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mutineer
2003-Aug-26, 07:42 PM
Two tumblers, I and II, contain equal volumes of pure liquids A and B respectively.
A spoonful of liquid A is taken from tumbler I and transferred to tumbler II.
The contents of tumbler II are stirred.
A precisely-equal spoonful of the contents of tumbler II is now transferred to tumbler I.

At the end of this operation, which of the liquids is the less contaminated?

(I’ll explain a few days from now why I selected this question.)

Glom
2003-Aug-26, 07:52 PM
I worked it out. I'll give a bit of time for people to vote before I explain it publicly.

SeanF
2003-Aug-26, 08:14 PM
Well, I voted, but I feel like I'm missing something here. I seem to remember hearing about this before, but the way the math is working out when I do it is not what I seem to remember from before.

At any rate, I voted the way my math works out, so when we're all done here I'll find out what I'm forgetting! :D

parejkoj
2003-Aug-26, 08:42 PM
Heh... that's kinda neat. when I did it in my head, it could have gone either way, but on paper it was pretty obvious (which probably means my answer was wrong...) :-?

mike alexander
2003-Aug-26, 08:49 PM
The correct answer is to choose the third door.

darkhunter
2003-Aug-26, 08:50 PM
My mind is going...I can feel it...I can feel it....

[Desperatly gathering the rement of a once great intelligence]

Old, half forgotten memories of a bology experiment (which probably had nothing to do with the question :D )

Now just the wait to see the right answer :lol:

Donnie B.
2003-Aug-26, 08:58 PM
I think it may be possible that the problem is not well specified. It may be necessary to know how much liquid was in each container to start with.

JimTKirk
2003-Aug-26, 08:59 PM
Come on now!

The true answer is...


There is no spoon! :lol: :lol: :lol:

Glom
2003-Aug-26, 09:00 PM
I think it may be possible that the problem is not well specified. It may be necessary to know how much liquid was in each container to start with.

We are told that there are equal volumes of each liquid in each tumbler.

mike alexander
2003-Aug-26, 09:17 PM
And the answer behind door three is: equal.

If you don't treat it as a Marilyn Vos Savant Parade problem but just as concentration problem in chemistry, it's easier.

Now, how much wood could a woodchuck chuck...

Glom
2003-Aug-26, 09:19 PM
Well, since Mike has revealed it, I'll post what I sent to mutineer.


I simplified things slightly. I assumed that each tumbler contained 1000ml of each liquid and a spoonful contained 1ml.

At first
Tumbler I contains 1000ml of Liquid A
Tumbler II contains 1000ml of Liquid B

After the spoonful is transferred from I to II
Tumbler I contains 999ml of Liquid A
Tumbler II contains 1000ml of Liquid B and 1ml of Liquid A resulting in 1001ml of total volume
Contamination due to Liquid A in Tumbler II is 1 part per 1001.

After the spoonful is transferred from II to I
Tumbler I contains 999+1/1001ml of Liquid A and 1000/1001 ml of Lquid B resulting in 1000ml of total volume
Contamination due to Liquid B in Tumbler I is 1000/1001 parts per 1000 or 1 part per 1001
Contamination due to Liquid A in Tumbler II is unchanged because a proportional quantity of both liquids should have been removed.

So contamination in each tumbler due to the minority liquid is 1 part per 1001. The contamination is equal.

mike alexander
2003-Aug-26, 09:34 PM
Glom&mutineer:

I didn't think I revealed it. Well, I didn't mean to reveal it.

How was I supposed to know there were so many chemists out there?

Oh, well... at least learn something from Feynman. Wasn't he the one who said if you don't understand the problem, recast it in a form you can understand?

