Each month, ERJ sets a weekly brainteaser, with questions of varying degrees of difficulty. Readers supplying the most accurate (and stylish) answers are then considered for the prestigious Brainiac of the Month title.
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 Danışmanlık, Eğitim 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
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:
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.
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.
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
10a + 20b = x = 10*0,06+20*0,12=0,6+2,4=3
A=( 4,8-30b)/20
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)
Let S be seal, and G be the gasket
20S+30G=4.8 Kg and 40S+50G=8.4 Kg
(40S+50G)-(20S+30G)=8.4-4.8=3.6 Kg.
10G=1.2 Kg so G=0.12 Kg
Gasket weight=0.12Kg. Seal weight=0.06Kg.
Result: 10S+20G=10x0.06+20x(0.12)=3.0Kg.