Quantitative Aptitude — Mensuration

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Quantitative Aptitude — Mensuration — Questions with Answers Open after you finish the quiz — all 22 questions, with answers and explanations.
Q1. A cone has a radius of 7 cm and a height of 24 cm. What is its volume? (Use π = 22/7)
  1. 1188 cm³
  2. 1344 cm³
  3. 1078 cm³
  4. 1232 cm³
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Answer: D. 1232 cm³
Volume of cone = (1/3)πr²h = (1/3) × (22/7) × (7 cm)² × 24 cm = (1/3) × 22 × 7 × 24 cm³ = 22 × 7 × 8 cm³ = 1232 cm³.
Q2. What is the ratio of the volumes of a cylinder, a cone, and a sphere, if they all have the same radius 'r' and the height of the cylinder and cone is equal to the diameter of the sphere (i.e., h = 2r)?
  1. 3:1:2
  2. 1:2:3
  3. 2:3:1
  4. 1:3:2
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Answer: A. 3:1:2
Let the radius be 'r' and height 'h = 2r'. Volume of Cylinder (Vc) = πr²h = πr²(2r) = 2πr³ Volume of Cone (Vco) = (1/3)πr²h = (1/3)πr²(2r) = (2/3)πr³ Volume of Sphere (Vs) = (4/3)πr³ Ratio Vc : Vco : Vs = 2πr³ : (2/3)πr³ : (4/3)πr³ Multiplying by 3/(2πr³) gives 3 : 1 : 2.
Q3. Observe the line chart showing the volume of a sphere as its radius increases. What trend does the chart illustrate regarding the relationship between the sphere's radius and its volume?
  1. Volume increases linearly with radius.
  2. Volume decreases as radius increases.
  3. Volume increases cubically with radius.
  4. Volume remains constant regardless of radius.
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Answer: C. Volume increases cubically with radius.
The formula for the volume of a sphere is V = (4/3)πr³. This shows that the volume is directly proportional to the cube of the radius (r³). The chart values (4.19, 33.5, 113, 268, 524) show a rapid, non-linear increase, consistent with a cubic relationship.
Q4. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the cylindrical part is 10 mm and the radius of the capsule is 3 mm. Find the total volume of the capsule. (Use π = 22/7)
  1. 108π mm³
  2. 144π mm³
  3. 90π mm³
  4. 126π mm³
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Answer: D. 126π mm³
Radius of capsule (r) = 3 mm. Length of cylindrical part (h) = 10 mm. Volume of cylindrical part = πr²h = π(3)²(10) = 90π mm³. Volume of two hemispheres = Volume of one sphere = (4/3)πr³ = (4/3)π(3)³ = (4/3)π(27) = 36π mm³. Total volume of capsule = Volume of cylindrical part + Volume of two hemispheres = 90π + 36π = 126π mm³.
Q5. A combined solid has a total volume of 1000 cm³. According to the pie chart showing the volume distribution, what is the volume contributed by the cylindrical part?
  1. 250 cm³
  2. 200 cm³
  3. 550 cm³
  4. 450 cm³
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Answer: C. 550 cm³
From the pie chart, the cylindrical part contributes 55% of the total volume. Volume of cylindrical part = 55% of 1000 cm³ = (55/100) × 1000 cm³ = 550 cm³.
Q6. An iron sphere of radius 12 cm is melted and recast into small spheres, each of radius 2 cm. How many small spheres can be formed?
  1. 36
  2. 72
  3. 108
  4. 216
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Answer: D. 216
Volume of large sphere = (4/3)π(12)³ cm³. Volume of one small sphere = (4/3)π(2)³ cm³. Number of small spheres = (Volume of large sphere) / (Volume of one small sphere) = [(4/3)π(12)³] / [(4/3)π(2)³] = (12/2)³ = 6³ = 216.
Q7. A metallic cuboid of dimensions 4 cm x 4 cm x 2 cm is melted and recast into a single sphere. Assuming π = 3, what is the diameter of the sphere?
  1. 4 cm
  2. 2 cm
  3. 6 cm
  4. 8 cm
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Answer: A. 4 cm
Volume of cuboid = length × breadth × height = 4 cm × 4 cm × 2 cm = 32 cm³. When recast, the volume remains the same. Volume of sphere = (4/3)πR³. Given V_sphere = 32 cm³ and π = 3. (4/3) × 3 × R³ = 32 4R³ = 32 R³ = 8 R = 2 cm. Diameter = 2R = 2 × 2 = 4 cm.
