6.4: Solve Simple Interest Applications
- Page ID
- 5029
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Use the simple interest formula
- Solve simple interest applications
Before you get started, take this readiness quiz.
- Solve 0.6y = 45. If you missed this problem, review Example 5.7.4.
- Solve \(\dfrac{n}{1.45}\) = 4.6. If you missed this problem, review Example 5.7.5.
Use the Simple Interest Formula
Do you know that banks pay you to let them keep your money? The money you put in the bank is called the principal, \(P\), and the bank pays you interest, \(I\). The interest is computed as a certain percent of the principal; called the rate of interest, \(r\). The rate of interest is usually expressed as a percent per year, and is calculated by using the decimal equivalent of the percent. The variable for time, t, represents the number of years the money is left in the account.
If an amount of money, \(P\), the principal, is invested for a period of t years at an annual interest rate r, the amount of interest, \(I\), earned is
\[I = Prt \nonumber\]
where
- \(I\) = interest
- \(P\) = principal
- \(r\) = rate
- \(t\) = time
Interest earned according to this formula is called simple interest.
The formula we use to calculate simple interest is I = Prt. To use the simple interest formula we substitute in the values for variables that are given, and then solve for the unknown variable. It may be helpful to organize the information by listing all four variables and filling in the given information.
Find the simple interest earned after 3 years on $500 at an interest rate of 6%.
Solution
Organize the given information in a list.
I = ?, P = $500, r = 6%, t = 3 years
We will use the simple interest formula to find the interest.
| Write the formula. | I = Prt |
| Substitute the given information. Remember to write the percent in decimal form. | I = (500)(0.06)(3) |
| Simplify. | I = 90 |
| Check your answer. Is $90 a reasonable interest earned on $500 in 3 years? | In 3 years the money earned 18%. If we rounded to 20%, the interest would have been 500(0.20) or $100. Yes, $90 is reasonable. |
| Write a complete sentence that answers the question. | The simple interest is $90. |
Find the simple interest earned after 4 years on $800 at an interest rate of 5%.
- Answer
-
$160
Find the simple interest earned after 2 years on $700 at an interest rate of 4%.
- Answer
-
$56
In the next example, we will use the simple interest formula to find the principal.
Find the principal invested if $178 interest was earned in 2 years at an interest rate of 4%.
Solution
Organize the given information in a list.
I = $178, P = ?, r = 4%, t = 2 years
We will use the simple interest formula to find the principal.
| Write the formula. | I = Prt |
| Substitute the given information. | 178 = P(0.04)(2) |
| Divide. | $$\dfrac{178}{0.08} = \dfrac{0.08P}{0.08}$$ |
| Simplify. | 2,225 = P |
| Check your answer. Is it reasonable that $2,225 would earn $178 in 2 years? | $$178 \stackrel{?}{=} 2,225 (0.04)(2)$$ |
| $$178 = 178 \; \checkmark$$ | |
| Write a complete sentence that answers the question. | The principal is $2,225. |
Find the principal invested if $495 interest was earned in 3 years at an interest rate of 6%.
- Answer
-
$2,750
Find the principal invested if $1,246 interest was earned in 5 years at an interest rate of 7%.
- Answer
-
$3,560
Now we will solve for the rate of interest.
Find the rate if a principal of $8,200 earned $3,772 interest in 4 years.
Solution
Organize the given information.
I = $3,772, P = $8,200, r = ?, t = 4 years
We will use the simple interest formula to find the rate.
| Write the formula. | I = Prt |
| Substitute the given information. | 3,772 = 8,200r(4) |
| Multiply. | 3,772 = 32,800r |
| Divide. | $$\dfrac{3,772}{32,800} = \dfrac{32,800r}{32,800}$$ |
| Simplify. | 0.115 = r |
| Write as a percent. | 11.5% = r |
| Check your answer. Is 11.5% a reasonable rate if $3,772 was earned in 4 years? | $$3,772 \stackrel{?}{=} 8,200 (0.115)(4)$$ |
| $$3,772 = 3,772 \; \checkmark$$ | |
| Write a complete sentence that answers the question. | The rate was 11.5%. |
Find the rate if a principal of $5,000 earned $1,350 interest in 6 years.
- Answer
-
4.5%
Find the rate if a principal of $9,000 earned $1,755 interest in 3 years.
- Answer
-
6.5%
Solve Simple Interest Applications
Applications with simple interest usually involve either investing money or borrowing money. To solve these applications, we continue to use the same strategy for applications that we have used earlier in this chapter. The only difference is that in place of translating to get an equation, we can use the simple interest formula.
We will start by solving a simple interest application to find the interest.
Nathaly deposited $12,500 in her bank account where it will earn 4% interest. How much interest will Nathaly earn in 5 years?
Solution
We are asked to find the Interest, I. Organize the given information in a list.
