Parallelogram 13 Year 10 12 Dec 2019Chrismaths Y10

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Noun: Parallelogram Pronunciation: /ˌparəˈlɛləɡram/

  1. a portmanteaux 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 week’s Parallelogram contains a WHOLE maths challenge paper – 25 glorious questions to stretch your brain over Chrismaths.

The time limit on maths challenges is 60 minutes, but it might take you up to 90 minutes to complete all the questions. However, you can tackle the challeges in one, two or three or more sessions. Just complete all the questions and hit the submit button before you go back to school in January. Bear in mind, the last 10 questions are tough. Also, in the real maths challenge paper, there is negative marking for tough questions, which means you lose marks for incorrect answers – don’t worry, in this Parallelogram you will not lose marks for wrong answers.

As this Parallelogram is longer than a typical Parallelogram, you can win a maximum of 400 (rather than 100) rewards points – this is a real opportunity to win an extra badge (or two).

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

Simon.

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

1.

3 marks

What is the value of 4002 - 2004?

  • 2004
  • 2002
  • 2000
  • 1998
  • 1996

4004 - 2004 = 2000, so 4002 - 2004 = 2000 - 2 = 1998.

2.

3 marks

You are told that 30 pupils have 25 different birthdays between them.

What is the largest number of pupils who could share the same birthday?

  • 2
  • 3
  • 4
  • 5
  • 6

There must be 25 pupils who all have different birthdays. If the remaining pupils all have the same birthday as one of these pupils, then 6 pupils will share the same birthday.

3.

3 marks

Four of these numbers can make two pairs so that each pair adds up to 98,765.

Which number is the odd one out?

  • 37,373
  • 45,678
  • 53,087
  • 61,392
  • 70,082

37,373 + 61,392 = 98,765, and 45,678 + 53,087 = 98,765, so 70,082 is the odd one out.

4.

3 marks

What is the value of a+b+c+d+e+f?

  • 360
  • 540
  • 720
  • 900
  • It depends on the triangle

The angles marked a°, c° and e° may be considered to be the exterior angles of the triangle, and therefore have a total of 360°. As b°=a°, d°=c° and f°=e° (all pairs of vertically opposite angles), b°+d°+f°=360°. So a°+b°+c°+d°+e°+f°=720°.

5.

3 marks

The sum of two numbers is 2. The difference between them is 4. What is their product?

  • -8
  • -3
  • 0
  • 3
  • 8

Let the numbers be x and y. Then x+y=2; xy=4. Adding these equations gives 2x=6, so x=3. Hence y=23=1 so the numbers are 3 and -1.

6.

4 marks

In Niatirb they use Cibara numerals. These are the same shape as normal Arabic numerals, but with the meanings in the opposite order. So "0" means "nine", "1" means "eight" and so on.

But they write their numbers from left to right and use arithmetic symbols just as we do. So, for example, they use 62 for the number we write as 37.

How do the inhabitants of Niatirb write the answer to the sum which they write as 837 + 742?

  • 419
  • 580
  • 1579
  • 5317
  • 8420

The sum is what we would write as 162 + 257 and this equals 419. However, in Niatirb it would be written 580.

7.

4 marks

Which of the following straight lines cuts the shaded area in half?

  • XA
  • XB
  • XC
  • XD
  • XE

The shaded area is a trapezium of area 123+7×5=25.

Line XD forms one side of a trapezium of area 12.5, since 121+4×5=12.5.

8.

4 marks

In March 2003, Welshman Tony Evans dropped a ball from an aircraft a mile above the Mojave desert to see if it would bounce. The ball was made from 6 million rubber bands, had a circumference of 14ft 8in, weighed 2600 pounds and took Mr Evans five years to build.

On average, roughly how many rubber bands did he add each day while building the ball?

  • 3
  • 33
  • 330
  • 3300
  • 33,000

The average number of rubber bands added each day was approximately 60000005×365600000018003300

9.

4 marks

The cuboids below all have the same volume. Which of them has the greatest surface area?

  • A)
  • B)
  • C)
  • D)
  • E)

The surface areas of the cuboids are: A 68; B 70; C 56; D 52; E 76.

10.

4 marks

What is the mean of 12, 13, 14 and 16?

