riz_gmat wrote:A solution contains 8 parts water for every 7 parts of lemonade syrup. How many parts of the solution should be removed and replaced with water so that the solution will now contain 40% lemonade syrup?
(A)1.5
(B)1.75
(C)2.14
(D)2.34
(E)2.64
OA is C
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 fractions or percentages.
Step 1: Convert any ratios to FRACTIONS.
W:
Here, water/total = 1/1.
S:
Since water:lemonade = 8:7, and 8+7=15, water/total = 8/15.
M:
Since the mixture is to be 40% lemonade, water/total = 60/100 = 3/5.
Step 2: Put the fractions over a COMMON DENOMINATOR.
W = 1/1 = 15/15.
S = 8/15.
M = 3/5 = 9/15.
Step 3: Plot the 3 numerators on a number line, with the numerators of W and S on the ends and the numerator for M in the middle.
W 15----------------9------------------8 S
Step 4: Calculate the distances between the numerators.
W 15-------
6--------9---------
1--------8 S
-
Step 5: 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 = 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, 1/7 of every 15 liters must come from the pure water:
(1/7) * 15 = 2.14.
The correct answer is
C.
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