**2.4 Solve Mixture and Uniform Motion Applications**

Topics covered in this section are:

- Solve coin word problems
- Solve ticket and stamp word problems
- Solve mixture word problems
- Solve uniform motion applications

**2.4.1 Solve Coin Word Problems**

Using algebra to find the number of nickels and pennies in a piggy bank may seem silly. You may wonder why we just don’t open the bank and count them. But this type of problem introduces us to some techniques that will be useful as we move forward in our study of mathematics.

If we have a pile of dimes, how would we determine its value? 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 $23$ 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 $23$ dimes, multiply $23$ by $\$0.10$ to get $\$2.30$.

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

$number \cdot value = total \ value$ $23 \cdot \$0.10 = \$2.30$ |

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 $number \cdot value = total \ value$

*number*is the number of coins*value*is the value of each coin*total value*is the total value of all the coins

If we had several types of coins, 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.

**Example 1**

Jesse has $\$3.02$ worth of pennies and nickels in his piggy bank. The number of nickels is three more than eight times the number of pennies. How many nickels and how many pennies does Jesse have?

**Solution**

Step 1. Read the problem.Determine the types of coins involved. Create a table. Write in the value of each type of coin. | pennies and nickels. Pennies are worth $\$0.01$. Nickels are with $\$0.05$. |

Step 2. Identify what we are looking for. | |

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. The number of nickels is three more than eight times the number of pennies. | Let $p=$ number of pennies. $8p+3=$ number of nickels. |

In the chart, multiply the number and the value to get the total value of each type of coin. | |

Step 4. Translate. Write the equation by adding the total value of all the types of coins. | $0.01p+0.05(8p+3)=3.02$ |

Step 5. Solve the equation. | $\begin{align*} 0.01p+0.40p+0.15&=3.02 \\ 0.41p+0.15&=3.02 \\ 0.41p&=2.87 \\ p&=7 \end{align*}$ $7$ pennies |

How many nickels? | $8p+3$ $8(7)+3$ $59$ nickels |

Step 6. Check the answer in the problem and make sure it makes sense.Jesse has $7$ pennies and $59$ nickels. Is the total value $\$3.02$? $\begin{align*} 7(0.01)+59(0.05)&=3.02 \\ 3.02&=3.02 \end{align*}$ |

The steps for solving a coin word problem are summarized below.

**HOW TO: Solve coin word problems.**

**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,” and “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.

**Identify**what you are looking for.**Name**what you 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.

**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.

**Solve**the equation using good algebra techniques.**Check**the answer in the problem and make sure it makes sense.**Answer**the question with a complete sentence.

**2.4.2 Solve Ticket and Stamp Word Problems**

Problems involving tickets or stamps are very much like coin problems. Each type of ticket and stamp has a value, just like each type of coin does. So to solve these problems, we will follow the same steps we used to solve coin problems.

**Example 2**

Danny paid $\$15.75$ for stamps. The number of $49$-cent stamps was five less than three times the number of $35$-cent stamps. How many $49$-cent stamps and how many $35$-cent stamps did Danny buy?

**Solution**

Step 1. Determine the types of stamps involved. | $49$-cent stamps and $35$-cent stamps |

Step 2. Identify we are looking for. | the number of $49$-cent stamps and the number of $35$-cent stamps |

Step 3. Write variable expressions to represent the number of each type of stamp. | Let $x=$ the number of $35$-cent stamps. |

“The number of 49-cent stamps was five less than three times the number of 35-cent stamps.” | $3x-5=$ the number of $49$-cent stamps |

Step 4. Write the equation from the total values. | $0.49(3x-5)+0.35x=15.75$ |

Step 5. Solve the equation. | $\begin{align*} 1.47x-2.45+0.35x&=15.75 \\ 1.82x-2.45&=15.75 \\ 1.82x&=18.2 \\ x&=10 \end{align*}$ $10$ $35$-cent stamps |

How many $49$-cent stamps? | $3x-5$ $3(10)-5$ $25 \ \ 49$-cent stamps |

Step 6. Check.$\begin{align*} 110(0.35)+25(0.49)&=15.75 \\ 3.50+12.25&=15.75 \\ 15.75&=15.75 \end{align*}$ | |

Step 7. Answer the question with a complete sentence. | Danny bought ten $35$-cent stamps and twenty-five $49$-cent stamps. |

In most of our examples so far, we have been told that one quantity is four more than twice the other, or something similar. In our next example, we have to relate the quantities in a different way.

