How many liters of a solution that is
10% alcohol by volume must be added
to 2 liters of a solution that is 50%
alcohol by volume to create a solution
that is 15% alcohol by volume?
A) 10
B) 12
C) 14
D) 16
E) 18
We can use ALLIGATION -- a great way to handle MIXTURE PROBLEMS.
Let X = the 10% solution and Y = the 50% solution.
Step 1: Plot the 3 percentages on a number line, with the percentages for X and Y on the ends and the percentage for the mixture in the middle.
X 10------------15------------50 Y
Step 2: Calculate the distances between the percentages.
X 10-----
5-----15-----
35-----50 Y
Step 3: Determine the ratio in the mixture.
The required ratio of X to Y is equal to the RECIPROCAL of the distances in red.
X:Y = 35:5 = 7:1.
Since X:Y = 7:1 = 14:2, 14 liters of X must be added to 2 liters of Y.
The correct answer is
C.
An alternate approach is to PLUG IN THE ANSWERS, which represent the amount of 10% solution that must be added to 2 liters of 50% solution.
Answer choice
C: 14 liters
Amount of alcohol in 14 liters of 10% solution = 10% of 14 = 1.4 liters.
Amount of alcohol in 2 liters of 50% solution = 50% of 2 = 1 liter.
(total alcohol)/(total volume) = (1.4 + 1)/(14 + 2) = (2.4)/16 = 24/160 = 3/20 = 15/100 = 15%.
Success!
The correct answer is
C.
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