1/3 common, so you get 1/3(3x^2 + 4x − 4), and only then split the middle term.
- Write the split line "1/3(3x^2 + 6x − 2x − 4)" and the factors "1/3(3x − 2)(x + 2)" on separate lines.
- End with the sentence "the zeroes of p(x) are 2/3 and −2".
Marks split as in the board’s published marking scheme
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 10sqrt(2) units.
Medha Ma’am
AskMedha’s AI teacher
Model answer · distance formula, circle centre
- 1.The values of 'a' are 5 and 3. The circle has centre C(2a, a − 7), passes through P(11, −9) and has diameter 10√2 units. Radius = 10√2 ÷ 2 = 5√2 units, and the radius is the distance CP.0.5 marks
- 2.By the distance formula, the distance between (x1, y1) and (x2, y2) is √[(x2 − x1)² + (y2 − y1)²]. CP² = (5√2)² = 50 (2a − 11)² + (a − 7 + 9)² = 500.5 marks
- 3.4a² − 44a + 121 + a² + 4a + 4 = 50 5a² − 40a + 75 = 0 Dividing by 5: a² − 8a + 15 = 00.5 marks
- 4.(a − 5)(a − 3) = 0 So a − 5 = 0 or a − 3 = 0, which gives a = 5 or a = 3.0.5 marks
- Halve the diameter first and write "Radius = 10√2 ÷ 2 = 5√2 units".
- Square the radius to get 50, then simplify to a² − 8a + 15 = 0.
- Write the factors (a − 5)(a − 3) = 0 and give both values, a = 5 or a = 3.
Marks split as in the board’s published marking scheme
(a) If triangle ABC ~ triangle 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 angle 1 = angle 2, show that triangle PQS ~ triangle TQR.
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.
Medha Ma’am
AskMedha’s AI teacher
Model answer · Tangent length, finding radius
- 1.Given: O is the centre of the circular ground, OP = 26 m, and PA = PB = 10 m are tangents from the external point P. Join OA. The tangent to a circle is perpendicular to the radius through the point of contact, so ∠OAP = 90°.0.5 marks
- 2.In right triangle OAP, by the Pythagoras Theorem, OP² = OA² + PA² (26)² = OA² + (10)² 676 = OA² + 1001 mark
- 3.OA² = 676 − 100 = 576 OA = √576 = 24 Therefore, the radius of the circular ground is 24 m.0.5 marks
- Write the reason for ∠OAP = 90°: "The tangent to a circle is perpendicular to the radius through the point of contact".
- Write the Pythagoras step as (26)² = OA² + (10)² in right triangle OAP before you solve it.
- End with the unit: "the radius of the circular ground is 24 m".
Marks split as in the board’s published marking scheme
(a) In the given figure, O is the centre of the circle and BCD is a tangent at point C. Prove that angle BAC + angle ACD = 90°. OR (b) Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
How the model answer begins
1. Given: ABCD is a quadrilateral circumscribing a circle with centre O. It touches AB, BC, CD and DA at P, Q, R and S. To prove: ∠AOB + ∠COD = 180° and ∠BOC + ∠AOD = 180°. Construction: Join OP, OQ, OR, OS, OA, OB, OC and…
The full answer from Medha Ma’am →(a) Prove that : tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = 1 + sec θ cosec θ OR (b) Prove that : (sin A + cos A)/(sin A − cos A) + (sin A − cos A)/(sin A + cos A) = 2/(2 sin^2 A − 1)
How the model answer begins
1. (a) We write the LHS in terms of sin θ and cos θ, using tan θ = sin θ/cos θ and cot θ = cos θ/sin θ. LHS = tan θ/(1 − cot θ) + cot θ/(1 − tan θ) = (sin θ/cos θ)/(1 − cos θ/sin θ) + (cos θ/sin θ)/(1 − sin θ/cos θ)
The full answer from Medha Ma’am →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.
How the model answer begins
1. The given points are A(5, –6) and B(–1, –4). Let the y-axis divide the line segment AB in the ratio k : 1 at the point P(0, y). By the section formula, the coordinates of the point which divides AB in the ratio k : 1…
The full answer from Medha Ma’am →Prove that 1/sqrt(5) is an irrational number.
How the model answer begins
1. Let us assume, to the contrary, that 1/√5 is rational. Then we can find integers p and q (q ≠ 0) such that 1/√5 = p/q, where p and q are coprime.
The full answer from Medha Ma’am →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 m^3 of air, find the height of the cylindrical part. (Use pi = 22/7).
How the model answer begins
1. The height of the cylindrical part of the room is 4 m, worked out below. Given: the room is a cylinder surmounted by a hemispherical dome, and it contains 1408/21 m³ of air. Let r be the radius of the hemisphere and of…
The full answer from Medha Ma’am →Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 2.
How the model answer begins
1. Two dice are thrown at the same time. Each die can show any one of the numbers 1, 2, 3, 4, 5, 6, so every outcome is an ordered pair (number on first die, number on second die). Number of possible outcomes = 6 × 6 = 36
The full answer from Medha Ma’am →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 ₹ 1,860 as the total annual interest. However, had he interchanged the amounts of investments in the two schemes, he would have received ₹ 20 more as annual interest. How much money did he invest in each scheme ?
How the model answer begins
1. Vijay invested ₹ 12,000 in scheme A and ₹ 10,000 in scheme B. The working is shown below. Given: scheme A gives 8% per annum and scheme B gives 9% per annum. The total annual interest is ₹ 1,860. Let Vijay invest ₹ x in…
The full answer from Medha Ma’am →(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 triangle ABC, if AD perpendicular BC and AD^2 = BD x DC, then prove that angle BAC = 90°.
How the model answer begins
1. Answering (b). Given: In triangle ABC, D is a point on side BC such that AD ⊥ BC and AD² = BD × DC. To prove: ∠BAC = 90°. Figure: triangle ABC with the perpendicular AD drawn from A to BC, and the right angles marked at…
The full answer from Medha Ma’am →(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. 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.
How the model answer begins
1. Answering option (b). Given: distance = 480 km. If the speed were 8 km/h less, the train would take 3 hours more. Let the uniform speed of the train be x km/h. Then the reduced speed is (x – 8) km/h.
The full answer from Medha Ma’am →Find the missing frequency 'f' in the following table, if the mean of the given data is 18. Hence find the mode. Daily Allowance: 11-13 (7), 13-15 (6), 15-17 (9), 17-19 (13), 19-21 (f), 21-23 (5), 23-25 (4)
How the model answer begins
1. Class marks xᵢ: 12, 14, 16, 18, 20, 22, 24; fᵢxᵢ: 84, 84, 144, 234, 20f, 110, 96.
The full answer from Medha Ma’am →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. 1 (ii) Determine the distance of the 8th round. 1 (iii) (a) Find the total distance run after completing all 10 rounds. 2 OR (iii) (b) If a runner completes only the first 6 rounds, what is the total distance run by the runner ? 2
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. 1 (ii) Find the length of the arc ACB. 1 (iii) (a) Find the area of each sector of the brooch. 2 OR (iii) (b) Find the total length of the silver wire used. 2
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'. 1 (ii) Find distance BC in terms of 'h'. 1 (iii) (a) Find the height CE of the lighthouse [Use sqrt(3) = 1.73] 2 OR (iii) (b) Find distance AE, if AC = 100 m. 2
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