ERJ Brainteaser Results 2019

List of Brainteaser questions over the past couple of months, along with the answers and readers supplying correct replies. The number of correct answers, as well as speed and quality of reply will help decide our Brainiac of the Month. 

 

APRIL 2019

 

Question 3: River crossing

A father, mother and their two children are crossing a river. Their boat can only take one adult or two children at a time. What is the minimum number of journeys needed to take the family of four across the river?

Answer: As per the first correct reply in from John Droogan, (see also table below from Jose Padron) it took them 9 journeys.

2 children across > 1 child back > 1 adult across > 1 child back > 2 children across > 1 child back > 1 adult across > 1 child back > (and finally) 2 children across.

Well done to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; David Mann, Polymer Business Development, France; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; John Bowen, consultant, Bromsgrove, UK; Jose Padron, spécialiste en développement de materiaux, Waterville TG Inc., Québec, Canada; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Bob Hall, senior director global marketing, Kraton Polymers LLC, Houston, Texas, USA.

 

Question 2: High five

Complete the following series

8,485, 8,516, 8,586, 8,611, _.

Answer: Apologies to those readers who spent time trying to work this one out as a mathematical series. There was, though, a strong clue in the title, as the figures represent the height above sea level of the world’s five highest mountains. Our answer, so, was 8848m for Mount Everest, known as Sagarmatha in Nepal and Chomolungma in Tibet.

Well done to peak performers: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany (thanks for the photo); David Mann, Polymer Business Development, France; Lars Linnemann, R&D manager, Genan A/S, Viborg, Denmark; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India (thanks for the table); Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; John Bowen, consultant, Bromsgrove, UK.

 

Question 1:  Next number

What number comes next in this series?

100, 500, 1, 50, 1000, 5, ???

Answer: The series derives from Roman numerals placed in alphabetical order:

C (100), D (500), I (1), L (50), M (1000), V (5) and then our answer X (10).

Congratulations maximus go to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Amparo Botella, Ismael Quesada SA, Spain; John Bowen, consultant, Bromsgrove, UK; David Mann, Polymer Business Development, France; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA.

 

MARCH 

 

The top prize this month goes to a clear winner and probably our leading specialist in the field of probability. Well earned congratulations go to: Michele Girardi of Scame Mastaf Spa in Italy, the new Brainiac of the Month.

 

Question 4: Degrees of difficulty

Add the next number to this series?

60, 90, 108, 120, 128.5714*, 135, 140, 144, 147.2727*, _

(*5th and 9th numbers are rounded to four decimal places)

Answer: Well done to: David Mann, Polymer Business Development, France (thanks for the drawing below); John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Bharat B Sharma, Sr VP Product Development & Technical Service (Elastomers), Reliance Industries Ltd, Vadodara, Gujarat, India; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Amparo Botella, Ismael Quesada SA, Spain; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, Larsen & Toubro Ltd, Kanchipuram, Tamil Nadu, India; Stephan Paischer, head of product management special products,  Semperit AG Holding, Vienna, Austria; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA.

As John Droogan noted, these are the angles inside a dodecagon: so, 10 x 180 / 12  = 150. We also liked Bharat B Sharma’s explanation (below) and similar approaches by Amparo Botella and Ramasubramanian P among others:

Sides Polygon          Sum of all angles      Interior angle l

3          Triangle                       180°                             60°

4          Quadrilateral               360°                             90°

5          Pentagon                    540°                             108°

6          Hexagon                     720°                             120°

7          Heptagon                    900°                             128.5714…°

8          Octagon                      1080°                           135°

9          Nonagon                     1260°                          140°

10        Decagon                     1440°                           144°

11        Hendecagon               1620°                           147.2727…°

12        Dodecagon                 1800°                           150

 

Question 3: Teacher teaser II

Question: A follow-on from last week’s ‘divisive question, we asked: There are 12 boys and 8 girls in the class. The teacher selects three children from the class at random to answer a question. What is the probability that the teacher chooses at least two boys?

Answer: A lot more clear-cut this week though only four correct answers suggest this type of question is proving a bit tricky for some. So extra well done to probably our best probability expert Michele Girardi, Scame Mastaf Spa, Suisio, Italy as well as to Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany, Bharat B Sharma, Sr V.P. product development & technical service (elastomers), Reliance Industries Ltd., India and P Ramasubramaniam , Larson & Toubro Ltd,  India, who provided the following detailed analysis.

 

Solution:

Probability of the teacher choosing at least two boys is the sum of probability of choosing all three boys and probability of choosing 2 boys and 1 girl.

Probability of choosing all 3 boys : P(BBB) = (12/20)*(11/19)*(10/18) = 0.193

Probability of choosing 2 boys and 1 girl: P(BBG,BGB,GBB) = (12/20)*(11/19)*(8/18) + (12/20)*(8/19)*(11/18)+(8/20)*(12/19)*(11/18) = (1056*3 )/6840 = 0.463

Probability of choosing at least 2 boys is : 0.193 + 0.463 = 0.656

Using formula:

P(at least 2 boys in a selection of 3) = (12C2 x 8C1 + 12C3 x 8C0)/20C3

= ((12×11/2) x (8/1) + (12x11x10/6)x1)/(20x19x18/6)

= (528 + 220)/1140 = 748/1140 = 0.656

 

Question 2: Teacher teaser I

In introducing herself. a new maths teacher tells her class that she has three children, before asking: if at least one of the children is a boy, what is the probability that she has three sons?

Answer: There were several different interpretations of this question, but the actual answer is 1/7 (or 14.3%), as all possible ways of the teacher having the three children must be considered – (ie BBB, BBG, BGB, GBB, GBG, GGB, BGG, but not GGG).

Extra well done to Michele Girardi, Scame Mastaf Spa, Suisio, Italy, who was the only one to get the correct answer, neatly explaining: There are 8 possible combinations of children, only one has no boys, and only one has 3 boys. So, the probability is 1/(8-1)

 

Tougher bonus question: Dividing time

If the digits 1 through to 9 are randomly arranged to make a number, what is the probability that that number is divisible by18?

Answer: 4/9 or 0.444. Well done to: Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Amparo Botella, Ismael Quesada SA, Spain; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, Larsen & Toubro Ltd, Kanchipuram, Tamil Nadu, India.

Again, Michele Girardi offered a detailed explanation: 18 = 2*3^2, so a number divisible by 18 must be divisible by 2 and 9. Numbers divisible by 2 must terminate by an even number, numbers are divisible by 9 if the sum of their digits is since: 1+2+3+4+5+6+7+8+9 = 45 (= 9×5), all the numbers created by this random arrangement are divisible by 9. Only the ones terminating by 2, 4, 6, 8, are also divisible by 2 and consequently by 18.  The probability of this event is 4/9 = 14.3%. Then the total number of permutations is 9!,there are 8! permutations for each of the 4 even final digits, so the probability is 4*8! /9! = 4/9.

 

Question 1: Factorial five, plus one

Complete the following set of factorials:

5! = 5 x 4 x 3 x 2 x 1=120, 4! = 4 × 3 × 2 × 1 = 24; 3! = ?; 2! = ?; 1! = ?; 0! = ?.

Answer:

5! = 5 x 4 x 3 x 2 x 1=120; 4! = 4 × 3 × 2 × 1 = 24; 3! = 3 x 2 x 1 = 6; 2! = 2 x 1 = 2; 1! = 1 x 1 = 1; 0! = 1

Well done to all the following readers who avoided tripping over the final step: realising that 0! is not zero: Andrew Beasley, product analyst, Hankook Tyre UK Ltd, UK; David Mann, Polymer Business Development, France; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; France Veillette, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada: John Bowen, consultant, Bromsgrove, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Ramasubramanian P, L&T Rubber Processing Machinery, India; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria.

