Brain Teasers ? or 50 Years On........... ?

Sorry frank. I've been very busy this past week and will be for the next 3 or 4 weeks, so I haven't really had time to give this much thought.

I visualise the arm and cartridge being "thrown" right/left with every revolution of the record. This would force the needle to  one side then the other of the spiral groove.

The amplitude of the oscillatlon will depend on the eccentricity of the hole, but not the frequency.

That's as far as I have got.

Hopefully others can see a way forward.................

The highest frequency will be when then distance between the stylus and centre hole is at it's greatest. The lowest will be the opposite side of the record. Assuming a constant rotational speed, the tangential velocity (which is proportional to frequency) is proportional to the distance from the hole. So...

Yes it is possible. I have constructed an elegant proof, but this margin is too small to contain it.

I might well have worded the question rather badly. Sincere apologies.

Fortunately it looks like sophiebear has got the right general picture. Moving from the Ellipse with its "power of 2", takes you towards a rectangle with its "power of Infinity".

Ok, I see it now. An ellipse, with the 1kHz signal in the groove travelling faster than it should when the true centre of the record is between the hole and the stylus, and travelling more slowly when the centre is remote from the stylus.

The perimeter of an ellipse is difficult to evaluate, but I can imagine someone figuring out the relationship between distance traced, speed and frequency change. At least a bit easier to imagine than Fermat's last Theory.

Nice one winky.

We started with "A" Levels and found a bit of difficulty with one ore two.

My "O" Level book has a few "Puzzles" at the end of some chapters, so if 15/16 year olds can do them, there should be no problems with the next three or four questions on this forum of life's successful souls

Robert has to saw a 10-metre long pole into 1-metre lengths. How long will it take him if he cuts one length every 3 minutes ?

A bag contains several discs, some red and some yellow. I have to take some out of the bag without looking. If I want to be sure that I pick at least four discs of the same colour, what is the least number of discs that I should take out of the bag ?

The rail journey from Ashfield to Beechgrove takes exactly 4 hours and trains leave each way on the hour and on the half hour. If you were on a train going from Ashfield to Beechgrove, how many trains going from Beechgrove to Ashfield would you pass during your journey ?

A coin and a die are tossed together. What is the probability of getting

1. a head on the coin and a six on the die,
2. a head on the coin or a six on the die (or both)
3. a head on the coin or a six on the die (but not both)

If it takes five men five days to plough five fields, how long should it take one man to plough one field, working at the same rate ?

As non of the luminaries has responded yet, I'll try some quick answers.

Roberts needs to make 9 cuts to get 10 segments as the last cut yields two segments. 9 cuts * 3 minutes = 27 minutes.

I don't get the second question. What do you mean by saying "to be sure"? More likely then not (>50%) or ...?

Lets say we look at the noon train from Ashfield. The train will meet the 8 a.m from B at departure and the the 4 p.m at Beechgrove station. Excluding these two, it will meet the 15 trains with departures from 8:30 a.m. to 15:30 p.m on its journey.

Assuming the coin toss can have only two outcomes, heads or tales, each has a chance of 1/2. For a six on the die, the chance is 1/6.

1. Head and six means both chances have to be multiplied -> 1/2 *1/6 = 1/12

2. Head or six or both means both chances can be added to each other -> 1/2 + 1/6 = 7/12

3. Head or six not both means the chances for head (1/2) and any number from 1 to 5 (5/6) and tales (1/2) and six (1/6) can be added -> 1/2 * 5/6 + 1/2 * 1/6= 1/2

Five days, each one needs five days for "his" field.

head/tails is 1/2;

6 or any mumber on die is 1/6

any combination of both is the same? So 1/12?

I get the following:

1) Head & Six probability is 1/2 * 1/6 = 1/12

2) Head or Six or both = 1/2 + 1/6 = 2/3

3) Head or Six but not both = 1/2 * 5/6 + 1/6 = 7/12

You can also calculate Head & Six but not both by subtracting (1) above from (2) ie 2/3 - 1/12 which again yield 7/12

2) odds for a head on a coin or a 6 (or both) is 1/2 + 1/12 = 7/12 (one combination of a 6 plus a head is included in the odds for a head)

3) odds for a head on the coin or a 6 (but not both) is therefore 1/2.

7 discs to be sure of at least 4 red or yellow discs.

I somehow managed to misunderstand the "four discs of the same colour" part . This is obviously the right answer.

Yep - I'm guilty of double-counting !

So I now get the same as others:

a) 1/2 * 1/6 = 1/12

b) 1/2 + 1/2 * 1/6 = 7/12

c) 1/2 * 5/6 + 1/2 * 1/6 = 1/2

Well, it looks like you've all been busy whilst I was in Norway.

Mulberry made a cracking start, then with great help from Lionel and sjbabbey and push from sophiebear you seem to have got them all sorted. great show.

staying with the "O" Level stuff one more time...............before moving back up to the "A" Level stuff......

A flat ring, (a better description might be a flat disc with a concentric hole) is formed by the area enclosed between two concentric rings.

A straight line ABC, cuts the outer circle at A and C and is tangential to the inner circle at B. AC is 200mm long.

What is the area of the flat ring (or the solid part of the flat disc with a concentric hole)

Don

That's a nice little problem. Hopefully a bit easier to solve than the goat puzzler ?

If the small hole radius is "r" and the outer ring radius is "R" then the area of the annular ring is given by:

Area = Pi * ( R^2 - r^2)

The lines BC, R and r all form a right-angle triangle. R forms the hypotenuse and the length of BC is 100 cm. Applying Pythagoras gives the following:

R ^ 2 = 100 ^2 + r^2 ;

Therefore (R^2 - r^2) = 100 ^2

So the area of the annular ring is Pi * 10,000 square cm

Hi Sophiebear,

Yes, alot easier than the tethered goat ! and you most certainly have the right concept, but.....................

I'll let you tidy up anarithmetical oopsie....starting with AC being 200mm long (not 200cm) - remember, this is your "O" Levels from 50 years ago, not this year. 50 years ago you had to get the whole thing right - arithmetic included, not just the difficult conceptual bit !

None-the-less, well done !!

Point taken Don....

But 50 years ago we would probably have been working in feet and inches !

Ok, now for the "A" Level equivalent of the previous problem.

A "bead" is made by drilling a straight hole through a solid sphere, passing through the centre of the sphere, ie along a diameter..

The length of the resulting hole in this "bead" is 6"

What is the volume of solid material in the resulting "bead"

Cheers

Don

PS. Marks are 5/10 for the correct arithmetic answer and 10/10 for the associated "proof"

Point taken - see the next one, ie the "bead" above - i've converted it to feet and inches especially for us old-timers

if the radius of the bead is 3 ins then on the basis that the volume would be 1/3Π r ˆ 3 i.e. 9Π cubic ins i.e. about 28.286 cu ins.

This doesn't take into account the volume of any material removed by the drill which cannot be calculated without knowing its width.

Nice try, but only 1/10 i'm afraid so far !!

Hint. the volume of a sphere is 4/3.∏. r^3.................. (not 1/3)

However, you are quite correct to discard the volume of any material removed by the drill, that is what turns a "sphere" into a "bead" and it is the volume of a "bead" that we are after.

How about a "bead" that started out as a 12" diameter sphere ? or a sphere that started out 40,000km in circumference (eg the Earth). What would be the volume left in such a "bead" whose "hole" was only 6" long.

5/10 for the numerical answer(s). 10/10 for the proof(s).