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 solid right circular cone of base diameter 14 cm and height 8 cm is melted to form a hollow sphere. If the external diameter of the sphere is 10 cm, find the internal diameter of the sphere.

  1. 3 cm

  2. 4 cm

  3. 6 cm

  4. 8 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Radius of base of cone is 7 cm and height is 8 cm. Volume of cone = 1/3 $\pi$ r2 h = 1/3$\pi$ (7)2 x 8 cm3 External radius of hollow sphere = 5 cm  Let internal radius be r cm. Volume of hollow sphere = 4/3 $\pi$ (R3 - r3) cm3 = 4/3 $\pi$ (53 - r3) cm3 Since volume of material used for cone will be the same as the volume of material used to make hollow sphere, 1/3 $\pi$ (7)2 8 = 4/3 $\pi$ (125 - r3) $\Rightarrow$ 98 = 125 - r 3 r3 = 125 - 98 = 27 r = 3 cm So, internal diameter of the sphere will be 2r = 2 x 3 = 6 cm.

Multiple choice

The radii of two circles are 6 cm and 8 cm. Find the radius of the circle with area equal to the sum of the areas of these circles.

  1. 20 cm

  2. 5 cm

  3. 2 cm

  4. 10 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Area of circle with radius 6 cm =  $\pi$(6)2 = 36$\pi$ cm2. Area of circle with radius 8 cm = $\pi$ (8)2 = 64$\pi$ cm2.  Total area of both circles = 36$\pi$ cm2 + 64$\pi$cm2 = 100$\pi$ cm2.  So, the radius of the new circle be R. $\pi$R2 = 100$\pi$cm2 R2 = 100 cm2 = R $\Rightarrow$ 10 cm. 

Multiple choice maths area of complex plane figures 2d and 3d figures

A square sheet of paper is converted into a cylinder by rolling it along its length. What is the ratio of the base radius to side of the square ?

  1. $\displaystyle \frac{1}{2\pi}$
  2. $\displaystyle \frac{\sqrt{2}}{\pi}$
  3. $\displaystyle \frac{1}{\sqrt{2\pi }}$
  4. $\displaystyle \frac{1}{\pi}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let length of each side of the square $=a$

Base radius of cylinder $=r$
Surface area of sheet $={ a }^{ 2 }$
Surface area of cylinder $=2\pi rh$
But height $h=a$, because the square sheet is rolled it along its length.
Thus, surface area of cylinder $=2\pi ra$
Therefore, ${ a }^{ 2 }=2\pi ra\Rightarrow a=2\pi r\Rightarrow \dfrac { r }{ a } =\dfrac { 1 }{ 2\pi  } $

Multiple choice maths surface area and volume of sphere solids surface area of a prism surface area of a prism and a pyramid volume of a sphere surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

From a solid sphere of radius $R$, a concentric solid sphere of radius $\dfrac{R}{2}$ is removed. The total surface area increases by

  1. $0\%$
  2. $25\%$
  3. $50\%$
  4. $75\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Solution:- (B) $25 \%$
Initial area of sphere $ \left( {A} _{1} \right) = 4 \pi {R}^{2}$
New area of sphere $\left( {A} _{2} \right) = 4 \pi {R}^{2} + 4 \pi {\left( \cfrac{R}{2} \right)}^{2} = 5 \pi {R}^{2}$
$\therefore$ Increase in area $= \cfrac{{A} _{2} - {A} _{1}}{{A} _{1}} \times 100$
$\Rightarrow$ Increase in area $= \cfrac{5 \pi {R}^{2} - 4 \pi {R}^{2}}{4 \pi {R}^{2}} \times 100 = 25 \%$
Hence the area will be increased by $25 \%$.
Multiple choice maths perimeter, area and volume surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

If a regular square pyramid has a base of side $8 cm$ and height of $30 cm$, then its volume is

  1. $120 cm^3.$
  2. $240 cm^3.$
  3. $640 cm^3.$
  4. $900 cm^3.$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given that:
Side $a=8\ cm$, Height $h=30\ cm$
As we know that
Volume of regular square pyramid 
$\Rightarrow a^2\dfrac{h}{3}$
$\Rightarrow 8^2\dfrac{30}{3}$
$\Rightarrow 64\times 10$
$\Rightarrow 640\ cm^3$
This is the required solution.
Multiple choice maths perimeter, area and volume surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

$VPQRS$ is rectangle based pyramid where $PQ = 30\ cm, QR = 20\ cm$ and volume is $2000\ {cm}^3$, then height (in cm) is

  1. $20$
  2. $40$
  3. $10$
  4. $30$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given : Length of base$(l)=30\ cm$, width of base$(h)=20\ cm$, Volume of pyramid$=2000\ cm^3$

Let $h$ be the height of the pyramid
We know that, volume of pyramid $=\dfrac{l\times w\times h}{3}$
$\implies 2000\ cm^3 = \dfrac{30 cm\times 20 cm\times h}{3}$
$\implies h=\dfrac{2000\times 3}{30\times 20} cm=10 cm$
Hence, height is $10 cm$.

Multiple choice maths measures and the circle surface area and volume of sphere surface area of a prism surface area of a prism and a pyramid

The base of a right prism is a square of perimeter 20 cm and its height is 30 cm. The volume of the prism is

  1. $700 cm^3$
  2. $750 cm^3$
  3. $800 cm^3$
  4. $850 cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, perimeter $=4a=20$cm
$\therefore a=5$ cm
Area $=a^2=25  cm^2$
Volume $=$ Area $\times$ Height
Volume $=25 \times 30$
Volume $=750  cm^3$

Multiple choice maths how many squares area of rectangular paths comparing areas spaces and boundaries - 2

The side of a square is 2 cm Semicircles are constructed on two sides of the square then the area of the whole figure is

  1. $ \displaystyle (4+\pi )cm^{2} $
  2. $ \displaystyle (4+4\pi )cm^{2} $
  3. $ \displaystyle 4\pi cm^{2} $
  4. $ \displaystyle 8\pi cm^{2} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The side of square is 2 cm 

Then area of square =$(2)^{2}=4 cm^{2}$
The Semicircle constructed on two side diameter 2 cm then radius =1 cm
Then area of one semicircle =$\frac{\pi r^{2}}{2}=\frac{\pi (1)^{2}}{2}=\frac{\pi }{2}$
Then  area of two semicircle=$2\times \frac{\pi }{2}=\pi $
So total area of whole figure=$(4+\pi )cm^{2}$

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

The radius of a sphere is r and radius of base of a cylinder is r and height is 2r. The ratio of their volumes will be-

  1. $2:3$
  2. $3:4$
  3. $4:3$
  4. $3:2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

The radius of the sphere $=r$

The radius of the cylinder $=r$

The height of the cylinder $=2r$

 

We know that the volume of the sphere

${{V} _{1}}=\dfrac{4}{3}\pi {{r}^{3}}$

 

We know that the volume of the cylinder

$ {{V} _{2}}=\pi {{r}^{2}}h $

$ {{V} _{2}}=\pi {{r}^{2}}\left( 2r \right) $

$ {{V} _{2}}=2\pi {{r}^{3}} $

 

Therefore, the required ratio

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{\dfrac{4}{3}\pi {{r}^{3}}}{2\pi {{r}^{3}}} $

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{2\pi {{r}^{3}}}{3\pi {{r}^{3}}} $

$ \dfrac{{{V} _{1}}}{{{V} _{2}}}=\dfrac{2}{3} $

$ {{V} _{1}}:{{V} _{2}}=2:3 $

 

Hence, this is the answer.

Multiple choice maths solids volume of a sphere surface area and volume of sphere surface areas and volumes surface areas and volumes of solids problems involving volume of combined solids application of surface area and volume of solids

A metallic hemispherical bowl is $0.25\;cm$ thick. The inside radius of the bowl is $5\;cm$. Find the volume of steel used in making the bowl.

  1. $43.25\;cm^3$
  2. $41.27\;cm^3$
  3. $42.25\;cm^3$
  4. $40.25\;cm^3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Volume of speed used $=$ $\dfrac { 2 }{ 3 } \pi \left( { r } _{ 1 }^{ 3 }-{ r } _{ 2 }^{ 3 } \right) $

${ r } _{ 1 }=5+0.25=5.25cm$
${ r } _{ 2 }=5cm$
$\therefore \quad $ Volume $=$ $\dfrac { 2 }{ 3 } \times \dfrac { 22 }{ 7 } \times \left( { \left( 5.25 \right)  }^{ 3 }-{ \left( 5 \right)  }^{ 3 } \right) $

                       $= \dfrac { 44 }{ 21 } \times \left( 144.70-125 \right) $

                       $= 41.27$ ${ cm }^{ 3 }$