These essential MCQ questions are selected from the most important topics in NTSE (National Talent Search Examination) 2026 Mathematics section for Class 10 students. Sections include Number System, Algebra, Geometry, Mensuration, Statistics, Probability and Trigonometry as per the NCERT Class 9 and Class 10 syllabus. For truely unlimited daily MCQ practice, visit Vooo AI Education.
📐 NTSE Mathematics
1The HCF of 36 and 48 is:
Answer: B — 12
HCF (Highest Common Factor) of 36 and 48: Factors of 36 = 1,2,3,4,6,9,12,18,36; Factors of 48 = 1,2,3,4,6,8,12,16,24,48. Highest common factor = 12. Alternatively: 36 = 2²×3², 48 = 2⁴×3. HCF = 2²×3 = 12. LCM = 2⁴×3² = 144. Note: HCF × LCM = 36 × 48 = 1728 = 12 × 144 ✓.
HCF (Highest Common Factor) of 36 and 48: Factors of 36 = 1,2,3,4,6,9,12,18,36; Factors of 48 = 1,2,3,4,6,8,12,16,24,48. Highest common factor = 12. Alternatively: 36 = 2²×3², 48 = 2⁴×3. HCF = 2²×3 = 12. LCM = 2⁴×3² = 144. Note: HCF × LCM = 36 × 48 = 1728 = 12 × 144 ✓.
2If a triangle has sides 3, 4 and 5, it is:
Answer: C — Right-angled
3² + 4² = 9 + 16 = 25 = 5². By the converse of Pythagoras theorem, if a² + b² = c², the triangle is right-angled. The (3, 4, 5) triplet is the smallest Pythagorean triple. Other Pythagorean triples: (5,12,13), (8,15,17), (7,24,25). A right-angled triangle has one angle = 90°.
3² + 4² = 9 + 16 = 25 = 5². By the converse of Pythagoras theorem, if a² + b² = c², the triangle is right-angled. The (3, 4, 5) triplet is the smallest Pythagorean triple. Other Pythagorean triples: (5,12,13), (8,15,17), (7,24,25). A right-angled triangle has one angle = 90°.
3The value of sin²θ + cos²θ equals:
Answer: C — 1
sin²θ + cos²θ = 1 is the fundamental Pythagorean trigonometric identity. Derived from: in a right triangle with hypotenuse r, sin θ = perpendicular/r, cos θ = base/r. So sin²θ + cos²θ = (p² + b²)/r² = r²/r² = 1. Other identities: 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ.
sin²θ + cos²θ = 1 is the fundamental Pythagorean trigonometric identity. Derived from: in a right triangle with hypotenuse r, sin θ = perpendicular/r, cos θ = base/r. So sin²θ + cos²θ = (p² + b²)/r² = r²/r² = 1. Other identities: 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ.
4The mean of 5 numbers is 20. If one number is removed, the mean becomes 15. What is the removed number?
Answer: A — 40
Sum of 5 numbers = 5 × 20 = 100. Sum of remaining 4 numbers = 4 × 15 = 60. Removed number = 100 - 60 = 40. Always remember: Mean = Sum/Count. To find the sum, multiply mean by count. This approach works for all mean-related problems.
Sum of 5 numbers = 5 × 20 = 100. Sum of remaining 4 numbers = 4 × 15 = 60. Removed number = 100 - 60 = 40. Always remember: Mean = Sum/Count. To find the sum, multiply mean by count. This approach works for all mean-related problems.
5The area of a circle with radius 7 cm is (use π = 22/7):
Answer: B — 154 cm²
Area of circle = πr² = 22/7 × 7 × 7 = 22 × 7 = 154 cm². Circumference = 2πr = 2 × 22/7 × 7 = 44 cm. Remember: Area uses r² while circumference uses r (not r²). Sector area = (θ/360°) × πr². These formulae are frequently used in NTSE mensuration questions.
Area of circle = πr² = 22/7 × 7 × 7 = 22 × 7 = 154 cm². Circumference = 2πr = 2 × 22/7 × 7 = 44 cm. Remember: Area uses r² while circumference uses r (not r²). Sector area = (θ/360°) × πr². These formulae are frequently used in NTSE mensuration questions.