Glom
2003-Aug-26, 09:36 PM
:oops:

mike alexander
2003-Aug-26, 09:54 PM
No :oops: please. I didn't mean for that to be taken wrong. From now on I'll be sure to use a :wink:

Besides, when I'm angry my rants tend to go on a lot longer. And the Vos Savant was extremely clever, don't you think? :wink:

What I will say is that I could not solve the problem until I recast it as a concentration problem. The old 'bag of tools' thing.

mutineer
2003-Aug-26, 11:17 PM
I think it may be possible that the problem is not well specified.
Sorry, Donnie. I guess it's too late now. I forgot to mention that the operation was carried out by a right-handed male, on a weekday, in the northern hemisphere. Also, that no overflow occurs, and that you are not allowed to use math to get the answer.


I didn't think I revealed it. Well, I didn't mean to reveal it.
Don't worry mike! So far as I'm concerned, you didn't reveal nuthin!


I simplified things slightly.
Sorry, Glom, your answer is wrong. It's the first sentence that's wrong.
The correct answer can be explained without math in a single sentence!
Anyone want to take up the challenge?


Heh... that's kinda neat. when I did it in my head, it could have gone either way, but on paper it was pretty obvious ...
Appreciated your answer, parejkoj! You really did get the answer, I think - i.e. the answer to why I found it a nice question.

I'll be back to say more.

Musashi
2003-Aug-26, 11:27 PM
Not a chemist....

Start with 1000 ml in each tumbler. Take 500ml of A and add to B. This makes:

Tumbler 1: 500ml of liquid A. Tumbler 2: 1000ml of liquid B and 500 ml of liquid A. Stir. I have to assume a good mixture, because I don't know how to do it otherwise. Now, take 500ml of solution AB and add back to Tumbler 1. I think it ends of like this:

The 500ml of AB transfered should be 1/3 A, or approx 167 ml, and 2/3 B or approx 333 ml. That gives a total of 667 ml A and 333 ml B in Tumbler 1.

That leaves 333 ml of A in tumbler 2. So, they are equally contaminated...

Darn, I shoulda writen it out first... I worked it out in my head and thought B would be more contaminated :oops: ... wish I could change my vote.

Normandy6644
2003-Aug-27, 12:17 PM
Um guys, I might be missing something, but in the setup of the problem it never says that either of the liquids are contaminated......could be just me though, I'm no chemist :D

cyswxman
2003-Aug-27, 12:58 PM
Intuitively, I think tumbler 1 is less contaminated since you are putting a tiny fraction of it's original liquid back in it.

mike alexander
2003-Aug-27, 04:06 PM
Oh. Of course. What an idiot (sel-referential).

If you pour all of A into B, mix well, then pour half back, the concentrations will be equal.

(I think Martin Gardner appeared to me in a dream...)

Musashi
2003-Aug-27, 04:41 PM
cyswxman, that's what I thought too, but I worked it out (see above), and I think that they are equally contaminated.

Glom
2003-Aug-27, 05:09 PM
Sorry, Glom, your answer is wrong. It's the first sentence that's wrong.
The correct answer can be explained without math in a single sentence!
Anyone want to take up the challenge?

Well then, what is it?

johnwitts
2003-Aug-27, 05:39 PM
I got it wrong...

Could the real answer be, 'Driving Miss Daisy?'

mutineer
2003-Aug-27, 08:07 PM
Well then, what is it?
Ah, you're an impatient fellow, Glom!
Apologies if I sounded impolite when I said your answer was wrong earlier - but really the sentence "I simplified things slightly" was not quite right! Using math is not the simple route.
I promise you that the correct solution can be explained in about thirty words of simple logic. (You might need to use a semicolon or two to keep it down to a single sentence.) Thing is, you need to come at the thing from a new direction - and suddenly it's a cinch! The correct solution is just unbelievably obvious.
I won't keep you waiting forever!

SeanF
2003-Aug-27, 08:42 PM
mutineer,

Glom's final answer was that "[t]he contamination is equal."

Are you saying that is wrong, or just that his solution wasn't really "simplified" because there's a simpler way to get it?

mutineer
2003-Aug-27, 11:01 PM
mutineer,

Glom's final answer was that "[t]he contamination is equal."

