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Practice · Class 9–10

Olympiad Practice Questions for Class 9–10

20 verified olympiad-style problems for Class 9 and Class 10 students, with instant solutions and explanations. Covers quadratics, coordinate geometry, trigonometry, arithmetic progressions, circle geometry. Free forever — no login, no ads.

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Question 1 of 20 · Class 9–10
If x² − 5x + 6 = 0, the roots are:
1 and 6
2 and 3
−2 and −3
−1 and −6
Answer: 2 and 3 — Factorise: (x − 2)(x − 3) = 0. So x = 2 or x = 3.
Question 2 of 20 · Class 9–10
The distance between points (3, 4) and (0, 0) is:
3
4
5
7
Answer: 5 — Distance = √(3² + 4²) = √(9 + 16) = √25 = 5.
Question 3 of 20 · Class 9–10
If sin θ = 4/5 and θ is acute, then cos θ =
3/5
4/5
5/3
5/4
Answer: 3/5 — cos²θ = 1 − sin²θ = 1 − 16/25 = 9/25. Since θ is acute, cos θ = 3/5.
Question 4 of 20 · Class 9–10
The HCF of 84 and 120 is:
6
12
24
36
Answer: 12 — 84 = 2² × 3 × 7; 120 = 2³ × 3 × 5. HCF = 2² × 3 = 12.
Question 5 of 20 · Class 9–10
A cylinder has radius 7 cm and height 10 cm. Its volume (π = 22/7) is:
154 cm³
440 cm³
1540 cm³
2200 cm³
Answer: 1540 cm³ — V = πr²h = (22/7) × 49 × 10 = 22 × 70 = 1540 cm³.
Question 6 of 20 · Class 9–10
The 15th term of the arithmetic progression 3, 7, 11, 15, … is:
55
59
63
67
Answer: 59 — a = 3, d = 4. T₁₅ = a + 14d = 3 + 56 = 59.
Question 7 of 20 · Class 9–10
If log₁₀ 2 = 0.301, then log₁₀ 8 is:
0.602
0.903
0.301
0.699
Answer: 0.903 — log 8 = log 2³ = 3 × log 2 = 3 × 0.301 = 0.903.
Question 8 of 20 · Class 9–10
The probability of drawing a king from a standard deck of 52 cards is:
1/13
1/26
1/52
4/13
Answer: 1/13 — There are 4 kings in 52 cards. Probability = 4/52 = 1/13.
Question 9 of 20 · Class 9–10
A tangent to a circle is perpendicular to the radius at the point of contact. If the radius is 8 cm and the tangent from an external point is 15 cm, the distance from the point to the centre is:
7 cm
17 cm
23 cm
√161 cm
Answer: 17 cm — Right triangle: distance² = 8² + 15² = 64 + 225 = 289. Distance = √289 = 17 cm.
Question 10 of 20 · Class 9–10
The mean of 5 numbers is 12. If a sixth number 18 is added, the new mean is:
13
14
15
16
Answer: 13 — Sum of 5 numbers = 60. New sum = 78. New mean = 78 ÷ 6 = 13.
Question 11 of 20 · Class 9–10
If the roots of x² − px + 12 = 0 are consecutive integers, then p =
5
6
7
8
Answer: 7 — Consecutive integers with product 12: 3 and 4. Sum = 7 = p.
Question 12 of 20 · Class 9–10
In an AP, the 5th term is 17 and the 12th term is 38. The common difference is:
2
3
4
5
Answer: 3 — T₁₂ − T₅ = 7d. So 38 − 17 = 21 = 7d, giving d = 3.
Question 13 of 20 · Class 9–10
If sin θ + cos θ = √2, then sin θ · cos θ =
0
1/2
√2/2
1
Answer: 1/2 — Square: (sinθ + cosθ)² = 2. Expand: sin²θ + 2sinθcosθ + cos²θ = 2. Since sin²θ + cos²θ = 1: 1 + 2sinθcosθ = 2. So sinθcosθ = 1/2.
Question 14 of 20 · Class 9–10
The distance between points (a, 0) and (0, b) is:
a + b
√(a² + b²)
√(a² − b²)
a² + b²
Answer: √(a² + b²) — Distance = √((a − 0)² + (0 − b)²) = √(a² + b²).
Question 15 of 20 · Class 9–10
In a circle, a chord of length 8 cm is 3 cm from the centre. The radius is:
4 cm
5 cm
√73 cm
√55 cm
Answer: 5 cm — The perpendicular from centre bisects the chord. Half-chord = 4 cm. By Pythagoras: r² = 3² + 4² = 25. Radius = 5 cm.
Question 16 of 20 · Class 9–10
A number is chosen at random from 1 to 100. Probability it is a perfect square is:
1/10
1/20
1/25
3/50
Answer: 1/10 — Perfect squares from 1 to 100: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 — that is 10 numbers. Probability = 10/100 = 1/10.
Question 17 of 20 · Class 9–10
If log₁₀ 2 = 0.301 and log₁₀ 3 = 0.477, then log₁₀ 6 =
0.176
0.699
0.778
0.903
Answer: 0.778 — log 6 = log(2 × 3) = log 2 + log 3 = 0.301 + 0.477 = 0.778.
Question 18 of 20 · Class 9–10
The volume of a cone with radius 6 cm and height 7 cm is: (π = 22/7)
176 cm³
264 cm³
352 cm³
792 cm³
Answer: 264 cm³ — V = (1/3)πr²h = (1/3)(22/7)(36)(7) = (1/3)(22)(36) = 22 × 12 = 264 cm³.
Question 19 of 20 · Class 9–10
The sum of first n odd natural numbers is:
n(n+1)/2
n²
n(n+1)
2n − 1
Answer: n² — 1 + 3 + 5 + … + (2n−1) = n². Standard identity, verified: n=3 gives 1+3+5 = 9 = 3² ✓.
Question 20 of 20 · Class 9–10
If a two-digit number is 4 times the sum of its digits, and reversing it gives a number 27 more, the original number is:
24
36
48
63
Answer: 36 — Let number = 10a + b. Then 10a + b = 4(a + b) → 6a = 3b → b = 2a. Also 10b + a = 10a + b + 27 → 9b − 9a = 27 → b − a = 3. Solving: b = 2a and b − a = 3 gives a = 3, b = 6. Number = 36.

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