# Rod Fractions

What fraction is the yellow rod of the orange rod?

Use this picture to help you. Note that it only uses orange and yellow rods.

You might like to use the interactivity further down this page to help you answer the following problems:

Using as many brown and red rods as you like, but no rods of any other colours, work out what fraction the red rod is of the brown one.

Using as many red and orange rods as you like, but no rods of any other colours, work out what fraction the red rod is of the orange one.

Can you find any other pairs of rods so that the length of the shorter rod compared with the longer rod is a fraction with 1 as its numerator?

Given an unlimited supply of any two differently coloured rods, can you find a general rule to work out what fraction the shorter rod is of the longer one?

You may like to explore this by using the pairs of colours suggested below as starting points.

Dark green and blue:

Pink and orange:

Light green and orange:

Why does your rule work?

Now, looking back at the different pairs of rods you have explored, can you find a way to express the longer rod as a fraction of the shorter rod?

Try using more than one of the longer rods.

Have a think about the above hint before looking at the next one.

How many rods of each of the two colours do you need to make two trains of the same length?

Pablo from International School of Brussels in Belgium and students from Holmgate Primary School found the fractions that some rods are of each other. This is Pablo's work:

(Click here to see a larger picture.)

... and this is Holmgate Primary School's work:

(Click here to see a larger picture.)

Adithya from PSBB KK Nagar Chennai in India made a list of pairs of rods where the length of the shorter rod compared with the longer rod is a fraction with 1 as its numerator:

(Red, Dark Green)

(Red, Light Green)

(Light Green,Dark Green)

(Light Green,Blue)

(Pink,Brown)

(White, Any Colour) 9 possibilities for 10 rods

Keshav from Colchester Royal Grammar School in the UK used a trick to work out how long the rods are compared to each other:

The trick to understanding how the different rods are linked to each other is by looking at their order on the right of the interactivity. At the top is the white rod, and at the bottom is the orange rod. To find the fraction that two rods make - or the number of times the smaller rod is needed to equal the size of the larger one - you need to know where the colour is in the column.

For example, the red rod is second in the column. The pink rod is fourth. If we put these into a fraction, we get $\frac24$, which equals $\frac12$. And, using the interactivity, we find that we do indeed need 2 reds to fill a pink rod. Another example is the green rod and the blue rod. The green is 3rd, and the blue is 9th. Writing this as a fraction, we get $\frac39 = \frac13,$ and we again find that we need 3 green rods to equal the length of a blue rod.

Scout from James Allen Girls School in the UK, and Adithya, both sent in images which show why Keshav's method works. Here is Adithya's image:

... and here is Scout's:

**Given an unlimited supply of any two differently coloured rods, can you find a general rule to work out what fraction the shorter rod is of the longer one?**

Carlos V from Kings College Alicante in Spain and Mahdi from Mahatma Gandhi International School in India both described this general method. This is Carlos' work:

First, you place a rod of one colour over the other rod. Then, continue adding rods until the length of the smaller rods is greater or equal to the larger rod.

If the length of the smaller rods is larger, add another large rod and continue adding smaller rods until they are equal. When it is equal, count how many smaller rods there are and then divide the amount of larger ones there are by the amount of smaller rods. That is the size of the smaller rod compared to the larger rod.

If you want to find the size of the larger rod compared to the smaller rod, divide the amount of smaller ones by the amount of larger rods.

Mahdi also sent in this method, which works if you know the length of the rods (you could measure them using white rods):

For any two rods of length '$x$' and '$y$', when you place $y$ number of rods of length $x$ straight in a line, it will make the same length as placing $x$ number of rods of length $y$ in a line. To make this a bit more clear, let us go through an example:

You have rods of length 4 and 7. So with the two rods of 4 and 7, place 7 such pink (4 unit) rods and 4 of the black (7 units).