Suppose Aniket sold a total of $100$ tickets. Each ticket was either an adult ticket or a child ticket. If he sold $20$ child tickets, how many adult tickets did he sell?

*Did you say “$80$”? How did you figure that out? Did you subtract $20$ from $100$?*

If he sold $45$ child tickets, how many adult tickets did he sell?

*Did you say “$55$”? How did you find it? By subtracting $45$ from $100$?*

Now, suppose Aniket sold *x* child tickets. Then how many adult tickets did he sell? To find out, we would follow the same logic we used above. In each case, we subtracted the number of child tickets from $100$ to get the number of adult tickets. We now do the same with $x$.

We have summarized this in the table.

We will apply this technique in the next example.

**Example 3**

A whale-watching ship had $40$ paying passengers on board. The total revenue collected from tickets was $\$1,196$. Full-fare passengers paid $\$32$ each and reduced-fare passengers paid $\$26$ each. How many full-fare passengers and how many reduced-fare passengers were on the ship?

**Solution**

Step 1. Determine the types of tickets involved. | full-fare tickets and reduced-fare tickets |

Step 2. Identify what we are looking for. | the number of full-fare tickets and reduced-fare tickets |

Step 3. Name. Represent the number of each type of ticket using variables. | Let $f=$ the number of full-fare tickets. $40-f =$ the number of reduced-fare tickets |

We know the total number of tickets sold was $40$. This means the number of reduced-fare tickets is $40$ less the number of full-fare tickets. Multiply the number times the value to get the total value of each type of ticket. | |

Step 4. Translate. Write the equation by adding the total values of each type of ticket. | $32f+26(40-f)=1,196$ |

Step 5. Solve the equation. | $\begin{align*} 32f+1,040f-26f&=1,196 \\ 6f&=156 \\ f&=26 \end{align*}$ full-fare tickets |

How many reduced-fare? | $40-f$ $40-26$ $14$ reduced-fare tickets |

Step 6. Check the answer.There were $26$ full-fare tickets at $\$32$ each and $14$ reduced-fare tickets at $\$26$ each. Is the total value $\$116$? $\begin{align*} 26 \cdot 32 =832& \\ 14 \cdot 26= 364& \\ \overline{1,196}& \end{align*}$ | |

Answer the question.Step 7. | They sold $26$ full-fare and $14$ reduced-fare tickets. |

**2.4.3 Solve Mixture Word Problems**

Now we’ll solve some more general applications of the mixture model. In mixture problems, we are often mixing two quantities, such as raisins and nuts, to create a mixture, such as trail mix. In our tables we will have a row for each item to be mixed as well as one for the final mixture.

**Example 4**

Henning is mixing raisins and nuts to make $25$ pounds of trail mix. Raisins cost $\$4.50$ a pound and nuts cost $\$8$ a pound. If Henning wants his cost for the trail mix to be $\$6.60$ a pound, how many pounds of raisins and how many pounds of nuts should he use?

**Solution**

Step 1. Determine what is being mixed. | The $25$ pounds of trail mix will come from mixing raisins and nuts. |

Step 2. Identify what we are looking for. | the number of pounds of raisins and nuts |

Step 3. Represent the number of each type of ticket using variables.As before, we fill in a chart to organize our information. We enter the price per pound for each item. We multiply the number times the value to get the total value. | Let $x=$ number of pounds of raisins. $25-x=$ number of pounds of nuts. |

Notice that the last column in the table gives the information for the total amount of the mixture. | |

Step 4. Translate into an equation. | The value of the raisins plus the value of the nuts will be the value of the trail mix. |

