M03-04

Do you have other questions of the same type??

Nevermind:
mixture-problems-with-best-and-easy-solutions-all-together-124644.html

mixture-problems-made-easy-49897.html

DS Mixture Problems to practice: search.php?search_id=tag&tag_id=43
PS Mixture Problems to practice: search.php?search_id=tag&tag_id=114

Hope it helps.

great question, looks like i need more practice on mixtures.

The term "replaced" is new to me and I wasn't sure how to approach this problem. Is this a term I might see on the GMAT? It seems like a science term that probably has a specific meaning, can anyone fill me in on what that is? Thanks! My assumption was that it means combining equal volumes.

Here replace means substituting some amount of mixture with equal amount of another mixture.

This helped me solve the question in a more intuitive way (according to me of course)
Assume Original Solution= 100 liters
Alcohol:Water = 50:50

Now we remove a certain amount from this solution which I define as 'x'
Alcohol:Water= 50 - (x/2) :50 - (x/2)

Next we add a solution that is 25% alcohol to it
Alcohol:Water= 50 - (x/2) + 25x/100 :50 - (x/2) + 75x/100

Simplify to get
Alcohol:Water= 50 - (x/4):50 + (x/4)

Now our target solution has Alcohol:Water = 30:70

So we can set either of our equations to get the answer
50 - (x/4) = 30 or
50 + (x/4) = 70

Solve for x to get 80

Allegation is probably the fastest way to do this for non-math geniuses.

The distance from the average of the original solution is 20 units and the distance from the new solution is 5 units.
Reducing to 4:1

So out of 5 total parts only 1/5th remains so 80% was replaced.

I always get excited with problems likes these! They're weighted heavily and categorized as difficult, but it seems to just be a weighted average problem in disguise. So if you know how to work with weighted averages, these types of problems are like free chicken!

We're given 3 solution concentrations to work with.

Starting: 50% (S)
Added solution: 25% (A)
Resultant solution: 30% (R)

Draw them out on a number line in increasing order, with the resultant solution in the middle (we put the resultant in the middle since it's an average of the other two solutions)

A-----R--------------------S
25-----30--------------------50

Now count the percentages in between the two.

From A to R: 5
From R to S: 20

Now, set them into a ratio. 20:5. Now simplify to 4:1.

**This means that 4 parts of the 25% solution were replaced for every 1 part of the existing solution. In other words, \(\frac{4}{5}\) replaced. (part over whole of 4:1 ratio) and \(\frac{4}{5}\) expands to \(\frac{80}{100}\) or 80%

I used to get tripped up on setting the ratio up backwards, at first glance it looks like it would be 5:20, since the space between A-R is 5, and R-S is 20, but think about it logically. The resultant (or average) solution is much closer to the "added" solution, meaning the resultant solution contains more of the added solution than the original solution, which is what brought "R" closer to "A" on our number line.

Once you get the hang of setting these weighted averages problems up like this, it becomes super easy.

Ans: E

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

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

RELATED VIDEO

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Let me jump in with my algebraic solution, hopefully this will help others as well!

so lets imagine, at first, we have \(x\) liters of solution of 50% strength, thus the alcohol content is
\(alcohol_1 = \frac{50}{100}x\)

by the end of the process, we have still the same amount \(x\) liters of solution of 30% strength. Notice that the process is replacing the some amount, thus the total amount (liter of solution) is unchanged
\(alcohol_2 = \frac{30}{100}x\)

So what do we need to do in order to achieve the 30% mix? We subtract some \(y\) amount of the50% solution, then add the same \(y\) amount of the new 25% solution
\((alcohol_1) + (replacement) = (alcohol_2)\)

\((\frac{50}{100}x) + (- \frac{50}{100}y + \frac{25}{100}y) = (\frac{30}{100}x)\)

Note that the equation above is valid since the actual volume of all the solution will actually not change ie \(x - y + y = x\)

some simplifications
\(50x - 25y = 30x\)
\(10x - 5y = 6x\)
\(4x = 5y \)

since the question stem ask about the fraction of the original solution that was replaced, or in other words:
\(\frac{y}{x} * 100 = \frac{4}{5} *100 = 80% \)

I think this is a high-quality question and I agree with explanation.