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A little light algebra to end the week

I thought I’d finish the week with a problem from my friends at ‘Quora’.. but then I found mostly quite difficult questions requiring University Maths! Yes I could follow (most) of it but it was all a little ‘off-piste’ for this diary,

So I have picked out a wordy problem that might look complicated at first sight, but we re

ally just need to turn it into a couple of equations…  Let’s see




Lets call our numbers a and b – That’s always a good start with Algebra.

The first equation gives us the sum. So

a + b = 26.

With the second equation, we need to decide which is the larger number. It doesn’t matter which be choose. Let’s say a is larger.

a =  5 + 2b  (a has a value 5 more than twice b)

This is where we can solve the simultaneous equations. I’ll make a video on all the ways to do this soon, but for now I am going to solve these ‘by substitution’

I am going to substitute into the first equation the expression for a in the second. In the () you will see I have replaced ‘a’ with what a = in the second

(5 + 2b) + b = 26

We can together together terms

5 + 3b = 26

Take 5 from both sides

3b = 21

And from this see that b = 7. And so a = 19 (Twice b plus 5).

As a final check 7 + 19 = 26

So there we have it the two numbers are 19 and 7


Reading the Question

My second post on avoiding and correcting errors seems a bit obvious! But it doesn’t stop many people, including myself from forgetting this ‘tip’ from time to time!

I’ll show this by an example from the exam paper I’ve just completed.

The first part I answered correctly – see if you can!

Its the second part I am writing about here.  Now I know why I made the mistake; I’d tutored students many times before on questions like this – or so I thought – and I wanted to show off!

I read the question as ‘What was the difference in Monthly average expenditure’ between the two years, and I have seen students answer this question before – when it was the question asked – by reading off all the data, adding all the numbers together and comparing.

And I usually tipped the more able students that they don’t have to do this – Its quicker to find the difference in each column and add these up. So that is what I did now….

Thinking you have seen the question before, like this, is a common reason for misreading – And its an error that is going to effect more able students, or at least ones with a good memory!

But this won’t help when the marks are added up.  Its the questions where you think ‘Oh, I know what to do with this one’ that you need to be most careful with, because they are the ones you are more likely to rush into without proper reading

OH, and have you spotted the important detail I missed in the question?

You will see I did too much work, and would have got no credit for it.











When the answer just doesn’t seem right

I’ve completed  a couple of Foundation exam papers this week, as a bit of forward planning for one of my students – and when I came too compare my answers with the official ones, I had made a couple of mistakes, and I”ll post about these over the weekend.

In this post, though, I’d like to share with you a question that I did get wrong initially, but where I spotted my own mistake.

The thing is – My answer just didn’t seem right.  And that’s what I’d like to share with you in this posts – It’s a very important skill with number questions; To be able to feel when your answer feels right.

This is the question






To make this comparison, you need to work out how much one biscuit cost.

For the 20 Biscuit tin the sum is  £1.50/20. Although this is from a calculator paper, I did this ‘long hand’ and got the answer 7.5p each.

I then did the calculation for the second tin, again ‘long hand/in my head’  and got the answer 5p per biscuit.

I don’t know how I did this – I made a mistake, and I never pretend I never make mistakes. I just wasn’t taking care.

But what I can do is think ‘Um, that doesn’t seem right. The second box costs a bit more than the first, and has slightly more biscuits.’

The answer just had to be about the same, not as different a 5p and 7.5p. That might seem like a big difference, but 7.5p is 50% more.

So I did the answer again, and found the cost per biscuit was also 7.5p each. The answer was that Nada was wrong, box 2 offers the same value, not better.

This might not work every time – If my wrong answer had been 7.4p per biscuit I may not have spotted my mistake.

But you would be surprised how often just thinking as you wrote your answer ‘does this make sense’, you can spot some basic errors

A little more on Higher Dimensions

Continuing yesterday’s theme of higher dimensions, in this post I look at a couple more 4 Dimension shapes and how they can be represented on a screen.

Thinking about how to show 4 dimensions its helpful to think of ways to show 3 dimensions on a 2D screen. One such way is to take ‘cross sections’.  For example, one way to show a cylinder is a series of circles.  Stack these together and you get a cylinder.

A cross section of a 4 Dimension shape will be a 3 Dimension shape,

Another way of thinking of this is to consider time as the forth dimension*.  Imagine seeing a sphere appearing as a dot, then growing to ‘full size’, then disappearing again at the same rate.  The diagram here would show stages of the process.

I’ve seen other representations of a ‘hypersphere’ but this is the one clearest to me.

How would a ‘Hypercube’ look like, using the same ‘cartoon technique?

(*Some people think this to be the case, through the 4-Dimension space-time physicists work with isn’t that simple, but it will do for this thought experiment).

