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ERJ Brainteaser - February

ERJ Brainteaser - February

2 Feb 2020

For the top award, we decided to look to those readers who battled/worked through the tricky bonus question run over weeks 2 and 3. Big congratulations, therefore, go to: Andrew Knox, Rubbond International; Michele Girardi, Scame Mastaf; Jose Padron, Waterville TG; Paul Knutson, Timken Belts, Michael Easton, Globus Group, the new joint holders of our Brainiac of the Month title.

In his garage, avid collector Pete has 200 types of tire, which he has arranged in seven piles.

Using the clues below, work out how many tires are in each pile.

Pile 1 + Pile 2 = 57 tires
Pile 2 + Pile 3 = 83 tires
Pile 3 + Pile 4 = 71 tires
Pile 4 + Pile 5 = 43 tires
Pile 5 + Pile 6 = 66 tires
Pile 6 + Pile 7 = 43 tires

BY Conti

Answer: This teaser proved popular with a good number of readers correctly answering: P1 contains 31 tires, P2 contains 26, P3 contains 57, P4 contains 14, P5 contains 29, P6 contains 37 and P7 contains 6 tires. Well done in order of reply to: Mirco Niklas, Process Technology, Freudenberg Technology Innovation SE & Co. KG Corporate R&D, Weinheim, Germany; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village, Kanchipuram, Tamil Nadu, India; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec; David Mann, key account manager, SPC Rubber Compounding, UK; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Michael Easton, sales and marketing director, Globus Group, Manchester, UK. John Bowen, consultant, Bromsgrove, UK; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Steve Wymark, technical manager, Fenner Dunlop Engineered Conveyor Solutions, Marfleet, Hull, UK; Daniel Willrich, redakteur, AutoRäderReifen-Gummibereifung, Hannover, Germany; Lionel Lesage, KSB (no other details supplied) Pushkar Raj Singh Shaktawat, ADM – OHP, Transportation, L&T Technology Services Ltd, Baroda, Gujarat, India.

If Fred leaves home and drives at a constant speed of 90km/hr, he will arrive at his office 5 minutes early. If the same journey is done travelling at 75km/hr, he will arrive six minutes late. How far is Fred’s home from his office?

Answer: A couple of wrong turns here and there but a good number of regular readers correctly worked out that Fred lives 82.5km from his office – quite a long commute. Well done to: John Bowen, consultant, Bromsgrove, UK: Michele Girardi, Scame Mastaf Spa, Suisio, Italy; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village, Kanchipuram, Tamil Nadu, India; Kirill Katsay, Omsk Carbon Group, Russia; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada.

Fred

As always, some neat working out, including Andrew Knox’s:
Distance travelled = speed x time.
If t is time in minutes, then distance D to the office is:
D = ((t-5)/60)x90, and
D = ((t+6)/60)x75
So, ((t-5)/60)x90 = ((t+6)/60)x75
Solving for t, t=60
D = (55/60)x90 (or (66/60)x75) = 82.5 km.

And we finally sorted that troublesome bonus question about the gap in the series: 12, 36 324, __, 52488:

12
12*(1+2) = 36
36*(3+6) = 324
324*(3+2+4) = 2916
2916*(2+9+1+6) = 52488

Extra well done to: Michele Girardi; Jose Padron; Paul Knutson (all mentioned above) and Michael Easton, sales and marketing director, Globus Group, Manchester.

Terry the terrier rolls a 30mm diameter rubber ball exactly three times around the perimeter of a room measuring 5 metres by 4 metres. To the nearest whole number, how many revolutions does the ball make?

Bonus question: Fill in the gap in the following series: 12, 36 324, __, 52388

Answer: As is often the case, this week’s Brainteaser was not exactly straightforward. As well as the official answer of 573 revolutions another valid answer 569 emerged. Guess it all depends on the definition of a ‘room’. In the 573 camp were: Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, USA; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd, Vedal Village, Kanchipuram, Tamil Nadu, India; Michael Easton, sales and marketing director, Globus Group, Manchester, UK; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec; while the 569-ers were John Bowen, consultant, Bromsgrove, UK: Jose Padron (again, see below); Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; and Michele Girardi, Scame Mastaf Spa, Suisio, Italy. Well done to all and everyone else who had a go..

Bonus Question: Fill in the gap in the following series: 12, 36 324, __, 52488

Continuing this week’s ‘not-straightforward’ theme there was an unfortunate typo in the final number of this series – 3 incorrectly keyed in, instead of 4. While there were a couple of correct answers, we have decided to roll this tricky teaser onto next week.

As was the case last week, we are extra grateful to Jose Padron for explaining:

There are three options to solve this:

Terry makes his way to the exact measures of the room
Terry is inside of the room; it is necessary to take in account the diameter of the ball
Terry is outside of the room; same way, taking in account the diameter of the ball.

Ball diam. = 0.03 m

From a general calculation: No of rev. = distance/(perimeter of the ball)

Case no 1

No of rev. = 3*(distance/(? * ball diam.)) = 3*18 / (3.1416*0.030) = 573 rev.

Case no 2

No of rev. = 3*((2*(long side-ball diam)+2(short side-ball diam))/(? * ball diam.)) = 3*(2*4.97+2*3.97) / (3.1416*0.03) = 569 rev.

Case no 3

No of rev. = 3*((2*(long side+ball diam)+2(short side+ball diam))/(? * ball diam.)) = 3*(2*5.03+2*4.03) / (3.1416*0.03) = 577 rev.

Come on Terry, good dog, my choice is 573 revolutions; the two others are somewhat extravagant.

On a shopping trip, Sheri bought 2kg of fish and 3kg of potatoes. She was accompanied by a friend who paid €4.20 for 1kg of fish and 1kg of potatoes. If the price of 1kg of fish was €2.30 more than 1kg of potatoes, how much did Sheri pay?

Answer: The lure of fish & chips seemed to attract quite a few new readers, but it was left down to our most seasoned Brainiacs to serve up the correct answer – €9.35. Well done, in order of reply, to: Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Ramasubramanian P, manager, marketing – mixer and LTKMPL products, rubber processing machinery, Larsen & Toubro Ltd; Paul Knutson, textile engineer, Timken Belts, Springfield, Missouri, US; Michele Girardi, Scame Mastaf Spa, Suisio, Italy; David Mann, key account manager, SPC Rubber Compounding, UK; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Hans-Bernd Lüchtefeld, marketing manager – automotive safety, Indorama Mobility Group, PHP Fibers GmbH, Obernburg, Germany. Many thanks to every else who had a go.

Some really neat working out sent in by everyone above, none more so than Jose Padron’s:

Jose Padron: Fish and chips