Step 5. Solve the equation. | $4.5x+8(25-x)=25(6.60)$ $-3.5x=-35$ $x=10$ pounds of raisins |

Find the number of pounds of nuts. | $25-x$ $25-10$ $15$ pounds of nuts |

Step 6. Check.$\begin{align*} 4.5(10)+8(15)&=25(6.60) \\ 45+120&=165 \\ 165&=165 \end{align*}$ | |

Step 7. Answer the question. | Henning mixed ten pounds of raisins with $15$ pounds of nuts. |

**2.4.4 Solve Uniform Motion Applications**

When you are driving down the interstate using your cruise control, the speed of your car stays the same—it is uniform. We call a problem in which the speed of an object is constant a uniform motion application. We will use the distance, rate, and time formula, $D=rt$, to compare two scenarios, such as two vehicles travelling at different rates or in opposite directions.

Our problem solving strategies will still apply here, but we will add to the first step. The first step will include drawing a diagram that shows what is happening in the example. Drawing the diagram helps us understand what is happening so that we will write an appropriate equation. Then we will make a table to organize the information, like we did for the coin, ticket, and stamp applications.

The steps are listed here for easy reference:

**HOW TO: Solve a uniform motion application.**

**Read**the problem. Make sure all the words and ideas are understood.- Draw a diagram to illustrate what is happening.
- Create a table to organize the information.
- Label the columns rate, time, distance.
- List the two scenarios.
- Write in the information you know.

**Identify**what you are looking for.**Name**what you are looking for. Choose a variable to represent that quantity.- Complete the chart.
- Use variable expressions to represent that quantity in each row.
- Multiply the rate times the time to get the distance.

**Translate**into an equation.- Restate the problem in one sentence with all the important information.
- Then, translate the sentence into an equation.

**Solve**the equation using good algebra techniques.**Check**the answer in the problem and make sure it makes sense.**Answer**the question with a complete sentence.

**Example 5**

Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne two hours to ride to the beach while it takes Dennis $1.5$ hours for the ride. Find the speed of both bikers.

**Solution**

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

- Draw a diagram to illustrate what it happening. Shown below is a sketch of what is happening in the example.
- Create a table to organize the information.
- Label the columns “Rate,” “Time,” and “Distance.”
- List the two scenarios.
- Write in the information you know.

**Step 2. Identify** what you are looking for.

You are asked to find the speed of both bikers. Notice that the distance formula uses the word “rate,” but it is more common to use “speed” when we talk about vehicles in everyday English.

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

- Complete the chart
- Use variable expressions to represent that quantity in each row.

We are looking for the speed of the bikers. Let’s let $r$ represent Wayne’s speed. Since Dennis’ speed is $7$ mph faster, we represent that as $r+7$

$r+7=$Dennis’ speed, $ \ $ $r=$ Wayne’s speed

Fill in the speeds into the chart.

- Multiply the rate times the time to get the distance.

**Step 4. Translate** into an equation.

- Restate the problem in one sentence with all the important information.
- Then, translate the sentence into an equation.

The equation to model this situation will come from the relation between the distances. Look at the diagram we drew above. How is the distance travelled by Dennis related to the distance travelled by Wayne?

Since both bikers leave from Riverside and travel to the beach, they travel the same distance. So we write:

**Step 5. Solve** the equation using algebra techniques.

Now solve this equation | $1.5(r+7)=2r$ $1.5r+10.5=2r$ $10.5=0.5r$ $21=r$ So Wayne’s speed is $21$ mph. |

Find Dennis’ speed | $r+7$ $21+7$ $28$ Dennis’ speed is $28$ mph. |

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

Dennis $28$ mph($1.5$ hours)=$42$ miles✓

Wayne $21$ mph($2$ hours)=$42$ miles✓

**Step 7. Answer** the question with a complete sentence.

Wayne rode at $21$ mph and Dennis rode at $28$ mph.

In Example 5, we had two bikers traveling the same distance. In the next example, two people drive toward each other until they meet.