Are you saying that is wrong, or just that his solution wasn't really "simplified" because there's a simpler way to get it?

The final answer was right. If you tackle to the IQ Question with math, you get the right answer (unless your arithmetic is at fault). I was hoping that people would treat it as a logic problem. BABBers go for a math approach far more readily than a "normal" population. It is entirely sensible and practical to use math - but less of a test of reasoning and mental agility (i.e. less of a test of IQ).

But Glom did not "simplify things slightly". It was that first sentence that I took objection to. Compared to the simple logical solution which I will finally present**, he "complicated things enormously".

**But I'd love someone else to come up with the perfect simple solution.

Normandy6644
2003-Aug-27, 11:26 PM
I'm sticking with my assertion that it was never stated that neither solution was said to be contaminated, therefore both have equal, and zero, contamination. :D

Musashi
2003-Aug-28, 12:14 AM
How about this. At the end of the experiment, both tumblers have the same amount of liquid they started with, so and missing A liquid in tumbler one is replaced by the same amount of B liquid. Therefore, they will have the same amount of contamination...

mutineer
2003-Aug-28, 12:49 AM
That's about it, Musashi!
Here is my own general explanation of the IQ Question:
-----------------------------------

There are two solutions to the IQ Question, each of which requires only a moment’s thought – but which produce contradictory answers.

(1) The first spoonful consists entirely of liquid A, but the second consists of a mixture – so the second produces less contamination. Therefore liquid A is the less contaminated.

(2) The two transfers leave equal volumes of liquid in each tumbler. The volumes of liquids A and B are equal. Thus, the amounts of contaminant in each tumbler must also be equal.

If both solutions occur to one, then one must look for an error in one of them. After further thought, one realises that (1) overlooks the fact that the second spoonful reduces the amount of contaminant in tumbler B. But (2) is logically flawless, and produces the correct answer.

I'll be back to say a few words more about why I selected the question.

mike alexander
2003-Aug-28, 01:15 AM
It is entirely sensible and practical to use math - but less of a test of reasoning and mental agility (i.e. less of a test of IQ).


Oh, poopy. (MST3K)

johnwitts
2003-Aug-28, 01:32 AM
OK...

Assume both tumblers contain 20 ml each. 1 spoon = 5 ml

Take 5 ml of A and mix with 20 ml of B. Mix becomes 25 ml, 4 parts B to 1 part A.

Take 5 ml of mix, which is 4 ml B and 1 ml A. Mix with 15 ml of remaining A.

Tumbler 'A' now has 16 ml A + 4 ml B.

Tumbler 'B' now has 16 ml B + 4 ml A.

Hence both have the same amount of cross contamination, and my origional assumption was incorrect. Can I have a re count?

mike alexander
2003-Aug-28, 01:59 AM
Hey everybody! Watch this! \:D/


Please explain how this is a test of IQ.




(Too bad there isn't an emoticon of an opening can of worms)

Normandy6644
2003-Aug-28, 03:09 AM
I was thinking that too.....more like a critical thinking test.

Jpax2003
2003-Aug-28, 05:59 AM
Two tumblers, I and II, contain equal volumes of pure liquids A and B respectively.
A spoonful of liquid A is taken from tumbler I and transferred to tumbler II.
The contents of tumbler II are stirred.
A precisely-equal spoonful of the contents of tumbler II is now transferred to tumbler I.

At the end of this operation, which of the liquids is the less contaminated?

(I’ll explain a few days from now why I selected this question.)

There are several ways of arriving at any of the 3 answers depending on the criteria. The answer is indeterminant because we must make assumptions about the question.

First, the both pure liquids could be the same liquid, therefore, no contaminant exists.

