swerve 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
We can PLUG IN THE ANSWERS, which represent the amount of Y.
Since the percentage for the mixture (35%) is closer to Y's percentage (30%) than to X's percentage (50%), Y must constitute MORE THAN 1/2 of the 16-liter mixture.
Thus, the amount of Y must be equal to more than 8 liters.
Eliminate A, B and C.
When the correct answer is plugged in, the mixture will be 35% alcohol.
D: Y= 10 liters, implying that X = 6 liters
Since Y is 30% alcohol, the amount of alcohol in 10 liters of Y = (3/10)(10) = 3.
Since X is 50% alcohol, the amount of alcohol in 6 liters of X = (1/2)(6) = 3.
Percentage of alcohol in the 16-liter mixture = alcohol/total = (3+3)/(16) = 6/16 = 3/8 = 37.5%.
Eliminate D.
The correct answer is
E.
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