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# 5.7: Simple Interest

Difficulty Level: At Grade Created by: CK-12

## Introduction

Student Council Investment

With a goal of boosting student attendance at football games, the student council has decided to invest a portion of their savings for middle school decorations. They figure that when the games are happening that they can decorate the middle school with balloons, banners and flyers.

“I think it will help make it a priority for students,” Jeremy said at the weekly student council meeting.

“It will be a lot of fun too. We could even host a pep rally to help get the kids charged up,” Candice suggested.

“We put 4000 in the bank in the sixth grade. Now that we are 8th\begin{align*}8^{th}\end{align*} graders that money has been sitting in the bank for two years at a 4% interest rate,” Jeremy explained. Candice began working out the math in her head. If they did put4000 in the bank for two years and they had a 4% interest rate, then there definitely is more money in there now. She began to complete the calculations in her head.

Do you have an idea how to figure this out? This problem involves principal, interest rates and time. This lesson will teach you all about calculating simple interest. Pay close attention and you will see this problem again at the end of the lesson.

What You Will Learn

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

• Use the simple interest formula 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 Formula I=Prt\begin{align*}\underline{I = Prt}\end{align*} to Find an Interest Rate

Money is a necessary part of everyday life, and as you get older, your relationship to money will change. In this lesson, we will explore some of the ways in which you will relate to money as you get older.

Saving money and making wise investments will be an important part of your financial planning. Part of making investing is earning interest. When you save money in the bank, the bank uses that money for its own investments. In return for using your money, the bank pays you a certain percent. This percent is your interest. Interest is the percent that a bank pays you for using keeping your money in their bank.

Banks compete with each other for your money because they want you to put your money in their bank. They try to give you the best “interest rate” that they can. This means that they will pay you a greater percentage than another bank to try to get your business. The greater the interest rate that they pay you; the more likely you are to invest your money with them in a savings account. The more money you save, the more they have to invest. They publish an interest rate r\begin{align*}r\end{align*} which tells you what percent they will pay you per year t\begin{align*}t\end{align*}. The principal, p\begin{align*}p\end{align*} is the amount of money that you have put into the bank.

You can use this information in the formula I=prt\begin{align*}I = prt\end{align*} in order to calculate the interest that you will earn on your principal p\begin{align*}p\end{align*}.

Take a few minutes and write this formula in your notebook.

Now let’s look at how we can use this formula to calculate interest.

Example

You invest 5,000 in a bank for 2 years at a 4% interest rate. What is the interest you have earned after this time? We start by looking at the given information. Then use the formula to calculate interest. p=5000,r=.04,t=2\begin{align*}p = 5000, r = .04, t = 2\end{align*}. Use the formula to calculate interest. III=prt=5000.042=400\begin{align*}I &= prt\\ I &= 5000 \cdot .04 \cdot 2\\ I &= 400\end{align*} The bank will pay you400 in interest over two years at that rate.

Example

An investor places 15,000 in a savings account that pays 4.5% interest. She will leave the money there for 6 years. What will her interest be? pIII=15000,r=.045,t=6=prt=15000.0456=4050\begin{align*}p &= 15000, r = .045, t = 6\\ I &= prt\\ I &= 15000 \cdot .045 \cdot 6\\ I &= 4050\end{align*} Her interest after 6 years will be4,050.

Now that you know how to calculate the interest, let’s look at figuring out the time.

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

Many investors may have specific goals—they want to earn a certain amount of interest on their investments. Because of this, they need to figure out the time that it takes to earn a certain amount of money. The formula I=prt\begin{align*}I = prt\end{align*} is an equation. We can use the Multiplication Property of Equations to solve for t\begin{align*}t\end{align*} if we know I,r\begin{align*}I, r\end{align*}, and p\begin{align*}p\end{align*}.

Let’s look at an example.