**Example 6**

Carina is driving from her home in Anaheim to Berkeley on the same day her brother is driving from Berkeley to Anaheim, so they decide to meet for lunch along the way in Buttonwillow. The distance from Anaheim to Berkeley is $395$ miles. It takes Carina three hours to get to Buttonwillow, while her brother drives four hours to get there. Carina’s average speed is $15$ miles per hour faster than her brother’s average speed. Find Carina’s and her brother’s average speeds.

**Solution**

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

- Draw a diagram to illustrate what it happening. Below shows a sketch of what is happening in the example.

- Draw a diagram to illustrate what it happening. Below shows a sketch of what is happening in the example.
- Create a table to organize the information.
- Label the columns rate, time, distance.
- List the two scenarios.
- Write in the information you know.

**Step 2. Identify** what we are looking for.

We are asked to find the average speeds of Carina and her brother.

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

- Complete the chart.
- Use variable expressions to represent that quantity in each row.

We are looking for their average speeds. Let’s let $r$ represent the average speed of Carina’s brother. Since Carina’s speed is $15$ mph faster, we represent that as $r+15$.

Fill in the speeds into the chart. - Multiply the rate times the time to get the distance.

**Step 4. Translate** into an equation.

- Restate the problem in one sentence with all the important information.
- Then, translate the sentence into an equation.

Again, we need to identify a relationship between the distances in order to write an equation. Look at the diagram we created above and notice the relationship between the distance Carina traveled and the distance her brother traveled.

The distance Carina traveled plus the distance her brother travel must add up to $395$ miles. So we write:

**Step 5. Solve** the equation using algebra techniques.

Now solve this equation | $3(r+15)+4r=395$ $3r+45+4r=395$ $7r+45=395$ $7r=350$ $r=50$ So Carina’s brother’s speed was $50$ mph. |

Carina’s speed is $r+15$ | $r+15$ $50+15$ $65$ So Carina’s speed was $65$ mph. |

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

Carina drove | $65$ mph ($3$ hours) $=$ | $195$ miles |

Her brother drove | $50$ mph ($4$ hours) $=$ | $200$ miles |

$\overline{395 \ miles}$ |

**Step 7. Answer** the question with a complete sentence.

Carina drove $65$ mph and her brother $50$ mph.

As you read the next example, think about the relationship of the distances traveled. Which of the previous two examples is more similar to this situation?

**Example 7**

Two truck drivers leave a rest area on the interstate at the same time. One truck travels east and the other one travels west. The truck traveling west travels at $70$ mph and the truck traveling east has an average speed of $60$ mph. How long will they travel before they are $325$ miles apart?

**Solution**

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

- Create a table to organize the information.
- Label the columns rate, time, distance.
- List the two scenarios.
- Write in the information you know.

**Step 2. Identify** what we are looking for.

We are asked to find the amount of time the trucks will travel until they are $325$ miles apart.

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

- Complete the chart.
- Use variable expressions to represent that quantity in each row.

We are looking for the time travelled. Both trucks will travel the same amount of time.

Let’s call the time*t*. Since their speeds are different, they will travel different distances. - Multiply the rate times the time to get the distance.

**Step 4. Translate** into an equation.

- Restate the problem in one sentence with all the important information.
- Then, translate the sentence into an equation.

We need to find a relation between the distances in order to write an equation. Looking at the diagram, what is the relationship between the distances each of the trucks will travel?

The distance travelled by the truck going west plus the distance travelled by the truck going east must add up to $325$ miles. So we write:

**Step 5. Solve** the equation using algebra techniques.

Now solve this equation | $\begin{align*} 70t+60t&=325 \\ 130t&=325\\ t&=2.5 \end{align*}$ |

So it will take the trucks $2.5$ hours to be $325$ miles apart.

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

Truck going West | $70$ mph ($2.5$ hours) $=$ | $175$ miles |

Truck going East | $60$ mph ($2.5$ hours) $=$ | $150$ miles |

$\overline {325 \ miles}$ |

**Step 7. Answer** the question with a complete sentence.

It will take the trucks $2.5$ hours to be $325$ miles apart.

