Parallelogram 13 Year 8 12 Dec 2019Chrismaths Y8

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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

Which of the following is closest to zero?

  • A) 6 + 5 + 4
  • B) 6 + 5 − 4
  • C) 6 + 5 × 4
  • D) 6 − 5 × 4
  • E) 6 × 5 ÷ 4

We work out the value of each expression in turn:

  • A: 6 + 5 + 4 = 11 + 4 = 15.
  • B: 6 + 5 − 4 = 11 − 4 = 7.
  • C: 6 + 5 × 4 = 6 + 20 = 26.
  • D: 6 − 5 × 4 = 6 − 20 = −14.
  • E: 6 × 5 ÷ 4 = 6 × 1.25 = 7.5.

We see that expression B has the value that is closest to zero.

2.

3 marks

What number is twenty-one less than sixty thousand?

  • A) 59,979
  • B) 59,981
  • C) 57,900
  • D) 40,001
  • E) 39,000

The direct method is to do the subtraction sum, as follows: 60,000 - 12 = 59,979.

Without doing the full subtraction we can see that the units digit of 60 000 − 21 is 9. In the context of the JMC you are entitled to assume that one of the given options is correct. As 59,979 is the only option with units digit 9, we deduce that this is the correct answer.

3.

3 marks

One lap of a standard running track is 400m.

How many laps does each athlete run in a 5000m race?

  • A) 4
  • B) 5
  • C) 8
  • D) 10
  • E) 1212

Since each lap has length 400m and each athlete runs 5000m, the number of laps that each athlete runs is:

5000400=504=1212.

4.

3 marks

In January 1859, an eight-year-old boy dropped a newly-hatched eel into a well in Sweden (apparently in order to keep the water free of insects). The eel, named Åle, finally died in August 2014.

How many years old was Åle when it died?

  • A) 135
  • B) 145
  • C) 155
  • D) 165
  • E) 175

To find Åle’s age we need to work out: 2014 − 1859 = 155.

So Åle’s age when it died was 155.

Alternatively, we can see that from 1859 to 1900 is 41 years, from 1900 to 2000 is 100 years, and from 2000 to 2014 is 14 years. So Åle’s age in years is 41 + 100 + 14, that is, 155.

5.

3 marks

What is the value of 125+0.25?

  • A) 0.29
  • B) 0.3
  • C) 0.35
  • D) 0.50
  • E) 0.65

We first convert the fraction into a decimal. We multiply both the numerator (the top) and the denominator (the bottom) by 4. This gives:

125=4100.

Therefore, expressed as a decimal, 125 is 0.04.

Hence 125+0.25=0.04+0.25=0.29.

6.

3 marks

Gill is now 28 years old and is a teacher of Mathematics at a school which has 600 pupils. There are 30 more girls than boys at the school.

How many girls are at Gill’s school?

  • A) 270
  • B) 300
  • C) 315
  • D) 330
  • E) 345

Suppose there were equal numbers of girls and boys at the school. Then there would be 300 girls and 300 boys. Each time we add one more girl and take away one boy, the difference between the number of girls and the number of boys goes up by 2, while the total number of pupils remains 600. So to have 30 more girls than boys, we would need to add 15 more girls and take away 15 boys. So a school with 600 pupils and 30 more girls than boys has 315 girls.

Alternatively, suppose that the school has g girls. Then, as there are 600 pupils altogether, there are 600g boys.

Since there are 30 more girls than boys: g600g=30.

This gives g600+g=30, that is, 2g600=30, and hence 2g=630.

Therefore g=315.

Hence, there are 315 girls at Gill’s school.

7.

3 marks

A distance of 8 km is approximately 5 miles.

Which of the following is closest to 1.2 km?

  • A) 0.75 miles
  • B) 1 mile
  • C) 1.2 miles
  • D) 1.6 miles
  • E) 1.9 miles

Since 8 km is approximately 5 miles, 1 km is approximately 58miles.

