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:

Two mixtures A and B contain milk and water in the ratios [#permalink]

Show Tags

02 Jan 2008, 07:16

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

73% (03:06) correct
27% (02:40) wrong based on 120 sessions

HideShow timer Statistics

Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?

Mixture B is a 5:4 ratio mixture of Milk:Water. In other words, for every 5 gallons of milk in B there are 4 gallons of water. To find out the actual number of gallons of milk in X gallons of mixture B there are a couple ways to go about it:

If we have 5 gallons of milk there are 4 gallons of water. So milk = 5/(5+4) or 5/9 is milk. You add 4 and 5 together to get the total volume of the mixture (milk + water) 90 gallons of mixture * 5/9 milk = 50 gallons of milk.

OR

You can see that 5:4 add up to 9 and that 90 is a multiple of 9. If 5:4 = 9 and 9*10 = 90 and we have 90 gallons of the mixture just multiply the ratio by 10. 5*10:4*10 = 50:40 = 50 gallons of milk

Let's say we have milk:water in a 1:9 ratio. We have 15 gallons of the mixture, how many gallons of milk?

1+9=10 so 1 gallon of milk for every 10 gallons of mixture 15/10 = 1.5 1*1.5:9*1.5 multiply each side by (15/10 or 1.5) 1.5:13.5 1.5+13.5 = 15 and 1.5 gallons of it is milk.

yea, figured it out. what u did was u took the total amount/total ratio = multiplier, in turn usnig the multilpier to find the respective amounts in the ratio.

Are we missing the final step? 2/7 x = 35 _________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Nope, we need 122.5 gallons of mixture A to be mixed in with mixture B to have 40% milk. The question is just asking how much A we need to mix in, but 35 gallons is indeed the amount of milk in the 122.5 gallons of Mixture A.

we're given that B has a total of 90 gallons. My first step is to figure out how much milk is in 90 gallons of B. Given the ratio 5:4 milk:water, i translate this into 5:9 milk:total.

90 gallons has 50 gallons milk and 40 water.

Now, since we are given info about milk, I put all numbers in terms of milk ... so the ratio of A goes from 2:5 milk:water to 2:7 milk:total. We want to find out how much of A we are going to add, so I set up as follows:

(50+2/7x) / (90+x) = 0.4 (or 40/100 or 4/10, whatever you want). X is the amount of A we will be adding.

I think, the easiest way to resolve this type of questions is making a simple table Component/Quantity of milk-water and then constructing an equation. For component A we will have 2/7A gallons of milk and 5/7 gallons of water. For component B we will have 5/9 gallons of milk and 4/9 of water. Therefore, we will have that:

0.4=(2/7A+5/9B)/(A+B) where B=90. You solve the equation and obtain A=35.

I think, the easiest way to resolve this type of questions is making a simple table Component/Quantity of milk-water and then constructing an equation. For component A we will have 2/7A gallons of milk and 5/7 gallons of water. For component B we will have 5/9 gallons of milk and 4/9 of water. Therefore, we will have that:

0.4=(2/7A+5/9B)/(A+B) where B=90. You solve the equation and obtain A=35.

but that answers the question of how much milk must be added, we are solving for how much of mixture A must be added. you can still get to the answer, it's just an extra step

I think, the easiest way to resolve this type of questions is making a simple table Component/Quantity of milk-water and then constructing an equation. For component A we will have 2/7A gallons of milk and 5/7 gallons of water. For component B we will have 5/9 gallons of milk and 4/9 of water. Therefore, we will have that:

0.4=(2/7A+5/9B)/(A+B) where B=90. You solve the equation and obtain A=35.

but that answers the question of how much milk must be added, we are solving for how much of mixture A must be added. you can still get to the answer, it's just an extra step

The question is about how much quantity of mixture A has to be added (".....How many gallons of A must be mixed with 90 gallons....") and that is what I have calculated.

I think, the easiest way to resolve this type of questions is making a simple table Component/Quantity of milk-water and then constructing an equation. For component A we will have 2/7A gallons of milk and 5/7 gallons of water. For component B we will have 5/9 gallons of milk and 4/9 of water. Therefore, we will have that:

0.4=(2/7A+5/9B)/(A+B) where B=90. You solve the equation and obtain A=35.

but that answers the question of how much milk must be added, we are solving for how much of mixture A must be added. you can still get to the answer, it's just an extra step

The question is about how much quantity of mixture A has to be added (".....How many gallons of A must be mixed with 90 gallons....") and that is what I have calculated.

0.4=(2/7A+5/9B)/(A+B) where B=90. You solve the equation and obtain A=35.

so you did set it up to get mixture A as your answer, but somewhere in your calculations you mixed up and found the amount of milk in mixture A instead (122.5*2/7) = 35.

