Parallelogram 13 Year 11 10 Dec 2020The Twelve Days of Christmas

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Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

  1. a portmanteau word combining parallel and telegram. A message sent each week by the Parallel Project to bright young mathematicians.

Season’s greetings. You will be glad to hear that I have packed lots of festive cheer into each question this week. Ho, ho, ho!

As you all move closer to your exams, you will be getting busier and busier with mocks and revision. With that in mind, this is the last Parallelogram for the rest of the academic year.

So, whether you started Parallel this year, or have been doing Parallel since Year 7, I hope that you have felt stretched, challenged, and are now considering continuing with mathematics (and science) next year and beyond. Have a wonderful Christmas break.

Cheerio, Simon

1. A Christmassy question based on an Intermediate Maths Challenge Problem

3 marks

1.1. On the seventh day of Christmas, my true love couldn’t find any swans a-swimming, so got me seven gold rings instead.

Only, my true love had them interlinked as shown in the diagram.

What is the smallest number of rings that need to be cut in order to separate all the rings?

Correct Solution: 3

The configuration includes three pairs of small rings. The rings in each pair are separate from the rings in the other two pairs so at least three rings will need to be cut. If the ring in each pair which is connected to the large ring is cut, then all of the rings can be separated. So the minimum number of rings which need to be cut is three.

2. A Christmassy question based on an Intermediate Maths Challenge Problem

4 marks

2.1 Father Christmas delivers presents to three neighbouring streets of houses.

After he delivers to the first street, he realises he is late and speeds up, covering the second street twice as quickly.

After the second street, he decides he still needs to deliver faster so covers the third street at maximum speed, one and half times quicker than he covered the second.

He takes ten minutes longer than he would do if he had covered all three streets at maximum speed.

How long does he take to deliver presents to all three streets?

  • 60 min
  • 42 min
  • 36 min
  • 30 min
  • 22 min
  • (Not answered)
Show Hint (–1 mark)
–1 mark

Let Father Christmas take x minutes to cover the last street. Then he takes 3x2 minutes to cover the second street and 3x minutes to cover the first street.

Show Hint (–1 mark)
–1 mark

Since the sum of these three times is 10 minutes more than if he had taken x minutes for each of the three streets, we have 3x+3x2+x=3x+10.

Show Hint (–1 mark)
–1 mark

The solution to 3x+3x2+x=3x+10 is x=4. Remember, this is the time taken to cover the last street.

Let Father Christmas take x minutes to cycle one mile. Then he takes 3x2 minutes to run one mile, and 3x minutes to walk one mile. Therefore: 3x+3x2+x=3x+10 is x=4.

Father Christmas takes 12 minutes to walk the first mile, 6 minutes to run the second mile and 4 minutes to cycle the third mile: a total time of 22 minutes.

3. The Twelve Days of Christmas

The next gift in this parallelogram is this video about the twelve days of Christmas. You’re welcome!

(If you have problems watching the video, right click to open it in a new window)

4 marks

3.1 The video introduced you to Pascal’s Triangle, in which every entry (apart from the 1s on either side) is the sum of the two above. What shape, familiar from a previous parallelogram, emerges if you shade all of the even numbers in Pascal's triangle (you may need to go to around the 9th row before you start seeing the pattern!)

  • A Borromean ring
  • A Sierpinski triangle
  • A Koch snowflake
  • A vortex
  • (Not answered)
Show Hint (–1 mark)
–1 mark

You can actually do this without any of the actual numbers in Pascal’s triangle, and instead just writing “O” for all the odd numbers, and “E” for all the even numbers, remembering that odd + odd = even, even + even = even, and even + odd = odd. The triangle starts off like this:

As you can see from this video, the shape that appears is a (very pixelated) Sierpinski triangle.

(If you have problems watching the video, right click to open it in a new window)

3 marks

3.2 Over the twelve days of Christmas, I receive 1 partridge on day 1 and for the remaining 11 days, so 12 partridges in total. I also receive 2 turtle doves on day 2 and for the remaining 10 days, so 22 turtle doves in total. But what item do I receive most of over the twelve days of Christmas?

  • Partridges
  • Geese
  • Swans
  • A tie between Geese and Swans
  • A tie between Partridges and Lords.
  • (Not answered)

It’s a tie between Geese and Swans. Watch this video for an explanation:

(If you have problems watching the video, right click to open it in a new window)

4. A Christmassy question based on an Intermediate Maths Challenge Problem

4 marks

4.1 Granny has made another of her special super-heavy Christmas cakes. At her Christmas party, five of the guests tried to guess the weight of the cake. Their guesses were 5040g, 5060g, 5110g, 5120g, and 5150g.

Actually, none of them was right. Only two were more than 30 grams out, and they were out by 70g and 90g.

What was the weight of the cake?

  • 5070g
  • 5080g
  • 5090g
  • 5110g
  • 5130g
  • (Not answered)
Show Hint (–1 mark)
–1 mark

The range of the guesses is just 110. This tells us that the two ‘wild’ guesses can’t be either side of the correct answer, as this would result in a range of 70 + 90 = 160.

Show Hint (–1 mark)
–1 mark

If the wild guesses were wildly high, they would be 5120g and 5150g. These can’t be 70g and 90g out because they don’t have a difference of 20g. So the ‘wild’ guesses must be 5040g and 5060g.

The guesses were 5040g, 5060g, 5110g, 5120g, 5150g. The range of these guesses is just 110g so the two 'wild' estimates must both be low or both high. Since the difference between 5040g and 5060g is 20g, 5040g is 90g out and 5060g is 70g out.

5. A Christmassy question based on an Intermediate Maths Challenge Problem

4 marks

5.1 Four busy elves, Al, Bertie, Chris and Di have made 150 toys between them. Al and Bertie have made 55 between them, and Al and Chris have made 65 between them.

What is the difference between the number of toys Al made, and the number of toys Di made?

  • 25
  • 30
  • 35
  • 40
  • 45
  • (Not answered)
Show Hint (–1 mark)
–1 mark

Letting Al’s, Bertie’s, Chris’ and Di’s totals be a, b, c and d, we have a+b=55 and a+b+c+d=150.

Subtracting one from the other we have c+d=95.

Show Hint (–1 mark)
–1 mark

a+c=65.

What expression will we get if we subtract a+c from c+d?

Show Hint (–1 mark)
–1 mark

c+da+c=da (The c’s cancel each other out) This is the difference between the number of toys Al made, and the number of toys Di made.

c+d=95, and a+c=65.

Al and Bertie have made 55 between them, so Chris and Di have made 150 - 55, that is 95, between them. As Al and Chris have made 65 between them, the difference between the amounts Al and Di have made is 95 – 65 = 30.

Before you hit the SUBMIT button, here are some quick reminders:

  • You will receive your score immediately, and collect your reward points.
  • You might earn a new badge...
  • Make sure you go through the solution sheet – it is massively important.
  • A score of less than 50% is ok – it means you can learn lots from your mistakes.
  • If you missed any earlier Parallelograms, make sure you go back and complete them. You can still earn reward points and badges by completing missed Parallelograms.
  • This was the last of our Parallelograms! So Long, and Thanks for All the Fish.

Cheerio, Simon.