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ERJ Brainteaser: October

ERJ Brainteaser: October

28 Oct 2022

For an excellent run of accurate answering as well as some expert analysis of our questions, the top award goes to Andrew Knox, new holder of the Brainiac of the Month title.

Question 4: Easy rider

A Tour de France competitor is cycling down a steep incline at constant acceleration, passing checkpoints A, B and C at a velocity of 6 m s−1, X m s−1, and 20 m s−1 respectively. If the time to taken to cycle from A to C is 35 seconds, what is the value of X and the distance between A and C?

Answer: This was quite a tricky one, so well done to everyone who had a go, especially our break-away pack: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Andrew Knox, Rubbond International, The Netherlands; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy.

Solutions

Question setter

To find the value of a:

a = (20-6)/35 = 14/35, therefore, a = 0.4m s−2

Then as t1/t2 = 4/3, 3t1 = 4t2, implying that t1 + t2 = 35.

This solves to give t1 = 20 and t2 = 15.

And as v = u + at, x = 6 + 0.4(20).

So, x = 14 m s-1

John Bowen

Rider passes A at 6m/s and 35 seconds later passes C at 20 m/s - an increase of 14 m/s, so acceleration, assumed uniform, is 14/35 m/s/s or 0.4m/s/s

We can now substitute in the equation of motion s = ut + 12ftsqd

So, s = 6 x 35 = 1/2 x 0.4 x 35 x 35 = 210 + 245 = 455m

To calculate the value of X, the velocity at B, we need either the time or distance to B, neither of which we have. If we call the time to B tB, then X = 6 + 0.4tB

Andrew Knox

If the time to taken to cycle from A to C is 35 seconds, what is the value of X and the distance between A and C?

Answer: Distance A-C is 455 m. (Relationships are v=a.t and v2=v1+a.t whereby acceleration a=speed increase/time = 14/35 m/s/s).

Point B however is undetermined, except that its distance from A is governed by the relationship d=((6+X)/2.t where t is the time to reach point B. If B happens to be half way between A and C, then speed X would for example be 13 m/s. But it can be anywhere along the route…

Michele Girardi

The speed X is 14.8 m/s and the distance AC is 455 m .

The equations of constant acceleration motion can be expressed as :

a = (V-V0)/t

V^2 -V0^2 = 2.a.s

naming :

a,x,c the speeds in A,B,C

g’ the acceleration, d the distance AC

it’s

a=6 c=20 t=35

g’= (c-a)/t= (20-6)/35 = 0,4

c^2 – a^2 = 2.g’.d

d = ( c^2-a^2)/2g’ = (400-36)/(2*0,4) = 455

in absence of information let's suppose that the checkpoints are equally spaced, at B the distance from A is d/2 and the speed is

x^2= a^2+2*g’*d/2 =36+2*0,4*455/2 = 218

x=sqrt(218)= 14.8

New (less tough) teaser on Monday

Question 3: Make my day!

?? December 2016

24 September 1944

16 November 1936

27 August 1728

Answer: The missing date is 14 December 2016 as neatly identified/worked out by a select set of readers (see solutions below). Well done, in order of correct reply, to: David Mann, key account manager, SPC Rubber Compounding, UK; Jonas Dispersyn, innovation platform leader – superior tire performance, NV Bekaert SA, Deerlijk, Belgium; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Stephan Paischer, head of product management and market intelligence, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, The Netherlands; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; and everyone else who had a go.

Solutions

David Mann

24 x 9 x 9 = 1944

16 x 11 x 11 = 1936

27 x 8 x 8 = 1728

14 x 12 x 12 = 2016

Amparo Botella

If you multiply the square of the month number by the day, you get the year (ie: 9*9*24=1944, 11*11*16=1936 and 8*8*27=1728)

then: 12*12*14=2016.

Jonas Dispersyn

The correlation between date (D), month (M) and year (Y) is Y = D x M²

So D = Y/M² = 2016/(12²) = 14

Michele Girardi

The dates' components are linked by the formula

day = year/month^2

day month year

27 8 1728 27 = 1728 / 64

16 11 1936 16 = 1936 / 121

24 9 1944 24 = 1944 / 81

14 12 2016 14 = 2016 / 144

Andrew Knox

Answer: 14 (14-12-2016)

(In each case day x (month)squared = the year)

Stephan Paischer

The answer is 14 December 2016.

The calculation pattern is as follow:

Year divided by day, and squareroot of the the result gives the month (as a figure from 1 to 12).

2016 / 14 = 144

Sqrt(144) = 12

12 = December

Question 2: My type of town

How might the following series begin?

