These 50 MCQ questions are specially designed for Math Olympiad competitions including IMO (International Mathematics Olympiad), NTSE, and school-level maths competitions 2026. Topics cover Number Theory, Algebra, Geometry, Logical Reasoning and Pattern Recognition. These questions are more challenging than standard school maths. Use Show Answer to understand step-by-step solutions. For unlimited daily maths MCQ practice, visit Vooo AI Education.
🔢 Number Theory (Q1–Q15)
1What is the sum of all prime numbers between 1 and 20?
Answer: C — 77
Primes between 1 and 20: 2, 3, 5, 7, 11, 13, 17, 19. Sum = 2+3+5+7+11+13+17+19 = 77.
Primes between 1 and 20: 2, 3, 5, 7, 11, 13, 17, 19. Sum = 2+3+5+7+11+13+17+19 = 77.
2How many factors does 36 have?
Answer: C — 9
36 = 2² × 3². Number of factors = (2+1)(2+1) = 9. Factors: 1,2,3,4,6,9,12,18,36.
36 = 2² × 3². Number of factors = (2+1)(2+1) = 9. Factors: 1,2,3,4,6,9,12,18,36.
3What is the remainder when 2¹⁰⁰ is divided by 3?
Answer: A — 1
2¹=2, 2²=4≡1(mod 3). So 2² ≡ 1 (mod 3). 2¹⁰⁰ = (2²)⁵⁰ ≡ 1⁵⁰ = 1 (mod 3). Remainder = 1.
2¹=2, 2²=4≡1(mod 3). So 2² ≡ 1 (mod 3). 2¹⁰⁰ = (2²)⁵⁰ ≡ 1⁵⁰ = 1 (mod 3). Remainder = 1.
4The LCM of two numbers is 180 and their HCF is 12. If one number is 36, what is the other?
Answer: C — 60
Product of two numbers = LCM × HCF = 180 × 12 = 2160. Other number = 2160 ÷ 36 = 60.
Product of two numbers = LCM × HCF = 180 × 12 = 2160. Other number = 2160 ÷ 36 = 60.
5Which is the smallest 4-digit perfect square?
Answer: B — 1024
32² = 1024. This is the smallest 4-digit perfect square since 31² = 961 (3 digits) and 32² = 1024 (4 digits).
32² = 1024. This is the smallest 4-digit perfect square since 31² = 961 (3 digits) and 32² = 1024 (4 digits).
6If a number is divisible by both 4 and 6, it is necessarily divisible by:
Answer: B — 12
LCM(4,6) = 12. So any number divisible by both 4 and 6 must be divisible by 12. Not necessarily 24 (e.g. 12 itself).
LCM(4,6) = 12. So any number divisible by both 4 and 6 must be divisible by 12. Not necessarily 24 (e.g. 12 itself).
7What is 1 + 2 + 3 + ... + 100?
Answer: C — 5050
Sum = n(n+1)/2 = 100×101/2 = 5050. This formula was famously discovered by Carl Friedrich Gauss as a child.
Sum = n(n+1)/2 = 100×101/2 = 5050. This formula was famously discovered by Carl Friedrich Gauss as a child.
8How many zeros are at the end of 100! (100 factorial)?
Answer: C — 24
Count factors of 5: ⌊100/5⌋ + ⌊100/25⌋ = 20 + 4 = 24. Each factor of 5 pairs with a factor of 2 to make a trailing zero.
Count factors of 5: ⌊100/5⌋ + ⌊100/25⌋ = 20 + 4 = 24. Each factor of 5 pairs with a factor of 2 to make a trailing zero.
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Start Free Practice →📐 Algebra & Equations (Q9–Q20)
9If x + y = 10 and xy = 21, what is x² + y²?
Answer: B — 58
x² + y² = (x+y)² - 2xy = 10² - 2(21) = 100 - 42 = 58.
x² + y² = (x+y)² - 2xy = 10² - 2(21) = 100 - 42 = 58.
10The value of (a-b)² + 2ab equals:
Answer: A — a² + b²
(a-b)² + 2ab = a² - 2ab + b² + 2ab = a² + b².
(a-b)² + 2ab = a² - 2ab + b² + 2ab = a² + b².
11If 2x - 3 = 11, what is the value of x?
Answer: C — 7
2x - 3 = 11 → 2x = 14 → x = 7.
2x - 3 = 11 → 2x = 14 → x = 7.
12The sum of three consecutive even numbers is 54. What is the largest?
Answer: C — 20
Let the numbers be n, n+2, n+4. Sum = 3n+6 = 54 → n = 16. Numbers are 16, 18, 20. Largest = 20.
