Solving Mixture Problems
One variable equations
The Problem
• A chemist has 20 ml of a solution that is
60% acid. How much 15% acid solution
must she add to get a solution that is 25%
acid?
Identify key words
• A chemist has 20 ml of a solution that is
60% acid. How much 15% acid solution
must she add to get a solution that is 25%
acid?
How much guides us to the question: How
much 15% solution?
To get guides us to where = will go.
A chemist has 20 ml of a solution that is 60% acid. How
much 15% acid solution must she add to get a solution that
is 25% acid?
• First, we draw a chart for mixtures:
Amount
First
Solution
Second
Solution
Final
Solution
Rate
Mixture
A chemist has 20 ml of a solution that is 60% acid. How
much 15% acid solution must she add to get a solution that
is 25% acid?
• We fill in the key information. Convert % to decimal:
Amount(ml) Rate (%)
First
Solution
20
.60
Second
Solution
Final
Solution
How much
.15
To get
.25
Mixture
A chemist has 20 ml of a solution that is 60% acid. How
much 15% acid solution must she add to get a solution that
is 25% acid?
•
The first column is added down.
20 + “how much” = “to get”
•
We have two unknowns! The question words are x, so
now we have 20 + x = “to get”
•
20 + x = “to get”
Because we have an equation, 20 + x means the same
amount as “to get.” Until you can use two variables,
you must use the 20 + x in place of “to get”
A chemist has 20 ml of a solution that is 60% acid. How
much 15% acid solution must she add to get a solution that
is 25% acid?
• Now we can fill in our table for the amounts:
Amount(ml) Rate %
First
Solution
20
.60
Second + X
Solution
.15
Final
= 20 + x
Solution
.25
Mixture
A chemist has 20 ml of a solution that is 60% acid. How
much 15% acid solution must she add to get a solution that
is 25% acid?
• Now we multiply each row across. It’s better to write the mixture as
“rate * amount” even though the columns are in the opposite order.
(remember, 3*4 is the same as 4*3).
Amount(ml) * Rate % = Mixture
First
Solution
20
* .60
= .60(20)
Second
Solution
X
* .15
= .15x
Final
Solution
(20 + x)
*
= .25(20 + x)
.25
A chemist has 20 ml of a solution that is 60% acid. How
much 15% acid solution must she add to get a solution that
is 25% acid?
• To get our final equation, we add down the Mixture column:
Amount (ml) Rate %
First
20
Solution
Second X
Solution
Final
(20 + x)
Solution
.60
Mixture
.60(20)
+
.15
.15x
=
.25
.25(20 + x)
A chemist has 20 ml of a solution that is 60% acid. How
much 15% acid solution must she add to get a solution that
is 25% acid?
•
We have our final equation!
.60(20) + .15x = .25(20 + x)
•
So simplify and solve:
.60(20) + .15x = .25(20 + x)
12 + .15x = .25(20 + x)
12 + .15x = 4 + .25x
12 -.10x = 4
-.10x = -8
X = 80
A chemist has 20 ml of a solution that is 60% acid. How
much 15% acid solution must she add to get a solution that
is 25% acid?
• Now we know that x = 80. Did we answer the question?
How much 15% acid solution must she add to get a
solution that is 25% acid?
Remember, “how much” was an amount measured in ml.
She must add 80 ml.
Other ways to determine amounts
The amounts column tends to be the difficult one in mixture
problems solved with only one variable. Here are the
usual patterns:
• Has 60 ml. + “how much” = “to get”
60 + x = (60 + x)
• “How much” + added to 60 ml = “to get” x + 60 = (60 + x)
The trickiest ones have the second unknown in the middle:
• “How much” + “added to” = to get 80 ml x + (80 - x) = 80
Remember to watch your question. You may need to
provide more than one answer!
There are other types of mixture problems that have slight
differences in their table set-ups:
• Financial interest mixtures have no final rate, and may not have a
final amount. “Interest earned” is the key phrase for the interest total
cell. Translation sentences are important in determining amounts.
Amount
Rate %
Interest
First
Investment
Second
Investment
Final/Total
May or may not have
Total Interest earned:
• Coins and mixed purchases (e.g. coffee and lattes) also don’t have
a final rate. Each rate is the value of the item: (e.g. nickels are .05,
a coffee may be 1.10). In each case a total value or cost is in the
final mixture cell. Again, translation sentences are important in
these.
Number
Rate
Value/Cost
First Item
Second
Item
Final/Total
May or may not have
Total value/cost:
• There are other mixture problems where a final rate is important.
These are usually determining the best price for a mixture of types of
nuts or types of flowers. In these cases, a final rate is provided, and
the problem is calculated like the solution mixtures: Add down in the
amount and value columns, multiply across on the items and final.
Amount
First Item
Second
Item
Final/Total
Rate
Value/Cost