We observe that the lengths are equal. Using this, we can calculate the ratio of both lengths easily. Intuitively, we can see why the statement in the first line is true because $y$ number of rods of length $x$ makes a total length of $xy$ and vice versa.

Aditya from Bishop Vesey's Grammar School in the UK also measured the rods and described this neat method which uses the Lowest Common Multiple (LCM):

The way one can work out what fraction the shorter rod is of the longer one is by essentially finding an LCM and simplifying the ratio into an $1:n$ form.

Firstly, to find the lowest common multiple between the two, you will need some sort of unit or measure to calculate with. I used the interactivity's grid squares and measured each square as $1$ unit. Using this, you can find an LCM between the lengths of the 2 rods. Using this, I then formed a ratio. For example, if we were to use the yellow and dark green rods, they both have a length of $5$ and $6$ units respectively and have an LCM of $30$. This signifies that if the two types of rods were placed consecutively under they both met at a column, there would be a total length of $30$.

This means that there are ($30\div5$) $6$ yellow rods and ($30\div6$) $5$ dark green rods. This allows us to form a ratio that there are $5$ dark green rods to every $6$ yellow rods. If put in the form of $1:n$, there would be $1$ green rod to every $1.2$ yellow rod - $1:1.2$

### Why do this problem?

Cuisenaire rods are the ideal context in which to give learners opportunities to express one quantity as a fraction of another (in this case lengths). There is no need to know the quantitative value of the rods' lengths, which means learners can focus on the comparisons, and the relationship between the lengths.### Possible approach

Ideally, learners would have real Cuisenaire rods to use, so that they can solve this problem practically as well as virtually. If they are not already familiar with Cuisenaire rods, it is essential to give them time to 'play' before having a go at this activity.Introduce the task by showing the second image of the two yellow rods and orange rod. Try not to say anything by way of explanation, simply ask, "What do you see? What do you wonder?". Give learners a few minutes of thinking time on their own before suggesting that they talk to a partner. Invite pairs to share their noticings, or wonderings, with the whole group, writing them up on the board without offering comment yourself. Encourage members of the class to respond to anything you have written. If it does not come up naturally during the conversation, ask, "What fraction is the yellow rod of the orange rod?".

At this point, introduce Cuisenaire - either the physical rods, or the interactivity. If you do not have real rods, it would be useful for students to have access to the interactivity in pairs, for example on a tablet or computer. (Throughout this activity, try to refer to the rods using their colours, rather than giving them any numerical values.) Show the image of the red rod and brown rod, and ask, "What fraction is the red rod of the brown rod?". Explain that learners can use as many brown and red rods as they like to find out, and give them a minute to work in pairs. You could invite a pair up to demonstrate what they did using the interactivity on the interactive whiteboard (IWB).

Set the class off on comparing the red and orange rods in a similar way, and share responses before giving them time to find other pairs of rods so that the length of the shorter rod compared with the longer rod is a fraction with 1 as its numerator. In a mini plenary, invite a pair of students to share their findings. Encourage learners to explain how they know they have found all the possible pairs of rods that satisfy this criteria.

Next, pose the question, "Given an unlimited supply of any two differently coloured rods, can you find a general rule to work out what fraction the shorter rod is of the longer one?". You could leave the images of the dark green/blue, pink/orange and light green/orange pairs on the board for learners, should they need a starting point. Allow some thinking and working time, then you may want to draw attention to some different ways of recording that you have observed.

In the final plenary, you could work on a full generalisation as a whole class. Invite a couple of pairs to use the IWB to share what they have done, or use a visualiser to capture them arranging real rods. You may find that some learners express the relationship between the rods in terms of a ratio, for example "For every three dark green rods, there are two blues", while others may be happy to go straight to a fractional relationship. Articulating

*why*the rule works is not necessarily straightforward, and you may need to help learners make the link between expressing the relationship as a ratio and as a fraction.

### Key questions

Tell me about what you're doing/what you've done.Have you tried using more than one of the longer rods?