If a portion of a half water/half alcohol mix is replaced with 25% alc

Question

If a fraction of a 50% alcohol solution is replaced with 25% alcohol solution, resulting in a final solution with 30% alcohol, what is the fraction of the original solution that was replaced?

A.  3\%  
B.  20\%  
C.  66\%  
D.  75\%  
E.  80\%

Bunuel · Accepted Answer

Official Solution:

If a fraction of a 50% alcohol solution is replaced with 25% alcohol solution, resulting in a final solution with 30% alcohol, what is the fraction of the original solution that was replaced?

A.  3\%  
B.  20\%  
C.  66\%  
D.  75\%  
E.  80\%

Let  x  be the fraction of the original solution that was replaced, and assume that the initial volume of the solution is 1 liter. Therefore,  x  liters of the 50% alcohol solution are replaced with  x  liters of the 25% alcohol solution.
 
After removing this fraction from the original solution, and before the replacement, the remaining volume of alcohol in the solution is  0.5(1-x)  liters. When  0.25x  liters of alcohol are added to the solution, the final volume of alcohol becomes  0.5(1-x)+0.25x  liters. We know that the resulting solution has an alcohol concentration of 30% or 0.3 liters of alcohol.
 
Thus, we have the equation  0.5(1-x)+0.25x=0.3 , which gives  x=0.8 . Therefore, 80% of the original solution was replaced.

Answer: E

gmat1011 · Answer

Hi Financier

It will work if you consider it as follows:

50% ---------------- 25%

------------30%---------

5%------------------20%

so ratio is 1:4 in final mixture

Earlier type 1 alcohol was 1

Now it is 1/5 ----> so loss of 4/5 = 80%...

BrentGMATPrepNow · Answer

When solving mixture questions, I find it useful to sketch the solutions with the ingredients SEPARATED.

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
 https://i.imgur.com/yt2KAw7.png 
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: 
 https://i.imgur.com/mNlQxXm.png

We're going to replace the x liters of missing solution with x liters of 25% alcohol solution
 https://i.imgur.com/pBODql1.png

When we add the volumes of alcohol and water, we get a final mixture that looks like this:
 https://i.imgur.com/JDsZfUL.png

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

RELATED VIDEO
 https://www.youtube.com/watch?v=7Or9gzROsjQ

EMPOWERgmatRichC · Answer

Hi All,

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. This is an example of a 'Weighted Average' question. Here's how we can set up that calculation using Algebra:

A = number of ounces of 50% solution 
B = number 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

GMATGuruNY · Answer

An alternate approach is to PLUG IN THE ANSWERS, which represent the proportion of 25% solution in the final mixture.
When the correct answer is plugged in, the final mixture = 30% alcohol.
Since the percentage in the final mixture -- 30% -- is closer to the percentage in the 25% solution than to the percentage in the original 50% solution, the 25% solution must constitute more than 1/2 of the final mixture.
Eliminate A and B.

D: 3/4 of the mixture = 25% alcohol, implying that 1/4 of the mixture = the original 50% alcohol 
Since there are 3 parts 25% alcohol for every 1 part 50% alcohol, we get:
Average percentage for every 4 parts =  \frac{(3*25)+(1*50)}{4} = \frac{125}{4} = 31.25 %
The resulting concentration is TOO HIGH.
Implication:
A greater proportion of the weaker solution -- the 25% alcohol solution -- is required.

E

E: 4/5 of the mixture = 25% alcohol, implying that 1/5 of the mixture = the original 50% alcohol 
Since there are 4 parts 25% alcohol for every 1 part 50% alcohol, we get:
Average percentage for every 5 parts =  \frac{(4*25)+(1*50)}{5} = \frac{150}{5} = 30 %
Success!

hemanthp · Answer

Yup. It is E indeed. 
- If V is volume of the mixture then V/2 is alc and V/2 is water. 
- Take Xml of the solution away (it takes X/2 alc with it). So the alc level now is (V-X)/2.
- Add X ml back but this solution only has X/4 alc. So new alc content = (V-X)/2 + X/4
- New alc content = 3V/10 as it is 30%.
Solving it gives X as 80%. 
More or less the same approach that Bunuel took.
Thank you,
Hemanth

KarishmaB · Answer

Question on Mixtures can be easily solved using weighted averages concept discussed here: https://gmatclub.com/forum/tough-ds-105651.html#p828579 I would also recommend that you go through the complete theory from some standard book if you are not comfortable. This question involves replacement rather than simply mixing two solutions hence it has one extra step at the end which I will discuss later. First of all consider, you have a solution with 50% alcohol. Part of it is removed and mixed with a 25% alcohol solution. In effect, this is similar to mixing two solutions, one of 25% alcohol and other of 50% alcohol. Ques1.jpg So 25% alcohol solution volume : 50% alcohol solution volume is 20:5 which is 4:1. Since out of 5 parts of 50% alcohol solution, 4 parts were removed to replace them with 4 parts 25% alcohol solution, the portion of 50% alcohol solution replaced was 4/5 = 80%. The question asks: what percentage of original alcohol was replaced? Since the solution is homogenous, it you replace 80% of it, 80% of the original amount of alcohol in the solution will be replaced. To understand this, lets say I have 100 liters of 50% alcohol solution. When I remove 80% of it, I remove 80 liters solution. In the solution I remove, I have 40 liters alcohol and 40 liters water. In the 20 liters solution that is remaining, 10 liters is alcohol and 10 liters is water. So amount of alcohol removed (and replaced) is 40/50 = 80%

jlgdr · Answer

Quick way Use Smart Numbers Give 100 for the initial amount Then you will have 50-0.25x = 30 x = 80 So % is 80/100 is 80% Hence E is the answer Hope it helps Cheers! J

wasario · Answer

Let's construct an equation to solve for this. You can either use a premade number to determine the amount of liquid we're dealing with (pick an easy number like 100), but for this example I'll just be using  s  to denote the amount of solution we have right now.

Currently, we have a 50% alcohol solution, and we're essentially taking out  x  amount of it, and putting in  x  amount of 20% alcohol solution to get a new solution that's 30% alcohol. We can construct the following equation to essentially figure out how much solution we need to remove to have the correct amount of alcohol:

0.5(s - x) + 0.25x = .3s

0.5(s - x)   is the amount of alcohol we have left after removing  x  amount of solution from the original mixture, and  0.25x   is the amount of alcohol we're putting back into the solution, and  .3s  is the total amount of alcohol we should have left in the end (we can do this because removing and adding back the same total amount of solution doesn't end up changing the net amount of solution we started and ended with). Now we can solve algebraically:

.5s - .5x + .25x = .3s 
 .2s - .25x = 0 
 .2s = .25x 
 x = .8s

Hence, we get that the amount we're removing is 80% of the original solution amount, thus our answer is  E

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If a portion of a half water/half alcohol mix is replaced with 25% alc

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