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Question

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Games | Cricket | Football | Hockey | Basketball |

No. of Students | 20 | 10 | 5 | 5 |

Represent the data using a circle graph.

Answer
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Here, ${\theta _C},{\theta _F},{\theta _H},{\theta _B}$ are the angles occupied by Cricket, Football, Hockey and Basketball respectively.

$\theta $ is the angle, which is used to draw the graph.

To find the $\theta $ we can use this formula.

$\theta = \dfrac{{{\text{Given}}\;{\text{Quantity}}}}{{{\text{Total}}\;{\text{Quantity}}}} \times 360^\circ $

Here, the given quantity is the number of students playing each game.

Total quantity is the total no of students and $360^\circ $ is the total angle of a circle.

According to the question:

Total no of students = 40.

No. of students who play Cricket = 20.

No. of students who play Football = 10.

No. of students who play Hockey = 5

No. of students who play Basketball = 5

Now, we need to find the angle subtended by each game on the circle.

$\therefore $ angle subtended by cricket = ${\theta _c}$

${\theta _c} = \dfrac{{{2}{0}}}{{{4}{0}}} \times 360^\circ = 180$

$\therefore $ angle subtended by football = $\theta r$

${\theta _F} = \dfrac{{1{0}}}{{4{0}}} \times 360^\circ = 90^\circ $

$\therefore $ angle subtended by Hockey = \[{\theta _H}\]

\[{\theta _H} = \dfrac{5}{{{4}{0}}} \times \mathop {{3}{6}0} = 45^\circ \]

$\therefore $ angle subtended by Basketball = ${\theta _B}$

\[{\theta _B} = \dfrac{5}{{{4}{0}}} \times \mathop {{3}{6}0}= 45^\circ \]

So now, we have all the data required to draw a circle graph.

$\therefore $ the graph will be like

${\theta _C}$ = Cricket

${\theta _F}$ = Football

${\theta _H}$ = Hockey

${\theta _B}$ = Basketball

We will 1st process the data. Also we could have made this by finding the percentage covered by each game.

Like the area covered by cricket = Ac.

Ac \[ = \dfrac{{{2}{0}}}{{4{0}}} \times \mathop {1{0}{0}} = 50\% \] area.

Similarly

AF = $\dfrac{{1{0}}}{{4{0}}} \times 100 = 25\% $ area.

AH = $\dfrac{5}{{\mathop {{4}{0}}\limits_2 }} \times \mathop {{1}{0}{0}} = \dfrac{{25}}{2} = 12.5\% $. area

AB = $\dfrac{5}{{40}} \times 100 = 12.5\% $ area