CBSE Class 10 Maths Standard 2026 Board Paper: Worked Solutions (Set 30/1/1)

Q11 mark

HCF of 960 and 432

Answer: (a) 48

1. 960 = 2⁶ × 3 × 5 and 432 = 2⁴ × 3³, so HCF = 2⁴ × 3 = 48.1 mark

Q21 mark

Is the number 2 prime, composite, both or neither?

Answer: (a) a prime number

1. 2 has exactly two factors, 1 and 2, so it is prime (the only even prime).1 mark

Q31 mark

Last digit of 6ⁿ for any natural number n

Answer: (b) 6

1. 6 × 6 = 36 ends in 6, and multiplying any number ending in 6 by 6 again ends in 6, so 6ⁿ always ends in 6.1 mark

Q41 mark

Number of zeroes of f(x) from its graph

Answer: (a) 0

1. The curve stays above the x-axis and never meets it, so f(x) has no zeroes.1 mark

Q51 mark

Pair of linear equations shown by coincident lines

Answer: (d) an infinite number of solutions

1. Coincident lines share every point, so the pair has infinitely many solutions.1 mark

Q61 mark

Common difference of the AP √2, 2√2, 3√2, …

Answer: (a) √2

1. d = 2√2 − √2 = √2.1 mark

Q71 mark

Similar triangles with 2AB = DE and BC = 8 cm: find EF

Answer: (d) 16 cm

1. The scale factor DE/AB = 2, so EF = 2 × BC = 16 cm.1 mark

Q81 mark

Where the mid-point of (5, −4) and (6, 4) lies

Answer: (a) x-axis

1. Mid-point = (11/2, 0). Its y-coordinate is 0, so it lies on the x-axis.1 mark

Q91 mark

cos θ when sin θ = a/b

Answer: (c) √(b² − a²)/b

1. Opposite = a, hypotenuse = b, so adjacent = √(b² − a²) and cos θ = √(b² − a²)/b.1 mark

Q101 mark

sin²A + 2cos²A when cos A = ½

Answer: (b) 5/4

1. cos²A = ¼, sin²A = ¾, so the value is ¾ + 2 × ¼ = 5/4.1 mark

Q111 mark

Angle of elevation of a 30 m tower from 10√3 m away

Answer: (d) 60°

1. tan θ = 30/(10√3) = √3, so θ = 60°.1 mark

Q121 mark

Angle between two tangents when ∠POQ = 120°

Answer: (a) 60°

1. In quadrilateral OPTQ the angles at P and Q are 90°, so ∠PTQ = 360° − 90° − 90° − 120° = 60°.1 mark

Q131 mark

Tangent PA with ∠POB = 125°: find ∠APO

Answer: (d) 35°

1. ∠OAP = 90° (tangent ⊥ radius). ∠POB is the exterior angle of triangle AOP, so ∠APO = 125° − 90° = 35°.1 mark

Q141 mark

Arc length: radius 21 cm, angle 60°

Answer: (a) 22 cm

1. Arc = (60/360) × 2 × (22/7) × 21 = 22 cm.1 mark

Q151 mark

Angle swept by the hour hand from 7:00 to 8:10

Answer: (c) 35°

1. The hour hand turns 30° per hour, i.e. ½° per minute. 70 minutes × ½° = 35°.1 mark

Q161 mark

Total surface area of a solid hemisphere of diameter 2d

Answer: (a) 3πd²

1. Radius = d, so TSA = 2πd² + πd² = 3πd².1 mark

Q171 mark

Median when mean = 12 and mode = 21

Answer: (c) 15

1. Mode = 3 Median − 2 Mean: 21 = 3M − 24, so M = 15.1 mark

Q181 mark

Probability of a number other than 3 on one die

Answer: (c) 5/6

1. 5 of the 6 equally likely outcomes are not 3, so P = 5/6.1 mark

Q191 mark

Assertion-reason: 53 Mondays in a leap year / non-leap year

Answer: (c) A is true, R is false

1. A leap year has 52 weeks and 2 extra days; 2 of the 7 possible pairs contain a Monday, so P = 2/7 (A true). A non-leap year has 1 extra day, so P = 1/7, not 5/7 (R false).1 mark