It is important to make sure that the units match when we use the distance rate and time formula. For instance, if the rate is in miles per hour, then the time must be in hours.

**Example 8**

When Naoko walks to school, it takes her $30$ minutes. If she rides her bike, it takes her $15$ minutes. Her speed is three miles per hour faster when she rides her bike than when she walks. What is her speed walking and her speed riding her bike?

**Solution**

First, we draw a diagram that represents the situation to help us see what is happening.

We are asked to find her speed walking and riding her bike. Let’s call her walking speed $r$. Since her biking speed is three miles per hour faster, we will call that speed $r+3$. We write the speeds in the chart.

The speed is in miles per hour, so we need to express the times in hours, too, in order for the units to be the same. Remember, $1$ hour is $60$ minutes. So:

$30$ minutes is $\frac{30}{60}$ or $\frac{1}{2}$ hour

$15$ minutes is $\frac{15}{60}$ or $\frac{1}{4}$ hour

We write the times in the chart.

Next, we multiply rate times time to fill in the distance column.

The equation will come from the fact that the distance from Naoko’s home to her school is the same whether she is walking or riding her bike.

So we say:

Translate to an equation. | distance walked $=$ distance covered by bike $\frac{1}{2}r=\frac{1}{4}(r+3)$ |

Solve this equation. | $\frac{1}{2}r=\frac{1}{4}(r+3)$ |

Clear the fractions by multiplying by the LCD of all the fractions in the equation. | $8\cdot \frac{1}{2}r=8 \cdot \frac{1}{4}(r+3)$ |

Simplify. | $4r=2(r+3)$ $4r=2r+6$ $2r=6$ $r=3$ $3$ mph walking speed $r+3$ biking speed $3+3$ $6$ |

Check. Walk $3$ mph ($0.5$ hour) $= 1.5$ miles Bike $6$ mph ($0.25$ hour) $= 1.5$ miles Yes, either way Naoko travels $1.5$ miles to school. |

Naoko’s walking speed is $3$ mph and her speed riding her bike is $6$ mph.

In the distance, rate and time formula, time represents the actual amount of elapsed time (in hours, minutes, etc.). If a problem gives us starting and ending times as clock times, we must find the elapsed time in order to use the formula.

**Example 9**

Cruz is training to compete in a triathlon. He left his house at 6:00 and ran until 7:30. Then he rode his bike until 9:45. He covered a total distance of $51$ miles. His speed when biking was $1.6$ times his speed when running. Find Cruz’s biking and running speeds.

**Solution**

A diagram will help us model this trip.

Next, we create a table to organize the information. We know the total distance is $51$ miles. We are looking for the rate of speed for each part of the trip. The rate while biking is $1.6$ times the rate of running. If we let $r =$ the rate running, then the rate biking is $1.6r$.

The times here are given as clock times. Cruz started from home at 6:00 a.m. and started biking at 7:30 a.m. So he spent $1.5$ hours running. Then he biked from 7:30 a.m until 9:45 a.m. So he spent $2.25$ hours biking.

Now, we multiply the rates by the times.

By looking at the diagram, we can see that the sum of the distance running and the distance biking is $51$ miles.

Translate into an equation. | distance running $+$ distance biking $= 51$ $1.5r + 2.25(1.6r) = 51$ |

Solve this equation. | $1.5r + 2.25(1.6r) = 51$ $1.5r+3.6r=51$ $5.1r=51$ $r=10$ mph running $1.6r$ biking speed $1.6(10)$ $16$ mph |

Check. Run $10$ mph ($1.5$ hours) $= 15$ mi Bike $16$ mph ($2.25$ hours) $= 36$ mi $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \overline{51 \ miles}$ |

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*Marecek, L., & Mathis, A. H. (2020). Solve Mixture and Uniform Motion Applications. In Intermediate Algebra 2e. OpenStax. https://openstax.org/books/intermediate-algebra-2e/pages/2-4-solve-mixture-and-uniform-motion-applications*.*License: CC BY 4.0. Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/2-introduction*