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 350,000 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:

Re: Mixture Problems - I collected from GMAT Club [#permalink]
11 Jan 2010, 00:44

I think the answer for 13 is B

From the question, we have below equation: x+y = 10(1)

From 1ST, y>4 > insufficent From 2ST, 3x + 5y <40 (3) Assume 3x+5y = 40 (2) From (1) & (2): we have x= 5, y=5 (3) < 40 when y<5, y= 4,3,2,1 x>5, x=6,7,8,9 therefore x>y. We don't need 1ST to answer this question.

Pls correct me, because I am not sure my solution is right. Thanks

Re: Mixture Problems - I collected from GMAT Club [#permalink]
02 Feb 2010, 17:43

samrus98 wrote:

tejal777 wrote:

Wanna confirm my answer for 2..(B)?

Yep, the answer for Q2 is B i.e. 2.5 => 3*0.25 = 0.75 gallons of white paint and 5*0.25 = 1.25 gallons of black paint are needed. But since only 0.5 and 1 gallon cans are available, 1 gallon of white paint and 1.5 gallon of black paint would be needed => 1 + 1.5 = 2.5

Thank you so much for your explanation. I am not too sure I am even understanding the question correctly. Why did you multiply the parts of paint by 0.25. Where does that number come from?

Re: Mixture Problems - I collected from GMAT Club [#permalink]
15 Mar 2010, 16:17

Wow, I am amazed yet again for GMAT clubbers! +1! _________________

My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own. I am the force; I can clear any obstacle before me or I can be lost in the maze. My choice; my responsibility; win or lose, only I hold the key to my destiny - Elaine Maxwell

Re: Mixture Problems - I collected from GMAT Club [#permalink]
31 Aug 2010, 22:54

Hi, Thanks a lot for this, I am really weak at mixture and work problems. I was planning on running through the OG and Kaplan looking for mixture problems. This saved me a lot of work!

Re: Mixture Problems - I collected from GMAT Club [#permalink]
11 Mar 2011, 04:35

8.Solution Y is 30 percent liquid X and 70 percent water. If 2 kilograms of water evaporate from 8 kilograms of solution Y and 2 kilograms of solution Y are added to the remaining 6 kilograms of liquid, what percent of this new solution is liquid X? (A) 30% (B) 33 1/3% (C) 37 1/2% (D) 40% (E) 50%

Step 1: Given: Y = 30% L + 70% W

Given: 8 kg solution. Therefore: 30% L = 2.4 kg Liquid X and 70% W = 5.6 kg Water.

Given: 2 kg water evaporate, therefore remaining water = 5.6 kg - 2kg = 3.6 kg water.

Now, total solution weight = 3.6 kg (water) + 2.4 kg (liquid X) = 6 kg Liquid (this is already stated in the problem)

Next step, the problem states that 2 kg solution Y are added to the 6 kg of liquid...Therefore in 2 kg of Solution Y, we have 30% Liquid X = 0.6 kg Liquid X AND 70% Water = 1.4 kg Water. (Add these up, they total to 2 kg).

The question asks, what percent of this NEW solution is liquid X?

So: Liquid X = Original solution + added solution = 2.4 kg + 0.6 kg = 3 kg in NEW solution.

New solution total = (Original + added) liquid X + (Original + added) water = (2.4+0.6) + (3.6+1.4) = 8 kg

Re: Mixture Problems - I collected from GMAT Club [#permalink]
11 Mar 2011, 08:13

15.Two different solutions of alcohol with respective proportions of alcohol to water of 3:1 and 2:3 were combined. What is the concentration of alcohol in the new solution if the first solution was 4 times the amount of the second solution?

Solution: Always start these problems off by assuming a smart number...assume 40 for the first solution here...the second solution is 10 (based on the question) now: 3x + x =40, therefore we have 30 units of alcohol and 10 units of water in the first solution. 2x + 3x = 10, therefore we have 4 units of alcohol to 6 units of water in the 2nd solution. Now adding alcohol up = 30 + 4 = 34. Percent of alcohol in new solution = alcohol / new total = (34/50) X 100 = 68 %

Re: Mixture Problems - I collected from GMAT Club [#permalink]
02 Jul 2011, 19:32

How much water (in grams) should be added to a 35%-solution of acid to obtain a 10%-solution?

1)There are 50 grams of the 35%-solution. 2)In the 35%-solution the ratio of acid to water is 7:13.

Using statement 1) we can solve the question.. in 50g of 35% solution we have 17.5g acid and 32.5g water. If we add x gram of water to the solution to obtain a 10% solution, the equation becomes 17.5 / (50+x) =10% ...Thus we can solve for x which is 125g.

Statement 2) talks about ratio and not the exact quantity..We will have multiple solutions for the same and hence not sufficient.

Re: Seed mixture X is 40 percent ryegrass and 60 + More Problems [#permalink]
28 Jan 2014, 23:27

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Hey, everyone. After a hectic orientation and a weeklong course, Managing Groups and Teams, I have finally settled into the core curriculum for Fall 1, and have thus found...

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

After I was accepted to Oxford I had an amazing opportunity to visit and meet a few fellow admitted students. We sat through a mock lecture, toured the business...