Mo2men wrote:A solution consists of only water and alcohol such that the ratio of alcohol to water in the solution is 7:3. How much amount of water should be added to the solution (in mililiters) so that the resulting solution contains 60% alcohol?
1.Total quantity of the resulting solution is 350 mililiters.
2.The original solution contains 10.5 mililiters of alcohol for every 4.5 mililiters of water.
Let W = the pure water, S = the original solution, and M = the mixture.
The following approach is called ALLIGATION -- a very good way to handle MIXTURE PROBLEMS.
Alligation can be performed only with PART TO WHOLE RELATIONSHIPS (percents or fractions).
Step 1: Convert any ratios to PERCENTS.
W:
Here, alcohol/total = 0%.
S:
Since alcohol/water = 7:3, and 7+3=10, alcohol/total = 7/10 = 70%.
M:
Here, alcohol/total = 60%.
Step 2: Plot the 3 percentages on a number line, with the percentages of W and S on the ends and the percentage for M in the middle.
W 0-----------------60-------------------70 S
Step 3: Calculate the distances between the percentages.
W 0-------
60--------60---------
10--------70 S
Step 4: Determine the ratio in the mixture.
The ratio of W to S in the mixture is equal to the RECIPROCAL of the distances in red.
W:S = 10:60 = 1:6.
Since W:S = 1:6, for every 7 liters of mixture, 1 liter of pure water must be combined with 6 liters of original solution.
Thus, the added pure water must constitute 1/7 of the final mixture.
Statement 1:
W = (1/7)(350) = 50mm.
SUFFICIENT.
Statement 2:
Thus, alcohol to water in the original solution = (10.5)/(4.5) = 21/9 = 7/3.
This information is given in the prompt.
INSUFFICIENT.
The correct answer is
A.
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