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Conversion Using Unit Analysis

Solve problems by converting units using unit analysis.

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Solve Problems Involving Rates and Unit Analysis
License: CC BY-NC 3.0

Joanne is training for the City to Surf marathon in Sydney, Australia. She knows she can run a 30 kilometer marathon in about 5 hours. How fast is Joanne going per minute?

In this concept, you will learn to solve problems involving rates and unit analysis.

Unit Rate

A rate often refers to speed or a rate could refer to the amount of money someone makes per hour.

When you talk about a unit rate, you look at comparing a rate to 1. An example of a unit rate would be 65 miles per hour. A key word when working with unit rate is the word “per”.

Let’s look at an example.

Jeff makes $150.00 an hour as a consultant. What is his rate per minute?

First, use units to convert hours to minutes.

\begin{align*}\begin{array}{rcl} \# \ \text{min} &=& 1 \ \text{hr} \times \frac{60 \ \text{min}}{1 \ \text{hr}} \\ \# \ \text{min} &=& 60 \ \text{min} \end{array}\end{align*} 

Next, convert \begin{align*}\frac{\$ 150}{\text{hr}}\end{align*} into \begin{align*}\frac{\$}{\text{min}}\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{\$150.00}{1 \ \text{hr}} &=& \frac{\$ 150.00}{60 \ \text{min}} \\ \frac{\$150.00}{ 60 \ \text{min} }&=& \$ 2.5 \end{array}\end{align*}

The answer is 2.5.

Jeff makes \begin{align*}\frac{\$ 2.50}{\text{min}}\end{align*}.

Unit analysis is when you look at how to measure individual units in different measurement amounts and it is used to convert units of measurement.

Let’s look at an example.

Juanita worked for 18 hours last week. She made $116.00 at the end of the week. Juanita was sure that her manager had made a mistake and that she should have made more money. Juanita makes $9.00 per hour. Did Juanita make the correct amount of money or was there a mistake?

First, find out how much Juanita should have made if she made.

\begin{align*}\begin{array}{rcl} \text{pay} &=& \frac{ \$9.00}{1 \text{ hr}} \times 18 \ \text{hr} \\ \text{pay} &=& \$162 \end{array}\end{align*}

Next, determine if Juanita was not paid correctly.

Juanita was paid $116 but should have been paid $162.

There was a mistake in her pay by $46.


Example 1

Earlier, you were given a problem about Joanne’s marathon training.

Joanne can run 30 km in 5 hours but wants to know her rate per minute.

First, use units to convert hours to minutes.

\begin{align*}\begin{array}{rcl} \# \ \text{min} &=& 5 \text{ hr} \times \frac{60 \ \text{min}}{1 \ \text{hr}}\\ \# \ \text{min} &=& 300\ \text{min} \end{array}\end{align*}

Next, convert 30 \begin{align*}\frac{\text{km}}{\text{hr}}\end{align*} into \begin{align*}\frac{\text{km}}{\text{min}}\end{align*}.

\begin{align*}\begin{array}{rcl} \frac{30 \ \text{km}}{1 \ \text{hr}} &=& \frac{30 \ \text{km}}{300 \ \text{min}} \\ \frac{30 \ \text{km}}{300 \ \text{min}} &=& 0.10 \end{array}\end{align*}

The answer is 0.10.

Joanne runs at a speed of \begin{align*}0.1 \frac{\text{km}}{\text{min}}\end{align*}.

Example 2

Jesse has a car that holds 14 gallons of gasoline. During the first week of the month, gasoline cost $2.75 per gallon. During the second week of the month, gasoline cost $2.50 per gallon. How much was the total cost for the 28 gallons of gasoline?

First, start by writing a variable expression to work on this problem. You know that the number of gallons of gasoline does not change. Therefore, let \begin{align*}x\end{align*} represent the number of gallons of gasoline.

\begin{align*}\text{Total cost} = 2.75x+2.50x\end{align*}

Next, substitute 14 for our variable \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} \text{Total cost} &=& 2.75(14)+2.50(14) \\ \text{Total cost} &=& 38.5+35 \\ \text{Total cost} &=& 73.5 \end{array} \end{align*}

The answer is 73.5.

Therefore, Jess paid $73.50 in gasoline for the two weeks.

