A Fistful of Brain Teasers

Don Atkinson posted:

Do the following bits of arithmetic and let me know what you think :-

2 + (1 x 9) = ?

3 + (12 x 9) = ?

4 + (123 x 9) = ?

5 + 91234 x 9) = ?

6 + etc

 

Belated but sincere thanks to ken c for the above teaser........ken didn’t incorporate the number “9” at the start of the 5 + (1234 x 9) line ! That was my error !

thanks ken.

Getting Home Early !

An office worker regularly arrives at his home station at exactly 5 pm (this is a fairy tale where the trains run EXACTLY to time !!!!)

His wife always meets him and drives him home (I already said it was a fairy tale!)

Today he took an earlier train and arrived at his home station at exactly 4 pm. It was a nice fine day today, so instead of telephoning, he set of to walk home, following the precise route his wife always takes and they met along the way. He got into the car and they drove home arriving 10 minutes earlier than usual.

Assuming the wife drove at a constant speed and that today she left home just in time to arrive at the station at 5 pm, can you figure out how long the office worker had been walking when his wife picked him up?

Eoink posted:

Is it 55 minutes?

It is indeed.......

......hopefully IB's endorsement increased your confidence a bit ?

Would either of you like to outline your reasoning, for the benefit of anyone watching from the "wings" and who might still be wondering how to figure it out ? (If not, I'll do it later.)

Time for wife to drive to station = t minutes
So she left at 5pm - t minutes
She was expecting to pick him up at 5pm and drive home, so returning home at 5pm + t minutes
Instead they returned home 10 minutes early at 5pm + t - 10 minutes
So the total journey time was 2t - 10 minutes instead of the usual 2t
She drove at constant speed, so she drove t - 5 minutes each way.
She left at 5pm - t minutes and met her husband t - 5 minutes later, thus 5pm - 5 minutes, or 4.55.
He left the station at 4pm, so had walked for 55 minutes.

That was a lot quicker to do than to try to explain

 

I did it in a slightly inelegant way, calculating times. And coming to paste this after typing out I see Eoink has beaten me - and I agree it take a lot longer to write it out in an understandable way than to do!

X = miles station to home
Y = miles walked to pick up point
Z = miles pick up point to home
A = car mpm (miles per min)
B = walk mpm

Time for Normal drive home (min) = X/A
Time for walk (min) = Y/B
Time for drive to or return from pickup (min) = Z/A

Today compared to normal:
Y/B + Z/A = 60 + X/A - 10
Y/B + Z/A - X/A = 50
- - - - -
Wife leaves home to gets home
Normal = 2X/A
today = 2Z/A
Today time is 10 min less, so
2X/A -10 = 2Z/A
-5 = Z/A - X/A
- - - - -
Substituting
Y/B -5 = 50
Y/B = 55

Eoink posted:

Time for wife to drive to station = t minutes
So she left at 5pm - t minutes
She was expecting to pick him up at 5pm and drive home, so returning home at 5pm + t minutes
Instead they returned home 10 minutes early at 5pm + t - 10 minutes
So the total journey time was 2t - 10 minutes instead of the usual 2t
She drove at constant speed, so she drove t - 5 minutes each way.
She left at 5pm - t minutes and met her husband t - 5 minutes later, thus 5pm - 5 minutes, or 4.55.
He left the station at 4pm, so had walked for 55 minutes.

That was a lot quicker to do than to try to explain

 

Nicely done !

And yes, it's nearly always a lot quicker to do these sums than it is to explain them

Innocent Bystander posted:

I did it in a slightly inelegant way, calculating times. And coming to paste this after typing out I see Eoink has beaten me - and I agree it take a lot longer to write it out in an understandable way than to do!

X = miles station to home
Y = miles walked to pick up point
Z = miles pick up point to home
A = car mpm (miles per min)
B = walk mpm

Time for Normal drive home (min) = X/A
Time for walk (min) = Y/B
Time for drive to or return from pickup (min) = Z/A

Today compared to normal:
Y/B + Z/A = 60 + X/A - 10
Y/B + Z/A - X/A = 50
- - - - -
Wife leaves home to gets home
Normal = 2X/A
today = 2Z/A
Today time is 10 min less, so
2X/A -10 = 2Z/A
-5 = Z/A - X/A
- - - - -
Substituting
Y/B -5 = 50
Y/B = 55

Again, nicely done.

Just shows......there is nearly always more than one way to solve a problem.

