[Guest]

Not sure if a similar problem has been posted before. I checked but didn't find any, so apologies in advance if I wasn't careful enough in the search.

There are two bottles each with a capacity of 20 liter (litre, if you are in US). One bottle contains 10 liter of water, the other contains 10 liter of wine. There are two things you can do: (a) pour fluid from either bottle onto the floor, and (b) pour fluid from one bottle into the other. The only constraint is that you must always have at least 10 liter of fluid in the wine bottle. The object is to dilute the wine in the wine bottle with water as much as possible. What is the maximum dilution you can achieve (what is the minimum possible percentage of wine in the fluid in the wine bottle)?

Edited October 8, 2009 by DeeGee

[Guest]

If you pour the wine out until there is only one drop left, then pour all of the water into that, you'll have it. However, not knowing the viscosity of the wine or the volume of one drop, a precise dilution ratio cannot be determined. (Actually, you'd get more dilution by simply pouring one drop from the wine into the water, so as not to lose a few drops of water stuck to the walls of its pitcher or through spillage.)

(EDIT) Ah, nevermind, I misread the constraint there.

Edited October 8, 2009 by Arcxjo

[Guest]

If you pour the wine out until there is only one drop left, then pour all of the water into that, you'll have it. However, not knowing the viscosity of the wine or the volume of one drop, a precise dilution ratio cannot be determined. (Actually, you'd get more dilution by simply pouring one drop from the wine into the water, so as not to lose a few drops of water stuck to the walls of its pitcher or through spillage.)

The OP states "The only constraint is that you must always have at least 10 liter of fluid in the wine bottle." So, one cannot pour the wine out until there is only one drop left. One must add to the wine first.

[Guest]

Since the puzzle has the condition that there must be 10 liters of wine in the wine bottle at all times, the only way to make any change would be to add water while removing wine.

I suspect with some artistic license that you could place the two bottles to one another, allow the liquid to flow from one to the other by making them perfectly horizontal and then the resultant mix would be 50/50 ultimately, once settled in a perfect world and have remained 10 liters throughout.

[Guest]

What if you poured all the water into the wine bottle waited tillthe water and wine seperated (water is heaver) and carefully poured the wine out that is floating on the top

[Guest]

What if you poured all the water into the wine bottle waited tillthe water and wine seperated (water is heaver) and carefully poured the wine out that is floating on the top

This solution has merit but I am pretty sure that since water is heavier than wine, it will mix when poured upon wine, not separate from the wine. However if wine is slowly poured upon water it is possible to float it over the water. Not sure in this scenario how that could occur and stay above 10 liters in the wine bottle

[plainglazed]

so far the most diluted I can get is just less than 44 1/2% wine.

[Guest]

This is the 1st time im using this site. Excuse me if i do something wrong. Since you cannot pour wine into water (level will drop below 10 lts), you have to pour water into wine. If you try this piecemeal, you will never achive the maximum dilution to 50% that you would get if you mixed the entire 10 ltr of water into wine. Therefore the answer can only be 50%

[Guest]

Think I got it...

I'm mixing up liter and tenths of liters...but principle and answer is the same, I used 1 liter of wine/water in a 2 liter bottle

Pour 1/10 of water into the wine, making it 10/11% wine, then dump out 1/11th of the new solution, repeat until you're out water and you get (10/11)^9 (can only do this 9 times, as the last bit of water dilutes the soltion once it's in and regardless of what is dumped out the ratio is the same.

if you do this with 1/100 of water you get (100/101)^99 ratio...

So:

(X /(X+1))^(X-1) as X-> infinity

e^-1 = ~36.788%

Outside the box, letting some of the alcohol in the wine evaporate then mix (keeping it at a liter with the water)...or covering the water and letting the wine evaporate as much as possible and keep it at a liter. but that's not gonna be an exact answer with the info provided

Edited October 8, 2009 by ajrettke

[Guest]

Add 1 liter of water to the wine, the liquid is now 10 parts wine to 1 part water or 90.90909% pure wine. Remove 1 liter of liquid from the bottle to get 10 liters of pure "Wineter" wich is 90.90909% wine. Add 1 liter of water to the Wineter and get 10 parts Wineter to 1 part water. Or 90.90909% Winter(90.90909% wine)is 82.64462% wine.

Wash rinse and repeat until all water is gone.

I came up with a liquid that contains 38.55432% wine and 61.44568% water

[plainglazed]

...where I just did it once but now integrating get just less than 42%. Add half the water to the wine, spill the mixture out until 10L remains, rinse and repeat?

