ERJ Brainteaser: October

Outside a cinema, a queue of 100 film-goers are each given a ticket with their seat number on it. The seat numbers are allocated based on queue-position, so that, for example, the 21st person in the line has seat number 21, the 32nd has seat number 32, the 77th has seat no 77 etc etc.

Unfortunately, the first person in the queue loses their ticket and on entering the cinema decides to sit in a random seat. If everyone else tries to sit in their allocated seat, what is probability that the final (100th) person in the queue will sit in seat number 100?

Answer: The probability was exactly 50%, as neatly worked out (see solutions below) by: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Rohit Kalé, distribution strategy manager, AMN/V/B2C/DIS, Michelin North America Inc., USA: Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; and everyone else who had a go.

Solutions

John Bowen
The sample space of the last queue member, consisting of all possible free seats, has only two elements. Either the seat intended for the first member is free or the last member’s seat is available. The reason is, that if another seat is not occupied, it was also not occupied when the member with that seat number entered the cinema. 
Hence it would be occupied by that member. Consequently, only the first and last members’ seats can be free. The queue would not be affected, in particular the behavior of the first 99 members would not be affected, if we switch the tickets of the first and last members. 
So, the last member’s seat will be free with the same probability as in the original scenario. The last member’s seat will be free in either the original or the adapted scenario. As the probabilities are equal and add up to one. The probability is a half.

Stephan Paischer 
The probability for the last person to get the assigned seat is 50%.
To solve this, we need to build up the approach from very low numbers of users.
Case 1: 2 users
The first user may chose one of two available seats, the second user needs to take the remaining one – the chance that it’s his assigned one is 50%.
1 – 2
R – R
W(2) – W(1)
Case 2: 3 users
Here there are 4 possible scenarios – 2 of which the 3rd user gets his assigned seat, 2 of which he doesn’t:
1 – 2 – 3           (R = right seat, W = wrong seat)
R – R – R
W(2) – W(1) – R
W(2) – W(3) – W(1)
W(3) – R – W(1)
For each number of users, the same pattern shows: in 50% of the cases, the last user will get his assigned seat, in 50% of the case he won’t.
This applies also for a large number such as 100 users.

Rohit Kalé 
As 98 seats would have been taken by the correct seat holders leaving only the one who lost the ticket and the final person with 2 seats.
So there is a possibility of the person with without ticket seating on the correct seat or occupying the seat of the final person. Hence it is 50%

Michele Girardi 
When 1 sits in A
- persons 2 to A-1 will sit correctly
- A will sit in B >A
- A+1 to B-1 will sit correctly
- B will sit in C>B
and so on.
At a certain point X will sit in 1 or 100. In the first case X+1 to 100 will sit correctly , in the second X+1 to 99 will too and 100 will sit in 1.
Since for every sequence A,B,C,...,X,1 there's a sequence A,B,C,...,X,100 and vice versa, there's a 50% probability that the final person in the queue will sit in the correct seat.

New teaser on Monday.

 

 

What is missing from this sequence and explain why.

Berillium, Carbon,  _,  Selenium?

Answer: A bit of uncertainty around this one and apologies as the question, perhaps, slightly favoured those with English as first language. Still, very well done to everyone who recognised the missing element as ‘gold’, just BeCAuSe: Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; David Mann, key account manager, SPC Rubber Compounding, UK; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; and everyone else who had a go.

New teaser on Monday.

 

The internal diameter at the bottom of a completely circular glass is 5cm and increases smoothly up to 10cm at the very top. If the internal height of the glass is 12cm, what volume of liquid will the glass hold when half full?

Answer: The answer is 275cc, but some of our expert Brainiacs managed to come up with interesting alternatives (see solutions below). Well done, in order of reply, to: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Rohit Kalé, distribution strategy manager, AMN/V/B2C/DIS, Michelin North America Inc., USA; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; and everyone else who had a go. 

Solutions

John Bowen

There are two possible answers here, depending on whether 'half full' means to half the volume of the glass, [which will be above half its height] or to a height of half its height or 6cm.  I shall give both solutions.

