Tag: maths

Questions Related to maths

If a circular grass lawn of $35\ m$ in radius has a path $7\ m$ wide running around it on the outside, then the area of the path is

maths circle measures area between two concentric circles the area of ring semicircle and ring

  1. $1450\ m^2$

  2. $1576\ m^2$

  3. $1694\ m^2$

  4. $3368\ m^2$

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Correct Option: C
Explanation:

Radius of bigger circle(with the path) = $35 + 7 = 42\ m.$
Thus area of the path $=$ Area of bigger circle $-$ Area of smaller circle
$\therefore$ Required area $= \pi (42)^2 - \pi (35)^2 = \dfrac{22}{7} \times (42 + 35)(42 - 35) = 22 \times 77 = 1694\ m^2$

A wire in the shape of an equilateral triangle encloses an area $s$ sq. cm  If the same wire is bent to form circle, the area of the circle will be

maths circle measures area between two concentric circles the area of ring semicircle and ring

  1. $\displaystyle \frac{\pi s^{2}}{9}$

  2. $\displaystyle \frac{3s^{2}}{\pi }$

  3. $\displaystyle \frac{3s}{\pi }$

  4. $\displaystyle \frac{3\sqrt{3}s}{\pi }$

Show answer
Correct Option: D
Explanation:

Area of equilateral triangle $= s$ sq.cm
$\Rightarrow  \dfrac{\sqrt3}{4} a^2 = s$, [where $a$, the side of equilateral triangle]
$\Rightarrow a= \sqrt{\dfrac{4s}{\sqrt3}}$
Now perimeter of equilateral triangle $ 3\times a =3 \times\sqrt{\dfrac{4s}{\sqrt 3}}$ cm 
Circumference of circle $=$ perimeter of equilateral triangle
$\Rightarrow 2\pi r= 3 \times\sqrt{\dfrac{4s}{\sqrt 3}}$, [where $r$ the radius of circle]
Solve the above expression for $r$, we get 
$r= \dfrac{3}{2\pi} \times \sqrt{\dfrac{4s}{\sqrt 3}}$
Area of circle $=\pi r^2 = \pi \times \left ( \dfrac{3}{2\pi} \times \sqrt{\dfrac{4s}{\sqrt 3}} \right )^2$
After simplification, we get
Area of circle $=\dfrac{3s\sqrt3}{\pi}$ sq.cm

A bicycle wheel has diameter 1m. If the bicycle travels one kilometer, then the number of revolutions the wheel make is.

maths circle measures area between two concentric circles the area of ring semicircle and ring

  1. $\dfrac {1}{\Pi }$

  2. $\dfrac {100}{\Pi }$

  3. $\dfrac {500}{\Pi }$

  4. $\dfrac {1000}{\Pi }$

Show answer
Correct Option: D
Explanation:

Let the number of revolution of the wheel is n.
Then,
n $\times$ circumference of wheel = Distance travelled by bicycle
$n \times  2\Pi  \times \frac {1}{2}=1$ kilometer 
$n \times  \Pi $=1000 meter
$n=\frac {1000}{\Pi }$

A dog is chained on a $6\ ft$ leash, fastened to the corner of a rectangular building. Calculate, about how much area does the dog have to move in.

maths circle measures area between two concentric circles the area of ring semicircle and ring

  1. $27\ ft^{2}$

  2. $36\ ft^{2}$

  3. $56.55\ ft^{2}$

  4. $84.82\ ft^{2}$

Show answer
Correct Option: D
Explanation:

A dog is chained on a $6$ ft leash to the corner of a rectangular building.
Since the building is rectangular, the dog is left with an angle of $360 - 90 = 270^o$ for it to roam around.
Also, the length of the leash will act as the radius of this sector.
$\therefore$ Area of the sector $= \cfrac{270}{360} \times \pi \times 6^2$
$= \cfrac{3}{4} \times \pi \times 36$
$= 84.82 \ \ ft^2$

The volume, V $cm^{2}$, of a hollow cylindrical pipe of length $l$ cm, outer radius R cm and inner radius r cm is given by the formula : $V\, =\, \pi\, (R^{2}\, -\, r^{2}).\, l$

Find r, if $V\, =\, 22,\, R\, =\, 2,\, l\, =\, 4$ and $\pi,\, 3\displaystyle \frac{1}{7}.$

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. 1.5

  2. 1.2

  3. 1.4

  4. 1.6

Show answer
Correct Option: A
Explanation:

Given $V= \pi \left ( R^{2}-r^{2} \right )l$

$V= \pi  R^{2}-\pi r^{2} l$

$\pi r^{2}l= \pi R^{2}l-V$


$ r^{2}= \dfrac{\pi R^{2}l-V}{\pi l}$

$\therefore  r= \sqrt{\dfrac{\pi R^{2}l-V}{\pi l}}$

Given $V=22 ,R=2 ,L=4 , \pi = 3\tfrac{1}{7}= \frac{22}{7}$

$\therefore r= \sqrt{\dfrac{\frac{22}{7}\times 4\times 4-22}{\dfrac{22}{7}\times4}}= \sqrt{\dfrac{352-154}{88}}= \sqrt{\dfrac{198}{88}}= \sqrt{\dfrac{9}{4}}= 1.5$

An iron pipe $20\space cm$ long has exterior diameter equal to $25\space cm$. If the thickness of the pipe is $1\space cm$, find the whole surface area of the pipe.

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. $3167\space cm^2$

  2. $3160\space cm^2$

  3. $3068\space cm^2$

  4. $3268\space cm^2$

Show answer
Correct Option: A
Explanation:

TSA of pipe $=$ $2\pi h(R+r)+2\pi ({ R }^{ 2 }-{ r }^{ 2 })$


                     $=$ $2\times \dfrac { 22 }{ 7 } \times 20(12.5+11.5)+2\times \dfrac { 22 }{ 7 } \left( { \left( 12.5 \right)  }^{ 2 }-{ \left( 11.5 \right)  }^{ 2 } \right) $


                     $=$ $\dfrac { 44\times 480 }{ 7 } +\dfrac { 44\times 24 }{ 7 } $

                    $ =$ $\dfrac { 21120 }{ 7 } +\dfrac { 1056 }{ 7 } =\dfrac { 22176 }{ 7 } $

                     $=$ $3167$ ${ cm }^{ 2 }$

The diameters of two cylinders are in the ratio of 2:1 and their volumes are equal. The ratio of their heights will be _________.

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. 1:6

  2. 1:2

  3. 1:4

  4. 3:4

Show answer
Correct Option: C
Explanation:
Let the diameter of the given two cylinders are $2x$ and $x$, 
so the radius of these cylinders are $x$ and $0.5 x$ respectively. 

Let the height of these cylinders are $ h _1$ and $ h _2$ respectively.

Given $ πx^2h _1=π(0.5x)^2h _2 $

$∴\dfrac{h _1}{h _2}=\dfrac{(0.5)^2}{1}=\dfrac 14$

$1:4.$

If the volume of a cylinder is $448\pi:cm^3$ and height 7 cm, its total surface area will be ______________.

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. $352:cm^2$

  2. $754.28:cm^2$

  3. $724.64:cm^2$

  4. $354:cm^2$

Show answer
Correct Option: B
Explanation:
Let the radius of cylinder is $r$ cm and height of this is $7$ cm.

Given, volume of cylinder
$=πr^2h$

$448π=πr^2×7⟹r2=64⟹r=8$ cm

∴  Required total surface area of cylinder
$=2πr(r+h)=2π×8(8+7)$

$=2×\dfrac {22} 7×8×15 $

$=754.28$ $cm^2$

200 wooden balls each of diameter 70 mm are to be painted Find the cost of painting these balls at 10 paise/$\displaystyle cm^{2}$

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. Rs.3080

  2. Rs.2771

  3. Rs.4000

  4. Rs.7000

Show answer
Correct Option: A
Explanation:

Total surface area of 200 balls of $\displaystyle \frac{35}{10}$ cm radius will be $\displaystyle 200\times 4\times \frac{22}{7}\times \frac{35}{10}\times \frac{35}{10}$ mm
The cost will be Rs $\displaystyle \frac{200\times 4\times 22\times 35\times 35}{7\times 10\times 10\times 100}$ which simplifies to Rs 3080

A rectangular paper of dimensions 6 cm and 3 cm is rolled to form a cylinder with height equal to the width of the paper, then its base radius is

maths area and volume of cylinder hollow cylinder volume of cylinder volume of cylinder and cone

  1. $ \displaystyle \frac{6}{\pi }cm $

  2. $ \displaystyle \frac{3}{2\pi }cm $

  3. $ \displaystyle \frac{6}{2\pi }cm $

  4. $ \displaystyle \frac{9}{2\pi }cm $

Show answer
Correct Option: C
Explanation:

The length of the rectangle become the circumference of the base of the cylinder 

$\therefore 2\pi r=6\Rightarrow r=\frac{6}{2\pi }$ cm