Example

Mrs. Duarte has $20,000 to invest. She wants to earn$10,000 in interest. She is considering a savings and loans bank that is offering her 5.6% interest per year. For how long will she have to leave her money in the bank in order to reach her goal of 10,000? Start by looking at the given information. I=10000,p=20000,r=.056\begin{align*}I = 10000, p = 20000, r = .056\end{align*} Solve for t\begin{align*}t\end{align*}. Next, we substitute the given values into the formula and solve the equation. I10000100001000011208.93=prt=20000.056t=1120t=1120t1120=t\begin{align*}I &= prt\\ 10000 &= 20000 \cdot .056 \cdot t\\ 10000 &= 1120t\\ \frac{10000}{1120} &= \frac{1120t}{1120}\\ 8.93 &= t\end{align*} She will have to leave her money in the bank for nearly 9 years. Exactly! We are using what we have learned about solving equations to figure out missing information regarding interest and banking. Example A bank is offering an interest rate of 4.75%. How long would it take to earn500 if you invested 12,000 in the bank? I=500,p=12000,r=.0475\begin{align*}I = 500, p = 12000, r = .0475\end{align*} Solve for t\begin{align*}t\end{align*}. I500500500570.88=prt=12000.0475t=570t=570t570=t\begin{align*}I &= prt\\ 500 &= 12000 \cdot .0475 \cdot t\\ 500 &= 570t\\ \frac{500}{570} &= \frac{570t}{570}\\ .88 &= t\end{align*} It would take .88 years or about 1012\begin{align*}10\frac{1}{2}\end{align*} months. Notice that we have a decimal part of a year here. We can estimate how many months that is based on the decimal. A whole would be represented by 12 months or one year. III. Use the Equation to Find the New Balance after a Given Time You can use the simple interest formula I=prt\begin{align*}I = prt\end{align*} to find any of the missing variables if you are given values of the others. We have used it to solve for I\begin{align*}I\end{align*} and t\begin{align*}t\end{align*}. Of course, once the bank pays you interest, your account balance grows. You start of with your principal p\begin{align*}p\end{align*} and then you add your interest I\begin{align*}I\end{align*}. Now let’s see how much a bank balance would be after a given time at a given interest rate. Example Jessica invests3,000 in a credit union at an interest rate of 3.9%. She leaves the money there for 5 years. What is her balance after that time?

To answer this question, we will need to do two things. First, we will need to figure out the amount of the interest. Then we can add this amount to the principal that Jessica first invested. This will give us the new balance.

First find the interest that she earned:

pIII=3000,r=.039,t=5=prt=3000.0395=585\begin{align*}p &= 3000, r = .039, t = 5\\ I &= prt\\ I &= 3000 \cdot .039 \cdot 5\\ I &= 585\end{align*}

She earned $585 in interest. Her principal was$3,000. How much does she have now?

585+3000=3585\begin{align*}585 + 3000 = 3585\end{align*}

She has 3,585. This is the new balance. Now let’s look at using this formula with some other real – world problems. IV. Solve Real – World Problems Involving Simple Interest Using the simple interest formula I=prt\begin{align*}I = prt\end{align*}, we can calculate the interest rate r\begin{align*}r\end{align*} if we are given the I,p\begin{align*}I, p\end{align*} and t\begin{align*}t\end{align*} values. As before, we will substitute the known values and then use inverse operations to find the missing value. Example A nurse put22,000 in the bank 15 years ago. She has earned 21,450 in interest—nearly as much as her initial investment. What was the interest rate that the bank was paying her? II214502145021450330000.065=21450,p=22000,t=15=prt=22000r15=330000r=330000r330000=r\begin{align*}I &= 21450, p = 22000, t = 15\\ I &= prt\\ 21450 &= 22000 \cdot r \cdot 15\\ 21450 &= 330000r\\ \frac{21450}{330000} &= \frac{330000r}{330000}\\ .065 &= r\end{align*} Because we are looking for a percent-an interest rate, we have to change the decimal to a percent. .065 = 6.5% The bank was paying 6.5%. Example A bank uses the same formula I=prt\begin{align*}I = prt\end{align*} to calculate the interest that you owe on a credit card. Normally, the interest rates are substantially higher than for a savings account. If you charge7,000 on a credit card and you bank charges you 15.9%, how much would you owe after a year?

First, look at the given information and then use the formula.

pIII=7000,r=.159,t=1=prt=7000×.159×1=1113\begin{align*}p &= 7000, r = .159, t = 1\\ I &= prt\\ I &= 7000 \times .159 \times 1\\ I &= 1113\end{align*}

Next, we add the interest to the balance. Because you owe on a credit card, this gives us the new balance that you need to pay.

