ERJ Brainteaser - April

Question 3: City series

Kabul, Tirana, Algiers, _?_...

Email your answer: correct replies on Friday.

12 Nov 2010; 9 Aug 2007; 6 May 2004; __?__.

Answer: As easy as 1,2,3 once you recognised that these dates can give the numerical sequence: 12, 11, 10 // 9, 8, 7 // 6, 5, 4 // 3, 2, 1 (or 3 Feb 2001). Well done to: Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; John Bowen, BSc [Hons], CEng, MIMMM, consultant, Bromsgrove, UK;  Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland; John Woods, technical consultant, Hanau Klein-Auheim, Germany; Hans-Bernd Luechtefeld, Consultant, Germany; Dr Kate Burns, senior technologist and regulatory officer, Prisma Colour Ltd, Birch Vale, High Peak, Derbyshire, UK; and (for their interesting alternative answers) Bharat B Sharma, technical director TWC Group (Techno Waxchem & Rajsha Chemicals), Kolkata, India; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain. 

SOLUTIONS

Andrew Knox

Answer: 3 Feb 2001

These dates can be abbreviated to read as 12-11-10, 9-8-7, 6-5-4, 3-2-1 (or 03-02-01 if preferred).

Taken together they form a simple sequence from 12 down to 1

John Bowen

Next in the series is 3rd February 2001

Each date is represented as follows: 12,11,10; 9,8,7; 6,5,4; so next will be 3,2,1

Kamila Staszewska

If the dates are presented in numerical format: 12/11/10, 09/08/07, 06/05/04 so the last missing date is 03/02/01 (3 Feb 2001).

John Woods

My proposed solution follows the following steps:

1. Transformation the dates into numerical form ,

12 November 2010 becomes 12.11 .10

9 August 2007 becomes 09.08 07

6 May 2004 ->  06.05.04

If we apply this and continue the spiral sequence downwards , it is logical to imagine that the next date is 03.02.01  , or the

3 February 2001 my proposal as the solution but I confess to being lazy and not looking up Saint’s days , Port Vale’s victories or other remote possibilities of an alternative solution.

Hans-Bernd Luechtefeld, Consultant

Days, months and years follow the row from 12 to 1, so the answer is 3 February 2001.

Dr Kate Burns

When written in dd/mm/yy format, the dates given are:

12/11/10

09/08/07

06/05/04

So the final one is 03/02/01 (counting down), which is 3rd February 2001.

ALTERNATIVELY

Bharat B Sharma

Answer - The next date would be 2 February 2001.

If we see closely, the trend is that be it date or month or year, everything is getting reduced by 3 (subtracting 3 years, 3 months, and 3 days v/s. earlier date).

If we reduce  

12 Nov 2010  If we reduce 3  years, 3 months, and 3 days, we get -- 9 Aug 2007

9 Aug 2007   If we reduce 3  years, 3 months, and 3 days, we get -- 6 May 2004

6 May 2004   If we reduce 3  years, 3 months, and 3 days, we get -- 2 Feb 2001

 Amparo Botella

The answer is 1-4-2001

As if you add the day, month and year, the result is a decrease of 9

12+11+2010 = 2033

9+8+2007 = 2024

6+5+2004 = 2015

As the year decreases in 3 years every time, the next year will be 2001, so the date must be:

1+4+2001 = 2006

This makes the sum of all factor the following series: 2033 -9 = 2024 – 9 = 2025 – 9 = 2006

New teaser on Monday.

 Moulded rubber seals are produced in pairs - one green, the other black - and transferred to containers holding more than 250 but less than 350 of the mouldings in total.

However, one seal has been lost from a container so that, if two seals were randomly drawn out, the probability of them being the same colour would be exactly 50%.

If there are more green seals than black seals, can you work out mathematically the colour of the missing seal?

Answer: Well done to everyone who had a go at this somewhat confusing (apologies) teaser – for starters, black would clearly be the colour of the missing seal. However, using very clever maths, Bharat B Sharma, technical director TWC Group (Techno Waxchem & Rajsha Chemicals), Kolkata, India was able to pinpoint the official answer. For their commendable replies, well done also to: Dr Kate Burns, senior technologist and regulatory officer, Prisma Colour Ltd, Birch Vale, High Peak, Derbyshire, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; and Kamila Staszewska, R&D / quality lead, Abcon Industrial Products Ltd, Cootehill, Co. Cavan, Ireland.

