# APR Teasers

On this page we will be adding some quiz questions that we post in our monthly client updates - every time we post a new teaser we will update the previous one with a solution, so look out for these each month.

**September 2021 **

*A standard 8x8 chessboard has each square marked with a different numeric value according to the following rule: for a given square, its value is the (arithmetic) mean of the other squares on the board from which a knight can reach the given square in one move. (Knights move in a standard "L"-shape.) For example, using standard algebraic chess notation, the value on the square a1 is equal to the mean of the values on squares b3 and c2.*

*If the square on the top left of the board (square a8) has a value of 21, what is the value of the square on the bottom right of the board (square h1)?*

As always, we will pay £50 to a charity of your choice as a prize to the first person providing the correct solution with their reasoning.

Please submit your entry to teasers@aprllp.com.

**Solution**

TBC

**August 2021 **

*At the Tokyo Olympics a gold medal weighed 556g, a silver medal weighed 550g and a bronze medal weighed 450g. *

*What was the total weight of all the medals awarded to athletes during the Olympics?*

As always, we will pay £50 to a charity of your choice as a prize to the first person providing the correct solution with their reasoning.

Please submit your entry to teasers@aprllp.com.

**Solution**

Answer = 1,240 kg

There were 781 gold medals, 770 silver medals and 850 bronze medals awarded.

*Source: https://olympics.com/tokyo-2020/olympic-games/en/results/all-sports/medalists.htm*

There were significantly more bronze medals awarded than golds and silvers because of sports like boxing awarding two bronze medals for each event.

The number of golds and silvers were not the same, partly because of the men’s high jump (where the gold medal was shared and there was no silver medal) and partly because of team sport events (football, rugby sevens, hockey etc) where the eventual gold medal winners used more players during the tournament, each of whom received a medal, than the eventual silver medal winners.

**July 2021 **

*A bag contains 100 marbles, and each marble is one of three different colours. *

*If you were to draw three marbles at random, the probability that you would get one of each colour is exactly 20 percent.*

*How many marbles of each colour are in the bag?*

As always, we will pay £50 to a charity of your choice as a prize to the first person providing the correct solution with their reasoning.

Please submit your entry to teasers@aprllp.com.

**Solution**

There are six different combinations of marbles which give one of each colour when drawn in that order:

x, y, z

x, z, y

y, x, z

y, z, x

z, x, y

z, y, x

The chance of the first one is x/100 (the probability the first marble is x-coloured) * y/99 (the probability the second marble is y-coloured, with only 99 balls left in the bag) * z/98 (the probability the third marble is z-coloured, with only 98 marbles left in the bag).

In fact, the probability will be the same for all six combinations: x*y*z / (100*99*98), with just the order of factors changing, which doesn’t matter.

So the total probability is 6*x*y*z / (100*99*98), and since we know this is 20%, we can simplify and get x*y*z = 32340.

So we need x+y+z=100 and x*y*z=32340, with x, y and z all whole numbers, and there are lots of routes to finding solutions here - whether you just check all combinations, or think about prime factors to limit the options you need to check. But however you get there, the only combination that fits is **21/35/44**.

**May 2021 **

Today, Anna is half the age of Chris.

8 years ago, Anna was half the age of Danielle.

9 years ago, Danielle was half the age of Ben.

When Anna was born, Ben was half as old as Anna is now.

How old will each person be when the sum of all of their ages is 65?

Please submit your entry to teasers@aprllp.com.

**Solution**

Anna = 12,

Danielle = 14

Ben = 17,

Chris = 22,

**April 2021 **

You have been given 31 dominoes, and a chessboard with 62 squares, where two diagonally opposite corners have been cut away. Each half of a domino is the same size as one of the chessboard squares. Can you find a way to lay the dominoes on the chessboard in such a way that each square is covered?

Please submit your entry to teasers@aprllp.com.

**Solution**

**March 2021 **

You're playing rock paper scissors with your friend. You've played together a lot, and you're quite good at guessing what move they will make: you get it right 50% of the time. You're playing a sudden death match: the first to win one round is the winner, and if you tie, you keep playing until someone wins. 'You can assume that an incorrect guess has an equal chance of resulting in either a draw or a loss'. What is the probability you win?

Please submit your entry to teasers@aprllp.com.

**Solution**

**February 2021**

Imagine you are a contestant on a game show. The host has randomly generated a number between 0 and 100 (this could be a whole number or a decimal) and asks you to guess a number lower than it. If you successfully guess a lower number, then you win the amount of your guess. If you guess a higher number, then you leave with nothing.

Assuming you want to maximise your expected earnings, what number should you guess?

Please submit your entry to teasers@aprllp.com. Look out for our next mailer to see if your reasoning holds up!

