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Saturn | Mercury | Silver | Emerald |

Lead | Ariane | Tungsten | Sea |

Soyuz | Lime | Apple | Sand |

Iron | Bottle | Step | Long March |

The four groups and their connections are:

Saturn, Ariane, Soyuz and Long March - Rockets

Mercury, Lead, Tungsten and Iron - Metals

Silver, Lime, Sand and Step - can all be preceded by Quick

Emerald, Sea, Apple and Bottle - Shades of Green

]]>Saturn | Mercury | Silver | Emerald |

Lead | Ariane | Tungsten | Sea |

Soyuz | Lime | Apple | Sand |

Iron | Bottle | Step | Long March |

As usual you can post your suggested answers as a comment on this website, reply to the post on Facebook, or retweet or reply on Twitter @quizmastershop.

Answer at 9.00 on Monday

]]>In this case you have six weights that are visibly identical. Three are light and three are heavy. All three light weights weigh the same, and all three heavy weights weigh the same.

Can you you determine which three weights are heavy using three weighings on a balance scales?

The table below shows the four different combinations, with an uppercase letter indicating a heavy weight, and a lowercase letter indicating a light weight.

1 | 2 | 3 | 4 |

a v B | a v B | a v b | a v b |

c v D | c v d | b v c | b v C |

e v F | d v E | c v D | D v E |

Which gives you the three heavy (and of course light) weights.

]]>In this case you have six weights that are visibly identical. Three are light and three are heavy. All three light weights weigh the same, and all three heavy weights weigh the same.

Can you you determine which three weights are heavy using three weighings on a balance scales?

As usual you can post your suggested answers as a comment on this website, reply to the post on Facebook, or retweet or reply on Twitter @quizmastershop.

Answer at 9.00 on Monday

]]>

A classic example of this is the dispute over the date of Twelfth Night – and we will be blogging about this next January. The correct answer is 5th January, but many, many people are insistent that it’s on the 6th. We’ve actually left this question out of our quizzes because it is so contentious.

*If you would like to receive regular Quiz Tips to your inbox click here*

The easiest and least awkward way to deal with this is to make it clear before you start the quiz that the answer on the sheet is the answer, even if it is wrong. Do not accept a web page that “shows” the correct answer, as you have no knowledge of the authenticity of the page. Stick to your guns.

We would recommend reading the questions and answers before the quiz to make sure that you are comfortable with both questions and answers.

]]>When he's put all the balls in position he notices that there is one missing. he then finds this ball on the floor.

What colour is this ball?

There are 15 Reds on a snooker table, each worth one point, plus the Yellow (2), Green (3), Brown (4), Blue (5), Pink (6) and Black (7), giving a total of 42 points.

The value of the balls in the second pocket is one higher than the first. The value of the balls in the third pocket is two higher than the first. And so on until the value of the balls in the sixth pocket is five higher than the first.

$$ 1 + 2 + 3 + 4 + 5 = 15 $$

So the value of the balls in the six pockets is

$$ 6x + 15 $$

where x is the value of the balls in the first pocket.

with successive values of x from zero this equates to 15, 21, 27, 33, 39 and 45. As 45 is higher than the total value of balls available (42) it must be 39. This leaves three points missing, and thus the missing ball is green.

The total points in each successive pocket is 4, 5, 6, 7, 8 and 9.

]]>When he's put all the balls in position he notices that there is one missing. he then finds this ball on the floor.

What colour is this ball?

As usual you can post your suggested answers as a comment on this website, reply to the post on Facebook, or retweet or reply on Twitter @quizmastershop.

Answer at 9.00 on Monday

]]>Can you you determine which weight in each coloured pair is the heavy weight using two weighings on a balance scales?

There are a few methods, but this one is quite elegant.

Firstly weigh a white weight and a blue weight against the other white weight and a red weight.

If this balances then the heavy white weight is with a light weight and vice versa. For the second weighing simply remove both white weights. This will reveal whether the blue or red weight is heavy, and the others can all be inferred from this.

If one side goes down then the light and heavy white weights are revealed, and red and blue weights are both light or both heavy. For the second weighing remove the white weights and weight the red weight against the other blue weight. This will reveal whether the original red and blue weights were both light or both heavy, and the others can be inferred from this.

]]>Can you you determine which weight in each coloured pair is the heavy weight using two weighings on a balance scales?

Answer at 9.00 on Monday

]]>For example, if you have the word STOP you could make

SOP by removing the T

STOPS by adding an S

STEP by changing the O to an E

POST by making an anagram

And of course there are many, many others.

Firstly, can you change LATE to REGAL in three steps.

LATE goes to LATER by adding an R

LATER goes to LAGER by changing the T to a G

LAGER goes to REGAL by anagramming

]]>For example, if you have the word STOP you could make

SOP by removing the T

STOPS by adding an S

STEP by changing the O to an E

POST by making an anagram

And of course there are many, many others.

Firstly, can you change LATE to REGAL in three steps.

And then can you devise any cunning word chains of your own and post them for others to solve. The best will get a mention on our social media.

As usual you can post your suggested answers and puzzles as a comment on this website, reply to the post on Facebook, or retweet or reply on Twitter @quizmastershop.

Answer at 9.00 on Monday

]]>Firstly consider a three-inch cube that is completely painted on all six faces. This cube is sawed into one-inch cubes. Some of the resulting one-inch cubes will have paint on three faces, some on two faces, some on one face, and some with no paint at all.

How many of each are there?

If this was easy enough, consider a four-inch cube. This is sawed into one-inch cubes. How many of these have paint on three, two, one and no faces?

You can keep going with this with bigger and bigger cubes, but perhaps it would be more fun to work out the general formula for each of the four types of one-inch cubes for an n-inch cube.

With the three-inch cube there will be 27 one-inch cubes, made up as follows:

The eight on the corners will have three painted faces.

The twelve on each edge will have two painted faces.

The six in the middle of the faces will have one painted face.

And the one in the middle will have no painted faces.

Moving onto the four-inch cube, there will be 64 one-inch cubes, made up as follows:

The eight on the corners will have three painted faces.

24, that is two on each edge, will have two painted faces.

24, that is four in the middle of the six faces will have one painted face.

And the eight in the cube in the middle will have no painted faces.

So, a generic solution . . .

There will always be eight corners so there will always be eight cubes with three painted faces.

There are twelve edges with (n-2) cubes on them, so there are 12(n-2) cubes with two painted faces.

There are six faces with (n-2)^2 cubes on them, so there are 6(n-2)^2 cubes with on painted face.

And there is a cube in the middle of the cube with (n-2)^3 one-inch cubes in it, so there are (n-2)^3 cubes with no painted faces.

]]>Firstly consider a three-inch cube that is completely painted on all six faces. This cube is sawed into one-inch cubes. Some of the resulting one-inch cubes will have paint on three faces, some on two faces, some on one face, and some with no paint at all.

How many of each are there?

If this was easy enough, consider a four-inch cube. This is sawed into one-inch cubes. How many of these have paint on three, two, one and no faces?

You can keep going with this with bigger and bigger cubes, but perhaps it would be more fun to work out the general formula for each of the four types of one-inch cubes for an n-inch cube.

Answer at 9.00 on Monday

]]>The first Grand Slam was won in 1908 and the most recent in 2018, but why were none won between 1932 and 1946 inclusive?

Simply, between those years France was excluded from the competition, and, as a Grand Slam requires wins against four other teams, there could be no Grand Slams.

]]>The first Grand Slam was won in 1908 and the most recent in 2018, but why were none won between 1932 and 1946 inclusive?

Answer at 9.00 on Monday

]]>Naturally this provoked some discussion and we all looked at the results with interest. Sure enough, Liverpool duly won.

A week later another email arrived with a prediction, and this too turned out to be correct. In fact our colleague received eight successive weekly emails, and all were correct.

By now we were all looking forward to the next email with feverish anticipation. However, the next one was different; same subject line, but the text read For this week's prediction, please send £100 to the following bank account.

However, were being scammed!

What is the scam?

We are not the only people receiving these emails. In fact a very large number of people are getting them.

All the scammers do is acquire, say, one million email addresses. The first week half get a prediction of Liverpool winning and half get a prediction of them not winning. Of course, half get a correct prediction, and only those get a second email. The half million remaining people are again split into two groups, with each half getting different predictions.

This continues with successive emails going only to those who have received correct predictions.

After eight weeks there are still about 4,000 people who have received eight correct predictions, but there is no guarantee that the ninth will be correct.

Would you have fallen for it?

]]>Naturally this provoked some discussion and we all looked at the results with interest. Sure enough, Liverpool duly won.

A week later another email arrived with a prediction, and this too turned out to be correct. In fact our colleague received eight successive weekly emails, and all were correct.

By now we were all looking forward to the next email with feverish anticipation. However, the next one was different; same subject line, but the text read For this week's prediction, please send £100 to the following bank account.

However, were being scammed!

What is the scam?

Answer at 9.00 on Monday

]]>Police research has shown that in these cases an eyewitness is correct 80% of the time. It is also known that 85% of the pupils at the school wear jumpers and not blazers, which has increased the chances of finding the culprit.

What is the probability of the thief wearing a blazer?

Let us assume that there are 100 pupils at the school, 85 of whom wear jumpers and 15 wear blazers, and they all steal something once.

For the 85 jumper-wearing thieves, the eyewitness will report that they were wearing a jumper 68 times (80% of 85) and will report they were wearing a blazer 17 times (20% of 85).

Likewise, for the 15 blazer-wearing thieves, the eyewitness will report that they were wearing a jumper 3 times (20% of 15) and will report they were wearing a blazer 12 times (80% of 15).

From this we can see that the eyewitness will report a blazer wearer 29 times, and that on 12 of those occasions the thief will be wearing a blazer. So there is a 12/29 chance or 41% that the thief was wearing a blazer.

]]>Police research has shown that in these cases an eyewitness is correct 80% of the time. It is also known that 85% of the pupils at the school wear jumpers and not blazers, which has increased the chances of finding the culprit.

What is the probability of the thief wearing a blazer?

Think carefully about this - it is harder than it appears.

Answer at 9.00 on Monday

]]>All you have to do is find a 4-digit number where the first digit is the number of zeroes in the number, the second digit is the number of ones, the third is the number of twos, and the fourth is the number of threes.

There are two of them to find.

And if you've found them, you can keep going and find a 5-digit number where the first digit is the number of zeroes, etc.

There are also one 7-digit number, one 8-digit number, one 9-digit number and one 10-digit number that you can find.

The numbers you are looking for are

1,210 and 2,020

21,200

3,211,000

42,101,000

521,001,000

and 6,210001,000

]]>All you have to do is find a 4-digit number where the first digit is the number of zeroes in the number, the second digit is the number of ones, the third is the number of twos, and the fourth is the number of threes.

There are two of them to find.

And if you've found them, you can keep going and find a 5-digit number where the first digit is the number of zeroes, etc.

There are also one 7-digit number, one 8-digit number, one 9-digit number and one 10-digit number that you can find.

Answer at 9.00 on Monday

]]>The men are shown five hats - three black and two white. The men are blindfolded, each has a hat placed on his head, they are tied in position so they can't move, and the blindfolds are removed. The first man can see both others. The second can only see the third. And the third can't see either of the other men. The unused hats have been hidden.

They are told that if any of them can say what colour hat they are wearing, they can all go free. But one incorrect answer means they are all dead men.

The first man says that he does not know.

The second man says that he does not know either.

The third man says that he does know.

How does he know, and what colour is his hat?

If the first man could see two white hats he would know that he was wearing a black hat. As he does not know, he must be able to see one or two black hats.

If the second man could see a white hat he would know that the black hat visible to the first man must be on his head. As he does not know, he must be looking at a black hat.

So the third man is wearing a black hat.

]]>The men are shown five hats - three black and two white. The men are blindfolded, each has a hat placed on his head, they are tied in position so they can't move, and the blindfolds are removed. The first man can see both others. The second can only see the third. And the third can't see either of the other men. The unused hats have been hidden.

They are told that if any of them can say what colour hat they are wearing, they can all go free. But one incorrect answer means they are all dead men.

The first man says that he does not know.

The second man says that he does not know either.

The third man says that he does know.

How does he know, and what colour is his hat?

Answer at 9.00 on Monday

]]>There used to be a festival called Christmastide which lasted twelve days and was followed by Epiphany, which celebrates the Wise Men visiting the baby Jesus. The first day of Christmas is Christmas Day and so the twelfth day is 5th January. So Twelfth Night is 5th January.

Somehow Twelfth Night and Epiphany have been conflated to be the same thing, but they are not.

]]>Over the twelve days of Christmas, who many gifts, in total, did my true love send to me?

That's it! Count the Partridge in a Pear Tree as one gift.

On the first day I receive one gift - a Partridge in a Pear Tree. On the second day I receive three gifts - two Turtle Doves and a Partridge in a Pear Tree. So this isn't 1+2+3 . . .+12

I receive twelve Partridges, one each day (12x1)

And I receive 22 Turtle Doves, two on each of eleven days (11x2)

And so on, until I receive twelve Drummers Drumming, all on the last day (1x12)

So we need to add 12, 22, 30, 36, 40, 42, 42, 40, 36, 30, 22 and 12, which equals 364.

]]>Over the twelve days of Christmas, who many gifts, in total, did my true love send to me?

That's it! Count the Partridge in a Pear Tree as one gift.

Answer at 9.00 on New Year's Eve

]]>There are nine pictures which each represent something and have one-word answers. The nine answers are all connected, have a Christmas theme and are in the correct order.

All you have to do is work out what each picture represents, and work out the connection between them.

And yes, pictures one and three are the same picture - the one picture represents two of the things!

In order the nine things are:

- Rudolf Nureyev (Rudolph)
- Usain Bolt (Dasher)
- Rudolf Nureyev (Dancer)
- Lipizzaner Horse at the Spanish Riding School (Prancer)
- Fox (Vixen)
- de Havilland DH 106 Comet (Comet)
- Cupid (Cupid)
- Donner Kebab meat (Donner)
- Blitzen - a Rick Riordan character (Blitzen)

Of course these are the nine reindeer.

]]>There are nine pictures which each represent something and have one-word answers. The nine answers are all connected, have a Christmas theme and are in the correct order.

All you have to do is work out what each picture represents, and work out the connection between them.

And yes, pictures one and three are the same picture - the one picture represents two of the things!

Answer at 9.00 on Monday

]]>He told us that of the 100 members, 90 liked red cabbage, 80 liked parsnips and 70 liked sprouts, and that he had to plate up the correct vegetables for each member.

We sympathised with his difficulty, and out of idle curiosity asked if any members didn't eat any vegetables. He replied that, commendably, very few did. And then knowing we like a puzzle he added cryptically:

19 times as many members eat all three vegetables as those that eat none, and the only people who eat only one vegetable all eat red cabbage.

Well that spoiled the evening, as we could not leave without working out how many people had three vegetables, two vegetables and one vegetable, and which do they eat.

If the ten members who don't eat red cabbage all eat parsnips there must be 70 people who eat both. If the 30 who don't eat red cabbage and parsnips all eat sprouts there must be at least 40 members who eat all three vegetables. The number who eat all three can't be more than 70 - the number who eat sprouts.

So we need to find a number divisible by 19 between 40 and 70, and 57 is the only one. There are 57 members who eat all three vegetables and only three who eat none.

Of the remaining 40 some eat one vegetable (red cabbage) and some eat two.

Let a = the number who eat red cabbage and parsnips, b = the number who eat red cabbage and sprouts, c = the number that eat parsnips and sprouts, and d = the number who eat red cabbage only.

$$ a + b + c + d = 40$$

Red cabbage eaters

$$ 57 + a + b + d = 90 $$

$$ a + b + d = 33 $$

Parsnip eaters

$$ 57 + a + c = 80 $$

$$ a + c = 23 $$

Sprouts eaters

$$ 57 + b + c = 70 $$

$$ b + c = 13 $$

And from these simultaneous equations

a = 16, b = 6, c = 7 and d = 11.

So 57 members eat all three vegetables, 16 eat red cabbage and parsnips, six eat red cabbage and sprouts, 7 eat parsnips and sprouts, eleven eat only red cabbage, and three eat no vegetables at all.

]]>