Question 4: Truck tire test
In a tire test, a truck is moving at a constant velocity of 15 m s−1 for X seconds. It then decelerates at a constant rate for X/2 seconds, reaching a velocity of 10 m s−1. It then immediately accelerates at a constant rate for 3X/2 seconds reaching a velocity of 20 m s−1. If the truck travels a total distance of 1312.5 m during the above test, what is the value of X?
Answer: Many tried but only a few top Brainiacs came up with the correct answer to this tricky teaser: 30 seconds (see Solutions below). Extra well done, so, to: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Andrew Knox, Rubbond International, The Netherlands; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; and Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; as well as to Jonas Dispersyn, innovation platform leader – superior tire performance, NV Bekaert SA, Deerlijk, Belgium, who was on the right track with his calculations.
To calculate this we need to determine the distance travelled in each phase:
Phase 1: 15 m/s for X sec, distance = 15X Phase 2: 15 m/s decelerating to 10 m/s over X/2 secs, distance =
[10.X/2 + 2.5.X/2]
Phase 3: !0 m/s accelerating to 20 m/s over 3X/2 secs, distance =
[10.3X/2 + 5.3X/2]
Adding these together, 43.75X = 1312.5, X = 30
Answer: 30 seconds. Average speed during deceleration is (15+10)/2 = 12.5 m/s and during acceleration (10+20)/2 = 15 m/s. So total distance travelled is (X x 15) = (X/2 x 12.5) + (3X/2 x 15) = 1312.5 metres whereby X = 30 and the total test time is 90 seconds.
I calculate the average speed for the deceleration period (from 15 to 10 m/sec, so 12,5 m/sec for x/2 sec) and the acceleration period (from 10 to 20 m/sec, so 15 m/sec for 3x/2 sec).
The distance is then the sum of the 3 products of speed and time:
D = v1 x t1 + v2 x t2 + v3 x t3
1312,5 = 15 x + 12,5 x/2 + 15 3x/2
X = 30 sec
Question 3: Lost & found
At a rubber factory, manager Simon finds a mislaid product batch: a bag containing a large number of seals.
To estimate the number of items in the batch, he randomly takes out 10 seals, marks them and then puts each back in the bag.
Simon then shakes the bag and randomly takes out 20 seals. He finds that three of these seals are marked. Work out an estimate for the total number of seals in the batch.
Answer: As shown by the clever solutions below, the best estimate is around the 67 mark. Very well done to: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Andrew Knox, Rubbond International, The Netherlands; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Jose Padron, laboratory analyst, Toyoda Gosei, Waterville, QC, Canada; John D Burrows, industry consultant, Luxembourg; and everyone else who had a go.
Batch size is 67.
We have 3 of 20 marked, and we know 10 in total were marked, so number in batch is [20 x 10]/3 = 67 Regards,
Answer: 67 (-ish)
The best estimate for the total number of seals "X" can be calculated as follows:
The ratio of marked seals to unmarked seals when the sample of 20 is taken is 10:(X-10)
Assuming everything is random this ratio will on average be the same as the sample, in which the ratio we are told is 3:17.
So 10/(X-10) = 3/17, so X-10 = 170/3 and so X = (170/3) +10 = 66.6667.
As it is assumed there is a whole number of seals in the bag, there would appear to be 67 seals in the bag.
However, when the 20 random seals are sampled, there can also only be a whole number of marked seals in the sample.
This means that on a subsequent sampling of 20 seals there is also a chance of selecting more or less than 3 marked seals.
Looked at the other way around, one would have to repeat the sampling a large number of times to get an accurate non-integer average to work out exactly how many seals are probably in this batch.
A much quicker way would simply be to count them all.
The estimated number of seals is 67, as the result of the proportion 10/x=3/20, x=200/3=67; about the confidence of the estimate, the statistical tables say that there is a 95% confidence that the number is between 26 and 312.
Probably a rough estimation of the filling level of the bag would yield better results with much less effort, considering that Simon had to:
- mix the seals, which is very difficult because of the friction of rubber
- inspect 100% to sort the remaining marked parts
- scrap them
The answer is 67 seals (66,66 to be precise).
Assuming an even distribution:
For the full batch: 10 out of x are marked -> 10/x
For the selected sample: 3 out of 20 are marked -> 3/20
So 10/x = 3/20 multiply by “x” and by “20”
200 = 3 x
X = 200/3
X = 66,66
The estimate for the number of seals in the bag is between 57 and 67
Let’s take the mark and recapture method.
According to the basic formula;
N = n*K/k
N= estimated population
n = marked specimens
K = recaptured specimens
k = marked specimens in recapture
N = 10*20/3 = 200/3 = 66.66 = 67 (rounded)
However, for smaller populations we can use this other formula:
N = [(n+1) * (K+1)/(k+1)]-1
N = [(11) * (21)/4]-1 = 56.75 = 57 (rounded)
So, estimation is between 57 and 67.
