SOLUTIONS

1. Solution:

The answer is not ‘about four inches’. In fact, the bookmarks are almost touching.

This picture shows how the two volumes look when they are placed on the shelf.

The bookmarks are placed at the front of Volume I and the back of Volume II. Since aardvark is near the front of Volume I and zebra is near the back of Volume II, the bookmarks are next to each other, separated by little more than the covers of the two volumes.

2. Solution:

Martin was wrong. The rolling coin makes a complete turn even though it has only travelled around half a circle. In the diagram, the quarter circle AB is the same as AC, so by the time B rolls round to C the coin will already be upside down after completing only half its journey. By the time point D rolls round to point E, the coin will be the right way up again.

3. Solution:

The missing word in this quiz is in fact the word ‘quiz’. The six words collectively use the 26 letters of the alphabet. (Collections of words that between them contain all the letters are known as pangrams.)

4. Solution:

She opened the Mixed tin. Suppose she took out a Rich Tea biscuit from that tin. Since the label was wrong, that must really be the Rich Tea tin. That means the tin labelled Digestive can only contain Mixed biscuits, and the one labelled Rich Tea must therefore contain the Digestive biscuits. Just the same logic applies if she removes a Digestive from the Mixed tin.

5. Solution:

Mr Jones wrapped the cue diagonally to make a package that, when measured end to end, was just under 1.5 metres in length.

Of course, when measured diagonally the parcel was as long as a cue, but the Post Office regulations did not say anything about diagonal measurements.

6. Solution:

This feat can be achieved by moving two matches as shown, so that the cherry appears inside the glass.

7. Solution:

It will take them three minutes. One pizza maker takes three minutes to cook one pizza, so in the same amount of time six pizza makers would make six pizzas.

8. Solution:

It is not knotted. Pull the right hand end and it first turns into this . . .

and then into this . . .

and finally becomes completely free.

9. Solution:

Maxine sent the letter. Every time the letter ‘x’ is typed, the daisy wheel shifts round by one. The letter ‘x’ is typed three times, the first time it produces an ‘x’, the second time a ‘y’ and the final time (in Maxine’s name at the end) a ‘z’.

The rest of the message reads:

Every time I type the letter x it starts to go funny. I must get it fixed when I come home. Lots of Love. Maxine.

10 Solution:

Mrs P put down £1 in small change (20 pence and 10 pence coins), from which the driver could see that she wanted a £1 ticket because otherwise she would have given him exactly 70p. Mrs Q put down a £1 coin, which could be used to pay for either type of ticket.

11. Solution:

Maria filled her can at the spot marked Z. Remember the old principle that the shortest distance between two points is a straight line? That principle applies here too – but it needs a mirror.

Imagine that the second half of her return journey, from Z to X, is reflected in the line of the river bank, so that the reflection of X is at P.

Then the total length of Maria’s journey is from Y to Z to P and this will be shortest when YZP is a straight line. In other words, when the angle between YZ and the river bank is the same as the angle between XZ and the river bank.

So Maria ran to the river bank and ‘bounced’ back to the picnic spot, like a snooker ball bouncing off a cushion.

12. Solution

It does not matter how fast she runs – even a world record sprint could not increase her average speed to 6 mph. This is because the Fun Run is 3 miles long, so if her overall speed was 6 mph she would do the whole run in 30 minutes. But she has already taken 30 minutes to run the first 2 miles at 4 mph, so she would have to run the final mile in 0 minutes! (This is a good illustration of the fact that averages can be quite complicated. You usually can’t average an average.)

13. Solution:

The sequence continues: TH TH TH TH. The letter pairs given in the puzzle are the endings of the ordinal numbers –firST, secoND, thiRD. TH then follows from fourTH to twentieTH.

14. Solution:

This is one of the earliest known examples of what is often called a ‘think of a number’ trick. Since it works for any number, let’s call that number ‘Blob’. Pick up ‘Blob’ beans in each hand. Now follow the instructions:

	Left hand	Right hand	Total
Start	Blob	Blob	Two Blobs
Four across	(Blob + 4)	(Blob – 4)	Still two Blobs
Cast away all in right	(Blob + 4)	0	Blob + 4
Do same no. from left	8*	0	8
Pick up five more	8	5	13

* (Blob + 4) take away (Blob – 4) equals 8, whatever ‘Blob’ is.

