## Sherlock Holmes Christmas Maths Challenge: Episode 5 Solution

Many congratulations to the **Episode 5** JOINT winners of our **Sherlock Holmes Maths Competition:**

**Year 10 Set 1 and Year 9 Set 1 – both from Canon Palmer School in Essex –**

for successfully working out that Holmes and Watson have been led in a figure-of-eight…

**For the answer to where Holmes and Watson must go next is:**

**…BACK TO 221B BAKER STREET!**

We’ll be sending the winners a **FREE Class set of Longman’s unique 2010 AQA Functional Maths Student Book (30 copies) ****and a tin of Cadbury’s Heroes.**

**DON’T MISS THE GRAND FINALE, Episode 6, to be published tomorrow (1st Dec.)**

We now know **WHO** stole the Devil’s Eye Ruby: Moriarty. And we know **WHY** he stole it: to build an EMP (Electro-Magnetic Pulse) weapon that could destroy all electronic devices within miles.

**BUT WHERE WILL HE USE IT, AND WHEN? See if you can solve the clues in Episode 6 tomorrow, and help Holmes & Watson to foil Moriarty’s plot…**

Meantime, here is the complete solution to Episode 5:

First part: The Science of deduction:

Working out the ratios

A=5 B=15 C=1 D=4 E=2 F=2 G=6 H=8 I=10 J=9 K=6

So there is J+F+G=17 MW going to Oxford Circus. So there is not enough power going to the trigger.We then worked out the remaining time on the song was 2mins 21 sec and that Jerry Rafferty sang a song called Baker Street which brought them to the location: 221B Baker Street.

## Free AQA Functional Maths for Black Friday!

Counter-intuitively, Black Friday is a good thing – US shops ‘go into the black’ after Thanksgiving, when everyone goes out shopping…

Try out this **free** Level 1 Functional Maths activity, from our **Longman AQA Functional Maths resources**:

And here are the **teacher notes**, including answers.

Let us know how your students get on!

## 4 Days to go: Holmes’ Christmas Challenge

For anyone that missed it, **EPISODE 5** of our **Sherlock Holmes Maths Competition **has been published.

**ENTRIES BY 5pm NEXT MONDAY, 29th NOVEMBER… **

**…will be eligible for this Episode’s prize: a FREE class set (30 copies) of Longman’s 2010 AQA Functional Maths Student Book and a tin of Cadbury’s Heroes.**

**This competition** is open to all UK Secondary Maths teachers, who can enter on their own behalf or on behalf of their GCSE class.

**GOOD LUCK!**

## After the AQA GCSE Maths Unit 1 exam, download FREE algebra tests for Unit 2

You’ve probably seen the recent Unit 1 exam paper for the new AQA GCSE Maths, with 50% of the marks coming from AO2 and AO3 questions.

Watch a short video about this big change from Series Editor Glyn Payne:

Our new resources ** Longman AQA GCSE Mathematics **provides a stack of practice for AO2 and AO3, as well as helping students secure their basic maths for AO1.

In particular, there are Practice Papers (for both modular and linear) in the ** Practice Books **that cover the full ability range: Foundation Sets, Middle Sets, Higher Sets plus the special ‘G to F’ and ‘A to A*’ Books. In addition, the **Assessment Pack** includes Chapter Tests and more Practice Papers.

Moving on to Unit 2 **Number and Algebra**, download **three Free Chapter Tests** for ‘Basic Algebra’:

**Higher sets**: (grades D to A*)

**Middle sets: **(grades E to B)

**Foundation sets: **(grades G to C)

And here are the **Answers **to all three tests – think we’ve got the right ones this time!

## Fibonacci Sequences – the Golden Rule about the Golden Ratio, which we just found out…

We just found out something new….

It turns out that** if you take the first ten terms of**** any Fibonacci sequence, the sum of those 10 terms is equal to the 7th term multiplied by 11.** Got it?

OK, let’s unpack that. Here’s the most basic Fibonacci sequence:

1,1,2,3,5,8,13,21,34,55

If you plug these into your head/calculator, you will find that the sum of these 10 terms is **143**.

Much quicker: take the 7th term (in this case 13), multiply that by 11 and you get, as if by magic… **143**.

And it works for any Fibonacci sequence. Let’s imagine one with bigger numbers starting with (to pick a number completely at random) 42:

42, 42, 84, 126, 210, 336, 546, 882, 1428, 2310

So we add all the terms up on a calculator and get **6,006**.

