Home » Archives for May 2020

Month: May 2020

A complicated word problem

Today I am going to be looking at a complicated word problem that an internet friend from Australia shared with me. (Apparently it was on ‘Reddit’ but I rarely look at that site)

Of course we need to use algebra to solve the problem, but forming the equations did take some thought.



The first equation doesn’t give to much difficulty. We let A’s current age be a, and B’s current age be b.

a = b + 15

Making the second equation

But what about the second sentence. That made my brain fry, just a little, when I first saw it.  And my heading is a bit inaccurate, because I started to make not one extra equation, but two extras.

And I am going to use an extra letter, c

Three different Times

There are 3 different times in the question. There is the current time, and we already have an equation for now.

Then there the time when ‘A is as old as B will be’

So at that time  B is ‘a’ years old .  The age of A at that point is described by the rest of the sentence, and we haven’t decoded that. But we do know the age of A then is 4 times something, so I am going to say A = 4c

We know 4c = a + 15  : Because at this different point of time, A is still 15 years older than B

We now need to decode the last part.

When B was c years old,  A = b/2 + 16  – sixteen years older than 1/2 of B’s current age

So using the 15 year old difference between the ages of A and B again

b/2 + 16 = c + 15

We can simplify this by taking 15 from both sides
c = b/2 + 1

I’m now going to substitute that expression for c into the earlier one

4(b/2 + 1) = a + 15
Simplify to give
2b + 4 = a + 15

so a = 2b – 11

We also know that a = b + 15
so 2b – 11 = b + 15, that gives us b = 26,  so a = 41



There is nothing particularly special about this complicated word problem, other than it was convoluted and took an extra step of algebra for me to see a solution.

For more Maths word problems, see here

Shapes and ratio : An exam question.

A question of shapes and ratio

I am going to answer a question today about shapes and ratio.

We would be getting to exam season now, if exams had not been cancelled this year. But I am going to post some exam questions with answers in this diary to keep me – and you – on your toes.

Try the question yourself before reading my answer. All questions will be from recent GCSE exams.


What does ‘Similar’ mean here?

A key work in this question is similar. This means the formulas for area and volume will be the same. The ratios between the height, length and any other measurement of the shape will be constant.

This means that if we take h, the height, as one measurement, and the ratios between the heights of A, B, and C as a:b:c, then the rations between the areas will be a2:b2:c2,  and the ratios between the volumes will be a3:b3:c3

See here for more information on the volume and areas of solid shapes


Working out the ratios

We have been given the Surface areas of A and B, so we can see the ratio is 4:25,  so the ratio a:b is 2:5 (by taking square roots)

We are given the ratio between the volumes of B and C as 27:64, so the ration b:c is 3:4 (by taking cube roots)

To find a ratio a:b:c using whole numbers, we need a LCM of 3 and 5, so that b can be the same in both. The LCM of 3 and 5 is 15, so a:b:c is 6:15:20

So the ratio of heingths a:c is 6:20, which can be simplified to 3:10.

Remember ratios work just like fractions – we get the simplest form by ‘cancelling’

Completing the square – What more can you do?

In my last post, which was a while ago, I looked at the three different ways to solve a Quadratic Equation.  The last of these, ‘completing the square’ may bot seem as obvious as the others, but in fact that is how the formula is found. Also there are questions in GCSE exams, especially on the Higher papers, that do guide the student through this method.

What is useful in the Completing the Square’ method is it can help find minimum and maximum values. You will learn another way of doing that later, called Calculus.

To find the minimum value of x2 + 4x – 1 we can complete the square by adding in 5, but that 5 must then be taken away.

x2 + 4x – 1 = x2 + 4x – 1 + 5 – 5 = (x + 2)2 – 5
We know that (x + 2)2 will be 0 when x = -2, but can never have a lower value.

This means the lowest value of x2 + 4x – 1 is -5 and this is when x = -2.  We can also show that is true by looking at the graph of this function.


This graph has been created by this website. Its called Desmos and looks like a state of the art way of creating graphs.

Go have a look at it!  There are beautiful ‘sliders’ that help you see how graphs change when the equation does.