Lesson Objectives
  • Demonstrate an understanding of how to translate phrases into algebraic expressions and equations
  • Demonstrate an understanding of the six-step method used for solving applications of linear equations
  • Learn how to solve mixture word problems

How to Solve Mixture Word Problems


In our last lesson, we reviewed the six-step method used to solve a word problem that involves a linear equation in one variable.

Six-step method for Solving Word Problems with Linear Equations in One Variable

  1. Read the problem and determine what you are asked to find
  2. Assign a variable to represent the unknown
    • If more than one unknown exists, we express the other unknowns in terms of this variable
  3. Write out an equation which describes the given situation
  4. Solve the equation
  5. State the answer using a nice clear sentence
  6. Check the result
    • We need to make sure the answer is reasonable. In other words, if asked how many students were on a bus, the answer shouldn't be (-4) as we can't have a negative amount of students on a bus.

Mixture Word Problems

Mixture word problems can be very frustrating. The key is to understand how to find the concentration of a particular substance within a given mixture. Let's start out with a simple example.
Example 1: Solve each word problem
An acid solution was made by mixing 5 gallons of an 80% acid solution and 7 gallons of a 50% acid solution. Find the concentration of acid in the new mixture.
For this problem, we don't need any variables. Let's think for a moment about how much acid is in each mixture:
Mixture (Gal) Acid % Pure Acid (Gal)
580%4
750%3.5
If we look at our table above, we can see that there are 4 gallons of pure acid in our 5-gallon mixture and 3.5 gallons of pure acid in our 7-gallon mixture. If we combine the two mixtures, we get a 12-gallon mixture which contains 7.5 gallons of pure acid. How can we find the % of acid for our new mixture? We divide the number of gallons of pure acid by the number of gallons of the total mixture:
7.5 ÷ 12 = 0.625
We can say our acid percentage is 62.5% for the mixture. The main thing is to understand that we can take the pure amount of a substance and divide it by the total mixture to find a concentration. Let's look at a more challenging mixture word problem.
Example 2: Solve each word problem
A local chemist made a solution that was 14% alcohol. He started out with 12 gallons of a 12% alcohol solution. He then added an unknown number of gallons of a 20% acid solution. How many gallons of the 20% acid solution did the chemist add to the initial alcohol solution?
Step 1) After reading the problem, it is clear that we want to find the number of gallons of the 20% acid solution that needs to be added to the 12% acid solution in order to obtain a solution that is 14% alcohol.
Step 2) let x = number of gallons of the 20% alcohol solution
Step 3) To write out an equation, let's first organize our information in a table:
Mixture (Gal) Alcohol % Pure Alcohol (Gal)
1212%1.44
x20%.2x
x + 1214%.14(x + 12)
From our table, we can see the number of gallons of pure acid from the 12% acid solution is 1.44 (12 • 0.12 = 1.44). Additionally, we represent the number of gallons of pure acid from our 20% acid solution as 0.2x. These two amounts must be equal to the amount of pure acid in the final solution, which is .14(x + 12).
(.14 » 14% acid concentration in the final solution)
(x + 12) » the number of gallons in the final solution
Let's set up our equation:
.14(x + 12) = .2x + 1.44
Step 4) Solve the equation
.14(x + 12) = .2x + 1.44
.14x + 1.68 = .2x + 1.44
Multiply each side by 100 to clear the decimals:
14x + 168 = 20x + 144
14x - 20x = 144 - 168
-6x = -24
x = -24 ÷ -6
x = 4
Step 5) Since x is the number of gallons of the 20% alcohol solution used, we can state our answer as:
The chemist used 4 gallons of the 20% alcohol solution.
Step 6) We can check our result by reading through the problem. We know the end result is a 16-gallon solution that is 14% alcohol. We can compare this to the sum of the alcohol from the two different solutions:
Solution 1 (12 gallons, 12% alcohol):
12 • .12 = 1.44
Solution 2 (4 gallons, 20% alcohol):
4 • .2 = .8
1.44 + 0.8 = 2.24
There are 16 gallons in the final solution with an alcohol percentage of .14, let's check to see if the number of gallons matches up:
16 • 0.14 = 2.24
2.24 = 2.24

Skills Check:

Example #1

Solve each word problem.

A metallurgist needs to make 12 oz. of an alloy containing 71% copper. She is going to melt and combine one metal that is 78% copper with another metal that is 66% copper. How much of each should she use?

Please choose the best answer.

A
2 oz. 78% copper, 10 oz. of 66% copper
B
5 oz. 78% copper, 7 oz. of 66% copper
C
3 oz. 78% copper, 9 oz. of 66% copper
D
3 oz. 78% copper, 9 oz. of 66% copper
E
11 oz. 78% copper, 1 oz. of 66% copper

Example #2

Solve each word problem.

3 ounces of Ryan's Premium Molasses was made by combining 2 ounces of cane molasses which costs $1 per ounce with 1 ounce of beet molasses which costs $4 per ounce. Find the cost per ounce of the mixture.

Please choose the best answer.

A
$2 per ounce
B
$3.25 per ounce
C
$2.75 per ounce
D
$2.25 per ounce
E
$1.75 per ounce

Example #3

Solve each word problem.

How many gallons of an 80% sugar solution must be mixed with 7 gallons of a 35% sugar solution to make a 59% sugar solution?

Please choose the best answer.

A
13 gallons
B
11 gallons
C
6 gallons
D
8 gallons
E
5 gallons
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