Q201 mark

Assertion-reason: zeroes of y² + 4y + 3 and the most zeroes a quadratic can have

Answer: (b) both true, R is not the correct explanation

1. y² + 4y + 3 = (y + 1)(y + 3), with zeroes −1 and −3, so A is true. R is true, but 'at most two' does not explain why this one has exactly two.1 mark

Q212 marks

1/α + 1/β for the zeroes of x² − 3x − 1

Answer: −3

1. α + β = 3 and αβ = −1.1 mark
2. 1/α + 1/β = (α + β)/(αβ) = 3/(−1)½ marks
3. = −3½ marks

Q222 marks

(A) DE ∥ BC with sides in x; find x. OR (B) similar triangles: find x and y

(A)

Answer: x = 4

1. DE ∥ BC, so by the Basic Proportionality Theorem AD/DB = AE/EC.1 mark
2. x/(x − 2) = (x + 2)/(x − 1)½ marks
3. x² − x = x² − 4, so x = 4.½ marks

(B)

Answer: x = 4.8 cm, y = 5 cm

1. △ABC ~ △XYZ gives AB/XY = BC/YZ = AC/XZ, i.e. 4/x = 6/7.2 = y/6.1 mark
2. x = 4 × 7.2/6 = 4.8 cm and y = 6 × 6/7.2 = 5 cm.½ + ½ mark

Q232 marks

Centre (x − 7, 2x), passes through (−9, 11), radius 5√2: find x

Answer: x = 3 or x = 5

1. Distance from the centre to (−9, 11) is the radius: √[(x − 7 + 9)² + (2x − 11)²] = 5√2.½ marks
2. (x + 2)² + (2x − 11)² = 50, so 5x² − 40x + 75 = 0, i.e. x² − 8x + 15 = 0.½ marks
3. (x − 3)(x − 5) = 0, so x = 3 or x = 5.½ + ½ mark

Q242 marks

(A) sin θ + cos θ when tan θ = 24/7. OR (B) a quotient when cot θ = 7/8

(A)

Answer: 31/25

1. Opposite 24, adjacent 7, hypotenuse √(576 + 49) = 25, so sin θ = 24/25 and cos θ = 7/25.1½ marks
2. sin θ + cos θ = 31/25.½ marks

(B)

Answer: 49/64

1. (1 + sin θ)(1 − sin θ)/[(1 + cos θ)(1 − cos θ)] = (1 − sin²θ)/(1 − cos²θ)1 mark
2. = cos²θ/sin²θ = cot²θ½ marks
3. = (7/8)² = 49/64½ marks

Q252 marks

Chord of the larger of two concentric circles (5 cm, 4 cm) touching the smaller

Answer: 6 cm

1. Draw the figure: chord AB of the larger circle touches the smaller circle at M.½ marks
2. OM ⊥ AB, so AM = √(5² − 4²) = 3 cm.1 mark
3. M is the mid-point of AB, so AB = 2 × 3 = 6 cm.½ marks

Q263 marks

Prove that √3 is irrational

Answer: Proof by contradiction

1. Suppose √3 is rational: √3 = p/q with q ≠ 0 and p, q co-prime.½ marks
2. Then 3q² = p², so 3 divides p², hence 3 divides p. … (i)1 mark
3. Write p = 3a: 9a² = 3q², so q² = 3a² and 3 divides q. … (ii)1 mark
4. (i) and (ii) make 3 a common factor of p and q, contradicting co-prime. So √3 is irrational.½ marks

Q273 marks

Ratio in which the x-axis divides the join of (−6, 5) and (−4, −1), and the point

Answer: 5 : 1, at (−13/3, 0)

1. Let P(x, 0) on the x-axis divide AB in the ratio k : 1.½ marks
2. P = ((−4k − 6)/(k + 1), (−k + 5)/(k + 1)).1 mark
3. The y-coordinate is 0: −k + 5 = 0, so k = 5.½ marks
4. The ratio is 5 : 1.½ marks
5. x = (−20 − 6)/6 = −13/3, so P = (−13/3, 0).½ marks

Q283 marks

(A) x = h + a cos θ, y = k + b sin θ: prove a sum of squares is 1. OR (B) prove a tan/sec identity equals 2 cosec A

(A)

Answer: Shown

1. (x − h)/a = cos θ1 mark
2. (y − k)/b = sin θ1 mark
3. LHS = cos²θ + sin²θ = 1 = RHS.1 mark

(B)