Hence 1.2 km is approximately 1.2×58 miles.

We have 1.2×58=1.2×58=68=0.75.

Therefore 1.2 km is approximately 0.75 miles.

Alternatively, since 8 km is approximately 5 miles, 1 mile is approximately 85 km, that is, 1.6 km.

It follows that 1.2 km is significantly less than 1 mile. So 0.75 miles is the correct option.

More precisely, it follows that 1.2 km is approximately 1.21.6 miles. Since:

1.21.6=1216=34=0.75,

it follows that 1.2 km is approximately 0.75 miles.

8.

3 marks

What is the value of

2+4+6+8+10+12+14+16+18+201+2+3+4+5+6+7+8+9+10?

  • A) 2
  • B) 10
  • C) 20
  • D) 40
  • E) 1,024

Each number in the numerator is a multiple of 2. If we take out this common factor, we obtain:

2+4+6+8+10+12+14+16+18+20=2×1+2+3+4+5+6+7+8+9+10.

It follows that:

2+4+6+8+10+12+14+16+18+201+2+3+4+5+6+7+8+9+10=2×1+2+3+4+5+6+7+8+9+101+2+3+4+5+6+7+8+9+10=2.

9.

3 marks

One of the three symbols +, −, × is inserted somewhere between the digits of 2016 to give a new number. For example, 20 − 16 gives 4.

How many of the following four numbers can be obtained in this way?

36; 195; 207; 320

  • A) 0
  • B) 1
  • C) 2
  • D) 3
  • E) 4

We see that:

  • 20 + 16 = 36.
  • 201 − 6 = 195.
  • 201 + 6 = 207.
  • 20 × 16 = 320.

So we can obtain all four of the given numbers.

10.

3 marks

A square is folded exactly in half and then in half again.

Which of the following could not be the resulting shape?

  • A)
  • B)
  • C)
  • D)
  • E)

To fold a figure exactly in half we need to fold it along a line of symmetry.

A square has four lines of symmetry, but these are of just two types. There are two lines of symmetry joining the midpoints of opposite edges, and two lines of symmetry joining opposite vertices (see figure 1).

When a square is folded along a line of symmetry joining the midpoints of edges, such as l, the resulting shape is a rectangle whose longer sides are twice as long as the shorter sides (see figure 2). When it is folded along a line of symmetry joining opposite vertices, such as m, the resulting shape is an isosceles right-angled triangle (see figure 3).

The rectangle shown in figure 2 has two lines of symmetry, n and p, joining midpoints of opposite edges. When the rectangle is folded along n, the resulting shape is a square with half the side length of the original square. This is the shape of option A.

When the rectangle is folded along p, the resulting shape is a rectangle whose longer sides are four times as long as the shorter sides. This is the shape of option B.

The isosceles right-angled triangle shown in figure 3 has one line of symmetry, labelled q. When the triangle is folded along this line the resulting shape is a smaller isosceles right-angled triangle. This is the shape of both options C and E.

We therefore see that the shapes of options A, B, C and E can be achieved. Since we have covered all the possible ways of folding the square in half and in half again, it follows that the shape that could not be achieved is D.

11.

3 marks

Which of the following statements is false?

  • A) 12 is a multiple of 2
  • B) 123 is a multiple of 3
  • C) 1234 is a multiple of 4
  • D) 12 345 is a multiple of 5
  • E) 123 456 is a multiple of 6

One method is to do the following divisions, to find out in which case the answer is not an integer.

  • A) 12 ÷ 2 = 6.
  • B) 123 ÷ 3 = 41.
  • C) 1,234 ÷ 4 = 308.5.
  • D) 12,345 ÷ 5 = 2,469.
  • E) 123,456 ÷ 6 = 20,576.

From these calculations we see that 1234 is not a multiple of 4. So the statement that is false is C.

