Applications of Linear Equations: Distance and Interest
Learning Objectives
Solve problems containing rates
Apply the steps for solving word problems to distance, rate, and time problems
Apply the steps for solving word problems to interest rate problems
Evaluate a formula using substitution
Rearrange formulas to isolate specific variables
Identify an unknown given a formula
Introduction
There is often a well-known formula or relationship that applies to a word problem. For example, if you were to plan a road trip, you would want to know how long it would take you to reach your destination. [latex]d=rt[/latex] is a well-known relationship that associates distance traveled, the rate at which you travel, and how long the travel takes.
Distance, Rate, and Time
If you know two of the quantities in the relationship [latex]d=rt[/latex], you can easily find the third using methods for solving linear equations. For example, if you know that you will be traveling on a road with a speed limit of [latex]30\frac{\text{ miles }}{\text{ hour }}[/latex] for 2 hours, you can find the distance you would travel by multiplying rate times time or [latex]\left(30\frac{\text{ miles }}{\text{ hour }}\right)\left(2\text{ hours }\right)=60\text{ miles }[/latex].
We can generalize this idea depending on what information we are given and what we are looking for. For example, if we need to find time, we could solve the [latex]d=rt[/latex] equation for t using division:
[latex]d=rt\\\frac{d}{r}=t[/latex]
Likewise, if we want to find rate, we can isolate r using division:
[latex]d=rt\\\frac{d}{t}=r[/latex]
In the following examples you will see how this formula is applied to answer questions about ultra marathon running.
Ann Trason
Ultra marathon running (defined as anything longer than 26.2 miles) is becoming very popular among women even though it remains a male-dominated niche sport. Ann Trason has broken twenty world records in her career. One such record was the American River 50-mile Endurance Run which begins in Sacramento, California, and ends in Auburn, California.[1] In 1993 Trason finished the run with a time of 6:09:08. The men’s record for the same course was set in 1994 by Tom Johnson who finished the course with a time of 5:33:21.[2]
In the next examples we will use the [latex]d=rt[/latex] formula to answer the following questions about the two runners.
What was each runner’s rate for their record-setting runs?
By the time Johnson had finished, how many more miles did Trason have to run?
How much further could Johnson have run if he had run as long as Trason?
What was each runner’s time for running one mile?
To make answering the questions easier, we will round the two runners’ times to 6 hours and 5.5 hours.
Example
What was each runner’s rate for their record-setting runs?
Show Solution
Read and Understand: We are looking for rate and we know distance and time, so we can use the idea: [latex]d=rt\\\frac{d}{t}=r[/latex]
Define and Translate: Because there are two runners, making a table to organize this information helps. Note how we keep units to help us keep track of what how all the terms are related to each other.
Runner
Distance =
(Rate )
(Time)
Trason
50 miles
r
6 hours
Johnson
50 miles
r
5.5 hours
Write and Solve:
Trason’s rate:
[latex]d=rt\\\frac{d}{t}=r[/latex]
[latex]\begin{array}{c}d=rt\\\\50\text{ miles }=\text{r}\left(6\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{8.33\text{ miles }}{\text{ hour }}\end{array}[/latex].
(rounded to two decimal places)
Johnson’s rate:
[latex]d=rt\\\frac{d}{t}=r[/latex]
[latex]\begin{array}{c}d=rt\\\\,\,\,\,\,\,\,50\text{ miles }=\text{r}\left(5.5\text{ hours }\right)\\\frac{50\text{ miles }}{6\text{ hours }}=\frac{9.1\text{ miles }}{\text{ hour }}\end{array}[/latex]
(rounded to two decimal places)
Check and Interpret:
We can fill in our table with this information.
Runner
Distance =
(Rate )
(Time)
Trason
50 miles
8.33 [latex]\frac{\text{ miles }}{\text{ hour }}[/latex]
6 hours
Johnson
50 miles
9.1 [latex]\frac{\text{ miles }}{\text{ hour }}[/latex]
5.5 hours
Now that we know each runner’s rate we can answer the second question.
Example
By the time Johnson had finished, how many more miles did Trason have to run?