1113+7000=8113\begin{align*}1113 + 7000 = 8113\end{align*}

You would owe $8,113. Now let’s go back to the dilemma in the introduction and apply what we have learned to this problem. ## Real-Life Example Completed Student Council Investment Here is the original problem once again. First, reread it. Then figure out the amount of the interest and the total amount in the student council bank account. There are two parts to your answer. With a goal of boosting student attendance at football games, the student council has decided to invest a portion of their savings for middle school decorations. They figure that when the games are happening that they can decorate the middle school with balloons, banners and flyers. “I think it will help make it a priority for students,” Jeremy said at the weekly student council meeting. “It will be a lot of fun too. We could even host a pep rally to help get the kids charged up,” Candice suggested. “We put$4000 in the bank in the sixth grade. Now that we are 8th\begin{align*}8^{th}\end{align*} graders that money has been sitting in the bank for two years at a 4% interest rate,” Jeremy explained.

Candice began working out the math in her head. If they did put $4000 in the bank for two years and they had a 4% interest rate, then there definitely is more money in there now. She began to complete the calculations in her head. Remember, there are two parts to your answer. Solution to Real – Life Example Now we need to figure out the interest and the final balance in the student council bank account. First, let’s find the amount of the interest. III=PRT=(4000)(.04)(2)=$320.00\begin{align*}I &= PRT\\ I &= (4000)(.04)(2)\\ I &= \320.00\end{align*}

Next, we add this to the original amount invested.

$4000+$320=4320.00\begin{align*}\4000 + \320 = \4320.00\end{align*} This is the new balance in the student council account. ## Time to Practice Directions: Use the simple interest formula I=prt\begin{align*}I = prt\end{align*} to solve for the Interest. 1. Find I\begin{align*}I\end{align*} if p=62,300,r=.0525,t=14\begin{align*}p = 62,300, r = .0525, t = 14\end{align*}. 2. Find I\begin{align*}I\end{align*} if p=9800,r=.028,t=9\begin{align*}p = 9800, r = .028, t = 9\end{align*}. 3. Find I\begin{align*}I\end{align*} if p=600,r=.05,t=8\begin{align*}p = \600, r = .05, t=8\end{align*}
4. Find I\begin{align*}I\end{align*} if p=2300,r=.06,t=12\begin{align*}p = \2300, r = .06, t=12\end{align*} 5. Find I\begin{align*}I\end{align*} if \begin{align*}p = \5500, r = .08, t=7\end{align*} 6. Find \begin{align*}I\end{align*} if \begin{align*}p = \400, r = .05\end{align*} and \begin{align*}t=5\end{align*} 7. Find \begin{align*}I\end{align*} if \begin{align*}p = \700, r = .03\end{align*} and \begin{align*}t=9\end{align*} 8. Find \begin{align*}I\end{align*} if \begin{align*}p = \500, r = .06\end{align*} and \begin{align*}t=12\end{align*} 9. Find \begin{align*}I\end{align*} if \begin{align*}p = \800, r = .09\end{align*} and \begin{align*}t=7\end{align*} 10. Find \begin{align*}I\end{align*} if \begin{align*}p = \950, r = .06\end{align*} and \begin{align*}t=4\end{align*} Directions: Find the new interest and then find the new balance with the given information. There are two steps to solving these problems. 1. \begin{align*}p = 43000, r = .0365, t = 11\end{align*} 2. \begin{align*}p = 7000, r = .079, t = 4\end{align*} 3. \begin{align*}p = 8000, r = .06, t = 3\end{align*} 4. \begin{align*}p = 18000, r = .04, t = 5\end{align*} 5. \begin{align*}p = 25000, r = .05, t = 3\end{align*} 6. \begin{align*}p = 3000, r = .05, t = 7\end{align*} 7. \begin{align*}p = 12000, r = .04, t = 5\end{align*} 8. \begin{align*}p = 9000, r = .06, t = 10\end{align*} 9. \begin{align*}p = 7500, r = .03, t = 8\end{align*} 10. \begin{align*}p = 27500, r = .04, t = 6\end{align*} Directions: Solve each problem. 1. Antonio receives a savings bond from his uncle for his birthday. It will be worth1,000 in \begin{align*}8\frac{1}{2}\end{align*} years but only cost his uncle $500. What was the interest rate paid on the bond? 2. Henrietta is thinking of investing a$10,000 inheritance in a savings account for 6 years that pays 4.3% interest. She owes $7,200 on a credit card which is charging her 11.9% interest. In how much time would the interest on the credit card be the same as the interest in the savings account after 6 years? 3. A mother wants to save money for her child’s education. Her child just turned 1 year old. She wants to have at least$80,000 in the bank by the time she is 18. She has placed \$35,000 in the bank at 6.5%. Will she reach her goal?

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