John D Burrows
The 20-seal sample (with 3 marked seals) should correspond to 30% (3/10) of the total population
This means a total population of 20/.3 = 66.66 seals in the bag
Let’s say 67 ………
Question 2: Square in the circle
A square is positioned inside a circle of area 49cm2. If each vertex of the square touches the circumference of the circle, what is the area of the square?
Answer: The answer to this tricky teaser is 31,19 cm², as so neatly worked out by a strong set of our top Brainiacs (see solutions below). Very well done, in order of correct reply, to: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Katharina Gottfried, technical manager, Arlanxeo Deutschland GmbH, Köln, Germany; David Mann, key account manager, SPC Rubber Compounding, UK; Andrew Knox, Rubbond International, The Netherlands; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Mehmet Koral, Erhardt-Leimer representative for Turkey, managing director, C&C Endüstriyel Danışmanlık, Eğitim ve Mümessillik Ltd, Göztepe, Istanbul, Turkey; Jose Padron, laboratory analyst, Toyoda Gosei, Waterville, QC, Canada; Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; and everyone else who had a go.
Area of the circle is pi.rsqd = 49
Area of the enclosed square is given by 2.rsqd [side is sq.rt [2r.sqd]
Ratio of ares circle:square is thus pi:2
So if the circle has an area of 49, the area of the square is 49x2/pi =
49 x 2 x 7/22 = 31.18 sq.cm.
Combining the 3 formulas
A circle = π*r²
a²+b²=c², with c=2r and a=b --> ½ a² = r²
A square = a²
it turns out
A circle = 49 cm² = π * ½ * a²
a² = 98 cm² / π ≈ 31,194 cm²
Let the circle have diameter 2R and the square have sides x
Area of the circle = πR2 = 49 sq cm
Now the square is made of 2 triangles with sides x,x, 2R
By Pythagoras 2x2=(2R)2=(2√(49/π))2
But x2 is the area of the square so the area is 98/π = 31.19 sq cm
Let the length of a diagonal through opposite vertices of the square be D, which is also the diameter D of the circle with radius r = D/2.
Let the length of each side of the square be x.
Then each half of the square is an equilateral triangle whose long side D2 = x2 +x2, so x = root(D2/2) (square root of (Dsquared/2))
Area of the circle is 49 cm2 = pie.r2 so r = root(49/pie) and D = 2.root(49/pie)
Area of square is x.x or x2 = D2/2 = ((2.root(49/pie))squared)/2 = 2.(49/pie) = 31.194 cm2
The answer is 31,19 cm².
From the surface of the circle, I calculate the radius of the circle.
A = 49
A = r² x pi
r = 3,949 cm
The radius of the circle multiplied by two also gives the diagonal of the square.
So I can apply Pytagoras’ rule: a² + b² = c²
Where a and b is the side of the square and c is the diagonal of the square.
a² + a² = d²
2 a² = d²
a = 5,585 cm
The surface of the square is a x a, so 31,19 cm².
We will work out the value of one edge of the square as X.
I will calculate the right answer correct to 3 significant figures.
The question gives us that the area of the circle is 49cm2, therefore we are able to utilize the equation for the area of this circle as πr2=49 (where r = radius of the circle).
We can now work out the radius of the circle by rearranging our equation:
r= √(49/π) = 3.9493...
As each vertex of the square touches the circumference of the circle, we can see that the diameter of the circle is equal to the diagonal length of the square.
We can calculate the diameter by doubling the radius, this gives us a value of 7.89865...
Next, we can use Pythagoras's theorem to calculate the value of X, we can do this as the diagonal line (which equals the diameter) cuts the square into two identical right-angled triangles.
As the diameter is the hypotenuse of these triangles, we can set up the equation:
a2+b2=(7.89865)2, and as a and b are both equal to the edge of the square x
As we know a length cannot be negative, we can state that X = 5.59 (this is the answer is rounded and correct to 3 sig figs).
The area of the square is 31.19 cm²
A circle = π*r²
Each vertex of square form a right triangle in the center.
Each side of the triangle is formed by the radius of the circle.
The long side of the right triangle represents the side of the square.
To obtain the value of the long side of the right triangle; we can use the Pythagorean theorem.
Where: a = b = r
solving for c² representing the area of the square,
c² = a²+b² = 2r² = 2*A circle/π = 2*49/3.1416 = 31.19
Question 1: Complete mystery
Complete the series:
31, 59, 90, 120, 151, 181…
Answer: The series is a cumulative total of days in the months of a year, starting in January, so: 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365. Very well done, in order of correct reply, to: Yoganand Nannapaneni, director, Mascot Systems Pvt. Ltd, Navi Mumbai, India; David Mann, key account manager, SPC Rubber Compounding, UK; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Stephan Paischer, head of product management special products, 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; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; 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.