CBSE Class 10 Maths Basic 2026 Board Paper: Worked Solutions (Set 430/1/1)

Q11 mark

LCM of a and b when their HCF is 1

Answer: (D) ab

1. HCF × LCM = a × b, and HCF = 1, so LCM = ab.1 mark

Q21 mark

What kind of number 3 + √2 is

Answer: (B) an irrational number

1. A rational number plus an irrational number is irrational.1 mark

Q31 mark

Discriminant of x² − 3x − 2 = 0

Answer: (B) 17

1. D = b² − 4ac = 9 − 4(1)(−2) = 17.1 mark

Q41 mark

x + 1/x = 3 as ax² + bx + c = 0: find a − b + c

Answer: (A) 5

1. Multiply by x: x² − 3x + 1 = 0, so a = 1, b = −3, c = 1 and a − b + c = 1 + 3 + 1 = 5.1 mark

Q51 mark

abscissa − ordinate for the point (3, −5)

Answer: (D) 8

1. 3 − (−5) = 8.1 mark

Q61 mark

Ratio in which a mid-point divides a segment

Answer: (C) 1 : 1

1. The mid-point is equally far from both ends.1 mark

Q71 mark

Which is not a similarity criterion for triangles

Answer: (D) RHS

1. AAA, SSS and SAS are similarity criteria; RHS is a congruence criterion.1 mark

Q81 mark

Find ∠P from two triangles with given sides and angles

Answer: (C) ∠P = 40°

1. QR/AB = 7.6/3.8 = 2, PR/AC = 6√3/(3√3) = 2, PQ/BC = 12/6 = 2, so △ABC ~ △RQP (SSS), with P matching C. ∠C = 180° − 80° − 60° = 40°, so ∠P = 40°.1 mark

Q91 mark

Tangent PA with OP = 10 cm and ∠APO = 30°: find AP

Answer: (D) 5√3 cm

1. ∠OAP = 90°, so AP = OP cos 30° = 10 × √3/2 = 5√3 cm.1 mark

Q101 mark

Which trigonometric statement is false

Answer: (C) sin 30° = cos 30°

1. sin 30° = ½ but cos 30° = √3/2. The other three are true (each side equals 1 or √2/2).1 mark

Q111 mark

Value of tan²A − 1/cos²A

Answer: (D) −1

1. 1/cos²A = sec²A = 1 + tan²A, so tan²A − sec²A = −1.1 mark

Q121 mark

Which marked angle is the angle of depression

Answer: (C) z

1. The angle of depression is measured at the observer, down from the horizontal line to the line of sight: that is z.1 mark

Q131 mark

Perimeter of the shaded sector in the figure

Answer: (C) l + 2r

1. The shaded sector is bounded by two radii and the arc of length l, so its perimeter is l + 2r (the chord a is inside the region, not on its boundary).1 mark

Q141 mark

Area of a quadrant : area of the circle

Answer: (C) 1 : 4

1. A quadrant is a quarter of the circle.1 mark

Q151 mark

Solid whose curved and total surface areas are equal

Answer: (D) Sphere

1. A sphere has no flat face, so its curved surface is its whole surface.1 mark

Q161 mark

Class mark of the median class of a grouped table

Answer: (D) 62.5

1. Cumulative frequencies: 2, 5, 12, 18, 24, 30. N/2 = 15 first falls in 55–70, so the class mark is (55 + 70)/2 = 62.5.1 mark

Q171 mark

Lower limit of the modal class (runs scored)

Answer: (B) 4000

1. The highest frequency, 10, is in 4000–5000.1 mark

Q181 mark

Which die event is a sure event

Answer: (D) a natural number < 7

1. Every face, 1 to 6, is a natural number less than 7, so the probability is 1.1 mark

Q191 mark

Assertion-reason: HCF is a factor of the LCM

Answer: (A) both true, R explains A

1. The HCF divides a (R), and a divides the LCM, so the HCF divides the LCM (A).1 mark

Q201 mark

Assertion-reason: p for which 4x + py + 8 = 0 and 2x + 2y + 2 = 0 is consistent

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

1. p = 4 makes all three ratios equal to 2 (infinitely many solutions), but any p ≠ 4 gives 4/2 ≠ p/2 and a unique solution, which is also consistent. So 'the value of p is 4' is false. R is a true statement. If you read 'consistent' as 'infinitely many solutions', you would choose (A): check your school's answer key.1 mark

