Quantitative Aptitude

Mensuration

2,507 Questions
All Direct Formula Application 1067 Shape Transformations 343 Mensuration Ratio Equations 277 Combined & Hollow Solids 247 Successive Percentage Changes 178 Inscribed & Circumscribed Figures 165 Cost Estimation Problems 100 Liquid Pour & Submergence 67 Unit Conversions 35 Path & Border Calculations 17 Time & Work Integration 7 Scale & Model Proportions 2

Mensuration Questions

Multiple choice

A cylindrical vessel with internal diameter $10\ cm$ and height $10.5\ cm$ is full of water. A solid cone of base diameter $7\ cm$ and height $6\ cm$ is completely immersed in water. Find the volume of water (i) displaced out of the cylinder (ii) left in the cylinder. (Take $\pi = \dfrac{22}{7}$)
  1. (i) $77\ cm^3$ and (ii) $748\ cm^3$
  2. (i) $70\ cm^3$ and (ii) $740\ cm^3$
  3. (i) $85\ cm^3$ and (ii) $786\ cm^3$
  4. (i) $88\ cm^3$ and (ii) $797\ cm^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume of cylinder = pi * r^2 * h = (22/7) * 5^2 * 10.5 = 825 cm^3. Volume of cone = (1/3) * pi * r^2 * h = (1/3) * (22/7) * 3.5^2 * 6 = 77 cm^3. The volume of water displaced is equal to the volume of the immersed object, which is 77 cm^3. The volume left is 825 - 77 = 748 cm^3.

Multiple choice

The volume and the total surface area of a solid right pyramid of height $4 cm$, and square base of side $6 cm$.
  1. $24cc, 86 sq. cm.$
  2. $48cc. 90 sq. cm.$
  3. $24cc, 80 sq. cm.$
  4. $48cc, 96 sq. cm.$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of pyramid = (1/3) * base_area * height = (1/3) * (6*6) * 4 = 48 cc. Slant height s = sqrt(height^2 + (side/2)^2) = sqrt(4^2 + 3^2) = 5. Lateral surface area = 4 * (1/2 * side * s) = 4 * (1/2 * 6 * 5) = 60. Total surface area = base_area + lateral_area = 36 + 60 = 96 sq cm.

Multiple choice

The radius of cylinder of maximum volume which can be inscribed in a right circular cone of radius $R$ and height $H$ (axis of cylinder and cone are same) is given by
  1. $\dfrac{R}{2}$
  2. $\dfrac{R}{3}$
  3. $\dfrac{2R}{3}$
  4. $\dfrac{2R}{5}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a cylinder of radius r and height h inscribed in a cone of radius R and height H, r/R = (H-h)/H. Volume V = pi * r^2 * h = pi * r^2 * H * (1 - r/R). Setting dV/dr = 0 gives r = 2R/3.

Multiple choice

A solid cylinder $8\ cm$ in diameter and $4\ cm$ in height is melted and $6$ spheres of equal sizes are made from the material. Then the diameter of each sphere is
  1. $4\ cm$
  2. $1\ cm$
  3. $1.2\ cm$
  4. $0.8\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume of cylinder = pi * r^2 * h = pi * 4^2 * 4 = 64 * pi. Volume of 6 spheres = 6 * (4/3) * pi * R^3 = 8 * pi * R^3. Setting them equal: 64 * pi = 8 * pi * R^3 => R^3 = 8 => R = 2. The diameter is 2 * R = 4 cm.

Multiple choice

The height of a cone and the radius of its base are respectively 9 and 3 cm. The cone is cut by a plane parallel to its base soas to divide it into two parts. The volume of frustum of cone is $44{ cm }^{ 3 }$, then the radius of upper circular of frustum is $\left{ Use\ \pi =\dfrac { 22 }{ 7 } \right} $.
  1. $\sqrt [ 3 ]{ 12 } cm$
  2. $\sqrt [ 3 ]{ 13 } cm$
  3. $\sqrt [ 3 ]{ 6 } cm$
  4. $\sqrt [ 3 ]{ 20 } cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of cone = (1/3)pi*r^2*h = (1/3)(22/7)*9*9 = 84.85. Frustum volume = 44. Remaining small cone volume = 84.85 - 44 = 40.85. Ratio of volumes = (r_small/r_large)^3 = 40.85/84.85. Solving for r_small gives the cube root of 13.

Multiple choice

A cylindrical vessel open at the top has a base diameter of 56 cm. If the total cost of painting the outer curved surface of the vessel is Rs 352 at the rate of Rs 0.2 per $100{ cm }^{ 2 },$ then the height of the vessel is
  1. $15 m$
  2. $10 m$
  3. $6 m$
  4. $12 m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Curved surface area (CSA) = 2 * pi * r * h. Diameter = 56 cm, so r = 28 cm. Cost = 352 Rs at 0.2 Rs per 100 cm^2 means total area = (352 / 0.2) * 100 = 176,000 cm^2. 2 * (22/7) * 28 * h = 176,000. 176 * h = 176,000, so h = 1000 cm = 10 m.