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 maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

Area of a sector having radius 12 cm and arc length 21 cm is

  1. 126 $cm^2$
  2. 252 $cm^2$
  3. 33 $cm^2$
  4. 45 $cm^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Arc Length : Perimeter = Area of Sector : Area of Circle

$21: 2\pi r = \; Area \; of \;  Sector : \pi r^2$

$21:24\pi = \; Area \; of \;  Sector :144\pi$

Area of Sector $= \dfrac{144 \pi *21}{24 \pi} = 126cm^2$

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments
The radius of a circle is $7 cm$, then area of the sector of this circle if the corresponding angle is $30^{\circ}$ is 
  1. $12.83 \,cm^2$
  2. $11.83 \,cm^2$
  3. $12.25 \,cm^2$

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

Area of a sector of a circle of radius '$r$' and angle $ = \dfrac { \theta  }{ 360 } \pi {r}^{2}$
Hence, area of the sector of the circle of  radius $ 7 $ cm and angle $ = \dfrac { 30 }{ 360 } \times \dfrac { 22 }{ 7 } \times 7 \times 7 = 12.83 \ \text{cm}^{2} $

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

The area of the sector of a circle whose radius is 6 m when the angle at the centre is $\displaystyle 42^{\circ}$ is 

  1. $\displaystyle 13.2\:m^{2}$
  2. $\displaystyle 14.2\:m^{2}$
  3. $\displaystyle 13.4\:m^{2}$
  4. $\displaystyle 14.4\:m^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Area of sector $\displaystyle =\frac{42}{360}\times \pi r^{2}$
$\displaystyle =\frac{42}{360}\times \frac{22}{7}\times6\times6=13.2m^{2}$

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

The area of the sector of a circle, whose radius is $6$ m when the angle at the centre is $42^0$, is

  1. $13.2$ sq. m
  2. $14.2$ sq. m
  3. $13.4$ sq.m
  4. $14.4$ sq. m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $\theta=42^0$, radius $=6$ m
Area of sector $=\, \displaystyle \frac {\theta}{360}\, \times\, \pi r^2$
$=\displaystyle \frac {42}{360}\, \times\, \displaystyle \frac {22}{7}\, \times\, 6\, \times\,6$
$ =\, 13.2$ sq. m
Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

Find the area of a sector with an arc length of $20 cm$ and a radius of $6 cm$.

  1. $20$ $cm^2$
  2. $40$ $cm^2$
  3. $60$ $cm^2$
  4. $80$ $cm^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Area of sector $=$ $\dfrac { Arc.length }{ 2\pi r } \times \pi { r }^{ 2 }$


                         $=$ $\dfrac { 20 }{ 2\pi r } \times \pi \times 6\times 6=60{ cm }^{ 2 }$

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

Find the area of a sector of a circle of radius $28$cm and central angle $45^0$.

  1. $616 cm^{2}$
  2. $308 cm^{2}$
  3. $508 cm^{2}$
  4. $154 cm^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Radius of sector $=28 cm$

Control angle $=45^{ o }$
Area of sector $=\cfrac { \theta  }{ 360° } \times \pi { r }^{ 2 }$
$=\cfrac { 45° }{ 360° } \times \cfrac { 22 }{ 7 } \times 28\times 28\ =308\quad { cm }^{ 2 }$

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

If a sector of a circle of diameter 21 cm subtends an angle of $120^{\circ}$ at the centre, then what is its area ? 

  1. $115.5 \ cm^2$.
  2. $84 \ cm^2$.
  3. $85.5 \ cm^2$.
  4. $78 \ cm^2$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Area of sector = $\cfrac{120}{360} \times \pi \times (\cfrac{21}{2})^2$

Thus area = $\cfrac{1}{3} \times \cfrac{22}{7} \times \cfrac{441}{4} = 115.5 cm^2$

Multiple choice maths circle measures area of a sector of a circle sector and arc of a circle area of sectors and segments

If the sector of a circle of diameter $14 cm$ subtends an angle of $30^{\circ}$ at the centre, then its area is

  1. $49 \pi$
  2. $\displaystyle \frac{49 \pi}{12}$
  3. $\displaystyle \frac{242}{3\pi}$
  4. $\displaystyle \frac{121}{3\pi}$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

Area of a sector $=\dfrac{\theta}{360^0} \times \pi r^2 =\dfrac{30}{360} \times \pi (7)^2 = \dfrac{49 \pi}{12}$


Also, 
$ \dfrac{121}{3\pi}=\dfrac{121 \times 7}{3 \times 22} = \dfrac{49 \times 22}{12 \times 7} = \dfrac{49 \pi}{12}$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The radius of the sphere is measured as $ \left( {10 \pm 0.02} \right)cm$. The error in the measurement of its volume is 

  1. $25.1 cc$
  2. $25.21 cc$
  3. $2.51 cc$
  4. $251.2 cc$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $r$ be the radius of the sphere.


