Question

The following table shows the favourite games of 40 students in a class.

Games	Cricket	Football	Hockey	Basketball
No. of Students	20	10	5	5

Represent the data using a circle graph.

Answer 1

Hint: First find the angle occupied by each game and then draw the graph.
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.

Complete step-by-step answer:
To find the $\theta$ we can use this formula.
$\theta ={\displaystyle \frac{\text{Given}\text{Quantity}}{\text{Total}\text{Quantity}}}\times {360}^{\circ }$
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 = ${\theta }_c$
${\theta }_c={\displaystyle \frac{20}{40}}\times {360}^{\circ }=180$
$\therefore$ angle subtended by football = $\theta r$
${\theta }_F={\displaystyle \frac{10}{40}}\times {360}^{\circ }={90}^{\circ }$
$\therefore$ angle subtended by Hockey = $${\theta }_H$$
$${\theta }_H={\displaystyle \frac{5}{40}}\times 360={45}^{\circ }$$
$\therefore$ angle subtended by Basketball = ${\theta }_B$
$${\theta }_B={\displaystyle \frac{5}{40}}\times 360={45}^{\circ }$$
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

Note: In the question where the graph is required to draw.
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 $$={\displaystyle \frac{20}{40}}\times 100=50\mathrm{\%}$$ area.
Similarly
AF = ${\displaystyle \frac{10}{40}}\times 100=25\mathrm{\%}$ area.
AH = ${\displaystyle \frac{5}{\underset{2}{40}}}\times 100={\displaystyle \frac{25}{2}}=12.5\mathrm{\%}$. area
AB = ${\displaystyle \frac{5}{40}}\times 100=12.5\mathrm{\%}$ area