1. Grade 10 Mixture Problems
A simple presentation by Mr. Agostini

2. The Problem
A chemistry teacher needs to make 20 L of 40% sulfuric acid solution. The two containers of acid solutions available contain 30% sulfuric acid and 50% sulfuric acid. How many litres of each solution must be mixed to make a container with 40% sulfuric acid solution.

3. The Idea
Did you know: 30% sulfuric acid contain is made up of 30% acid and the rest is water to dilute the solution. So in a 100 L container, 30L is acid and 70 L is water.
You are going to mix the containers together to obtain 20 litres of a solution that 40% of it will be sulfuric acid. How many litres from each container will we need to use to make this mixture.
If you had to guess, what would you say: Recall 30% acid in one container and 50% in the other. You need 40% in the mixture. How many litres from each mixture do you think we would need.

4. Defining your Variables First
Define your variables first: (Remember the last sentence in the question usually tells us what we are looking for)
Let x be the amount in L that is poured from container 1 at 30% acid solution
Let y be the amount in L that is poured from container 2 at 50% acid solution

5. Creating your two equations: Equation 1
Equation 1 is usually build on the amount of liquid (in this case) that is required. We need 20 L.
Therefore: x amount is being poured in from container 1 and y amount is being poured in from container 2, to make 20 L. Their sum is 20 then.
So the equation is:
x + y = 20 (Equ: 1)

6. Creating your two equations: Equation 2
We create equation 2 from the amount of pure acid that is needed from each container and how much pure acid is in the 20L container. (Recall: 40% of 20 L of pure acid is required)
The amount(L) of pure acid from container 1 will be 30% of x amount(L) poured into the beaker or 0.30x.
The amount(L) of pure acid from container 2 will be 50% of y amount(L) poured into the beaker or 0.50y
Therefore, we combine 30% of x with 50% of y to obtain 40% of 20 L
The second equation is:
0.30x y = 0.40(20) (Equ: 2)

7. So what do we have: Let x = amount in L of 30% sulphuric acid soluton.
Let y = amount in L of 50% sulphuric acid soluton.
(1) x + y = 20
(2) 0.30x y = 0.40(20) Or
(2) 0.3x + 0.5y = 8

8. When Solving you can use any method you wish. Yay!!!!!
Elimination Method:
( x -3)
( x 10)
-3x – 3y = -60
3x + 5y = 80
Back sub into x + y = 20
x + 10 = 20
x = 20 – 10
x = 10
Therefore 10 L of container 1 (at 30% acid solution) and 10 L of container 2 (at 50% acid solution) is needed to make 20 L at 40% acid solution.
x + y = 20
0.3x + 0.5y = 8 +
2y = 20
y = 10

9. When Solving you can use any method you wish. Yay!!!!!
Substitution Method:
(1) x + y = 20
Back sub into x + y = 20
(2) 0.3x + 0.5y = 8
10 + y = 20
y = 20 – x sub into (2)
y =
0.3x + 0.5(20 – x) = 8
y = 10
0.3x + 10 – 0.5x = 8
Therefore 10 L of container 1 (at 30% acid solution) and 10 L of container 2 (at 50% acid solution) is needed to make 20 L at 40% acid solution.
0.3x – 0.5x =
– 0.2x = - 2
x = 10