1. Chemistry
Mixture
Problems

2. Let x = the mL of the 18% solution Let y = the mL of the 78% solution
A chemist needs 150 milliliters of a 50% saline solution but has only 18% and 78% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.
We’re asked to find the amount used of each of the solutions she has on hand, so …
Let x = the mL of the 18% solution
Let y = the mL of the 78% solution

3. Our first equation is x + y = 150
Now we need to find out two relationships between these amounts:
1) A chemist needs 150 milliliters of solution.
The amount of 18% solution + the amount of 78% solution = 150 mL
Our first equation is x + y = 150

4. Our second equation is .18x + .78y = 75
We need one more relationship between the amounts:
2) A chemist needs 150 milliliters of a 50% saline solution.
That means that she needs 150(.05) or 75 mL of salt in her solution
Amount of salt in the 18% solution = .18x
Amount of salt in the 78% solution = .78y
Our second equation is .18x + .78y = 75

5. Now we have two equations in two unknowns
x + y = 150
.18x + .78y = 75
I would solve this by elimination
.18x + .78y = 75 multiply by 100 to clear the decimals
18x + 78y = 7500

6. Multiply the top equation by -18 to clear the x column:
(I picked 18 because it’s smaller than 78 - the numbers will be easier to work with)
x + y = 150
18x + 78y = 7500
-18x - 18y = -2700
Now just add the equations together. The x’s will add up to zero and disappear!
60y = 4800
y = 80
She needs 80ml of 78% solution.

7. Now let’s find the amount of 18% solution and check our work
x + y = 150
y = 80
x + 80 = 150
x = 70
Now we have 80 mL of 78% solution and 70 mL of 18% solution.
Does this fit our story?
Does 80 at 78% + 70 at 18% = 75?
80(.78) + 70(.18)
= 75
We have a winner!
80 ml of 78% solution and 70 mL of 18% solution