Among the many neat explanations provided were Andrew Beasley’s concise definition of n! as n! = n*(n-1)! for n > 0 and this longer one from Ramasubramanian P:

Mathematically, factorial of a positive integer n is the product of all the integers less than or equal to n and greater than or equal to 1. The significance of ‘factorial’ is that it represents the number of ways ‘n’ elements of a set can be distinctly arranged in a sequence. For example, for a set with three elements, say A,B,C the number of ways this set can be distinctly arranged is 6: ABC, ACB, BCA, BAC, CAB, CBA. Similarly, two elements can be arranged in two distinct ways. A single element can be arranged in only one distinct way.  And a set with no elements can be arranged in one distinct way, which is again a NULL set.  Hence 0! = 1 too.

 

FEBRUARY

 

Great to have a first-time winner of our top award and so well deserved for managing an extra tricky series of teasers this month, including Question 1, which seemed to throw so many of our top players. Huge congratulations go to Ramasubramanian P of Larsen & Toubro in India, our new Brainiac of the Month.

 

Question 4: Country file

Starting in Turkey and moving west across Europe, find the missing city in the following sequence:

Ankara, _, Oslo, Andorra la Vella.

Answer: These are the names (in English) of capital cities beginning and ending with the same letter. The missing city, so, was Warsaw.

Expertly well done to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; David Mann, Polymer Business Development, France; Ramasubramanian P, L&T Rubber Processing Machinery, Larsen & Toubro Ltd, Tamil Nadu, India. A special mention also to Michele Girardi Scame Mastaf Spa, Suisio, Italy for his help in pointing out a slight blip with the question on Monday.

Bonus question: More soup

As we had two valid answers for Question 3, we ran this on into this week, asking readers to identify what came next in the ‘officially’ correct sequence ABC, BCE, CEH, EHM, _

Answer: HMU came next in the sequence, as David Mann explained: gaps of 5 and 8 according to the Fibonnaci series.

Well done to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; David Mann, Polymer Business Development, France; Ramasubramanian P, L&T Rubber Processing Machinery, Larsen & Toubro Ltd, Tamil Nadu, India; Amparo Botella, Ismael Quesada SA, Spain. A special mention also to Hans-Bernd Lüchtefeld of PHP Fibers GmbH in Germany, who came quite close with his answer (pictured across).

 

Question 3: Alphabet soup

Find what comes next in the following sequence:

ABC, BCE, CEH, _

Answer: As sometimes happens, a valid, alternative answer has emerged for this week’s Brainteaser. So well done to the following readers who worked out the ‘officially’ correct sequence: ABC, BCE, CEH, EHM: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; John Bowen, consultant, Bromsgrove, UK; Michael Easton, sales & marketing director, Globus Group, Manchester, UK; Michele Girardi, Scame Mastaf Spa, Suisio, Italy;

And, equally well done to readers who provided the alternative valid answer EHL: Amparo Botella, Ismael Quesada SA, Spain; Bharat B Sharma, Sr V.P. product development & technical service (elastomers), Reliance Industries Ltd, Vadodara Manufacturing Division, Petrochemicals, Vadodara, Gujarat, India; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Ramasubramanian P, L&T Rubber Processing Machinery, Larsen & Toubro Ltd, Tamil Nadu, India; Varun Sureka, Hartex Rubber Pvt. Ltd, Hyderabad, India.

This sequence was explained concisely by Amparo Botella: ABC, BC(D)E, CE(FG)H, EH(IJK)L

 

Question 2: Flight time

A projectile is fired at a horizontal velocity of 150 metres/second. It hits a target at a horizontal distance of 30 metres. Assuming negligible air resistance effect, what is a) the flight time of the projectile? and b) its vertical displacement?

Answers: Well done to the following readers who answered part (a) correctly ie 0.2 seconds: John Bowen, consultant, Bromsgrove, UK; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; P. Ramasubramanian, Larsen & Toubro Ltd, Rubber Processing Machinery, Tamil Nadu, India: David Mann, Polymer Business Development, France; Olaf Mayer-Mader, Abteilung BA-M, product management and business development PA, Festo AG & Co. KG, Denkendorf, Germany.