[Guest]

You only have to keep 10 liters in the wine bottle, not the water bottle. So pour 1 drop of water into the wine bottle. Now it has 10 liters plus 1 drop. Shake the crap out of it to mix, then pour one drop back into the water bottle. Still 10 liters in the wine bottle and the other bottle is filled with some weak drink indeed,

[Guest]

You only have to keep 10 liters in the wine bottle, not the water bottle. So pour 1 drop of water into the wine bottle. Now it has 10 liters plus 1 drop. Shake the crap out of it to mix, then pour one drop back into the water bottle. Still 10 liters in the wine bottle and the other bottle is filled with some weak drink indeed,

Good try but he was looking for the dilution in the wine bottle.

"(what is the minimum possible percentage of wine in the fluid in the wine bottle)? "

I would have to agree with ajrettke and his answer using limits.

Another way to look at this is that you want to remove as much of the wine as possible. By adding smaller amounts of water and removing the same water, the percentage of wine (during each fill and drain event) gets higher as the volumes approach the limit of zero.

Edited October 8, 2009 by twhedge

[Guest]

Please tell me if this is right.

It would appear that pouring small amounts of water at a time and then reducing the current mixture back to 1 liter gives a better water/wine ratio.

So. Let W be wine and T for water.

d is the amount of water we pour each time.

At the start we have: 10W

Then pour d amount of water: 10W + dT

Then pour empty to 1 liter: (10W + dT) / (10+d)

Repeat this process to what we have now: [ ((10W + dT) / (10+d)) + dT ] / (10W + dT)

Now unless I'm mistaken this generates the series:

[ 10W + dT + dT(10+d) + dT(10+d)^2 + dT(10+d)^3 + ... + dT(10+d)^(n-1) ] / (10W + dT)^n

where n = 10/d (the number of times we have to repeat the process.

So at the end we will have 10/(10W + dT)^n amount of water,

and dT + dT(10+d) + dT(10+d)^2 + dT(10+d)^3 + ... + dT(10+d)^(n-1) ] / (10W + dT)^n

We can cancel the denominator for both these. So all we need to do is find the value as d approaches 0 and n approaches infinity.

I will do this in soon. (I have to eat now) .

[Guest]

Right my last post is mostly nonsense. I'll will make a revised post shortly.

[Guest]

I'm going to make a more general formula this time.

So. Let W be wine and T for water.

d is the amount of water we pour each time.

A is our starting amount of liters of each.

At the start we have: AW

Then pour d amount of water: AW + dT

Then pour empty to 1 liter: A*(AW + dT) / (A+d)

Repeat this process to what we have now: (A*(A*(AW + dT) / (A+d)) + dT) / (A + d)

So this becomes the series:

{[A^n * AW] + [A^n * dT] + [(A+d) * A^(n-1) * dT] + ... + [(A+d)^(n-1) * A^1 * dT]}/(A+d)^n

Giving amount of wine: A^n * AW/(A+d)^n

Giving amount of water: dT[A^n + ((A+d) * A^(n-1)) + ... + ((A+d)^(n-1) * A]/(A+d)^n

where n = A/d (The number of times we must do the process)

Now we need to find the amounts as d approaches 0 and n approaches infinity.

I know I haven't had anywhere near as much maths education as some members of this board so I may very well get this next bit wrong.

We know that the amount of wine and water must total A. Also the amount of wine must be proportional to the starting amount.

The amount of wine we have is (1^n)/(1+d)^n

d is A/n: (1^n)/(1+1/n)^n

Now as d approaches 0.

The numerator is 1.

Fortunately I know that the denominator is e. So when A is 1 we have 1/e left. As this is proportional to A, We will generally have A/e amount of wine left.

This gives 1/e wine of the total solution.

[Guest]

Good work ajrettke and psychic mind. That is the correct answer. I did it the same way as you psychic mind, but didnt "see the e" and had to take a differential to maximise. Nice thinking that.

As for ajrettke's solution, I think there is a mistake in taking (X /(X+1))^(X-1) whereas it should really be

(X /(X+1))^(X) but you get the same answer as you are taking X -> infinity.

[Guest]

This can be done in a multitude of ways, but you can add about a liter of water or any smaller amount. After doing so pour out ome of the liquid in the wine contianer being careful to keep at least 10 liters in it. Do this repetedly until the water in the water container is gone.

[Guest]

The puzzle is solved. I love the way my college mathematics come into play in these puzzles.