The glass is a truncated cone: the full cone has a height of 24cm and a base radius of 5cm. Volume is given by Pi.rSqd.h/3 [where Pi =22/7, r = base radius, h = height, volume = 628cc

To calculate the volume of the glass we need to calculate the volume of the truncated smaller cone and deduct it from this: r = 2.5, h = 12, volume = 78.5cc which gives the volume of our glass as 549.5cc, so if half filled by volume it will contain 274.75cc.   This however will fill it above 6cm or half-way up.

If we fill the glass to half its height, the whole cone volume will be 265cc [r = 3.75, h = 18] and to calculate the volume in our glass we deduct the truncated portion, 78.5 to give a volume of 186.5cc to the 6cm height.

Andrew Knox

Volume of the glass (a truncated cone) is Pi/3*h*(r2 + r*R + R2) = 3.1415926/3 x 12 x (2.5squared + 2.5x5 + 5 squared) = 550 cc (cubic centimetres).

Question then is when is the glass half full?

James Bond would probably shoot the barman if his Martini glass was only filled to half of its height. Surprisingly a conical glass is in fact half full when filled to 79.37% of its height.

So, the answer is: A glass is half full when it contains half the volume it can potentially hold - in the case of this truncated cone, when containing  550/2 = 275 cc.

Amparo Botella

V=((h*?)/3)*(R2+r2+R+r)

V=((12*3,1425)/3)*( 52+2,52+5+2,5)

V=12,57*38,75

V= 487,0875 cm3 – full glass

Half glass= 243,54375 cm3

Michele Girardi 

Assuming that "half full" means "full at half height", the answer is 186.5 cm^3 , otherwise the half volume is 274.9 cm^3

Solution 1

The volume of a cone is base surface x height / 3 The first thing to do is to define  the cone that originates the glass by truncation .

The extra height x is given by the equation (12+x):x=10:5, that yields x = 12 .

The volume of the cone obtained subtracting the glass is 5^2/4*PI*12/3 = 78.54

The cone obtained adding the half full glass has a height of 12+6 = 18 , its base is (5+10)/2 = 7.5 and the volume is  (7.5^2)/4*PI*18/3 = 265.07.

The liquid content of the glass half full is given by the difference: 265.07-78.54= 186.5

The volume of the full glass is (10^2)/4*PI*24/3 - 78.54 = 549.78 , the half is 274,9

Solution 2

Considering a cross section of the glass, the volume is given by the integral  x^2*PI on dy .

Where x ranges from 2.5 to (5+2,5)/2= 3.75 (radius of the circle of the top of the liquid) on the edge line it's  y = (x-2,5)*12/2.5 and   dy = 12/2.5*dx

The volume is the integral of  x^2*PI*12/2.5 in dx   with x between 2,5 and 3,75

= 12/2.5*PI*[x^3/3]

= 12/2.5*PI/3*(3.75^3-2.5^3)= 186.5

Rohit Kalé 

There are two answers depending upon how you look at half full.

If you go purely by half volume then it is 225cm3

If you go half of the dept of the glass then it is 164cm3

Jose Padron

Hello, volume of the half-full glass is 274.89 cm³ (or 186.53 cm³ see second interpretation)

According to the calculation

Diameter at the bottom = 5 cm, thus radius; r = 2.5 cm

Diameter at the top = 10 cm. thus, radius R = 5 cm

Height to the top; h = 12 cm

The formula for a truncated cone volume is:   1/3*?*h*(r^2+(r*R)+R^2)

Entering each data; truncated cone volume = 1/3*?*12*(2.5²+(2.5*5)+5^2) 

Volume = 549.78 ml

The half-full glass is V/2 = 549.78/2 = 274.89 cm³ = 9.67 UK fluid oz. = 0.484 UK pint

To complicate a little more. Here is another interpretation coming from an intense discussion around a glass of beer.

