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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Mixture Problem
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prachi18oct
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GMATGuruNY
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I posted three different approaches -- alligation, plugging in the answers, and algebra -- here:prachi18oct 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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Hi prachi18oct,
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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Abhijit K
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Assume we had a total of 10 litres.
5 ltrs of alcohol and 5 litres of water.Let the total ltrs taken out of the solution be 2x. Therefore x ltrs of alcohol are removed and x ltrs of water is removed.
New solution added again is 2x. Therefore alcohol content of the new solution is 0.25*2x=0.5x
Equation:-
(5-x+0.5x)/10=3/10
5-0.5x=3
0.5x=2
x=4
Original alcohol was 5 ltrs. Alcohol replaced is 4 ltrs. Therefore alcohol % replaced = 4/5= 80%
Hence, correct answer choice is E
5 ltrs of alcohol and 5 litres of water.Let the total ltrs taken out of the solution be 2x. Therefore x ltrs of alcohol are removed and x ltrs of water is removed.
New solution added again is 2x. Therefore alcohol content of the new solution is 0.25*2x=0.5x
Equation:-
(5-x+0.5x)/10=3/10
5-0.5x=3
0.5x=2
x=4
Original alcohol was 5 ltrs. Alcohol replaced is 4 ltrs. Therefore alcohol % replaced = 4/5= 80%
Hence, correct answer choice is E
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Another way to PLUG IN THE ANSWERS.prachi18oct 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
D: 75%
Here, 3/4 of the 50% solution is replaced with 25% solution.
Since there are 3 parts 25% solution for every 1 part 50% solution, we get:
Average percentage per 4 parts = (3*25 + 1*50)/4 = 125/4 ≈ 31%.
The resulting percentage is too high.
Implication:
More of the replacement solution -- which has a lower percentage of alcohol -- is required.
The correct answer is E.
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Followed here and elsewhere by over 1900 test-takers.
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.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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