Certainly! Here are 50 MCQs of Maths for Class 9 with answers:
1. The value of 2x+3=112x + 3 = 112x+3=11 is:
- A) 4
- B) 5
- C) 6
- D) 7
Answer: B) 4
Solving 2x+3=112x + 3 = 112x+3=11, we get 2x=82x = 82x=8, so x=4x = 4x=4.
2. The LCM of 12 and 15 is:
- A) 30
- B) 60
- C) 120
- D) 45
Answer: B) 60
The least common multiple (LCM) of 12 and 15 is 60.
3. The HCF of 16 and 24 is:
- A) 4
- B) 8
- C) 6
- D) 12
Answer: B) 8
The highest common factor (HCF) of 16 and 24 is 8.
4. If the ratio of two numbers is 2:3, then their sum is:
- A) 5
- B) 6
- C) 8
- D) 9
Answer: B) 6
The sum of two numbers in a ratio of 2:3 is 2+3=52 + 3 = 52+3=5.
5. The solution of the linear equation 3x−4=5x+63x – 4 = 5x + 63x−4=5x+6 is:
- A) 5
- B) -5
- C) 10
- D) -10
Answer: B) -5
Solving 3x−4=5x+63x – 4 = 5x + 63x−4=5x+6, we get −2x=10-2x = 10−2x=10, so x=−5x = -5x=−5.
6. The area of a rectangle with length 8 cm and width 5 cm is:
- A) 30 cm²
- B) 40 cm²
- C) 35 cm²
- D) 45 cm²
Answer: B) 40 cm²
The area of a rectangle is given by Length×Width=8×5=40\text{Length} \times \text{Width} = 8 \times 5 = 40Length×Width=8×5=40 cm².
7. The perimeter of a square with side 4 cm is:
- A) 16 cm
- B) 12 cm
- C) 8 cm
- D) 10 cm
Answer: A) 16 cm
The perimeter of a square is 4×side=4×4=164 \times \text{side} = 4 \times 4 = 164×side=4×4=16 cm.
8. The square root of 81 is:
- A) 9
- B) 8
- C) 7
- D) 10
Answer: A) 9
The square root of 81 is 81=9\sqrt{81} = 981=9.
9. The volume of a cube with side 3 cm is:
- A) 27 cm³
- B) 9 cm³
- C) 12 cm³
- D) 15 cm³
Answer: A) 27 cm³
The volume of a cube is side3=33=27\text{side}^3 = 3^3 = 27side3=33=27 cm³.
10. The perimeter of a triangle with sides 3 cm, 4 cm, and 5 cm is:
- A) 10 cm
- B) 12 cm
- C) 14 cm
- D) 15 cm
Answer: B) 12 cm
The perimeter of a triangle is the sum of its sides: 3+4+5=123 + 4 + 5 = 123+4+5=12 cm.
11. The sum of the interior angles of a triangle is:
- A) 180°
- B) 360°
- C) 90°
- D) 270°
Answer: A) 180°
The sum of the interior angles of any triangle is always 180°.
12. The perimeter of a circle is called:
- A) Diameter
- B) Circumference
- C) Area
- D) Radius
Answer: B) Circumference
The perimeter of a circle is called the circumference.
13. The value of cos60∘\cos 60^\circcos60∘ is:
- A) 1
- B) 12\frac{1}{2}21
- C) 22\frac{\sqrt{2}}{2}22
- D) 0
Answer: B) 12\frac{1}{2}21
By definition, cos60∘=12\cos 60^\circ = \frac{1}{2}cos60∘=21.
14. The ratio of the areas of two similar triangles is:
- A) Equal to the ratio of their sides
- B) Equal to the square of the ratio of their sides
- C) Equal to the sum of their sides
- D) None of the above
Answer: B) Equal to the square of the ratio of their sides
For two similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides.
15. The sum of 2x and 3x is:
- A) 5x
- B) 6x
- C) 7x
- D) 4x
Answer: A) 5x
The sum of 2x2x2x and 3x3x3x is 5x5x5x.
16. A line parallel to the x-axis has the equation:
- A) y=0y = 0y=0
- B) x=0x = 0x=0
- C) y=ky = ky=k
- D) x=kx = kx=k
Answer: C) y=ky = ky=k
A line parallel to the x-axis is of the form y=ky = ky=k, where kkk is a constant.
17. The distance between the points (3, 4) and (6, 8) is:
- A) 5
- B) 4
- C) 3
- D) 7
Answer: A) 5
Using the distance formula (x2−x1)2+(y2−y1)2\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}(x2−x1)2+(y2−y1)2, we get (6−3)2+(8−4)2=9+16=25=5\sqrt{(6-3)^2 + (8-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5(6−3)2+(8−4)2=9+16=25=5.
18. The area of a triangle with base 10 cm and height 8 cm is:
- A) 40 cm²
- B) 50 cm²
- C) 30 cm²
- D) 20 cm²
Answer: B) 50 cm²
The area of a triangle is 12×base×height=12×10×8=50\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 8 = 5021×base×height=21×10×8=50 cm².
19. The value of 9+3×29 + 3 \times 29+3×2 is:
- A) 15
- B) 18
- C) 21
- D) 17
Answer: A) 15
According to the order of operations (BODMAS), first multiply 3×2=63 \times 2 = 63×2=6, then add 9+6=159 + 6 = 159+6=15.
20. The value of 2(3x+4)=122(3x + 4) = 122(3x+4)=12 is:
- A) 1
- B) 2
- C) 3
- D) 4
Answer: C) 3
Solving 2(3x+4)=122(3x + 4) = 122(3x+4)=12, we get 6x+8=126x + 8 = 126x+8=12, then 6x=46x = 46x=4, so x=46=23x = \frac{4}{6} = \frac{2}{3}x=64=32.
21. The area of a sector with angle 60° and radius 10 cm is:
- A) 50π cm²
- B) 25π cm²
- C) 10π cm²
- D) 20π cm²
Answer: B) 25π cm²
The area of a sector is θ360×πr2\frac{\theta}{360} \times \pi r^2360θ×πr2. Substituting θ=60\theta = 60θ=60, and r=10r = 10r=10, we get 60360×π×102=25π\frac{60}{360} \times \pi \times 10^2 = 25\pi36060×π×102=25π cm².
22. The probability of getting an even number when a die is rolled is:
- A) 12\frac{1}{2}21
- B) 13\frac{1}{3}31
- C) 23\frac{2}{3}32
- D) 16\frac{1}{6}61
Answer: A) 12\frac{1}{2}21
There are 3 even numbers (2, 4, 6) out of 6 possible outcomes, so the probability is 36=12\frac{3}{6} = \frac{1}{2}63=21.
23. The midpoint of the points (4, 6) and (8, 10) is:
- A) (6, 8)
- B) (5, 7)
- C) (4, 7)
- D) (5, 8)
Answer: A) (6, 8)
The midpoint is (x1+x22,y1+y22)=(4+82,6+102)=(6,8)\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{4+8}{2}, \frac{6+10}{2} \right) = (6, 8)(2×1+x2,2y1+y2)=(24+8,26+10)=(6,8).
24. The value of (x+1)(x−1)(x + 1)(x – 1)(x+1)(x−1) is:
- A) x2+1x^2 + 1×2+1
- B) x2−1x^2 – 1×2−1
- C) x2+2x^2 + 2×2+2
- D) x2−2x^2 – 2×2−2
Answer: B) x2−1x^2 – 1×2−1
Using the identity (a+b)(a−b)=a2−b2(a + b)(a – b) = a^2 – b^2(a+b)(a−b)=a2−b2, we get (x+1)(x−1)=x2−1(x + 1)(x – 1) = x^2 – 1(x+1)(x−1)=x2−1.
25. If a line intersects the x-axis at (2, 0) and y-axis at (0, 3), the equation of the line is:
- A) x2+y3=1\frac{x}{2} + \frac{y}{3} = 12x+3y=1
- B) x3+y2=1\frac{x}{3} + \frac{y}{2} = 13x+2y=1
- C) x2−y3=1\frac{x}{2} – \frac{y}{3} = 12x−3y=1
- D) x3−y2=1\frac{x}{3} – \frac{y}{2} = 13x−2y=1
Answer: A) x2+y3=1\frac{x}{2} + \frac{y}{3} = 12x+3y=1
This is the standard form of the equation of a line intercepting the x-axis at 2 and the y-axis at 3.
26. The value of 4(3x+5)=324(3x + 5) = 324(3x+5)=32 is:
- A) 2
- B) 3
- C) 5
- D) 4
Answer: A) 2
Simplifying the equation: 4(3x+5)=324(3x + 5) = 324(3x+5)=32, we get 12x+20=3212x + 20 = 3212x+20=32, then 12x=1212x = 1212x=12, so x=2x = 2x=2.
27. The probability of drawing a red card from a standard deck of 52 cards is:
- A) 12\frac{1}{2}21
- B) 14\frac{1}{4}41
- C) 2652\frac{26}{52}5226
- D) 113\frac{1}{13}131
Answer: A) 12\frac{1}{2}21
There are 26 red cards (13 diamonds and 13 hearts) in a deck of 52 cards, so the probability is 2652=12\frac{26}{52} = \frac{1}{2}5226=21.
28. The distance between the points (2, 3) and (5, 7) is:
- A) 5
- B) 10\sqrt{10}10
- C) 34\sqrt{34}34
- D) 4
Answer: C) 34\sqrt{34}34
Using the distance formula, the distance is (5−2)2+(7−3)2=9+16=34\sqrt{(5 – 2)^2 + (7 – 3)^2} = \sqrt{9 + 16} = \sqrt{34}(5−2)2+(7−3)2=9+16=34.
29. The median of the numbers 3, 7, 5, 2, 6 is:
- A) 5
- B) 6
- C) 4
- D) 3
Answer: A) 5
Arranging the numbers in ascending order: 2, 3, 5, 6, 7. The middle value (median) is 5.
30. The perimeter of a semicircle with radius 7 cm is:
- A) 222222 cm
- B) 141414 cm
- C) 212121 cm
- D) 101010 cm
Answer: A) 222222 cm
The perimeter of a semicircle is πr+2r\pi r + 2rπr+2r. Substituting r=7r = 7r=7, we get π(7)+2(7)≈22\pi(7) + 2(7) \approx 22π(7)+2(7)≈22 cm.
31. If the roots of the quadratic equation x2−5x+6=0x^2 – 5x + 6 = 0x2−5x+6=0 are:
- A) 2, 3
- B) 1, 6
- C) 3, 4
- D) 0, 6
Answer: A) 2, 3
The equation can be factored as (x−2)(x−3)=0(x – 2)(x – 3) = 0(x−2)(x−3)=0, so the roots are x=2x = 2x=2 and x=3x = 3x=3.
32. The sum of the angles of a quadrilateral is:
- A) 360°
- B) 180°
- C) 90°
- D) 270°
Answer: A) 360°
The sum of the interior angles of any quadrilateral is always 360°.
33. The height of a cone is 12 cm, and its radius is 5 cm. The volume of the cone is:
- A) 100π100\pi100π cm³
- B) 200π200\pi200π cm³
- C) 300π300\pi300π cm³
- D) 150π150\pi150π cm³
Answer: A) 100π100\pi100π cm³
The volume of a cone is 13πr2h\frac{1}{3} \pi r^2 h31πr2h. Substituting r=5r = 5r=5 and h=12h = 12h=12, we get 13π(52)(12)=100π\frac{1}{3} \pi (5^2)(12) = 100\pi31π(52)(12)=100π cm³.
34. The equation of a straight line with slope 2 and passing through the point (0, 3) is:
- A) y=2x+3y = 2x + 3y=2x+3
- B) y=3x+2y = 3x + 2y=3x+2
- C) y=x+3y = x + 3y=x+3
- D) y=2x−3y = 2x – 3y=2x−3
Answer: A) y=2x+3y = 2x + 3y=2x+3
The equation of a line is y=mx+cy = mx + cy=mx+c, where mmm is the slope and ccc is the y-intercept. Since the slope is 2 and the line passes through (0, 3), the equation is y=2x+3y = 2x + 3y=2x+3.
35. The ratio of the areas of two squares is 25:36. The ratio of their sides is:
- A) 5:6
- B) 25:36
- C) 6:5
- D) 3:4
Answer: A) 5:6
The ratio of the areas of two similar squares is the square of the ratio of their sides. Therefore, 25:36=5:6\sqrt{25} : \sqrt{36} = 5:625:36=5:6.
36. The surface area of a cube with side 4 cm is:
- A) 48 cm²
- B) 64 cm²
- C) 96 cm²
- D) 128 cm²
Answer: A) 48 cm²
The surface area of a cube is 6×side2=6×42=6×16=486 \times \text{side}^2 = 6 \times 4^2 = 6 \times 16 = 486×side2=6×42=6×16=48 cm².
37. The sum of the first 20 natural numbers is:
- A) 200
- B) 210
- C) 220
- D) 230
Answer: B) 210
The sum of the first nnn natural numbers is n(n+1)2\frac{n(n + 1)}{2}2n(n+1). For n=20n = 20n=20, the sum is 20(20+1)2=20×212=210\frac{20(20 + 1)}{2} = \frac{20 \times 21}{2} = 210220(20+1)=220×21=210.
38. The value of (2x+3)−(x−4)(2x + 3) – (x – 4)(2x+3)−(x−4) is:
- A) x+7x + 7x+7
- B) x−1x – 1x−1
- C) x+1x + 1x+1
- D) x−7x – 7x−7
Answer: A) x+7x + 7x+7
Simplifying the expression: (2x+3)−(x−4)=2x+3−x+4=x+7(2x + 3) – (x – 4) = 2x + 3 – x + 4 = x + 7(2x+3)−(x−4)=2x+3−x+4=x+7.
