Two Buckets

[LaurV]

Originally Posted by Uncwilly

The problem is, each bucket is exactly 1 gallon in size.

But you have an additional cup

[retina]

Two gals gallons and one cup.

[MattcAnderson]

Thanks for all the good replies, club.

Right answers go to Retina, LaurV, and Mawn.

The answer is c) same amount of water in the alcohol as alcohol in the water. This is because both buckets contain a gallon, and any fluid that is not alcohol is water and vice versa.


Cheers,
Matt

[retina]

Originally Posted by MattcAnderson

Right answers go to Retina, LaurV, and Mawn.

Hmm, well apparently not as the follow up discussion would tend to indicate some other factors are at play here.

[science_man_88]

Originally Posted by retina

Hmm, well apparently not as the follow up discussion would tend to indicate some other factors are at play here.

I pointed out to the OP through PM that it matters if it's v/v, w/v, or w/w measured.

[MattcAnderson]

The question is this.
First you have a bucket with a gallon of water and a gallon of alcohol.
Then you take a cup of alcohol and pour it into the water.
Then you take the same cup and pour a gallon of the mixture back into the alcohol.
Why is there the same ammount of alcohol in the water as water in the alcohol?
Let A be the ammount of alcohol in bucket 1.
Let G be one gallon
Let C be the cup volume
After the first cup action, the bucket 1 will contain
A = G - C
Let W be the ammount of water in bucket 2
After the first cup action, bucket 2 will contain
W = G + C

okay this is a bad way to start. Lets try new letters
Let W be a gallon of water
let A be a gallon of Alcohol
let 'a' be a cup of alcohol.
the 3 ammounts we have are
Bucket 1 | Bucket 2
A W
A-a W+a
The third step gets hairy, but know this.
At the end, both buckets have the same volume of liquid
Whatever is not Alcohol in either bucket is water.
So bucket 1 conatins A-a+(a1 + w1) and by deduction
bucket 2 will contain W+a - ( a1 + w1).
Rearanging terms gives
Bucket one has in the end, A-a+a1+w1
Bucket 2 has also, W+w1+a-a1
This does not explain it well, but the book says the volumes of liquid are the same.
The title of the chapter is the fine frustrating art of numbermanship.
Regards,
Matt A
8/19/2014

[LaurV]

You are making it extremely complicate. The right "motivation" is illustrated very well in post #6 by The Mawn, you have some quantity q of liquid one and some quantity q of liquid 2. No matter how you mix them in two recipients R1 and R2, you can repeat your "cup-related" activity a billion times, masturbation, as Chris would say, but if you end up with the same quantity q of mixture in both recipients a and b, then you have in recipient R1 a quantity x of liquid 1 and a quantity q-x of liquid 2. And because you only had a total of q for each, then it is a must that you have in the second recipient a quantity q-x of liquid 1 and a quantity of q-(q-x)=x of liquid 2.

So, always the concentration of liquid 1 in liquid 2 in the first recipient is x/q, which is the same as the concentration of the liquid 2 in the liquid 1 in the second recipient.

And this is the solution given by the Mensa and HIQS tests too.

What the people here are pointing out is that in real life, this is not true, because mixing two liquids (water and alcohol here) will very seldom end up with the sum of volumes. Example, mixing one gallon of ethanol with one gallon of water will give you a total volume of 1.92 gallons (and not 2.0 gallons as expected). And why this happens is because the molecules of the two liquids are different in size, and go "in between". Imagine having a bucket of sand and a bucket of small rocks. When you mix them together, you will get only a bucket and half, because the sand goes between the rocks, occupying space which was empty before in the rocks bucket, without overflowing the bucket.

So, in real life, the same cup can hold "more" mixture of liquids, than it can hold from every liquid apart. It is like your cup is "growing up". Assume you have a very minuscule cup which can only hold 100 molecules of liquid 1, and - because the molecules of liquid 2 are smaller - it can hold 150 molecules of liquid 2. You would expect that if you fill the cup with a fifty-fifty percent mixture (or fill half of the cup with liquid 1 and another half with liquid2), to have 50 molecules of liquid 1 and 75 molecules of liquid 2 in the cup. But in reality, this is always false. If your liquids repel each other, you may have only 45 molecules of liquid 1, and 68 of liquid 2. In the given case of water and alcohol, you would expect a 1.92/2 volume decrease, so your cup will hold 52 molecules of liquid 1 and 78 molecules of liquid 2. Remark I didn't say water and alcohol, I only used the number for proportion, because, in fact, water and alcohol molecules do not have the size proportion given above.

But you got the idea. My question was: does this "volume shrinking" influences the result, assuming you end up with the same volume in the recipients? And if so, how? [edit: and of course, I understand that this is a new puzzle!]

[wblipp]

Originally Posted by MattcAnderson

After the first cup action, bucket 2 will contain
W = G + C

As has been pointed out many times in this thread, this is wrong.

I don't think that anybody is confused about the answer if this approximation is used.

[pdazzl]

I wrote it out in long hand fraction and got that they are the same (polar opposites).

Water bucket contains:
256/17 cups water
16/17 cups alcohol

Alcohol bucket contains:
256/17 cups of alcohol
16/17 cups of water


Also, the question didn't state the bucket was of one gallon size, only that it contained a gallon of each. What if each bucket could only contain say 16.5 cups and each pour resulted in half a cup spilled? :) [Yes, the second pour would not cause it to overflow but it makes for an interesting problem]

[wblipp]

The real world answer is more water in the alcohol bucket.

Beginning Bucket 1:
16 cups of water
0 cups of alcohol
Total volume 16 cups

Beginning Bucket 2:
0 cups of water
16 cups of alcohol
Total volume 16 cups

Step 1 pours 1 cup from Bucket 2 to Bucket 1. This cup contains 1 cup of alcohol. It mixes in Bucket 1, resulting in less that 17 total cups.

Step 1 Bucket 1:
16 cups of water
1 cups of alcohol
Total volume less than 17 cups

Step 1 Bucket 2:
0 cups of water
15 cups of alcohol
Total volume 15 cups

Step 2 pours one cup from Bucket 1 to Bucket 2. The water and alcohol are mixed 16:1, so the cup will have water and alcohol in that mixture. But you will get more than 1/17 of cup of alcohol plus 16/17 of a cup of water because that mixture has a volume of less than 1 cup. Let this additional amount be a factor of (1+a) - we could look up a, but we know it is a small positive number. The second pour will move (1+a)/17 cups of alcohol and (16+16a)/17 cups of water.

Step 2 Bucket 1:
(256-16a)/17 cups of water
(16-a)/17 cups of alcohol
Total volume less than 16 cups

Step 2 Bucket 2:
(16+16a)/17 cups of water
(256+a)/17 cups of alcohol
Total volume less than 16 cups

a is a positive number, so 16-a < 16+16a, so there is more alcohol in the water bucket.

Note that a=0 gives the expected solution of equal results.

[Uncwilly]

I still protest that you all may be wrong. The OP did not state the volume of the buckets. More information is needed. If volume of the buckets is less than a gallon and a cup, the size becomes a major factor.