Second, we do not know the nature of the contaminant. If it is soluble in both or either liquids, we can use mathematics to answer the equation. If the contaminant sinks in one or both and samples were taken from the top of the column, no contaminant would be transferred. If the liquids are pure and without solutes, then do we consider the liquids to be contaminants on their own? If they are contaminants to each other, are these liquids soluble together? If oil and water were the pure liquids, which didn't mix, would you still consider it a contaminant? The spoon may be able, hypothetically, to remove the entire portion of Liquid A from Liquid B, depending on the nature of the liquids.

Third, depending on the nature of the liquds, we could get variable answers. Are we measuring contamination by %Volume or %Mass. Density would play a key factor here.

Fourth, what is the size of the spoon? If the spoon is large enough to hold the entire contents of liquid A and B then they would be co-mingled to the point of similarity.

I think a test of IQ should not be answering questions but formulating them. We all know that all lines of inquiry can be broken into simple yes-no questions. "It is the question that drives us, Neo."

There are several possible tricks, but the semantics of the argument provide a pretty straight-forward logic. We must assume that there actually are liquids and no tricks about fluid gasses. We must assume that the contents are greater than zero, or else we have a divide-by-zero error. We must assume that the Tumbler I has volume V of Liquid A and that Tumbler II has volume V of Liquid B... therefore we do not assume that the statement means Tumbler I has 50/50 mix of Liquids A & B to start. We must assume that that the Tumbler has a capacity of Volume T and the Spoon has a Volume S1 and holds a Volume U and that T>SUV. (if T<SUV then contents spill out giving skewed results) We must assume that Liquid A and Liquid B are not the same liquid. We must assume that the spoon holds a volume of liquid less than Volume V. We must assume that the spoon is used in a proper manner (e.g. it is not inverted so as not to hold liquid, but even then viscosity and cling would allow transport, same with a slotted spoon). We must assume that the liquids are soluble or mixable, elsewise stirring in unnecessary (we must assume the questioner is not an idiot). We must assume that the liquids themselves are the contaminant, as no other matter are mentioned in the statement. We must assume that temperature is constant, there is no change in Volume V through vaporation, that the liquids are non-reactive. We must assume that...

The Answer is that people make a lot of assumptions when answering (or constructing) what appear to be simple or basic questions.

SAMU
2003-Aug-28, 07:26 AM
Et al, see above

Well said.

jokergirl
2003-Aug-28, 08:46 AM
I knew that one already, so I exclude myself from this poll.

;)

mutineer
2003-Aug-28, 09:54 AM
The Answer is that people make a lot of assumptions when answering (or constructing) what appear to be simple or basic questions.
You are quite right, of course!
Actually, if the amounts of both liquids are negative and the spoon is larger than the tumbler, you can still get the right answer if you work with signed binary variables.
Real problem is, I forgot to mention it was raining at the time of the experiment ...

Kaptain K
2003-Aug-28, 11:17 AM
I knew that one already, so I exclude myself from this poll.

;)
Me too! 8)

mutineer
2003-Aug-28, 04:14 PM
I said at the outset that I would be back to explain why I selected this question.

Well, all of you who got the wrong answer, give yourself a slap on the back. What makes this IQ Question such a gem (so intriguing) is that getting it right shows a NEGATIVE CORRELATION with IQ!


Please explain how this is a test of IQ.So your skepticism is well justified, mike!

Intelligent people RELIABLY get this IQ Question WRONG … whereas subjects of lower intelligence produce more varied answers, and are more likely to go for the “equal contamination” answer.

The intelligent mind grasps at once that there is an inequality involved. The first spoonful is pure; the second is not. The intelligent mind concludes that liquid A is the less contaminated. The intelligent mind just cannot resist going there! To no one does the thought occur: “Only the second spoonful takes away any contamination, so tumbler B must be the less contaminated.”

