The post A complicated word problem appeared first on Tackle Maths.

]]>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

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

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

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]]>The post Shapes and ratio : An exam question. appeared first on Tackle Maths.

]]>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.

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 a^{2}:b^{2}:c^{2}, and the ratios between the volumes will be a^{3}:b^{3}:c^{3}

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

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’

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]]>The post Completing the square – What more can you do? appeared first on Tackle Maths.

]]>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 x^{2} + 4x – 1 we can complete the square by adding in 5, but that 5 must then be taken away.

x^{2} + 4x – 1 = x^{2} + 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 x^{2} + 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.

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]]>The post The three different ways to solve a quadratic appeared first on Tackle Maths.

]]>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 x^{2} – 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 x^{2} – 8x + 16

So how can we get from x^{2} – 8x + 15 = 0 to x^{2} – 8x + 16?

By adding 1 to each side to give x^{2} – 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.

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]]>The post Constructions with compass and straight edge appeared first on Tackle Maths.

]]>I do find constructions with compass and straight edge is one topic that is easiest to tutor by watching videos together

I know how to do them, but when I have previously taught large groups , I have let others do the demonstration by using YouTube Videos!

In fact, I am going to do the same in this post!! But first let me explain some history, because constructions with a compass and straight edge have a place in ancient history. It was one of those things the Greeks of antiquity liked to do. In fact it was they who chose the rules.

So what exactly are these ‘Constructions’? Basically they are exercises in Geometry. There are a few of them but here we will look at three.

The first asks use to bisect a given line. Bisect means ‘divide into 2 equal parts. (Note all these drawings are example sketches. They have not been created by constriction. See the videos below on how to do those.)

Then draw a perpendicular line at a chosen point on a line.

And then, Bisect an angle

You might be thinking – well that looks easy enough; just use my ruler and protractor. But that is because I haven’t shared the major catch; You can only use a pair of compasses and a straight edge. The rules say specifically a straight edge – You can use a ruler, for this BUT you have to ignore the markings! This is NOT a measuring exercise.

Why these restrictrictions? Were those ancient Greeks being just plain awkward? Well maybe, but also, they may not have had a protractor, but mostly this is an exercise in how things fit together.

This is the point where I hand over to other experts…..

Here are some more videos showing these constructions

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]]>The post Words to know for Maths Questions II – Prove, Show, Hence appeared first on Tackle Maths.

]]>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.

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

answer to the first.

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.

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]]>The post Christmas Lectures 2019 appeared first on Tackle Maths.

]]>a maths theme, and the lecturer was to be one of my favourite YouTube Mathematician, Dr Hannah Fry.

Well, there are ways in which they can’t! For me they were an intrinsic part of visiting my Grandparents after Christmas, and of course, I’ll never watch them as part of the target audience again. There are now only three lectures each year, compared to six in my youth.

I give that as background, so that my initial impression makes more sense, which was that this year’s lectures seemed to cover a lot of ground. There was a dizzying pace from one example, experiment or set piece to the next, with the Mathematics explained only at a surface level.

Rewatching for this post, I found I enjoyed the lectures more second time round. Dr Fry’s purpose for her talks was, I think, to show what Maths can do rather than how it could do it. I think the lecturers in my Christmas memories went deeper, but perhaps that was a trick of perspective.

The subtitle of this year’s talks was ‘The Hidden Power of Maths, and if my concern was that a lot of the Maths did stay hidden, Dr Fry did give a lot of examples of how maths effects our lives today.

I think her greatest success was to encourage her audience to appreciate that Maths isn’t just about numbers. As she says early in the first lecture, “Maths can offer a new way of looking at almost anything.” She gives many interesting examples of where Maths can be used. Like Crowd control, weather forecasting, disease control and football tactics. That was all in the first lecture.

In the middle episode, Fry made a good job of explaining the idea of an algorithm. They are good at solving a Rubik’s cube. They are less good at recognising a dog as a dog, and in making a cup of tea. Instructions to a machine have to be very specific in definitions and what to include

I thought the final talk was the better for not trying to cover so many examples. It opened with a spectacular cycling stunt, and closed with some examples of how maths can be used to fool us with fake news and fake music. I confess I found it very hard to tell the fake music from the real.

Despite my reservations on the breakneck speed, I would still recommend The Royal Institution Christmas lectures to any Maths student. They are available on YouTube.

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]]>The post Words to know for Maths Questions I – Exact, Sketch, Plot appeared first on Tackle Maths.

]]>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.

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)

The diagram below shows a sketch of the graph of f(x) = 1/(1+x^{2})

In sketching this function, the important things to note are

i) Its always positive

ii) It will have a maximum when 1+x^{2} 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

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

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]]>The post Quadratic Roots : An A-level question appeared first on Tackle Maths.

]]>*The equation kx ^{2} + 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

b^{2} – 4ac > 0 where the equation is ax^{2} + bx + c = 0

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

16k^{2} – 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

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 x^{2} + 3x + 3 = 0

If we factorise out the 3/4 we get

1/4 (3x^{2} + 12x + 12) = 0

3/4(x^{2} + 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.

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]]>The post Problem solving with Similar Triangles appeared first on Tackle Maths.

]]>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

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 cm ^{2}**

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 cm^{2}

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