Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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%

OA explanation: Find the difference between the original and final amounts of alcohol. Suppose we had 1 liter or 1 gallon of the initial solution, and has been replaced. We can build an equation using this information:

How do you solve this? I don't get how the OA came up with the equation below. thanks

Here is the how I solved the problem Consider the quantity of solution as 1 litter. It is given that 50% of the solution was alcohol. Let X be the amount of alcohol removed from the solution.

The equation to remove X amount of alcohol from the 50% alcohol solution is 0.5 litters alcohol - 0.5 * X litters alcohol

The X litters removed from the solution is replaced by 25% alcohol solution. 0.5 litters alcohol - 0.5 * X litters alcohol + 0.25 * X litters alcohol

The resulting solution contains 30% alcohol. Therefore the final equation is 0.5 litters alcohol - 0.5 * X litters alcohol + 0.25 * X litters alcohol = 0.30 litters of alcohol.

If you solve for the above equation, you will get X as 0.8, which is 80%

Q: 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%

Thanks in Advance -Amit

b) 20% solution1 is 50% alcohol and solution2 is 25% alcohol. to get 30% alcohol solution we need solution1/ solution2 in ratio 1:4

Let x be removed out of the first solution of 50% alcohol and x be added from the other solution of 25% alcohol to make a wholesome (a peg ??) 30% alcohol solution ( I don't take it with water though!!!)

Anyways, in equation form it is 50(1-x) + 25x = 30, which gives x = 4/5. so 80% quantity of the first solution is replaced with 80% quantity of second solution to get a wholesome 30% solution.
_________________

Hi, Thanks for the reply. But the OA is E(80%). Could you please justify? -Amit

ohk...read the q wrong...since we have 100% of original solution [50% alcohol] and after mixing it is only 20% so the 1st solution is replaced by 100 - 20 = 80%

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%

OA explanation: Find the difference between the original and final amounts of alcohol. Suppose we had 1 liter or 1 gallon of the initial solution, and has been replaced. We can build an equation using this information:

How do you solve this? I don't get how the OA came up with the equation below. thanks

Let x - amount of total mixture y - amount of mixture replaced with 25% alcohol

Now, 0.5(x-y) + 0.25y = 0.3x 0.5(x-y) represents 50% alcohol quantity in new mixture 0.25y represents 25% alcohol quantity in new mixture 0.3x represents total alcohol quantity in new mixture

solving y = (4/5)x hence y is 80% of x that means E.

This question was also posted in PS subforum. Below is my solution from there.

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%

Question can be solved algebraically or using allegation method.

Algebraic approach:

Initial solution is "half water/half alcohol mix" means it's 50% (0.5) alcohol solution.

Let the portion replaced be \(x\) and the volume of initial solution be 1 unit.

Then the amount of alcohol after removal of a portion will be \(0.5(1-x)\) and the amount of alcohol added will be \(0.25x\), so total amount of alcohol will be \(0.5(1-x)+0.25x\). On the other hand as in the end 30% alcohol solution was obtained then the amount of alcohol in the end was \(0.3*1\).

if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced? you are correct

Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal % of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistake in my logic .

if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced? you are correct

Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal % of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistake in my logic .

Both questions would give 80% because the ratio was 1:1,

So, replacing 80% of the solution means replacing 80% of the alcohol.

In the red-part, why did you further cut down the alcohol content to half? It should be:

if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced? you are correct

Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal % of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistake in my logic .

Both questions would give 80% because the ratio was 1:1,

So, replacing 80% of the solution means replacing 80% of the alcohol.

In the red-part, why did you further cut down the alcohol content to half? It should be:

0.8*0.5=0.4 g of alcohol replaced.

0.4 is 80% of 0.5

0.8*0.5 gal of alcohol was removed from the original solution (0.8 gal of original mix was removed ) and replaced with 0.8*0.25 gal of alcohol ( 0.8 gal of new mix added whose alcohol conc is 25%) from the addition of the new mixture.

The question asks what % of original alcohol ( 0.4 gal ) was replaced ( 0.2 gal ) .

Even if you would consider the question to mean what % of original alcohol ( 0.4 gal ) was replaced with water ( 0.2 gal ) ; even then the ans would be 40%.

0.8*0.5 gal of alcohol was removed from the original solution (0.8 gal of original mix was removed ) and replaced with 0.8*0.25 gal of alcohol ( 0.8 gal of new mix added whose alcohol conc is 25%) from the addition of the new mixture.

The question asks what % of original alcohol ( 0.4 gal ) was replaced ( 0.2 gal ) .

Even if you would consider the question to mean what % of original alcohol ( 0.4 gal ) was replaced with water ( 0.2 gal ) ; even then the ans would be 40%.

Again , open to discussion .

I'm going to do this backwards:

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?

1 G mix 0.5 g Original Alcohol 0.5 g Original Water ***********

Answer: 80% of original alcohol was replaced.

0.4g of alcohol from 0.5g of original alcohol was replaced, (80% of original alcohol replaced- removed AND something else was added), with 0.1g Alcohol+0.3g Water.

Also, 0.4g of water from 0.5g of original water was replaced, (80% of original water replaced- removed AND something else was added), again with 0.1g Alcohol+0.3g Water

So, A+W 0.5+0.5

Removed: 0.4A+0.4W

Remaining 0.1A+0.1W

Added (Placed back for alcohol)+(Placed back for water) (0.1A+0.3W)+(0.1A+ 0.3W)

Total Content: 0.1A+0.1W+0.1A+0.3W+ 0.1A+ 0.3W 0.3A+0.7W

I don't see any problem with the wordings.
_________________

Q : what percentage of the [highlight]original alcohol[/highlight] was replaced? 80% would have been correct if the question asked : what percentage of the [highlight]original mixture[/highlight] was replaced?

In this case , 1 gal alcohol mix => original alcohol = 0.5 gal ; replaced alcohol = 0.8gal*0.25% = 0.2 gal % of original alcohol replaced = (0.2gal)/(0.5gal) *100 = 40%

There might be mistake in this question .point out any mistakes in my logic .

I also arrived at the same answer. I think the question should as "what % of original mixture was replaced" or 40% should be one of the choices.