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Study Guides > ALGEBRA / TRIG I

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
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. d=rtd=rt 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 d=rtd=rt, 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 30 miles  hour 30\frac{\text{ miles }}{\text{ hour }} for 22 hours, you can find the distance you would travel by multiplying rate times time or (30 miles  hour )(2 hours )=60 miles \left(30\frac{\text{ miles }}{\text{ hour }}\right)\left(2\text{ hours }\right)=60\text{ miles }.

Try It

[ohm_question]184997[/ohm_question]
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 d=rtd=rt equation for tt using division:

d=rtd=rt

dr=t\frac{d}{r}=t

Likewise, if we want to find rate, we can isolate rr using division:

d=rtd=rt

dt=r\frac{d}{t}=r

In the following examples, you will see how this formula is applied to answer questions about ultra marathon running.
Ann Trason Ann Trason
Ultra marathon running (defined as anything longer than 26.226.2 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 5050-mile Endurance Run, which begins in Sacramento, California, and ends in Auburn, California.[footnote]"Ann Trason." Wikipedia. Accessed May 05, 2016. https://en.wikipedia.org/wiki/Ann_Trason.[/footnote] In 1993, Trason finished the run with a time of 6:09:086:09:08.  The men's record for the same course was set in 1994 by Tom Johnson, who finished the course with a time of  5:33:215:33:21.[footnote] "American River 5050 Mile Endurance Run." Wikipedia. Accessed May 05, 2016. https://en.wikipedia.org/wiki/American_River_50_Mile_Endurance_Run.[/footnote] In the next examples, we will use the d=rtd=rt formula to answer the following questions about the two runners.
  1. What was each runner's rate for their record-setting runs?
  2. By the time Johnson had finished, how many more miles did Trason have to run?
  3. How much further could Johnson have run if he had run as long as Trason?
  4. What was each runner's time for running one mile?
To make answering the questions easier, we will round the two runners' times to 66 hours and 5.55.5 hours.  

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: d=rtd=rt, dt=r\frac{d}{t}=r. 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 5050 miles rr 66 hours
Johnson 5050 miles rr  5.55.5 hours
Write and Solve: Trason's rate:

d=rtd=rt

50 miles =r(6 hours)50\text{ miles }=r\left(6\text{ hours}\right)

50=6r50=6r

r≈8.33r\approx 8.33

Johnson's rate:

r=dtr=\frac{d}{t}

r=50 miles5.5 hoursr=\frac{50\text{ miles}}{5.5\text{ hours}}

r=505.5r=\frac{50}{5.5}

r≈9.10r\approx 9.10

Check and Interpret: We can fill in our table with this information.
Runner Distance = (Rate ) (Time)
Trason 5050 miles 8.338.33  miles  hour \frac{\text{ miles }}{\text{ hour }} 66 hours
Johnson 5050 miles 9.19.1  miles  hour \frac{\text{ miles }}{\text{ hour }} 5.55.5 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?

Answer: Here is the table we created for reference:

Runner Distance = (Rate ) (Time)
Trason 5050 miles 8.338.33  miles  hour \frac{\text{ miles }}{\text{ hour }} 66 hours
Johnson 5050 miles 9.19.1  miles  hour \frac{\text{ miles }}{\text{ hour }}  5.55.5 hours
Read and Understand: We are looking for how many miles Trason still had on the trail when Johnson had finished after 5.55.5 hours. This is a distance, and we know rate and time. Define and Translate: We can use the formula d=rtd=rt again. This time the unknown is dd, and the time Trason had run is 5.55.5 hours. Write and Solve:

d=rtd=8.33 miles  hour (5.5 hours)d=45.82 miles\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}

Check and Interpret: Have we answered the question? We were asked to find how many more miles she had to run after 5.55.5 hours.  What we have found is how long she had run after 5.55.5 hours. We need to subtract d=45.82 miles d=45.82\text{ miles } from the total distance of the course.

50 miles −45.82 miles =4.18 miles 50\text{ miles }-45.82\text{ miles }=4.18\text{ miles }

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 for the same amount of time as Trason?

Answer: Here is the table we created for reference:

Runner Distance = (Rate ) (Time)
Trason 5050 miles 8.338.33  miles  hour \frac{\text{ miles }}{\text{ hour }} 66 hours
Johnson 5050 miles 9.19.1  miles  hour \frac{\text{ miles }}{\text{ hour }} 5.55.5 hours
Read and Understand: The word further implies we are looking for a distance. Define and Translate: We can use the formula d=rtd=rt again. This time the unknown is dd, the time is 66 hours, and Johnson's rate is 9.1 miles  hour 9.1\frac{\text{ miles }}{\text{ hour }} Write and Solve:

d=rtd=9.1 miles  hour (6 hours )d=54.6 miles \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}.

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 9.1 miles  hour 9.1\frac{\text{ miles }}{\text{ hour }} for 66 hours. Johnson would have run 54.654.6 miles, so that's 4.64.6 more miles than he ran during 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?

Answer: Here is the table we created for reference:

Runner Distance = (Rate ) (Time)
Trason 5050 miles 8.338.33  miles  hour \frac{\text{ miles }}{\text{ hour }} 66 hours
Johnson 5050 miles 9.19.1  miles  hour \frac{\text{ miles }}{\text{ hour }} 5.55.5 hours
Read and Understand: we are looking for time, and this time our distance has changed from  5050 miles to 11 mile, so we can use

d=rtdr=td=rt\\\frac{d}{r}=t

Define and Translate: we can use the formula d=rtd=rt again. This time the unknown is tt, the distance is 11 mile, and we know each runner's rate. It may help to create a new table:
Runner Distance = (Rate ) (Time)
Trason 11 mile 8.338.33  miles  hour \frac{\text{ miles }}{\text{ hour }} tt hours
Johnson 11 mile 9.19.1  miles  hour \frac{\text{ miles }}{\text{ hour }} tt hours
Write and Solve: Trason: We will need to divide to isolate time.

d=rt1 mile =8.33 miles  hour (t hours )1 mile 8.33 miles  hour =t hours 0.12 hours =t\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}.