Alternatively, we have the following tests for whether a positive integer is a multiple of n, for small values of n:

  • n) Test for being a multiple of n
    1. The units digit is even.
    1. The sum of the digits is a multiple of 3.
    1. The tens and units digits taken together make an integer which is a multiple of 4.
    1. The units digit is 0 or 5.
    1. A multiple of 2 and a multiple of 3.

We apply these tests to the statements given in the options:

  • A) The units digit of 12 is 2, which is even. So 12 is a multiple of 2.
  • B) The sum of the digits of 123 is 1 + 2 + 3 = 6. This is a multiple of 3. So 123 is a multiple of 3.
  • C) The tens and units digits make 34. This is not a multiple of 4. So 1,234 is not a multiple of 4.
  • D) The units digit is 5. So 12,345 is a multiple of 5.
  • E) 123,456 has units digit 6 and the sum of its digits is 21. So it is a multiple of 2 and 3, and hence of 6.

Therefore it is statement C that is false.

12.

3 marks

The musical Rent contains a song that starts “Five hundred and twenty five thousand six hundred minutes”.

Which of the following is closest to this length of time?

  • A) a week
  • B) a year
  • C) a decade
  • D) a century
  • E) a millennium

The song refers to 525,600 minutes. Since there are 60 minutes in an hour, 525,600 minutes is:

52560060 hours, that is, 8760 hours.

Since there are 24 hours in one day, 8760 hours is:

876024 days, that is, 365 days.

It follows that of the given options a year is closest to the length of time corresponding to the number of seconds mentioned in the song.

Note A calendar year has 365 days in a normal year, but 366 days in a leap year such as 2016.

Astronomers use several different definitions of a year which differ a little from each other in length. One common definition is the sidereal year which is the time it takes the earth to complete one revolution about the sun, relative to the distant stars. The sidereal year has length 365.256363004 days.

Because this is not an exact number of days, a calendar with the same number of days in each year would not be synchronised with the position of the earth relative to the sun and so it would get out of step with the seasons.

Different cultures have solved this problem in different ways. In the most commonly used calendar, introduced by Pope Gregory XIII in 1582, the rules for the length of a year are:

A year is a leap year if its number is a multiple of 4, except that if the number is a multiple of 100 but not of 400, then it is not a leap year. A standard year has 365 days. A leap year has 366 days.

For example, 2016 is a multiple of 4. Therefore the current year is a leap year. 2017 is not a multiple of 4, so next year will not be a leap year. 2000 is a multiple of 100 and also of 400. So the year 2000 was a leap year. However, 2100 is divisible by 100 by not by 400. So the year 2100 will not be a leap year.

13.

3 marks

The diagram shows five circles placed at the corners of a pentagon. The numbers 1, 2, 3, 4, 5 are placed in the circles shown, one in each, so that the numbers in adjacent circles always differ by more than 1.

What is the sum of the numbers in the two circles adjacent to the circle which contains the number 5?

  • A) 3
  • B) 4
  • C) 5
  • D) 6
  • E) 7

The number 3 cannot be placed next to either 2 or 4. So it will be between 1 and 5. Therefore 5 is next to 3, and not next to 1. Also, 5 cannot be next to 4. Therefore 5 is next to 2. Therefore the sum of the numbers next to 5 is 3 + 2, that is, 5.

14.

3 marks

In the diagram, AB=AC and D is a point on AC such that BD=BC. Angle BAC is 40°.

What is angle ABD?

  • A) 15°
  • B) 20°
  • C) 25°
  • D) 30°
  • E) 35°

Since AB=AC, the triangle ABC is isosceles. Therefore ACB=CBA. Since the angles in a triangle add up to 180°, we have BAC+ACB+CBA = 180°. Hence 40° + 2ACB = 180°.

It follows that 2ACB = 180° − 40° = 140°. Therefore ACB = 70°.

Since BD=BC, the triangle BCD is isosceles. Therefore BDC=DCB. Since DCB is the same as ACB, it follows that BDC=ACB = 70°.