Now, much quicker, jot down the 7th term (here: 546), multiply it by 11 and, hey presto: 546×11=**6,006**

It’s a trick you can cheerfully teach to your maths students – get them to take it home and impress their parents. Something along the lines: “hey, dad/mum, I bet I can add up a Fibonacci sequence faster in my head than you can using a calculator”.

Our thanks to the **Republic of Maths** blog that brought this to our attention, via **this video link**.

**One final note:** the covers on our (as in: Longman’s) **AQA GCSE Maths Student Books **this year were inspired by spirals. A spiral galaxy on our **Foundation sets book**, a spiral staircase on our **Higher sets book**, and a **chameleon’s tail **on our **Middle sets book **(someone must’ve been wondering what this image was?) –

## EPISODE 5: Sherlock Holmes Maths Competition – FREE Christmas Maths challenge to download

**SHERLOCK HOLMES AND THE MYSTERY OF THE DEVIL’S EYE**

**Episode 5 – the Christmas Lights at O X F O R D C I R C U S**

**The story so far… **Called to investigate the disappearance from Harrods of the world’s largest ruby, the **Devil’s Eye, **Holmes and Watson follow a trail from there to the London Eye to the tunnels under the Thames. They emerge from these on Guy Fawkes night to be greeted by fireworks above Borough Market, where Holmes finally discovers in **Episode 4** **WHO**** has stolen the ruby and WHY… **

His old nemesis, **Moriarty**, plans to use the ruby in a laser to create **the world’s most powerful EMP (Electro-Magnetic Pulse) weapon**. A weapon so powerful it could knock out every electronic device in, *or above, *London – including all mobile phones, hospital equipment, and aircraft guidance systems… To be activated the device needs two things: (i) a vast metal antenna of some shape, (ii) a huge source of electric power. **Wherever will Moriarty find those? **

Ably assisted by **the Maths department at the Littlehampton Academy**, Holmes and Watson solve the code and arrive at… **O X F O R D C I R C U S, where the Christmas Lights are about to be switched on:**

And here’s the accompanying **worksheet**:

**WHAT TO DO: **

**DOWNLOAD THE PDFs, SOLVE THE CLUES, AND SUBMIT YOUR ANSWERS TO: ****SherlockH_221B@yahoo.co.uk**** **

**BY MONDAY 29th NOVEMBER.**

The winning entry will be drawn at random from all correct answers, and will receive a **FREE Class set of Longman’s unique 2010 AQA Functional Maths Student Book (30 copies) ****and a tin of Cadbury’s Heroes.**

**If you’re NEW to this competition…**

View **this page of our blog** to keep up to date with all the action. The competition is free to enter, and you can join in any Round/Episode. All the page references in the Episode point to maths support in Longman’s** AQA GCSE Middle sets Student Book**. All new and existing subscribers are entitled to receive a single free copy of this book – **request your copy by leaving a comment on this post.**

**GOT ANY QUESTIONS?** Just leave a comment on this post and we’ll get straight back to you.

## Sherlock Holmes GCSE Maths Challenge – episode 4 solution

Many congratulations to the **Episode 4** winners of our **Sherlock Holmes maths competition:**

**Jonathon Cox at The Littlehampton Academy**

for successfully working out where Holmes and Watson must go next.

**And the answer is:**

**OXFORD CIRCUS.**

We’ll be sending the winners a **FREE Class set of Longman’s unique AQA A-A* and G-F Practice Books (15 copies of each) ****and a tin of Cadbury’s Heroes.**

Coming up next… a spot of Christmas shopping in Episode 5 on Wednesday.

Here is the **winning solution to episode 4:**

The solution to how far away Moriarty is from the tunnel.

The path of the firework is y=300+44x-x^2. This is a quadratic so using the quadratic formula to find x when y=0

x= (-b +/- square root (b^2 – 4ac))/(2a)

where a=-1, b=44 and c=300

x = (-44 +/- square root (44^2 – 4X-1X300))/(2X-1)

x = (-44 +/- square root (1936 + 1200))/(-2)

x = (-44 +/- square root (3136))/(-2)

x = (-44 +/- 56)/(-2)

Either x = (-44 – 56)/(-2) = -100/-2 = +50 (where the tunnel is)

or x = (-44 + 56)/(-2) = 12/-2 = -6 (where Moriarty must have fired the firework from)

So 50 – -6 = 56 metres away from the tunnel

Moriarty is 56 metres awayThe Code.

Using the Coordinates of the Tate Modern (34, 14) and the code FG DG.

Using a shifted substitution D=1, E=2, F=3 etc

The Coordinates generated from substitution of CK EE gives (08, 22) which give the coordinates for Oxford Circus

Moriarty’s Target is Oxford Circus

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