# ERJ Brainteaser: January 2024

26 Jan 2024

For his excellent answers to all four questions, especially problematic Q3, it's big congratulations to** David Mann**, this year's first** Brainiac of the Month**.

Question 4: **Numbers and letters**

5 (IV) , 6 (IX), 7(V), 8 (?)

*Clues: fIVe, sIX...*

**Answer**: You either saw it or you didn’t: in brackets were roman numerals when the numbers in this sequence were written out – a making the answer **I** (8). Maximus marks to: **John Bowen**, rubber & tire industry consultant, Bromsgrove, UK; **John Coleman**, membership manager, Circol ELT, Dublin, Ireland; **David Mann**, Polymer Business Development (retired); **Stephan Paischer**, head of product management and market intelligence, Semperit AG Holding, Vienna, Austria; **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands; **Amparo Botella**, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; **Suman Dhar**, deputy general manager, international business development, Zinc BU, Rubamin, Gorwa, Vadodara, Gujarat, India; **Michele Girardi**, quality manager, Scame Mastaf SpA, Suisio, Italy; and everyone else who had a go.

New teaser on Monday

Question 3: **Missing metal**

CaMg, CaS, CaCa (CaSc), Ca??

*Apologies for typo Sc previously given as Cs*

*Clues: London…*

**Answer**: It did not help that there was initially a typo - apologies, at least that shows our questions are hand-made - in this obscure-looking teaser. So extra well done for decoding the answer Cr (see Solutions below) to: **David Mann**, Polymer Business Development (retired), UK; **Michele Girardi**, quality manager, Scame Mastaf SpA, Suisio, Italy; **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands; **John Coleman**, membership manager, Circol ELT, Dublin, Ireland; **Stephan Paischer**, head of product management and market intelligence, Semperit AG Holding, Vienna, Austria; **John Bowen**, rubber & tire industry consultant, Bromsgrove, UK; **Sudi Sudarshan**, strategic market & technology insights lead, Bridgestone Americas Inc., USA; and everyone else who had a go.

SOLUTIONS

**David Mann**

*First stab: Olympic years expressed in atomic numbers?*

*CaMg = 2012*

*CaS = 2016*

*CaCa = 2020*

*So 2024 = CaCr*

**Michele Girardi**

*Cesium should be substituted by Chromium. Then substituting the letters with the atomic numbers, one gets the sequence 2020, 2016, 2020, 2024, 2028*

**Andrew Knox**

*If Sc this then refers to the Olympic Games then the answer is Cr for 2024.*

**John Coleman**

*Initially, I was rattling my brain to fit Cesium (Cs, 55)) into the equation, but after seeing the typo in the question this morning things became much clearer…*

*Ca(20)Mg(12) – 2012 Olympics in London.*

*Ca(20)S(16) – 2016 Olympics in Rio.*

*Ca(20)Ca(20) (Ca(20)Sc(21)) – 2020 Olympics in Tokyo (covid-delayed until 2021).*

*So, the next Olympics are this year in Paris, therefore the atomic number in question is 24 = Cr (Chromium).*

**John Bowen**

*Missing metal is Chromium, Cr*

*Atomic Numbers of metals:*

*CaMg = 2012*

*CaS = 2020*

*CaCa = 2020 [CaSc = 2021]*

*CaCr = 2024*

*These refer to years in which Olympic Games were held [or not in the case of 2020 - replaced by 2021 due to covid]*

*Next is 2024, CaCr, in Paris.*

**Stephan Paischer**

*One answer could be Cr (chrome).*

*If I take the periodic number of the single elements, f.e. Ca 20, Mg 12, S 16, then Cr gives 24, which gives the following series:*

*CaMg 2012*

*CaS 2016*

*CaCa 2020*

*CaCr 2024*

*This could be a series of Summer Olympic games (London, Rio, Tokyo, Paris).*

**Sudi Sudarshan**

*Thank you for your correction from (CaCs) to (CaSc) as that threw me off! 2055 didn’t make any sense.*

*Solution: The members of the set are the years in which the Summer Olympics were held, by translating each element’s atomic symbol to its atomic number and concatenating the two numbers.*

*So CaMg = 2012, CaS = 2016, CaCa = 2020 (CaSc) = (2021) - of course the originally scheduled Tokyo Olympics were postponed from 2020 to 2021 due to the COVID pandemic.*

