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# 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

Conclusion

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.

# The three different ways to solve a quadratic

In this post I’ll look at the three different ways to solve a quadratic equation

Oh yes, Sorry, first post for a while. I spent much of February making videos for my Facebook Group  (And yes, that’s a link to it!)

Let’s consider the equation  x2 – 8x + 15 = 0

The first way we learn is to factorise, but spotting the factorisation isn’t always is easy or straightforward.

If you can’t see an immediate factorisation, you might want to reach for the formula, The formula has the advantages that it can always be applied  – and can tell you quite quickly if there are real* solutions. Its a good idea to work out the square root part first, as this will tell you if there are solutions and how many.

In this case : 64 – 4 x 15 x 1 =   64 –  60 = 4.  The square root of 4 is 2.

(8 + 2)//2  = 5 and (8 – 2)/2 = 3

The fact that the solutions are integers suggests that the factorisation does exist, and we could have spotted it if we had looked a bit longer.

What numbers multiply to make +15?   1 & 15 and 3 & 5; and the negatives.   Our numbers need to add to -8, so the ‘answer’  is -3 and -5.

The factorisation is (x – 3)(x – 5), and from that we can see the roots are 3 and 5.  Roots is just a fancy mathematician way of saying the solution.

The last method is called ‘Completing the square’, which isn’t immediately obvious but is actually the way the formula can be found.

We take the number before the x  (called the co-efficient) and halve it – In this case that gives us -4, which we write like this

(x – 4)2 – which can be expanded to x2 – 8x + 16

So how can we get from x2 – 8x + 15 = 0  to x2 – 8x + 16?

By adding 1 to each side to give x2 – 8x + 16 = 1,  or (x – 4)2 = 1

Taking the square root of both sides gives  x – 4 = +-1.
Add 4 to both sides gives  x = 3 and x = 5

We get the same answer for the three different ways to solve a quadratic – but it would be awkward if that were not so!

For help on this and other Maths topics, see my current deals on Lockdown tuition.

# Words to know for Maths Questions II – Prove, Show, Hence

### Words to know for Maths questions

The post before last, I gave a few words to know for Maths Questions, and today I carry on with that.

The first post was about words that describe the answers we should give.  Today’s words are about how to tackle the question.

Questions that use the words like prove and show that gives you the end point – the answer – in the question. The requirement then is to show understanding of how to get to the answer.

We are all told that it is important to ‘show our working’ but in this type of question a prove question, its the ‘working’ that is the point.  You need to show each step of the process of getting to the answer. Its fair to say that ‘Mathematical Proof’  is a rigorous process.

The further you go in Mathematics, the more important ‘Proof’ is compared to ‘doing sums’  – especially in what is called ‘Pure Mathematics’.  This was we get results, or answers, that can be reused many times in other questions.    I’ll write another Post on Proofs later but here is a link to some examples of proofs.

## Two more useful words

The other words I want to mention today are Hence and Otherwise. Hence is a very helpful word because if you heed it, you will save time in answering a question.

A simple example would be a question like this

The question here says, solve the second part using your

I think with this question, we would do it this way anyway, but the word Hence is more than a big clue, it is an instruction to use the first part.

The alternative is Hence or Otherwise.  Questions with this wording also give a big clue that the quick way to answer is to use an earlier part of the question, but the otherwise lets the student find an alternative way.

My advice would be to look for a way to use the work you have already done. Its probably going to be quicker. Only go for otherwise if you really can’t see a way of doing this.

# Words to know for Maths Questions I – Exact, Sketch, Plot

### Three Keywords : Exact  Sketch and Plot

Exact  Sketch and Plot are three words you may see in exam questions that have specific meanings.

This is the first of 2 or 3 posts I intend to make this week about words you may see in an exam question and the specific meanings of these words. I might run a quiz on these words in a few weeks time.

## Exact

One of my students likes to reach for his calculator to given a numerical answer to questions – but this is what you should not do when a question asks for an exact answer.

In an exact answer questions,  symbols like √2,  e, and π are exact descriptions of values. 1.414,  2.718 and 3.142 are approximations.

A question that asks you to find an exact value, and you reach in your calculations  π + e, then leave this as the answer. Do not write the answer as 5.860  (though that is correct to 3 s.f, its not exact)

Sketch

If a questions ask for a sketch of the graph of a function, it is not asking you to find the values over a range, and plot each point. The idea behind a ‘Sketch’ question is to give a general idea of the shape of the graph.  Certain points of interest of the graph are important in a sjetch

i) Points where the axes are crossed
ii) Turning points and points of inflection – where the gradient is 0
iii) Asymptotes – lines to which the curve is getting close to but not touching
iv) An idea of what the function does outside the range; Gets bigger/smaller, tends to 0, repeats (as in sine curve)

## An Example

The diagram below shows a sketch of the graph of f(x) = 1/(1+x2)

In sketching this function, the important things to note are
i) Its always positive
ii) It will have a maximum when 1+x2 is minimum (when x=0). This will also be the pint the y axis is crossed
iii) To the left and right the line will get closer and closer to the x axis.

