Practice Solving Mixture Word Problems

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.

Manipulative Mathematics

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

\(\text{number ⋅ value = total value}\)

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.

\(\text{number ⋅ value = total value}\)

\(17 \cdot \ $ 0.10 = \ $ 1.70\)

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.

Type
Number · Value ($) = Total Value ($)


Quarters
14
0.25
3.50
Dimes
17
0.10
1.70
Nickels
21
0.05
1.05
Pennies
39
0.01
0.39



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.

Example 3.26

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.

    Type

    Number • Value($) = Total Value(S)

    Dimes

     

    \(0.10\)

     

    Nickels

     

    \(0.05\)

     

     

    \(2.25\)

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.

\(\begin{aligned} \text { Let } d &=\text { number of dimes. } \\ d+9 &=\text { number of nickels } \end{aligned}\)

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

Type

Number • Value($) = Total Value($)

Dimes

\(d\)

\(0.10\)

 

Nickels

\(d+ 9\)

\(0.05\)

 

 

\(2.25\)



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 number⋅value=totalvalue. See how this is done in the table below.

Type

Number • Value(S) = Total Value(S)

Dimes

\(d\)

\(0.10\)

\(0.10d\)

Nickels

\(d + 9\)

\(0.05\)

\(0.05(d + 9) \)

 

\(2.25\)


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.


Translate to an equation.
\( \underbrace{\text{Value of dimes }}_{0.10d} \underbrace{+}_{+} \underbrace{\text{value of nickels }}_{0.05(d+9)} \underbrace{=}_{=} \underbrace{\text{total value of coins }}_{2.25} \)


Step 5. Solve the equation using good algebra techniques.

Now solve this equation. \(0.10 d+0.05(d+9)=2.25\)
Distribute. \(0.10 d+0.05 d+0.45=2.25\)
Combine like terms. \(0.15 d+0.45=2.25\)
Subtract 0.45 from each side. \(0.15 d=1.80\)
Divide. \(d=12\)
So there are 12 dimes.  
The number of nickels is d+9. \(d+9\)
  \(12+9\)
  \(21\)


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 \(21(0.05)=\frac{1.05}{\$ 2.25}\text{✓}\)

Table 3.7

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.

adalberto has $2.25 in dimes and nickles in his pocket, he has nine more nickles than dimes. how many of each type does he ha

Try It 3.51

Michaela has $2.05 in dimes and nickels in her change purse. She has seven more dimes than nickels. How many coins of each type does she have?

Try It 3.52

Liliana has $2.10 in nickels and quarters in her backpack. She has 12 more nickels than quarters. How many coins of each type does she have?

How To

Solve Coin Word Problems.

Step 1. 

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

  • Determine the types of coins involved.
  • Create a table to organize the information.
  • 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.

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

Step 2. Identify what we are looking for.

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.

Step 4. Translate into an equation.

It may be helpful to restate the problem in one sentence with all the important information. Then, translate the sentence into an equation.

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

Step 5. Solve the equation using good algebra techniques.

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

Step 7. Answer the question with a complete sentence.

Example 3.27

Maria has $2.43 in quarters and pennies in her wallet. She has twice as many pennies as quarters. How many coins of each type does she have?

Solution

Step 1. Read the problem.

Determine the types of coins involved.

We know that Maria has quarters and pennies.

Create a table to organize the information.

  • 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.

Type

Number • Value($) = Total Value(S)

Quarters

 

\(0.25\)

 

Pennies

 

\(0.01\)

 

 

\(2.43\)

 

Step 2. Identify what you are looking for.

  • We are looking for the number of quarters and pennies.

Step 3. Name. Represent the number of quarters and pennies using variables.

  • We know Maria has twice as many pennies as quarters. The number of pennies is defined in terms of quarters.
  • Let q represent the number of quarters.
  • Then the number of pennies is \(2q\).

Type

Number • Value(S) = Total Value(S)

Quarters

\(q\)

\(0.25\)

 

Pennies

\(2q\)

\(0.01\)

 

 

\(2.43\)


Multiply the 'number' and the 'value' to get the 'total value' of each type of coin.