Answer: Shown

1. Write tan A = sin A/cos A and sec A = 1/cos A in both fractions.1 mark
2. LHS = sin A/(cos A + 1) − sin A/(cos A − 1)½ marks
3. = sin A × (−2)/(cos²A − 1) = sin A × (−2)/(−sin²A)1 mark
4. = 2/sin A = 2 cosec A = RHS.½ marks

Q293 marks

(A) Incircle radius of a right triangle with legs 4 cm and 3 cm. OR (B) prove PM = ½(PQ + QR + PR) for an escribed circle

(A)

Answer: 1 cm

1. AC = √(3² + 4²) = 5 cm.½ marks
2. Let the tangent lengths from B be BD = BE = x.½ marks
3. Then AD = AF = 4 − x and CE = CF = 3 − x (equal tangents).½ marks
4. AF + CF = AC: 7 − 2x = 5, so x = 1.1 mark
5. With ∠B = 90°, OEBD is a square, so the radius = x = 1 cm.½ marks

(B)

Answer: Shown

1. Tangents from an external point are equal: PM = PN, QS = QM, RS = RN.1½ marks
2. PM + PN = (PQ + QM) + (PR + RN).½ marks
3. 2PM = PQ + QS + RS + PR½ marks
4. = PQ + QR + PR, so PM = ½(PQ + QR + PR).½ marks

Q303 marks

Volume of a cylinder with hemispherical ends: total height 20 cm, diameter 7 cm

Answer: 4081/6 cm³ = 680⅙ cm³

1. Radius = 7/2 cm.½ marks
2. Cylinder height = 20 − 7/2 − 7/2 = 13 cm.½ marks
3. Volume = πr²h + 2 × (2/3)πr³ = (22/7)(7/2)²(13) + (4/3)(22/7)(7/2)³1 mark
4. = 38.5 × (13 + 14/3) = 4081/6 = 680⅙ cm³.1 mark

Q313 marks

Two dice: list outcomes; P(same number), P(different numbers)

Answer: 36 outcomes; 1/6 and 5/6

1. The outcomes are the 36 ordered pairs (1, 1), (1, 2), …, (6, 6).1 mark
2. Same number: (1, 1), (2, 2), …, (6, 6), so P = 6/36 = 1/6.1 mark
3. Different numbers: P = 1 − 1/6 = 5/6.1 mark

Q325 marks

Graph 2x − 3y + 6 = 0, 2x + 3y − 18 = 0 and x = 0; vertices and area of the triangle

Answer: (0, 2), (3, 4), (0, 6); area 6 square units

1. 2x − 3y + 6 = 0 through (0, 2), (3, 4), (−3, 0): plot and draw it.1 mark
2. 2x + 3y − 18 = 0 through (0, 6), (3, 4), (9, 0): plot and draw it.1 mark
3. x = 0 is the y-axis.½ marks
4. Vertices: A(0, 2), B(3, 4), C(0, 6).1½ marks
5. Base AC = 4 on the y-axis, height = 3 (x-coordinate of B): area = ½ × 4 × 3 = 6 square units.1 mark

Q335 marks

(A) Speeds of two trains over 200 km, one hour apart. OR (B) two squares: areas sum 640 m², perimeters differ by 64 m

(A)

Answer: 50 km/h and 40 km/h

1. Let the faster speed be x km/h, so the slower is (x − 10) km/h: 200/(x − 10) − 200/x = 1.2 marks
2. 2000 = x(x − 10), so x² − 10x − 2000 = 0.1 mark
3. (x − 50)(x + 40) = 0.1 mark
4. x = 50 (x = −40 is rejected).½ marks
5. Faster train 50 km/h, slower train 40 km/h.½ marks

(B)

Answer: 24 m and 8 m

1. Sides x m and y m (x > y): x² + y² = 640.1 mark
2. 4x − 4y = 64, so y = x − 16.1 mark
3. x² + (x − 16)² = 640 gives x² − 16x − 192 = 0.1 mark
4. (x − 24)(x + 8) = 0.1 mark
5. x = 24 (x = −8 is rejected).½ marks
6. y = 8: the sides are 24 m and 8 m.½ marks

Q345 marks

(A) State and prove the Basic Proportionality Theorem. OR (B) medians of similar triangles: prove two pairs of triangles similar

(A)