And here is my sketch of the function.  I should mention here I find it much easier to sketch graphs with hand pencil and paper than I do with a mouse. This isn’t great but gives the idea ## Plot

In an exam question on graphs, plot is asking for the opposite of sketch.  This is where you are expected to work out values of the function, plot them on the graph and join up.  Draw has a similar meaning.

Next Post

In the next post I will give some of the words that signify a ‘proof’ type question, and what each one means

# Quadratic Roots : An A-level question

In this post I am going to be looking at question given to my A-Level student in a recent test. We need to be considering how to find when there are no quadratic roots using a fact I considered in a recent video on my YouTube Channel

### The Question we are looking at is

The equation kx2 + 4kx + 3 = 0 has no real roots

Show that 0 <= k < 3/4

As I give in the video, the clue to finding out when there can be no real roots lies in the formula for solving quadratic equations. The part in a square root sign must be > 0

b2 – 4ac > 0   where the equation is ax2 + bx + c = 0

### Form another inequality

In this question, the a and b are given in terms of another unknown, k.  but we can now form an inequality statement

(4k)2 – 4 x k x 3 < 0

Which we can reform as 16k2 – 12k < 0

and factorise to

3k(4k – 3) < 0.

The graph of this new quadratic is shown here.  Because the k2 is positive, the curve is a smiley face, not a glum face.

This means that the part where the graph is negative

is between the roots

This means we know that 0 < k < 3/4

### Boundary cases

But the interesting cases are where k = 0 and k = 3/4.  We have to be careful here because its not clear what these values mean for our original equation at these values. We need to check back.

If k = 3/4, the original equation is

3/4 x2 + 3x + 3 = 0

If we factorise out the 3/4 we get

1/4 (3x2 + 12x + 12) = 0
3/4(x2 + 4x + 4) = 0

3/4 (X+2)2 =0

And as we can see, -2 is a root of this equation, but there are no others.  so k=3/4 does NOT meet the criteria of there being no real roots but does describe the special case of there being exactly 1 root.

One might think that k=0 will also give a 1 root case. Let’s see.

Setting k=0 and the original equation becomes

3 = 0   –  which doesn’t seem to make sense. It certainly has no solution for x; For all x its an equation that will never be equal.

So k=0 also meets the required condition that the equations has no roots, or solutions

So we have shown that if k meets the criteria 0<= k < 3/4,  the equation has no roots.

# Problem solving with Similar Triangles

This  posts uses Similar triangles to solve a question that was posted in a Facebook group for tutors.   I thought it was fun so I’m sharing it here.

My answer uses the idea of Similar Triangles. If you are not sure what they are, click here.

My first thought on seeing the question was that similar triangles were involved but I didn’t realise they were studied before 11.

Anyway, here is the question. There are two parts but the method is the same in both. You can just look at one of the lower triangle and the overlap. The whole of the other triangle can be disregarded

### Solution to similar triangles question

The key is seeing that the shaded area is ‘similar’ to the lower triangle (or the other one).  We can show they are similar because all the lines are parallel, so the angles are the same – corresponding angles

Both Triangles have a vertical line and a horizontal line, so those two pairs are parallel, and the two hypotenuses are ‘parallel’ as they are part of the same line.

As they are similar, the sides must be in the same ratio.  The ratio between them can be seen most clearly by comparing the lengths of the bases….  6 for the bottom triangle, 1 for the overlap

So the ration is 6:1. The lower triangle has height of 4,  so the height is 4

4/6 * 1 – or 2/3 a square.

The area is 1/2 * bh – so that makes 1/2 * 1 * 2/3  = 1/3 cm2

### The second Triangle

The method for the second triangle is the same.  here the ratio is 7:3 and the height of the lower triangle is 5cm.

Using the ratios we can see the height of the overlap is 3/7 * 5 =15/7

The area is 1/2 * 3 * 15/7 = 45/14 cm2

# How to solve a Sudoku using Set Theory

In this entry I am going to look at ow to solve a Sudoku puzzle using Sets and Venn diagrams

### What is a Sudoku Puzzle

Sudoku puzzles have now been around for over half my life, and sometimes I’ll have a go at one. They are not my favourite sort of puzzle but they divert the mind for a while.