Before we move off 4-Dimension shapes, I’d like share one of my favourite shapes – This can only exist if we have 4-Dimensions (The closest 3 dimension idea is a mobius strip)

In the picture, it looks like the bottle goes ‘through itself’ but in the ‘4 dimensions’ this would not be the case.  Rather like if we want to het past a wall we step over it, using the third  dimension that a creature who knew only two dimensions could not


Living on a higher dimension…

… is something I am sure we’d all like to do; Well if I meant a level of unbound wisdom; But that’s enough of existential ponderings. This is a Maths Diary so I probably mean Dimension, as in shapes!

Maths students study shapes that have 2 or 3 dimensions …  but the underlying maths regarding shape can be extended to more dimensions

For example, we all know the area of a square is  L2 where L is the length of one of its sides   The volume of a cube is L3, again where L is the length of one of the sides.


So what does L4 represent. Its definitely something we can write down…  but does it have a meaning.  Its fair to say that, by extension this  would be the ‘amount of space’ occupied by an equal-lengthed shape in 4 Dimensions!

The difficulty lies in trying to relate that to what we know, as we don’t know 4 dimensions. A 4-dimension cube is often called a hypercube, and below is representation of one; But here we are somewhat up against it because we are trying to show 4 dimensions, using just a 2-dimension screen.  I’ve seen various ways of doing this, and the way I’m showing here is the clearest to me… Think of the ‘cube inside’ as being smaller only because its further away. Its really the same size.




To an extent we also had this problem in drawing the cube, as shown above, as there we were trying to show three dimensions on a two dimension screen.  That was only a ‘gap’ of one extra dimension though, and we are familiar with what a cube looks like.

In my next post I will continue with this theme, and consider how we can show other 4 dimension shapes in two dimensions.



Is BODMAS for Life (Or just for Christmas)?

OK so its a bit eccentric having a post about Christmas at the start of August, but I liked the title so I am sticking to it

The question is –  Does the rule for order of calculation, BODMAS  always apply?

Well the real answer is ‘yes’ – but should the rules should be bent sometimes?

This idea started with a post I saw on Facebook – I gave my initial answer yesterday, and this morning I had to admit I got it wrong – following BODMAS to the letter.

The question is, simply, what is the value of



Add 2 + 2 to get 4; multiply by the 2 outside the bracket and get 8;  then 8 divided by 8…  we get the answer 1.

Actually, by BODMAS rules the divide should come before the multiply (D before M)   so it should be 8 divided by 2 (Giving 4)….  4 x 4 = 16

That is probably the ‘Correct’ answer and I had to accept I was wrong  – and there is nothing bad about accepting one is wrong sometimes


But I still feel somewhat attached to my original answer!  To me,  2(2+2)  LOOKS like a single unit for calculation. If the x sign had been there between the first 2 and the ( , as below, I don’t think I’d have made the same mistake


For me, the () is such a powerful sign,  I see any digit next to it as ‘belonging’ to it, and hence how I did that calculation in the way I did. I can’t claim that is the official rule; just the way I read it.

So my recommendation is, BODMAS rules as they are, if you want to communicate a calculation, if there is any doubt on what you mean, include extra brackets to avoid confusion


Re-arranging the furniture

In my last post I gave a formula for finding the temperature in Fahrenheit if we know the temperature in Celsius.

The reverse can be found with a simple rearranging of the elements – indeed I believe I wrote about this in my diary before

F = 9/5 * C – 32

F + 32 = C * 9/5

C = (F – 32) * 5/9

This is a subject on which I helped a student once after his Science tutor said he wasn’t getting it. After working on it with me, the Science teacher noted the improvement.

In fact a study of maths does help with a study of the Sciences, execially Physics

This has been a short post, tidying up the last one and with a bit of a repeat  – In August I plan to feature a ‘Mathematician of the week’ so keep readiing!

Oooh what a scorcher!

Well,its hot here in the UK, and that may not make people think immediately of maths! But its a subject useful

for every occasion.

So today I am writing about ‘temperature’ and how it helps us understand Maths.

Temperature is the sample I often use when explaining negative numbers to students.  It is one example were minus numbers are used in everyday life – but perhaps not in this weather!

Another time temperature is useful when I tutor, is using formulas and rearranging formulas. I think I may have written about this in this diary before, but there is no harm in revising the ideas.

As you know there are two temperature scales in common use; Celsius and Fahrenheit, both named after the scientist who invented them.

(Since these are usually abbreviated to  ͦ C  and  ͦ F some people think the first of these is called Centigrade. This is understandable because it started being used widely about the same time we started using metric measurements like centimetres – But its more accurate to use the inventors name; the initial is just co-incidence)

These days both scales are defined by the freezing and boiling points of water – 0 and 100 in Celsius; 32 and 212 in Fahrenheit. Quite why these names are chosen seems very strange, but its likely that wasn’t how Mr Fahrenheit came up with his scale.

We can use these numbers to find the formula to convert  ͦ C  to  ͦ F

Lets say this formula is F(c) = ac + b – It is a linear function, we do know that

F(0) = a x 0 + b = 32 so b = 32

F(100) = a x 100 + 32 = 212

From this we can get 100a = 180  (I’ve simplified the a and taken 32 from both sides

This gives a = 9/5  (180/100 simplified)

So F = 9c/5 + 32 – which is the familiar formula, obtained from the fixed point and a spot of function algebra

Taking our Polygon formula to the limit

In yesterday’s post I considered how we can find the area of a many sided regular polygon given the number of sides and the length of each one.  We found the formula

For this post I’m going to add Circumference into the mix; all the way round the shape, the length L, S times

C = L x S

In the formula, we are now going to replace the L with a C, since the more sides we have, L is going to get small, and C is an easier thing to measure.

L = C/S is just a rearrangement of the formula above which we can use to replace L in our area formula

This gives a formula with C in of

A = C2/4STan(180/S)

We now have moved all mention of S to the bottom of our formula.

Now if C is kept the same but S gets bigger and bigger, what does that mean for our formula?

As S gets bigger. 180/S will get closer to 0, and so too will Tan(180/S), S will obviously get bigger, so what does that mean for  S x Tan(180/S).

Now this is where I am going to cheat a little; It may be possible using mathematical techniques to see what happens as S gets bigger…  but I am just going to plug some numbers in

When S = 10,  S x Tan(180/S) = 3.249
When S = 50, S x Tan(180/S) = 3.146
When S = 100, S x Tan(180/S) = 3.143
When S = 1000, S x Tan(180/S) = 3.142

S has to get quite big before the pattern is clear, but it seems that
S x Tan(180/S) is getting close to a very familiar number, π.

So it seems for very large values of S

A = C2/4π

Remember that C = 2πr2 so C2 = 4π2r2

So, A = 4π2r2/4π

The 4π on the bottom cancels with elements on the top and we are left with

A = πr2

Which is the familiar area of a circle! And if you think about polygons with many, many sides you will see they are very close to being circles.

This is why I love maths! Everything fits together!

Area of a polygon

I meant to post this last week, but I could see after my first draft I’d made a mistake in the Maths!  Yes it happens……

Anyway, many of my posts have addressed specific exam questions but sometimes I just get a bug about something .. and then I think,  why have I never thought about that before!!

I don’t know of any maths course this is on, but hey, I just wanted to know, as I wrote my last two posts….   Is there a formula for the area of a regular Polygon!

If it is 3 sided or 4 sided – a triangle and a square – then we know the formula for area, but I was thinking – what about a formula that works for any regular polygon – That is to say, one with all the sides the same.





Here is a polygon and lets say the length of all the sides is L.  You can count the sides here and see there are 8 – this is an octagon  – but let this represent ‘any polygon’ with a number of sides S.

What I will be looking for is a formula where A = something with L and S in, as they are the two ‘properties’ of the polygon that might change in our ‘general’ polygon

The art of finding the area of any unfamiliar shape is to divide it into shapes for which you know how to find the area.  Any polygon can be divided into triangles by drawing lives from each corner into the middle.

How many triangles?  S  – One for each side of the polygon.


What is the area of each of these triangles?  For this we need to know the base and the height; The base is L. To find the height we will need to use some trigonometry. Half of the triangle is a right angled triangle. I am going to use the angle at the top of the half triangle, which I have labelled x

The full angle at the top is 1/S of the 360 degrees at the centre  =

360/S.   x is 1/2 of this  = 180/S

Tan(x) = L/2  /  h  (Opposite/Adjacent where the ‘opposite’ is half the base  and h is what we will call the height, for now.

Add in some detail and rearrange

Tan(180/S) = L/2h
Rearrange again to give h = L/2Tan(180/S).

Area of the small triangle is (using 1/2 x b x h)   = L2/4Tan(180/S).

This is only 1 triangle out of S triangles, so the formula for the whole polygon is

Now, as I said at the start, I worked out this formula a week ago, bit I wanted to check my work because it looked a bit cumbersome….. But it does seem to stand up, and I’ll show you how.

There are two polygons for which we know another formula; The Triangle and the square (S = 3 and S = 4).  Lets see what happens if we make S= 4

A =  L2 x 4 / 4Tan(180/4)

Tan(180/4) = Tan(45) = 1 – so we cand leave this term out. Also the 4 and 4 can cancel, and we get

A =  L2 – the simple and familiar formula for the area of a square!

Checking this for the triangle is more tricky because we to find the ‘normal’ area for a regular 3-polygon – or as we usually call it, an equilateral triangle.  For this we need our old friend, Pythagoras’ Theorem.

Using the theorem in half the triangle, we get

(L/2)2 + H2 = L2

H2 =  L2 – L2/4  = 3L2/4

H = √3L/2

so A = √3L2/4

Now lets see what we get from our polygon formula with S=3.

Tan(180/3) = Tan 60 = √3

A = (L2 x 3)/(4 x √3)

Remember that 3/√3 = √3

so A = √3L2/4

The same as with the direct method!

That, I claim, justifies my formula.  If anyone can think of an alternative way of finding the area of a pentagon or hexagon, then the formula can be checked for these shapes too.

Now, interesting things happen to our formula if S gets bigger and bigger, but that will have to wait for another post!