# 6.21: Simple Interest

**Basic**Created by: CK-12

**Practice**Percent Equations

A payday loan gives workers a cash advance on their paycheck. The borrower promises to repay the loan when he or she receives his or her next paycheck. In exchange for the cash advance, the lender charges a high fee. Logan finds himself in a position where if he doesn’t pay his phone bill, his phone will be shut off. It is four weeks until his next paycheck and he doesn’t have enough money to pay his bill. So he takes out a payday loan for $150. The lender charges a simple interest rate of 400%. How much interest is Logan charged when he repays it in four weeks?

In this concept, you will learn to solve real-world problems involving simple interest.

### Solving Problems Involving Simple Interest

You can use the simple interest equation to calculate other unknowns besides the interest rate, like the amount of time it takes to earn a specific amount of interest.

Let’s take a look at an example.

Ben deposited $1,200 in a certificate of deposit (CD) at an interest rate of 5.5%. He earned $198 in simple interest. How long was the CD for?

First, write the equation.

\begin{align*}I=Prt\end{align*}

Next, substitute the given values.

\begin{align*}\$198=\$1,200\times 5.5\%\times t\end{align*}

Next, write the percent as a decimal.

\begin{align*}\$198=\$1,200\times 0.055\times t\end{align*}

Next, simplify.

\begin{align*}\$198=\$66\times t\end{align*}

Then, solve for \begin{align*}t\end{align*}

\begin{align*}3=t\end{align*}

The answer is the CD was for 3 years.

You can also use the simple interest equation to figure out the amount of the interest.

Let’s look at another example.

Troy deposited $400 into his savings account. How much interest will he receive at the end of one year if the interest rate is 3%?

First, write the equation.

\begin{align*}I=Prt\end{align*}

Next, substitute the given values.

\begin{align*}=\$400\times 3\% \times 1\end{align*}

Next, write the percent as a decimal.

\begin{align*}=\$400\times 0.03\times 1
\end{align*}

Next, simplify and solve for \begin{align*}I\end{align*}

\begin{align*}=\$12\end{align*}

The answer is $12.

Troy will receive $12 in interest at the end of one year.

Let’s take a look at another example.

Courtney borrowed $7,500 for 4 years at an annual interest rate of 8%. How much interest will she pay on the loan?

First, write the equation.

\begin{align*}I=Prt\end{align*}

Next, substitute the given values.

\begin{align*}=\$7,500\times 8\%\times 4\end{align*}

Next, write the percent as a decimal.

\begin{align*}=\$7,500\times 0.08\times 4\end{align*}

Next, simplify and solve for \begin{align*}I\end{align*}

\begin{align*}=\$2,400\end{align*}

The answer is $2,400.

Courtney will pay $2,400 interest on the loan.

Once you know the interest, you can go back and add it to the principal.

### Examples

#### Example 1

Earlier, you were given a problem about Logan and his payday loan.

If the lender charges a simple interest rate of 400% on Logan’s $150 loan, how much interest is Logan charged when he repays it in four weeks?

First, the loan is for four weeks, so convert this to years. There are 52 weeks in a year.

\begin{align*}t=\frac{4}{52}=0.0769 \text{ years}\end{align*}

Next, use the simple interest equation.

\begin{align*}I=Prt\end{align*}

Next, substitute the given values.

\begin{align*}=\$150\times 400\%\times 0.0769\end{align*}

Next, write the percent as a decimal.

\begin{align*}=\$150\times 4.00\times 0.0769\end{align*}

Then, solve for \begin{align*}I\end{align*}

\begin{align*}=\$46.14\end{align*}

The answer is Logan pays $46.14 in interest.

#### Example 2

Joanna borrowed $500 at an interest rate of 8%. At the end of the loan period, she had to pay back $530. How long was the loan for?

The amount to be repaid includes the principal plus the interest. Subtract the principal from the amount to be repaid to find the amount of the interest.

\begin{align*}\$530-\$500=\$30\end{align*}

First, write the equation.

\begin{align*}I=Prt\end{align*}

Next, substitute the given values.

\begin{align*}\$30=\$500\times 8\%\times t\end{align*}

Next, write the percent as a decimal.

\begin{align*}\$30=\$500\times 0.08\times t\end{align*}

Next, simplify.

\begin{align*}\$30=\$40\times t\end{align*}

Then, solve for \begin{align*}t\end{align*}

\begin{align*}\frac{30}{40} = \frac{3}{4} = t\end{align*}

Three-quarters of a year is three-quarters of 12 months, which is 9 months.

The answer is the loan was for 9 months.

#### Example 3

John earned $96.00 on a $1,200 deposit at 2% interest rate. How long did he invest?

First, write the equation.

\begin{align*}I=Prt\end{align*}

Next, substitute the given values.

\begin{align*}\$96=\$1,200\times 2\%\times t\end{align*}

Next, write the percent as a decimal.

\begin{align*}\$96=\$1,200\times 0.02\times t\end{align*}

Next, simplify.

\begin{align*}\$96=\$24\times t\end{align*}

Then, solve for \begin{align*}t\end{align*}

\begin{align*}\frac{96}{24}=4=t\end{align*}

The answer is 4 years.

#### Example 4

Karen paid $48.00 on $600.00 at a 4% interest rate. How long did she have the loan?

First, write the equation.

\begin{align*}I=Prt\end{align*}

Next, substitute the given values.

\begin{align*}\$48=\$600\times 4\%\times t\end{align*}

Next, write the percent as a decimal.

\begin{align*}\$48=\$600\times 0.04\times t\end{align*}

Next, simplify.

\begin{align*}\$48=\$24\times t\end{align*}

Then, solve for \begin{align*}t\end{align*}

\begin{align*}\frac{48}{24}=2=t\end{align*}

The answer is 2 years.

#### Example 5

Eric earned $130.00 on a $2,000 deposit at a 1% interest rate. How long did he invest?

First, write the equation.

\begin{align*}I=Prt\end{align*}

Next, substitute the given values.

\begin{align*}\$130=\$2,000 \times 1\% \times t\end{align*}

Next, write the percent as a decimal.

\begin{align*}\$130=\$2,000 0.01\times t\end{align*}

Next, simplify.

\begin{align*}\$130 = \$20 \times t\end{align*}

Then, solve for \begin{align*}t\end{align*}

\begin{align*}\frac{130}{20} = 6.5=t\end{align*}

The answer is 6.5 years.

### Review

Calculate the simple interest for each amount.

- $1,500.00 at 3% for 1 year
- $2,300.00 at 2% for 2 years
- $500.00 at 4% for 2 years
- $2,500.00 at 5% for 5 years
- $1,500.00 at 11% for 2 years
- $3,500 at 3% for 5 years
- $3,500 at 4% for 15 years
- $2,300 at 2% for 3 years
- $5,500 at 5% for 6.5 years
- $12,000 at 4% for 5 years

Find the length of time for each loan.

- Principal: $1,250; interest rate: 6%; simple interest: $300
- Principal: $4,800; interest rate: 7.5%; simple interest: $900

Solve each problem.

- Juan invested $5,000 in an account that pays 5% interest. If interest is paid 4 times a year, how much is each interest payment?
- Sophie put $330 in a savings account at a simple interest rate of 4% per year. Avi put $290 in a savings account at a simple interest rate of 5% per year. Who will have earned more interest after 2 years? How much more?
- How much did Madi is on invested in a certificate of deposit for 4 years at a 6% interest rate. At the end of the 4 years, the value of the certificate of deposit was $3,100.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 6.21.

### Resources

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Compound interest

Compound interest refers to interest earned on the total amount at the time it is compounded, including previously earned interest.future value

In the context of earning interest, future value stands for the amount in the account at some future time .Interest

Interest is a percentage of lent or borrowed money. Interest is calculated and accrued regularly at a specified rate.Interest Rate

The interest rate is the percentage at which interest accrues.present value

In the context of earning interest, present value stands for the amount in the account at time 0.Principal

The principal is the amount of the original loan or original deposit.Simple Interest

Simple interest is interest calculated on the original principal only. It is calculated by finding the product of the the principal, the rate, and the time.### Image Attributions

In this concept, you will solve real-world problems involving simple interest.

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