Question 5: **Track & trace**

A manufacturer could place 8 tractor tires or 10 truck tires into a container for shipping. In one shipment, he sent a total of 96 containers. If there were more tractor tires than truck tires, how many tires did he ship?

**Answer**: Acknowledging the point about mixed containers (see impressive Solutions below), the answer to this tricky teaser was 852 & 768. Very well done to: **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands; **Michele Girardi**, Scame Mastaf Spa, Suisio, Italy; **Trevor Elison**, manufacturing engineer, Mountville Rubber Co., LaGrange, Georgia, USA; **David Mann**, key account manager, SPC Rubber Compounding, UK; **Dr Katharina Gottfried**, TSAD manager, Arlanxeo Deutschland GmbH, TSAD EMEA II, Koeln, Germany; and everyone else who had a go.

*Solutions:*

*Andrew Knox*

*To comply with the stipulations stated here, it is assumed that mixed containers are not allowed.*

*We are also not told that the manufacturer is attempting to maximise the number of tyres shipped. There is therefore not a unique answer to this question.*

*So if there are X containers of 8 tractor tyres and Y containers of 10 truck tyres, then X + Y = 96 and 8X > 10Y. These criteria are met by X is 54 or more.*

*If X = 54, total tyres shipped is (8 x 54) + (10 x 42) = 432 + 420 = 852 tyres.*

*If all 96 containers contain only tractor tyres, 768 tyres in total will be shipped.*

*For all other combinations of X > 54, the total number of tyres shipped will go down as X increases, but the criteria that there are more tractor tyre than truck tyres shipped will continue to be met.*

* *

*Trevor Elison*

*Between 768 and 852 Tires Shipped*

*Containers_Tractor = C_F*

*Containers_Truck = C_T*

*C_F + C_T = 96*

*Rearranging*

*C_F = 96 – C_T*

*8*C_F > 10*C_T*

*Substituting*

*8(96 – C_T) – 10*C_T > 0*

*768 – 18*C_T > 0*

*768 > 18*C_T*

*42.67 > C_T*

*C_T = 42, 41 . . . 0*

*C_F = 96 – C_T = 54, 55, . . . 96*

*When C_T = 42, 420 Truck Tires and 432 Tractor Tires Shipped = 852 Total Tires*

*When C_T = 0, 0 Truck Tires and 768 Tractor Tires Shipped = 768 Total Tires*

* *

*Michele Girardi*

*The total number of tires shipped is between 768 and 852 .*

*Calling x the number of tractor tire containers, the number of tractor tires is 8x, of truck tires is (96-x)*10 .*

* More tractor tires than truck tires means*

*8x > (96-x)*10*

*8x > 960 -10x*

*18x > 960*

*x > 960/18 = 53.8*

*x >= 54*

*corresponding to 54*8+(96-54)*10 = 852 tires*

*increasing the tractor tires to the maximum, for x = 96 the total is 96*8 = 768*

*So, the total number of tires is between 768 and 852*

* *

*David Mann*

*For the numbers to be about equal, the containers would be in the ratio (10/18):(8/18), making 53.33:42.67 for 96 containers.*

*To have just more tractor tyres, the numbers would be:*

*54 tractor containers = 432 tyres*

*42 truck tyre containers = 420 tyres*

*So 852 in all.*

* *

*Dr Katharina Gottfried*

*With x = tractor and y = truck:*

*8x>10y*

*x+y=96*

*Solving the equations:*

*x>53,3333*

*y<42,6667*

*With 54 containers full with tractor tires (=432) and 42 containers full with truck tires (=420), in total 852 tires has been shipped.*

*Going to next container pairs (55 tractors + 41 trucks; 56 tractors + 40 trucks) the total sum of tires is -2 each (850, 848, …).*

New teaser on Monday

Question 4:** Keep it country II**

Fill in the gap in this sequence:

… Switzerland, Austria, __, Denmark, Norway, Poland…

Clues: 00

**Answer**: A technical blip with this one (apologies) prevented us adding Sweden (46) as an additional clue – so that can also be accepted as well as the official answer **UK (44)**. Very well done to the highly select set of Brianiacs who recognised this sequence: **Michele Girardi**, Scame Mastaf Spa, Suisio, Italy; **John Bowen**, rubber industry consultant, Bromsgrove, Worcs, UK; **Serafina Vulcano**, Martinello Articoli Tecnici SpA, Italy; **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands; **Jose Padron**, material development specialist, Waterville TG Inc. Waterville, Québec, Canada: and everyone else who had a go.

Solutions

*Michele Girardi*

*These countries are ordered by international dialling code ; Switzerland (41), Austria (43), __, Denmark(45), Norway(47), Poland(48) .The missing one is UK (44)*

*John Bowen*

*Missing one is United Kingdom:*

*Dialing codes are*

*Switzerland 41*

*Austria 43*

*UK 44*

*Denmark 45*

*Norway 47*

*Poland 48*

* *

*Andrew Knox*

*History repeats itself here with a list of European country dialling codes in ascending order, i.e. … Switzerland (41), Austria (43), __, Denmark (45), Norway (47), Poland (48)…*

*The gap is therefore filled by UK (44).*

*One wonders however why Sweden (46) is missing in this sequence?*

*I still quite like [another] answer, however, that this is a list of the founder members in of the European Free Trade Association (EFTA) – the missing countries in 1960 being Portugal, Sweden and the United Kingdom.*

*So, the answer could be Portugal, Sweden and the United Kingdom. *

*Jose Padron*

Question 3: **Particle science**

Particle X passes through point P with velocity 2.8 m s−1 and constant acceleration 0.12 m s−2. Three seconds later particle Y passes through P with velocity 2.4 m s−1 and constant acceleration 0.2 m s−2. What is the distance from P when the two particles meet?

*Clues: Clues: This equation of motion might help: s = ut + 0.5 (a x t squared)*

(Where s = distance, u + initial velocity, v= velocity at time t, a = acceleration and t= time.)

**Answer**: Our questions usually garner a handful of replies within a couple of hours. Not so with this one, which held out for almost a day before the first correct answer arrived in. Extra well done so, in order of reply, to: **Michele Girardi**, quality manager, Scame Mastaf Spa, Suisio, Italy; **Trevor Elison**, manufacturing engineer, Mountville Rubber Co., LaGrange, Georgia, USA; **Andrew Knox**, Rubbond International, The Netherlands.

Solutions

*Michele Girardi*

*The equation of motion applicable when the origin of time is zero are *

*S=S0+V0*t+1/2*a*t^2*

*V=V0+a*t*

*After 3 seconds , X will have reached the position and speed *

*Sx = 2.8*3+1/2*0.12*3^2 = 4.54*

*Vx= 2.8+0.12*3=3.16*

*Let's take t=0 when Y is in P , the equations of motion are *

*X : S = 4.54+3.16*t+1/2*0.12*t^2*

*Y: S = 2.4*t+1/2*0.2*t^2*

*the time at which the particles will have the same position is given by *

*4.54+3.6*t+1/2*0.12*t^2=0+2,4*t+1/2*0.2^t^2*

*whose meaningful solution is t=27.2s , corresponding to a distance of 139.4m*

*Trevor Elison *

*Particles meet 140m from point P*

*Assume both particles are on the same path.*

*Assume both accelerations are measured with the same accuracy. (.2= .20 m/s^2)*

* Solve for s_x at 3.0 seconds = 8.94m*
* Solve for v_x at 3.0 seconds = 3.16m/s (v_x,3 = u_x + a_x*t)*
* Set s_x,f = s_y,f at time final (t_f)*
* Rearrange to format ax^2 + bx + c = 0*