And even more well done to those readers who answered the more tricky part (b) correctly ie 0.196 metres (0.2m also accepted): John Bowen; Thierry Montcalm and David Mann.

John Bowen’s solution neatly shows how it’s done:

a] Flight time = 30/150 = 0.2 secs

b] Vertical displacement: we use s = ut + 1/2.f.tsqd = 0 [fired horizontally so no initial vertical component] + 1/2 * 9.8*0.2*0.2 [f = acceleration due to gravity, 9.8m/sec/sec] = 0.196 metres.

 

 

Question 1: City sequence

Fill in the missing cities in the following sequence:

Reykjavík, _, Oslo Montevideo , _, Wellington.

Answer: Just two correct replies, so extra well done to: P Ramasubramanian, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village, Kanchipuram, Tamil Nadu, India; and Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan. Our other Brainiacs really do need to get out more.

As P Ramasubramanian nicely explained: This is a sequence of national capital cities sorted by latitudes.  Northernmost national capital is Reykjavic (capital of Iceland @ 64.14 N) and the southernmost national capital is Wellington (capital of New Zealand @ -41.28 S).

So, the sequence is:

Reykjavic – capital of Iceland (64.14 N), Helsinki – capital of Finland (60.17 N), Oslo – capital of Norway (59.9 N) …  Montevideo – capital of Uruguay (34.88 S), Canberra – capital of Australia (35.3 S), Wellington – capital of New Zealand (41.28 S).

 

JANUARY

 

When the going got tough, Brainiacs around the world really got going, and this month’s award could have gone to three or four top contestants. Both for dealing so expertly with all four questions, I am sure everyone will join in congratulating JOHN BOWEN, our first BRAINIAC OF THE MONTH for 2019

 

Question 4: Next number B

Find the next number in the following series:

123, 354, 897, 1875, 10626, ?

Answer: Many thanks to all our readers who had a go at this one – not everyone got to the correct answer, 16887. Extra well done, so, in order of reply to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Bharat B Sharma, Sr VP product development & technical service (elastomers), Reliance Industries Ltd, Gujarat, India; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village, Kanchipuram, Tamil Nadu, India; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; John Bowen, consultant, Bromsgrove, UK: Andrew Beasley, product analyst, Hankook Tyre UK Ltd, UK; Michael Easton, sales and marketing director, Globus Group, Manchester, UK; Amparo Botella, Ismael Quesada SA, Spain; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan.

Lots of neat workings out, but here’s John Droogan’s reply – within minutes of the Monday send-out – to show how it’s done:

123 > 231 then add 123 gives 354

354 > 543 then add 345 gives 897

897 > 978 then add 897 gives 1875

1875 >8751 then add 1875 gives 10626

10626 > 06261 then add 10626 gives 16887.

 

Question 3: Football focus

In tests for the next World Cup, a football is modelled as a particle and air resistance is ignored. A player kicks the ball from a height of 0.6m above the ground, propelling it vertically upwards at a speed of 10.5m/s. From the modelling, what is the greatest height above the ground reached by the ball, and calculate the length of time the ball is more than 2m above the ground.

Answer: Said it might get tougher and it did with only three correct replies to this week’s teaser, from: John Bowen, consultant, Bromsgrove, UK: Bharat B Sharma, Sr VP product development & technical service (elastomers), Reliance Industries Ltd, Gujarat, India; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA. Their workings out of the correct answers 6.22 metres and 1.86 seconds are each given below:

John Bowen

This problem uses the basic Laws of Motion:

i] To calculate height reached – ie when the ball stops in its vertical trajectory we use

v sqd – u sqd = 2fs, where f = acceleration due to gravity [ in this case -9.81m/sec/sec] and s = distance, u = initial velocity, v = final velocity, so

0 – 110.25 = -2*9.81*s

s = 5.62, so total height = 5.62 + 0.6 = 6.22m

ii] to determine the time above 2m we need to solve a quadratic of the form s = ut + 1/2f*t sqd where t = time in seconds; s = 1.4 [as it starts at a height of 0.6m]

so 1.4 = 10.5*t – 1/2*9.81*t sqd

solving for t gives values of 1.99 and 0.143, so the ball is above 2m height for 1.85 seconds

Greatest height above the ground reached by the ball = 6.225m

Total time the ball is more than 2m above the ground.= 1.857 Sec. (gravity considered as 9.8m/sq.second)

Bharat B Sharma  

1  Initial height = 0.6                                                               

v2 = u2 + 2as          v=0                                                      

u=10.5m/s                                           

(10.5)2= 2×9.8xs     a= (-9.8m/sec2)                                             

s= (10.5×10.5/2×9.8 )+0.5 m(Initial height)                                     

s=5.625m+0.6m   =  6.225m                                                                            

2. Time require to keep ball above 2 m from ground (upward 2m to 6.225m and return)

Distance covered  4.225 m upward and 4.225 m down ward                               

Going up  time        Final speed = 0, s= 4.225 m, a =-9.8m/sq.sec==                   

Initial speed at 2 m = sq root of 2×9.8×4.225= sq rt of 82.81 m/sec

                    =9.1 m/sec                                                     

Time taken to trave 4.225 m= (V=u+at) =0=9.1-9.8t)                      

                    = time = 9.1/9.8= 0.9286 sec                                               

Going down = distance covered 4.225m, Initial speed is 0 and a=9.8m/sec2

v sqare=u sq+2as    u=0, a=9.8m/sq.sec, s=4.225)                         

v sq=  82.81                                         

v        9.1m/sec                                             

Time for return journey will also be 0.9286sec                                

Total time the ball will be above 2 meter height =2*0.92857=  1.857 sec.

Paul Knutson

Greatest height = 6.22 m.  Time above 2 m is 1.86s.  All derived from formula y=v0*t-.5*g*t^2

 

Question 2: Next number A

 Find the next number in the following sequence: 10, 9, 17, 50, 199, _

Answer: Clues were offered but not really needed as correct replies rolled in, including Michele Girardi’s a(i+1)=a(i)*i-1 making the sequence: 1,10; 2,9; 3,17; 4,50; 5,199; 6, 994. Well done in order of reply to: John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Andrew Beasley, product analyst, Hankook Tyre UK Ltd, UK; John Bowen, consultant, Bromsgrove, UK: Dave Stuckey, Dow, South Wales, UK; Hans-Bernd Lüchtefeld, market research & communication manager, PHP Fibers GmbH, Obernburg, Germany; Bharat B Sharma, Sr VP product development & technical service (elastomers), Reliance Industries Ltd, Gujarat, India; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Michael Easton, sales and marketing director, Globus Group, Manchester, UK; Jon Cutler, materials development manager, Trelleborg Sealing Solutions, Tewkesbury, UK; Martin Jones, customer service manager, network logistics & transport, DHL Supply Chain UK; Amparo Botella, Ismael Quesada SA, Spain; Thierry Montcalm, R&D manager, Soucy Techno Inc., Canada; Trey Thies, growth strategist, engineered performance products, Milliken & Co., Spartanburg, South Carolina, USA; Yuichi (Joe) Sano, Sumitomo Electric Industries Ltd, Itami, Japan; David Mann, manager rubber technology, SI Group, Béthune, France.

 

Question 1: Rubber connect

Find the rubber connection in: England, Peru, Denmark, Malaysia.

Answer: A relatively easy one to start the year, as long you were not thrown off track by ‘Malaysia’ – things might get tougher as we go along. Well done to the following readers who identified EPDM in the country names: David Mann, manager rubber technology, SI Group, Béthune, France; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; John Droogan, advanced polymers and composites, MegaChem (UK) Ltd, Caldicot, Monmouthshire, UK; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; John Bowen, consultant, Bromsgrove, UK: France Veillette, chef environnement, usine de Joliette, Bridgestone Canada Inc. Canada; Thierry Montcalm, R&D and innovation manager, Soucy Techno, Canada; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands.

 

Brainteaser Awards and Results 2018