SOLUTIONS

Bharat B Sharma

Suppose – G is Green seals and B is Black Seals.

                Total seal in container is T = G+B  and probability (in drawing two seals) is each 0.5 (50%) for Black and Green

                {G(G-1)+B(B-1)}/{T(T-1)} = ½

                2{G(G-1)+B(B-1)}} = T(T-1)

2G2 – 2G + B2- 2B = (G+B)2 – (G+B)

                G2 – 2GB + B2=G+B

                (G-B)2 = T  -- Remaining seals must be a perfect square

Since one of the seals is lost, so the total will not be even (it will be odd).

Squares between 250 and 250 are squares of 16, 17, and 18 and only square of 17 is odd (289) (others are 256 and 324).

With 289 is only odd number, the total that would have been including the missing seal – should have been 289+1=290 (or 145 each of black and green seal).

We have two equations now:

G+B =289

G-B =17

Therefore 2G = 289+17    or G=153

And B= 289-153 or B= 136

Given that only one ball is lost during filling of container , we should be having originally 153 pairs of Green and Black. We have two scenario-

#1 If green is lost, meaning Green = 152 , making black as 153 (Black >Green).

#2 If Black is lost, meaning Black = 152 , making Green as 153 (Green > Black)

In our solution, Green seals are more (153) than Black (136)  (2nd condition)

Therefore, scenario #2 is more mathematically valid meaning

A BLACK SEAL WAS LOST DURING TRANSFERRING IN CONTAINER!!

Dr Kate Burns

If the seals are produced in pairs, one is missing and there are more green seals, then the missing seal must be black.

To go a step further, I set out to prove mathematically that in this case the probability of picking two of the same colour was 50%.

If the number of green seals (ng) is one greater than the number of black seals (nb), ng=nb+1. The total number of seals is 2nb+1

Probability of picking two green seals P(g,g) = nb+1/(2nb+1) * nb/2nb = (nb+1) / 2(2nb+1)

Probability of picking two black seals P(b,b) = nb/(2nb+1) * nb-1/2nb = (nb-1)/2(2nb+1)

Probability of picking the same colour twice = P(g,g) + P(b,b) = (nb+1)/ 2(2nb+1) +  (nb-1)/2(2nb+1) = 2nb/ 2(2nb+1)

This is close to 50%, but is not exactly 50% as stated in the question. As the number of total mouldings is between 250 and 350, the actual chance of both seals being the same colour ranges from 49.801% to 49.857%.

Andrew Knox

Answer: If the seals are produced in pairs, and only one seal from this container has been lost, and given there are more green seals than black, then the colour of the lost seal must be black.

In order for two seals to be withdrawn from this container and the chance to be EXACTLY 50% that they are of the same colour, the first of the two seals selected must be the opposite colour to the lost seal, i.e. the first to be selected must be green, leaving an equal number of both colours before the second seal is withdrawn.

The statement "However, one seal has been lost from a container so that, if two seals were randomly drawn out, the probability of them being the same colour would be exactly 50%" is however not true. This is only the case if the first seal removed is green. If the first seal removed is black, there will be two more green seals remaining n the container than black seals, whereby the chance the second seal selected also being black will be only (n-2)/(2n-2) x 100 percent, where n is the original number of black or green seals that should have been in the container before one was lost.

Kamila Staszewska

There are more green and black so G>B

P of drawing two seals of the same colour  = 50%

P=((G(G-1))/2 + (B(B-1))/2)/((G+B)*(G+B-1)/2)=1/2

Solving this will give (G-B)^2= G+B

The difference d forces G-B to be positive whole number d=G-B

If the green is missing then the original number is G+1 and B

Original difference: (G+1)-B=(G-B)+1=d+1

Since d>0, d+1>=2 which is not possible from losing only ONE seal.

If the black  is missing then the original numbers were G and B+1

Putting in the original difference equation: G-(B+1)=(G-B)-1=d-1

For d to be positive whole number: d-1>=0

So G>=B+1-1=B which is possible so the missing seal was black.

New (more straightforward) teaser on return from Easter holiday on 7 April.