marcodonzelli wrote:
How much water should be added to 10 liters of a 20%-solution of alcohol to reduce the concentration of alcohol in the solution by 75% ?
A. 25 liters
B. 27 liters
C. 30 liters
D. 32 liters
E. 35 liters
One iteration of a weighted average equation here is
(Concentration of A)(Vol A) + (Concen of B)(Vol B) = (Desired Concen)(Total vol = A+B)
This question is a bit unusual.
First, often, the desired concentration is given. Second, it might be hard to discern which value to reduce by 75 percent. In the equation above for this question,
A = 10 liters of soltion with 20% (=.20) alcohol
B = x liters of water with 0% (=0.0) alcohol
If water is zero percent alcohol, its percentage can't be reduced any further. The question therefore asks that 20 percent concentration be reduced by 75 percent:
(.20)(.25) = .05. The latter is the desired concentration of resultant solution.
In the equation, you don't have to insert second term on LHS. Doing so just makes clear that you
are adding x, but its concentration of alcohol has no weight. So
(.20)10 + (0)x = (.05)(10 + x)
2 + 0x = .5 + .05x
1.5 = .05x
\(\frac{1.5}{.05}\)= x
x = 30
Answer C
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