# Mediant madness

## Problem

Kyle got 75% on section A of his maths exam, but he only got 35% on section B.

He says this means he should score 55% overall, but his teacher said he only scored 50% overall! How can this be?

Once you have had a chance to think about it, click below to see what he scored on each section.

In section A, there were 12 questions and Kyle got 9 correct.

In section B, there were 20 questions and Kyle got 7 correct.

Overall, there were 32 questions and Kyle got 16 correct.

As section A had fewer questions than section B, the score in section B is given more weighting.

The difference in overall scores is because Kyle worked out the

*average*of his two scores, but his teacher worked out the

*mediant*.

Given two fractions $\frac{a}{c}$ and $\frac{b}{d}$, the mediant is defined as $\frac{a+b}{c+d}$.

Do you agree that Kyle only deserved 50%?

Here is an interactive diagram for you to explore the properties of the mediant of two fractions.

Can you find some values of $a, b, c$ and $d$ so that Kyle and his teacher both get the same value for the overall score (that is, the average and the mediant are the same)?

The pass mark for an exam is 50%. If I scored 25% on the first $n$ questions, under what circumstances can I still pass the exam?

Is it true that the mediant always lies between the two fractions? How do you know?

## Getting Started

If Kyle scored 75% on section A, how many questions might there have been?

How many did he get right in each case?

In section B, he scored 35%, how many questions might there have been?

How many did he get right in each case?

Do any of your examples give a total score of 55% when you add together Kyle's marks for the two sections?

## Student Solutions

Pradeep from Peak School in Hong Kong and Vicente José from Kings College Alicante in Spain gave their opinions about the dispute. Pradeep said:

75% on Section 1

35% on Section 2

The mean of these two gives 55% so technically Kyle is right but if you take the overall score of both exams you get 50% so the teacher is right too.

This problems answer depends on how you choose to grade either you take the mean of the scores or take the overall scores.

Vicente José said:

The first question asked in the problem is if we agree with that Kyle only deserves a 50% in her exam. I do agree she only deserves a 50% in her exam as she must have the combined mean, mediant, of her two exams depending on the number of marks awarded on each in order to give more weighting to the one with more marks.

Oliver from The Priory Academy LSST in England and Clara from Bangkok Patana School in Thailand found some examples of when Kyle and the teacher would agree about the overall score. This is Clara's work:

The average and the median marks are the same when the both fractions / percentages are equal. The tests scores where both Kyle and his teacher would agree on his test score would be:

$\frac0{12}$ and $\frac0{20} =0\%$

$\frac3{12}$ and $\frac5{20} =25\%$

$\frac6{12}$ and $\frac{10}{20} =50\%$

$\frac9{12}$ and $\frac{15}{20} = 75\%$

$\frac{12}{12}$ and $\frac{20}{20} = 100\%$

Siddhant from Indus International School of Bangalore in India found some examples of pairs of fractions ($\frac ac $ and $\frac bd$) for which the mean is equal to the mediant.

When I started solving this problem, I realized that there were several, unique solutions to the first part of the problem. For example, one solution I got was:

a = 2

b = 1

c = 8

d = 4

In this case, the averages and the mediant are both 25%.

Another example could be:

a = 4

b = 2

c = 8

d = 4

In this case, the averages and the mediant are both 50%.

What do Siddhant's fractions have in common with what Clara and Oliver noticed?

Vicente José found another way to generate a pair of fractions which have the mediant equal to the mean:

For the mean, $\dfrac{\text{grade A + grade B}}2$, to be the same as the mediant, there are two combinations possible.

1) $c=d$ and so there are the same number of marks in each test and therefore the same weighting is taken on each and so the mean = mediant

Eg. $\frac{25}{50}$ and $\frac{40}{50}$

Mediant: $\frac{25+40}{50+50} =\frac{65}{100} = 65\%$

Mean: $( \frac{25}{50} + \frac{40}{50} ) \div2 = 1.3\div2 = 65\%$

2) $\frac ab=\frac cd$ if both fractions are equal the mediant and mean will be the same as the mean and mediant will both be $\frac ab$ which is also $\frac cd$ (as we had already seen).**The pass mark for an exam is 50%. If I scored 25% on the first $n$ questions, under what circumstances can I still pass the exam?**

Clara started by looking at different ways to get 25%:

The only results you can get to get $25\%$ are:

$\frac14, \frac28, \frac3{12}, \frac4{16}, \frac5{20},$ and so on.