First, I calculated the figures considering a half-full glass as the half of the total volume.

However, for those who think a half-full glass is the volume at the middle of the total height, not completely true, because often our eyes and brain can misunderstand this situation.

The volume at the half of the height will be; using the same formula;

r = 2.5 cm

R = 3.75 cm (radius at half of the conical sector)

h = 6 cm (half of the height)

Entering each data; truncated cone volume = 1/3*?*6*(2.5²+(2.5*3.5)+3.5^2) = 186.53 cm³

Latter situation leads us to have less beer in our glass.

New teaser on Monday.

 

Which number completes the following?

31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, _.

Answer: This one tripped up quite a few readers, but not the following select group, who quickly worked out the answer as 365 (solutions below). Well done, in order of correct reply, to: David Mann, key account manager, SPC Rubber Compounding, UK; Rohit Kalé, distribution strategy manager, AMN/V/B2C/DIS, Michelin North America Inc., USA; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; and everyone else who had a go.

Solutions

David Mann

These are the number of days at the end of each month, so the last number is 365.

Andrew Knox

This series is the number of days in each month of the year added together to make the following number in the series.
(Clue lies in calculating the delta's, being: 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, whereby 28 appears to be unusual, as do the two 31's in sequence).

Rohit Kalé 

The answer is based on the cumulative numbers sequence of the number days in a month in a normal year. (Not a leap year)

Jan    31    31
Feb    28    59
Mar    31    90
Apr    30    120
May    31    151
Jun    30    181
Jul    31    212
Aug    31    243
Sep    30    273
Oct    31    304
Nov    30    334
Dec    31    365

Jose Padron

Hello, the answer is 365, the number of days in a non-bissextile year
This 11-series is the accumulation of the number of days each month.
This series is only valid in these years where February has 28 days.
month    Number of days        subtotal        total
January    31                31
February    28    +    31    =    59
March    31    +    59    =    90
April    30    +    90    =    120
May    31    +    120    =    151
June    30    +    151    =    181
July    31    +    181    =    212
August    31    +    212    =    243
September    30    +    243    =   273
October    31    +    273    =    304
November    30    +    304    =    334
December    31    +    334    =    365
 

John Bowen

The final number is 365 - the numbers given are the running total of the days of each month through the year [not a leap year] so the final number is the number of days total in a normal year with February having 28 days.

Amparo Botella

The series corresponds to the increase of number of days of a month considering February as 28.
January – 31
January+Feburary – 59
January+February+March – 90
January+February+March+April – 120
...............
January+February+March+April+May+June+July+August+September+October+November+December – 365

Michele Girardi

The objects of  the series are the days elapsed since the beginning of the year , the last day of the month
month   Days in month    Last day of month
       1               31               31
       2               28               59
       3               31               90
       4               30              120
       5               31              151
       6               30              181
       7               31              212
       8               31              243
       9               30              273
      10               31             304
      11               30             334
      12               31             365

 

New teaser on Monday.

 

Can you find the special numbers that meet the following criteria? 

1. They are all between 300 and 600.
2. The middle digit is odd and the other digits are even.
3. The sum of the digits is 13.

 

Answer: 418, 436, 454, 472, 490, well done to John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Amparo Botella, Ismael Quesada SA, Spain; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Giorgio Marzari, sales director, RDC srl, Bareggio (Mi), Italy; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada and Rohit Kalé, distribution strategy manager, AMN/V/B2C/DIS, Michelin North America Inc., USA.

Shedding extra light on the subject:

 

Stephan Paischer:

The answer is 418, 436, 454, 472 and 490.

1 – as per rule 1 and 2, only numbers in the 400-range and the number 600 is suitable (the 300- and 500-range is not suitable as the first digit is odd).
2 – as per rule 2, only numbers in the 410 / 430 / 450 / 470 / 490 range are suitable (as the second digit needs to be odd).
3 – among these 5 ranges, only one number each is suitable – the one giving 13 as a sum of digits.

Rohit Kalé

And Jose Padron