Mathematics is an integral part of a student’s curriculum in Class 10, as it lays a foundation for further education. It helps develop practical skills in solving problems. Thus, the concepts covered in the Class 10 Maths exam are essential to help students get ready for the board exams and various competitive entrance tests. Algebra, geometry, trigonometry, statistics, and probability are some of the vital components of the curriculum. Practicing multiple-choice questions (MCQs) is an effective strategy to help students assess their understanding and strengthen their problem-solving abilities. This post provides 50 MCQs from various topics in Class 10 Maths to test your knowledge and enhance exam readiness.
MCQs for Class 10 Maths
1. The value of 9\sqrt{9}9 is:
- A) 3
- B) 9
- C) 6
- D) 4
Answer: A) 3
The square root of 9 is 3, as 32=93^2 = 932=9.
2. If the length of a rectangle is 5 cm and its width is 3 cm, then its area is:
- A) 8 cm²
- B) 15 cm²
- C) 20 cm²
- D) 18 cm²
Answer: B) 15 cm²
The area of a rectangle is given by Length×Width=5×3=15\text{Length} \times \text{Width} = 5 \times 3 = 15Length×Width=5×3=15 cm².
3. The perimeter of a square with side length 6 cm is:
- A) 24 cm
- B) 12 cm
- C) 36 cm
- D) 18 cm
Answer: B) 24 cm
The perimeter of a square is given by 4×side=4×6=244 \times \text{side} = 4 \times 6 = 244×side=4×6=24 cm.
4. If the angles of a triangle are in the ratio 2:3:4, then the angles are:
- A) 40°, 60°, 80°
- B) 30°, 45°, 105°
- C) 36°, 54°, 90°
- D) 20°, 30°, 130°
Answer: A) 40°, 60°, 80°
The sum of the angles in a triangle is 180°. The ratio is 2x:3x:4x2x : 3x : 4x2x:3x:4x, solving gives x=20x = 20x=20, so the angles are 40°, 60°, and 80°.
5. The distance between two points (2,3)(2, 3)(2,3) and (5,7)(5, 7)(5,7) is:
- A) 25\sqrt{25}25
- B) 18\sqrt{18}18
- C) 13\sqrt{13}13
- D) 17\sqrt{17}17
Answer: D) 17\sqrt{17}17
Using the distance formula, (x2−x1)2+(y2−y1)2=(5−2)2+(7−3)2=9+16=17\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} = \sqrt{(5 – 2)^2 + (7 – 3)^2} = \sqrt{9 + 16} = \sqrt{17}(x2−x1)2+(y2−y1)2=(5−2)2+(7−3)2=9+16=17.
6. The value of cos0∘\cos 0^\circcos0∘ is:
- A) 1
- B) 0
- C) ∞\infty∞
- D) -1
Answer: A) 1
By definition, cos0∘=1\cos 0^\circ = 1cos0∘=1.
7. The solution of the equation x2−16=0x^2 – 16 = 0x2−16=0 is:
- A) x=4x = 4x=4
- B) x=−4x = -4x=−4
- C) x=4x = 4x=4 and x=−4x = -4x=−4
- D) x=16x = 16x=16
Answer: C) x=4x = 4x=4 and x=−4x = -4x=−4
Solving the equation x2=16x^2 = 16×2=16, we get x=4x = 4x=4 or x=−4x = -4x=−4.
8. The LCM of 12 and 18 is:
- A) 36
- B) 72
- C) 24
- D) 18
Answer: A) 36
The LCM of 12 and 18 is 36, as it is the smallest number divisible by both 12 and 18.
9. The sum of the angles in a quadrilateral is:
- A) 180°
- B) 360°
- C) 90°
- D) 270°
Answer: B) 360°
The sum of the interior angles of any quadrilateral is always 360°.
10. The value of 35+25\frac{3}{5} + \frac{2}{5}53+52 is:
- A) 55\frac{5}{5}55
- B) 65\frac{6}{5}56
- C) 15\frac{1}{5}51
- D) 75\frac{7}{5}57
Answer: D) 75\frac{7}{5}57
The sum of fractions with the same denominator is 3+25=75\frac{3 + 2}{5} = \frac{7}{5}53+2=57.