Answer: Statement and proof

1. Statement: a line drawn parallel to one side of a triangle, meeting the other two sides at distinct points, divides those sides in the same ratio.1 mark
2. Given △ABC with DE ∥ BC, D on AB, E on AC. To prove AD/DB = AE/EC. Construction: join BE and CD; draw EM ⊥ AB and DN ⊥ AC. Figure.2 marks
3. ar(ADE)/ar(BDE) = (½·AD·EM)/(½·DB·EM) = AD/DB and ar(ADE)/ar(CDE) = AE/EC. △BDE and △CDE stand on the same base DE between the same parallels, so their areas are equal. Hence AD/DB = AE/EC.2 marks

(B)

Answer: Shown

1. (i) △ABC ~ △PQR gives ∠A = ∠P.½ marks
2. AB/PQ = AC/PR, and AB = 2AM, PQ = 2PN (CM and RN are medians), so AM/PN = AC/PR.1½ marks
3. So △AMC ~ △PNR (SAS similarity).½ marks
4. (ii) △ABC ~ △PQR gives ∠B = ∠Q.½ marks
5. AB/PQ = BC/QR with AB = 2MB, PQ = 2NQ, so MB/NQ = BC/QR.1½ marks
6. So △CMB ~ △RNQ (SAS similarity).½ marks

Q355 marks

Mean 35, total frequency 25: find the missing frequencies x and y

Answer: x = 3, y = 5

1. Table with class marks 5, 15, …, 65 and fᵢxᵢ: Σfᵢ = 17 + x + y and Σfᵢxᵢ = 605 + 15x + 45y.2 marks
2. Σfᵢ = 25, so x + y = 8. … (i)½ marks
3. Mean = (605 + 15x + 45y)/25 = 35.½ marks
4. 15x + 45y = 270, so x + 3y = 18. … (ii)1 mark
5. Solving (i) and (ii): x = 3 and y = 5.½ + ½ mark

Q364 marks

Case study: potato race (bucket 5 m from the first potato, potatoes 3 m apart, 10 potatoes)

(i)

Answer: 10 m

1. To the first potato and back: 5 + 5 = 10 m.1 mark

(ii)

Answer: 16 m

1. The second potato is 8 m away: 8 + 8 = 16 m.1 mark

(iii)(a)

Answer: 370 m

1. The distances 10, 16, 22, … form an AP with a = 10, d = 6.1 mark
2. S₁₀ = (10/2)[2 × 10 + 9 × 6] = 5 × 74 = 370 m.1 mark

(iii)(b)

Answer: 74 seconds

1. AP with a = 10, d = 6.½ marks
2. Total distance S₁₀ = 5 × 74 = 370 m.1 mark
3. Time = 370 ÷ 5 = 74 s.½ marks

Q374 marks

Case study: radio tower in two sections, wires from a point 6 m away (elevations 30° and 60°)

(i)

Answer: OB = 4√3 m

1. cos 30° = OP/OB: √3/2 = 6/OB.½ marks
2. OB = 12/√3 = 4√3 m.½ marks

(ii)

Answer: OA = 12 m

1. cos 60° = OP/OA: ½ = 6/OA.½ marks
2. OA = 12 m.½ marks

(iii)(a)

Answer: AB = 4√3 m

1. tan 30° = BP/6, so BP = 2√3 m.1 mark
2. tan 60° = AP/6, so AP = 6√3 m.½ marks
3. AB = AP − BP = 4√3 m.½ marks

(iii)(b)

Answer: 6√3 m²

1. tan 30° = BP/6, so BP = 2√3 m.1 mark
2. Area of △OPB = ½ × BP × OP = ½ × 2√3 × 6 = 6√3 m².1 mark

Q384 marks

Case study: silver brooch, circle of diameter 35 cm with 5 diameters (10 sectors)

(i)

Answer: 17.5 cm

1. r = 35/2 = 17.5 cm.1 mark

(ii)

Answer: 110 cm

1. Circumference = 2 × (22/7) × (35/2) = 110 cm.1 mark

(iii)(a)

Answer: 285 cm

1. Wire = circumference + 5 diameters = 110 + 5 × 351 mark
2. = 285 cm.1 mark

(iii)(b)

Answer: 385/4 = 96.25 cm²

1. Each sector has angle 360°/10 = 36°.1 mark
2. Area = (36/360) × (22/7) × (35/2)² = 385/4 = 96.25 cm².1 mark