Parallelogram 14 Level 2 7 Dec 2023Chrismaths - Part 1

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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.

It’s nearly Christmas. Not very many people realise that the original name for this holiday was Chrismaths. For centuries, children traditionally completed mathematics problems during the festive period. The amount of effort that children put into their mathematics problems allowed Santa to decide who had been naughty and who had been nice.

With this in mind, this next few week’s of Parallelogram contain a WHOLE maths challenge paper, to stretch your brain over Chrismaths.

Good luck and happy Chrismaths. Don’t eat too many mince pie charts!


PS: I want to say thank you to the UK Mathematics Trust, who own the copyright to these questions.


3 marks

Which of the following calculations gives the largest answer?

  • A) 1 − 2 + 3 + 4
  • B) 1 + 2 − 3 + 4
  • C) 1 + 2 + 3 − 4
  • D) 1 + 2 − 3 − 4
  • E) 1 − 2 − 3 + 4
  • (Not answered)

Since the numbers are the same in each sum, the largest answer results when the amount subtracted is the smallest. In option A the number 2 is subtracted, in option B it is 3, in option C it is 4, in option D both 3 and 4 are subtracted, and in option E both 2 and 3. So A gives the largest answer.

Alternatively, we can see directly that the sums have the following answers:

  • A: 1 − 2 + 3 + 4 = 6
  • B: 1 + 2 − 3 + 4 = 4
  • C: 1 + 2 + 3 − 4 = 2
  • D: 1 + 2 − 3 − 4 = −4
  • E: 1 − 2 − 3 + 4 = 0

and therefore that A gives the largest answer.


3 marks

What is the value of x in this triangle?

  • A) 45
  • B) 50
  • C) 55
  • D) 60
  • E) 65
  • (Not answered)

The angle marked on the diagram as y° and the angle that is 110° are angles on a straight line. Therefore their sum is 180°.

It follows that y°=70°. Therefore, by the Exterior Angle Theorem 120°=x°+y°=x°+70°.

It follows that x=50.


3 marks

What is the smallest prime number that is the sum of three different prime numbers?

  • A) 11
  • B) 15
  • C) 17
  • D) 19
  • E) 23
  • (Not answered)

Consider three different prime numbers which include 2, say the prime numbers 2, p and q. Then p and q will both be odd numbers, and therefore 2+p+q will be an even number greater than 2 and so cannot be a prime number. So, if we seek prime numbers that are sums of three different prime numbers, we need only consider sums of three different odd prime numbers.

The three smallest odd prime numbers are 3, 5 and 7, but their sum is 15 which is not prime. If we replace 7 by the next odd prime, 11, we have three odd primes with sum 3 + 5 + 11 = 19, which is a prime number.

We cannot obtain a smaller prime number as a sum using 3 and two other odd prime numbers. If we do not include 3, the smallest sum of three odd prime numbers that we can obtain is 5 + 7 + 11 = 23 which is greater than 19.

We can therefore deduce that 19 is the smallest prime number which is the sum of three different prime numbers.


4 marks

The diagram shows a square inside an equilateral triangle.

What is the value of x+y?

  • A) 105
  • B) 120
  • C) 135
  • D) 150
  • E) 165
  • (Not answered)

We let P,Q,R,S and T be the points shown in the diagram. We also let QRP=p° and TRS=q°.

Because it is an angle of a square, PRT=90°. Because they are angles of an equilateral triangle PQR=RST=60°.

Because the angles of a triangle have sum 180°, from triangle PQR we have x+p+60=180 and from triangle TRS,y+q+60=180. Therefore x+p=120 and y+q=120.

Because QRP, PRT and TRS are angles on a straight line, p+q+90=180 and therefore p+q=90. It follows that x+y=x+p+y+qp+q=120+12090=150.

There is a quick method that it is all right to use in the context of the JMC, but which would not be acceptable if you had to give a full solution with detailed reasons.

We have already shown that p+q=90. Since the question does not give us individual values for p and q, we can assume that the answer is independent of their actual values. So, for simplicity, we assume that p=q=45. Therefore in each of the triangles PQR and RST one of the angles is 60° and one is 45°. Therefore, because the sum of the angles in a triangle is 180°, both x and y are equal to 1806045=75. We conclude that x+y=75+75=150.


5 marks

Gill is now 27 and has moved into a new flat. She has four pictures to hang in a horizontal row on a wall which is 4800mm wide. The pictures are identical in size and are 420mm wide.

Gill hangs the first two pictures so that one is on the extreme left of the wall and one is on the extreme right of the wall. She wants to hang the remaining two pictures so that all four pictures are equally spaced.

How far should Gill place the centre of each of the two remaining pictures from a vertical line down the centre of the wall?

  • A) 210 mm
  • B) 520 mm
  • C) 730 mm
  • D) 840 mm
  • E) 1040 mm
  • (Not answered)

Let the three gaps between the pictures each be g mm wide.

Since each of the four pictures is 420mm wide and the wall is 4800mm wide:

and therefore:

It follows that:

The distance between the centres of the middle two pictures is equal to the width of one picture and the width of the gap, that is, in mm, 420+g=420+1040=1460.

The distance between the centre of one of these pictures and the centre line is half this distance. Therefore the required distance is, in mm, 12×1460=730.

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... if not, then maybe next week.
  • 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.
  • The next Parallelogram is next week, at 3pm on Thursday.
  • Finally, 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.

Cheerio, Simon.