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

The volume of the frustum of a cone is 7,740 cm$^3$. The bottom and top radius are 14 cm and 7 cm. Find its height. (Use $\pi$ = 22/7). Round off your answer to nearest whole number.

  1. 21 cm

  2. 22 cm

  3. 23 cm

  4. 24 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of frustum = (1/3) * pi * h * (R^2 + r^2 + R*r). 7740 = (1/3) * (22/7) * h * (14^2 + 7^2 + 14*7). 7740 = (1/3) * (22/7) * h * (196 + 49 + 98) = (1/3) * (22/7) * h * 343. 7740 = 22 * h * (343/21) = 22 * h * 16.333. 7740 = 359.33 * h. h = 21.54, which rounds to 22.

Multiple choice

The height of the frustum cone is $18$ in. The area of the top and bottom circle of the cone is $4$ in$^2$ and $16$ in$^2$ respectively. Find its volume.
  1. $168$ in$^3$
  2. $158$ in$^3$
  3. $146$ in$^3$
  4. $148$ in$^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume of a frustum = (1/3)pi*h(R^2 + r^2 + R*r). Here, pi*R^2 = 16 and pi*r^2 = 4. So R^2 = 16/pi and r^2 = 4/pi. R = 4/sqrt(pi), r = 2/sqrt(pi). Volume = (1/3)h(Area1 + Area2 + sqrt(Area1*Area2)). Volume = (1/3)18(16 + 4 + sqrt(16*4)) = 6*(20 + 8) = 6*28 = 168.

Multiple choice

The volume of frustum of a cone is 3,600 cm$^3$. The area of the bottom circle of the cone is 4 cm$^2$ and the area of the top circle of the cone is 16 cm$^2$. Find its height.

  1. 385.7 cm

  2. 375.7 cm

  3. 365.7 cm

  4. 355.7 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Volume of frustum V = (1/3) * pi * h * (R^2 + r^2 + R*r). Given areas A1 = pi*R^2 = 16 and A2 = pi*r^2 = 4, so R = 4/sqrt(pi) and r = 2/sqrt(pi). Substituting these into the volume formula leads to h = 3600 / (1/3 * (16 + 4 + sqrt(16*4))) = 3600 / (1/3 * 28) = 385.7 cm.

Multiple choice

Calculate the total surface area of a frustum of cone whose upper base radius is 15 m and lower base radius is 5 m and its height is 15 m. (Use $\pi$ = 3).
  1. 1,931.66 m$^2$
  2. 1,831.66 m$^2$
  3. 1,731.66 m$^2$
  4. 1,631.66 m$^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total surface area = pi * (R^2 + r^2 + (R+r)*l). l = sqrt(h^2 + (R-r)^2) = sqrt(15^2 + 10^2) = sqrt(325) = 18.027. Using pi = 3: Area = 3 * (15^2 + 5^2 + (15+5)*18.027) = 3 * (225 + 25 + 360.54) = 3 * 610.54 = 1831.62. Matches option B.

Multiple choice

The volume of frustum of a cone is 180.27 ft$^3$. The area of the bottom circle of the cone is 27 ft$^2$ and the area of the top circle of the cone is 12 ft$^2$. Find its height.

  1. 9.28 ft

  2. 9.38 ft

  3. 9.58 ft

  4. 9.48 ft
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The volume of a frustum of a cone is given by V = (1/3) * h * (A1 + A2 + sqrt(A1 * A2)), where A1 and A2 are the areas of the base and top circles. Substituting the given values, we get 180.27 = (1/3) * h * (27 + 12 + sqrt(27 * 12)) = 19 * h. Solving for h gives h = 180.27 / 19 = 9.48 ft.

Multiple choice

The frustum cone curved surface area is 600 $mm^3$. The upper and lower circle area of a cone is 25 and 35 mm. Find the total surface area.
  1. 660 $mm^3$
  2. 650 $mm^3$
  3. 560 $mm^3$
  4. 660 $mm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The total surface area of a frustum is the sum of the curved surface area and the areas of the two circular bases. Total Area = 600 + 25 + 35 = 660. The units in the options are inconsistent, but 660 is the correct numerical value.

Multiple choice

The lateral area of the frustum of a right circular cone is calculated by using the formula _________________.
  1. $\displaystyle \frac{1}{2} (2\pi R + 2\pi r) l$
  2. $\displaystyle \frac{1}{2} (2\pi R - 2\pi r) l$
  3. $\displaystyle \frac{1}{2} (2\pi R + 2\pi r) $
  4. $\displaystyle \frac{1}{3} (2\pi R + 2\pi r) l$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The lateral area of a frustum is the average of the circumferences of the two bases multiplied by the slant height: (2*pi*R + 2*pi*r)/2 * l = pi(R+r)l.