I = ?, P = $12,500, r = 4%, t = 5 years
| Write the formula. | I = Prt |
| Substitute the given information. | I = (12,500)(0.04)(5) |
| Simplify. | I = 2,500 |
| Check your answer. Is $2,500 a reasonable interest on $12,500 over 5 years? | At 4% interest per year, in 5 years the interest would be 20% of the principal. Is 20% of $12,500 equal to $2,500? Yes. |
| Write a complete sentence that answers the question. | The interest is $2,500. |
Areli invested a principal of $950 in her bank account with interest rate 3%. How much interest did she earn in 5 years?
- Answer
-
$142.50
Susana invested a principal of $36,000 in her bank account with interest rate 6.5%. How much interest did she earn in 3 years?
- Answer
-
$7,020
There may be times when you know the amount of interest earned on a given principal over a certain length of time, but you don't know the rate. For instance, this might happen when family members lend or borrow money among themselves instead of dealing with a bank. In the next example, we'll show how to solve for the rate.
Loren lent his brother $3,000 to help him buy a car. In 4 years his brother paid him back the $3,000 plus $660 in interest. What was the rate of interest?
Solution
We are asked to find the rate of interest, r. Organize the given information.
I = 660, P = $3,000, r = ?, t = 4 years
| Write the formula. | I = Prt |
| Substitute the given information. | 660 = (3,000)r(4) |
| Multiply. | 660 = (12,000)r |
| Divide. | $$\dfrac{660}{12,000} = \dfrac{(12,000)r}{12,000}$$ |
| Simplify. | 0.055 = r |
| Change to percent form. | 5.5% = r |
| Check your answer. Is 5.5% a reasonable interest rate to pay your brother? | $$660 \stackrel{?}{=} (3,000)(0.055)(4)$$ |
| $$660 = 660 \; \checkmark$$ | |
| Write a complete sentence that answers the question. | The rate of interest was 5.5%. |
Jim lent his sister $5,000 to help her buy a house. In 3 years, she paid him the $5,000, plus $900 interest. What was the rate of interest?
- Answer
-
6%
Hang borrowed $7,500 from her parents to pay her tuition. In 5 years, she paid them $1,500 interest in addition to the $7,500 she borrowed. What was the rate of interest?
- Answer
-
4%
There may be times when you take a loan for a large purchase and the amount of the principal is not clear. This might happen, for instance, in making a car purchase when the dealer adds the cost of a warranty to the price of the car. In the next example, we will solve a simple interest application for the principal.
Eduardo noticed that his new car loan papers stated that with an interest rate of 7.5%, he would pay $6,596.25 in interest over 5 years. How much did he borrow to pay for his car?
Solution
We are asked to find the principal, P. Organize the given information.
I = 6,596.25, P = ?, r = 7.5%, t = 5 years
| Write the formula. | I = Prt |
| Substitute the given information. | 6,596.25 = P(0.075)(5) |
| Multiply. | 6,596.25 = 0.375P |
| Divide. | $$\dfrac{6,596.25}{0.375} = \dfrac{0.375P}{0.375}$$ |
| Simplify. | 17,590 = P |
| Check your answer. Is $17,590 a reasonable amount to borrow to buy a car? | $$6,596.25 \stackrel{?}{=} (17,590)(0.075)(5)$$ |
| $$6,596.25 = 6,596.25 \; \checkmark$$ | |
| Write a complete sentence that answers the question. | The amount borrowed was $17,590. |
Sean's new car loan statement said he would pay $4,866.25 in interest from an interest rate of 8.5% over 5 years. How much did he borrow to buy his new car?
- Answer
-
$11,450
In 5 years, Gloria's bank account earned $2,400 interest at 5%. How much had she deposited in the account?
- Answer
-
$9,600
In the simple interest formula, the rate of interest is given as an annual rate, the rate for one year. So the units of time must be in years. If the time is given in months, we convert it to years.
Caroline got $900 as graduation gifts and invested it in a 10-month certificate of deposit that earned 2.1% interest. How much interest did this investment earn?
Solution
We are asked to find the interest, I. Organize the given information.
I = ?, P = $900, r = 2.1%, t = 10 months
| Write the formula. | I = Prt |
| Substitute the given information, converting 10 months to \(\dfrac{10}{12}\) of a year. | $$I = \$900 (0.021) \left(\dfrac{10}{12}\right)$$ |
| Multiply. | I = 15.75 |
| Check your answer. Is $15.75 a reasonable amount of interest? | If Caroline had invested the $900 for a full year at 2% interest, the amount of interest would have been $18. Yes, $15.75 is reasonable. |
| Write a complete sentence that answers the question. | The interest earned was $15.75. |
Adriana invested $4,500 for 8 months in an account that paid 1.9% interest. How much interest did she earn?
- Answer
-
$57.00
Milton invested $2,460 for 20 months in an account that paid 3.5% interest How much interest did he earn?