*The next scheduled Summer Olympics are in 2024 in Paris. CaXX = 2024 The element with atomic number 24 is Chromium (Cr).*

Question 2: **Rush hour**

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

**Answer**: As was pointed out Lucy has quite a commute at **82.5km** - maybe she works from home on some days of the week. Very well done to: **Antonella Pagliarulo**, product development manager, performance polymer solutions,Thomas Swan & Co. Ltd, Consett, Co. Durham, UK; **John Bowen**, rubber & tire industry consultant, Bromsgrove, UK; **Stephan Paischer**, head of product management and market intelligence, Semperit AG Holding, Vienna, Austria; **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands; **Michele Girardi**, quality manager, Scame Mastaf SpA, Suisio, Italy; **Sudi Sudarshan**, strategic market & technology insights lead, Bridgestone Americas Inc., USA; **Peter D. Talbot**, research scientist, Chem-Trend LP, Howell, Michigan, USA; **David Mann**, Polymer Business Development (retired), UK; and (for coming close) **Amparo Botella**, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain.

SOLUTIONS

**Antonella Pagliarulo**

*The distance home-work is given by speed x time. *

*The speed we know in both cases, time is x-5min and x+6min, respectively.*

*The solution is given by solving the system of equations:*

*First case: 90km/60min · (x – 5min) = y km*

*Second case: 75km/60min · (x + 6min) = y km*

*The result is 82.5 km*

**John Bowen**

*Lucy's home is 82.5 km from her office.*

*Calculation:*

*We need to set up two simultaneous equations.*

*Lwt distance = d, and time = t hours*

*Then at 90 kph, d/90 = t - 1/12*

*and at 75 kph d/75 = t + 1/10*

*So converting times from hours to minutes then multiplying by 90 these become d = 90t - 7.5 --- Eqn 1*

*and 1.2d = 90t + 9 ---Eqn 2*

*Subtracting 1 from 2 gives us*

*0,2d = 16.5*

*So d = 82.5km*

*CHECK:*

*At 90km/hr, time = 82.5/90 hr = 55 minutes*

*and at 75 km/hr, time = 82.5/75 = 66 minutes*

*as my old maths master would say, 'QED'*

**Stephan Paischer**

*If I define the travel time as x for speed of 90 km/h, and to (x+11) for speed of 75 km/h, and then calculate the distance by multiplying speed (in km/h) with travel time (x/60, in hrs):*

*(90 * x) / 60 = ((75 * (x+11) / 60*

*X = 55*

*Travel times are 55 mins for 90 km/h, and 66 mins for 75 km/h.*

*The distance is 90 * 55 / 60 = 82,5 km, or 75 * 66 / 60 = 82,5 km.*

**Andrew Knox**

*90 kph = 1.5 km/min, 75 kph = 1.25 km/min*

*Distance to office D = 1.5(t-5) = 1.25(t+6)*

*So, solving for t, t = 60, and D = 90 x 55/60 = 82.5 km*

*(Similarly D = 75 x 66/60 = 82.5 km)*

**Michele Girardi**

*t = s*(1/v)*

*t1 = s*(1/v1) t2 =s*(1/v2)*

*t2-t1 = s*(1/v2-1/v1)*

*11/60 =s*(1/75-1/90)*

*s = 82.5*

**Sudi Sudarshan**

*Let x be the distance in km from Lucy’s home to her office and T the time in minutes taken to reach the office on time.*

*Time taken in minutes at a speed of 90 km/h = x*60/90 =2x/3 = T-5*

*Time taken in minutes at a speed of 75 km/h = x*60/75 = 4x/5 = T+6*

*Subtracting the first equation from the second: 4x/5 – 2x/3 = (T+6)-(T-5)*

*2x/15 = 11 or x = 15*11/2 = 82.5 km*

*That’s quite a commute!*

*Bonus: Time taken to arrive on time = 2*x/3 + 5 = 2*82.5/3 + 5 = 55+5 = 60 Minutes or 1 hour*

**David Mann**

*This took more thought than I expected. Schoolboy error, forgetting to change the units from Km / h to Km / minute!*

*Speed = distance / time*

*Let the speeds be S1 (90 km / h ) and S2 (75 km /h ), the times T1 and T2, and the distance D.*