$\Rightarrow$  $r=10$


Error in the measurement of radius $=\Delta r$

$\therefore$  $\Delta r=0.02\,m$

$\Rightarrow$  Volume of the sphere $(V)=\dfrac{4}{3}\pi r^3$

We need to find error in calculating the volume that is $\Delta V$

$\Delta V=\dfrac{dv}{dr}\times \Delta r$

         $=\dfrac{d\left(\dfrac{4}{3}\pi r^3\right)}{dr}\times \Delta r$

         $=\dfrac{4}{3}\pi\dfrac{d(r^3)}{dr}\times \Delta r$

         $=\dfrac{4}{3}\pi(3r^2)\times (0.0.2)$

         $=4\pi r^2\times 0.02$

         $=4\times 3.14\times (10)^3\times 0.02$

         $=251.2\,cm^3$ i.e. $251.2\,cc$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The height of a cylinder is equal to the radius. If an error of $\alpha$ % is made in the height, then percentage error in its volume is

  1. $\alpha$ %
  2. $2\alpha$ %
  3. $3\alpha$ %

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Volume of cylinder $V= \pi { r }^{ 2 }h$
Since, $h=r$
$V=\pi {h}^{3}$ 
$\displaystyle \dfrac{dV}{dh}=3\pi h^{2}$
Given, percentage error in measuring height $=\alpha$%
$\Rightarrow \displaystyle \dfrac { \Delta h }{ h } =\dfrac { \alpha }{ 100 } $
$\Rightarrow \displaystyle  { \Delta h }=\dfrac {\alpha h }{ 100 } $
Now, approximate error in measuring V$\displaystyle =dV= (\dfrac{dV}{dh}){ \Delta h}$
                                          $\displaystyle = \dfrac{3\alpha }{100} {\pi h^{3}} =3\alpha$% of V
Percentage error in measuring $V =3\alpha$%
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If there is an error of $0.01 cm$ in the diameter of a sphere then percentage error in surface area when the radius $= 5 cm$, is

  1. $0.005\%$
  2. $0.05\%$
  3. $0.1\%$
  4. $0.2\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Surface area of sphere $S=4\pi r^{2}$
$\displaystyle S=\pi D^{2}$
$\Rightarrow S=100\pi$
Also, $ \displaystyle \frac{dS}{dD}=2\pi D=20\pi$
Approximate error in S is $\displaystyle dS=(\frac{dS}{dD})\Delta D$
                                         $ =20\pi (0.01)$
                                          $=\dfrac{1}{500} S$
                                           $=0.2$% of S
Percentage error in $S=0.2%$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If errors of $1\%$ each are made in the base radius and height of a cylinder, then the percentage error in its volume is

  1. $1\%$
  2. $2\%$
  3. $3\%$

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

Given, percentage error in r is 1%
$\Rightarrow \displaystyle \frac{\Delta r}{r}=\frac{1}{100}$

$\Rightarrow \displaystyle \Delta r=\frac{r}{100}$
Also given, percentage error in h is 1%
$\Rightarrow \displaystyle \frac{\Delta h}{h}=\frac{1}{100}$

$\Rightarrow \displaystyle \Delta h=\frac{h}{100}$
Now, volume of cylinder $V=\pi r^{2}h$
$\Delta V=\pi [r^{2}\Delta h+2rh\Delta r]$
$\displaystyle \Delta V=\pi[r^{2}\frac{h}{100}+2rh\frac{r}{100}]$

$\displaystyle\Delta V=\pi r^{2}h[\frac{3}{100}]$
$\Rightarrow\displaystyle\frac{\Delta V}{V}=\frac{3}{100}$
Percentage error in V is 3%
 

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If the radius of a sphere is measured as $9 \ cm$ with an error of $ 0.03 \ cm$ then, find the approximate error in calculating its volume.

  1. $\displaystyle 9.72\pi\:\: cm^{3}$
  2. $\displaystyle 7.92\pi\:\: cm^{3}$
  3. $\displaystyle 8.72\pi\:\: cm^{3}$

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

Given, $r=9 cm, \Delta r=0.03cm$

We know Volume of sphere with radius 'r' is $V=\cfrac{4}{3}\pi r^3$
$\therefore \Delta V=4\pi r^2\Delta r$
$\Rightarrow \Delta V=4\pi\times 81\times .03=9.72\pi  cm^3 $(using given values)

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

If the error committed in measuring the radius of the circle is $0.05\%$, then the corresponding error in calculating the area is:

  1. $0.05\%$
  2. $0.025\%$
  3. $0.25\%$
  4. $0.1\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac { dr }{ r } =0.05\Rightarrow dr=(0.05)r$

Area of circle $=\pi r^2$
$\ A=\pi r^{ 2 }\Rightarrow \dfrac { dA }{ dr } =2\pi r\ dA=2\pi rdr\Rightarrow \dfrac { dA }{ A } =\dfrac { 2\pi rdr }{ \pi r^{ 2 } } =\dfrac { 2dr }{ r } \ \therefore \dfrac { dA }{ A } =2(0.05)^{ 2 }=0.1$
$\therefore$ Corresponding error in area $= 0.1\%$