Parallelogram 24 Year 8 25 Mar 2021Easter challenges

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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.
  • Tackle each Parallelogram in one go. Don’t get distracted.
  • Finish before the end of Easter holidays.
  • Your score & answer sheet will appear immediately after you hit SUBMIT.
  • Don’t worry if you score less than 50%, because it means you will learn something new when you check the solutions.

Junior Maths Challenge (UKMT) Mock

Some of you will be doing the UKMT’s Junior Maths Challenge on 30th April, so this Parallelogram consists of 25 Junior Maths Challenge questions – some easy, some tricky. This is an opportunity to get some practice before the real JMC. And, if you are not doing the JMC, then it is an excellent way to develop your maths skills.

The real JMC allows you one hour to attempt 25 questions, so in theory you have roughly sixty minutes to complete these 25 questions. However, don’t rush, don’t guess and don’t worry if you need longer than sixty minutes.

Here are a few things to bear in mind when you do the real JMC:

  • you may not use a calculator.
  • the first half of the questions are not too tricky (also true below).
  • the second half of the questions are really tricky (also true below).
  • you will lose marks for wrong answers in the second half of the questions (not true below).

So don’t take wild guesses for the second half of the questions in the real JMC, and don’t take wild guesses for trickier questions below. However, if you have a rough idea of the answer or have narrowed it down to a couple of possible answers, then it is worth picking an answer.

Make sure you have a paper and pencil to hand for sketches and calculations.

Because this Parallelogram is longer and tougher than a typical Parallelogram, we will be awarding TREBLE POINTS for this week's answers! So there's even more reason to give this one your best shot.

Finally, afterwards, make sure you go through all the answers and see where you went right and where you went wrong. This is 100% the best way to improve your maths skills.

The next Parallelogram will contain our usual mix of videos, humour, puzzles and problems.

1.

5 marks

1.1 What is the smallest four-digit positive integer which has four different digits?

  • 1032
  • 2012
  • 1021
  • 1234
  • 1023

Here it is easy just to check the options that are given. A, D and E are the only options in which all four digits are different. Of these, clearly, E is the smallest.

For a complete solution we need to give an argument to show that 1023 really is the smallest four digit positive integer with four different digits. It is easy to do this.

To get the smallest possible number we must use the four smallest digits, 0, 1, 2 and 3. A four digit number cannot begin with a 0. So we must put the next smallest digit, 1, in the thousands place, as a four-digit number beginning with 2 or 3 is larger than one beginning with a 1. For similar reason the hundreds digit must be the smallest remaining digit, 0. Similarly the tens digit must be 2 and the units digit must be 3.

So the required number is 1023.

2.

5 marks

2.1 How many of the six faces of a die have fewer than three lines of symmetry?

  • 2
  • 3
  • 4
  • 5
  • 6

Each of faces 1, 4 and 5 has four axes of symmetry, whilst each of faces 2, 3 and 6 has two axes of symmetry only.

3.

5 marks

3.1 Which of the following has exactly one factor other than 1 and itself?

  • 6
  • 8
  • 13
  • 19
  • 25

The factors of 6 are 1, 2, 3 and 6; the factors of 8 are 1, 2, 4 and 8; the factors of 13 are 1 and 13; the factors of 19 are 1 and 19; and the factors of 25 are 1, 5 and 25. We see from this that, of the numbers we are given as options, only 25 has exactly one factor other than 1 and itself. It is worth noticing that 25 is the only square number.

4.

5 marks

4.1 Which of the following points is not at a distance of 1 unit from the origin?

  • (0, 1)
  • (1, 0)
  • (0, -1)
  • (-1, 0)
  • (1, 1)

Each of points (0, 1), (1, 0), (0, -1) and (-1, 0) is 1 unit from the origin, but the point (1, 1) is at a distance of 2 units from the origin.

5.

5 marks

5.1 One of the mascots for the 2012 Olympic Games is called "Wenlock" because the town of Wenlock in Shropshire first held the Wenlock Olympian Games in 1850. How many years before the 2012 Olympics was that?

  • 62
  • 152
  • 158
  • 162
  • 172

All you have to do is a subtraction: 2012 - 1850 = 162.

6.

5 marks

6.1 Each square in the figure is 1 unit by 1 unit.

What is the area of triangle ABM (in square units)?

  • 4
  • 4.5
  • 5
  • 5.5
  • 6

Triangle ABM has base 3 units and height 3 units, so its area is 12×3×3 units², that is 412 units².

7.

5 marks

7.1 A small ink cartridge has enough ink to print 600 pages. Three small cartridges can print as many pages as two medium cartridges. Three medium cartridges can print as many pages as two large cartridges. How many pages can be printed using a large cartridge?

  • 1200
  • 1350
  • 1800
  • 2400
  • 5400

Three small cartridges have enough ink for 3×600=1800 pages. So 1800 is the number of pages that two medium cartridges can print. Hence one medium cartridge can print 12 (1800) = 900 pages. So three medium cartridges have enough ink for 3×900=2700 pages. So 2700 is the number of pages that two large cartridges can print. Hence one large cartridge can print 12 (2700) = 1350 pages.

An alternative, algebraic method, is to let m be the number of pages that a medium cartridge can print, and l be the number of pages that a large cartridge can print. From the information we are given we have the equations 3×600=2m and 3m=2l.

From these we can deduce that l = 32 m = 32 × 3×6002 =9×150=1350.

8.

5 marks

8.1 The figure on the right shows an arrangement of ten square tiles.

Which labelled tile could be removed, but still leave the length of the perimeter unchanged?

  • A
  • B
  • C
  • D
  • E

Removing tile A or tile B or tile D has the effect of reducing the perimeter by a distance equal to twice the side of one tile, whilst removing tile C increases the perimeter by that same distance.

Removing tile E, however, leaves the length of the perimeter unchanged.

9.

5 marks

9.1 The diagram on the right shows the positions of four people (each marked x) in an Art Gallery. In the middle of the room is a stone column. Ali can see none of the other three people. Bea can see only Caz. Caz can see Bea and Dan. Dan can only see Caz.

Who is at position p?

  • Ali
  • Bea
  • Caz
  • Dan
  • More information needed.

The person in position p is the only one who can see two of the others. We are told that Caz can see Bea and Dan but everyone else can see just one other person or no-one. So it must be Caz who is at position P.

We can also see that Ali, who can see no-one, must be in the position shown. Bea and Dan must be in the other two positions, but the information we are given doesn't enable us to work out which is where.

10.

5 marks

10.1 The diagram shows three squares of the same size. What is the value of x?

  • 105
  • 120
  • 135
  • 150
  • 165

The triangle in the centre of the diagram is equilateral since each of its sides is equal in length to the side of one of the squares. The sum of the angles at a point is 360°, so x=36090+90+60=120.

11.

5 marks

11.1 In the diagram shown, all the angles are right angles and all the sides are of length 1 unit, 2 units or 3 units.

What, in square units, is the area of the shaded region?

  • 22
  • 24
  • 26
  • 28
  • 30

Divide the whole figure into horizontal strips of height 1 unit: its area is (3 + 6 + 8 + 8 + 8 + 6 + 3) units2 = 42 units2. Similarly, the unshaded area is (1 + 4 + 6 + 4 + 1) units2 = 16 units2. So the shaded area is 26 units2.

Alternative solution: notice that if the inner polygon is moved a little, the answer remains the same – because it is just the difference between the areas of the two polygons. So, although we are not told it, we may assume that the inner one is so positioned that the outer shaded area can be split neatly into 1 by 1 squares – and there are 26 of these.

12.

5 marks

12.1 The six-member squad for the Ladybirds five-a-side team consists of a 2-spot ladybird, a 10-spot, a 14-spot, an 18-spot, a 24-spot and a pine ladybird (on the bench). The average number of spots for members of the squad is 12. How many spots does the pine ladybird have?

  • 4
  • 5
  • 6
  • 7
  • 8

The total number of spots which the six ladybirds have is 6×12=72. So the number of spots which the pine ladybird has is 722+10+14+18+24=4.

13.

5 marks

13.1 In the multiplication grid on the right, the input factors (in the first row and the first column) are all missing and only some of the products within the table have been given.

What is the value of A + B + C + D + E ?

  • 132
  • 145
  • 161
  • 178
  • 193

Let the input factors be p,q,r,s and t along the top row, and v,w,x,y and z in the first column, as shown in the grid on the right.

We see that wp=15 and xp=18, so that p is a divisor of both 15 and 18, and hence is either 1 or 3. However, if p=1, then it would follow that w=15. But this is not possible as wr=40. We deduce that p=3. It follows that w=5 and x=6. Since w=5 and wr=40, we have that r=8.

Therefore, as zr=56, z=7. Also, as x=6 and xs=60, s=10. Since vs=20, it follows that v=2. Then, as vq=10, q=5. Hence, as yq=20, we have that y=4 . Finally, as yt=24, we deduce that t=6.

This enables us to complete the table, as shown on the left. (Though, really, we need only to calculate the diagonal entries that we have shown in bold.)

Therefore, we have A+B+C+D+E=6=25+48+40+42=161.

Next up, we have the trickier questions.

14.

6 marks

14.1 Talulah plants 60 tulip bulbs. When they flower, she notes that half are yellow; one third of those which are not yellow are red; and one quarter of those which are neither yellow nor red are pink. The remainder are white.

What fraction of the tulips are white?

  • 124
  • 112
  • 16
  • 15
  • 14

Since half the 60 bulbs are yellow, 30 are yellow and 30 are not yellow. One third of the 30 bulbs that are not yellow are red, so 10 are red and 20 are neither yellow nor red. One quarter of the 20 that are neither yellow nor red, are pink. So 5 bulbs are pink.

This leaves 15 bulbs which are neither yellow, nor red nor pink. So there are 15 white bulbs. Therefore the fraction of bulbs that are white is 1560, that is, 14.

15.

6 marks

15.1 Mathias is given a grid of twelve small squares.

He is asked to shade grey exactly four of the small squares so that his grid has two lines of reflection symmetry.

How many different grids could he produce?

  • 2
  • 3
  • 4
  • 5
  • 6

The two lines of reflection symmetry of the grid are shown in the figure on the right as broken lines.

We see from this that if the grid with some of the small squares shaded grey has both these lines of reflection symmetry, then all the four squares labelled P must be the same colour. Similarly, all the four squares labelled Q must be the same colour, the two squares labelled R must be the same colour, and the two squares labelled S must be the same colour.

It follows that there are only three ways in which Mathias can shade exactly four of the small squares of the grid grey so that the result has two lines of reflection symmetry. These are:

(i) shade grey all the squares labelled P, and no others,
(ii) shade grey all the squares labelled Q, and no others,
and (iii) shade grey all the squares labelled R and all those labelled S, and no others.

The three different grids that Mathias could produce are shown in the figure below.

16.

6 marks

16.1 The lengths, in cm, of the sides of the equilateral triangle PQR are as shown.

Which of the following could not be the values of x and y?

  • (18, 12)
  • (15, 10)
  • (12, 8)
  • (10, 6)
  • (3, 2)

As triangle PQR is equilateral, x+2y=3xy=5yx. Equating any two of these expressions gives 2x=3y.

The only pair of given values which does not satisfy this equation is x=10, y=6.

17.

6 marks

17.1 The diagram is a “map” of Jo’s local rail network, where the dots represent stations and the lines are routes.

Jo wants to visit all the stations, travelling only by train, starting at any station and ending at any station, with no restrictions on which routes are taken.

What is the smallest number of stations that Jo must go to more than once?

  • 1
  • 2
  • 3
  • 4
  • 5

We label the stations as shown in the diagram on the right so that we can refer to them. Jo can visit the stations in the order:

O,P,Q,R,S,R,T,X,Y,Z,Y,X,W,V,U.

In this way she visits all the stations and goes through just three of them, R, Y and X more than once.

It remains only to show that Jo cannot visit all the stations without going through at least three of them more than once.

Jo must go through the junction stations R and X more than once. If Jo does not start or finish at O, she will have to go through P more than once. If she does not start or finish at U she will have to go through V more than once. If she starts at O and finishes at U, or vice versa, she will have to go through Y more than once. It follows that Jo has to go through at least three stations more than once. So this is the smallest number of stations Jo must go through more than once.

18.

6 marks

18.1 In rectangle PQRS, the ratio of PSQ to PQS is 1:5 . What is the size of QSR?

  • 15°
  • 18°
  • 45°
  • 72°
  • 75°

Let QSR =x°. Since PS is parallel to QR, the alternate angles PQS and QSR are equal. So PQS =x°. Therefore, as PSQ:PQS =1:5, PSQ =15x°. Therefore, from the right angled triangle PQS we deduce that x+15x+90=180 and so 65x=90. Therefore x=56×90=75.

So QSR is 75°.

19.

6 marks

19.1 Jack and Jill played a game for two people. In each game, the winner was awarded 2 points and the loser 1 point. No games were drawn. Jack won exactly 4 games and Jill had a final score of 10 points. How many games did they play?

  • 5
  • 6
  • 7
  • 8
  • impossible to determine

Jack won exactly 4 games. So Jill lost 4 games and gained 4 points from these losses. Jill gained 10 points altogether and so gained 10 − 4 = 6 points from her wins. Since there are 2 points for a win, Jill won 3 games. So Jack won 4 games, Jill won 3 games and altogether 4 + 3 = 7 games were played.

20.

6 marks

20.1 Dominic wants to place the six dominoes above in a hexagonal ring so that, for every pair of adjacent dominoes, the numbers of pips match. The ring on the right indicates how one adjacent pair match.

In a completed ring, how many of the other five dominoes can he definitely not place adjacent to ?

  • 1
  • 2
  • 3
  • 4
  • 5

There are only two 1-pip dominoes among the six Dominic has. These must therefore be adjacent. Likewise for the two 2-pips, the two 6-pips and the two 4-pips. So the ring must include the three adjacent dominoes:

and also the three adjacent dominoes:

The ring may now be completed two ways, i.e., by connecting top-left 5-pip to bottom-left 5-pip and the top-right 5-pip to bottom-right 5-pip … or by connecting top-left 5-pip to bottom-right 5-pip and the top-right 5-pip to bottom-left 5-pip.

We therefore see that Dominic can create two different rings of six dominoes, as shown below.

We now see that there are just 2 dominoes that the [1:5] domino cannot be adjacent to, namely the domino [2:5] and, of course, the domino [4:6].

21.

6 marks

21.1 Pablo’s teacher has given him 27 identical white cubes.

She asks him to paint some of the faces of these cubes grey and then stack the cubes so that they appear as shown.

What is the largest possible number of the individual white cubes which Pablo can leave with no faces painted grey?

  • 8
  • 12
  • 14
  • 15
  • 16

We can see 19 of the 27 cubes. Of these there are 12 which we can see have at least one grey face. The remaining cubes could have all their faces white. So the maximum number of cubes that could be all white is 27 − 12 = 15.

22.

6 marks

22.1 Peter wrote a list of all the numbers that could be produced by changing one digit of the number 200. How many of the numbers in Peter's list are prime?

  • 0
  • 1
  • 2
  • 3
  • 4

If the hundreds or tens digit of 200 is changed, but the units digit is unchanged, the resulting number is, like 200, a multiple of 10, and so cannot be prime. So we need only consider the 9 numbers that we can get by changing the units digit. Of these, we can see immediately that 202, 204, 206 and 208 are all divisible by 2 and 205 is divisible by 5. So none of them is prime.

This just leaves 201, 203, 207 and 209. Now:
201=3×67,
203=7×29,
207=3×3×23 and
209=11×19.

None of them are prime, so none of the numbers in Peter's list are prime.

23.

6 marks

23.1 Sam wants to complete the diagram so that each of the nine circles contains one of the digits from 1 to 9 inclusive and each contains a different digit.

Also, the digits in each of the three lines of four circles must have the same total.

What is this total?

  • 17
  • 18
  • 19
  • 20
  • 21

We suppose that x, a, b, c, d, e, and f are the numbers in the circles, as shown in the diagram. These numbers are the digits 1, 3, 4, 6, 7, 8, and 9, in some order.

Let the common total of the digits in the three lines be k.

Then: x+a+b+5+x+e+f+2+5+c+d+2=3k,

that is, x+a+b+c+d+e+f+x+14=3k.

Now x+a+b+c+d+e+f=1+3+4+6+7+8+9=38

and therefore

52+x=3k.

It follows that 52+x is divisible by 3. Now x is one of 1, 3, 4, 6, 7, 8, 9, and the only one of these possible values for which 52 + x is divisible by 3 is x = 8. This gives 52 + x = 60.

Hence 60=3k, and so k=20.

In the context of the JMC, we can stop here. But to give a full mathematical solution we should satisfy ourselves that it is possible to place the digits in the circles so that each line adds up to 20.

One way this can be done is shown in the diagram opposite.

24.

6 marks

24.1 The interior angles of a triangle are 5x+3y°, 3x+20° and 10y+30°, where x, y are positive integers.

What is the value of x+y?

  • 15
  • 14
  • 13
  • 12
  • 11

Crucially, the question said that x and y are positive integers.

The angles of a triangle add up to 180°. So 5x+3y+3x+20+10y+30=180. This gives 8x+13y+50=180. Therefore 8x+13y=130, and so

8x=13013y=1310y.

It follows that 8x is a multiple of 13 and hence, as 8 and 13 have no common factors, it must be that x is a multiple of 13. So x=13a, where a is a positive integer. It then follows that

813a=1310y

and so, cancelling the factor 13

8a=10y

and hence 10y is a positive number which is a multiple of 8. The only positive value of y for which 10y>0 and 10y is a multiple of 8, is y=2, which gives a=1 and hence x=13. So x+y=13+2=15.

25.

6 marks

25.1 A die has the shape of a regular tetrahedron, with the four faces having 1, 2, 3 and 4 pips. The die is placed with 4 pips ‘face down’ in one corner of the triangular grid shown, so that the face with 4 pips precisely covers the triangle marked with 4 pips.

The die is now ‘rolled’, by rotating about an edge without slipping, so that 1 pip is face down. It is rolled again, so that 2 pips are face down, as indicated. The rolling continues until the die rests on the shaded triangle in the opposite corner of the grid.

How many pips are now face down?

  • 1
  • 2
  • 3
  • 4
  • it depends on the route taken

We first consider what happens if one vertex of the tetrahedral die remains fixed on the grid and the die rolls about this fixed point. In doing this the die covers in succession six faces making up a hexagon. The faces adjacent to the fixed vertex are each face down twice in this hexagon. If the die is rolled about any of the edges on the perimeter of the hexagon, the remaining face is then face down. This is shown in the diagram on the right for the case where the faces with 1, 2 and 3 pips are adjacent to the fixed vertex, and the remaining face has 4 pips on it.

The pattern shown above can be extended to the whole of the given triangular grid, as shown in the diagram on the right. (Indeed, this pattern could be extended to a triangular grid filling the entire plane.)

For any triangle in the grid, the face then face down together with the two preceding faces form part of a hexagon, and determine the orientation of the die. (Here, by the orientation of the die, we mean the positions of all four of its faces.) The die can roll to this triangle across any of its edges. However, the pattern implies that the same face is face down in all these cases. In particular, the face with 1 pip is face down when the die reaches the shaded triangle.

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 after the Easter holidays.
  • 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.