11. The graph of the equation y=x2−4x+3y = x^2 – 4x + 3y=x2−4x+3 is:
- A) A straight line
- B) A parabola opening upwards
- C) A parabola opening downwards
- D) A hyperbola
Answer: B) A parabola opening upwards
Since the coefficient of x2x^2×2 is positive, the graph is a parabola opening upwards.
12. The value of sin90∘\sin 90^\circsin90∘ is:
- A) 0
- B) 1
- C) ∞\infty∞
- D) -1
Answer: B) 1
By definition, sin90∘=1\sin 90^\circ = 1sin90∘=1.
13. The area of a circle with radius 7 cm is:
- A) 49π cm²
- B) 14π cm²
- C) 28π cm²
- D) 35π cm²
Answer: A) 49π cm²
The area of a circle is πr2\pi r^2πr2, so the area is π×72=49π\pi \times 7^2 = 49\piπ×72=49π cm².
14. The value of 2x+3=112x + 3 = 112x+3=11 is:
- A) x=4x = 4x=4
- B) x=3x = 3x=3
- C) x=5x = 5x=5
- D) x=2x = 2x=2
Answer: A) x=4x = 4x=4
Solving 2x=82x = 82x=8, we get x=4x = 4x=4.
15. The volume of a cube with side length 5 cm is:
- A) 25 cm³
- B) 125 cm³
- C) 20 cm³
- D) 100 cm³
Answer: B) 125 cm³
The volume of a cube is side3=53=125\text{side}^3 = 5^3 = 125side3=53=125 cm³.
16. The quadratic formula is used to find the roots of:
- A) Linear equations
- B) Exponential equations
- C) Polynomial equations of degree 2
- D) Logarithmic equations
Answer: C) Polynomial equations of degree 2
The quadratic formula is used to find the roots of quadratic equations ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0.
17. The solution of the system of equations 2x+y=72x + y = 72x+y=7 and x−y=1x – y = 1x−y=1 is:
- A) x=3,y=2x = 3, y = 2x=3,y=2
- B) x=4,y=1x = 4, y = 1x=4,y=1
- C) x=2,y=3x = 2, y = 3x=2,y=3
- D) x=5,y=0x = 5, y = 0x=5,y=0
Answer: A) x=3,y=2x = 3, y = 2x=3,y=2
Solving the system of equations gives x=3x = 3x=3 and y=2y = 2y=2.
18. The discriminant of the quadratic equation x2+4x+3=0x^2 + 4x + 3 = 0x2+4x+3=0 is:
- A) 16
- B) 8
- C) 12
- D) 4
Answer: A) 16
The discriminant of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 is Δ=b2−4ac\Delta = b^2 – 4acΔ=b2−4ac. Here, Δ=42−4×1×3=16−12=4\Delta = 4^2 – 4 \times 1 \times 3 = 16 – 12 = 4Δ=42−4×1×3=16−12=4.
19. The number of diagonals in a polygon with 5 sides is:
- A) 5
- B) 10
- C) 7
- D) 8
Answer: C) 7
The number of diagonals in a polygon with nnn sides is given by n(n−3)2\frac{n(n-3)}{2}2n(n−3). For n=5n = 5n=5, we get 5(5−3)2=7\frac{5(5-3)}{2} = 725(5−3)=7.
20. The ratio of the areas of two squares whose side lengths are 3 cm and 5 cm respectively is:
- A) 9:25
- B) 25:9
- C) 3:5
- D) 5:3
Answer: A) 9:25
The ratio of the areas of two squares is the square of the ratio of their side lengths. (35)2=925\left( \frac{3}{5} \right)^2 = \frac{9}{25}(53)2=259.
21. The sum of the interior angles of a hexagon is:
- A) 720°
- B) 1080°
- C) 360°
- D) 540°
Answer: A) 720°
The sum of the interior angles of a polygon with nnn sides is given by 180(n−2)180(n-2)180(n−2). For a hexagon, 180(6−2)=720°180(6-2) = 720°180(6−2)=720°.