  • 15
  • 115
  • 512
  • 724
  • 516

The four fractions total 54, so their mean is 54÷4=516.

11.

4 marks

The diagram shows a square board in which strips of white squares alternate with strips of black and white squares. A larger board, constructed in the same way, has 49 black squares.

How many white squares are there on the larger board?

  • 176
  • 196
  • 245
  • 289
  • 392

The 49 black squares will be in a 7 × 7 formation, so the board will measure 15 × 15 squares. Hence the number of white squares = 225 - 49 = 176.

12.

4 marks

This figure is made from a straight line 16cm long and two quarter circles, one with its centre at the midpoint of the straight line.

What is the area of the figure (in cm2)?

  • 64
  • 16π
  • 32 + 16π
  • 32π
  • 16 + 8π

As the diagram shows, the figure may be cut into two parts which fit together to form a square measuring 8cm × 8cm.

13.

4 marks

Four of these points lie on a single straight line. Which is the odd one out?

  • (−3, −3)
  • (−2, −1)
  • (2, 5)
  • (4, 11)
  • (5, 13)

Points (−3, −3), (−2, −1), (4, 11) and (5, 13) all lie on the line whose equation is y=2x+3, but (2, 5) does not lie on this line.

14.

4 marks

In this addition sum, each letter represents a different non-zero digit.

What is the value of a+w+a+y?

  • 13
  • 15
  • 16
  • 17
  • 18

We note first that y = 5 is the only non-zero digit which, when it is multiplied by 3, has itself as the units digit. So there is a carry of 1 into the tens column. We note also that a = 1 or a = 2 as “fly” < 1000 and therefore 3 × “fly” < 3000.

We now need 3 × l + 1 to end in either 1 or 2 and the only possibility is l = 7, giving a = 2 with a carry of 2 into the hundreds column.

As a = 2, f must be at least 6. However, if f = 6 then w = 0 which is not allowed. Also the letters represent different digits, so f ≠ 7 and we can also deduce that f ≠ 9 since f = 9 would make w = 9.

Hence f = 8, making w = 6 and the letters represent 875 × 3 = 2625.

15.

4 marks

Only one of these triangles can actually be made.

Which is it?

  • A
  • B
  • C
  • D
  • E

As triangle A has two equal sides, it should have two equal angles, but its angles are 25°, 110° and 45° so it is impossible to make.

In a triangle, the smallest angle lies opposite the shortest side which makes B impossible since 20° is the smallest angle, but 5 cm is not the shortest side.

Triangle C is impossible as 42+7282, so it does not obey Pythagoras' Theorem.

The longest side of a triangle must be shorter than the sum of the other two sides, but this is not the case in triangle E, so it cannot be made either.

Triangle D, however, is obtained by cutting an equilateral triangle of side 6cm in half along an axis of symmetry and so can certainly be made.

16.

4 marks

If the pattern shown is continued, what number will appear directly below 400?

  • 438
  • 439
  • 440
  • 441
  • 442

Note that the number at the end of the nth row is n2, so 400 will lie at the end of the 20th row.

The row below will end in 212, ie 441, so the number directly below 400 will be 440.

17.

4 marks

A, B, C, D, E, P and Q are points on the number line as shown.

One of the points represents the product of the numbers presented by P and Q. Which is it?

  • A
  • B
  • C
  • D
  • E

As P and Q both lie between 0 and 1, their product will be greater than 0 but smaller than P and smaller than Q. Of the options available, only B satisfies these conditions.

Furthermore, its position is correct since P is approximately equal to 12, which means that the product of P and Q lies approximately halfway between 0 and Q.

18.

4 marks

In the triangle PQR, there is a right angle at Q and angle QPR is 60°.

The bisector of the angle QPR meets QR and S, as shown.

What is the ratio of QS : SR?

  • 1 : 1
  • 1 : 2
  • 1 : (3 - 3)
  • 1 : 3
  • 1 : 2

Angle PRS = 30°, so triangle PRS is isosceles with SP = SR.

Hence QSSR=QSSP=12 as PQS is half of an equilateral triangle.

Alternatively, we can use the angle bisector theorem: QSSR=PQPR=12as PQR is also half of an equilateral triangle.

19.

