In my last post I gave the following exam question which I had looked at with two of my students

And now, as promised, the answer.

I’ve added a few coloured lines to help me talk this through.

The width of the square is the width of two of the circles. I have marked this with the two parallel grey lines of different shades. As we are told each of the circles has a radius of 24cm, that makes the width of the square 4 x this – 96cm!

The height of the rectangle is a bit less, because there is an overlap of the circle diameters in that direction.

We do know the height includes two radiuses* – shown as the parallel red/pink lines. But there is also the height of the green triangle, shown in blue on the height of the rectangle.

We can find that using Pythagoras Theorem – By splitting the green triangle in half, we have two right angled triangle with sides 48cm (the hypotenuse) and 24cm…   so h² + 24² = 48².

h = sqrt(1728) = 41.57cm (to 4 s.f.)**

It can be easy after doing the most complex part of a calculation to forget we haven’t yet finished the question. Let’s not fall into that trap. We still need to find the total height of the rectangle, and thus the area of the rectangle.

Total height is 41.57 + 2 x 24 = 89.57cm

Area = 89.57 x 96 = 8600cm² (To 3 s.f)

A couple of notes about my solution
* Yes, the official plural is radii – from the Latin!  I always think that looks a bit old fashioned so I tend to say radiuses!
**Its often a judgement call how one rounds ones answer. Exam questions will sometimes say (Round to 3 s.f. or 1d.p) but if they don’t, show what rounding you have done. It is not appropriate to give more figures than any rounding given in the data provided….  BUT  if you know you are going to round to 3 s.f. at the end, keep one extra figure in the intermediate numbers. That’s why I use 4.s.f for h, even though I know I am going to give a less precise answer at the end

Today I’d like to share with you an exam question that stumped me for a while a week ago.

 To find the area of the rectangle we need to find the length of the rectangle and its height.

The Length isn’t too hard, but can you see how to find the height before I post the answer here on Thursday?

Growing up , Loving maths, I found out a few things for myself, and I’ll post about one of those next time.

But this is a simple rule with numbers my Dad showed me.  Try adding up the odd numbers one by one, and see what you get

1  = 1

1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25

You will recognise these numbers – all the squares!  And this is the secret behind a type of questions I’ve seen on exam papers recently

There are a lot of different sequence questions, and its best not to leap in until you have really recognised. But if nothing appears to you, a next step is to look at how the numbers are increasing. Actually write the numbers in the gaps. If you worked them out in your head you might not see the pattern.

Here the gaps are :  7   11  15  19

And you will see straight away they are increasing steadily – By 4 each time.  – And when you see this – that the second level of differences are constant – you will know you have a ‘quadratic’ sequence.

That is to say, the expression in terms of n will be of the form

an² + bn + c   – where b and c could be 0 but a is not (If a was 0, the first differences would be constant)

If you play with these sequences for long enough, you will find something very interesting but for this posts I’ll let you in on the secret

To find what a should be, take half of the ‘constant’ you find by following the differences. In this case that was 4, so a = 2.

So our sequence rule starts 2n²

We are now going to write our sequence again, and underneath, write double the square numbers

4    11   22  37   56
2     8    18  32    50

Takes away the bottom from from the top, you get

2  3  4  5  6

or n + 1 for each position

This means our full answer is

2n² + n + 1

A contact of mine on Facebook kindly provided me with a set of harder problem style questions recently for one of my more capable students…  These are questions where you can’t just apply the maths you know – You have to think a bit.

I’m not going to use them all for blog posts, but this was an interesting one I think.

First thought – How can we tell that? Your calculator isn’t going to help; Numbers that big are not going to give you the last digit.

So if we can’t tell directly, what can we tell? Actually, we need to start playing, and playing with numbers is something I love to do.

Let’s start by looking at what powers of 4 are like – That’s going to help us

You don’t have to go very far before seeing that every second number – every power of 4 to an odd number – end in a 4.

Continue that pattern on and we can see 4 to power 333 is going to end in 4.

 

 

 

 

What about powers of 3  –  and here I did need my calculator – 3  9  27  81  243   729    2187  6561

Every fourth number in this list 4th, 8th – and indeed the one before the 3 would be 1, 3^ 0) ends in a 1.

All the numbers divisible by 4 in fact.  So we can say for sure that 3 to the power 444 end in a 1.

And a number ending in a 4 plus a number ending in a 1 will end in a 5.

WE have shown what we were asked to show, and we didn’t have to work out the whole number

 

 

 

Ahh  I thought I’d posted the answer to my last post the day after, and now I see I didn’t. Whoops

In my last post I described a question where I couldn’t see the answer.