I am inclined to think that however high up the IQ scale you go, people’s minds tend to work the same way on this one. Only when the question is highlighted (e.g. by being selected to go on a BB), is suspicion aroused that one should be wary of ones first thoughts. One should think it over again. Or at least check it out with math. I would guess that most people who got the right answer applied figures to the problem – clearly a sensible thing to do, glom and mike. But second best to reasoning out the right answer without math. Congrats to anyone who did so!

//////////////////

Great fun can be had by having an accomplice introduce this question at a casual social gathering. Every intelligent person will immediately grasp the wrong answer. Play dumb and go for the right answer! Listen to their explanations of how you have failed to spot the solution. Look a bit confused as they make ever more desperate attempts to make you see sense. Take the line that: “Well, I don’t know. There seems to be a sort of mutuality about the whole thing. One spoonful goes one way. One goes the other. I’d kinda guess there was equal contamination on both sides.” Eventually, as they grow increasingly exasperated with your obtuseness, and raise their eyes to the ceiling, someone will reach for a pen and paper and try working it out with figures to convince you.

Enjoy the denouement!

snowcelt
2003-Aug-28, 04:18 PM
Oh sure, keep the morons in suspense. What is the offical answer mutineer?

SeanF
2003-Aug-28, 04:45 PM
Oh sure, keep the morons in suspense. What is the offical answer mutineer?



mutineer,

Glom's final answer was that "[t]he contamination is equal."

Are you saying that is wrong, or just that his solution wasn't really "simplified" because there's a simpler way to get it?

The final answer was right.

snowcelt
2003-Aug-28, 05:09 PM
I had heard this one years ago and it always escaped me. I always thought there was some kind of trick. Still looking hard for the trick. I must be one of the lower iq types. Just do not get it.

If I have 2 100 ml containers labeled a and b.
I remove 10ml from a and place in b leaving 90ml of pure liquid in a and 110 ml of contaminated liquid in b.
a is 100% pure.
b is 90.909090...% b and 9.1010...% a
How in the heck can one do the math on this?
I can not see this through.
With this example, walk me through please.

weatherc
2003-Aug-28, 06:14 PM
There is a way to show this graphically that shows how this can work. Take two beakers of liquid, each with 20 parts of liquid:

AAAAA BBBBB
AAAAA BBBBB
AAAAA BBBBB
AAAAA BBBBB

Now, take five parts of Liquid A and place it on top of Liquid B:

-------- AAAAA
-------- BBBBB
AAAAA BBBBB
AAAAA BBBBB
AAAAA BBBBB

When Liquid B is mixed with the parts from Liquid A, the result can be represented in the following manner:

-------- BBBBA
-------- BBBBA
AAAAA BBBBA
AAAAA BBBBA
AAAAA BBBBA

Now, skim the top five parts off of Liquid B (which will include 4 parts B + 1 part A), and place it on top of Liquid A:

BBBBA BBBBA
AAAAA BBBBA
AAAAA BBBBA
AAAAA BBBBA

The mixture in Liquid B can be rearranged in the following manner to show the final result, which is equal contamination:

BBBBA AAAAB
AAAAA BBBBB
AAAAA BBBBB
AAAAA BBBBB

There. No messing around with any scary percentages or math to get the final answer (which is good, because I'm bad at math). :D

snowcelt
2003-Aug-28, 06:44 PM
thankyou very much. :oops:

weatherc
2003-Aug-28, 06:52 PM
snowcelt,

Nothing to be embarrassed about. I couldn't figure it out either until I started thinking visually instead of mathematically.

mike alexander
2003-Aug-28, 07:03 PM
I've always been highly skeptical of the whole IQ thing in the first place. Not because I don't think some people are smarter than others (whatever that means), but because it tends to measure what the tests are testing for (which sounds tautological, but I guess that's close to my thinking).

One place I see a deficit is failure to link the more abstract IQ concepts to demonstrable survival value; I'm thinking in evolutionary terms here. Smart has to be relative to the local environment. As a deliberate exaggeration, surviving in a departmental colloquium and in the Kalahari desert may require very different sorts of smarts.

