Problem C and Solution

An Average Quiz

For a recent quiz about averages, the following information is known:

There were three questions on the quiz.

Each question was worth \(5\) marks.

Each question was marked right or wrong (no part marks).

\(30\%\) of the students got all \(3\) questions correct.

\(40\%\) of the students got exactly \(2\) questions correct.

\(25\%\) of the students got exactly \(1\) question correct.

\(5\%\) of the students got no question correct.

Determine the overall class average for this quiz.

**Solution 1**

To determine the average, we must determine the sum of all the marks and divide by the number of students. We will use the information given for a class of \(100\) students.

Since \(30\%\) of the students got all \(3\) questions correct, \(30\) students each scored \(15\) marks and earned a total of \(30\times 15=450\) marks.

Since \(40\%\) of the students got exactly \(2\) questions correct, \(40\) students each scored \(10\) marks and earned a total of \(40\times 10=400\) marks.

Since \(25\%\) of the students got exactly \(1\) question correct, \(25\) students each scored \(5\) marks and earned a total of \(25\times 5=125\) marks.

Since \(5\%\) of the students got no questions correct, \(5\) students scored \(0\) marks and earned a total of \(5\times 0=0\) marks.

The total number of marks earned by the \(100\) students was \(450+400+125+0=975\).

The average mark on the quiz was then \(975\div 100=9.75\) out of \(15\), or \(65\%\).

**Solution 2**

To determine an average, we must determine the total of all the marks and divide by the number of students.

Let \(n\) represent the number of students who wrote the quiz where \(n\) is a positive integer.

Since \(30\%\) of the students got all \(3\) questions correct, \(0.30n\) students each scored \(15\) marks and earned a total of \(0.30n\times 15=4.5n\) marks.

Since \(40\%\) of the students got exactly \(2\) questions correct, \(0.40n\) students each scored \(10\) marks and earned a total of \(0.40n\times 10=4n\) marks.

Since \(25\%\) of the students got exactly \(1\) question correct, \(0.25n\) students each scored \(5\) marks and earned a total of \(0.25n\times 5=1.25n\) marks.

Since \(5\%\) of the students got no questions correct, \(0.05n\) students scored \(0\) marks and earned a total of \(0.05n\times 0=0\) marks.

The total number of marks earned by the \(n\) students was \(4.5n+4n+1.25n+0=9.75n\).

The average mark on the quiz was then \(\frac{9.75n}{n}=9.75\) (since \(n\neq 0\)) out of \(15\), or \(65\%\).