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%3%
B. 20%20%
C. 66%66%
D. 75%75%
E. 80%
[spoiler]OA: E[/spoiler]
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Mixture problem unclear
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GMATGuruNY
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The following approach is called alligation -- a very good way to handle MIXTURE PROBLEMS.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
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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Hi Mo2men,
This is an example of a 'Weighted Average' question.
We're told that we're going to mix a certain amount of a 50% alcohol solution with a certain amount of a 25% alcohol solution and end up with a 30% alcohol solution. Here's how we can set up that calculation using Algebra:
A = # of ounces of 50% solution
B = # of ounces of 25% solution
A+B = total ounces of the mixed solution
(.5A + .25B)/(A+B) = .3
.5A + .25B = .3A + .3B
.2A = .05B
20A = 5B
4A = B
A/B = 1/4
This means that for every 1 ounce of solution A, we have 4 ounces of solution B.
To answer the question that's asked, imagine that you have 5 ounces of solution A. We're told to REPLACE 4 ounces of it with 4 ounces of solution B (thus, we'll end up with the 30% mixture that we're after).
Since 4 ounces of the 5 ounces were replaced, 4/5 of the original alcohol and 4/5 of the original water was replaced.
4/5 = 80%
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
This is an example of a 'Weighted Average' question.
We're told that we're going to mix a certain amount of a 50% alcohol solution with a certain amount of a 25% alcohol solution and end up with a 30% alcohol solution. Here's how we can set up that calculation using Algebra:
A = # of ounces of 50% solution
B = # of ounces of 25% solution
A+B = total ounces of the mixed solution
(.5A + .25B)/(A+B) = .3
.5A + .25B = .3A + .3B
.2A = .05B
20A = 5B
4A = B
A/B = 1/4
This means that for every 1 ounce of solution A, we have 4 ounces of solution B.
To answer the question that's asked, imagine that you have 5 ounces of solution A. We're told to REPLACE 4 ounces of it with 4 ounces of solution B (thus, we'll end up with the 30% mixture that we're after).
Since 4 ounces of the 5 ounces were replaced, 4/5 of the original alcohol and 4/5 of the original water was replaced.
4/5 = 80%
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
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Brent@GMATPrepNow
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When solving mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED.Mo2men 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%3%
B. 20%20%
C. 66%66%
D. 75%75%
E. 80%
[spoiler]OA: E[/spoiler]
Since we're asked to find a PERCENTAGE, we can assign a "nice" value to the original volume.
So, let's say we start with 100 liters
As you can see, 50 liters is alcohol and 50 liters is water
Now let's remove x liters of the mixture from the container.
Half of removed x liters will be alcohol, half of the removed x liters will be water
In other words, we're removing 0.5x liters of alcohol, and 0.5x liters of water.
So, the resulting mixture looks like this:
We're going to replace the x liters of missing solution with x liters of 25% alcohol solution
When we add the volumes of alcohol and water, we get a final mixture that looks like this:
Finally, we want the final mixture to be 30% alcohol.
In other words, we want: (volume of alcohol)/(total volume of mixture) = 30/100 (aka 30%)
We get: (50-0.25x)/100 = 30/100
Cross multiple: (50-0.25x)(100) = (100)(30)
Simplify: 5000 - 25x = 3000
Solve to get: x = 80
So, 80 liters were originally removed
Answer: E
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
