# Simple question about circumference of circle

Q: The physical education teacher asked to one classroom, by vote, choose a sport between volleyball, basketball and football, to practice in class the following week.

pie chart:

The segment AB, which is the diameter of the pie chart, measuring 12cm, and the length of the smaller arc $\stackrel \frown{AC}$ is $\frac{3\pi}{2}$ cm.

a) which was the sport most votes? how many students chose this sport?

b) how many students participated in the voting?

c) How many students voted in football?

the most important for me is understand how to solve this problem and any other question similar to this.

• It doesn't seem like you've been given enough information to (fully) answer any of the questions. I think you have to know how many students are in the class. Otherwise, the best you can do is give proportions. Jun 6, 2013 at 23:03

To answer some of your questions, we need to know how many students voted.

However, you have all the information to determine what percentage of students who voted, voted for each sport.

A circle with diameter $12$ cm has a circumference of $C = 12 \pi$. $\;\stackrel \frown{AB} = 6 \pi$, or $\dfrac 12\;$ the circumference of the circle.

What fraction of $12\pi$ is $\stackrel \frown{AC} = \frac{3\pi}{2}$? $$\dfrac{3\pi/2}{12\pi} = \frac{3}{24} = \frac 18$$

What must the length of $\stackrel \frown{BC}$ given that $6 \pi + 3\pi/2 + \stackrel \frown{BC} = 12\pi$? We must have that the length of $\stackrel\frown{BC} = 9\pi/2 cm$. And $\dfrac{9 \pi/2}{12 \pi} = \dfrac 9{24} = \dfrac 38$ of the circumference of the entire circle.

We can't know, given the information provided, how many students voted. At any rate, of those that did vote, you can determine the percentage or fraction of that total that voted for each sport, respectively.

Now, you have that $\frac 12$ or $50\%$ of all students voted to play volleyball, $\frac 18$ or $12.5\%$ of all students voted to play basketball, and $\frac 38$ or $37.5\%$ of the students voted to play football.

Edit/Update: Okay, so now we know that basketball received $4$ votes. This is sufficient to compute the number of students who voted, and the number of students who voted for any given sport.

So we'll let $X$ be the number of students voting.
Then $\dfrac 1 8\times X = 4 \implies X = 32$ students voted.

\begin{align} \hline \\ \text{Students voting for basketball}: \dfrac 18 \times 32 & = \;\;4 \\ \text{Students voting for football}: \dfrac 38 \times 32 & = 12 \\ \text{Students voting for volleyball}: \dfrac 12 \times 32 &= 16 \\ \hline \\ \text{Total number of students voting}:\quad\qquad\; & \phantom{ = } 32 \\ \end{align}

• sorry, I'm dumb, messed up the end of question, edited, now the question is complete, sorry again Jun 7, 2013 at 1:16
• yes, it's very clearly for me, thanks Jun 7, 2013 at 1:53
• Great, you're welcome! Jun 7, 2013 at 1:54

Note that you do not have enough information to provide numerical answers to any of the questions asked. Let's assume that $n$ students can vote. What fraction of these students voted for each sport? @amWhy gives some good hints how to answer this question.

You also know that each piece of the pie must represent a whole number. (It doesn't make any sense to say 1.5 students voted for football.) With that in mind, what is the minimum number of students that participated in voting? What other possible numbers of students can satisfy the information given?