24 March 2017

In this post we’re going to look at a question that combines a number of topics – ratio, time and common multiples – taken from a sample paper sent out by one of the most competitive schools in London.

*In this series we look at eleven of the most difficult questions from the 11+ maths exam – and show you how to do them.*

*The first part of our series, about cows, chickens and simultaneous equations, is available here. *

In the first part of this series we looked at simultaneous equations, and how expressing information using variables is an essential skill for any 11+ candidate. Today we’re going to look at a question that combines a number of topics – ratio, time and common multiples – taken from a sample paper sent out by one of the most competitive schools in London. Let’s get started!

Mrs Varma has a watch (a 12-hour watch) which gains 3 minutes every 2 hours.

- She sets her watch to the correct time at noon on 1
^{st}January. If she doesn’t reset it, when will it next show the correct time?

Ok, so let’s begin – as we always should – by establishing exactly what it is that we know.

We know that:

- Mrs Varma’s watch is a 12-hour watch.
- It is gradually getting
*faster*, by three minutes every two hours. - She is setting it at noon on 1
^{st}of January. It is going to show the correct time at some point in the future.

So where to begin? Firstly, and this is a good trick for any question about time at 11+, establish a *constant*. By that I mean establish something that doesn’t change.

In this case, our constant is going to be the *real time. *Mrs Varma’s watch is not going to show the real time. It’s going fast, and so as soon as she starts it it gradually pulls further and further away from the real time. Let’s have a look at this:

The question is, if Mrs Varma’s watch is gradually getting more and more wrong, how can it show the right time?

The answer lies in the 12-hour clock. If you look at the time on a 12-hour clock, you can’t say if it’s AM or PM. You know only that it’s 3 o’clock, for example. And so, if Mrs Varma’s watch gets so far away from the real time that it’s *12 hours *wrong, it will be showing the same time as the correct time!

That might be a little hard to understand. Perhaps this will make it easier:

So how do express this without drawing loads of clocks? Well, the answer is ratio.

**As is always the case with 11+ maths, we don’t mix units. That means that if we are dealing with hours or minutes, we always go with minutes. **

Remember! When Mrs Varma’s watch adds 12 hours it will tell the right time.

We know that 2 hours = 120 minutes.

Therefore every 120 minutes Mrs Varma’s watch adds 3 minutes.

That means that every 1200 minutes it adds 30 minutes!

That means that every 2400 minutes it adds 1 hour!

That means that every (2400 x 12) or 28,800 minutes it adds 12 hours.

Which means that Mrs Varma’s watch will be exactly 12 hours wrong, and so will be telling the right time, after 28,800 minutes.

We divide 28,800 minutes by 60 (to convert minutes into hours) and we get 480. 480 divided by 24 (to convert hours into days) is 20 days. So Mrs Varma’s watch will tell the correct time *exactly 20 days *after our project began, at noon on the 21^{st} January.

Now for the next bit…

b) Mrs Eddison’s watch (also a 12 hour watch) loses 5 minutes every 2 hours. She also sets her watch to the correct time at noon on 1^{st} When will our two watches show the same time?

Ok, so while Mrs Varma’s watch goes fast, Mrs Eddison’s watch is going slow. We’ve established that ‘the same time’ means 12 hours apart from one another. So how are we going to find out when the watches are going to be 12 hours apart?

Well, we know that Mrs Varma’s watch gains 3 minutes every two hours. And we know that Mrs Eddison’s watch loses 5 minutes every two hours. That means that after two hours the watches will be 8 minutes apart from one another. So let’s do some more ratio!

Remember – 12 hours is 60 x 12 minutes, or 720 minutes.

So we can see that the two watches will be 720 minutes, or 12 hours, different after 180 hours. So how many days is that?

180 divided by 24 = 7.5, which means that the watches will say the same time after 7 and a half days. That is at midnight, or 00:00 on the 9^{th} January.

Ok, now on to the final bit!

c) When will our watches next show the same, correct time?

In order to do this final part we need to work out how often each watch shows the correct time.

We already know that Mrs Varma’s watch shows the correct time every 20 days – we learned that in the first question.

We also need to know how often Mrs Eddison’s watch show’s the correct time.

So, let’s do some ratio for Mrs Eddison as we did for Mrs Varma. We’re looking to see how long it takes Mrs Eddison’s watch to get 12 hours (720 minutes) away from the real time.

So we can see that Mrs Eddison’s watch tells the correct time every 12 days.

We know that Mrs Varma’s watch tells the correct time every 20 days. So what we need to do is find the **lowest common multiple** of 12 and 20. The answer is 60. The watches will both tell the correct time every 60 days. 60 days after 1^{st} January is **noon **on the 2^{nd }March, and so that’s our answer.

Not easy! But this question demonstrates the power of knowing how to do ratio. When it comes to the 11+, good knowledge of ratio is hugely important. Until next time!