SOLUTION: Bottle A contains some water. Bottle B contains an equal amount of pure alcohol. First 1/3 of the water in A is poured into B. Stir Well. Then 1/3 of the mixture in B is poured int
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Question 589674: Bottle A contains some water. Bottle B contains an equal amount of pure alcohol. First 1/3 of the water in A is poured into B. Stir Well. Then 1/3 of the mixture in B is poured into A. Stir Well. Again 1/3 of the mixture in A is poured into B. Stir Well. Finally the correct amount of mixture is poured from B into A so the both bottles contains the same amount of mixture. In what ratio is the pure alcohol divided between A and B now?
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Call the amount of water in A at the start unit
Call the amount of alcohol in B at the start unit
----------
After the 1st step,
A has units of water in it
B has units of water plus unit of alcohol
----------
After the 2nd step,
units of water from B is added to A, and
units of alcohol from B is added to A
Now A contains units of water
and units of alcohol
Now B contains units of water
and units of alcohol
---------
Note that after each step the total units of both water and alcohol must = 1
---------
After the next step:
units of water from A is added to B
units of alcohol from A is added to B
Now A contains units of water
and units of alcohol
Now B contains units of water
and units of alcohol
---------
check: water = and
alcohol =
OK
---------
After the next step:
Some fraction of the mixture in B is poured into A
Let that fraction =
Now B contains units of water
and units of alcohol
Now A contains units of water
and units of alcohol
----------
The mixture in A now = the mixture in B, so
----------
The units of alcohol in A is now
The units of alcohol in B is now
The ratio of alcohol is
A/B = 13/14
----------
Unless I goofed somewhere
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