Last Updated: 09 Dec, 2022

Some 11+ questions can frighten even the best-prepared candidates (and their parents!) but help is at hand in UK Study Centre's new blog series.

There’s an anecdote that UK Study Centre tutors hear quite frequently on first meeting the parents of their tutees: “I didn’t think we’d need a tutor, after all I consider myself pretty good at maths, but then I looked at the questions…”

Ah yes, 11+ maths – the point at which Mum and Dad begin to realise that it has been a *long *time since school and that there’s quite a lot of dust in the attic. Luckily help is at hand. Here at UK Study Centre we’re used to helping pupils through the 11+ entrance process and are very familiar with the kind of brain-muddling maths problems that tend to appear in the papers, so we thought it might be helpful to lend some support to parents and pupils in need. In this blog series we’re going to look at eleven maths questions from 11+ papers that can perturb even the hardiest of candidates, and show you, simply, clearly and succinctly, how to understand them. If you can understand and solve these questions, knowing exactly how they work, it’s unlikely that you’re going to be fazed by much else in the 11+. Without further ado then, let's get started on the first of our eleven 11+ questions!

Question: A farmer owns cows and chickens. Altogether the animals have 30 heads and 100 legs. How many cows are there?

Ok, so let’s get to work!

First of all, from the information given (and with a bit of thought) we can deduce the following:

- We know that if there are 30 heads altogether then our farmer owns 30 animals, since chickens and cows only have one head
- We know that the 100 legs consist of all of the legs of the cows (which have four legs each) and all of the legs of the chickens (which have two legs each). From that we know
*that some multiple of 4, when added to some multiple of 2, will equal 100.*

So what can we put down on paper? First of all, let’s call the cows, and by that we mean the number of cows (this is very important) *x, *and the number of chickens *y. *

So we can say that all of the cows (or the number of the cows) or *x, *added to all of the chickens (or the number of chickens, or *y, *will equal 30.

*x *+ *y = 30 *

OK, so we’ve found a way to represent the numbers of the cows and the chickens. The issue is, how are we going to represent their legs?

**One of the key skills that 11+ candidates need to master is representing information in terms of other information. **

If the cows are called *x *, then what are their legs called?

Let’s imagine that there is only one cow. There is 1 of him. But he has 4 legs. That means the ratio of him to his legs is 1:4. So we can represent his legs by saying ‘the number of him multiplied by 4’.

Another way of thinking of this is to image a room full of people. If we want to represent the legs of the people in the room ,we would say ‘the number of the people in the room multiplied by 2’.

Tutor’s Tip: Practise discussing with your child the name of things ‘in terms of’ the names of other things. Say, for example, that you are 31 years older than your child. Ask your child – for the sake of example, let’s call him Zinedine, to not call you by your name, but to call you ‘Zinedine + 30’. Or if the cat (Boris) is five years younger than Zinedine, call the cat ‘Zinedine – 5’…

And so, if *x *is the number of cows, then 4 x *x, *or 4*x is* the number of legs, because for every cow there are four legs.

Using the same logic, we can see that the chickens’ legs can be represented as 2*y. *

So now we have another equation:

4*x + *2*y = *100 (that is, all of the legs of the cows, added to all the legs of the chickens)

So we know that:

*x *+ *y = *30

and

4*x + *2*y = *100

So where do we go now? Well, here’s the big tip: **When faced with simultaneous equations, the first step should always be to get rid of one of the terms. **

You can do this by using the following method:

- Firstly line up your equations with the larger one on top of the smaller one thus:

4*x + *2*y = *100* *

*x *+ *y = *30

- The next step is to
**make one of the terms the same.**Here we can see that we have 2*y*in the upper equation and*y*in the lower equation. Why don’t we make that lower equation into 2*y*as well?

We do that by multiplying **the whole of the lower equation by 2. **

*2 ( x *+ *y = *30) becomes 2*x + *2*y = *60.

So now we have two equations that have like terms:

4*x + *2*y = *100

2*x + *2*y = *60

Because two of the terms – in this case the number of *y* – are the same, we can subtract one equation from the other thus:

We do this to remove the *y *from the equation, so that we only have to deal with the *x. *

So we’re left with 2*x = 40* which we can solve to be *x = *20.

Now, if we go back to what *x *actually means, we can see that *x * is the number of cows…

And so if there are 30 animals in total, and there are 20 cows, that means that there are 10 chickens!

SOLVED. Congratulations.

The big thing to take away from this question is that we can use a variable, *x, *to represent a *certain number of something.* There are other 11+ questions in sample papers where a certain number of something is involved, e.g a certain number of ladies have lost their gloves, or there are a certain number of coins that are worth a similar amount etc.

We also learned that we can represent a number (in fact, anything really) *in terms of* something else.