If you do Sudoku, you may not realise it, but you are doing a problem in 3 dimensions. Each number has to be unique in three directions – in each column, in each row, in each box. So that is one I did a few days ago – and yes, I did finish it!

### Venn Diagrams

But before I describe how I went about that, lets track back and look about something I learnt in about Year 7 (or ‘First year seniors’ as we called it in my day!)

Actually I did ask my A-level student from 2 years ago about sets and Venn Diagrams – and he said he hadn’t learnt about them. But there are definitely questions about them on GCSE exams now.

Venn Diagrams were the invention of English Mathematician John Venn who was working about 100 years ago. They are the clearest way to show sets and how they relate to each other.

Let’s just step back one step first though. What is a set?

A set is just a collection of things. Of letters, of people, of cats… even of sets!  Let’s say set A is the ‘the set of all the vowels in the alphabet’,   B is the set of all the letters in the word FACE This is a Venn Diagram showing set A and set B. The place to look is where the two circles overlap. In that space I have written the letters A and E because  they belong to both sets.

I said that the ‘things’ in the sets could themselves be sets themselves. That might sound like a strange thing to say.  But lets say C is the set of all the elephants in the world and set D which is ‘the set of all sets of animals!’.   Then set C would be in set D!

### How to solve a Sudoku puzzle – using Sets!

Let’s get back to Sudoku. How can set theory help?

When I look on what number I can put into an empty square – let’s say the square in the middle of the second row

In this Venn Diagram, I’m defining the set ‘Row’ to be ‘all numbers that don’t yet appear in row 2. Likewise, the sets ‘Column’ and ‘Square’ are all the numbers that don’t yet appear in the 5th column and the top-middle square.

NOTE: I’ve said numbers are members if they DON’T already appear on the Row, column and square. That is because that is what qualifies them as the right number for the square.

The number 1 is already in the row, column and square.  But we can put the number 2 on the diagram. Its not in the column yet. If I consider number 3 to 9 in turn and add them to my Venn Diagram, I get this The right number to put in the box needs to qualify in all three ways. It needs to be in all three sets.   The middle of the Venn Diagram, where all three sets (i.e. the circles) intersect.   We actually have two numbers in that space. 6 and 8.  Actually that means we don’t yet know what number to put in this square. More of the puzzle needs to be solved before this box is.

Do I draw Venn Diagrams  for every blank square? Well, Ok, I draw them in my head, but my thinking follows the same line

I am solving a Sudoku puzzle using ‘Set Theory’

# The next number in a sequence

You probably saw questions ‘What is the next number in the sequence?’   quite early in studying numbers.

### Next number in a sequence example

2  4   6   8   10  ..  what is the next number?

Let’s just ignore the people who can see the answer straight off,  and look for a method.

Look for the GAPS between each number. That is always the best start. In this example, the gap between each number is 2 – To put it another way, the numbers are going UP by 2 each time.

Once we have spotted that, we can move on through the sequence, adding two onto the last number each time.

… 12  14  16

Where we can get clever though is trying to find a general term in this sequence.  This is a step up in effort, for sure.

What we do here is give all the numbers in a sequence a ‘place in the  sequence’   which we do, in true algebra fashion, by using a letter. Its normal to use n in sequences.  Questions will usually ask ‘find the nth’ term?’

We say we the first term in the sequence is n=1,  then n=2 for the second, and so on.  The general term then uses n in a formula. Let’s see how its done

### Finding the nth term

The first thing is to see the gap, as we saw before. In the first example, the gap was 2

1  4  7 10 13

In this sequence the gap is 3

So we start our nth term formula by putting this gap number in front of the n.   2n for the first example,  3n for this example.

So – does ‘3n’ give the sequence we’ve been asked to investigate?

No, because that sequence is

3  6  9  12 15 ..

But we can compare the two sequences – something we do a lot when looking for nth term formulas – and see our sequence is 2 less for eeach term than 3n: 3 6 9 …

so we have our formula   3n – 2

We can check, say, the 5th term  –   5 x 3 – 2 = 15 – 2 = 13. That matches what we were given. We can now confidently predict the 100th term

100 x 3 – 2 = 200 – 2 = 298

### Is the number in the sequence?

Another common question is – is 100 in the sequence 1 4 7 10….

This question has not asked you to find the nth term – but that is the route to finding the answer.

We have already found the nth term for this sequence. This means we need to find n where

3n – 2 = 100.

We start to solve this like an equation, by taking 2 from both sides

3n = 98.

Now we hit a hitch.  n is not going to be a whole number because 3 is not a factor of 98.  In sequences we are only interested in cases where n IS a whole number.

from this we can say that 100 is NOT in the sequence because there is no n where 3n – 2 = 100

### More Practice

For more practice, see this website