Linear Equation Solution Mixing Problem (mixing to get given % acid solution)

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spaceshowfeature1

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The Problem is: Example 11 We have a 40% acid solution and we want 75 liters of a 15% acid solution. How much water should we put into the 40% solution to do this?

I try to solve it by plugging it into the standard formula, but the question is very vague, and full of percentages that dont regester. I read the problem over and over, and I never even get to the first step. The words don’t regester anything in my brain. Please help me.

Also, thanks to everyone who helped me on the last problem.
 
spaceshowfeature1 said:
The Problem is: Example 11 We have a 40% acid solution and we want 75 liters of a 15% acid solution. How much water should we put into the 40% solution to do this?

I try to solve it by plugging it into the standard formula, but the question is very vague, and full of percentages that dont regester. I read the problem over and over, and I never even get to the first step. The words don’t regester anything in my brain. Please help me.

Also, thanks to everyone who helped me on the last problem.
Click to expand...

Go ahead and show what your "standard formula" is, and how you plugged in, so we can see if you are partly right. The more you show, the better, even if it's wrong.

It can be helpful to paraphrase or diagram a problem when you are confused by it. Here's my paraphrase:

We are starting with two things: a 40% acid solution, and pure water. We want to mix together some amount of each to make 75 liters of 15% acid solution. How much of each?

I notice that it only asks about the amount of water, but not about the amount of 40% solution. We can find both, which will allow us to check our solution.

So I can repeat the paraphrase, sticking in two variables for the unknowns:

We are starting with two things: a 40% acid solution, and pure water. We want to mix together x liters of the former and y liters of the latter to make 75 liters of 15% acid solution. What are x and y?

(If you only know how to work with one variable, we can deal with that -- that is one reason we want to see your work, so we can know what you expect to use.)
 
Dr.Peterson said:
Go ahead and show what your "standard formula" is, and how you plugged in, so we can see if you are partly right. The more you show, the better, even if it's wrong.

It can be helpful to paraphrase or diagram a problem when you are confused by it. Here's my paraphrase:

We are starting with two things: a 40% acid solution, and pure water. We want to mix together some amount of each to make 75 liters of 15% acid solution. How much of each?

I notice that it only asks about the amount of water, but not about the amount of 40% solution. We can find both, which will allow us to check our solution.

So I can repeat the paraphrase, sticking in two variables for the unknowns:

We are starting with two things: a 40% acid solution, and pure water. We want to mix together x liters of the former and y liters of the latter to make 75 liters of 15% acid solution. What are x and y?

(If you only know how to work with one variable, we can deal with that -- that is one reason we want to see your work, so we can know what you expect to use.)
Click to expand...
The problem is single variabled. The standard formula is secondary liquid in water= percent of solution x volume of solution.
 
w = water

Li'l diagram:
w..........0%
75-w....40%
--------------
75.......15%

40(75-w) = 15(75)
Solve for w

HOKAY?
 
spaceshowfeature1 said:
The problem is single variabled. The standard formula is secondary liquid in water= percent of solution x volume of solution.
Click to expand...

Variables are actually part of the solution method, not the problem; you're saying that you are familiar only with using one variable, so you want to use that method. That's fine.

You can see an example of this sort of problem here. They first use two variables, x and y, but then point out that if we know x, then we can find y by subtracting x from the total, so we can use "total - x" in place of y, and now it is a one-variable problem. That will work here.

Here is my modified paraphrase, using the fact that the total is 75 liters:

We are starting with two things: a 40% acid solution, and pure water. We want to mix together x liters of 40% acid solution and (75-x) liters of pure water (which is 0% acid!) to make 75 liters of 15% acid solution. What are x and y?​

See what you can do with it.
 
Denis said:
w = water

Li'l diagram:
w..........0%
75-w....40%
--------------
75.......15%

40(75-w) = 15(75)
Solve for w

HOKAY?
Click to expand...

As in the Purplemath example, I find it valuable (both for communicating ideas to a student, and for the student knowing what he is doing) if we label everything clearly. The variable w isn't "water"; it's the number of liters of water. And although you know what each row and column in the table means, others won't. Here's my version of your table (borrowed from stapel):

liters sol'npercent acidliters acid
40% sol'n75 – w0.400.40(75 – w)
0% sol'nw00
mixture75
0.15(0.15)(75)

All the same stuff, but now clearly labeled.
 
Dr.Peterson said:
As in the Purplemath example, I find it valuable (both for communicating ideas to a student, and for the student knowing what he is doing) if we label everything clearly. The variable w isn't "water"; it's the number of liters of water. And although you know what each row and column in the table means, others won't. Here's my version of your table (borrowed from stapel):

liters sol'npercent acidliters acid
40% sol'n75 – w0.400.40(75 – w)
0% sol'nw00
mixture750.15(0.15)(75)

All the same stuff, but now clearly labeled.
Click to expand...
Yes! I used this type of table to figure out the d=rt Problems. I still have one question: Why are the liters of the solution 75-w?
 
spaceshowfeature1 said:
Yes! I used this type of table to figure out the d=rt Problems. I still have one question: Why are the liters of the solution 75-w?
Click to expand...

You want 75 liters of mixture (the bottom row of the table). That includes w liters of pure water. The rest of the mixture is the total minus the part: 75 - w.

Read the Purplemath page I referred to again, because that explains it.

This "total minus part" is found in very many problems involving mixtures, whether of solutions, distances, prices, investments, or whatever. It's definitely worth being familiar with.
 
Dr.Peterson said:
You want 75 liters of mixture (the bottom row of the table). That includes w liters of pure water. The rest of the mixture is the total minus the part: 75 - w.

Read the Purplemath page I referred to again, because that explains it.

This "total minus part" is found in very many problems involving mixtures, whether of solutions, distances, prices, investments, or whatever. It's definitely worth being familiar with.
Click to expand...
Thanks again!
 
Generally speaking:

QTY.....%

a.... @ u
b.... @ v
----------
a+b @ w

(au + bv)/(a+b) = w
 
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