Mixture Replacement Problem

Question

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) 20%
(C) 66%
(D) 75%
(E) 80%

Would someone please help me with this? I really don't know how to solve this.

Thank you in advance.

GMATGuruNY · Answer

Yvette726 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) 20%
(C) 66%
(D) 75%
(E) 80%

Would someone please help me with this? I really don't know how to solve this.

Thank you in advance.

Alcohol percentage in the original solution: 50%.
Alcohol percentage in the replacement solution: 25%.
Alcohol percentage in the mixture: 30%.

Let O = the original solution and R = the replacement solution.
The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the percentages for O and R on the ends and the percentage for the mixture in the middle.
O 50%-------------30%-----------25% R

Step 2: Calculate the distances between the percentages.
O 50%-----20-----30%-----5-----25% R

Step 3: Determine the ratio in the mixture.
The required rate of O to R is the RECIPROCAL of the distances in red.
O:R = 5:20 = 1:4.

Since O:R = 1:4, of every 5 liters of total mixture, O=1 liter and R=4 liters.
Thus, R/total = 4/5 = 80%.
Since R constitutes 80% of the resulting mixture, 80% of the original solution is replaced by R.
Implication:
80% of the original amount of alcohol -- as well as 80% of the original amount of water -- is replaced by R.

The correct answer is E.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html

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Yvette726 · Answer

GMATGuruNY wrote:Yvette726 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) 20% (C) 66% (D) 75% (E) 80% Would someone please help me with this? I really don't know how to solve this. Thank you in advance. Alcohol percentage in the original solution: 50%. Alcohol percentage in the replacement solution: 25%. Alcohol percentage in the mixture: 30%. Let O = the original solution and R = the replacement solution. The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS. Step 1: Plot the 3 percentages on a number line, with the percentages for O and R on the ends and the percentage for the mixture in the middle. O 50%-------------30%-----------25% R Step 2: Calculate the distances between the percentages. O 50%-----20-----30%-----5-----25% R Step 3: Determine the ratio in the mixture. The required rate of O to R is the RECIPROCAL of the distances in red. O:R = 5:20 = 1:4. Since O:R = 1:4, of every 5 liters of total mixture, O=1 liter and R=4 liters. Thus, R/total = 4/5 = 80%. Since R constitutes 80% of the resulting mixture, 80% of the original solution is replaced by R. Implication: 80% of the original amount of alcohol -- as well as 80% of the original amount of water -- is replaced by R. The correct answer is E. For two similar problems, check here: https://www.beatthegmat.com/ratios-fract ... 15365.html Thank you Mitch! I still have trouble understanding this method, but it does seem to be very efficient in solving this type of problems. I will keep trying!

GMATGuruNY · Answer

Yvette726 wrote: Thank you Mitch! I still have trouble understanding this method, but it does seem to be very efficient in solving this type of problems. I will keep trying!

An alternate approach:

Let O = the original solution and R = the replacement solution.
We can PLUG IN THE ANSWERS, which represent the percentage of R in the mixture.
When the correct answer choice is plugged in, the percentage of alcohol in the mixture will be 30%.
Since the percentage of alcohol in the mixture (30%) is VERY close to the percentage of alcohol in R (25%), R must constitute MOST of the mixture.

E: R=80 liters, O=20 liters, for a total of 100 liters
Since R is 25% alcohol, the amount of alcohol in R = .25(80) = 20 liters.
Since O is 50% alcohol, the amount of alcohol in O = .5(20) = 10 liters.
(total alcohol)/(total volume) = (20+10)/100 = 30%.
Success!

The correct answer is E.

Rich.C@EMPOWERgmat.com · Answer

Hi Yvette726,

This mixture question is more complex than normal for the GMAT (and as such is NOT typical), but you can still solve it with the part/whole formula and some algebra. Here's how:

The question doesn't tell us the actual amount of mix that we have, so I'm going to choose a total of 10 gallons.

Total = 10 gallons.

We're told that it's half water and half alcohol, so

Water = 5 gallons
Alcohol = 5 gallons

Next, we're told that we're going to replace some of this original mixture with a new mixture, which is 25% alcohol, and the end mixture will be 30% alcohol. (Note: The current mixture is 50% alcohol and when we "take some out", we're taking out an equal amount of water and alcohol).

By "replacing" liquid, we will still have a total of 10 gallons. Here's how the part/whole formula comes into play:

Original: Alcohol/Total = 5/10 = .5
After replacement = X/10 = .3
X = the new mixture of 50% alcohol and 25% alcohol

Now, here's how we figure out X:

First, the weighted average formula:

A = gallons that are 50% alcohol
B = gallons that are 25% alcohol

(.5A + .25B)/10 = .3
A + B = 10

We now have a "system" of equations. You can solve this using either "substitution" or "combination"
.5A + .25B = 3
5A + 2.5B = 30

B = 10 - A

Now, substitute:

5A + 2.5(10 - A) = 30
5A + 25 - 2.5A = 30
2.5A = 5
A = 2

Plug back in:
B = 8

Since we're interested in what fraction was replaced, we need B and the total:

8/10 = 80%

GMAT assassins aren't born, they're made,
Rich

Contact Rich at [email protected]

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