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Mixture

by akhilsuhag » Fri Aug 29, 2014 2:44 pm
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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by GMATGuruNY » Fri Aug 29, 2014 7:41 pm
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
The following approach is called alligation -- a very good way to handle MIXTURE PROBLEMS.
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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by j_shreyans » Fri Aug 29, 2014 10:14 pm
Hi ,

Is there any other approach to solve this question?

Thanks ,

Shreyans
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by GMATGuruNY » Sat Aug 30, 2014 3:01 am
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
Alternate approach 1: PLUG IN THE ANSWERS
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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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

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by GMATinsight » Sat Aug 30, 2014 3:13 am
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
Answer: Option E
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