### Let’s Think About It

Nisi entered a local triathlon and won the grand prize of $4000. She decides to put all $4000 in the bank and leave it untouched for two years until she finishes high school. The bank offered her a 4% interest rate. After the two years is up, how much money will Nisi have in the bank?

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

### Guidance

Saving money and making wise investments will be an important part of your financial planning. Part of 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. **Interest** is the percent that a bank pays you for keeping your money in their bank.

All banks have an interest rate \begin{align*}r\end{align*}

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

Now let’s look at an example using this formula to calculate interest.

You invest $5,000 in a bank for 2 years at a 3% interest rate. What is the interest you have earned after this time?

First, using the formula for simple interest, fill in all of the numbers you know.

\begin{align*}\begin{array}{rcl}
p &=& \$ 5000 \\
r &=& 3 \% \ \text{or} \ 0.03 \\
t &=& 2 \ \text{years}
\end{array}\end{align*}

\begin{align*}\begin{array}{rcl} I &=& PRT \\ I &=& (5000)(0.03)(2) \end{array}\end{align*}

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

\begin{align*}\begin{array}{rcl} I &=& (5000)(0.03)(2) \\ I &=& 300 \end{array}\end{align*}

The answer is 300.

You have earned $300 in interest in the two years.

Let’s look at another example.

Mrs. Payne 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?

First, using the formula for simple interest, fill in all of the numbers you know.

\begin{align*}\begin{array}{rcl}
I &=& 10000 \\
p &=& \$ 20000 \\
r &=& 5.6 \% \ \text{or} \ 0.056
\end{array}\end{align*}

\begin{align*}\begin{array}{rcl} I &=& PRT \\ 10000 &=& (20000)(0.056)(t) \\ 10000 &=& 1120 \ t \end{array}\end{align*}

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

\begin{align*}\begin{array}{rcl} 10000 &=& \ 1120 \ t \\ \frac{10000}{1120} &=& \frac{1120 \ t}{1120} \\ t &=& 8.93 \end{array}\end{align*}

The answer is 8.93.

Mrs. Payne will have to leave her money in the bank for almost 9 years.

You can use the simple interest formula \begin{align*}(I = PRT)\end{align*}

Let’s try another example.

Jessica invests $3,000 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?

First, using the formula for simple interest, fill in all of the numbers you know.

\begin{align*}\begin{array}{rcl}
p &=& \$ 3000 \\
r &=& 3.9 \% \ \text{or} \ 0.039 \\
t &=& 5 \ \text{years}
\end{array}\end{align*}

\begin{align*}\begin{array}{rcl} I &=& PRT \\ I &=& (3000)(0.039)(5) \end{array}\end{align*}

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

\begin{align*}\begin{array}{rcl} I &=& (3000)(0.039)(5) \\ I &=& 585 \end{array}\end{align*}

Then, to find the bank balance, add the interest earned to the amount invested.

\begin{align*}\begin{array}{rcl} \text{balance} &=& 3000 + 585 \\ &=& 3585 \end{array}\end{align*}

The answer is 3585.

Jessica’s balance will be $3585 in the five years.

### Guided Practice

A nurse put $22,000 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?

First, using the formula for simple interest, fill in all of the numbers you know.

\begin{align*}\begin{array}{rcl} I &=& \$ 21450 \\ p &=& \$ 22000 \\ t &=& 15 \ \text{years} \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} I &=& PRT \\ 21450 &=& (22000)(r)(15) \\ 21450 &=& 330000 \ r \end{array}\end{align*}

Next, solve for \begin{align*}r\end{align*} by dividing by 330000.

\begin{align*}\begin{array}{rcl} 21450 &=& \ 330000 \ r \\ \frac{21450}{330000} &=& \frac{330000 \ r}{330000} \\ r &=& 0.065 \end{array}\end{align*}

Then, multiply the value of \begin{align*}r\end{align*} by 100 since interest rates are reported as percentages.

\begin{align*}\begin{array}{rcl} r &=& 0.065 \times 100 \\ r &=& 65 \% \end{array}\end{align*}

The answer is 6.5%.

The nurse invested at 6.5% interest.

### Examples

#### Example 1

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?

First, using the formula for simple interest, fill in all of the numbers you know.

\begin{align*}\begin{array}{rcl} p &=& \$ 15000 \\ r &=& 4.5 \% \ \text{or} \ 0.045 \\ t &=& 6 \ \text{years} \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} I &=& PRT \\ I &=& (15000)(0.045)(6) \end{array}\end{align*}

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

\begin{align*}\begin{array}{rcl} I &=& (15000)(0.045)(6) \\ I &=& 4050 \end{array}\end{align*}

The answer is 4050.

The investor earned $4050 in interest in the six years.

#### Example 2

A bank is offering an interest rate of 4.75%. How long would it take to earn $500 if you invested $12,000 in the bank?