This is a pattern where the numerator increases by $1$ and the denominator increases by $4$ each time. All of the fractions can be simplified to $\frac14.$

Siddhant began using algebra, but was actually using the sum, not the mediant:

Pass Mark $= 50\%$

$25\%$ of $n$ questions $= 0.25n$

$$\begin{align}&\tfrac{0.25n}n +\tfrac bd \ge 0.5\\

\Rightarrow & \tfrac{0.25n}n = 0.5 - \tfrac bd\\

\Rightarrow &0.25 = 0.5 = -\tfrac bd\\

\Rightarrow &-0.25 = -\tfrac bd\\

\Rightarrow&\tfrac bd = 0.25\end{align}$$

The only way that he will pass is if he scores greater than or equal to $25\%$ in his remaining questions.

Vicente José also used algebra:

If $\frac ab = 0.25$ and $b=n$ then $\frac ab = \frac{0.25n}{n}$

If mediant is $\frac{a+b}{c+d}$ and $\frac ab = \frac{0.25n}{n}, \frac cd = \frac yx$ then $\frac{0.25n+y}{n+x}$

If the pass grade is $50\%$ (or $0.5$) then $$\frac{0.25n+y}{n+x}=0.5$$

Multiply by $n+x$: $$0.25n + y = 0.5(n+x)$$ Multiply by $2$: $$0.5n + 2y = n+x$$ Subtract $0.5n$: $$2y = x + 0.5n$$ Divide by two: $$y=0.5x+0.25n$$

The final equation is a linear equation with three unknowns $y, x$ and $n.$ $y$ and $x$ are representative of the next grade obtained, $\frac yx$; and $n$ the number of marks in the first grade.

Vicente José used this to create a graph:

We will then make this an inequality of $y>0.5x+0.25n$, to get all possible answers to make her average score $50\%,$ and we will do another inequality, $y\le x,$ in order to get the real answers as it is impossible to achieve more marks than can be awarded. The area made by the intersection of these will be all possible answers of $\frac cd$ or $\frac yx$ to get a $50\%$ or above.

In the image of the graph you can see the possible answers when the first test was out of $20$ marks.

**Is it true that the mediant always lies between the two fractions? How do you know?**

This is Vicente José's explanation, with a diagram added, which is taken from the vector graph in the question:

This answer is going to be answered in relation to the vector graph given. We know this as when this is put in vector quantities as shown in the question and we add these up, the angle made by the final point to the origin of the vectors and a horizontal imaginary axis will always lie between the first and second vector/grade.

## Teachers' Resources

### Why do this problem?

This problem brings together opportunities to think about the familiar topics of fractions, averages, and gradient. We hope unexpected result in the context of averaging test scores will provoke students to be curious about the mediant of two fractions.### Possible approach

"Kyle got 75% on section A of his maths exam, but he only got 35% on section B.He says this means he should score 55% overall, but his teacher said he only scored 50% overall! How can this be?"

Give students some time to think about this question, and then discuss their thoughts in pairs or small groups. Then bring the class together - "Who is right, Kyle or his teacher?"

*"We don't have enough information because we don't know if the two sections had equal weighting"*

Next, share the extra information from the problem:

In section A, there were 12 questions and Kyle got 9 correct.

In section B, there were 20 questions and Kyle got 7 correct.

Overall, there were 32 questions and Kyle got 16 correct.

The difference in overall scores is because Kyle worked out the

*average*of his two scores, but his teacher worked out the

*mediant*. Define the mediant:

"Given two fractions $\frac{a}{c}$ and $\frac{b}{d}$, the mediant is defined as $\frac{a+b}{c+d}$."

Then share the interactive diagram below. If students have access to computers or tablets, give them some time to explore the diagram. Alternatively, they could use squared paper to create their own examples.

Here are three questions that students could explore:

Can you find some values of $a, b, c$ and $d$ so that Kyle and his teacher both get the same value for the overall score (that is, the average and the mediant are the same)?

The pass mark for an exam is 50%. If I scored 25% on the first $n$ questions, under what circumstances can I still pass the exam?

Is it true that the mediant always lies between the two fractions? How do you know?

###

Key questions

What aspect of the diagram represents the fractions $\frac{a}{c}$ and $\frac{b}{d}$?How does the gradient of the line joining $(0,0)$ to $((c+d),(a+b))$ compare to the other gradients?