[superprismatic]

4 liters of wine is removed from a wine cask and replaced with

4 liters of water. The wine in the cask is thus diluted. This

procedure of removing 4 liters from the cask and replacing it

with 4 liters of water is done two more times. The diluted

wine left in the cask is now .512 the strength of what it was

initially. How much wine did the cask originally hold?

[preflop]

20 Liters

[Guest]

0.512^(1/4) = .845

4/(1-0.845897011) = 25.9566672 liters

check my work:

(26*22/26*22/26*22/26*22/26)/26 = 0.512622107 is about .512 strength

26 liters

but the answer's not exact, so it could be way off.

[superprismatic]

0.512^(1/4) = .845

4/(1-0.845897011) = 25.9566672 liters

check my work:

(26*22/26*22/26*22/26*22/26)/26 = 0.512622107 is about .512 strength

26 liters

but the answer's not exact, so it could be way off.

Hi, pengwen! Welcome. Please explain the individual steps leading to your answer. The fun of this site lay in how people get to their solutions.

[Guest]

20

After first transfer, strength of wine = (X-4)/X , where X is the volume of cask.

After 2nd transfer strength = [(X-4)/X]^2

After 3rd transfer strength = [(X-4)/X]^3 which is =.512

Solving for X gives volume of cask.

Edited August 18, 2011 by superprismatic
Please use spoilers so others don't accidentally see what they may want to work out themselves.

[Guest]

@ skpoudyal

Thks for the simplified 3rd degree polynomial. Got stuck in reducing that.

Resorted to goal seek in excel to get answer. Hope that is not against the law.

[Guest]

Instead of dealing with a third degree polynomial, I chose to put the formulas into a spreadsheet. At each step, calculated amount of water removed and amount of wine removed. Then proportion of wine after adding 4 liters more of water. Cell A1 was my original guess. From here it is simple to plug in different options into cell A1 until results give the desired .512. Solution at which I arrived doing it this way

20

[dark_magician_92]

20 litres.

eqn is : x*(1-4/x)^3=0.512x

(1-4/x)^3=0.512=0.8^3

1-4/x=0.8

x=20

[Guest]

Whupps, I thought the cask was diluted 4 times, not 3.

I figured that 0.512 is the dilution quotient times itself a few times. So: cube root of 0.512 = 0.8

The 0.2 (1-0.8) is then the water part of the wine. To find the amount of wine initially: 4/(0.2) = 20

With that process, you can have the equation: CaskSize = 4/(1-0.512^(1/TimesDiluted))

1: 8.196721311

2: 14.06181765

*3: 20*

4: 25.95666716

5: 31.9207495

Hope that covers it all.