4 marks

Three rectangular-shaped holes have been drilled passing all the way through a solid 3 x 4 x 5 cuboid. The diagrams show the front, side and top views of the resulting block.

What fraction of the original cuboid remains?

  • 1330
  • 715
  • 12
  • 815
  • 1730

Consider the cuboid to be made up of 60 unit cubes. The front and side views show that top and bottom layers consist of the same number of cubes and from the top view we see that this number is 14. The front and side views indicate that only the 4 corner cubes remain in the middle layer, so the total number of cubes remaining is 2 × 14 + 4 = 32. The required fraction, therefore, is 3260=815.

20.

4 marks

What is the largest power of 2 that divides 12721?

  • 21
  • 27
  • 28
  • 263
  • 2127

Using the formula for the difference of two squares:

12721=127212=127+11271=128×126=27×2×63=28×32×7.

21.

5 marks

A square is divided into four congruent rectangles and a smaller square, as shown. The diagram is not to scale.

The area of the small square is 14 of the area of the whole square.

What is the ratio of the length of a short side of one of the rectangles to the length of a long side?

  • 1 : 2
  • 1 : 3
  • 1 : 2
  • 1 : 3
  • 1 : 4


Let the length of a short side of a rectangle be x and the length of a long side be y. Then the whole square has side of length y+x, whilst the small square has side of length yx.

As the area of the whole square is four times the area of the small square, the length of the side of the whole square is twice the length of the side of the small square.

Therefore y+x=2yx i.e. y=3x so x:y=1:3.

22.

5 marks

In a maths exam with N questions, you score m marks for a correct answer to each of the first q questions, and m+2 marks for a correct answer to each of the remaining questions.

What is the maximum possible score?

  • m+2N2q
  • Nm
  • mp+m+2q
  • Nm+1
  • Nm+qm+2

If you answer all questions correctly, you receive m marks for each of the N questions plus an extra 2 marks for the last Nq questions.

So the maximum possible score =m×N+2×Nq=mN+2N2q=m+2N2q.

23.

5 marks

In the diagram, the letter S is made from two arcs, KL and MN, which are each five-eights of the circumference of a circle of radius 1, and the line segment LM, which is tangent to both circles.

At points K and N, common tangents to the circles touch one of the circles.

What is the length LM?

  • 32
  • 32
  • 2
  • 322
  • 1+2


Let the centres of the circles be J and O and let NI be the common tangent shown.

Let P be the point of intersection of JO and LM.

As arc KL is 5/8 of the circumference of the top circle, IJL = 45°.

Consider quadrilateral IJON: sides IJ and NO and are radii, so they are both of unit length and they are both perpendicular to tangent NI. So IJON is a rectangle.

Hence IJO=90° and LJP=90°45°=45°. In triangle JLP, JPL=180°90°+45°=45°.

So triangle JLP is isosceles and LP=LJ= 1 unit.

By a similar argument, it may be shown that PM is also of length 1 unit, so LM=LP+PM= 2 units.

24.

5 marks

If p, q and pq are all positive integers, which of the following is least?

  • q2p2
  • p2q2
  • qp
  • qp
  • pq

We may deduce that p>q, so, as p and q are both positive, p2q2 > pq > 1.

We may also deduce that 0 < q2p2 < qp < qp < 1.

Hence q2p2 is the least of the five numbers.

25.

5 marks


The diagram shows a square with two lines from a corner to the middle of an opposite side.

The rectangle fits exactly inside these two lines and the square itself.

What fraction of the square is occupied by the shaded rectangle?

  • A) 13
  • B) 25
  • C) 310
  • D) 12
  • E) 38

Triangles ABG, GFE and EFA are similar, so AF:FE=EF:FG=GB:BA=1:2.

Thus if AF=a, then FE=2a, FG=4a, and the shaded area is 8a2.

By Pythagoras' Theorem, AE=5a, so AD=25a and the area of the square is 20a2.

Thus the required fraction is 820, which is 25.

Alternatively, as we have shown that FG=4AF, we can divide parallelogram AGCE into 10 congruent triangles, 8 of which make up rectangle EFGH.

So the area of the rectangle is 45 of the area of the parallelogram, which in turn is half the area of square ABCD.

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 will appear on Thursday 10 January, at 3pm.
  • 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.