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This problem trains for: AMC-8, AMC-10, AIME, Math Kangaroo 7-8, Math Kangaroo 9-10.
A solution of alcohol in water is 60% concentrated. It is mixed with a solution of alcohol in water that is 40% to obtain a solution that is 46% concentrated. If the amount of the first solution is doubled and the amount used from the second solution is tripled, what is the concentration of the resulted mixture?
Denote the initial amounts of the two solutions by m and n, respectively. Since the mixing process does not involve a reaction, the total quantity of alcohol remains unchanged (invariant). Write an equation that sets the total quantity of alcohol invariant before and after the mixing operation:
The second mixture consists of the amounts 2m and 3n, respectively. Write again the conservation of the total amount of alcohol:
We have to eliminate the variables carefully since we have three unknowns and only two equations. But, since each term contains either a factor of m or one of n, we can reduce the number of variables to only two by using the change of variable:
Divide the first equation by n:
and use the change of variable:
and solve for y:
Now substitute this in the second equation: