# How to help every child understand ratio – method 2

krisboulton
25th September 2015 at 17:40

In this post I shared the box method for solving ratio problems.  The second method I hinted at is the bar model.  Although not a new tool, it has fallen out of our institutional consciousness.  It is not quite as straight forward as the box method, however, it is more versatile, being useful for a range of problems that would otherwise typically require an algebraic model.  Though still not nearly as powerful as algebra, in terms of its versatility and its simplicity, its visual nature means it is much easier for all young children to comprehend, making it a useful bridging step up to full algebraic modelling in years 9 onwards.

Let’s look at the same three problems as last time, but this time, modelled as a bar model.  The three questions were:

1. Jack and Jill share £28 in the ratio 5:2, how much does Jack receive?
2. Jack and Jill share some money in the ratio 5:2.  Jack receives £15, how much does Jill receive?
3. Jack and Jill share some money in the ratio 7:3.  Jack gives £20 of his money to Jill, so now they have the same amount each.  How much money do they have altogether?​

Question 1

The setup is actually very similar to that of the box method.  As noted, the box method might be more intuitive at first, since it directly maps on to the orientation of the ratio notation, however that model is useful for ratio problems only; the advantage of taking more time and effort over bar modelling is that it can eventually be used for problems with fractions and for solving simple linear equations as well.

With the model set up, as with the box method, we look at how many pieces represent £28.  There are 7 of them, so we divide the £28 between the 7 pieces, and see that each piece must be worth £4.

From this we can now easily see that Jack received £20.

Question 2

Notice that, as with the box method, all that really changes is where the information and question marks go.

This time, the £15 is being split between 5 pieces, making £3 per piece.

And so we can now see clearly that Jill receives £6.

Question 3

As a reminder, the algebraic model for this problem looks like this:

A child would require a reasonably advanced level of algebraic rearrangement just to be able to solve this equation, and the ability needed to actually construct the model in the first place is even greater still!  Yet, with the bar model, even young children can cope easily with a problem of this kind:

Before

We can see from the model that Jack needed to pass two ‘pieces’ over to Jill, for them to have an equal number, and that those two were worth £20.  So, each piece must be worth £10.  There are ten pieces altogether, therefore they had £100 between them.

As with the box method, bar modelling has limitations in terms of the kinds of numbers it can comfortably deal with; had the problem above been in the ratio 8:3 rather than 7:3, it wouldn’t have been quite so straight forward.  It is still possible to use bar models to solve a problem like that – split the boxes into a great number of smaller pieces – but eventually the power of algebra to bat aside any variety of problem of number values rather sucks the purpose out of making such an effort.  The box method and the bar model are excellent visual intermediary steps for younger year groups (primary and Year 7-8, maybe) as well as for pupils with lower prior attainment, while crossing the bridge to fully abstract algebraic modelling.

For more information, there is a useful series of YouTube videos here, and an excellent, thorough, guide to using bar models here.