6Factorise: x² - 5x + 6
Answer: B — (x-2)(x-3)
To factorise x² - 5x + 6: find two numbers that multiply to +6 and add to -5. Those numbers are -2 and -3: (-2)×(-3) = +6, (-2)+(-3) = -5. So x² - 5x + 6 = (x-2)(x-3). Verification: (x-2)(x-3) = x² - 3x - 2x + 6 = x² - 5x + 6 ✓.
To factorise x² - 5x + 6: find two numbers that multiply to +6 and add to -5. Those numbers are -2 and -3: (-2)×(-3) = +6, (-2)+(-3) = -5. So x² - 5x + 6 = (x-2)(x-3). Verification: (x-2)(x-3) = x² - 3x - 2x + 6 = x² - 5x + 6 ✓.
7The probability of getting a head when a fair coin is tossed is:
Answer: C — 1/2
Probability = Favourable outcomes / Total outcomes = 1/2 (one head outcome out of two possible outcomes: H or T). For two tosses: P(both heads) = 1/2 × 1/2 = 1/4. P(at least one head) = 1 - P(no heads) = 1 - 1/4 = 3/4. Probability always lies between 0 (impossible) and 1 (certain).
Probability = Favourable outcomes / Total outcomes = 1/2 (one head outcome out of two possible outcomes: H or T). For two tosses: P(both heads) = 1/2 × 1/2 = 1/4. P(at least one head) = 1 - P(no heads) = 1 - 1/4 = 3/4. Probability always lies between 0 (impossible) and 1 (certain).
8If 3x + 5 = 20, then x =
Answer: C — 5
3x + 5 = 20 → 3x = 20 - 5 = 15 → x = 15/3 = 5. Linear equations: always perform the same operation on both sides. Verification: 3(5) + 5 = 15 + 5 = 20 ✓. For simultaneous equations (two variables), use substitution or elimination method.
3x + 5 = 20 → 3x = 20 - 5 = 15 → x = 15/3 = 5. Linear equations: always perform the same operation on both sides. Verification: 3(5) + 5 = 15 + 5 = 20 ✓. For simultaneous equations (two variables), use substitution or elimination method.
9The number of diagonals in a hexagon is:
Answer: B — 9
Number of diagonals in a polygon = n(n-3)/2, where n = number of sides. For hexagon (n=6): 6(6-3)/2 = 6×3/2 = 9 diagonals. For other polygons: Triangle (3 sides) = 0; Quadrilateral (4) = 2; Pentagon (5) = 5; Hexagon (6) = 9; Octagon (8) = 20. This formula is important for NTSE geometry.
Number of diagonals in a polygon = n(n-3)/2, where n = number of sides. For hexagon (n=6): 6(6-3)/2 = 6×3/2 = 9 diagonals. For other polygons: Triangle (3 sides) = 0; Quadrilateral (4) = 2; Pentagon (5) = 5; Hexagon (6) = 9; Octagon (8) = 20. This formula is important for NTSE geometry.
10A number is divisible by 11 if:
Answer: C — Difference of sum of alternate digits is divisible by 11 (or zero)
Divisibility rule for 11: subtract the sum of digits at even positions from sum of digits at odd positions; if the result is 0 or divisible by 11, the number is divisible by 11. Example: 121: (1+1) - 2 = 0. 1331: (1+3) - (3+1) = 4-4 = 0 ✓. Other rules: divisible by 2 (even); by 3 (digit sum divisible by 3); by 9 (digit sum divisible by 9).
Divisibility rule for 11: subtract the sum of digits at even positions from sum of digits at odd positions; if the result is 0 or divisible by 11, the number is divisible by 11. Example: 121: (1+1) - 2 = 0. 1331: (1+3) - (3+1) = 4-4 = 0 ✓. Other rules: divisible by 2 (even); by 3 (digit sum divisible by 3); by 9 (digit sum divisible by 9).
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