Hopefully anybody browsing and following both yourself and Eoin, will now be able to say "ah! now I understand !"

Reasoning: from the question, the volume of the bead solid is same regardless of diameter of original sphere - if  large diameter, the hole will be almost as large diameter to get the 6in hole height, making the bead very large but with thin wall. If small, the bead will be smaller, but thicker. If hole infinitesimally small, the sphere could only be 6” diameter, hence volume of the solid is volume of the 6in sphere, 4/3 πr³

Don Atkinson posted:
Beachcomber posted:

What is the diameter of the hole?

Drilling a hole through a sphere creates a bead.

The length of the hole through the resultant bead is 6".

 

Yes. I know the length of the hole is about 6" (slightly less at the edges of the hole).  But I don't see any information which tells me what diameter the hole is.  If it is 6" then nothing is left of the bead.  If it is .00001 inches then as near as dammit the bead is pretty much the same as it was.

I must be missing something - either the diameter of the drill bit used to drill the hole or something else...

Beachcomber posted:
Don Atkinson posted:
Beachcomber posted:

What is the diameter of the hole?

Drilling a hole through a sphere creates a bead.

The length of the hole through the resultant bead is 6".

 

Yes. I know the length of the hole is about 6" (slightly less at the edges of the hole).  But I don't see any information which tells me what diameter the hole is.  If it is 6" then nothing is left of the bead.  If it is .00001 inches then as near as dammit the bead is pretty much the same as it was.

I must be missing something - either the diameter of the drill bit used to drill the hole or something else...

From Don’s drawing it is the edges of the hole that are 6” long, not the centre. My assumption was that the hole can be any diameter, but with the fixed height the volume of the bead surrounding would be the same, as I tried to say in explaining my splution. The assumption was purely through Don’s pointedly giving only the one dimension.

Ah, - I see - so if the length of the hole is 6 inches long at the edges of the hole then... but wait, he said 6 inches along the diameter, so that was why I assumed that the sphere is 6 inches in diameter.  If we are saying that it is 6 inches long at the edges of the hole, then things get awfully complicated.  We are now talking about a chord that is 6 inches long.  But any size of sphere can have parallel chords that are 6 inches long.  Are you saying that the segment created by a chord that is 6 inches long is the same for all spheres (greater than 6 inches in diameter)?  

I think we have three volumes. We have a cylinder through the sphere of radius r, and volume 6 PI r^2. We have two segments of the spehere, one at each end of the cylinder which are missing, radius of r and height h. We have a sphere of diameter 6 + 2*h, so volume 4/3 PI (3+h)^2. I can't immediately see how to work out h from r and 6, if I could I could solve the three volumes and I strongly suspect as IB does that given the phrasing of the question they'll be independent of r as the terms will cancel.

Ah, I'm looking at as a sphere, hollowed out by a cylinder and with the cap at each end of the cylinder removed. So we have the volume of the sphere, from which we subtract the volume of the cylinder and the volume of the 2 caps. I'm pretty sure that if I solve for these, the terms with the radius of the cylinder will cancel out, and we'll end up with the volume of a 6 inch sphere, but my near-40 year old A-Level maths is not coming back to me.

All three of you, IB, Beachcomber and Eoin are well on the way to success.

However, you are all basing your projected outcome on either TRUST or assumptions.

I’m somewhat flattered that you trust that I have provided sufficient information that you can assume that the radius of the sphere is irrelevant. 

Well, you can! It doesn’t matter whether the sphere is 6” (the smallest), 12” (say) or 8,000 miles (the world). The volume of the residual “bead” or “ring” is the same. Almost unbelievable !

However, as I say, this is based on trust. Eoin has a way forward to prove that R is irrelevant. But there are other ways.

Eoink posted:

Ah, I'm looking at as a sphere, hollowed out by a cylinder and with the cap at each end of the cylinder removed. So we have the volume of the sphere, from which we subtract the volume of the cylinder and the volume of the 2 caps. I'm pretty sure that if I solve for these, the terms with the radius of the cylinder will cancel out, and we'll end up with the volume of a 6 inch sphere, but my near-40 year old A-Level maths is not coming back to me.

Hi Eoin,

Twenty years ago, I would have reached for my school maths homework book, but these days it’s much easier, and just as valid, to “google” ........

volume of a sphere

volume of a cylinder

volume of a spherical dome

my guess is that it’s only the dome that the old grey matter needs to refresh.