By the External Angle Theorem, BDC=BAD+ABD. That is, 70° = 40° + ABD. It follows that ABD = 30°.

15.

3 marks

How many of these four expressions are perfect squares?

  • 13 + 23

  • 13 + 23 + 33

  • 13 + 23 + 33 + 43

  • 13 + 23 + 33 + 43 + 53

  • A) 0

  • B) 1

  • C) 2

  • D) 3

  • E) 4

We find the value of each expression to see which of them are squares. After we have found the value of the one expression, we can find the value of the next expression by adding the extra term to the sum we have already calculated. In this way, we obtain:

  • 13 + 23 = 1 + 8 = 9 = 32.
  • 13 + 23 + 33 = 9 + 33 = 9 + 27 = 36 = 62.
  • 13 + 23 + 33 + 43 = 36 + 43 = 36 + 64 = 100 = 102.
  • 13 + 23 + 33 + 43 + 53 = 100 + 53 = 100 + 125 = 225 = 152.

We now see that all four of the expressions are perfect squares.

Note The first cube 13 is, of course, also a square since 13 = 12. We see from the above calculations that the sum of the first two cubes is square, and so also are the sums of the first three cubes, the first four cubes and the first five cubes. Just from these five examples we cannot deduce that the sum of the first n cubes is always a square, but it looks as though it might be true.

You should not be satisfied until you have found either a proof that this is always true, or a counterexample to show that it is false for at least one positive integer n.

16.

4 marks

Each of the nine small squares in this grid can be coloured completely black or completely white.

What is the largest number of squares that can be coloured black so that the design created has rotational symmetry of order 2, but no lines of symmetry?

  • A) 4
  • B) 5
  • C) 6
  • D) 7
  • E) 8

We note first that the colour of the centre square does not affect the symmetries of the design. Since we are looking for the largest number of squares that can be coloured black, we need only consider colourings in which this square is black.

To have a colouring which has rotational symmetry of order 2, each pair of opposite squares, that is, the pairs labelled p,q,r and s in the diagram on the left, must be the same colour.

The second diagram above shows that, if all four of these pairs are black, then the design has four lines of symmetry. The third and fourth diagrams show that, if one of these pairs is coloured white and the other three pairs are black, then the design has two lines of symmetry.

The diagram on the right shows, however, that we can create a design with no lines of symmetry by colouring the appropriate two pairs white and the other two pairs black. It can be seen that the design in this diagram has rotational symmetry of order 2.

This shows that the largest number of squares that can be coloured black to make a design with rotational symmetry of order 2, but no lines of symmetry, is 5.

17.

4 marks

In a group of 48 children, the ratio of boys to girls is 3 : 5.

How many boys must join the group to make the ratio of boys to girls 5 : 3?

  • A) 48
  • B) 40
  • C) 32
  • D) 24
  • E) 8

Since the ratio of boys to girls is 3 : 5, the proportion of boys in the group is 38. Therefore the number of boys is:

38×48=18.

It follows that the number of girls in the group is 4818=30.

The ratio 5 : 3 is the same as the ratio 50 : 30. So in a group which has 30 girls, in which the ratio of boys to girls is 5 : 3, there will be 50 boys. The number of boys is 18. To make this up to 50, the number of boys that need to join the group is 5018=32.

18.

4 marks

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

What digit does X represent?

  • A) 1
  • B) 3
  • C) 5
  • D) 7
  • E) 9

From the units column we see that either E+E=S, or there is a carry, in which case E+E=10+S. However, from the tens column, we deduce that E+ES. So there is a carry from the units column to the tens column.

Therefore, from the tens column we see that: 1+E+E=10+E.

If we subtract E+1 from both sides of this equation, we deduce that E=9. It now follows from the units column that S=8. Hence the sum is

It follows that X represents 7.

19.

4 marks

Three boxes under my stairs contain apples or pears or both. Each box contains the same number of pieces of fruit. The first box contains all twelve of the apples and one-ninth of the pears.

How many pieces of fruit are there in each box?

  • A) 14
  • B) 16
  • C) 18
  • D) 20
  • E) 36

The first box contains one-ninth of the pears. Therefore, the other two boxes contain, between them, eight-ninths of the pears. So, since they contain the same number of pieces of fruit, but no apples, they each contain four-ninths of the pears.

So the difference between the number of pears in each of the second and third boxes and the number of pears in the first box is equal to three-ninths of the number of pears, that is, to one-third of the number of pears.

Therefore, as the first box holds the same number of pieces of fruit as the other two boxes, the 12 apples in the first box match the extra one-third of the number of pears in the other two boxes. Since one-third of the pears amount to 12 pears, it follows that altogether there are 36 pears.

Hence the total number of pieces of fruit is 48. Since these are equally divided between the boxes, each of the three boxes contains 16 pieces of fruit.

Alternatively, Since the first box contains one-ninth of the pears, we avoid fractions by supposing that there are 9p pears. It follows that there are p pears in the first box and altogether 8p pears in the other boxes. As they contain equal numbers of pieces of fruit, there are 4p pears in each of these boxes.

Hence there are 12+p pieces of fruit in the first box, and 4p in each of the other two boxes.

Since each box contains the same number of pieces of fruit: 12+p=4p, and hence 12=3p, from which it follows that p=4.

We have seen that there are 12+p pieces of fruit in the first box. Since p=4, we deduce that the number of pieces of fruit in each box is 16.

20.

4 marks

A cyclic quadrilateral has all four vertices on the circumference of a circle. Brahmagupta (598–670AD) gave the following formula for the area, A, of a cyclic quadrilateral whose edges have lengths a,b,c,d:

where s is half of the perimeter of the quadrilateral.

What is the area of the cyclic quadrilateral with sides of length 4 cm, 5 cm, 7 cm and 10 cm?

  • A) 6 cm2
  • B) 13 cm2
  • C) 26 cm2
  • D) 30 cm2
  • E) 36 cm2

In Brahmagupta’s formula, we have a=4,b=5,c=7 and d=10.

The total length of the perimeter, in cm, is 4 + 5 + 7 + 10 = 26. Therefore s=26÷2=13.

It follows that:

  • s − a = 13 − 4 = 9,
  • s − b = 13 − 5 = 8,
  • s − c = 13 − 7 = 6, and
  • s − d = 13 − 10 = 3.

Hence: sasbscsd=9×8×6×3=9×144.

Therefore:

Therefore the area of the cyclic quadrilateral is 36 cm2.

21.

5 marks

The diagram shows a pentagon drawn on a square grid. All vertices of the pentagon and triangle are grid points.

What fraction of the area of the pentagon is shaded?

  • A) 27
  • B) 13
  • C) 25
  • D) 14
  • E) 29

We assume that the small squares making up the grid have a side length of 1 unit.

The pentagon consists of a 6 × 6 square, with half of a 3 × 3 square removed. Hence the area of the pentagon, in square units, is:

6×6123×3=3692=632.

The area of a triangle is half the base multiplied by the height. The shaded triangle has a base of length 3 units and height 6 units. Therefore, its area, in square units, is:

123×6=9.

It follows that the fraction of the area of the pentagon that is shaded is:

9632=9×263=27.

22.

5 marks

Four copies of the triangle shown are joined together, without gaps or overlaps, to make a parallelogram.

What is the largest possible perimeter of the parallelogram?

  • A) 46 cm
  • B) 52 cm
  • C) 58 cm
  • D) 62 cm
  • E) 76 cm


The four triangles between them have 12 edges. When they are fitted together there must be at least 3 places where two of the triangles meet along an edge. At each of these places 2 of the 12 edges form interior edges of the parallelogram. So the perimeter of the parallelogram will be made up of at most 123×2=6 of the edges of the triangles. So, if we can arrange the triangles so that the perimeter is made up of all four edges of length 13 cm, and two edges of length 12 cm, this would give us the largest possible perimeter.

The diagram shows that such an arrangement is possible. So the length of the longest possible perimeter is 4 × 13 cm + 2 × 12 cm = 76 cm.

23.

5 marks

The diagram shows the first few squares of a ‘spiral’ sequence of squares. All but the first three squares have been numbered.

After the first six squares, the sequence is continued by placing the next square alongside three existing squares – the largest existing square and two others.

The three smallest squares have sides of length 1.

What is the side length of the 12th square?

  • A) 153
  • B) 123
  • C) 83
  • D) 53
  • E) 13

We see from the diagram in the question that the side length of the 4th square is 3, and the side length of the 5th square is 4.

Also, the side length of the 6th square is the sum of the side lengths of the 4th and 5th squares. So the 6th square has side length 3 + 4 = 7. Similarly the side length of the 7th square is the sum of the side lengths of the 5th and 6th squares. So the 7th square has side length 4 + 7 = 11.

The sequence of side lengths continues in this way. After the first three numbers, each number in this sequence is the sum of the previous two numbers. Therefore the first 12 numbers in the sequence are:

1, 1, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123.

We therefore see that the side length of the 12th square is 123.

24.

5 marks

Part of a wall is to be decorated with a row of four square tiles.

Three different colours of tiles are available and there are at least two tiles of each colour available. Tiles of all three colours must be used.

In how many ways can the row of four tiles be chosen?

  • A) 12
  • B) 18
  • C) 24
  • D) 36
  • E) 48

We will suppose that the colours of the tiles that are available are red, blue and green.

We need to count the number of different ways in which we can choose to places the tiles, so that we use at least one tile of each of the three colours. To do this we will need two tiles of one colour and one tile of each of the other two colours.

The method we use is to split our choice of how to arrange the tiles into a sequence of independent choices, and to count the number of different ways in which we can make these choices.

The first choice is to decide the colour of the two tiles with the same colour. There are 3 ways to choose this colour. Once we have made this choice, we have no further choice of colours as we must have one tile of each of the two other colours.

We now need to count the number of different ways to place the four tiles in a row after we have chosen their colours.

Suppose, for the sake of argument, that we have decided to have two red tiles and hence one blue tile and one green tile.

We can decide how to place these tiles by first choosing the place for the blue tile, and then the place for the green tile. Then there will be no further choice, as we will need to put the two red tiles in the two remaining places.

There are 4 choices for the position of the blue tile.

Then, whichever position we choose for the blue tile, there will remain 3 choices of position for the green tile.

We thus see that placing the tiles involves three independent choices and that the number of possibilities for these choices are 3, 4 and 3 respectively.

Each of the 3 choices of colour shared by two of the tiles may be combined with any of the 4 choices of position for the tile of the second colour. So there are 3 × 4 ways to combine these choices.

Each of these 3 × 4 ways of making the first two choices may be combined with any of the 3 choices of position for the tile of the third colour.

Therefore the total number of different ways of combining these choices is 3 × 4 × 3 = 36. Hence there are 36 different ways in which the row of four tiles can be chosen.

25.

5 marks

Beatrix places dominoes on a 5 × 5 board, either horizontally or vertically, so that each domino covers two small squares.

She stops when she cannot place another domino, as in the example shown in the diagram.

When Beatrix stops, what is the largest possible number of squares that may be still uncovered?

  • A) 4
  • B) 5
  • C) 6
  • D) 7
  • E) 8

The board has 25 squares. Each domino covers two squares. So, however many dominoes are placed on the board, the total number of squares that are covered will be even. So the number of uncovered squares will be odd.

In particular, it is not possible to have 8 uncovered squares. The diagram shows that it is possible for Beatrice to place 9 dominoes on the board in such a way that she cannot place another domino. This leaves 7 uncovered squares.

As this is the largest odd number given as one of the options, we conclude that D is the correct option.

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

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Cheerio, Simon.