SOLUTION: . Ulysses made two alcohol solutions of different concentrations. When he mixed 4 liters of the first solution with 6 liters if the second, he obtained 35% solution of alcohol. But
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Question 1200676: . Ulysses made two alcohol solutions of different concentrations. When he mixed 4 liters of the first solution with 6 liters if the second, he obtained 35% solution of alcohol. But when he mixed 6 liters of the first solution with 4 liters of the second, he obtained a 50% solution of alcohol. What would be the percentage of alcohol if he mixed 2 liters of first solution with 8 liters of the second?
A. 10%
B. 20%
C. 25%
D. 72.5% Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Ulysses made two alcohol solutions of different concentrations.
When he mixed 4 liters of the first solution with 6 liters if the second,
he obtained 35% solution of alcohol.
But when he mixed 6 liters of the first solution with 4 liters of the second,
he obtained a 50% solution of alcohol.
What would be the percentage of alcohol if he mixed 2 liters of first solution
with 8 liters of the second?
~~~~~~~~~~~~~~
Let x be the concentration of the alcohol in the first solution,
expressed as a decimal number.
Let x be the concentration of the alcohol in the second solution.
Write equations for the concentrations as you read the problem
4x + 6y = 0.35(4+6), or 4x + 6y = 3.5, (1)
6x + 4y = 0.5*(6+4), or 6x + 4y = 5. (2)
To solve equations (1), (2), multiply eq(1) by 3; multiply eq(2) by 2. You will get
12x + 18y = 10.5 (1')
12x + 8y = 10 (2')
Subtract eq(2') from eq(1')
10y = 0.5 ---> y = 0.05.
Then from equation (2)
6x = 5 - 4*0.05 = 4.8 ---> x = 4.8/6 = 0.8.
Thus the concentrations of the original mixtures are x= 0.8, y= 0.05.
Then the concentration of the desired mixture is
C = = = = 0.2 = 20%. ANSWER
