8 higher-order questions on the Class 10 syllabus, set out like a 25-mark, 1-hour paper. Try it in one sitting, then open each worked solution to mark your work.
Our own practice paper, not a CBSE paper. CBSE has not yet published a sample paper, blueprint or Class 10 syllabus for Mathematics Advanced, so we wrote these questions ourselves on the current Class 10 chapters, at the higher-order (HOTS) level CBSE says the paper will have. What CBSE has said about the paper.
Find the smallest natural number that leaves a remainder of 7 when divided by 12, by 15 and by 20, and is also divisible by 11.
Answer: 187
LCM(12, 15, 20) = 2² × 3 × 5 = 60, so the number is 60k + 7 for a whole number k. [1]
Try k = 0, 1, 2, 3, …: 7, 67, 127, 187. 7 = 11 × 0 + 7, 67 = 11 × 6 + 1, 127 = 11 × 11 + 6, 187 = 11 × 17. The smallest is 187. [1]
Check: 187 = 12 × 15 + 7 = 15 × 12 + 7 = 20 × 9 + 7.
Find the value(s) of k for which the pair of equations kx + 3y = k − 3 and 12x + ky = k has no solution.
Answer: k = −6
No solution needs k/12 = 3/k ≠ (k − 3)/k. From k/12 = 3/k: k² = 36, so k = 6 or −6. [1]
If k = 6: 3/k = 1/2 and (k − 3)/k = 3/6 = 1/2. All three ratios are equal, so the lines coincide (infinitely many solutions): reject.
If k = −6: 3/k = −1/2 but (k − 3)/k = −9/−6 = 3/2, so the lines are parallel. Answer: k = −6. [1]
θ is an acute angle and sin θ + cos θ = √2 cos θ. Prove that cos θ − sin θ = √2 sin θ.
Answer: Proof (cos θ = (√2 + 1) sin θ)
From the given equation, sin θ = (√2 − 1) cos θ. [½]
So cos θ = sin θ/(√2 − 1) = sin θ(√2 + 1)/((√2 − 1)(√2 + 1)) = (√2 + 1) sin θ, since (√2 − 1)(√2 + 1) = 2 − 1 = 1. [1]
Hence cos θ − sin θ = (√2 + 1) sin θ − sin θ = √2 sin θ. [½]
In triangle ABC, D is a point on AB and E is a point on AC with DE ∥ BC. F is a point on AD with FE ∥ DC.
Answer: (a) proof; (b) AD = 6 cm, DB = 3 cm
(a) In triangle ABC, DE ∥ BC, so by the Basic Proportionality Theorem AD/DB = AE/EC. In triangle ADC, FE ∥ DC, so AF/FD = AE/EC. [1]
Therefore AF/FD = AD/DB. Inverting and adding 1 to both sides: (FD + AF)/AF = (DB + AD)/AD, that is AD/AF = AB/AD, so AD² = AF × AB. [1]
(b) AD² = 4 × 9 = 36, so AD = 6 cm and DB = 9 − 6 = 3 cm. [1]
a, b and c are real numbers with a ≠ b. The quadratic equation (a − b)x² + (b − c)x + (c − a) = 0 has equal roots. Prove that 2a = b + c.
Answer: Proof (discriminant = ((a − b) − (c − a))²)
Equal roots means the discriminant is zero: (b − c)² − 4(a − b)(c − a) = 0. [1]
Put u = a − b and w = c − a. Then b − c = −(u + w), and the condition becomes (u + w)² − 4uw = (u − w)² = 0. [1]
So u = w: a − b = c − a, which gives 2a = b + c. [1]
A quicker route: the coefficients add up to 0, so x = 1 is a root. With equal roots both roots are 1, so their product (c − a)/(a − b) = 1, which gives the same result.
The median of the distribution below is 28 and the total frequency is 60. Find x and y.
| Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Frequency | 5 | x | 20 | 15 | y |
Answer: x = 9, y = 11
N = 60, so N/2 = 30. The median 28 lies in the class 20–30, so l = 20, h = 10, f = 20 and cf = 5 + x. [1]
28 = 20 + [(30 − (5 + x))/20] × 10, so 8 = (25 − x)/2 and x = 9. [1]
5 + 9 + 20 + 15 + y = 60 gives y = 11. [1]
Check: the cumulative frequencies are 5, 14, 34, 49, 60, and 30 falls between 14 and 34, so 20–30 really is the median class.
Three circles, each of radius 7 cm, are drawn in a plane so that each circle touches the other two from outside.
Answer: (a) side 14 cm; (b) (49√3 − 77) cm² ≈ 7.77 cm²; (c) 22 cm
(a) When two circles touch from outside, the distance between their centres is the sum of the radii: 7 + 7 = 14 cm. All three distances are 14 cm, so the centres form an equilateral triangle of side 14 cm. [1]
(b) Area of the triangle = (√3/4) × 14² = 49√3 cm². [1]
Each angle of the triangle is 60°, so inside the triangle each circle covers a sector of 60°. The three sectors together: 3 × (60/360) × (22/7) × 7² = (1/2) × 154 = 77 cm². [1]
Enclosed area = (49√3 − 77) cm² ≈ 49 × 1.73 − 77 = 84.77 − 77 = 7.77 cm². [1]
(c) The boundary is three arcs, each 60° of a circle of radius 7 cm: 3 × (60/360) × 2 × (22/7) × 7 = 22 cm. [1]
Answer: (a) proof; (b) AF = 7 cm, BD = 5 cm, CE = 3 cm; (c) s = 15, 15 − 8 = 7
(a) Let PA and PB be tangents from P to a circle with centre O, touching it at A and B. Join OA, OB and OP. OA ⊥ PA and OB ⊥ PB (a tangent is perpendicular to the radius at the point of contact). In right triangles OAP and OBP: OA = OB (radii) and OP is common, so the triangles are congruent (RHS) and PA = PB. [2]
(b) By part (a), let AF = AE = x, BF = BD = y, CD = CE = z. Then x + y = 12, y + z = 8, z + x = 10. Adding: 2(x + y + z) = 30, so x + y + z = 15. [1]
Hence AF = x = 15 − 8 = 7 cm, BD = y = 15 − 10 = 5 cm, CE = z = 15 − 12 = 3 cm. [1]
(c) In general 2(x + y + z) = AB + BC + CA = 2s, so x + y + z = s and AF = x = s − (y + z) = s − BC. Here s = 15 and 15 − 8 = 7 = AF. [1]
Practice paper 1 · How the Maths Advanced paper works · Class 10 chapter notes and questions · Class 10 sample papers (80 marks)
Written by CBSE Math Revision on 4 October 2026. Every answer was worked twice: once by hand and once by a computer check. Spotted a mistake? Tell us.