<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 6.7: Simple Interest

Difficulty Level: At Grade Created by: CK-12

## Introduction

The Loan

Since Taylor has been working at the candy store, she has had her eye on a new bike. She has saved $56.00, but the bike is being sold for$156.00.

Taylor wouldn’t mind waiting, except for the fact that several of her friends are going to go on a big bike ride on the weekend and she wants to go with them. She has decided to ask her brother if she can borrow the money.

The one catch is that he wants to charge her interest. Her brother will loan her the $100.00, but he wants to charge her 15% interest per month. “I don’t think that is very fair,” Taylor tells him when he presents the deal. “It isn’t that much more that you have to pay back.” Taylor disagrees. How much is 15% interest on$100.00?

This lesson will teach you all about simple interest and about how to calculate simple interest. By the end of the lesson you will understand how to figure out the interest on Taylor's loan.

What You Will Learn

In this lesson, you will learn how to use the following skills.

• Use the simple interest equation I=Prt\begin{align*}I = Prt\end{align*} to find an interest rate.
• Use the equation to find the time required to earn a given amount.
• Use the equation to find the new balance after a given time.
• Solve real-world problems involving simple interest.

Teaching Time

I. Use the Simple Interest Equation I=Prt\begin{align*}I=Prt\end{align*} to find an Interest Rate

This lesson is all about borrowing money, paying money back and the fees associated with borrowing money. This is where you will learn all about interest.

What is interest?

Interest is a charge for money that is borrowed. When you borrow money, you pay the lender interest for borrowing the money. When you deposit money into a saving account at a bank, the bank pays you interest since the bank is borrowing money from you. So interest can be something that you have to pay or that is paid to you.

The amount of money that is invested or borrowed is called the principal. If you borrow 500.00 this is the principal. It is the initial amount before any interest is added on. The rate of interest is the percent charged or earned. We also have to consider the time in years that the money is borrowed or deposited when computing interest. How do we calculate interest? We can calculate interest by using an equation. Here is the equation. InterestI=Principal×rate×time=Prt\begin{align*}\text{Interest} & = \text{Principal} \times \text{rate} \times \text{time}\\ I & = Prt\end{align*} Take a few minutes to write this equation down in your notebook. If you know the amount of interest, the principal, and the time, you can find the interest rate. Example If you deposit2,000 at USA Savings Bank, at the end of 2 years you will have received $240 in simple interest. What is the interest rate at USA Savings Bank? I=Prt$240=$2,000×r×2 Substitute values.$240=4,000×rSimplify.0.06=r Solve for r.\begin{align*}&I = Prt\\ &\ 240 = \ 2,000 \times r \times 2 \qquad \ \leftarrow \text{Substitute values.}\\ &\ 240 = \ 4,000 \times r \qquad \qquad \leftarrow \text{Simplify.}\\ &0.06 = r \qquad \qquad \qquad \qquad \ \leftarrow \text{Solve for}\ r.\end{align*} Change 0.06 to the percent 6%. The interest rate is 6% per year. Example If you borrow3,600 from USA Savings Bank for 18 months, at the end of the 18 months you will repay $4,059 to the bank. What is the interest rate for this loan? 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.$4,059$3,600=$459\begin{align*}\ 4,059 - \ 3,600 = \ 459\end{align*}

Since there are 12 months in a year, 18 months is 1812\begin{align*}\frac{18}{12}\end{align*} or 112\begin{align*}1 \frac{1}{2}\end{align*} years.

I=Prt$459=$3,600×r×112 Substitute values.$459=$5,400×r  Simplify.0.085=r  Solve for r.\begin{align*}&I = Prt\\ &\ 459 = \ 3,600 \times r \times 1 \frac{1}{2} \qquad \ \leftarrow \text{Substitute values.}\\ &\ 459 = \ 5,400 \times r \qquad \qquad \ \ \leftarrow \text{Simplify.}\\ &0.085 = r \qquad \qquad \qquad \qquad \ \ \leftarrow \text{Solve for}\ r.\end{align*}

Change 0.085 to the percent 8.5%.

The interest rate is 8.5% per year.

6S. Lesson Exercises

Find the interest rate for each problem.