Q8. A circular racing track has been developed in a field. If the difference between the outer circumference and the inner circumference of the racing track is 33 m, then find the width of the track (in m). (Use pi = 22/7)
  1. 5 whole 1/4
  2. 5 whole 3/4
  3. 5 whole 1/5
  4. 4 whole 3/4
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Answer: A. 5 whole 1/4
Let the outer radius be R and inner radius be r. The width of the track is (R - r). The difference in circumferences is 2πR - 2πr = 2π(R - r) = 33. Substituting π = 22/7, we get 2 * (22/7) * (R - r) = 33. Therefore, 44/7 * (R - r) = 33, which gives (R - r) = (33 * 7) / 44 = 21/4 = 5 whole 1/4 m.
Q9. The length of each edge of a cube is 2.6 cm. What is the total surface area (in cm squared) of the cube?
  1. 39.96
  2. 40.56
  3. 40.76
  4. 40.36
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Answer: B. 40.56
The total surface area of a cube is given by the formula 6 * (side squared). Here, the side of the cube is 2.6 cm. So, total surface area = 6 * (2.6^2) = 6 * 6.76 = 40.56 cm squared.
Q10. A solid metallic sphere of radius 3 cm is melted and drawn into a wire of thickness 4 mm. What is the length of the wire (in m)?
  1. 9.25
  2. 7.5
  3. 9
  4. 8
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Answer: C. 9
Volume of the metallic sphere = Volume of the cylindrical wire. Radius of the sphere = 3 cm. Thickness of the wire means its diameter = 4 mm, so its radius = 2 mm = 0.2 cm. Volume of sphere = (4/3) * π * (radius cubed) = (4/3) * π * (3^3) = 36π. Volume of wire = π * (radius squared) * height = π * (0.2^2) * h = 0.04πh. Equating both volumes: 36π = 0.04πh, which gives h = 36 / 0.04 = 900 cm = 9 m.
Q11. The area of a rectangle is 453.6 m squared. If its length is 27m, then what is the perimeter of the rectangle?
  1. 86.6 m
  2. 87.6 m
  3. 85.4 m
  4. 88.8 m
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Answer: B. 87.6 m
Area of a rectangle = length * breadth. Given, area = 453.6 m squared and length = 27 m. So, 27 * breadth = 453.6, which gives breadth = 453.6 / 27 = 16.8 m. The perimeter of a rectangle = 2 * (length + breadth) = 2 * (27 + 16.8) = 2 * 43.8 = 87.6 m.
Q12. The ratio of the length, width and height of a cuboid is 4 : 3 : 5 and the sum of the lengths of all its edges is 144 cm. Find the total surface area of the cuboid.
  1. 1620 cm squared
  2. 1026 cm squared
  3. 756 cm squared
  4. 846 cm squared
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Answer: D. 846 cm squared
Let the length, width, and height of the cuboid be 4x, 3x, and 5x respectively. A cuboid has 12 edges (4 lengths, 4 widths, 4 heights). The sum of all edges = 4 * (length + width + height) = 144. So, 4 * (4x + 3x + 5x) = 144 => 4 * 12x = 144 => 48x = 144 => x = 3. Therefore, length = 12 cm, width = 9 cm, and height = 15 cm. Total surface area of a cuboid = 2 * (length * width + width * height + height * length) = 2 * (12 * 9 + 9 * 15 + 15 * 12) = 2 * (108 + 135 + 180) = 2 * 423 = 846 cm squared.
Q13. The circumference of a circle is given as 308 m. What is the area of the circle? [Use pi = 22/7]
  1. 7646 m squared
  2. 7546 m squared
  3. 7556 m squared
  4. 7446 m squared
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Answer: B. 7546 m squared
Circumference of a circle = 2 * π * r = 308. Substituting π = 22/7, we get 2 * (22/7) * r = 308 => 44/7 * r = 308 => r = (308 * 7) / 44 = 49 m. Area of the circle = π * (radius squared) = (22/7) * 49 * 49 = 22 * 7 * 49 = 154 * 49 = 7546 m squared.
Q14. The outer radius of a spherical shell is 9 cm and the thickness of the shell is 1 cm. Find the volume of the metal used for the shell (in cubic cm). (Use pi = 22/7)
  1. 912 whole 2/3
  2. 915 whole 1/3
  3. 909 whole 1/3
  4. 909 whole 2/5
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Answer: C. 909 whole 1/3
Outer radius (R) = 9 cm. Thickness = 1 cm, so inner radius (r) = 9 - 1 = 8 cm. Volume of the metal in a spherical shell = (4/3) * π * (R cubed - r cubed) = (4/3) * (22/7) * (9^3 - 8^3) = 88/21 * (729 - 512) = 88/21 * 217 = (88 * 31) / 3 = 2728 / 3 = 909 whole 1/3 cubic cm.
Q15. The radius of the base of a conical tent is 9m and its height is 12 m, find the cost of the material needed to make it if it costs Rupees 100 per pi m squared.