In fact, this use of a symbol (‘Blob’) to represent any number is an example of algebra. Instead of calling it ‘Blob’, mathematicians usually prefer ‘x’.

15. Solution:

Make an ordinary knot in a strip of paper, and very carefully flatten the knot, keeping it tight as you do so. It will form a regular pentagon.

16. Solution:

The man took the goat across, leaving the wolf and cabbages behind. Then he returned and took the wolf across and brought the goat back. He then rowed across with the cabbages. Finally, he returned to pick up the goat and take it across, making a total of seven crossings.

17. Solution:

After 280 seconds, the Big Wheel would have gone round exactly 28 times, the Rotator would have gone round exactly 35 times, and the Twirler would have gone round exactly 40 times.

This happens because 280 is exactly divisible by 10, 8 and 7; it is also the smallest number with that property. So they were in line again after 280 seconds, or just under 5 minutes.

18. Solution:

Move the bottom-left toothpick as shown. You now have a mirror image of the original deer.

19. Solution:

Mum is 49. Weekends account for two sevenths of the week, so 35 is five sevenths of mum’s age. Divide 35 by 5 and multiply by 7 to get 49.

20. Solution:

Dougal’s original pocket money was £6 per week. Compare the two ways of increasing his pocket money. This tells you that one third of the original, plus £1, is equal to one quarter of the original, plus £1.50.

The difference between one third and one quarter is one twelfth, so the extra 50p is one twelfth of the original pocket money, which must therefore have been 50p x 12 = £6.

21. Solution:

The carpenter made this cut through the original plank, as in the first figure, and then rearranged the pieces as in the second illustration.

22. Solution:

Forty. (In French there are several numbers: deux, cinq, dix and cent.)

23. Solution:

The trains up and down the line arrived every ten minutes, but the down train was always two minutes after the up train. So Jessica was four times as likely to arrive in the eight minutes before the up train arrived as she was to arrive in the mere two minutes before the down train arrived.

24. Solution:

The word that becomes decoded is ‘DECODED’: it is unchanged when you look at it upside down in the mirror (have a look and see!). The reason is that all the letters in that word are symmetrical top and bottom. All the other words contain at least one letter that does not have this line of symmetry.

25. Solution:

Debbie is a small girl going to school and back. When she enters the lift in the morning she can reach the bottom button marked ‘GROUND’ but, on returning, she cannot reach any button higher than ‘FLOOR 5’. Only if an adult is getting into the lift at the same time can she ask them to press ‘FLOOR 16’ for her and go all the way up. (This puzzle has traditionally been presented as being a dwarf in the lift, but it has always seemed absurd even in the fantasy world of puzzles that it never occurred to the dwarf to either (a) take a stick, or (b) move apartment.)

26. Solution:

One way is to start by going diagonally down from point A to B, then across to C, etc.

27. Solution:

No, he wasn’t. He handed over his £10, and Janet said, ‘Thanks,’ and started to walk away. Her brother said, ‘Hey, if you don’t give me £15, you’ve lost the bet!’ ‘Oh, yes, so I have,’ said Janet. ‘So I owe you five pounds.’ She handed £5 over and went off to the cinema with £10 in her pocket and a smile on her face.

28. Solution:

When he walks up the escalator, Mr Watson takes 20 seconds to reach the top. In that time he has walked up 40 steps (two steps per second). If he stands still, in 20 seconds he only rises by half the height of the escalator. So the height of the escalator, H, is 40 steps plus half of its height, H, which means the escalator must be 80 steps high.

(This is similar reasoning to the old puzzle ‘If a brick weighs one pound plus half a brick, how much does a brick weigh?’ – to which the answer is ‘two pounds’.)

29. Solution:

These are the moves the trains should have made.

30. Solution:

The ladder was attached to the side of the ship, so the ladder and the ship would rise together as the tide came in.

31. Solution:

She won 5 prizes.

There were only four girls involved, and six differences between them were given. Therefore every possible difference between the four girls was named and we are looking for four numbers whose six differences are 1, 2, 3, 4, 5 and 6. The differences between numbers don’t change if all the numbers are increased or decreased by the same amount, so we can start by assuming that the first number is 1, in which case the largest number (to give a maximum difference of 6) must be 7.

1 __ __ 7

We have to fill in the two middle numbers so that the six differences are the numbers 1 to 6. There are just two ways to do this:

1 2 5 7 or 1 3 6 7

The total number of prizes won in the first case is 1+2+5+7 = 15, which can be raised to 27 if we add 3 prizes to each girl’s total, making their totals 4, 5, 8 and 10. The total of the second set is, however, 17, which cannot be raised to 27 by increasing each number by the same amount. Therefore 4, 5, 8, 10 are their numbers of prizes, and Bernice, who won 1 more than another girl, won 5 prizes.

32. Solution

In a way they are all right! Rivers make up part of the Spain–Portugal border. A mathematician called Hugo Steinhaus pointed out in 1954 that the length of a river depends on the size of the ruler that you use to measure it. In particular, he pointed out, if you use a small ruler and trace its banks in and out of every little creek and inlet, every twist and turn, then you will end up with a total length far in excess of the length calculated from a map, or given in geography books.

So when the Spanish claimed years ago (yes, they really did) that their boundary with Portugal was 987 km, they were probably correct, by the scale they used to measure it, and the Portuguese may well have been correct when they claimed about the same time that it was 1,214 km. Pamela is most likely to be wrong, because if she really tried to get down on her hands and knees and measured the length of the border with a school ruler she would end up with a total length far more than 2,000 km.

33. Solution:

No, it is not possible! However you place the 31 dominoes, there will always be two squares left that are not adjacent to each other, as the diagram overleaf illustrates. Trial and error will also show that the two squares that remain are always both black squares. This is a clue to the explanation. When a domino is placed on the board it covers two adjacent squares – one white square and one black square. However many dominoes you place, they will always cover equal numbers of black and white squares.

But the board with the corners missing has 32 black squares and only 30 white squares so, after placing 30 dominoes, two black squares must remain, and these can never be covered by the 31st domino.

34. Solution:

The simplest way to get the two columns to add to the same total is to invert the number 9, like this:

There are other ways, too – for example, pick up the 5 card and place it on top of the 7, so that both columns add to 19.

35. Solution:

Watson was wearing a white handkerchief. If he had been wearing the blue handkerchief, Mrs Hudson would immediately have known that hers was one of the two white handkerchiefs. The fact that she didn’t know means that she saw a white handkerchief on Watson’s head.

This type of puzzle illustrates the interesting principle that sometimes you can make useful deductions from apparently ‘no’ information. There is a famous example from a real Sherlock Holmes story called ‘Silver Blaze’. Inspector Gregory asks whether there is anything which Holmes wishes to draw to his attention.

‘To the curious incident of the dog in the night time,’ [replied Holmes]. ‘But the dog did nothing in the night time.’ ‘That was the curious incident,’ remarked Sherlock Holmes.

36. Solution:

Number the squares from 1 to 7, as in this illustration.

Then you need to make 6 slides and 9 jumps, a total of 15 moves. This is how it is done: 3–4; 5–3; 6–5; 4–6; 2–4; 1–2; 3–1; 5–3; 7–5; 6–7; 4–6; 2–4; 3–2; 5–3; 4–5.

37. Solution:

Mack’s strategy is to turn over the first piece of paper, then swap to the second piece. If the second number is higher than the first, he sticks. If it is lower, he swaps to the third.

If we call the three numbers L(lowest), M(iddle) and H(ighest), then there are six possible combinations that Mack could turn over, and in three of them he wins:

1. L M H (loses)

2. L H M (wins)

3. M H L (wins)

4. M L H (wins)

5. H L M (loses)

6. H M L (loses)

38. Solution:

Moving these three matches leaves four small squares and the large outer square.

39. Solution:

Yes. There is only one position in which the little hand is pointing in the correct direction. If it is quarter past the hour, the little hand should be quarter of the way between the hour marks, if it is twenty past the hand should be a third of the way, and so on. Since the hand is exactly midway between two of the hour symbols, it is half past the hour . . . which makes it 1:30.

40. Solution:

Jack saw that Jill’s face was dirty and assumed that his face was also dirty, so he went to wash it. Jill did not see anything on Jack’s face, which was in fact clean, and did not realise that there was anything on her face, so she did not go to wash it.

41. Solution:

Bob only needs to go to the other end and back once, and he only needs to make two connections!

If we call the four cables A, B, C, D at one end, and 1, 2, 3, 4 at the other, then Bob first connects A and B together. At the far end he labels the cables 1, 2, 3 and 4. He then tests until he finds a pair – let’s say 2 and 4 – which form a circuit. He now connects one of that pair to another unused cable – say 2 to 3 – and goes back to the other end of the tunnel, disconnects A and B and again looks for the pair which form a circuit. One of these will be A or B (the other end of 2), the other will be C or D (the other end of 3). Let’s suppose the pair is A and C. He now has all the information he needs. In this example:

A is 2, so B is 4, C is 3 and D must be 1.

Of course this works whichever combinations turn out to form the circuits.

42. Solution:

Place your fingers on the two heads in the bottom row and slide them round to a position exactly above the two tails in the top row. Then, keeping your fingers firmly on the original two coins, push them down, so that the two columns of coins move down far enough to make a square again.

43. Solution:

If the sons divided the land like this, they would each receive an equal-sized L-shaped piece of land.

44. Solution:

Josie called ‘eight’, making the running total up to 22. Whatever Becky calls on her turn, Josie always chooses a number that makes their combined total up to 11.

When they have both called out nine numbers the total will be 99, and Becky will now be forced to take the total up to at least 100.

45. Solution:

Bill is wrong, he is just as likely to pull out a single tea bag as a double. To empty a jar of each of its 40 tea bags, Bill must first pull out the double tea bag (once) and then later pull out the remaining single tea bag (once), so he will pull out 20 doubles and 20 singles. If for any reason he is more likely to pull out doubles at the start of the jar, this means that at the end of the jar he is more likely to pull out the singles that he has created. Since he can’t remember how full the jar was, the answer must be 50-50.

46. Solution:

Tom was planning to pick up the last but one and pour its contents into the second tumbler, then replace the empty glass in its original position.

47. Solution:

He should have broken all three links in one piece of chain, leaving four complete pieces of chain, which he should then have joined together with the three broken links.

48. Solution:

The letters are the first letters of the numbers one to eight, so the next letter should be n for nine.

49. Solution:

Although he was actually away for only 30 days, he experienced 31 nights: one for each day, plus one for his eastward circumnavigation of the world. When you travel eastwards the sun goes overhead faster so the days and nights get shorter. In Jules Verne’s Around the World in 80 Days, the plot hinges on the fact that the hero Phileas Fogg experiences 81 ‘days’ on his journey, even though he is away for only 80 days.

What if you were to fly west? Your days would now take longer. Since the circumference of the Earth is about 24,000 miles, if you flew westwards around the equator at about 1,000 mph your nights could last an eternity. Any faster and the sun would start setting in the east!

50. Solution:

This is the new pool. The trees remain in place at the middle points of each side, instead of being at the corners.

51. Solution:

The object is a cube, seen from the three directions shown in the second illustration.

52. Solution:

In order, the answers are: a coffin; your breath; your name; fire; and footsteps.

53. Solution:

Six! To realise why rearranging the pieces will not save any cuts, think of the small cube in the middle of the cube on the table. It has six separate faces, each of which has to be made by a separate saw cut.

54. Solution:

This is the simplest and prettiest solution:

(Mind you, you can probably imagine the arguments about who got the more crumbly middle pieces!)

55. Solution:

The use of a ruler or any straight edge will tell you that the line on the right is actually the continuation of the top line, even though the one on the left might appear to be correct on first impression.

56. Solution:

Four people (including Brian) went to the meeting.

159 is the product of the number of people Brian invited and the number of pages in each document. The only two numbers which, when multiplied together, give 159 are 3 and 53, which are known as its prime factors. So the number of people given documents was either 159, 53 or 3. It is extremely unlikely that Stephen would have had time to distribute 53 documents to 53 people (let alone 159) in the few minutes available, so he must have distributed three copies of the document, and each copy had 53 pages in it. Brian kept the original, making four documents for the four at the meeting.

57. Solution:

FOUR into FIVE: can be done in seven steps, for example:

FOUR POUR POUT PORT PORE FORE FIRE FIVE

Using common words ONE into TWO can be done in ten at most:

ONE ODE ODD ADD AID LID LIP TIP TOP TOO TWO

58. Solution:

More than half fill the vase and tip it so that the water runs out until it reaches from the lip of the vase to the edge of the base, as in the first figure.

Now the vase is half full. Tilt the vase upright and mark the level with the pen to show how far up the side the water comes. Finally tip out the water a little at a time until, when tilted, the water goes from the mark on the side to the bottom edge.

59. Solution

The proportion of males on the island was still half! The King’s policy had no effect at all on the ratio of boy babies to girl babies. Telling parents to stop having children when their first girl is born is the same as asking them to stop tossing a coin when the first head appears. On average, there will still be 50% boys and 50% girls. (If you don’t believe us, toss a coin to simulate the King’s rule.)

60. Solution:

The ball costs 50p (and the bat costs £10.50).

61. Solution:

The trick is very simple: you move onto the long diagonal, from 1 to 2. Your opponent then has to move off the diagonal, allowing you at your second turn to move back onto the diagonal . . . and all the time you are getting nearer to END, which you will be the one to occupy, because it is on the long diagonal.

62. Solution

They do not move closer together or further apart.

To see why this is so, imagine that there is a block of wood between them. The right-hand bolt is being screwed as if it were screwing into the block, so the wooden block is being pulled towards the right.

The left-hand bolt, however, is being turned as if it were being unscrewed from the same wooden block. In other words, it is releasing it to allow it to move to the right. So the two bolts do not move relative to each other (apart from their separate rotations), and only the imaginary block moves. Take away the imaginary block . . .

63. Solution:

The reason is simply that the birds spend much longer away from the nest (over 14 minutes) than they do in the nest within sight of Edward Spinks (under 30 seconds), so the chance is more than 29:1 that the first sighting will be a bird flying back to the nest. Since he has only been watching for a week, it is no surprise that so far Mr Spinks’ first sightings have always been arrivals. (If Mr Spinks started watching before dawn, of course, he would see a bird leave first.)

64. Solution:

Since all of the statements contradict each other, at least three of them must be false. In fact number 3 is true and the others are false.

65. Solution:

Their conversation included two sentences that include every letter of the alphabet (known as pangrams). These were:

‘Pack my box with five dozen liquor jugs’, which does it in 32 letters.

‘Quick Baz, get my woven flax jodhpurs’, which does it in 30, and is one of the shortest coherent sentences in the English language that uses all 26 letters.

‘A quick brown fox jumped over that lazy dog’, would contain every letter if it said ‘jumps’ instead of ‘jumped’.

66. Solution:

You cannot float across on top of the planks because then you would certainly be in danger of falling into the water. Instead, arrange the planks at one corner of the moat, as in the diagram.

With this arrangement, two planks as short as 2.9 metres can be used to cross a 3-metre moat, allowing for a little overlap where the planks rest on each other or on the bank.

67. Solution:

You cannot write ‘one’ because then there are two. But if you write two, then there is only one! Lots of words would fit, for example ‘small’, ‘minimal’ or ‘finite’.

68. Solution:

The difference between 30-all and deuce in tennis is purely psychological: the two scores are exactly equivalent, since to win a game in either situation one player needs to win two points in a row. Whenever a game reaches 30-all the umpire could therefore always call deuce.

Love all, however, is not the same as 30-all. In all serious tennis matches the server is more likely to win a point than the receiver, but it takes a bit of maths (or a lot of experience of playing tennis) to prove that it is more of an advantage to the server (Tuttle) if the score is love all than if it is 30-all. (In fact, if the server’s chance of winning any point is always 60%, then his chance of winning the game from love-all is about 74%, while his chance of winning from 30-all falls to 69%.)

69. Solution:

The Dippleton v Bimpley game has not been played yet – in fact it is the only game left to be played in the league.

When the league finishes, there will have been six games (AB, AC AD BC BD and CD). Each game produces two points (either two for the winner, or one for each team in a draw) so the total number of points in the league so far (4+3+2+1=10) is double the number of games that have been played (5).

Chinfield has four points, which it can only have got by winning two matches 1-0 and losing one 1-0; or by winning once and drawing twice. Either way, Chinfield has played three matches. And because Argleford has conceded as many goals as they scored, they much also have played three matches to have three points (either winning one, losing one and drawing one; or by drawing all three matches).

So Argleford and Chinfield have both played all their three matches, which means the sixth game that hasn’t been played yet must by Dippleton v. Bimpley.

(With a bit more work, you can find what all the results have been so far: AvB 0-0; AvC 1-1; AvD 1-1; CvB 1-0; CvD 0-0.)

70. Solution:

It was 1 January. In any year, Christmas Day is 358 (or 359) days later than New Year’s Day, and since neither 358 nor 359 is exactly divisible by seven, the day of the week is always different on these two dates. Damien’s birthday is on 31 December, which explains how his birthday can appear to leap by so much. And Janet (who may not be present at this gathering) lives somewhere in the southern hemisphere.

71. Solution:

Thirty-four is the inevitable number of breaks.

Barry starts with one piece of chocolate. Every time he makes a break, he creates one extra piece of chocolate. This is true whether he breaks a small piece into two, or a large rectangle into two parts. Since he ends up with 35 pieces, he must have made 34 breaks.

72. Solution:

To your left. East is to the right when you travel North, but if you are travelling South (which you must be doing if you are leaving the North Pole), East is to the left.

73. Solution:

Potatoes. This has nothing to do with vegetables, though there may be some exotic explanations people try to use about how easy it is to ‘scatter’ different objects. The explanation is very simple. The words in the sentence have increasing numbers of letters, 1, 2, 3, 4, etc. The gap is a word with 8 letters, of which only potatoes fits from the five supplied.

74. Solution

The two containers are simply cylinders, so you can half fill either one exactly by tilting it and filling it so that the water just reaches the bottom edge and the top edge of the cylinder, as in this sketch.

So you can fill each container once, then half fill each once, to transfer a total of 3 litres + 7 litres + 1.5 litres + 3.5 litres = 15 litres in four moves.

75. Solution

The surprising answer is that the ball would rebound at 120 mph. Its speed relative to the train before impact is 20 + 50 = 70 mph, so the ball will rebound at 70 mph relative to the train. But the train itself is travelling at 50 mph, so relative to Craig – and the ground he is standing on – the ball is travelling at a phenomenal 70 + 50 mph. While superballs this bouncy don’t exist, this huge amount of acceleration can be observed even by dropping an ordinary bouncy ball onto something moving upwards (like a tennis racket).

76. Solution:

The number opposite 4 is in fact 6. Most people say 3, but this is not an ordinary die because it has two 6s and no 5. To see why, look at the 2 and 6 in the first two views. In one the 2 slopes to the left, in the other it slopes to the right, so there must be different faces involved. If it is the same 6 in both views then 3 and 4 end up on the same face, so the two 6s must be different. Here is what the whole cube looks like unfolded:

77. Solution:

There are still 100 olives in each dish, so there must be exactly the same number of black olives in the green dish as there are green olives in the black dish. If this was not true then you would have created olives out of nowhere.

20 green olives were transferred to the black dish. ‘N’ of those green olives were then trasnferred back to the green dish, leaving ‘20 minus N’ green olives in the black dish. Exactly the same number of black olives, ‘20 minus N’, went to the green dish.

	Black dish	Green dish
Start	100 Black	100 Green
After swap 1	100 B + 20 G	80 G
After swap 2	100 B – 3 B = 97 B 20 G – 17 G = 3 G	= 3 B 80 G + 17 G = 97 G

The best way to prove it to yourself is to take a real bowl of olives and try it out.

78. Solution:

The plots were joined like this. Mr Brown’s path was made to wind in and out of the others, and is more than twice as long and twice as expensive.

79. Solution:

One thing James Bond will not die of is freezing. All that a fridge does is transfer heat from inside to outside, and it uses a lot of electricity to do this. Since the fridge’s outside is in the room, the room will in fact slowly warm up. So Bond is at risk of dying of dehydration, suffocation or overheating!

80. Solution:

On the blocks were the letters T W E L V E O N E, which, when rearranged, form the words E L E V E N T W O.

81. Solution:

The label is now on the outside, at the front.

82. Solution:

The writer of the message has a special number and he has, of course, added this number to each of the numbers in the message, since that is the rule of the club. If his number was bigger than 4, it would fall outside the range of numbers that he says members have; if his number was smaller than 4, then it would be impossible for there to be other members with smaller numbers than his. So his number is 4, which means there are 6 members in the club, all with numbers between 2 and 5, two of whom have a smaller number than him, and one with a larger number. All of the numbers add up to 23. The only possible combination of six numbers that fits is 3, 3, 4, 4, 4, 5.

83. Solution:

Arguably there was nothing wrong with her reasoning. This is a version of a well known paradox called ‘The unexpected hanging’, which has been discussed endlessly by many authors. Here is one way in which you might explain the paradox: it is illogical to tell someone when they will have a surprise. The story could be cut down to Andrew saying: ‘Jenny, I’m going to give you a surprise this evening. I will take you for a meal at the Tandoori.’ She might logically say ‘that’s not a surprise because you’ve told me’ so that when he then actually does take her for the meal, she is slightly surprised.

84. Solution:

Here is the simplest way to place the four coins.

Below is how the five coins are placed. One coin has to be much smaller than the other four. The leaning coins are unlikely to stay in place by themselves – you will have to hold them in place – but that isn’t disallowed by the conditions of the puzzle.

85. Solution:

The cheetah won again! They ran at the same speeds, so that when the leopard had run 90 paces, the cheetah had run 100 and caught up with the leopard. That meant they still had 10 more paces to run to the stream, by which time the cheetah would have nosed ahead.

86. Solution:

The volumes that he does not move will stay in their original order on the shelf. Therefore they must already be in the correct order. So all he has to do is to leave as many volumes as possible which are already in ascending order from left to right. The most he can pick out is four, and he can do this in four ways: 3 4 9 10; 3 5 9 10; 3 4 6 8; 3 5 6 8.

It makes no difference which sequence he chooses. He still has to remove and replace each of the remaining six volumes.

(There is a far harder question, which does not have a simple solution. Since it takes a real effort to push several heavy volumes along the shelf in order to replace the volume he has removed, does the order in which he removes and replaces these six volumes make any difference? The short answer is yes.)

87. Solution:

This is the object, in perspective.

88. Solution:

Yes. The fastest they can do it is seven minutes. Mike puts polish on the shoes of two of the men while Merv puts it on the third man’s shoes. Merv then shines Mike’s two men while Mike shines Merv’s.

Mike: Polish Mr A; Polish Mr B; Shine Mr C (5 minutes)

Merv: Polish Mr C; Shine Mr A; Shine Mr B (7 minutes)

89. Solution:

Old MacDonald’s collection of animals consisted of:

PIG (piglet)

KID (goat)

OX (oxen)

SHEEP (sheep)

It is slightly unusual to have an ox on the same farm as a pig, sheep and kid, but that’s the way it was. There are some more obscure alternatives, such as DEER.

90. Solution:

He turned right instead (naturally, what else could he do?) like this:

No one asked about how long his curious journey might have been, but the answer would look something like the above. He cycles along the sides of a polygon that can be drawn round the ground, and at each corner, he pirouettes to the right.

91. Solution

Mrs Smith will overtake all the buses which were occupying positions on her route at the moment she set out, except for those that leave her route during the one hour she is driving along it. She will pass coming in the other direction the same original number of buses plus all the buses which have entered her route during the same one hour. The original number of buses on the route must therefore be 12, with 4 buses leaving and 4 buses joining during her one hour drive, which means that the buses on average leave every 15 minutes.

92. Solution:

The three longest common words that use letters out of QWERTYUIOP are PROPRIETOR, PERPETUITY and (surprisingly) TYPEWRITER. All three of these words appeared in the puzzle.

93. Solution:

Charlie Snod did it.

Suppose the first statement is false. That would mean that ‘this statement and the one after the first true statement are both true’, which is a contradiction since we said the first was false. So the first statement must be true. This means statement 2 is false, which means statement 3 is true. So (from 2 being false) statement 5 is false. If 5 is false then less than half of the statements are true, and the only way for this to happen is if statements 4 and 6 are also false. So from statement 4, Charlie Snod must be guilty (and Henry van Eyck is not a Dutchman).

94. Solution:

The traditional answer to this is that, since the secretary has established that the third letter is T, there is no reason for him or her to know what it stands for.

However, it is clear from their conversation that they have a bad telephone line. T for Tummy sounds very like D for Dummy, so it is very natural for the secretary to ask the caller to repeat the T (although a better response would have been ‘Sorry, is that T for Tango?’).

95. Solution

Any line through X will divide the bottom row of three buns fairly, and any line through Y will divide the top two fairly, so the line XY is one solution to their problem. Similarly, any line through A bisects the left-hand bun and any line through B (the midpoint of the group) bisects the block of four buns on the right, so the other solution is the line AB (see overleaf).

96. Solution:

Gary had heard this trick before and knew that the secret was to put a mat exactly in the middle of the table. Then wherever Marian put a mat, he would put one symmetrically opposite, so that when it became impossible to place a mat it would be Marian who reached this situation first. Unfortunately, in practice it is impossible to place a beer mat exactly at the centre of a large rectangular table without the use of a ruler or a length of string, so if you try this trick, make the bet a small one!

97. Solution:

It can be done in five moves, by inverting (for example) 1 2 3, 4 5 6, 7 8 9, 8 9 10, and finally 8, 9 and 11. Glasses 8 and 9, having each been inverted three times, are finally the correct way up, like the rest.

There are, of course, many other combinations you could use to achieve it in five moves.

98. Solution:

You have to turn over two cards – the lolly card and the vanilla card. If the lolly card doesn’t have chocolate on the other side, then Alf is wrong. And if the vanilla card has a lolly on the other side, then he is also wrong. But it doesn’t matter what is on the other side of the other two cards: there could be any ice cream on the back of a chocolate card, since what Alf said did not exclude this possibility!

99. Solution:

Charles’ numbers were not random. As he wrote down his numbers, he made sure that each digit in his first number when added to each digit of his father’s first number always added to 9, and the same for the second numbers.

Dad’s first number	7 2 5 8 3 9 1
	+
Charles’ first number	2 7 4 1 6 0 8 = 9 9 9 9 9 9 9

That means that the first four numbers would always be equal to 9,999,999 + 9,999,999 = 19,999,998. Then all he had to do to add the five numbers was take dad’s third number, add 20,000,000 and take away two – a very simple calculation, which can be done by putting a 2 in front of dad’s third number and subtracting two from the end of it.

100. Solution:

The right way round. The easiest way to demonstrate this to yourself is to write on a piece of clear plastic, and look through it at a mirror.

101. Solution:

In order to survive, Sybil needs to run fast enough to remain stationary relative to the wheels. The centre of the wheel is moving forward at 10 mph, but the top of the caterpillar track is moving at 20 mph (while the bottom of the track in contact with the ground is stationary).

So Sybil needs to run at 10 mph towards the back of the tank, in order to be moving forwards at the same speed as the tank. If she can manage more than 10 mph then she will move steadily towards the back of the moving tank and eventually fall off.

102. Solution:

Adam and Ben get the same amount of crust. There is a neat way to prove this. Imagine that the ‘vertical’ cut is moved to the right (see the dotted line). The length of Adam’s top-left crust has increased by the portion indicated by the heavy lines, but his other piece bottom-right has shortened by the same amount. So moving the cut to the right has no effect on Adam’s total crust. The same will be true if you move the horizontal cut up or down. So, in order to compare how much crust Adam and Ben get, we can move mum’s cut so that the two lines pass through the centre of the circle. It then becomes obvious that Adam and Ben got the same amound of crust as each other (two quarters, or half, each).

103. Solution:

This is how it is done. It is not easy to visualise, so we have shown the original cube, and also exploded it to show the three separate (identical) pieces.

This solution comes from the fact that when you look at a cube along any of its long diagonals, its outline is the shape of a hexagon. The three cuts are across the opposite sides of the hexagon.

104. Solution

The most valuable envelope contains £37. In order to make every combination from £1 to £63, Harry has to put the following amounts in the first six envelopes:

£1   £2   £4   £8   £16   £32

That leaves £37 to go into the seventh envelope. To make values between £64 and £100, he hands over £37 envelopes plus the combinations of the other six envelopes that make £27, £28, £29... all the way to £63.

105. Solution:

(c.) Exactly 50-50. Since you have 4 and I have 3 coins, it is certain that you have either tossed more heads than I have, or more tails. (You can’t have done both.) Since these are the only two options, and since there is nothing that favours a coin to end as tails rather than heads, the chance that you toss more heads than me must be 50-50 – a simple answer to what at first sight looks like it might be a difficult probability question.