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Square peg, round hole

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.

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





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


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 leftn 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 ration:

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


Substituting for leftn 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 ration = 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!


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


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):

  1. Send 2 + 1 across first. Total time = 2 mins
  2. 2 returns with the torch. Total time = 2+2 = 4 mins
  3. Send 10 + 5 across . Total time = 4 + 10 = 14 mins
  4. 1 returns with the torch. Total time = 14 +1 = 15 mins
  5. 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


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


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



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 2020Puzzle cross

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


D. 6 - as demonstrated below.

Teach First Futures SchemeFebruary 2020

What is the next number in the sequence?

1, 11, 21, 1211, 111221, …

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


C. 312211

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

So with a starting point of 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


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.

Fizz Buzz

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.

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

Excel 2019 Solution - 58 characters


Pre 2019 Excel Solution - 61 characters


Tennis Teaser

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.)


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 

Square peg round hold

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?


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.