Top 50 MCQs – CBSE Class 10 Maths (2026)
Top 50 MCQs – CBSE Class 10 Maths (2026)
1. If HCF(336, 54) = 6, then LCM(336, 54) is:
✅ Answer: (a) 3024
We use \( \text{HCF} \times \text{LCM} = a \times b \). So, \( \text{LCM} = \frac{336 \times 54}{6} = 3024 \).
2. The zeroes of the polynomial \( x^2 - 3 \) are:
✅ Answer: (a) \( \sqrt{3}, -\sqrt{3} \)
Solve \( x^2 - 3 = 0 \Rightarrow x = \pm \sqrt{3} \).
3. The pair of equations \( 2x + 3y = 5 \) and \( 4x + 6y = 10 \) has:
✅ Answer: (c) Infinitely many solutions
The second equation is exactly twice the first ⇒ coincident lines.
4. In an AP, if \( a = 5 \), \( d = 3 \), then \( a_{10} \) is:
✅ Answer: (b) 32
\( a_n = a + (n-1)d = 5 + 9 \times 3 = 32 \).
5. If \( \sin A = \frac{1}{2} \), then \( \cos A = \)?
✅ Answer: (a) \( \frac{\sqrt{3}}{2} \)
If \( \sin A = \frac{1}{2} \), then \( A = 30^\circ \), so \( \cos 30^\circ = \frac{\sqrt{3}}{2} \).
6. The distance between points (2, 3) and (5, 7) is:
✅ Answer: (a) 5 units
Distance = \( \sqrt{(5-2)^2 + (7-3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).
7. A tangent PQ at point P of a circle of radius 5 cm meets a line through the centre O at Q such that OQ = 13 cm. Length PQ is:
✅ Answer: (a) 12 cm
OP ⊥ PQ ⇒ \( OP^2 + PQ^2 = OQ^2 \) ⇒ \( 5^2 + PQ^2 = 13^2 \) ⇒ \( PQ = \sqrt{169 - 25} = \sqrt{144} = 12 \).
8. If the probability of winning a game is 0.6, then the probability of losing it is:
✅ Answer: (b) 0.4
P(lose) = 1 – P(win) = 1 – 0.6 = 0.4.
9. The median of the data: 13, 15, 16, 18, 20 is:
✅ Answer: (a) 16
Odd number of terms ⇒ median = middle term = 3rd term = 16.
10. The volume of a sphere of radius 3 cm is:
✅ Answer: (a) \( 36\pi \text{cm}^3 \)
Volume = \( \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (27) = 36\pi \).
11. The decimal expansion of \( \frac{17}{8} \) will terminate after how many places?
✅ Answer: (c) 3
\( \frac{17}{8} = \frac{17}{2^3} \). Denominator is of the form \( 2^m5^n \) ⇒ terminates after max(m,n) = 3 places.
12. If one zero of \( 2x^2 - 3x + k \) is reciprocal of the other, then \( k = \)?
✅ Answer: (a) 2
Let zeroes be \( \alpha \) and \( \frac{1}{\alpha} \). Then product = \( \alpha \cdot \frac{1}{\alpha} = 1 = \frac{k}{2} \) ⇒ \( k = 2 \).
13. The value of \( \tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ \) is:
✅ Answer: (b) 1
Use \( \tan(90^\circ - \theta) = \cot \theta \). So, \( \tan 48^\circ = \cot 42^\circ \), \( \tan 23^\circ = \cot 67^\circ \). Product = \( \cot 42^\circ \tan 42^\circ \cdot \cot 67^\circ \tan 67^\circ = 1 \cdot 1 = 1 \).
14. The ratio of the areas of two similar triangles is 9 : 16. The ratio of their corresponding sides is:
✅ Answer: (a) 3 : 4
Area ratio = (side ratio)\(^2\) ⇒ side ratio = \( \sqrt{9/16} = 3/4 \).