22. If the base of a triangle is 8 cm and the height is 5 cm, then its area is:
- A) 10 cm²
- B) 20 cm²
- C) 40 cm²
- D) 30 cm²
Answer: B) 20 cm²
The area of a triangle is 12×base×height=12×8×5=20\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 5 = 2021×base×height=21×8×5=20 cm².
23. The mean of the following numbers: 4, 7, 10, 15, 18 is:
- A) 10
- B) 11
- C) 12
- D) 13
Answer: B) 11
The mean is calculated as the sum of the numbers divided by the number of terms. 4+7+10+15+185=545=10.8\frac{4 + 7 + 10 + 15 + 18}{5} = \frac{54}{5} = 10.854+7+10+15+18=554=10.8, rounded to 11.
24. The slope of the line 2x−3y=62x – 3y = 62x−3y=6 is:
- A) 2
- B) 23\frac{2}{3}32
- C) -2
- D) 32\frac{3}{2}23
Answer: B) 23\frac{2}{3}32
Rewriting the equation in slope-intercept form y=mx+by = mx + by=mx+b, we get y=23x−2y = \frac{2}{3}x – 2y=32x−2. So the slope mmm is 23\frac{2}{3}32.
25. If the radius of a circle is 10 cm, then its circumference is:
- A) 20π cm
- B) 10π cm
- C) 100π cm
- D) 50π cm
Answer: A) 20π cm
The circumference of a circle is given by 2πr2\pi r2πr. Substituting r=10r = 10r=10, we get 2π×10=20π2\pi \times 10 = 20\pi2π×10=20π cm.
26. The value of 34×25\frac{3}{4} \times \frac{2}{5}43×52 is:
- A) 15\frac{1}{5}51
- B) 620\frac{6}{20}206
- C) 320\frac{3}{20}203
- D) 520\frac{5}{20}205
Answer: C) 620\frac{6}{20}206
Multiplying the fractions: 34×25=620\frac{3}{4} \times \frac{2}{5} = \frac{6}{20}43×52=206.
27. If the perimeter of a rectangle is 40 cm and the length is 12 cm, then its width is:
- A) 8 cm
- B) 10 cm
- C) 6 cm
- D) 5 cm
Answer: A) 8 cm
The perimeter of a rectangle is 2×(Length+Width)2 \times (\text{Length} + \text{Width})2×(Length+Width). Let width = www, then 2(12+w)=402(12 + w) = 402(12+w)=40, solving gives w=8w = 8w=8 cm.
28. The product of the roots of the equation x2−5x+6=0x^2 – 5x + 6 = 0x2−5x+6=0 is:
- A) 5
- B) 6
- C) -6
- D) 1
Answer: B) 6
The product of the roots of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0 is given by ca\frac{c}{a}ac. Here c=6c = 6c=6 and a=1a = 1a=1, so the product is 6.
29. The roots of the equation x2+2x+1=0x^2 + 2x + 1 = 0x2+2x+1=0 are:
- A) 1
- B) 0
- C) -1
- D) -2
Answer: C) -1
This is a perfect square equation, and its roots are x=−1x = -1x=−1.
30. The volume of a cylinder with radius 3 cm and height 4 cm is:
- A) 36π cm³
- B) 12π cm³
- C) 9π cm³
- D) 48π cm³
Answer: A) 36π cm³
The volume of a cylinder is given by πr2h\pi r^2 hπr2h. Substituting r=3r = 3r=3 and h=4h = 4h=4, we get π×32×4=36π\pi \times 3^2 \times 4 = 36\piπ×32×4=36π cm³.
31. The roots of the equation x2+4x+5=0x^2 + 4x + 5 = 0x2+4x+5=0 are:
- A) Real and distinct
- B) Real and equal
- C) Imaginary
- D) Complex
Answer: C) Imaginary
The discriminant Δ=b2−4ac=42−4×1×5=16−20=−4\Delta = b^2 – 4ac = 4^2 – 4 \times 1 \times 5 = 16 – 20 = -4Δ=b2−4ac=42−4×1×5=16−20=−4, so the roots are imaginary.