0.120.12  hours is about  7.27.2 minutes, so Trason's time for running one mile was about 7.27.2 minutes. WOW! She did that for  66 hours! Johnson: We will need to divide to isolate time.

d=rt1 mile =9.1 miles  hour (t hours )1 mile 9.1 miles  hour =t hours 0.11 hours =t\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}.

0.110.11 hours is about 6.66.6 minutes, so Johnson's time for running one mile was about  6.66.6 minutes. WOW! He did that for 5.55.5 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  5050-mile course.  Yes, we answered our question. Trason's mile time was 7.2minutesmile7.2\frac{\text{minutes}}{\text{mile}} and Johnsons' mile time was 6.6minutesmile6.6\frac{\text{minutes}}{\text{mile}}

In the following video, we show another example of answering many rate questions given distance and time. https://youtu.be/3WLp5mY1FhU

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 PP), which then grows slowly according to the interest rate (rr, measured in percent) and the length of time (tt, usually measured in months) that the money stays in the account. The amount earned over time is called the interest (II), which is then given to the customer.
CautionBeware! 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 I=P ⋅ r ⋅ t\displaystyle I=P\,\cdot \,r\,\cdot \,t 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 tt using division:

            I=P ⋅ r ⋅ tIP ⋅ r=P⋅ r ⋅ t P ⋅ r           t=I r ⋅ t\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}

Below is a table showing the result of solving for each individual variable in the formula.
Solve For Result
II I=P ⋅ r ⋅ tI=P\,\cdot \,r\,\cdot \,t
PP P=Ir ⋅ t{P}=\frac{I}{{r}\,\cdot \,t}
rr r=IP ⋅ t{r}=\frac{I}{{P}\,\cdot \,t}
tt  t=IP ⋅ r{t}=\frac{I}{{P}\,\cdot \,r}
In the next examples, we will show how to substitute given values into the simple interest formula, and decipher which variable to solve for.

Example

If a customer deposits a principal of $2000 at a monthly rate of  0.7%0.7\%, what is the total amount that she has after 2424 months?

Answer: Substitute in the values given for the Principal, Rate, and Time.

I=P ⋅ r ⋅ tI=2000⋅0.7%⋅24\displaystyle\begin{array}{l}I=P\,\cdot \,r\,\cdot \,t\\I=2000\cdot 0.7\%\cdot 24\end{array}

Rewrite  0.7%0.7\% as the decimal  0.0070.007, then multiply.

I=2000⋅0.007⋅24I=336\begin{array}{l}I=2000\cdot 0.007\cdot 24\\I=336\end{array}

Add the interest and the original principal amount to get the total amount in her account.

2000+336=2336 \displaystyle 2000+336=2336

She has  $2336 after  2424 months.

The following video shows another example of finding an account balance after a given amount of time, principal invested, and a rate. https://youtu.be/XkGgEEMR_00 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 3.5%3.5\% monthly interest for 33 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 I=P ⋅ r ⋅ tI=P\,\cdot \,r\,\cdot \,t Define and Translate: we know PP, rr, and tt so we can use substitution. rr  =0.0350.035, PP =  $600, and tt =33 years. We have to be careful, because rr is in months, and tt is in years.  We need to change tt into months, because we can't change the rate—it is set by the bank.

t=3 years ⋅12 months year =36 months {t}=3\text{ years }\cdot12\frac{\text{ months }}{\text {year }}=36\text{ months }

Write and Solve: Substitute the given values into the formula.

I=P ⋅ r ⋅ tI=600 ⋅ 0.035 ⋅ 36I=756\begin{array}{l} I=P\,\cdot \,r\,\cdot \,t\\\\I=600\,\cdot \,0.035\,\cdot \,36\\\\{I}=756\end{array}

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 $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. https://youtu.be/mRV5ljj32Rg

Example

After  1010 years, Jodi's account balance has earned  $1080 in interest. The rate on the account is  0.09%0.09\% 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 II =  $1080, rr =0.0090.009, and tt =1010 years, so we can use P=Ir ⋅ t{P}=\frac{I}{{r}\,\cdot \,t} 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.

t=10 years ⋅12 months  year =120 months {t}=10\text{ years }\cdot12\frac{\text{ months }}{\text{ year }}=120\text{ months }

Write and Solve: Substitute the given values into the formula

P=IR ⋅ TP=10800.009 ⋅ 120P=10801.08=1000\begin{array}{l}{P}=\frac{I}{{R}\,\cdot \,T}\\\\{P}=\frac{1080}{{0.009}\,\cdot \,120}\\\\{P}=\frac{1080}{1.08}=1000\end{array}

Check and Interpret: We were asked to find the principal given the amount of interest earned on an account.  If we substitute PP =$1000 into the formula I=P ⋅ r ⋅ tI=P\,\cdot \,r\,\cdot \,t we get

I=1000 ⋅ 0.009 ⋅ 120I=1080I=1000\,\cdot \,0.009\,\cdot \,120\\I=1080

Our solution checks out. Jodi invested  $1000.

The last video shows another example of finding the principal amount invested based on simple interest. https://youtu.be/vbMqN6lVoOM

Try it

[ohm_question]196954[/ohm_question]
In the next section, we will apply our problem-solving method to problems involving dimensions of geometric shapes.

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