But it’s surprising how nicely the various terms cancel out leaving a volume devoid of R.

I’ll post my solutions tomorrow. One as outlined above, the other based on integration. You are on the right track.

Eoink posted:

Thanks Don, my brain’s a bit frazzled after one of “those days” at work, I’ll try to have a look tomorrow. 

There’s no rush Eoin, these brain teasers are meant to be enjoyable and a distraction from the pressures of work.

The ref to Teresa May etc was simply topical melodrama 


 

Integation Solution JPEG

This is my "integration" solution.

I hope it is readable, but I will post a typed-up version later if necessary.

This satisfies Putin's IB's challenge above to "...provide proof that there is a solution"

I await an alternative solution, hint - the one that Eoin outlined above would do nicely and is dead easy to do.

Don Atkinson posted:

Integation Solution JPEG

This is my "integration" solution.

I hope it is readable, but I will post a typed-up version later if necessary.

This satisfies Putin's IB's challenge above to "...provide proof that there is a solution"

I await an alternative solution, hint - the one that Eoin outlined above would do nicely and is dead easy to do.

It’s easy when you solve the right way Don, my first pass was 3 sheets of A4.

We have a sphere drilled through. The sphere is thus a bead made up of a sphere minus a cylinder through the sphere and minus two caps at each end. Let us assign the drilled hole an arbitrary radius of r. Let us say the caps have height h. (I’m known for originality in unit naming.)

Looking at Don’s picture above, we can easily see that the sphere has a diameter of 6+2*h, the length of the cylinder plus the height of the two caps. So it has radius 3+h.

The volume of the sphere is 4/3 Pi rsphere^3 = 4/3 pi (3+h) ^ 3 =

4/3*pi*h^3 + 12*pi*h^2 + 36*pi*h + 36*pi

——————————-

Google tells me that rsphere = (r^2+h^2)/2*h = 3+h which resolves to r^2 = 6*h + h^2

_____________

The volume of the cylinder is 6*pi*r^2 = 6*pi*(6*h + h^2) = 36*pi*h + 6*pi*h^2

Again from Google, the volume of a cap is pi/3*(3*r^2*h + h^3), substitute for r^2 and the volume of 2 caps is 6*pi*h^2 + 4*pi/3*h^3

Subtract the volume of the cylinder and the two caps from the volume of the sphere, all the h terms cancel out and you have 36*pi^2.

I can’t help thinking my 16 year old self would have seen this as a light workout before starting the hard bit of A’Level maths, not any more.

Great question, thanks Don.

Don Atkinson posted:

Well done Eoin.

quite satisfying when you get a result like that 

Thanks Don, very satisfying, especially after I realised that resolving r and h for r and not for h made the equations about half the length.  Thanks again, a good mental workout.

Boring Sphere Geometry Sol JPEG

My version of the "geometry" solution.

Some of the intermediate maths is omitted eg

......in the line starting "2 x DOMES" the second equation is obtained by substituting h = R - 3 (see diagram) then expanding.

.....And in the line starting "CYLINDER", r² is substituted by R² - 9 (see diagram)

Perhaps the purists would have us derive the formulae for a sphere, a dome and a cylinder, but then the previous "integration" method is probably easier

 Anyway, I think the concept of a 6" long hole through the Earth leaving a bead the volume as a 6" long hole through a 12" globe is somewhat surprising !

Chultenham Tin Cup JPEG

Chultenham Tin Cup.

My donkey (and donkey jockey) were running in some fancy race yesterday.

The diagram shows part of the “field of play”. Both the three mile wide turf and two mile wide sand extend well to the left and right of the field of play.

My donkey and rider, in competition with others, simply had to cross the field of play from the “Start” to “Finish” by the quickest route possible. It is important to note that all the donkeys travelled at the same speed as each other and could cross the turf at twice the speed they could travel across sand.

Being (slightly) more intelligent than my donkey and my donkey jockey, I felt my role in this race was to figure out the fastest route that I should encourage my donkey and jockey to take – if we all took the same route it would be a dead heat !

The straight line from start to finish didn’t seem to me to offer the fastest route and that was clearly true because we beat all those donkeys that took the straight line.

We also happened to beat all the other donkeys as well.

What route did I choose ?

PS: this one struck me as being a bit more difficult than some of the other teasers. It might be worth sharing your various thoughts...........rather than having a race to see who's first ! (pun intended !)

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