Check your answer 35 to see for yourself ((90*5/9)+(35*2/7))/90+35 (50+10)/125 60/125 = 48%

Now try it with 122.5 ((90*5/9)+(122.5*2/7)/90+122.5 (50+35)/212.5 85/212.5 = 40%

I think, the easiest way to resolve this type of questions is making a simple table Component/Quantity of milk-water and then constructing an equation. For component A we will have 2/7A gallons of milk and 5/7 gallons of water. For component B we will have 5/9 gallons of milk and 4/9 of water. Therefore, we will have that:

0.4=(2/7A+5/9B)/(A+B) where B=90. You solve the equation and obtain A=35.

Sorry, I was wrong.... A=122.5; I do not know where I got lost

I think, the easiest way to resolve this type of questions is making a simple table Component/Quantity of milk-water and then constructing an equation. For component A we will have 2/7A gallons of milk and 5/7 gallons of water. For component B we will have 5/9 gallons of milk and 4/9 of water. Therefore, we will have that:

0.4=(2/7A+5/9B)/(A+B) where B=90. You solve the equation and obtain A=35.

Sorry, I was wrong.... A=122.5; I do not know where I got lost

No worries, your equation was set up perfectly. Try working it through again and see if you can spot your mistake. If you still can't find it you're welcome to post it up here and we'll be happy to take a look at it for you.

Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?

How can I do this without using the ratio approach?

As you may have seen in a previous post, I like to shy away from algebra if there is some other framework I can use instead. For mixtures and solutions, I simply draw this box, fill in what is given, and then make everything work out to the bottom right corner.

The rows represent the solutions being mixed. The first solution first, the added solution second, and the resulting solution third. The columns represent the parts of the solution. The first (left) column is the total solution, the middle represents the percentage of the "stuff" (in this case, milk, but could be acid, salt, alcohol, etc), and the third column is the total "stuff" in the solution.

We multiply the first column by the second column to get the third column. When we mix them together, the total amounts mix, as do the amounts of the "stuff", so we add down. Note that the middle column doesn't add - it's just there to get us from the left to the right.

So here is how I applied it to this problem. We know how much of B there is, and we are asked how much of A is mixed with it to get 40% milk all together. Follow the chart, and you see that the bottom row multiplies to the right box, and the right column adds to the right box, so we have the right box from two directions. That's how we solve.

50 + 2/7x = .4(90 + x) x = 122.5

Check out some earlier posts I made on this topic.

No of gallon milk from A+No of gallon milk from B = (total gallon of mixture)*40/100(40% is milk in tat) Since we are comparing only the milk so both RHS and LHS should be milk.

I found Ian7777's all posts very clear, detail and useful. More than that his attitude is humble and highly positive. I always enjoy reading his posts and learned a lot from him.

Thanks Ian.

GT

ian7777 wrote:

bmwhype2 wrote:

Two mixtures A and B contain milk and water in the ratios 2:5 and 5:4 respectively. How many gallons of A must be mixed with 90 gallons of B so that the resultant mixture contains 40% milk?

How can I do this without using the ratio approach?

As you may have seen in a previous post, I like to shy away from algebra if there is some other framework I can use instead. For mixtures and solutions, I simply draw this box, fill in what is given, and then make everything work out to the bottom right corner.

The rows represent the solutions being mixed. The first solution first, the added solution second, and the resulting solution third. The columns represent the parts of the solution. The first (left) column is the total solution, the middle represents the percentage of the "stuff" (in this case, milk, but could be acid, salt, alcohol, etc), and the third column is the total "stuff" in the solution.

We multiply the first column by the second column to get the third column. When we mix them together, the total amounts mix, as do the amounts of the "stuff", so we add down. Note that the middle column doesn't add - it's just there to get us from the left to the right.

So here is how I applied it to this problem. We know how much of B there is, and we are asked how much of A is mixed with it to get 40% milk all together. Follow the chart, and you see that the bottom row multiplies to the right box, and the right column adds to the right box, so we have the right box from two directions. That's how we solve.

50 + 2/7x = .4(90 + x) x = 122.5

Check out some earlier posts I made on this topic.

Re: Two mixtures A and B contain milk and water in the ratios [#permalink]

Show Tags

22 Nov 2014, 13:24

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. _________________

Last year when I attended a session of Chicago’s Booth Live , I felt pretty out of place. I was surrounded by professionals from all over the world from major...

I recently returned from attending the London Business School Admits Weekend held last week. Let me just say upfront - for those who are planning to apply for the...