39. The distance between the points (1, 2) and (4, 6) is:
- A) 3
- B) 5
- C) 10\sqrt{10}10
- D) 14\sqrt{14}14
Answer: B) 5
Using the distance formula (x2−x1)2+(y2−y1)2\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}(x2−x1)2+(y2−y1)2, the distance is (4−1)2+(6−2)2=9+16=5\sqrt{(4 – 1)^2 + (6 – 2)^2} = \sqrt{9 + 16} = 5(4−1)2+(6−2)2=9+16=5.
40. If the angle between two intersecting lines is 90°, then they are:
- A) Parallel
- B) Perpendicular
- C) Oblique
- D) Skew
Answer: B) Perpendicular
Two lines that intersect at an angle of 90° are called perpendicular lines.
41. The value of sin30∘\sin 30^\circsin30∘ is:
- A) 1
- B) 12\frac{1}{2}21
- C) 22\frac{\sqrt{2}}{2}22
- D) 32\frac{\sqrt{3}}{2}23
Answer: B) 12\frac{1}{2}21
From trigonometric tables, sin30∘=12\sin 30^\circ = \frac{1}{2}sin30∘=21.
42. The quadratic equation x2−5x+6=0x^2 – 5x + 6 = 0x2−5x+6=0 has:
- A) 2 real roots
- B) 2 complex roots
- C) 1 real root
- D) No roots
Answer: A) 2 real roots
The quadratic equation can be factored as (x−2)(x−3)=0(x – 2)(x – 3) = 0(x−2)(x−3)=0, so it has 2 real roots, x=2x = 2x=2 and x=3x = 3x=3.
43. The area of a circle with radius 7 cm is:
- A) 49π49\pi49π cm²
- B) 14π14\pi14π cm²
- C) 494949 cm²
- D) 22π22\pi22π cm²
Answer: A) 49π49\pi49π cm²
The area of a circle is πr2\pi r^2πr2. Substituting r=7r = 7r=7, we get π×72=49π\pi \times 7^2 = 49\piπ×72=49π cm².
44. The value of 3x−7=2x+53x – 7 = 2x + 53x−7=2x+5 is:
- A) 12
- B) 15
- C) 5
- D) 2
Answer: C) 12
Solving 3x−7=2x+53x – 7 = 2x + 53x−7=2x+5, we get x=12x = 12x=12.
45. The length of the diagonal of a rectangle with length 6 cm and width 8 cm is:
- A) 10 cm
- B) 12 cm
- C) 14 cm
- D) 16 cm
Answer: A) 10 cm
Using the Pythagorean theorem, the diagonal is 62+82=36+64=100=10\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 1062+82=36+64=100=10 cm.
46. The length of the altitude of a triangle with area 30 cm² and base 5 cm is:
- A) 6 cm
- B) 7 cm
- C) 5 cm
- D) 8 cm
Answer: A) 6 cm
The area of a triangle is 12×base×height\frac{1}{2} \times \text{base} \times \text{height}21×base×height. Substituting area=30\text{area} = 30area=30, base=5\text{base} = 5base=5, we get 30=12×5×h30 = \frac{1}{2} \times 5 \times h30=21×5×h, so h=6h = 6h=6 cm.
47. The value of (x+2)2−(x−2)2(x + 2)^2 – (x – 2)^2(x+2)2−(x−2)2 is:
- A) 4x4x4x
- B) 444
- C) x2+4x^2 + 4×2+4
- D) 4x+44x + 44x+4
Answer: A) 4x4x4x
Using the difference of squares formula, (x+2)2−(x−2)2=4x(x + 2)^2 – (x – 2)^2 = 4x(x+2)2−(x−2)2=4x.
48. The volume of a sphere with radius 5 cm is:
- A) 500π500\pi500π cm³
- B) 5003π\frac{500}{3} \pi3500π cm³
- C) 5004π\frac{500}{4} \pi4500π cm³
- D) 43πr3\frac{4}{3} \pi r^334πr3
Answer: B) 5003π\frac{500}{3} \pi3500π cm³
The volume of a sphere is 43πr3\frac{4}{3} \pi r^334πr3. Substituting r=5r = 5r=5, we get 43π(5)3=5003π\frac{4}{3} \pi (5)^3 = \frac{500}{3} \pi34π(5)3=3500π cm³.
49. If cosA=0.5\cos A = 0.5cosA=0.5, then AAA is:
- A) 30°
- B) 45°
- C) 60°
- D) 90°
Answer: C) 60°
The cosine of 60° is 0.50.50.5.
50. The value of 3x+7=10x−53x + 7 = 10x – 53x+7=10x−5 is:
- A) 4
- B) 2
- C) -2
- D) -4
Answer: C) -2
Solving 3x+7=10x−53x + 7 = 10x – 53x+7=10x−5, we get −7x=−12-7x = -12−7x=−12, so x=−2x = -2x=−2.
I hope these MCQs help you in your preparation! Let me know if you’d like more or need any clarifications.
Also Read: MCQs of Maths for Class 8
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