Problems Involving Formulas I
Learning Outcomes
- Solve a formula for a specific variable
- Use the distance, rate, and time formula
- Apply for the steps for solving word problems to interest rate problems
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 [latex]2[/latex] 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].Try It
[ohm_question]184997[/ohm_question][latex]d=rt[/latex]
[latex]\frac{d}{r}=t[/latex]
Likewise, if we want to find rate, we can isolate [latex]r[/latex] using division:[latex]d=rt[/latex]
[latex]\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- 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?
Example
What was each runner's rate for their record-setting runs? Round to two decimal places.Answer: Read and Understand: We are looking for rate and we know distance and time, so we can use the idea: [latex]d=rt[/latex], [latex]\frac{d}{t}=r[/latex]. Let's solve one runner's rate using the original formula, and one using the rearranged formula. 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 how all the terms are related to each other.
| Runner | Distance = | (Rate ) | (Time) |
|---|---|---|---|
| Trason | [latex]50[/latex] miles | [latex]r[/latex] | [latex]6[/latex] hours |
| Johnson | [latex]50[/latex] miles | [latex]r[/latex] | [latex]5.5[/latex] hours |
[latex]d=rt[/latex]
[latex]50\text{ miles }=r\left(6\text{ hours}\right)[/latex]
[latex]50=6r[/latex]
[latex]r\approx 8.33[/latex]
Johnson's rate:[latex]r=\frac{d}{t}[/latex]
[latex]r=\frac{50\text{ miles}}{5.5\text{ hours}}[/latex]
[latex]r=\frac{50}{5.5}[/latex]
[latex]r\approx 9.10[/latex]
Check and Interpret: We can fill in our table with this information.| Runner | Distance = | (Rate ) | (Time) |
|---|---|---|---|
| Trason | [latex]50[/latex] miles | [latex]8.33[/latex] [latex]\frac{\text{ miles }}{\text{ hour }}[/latex] | [latex]6[/latex] hours |
| Johnson | [latex]50[/latex] miles | [latex]9.1[/latex] [latex]\frac{\text{ miles }}{\text{ hour }}[/latex] | [latex]5.5[/latex] hours |
Example
By the time Johnson had finished, how many more miles did Trason have to run?Answer: Here is the table we created for reference:
| Runner | Distance = | (Rate ) | (Time) |
|---|---|---|---|
| Trason | [latex]50[/latex] miles | [latex]8.33[/latex] [latex]\frac{\text{ miles }}{\text{ hour }}[/latex] | [latex]6[/latex] hours |
| Johnson | [latex]50[/latex] miles | [latex]9.1[/latex] [latex]\frac{\text{ miles }}{\text{ hour }}[/latex] | [latex]5.5[/latex] hours |
[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 [latex]5.5[/latex] hours. What we have found is how long she had run after [latex]5.5[/latex] 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 }=4.18\text{ miles }[/latex]
Examples
How much further could Johnson have run if he had run for the same amount of time as Trason?Answer: Here is the table we created for reference:
| Runner | Distance = | (Rate ) | (Time) |
|---|---|---|---|
| Trason | [latex]50[/latex] miles | [latex]8.33[/latex] [latex]\frac{\text{ miles }}{\text{ hour }}[/latex] | [latex]6[/latex] hours |
| Johnson | [latex]50[/latex] miles | [latex]9.1[/latex] [latex]\frac{\text{ miles }}{\text{ hour }}[/latex] | [latex]5.5[/latex] hours |
[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 [latex]6[/latex] hours. Johnson would have run [latex]54.6[/latex] miles, so that's [latex]4.6[/latex] more miles than he ran during the race.Example
What was each runner's time for running one mile?Answer: Here is the table we created for reference:
| Runner | Distance = | (Rate ) | (Time) |
|---|---|---|---|
| Trason | [latex]50[/latex] miles | [latex]8.33[/latex] [latex]\frac{\text{ miles }}{\text{ hour }}[/latex] | [latex]6[/latex] hours |
| Johnson | [latex]50[/latex] miles | [latex]9.1[/latex] [latex]\frac{\text{ miles }}{\text{ hour }}[/latex] | [latex]5.5[/latex] hours |
[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 [latex]t[/latex], the distance is [latex]1[/latex] mile, and we know each runner's rate. It may help to create a new table:| Runner | Distance = | (Rate ) | (Time) |
|---|---|---|---|
| Trason | [latex]1[/latex] mile | [latex]8.33[/latex] [latex]\frac{\text{ miles }}{\text{ hour }}[/latex] | [latex]t[/latex] hours |
| Johnson | [latex]1[/latex] mile | [latex]9.1[/latex] [latex]\frac{\text{ miles }}{\text{ hour }}[/latex] | [latex]t[/latex] hours |
[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].
[latex]0.12[/latex] hours is about [latex]7.2[/latex] minutes, so Trason's time for running one mile was about [latex]7.2[/latex] minutes. WOW! She did that for [latex]6[/latex] 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].