Q212 marks

Solve x/2 + 2y/3 = −1 and x − y/3 = 3

Answer: x = 2, y = −3

1. Clear fractions: 3x + 4y = −6 and 3x − y = 9.½ marks
2. Subtract: 5y = −15, so y = −3.1 mark
3. Then 3x = 9 + y = 6, so x = 2.½ marks

Q222 marks

(a) PQ ∥ RS: prove △POQ ~ △SOR. OR (b) △OSR ~ △OQP with given angles: find ∠OSR and ∠OQP

(a)

Answer: Proved (AA)

1. PQ ∥ RS with transversal PS: ∠QPO = ∠RSO (alternate angles). With transversal QR: ∠PQO = ∠SRO.1 mark
2. So △POQ ~ △SOR by AA similarity (∠POQ = ∠SOR as vertically opposite angles also agrees).1 mark

(b)

Answer: ∠OSR = 55°, ∠OQP = 55°

1. ∠ROQ is an exterior angle of △SOR: ∠OSR = 125° − 70° = 55°.1 mark
2. △OSR ~ △OQP, so ∠OQP = ∠OSR = 55°.1 mark

Q232 marks

Chord of the larger of two concentric circles (10 cm, 6 cm) touching the smaller

Answer: 16 cm

1. The radius to the point of contact M is perpendicular to the chord AB, so AM = √(10² − 6²) = 8 cm.1½ marks
2. AB = 2 × 8 = 16 cm.½ marks

Q242 marks

(a) A and B when tan(A + B) = 1 and tan(A − B) = 1/√3. OR (b) prove tan 45° = 1 geometrically

(a)

Answer: A = 37½°, B = 7½°

1. A + B = 45° and A − B = 30°.1 mark
2. Adding: 2A = 75°, so A = 37½° and B = 7½°.1 mark

(b)

Answer: Shown

1. Take △ABC with ∠B = 90° and ∠A = 45°. Then ∠C = 45°, so the triangle is isosceles and BC = AB.1 mark
2. tan 45° = tan A = BC/AB = 1.1 mark

Q252 marks

Minor segment area: diameter 20 cm, chord subtends 60° (π = 3.14, √3 = 1.73)

Answer: ≈ 9.08 cm²

1. r = 10 cm. Sector area = (60/360) × 3.14 × 100 ≈ 52.33 cm².1 mark
2. The triangle is equilateral: area = (√3/4) × 100 = 43.25 cm². Segment = 52.33 − 43.25 ≈ 9.08 cm².1 mark

Q263 marks

(a) Prove √3 irrational. OR (b) complete a factor tree for x

(a)

Answer: Proof by contradiction

1. Suppose √3 = p/q with q ≠ 0 and p, q co-prime. Then 3q² = p², so 3 divides p.1 mark
2. Write p = 3c: q² = 3c², so 3 divides q.1 mark
3. 3 is then a common factor of p and q, a contradiction. So √3 is irrational.1 mark

(b)

Answer: a = 3, b = 7, y = 420, x = 840 = 2³ × 3 × 5 × 7

1. 210 = a × 70, so a = 3; 35 = 5 × b, so b = 7.1 mark
2. y = 2 × 210 = 420 and x = 2 × 420 = 840.1 mark
3. x = 2 × 2 × 2 × 3 × 5 × 7 = 2³ × 3 × 5 × 7.1 mark

Q273 marks

Quadratic polynomial with sum of zeroes 0 and product −9; find its zeroes

Answer: x² − 9; zeroes 3 and −3

1. p(x) = x² − (sum)x + (product) = x² − 0x − 9.1½ marks
2. x² − 9 = (x − 3)(x + 3).1 mark
3. Zeroes: 3 and −3.½ marks

