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4.8: Formulas and Unit Analysis

Created by: CK-12

Introduction

The Cost of a Climb

“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 ratios and proportions to solve this. By thinking of one trip as a unit, we can look at the proportion and solve for the correct amount of money. Use this lesson to learn even more about how to work with proportions and then solve this problem at the end of the lesson.

What You Will Learn

In this lesson, you will learn how to complete the following tasks using different skills.

  • Solve various rate problems using given formulas and conversions of units of measure.
  • Use unit analysis to check results.
  • Write and evaluate variable expressions involving change in measurable units over time, and confirm by unit analysis.
  • Solve real-world problems involving formulas for rates and other derived measurements suitable for unit analysis.

Teaching Time

I. Solve Various Rate Problems using Given Formulas and Conversions of Units of Measure

Think back a few lessons ago when we talked about rates and unit rates. When we work with a rate we examine 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. 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. Let’s look at an example.

Example

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. 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 a proportion.

\frac{\$ 150}{1 \ hour} = \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.

Example

If William drives at a rate of 55 miles per hour (mph), how far will he travel in 3.5 hours?

This problem requires a formula. If you think back, you will remember that when we work with distance, speed and time, we can use the following formula.

d = rt

To solve this problem, we substitute the known values for the variables in the formula. The rate is 55 miles per hour, and the time is 3.5 hours. Be sure to include the units when substituting.

d &= rt\\d &= \left( \frac{55 \ miles}{1 \ hour}\right)(3.5 \ hours)

The unit of hours cancels out.

d = \left( \frac{55 \ miles}{1 \ hour}\right)(3.5 \ hours) = 192.5 \ miles

William will travel 192.5 miles in 3.5 hours.

Here is a new formula that you will use as you work on higher levels of math and science. It is the formula for velocity. Let’s look at an example that uses this formula.

Example

The formula for the velocity of an object is v=\frac{d}{t}, where v represents velocity, d represents distance, and t represents time. If an object travels 1000 meters in 2 seconds, what is its velocity in kilometers per second?

First, substitute the known values for the variables in the formula. The distance is 1000 meters and the time is 2 seconds.

v = \frac{d}{t} = \frac{1000 \ meters}{2 \ seconds} = 500 \ meters \ per \ second

So the velocity of the object is 500 meters per second. Now use unit analysis to find the velocity in kilometers per second. Notice that we use the measurement that you learned in the last lesson that 1000 meters is equal to 1 kilometer.

\frac{500 \ meters}{1 \ second} \times \frac{1 \ kilometer}{1000 \ meters}

The meters in the first ratio cancels out the meters in the second ratio and you are left with kilometers per second.

\frac{500 \ kilometers}{1000 \ seconds} = \frac{1}{2} \ kilometer \ per \ second

The object has a velocity of 0.5 kilometer per second.

II. Use Unit Analysis to Check Results

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.

Let’s look at an example.

Example

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}

If Juanita makes $9.00 per hour, we can multiply 18 \times 9. We have an answer of $162.00. Juanita was only paid $116.00, so there definitely was an error in her payment.

Example

The Travis family went to Florida on their vacation. They decided to drive verses fly so that they could save some money. The family figured that they traveled 300 miles in the first part of the day. They drove for 5 hours. If they were traveling 60 mph, would their distance and time be accurate?

To solve this problem, we can check our work using the formula d = rt. Once we have the formula, we can substitute the given values into the problem. If the quantity on the left side of the equation is equal to the quantity on the right side of the equation, then the values are equal and our work is accurate.

Let’s sort this out.

d &= rt\\300 \ miles &= (60)(5)\\300 &= 300

The calculations that the Travis family made were correct.

III. Write and Evaluate Variable Expressions Involving Change in Measurable Units over Time, and Confirm by Unit Analysis

Sometimes, we will have a value that changes over time. We can use variable expressions and unit analysis to work with values that change. An example of this would be gasoline prices. During a month, the price of one gallon of gasoline can vary greatly. When this happens, the price to fill up a tank of gas can change week to week. The number of gallons in the tank does not change, but the cost of the fill-up does change.

Let’s look at an example.

Example

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?

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.

IV. Solve Real – World Problems Involving Formulas for Rates and

Other Derived Measurements Suitable for Unit Analysis

Money exchange is a way that we use unit analysis in everyday life. If you have ever traveled to a foreign country where you needed to exchange your money for money from another country, then you have used unit analysis. Let’s look at an example.

Example

Tanya is traveling to Mexico. She wants to exchange $500 for Mexican pesos. By looking at the current exchange rate, she sees that 12.5 pesos is equal to 1 dollar. How many pesos will Tanya receive when she exchanges her money?

First, think about the units. Our units are in pesos. We know that we want to exchange $500.00 for pesos. Let’s start with the unit rate.

\frac{12.5}{1 \ dollar}

We are converting $500.00 so we can multiply that times our unit exchange rate.

\$ 500 \left( \frac{12.5}{1} \right)

Tanya will receive 6250 pesos for her $500.00.

Real-Life Example Completed

The Cost of a Climb

Here is the original problem once again. Reread it and then write a proportion to solve the problem. There are two parts to your answer.

“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.

Remember, write a proportion and solve for the amount of money that it would have cost Apa Sherpa to climb Mount Everest 19 times.

Solution to Real – Life Example

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.

Vocabulary

Here are the vocabulary words that are found in this lesson.

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.

Time to Practice

Directions: Use what you have learned to solve each problem.

  1. Peter runs at a rate of 10 kilometers per hour. What is his rate in centimeters per second, to the nearest centimeter?
  2. A cheetah can run at a speed of 60 miles per hour. What is this rate in feet per second?
  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 minutes 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.?
  16. If the mass of an object is 9 kilograms and the volume is 40 cu. cm, what is the density of the object in grams per cubic centimeter?

Directions: Write six different problems and solve them. Use money exchange, time exchange, the velocity formula and the distance formula in your problems.

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Jan 14, 2013

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Mar 17, 2014
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CK.MAT.ENG.SE.1.Middle-School-Math-Grade-8.4.8

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