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.

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:

Re: PS: Mixture Problem [#permalink]
11 Feb 2011, 23:55

One may solve it by alligation as well.

50 30 \ / 40 / \ 10 10

50 - Percent of acid in first solution 30 - Percent of acid in second solution 40 - Percent of acid in mixture of both LHS 10 -> Percent of acid in mixture of both - Percent of acid in second solution = 40-30 -> Portion of the second in the mixture RHS 10 -> Percent of acid in first solution - Percent of acid in mixture of both = 50-40 ->Portion of the first in the mixture

\frac{Portion \quad of \quad second}{Portion \quad of \quad first}=\frac{10}{10}=\frac{1}{1} \frac{Portion \quad of \quad first}{Total \quad mixture}=\frac{1}{1+1}=\frac{1}{2}

If the new mixture contains 1/2 of the first solution; then half of the first solution must have been replaced with second solution.

Re: PS: Mixture Problem [#permalink]
28 Jun 2011, 13:05

imerial wrote:

Seems like it's easier just to solve it straight. (30%+50%)/2=40%. So the amount replaced was 50-30 or 20%. Or am I missing something?

I don't think this is correct.

You have a 50% solution and a 30% solution. You are replacing part of the 50% solution with the 30% solution to make a 40% solution.

There are lots of different approaches to solve the problem.

Intuitively -- you know that 40% is in between 30% and 50% -- so 1/2 of the 50% solution will need to be replaced.

You can use algebra and solve this or use the tables. You can also use the "allegation" method -- which is quite useful in these problems. I don't quite follow how you got 20%.

Re: PS: Mixture Problem [#permalink]
28 Jun 2011, 19:25

Sorry, the 20% was a typo. My thought process was:

1) We have a 50% solution. 2) We want a 40% solution. 3) We have at our disposal a 30% solution. 4) 40 is the average of 50 and 30, so the solution needs to be half 50 and half 30. 5) Therefore, we replace half the 50% solution with 30% solution. 6) Now instead of being all 50 we're half 50 half 30. That means we replaced half the solution.

Does that work or have I misunderstood? I saw all the complex math in the answers and got confused as to why it was necessary.

Re: PS: Mixture Problem [#permalink]
28 Jun 2011, 20:13

imerial wrote:

Sorry, the 20% was a typo. My thought process was:

1) We have a 50% solution. 2) We want a 40% solution. 3) We have at our disposal a 30% solution. 4) 40 is the average of 50 and 30, so the solution needs to be half 50 and half 30. 5) Therefore, we replace half the 50% solution with 30% solution. 6) Now instead of being all 50 we're half 50 half 30. That means we replaced half the solution.

Does that work or have I misunderstood? I saw all the complex math in the answers and got confused as to why it was necessary.

Re: PS: Mixture Problem [#permalink]
07 Nov 2011, 02:24

lonewolf wrote:

Hermione wrote:

Does anyone have a quick and easy way of solving this problem? I had to pick from the answer choices but I want to know how to solve it outright.

Some part of the 50% solution of acid was replaced with the equal amount of 30% solution of acid. As a result, 40% solution f acid was obtained. what part of the original solution was replaced?

1/5 1/4 1/2 3/4 4/5

Let X = Part of acid solution replaced, then 1-X will be the parts not replaced.