Example 3

Ten apples cost $3.99. What is the cost for one apple?

First, set up a proportion.

\begin{align*} \frac{\$3.99}{10 \ \text{apples}} = \frac{x}{1 \ \text{apple}}\end{align*}

Next, cross multiply.

\begin{align*} \begin{array}{rcl} \frac{3.99}{10} &=& \frac{x}{1} \\ 10 x &=& 1 \times 3.99 \\ 10x &=& 3.99 \end{array}\end{align*}

Then, divide both sides by 10 to solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} 10x &=& 3.99 \\ \frac{10x}{10} &=& \frac{3.99}{10} \\ x &=& 0.399 \end{array}\end{align*}

The answer is 0.399.

One apple costs $0.40 or 40¢.

Example 4

Fifteen gallons of gasoline costs $45.00. How much is it per gallon?

First, set up a proportion.

\begin{align*}\frac{\$45.00}{15 \ \text{gallons}} = \frac{x}{1 \ \text{gallon}}\end{align*} 

Next, cross multiply.

\begin{align*}\begin{array}{rcl} \frac{45}{15} &=& \frac{x}{1} \\ 15 x &=& 1 \times 45 \\ 15x &=& 45 \end{array}\end{align*}

Then, divide both sides by 15 to solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} 15x &=& 45 \\ \frac{15x}{15} &=& \frac{45}{15} \\ x &=& 3 \end{array}\end{align*}

The answer is 3.

One gallon of gas costs $3.00.

Example 5

Two tickets to a ballgame costs $111.50. What is the cost for one ticket?

First, set up a proportion.

\begin{align*} \frac{\$111.50}{2 \ \text{tickets}} = \frac{x}{1 \ \text{ticket}}\end{align*}

Next, cross multiply.

\begin{align*}\begin{array}{rcl} \frac{111.50}{2} &=& \frac{x}{1} \\ 2x &=& 1 \times 111.50 \\ 2x &=& 111.50 \\ \end{array}\end{align*}

Then, divide both sides by 2 to solve for \begin{align*}x\end{align*}.

\begin{align*}\begin{array}{rcl} 2x &=& 111.50 \\ \frac{2x}{2} &=& \frac{111.50}{2} \\ x &=& 55.75 \end{array}\end{align*}

The answer is 55.75.

One ticket to the game is $55.75


Use what you have learned to solve each problem.

  1. Peter runs at a rate of 10 kilometers per hour. How many kilometers will he cover in 8 hours?
  2. A cheetah can run at a speed of 60 miles per hour. What is his distance after 6 hours?
  3. What is the distance formula?
  4. If a car travels at a rate of 65 miles per hour for 30 minutes, how far will it travel?
  5. A train travels at a rate of 50 miles per hour. If it needs to travel 320 miles, how many minutes will it take?
  6. A car travels 65 mph for 12 hours. How many miles will it travel?
  7. A bus traveled 300 miles at an average speed of 50 miles per hour. How long did this trip take the bus?
  8. A car traveled at an average speed of 40 miles per hour through a construction zone. If the car traveled 20 miles at this rate, how many hours did it take to travel the 20 miles?
  9. What is velocity?
  10. What is the formula for velocity?
  11. What is the velocity of an object that travels 500 miles in 2.5 hours?
  12. If an object has a velocity of 125 miles per hour, how long will it take to travel 4,375 miles?
  13. If an object has a velocity of 7 kilometers per minute, how far will it travel in 2 hours?
  14. If an object has a velocity of 4 meters per second, how many kilometers will it travel in 2 days?
  15. The formula for density is \begin{align*}D=\frac{m}{v}\end{align*} where \begin{align*}D\end{align*} represents the density of an object, \begin{align*}m\end{align*} represents the mass of the object, and \begin{align*}v\end{align*} represents the volume of the object. What is the density of a brick that weighs 9 pounds and has a volume of \begin{align*}36 \ cu.\ in.\end{align*}?

Review (Answers)

To see the Review answers, open this PDF file and look for section 4.17. 


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Rate A rate is a special kind of ratio that compares two quantities.
Unit Analysis Unit analysis is a method of converting units of measurement by using ratios and proportions.
Unit Rate A unit rate is a ratio that compares a quantity to one. The word “per” is a key word with unit rates.

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