Mixture Problem

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prachi18oct: Master | Next Rank: 500 Posts; Posts: 269; Joined: Sun Apr 27, 2014 10:33 pmFollowed by: 5 members; Thanked: 8 times

by prachi18oct » Fri Feb 20, 2015 11:42 am

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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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Feb 20, 2015 11:56 am

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

I posted three different approaches -- alligation, plugging in the answers, and algebra -- here:

https://www.beatthegmat.com/mixture-t278877.html

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by [email protected] » Fri Feb 20, 2015 4:34 pm

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

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Abhijit K: Master | Next Rank: 500 Posts; Posts: 110; Joined: Mon Feb 09, 2015 1:52 amFollowed by: 4 members

by Abhijit K » Sat Feb 21, 2015 2:11 am

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

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Feb 21, 2015 4:41 am

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

Another way to PLUG IN THE ANSWERS.

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 â‰ˆ 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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