Show Solution
Here is the table we created for reference:
Runner
Distance =
(Rate )
(Time)
Trason
50 miles
8.33 [latex]\frac{\text{ miles }}{\text{ hour }}[/latex]
6 hours
Johnson
50 miles
9.1 [latex]\frac{\text{ miles }}{\text{ hour }}[/latex]
5.5 hours
Read and Understand: We are looking for how many miles Trason still had on the trail when Johnson had finished after 5.5 hours. This is a distance, and we know rate and time.
Define and Translate: We can use the formula [latex]d=rt[/latex] again. This time the unknown is d, and the time Trason had run is 5.5 hours.
Write and Solve:
[latex]\begin{array}{l}d=rt\\\\d=8.33\frac{\text{ miles }}{\text{ hour }}\left(5.5\text{ hours }\right)\\\\d=45.82\text{ miles }\end{array}[/latex].
Check and Interpret:
Have we answered the question? We were asked to find how many more miles she had to run after 5.5 hours. What we have found is how long she had run after 5.5 hours. We need to subtract [latex]d=45.82\text{ miles }[/latex] from the total distance of the course.
[latex]50\text{ miles }-45.82\text{ miles }=1.48\text{ miles }[/latex]
The third question is similar to the second. Now that we know each runner’s rate, we can answer questions about individual distances or times.
Examples
How much further could Johnson have run if he had run as long as Trason?
Show Solution
Here is the table we created for reference:
Runner
Distance =
(Rate )
(Time)
Trason
50 miles
8.33 [latex]\frac{\text{ miles }}{\text{ hour }}[/latex]
6 hours
Johnson
50 miles
9.1 [latex]\frac{\text{ miles }}{\text{ hour }}[/latex]
5.5 hours
Read and Understand: The word further implies we are looking for a distance.
Define and Translate: We can use the formula [latex]d=rt[/latex] again. This time the unknown is d, the time is 6 hours, and Johnson’s rate is [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex]
Write and Solve:
[latex]\begin{array}{l}d=rt\\\\d=9.1\frac{\text{ miles }}{\text{ hour }}\left(6\text{ hours }\right)\\\\d=54.6\text{ miles }\end{array}[/latex].
Check and Interpret:
Have we answered the question? We were asked to find how many more miles Johnson would have run if he had run at his rate of [latex]9.1\frac{\text{ miles }}{\text{ hour }}[/latex] for 6 hours.
Johnson would have run 54.6 miles, so that’s 4.6 more miles than than he ran for the race.
Now we will tackle the last question where we are asked to find a time for each runner.
Example
What was each runner’s time for running one mile?
Show Solution
Here is the table we created for reference:
Runner
Distance =
(Rate )
(Time)
Trason
50 miles
8.33 [latex]\frac{\text{ miles }}{\text{ hour }}[/latex]
6 hours
Johnson
50 miles
9.1 [latex]\frac{\text{ miles }}{\text{ hour }}[/latex]
5.5 hours
Read and Understand: we are looking for time, and this time our distance has changed from 50 miles to 1 mile, so we can use
[latex]d=rt\\\frac{d}{r}=t[/latex]
Define and Translate: we can use the formula [latex]d=rt[/latex] again. This time the unknown is t, the distance is 1 mile, and we know each runner’s rate. It may help to create a new table:
Runner
Distance =
(Rate )
(Time)
Trason
1 mile
8.33 [latex]\frac{\text{ miles }}{\text{ hour }}[/latex]
t hours
Johnson
1 mile
9.1 [latex]\frac{\text{ miles }}{\text{ hour }}[/latex]
t hours
Write and Solve:
Trason:
We will need to divide to isolate time.
[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=8.33\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{8.33\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.12\text{ hours }=t\end{array}[/latex].
0.12 hours is about 7.2 minutes, so Trason’s time for running one mile was about 7.2 minutes. WOW! She did that for 6 hours!
Johnson:
We will need to divide to isolate time.
[latex]\begin{array}{c}d=rt\\\\1\text{ mile }=9.1\frac{\text{ miles }}{\text{ hour }}\left(t\text{ hours }\right)\\\\\frac{1\text{ mile }}{\frac{9.1\text{ miles }}{\text{ hour }}}=t\text{ hours }\\\\0.11\text{ hours }=t\end{array}[/latex].
0.11 hours is about 6.6 minutes, so Johnson’s time for running one mile was about 6.6 minutes. WOW! He did that for 5.5 hours!
Check and Interpret:
Have we answered the question? We were asked to find how long it took each runner to run one mile given the rate at which they ran the whole 50-mile course. Yes, we answered our question.
Trason’s mile time was [latex]7.2\frac{\text{minutes}}{\text{mile}}[/latex] and Johnsons’ mile time was [latex]6.6\frac{\text{minutes}}{\text{mile}}[/latex]
In the following video, we show another example of answering many rate questions given distance and time.
Simple Interest
In order to entice customers to invest their money, many banks will offer interest-bearing accounts. The accounts work like this: a customer deposits a certain amount of money (called the Principal, or P), which then grows slowly according to the interest rate (R, measured in percent) and the length of time (T, usually measured in months) that the money stays in the account. The amount earned over time is called the interest (I), which is then given to the customer.
Beware! Interest rates are commonly given as yearly rates, but can also be monthly, quarterly, bimonthly, or even some custom amount of time. It is important that the units of time and the units of the interest rate match. You will see why this matters in a later example.
The simplest way to calculate interest earned on an account is through the formula [latex]\displaystyle I=P\,\cdot \,R\,\cdot \,T[/latex].
If we know any of the three amounts related to this equation, we can find the fourth. For example, if we want to find the time it will take to accrue a specific amount of interest, we can solve for T using division:
Add the interest and the original principal amount to get the total amount in her account.
[latex] \displaystyle 2000+336=2336[/latex]
She has $2336 after 24 months.
The following video shows another example of finding an account balance after a given amount of time, principal invested, and a rate.
In the following example you will see why it is important to make sure the units of the interest rate match the units of time when using the simple interest formula.
Example
Alex invests $600 at 3.25% monthly interest for 3 years. What amount of interest has Alex earned?
Show Solution
Read and Understand: The question asks for an amount, so we can substitute what we are given into the simple interest formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex]
Define and Translate: we know P, R, and T so we can use substitution. R = 0.0325, P = $600, and T = 3 years. We have to be careful! R is in months, and T is in years. We need to change T into months because we can’t change the rate—it is set by the bank.
[latex]{T}=3\text{ years }\cdot12\frac{\text{ months }}{ year }=36\text{ months }[/latex]
We were asked what amount Alex earned, which is the amount provided by the formula. In the previous example we were asked the total amount in the account, which included the principal and interest earned.
Alex has earned $756.
In the following video we show another example of how to find the amount of interest earned after an investment has been sitting for a given monthly interest.
Example
After 10 years, Jodi’s account balance has earned $1080 in interest. The rate on the account is 0.09% monthly. What was the original amount she invested in the account?
Show Solution
Read and Understand: The question asks for the original amount invested, the principal. We are given a length of time in years, and an interest rate in months, and the amount of interest earned.
Define and Translate: we know I = $1080, R = 0.009, and T = 10 years so we can use [latex]{P}=\frac{I}{{R}\,\cdot \,T}[/latex]
We also need to make sure the units on the interest rate and the length of time match, and they do not. We need to change time into months again.
[latex]{T}=10\text{ years }\cdot12\frac{\text{ months }}{ year }=120\text{ months }[/latex]
We were asked to find the principal given the amount of interest earned on an account. If we substitute P = $1000 into the formula [latex]I=P\,\cdot \,R\,\cdot \,T[/latex] we get
Problem Solving Using Distance, Rate, Time (Running). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: https://youtu.be/3WLp5mY1FhU. License: CC BY: Attribution
Simple Interest - Determine Account Balance (Monthly Interest). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: https://youtu.be/XkGgEEMR_00. License: CC BY: Attribution
Simple Interest - Determine Interest Balance (Monthly Interest). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: https://youtu.be/mRV5ljj32Rg. License: CC BY: Attribution
Simple Interest - Determine Principal Balance (Monthly Interest). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. Located at: https://youtu.be/vbMqN6lVoOM. License: CC BY: Attribution
Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. Located at: http://nrocnetwork.org/dm-opentext. License: CC BY: Attribution
Beware! Interest rates are commonly given as yearly rates, but can also be monthly, quarterly, bimonthly, or even some custom amount of time. It is important that the units of time and the units of the interest rate match. You will see why this matters in a later example.