SOLUTION: Could I please have a chart on how to set this up? I know that it's similar to an interest chart, but that's all. A 20% acid solution is mixed with a 70% acid solution to get 50

Question 580075: Could I please have a chart on how to set this up? I know that it's similar to an interest chart, but that's all.
A 20% acid solution is mixed with a 70% acid solution to get 50 liters of a 40% solution.
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
Why use a chart? If you use a chart, it becomes a recipe for solving such problems without understanding what you are doing. Instead let's analyze the information that you are given and see if we can't think our way through this problem.
.
By reading the problem we know that we have to find out how much volume of a 20% solution we are going to mix with an unknown volume of 70% solution. So we have to find two unknown volumes. Let's call the unknown volume of the 20% solution T and the unknown volume of the 70% solution S. (T stands for 20% and S stands for 70%).
.
The problem says that when we add these two volumes together we end up with 50 liters of a mixed solution. So let's set up an equation for that. When we mix the two unknown volumes we are adding them together, and we want to have the answer be 50 liters. In equation form this is:
.
T + S = 50
.
Since we have two unknowns (T and S) we know that we need two equations to solve for these two unknowns. We already have one of the equations. Now we need to look for a second one.
.
Let's think about how much acid we have in each of the volumes. In T (the volume of the 20% solution) 20% (or 0.2) of it is acid. So if we multiply 0.2 times T that answer should equal the volume of acid in that solution. Similarly, in S (the volume of the 70% solution) 70% (or 0.7) of it is acid. So if we multiply 0.7 times T that answer should equal the volume of acid in that solution. How much of the mixture should be acid? We are told that we are to end up with 50 liters of the mixture, and 40% (or 0.4) of that mixture should be acid. For this we can multiply 0.4 times 50 and get an answer of 20. That volume of 20 liters is to be the volume of the acid in the mixture.
.
So now we can write our second equation that we need. We can add the volume of the acid in the two original solutions and set it equal to the volume of acid in the mixture. In equation form this becomes:
.
0.2T + 0.7S = 20
.
We have now reduced the problem to an algebraic exercise, namely solving for two equations that each have two unknowns. Our two equations are:
.
T + S = 50 and
0.2T + 0.7S = 20
.
You probably already know that we can solve two equations with two unknowns by several methods. For example, we can use variable elimination or substitution. (You will learn some other methods later.)
.
Let's use substitution. From the first equation, we can subtract S from both sides and get:
.
T = 50 - S
.
Then we can go to the second equation and substitute 50 - S in place of T. When we do that substitution, the second equation becomes:
.
0.2(50 - S) + 0.7S = 20
.
We then do the distributed multiplication of the first term by multiplying 0.2 times each of the quantities in the parentheses to get:
.
10 - 0.2S + 0.7S = 20
.
Then we subtract 10 from both sides of the equation to get rid of the 10 on the left side, and the equation we are then left with is:
.
-0.2S + 0.7S = 10
.
By combining the two terms on the left side we have:
.
0.5S = 10
.
and finally we divide both sides by 0.5 and the equation becomes:
.
S = 20
.
This tells us that we need 20 liters of S (the 70% solution) in the mixture. And since the mixture is to be 50 liters, we know that the other 30 liters must come from the 20% solution (T).
.
That's the answer --- 30 liters of 20% solution get mixed with 20 liters of the 70% solution to produce 50 liters of a 40% solution.
.
Hope this helps you to understand how you can think your way through problems such as this one. Just remember that in mixture problems involving two unknowns you need to have two independent equations. (Mixing coffee or nuts of two different prices, or acids of different strengths are common examples).
RELATED QUESTIONS
"At noon, a pump is turned on to fill an empty pool. Normally, the pump would fill the... (answered by josgarithmetic,stanbon,ikleyn)

I submitted a problem earlier today and the response definitely helped, but I am stumped... (answered by ptaylor)

1. There are 40 dimes and quarters in the drawer. Peggy counted them and found that their (answered by jim_thompson5910)

A chemist wishes to mix some pure acid with a 10% solution to obtain 60 liters of a 20%... (answered by Mathtut)

Okay, I am taking an online Math class and have come across all of the types of work... (answered by stanbon)

I have been working on this math problem and I got an answer, but I am not sure if it is... (answered by stanbon,solver91311)

Elaan paddled out to check on her crab traps. Going to the traps she caught a falling... (answered by mananth)

Hello Can someone please help me with this. Mr. Doodle�s grade distribution over the... (answered by solver91311,robertb)

My teacher went over these problems in class today. I know that I would set up a chart... (answered by stanbon)