*T1 = (D/S1)*60*

*T2 = (D/S2)*60*

*T2 - T1 = 11*

*so*

*(D/75)*60 - (D/90)*60 = 11*

*Multiply by 450*

*6D- 5D = 450 * 11 / 60*

*D = 82.5 km*

**Peter D. Talbot**

*Let x = distance of Lucy’s home from work*

*Let t = time taken to travel from home to office in hr. NOTE: 5 min = 0.0833 hr; 10 min = 0.1 hr.*

*x/90 = (t - 0.833), x = 90(t – 0.833)*

*x/75 = (t + 0.1), x = 75(t + 0.1)*

*Therefore, solving for t,*

*90t – 15 = 75t + 7.5, *

*t = 1*

*Therefore, solving for x,*

* x = 90 – 7.5 = 82.5km*

*Alternatively,*

*x = 75 + 7.5 = 82.5km*

*Further checking answers,*

*82.5km @ 90km/hr = 0.9167 hr = 55 minutes*

*82.km @ 75km/hr = 1.1 hr = 66 minutes*

**Amparo Botella**

*The distance between Lucy home and the office is around 81 km.*

*Using the formula of distance=speed * time, and converting minutes into hours:*

*D=90*(T-0,08)*

*D=75*(T+0,1)*

*90T-7,2=75T+7,5*

*15T=14,7*

*T=0,98*

*D=90*(0,98-0,08)=81*

*D=75*(0,98+0,1)=81*

Question 1: **Tricky triangle**

What are the dimensions of the square in this triangle, if the length of the horizontal base is 4cm, the vertical side is 3cm and the hypotenuse is 5cm?

**Answer**: Quite a tester after the holiday break, so very well done (Solutions below) to: **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands; **David Mann**, Polymer Business Development (retired), UK; **Amparo Botella**, procurement & quality, Ismael Quesada SA, Elche, Alicante, Spain; **Alfred Johnson**, project specialist, AlMailem Tires, Kuwait; **Sudi Sudarshan**, strategic market & technology insights lead, Bridgestone Americas Inc., USA; and everyone else who took off their party hat to have a go.

SOLUTIONS

**Andrew Knox**

*The sides of the square are each approx. 1.7142857 (12/7).*

*Let the sides of the square be x, then the area of the square is x2, and the areas of the two small triangles remaining are x.(3-x)/2 and x.(4-x)/2.*

*We know the area of the outer triangle is 3.4/2 = 6*

*So, x2 + x.(3-x)/2 + x.(4-x)/2 = 6, or, simplified, 7/2.x = 6 so x = approx 1.7142857*

**David Mann**

*Consider the triangle with base A2 *

*Height is O2 *

*O2 = A1 (sides of a square) *

*As the big triangle is base 4, height 3, then O2 = A2 x 0.75 *

*But O2 = A1 so A1=A2 x 0.75 *

*A2 = A1 / 0.75 *

*A1 + A2 = 4 *

*So A1 + (A1 / 0.75) = 4 *

*Multiply by 0.75 *

*1.75 A1 = 3 *

*A1 = 1.714 *

*The square has sides of 1.714.*

**Amparo Botella**

*The result of the first brainteaser of the year is: each side of the square is 1,714 cm.*

* *

* *

* *

*A= 3 A=x+(3-x)*

*B= 4 B=x-(4-x)*

*(3-x)*x + (4-x)*x + x2 = 6 (el area de los dos triangulos pequeños + el area del cuadrado es igual al area del triangulo grande que es 3*4/2=6)*

* 2 2*

*3x-x2 + 4x-x2 + x2 = 6*

* 2 2*

*3x-x2+4x-x2+2x2=12*

*7x-2x2+2x2=12*

*7x=12*

*X=12/7= 1.714*

**Alfred Johnson**

*The sides of the square will be [around 1.8 cm] each. I couldn't solve it with the traditional trigonometry, however, MS powerpoint came for help...!*

**Sudi Sudarshan**

*My answer: 12/7 cm x 12/7 cm*

*Solution:*

*Let x be the side of the square.*

* Based on proportionality of triangles,*

* x/(3-x) = 4/3 *

*3x = 4(3-x) = 12 – 4x*

*7x = 12*

*X = 12/7*