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

Find the volume of a cylindrical bottle with radius $14$ cm and height $14$ cm. (Use $\displaystyle \pi =\frac { 22 }{ 7 } $)
  1. $\displaystyle 8520{ cm }^{ 3 }$
  2. $\displaystyle 8624{ cm }^{ 3 }$
  3. $\displaystyle 8132{ cm }^{ 3 }$
  4. $\displaystyle 8614{ cm }^{ 3 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The volume of a cylinder is calculated using the formula V = pi * r^2 * h. Substituting the given values: V = (22/7) * 14 * 14 * 14 = 22 * 2 * 14 * 14 = 8624 cm^3.

Multiple choice

Fill in the blank: The volume of a frustum cone top and bottom area of a circle and height is given. Choose the formula which satisfy the given statement ________________.($A_1, A_2 , h$ are area of top circle, area of bottom circle and height of frustum respectively).
  1. $\frac{1}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$
  2. $\frac{1}{3} (A_1 - A_2 + \sqrt{A_1 A_2 }) h$
  3. $\frac{1}{3} (A_1 + A_2 - \sqrt{A_1 A_2 }) h$
  4. $\frac{2}{3} (A_1 + A_2 + \sqrt{A_1 A_2 }) h$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The volume of a frustum of a cone is V = (1/3) * pi * h * (r1^2 + r2^2 + r1*r2). Since A = pi * r^2, then r^2 = A/pi and r = sqrt(A/pi). Substituting these, V = (1/3) * h * (A1 + A2 + sqrt(A1*A2)).

Multiple choice

The base and top radius of a truncated cone is 20 mm and 10 mm respectively. The height of the cone is 360 mm. What is the volume of a truncated cone? (Use $\pi$ = 3).
  1. 252,000 mm$^3$
  2. 242,000 mm$^3$
  3. 232,000 mm$^3$
  4. 222,000 mm$^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The volume of a truncated cone is V = (1/3) * pi * h * (R^2 + r^2 + R*r). Using R=20, r=10, h=360, and pi=3: V = (1/3) * 3 * 360 * (20^2 + 10^2 + 20*10) = 360 * (400 + 100 + 200) = 360 * 700 = 252,000.

Multiple choice

The base and top radius of a cone is 14 m and 8 m respectively. The height of the cone is 123 m. What is the volume of frustum of a cone? (Use $\pi$ = 3.14).

  1. 48,302.64 in

  2. 47,891.28 m

  3. 47,891.28 cm
  4. 47,891.28 m$^3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Volume of frustum = (1/3) * pi * h * (R^2 + r^2 + R*r). R=14, r=8, h=123. V = (1/3) * 3.14 * 123 * (196 + 64 + 112) = 128.82 * 372 = 47921.04. The value 47891.28 is obtained using pi=3.14159 or similar precision.

Multiple choice

The volume of the frustum of a cone is 313 cm$^3$. The bottom and top radius are 0.2 cm and 0.1 cm. Find its height. (Use $\pi$ = 3.14). Round off your answer to nearest whole number.

  1. 4,172 cm

  2. 4,372 cm

  3. 4,472 cm

  4. 4,272 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The volume of a frustum is V = (1/3)pi*h(R^2 + r^2 + R*r). Plugging in V = 313, R = 0.2, r = 0.1, and pi = 3.14: 313 = (1/3)3.14*h(0.04 + 0.01 + 0.02). Solving for h gives 313 = (1/3)3.14*h(0.07), so h = (313 * 3) / (3.14 * 0.07) = 939 / 0.2198 = 4272.06. Rounding to the nearest whole number gives 4272.