ERJ Brainteaser: November

In this diagram, the cylindrical can is on a horizontal surface and AC is a diameter of the base. Also, B is on the top edge of the can and BC is vertical. If the radius of the base is 5cm and the volume is 1178cm3, can you work out the angle between the rod and the base of the can?

Answer: This tin can was not everyone’s cup of tea with only a select handful of readers correctly working out the answer at 56,31°. Extra well done, so, to: Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Jose Padron, material development specialist, Waterville TG Inc. Waterville, Québec, Canada; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands. Honourable mentions also to Simon Robinson, editor, Urethanes Technology International, London, UK and Peter McKenzie Vass, technical director, Kerry Abrasives. Listowel, Co. Kerry, Ireland, who were on the right track.    

Solutions

Stephan Paischer
Tan(alpha) = 15/10 (10 being the diameter on the base and 15 the height of the can) = 56,31°.

Michele Girardi
Volume = (1/2AC)^2*PI*BC
1178=5^2*PI*BC     leading to BC = 15
tg(BAC)=BC/AC=15/10=1.5
angle BAC= inv tg(1.5)= 56.3

Jose Padron
With a volume of 1178 cm³ and radius 5 cm, we can get height of the can.
V= A*H; H=V/A and A= PI*r²
H= 1178 cm³/(PI*5²) = 14.999 cm = BC
Diam. Base = 2r = 2*5 = 10 cm = AC
Thus; Tangent of angle BAC = BC/AC = 14.999/10 = 1.4999
Angle between rod and base = Arc Tan (1.4999) = 56.31

Andrew Knox
Diameter base A-C is 10 cm, area of base s (Pi)*r2 = 3.1415926*25 = 78.54 cm2 If volume is 1178 cm2, then height B-C is 1178/78.54 = 14.99876 cm
Angle between A-B and A-C is deduced from tan (a) = B-C/ A-C = 15/10 = 1.499876
From tables angle (a) is approx. 56,4 degrees.

 
New teaser on Monday

What characteristic do these three 12-digit numbers share with each other, but with no other 12-digit number?

100307124369
111824028801
433800063225

Clues: 316713, 334401, 656635

Bonus question:

Fill in the blanks using +, -, *, or / to make this equation valid. You may also add parentheses if necessary.

5 _ 5 _ 5 _ 5 _ 5 = 24

Answer: The solution to the main question is that they are all square numbers (see below), while the more straightforward extra question could be worked out as ((5x5x5)-5)/5 = 24. Well done to: Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Jose Padron, material development specialist, Waterville TG Inc. Waterville, Québec, Canada; Karl Heinz Sandholzer, product management vice president, Saurer Technologies GmbH & Co. KG, Twisting Solutionsm, Kempten, Germany.

Solution

They are all square numbers:

100307124369 = 316713^2
111824028801 = 334401^2
433800063225 = 656635^2

The sum of their digits are square numbers:

1+0+0+3+0+7+1+2+4+3+6+9 = 36 = 6^2
1+1+1+8+2+4+0+2+8+8+0+1 = 36 = 6^2
4+3+3+8+0+0+0+6+3+2+2+5 = 36 = 6^2

The sum of their digit pairs are square numbers:

10+03+07+12+43+69 = 144 = 12^2
11+18+24+02+88+01 = 144 = 12^2
43+38+00+06+32+25 = 144 = 12^2

The sum of their digit triplets are square numbers:

100+307+124+369 = 900 = 30^2
111+824+028+801 = 1764 = 42^2
433+800+063+225 = 1521 = 39^2

The sum of their digit quadruplets are square numbers:

1003+0712+4369 = 6084 = 78^2
1118+2402+8801 = 12321 = 111^2
4338+0006+3225 = 7569 = 87^2

The sum of their digit sextuplets are square numbers:

100307+124369 = 224676 = 474^2
111824+028801 = 140625 = 375^2
433800+063225 = 497025 = 705^2

New teaser on Monday

There is a large number of rubber seals in a box and Andy has to work out an estimate for the number. So, he takes out at random 10 seals and puts a mark on each. He then puts each seal back in the box and shakes it before randomly taking out 20 seals. If Andy finds a mark on three of the seals, work out an estimate for the total number of  cubes (sorry: should have said ‘seals’] in the bag.

Answer: Well done to everyone who worked out the answer as 67 (solution below). Extra marks to those readers who kindly did not mention the ‘cube’ error (see above) and to Andrew Knox for pointing out that it would be quicker for Andy just to count the seals.

This one tripped up a fair number of readers, but not: David Mann, key account manager, SPC Rubber Compounding, UK; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Jose Padron, material development specialist, Waterville TG Inc. Waterville, Québec, Canada. 

Solutions:

David Mann
If 10 out of x are marked and 3 out of 20 are marked, then 10/x = 3/20; so there are about 67 seals in the box.

Stephan Paischer 
If 10 out of x are marked and 3 out of 20 are marked, then 10/x = 3/20; so there are about 67 seals in the box.

John Bowen
Of 20 seals removed, 3 are marked, so to retrieve all 10 marked seals will require [10/3]  *20 = 67 seals in total [rounded up from 66.67] Andy has 67 seals in the box.

Jose Padron
If Andy takes 3 marked parts in 20 parts this is 30% of the total parts because he marked 10 of them. So, if 20 parts = 30% of total parts. Total parts = 20*100/.30 = 66.66 ~ 67 parts.

Michele Girardi
Answer is 67, which is the solution of the equation 3/20 = 10/x With a  statistical software we can estimate  that there's 95% confidence that the  proportion of marked seals is  between  0,032071 and 0,378927, consequently the number of seals is between 26 ( 10/0.378927)  and 312 ( 10/0.032071).

Andrew Knox
Answer: No idea how many cubes there might be in the bag, but the no. of seals in the box are likely so be about 67.
If X is the total no. of seals, then the proportion of marked seals added and marked seals sampled should be the same. So, 10/X = 3/20, or X = (10 x 20)/3 = 66.66666.
However, there is a fairly high chance Andy could have found only 2 marked seals in the sample, in which case the estimated no. of seals would be 100, or 4 marked seals, in which case the estimated no. would be 50 seals.
So, Andy's method of estimating the no. of seals in the box is not a good one! In fact he would be better off just counting them.

What is the area of the shaded section in this rectangle?

Answer: Not so tricky, once you drew verttical lines to break the rectangle into four smaller ones: showing that exactly half or 21 sq cm of the total area was shaded. Well done, in order of sending in the correct reply,  to: John Bowen, consultant, Bromsgrove, UK; Carlo Nolli, PMG Compounds; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; David Mann, key account manager, SPC Rubber Compounding, UK; Andrew Knox, Rubbond International, Ohé en Laak, The Netherlands; Jose Padron, material development specialist, Waterville TG Inc., Waterville, Québec, Canada; Daniel Willrich, redakteur, AutoRäderReifen-Gummibereifung, Hannover, Germany; Michele Girardi, Scame  Mastaf Spa, Suisio, Italy; Tom Bell, S & T Group LLC, marketing & business services in elastomers, Houston, Texas, USA - and everyone else who had a go.