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4.17: Solve Problems Involving Rates and Unit Analysis

Difficulty Level: At Grade Created by: CK-12
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Practice Conversion Using Unit Analysis
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“I would LOVE to climb Mount Everest!” Josh exclaimed at breakfast one morning.

“Really?” his Dad said smiling. “Well son, you had better start saving now.”

Josh looked up from his oatmeal with a puzzled look on his face.

“What makes you say that?” Josh asked.

“What makes me say that is that the going rate for one climb on Everest is about $60,000. That’s what makes me say that,” his Dad explained taking a sip of his coffee. “Really? Wow! I had no idea,” Josh said. “Well, I guess I’ll just have to make a lot of money!” Josh said leaving the table. He kept thinking about what his Dad had said all the way to school. Sixty thousand dollars was a lot of money to climb a mountain, but what really amazed Josh was thinking about the numbers of people who had climbed the mountain more than once. When he got to class, he looked up in his book that Apa Sherpa a man from Nepal had successfully climbed Everest 19 times. Now he was often a guide who was paid, but still, Josh couldn’t help thinking about how much money Apa Sherpa would have spent if he had paid to climb Everest 19 times at the rate his father spoke about. How much would it have cost? We can use units, ratios and proportions to solve this problem. By thinking of one trip as a unit, we can look at the proportion and solve for the correct amount of money. Guidance A rate refers to speed or a rate can refer to the amount of money someone makes per hour. When we talk about a unit rate, we look at comparing a rate to 1, or how much it would take for 1 of something. It could be one apple, one mile, one gallon. We are comparing a quantity to one. A key word when working with unit rate is the word “per”. We can use ratios and proportions to solve problems involving rates and unit rates. Jeff makes$150.00 an hour as a consultant. What is his rate per minute?

To figure this out, we have to think about the unit rate that Jeff is paid as a consultant. You will see that we have “an hour” written into the problem. This is the unit rate. Also notice that the would "per" is used in the problem.

Let’s write the unit rate as a ratio compared to 1.

$\frac{\ 150.00}{1}$

Next, we need to think about what the problem is asking for. It is asking for his rate per minute. The given information is in hours, so we need to write a ratio that compares hours to minutes.

$\frac{1 \ hour}{60 \ minutes}$

Now we can write an expression combining the two ratios.

$\frac{\ 150}{1 \ hour} \cdot \frac{1 \ hour}{60 \ minutes}$

That’s a great question. We don’t compare hours to hours because we aren’t comparing hours. We are comparing money to hours and we need to figure out the rate of money per minute. You always have to think about what is being compared when working with proportions.

Next, we can solve. Notice that because 1 hour is diagonal from 1 hour, we can cross cancel the hours. That leaves us with a ratio that compares money to minutes.

$\frac{\ 150}{60 \ minutes}$

This also helped us to convert hours to minutes making it easier to figure out the answer to the problem. Now we can divide to figure out our answer.

Jeff makes $2.50 per minute. What is unit analysis? Unit analysis is when we look at how to measure individual units in different measurement amounts and it is used to convert units of measurement. When we use unit analysis, we convert different measurement units by comparing the units using ratios and proportions. Unit analysis is very helpful when checking results. Take a look at this dilemma. Juanita worked for 18 hours. She made$116.00 at the end of her shift. 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? To work on this problem, we can use unit analysis. Let’s start by writing a ratio to compare how much Juanita made for the hours worked. $\frac{18 \ hours}{\ 116.00}$ Next, we can use her hourly rate to work with. She makes$9.00 per hour.

$\frac{\ 9}{1 \ hour}$

Solution: .39

Example B

Fifteen gallons of gasoline costs $45.00. How much is it per gallon? Solution:$3.00

Example C

Two tickets to a ballgame costs $111.50. What is the cost for one ticket? Solution:$55.75

Now let's go back to the dilemma from the beginning of the Concept.

Now let’s look at solving this problem.

We know that it costs $60,000 for 1 trip up Mount Everest. $\frac{\ 60,000}{1} &= \frac{x}{19}\\x &= \ 1,140,000$ We can also use unit analysis to solve this problem.$60,000 dollars $\left( \frac{19}{x \ dollars}\right)$

$60,000 \times 19 = \ 1,140,000$ is the cost of the nineteen trips.

This is our solution.

Vocabulary

Rate
a unit that is in relationship with another unit. It could be a pay rate or a rate of speed. It is a unit that is measured.
Unit Rate
a rate compared to 1. A pay rate would be an amount of money per hour. A gasoline unit rate would be the amount of money for one gallon of gasoline.
Unit Analysis
is a method of converting different units of measurement by using ratios and proportions to compare and convert the units.

Guided Practice

Here is one for you to try on your own.

Solve and then check using unit analysis.

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?

Solution

Let’s start by writing a variable expression to work on this problem. We know that the number of gallons of gasoline does not change. That can be our variable.

$x =$ number of gallons of gasoline

The other parts of the expression include the different prices for the gasoline.

$2.75x+2.50x$

This expression will help us to determine how much money Jesse spent on 28 gallons of gasoline. Each full tank is 14 gallons. We can substitute 14 for our variable $x$ .

$2.75(14)&+ 2.50(14)\\\ 38.50 &+ \ 35.00$

The total amount of money spent was $73.50. We can check our work by using unit analysis. $\frac{2.75}{1 \ gallon} &= \frac{x}{14 \ gallons} = \ 38.50\\\frac{2.50}{1 \ gallon} & = \frac{x}{14 \ gallons} = \ 35.00$ The sum of the money spent was$73.50.

Practice

Directions: 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 $D = \frac{m}{v}$ where $D$ represents the density of an object, $m$ represents the mass of the object, and $v$ represents the volume of the object. What is the density of a brick that weighs 9 pounds and has a volume of 36 cu. in.?

Vocabulary Language: English

Rate

Rate

A rate is a special kind of ratio that compares two quantities.
Unit Analysis

Unit Analysis

Unit analysis is a method of converting units of measurement by using ratios and proportions.
Unit Rate

Unit Rate

A unit rate is a ratio that compares a quantity to one. The word “per” is a key word with unit rates.

Jan 23, 2013

Apr 17, 2015