As to the problem at hand, I find it interesting that when I got bollixed up in the abstract logic of it I recast it in a concrete problem using the mathematical tools I have. Once I had a specific answer I could think more generally by trying an extreme (pour all of A into B and mix). Once I had two examples that gave the same answer I could then go back and 'construct the logic' of it, but it was post-hoc because I pretty much knew what the answer had to be. The problem told me more about the way I think than how 'intelligent' I am. Which is more interesting, anyway.

So I am more than mollified. Actually a fun run, mutineer. Got any more?

(Others can also pick up a copy of Martin Gardner's book, 'Aha! Insight!' which collects a bunch of problems of this type)

SeanF
2003-Aug-28, 07:14 PM
weatherc's graphic format only really works, though, if the number of columns and rows are equal (or factors, I think), so it doesn't necessarily apply to all possible amounts.

I look at it this way. At the end of the experiment, you're going to have tumbler 1 with some A and some B, and tumbler 2 with some A and some B, right?

Now, we know that the A in tumbler 1 combined with the A in tumbler 2 would give us the original total amount of A. We also know, since we ended up with the same amount of liquid in both tumblers, that the A in tumbler 1 combined with the B in tumbler 1 is equal to the original total amount of A. Logically, if A1 + B1 = A1 + A2, then B1 = A2.

Therefore, we conclude that the A in tumbler 2 must be equal to the B in tumbler 1, and the contamination is equal.

mutineer
2003-Aug-28, 08:15 PM
Actually a fun run, mutineer. Got any more?
Thanks, mike! I've been racking my brains - but this one's a hard act to follow. I can't think of another that compares.

Anyone else want to give us an "IQ Question II"?

mike alexander
2003-Aug-28, 11:36 PM
Hmmm...this is an old one, so if everybody comes back with... THAT'S AN OLD ONE... what can I say?

You are handed two identical-looking slugs of metal that are clinging to each other. One is a magnet, the other is not. Determine which one is the magnet.

You may separate them and bring them back together, but only ONCE.

johnwitts
2003-Aug-28, 11:47 PM
Magnets can be easily identified using a fridge.

(Other than that, who knows?)

frenat
2003-Aug-29, 12:28 AM
Separate them and move them individually close to your monitor. If the monitor display changes shapes and colors that one is the magnet.

johnwitts
2003-Aug-29, 12:34 AM
Separate them and move them individually close to your monitor. If the monitor display changes shapes and colors that one is the magnet.

Then, go out and buy a new monitor...

Grey
2003-Aug-29, 02:31 AM
You are handed two identical-looking slugs of metal that are clinging to each other. One is a magnet, the other is not. Determine which one is the magnet.

You may separate them and bring them back together, but only ONCE.
I assume the challenge is to do this without recourse to other materials, or the problem becomes much too easy. I've sent Mike my solution, but I'll give the rest of you a chance to think about it a while longer. :)

Jpax2003
2003-Aug-29, 06:16 AM
You are handed two identical-looking slugs of metal that are clinging to each other. One is a magnet, the other is not. Determine which one is the magnet.

You may separate them and bring them back together, but only ONCE.

Well, you could wait to see which one will end up on a late night infomercial with some british accent promoting it for it's health benefits...

or

Assume that the slugs of metal are machined in normal magnet shapes... probably a short cylinder. rotate one of the cylinders 90 degrees from the other so that it is perpendicular. When you hold slug A face-forward and bring it toward slug B, you will be able to tell which is the magnet.

A is held still and approaches, B is perpendicular:
If B immediately clings to the face of A, then A is the magnet.
If B freely rotates 90 degrees then clings (face-to-face) to A, then B is the magnet.

The flat face of the cylinder in this illustration is the pole N or S, and the magnetic lines of force will try to orient one of the poles toward the other metallic object.