When I took the question away and studied it, I realised I hadn’t read the question in full and it wasn’t my small doubts on Circle Theorems that were the issue.

However good you get at Maths there will always be some thinks you spot quicker than others – but in all cases, reading the quesion is so important!

The line I have underlined tells me that < CBA is x degrees – I was trying to think of another Circle Theorem that told me that! The alternate angle theorem tells us that <BAC is x degrees.  Once we know the opposite angles are the same, so are the sides, so AC = BC

Here is a thing I always say to my students…  make sure you read the question ..

And here is a confession..  though I can do algebra and number skills in my head, if I ever have to stop and think, its on the Geometry and Angles questions.

So yesterday was with one of my brightest students, and we’d finished the main activity for the day so I brought out my reserve exam questions…  and the first one was about the Alternative Angle Theorem.

I know this theorem well and I know I know it… but I don’t

always believe I know I know it!  And yesterday I looked at the question and my student was looking at me expectantly…   Later in the evening I realised the main problem was I hadn’t read the question!  Doh, as Homer Simpson would say

 

 

Let’s have a look at the question – and how much easier it was once I’d read ALL the information!

I could immediately see that this was an Alternative Angle question – that is a picture of exactly when it applies.  But ‘Prove AC = BC’ – I couldn’t see how to use the theorem to prove that. There didn’t seem to be enough information in the DIAGRAM!! Its where my moment of confidence crisis kicked in and I’m sharing this with you to show it happens to tutors too (especially when watched by eager students). The way out if this happens to you is to sit back, stay calm – possibly come back to this question later – and read ALL the information given

 

So who can spot what I didn’t read!!  The answer to this – and the whole question – will be given tomorrow

For this week’s posts I’m dipping into the weird land of Quora. As I may have mentioned before the Maths questions on this Q&A based social media site can be weird and can be trivial, like what is the LCM of 2 and 3. 

That is 6 by the way but I’m not going to dedicate a whole post to that.

Others today ask ‘Why do Negative numbers exist’ which is perhaps deeper than the questioner intended, and ‘What are the 100 ways of asking a woman for her number’, which is perhaps only loosely a Maths question. Though if this can be interpreted as her favourite number (not her phone number which I suspect was the idea) then it could be a more interesting one. What does someone’s favourite number say about them.  Have I posted about my favourite number before? I should do that sometime.

There are 5 numbers above and 5 below. – That’s 2 times 5.  Plus the two numbers on the line.

 So that is 2 x 5 + 2 = 12.

I like to generalise these things and look for patterns, and that means , yes, algebra.

So what if the number of balls above the line is a.

The total number of balls is  2a + 2  (Times 2 then add 2)

Now lets think about the problem we are solving. There are 11 balls between the 8^th and 20^th (Count them,  9^th, 10^th, 11^th…. 19^th)
use our formula  – 2 x 11 + 2 = 24.

There are 24 balls in our circle.  We have solved the problem but, better than that, we can answer quickly even if the question setter changed the numbers!

I like that approach; a bit more work and you can answer so many similar questions

So its the gift on days 6 and 7 that she gets most of; The geese a-laying and the swans a-swimming; 42 of each..  Lets hope she has a large pond 🙂

What does The twelve days ‘True Love’ get most of?

The twelve drummers drumming of course!

But is that really true because she only gets 12 drummers on the last day, on that one day. That’s a total number of 12 drummers.

She gets 11 pipers piping on the last two days, a total of 22 pipers!  So that is the most, yeah, she gets most pipers?

Work back through the twelve days and see what she gets most of.

Answer in the next post, later this week.

Should I be posting more ‘seasonal problems’  Ha  maybe….   I’m a bit of a scrooge  really,  try not tow think of Christmas until December reaches double figures in the date

But here is a problem posted in one of my Facebook groups which I thought I’d share with my diary

Might be the way I think, but I automatically think of Algebra when I see this problem

There are two things that we don’t know,  the weight of the box and the weight of a chocolate.  OK, we have only been asked for the weight of the box, not a chocolate but the weight of each chocolate is there in the problem, and I always like to add extra, especially if its chocolate!

Let b be the weight of the box, c the weight of 1 chocolate

b + 8c = 280g  : Eq1
b + 5c = 199g  : Eq2

Subtract Eq 2 from Eq 1 gived

3c = 81g   – so c = 27g

So from Eq 2  b + 135g  so c = 64g

I always like to use the other equation as a check when solving simultaneous equations

64 + 8 x 27 + 280g  as required by the check