Type

Number  •  Value(S) = Total Value(S)

Quarters

\(q\)

\(0.25\)

\(0.25q\)

Pennies

\(2q\)

\(0.01\)

\(0.01(2 q) \)

 

\(2.43\)


Step 4. Translate. Write the equation by adding the 'total value' of all the types of coins.

Step 5. Solve the equation. \(0.25 q+0.01(2 q)=2.43\)
Multiply. \(0.25 q+0.02 q=2.43\)
Combine like terms. \(0.27 q=2.43\)
Divide by 0.27. \(q=9 \, \text{quarters}\) 
The number of pennies is 2q. \(2 q\)
\(2 \cdot 9\)
\(18 \, \text{pennies} \)
Step 6. Check the answer in the problem.  
Maria has 9 quarters and 18 pennies. Does this make $2.43?  
\begin{array}{lrl}
\text { 9 quarters } & 9(0.25) & =2.25 \\
\text { 18 pennies } & 18(0.01) & =\frac{0.18}{\$ 2.43 \text{✓}} \\
\text { Total } & &
\end{array}
 
Step 7. Answer the question. Maria has nine quarters and eighteen pennies.

Table 3.8

Try It 3.53

Sumanta has $4.20 in nickels and dimes in her piggy bank. She has twice as many nickels as dimes. How many coins of each type does she have?

Try It 3.54

Alison has three times as many dimes as quarters in her purse. She has $9.35 altogether. How many coins of each type does she have?

In the next example, we'll show only the completed table - remember the steps we take to fill in the table.

Example 3.28

Danny has $2.14 worth of pennies and nickels in his piggy bank. The number of nickels is two more than ten times the number of pennies. How many nickels and how many pennies does Danny have?

Solution
Step 1. Read the problem.  
Determine the types of coins involved. pennies and nickels
Create a table.  
Write in the value of each type of coin. Pennies are worth $0.01.
Nickels are worth $0.05.
Step 2. Identify what we are looking for. the number of pennies and nickels
Step 3. Name. Represent the number of each type of coin using variables.  
The number of nickels is defined in terms of the number of pennies, so start with pennies. Let \(p\)= number of pennies.
The number of nickels is two more than ten times the number of pennies. \(10p+2\)= number of nickels.
Multiply the number and the value to get the total value of each type of coin.  

Type

Number • Value ($) = Total Value ($)

pennies

\(p\)

\(0.01\)

\(0.01p\)

nickels

\(10p + 2\)

\(0.05\)

\(0.05(10p + 2) \)

 

\($2.14\)

Step 4. Translate. Write the equation by adding the total value of all the types of coins. \(0.01 p+0.05(10 p+2)=2.14\)
Step 5. Solve the equation. \(\begin{aligned} 0.01 p+0.50 p+0.10 &=2.14 \\ 0.51 p+0.10 &=2.14 \\ 0.51 p &=2.04 \\ p &=4 \text { pennies } \end{aligned}\)
How many nickels? \begin{align}
\begin{array}{r}
10 p+2 \\
10(4)+2 \\
42 \text { nickels }
\end{array}
\end{align}
Step 6. Check the answer in the problem and make sure it makes sense
Danny has four pennies and 42 nickels.
Is the total value $2.14?
\begin{align}
\begin{aligned}
4(0.01)+42(0.05) & \stackrel{0}{=} 2.14 \\
2.14 &=2.14 \sqrt{ }
\end{aligned}
\end{align}
 
Step 7. Answer the question. Danny has four pennies and 42 nickels.
Try It 3.55

Jesse has $6.55 worth of quarters and nickels in his pocket. The number of nickels is five more than two times the number of quarters. How many nickels and how many quarters does Jesse have?

Try It 3.56

Elane has $7.00 total in dimes and nickels in her coin jar. The number of dimes that Elane has is seven less than three times the number of nickels. How many of each coin does Elane have?