Week 13Chrismaths Y7

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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 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

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.

2.

3 marks

It has just turned 22:22. How many minutes are there until midnight?

  • A) 178
  • B) 138
  • C) 128
  • D) 108
  • E) 98

There are 60 − 22 = 38 minutes from 22:22 to 23:00 and then a further 60 minutes to midnight.

Since 38 + 60 = 98, there are 98 minutes to midnight.

3.

3 marks

What is the value of 123451+2+3+4+5?

  • A) 1
  • B) 8
  • C) 678
  • D) 823
  • E) 12,359

In the context of the JMC we can answer the question by eliminating the options that cannot be correct without having to do a division sum to evaluate the fraction given in this question.

Since 12,345 and 1 + 2 + 3 + 4 + 5 = 15 are both odd numbers, 123451+2+3+4+5 must also be odd. So only options A, D and E could be correct.

However, it is clear that 12345151 and 123451512359

This rules out options A and E and leaves D as the correct option.

4.

3 marks

In this partly completed pyramid, each rectangle is to be filled with the sum of the two numbers in the rectangles immediately below it.

What number should replace x?

  • A) 3
  • B) 4
  • C) 5
  • D) 7
  • E) 12

We have used p,q,r,s and t for the numbers in certain of the rectangles as shown in the diagram.

We now repeatedly use the fact that the number in each rectangle in the first four rows is the sum of the numbers in the rectangles immediately below it. This enables us to work out the values of p,q,r,s and t.

Applied to the top rectangle, this gives 105=p+47. It follows that p=10547=58.

Then, as p=31+q, we have 58=31+q. Therefore q=27. Next, from 47=q+r, we deduce that 47=27+r. This gives r=20.

Next, from r=13+s, we have 20=13+s. Hence s=7.

We also have 13=9+t. Therefore t=4. Finally, s=t+x. Therefore 7=4+x.

We can now conclude that x=3.

5.

3 marks

The difference between 13 of a certain number and 14 of the same number is 3.

What is that number?

  • A) 24
  • B) 36
  • C) 48
  • D) 60
  • E) 72

Since 1314=4312=112

it follows that 112 th of the number is 3. Therefore the number is 12×3=36.

6.

3 marks

What is the value of x in this triangle?

  • A) 45
  • B) 50
  • C) 55
  • D) 60
  • E) 65

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.

7.

3 marks

The result of the calculation 123,456,789 × 8 is almost the same as 987,654,321 except that two of the digits are in a different order.

What is the sum of these two digits?

  • A) 3
  • B) 7
  • C) 9
  • D) 15
  • E) 17

In the context of the JMC we are entitled to assume the truth of the statement that 123456789×8 is obtained by interchanging two of the digits of 987654321. This leads to a quick way to answer the question without the need for a lot of arithmetic.

Because 9×8=72, the units digit of 123456789×8 is a 2. Starting from 987654321, to obtain a 2 as the units digit we need to interchange the digits 1 and 2. So these are the two digits which are in a different order in the answer to the calculation. Now comes the easy bit, 1+2=3.

Alternatively, if we cannot take the statement in the question on trust, the only thing to do is to actually multiply 123456789 by 8. If you do this you will see that the answer is 987654312. It follows that it is the digits 1 and 2 that need to be interchanged.

Note that, in fact, as soon as we get as far as working out that 89×8=712 we can deduce that the digits 1 and 2 need to be interchanged. It is, however, necessary to do the whole sum to check that all the other digits are in the right order.

Note The number 123456789×8 is a multiple of 8. We have the following test for whether a number is a multiple of 8.

An integer is a multiple of 8, if, and only if, its last three digits form a number which is a multiple of 8.

Since 321 is not a multiple of 8, this shows immediately that 987654321 is not equal to 123456789×8 .

8.

3 marks

Which of the following has the same remainder when it is divided by 2 as when it is divided by 3?

  • A) 3
  • B) 5
  • C) 7
  • D) 9
  • E) 11

The answer may be found by trying the options in turn. In this way we find that 7 has remainder 1 when divided by 2 and when divided by 3.

Alternatively, we see that all the numbers given as options are odd and so each has remainder 1 when divided by 2. So the correct option will be a number which also gives remainder 1 when divided by 3.

It is easy to see that, of the given options, only 7 meets this requirement.

9.

3 marks

According to a newspaper report, “A 63-year-old man has rowed around the world without leaving his living room.”

He clocked up 25,048 miles on a rowing machine that he received for his 50th birthday.

Roughly how many miles per year has he rowed since he was given the machine?

  • A) 200
  • B) 500
  • C) 1,000
  • D) 2,000
  • E) 4,000

The man is now 63 years old and was given the rowing machine for his 50th birthday. So he has had the rowing machine for 13 years and, possibly, a few months. Therefore the average number of miles per year that he has rowed is roughly

25048132600013=2000.

Therefore, 2000 is roughly the number of miles per year that the man has rowed.

10.

3 marks

In the expression 1 □ 2 □ 3 □ 4 each □ is to be replaced by either + or ×.

What is the largest value of all the expressions that can be obtained in this way?

  • A) 10
  • B) 14
  • C) 15
  • D) 24
  • E) 25

In general, we obtain a larger number by multiplying two positive integers together than by adding them. The only exceptions are when one of the positive integers is 1, because 1+n>1×n, and when both are 2, because 2+2=2×2.

Therefore to obtain the largest possible value we need to replace the first □ with + but the other two with ×. This leads us to the answer 1+2×3×4=1+24=25.

11.

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

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.

12.

3 marks

A fish weighs a total of 2 kg plus a third of its own weight.

What is the weight of the fish in kg?

  • A) 213
  • B) 3
  • C) 4
  • D) 6
  • E) 8

Since the fish weighs 2 kg plus one third of its weight, 2 kg is two thirds of its weight. Therefore one third of its weight is 1 kg, and so the total weight of the fish is 2 kg + 1 kg = 3 kg.

Alternatively, we can also solve this problem using algebra.

We let x be the weight of the fish in kg. Now we use the information in the question to create an equation involving x that we can solve.

Because the fish weighs 2 kg plus one third of its weight, x=2+13x

It follows that x13x=2, and hence, 23x=2.

Because 32×23=1, we multiply both sides of this equation by 32. In this way we deduce that x=32×2=3.

13.

3 marks

In the figure shown, each line joining two numbers is to be labelled with the sum of the two numbers that are at its end points.

How many of these labels are multiples of 3?

  • A) 10
  • B) 9
  • C) 8
  • D) 7
  • E) 6

The figure consists of a regular octagon whose vertices are labelled with the positive integers from 1 to 8 inclusive. There is a line joining each pair of the vertices. It follows that the number of labels which are multiples of 3 is equal to the number of pairs of distinct integers in the range from 1 to 8, inclusive, whose sum is a multiple of 3.

The following table shows all multiples of 3 that can be a label, and for each multiple, the different ways of writing it as a sum of two distinct integers in the range from 1 to 8. We do not need to go beyond 15, as the largest integer that can appear as a label is 7 + 8 = 15.

It follows that the number of labels which are multiples of 3 is 1 + 2 + 4 + 2 + 1 = 10.

14.

3 marks

Digits on a calculator are expressed by a number of horizontal and vertical illuminated bars. The digits and the bars which represent them are shown in the diagram.

How many digits are both prime and represented by a prime number of illuminated bars?

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

The digits that are primes are 2, 3, 5 and 7 (It is important to remember that 1 is not a prime number). The numbers of illuminated bars used to represent them are:

2: 5 bars; 3: 5 bars; 5: 5 bars; 7: 3 bars.

We see that each of them is represented by a prime number of bars. So there are 4 of the digits with the required property.

15.

3 marks

Which of the following is divisible by all of the integers from 1 to 10 inclusive?

  • A) 23 × 34
  • B) 34 × 45
  • C) 45 × 56
  • D) 56 × 67
  • E) 67 × 78

It is easy to rule out four of the options using the fact that a product of integers is divisible by a prime number p if, and only if, at least one of the integers making up the product is divisible by p.

Using this we see that, since neither 23 nor 34 is a multiple of 3, it follows that 23 × 34 is not a multiple of 3, since neither 34 nor 45 is a multiple of 7, 34 × 45 is not a multiple of 7, since neither 56 nor 67 is a multiple of 5, 56 × 67 is not a multiple of 5, and, similarly, 67 × 78 is not a multiple of 5.

This rules out the options A, B, D and E. In the context of the JMC this is enough for us to be able to conclude that option C is the correct answer.

However, for a full solution, we would need to check directly that 45 × 56 is divisible by all the integers from 1 to 10. This is straightforward.

Every integer is divisible by 1. Since 45 is divisible by 3, 5 and 9, so also is 45 × 56. Since 56 is divisible by 2, 4, 7 and 8, so also is 45 × 56. Since 45 × 56 is divisible both by 2 and by 3, which have no common factor, it is also divisible by 6. Since 45 × 56 is divisible by both 2 and by 5, which have no common factor, it is also divisible by 10.

Therefore 45 × 56 is divisible by all the integers from 1 to 10, inclusive.

16.

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

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.

17.

4 marks
  • Knave of Hearts: “I stole the tarts.”
  • Knave of Clubs: “The Knave of Hearts is lying. ”
  • Knave of Diamonds: “The Knave of Clubs is lying.”
  • Knave of Spades “The Knave of Diamonds is lying.”

How many of the four Knaves were telling the truth?

  • A) 1
  • B) 2
  • C) 3
  • D) 4
  • E) more information needed

Either the Knave of Hearts stole the tarts or he is innocent.

If the Knave of Hearts stole the tarts, he was telling the truth. So the Knave of Clubs was lying. Hence the Knave of Diamonds was telling the truth. Therefore the Knave of Spades was lying. So in this case two of the four Knaves were lying.

If the Knave of Hearts did not steal the tarts, he was lying. So the Knave of Clubs was telling the truth. Hence the Knave of Diamonds was lying. Therefore the Knave of Spades was telling the truth. So also in this case two of the four Knaves were lying.

We cannot tell from the information given whether or not the Knave of Hearts stole the tarts. But, as we have seen, we can be sure that, whether he stole them or not, two of the Knaves were telling the truth and two were lying.

18.

4 marks

Each of the fractions 263718459 and 527436918 uses the digits 1 to 9 exactly once.

The first fraction simplifies to 17.

What is the simplified form of the second fraction?

  • A) 18
  • B) 17
  • C) 534
  • D) 961
  • E) 27

We are told in the question that 263718459=17.

Now note that 5274=2×2637 and 36918=2×18459. It follows that, by cancelling the common factor 2 in the numerator and the denominator,

527436918=2×26372×18459=263718459=17.

If you do not spot the quick method used above, there is nothing for it but to try out the options in turn.

We first consider option A. By cross multiplication we have

527436918=185274×8=36918×1.

We can see that the equation 5274×8=36918×1 cannot be correct just by looking at the units digits on the two sides of the equation. On the left hand side 4 × 8 gives a units digit of 2, but on the right hand side 8 × 1 gives a units digit of 8. So the equation is not correct.

We deduce that

52743691818

and so option A is not the correct one.

Next we look at option B. Using cross multiplication again, we have

527436918=175274×7=36918×1.

It is straightforward to check that the equation 5274×7=36918 is true. Therefore option B is correct.

19.

4 marks

One of the following cubes is the smallest cube that can be written as the sum of three positive cubes.

Which is it?

  • A) 27
  • B) 64
  • C) 125
  • D) 216
  • E) 512

The positive cubes are the numbers in the sequence 1, 8, 27, 64, 125, 216, 343, ... .

It is straightforward to check that none of the first fives cubes in this sequence is the sum of three smaller positive cubes.

For example, as 27 + 27 + 27 = 81, and 81 < 125, any three cubes with sum 125 must include 64 at least once. The three cubes couldn’t include 64 twice because 64 + 64 > 125. However, if we had p+q+64=125, where p and q are positive cubes which are smaller then 64, then p+q=12564=61, which is impossible as the only values p and q can take are 1, 8 and 27. So 125 is not the sum of three positive cubes.

However, 216 = 27 + 64 + 125, and so 216 is the sum of three positive cubes and so is smallest cube that can be written as the sum of three positive cubes.

20.

4 marks

The diagram shows a pyramid made up of 30 cubes, each measuring 1m × 1m × 1 m.

What is the total surface area of the whole pyramid (including its base)?

  • A) 30 m2
  • B) 62 m2
  • C) 72 m2
  • D) 152 m2
  • E) 180 m2

The surface area that can be seen by looking up from below the pyramid is that of 4 × 4 = 16 squares each measuring 1m × 1 m. So the surface area of the base is 16 m2.

The view looking down from above the pyramid is shown in the diagram. The surface area that can be seen is made up of a complete 1m × 1m square, some three-quarter squares and some half squares. Without adding these up, we can see that the surface area is also 16 m2.

The view from each of the four sides is made up of a pyramid of 10 squares whose surface area is 10 m2.

Therefore, the total surface area is 2 × 16 m2 + 4 × 10 m2 = 72 m2.

21.

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

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:
4×420+3g=4800

and therefore:
3g=48004×420=48001680=3120.

It follows that:
g=13×3120=1040.

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.

22.

5 marks

The diagram shows a shaded region inside a regular hexagon.

The shaded region is divided into equilateral triangles.

What fraction of the area of the hexagon is shaded?

  • A) 38
  • B) 25
  • C) 37
  • D) 512
  • E) 12

We form a complete grid inside the hexagon, as shown in the figure.

In this way the hexagon is divided up into a number of congruent equilateral triangles and, around the edge, some triangles each congruent to half of the equilateral triangles.

We could now use the grid to work out the shaded and unshaded areas in terms of the areas of the equilateral triangles, and hence work out which fraction of the area of the hexagon is shaded.

It is a little easier to exploit the sixfold symmetry of the figure and just work out the fraction of the area surrounded by the heavy lines that is shaded.

We see that in this part of the hexagon there are six shaded equilateral triangles, four unshaded equilateral triangles and four unshaded triangles whose areas are each half that of the equilateral triangles. So the unshaded area is equal to that of six of the equilateral triangles. It follows that the shaded area is equal to the unshaded area.

We conclude that the fraction of the hexagon that is shaded is 12.

23.

5 marks

The diagram shows four shaded glass squares, with areas of 1 cm2, 4 cm2, 9 cm2 and 16 cm2, placed in the corners of a rectangle. The largest square overlaps two others. The area of the region inside the rectangle but not covered by any square (shown unshaded) is 1.5 cm2.

What is the area of the region where the squares overlap (shown dark grey)?

  • A) 2.5 cm2
  • B) 3 cm2
  • C) 3.5 cm2
  • D) 4 cm2
  • E) 4.5 cm2


The squares in the corners of the large rectangle are of sizes 1 cm × 1 cm, 2 cm × 2 cm, 3 cm × 3 cm and 4 cm × 4 cm.

The white rectangle has width 1 cm and area 1.5 cm2. It follows that it has height 1.5 cm.

We can now deduce that the large rectangle has height 5.5 cm and width 5 cm, and hence that the lengths are, in cm, as shown in the diagram.

We therefore see that the region shown dark grey is made up of two rectangles, one with width 2 cm and height 1.5 cm, and the other with width 2 cm and height 0.5 cm.

Therefore the area of this region is (2 × 1.5) cm2 + (2 × 0.5) cm2 = 3 cm2 + 1 cm2 = 4 cm2.

Once we have shown, as above, that the large rectangle has height 5.5 cm and width 5 cm, we can find the area of the overlap without finding the dimensions of the two rectangles that make it up. Instead we can give an argument just in terms of areas, as follows.

The area of the large rectangle is 5.5 cm × 5 cm = 27.5 cm2. Since the area not covered by any of the squares is 1.5 cm2, the area of the large rectangle covered by the squares is 27.5 cm2 − 1.5 cm2 = 26 cm2.

The total area of the squares is 1 cm2 + 4 cm2 + 9 cm2 + 16 cm2 = 30 cm2.

The difference between these two areas is accounted for by the overlap. Therefore the area of the overlap is 30 cm2 − 26 cm2 = 4 cm2.

24.

5 marks

A palindromic number is a number which reads the same when the order of its digits is reversed.

What is the difference between the largest and smallest five-digit palindromic numbers that are both multiples of 45?

  • A) 9,180
  • B) 9,090
  • C) 9,000
  • D) 8,910
  • E) 8,190

We use the notation abcde for the number which is represented by the digits a,b,c,d and e when expressed using the standard base 10. Using this notation we can write a five-digit palindromic number as abcba, where a,b and c are digits.

Since 45 = 5 × 9, and 5 and 9 have no common factors, the five-digit palindromic number abcbais a multiple of 45 if, and only if, it is a multiple of both 5 and 9.

A number is a multiple of 5 if, and only if, its units digit is 0 or 5. Here the units digit a cannot be 0, since otherwise abcba would not be a five-digit number. We deduce that a is 5. Thus a five-digit palindromic number which is divisible by 5 is of the form 5bcb5, where band c are digits.

A number is a multiple of 9 if, and only if, the sum of its digits is a multiple of 9.

The smallest five-digit palindromic number that is a multiple of 45 has the form 50c05, corresponding to taking b to be 0. The digit sum of50c05 is 10+c. For this to be a multiple of 9 we need to take the digit c to be 8. Therefore 50805, with digit sum 18, is the smallest five-digit palindromic number which is a multiple of 45.

The largest five-digit palindromic number that is a multiple of 45 has the form 59c95, corresponding to taking b to be 9. The digit sum of59c95 is 28+c. For this to be a multiple of 9 we need to take the digitc to be 8. Therefore 59895, with digit sum 36, is the largest five-digit palindromic number which is a multiple of 45.

The difference between these two numbers is 5989550805=9090.

25.

5 marks

The four straight lines in the diagram are such that VU=VW. The sizes of UXZ, VYZ and VZX are x°,y°and z°.

Which of the following equations gives x in terms of y and z?

  • A) x=yz
  • B) x=180yz
  • C) x=yz2
  • D) x=y+z90
  • E) x=yz2

Because VU=VW, the triangle VUW is isosceles and so VUW and VWU are equal.

Because they are vertically opposite VWU and YWX are equal.

Therefore VUW, VWU and YWX are all equal. We let the size of these three angles be t°, as marked on the diagram.

We now apply the Exterior Angle Theorem in turn to the exterior angle, VUW of triangle UZX and the exterior angle ZYW of triangle WYX. This gives t=z+x and y=t+x.

Therefore, using the first equation to substitute z+x for t in the second equation, we have y=z+x+x=z+2x.

Hence, by rearranging the last equation, 2x=yz.

By dividing both sides of the last equation by 2, we conclude that x=yz2.

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