  1. Rupees 14,500
  2. Rupees 13,000
  3. Rupees 15,000
  4. Rupees 13,500
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Answer: D. Rupees 13,500
Radius (r) = 9 m, Height (h) = 12 m. Slant height (l) = square root of (r squared + h squared) = square root of (9 squared + 12 squared) = square root of (81 + 144) = square root of 225 = 15 m. The material needed for a conical tent corresponds to its curved surface area = π * r * l = π * 9 * 15 = 135π m squared. Cost of the material = Area * Rate = 135π * (100 / π) = 135 * 100 = Rupees 13,500.
Q16. A right circular cone is surmounted on a hemisphere. Base radius of the cone is equal to radius of the hemisphere. The diameter of the hemisphere is 12 cm while the height of the cone is 8 cm. Find the cost of painting the compound object if it costs Rupees 25 to paint pi cm squared.
  1. Rupees 10,371
  2. Rupees 3,300
  3. Rupees 6,930
  4. Rupees 4,400
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Answer: B. Rupees 3,300
Diameter of hemisphere = 12 cm, so radius (r) = 6 cm. Height of cone (h) = 8 cm. Slant height of cone (l) = square root of (6 squared + 8 squared) = square root of 100 = 10 cm. Total surface area of the compound object = Curved surface area of cone + Curved surface area of hemisphere = π * r * l + 2 * π * (r squared) = π * 6 * 10 + 2 * π * (6 squared) = 60π + 72π = 132π cm squared. Cost of painting = 132 * 25 = Rupees 3,300.
Q17. A semicircle has been drawn on the length of a rectangle. The area of the shaded region in the figure is: (Assume the semicircle is removed from the rectangle, length = 14 cm, breadth = 10 cm)
  1. 77 cm squared
  2. 14 cm squared
  3. 129 cm squared
  4. 63 cm squared
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Answer: D. 63 cm squared
Length of the rectangle = 14 cm, Breadth = 10 cm. Total area of the rectangle = length * breadth = 14 * 10 = 140 cm squared. A semicircle is drawn on the length, so its diameter is 14 cm, meaning its radius (r) = 7 cm. Area of the semicircle = (π * r squared) / 2 = (22/7 * 7 * 7) / 2 = 77 cm squared. Area of the shaded region = Area of rectangle - Area of semicircle = 140 - 77 = 63 cm squared.
Q18. The area of four side walls of a cubical box is 36 cm squared. Its edge is:
  1. square root of 6 cm
  2. 3 cm
  3. 9 cm
  4. 6 cm
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Answer: B. 3 cm
The area of the four side walls of a cube (lateral surface area) is given by 4 * (edge squared). Given, 4 * (edge squared) = 36. Therefore, edge squared = 36 / 4 = 9. Taking the square root, edge = square root of 9 = 3 cm.
Q19. If length of each side of a cube is doubled, then its volume becomes how many times the original volume?
  1. becomes 8 times
  2. is doubled
  3. becomes 9 times
  4. becomes 6 times
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Answer: A. becomes 8 times
Let the initial side of the cube be x. Original volume = x cubed. When the side is doubled, the new side becomes 2x. New volume = (2x) cubed = 8 * (x cubed). Therefore, the new volume becomes 8 times the original volume.
Q20. The volume of a hemisphere is 19404 cm cubed. Its radius is: (Use pi = 22/7)
  1. 30 cm
  2. 19 cm
  3. 20 cm
  4. 21 cm
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Answer: D. 21 cm
Volume of a hemisphere = (2/3) * π * (radius cubed). Given, (2/3) * (22/7) * (r cubed) = 19404. This gives 44/21 * (r cubed) = 19404 => r cubed = (19404 * 21) / 44 = 441 * 21 = 21 * 21 * 21. Taking the cube root, r = 21 cm.
Q21. The interior angle of a regular polygon is 108 degrees. The number of the sides of the polygon is:
  1. 5
  2. 360
  3. 15
  4. 108
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Answer: A. 5
Each interior angle of a regular polygon = 180 minus (each exterior angle). Given interior angle = 108 degrees, so each exterior angle = 180 - 108 = 72 degrees. The number of sides of a regular polygon = 360 divided by each exterior angle = 360 / 72 = 5 sides.
Q22. If the base of cylinder is the same as that of a cone, and the height of the cylinder is also the same as that of the cone, then find the ratio of the volumes of the cylinder and the cone.
  1. 1 : 3
  2. 3 : 2
  3. 3 : 1
  4. 2 : 3
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Answer: C. 3 : 1
Let the common radius be r and common height be h. Volume of cylinder = π * (r squared) * h. Volume of cone = 1/3 * π * (r squared) * h. The ratio of their volumes = [π * (r squared) * h] divided by [1/3 * π * (r squared) * h] = 1 divided by (1/3) = 3 : 1.
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