[latex]0.11[/latex] hours is about [latex]6.6[/latex] minutes, so Johnson's time for running one mile was about [latex]6.6[/latex] minutes. WOW! He did that for [latex]5.5[/latex] 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 [latex]50[/latex]-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]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 [latex]P[/latex]), which then grows slowly according to the interest rate ([latex]r[/latex], measured in percent) and the length of time ([latex]t[/latex], usually measured in months) that the money stays in the account. The amount earned over time is called the interest ([latex]I[/latex]), 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.[latex]\displaystyle\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,I=P\,\cdot \,r\,\cdot \,t\\\\ \frac{I}{{P}\,\cdot \,r}=\frac{P\cdot\,r\,\cdot \,t}{\,P\,\cdot \,r}\\\\\,\,\,\,\,\,\,\,\,\,\,{t}=\frac{I}{\,r\,\cdot \,t}\end{array}[/latex]
Below is a table showing the result of solving for each individual variable in the formula.| Solve For | Result |
|---|---|
| [latex]I[/latex] | [latex]I=P\,\cdot \,r\,\cdot \,t[/latex] |
| [latex]P[/latex] | [latex]{P}=\frac{I}{{r}\,\cdot \,t}[/latex] |
| [latex]r[/latex] | [latex]{r}=\frac{I}{{P}\,\cdot \,t}[/latex] |
| [latex]t[/latex] | [latex]{t}=\frac{I}{{P}\,\cdot \,r}[/latex] |
Example
If a customer deposits a principal of [latex]$2000[/latex] at a monthly rate of [latex]0.7\%[/latex], what is the total amount that she has after [latex]24[/latex] months?Answer: Substitute in the values given for the Principal, Rate, and Time.
[latex]\displaystyle\begin{array}{l}I=P\,\cdot \,r\,\cdot \,t\\I=2000\cdot 0.7\%\cdot 24\end{array}[/latex]
Rewrite [latex]0.7\%[/latex] as the decimal [latex]0.007[/latex], then multiply.[latex]\begin{array}{l}I=2000\cdot 0.007\cdot 24\\I=336\end{array}[/latex]
Add the interest and the original principal amount to get the total amount in her account.[latex] \displaystyle 2000+336=2336[/latex]
She has [latex]$2336[/latex] after [latex]24[/latex] months.Example
Alex invests [latex]$600[/latex] at [latex]3.5\%[/latex] monthly interest for [latex]3[/latex] years. What amount of interest has Alex earned?Answer: 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 [latex]P[/latex], [latex]r[/latex], and [latex]t[/latex] so we can use substitution. [latex]r[/latex] =[latex]0.035[/latex], [latex]P[/latex] = [latex]$600[/latex], and [latex]t[/latex] =[latex]3[/latex] years. We have to be careful, because [latex]r[/latex] is in months, and [latex]t[/latex] is in years. We need to change [latex]t[/latex] into months, because we can't change the rate—it is set by the bank.
[latex]{t}=3\text{ years }\cdot12\frac{\text{ months }}{\text {year }}=36\text{ months }[/latex]
Write and Solve: Substitute the given values into the formula.[latex]\begin{array}{l} I=P\,\cdot \,r\,\cdot \,t\\\\I=600\,\cdot \,0.035\,\cdot \,36\\\\{I}=756\end{array}[/latex]
Check and Interpret: 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 [latex]$756[/latex].Example
After [latex]10[/latex] years, Jodi's account balance has earned [latex]$1080[/latex] in interest. The rate on the account is [latex]0.09\%[/latex] monthly. What was the original amount she invested in the account?Answer: 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 [latex]I[/latex] = [latex]$1080[/latex], [latex]r[/latex] =[latex]0.009[/latex], and [latex]t[/latex] =[latex]10[/latex] 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 }}{\text{ year }}=120\text{ months }[/latex]
Write and Solve: Substitute the given values into the formula[latex]\begin{array}{l}{P}=\frac{I}{{R}\,\cdot \,T}\\\\{P}=\frac{1080}{{0.009}\,\cdot \,120}\\\\{P}=\frac{1080}{1.08}=1000\end{array}[/latex]
Check and Interpret: We were asked to find the principal given the amount of interest earned on an account. If we substitute [latex]P[/latex] =[latex]$1000[/latex] into the formula [latex]I=P\,\cdot \,r\,\cdot \,t[/latex] we get[latex]I=1000\,\cdot \,0.009\,\cdot \,120\\I=1080[/latex]
Our solution checks out. Jodi invested [latex]$1000[/latex].Try it
[ohm_question]196954[/ohm_question]Contribute!
Licenses & Attributions
CC licensed content, Original
- Image: Ann Trason Trail Running. Authored by: Lumen Learning. License: CC BY: Attribution.
- Rates. Provided by: Lumen Learning License: CC BY: Attribution.
- Problem Solving Using Distance, Rate, Time (Running). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Simple Interest - Determine Account Balance (Monthly Interest). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Simple Interest - Determine Interest Balance (Monthly Interest). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Simple Interest - Determine Principal Balance (Monthly Interest). Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
CC licensed content, Shared previously
- Ann Trason. Provided by: Wikipedia Located at: https://en.wikipedia.org/wiki/Ann_Trason. License: CC BY-SA: Attribution-ShareAlike.
- American River 50 Mile Endurance Run. Provided by: Wikipedia Located at: https://en.wikipedia.org/wiki/American_River_50_Mile_Endurance_Run. License: CC BY-SA: Attribution-ShareAlike.
- Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