*a = (a_y-a_x)/2, b = (u_y – v_x,3), c = -s_x,3 (.04, -.76, -8.94)*

* Solve for t_f using the quadratic formula = 27.22s (from the time Y passes point P)*
* Substitute t_f into either distance equation (at the time Y passes point P) to obtain distance where the two particles meet*
* 139.3659m. Round for significant figures = 140m** *

*Andrew Knox*

*Distance from point P when the two particles meet (assuming they are travelling in exactly the same direction!) is approx. 139.27 m, time elapsed for particle Y is approx 27.2 s after passing point P.*

*Calculation: Distance travelled by particle X in its 3 second "head start" is v.t + (1/2) a.t.t, so distance = 2.8.t + 0.06.t.t or 8.94 m.*

*Its velocity after 3 seconds is 2.8 + 0.12x3 = 3.16 m/s.*

*So the distance travelled from point P by X, from the moment Y passes point P, is Dx = 8.94 + 3.16T + 0.06 t.t*

*This needs to equate to the distance travelled from point P by particle Y which is Dy = 2.4.t + 0.1.t.t.*

*The two particles meet when Dx = Dy, so solve for t = ca. 27.2 s, and Dx & Dy are then both equal ca. 139.27 m.*

New, slightly easier, teaser on Monday.

Question 2: **What comes next?**

08 Friday, 12 Wednesday, 16 Monday, 20 Saturday, ?...

**Answer**: 24 Thursday ie the next ‘leap day’ on the calendar, 29th February 2024 – though some of our Brainiacs avoided all that and went straight down the maths route. Very well done, in order of reply, to: **John Bowen**, rubber industry consultant, Bromsgrove, Worcs, UK; **Andrew Knox**, Rubbond International, Ohé en Laak, The Netherlands; **Stephan Paischer**, head of product management special products, Semperit AG Holding, Vienna; **David Mann**, key account manager, SPC Rubber Compounding, UK; **France Veillette**, chef environnement, Usine de Joliette, Bridgestone Canada Inc., Canada; **Suman Dhar**, deputy general manager- international business at zinc chemicals producer Rubamin Private Ltd, Vadodara, Gujarat, India; **Michele Girardi**, Scame Mastaf Spa, Suisio, Italy; **John Coleman**, membership manager, Circol ELT, Dublin, Republic of Ireland. And, an extra well done to those – you know who you are! – who took the leap-year route.

New teaser on Monday.

Question 1: **Percentage play**

Complete this sequence:

O = 10%, ONE H = 40%, ONE HUND = 70%, ___.

**Answer**: You either got this one quite quickly or, perhaps, went on a ‘wild goose chase’. As explained nicely by David Mann and Michele Girardi (see Solutions below) below, the answer is ONE HUNDRED = 100%. Very well done, in order of reply, to: **Stephan Paischer**, head of product management and market intelligence, Semperit AG Holding, Vienna, Austria; **David Mann**, key account manager, SPC Rubber Compounding, UK; **John Bowen**, rubber industry consultant, Bromsgrove, Worcs, UK; **Andrew Knox**, Rubbond International, The Netherlands; **Michele Girardi**, quality manager, Scame Mastaf Spa, Suisio, Italy; **Suman Dhar**, deputy general manager- international business at zinc chemicals producer Rubamin Private Ltd, Vadodara, Gujarat, India; **John Coleman**, membership manager, Circol ELT, Dublin, Republic of Ireland; **Serafina Vulcano**, Martinello Articoli Tecnici SpA, Italy; and everyone else who had a go.

Solutions

*David Mann*

* The sequence is*

*O*

*ONE*

*ONE HUND*

*ONE HUNDRED*

*and the percentages are the number of letters divided by the number of letters in ONE HUNDRED.*

*So the next term is ONE HUNDRED = 100%*

*Michele Girardi*

*Hello everybody, the answer is: ONE HUNDRED=100%, as can be seen from the following sequence: *

*O 1 letter 10%*

*ONE H 4 letters 40%*

*ONE HUND 7 letters 70%*

*ONE HUNDRED 10 letters 100%*