I hope that's the answer.

mutineer
2003-Aug-29, 09:05 AM
I'm not sure about this "only ONCE" thing.
I don't believe you have to bring the two slugs into actual contact at all - just within probing distance.
Like Grey, I've sent mike my solution.

mike alexander
2003-Sep-02, 01:15 AM
Here's an answer to the magnet question, which several posters have already gotten:

Two identical-looking metal bars stuck together, only one is a permanent magnet. Figure out which one is the magnet. You may manipulate them but you can only take them apart and put them back together once.

Pull them apart and place the face of one against the midpoint of the other one. If they stick together, the one touching the midpoint is the magnet, otherwise it is the other one.

This depends on knowing that the magnetic attraction at the center of a bar magnet is (ideally) zero. In practice it is very close to it (realizing that in thought problems like this everything is idealized).

Mutineer wrote to ask if there is some way to solve the problem without specific knowledge of how the dipole in a bar magnet works (apologies for severely truncating your missive, M). Since I first heard this one many years ago, I had no idea. Part way through my reply to him I realized there might be a way, as long as you know how a bar magnet attracts things. As so:

Lay a bar magnet on the table in front of you. Bring a chunk of iron toward the right magnet face: it will be pulled to the left. Now bring the iron chunk toward the left magnet face and it will be pulled to the right. Logically, there must be a place in the magnet where the direction of pull reverses, leaving no net pull on the iron. In the abscence of any other factors, it is likewise logical to assume this happens at the center of symmetry of the magnet. Continuing, bringing a chunk of iron to the center of the magnet should result in no attraction of the iron. (again, the ideal magnet and the ideal, probably dimensionless, chunk of iron). The choice then becomes trivial.
](*,)

wedgebert
2003-Sep-02, 02:59 AM
I was just going to break one of the slugs into little pieces and see how they react with the other slug.

mike alexander
2003-Sep-02, 03:07 AM
I was just going to break one of the slugs into little pieces and see how they react with the other slug


Hmmm. If you break the MAGNET in two, you'll end up with two magnets, which you could then test for like poles. If you break the other one, there won't be any poles.

Separating with a vengeance.

wedgebert
2003-Sep-02, 03:43 AM
I was just going to break one of the slugs into little pieces and see how they react with the other slug


Hmmm. If you break the MAGNET in two, you'll end up with two magnets, which you could then test for like poles. If you break the other one, there won't be any poles.

Separating with a vengeance.

Well there you go, problem solved. If the two halves attract/repel each other, that one was the magnet. Otherwise the other one is the magnet.

Jpax2003
2003-Sep-02, 09:40 PM
I was just going to break one of the slugs into little pieces and see how they react with the other slug


Hmmm. If you break the MAGNET in two, you'll end up with two magnets, which you could then test for like poles. If you break the other one, there won't be any poles.

Separating with a vengeance.

I thought we had to do this without recourse to other materials. How to you propose to break the metal magnet to pieces... with bare hands?

If you can break the magnet with bare hands you are probably a superman from the planet Krypton and, therefore, you have laser vision which may be affected by magnets, so you wouldn't need to break them anyways.

mike alexander
2003-Sep-02, 10:17 PM
Dang! You probably saw the red S hrough my cheap shirts.

Yes, I put restrictions on the original thought problem. But isn't it fun to see people looking for other ways to solve it?

frenat
2003-Sep-03, 01:28 AM
Actually, some magnets are easily broken. It just depends on how thick the magnet is.

The Supreme Canuck
2003-Sep-03, 01:35 AM
Magnets can be fragile:

http://www.scitoys.com/scitoys/scitoys/magnets/gauss.html


It is very important that you keep the magnets from jumping together. They are made of a brittle sintered material that shatters like a ceramic.

https://4ua.com/scitoys/cgi-bin/shop.exe?page=magnet_desc.html&QX=0a&SID=169597267 4&something_to_remember_me_by_an_bye_and_bye&


CAUTION: Neodymium magnets are so powerful that they will break if allowed to jump together.