___, Washington, Edinburgh, Rome, Tbilisi, Yaoundé.

Answer: This is about noticing that the first letter in the name of each city shows QWERTY – as on keyboards – once, of course, Q is added. While most readers went for Quito, Tim Clayfield’s Qingdao was equally valid, as (currently) not all cities in the series are country capitals. Very well done, in order of reply, to: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Stephan Paischer, head of product management and market intelligence, Semperit AG Holding, Vienna, Austria; Andrew Knox, Rubbond International, The Netherlands; Frank Bloemendaal, research & development, Polycomp, Vorden, The Netherlands; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Jose Padron, laboratory analyst, Toyoda Gosei, Waterville, QC, Canada; Hans-Bernd Lüchtefeld, marketing manager, Indorama Mobility Group, Germany; Tim Clayfield, Interface Polymers, Loughborough, UK; Yoganand Nannapaneni, director, Mascot Systems Pvt. Ltd, Navi Mumbai, India; and everyone else who had a go.

New teaser on Monday.

Question 1: Crate expectations

At a car assembly factory, the total weight of a crate, 20 seals and 30 gaskets equals 4.8kg. The total weight of the crate, 40 seals and 50 gaskets equals 8.4kg. What is the total weight of the crate, 10 seals and 20 gaskets?

Answer: This tricky teaser generated a range of impressive solutions (see below) to get to the correct answer 3kg. Very well done to: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Andrew Knox, Rubbond International, The Netherlands; Stephan Paischer, head of product management and market intelligence, Semperit AG Holding, Vienna, Austria; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Jonas Dispersyn, innovation platform leader – superior tire performance, NV Bekaert SA, Deerlijk, Belgium; Mehmet Koral, Erhardt-Leimer representative for Turkey, managing director, C&C Endüstriyel Dan??manl?k, E?itim ve Mümessillik Ltd, Göztepe, Istanbul, Turkey; and everyone else who had a go.

Solutions

John Bowen

To solve this one we need to set up some simultaneous equations:

Let weight of crate = C

Weight of a seal = S

Weight of a gasket = G

We are told C + 20S + 30G = 4.8kg-----------Eqn 1

and C + 40S + 50G = 3.6kg-----------Eqn 2

Subtract 2 from 1: 20S + 20G = 3.6kg

so 10S + 20G = 1.8kg-------Eqn3

Subbtract 3 from 1 gives us

C + 10S + 20G = 3.0kg

This is digging up the old school algebra!

Andrew Knox

There is in fact one too few equations here to deduce the correct answer - some trial and error is required, and there could be more than one solution.

Let the crate itself weigh "C" grams, the seals "s" grams each and the gaskets "g" grams each (a bad choice of letters in retrospect...).

We then have as given:

C + 20s + 30g = 4800, and

C + 40s + 50g = 8400

Subtracting we get 20s + 20g = 3600, so s = 180 - g

Substituting for s in C + 20s + 30g = 4800 gives C = 4800 - (3600 - 20g) - 30g or C = 1200 - 10g

Substituting for s in C + 40s + 50g = 8400 gives the same result.

So we have 3 variables and only two equations which means we cannot solve the problem by deduction alone and must resort to trial and error.

By chance s = g = 90 works, by which the crate C would weigh 300 grams.

So, to answer the question, the total weight of the crate, 10 seals and 20 gaskets will be C + 900 + 1800 = 3000 grams or 3.0 kg.

But, s = 75 and g = 105 also works, whereby the crate would weigh 150 g.

In that case the total weight of the crate, 10 seals and 20 gaskets will be 150 + 750 + 2100 = 3000 grams or 3.0 kg as well.

Stephan Paischer

To solve the equation completely (ie the individual weight of the crate, a seal and a gasket) is not possible as there is not enough input.

But by subtracting the two weights, you can get the weight for the difference of the two cases, i.e. the weight for 20 seals and 20 gaskets.

20 x + 20 y = 3,6

10 x + 10 y = 1,8

So, the value of 1,8 kg is equivalent to 10 seals and 10 gaskets.

The requested value (crate + 10 seals + 20 gaskets) is one step (10 seals and 10 gaskets) below the first input equations (crate + 20 seals + 30 gaskets). So I reduce the weight of the first input equation (4,8 kg) by the step value (1,8 kg) to get the answer.

4,8 – 1,8 = 3,0 kg

Michele Girardi

The data are :

20S + 30G = 4,8

40S + 50G = 8,4

Multiplying by 2 and -1, summing

10G = 4.8*2 - 8.4 = 1.2 and 30G = 1.2*3 = 3.6

substituting in the first

20S + 3.6 = 4.8 20S = 1.2

Finally:

10S + 20G = 1.2/2 +1.2*2 = 0,6+2.4 = 3

Amparo Botella