notrandom wrote:
A certain quantity of 40% concentration solution is replaced with 25% concentration solution such that the concentration of the combined amount is 35%.
What's the ratio of the amount of solution that was replaced to the amount of solution that was not replaced?
A) 1:3
B) 1:2
C) 2:3
D) 2:1
E) 3:1
We let m = the amount of solution with 40% concentration and n = the amount of solution with 25% concentration.
Originally we have 0.4m concentration. Since the amount n is removed from m, 0.4n concentration is removed. Since the original amount n is replaced a new amount n, the total amount is still m, however, the 0.4n concentration that is removed is only replaced with 0.25n concentration. Thus we have:
(0.4m - 0.4n + 0.25n)/(m - n + n) = 0.35
(0.4m - 0.15n)/m = 0.35
0.4m - 0.15n = 0.35m
0.05m = 0.15n
5m = 15n
m = 3n
Since m = 3n and amount n is replaced, thus amount 2n is not replaced. Therefore, the ratio is 1/2.
Alternate Solution:
Let x = the original amount of 40% concentration and y = the amount that was taken out and replaced. Note that we started with x liters, and we ended with the same amount (because it was replaced).
Thus, we start with x liters of 40% solution, we take out y liters of 40% solution, and then add back y liters of 25% solution. The result is x liters of 35% solution. Converting the percents to decimals, we can create the equation:
0.40x – 0.40y + 0.25y = 0.35x
.05x = 0.15y
5x = 15y
x = 3y
The question is to determine the ratio of the solution that was replaced to the amount of solution that was not replaced. The amount of solution that was replaced is y, and the amount not replaced is (x – y). Thus, by substituting 3y for x, the ratio can be expressed as:
y/(x – y) = y/(3y – y) = y/2y = 1/2
Answer: B
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