Q283 marks

(a) Solve x + 3y = 6 and 2x − 3y = 12 graphically. OR (b) complementary angles in the ratio 1 : 2

(a)

Answer: x = 6, y = 0

1. x + 3y = 6 passes through (0, 2) and (6, 0); 2x − 3y = 12 passes through (0, −4) and (6, 0).1 mark
2. Draw both lines on one grid.1 mark
3. They meet at (6, 0), so x = 6, y = 0.1 mark

(b)

Answer: x = 30°, y = 60°

1. Complementary: x + y = 90. Ratio 1 : 2: y = 2x.1½ marks
2. 3x = 90, so x = 30.1 mark
3. y = 60: the angles are 30° and 60°.½ marks

Q293 marks

Prove a rectangle circumscribing a circle is a square

Answer: Proof

1. Let ABCD touch the circle at P, Q, R, S on AB, BC, CD, DA. Tangents from a point are equal: AP = AS, BP = BQ, CR = CQ, DR = DS.1 mark
2. Adding: AB + CD = AD + BC.1 mark
3. In a rectangle AB = CD and AD = BC, so 2AB = 2BC, i.e. AB = BC: all sides are equal and ABCD is a square.1 mark

Q303 marks

Prove (1 + cot²A)/(1 + tan²A) = [(1 − cot A)/(1 − tan A)]²

Answer: Shown

1. LHS = cosec²A/sec²A = cos²A/sin²A = cot²A.1 mark
2. (1 − cot A)/(1 − tan A) = (1 − 1/tan A)/(1 − tan A) = [(tan A − 1)/tan A]/(1 − tan A) = −1/tan A.1½ marks
3. RHS = (−cot A)² = cot²A = LHS.½ marks

Q313 marks

Pens: P(customer does not buy) from 200 pens; P(buys) after mixing in another 100

Answer: 1/10, then 13/15

1. Defective pens = 200 − 180 = 20, so P(not buy) = 20/200 = 1/10.1½ marks
2. Mixed lot: 300 pens, 180 + 80 = 260 good.½ marks
3. P(buy) = 260/300 = 13/15.1 mark

Q325 marks

(a) Squares differ by 180; smaller² = 8 × greater: find the numbers. OR (b) k for equal roots of 2x² + kx + 3 = 0, and the roots

(a)

Answer: 18 and 12

1. Greater x, smaller y: x² − y² = 180 and y² = 8x.1 mark
2. x² − 8x − 180 = 0.1 mark
3. (x − 18)(x + 10) = 0, so x = 18 (x = −10 is rejected: the numbers are positive).1½ marks
4. y² = 8 × 18 = 144, so y = 12.1 mark
5. The numbers are 18 and 12.½ marks

(b)

Answer: k = ±2√6; roots −√6/2, −√6/2 or √6/2, √6/2

1. Real and equal roots: D = k² − 4(2)(3) = 0.1 mark
2. k² = 24, so k = ±2√6.1 mark
3. Equal roots are x = −k/(2a) = −k/4.1 mark
4. k = 2√6: x = −√6/2 (twice).1 mark
5. k = −2√6: x = √6/2 (twice).1 mark

Q335 marks

State BPT; use it to prove ABCD is a trapezium when AO/BO = CO/DO

Answer: Statement and proof

1. BPT: a line drawn parallel to one side of a triangle, meeting the other two sides at distinct points, divides them in the same ratio.1 mark
2. Construction: through O draw OE ∥ AB, meeting AD at E. Figure.1 mark
3. In △ABD, OE ∥ AB, so by BPT DE/EA = DO/OB.1 mark
4. Given AO/BO = CO/DO, so CO/AO = DO/BO = DE/EA. In △ADC, E and O divide DA and CA in the same ratio, so EO ∥ DC (converse of BPT).1½ marks
5. AB ∥ EO and EO ∥ DC give AB ∥ DC, so ABCD is a trapezium.½ marks

