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Two mixtures A and B contain milk and water in the ratios [#permalink]
02 Jan 2008, 06:16

00:00

A

B

C

D

E

Difficulty:

35% (medium)

Question Stats:

50% (02:40) correct
50% (02:17) wrong based on 8 sessions

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?

Re: rates - milk and water [#permalink]
03 Jan 2008, 11:07

3

This post received KUDOS

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: rates - milk and water [#permalink]
02 Jan 2008, 08:22

1

This post received KUDOS

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.

Re: rates - milk and water [#permalink]
02 Jan 2008, 08:05

eschn3am wrote:

Mixture A: 2:5 ratio of Milk:Water Mixture B: 5:4 ratio of Milk:water

90 gallons mixture B with a ratio of 5:4 = 50 gallons of milk and 40 gallons of water 90/(5+4) = 10. 10*5 = 50 and 10*4 = 40. (50+(2/7x))/(90+x) = .40 50+(2/7x)=.4(90+x) 50+(2/7x)=(2/5x)+36 50+(10/35x)=(14/35x) +36 4/35x=14 4x=14*35 4x=490 x=122.5 gallons

that's how you would solve it out step-by-step mathematically. Hope that helps, let me know if you have any questions

please explain the reasoning.
_________________

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

Re: rates - milk and water [#permalink]
02 Jan 2008, 08:31

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

Re: rates - milk and water [#permalink]
02 Jan 2008, 08:42

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.

Re: rates - milk and water [#permalink]
02 Jan 2008, 11:13

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.

Re: rates - milk and water [#permalink]
03 Jan 2008, 04:16

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.

Re: rates - milk and water [#permalink]
03 Jan 2008, 04:50

automan wrote:

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

Re: rates - milk and water [#permalink]
03 Jan 2008, 05:18

eschn3am wrote:

automan wrote:

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.

Re: rates - milk and water [#permalink]
03 Jan 2008, 05:27

automan wrote:

eschn3am wrote:

automan wrote:

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%

Re: rates - milk and water [#permalink]
03 Jan 2008, 09:14

automan wrote:

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

Re: rates - milk and water [#permalink]
03 Jan 2008, 09:22

automan wrote:

automan wrote:

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.

Re: rates - milk and water [#permalink]
17 Aug 2009, 23:38

Equation should be like this

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.

Re: rates - milk and water [#permalink]
18 Aug 2009, 09:38

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.