These essential MCQ questions are selected from the most important topics in CBSE Class 10 Mathematics 2026 as per the latest NCERT syllabus. Sections include Real Numbers, Polynomials, Pair of Linear Equations, Quadratic Equations, Triangles, Introduction to Trigonometry, and Statistics & Probability. For truly unlimited daily MCQ practice, visit Vooo AI Education.
📐 CBSE Class 10 Mathematics
1The HCF of 12 and 18 is:
Answer: B — 6
HCF (Highest Common Factor) is the largest number that divides both numbers. 12 = 2² × 3; 18 = 2 × 3². HCF = 2¹ × 3¹ = 6. LCM = 2² × 3² = 36. Key relation: HCF × LCM = Product of two numbers → 6 × 36 = 216 = 12 × 18 ✓. The Fundamental Theorem of Arithmetic states that every integer > 1 can be uniquely expressed as a product of prime factors.
HCF (Highest Common Factor) is the largest number that divides both numbers. 12 = 2² × 3; 18 = 2 × 3². HCF = 2¹ × 3¹ = 6. LCM = 2² × 3² = 36. Key relation: HCF × LCM = Product of two numbers → 6 × 36 = 216 = 12 × 18 ✓. The Fundamental Theorem of Arithmetic states that every integer > 1 can be uniquely expressed as a product of prime factors.
2The zeroes of the polynomial p(x) = x² - 5x + 6 are:
Answer: B — 2 and 3
Factorising x² - 5x + 6: find two numbers whose product = 6 and sum = -5. Those are -2 and -3. So x² - 5x + 6 = (x-2)(x-3) = 0 → x = 2 or x = 3. Verification: Sum of zeroes = 2+3 = 5 = -(-5)/1 ✓; Product of zeroes = 2×3 = 6 = 6/1 ✓. For ax²+bx+c: sum of zeroes = -b/a; product of zeroes = c/a.
Factorising x² - 5x + 6: find two numbers whose product = 6 and sum = -5. Those are -2 and -3. So x² - 5x + 6 = (x-2)(x-3) = 0 → x = 2 or x = 3. Verification: Sum of zeroes = 2+3 = 5 = -(-5)/1 ✓; Product of zeroes = 2×3 = 6 = 6/1 ✓. For ax²+bx+c: sum of zeroes = -b/a; product of zeroes = c/a.
3The discriminant of the quadratic equation 2x² - 4x + 2 = 0 is:
Answer: A — 0
Discriminant D = b² - 4ac. Here a=2, b=-4, c=2. D = (-4)² - 4(2)(2) = 16 - 16 = 0. When D = 0: two equal real roots (x = -b/2a = 4/4 = 1). When D > 0: two distinct real roots. When D < 0: no real roots (complex roots). Here both roots = 1. This equation can be simplified to x² - 2x + 1 = 0 → (x-1)² = 0 → x = 1 (repeated root).
Discriminant D = b² - 4ac. Here a=2, b=-4, c=2. D = (-4)² - 4(2)(2) = 16 - 16 = 0. When D = 0: two equal real roots (x = -b/2a = 4/4 = 1). When D > 0: two distinct real roots. When D < 0: no real roots (complex roots). Here both roots = 1. This equation can be simplified to x² - 2x + 1 = 0 → (x-1)² = 0 → x = 1 (repeated root).
4If two triangles are similar, then their corresponding sides are:
Answer: B — Proportional
In similar triangles, all corresponding angles are equal and corresponding sides are in proportion (ratio). AA (Angle-Angle), SSS (Side-Side-Side), and SAS (Side-Angle-Side) are criteria for triangle similarity. Congruent triangles have equal corresponding sides AND equal angles. The ratio of areas of similar triangles = square of the ratio of their corresponding sides. Basic Proportionality Theorem (Thales' Theorem) is fundamental here.
In similar triangles, all corresponding angles are equal and corresponding sides are in proportion (ratio). AA (Angle-Angle), SSS (Side-Side-Side), and SAS (Side-Angle-Side) are criteria for triangle similarity. Congruent triangles have equal corresponding sides AND equal angles. The ratio of areas of similar triangles = square of the ratio of their corresponding sides. Basic Proportionality Theorem (Thales' Theorem) is fundamental here.
5The value of sin²θ + cos²θ is:
Answer: B — 1
sin²θ + cos²θ = 1 is the most fundamental trigonometric identity. Other key identities: 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ. These are derived from Pythagoras theorem. Standard values to memorise: sin 0°=0, sin 30°=1/2, sin 45°=1/√2, sin 60°=√3/2, sin 90°=1. cos values are the reverse. tan = sin/cos. These identities are used extensively in Class 10 proofs.
sin²θ + cos²θ = 1 is the most fundamental trigonometric identity. Other key identities: 1 + tan²θ = sec²θ; 1 + cot²θ = cosec²θ. These are derived from Pythagoras theorem. Standard values to memorise: sin 0°=0, sin 30°=1/2, sin 45°=1/√2, sin 60°=√3/2, sin 90°=1. cos values are the reverse. tan = sin/cos. These identities are used extensively in Class 10 proofs.