1. Jesse borrowed $500.00. At the end of the year he paid back$50.00 in interest. What was the interest rate?
2. Karen earned $200.00 in two years of simple interest on her initial investment of$400.00. What was the annual (yearly) interest rate?

Take a few minutes to check your work with a friend.

II. Use the Equation to Find the Time Required to Earn a Given Amount

Now that you know how to find the interest rate, we can use the equation to calculate the amount of time it takes to earn a specific amount of interest.

Example

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

I=Prt$198=$1,200×5.5%×tSubstitute values.$198=$1,200×0.055×tWrite the percent as a decimal.198=66×t Simplify.3=tSolve for t.\begin{align*}& I = Prt\\ & \ 198 = \ 1,200 \times 5.5 \% \times t \qquad \leftarrow \text{Substitute values.}\\ & \ 198 = \ 1,200 \times 0.055 \times t \qquad \leftarrow \text{Write the percent as a decimal.}\\ & \ 198 = 66 \times t \qquad \quad \qquad \qquad \ \leftarrow \text{Simplify.}\\ & 3 = t \qquad \qquad \qquad \qquad \qquad \quad \leftarrow \text{Solve for}\ t.\end{align*} The CD was for 3 years. Example Joanna borrowed500 at an annual 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.$530$500=$30\begin{align*}\ 530 - \ 500 = \ 30\end{align*}

I=Prt$30=$500×8%×t  Substitute values.$30=$500×0.08×t Write the percent as a decimal.$30=$40×tSimplify.34=t  Solve for t.\begin{align*}& I = Prt\\ & \ 30 = \ 500 \times 8 \% \times t \qquad \ \ \leftarrow \text{Substitute values.}\\ & \ 30 = \ 500 \times 0.08 \times t \ \quad \quad \leftarrow \text{Write the percent as a decimal.}\\ & \ 30 = \ 40 \times t \qquad \qquad \qquad \leftarrow \text{Simplify.}\\ & \frac{3}{4} = t \qquad \qquad \qquad \qquad \quad \ \ \leftarrow \text{Solve for}\ t.\end{align*}

34\begin{align*}\frac{3}{4}\end{align*} of a year is 34\begin{align*}\frac{3}{4}\end{align*} of 12 months, or 9 months.

The loan was for 9 months.

6T. Lesson Exercises

Find the amount of time for each problem.

1. John earned $124.00 on a$1200 deposit at 2% annual interest rate.
2. Karen paid $25.00 on$600.00 at a 4% annual interest rate.

Take a few minutes to check your work with a friend.

III. Use the Equation to Find the New Balance After a Given Time

We can also use the equation to figure out the amount of the interest. Let’s look at a few examples.

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% per year? First we write the equation. Then we substitute the given values and solve. I=Prt=$400×3%×1Substitute values.=$400×0.03×1 Write the percent as a decimal.=$12  Solve for I.\begin{align*}I & = Prt\\ & = \ 400 \times 3 \% \times 1 \qquad \leftarrow \text{Substitute values.}\\ & = \ 400 \times 0.03 \times 1 \ \quad \leftarrow \text{Write the percent as a decimal.}\\ & = \ 12 \ \qquad \qquad \qquad \ \leftarrow \text{Solve for}\ I.\end{align*}

Troy will receive $12 in interest at the end of one year. Example Courtney borrowed$7,500 for 4 years at an annual interest rate of 8%. How much interest will she pay on the loan?

I=Prt=$7,500×8%×4Substitute values.=$7,500×0.08×4 Write the percent as a decimal.=2,400 Solve for I.\begin{align*}& I = Prt\\ & = \ 7,500 \times 8 \% \times 4 \qquad \leftarrow \text{Substitute values.}\\ & = \ 7,500 \times 0.08 \times 4 \ \quad \leftarrow \text{Write the percent as a decimal.}\\ & = \ 2,400 \qquad \qquad \qquad \ \leftarrow \text{Solve for}\ I.\end{align*} Courtney will pay2,400 interest on the loan.

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

Look at the first example.

Example

Troy deposited $400 into his savings account. We know that at the end of the year he received$12.00 in interest. We can add the amounts together to find the new balance.

400+12=412.00\begin{align*}400 + 12 = \ 412.00\end{align*} The new balance is412.00.

In the second example, Courtney borrowed $7,500 and will pay$2400 interest on the loan. Here is the total she will pay to pay back the loan. We add the interest and the Principal.