**Solution**

To maximise your expected winnings, you should guess 50.

Suppose you make a guess of X. Then the probability that the host’s number is higher than your guess is (100-X)/100. The product X*(100-X)/100 gives your expected earnings and this is maximised when X=50.

**January 2021**

Three students decide to play a round-robin pool tournament, where the winner stays on and the loser waits their turn to play again.

At the end of the evening, they count the number of games that each of them played:

Ash played 9,

Beth played 13,

Craig played 16.

Who lost the second game?

**Solution**

Ash

The total number of games played is (9 + 13 + 16) / 2 = 19.

If a player loses every game they play, then they play alternate games. So the minimum number of games it is possible to have played is 9: games 2, 4, 6, … , 18. As Ash only played 9 games, they must have played in each even numbered game and lost them all.

**November 2020**

A Nations cup group consists of 4 teams who all play each other twice. The teams finishing first and second after all the matches have been played qualify for the knockout stage of the competition. 3 points are awarded for a win and 1 for a draw (0 for a loss). In the case of a points tie a scale of other metrics are used to decide final positions such that no team has the same final position in the group.

What is:

a) the lowest number of points a team can get and still qualify for the knockout stage?

b) the highest number of points a team can get and not qualify for the knockout stage?

A. 4 & 11 B. 5 & 11 C. 6 & 11 D. 4 & 12 E. 5 & 12

**Solution**

Option D, 4 points & 12 points.

Scenario/reasoning for a) = 4:

One team wins all six of its matches. The three other teams all draw with each other in all of their matches, thereby all finishing on 4 points. One of these teams will finish second, based on head-to-head away goals etc.

Scenario/reasoning for b) = 12:

One team loses all six of its matches. The three other teams all beat each other once and lose to each other once, thereby all finishing on 12 points. One of these teams will finish third, based on head-to-head goal difference etc.

**October 2020**

Cari and Morgan play rock-paper-scissors 10 times. You know that Cari uses rock three times, scissors six times, and paper once. Morgan uses rock twice, scissors four times, and paper four times. You know there are no ties in all 10 games but the order of games is unknown.

Who wins and by how many games?

A. Morgan by 6 B. Morgan by 3 C. It is a draw D. Cari by 2 E. Cari by 4

**Solution**

Cari by 4. As there are no ties we know the six times Cari plays scissors must be against non scissor plays from Morgan. So Cari plays scissors against rock twice and paper four times putting her two games ahead. All remaining 4 games Morgan must now play scissors against Cari who will rock three times and paper once giving Cari another three wins and one loss. This leads to a total net win of 4 games by Cari.

Cari | Morgan | Winner |

R | S | C |

R | S | C |

R | S | C |

S | R | M |

S | R | M |

S | P | C |

S | P | C |

S | P | C |

S | P | C |

P | S | M |

**September 2020**

100 people are lined up waiting for a slice of pizza they will all be sharing. The first person takes 1% of the pizza. The second then takes person then takes 2% of what is remaining. The third takes 3% of what is remaining and this continues with the nth person always taking n% of the remaining pizza.

Which number in the line will get the largest slice of pizza?

A. 1 B. 3 C. 5 D. 10 E. 25

**Solution**

D. 10

*For each nth person the size of slice they receive can be shown as:*

*\(slice_{n} = \frac{n}{100}\ left_{n-1}\)*

*Where left _{n} is the amount of pizza left after the nth person takes their slice and can be shown as:*

*\(left_{n} = left_{n-1}\frac{100-n}{100}\ \)*

*Now consider the following ratio of the slice size received by the nth and (n-1)th person denoted ratio _{n}:*

*\(ratio_{n} = \frac{\frac{n}{100}\ left_{n}}{\frac{n-1}{100}\ left_{n-1}}\ \)*

*Substituting for left _{n} leaves:*

*\(ratio_{n} = \frac{\frac{n}{100}\ left_{n-1} \frac{100-n}{100}\ }{\frac{n-1}{100}\ left_{n-1}}\ \)*

*\(ratio_{n} = \frac{n(100-n)}{(n-1)100}\ \)*

*To maximise for n we now set ratio _{n }= 1 and solve:*

*\(100(n-1) = 100n - n^{2} \)*

*\(n^{2} = 100 \)*

*\(n = 10 \)*

**August 2020**

The APR partners have decided to have new office name plates printed, in order to further increase the respect in which they are held by APR staff and allow them to spend a few hours imagining that they will get back to their offices this year.

The company providing the name plates charges nothing for the plate itself, and instead charges according to the letters being printed. The following quotes have been obtained:

Roger Austin: £19

Gary Heslop: £15

Tim Nash: £11

How much will the company charge if Chris Bryce decides to purchase a name plate (from his own funds, of course!)?

A. £12 B. £14 C. £15 D. £17 E. £0. Chris should not have a name plate!

**Solution**

B. £14

*The sign maker charges £1 for consonants, £2 for a vowel and applies a double charge for capital letters.*

*This can be solved by constructing simultaneous equations using the information known about the 3 partner name plates.*

**July 2020**

Four people are on one side of a weak bridge at night and need to get across. The bridge will only support the weight of two people at a time and a torch is needed to cross safely, but the group only has a single torch. Each person takes a different amount of time to cross - 10 minutes, 5 minutes, 2 minutes and 1 minute respectively - and any pair crossing the bridge must travel at the speed of the slower person.

What is the shortest time in which all four can get from one side of the bridge to the other?

A. Not possible B. 48 mins C. 21 mins D. 19 mins E. 17 mins

**Solution**

E. 17 mins

*The key is that each pair can only cross at the speed of the slowest individual so finding a way for the two slowest members to only have cross once and at the same time will improve efficiency. Hence the most efficient way to get all individuals across is as follows (all individuals are denoted by the time it takes them to cross):*

*Send 2 + 1 across first. Total time = 2 mins**2 returns with the torch. Total time = 2+2 = 4 mins**Send 10 + 5 across . Total time = 4 + 10 = 14 mins**1 returns with the torch. Total time = 14 +1 = 15 mins**2 +1 cross again so all individuals are now on the correct side. Total time = 15 + 2 =***17 mins**.

**June 2020**

100 people walk down a corridor with 100 lights, all initially off. Each person flicks all switches that their number divides. E.g. first person flicks all switches, number 3 flicks 3,6,9,...,99 and the 37th person flicks just 37 and 74.

How many switches are now in the on position after all the people have walked through?

A. 0 B. 10 C. 25 D. 100 E. More information is required

**Solution**

B. 10

*Each numbered switch will be flipped a number of times equal to its number of factors. As all switches begin in the off position only those that are flipped an odd number of times will be in an on position. The only numbers that have an odd number of integer factors are square numbers so only square numbered switches will be on at the end. From 1-100 there are 10 square numbers so 10 switches will be in the on position once everyone has passed through.*

**May 2020**

A wholesale fruit factory produces only boxes of apples, boxes of oranges or boxes of mixed apples and oranges. These are labelled by an automated machine however an error has been introduced in the label machines code and now means that every box has been labelled incorrectly.

If you have one of each box which box should you remove a single piece of fruit from to allow you to relabel all boxes correctly.

A. Apples & Oranges B. Apples C. Oranges D. Any of them E. None of them

**Solution**

*A. Apples and Oranges*

*The key is that all boxes are incorrectly labelled.*

*This means that the box labelled apples and oranges cannot contain mixed fruit. Hence if you remove an apple from this box you know this box should be labelled apples and if you remove an orange it should be labelled oranges.*

*Relabelling this box means now have a box without a label (that you just removed to put on the previously labelled mixed fruit) and a box with a label that you know is incorrect. So, move the label from the box you know is incorrect (the only box with the original label applied by the broken machine) to the box without a label then place the mixed fruit label on the remaining unlabelled box and they will all be correctly labelled.*

**April 2020**

For a bar of chocolate comprising of 30 squares (a large 6 by 5 bar of your favourite brand say) how many times would you need to break the bar to make it into 30 individual squares?

A. 11 B.15 C. 29 D. 30 E. 31

**Solution**

*C.29*

*Any time the bar is broken it only creates a single extra piece hence for any chocolate bar of n squares it will always take n-1 breaks to break it down into it's individual squares.*

**March 2020**

Here is an ordinary cross. Drawing only 2 straight lines what is the greatest number of pieces you can divide the cross into?

A. 3 B.4 C. 5 D. 6 E. 8

**Solution**

*D. 6 - as demonstrated below.*

**February 2020**

What is the next number in the sequence?

1, 11, 21, 1211, 111221, …

A. 1122221 B. 1212221 C. 312211 D. 1331212 E. 22

**Solution**

*C. 312211*

*Each number is a description of the digits which make up the previous number.*

*So with a starting point of 1:*

*1*

*1 has one 1, so we get 11*

*11 has two 1s, so we get 21*

*21 has one 2 followed by one 1, so we get 1211*

*1211 has one 1 followed by one 2 followed by two 1s, so we get 111221*

*111221 has three 1s followed by two 2s followed by one 1, so we get 312211*

**January 2020**

The first six volumes of Encyclopaedia of Mathematics are arranged in order on my shelf from left to right. The six volumes contain mathematicians with surnames beginning; A-Ba, Be-Ca, Ce-Ei, Ek-Fe, Fee-Fi and, Fo-Fum respectively. If one ignores the covers, which of the following encyclopaedia entries could be on the page ‘next to’ the page with the entry for Einstein as they sit on my shelf?

A. Abel B. Bernoulli C. Cantor D. Euler E. Fibonacci

**Solution**

*B. Bernoulli*

*Ignoring mathematicians who have been omitted from this problem and skating over the fact that calling Einstein a mathematician has offended some readers' sensibilities, Einstein could be last entry in volume 3 (surnames Ce-Ei). As the book sits on the shelf this would make his page the left most page.*

*This would make him 'next to' the first page of volume 2 (surnames Be-CA) as this is the right-most as it sits on the shelf. Of the options given only Bernoulli could be the first entry in this volume meaning he could be 'next to' Einstein.*

**December 2019**

Create the shortest possible formula that can be copied from cell A1 downwards to replicate the game "FizzBuzz".

The game works like this:

- The players count up from 1.
- Each multiple of 3 is replaced with the word "Fizz".
- Each multiple of 5 is replaced with the word “Buzz”.
- If a number is both a multiple of 3 and a multiple of 5, then "FizzBuzz" should be used.

Other rules:

- No VBA allowed.
- Your formula has to be written into the cell - no hiding it in a named range or similar.
- That means no row numbers in a separate column, no inputs in fixed cells, no range names, no variables, no formulae in conditional formatting or text formatting, etc.
- Any Excel version allowed, but no add-in or other non-standard functions.
- Array formulae are fine, but the curly braces add 2 to your character count.

**Solution**

*There are two equally valid solutions for this teaser depending on the version of Excel available to you:*

*Excel 2019 Solution - 58 characters*

*SWITCH(GCD(ROW(),15),1,ROW(),3,"Fizz",5,"Buzz","FizzBuzz")*

*Pre 2019 Excel Solution - 61 characters*

*CHOOSE(MOD(GCD(ROW(),15),11),ROW(),,"Fizz","FizzBuzz","Buzz")*

**November 2019**

A tennis player has a 60% chance of winning any given point in her service game. Assuming that all points are independent of each other, what is the probability that she will win the service game?

(The scoring system in tennis is such that a player must score a minimum of 4 points and be at least 2 clear points ahead of their opponent in order to win a game.)

**Solution**

*P(win) = P(win without reaching deuce) + P(reach deuce and then win)*

* = P(win without reaching deuce) + P(reach deuce)*P(win from deuce)*

*P(win without reaching deuce) = P(win at least 4 out of first 6 points)*

* = P(win 4 out of 6) + P(win 5 out of 6) + P(win 6 out of 6)*

* = (6*5/2!)*(0.4^2)*(0.6^4) + 6*0.4*(0.6^5) + 0.6^6*

* = 0.31104 + 0.186624 + 0.046656*

* = 0.54432*

*P(reach deuce) = P(win exactly 3 out of first 6 points)*

* = (6*5*4/3!)*(0.4^3)*(0.6^3)*

* = 0.27648*

*To calculate P(win from deuce), define the function F as follows:*

*F(0) = win from deuce*

*F(1) = win from advantage*

*F(-1) = win from opponent’s advantage*

*Then F(0) = 0.6*F(1) + 0.4*F(-1)*

* = 0.6*(0.6 + 0.4*F(0)) + 0.4*(0.6*F(0) + 0)*

* = 0.36 + 0.48*F(0)*

*Therefore F(0) = 0.36/(1 – 0.48) = 9/13*

*Hence, P(win) = 0.54432 + 0.27648*9/13*

* = 0.7357*

**October 2019**

One day a jailer brought three prisoners out and had them stand in a line, one behind the other. He then showed them five hats (three red and two white) which he hid in a bag. From these five hats, he selected three and put one on each of the prisoners’ heads. None of the men could see what colour hat he himself wore, but could see the colour of any hats in front of him.

Starting from the back of the line, the jailer asks the first man what colour his hat is, offering him his freedom if he is correct but doubling his sentence if he gets it wrong; he declines to answer. The jailer then asks the second man the colour of his hat, making the same offer; he also declines to answer.

What colour is the third man’s hat?

**Solution**

*The answer is red.*

*Assuming all prisoners are logical and only answer if they are certain.*

*Starting at the back the last prisoner can see the other two hats. If they are both white then he will know he is wearing red. As he does not answer at least one of the front two prisoners must be wearing a red hat.*

*The second prisoner now knows that at least one of himself or the prisoner in front is wearing a red hat. If he can see a white hat on the front prisoner then he will deduce his is red and answer.*

*Given the second prisoner does not answer the third prisoner knows his hat is not white and so must be red.*