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 paper is in the form of a rectangle $ABCD$ where $AB = 22\ cm$ and $BC = 14\ cm$. A semi-circular portion with $BC$ as diameter is cut off. Find the area of the remaining part. $\left(Use\ \displaystyle \pi =\dfrac{22}{7} \right)$
  1. $130$ $\displaystyle cm^{2} $
  2. $231$ $\displaystyle cm^{2} $
  3. $200$ $\displaystyle cm^{2} $
  4. $441$ $\displaystyle cm^{2} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The area of the rectangle is 22 * 14 = 308 cm^2. The semi-circle has a diameter of 14 cm, so the radius is 7 cm; its area is (1/2) * (22/7) * 7 * 7 = 77 cm^2. The remaining area is 308 - 77 = 231 cm^2.

Multiple choice

Find the volume of a shape obtained by placing a cone onto the circular end of hemisphere. The hemisphere has a radius $5$ cm and height of the cone is $4$ cm.
  1. $266.5\ cm^{3}$
  2. $366.5\ cm^{3}$
  3. $456.5\ cm^{3}$
  4. $576.5\ cm^{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume = Volume of hemisphere + Volume of cone. V = (2/3)pi*r^3 + (1/3)pi*r^2*h. r=5, h=4. V = (2/3)pi(125) + (1/3)pi(25)(4) = (250/3)pi + (100/3)pi = (350/3)pi. (350/3) * 3.14159 = 366.519.

Multiple choice

Consider two cylinders($C_1$ and $C_2$), in which $C_1$'s height is double that of the height of other cylinder $C_2$ and their radius is half of the radius of $C_2$. Then find the volumes of these two cylinders and compare them.
  1. Volume of $C_1 > $ Volume of $C_2$
  2. Volume of $C_1=2 \times$ Volume of $C_2$
  3. Volume of $C_1= $ Volume of $C_2$
  4. Volume of $C_1=\dfrac{1}{2}\times $ Volume of $C_2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

V1 = pi * (r/2)^2 * (2h) = pi * (r^2/4) * 2h = 0.5 * pi * r^2 * h. V2 = pi * r^2 * h. Thus V1 = 0.5 * V2.

Multiple choice

A right angle triangle with perpendicular and base as $8$ and $6$ $cm$ respectively, is rotated about its longest side. Find the volume of the solid generated.
  1. $50.28\pi$
  2. $150.85\pi$
  3. $76.8\pi$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Rotating a right triangle about its hypotenuse creates two cones sharing a base. Hypotenuse = sqrt(8^2 + 6^2) = 10. The altitude to the hypotenuse is h = (8 * 6) / 10 = 4.8. The radius of the common base is 4.8. Volume = (1/3) * pi * r^2 * h1 + (1/3) * pi * r^2 * h2 = (1/3) * pi * r^2 * (h1 + h2) = (1/3) * pi * (4.8)^2 * 10 = 76.8 * pi.

Multiple choice

A iron rod is in the form of a square enclosing an area of $22{cm}^{2}$. If the same iron rod is bent into a circle, then find the area of that circle?
  1. $24{cm}^{2}$
  2. $26{cm}^{2}$
  3. $28{cm}^{2}$
  4. $30{cm}^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A square with area 22 has side length sqrt(22). The perimeter is 4 * sqrt(22). When bent into a circle, the circumference 2 * pi * r = 4 * sqrt(22), so r = 2 * sqrt(22) / pi. Area = pi * r^2 = pi * (4 * 22 / pi^2) = 88 / pi. Using pi approx 3.14, 88 / 3.14 approx 28.02.

Multiple choice

The shape of a solid is a cylinder surmounted by a cone. If the volume of the solid is $40656\space cm^3$, the diameter of the base is $42\space cm$ and the height of the cylinder is $20\space cm$, find the slant height of the conical portion.
  1. $45\space cm$
  2. $35\space cm$
  3. $40\space cm$
  4. $50\space cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume = pi * r^2 * h_cyl + (1/3) * pi * r^2 * h_cone. r = 21, h_cyl = 20. Volume = pi * 21^2 * 20 + (1/3) * pi * 21^2 * h_cone = 40656. Using pi = 22/7, 8820 * 22/7 + 147 * 22/7 * h_cone = 40656. 27720 + 462 * h_cone = 40656. 462 * h_cone = 12936, h_cone = 28. Slant height = sqrt(r^2 + h_cone^2) = sqrt(21^2 + 28^2) = sqrt(441 + 784) = sqrt(1225) = 35.

Multiple choice

A company makes a metallic block having square bases with volume $640$ $cm^3$, and height $10$ cm. If Robin wants to paint the block with the paints at the rate of Rs $15$ per $cm^2$, then find the cost of painting Robin has to pay.
  1. Rs $5680$
  2. Rs $6595$
  3. Rs $6720$
  4. Rs $6690$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Volume = base_area * height = 640. base_area = 640 / 10 = 64. Since base is square, side = 8. Surface area to paint = 2 * base_area + perimeter * height = 2 * 64 + (4 * 8) * 10 = 128 + 320 = 448. Cost = 448 * 15 = 6720.

Multiple choice

A man running round a race course notes that the sum of the distance of two flag posts from him is always $10\ m$ and the distance between the flag posts is $8\ m$. The area of the path he encloses in square metres is:
  1. $15 \pi$
  2. $20 \pi$
  3. $27 \pi$
  4. $30 \pi$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The path is an ellipse where the sum of distances to foci is 2a = 10 (a=5) and distance between foci is 2ae = 8 (ae=4). Thus e = 0.8. b^2 = a^2(1-e^2) = 25(1-0.64) = 25(0.36) = 9. Area = pi * a * b = pi * 5 * 3 = 15 pi.

Multiple choice

Some liquid is filled in a cylindrical vessel of radius $R$. Let ${ F }{ 1 }$ be the force applied by the liquid on the bottom of the cylinder. Now the same liquid is poured into a vessel of uniform square cross-section of side $R$. Let ${ F }{ 2 }$ be the force applied by the liquid on the bottom of this new vessel. (Neglect atmosphere pressure). Then
  1. ${ F }_{ 1 }=\pi { F }_{ 2 }$
  2. ${ F }_{ 1 }=\cfrac { { F }_{ 2 } }{ \pi } $
  3. ${ F }_{ 1 }=\sqrt { \pi { F }_{ 2 }}$
  4. ${ F }_{ 1 }={ F }_{ 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The force applied by the liquid on the bottom of a vessel is equal to the weight of the liquid column above it (Pressure * Area = Density * g * Height * Area). Since the volume and density are the same, the weight of the liquid is the same in both vessels, resulting in the same force on the bottom.