Q345 marks

(a) Volume of a cone-on-hemisphere toy (r = 5 cm, cone height = diameter). OR (b) cube topped by a hemisphere of radius 3.5 cm

(a)

Answer: 500π/3 cm³ = 11000/21 cm³ (π = 22/7)

1. r = 5 cm; cone height = diameter = 10 cm.1 mark
2. Cone volume = (1/3)π(5²)(10) = 250π/3 cm³.1½ marks
3. Hemisphere volume = (2/3)π(5³) = 250π/3 cm³.1½ marks
4. Toy = 500π/3 = 500 × 22/21 = 11000/21 cm³ (with π = 22/7).1 mark

(b)

Answer: Edge 7 cm; TSA 332.5 cm²

1. The hemisphere's base must fit on a face, so the least edge = diameter = 7 cm.1 mark
2. TSA = 6 faces − base circle covered + curved hemisphere = 6a² − πr² + 2πr² = 6a² + πr².2 marks
3. = 6 × 49 + (22/7) × 3.5²1 mark
4. = 294 + 38.5 = 332.5 cm².1 mark

Q355 marks

Mean lifetime of 200 electrical components (grouped data)

Answer: 61 hours

1. Class marks xᵢ: 10, 30, 50, 70, 90, 110.1 mark
2. fᵢxᵢ: 100, 1050, 2500, 4200, 2700, 1650.2 marks
3. Σfᵢ = 200 and Σfᵢxᵢ = 12200.1 mark
4. Mean = 12200/200 = 61 hours.1 mark

Q364 marks

Case study: fireman's ladder to a 15 m roof at 60°, then at 30° from the opposite wall

(i)

Answer: 10√3 m

1. sin 60° = 15/L, so L = 15 × 2/√3 = 10√3 m.1 mark

(ii)

Answer: 5√3 m

1. tan 60° = 15/d, so d = 15/√3 = 5√3 m.1 mark

(iii)(a)

Answer: 15√3 m

1. Diagram: building of height 15 m, ladder from the top of the building down to the foot of the opposite wall, 30° with the road.1 mark
2. tan 30° = 15/w, so the road width w = 15√3 m.1 mark

(iii)(b)

Answer: 30 m

1. sin 30° = 15/L.1 mark
2. L = 15 ÷ ½ = 30 m.1 mark

Q374 marks

Case study: spiral of semicircles (radii 50, 100, 150 cm, …) with 10, 20, 30 … saplings

(i)

Answer: 650 cm

1. Radii form an AP with a = 50, d = 50: a₁₃ = 50 + 12 × 50 = 650 cm.1 mark

(ii)

Answer: n = 10

1. 50 + (n − 1)50 = 500, so 50n = 500 and n = 10.1 mark

(iii)(a)

Answer: 660 saplings

1. Saplings 10, 20, 30, …: S₁₁ = (11/2)[2 × 10 + 10 × 10]1 mark
2. = (11/2) × 120 = 660.1 mark

(iii)(b)

Answer: Till the 9th spiral

1. Sₙ = (n/2)[20 + 10(n − 1)] = 5n(n + 1) = 450.1 mark
2. n² + n − 90 = 0, so (n + 10)(n − 9) = 0 and n = 9.1 mark

Q384 marks

Case study: circular park with gates A(10, 20) and B(50, 50); fountains trisect AB

(i)

Answer: C(30, 35)

1. AB is a diameter, so C is its mid-point: ((10 + 50)/2, (20 + 50)/2) = (30, 35).1 mark

(ii)

Answer: 25 units

1. AB = √(40² + 30²) = 50, so the radius = 25.1 mark

(iii)(a)

Answer: P(70/3, 30)

1. P divides AB in the ratio 1 : 2.1 mark
2. P = ((1 × 50 + 2 × 10)/3, (1 × 50 + 2 × 20)/3) = (70/3, 30).1 mark

(iii)(b)

Answer: AQ = 100/3 units

1. Q divides AB in the ratio 2 : 1, so AQ = (2/3) AB.1 mark
2. AQ = (2/3) × 50 = 100/3 units.1 mark