If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
A) 3
B) 12
C) 64
D) 75
E) 80
Mixture
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- akhilsuhag
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The following approach is called alligation -- a very good way to handle MIXTURE PROBLEMS.akhilsuhag wrote:If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
A) 3
B) 12
C) 64
D) 75
E) 80
Let F = the first solution and R = the replacement solution.
Step 1: Plot the 3 percentages on a number line, with the percentage for F and R (50% and 25%) on the ends and the percentage for the mixture (30%) in the middle.
F 50--------30-------25 R
Step 2: Calculate the distances between the percentages.
F 50---20---30---5---25 R
Step 3: Determine the ratio in the mixture.
The ratio of F to R in the mixture is the RECIPROCAL of the distances in red.
F:R = 5:20 = 1:4.
Since F:R = 1:4, of every 5 liters, 1 liter is F and 4 liters are R.
Thus:
R/Total = 4/5 = 80%.
The correct answer is E.
More practice with alligation:
https://www.beatthegmat.com/ratios-fract ... 15365.html
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Alternate approach 1: PLUG IN THE ANSWERSakhilsuhag wrote:If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
A) 3
B) 12
C) 64
D) 75
E) 80
The answer choices represent the percentage of replacement solution in the mixture.
When the correct answer choice is plugged in, the percentage of alcohol in the mixture will be 30%.
Answer choice D: 75%
Let the replacement solution = 75 liters and the original solution = 25 liters.
Amount of alcohol in the 25% replacement solution = (0.25)(75) = (75/4) = 18.75.
Amount of alcohol in the 50% original solution = (0.5)(25) = 12.5.
Resulting percentage:
(total alcohol)/(total volume) = (18.75 + 12.5)/(75+25) = 31.25%.
The percentage of alcohol is too high.
Eliminate D.
To reduce the percentage of alcohol in the mixture, MORE of the replacement solution -- which has a lower percentage of alcohol -- is needed.
Thus, the correct answer must be GREATER than 75.
The correct answer is E.
Alternate approach 2: Algebra
Let F = the first solution and R = the replacement solution
Percentage of alcohol in F = 0.5F.
Percentage of alcohol in R = 0.25R.
Percentage of alcohol in the mixture of F and R = 0.3(F+R).
Thus:
0.5F + 0.25R = 0.3(F+R)
50F + 25R = 30(F+R)
50F + 25R = 30F + 30R
20F = 5R
4F = R.
If F=1, then R=4, implying that R constitutes 4 of every 5 liters.
Thus:
R/total = 4/5 = 80%.
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Answer: Option Eakhilsuhag wrote:If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
A) 3
B) 12
C) 64
D) 75
E) 80
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