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.
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
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
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…
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
a,x,c the speeds in A,B,C
g’ the acceleration, d the distance AC
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
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.
24 x 9 x 9 = 1944
16 x 11 x 11 = 1936
27 x 8 x 8 = 1728
14 x 12 x 12 = 2016
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)
The correlation between date (D), month (M) and year (Y) is Y = D x M²
So D = Y/M² = 2016/(12²) = 14
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
Answer: 14 (14-12-2016)
(In each case day x (month)squared = the year)
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.
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!
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.
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
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
10S + 20G = 1.2/2 +1.2*2 = 0,6+2.4 = 3
We have the formula:
20a + 30b = 4,8
40a + 50b = 8,4
10a + 20b = x = 10*0,06+20*0,12=0,6+2,4=3
9,6 -60b + 50b=8,4
The number we are looking for is 3.
Let use C for crate, S for seals and G for gaskets.
We have then following two equations
C+20S+30G=4.8 (1) and C+40S+50G=8.4 (2)
And we are looking for C+10S+20G
Or a little bit rearranged (C+10G)+(10S+10G) (3)
Subtracting the two known equation from each other
0C+20S+20G=3.6 or 10S+10G=1.8 (4) or S=0.18-G (5)
(5) in (2)
(4) and (5) in (6)
Final answer: 3kg
Let S be seal, and G be the gasket
20S+30G=4.8 Kg and 40S+50G=8.4 Kg
10G=1.2 Kg so G=0.12 Kg
Gasket weight=0.12Kg. Seal weight=0.06Kg.