ERJ Brainteaser: April
29 Apr 2022
As regular readers know, our top award sometimes goes to a best newcomer (or relative newcomer). Our congratulations go to Katharina Gottfried of Arlanxeo Deutschland, our new Brainiac of the Month.
Question 4: Letters & numbers
H + O = 9
B + C = ?
N + S = 23
Answer: ‘Elementary my dear readers’, as this not too difficult teaser is based on the atomic number of elements, So:
H [Hydrogen] =1, O [Oxygen] =8 so H + O = 9
N [Nitrogen] = 7, S [Sulphur] =16 so N + S = 23, so:
B [Boron] = 5, C [Carbon] = 6 so B+C =11
Well done in order of reply to: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, The Netherlands; Jose Padron, laboratory analyst, Waterville TG Inc., Waterville, Québec, Canada; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; and everyone else who had a go.
New teaser on Tuesday.
Question 3 : Country couples
Australia + Austria = 104
Chile + China = ?
Hungary + India = 127
Answer: THe answer as first supplied first by Andrew Knox is 142, ie Chile (56) + China (86) = 142. These are the international telephone dialling codes for each country added together, so Australia (61) + Austria (43) = 104, etc.
Very well done, in order of reply to: Andrew Knox, Rubbond International, The Netherlands; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Jose Padron, laboratory analyst, Waterville TG Inc., Waterville, Québec, Canada; Katharina Gottfried, Arlanxeo Deutschland GmbH, TSAD EMEA II, Köln, Germany; and everyone else who had a go.
Question 2: Building blokes
On a construction site, Alan can build a wall in nine hours, Brian in six hours and Colin in four and a half hours. How long will it take all three to build the wall if they all work together?
Answer: Back on a steady footing this week, with no dispute over the correct answer: 2 hours. Very well done to: Katharina Gottfried, Arlanxeo Deutschland GmbH, TSAD EMEA II, Köln, Germany; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Andrew Knox, Rubbond International, The Netherlands; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria.
Solutions
Katharina Gottfried
Together they would need 2 hours to build a wall (1/x = 1/9 + 1/6 + 1/4,5)
John Bowen
We need to consider the rate of work:
Alan builds 0.1111 walls/hr
Brian build 0.1666 walls/hr
Colin 0.2222 walls/hr
So working together, and assuming they do not fall over each other, they will produce 0.5 walls/hr so it will take 2 hours to build the wall.
Andrew Knox.
In 1 hour, Alan will build 1/9 th of the wall, Brian 1/6 th (or 1.5/9 th) and Colin 1/4.5 or 2/9 th (together = 4.5/9, or 1/2).
Adding these together, when all three work on the wall, they will complete 1/9 + 1.5/9 + 2/9 = 4.5/9 or 1/2 of the wall in 1 hour, and so will complete the whole wall in 1/0.5 = 2 hours.
Michele Girardi
The speed in walls/hour of the 3 are : 1/9, 1/6, 2/9
The combined speed is 1/9+1/6+2/9 = (6+9+12)/54 = 27/54 = 1/2
The time in hours for one wall is ( 1 / (1/2) ) = 2
Stephan Paischer
Alan: in 1 hr, he finishes 1/9th of the wall.
Brain: in 1 hr -> 1/6th
Colin: in 1 hr -> 1(4.5)th = 2/9th of the wall.
1/9 + 1/6 + 2/9 = 0.5
This means in 1 hr, they finish half of the wall. So in 2 hrs, the finish the complete wall.
New teaser on Tuesday
Question 1: Pony puzzle
A farmer has two horses in a stable: a black one and a white one.
One is male. What is the probability that both horses are male?
Answer: While 50% was a recurring reply, it’s extremely well done to John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK, who hit on the exact logic behind this question: The probability that both are male is 1/3. We already have one male, so the possible outcomes are now : MM, MF, FM, [FF does not now exist as one we know to be M]. So, the probability of MM is 1/3.
But, as happens now & then, disagreement arises with the official answers and/or the logic applied to solutions. Having considered the points made below, we also accept 50% as a (maybe even more) valid answer to this teaser: So well done also to: Andrew Knox, Rubbond International, The Netherlands; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria.
50/50 solutions
Amparo Botella
If we talk about having 2 males, the reply FM and MF is the same one. Is one female and one male, it does not matter the way you write it. So this reply is not right. We are talking about a permutation where the order of the factors does not matter, we are just talking about having two males, it the order does not matter, FM and MF is exactly the same, so the probability of MM is 1/2.
Andrew Knox
The four possible outcomes given in the solution would be correct if nothing were known about the genders of the two horses in the stable, and they were led out one after the other. The outcome then would indeed be one of the following four: MM, MF, FM, FF, each with a probability of 1/4.
In this instance however we know that one horse is male. This leads to the other horse necessarily being either male or female, and we assume with equal probability. Furthermore, the gender of the "other horse" is totally independent of the gender of its male stablemate.
Stating that FM is still a possibility, which it isn't if the "first horse" is known to be male.
If we were simplify matters by taking out one of the two horses from the stable and noting it is male, then there is only one horse left in the stable, of unknown gender.
There are then only two possible outcomes: MM, MF, each with a probability of 1/2.
The corollary of the solution given is that the chance of the "other horse" being female is 2/3. Taken to the roulette wheel, if say betting on red or black, if black came up on the first spin, your solution would suggest there is twice the chance of the next spin being red than of its being black, and we know that is not true.
I have to add that the answer given for the question about the pencils falling behind the desk was similarly flawed, for similar reasons.
The difficulties appear to arise when statistical probabilities are mixed up with logic.