6The mean of 5 numbers is 20. If one number is removed and the mean becomes 18, the removed number is:
Answer: C — 28
Sum of 5 numbers = 5 × 20 = 100. After removing one number, sum of 4 numbers = 4 × 18 = 72. Removed number = 100 - 72 = 28. Mean = Sum of observations / Number of observations. This is a direct application of the mean formula. In statistics, Class 10 also covers median (middle value) and mode (most frequent value) for grouped and ungrouped data.
Sum of 5 numbers = 5 × 20 = 100. After removing one number, sum of 4 numbers = 4 × 18 = 72. Removed number = 100 - 72 = 28. Mean = Sum of observations / Number of observations. This is a direct application of the mean formula. In statistics, Class 10 also covers median (middle value) and mode (most frequent value) for grouped and ungrouped data.
7The nth term of an AP is given by aₙ = 3n + 2. Its common difference is:
Answer: C — 3
aₙ = 3n + 2. Common difference d = aₙ₊₁ - aₙ = [3(n+1)+2] - [3n+2] = 3n+5 - 3n-2 = 3. First term a₁ = 3(1)+2 = 5. Second term a₂ = 3(2)+2 = 8. d = 8-5 = 3 ✓. In general, if aₙ = an + b (linear in n), the common difference = coefficient of n = a. The first term = a + b. This is a direct and important observation for Class 10 AP problems.
aₙ = 3n + 2. Common difference d = aₙ₊₁ - aₙ = [3(n+1)+2] - [3n+2] = 3n+5 - 3n-2 = 3. First term a₁ = 3(1)+2 = 5. Second term a₂ = 3(2)+2 = 8. d = 8-5 = 3 ✓. In general, if aₙ = an + b (linear in n), the common difference = coefficient of n = a. The first term = a + b. This is a direct and important observation for Class 10 AP problems.
8A tangent to a circle makes an angle of __ with the radius at the point of tangency.
Answer: D — 90°
A tangent to a circle at any point is perpendicular to the radius drawn to that point. This is a fundamental theorem of circles in Class 10. Consequences: (1) From an external point, two tangents drawn to a circle are equal in length. (2) The line joining the centre to the external point bisects the angle between the two tangents. These properties are used extensively in Class 10 geometry problems involving tangents.
A tangent to a circle at any point is perpendicular to the radius drawn to that point. This is a fundamental theorem of circles in Class 10. Consequences: (1) From an external point, two tangents drawn to a circle are equal in length. (2) The line joining the centre to the external point bisects the angle between the two tangents. These properties are used extensively in Class 10 geometry problems involving tangents.
9The volume of a sphere of radius r is:
Answer: C — (4/3)πr³
Volume of sphere = (4/3)πr³. Surface area of sphere = 4πr². Volume of hemisphere = (2/3)πr³. Surface area of hemisphere (curved) = 2πr². Other key formulas: Volume of cylinder = πr²h; Volume of cone = (1/3)πr²h; Volume of cube = a³; Volume of cuboid = l×b×h. Surface area of sphere is the area of the outer surface — useful when coating or painting a spherical object.
Volume of sphere = (4/3)πr³. Surface area of sphere = 4πr². Volume of hemisphere = (2/3)πr³. Surface area of hemisphere (curved) = 2πr². Other key formulas: Volume of cylinder = πr²h; Volume of cone = (1/3)πr²h; Volume of cube = a³; Volume of cuboid = l×b×h. Surface area of sphere is the area of the outer surface — useful when coating or painting a spherical object.
10A card is drawn from a well-shuffled deck of 52 cards. The probability of drawing a king is:
Answer: B — 1/13
A standard deck has 52 cards with 4 kings (one per suit: hearts, diamonds, clubs, spades). P(king) = 4/52 = 1/13. Probability = Favourable outcomes / Total outcomes. Key facts: 52 cards = 4 suits × 13 cards. Each suit has Ace, 2-10, Jack, Queen, King. P(face card) = 12/52 = 3/13. P(red card) = 26/52 = 1/2. P(ace) = 4/52 = 1/13. Probability always lies between 0 and 1 inclusive.
A standard deck has 52 cards with 4 kings (one per suit: hearts, diamonds, clubs, spades). P(king) = 4/52 = 1/13. Probability = Favourable outcomes / Total outcomes. Key facts: 52 cards = 4 suits × 13 cards. Each suit has Ace, 2-10, Jack, Queen, King. P(face card) = 12/52 = 3/13. P(red card) = 26/52 = 1/2. P(ace) = 4/52 = 1/13. Probability always lies between 0 and 1 inclusive.
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