32. The sum of the first 20 natural numbers is:
- A) 210
- B) 190
- C) 100
- D) 300
Answer: A) 210
The sum of the first nnn natural numbers is given by n(n+1)2\frac{n(n+1)}{2}2n(n+1). For n=20n = 20n=20, we get 20(20+1)2=210\frac{20(20+1)}{2} = 210220(20+1)=210.
33. The area of a sector of a circle with central angle 60° and radius 6 cm is:
- A) 6π cm²
- B) 12π cm²
- C) 9π cm²
- D) 3π cm²
Answer: D) 3π cm²
The area of a sector is given by θ360×πr2\frac{\theta}{360} \times \pi r^2360θ×πr2. Substituting θ=60\theta = 60θ=60, and r=6r = 6r=6, we get 60360×π×62=3π\frac{60}{360} \times \pi \times 6^2 = 3\pi36060×π×62=3π cm².
34. The number of sides of a regular polygon if each interior angle is 120° is:
- A) 6
- B) 8
- C) 10
- D) 12
Answer: A) 6
The formula for the interior angle of a regular polygon is (n−2)×180n=120\frac{(n-2) \times 180}{n} = 120n(n−2)×180=120, solving gives n=6n = 6n=6.
35. The value of tan45∘\tan 45^\circtan45∘ is:
- A) 0
- B) 1
- C) ∞\infty∞
- D) -1
Answer: B) 1
By definition, tan45∘=1\tan 45^\circ = 1tan45∘=1.
36. The solution of the equation 3x+5=2x+83x + 5 = 2x + 83x+5=2x+8 is:
- A) x=1x = 1x=1
- B) x=3x = 3x=3
- C) x=2x = 2x=2
- D) x=4x = 4x=4
Answer: D) x=4x = 4x=4
Solving the equation, we get 3x−2x=8−53x – 2x = 8 – 53x−2x=8−5, so x=4x = 4x=4.
37. If the base of a triangle is 10 cm and its height is 6 cm, the area of the triangle is:
- A) 60 cm²
- B) 30 cm²
- C) 20 cm²
- D) 15 cm²
Answer: B) 30 cm²
The area of a triangle is given by 12×base×height=12×10×6=30\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 6 = 3021×base×height=21×10×6=30 cm².
38. The value of 78+14\frac{7}{8} + \frac{1}{4}87+41 is:
- A) 58\frac{5}{8}85
- B) 98\frac{9}{8}89
- C) 34\frac{3}{4}43
- D) 74\frac{7}{4}47
Answer: B) 98\frac{9}{8}89
The sum of the fractions is 78+28=98\frac{7}{8} + \frac{2}{8} = \frac{9}{8}87+82=89.
39. The value of sin30∘\sin 30^\circsin30∘ is:
- A) 32\frac{\sqrt{3}}{2}23
- B) 12\frac{1}{2}21
- C) 22\frac{\sqrt{2}}{2}22
- D) 1
Answer: B) 12\frac{1}{2}21
By definition, sin30∘=12\sin 30^\circ = \frac{1}{2}sin30∘=21.
40. The area of a right-angled triangle with base 5 cm and height 12 cm is:
- A) 30 cm²
- B) 25 cm²
- C) 60 cm²
- D) 50 cm²
Answer: A) 30 cm²
The area of a right-angled triangle is 12×base×height=12×5×12=30\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 3021×base×height=21×5×12=30 cm².
I hope these additional MCQs help you in your studies! Let me know if you need further practice or assistance in any class 10 maths problem.
Also Read: MCQs of Maths for Class 8
You may also like to read this: Comprehensive MCQs on Percentages: Test Your Skills with Basic to Advanced Questions
As a seasoned content writer specialized in the fitness and health niche, Arun Bhagat has always wanted to promote wellness. After gaining proper certification as a gym trainer with in-depth knowledge of virtually all the information related to it, he exercised his flair for writing interesting, informative content to advise readers on their healthier lifestyle. His topics range from workout routines, nutrition, and mental health to strategies on how to be more fit in general. His writing is informative but inspiring for people to achieve their wellness goals as well. Arun is committed to equipping those he reaches with the insights and knowledge gained through fitness.