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3.3 Solve mixture applications

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By the end of this section, you will be able to:
  • Solve coin word problems
  • Solve ticket and stamp word problems
  • Solve mixture word problems
  • Use the mixture model to solve investment problems using simple interest

Before you get started, take this readiness quiz.

  1. Multiply: 14(0.25).
    If you missed this problem, review [link] .
  2. Solve: 0.25 x + 0.10 ( x + 4 ) = 2.5 .
    If you missed this problem, review [link] .
  3. The number of dimes is three more than the number of quarters. Let q represent the number of quarters. Write an expression for the number of dimes.
    If you missed this problem, review [link] .

Solve coin word problems

In mixture problems    , we will have two or more items with different values to combine together. The mixture model is used by grocers and bartenders to make sure they set fair prices for the products they sell. Many other professionals, like chemists, investment bankers, and landscapers also use the mixture model.

Doing the Manipulative Mathematics activity Coin Lab will help you develop a better understanding of mixture word problems.

We will start by looking at an application everyone is familiar with—money!

Imagine that we take a handful of coins from a pocket or purse and place them on a desk. How would we determine the value of that pile of coins? If we can form a step-by-step plan for finding the total value of the coins, it will help us as we begin solving coin word problems.

So what would we do? To get some order to the mess of coins, we could separate the coins into piles according to their value. Quarters would go with quarters, dimes with dimes, nickels with nickels, and so on. To get the total value of all the coins, we would add the total value of each pile.

Piles of pennies, nickels, dimes, and quarters

How would we determine the value of each pile? Think about the dime pile—how much is it worth? If we count the number of dimes, we’ll know how many we have—the number of dimes.

But this does not tell us the value of all the dimes. Say we counted 17 dimes, how much are they worth? Each dime is worth $0.10—that is the value of one dime. To find the total value of the pile of 17 dimes, multiply 17 by $0.10 to get $1.70. This is the total value of all 17 dimes. This method leads to the following model.

Total value of coins

For the same type of coin, the total value of a number of coins is found by using the model

n u m b e r · v a l u e = t o t a l v a l u e

where
     number is the number of coins

     value is the value of each coin

     total value is the total value of all the coins

The number of dimes times the value of each dime equals the total value of the dimes.

n u m b e r · v a l u e = t o t a l v a l u e 17 · $0.10 = $ 1.70

We could continue this process for each type of coin, and then we would know the total value of each type of coin. To get the total value of all the coins, add the total value of each type of coin.

Let’s look at a specific case. Suppose there are 14 quarters, 17 dimes, 21 nickels, and 39 pennies.

This table has five rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Quarters, 14, 0.25, and 3.50. The third row reads Dimes, 17, 0.10, and 1.70. The fourth row reads Nickels, 21, 0.05, and 1.05. The fifth row reads Pennies, 39, 0.01, and 0.39. The extra cell reads 6.64.

The total value of all the coins is $6.64.

Notice how the chart helps organize all the information! Let’s see how we use this method to solve a coin word problem.

Adalberto has $2.25 in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type of coin does he have?

Solution

Step 1. Read the problem. Make sure all the words and ideas are understood.

  • Determine the types of coins involved.
    Think about the strategy we used to find the value of the handful of coins. The first thing we need is to notice what types of coins are involved. Adalberto has dimes and nickels.
  • Create a table to organize the information. See chart below.
    • Label the columns “type,” “number,” “value,” “total value.”
    • List the types of coins.
    • Write in the value of each type of coin.
    • Write in the total value of all the coins.
    We can work this problem all in cents or in dollars. Here we will do it in dollars and put in the dollar sign ($) in the table as a reminder.
    The value of a dime is $0.10 and the value of a nickel is $0.05. The total value of all the coins is $2.25. The table below shows this information.
    This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Dimes, blank, 0.10, and blank. The third row reads Nickels, blank, 0.05, and blank. The extra cell reads 2.25.

Step 2. Identify what we are looking for.

  • We are asked to find the number of dimes and nickels Adalberto has.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

  • Use variable expressions to represent the number of each type of coin and write them in the table.
  • Multiply the number times the value to get the total value of each type of coin.

Next we counted the number of each type of coin. In this problem we cannot count each type of coin—that is what you are looking for—but we have a clue. There are nine more nickels than dimes. The number of nickels is nine more than the number of dimes.

Let d = number of dimes. d + 9 = number of nickels

Fill in the “number” column in the table to help get everything organized.

This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Dimes, d, 0.10, and blank. The third row reads Nickels, d plus 9, 0.05, and blank. The extra cell reads 2.25.

Now we have all the information we need from the problem!

We multiply the number times the value to get the total value of each type of coin. While we do not know the actual number, we do have an expression to represent it.

And so now multiply n u m b e r · v a l u e = t o t a l v a l u e . See how this is done in the table below.

This table has three rows and four columns with an extra cell at the bottom of the fourth column. The top row is a header row that reads from left to right Type, Number, Value ($), and Total Value ($). The second row reads Dimes, d, 0.10, and 0.10d. The third row reads Nickels, d plus 9, 0.05, and 0.05 times the quantity (d plus 9). The extra cell reads 2.25.

Notice that we made the heading of the table show the model.

Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence. Translate the English sentence into an algebraic equation.

Write the equation by adding the total values of all the types of coins.

The sentence, “value of dimes plus value of nickels equals total value of coins,” can be translated to an equation. Translate “value of dimes” to 0.10d, translate “value of nickles” to 0.05d, and translate “total value of coins” to 2.25. The full equation is 0.10d plus 0.05 times the quantity d plus 9 equals 2.25.

Step 5. Solve the equation using good algebra techniques.

Now solve this equation. .
Distribute. .
Combine like terms. .
Subtract 0.45 from each side. .
Divide. .
So there are 12 dimes.
The number of nickels is d + 9 . .
.
21

Step 6. Check the answer in the problem and make sure it makes sense.

Does this check?

12 dimes 12 ( 0.10 ) = 1.20 21 nickels 21 ( 0.05 ) = 1.05 ____ $2.25 ✓

Step 7. Answer the question with a complete sentence.

  • Adalberto has twelve dimes and twenty-one nickels.

If this were a homework exercise, our work might look like the following.

A homework assignment written on lined loose leaf paper. The assignment reads: “Adalberto has 2 dollars and 25 cents in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type does he have?” Below this is a table. The first row of the table is a header row, and each cell names the column or columns below it. The first cell from the left is named “Type.” The second cell contains the equation “Number” times “Value” equals “Total Value,” with one column corresponding to “Number,” one column corresponding to “Value,” and one column corresponding to total value. Hence the content of the “Number” column times the content of the “Value” column equals the content of the “Total Value” column. In the second row of the table, the “Type” column contains “Dimes,” the “Number” column contains d, the “Value” column contains 0.10, and the “Total Value” column contains 0.10d. In the third row of the table, the “Type” column contains “Nickels,” the “Number” column contains d plus 9, the “Value column contains 0.05, and the “Total Value” column contains 0.05 times the quantity d plus 9. One row down, the “Total Value” column contains one more cell, which contains 2.25. Below the table is the equation 0.10d plus 0.05d plus 0.45 equals 2.25. Below this is 0.15d plus 0.45 equals 2.25. Below this is 0.15d equals 1.80. To the right is d plus 9, which translates to 12 plus 9, or 21 nickels. To the right of this is the checking stage, where we see if 12 dimes and 21 nickels amount to 2 dollars and 25 cents. 12 times 0.10 equals 1.20, and 21 times (0.05) equals 1.05. 1.20 plus 1.05 equals 2.25.

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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