15. If the roots of \( x^2 + 4x + k = 0 \) are real and equal, then \( k = \)?
✅ Answer: (a) 4
Discriminant \( D = b^2 - 4ac = 16 - 4k = 0 \) ⇒ \( k = 4 \).
16. The coordinates of the point which divides the line segment joining (4, –3) and (8, 5) in the ratio 3:1 internally are:
✅ Answer: (a) (7, 3)
Section formula: \( \left( \frac{3 \cdot 8 + 1 \cdot 4}{4}, \frac{3 \cdot 5 + 1 \cdot (-3)}{4} \right) = \left( \frac{28}{4}, \frac{12}{4} \right) = (7, 3) \).
17. The value of \( \frac{1 + \tan^2 A}{1 + \cot^2 A} \) is:
✅ Answer: (b) \( \tan^2 A \)
Numerator = \( \sec^2 A \), denominator = \( \csc^2 A \). So, \( \frac{\sec^2 A}{\csc^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \tan^2 A \).
18. A die is thrown once. The probability of getting a prime number is:
✅ Answer: (a) \( \frac{1}{2} \)
Prime numbers on a die: 2, 3, 5 ⇒ 3 outcomes. Total = 6 ⇒ P = \( \frac{3}{6} = \frac{1}{2} \).
19. The sum of first 10 natural numbers is:
✅ Answer: (b) 55
Sum = \( \frac{n(n+1)}{2} = \frac{10 \times 11}{2} = 55 \).
20. The angle of elevation of the top of a tower from a point 30 m away from its foot is \( 30^\circ \). Height of the tower is:
✅ Answer: (c) \( \frac{30}{\sqrt{3}} \) m
\( \tan 30^\circ = \frac{h}{30} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{30} \Rightarrow h = \frac{30}{\sqrt{3}} \).
21. The product of the zeroes of \( 3x^2 - 5x + 2 \) is:
✅ Answer: (a) \( \frac{2}{3} \)
Product = \( \frac{c}{a} = \frac{2}{3} \).
22. If \( \cos \theta = \frac{12}{13} \), then \( \sin \theta = \)?
✅ Answer: (a) \( \frac{5}{13} \)
\( \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13} \).
23. The perimeter of a quadrant of a circle of radius 7 cm is:
✅ Answer: (a) 25 cm
Perimeter = \( \frac{1}{4}(2\pi r) + 2r = \frac{1}{2} \pi r + 2r = \frac{22}{7} \cdot \frac{7}{2} + 14 = 11 + 14 = 25 \) cm.
24. The mode of 5, 7, 6, 5, 9, 5, 8 is:
✅ Answer: (a) 5
5 occurs most frequently (3 times).
25. The value of \( \sin^2 \theta + \cos^2 \theta = \)?
✅ Answer: (b) 1
Fundamental identity.
26. The discriminant of \( 2x^2 - 4x + 3 = 0 \) is:
✅ Answer: (b) -8
\( D = (-4)^2 - 4(2)(3) = 16 - 24 = -8 \).
27. The 10th term of the AP: 2, 7, 12, ... is:
✅ Answer: (b) 47
\( a = 2, d = 5 \), \( a_{10} = 2 + 9 \times 5 = 47 \).
28. If two tangents inclined at \( 60^\circ \) are drawn to a circle of radius 3 cm, then length of each tangent is:
✅ Answer: (a) \( 3\sqrt{3} \) cm
Angle between tangents = \( 60^\circ \) ⇒ angle between radius and tangent = \( 30^\circ \). So, \( \tan 30^\circ = \frac{3}{l} \Rightarrow l = \frac{3}{\tan 30^\circ} = 3\sqrt{3} \).
29. The mean of first five multiples of 3 is:
✅ Answer: (a) 9
Numbers: 3, 6, 9, 12, 15. Mean = \( \frac{45}{5} = 9 \).
30. The value of \( \cot (90^\circ - \theta) = \)?
✅ Answer: (a) \( \tan \theta \)
Co-function identity.
31. The HCF of two consecutive even numbers is:
✅ Answer: (b) 2
Example: 4 and 6 → HCF = 2.
32. The graph of \( y = x^2 - 4 \) intersects x-axis at:
✅ Answer: (a) (2, 0) and (–2, 0)
Set \( y = 0 \): \( x^2 = 4 \Rightarrow x = \pm 2 \).
33. The area of a circle is \( 154 \text{cm}^2 \). Its diameter is:
✅ Answer: (b) 14 cm
\( \pi r^2 = 154 \Rightarrow r^2 = 49 \Rightarrow r = 7 \Rightarrow d = 14 \).
34. If \( P(E) = 0.05 \), then \( P(\text{not } E) = \)?
✅ Answer: (a) 0.95
\( P(\text{not } E) = 1 - P(E) = 0.95 \).
35. The common difference of the AP whose nth term is \( 3n + 5 \) is:
✅ Answer: (a) 3
\( a_n = 3n + 5 \), so \( a_1 = 8, a_2 = 11 \), d = 3.
36. The value of \( \frac{\tan 25^\circ}{\cot 65^\circ} = \)?
✅ Answer: (b) 1
\( \cot 65^\circ = \tan(90^\circ - 65^\circ) = \tan 25^\circ \), so ratio = 1.
37. The curved surface area of a hemisphere of radius r is:
✅ Answer: (a) \( 2\pi r^2 \)
CSA of hemisphere = half of sphere = \( \frac{1}{2} \times 4\pi r^2 = 2\pi r^2 \).
38. The pair of equations \( x + 2y + 5 = 0 \) and \( -3x - 6y + 1 = 0 \) has:
✅ Answer: (b) No solution
Ratios: \( \frac{1}{-3} = \frac{2}{-6} \ne \frac{5}{1} \) ⇒ parallel lines.
39. The probability of an impossible event is:
✅ Answer: (b) 0
40. If \( \alpha, \beta \) are roots of \( x^2 - 5x + 6 = 0 \), then \( \alpha + \beta = \)?
✅ Answer: (a) 5
Sum of roots = \( -\frac{b}{a} = 5 \).
41. The distance of point (–3, 4) from origin is:
✅ Answer: (a) 5
\( \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = 5 \).
42. The value of \( \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ = \)?
✅ Answer: (b) 1
This is \( \sin(A+B) = \sin 90^\circ = 1 \).
43. The total surface area of a cube of side 5 cm is:
✅ Answer: (b) 150 cm²
TSA = \( 6a^2 = 6 \times 25 = 150 \).
44. The empirical relationship between mean, median, and mode is:
✅ Answer: (a) Mode = 3 Median – 2 Mean
45. The value of \( \cos 0^\circ + \sin 90^\circ = \)?
✅ Answer: (c) 2
\( \cos 0^\circ = 1, \sin 90^\circ = 1 \), sum = 2.
46. The number of tangents that can be drawn from an external point to a circle is:
✅ Answer: (b) 2
47. The mid-point of line segment joining (–1, 7) and (4, –3) is:
✅ Answer: (a) (1.5, 2)
Midpoint = \( \left( \frac{-1+4}{2}, \frac{7 + (-3)}{2} \right) = (1.5, 2) \).
48. The value of \( \frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} = \)?
✅ Answer: (a) 0
\( \tan 45^\circ = 1 \), so \( \frac{1 - 1}{1 + 1} = 0 \).
49. The probability of getting a number less than 3 when a die is thrown is:
✅ Answer: (a) \( \frac{1}{3} \)
Favorable: 1, 2 ⇒ 2 outcomes. P = \( \frac{2}{6} = \frac{1}{3} \).
50. The height of a cone with radius 3 cm and volume \( 12\pi \text{cm}^3 \) is:
✅ Answer: (a) 4 cm
Volume = \( \frac{1}{3}\pi r^2 h = 12\pi \Rightarrow \frac{1}{3} \pi (9) h = 12\pi \Rightarrow 3h = 12 \Rightarrow h = 4 \).

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