Let the numbers be n, n+2, n+4. Sum = 3n+6 = 54 → n = 16. Numbers are 16, 18, 20. Largest = 20.
13If x² = 144, what are the possible values of x?
Answer: B — 12 and -12
x² = 144 means x = ±√144 = ±12. Both 12² and (-12)² equal 144.
x² = 144 means x = ±√144 = ±12. Both 12² and (-12)² equal 144.
14What is the value of 3³ + 4²?
Answer: C — 43
3³ = 27 and 4² = 16. So 27 + 16 = 43.
3³ = 27 and 4² = 16. So 27 + 16 = 43.
📏 Geometry & Mensuration (Q15–Q25)
15The sum of interior angles of a hexagon is:
Answer: B — 720°
Sum of interior angles = (n-2) × 180° = (6-2) × 180° = 4 × 180° = 720°.
Sum of interior angles = (n-2) × 180° = (6-2) × 180° = 4 × 180° = 720°.
16If the radius of a circle is doubled, its area becomes:
Answer: C — 4 times
Area = πr². If r becomes 2r, area = π(2r)² = 4πr². So area becomes 4 times the original.
Area = πr². If r becomes 2r, area = π(2r)² = 4πr². So area becomes 4 times the original.
17In a right triangle, if the two shorter sides are 3 cm and 4 cm, the hypotenuse is:
Answer: B — 5 cm
Pythagorean theorem: h² = 3² + 4² = 9 + 16 = 25. h = 5 cm. The 3-4-5 triangle is a classic Pythagorean triple.
Pythagorean theorem: h² = 3² + 4² = 9 + 16 = 25. h = 5 cm. The 3-4-5 triangle is a classic Pythagorean triple.
18The volume of a cube with side 4 cm is:
Answer: C — 64 cm³
Volume of cube = side³ = 4³ = 4×4×4 = 64 cm³.
Volume of cube = side³ = 4³ = 4×4×4 = 64 cm³.
19Two angles of a triangle are 60° and 70°. What is the third angle?
Answer: B — 50°
Sum of angles in a triangle = 180°. Third angle = 180° - 60° - 70° = 50°.
Sum of angles in a triangle = 180°. Third angle = 180° - 60° - 70° = 50°.
20The diagonal of a square with side 5 cm is:
Answer: C — 5√2 cm
Diagonal of square = side × √2 = 5√2 ≈ 7.07 cm. Using Pythagoras: d² = 5² + 5² = 50, d = √50 = 5√2.
Diagonal of square = side × √2 = 5√2 ≈ 7.07 cm. Using Pythagoras: d² = 5² + 5² = 50, d = √50 = 5√2.
🧩 Logical Reasoning & Patterns (Q21–Q30)
21What comes next? 1, 1, 2, 3, 5, 8, 13, __
Answer: C — 21
This is the Fibonacci sequence where each number is the sum of the two preceding ones. 8 + 13 = 21.
This is the Fibonacci sequence where each number is the sum of the two preceding ones. 8 + 13 = 21.
22A clock shows 3:15. What is the angle between the hour and minute hands?
Answer: A — 7.5°
At 3:15, minute hand is at 90°. Hour hand moves 0.5° per minute, so at 3:15 it is at 90° + 15×0.5° = 97.5°. Angle = 97.5° - 90° = 7.5°.
At 3:15, minute hand is at 90°. Hour hand moves 0.5° per minute, so at 3:15 it is at 90° + 15×0.5° = 97.5°. Angle = 97.5° - 90° = 7.5°.
23If you arrange the letters of MATHS in alphabetical order, which letter comes in the middle?
Answer: C — M
MATHS alphabetically: A, H, M, S, T. The middle (3rd) letter is M.
MATHS alphabetically: A, H, M, S, T. The middle (3rd) letter is M.
24How many triangles are in a regular hexagon divided into 6 equal triangles?
Answer: C — 12
6 small triangles + 6 larger triangles formed by adjacent pairs = 12 triangles total. Counting all possible triangles: individual (6) + two-triangle combos (6) = 12.
6 small triangles + 6 larger triangles formed by adjacent pairs = 12 triangles total. Counting all possible triangles: individual (6) + two-triangle combos (6) = 12.
25In a group of 30 students, 18 play cricket, 15 play football and 5 play both. How many play neither?
Answer: B — 2
By inclusion-exclusion: Cricket ∪ Football = 18 + 15 - 5 = 28. Neither = 30 - 28 = 2.
By inclusion-exclusion: Cricket ∪ Football = 18 + 15 - 5 = 28. Neither = 30 - 28 = 2.
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