Bunuel wrote:
Solution X, which is 50% alcohol, is combined with solution Y, which is 30% alcohol, to form 16 liters of a new solution that is 35% alcohol. How much of solution Y is used?
A. 4 liters
B. 6 liters
C. 8 liters
D. 10 liters
E. 12 liters
When solving mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED:
Since we want to determine the volume of solution Y needed, let's...
Let y = volume (in liters) of solution Y needed
This means 16 - y = volume (in liters) of solution X needed (since the combined volume of both amounts is 16 liters)
So, we get:
Now let's determine the
volume of alcohol in each container.
Solution Y is 30% alcohol. We have y liters of solution Y.
So, the volume of alcohol =
0.3ySolution X is 50% alcohol. We have 16 - y liters of solution X.
So, the volume of alcohol = 0.5(16 - y) =
8 - 0.5yThe combined solution is 35% alcohol. There are 16 liters of this solution.
So, the volume of alcohol = 0.35(16) =
5.6So, our sketch looks like this:
At this point, we can focus on the volume of alcohol in each container.
We know that: (volume of alcohol in 1st container) + (volume of alcohol in 2nd container) = volume of alcohol in combined solution.
In other words:
0.3y + (
8 - 0.5y) =
5.6Simplify: 8 - 0.2y = 5.6
Subtract 8 from both sides to get: -0.2y = -2.4
Solve: y = 12
Answer: E
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