7500+2400=9,900\begin{align*}7500 + 2400 = \ 9,900\end{align*} The new balance is9,900.

IV. Solve Real-World Problems Involving Simple Interest

Using simple interest is something that we do in real life all the time. This equation that you have learned is very important. Notice in Courtney’s example that she will pay back a lot of money in interest on the loan. This is always something that you have to consider when you borrow money. There is a lot of extra money that you end up paying because of the interest. That is why it is such a big deal when you hear people talk about a low interest rate.

Example

A car loan in the amount of $9,000 at an annual rate of 7% for 3 years is to be repaid in 36 monthly installments, including principal and interest. How much is each payment? The amount to be repaid includes the principal plus the interest. First find the amount of interest. I=Prt=$9,000×7%×3Substitute values.=$9,000×0.07×3 Write the percent as a decimal.=$1,890 Solve for I.\begin{align*}I & = Prt\\ & = \ 9,000 \times 7 \% \times 3 \qquad \leftarrow \text{Substitute values.}\\ & = \ 9,000 \times 0.07 \times 3 \ \ \quad \leftarrow \text{Write the percent as a decimal.}\\ & = \ 1,890 \qquad \qquad \qquad \ \leftarrow \text{Solve for}\ I.\end{align*}

Since the interest is $1,890, the amount to be repaid is$9,000+1,890\begin{align*}\9,000 + \1,890\end{align*}, or10,890.

Divide $10,890 by 36 to find the amount of each monthly payment.$10,890÷36=302.50\begin{align*}\ 10,890 \div 36 = \ 302.50\end{align*} Each payment is302.50.

## Real Life Example Completed

The Loan

Here is the original problem once again. Reread it and underline any important information. Then figure out the interest.

Since Taylor has been working at the candy store, she has had her eye on a new bike. She has saved $56.00, but the bike is being sold for$156.00.

Taylor wouldn’t mind waiting, except for the fact that several of her friends are going to go on a big bike ride on the weekend and she wants to go with them. She has decided to ask her brother if she can borrow the money.

The one catch is that he wants to charge her interest. Her brother will loan her the $100.00, but he wants to charge her 15% interest per month. “I don’t think that is very fair,” Taylor tells him when he presents the deal. “It isn’t that much more that you have to pay back.” Taylor disagrees. How much is 15% interest on$100.00?

To figure this out we multiply the 15% times $100.00. .15×100=$15.00\begin{align*}.15 \times 100 = \ 15.00\end{align*} per month.

If it takes her 3 months to pay him back, she will have to pay $45.00 in interest. “No thanks,” Taylor says to her brother. “That isn’t a very good deal. I am going to ask Dad for a loan. I am sure that he won’t charge me interest.” Taylor grinned at her brother and went off to find her Dad. ## Vocabulary Interest the amount of money added to a loan or to a deposit based on the amount of principal, an interest rate, and the amount of time for which the interest is paid. Principal the original amount of money borrowed or invested Interest Rate the percent that is being given for an investment or for a loan. ## Technology Integration ## Time to Practice Directions: Find the simple interest on each amount (all interest rates are per year). 1.$500.00 at 4% for 2 years

2. $200.00 at 5% for 3 years 3.$5000.00 at 2% for 2 years

4. $600.00 at 10% for 1 year 5.$1200.00 at 4% for 2 years

6. $1500.00 at 3% for 1 year 7.$2300.00 at 2% for 2 years

8. $500.00 at 4% for 2 years 9.$2500.00 at 5% for 5 years

10. $1500.00 at 11% for 2 years Directions: Find the interest rate for each loan. 11. principal:$2,500; time: 2 years; simple interest: $450 12. principal:$5,600; time: 9 months; simple interest: $357 Directions: Find the length of time for each loan. 13. principal:$1,250; interest rate: 6%; simple interest: $300 14. principal:$4,800; interest rate: 7.5%; simple interest: $900 Directions: Solve each problem. 15. Juan invested$5,000 in an account that pays 5% interest per year. If interest is split into four payments per year, how much is each interest payment?

16. 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?

17. Madison 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. How much did Madison deposit originally?

Show Hide Details
Description
Tags:
Subjects:
Search Keywords: