[Guest]

You and 9 other friends are preparing a party at your house. As a welcoming drink you filled a 20 liter bowl with sweet, completely non-alcoholic juice. You then dipped a 1/2 liter can into the bowl, drank the contents and poured from one of the many bottles of 40% vodka into the bowl to replace the 1/2 liter juice you took. Remember, this is just a welcoming drink... no need to get carried away.

A short while later one of your friends came to the bowl of the welcoming drink, and decided to spice it up a little. She took 1/2 a liter from the bowl, drank it, and replaced it with the vodka.

During the preparation of the party each of your friends had the same brilliant idea, so that before the guests arrived each of your 9 friends and yourself had replaced 1/2 a liter of the contents of the bowl with vodka.

Assuming that it was always done using perfect measures, and the liquid in the bowl would always be perfectly blended, what was the alcohol percentage (with reasonable precision) of the welcoming drink at the time of the guests' arrival?

[Guest]

9.975% I don't really feel like writing out my full solution since I did it in Excel. If others agree, I will let someone else post a solution.

[Guest]

a mixture at party time that was 8.03% alcohol by volume.

This is due to 10 additions of 1/2 liter of 40% vodka, or 0.2 liters per addition. Meanwhile, each friend actually reduced the alcohol slightly when they tested their 1/2 liter. In the end, it was about 1.6 liters of alcohol in the 20 liter mixture.

[Guest]

Neither of you can be right. I think it should be a little more than 9%. It should definitely be between 9% and 10%, because time you add 1/2 liter of 40% alcohol, that's adding 1% alcohol to the whole 20 liters. Each of the 9 friends also takes out progressively more of the existing alcohol out of the drink, but no one ever takes out 1% because that would mean that the alcohol content was already at 40% in the punch.

The host adds 1%. The first friend takes out 1/40 of 1 %, or .025%, and adds 1%, so there's 1.975% after the host and one friend. The next friend takes out 1/40 of 1.975% (which is a little less than 0.05%) and adds 1%.

If we round up every time and say that after the 1st friend there's 2%, then after the 4th friend, there's 5%, and so the 5th friend takes out 1/40 of 5% or .125% and adds 1%. If we assume that the average taken out by each friend is that .125%, then the 9 friends take out 1.125% and add 9%, putting the total alcohol at 8.875%, which rounds to 9. That's an overestimate of what gets taken out, though, so my guess is that the actual content is a little more than 9%.

[Guest]

I'm not sure what I missed that first time around, but now I am getting 8.94682%

Edited April 13, 2009 by wafflechip

[Guest]

I too counted on each friend drinking a ratio of juice/alcohol (with the ratio changing each time), but after 10 total additions, I ended up with 4.705996 liters of alc in the mix- for a percentage by volume of 11.765% Did I miss something?

BTW, this is my first post- got tired of being a "guest" I guess!

[Guest]

8.94691986469167633056640625% alcohol

Below are my numbers. The x/20 is the alcohol in the 20 liter bowl. The following number is the amount of alcohol taken from the bowl (1/40th of the current alcohol). So to get the next number I took .5 -.0125 + .5 = .9875. And so on.

.5/20

.0125

.9875/20

.02462

1.462875/20

.036571875

1.926303125/20

0.048157578125

2.378145546875/20

0.059453638671875

2.818691908203125/20

0.070467297705078125

3.248224610498046875/20

0.081205615262451171875

3.667018995235595703125/20

0.091675474880889892578125

4.075343520354705810546875/20

0.101883588008867645263671875

4.473459932345838165283203125/20

With this last numberI divided the 4.473459932345838165283203125/20 and got 0.22367299661729190826416015625. Since the alcohol is only 40%,I multiplied this by .4 and got: 0.0894691986469167633056640625

[Guest]

Dan, It looks like you used ~50% Alcohol by volume instead of 40%? At least, I'm in your neighborhood if I change my math.

I, too, got 8.946% using this formula:

ounces of booze previously in drink-(% of ounces of booze previosuly in drink times ounces of juice removed)+new ounces of booze added

[Guest]

Forgot to actually take out the alcohol each friend drank- Now after 10 additions i've got 4.411991 liters of Vodka, and Vodka Being 40% Alc/vol, that 1.764797 liters alcohol for 4.412% total.

Probably still wrong, but i feel better about it.

[Prof. Templeton]

8.94691986469167633056640625% alcohol

Percentage of alcohol and liters of Vodka

39¹⁰/40⁹ = 31.053 liters of punch left and 8.947 liters of vodka. About 22.37% vodka after 10 replacements and 8.947% alcohol total.

[Guest]

Quite a few correct answers. I was hoping for some elaboration of principles rather than excel results, but I guess I have to ask for it specifically then.

I also guess that this was a bit more of a math task than a logic puzzle. This was not intended.

Anyway...

Initially there is 20 liter of pure non-alcoholic juice in the bowl.

Friend 1 removes 2.5% of the liquid replacing it with vodka = 20 * 0.975 = 19.5 liter juice left

Friend 2 removes 2.5% of the liquid replacing it with vodka = 20 * 0.975 * 0.975 liter juice left

Friend n removes 2.5% of the liquid replacing it with vodka = 20 * 0.975^n liter juice left

10 people each replacing 1/2 liter from the bowl with vodka results in a bowl with 20 * 0.975^10 = 15.527 liter juice and 4.473 liter vodka.

So the bowl contains 22.367% 40%-vodka for a total alcohol percentage of 8.947

Edited April 13, 2009 by uhre

[Guest]

% Of Alcohol = 1 - { (20*0.975^10) + [(0.6)*(0.5)(0.975^10 - 1)/(0.975 - 1)] }

My answer is 8.94%

[Prof. Templeton]

My previous post confused liters with half liters, but the percentage remains the same.

[Guest]

I take it we assume we're ignoring

the apparent specific volume of alcohol in aqueous mixtures? If I have 170 proof alcohol and add water to 1 liter of alcohol to fill a 2 liter container, my proof is closer to 105 instead of the anticipated 85. Figuring in the sugars of the juice, this could create a different applied proof in as little as an hour or two.

I'm waiting on the equations from a friend, but I take it we're not supposed to think about this, right?

[Guest]

Quite a few correct answers. I was hoping for some elaboration of principles rather than excel results, but I guess I have to ask for it specifically then.

I also guess that this was a bit more of a math task than a logic puzzle. This was not intended.

Anyway...

Initially there is 20 liter of pure non-alcoholic juice in the bowl.

Friend 1 removes 2.5% of the liquid replacing it with vodka = 20 * 0.975 = 19.5 liter juice left

Friend 2 removes 2.5% of the liquid replacing it with vodka = 20 * 0.975 * 0.975 liter juice left

Friend n removes 2.5% of the liquid replacing it with vodka = 20 * 0.975^n liter juice left

10 people each replacing 1/2 liter from the bowl with vodka results in a bowl with 20 * 0.975^10 = 15.527 liter juice and 4.473 liter vodka.

So the bowl contains 22.367% 40%-vodka for a total alcohol percentage of 8.947

These answers are too high. It appears many of you forgot a concept:

This answer exagerates the amount of alcohol in the punch after ten people. It assumes that only punch is removed, and not a mixture of alchol and punch when the .5L cup is taked out. We are assuming the liquid mixes completely and instantly, so we must account for the withdrawel of a certain percentage of alcohol with each person. IN this case it will be (.5L/20L)*(Current Alcohol Level in Punch). After accounting for the simultaneous subtracting of alchol while adding alcohol I got an alcohol percentaage of %8.53, assuming the alcohol is inert. Observe that the remaining alcohol should approach 40% if the party is huge and an infinite number of people show up. See the attached excel file.

Punch__Drunk.xls

Edited April 13, 2009 by hungryjeff

[Prof. Templeton]

I take it we assume we're ignoring

the apparent specific volume of alcohol in aqueous mixtures? If I have 170 proof alcohol and add water to 1 liter of alcohol to fill a 2 liter container, my proof is closer to 105 instead of the anticipated 85. Figuring in the sugars of the juice, this could create a different applied proof in as little as an hour or two.

I'm waiting on the equations from a friend, but I take it we're not supposed to think about this, right?

The OP says nothing about proof, only percentage of alcohol.

It appears many of you forgot a step:

This answer exagerates the amount of alcohol in the punch after ten people. It assumes that only punch is removed, and not a mixture of alchol and punch when the .5L cup is taked out. We are assuming the liquid mixes completely and instantly, so we must account for the withdrawel of a certain percentage of alcohol with each person. IN this case it will be (.5L/20L)*(Current Alcohol Level in Punch). After accounting for the simultaneous subtracting of alchol while adding alcohol I got an alcohol percentaage of %8.53, assuming the alcohol is inert. Observe that the remaining alcohol should approach 40% if the party is huge and an infinite number of people show up.

Most of the answers do indeed take into account that the mixture is no longer just punch after the first person replaces his 1/2 liter.

[Guest]

I take it we assume we're ignoring

the apparent specific volume of alcohol in aqueous mixtures? If I have 170 proof alcohol and add water to 1 liter of alcohol to fill a 2 liter container, my proof is closer to 105 instead of the anticipated 85. Figuring in the sugars of the juice, this could create a different applied proof in as little as an hour or two.

I'm waiting on the equations from a friend, but I take it we're not supposed to think about this, right?

That is correct. The puzzle was merely constructed like this to present a problem of dealing with the relative change of 2 intermixed substances in an interesting way.

[Guest]

The OP says nothing about proof, only percentage of alcohol.

Most of the answers do indeed take into account that the mixture is no longer just punch after the first person replaces his 1/2 liter.

Not correctly and completely.

20*(.975^n) assumes pure punch is subtracted in each iteration. You cannot subtract this from 20L to get an accurate volume of alcohol in the punch after 10 people. Check my spreadsheet. Is it wrong?

[Guest]

Not correctly and completely.

20*(.975^n) assumes pure punch is subtracted in each iteration. You cannot subtract this from 20L to get an accurate volume of alcohol in the punch after 10 people. Check my spreadsheet. Is it wrong?

You are correct that 20* 0.975^n assumes pure punch being subtracted - but it is a deminshing amount of punch - ie. 0.5 liter, 0.4875 liter, 0.4753152 liter, etc - the rest up till .5 liter must then be vodka.

Regarding your excel sheet - it seems you agree with the majority. At the row for 10 people you have 8.95 which is exactly (well, rounded to 2 decimals) what the majority in here concluded.

Edited April 13, 2009 by uhre

[Guest]

I take it we assume we're ignoring

the apparent specific volume of alcohol in aqueous mixtures? If I have 170 proof alcohol and add water to 1 liter of alcohol to fill a 2 liter container, my proof is closer to 105 instead of the anticipated 85. Figuring in the sugars of the juice, this could create a different applied proof in as little as an hour or two.

You are confusing ABW and ABV. Alcoholic proof is simply ABV*2. Why? I have no clue, but the mixture you describe will be exactly 42.5% ABV, or 85 proof.

[Guest]

10%-This is because there are 40 1/2 liters of juice in the bottle. If 40% vodka was added, the 40 1/2 liters diluted the vodka to adding 1% alcohol to the juice. Then that happened 9 more times, adding up to 10%. I know I'm wrong, probably.