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Re: Three friends A, B and C play a game in a pub. The rules are simple. [#permalink]
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Step 1: A vs B:
Since A has a higher percentage of alcohol than B, A has to give B 200 ml of wine. Considering that A has a 60% strength mixture, B will end up having 800 ml of wine that has 604

Step 2: B vs C:
B has a higher percentage of alcohol than C, so B has to give 200 ml of alcohol to C now. C will end up having wine that has 514 = 50.25%

Step 3: A gives C 200 ml of wine. We see that computing percentages is going to cumbersome. So, let us just forget the numbers and think about how the game pans out.

Step 4: A would give B 200 ml of wine. We note straight away that A will always give the wine. A had the highest percentage alcohol to start with. And no mixture can reach 60% alcohol levels. So, A will give all his wine away by this round.

Post this, we notice that, between B and C whoever has higher percentage alcohol will keep giving wine to the other. Eventually, all the wine will be with just one person. At the rate of 3 minutes per round, the game, the game will definitely be over before 1 hour.

So, the concentration of alcohol in the wine D drinks is 603 = 52.67%.
GMAT Club Bot
Re: Three friends A, B and C play a game in a pub. The rules are simple. [#permalink]
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