Deepthi Subbu wrote:Solution A contains 18 ounces of alcohol and 32 ounces of water. Solution B contains 72 ounces of water and 8 ounces of alcohol. What percent of Solution B must be added to
Solution A so that the resulting mixture contains 5/7 water?
Here's how to solve with ALLIGATION -- a very good way to handle MIXTURE PROBLEMS.
Alligation can be performed only with fractions or percentages.
Thus, the ratios here must be converted.
Step 1: Convert the ratios to FRACTIONS.
A:
Since alcohol:water = 18:32, and 18+32=50, alcohol/total = 18/50 = 9/25.
B:
Since alcohol:water = 8:72, and 8+72=80, alcohol/total = 8/80 = 1/10.
Mixture:
Since water/total = 5/7, alcohol/total= 2/7.
Step 2: Put the fractions over a COMMON DENOMINATOR.
A = 9/25 = 126/350.
B = 1/10 = 35/350.
Mixture = 2/7 = 100/350.
Step 3: Plot the 3 numerators on a number line, with the numerators of A and B (126 and 35) on the ends and the mixture's numerator (100) in the middle.
A 126---------------100---------------35 B
Step 4: Calculate the distances between the numerators.
A 126------
26------100------
65------35 B
Step 5: Determine the ratio in the mixture.
The ratio of A to B in the mixture is the RECIPROCAL of the distances in red.
A:B = 65:26 = 5:2.
Since A:B = 5:2 = 50:20, and A's actual volume = 50 ounces, the mixture must be composed of 50 ounces of A and 20 ounces of B.
Thus:
(Amount of B in the mixture)/(total volume of B) = 20/80 = 1/4 = 25%.
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