- Answer
-
$143.50
Practice Makes Perfect
Use the Simple Interest Formula
In the following exercises, use the simple interest formula to fill in the missing information.
| Interest | Principal | Rate | Time (years) |
|---|---|---|---|
| $1200 | 3% | 5 |
| Interest | Principal | Rate | Time (years) |
|---|---|---|---|
| $1500 | 2% | 4 |
| Interest | Principal | Rate | Time (years) |
|---|---|---|---|
| $4410 | 4.5% | 7 |
| Interest | Principal | Rate | Time (years) |
|---|---|---|---|
| $2212 | 3.2% | 6 |
| Interest | Principal | Rate | Time (years) |
|---|---|---|---|
| $577.08 | $4580 | 2 |
| Interest | Principal | Rate | Time (years) |
|---|---|---|---|
| $528.12 | $3260 | 3 |
In the following exercises, solve the problem using the simple interest formula.
- Find the simple interest earned after 5 years on $600 at an interest rate of 3%.
- Find the simple interest earned after 4 years on $900 at an interest rate of 6%.
- Find the simple interest earned after 2 years on $8,950 at an interest rate of 3.24%.
- Find the simple interest earned after 3 years on $6,510 at an interest rate of 2.85%.
- Find the simple interest earned after 8 years on $15,500 at an interest rate of 11.425%.
- Find the simple interest earned after 6 years on $23,900 at an interest rate of 12.175%.
- Find the principal invested if $656 interest was earned in 5 years at an interest rate of 4%.
- Find the principal invested if $177 interest was earned in 2 years at an interest rate of 3%.
- Find the principal invested if $70.95 interest was earned in 3 years at an interest rate of 2.75%.
- Find the principal invested if $636.84 interest was earned in 6 years at an interest rate of 4.35%.
- Find the principal invested if $15,222.57 interest was earned in 6 years at an interest rate of 10.28%.
- Find the principal invested if $10,953.70 interest was earned in 5 years at an interest rate of 11.04%.
- Find the rate if a principal of $5,400 earned $432 interest in 2 years.
- Find the rate if a principal of $2,600 earned $468 interest in 6 years.
- Find the rate if a principal of $11,000 earned $1,815 interest in 3 years.
- Find the rate if a principal of $8,500 earned $3,230 interest in 4 years.
Solve Simple Interest Applications
In the following exercises, solve the problem using the simple interest formula.
- Casey deposited $1,450 in a bank account with interest rate 4%. How much interest was earned in 2 years?
- Terrence deposited $5,720 in a bank account with interest rate 6%. How much interest was earned in 4 years?
- Robin deposited $31,000 in a bank account with interest rate 5.2%. How much interest was earned in 3 years?
- Carleen deposited $16,400 in a bank account with interest rate 3.9%. How much interest was earned in 8 years?
- Hilaria borrowed $8,000 from her grandfather to pay for college. Five years later, she paid him back the $8,000, plus $1,200 interest. What was the rate of interest?
- Kenneth lent his niece $1,200 to buy a computer. Two years later, she paid him back the $1,200, plus $96 interest. What was the rate of interest?
- Lebron lent his daughter $20,000 to help her buy a condominium. When she sold the condominium four years later, she paid him the $20,000, plus $3,000 interest. What was the rate of interest?
- Pablo borrowed $50,000 to start a business. Three years later, he repaid the $50,000, plus $9,375 interest. What was the rate of interest?
- In 10 years, a bank account that paid 5.25% earned $18,375 interest. What was the principal of the account?
- In 25 years, a bond that paid 4.75% earned $2,375 interest. What was the principal of the bond?
- Joshua's computer loan statement said he would pay $1,244.34 in interest for a 3 year loan at 12.4%. How much did Joshua borrow to buy the computer?
- Margaret's car loan statement said she would pay $7,683.20 in interest for a 5 year loan at 9.8%. How much did Margaret borrow to buy the car?
- Caitlin invested $8,200 in an 18-month certificate of deposit paying 2.7% interest. How much interest did she earn form this investment?
- Diego invested $6,100 in a 9-month certificate of deposit paying 1.8% interest. How much interest did he earn form this investment?
- Airin borrowed $3,900 from her parents for the down payment on a car and promised to pay them back in 15 months at a 4% rate of interest. How much interest did she owe her parents?
- Yuta borrowed $840 from his brother to pay for his textbooks and promised to pay him back in 5 months at a 6% rate of interest. How much interest did Yuta owe his brother?
Everyday Math
- Interest on savings Find the interest rate your local bank pays on savings accounts.
- What is the interest rate?
- Calculate the amount of interest you would earn on a principal of $8,000 for 5 years.
- Interest on a loan Find the interest rate your local bank charges for a car loan.
- What is the interest rate?
- Calculate the amount of interest you would pay on a loan of $8,000 for 5 years.
Writing Exercises
- Why do banks pay interest on money deposited in savings accounts?
- Why do banks charge interest for lending money?
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) On a scale of 1–10, how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?
Contributors and Attributions
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at http://cnx.org/contents/fd53eae1-fa2...49835c3c@5.191."