First, using the formula for simple interest, fill in all of the numbers you know.

\begin{align*}\begin{array}{rcl} I &=& \$ 500 \\ p &=& \$ 12000 \\ r &=& 4.75 \% \ \text{or} \ 0.0475 \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} I &=& PRT \\ 500 &=& (12000)(0.0475)(t) \\ 500 &=& 570 \ t \end{array}\end{align*}

Next, solve for \begin{align*}t\end{align*} by dividing both sides by 570.

\begin{align*}\begin{array}{rcl} 500 &=& \ 570 \ t \\ \frac{500}{570} &=& \frac{570 \ t}{570} \\ t &=& 0.877 \end{array}\end{align*}

The answer is 0.88.

It would take 0.88 years or 10.6 months to gain $500 in interest.

#### Example 3

If you charge $7,000 on a credit card and you bank charges you 15.9%, how much would you owe after a year?

First, using the formula for simple interest, fill in all of the numbers you know.

\begin{align*}\begin{array}{rcl} p &=& \$ 7000 \\ r &=& 15.9 \% \ \text{or} \ 0.159 \\ t &=& 1 \ \text{year} \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} I &=& PRT \\ I &=& (7000)(0.159)(1) \end{array}\end{align*}

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

\begin{align*}\begin{array}{rcl} I &=& (7000)(0.159)(1) \\ I &=& 1113 \end{array}\end{align*}

Then, to find the total owing after one year, add the interest earned to the amount borrowed.

\begin{align*}\begin{array}{rcl} \text{amount owing} &=& 7000 + 1113 \\ &=& 8113 \end{array}\end{align*}

The answer is 8113.

You would owe $8113 after one year.

### Follow Up

Nisi left her $4000 prize money in the bank for two years at 4% interest. She needs to figure out how much money she now has.

First, using the formula for simple interest, fill in all of the numbers you know.

\begin{align*}\begin{array}{rcl} p &=& \$ 4000 \\ r &=& 4.0 \% \ \text{or} \ 0.04 \\ t &=& 2 \ \text{years} \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} I &=& PRT \\ I &=& (4000)(0.09)(2) \end{array}\end{align*}

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

\begin{align*}\begin{array}{rcl} I &=& (4000)(0.04)(2) \\ I &=& 320 \end{array}\end{align*}

Then, to find the total amount in the bank, add the interest earned to the original amount.

\begin{align*}\begin{array}{rcl} \text{balance} &=& 4000 + 320 \\ &=& 4320 \end{array}\end{align*}

The answer is 4320. After two years, Nisi will have $4320.

### Video Review

https://www.youtube.com/watch?v=r3-lyBGlJ98&safe=active

### Explore More

Use the simple interest formula \begin{align*}I = PRT\end{align*} to solve for the Interest.

1. Find \begin{align*}I\end{align*} if \begin{align*}p = 62, 300, \ r = 0.0525, \ t = 14\end{align*}.

2. Find \begin{align*}I\end{align*} if \begin{align*}p = 9800, \ r = 0.028, \ t = 9\end{align*}.

3. Find \begin{align*}I\end{align*} if \begin{align*}p = $600, \ r = 0.05, \ t = 8\end{align*}

4. Find \begin{align*}I\end{align*} if \begin{align*}p = $2300, \ r = 0.06, \ t = 12\end{align*}

5. Find \begin{align*}I\end{align*} if \begin{align*}p = $5500, \ r = 0.08, \ t = 7\end{align*}

6. Find \begin{align*}I\end{align*} if \begin{align*} p = $400, \ r = 0.05\end{align*}, \begin{align*}t = 57\end{align*}.

7. Find \begin{align*}I\end{align*} if \begin{align*}p = $700, \ r = 0.03\end{align*}, \begin{align*}t = 9\end{align*}

8. Find \begin{align*}I\end{align*} if \begin{align*}p = $500, \ r = 0.06\end{align*}, \begin{align*}t = 12\end{align*}

9. Find \begin{align*}I\end{align*} if \begin{align*}p = $800, \ r = 0.09\end{align*}, \begin{align*}t = 7\end{align*}

10. Find \begin{align*}I\end{align*} if \begin{align*}p = $950, \ r = 0.06\end{align*}, \begin{align*}t = 4\end{align*}

Find the new interest and then find the new balance with the given information. There are two steps to solving these problems.

11. \begin{align*}p = 43000, \ r = 0.0365, \ t = 11\end{align*}

12. \begin{align*}p = 7000, \ r = 0.079, \ t = 4\end{align*}

13. \begin{align*}p = 8000, \ r = 0.06, \ t = 3\end{align*}

14. \begin{align*}p = 18000, \ r = 0.04, \ t = 5\end{align*}

15. \begin{align*}p = 25000, \ r = 0.05, \ t = 3\end{align*}

16. \begin{align*}p = 3000, \ r = 0.05, \ t = 7\end{align*}

17. \begin{align*}p = 12000, \ r = 0.04, \ t = 5\end{align*}

18. \begin{align*}p = 9000, \ r = 0.06, \ t = 10\end{align*}

19. \begin{align*}p = 7500, \ r = 0.03, \ t = 8\end{align*}

20. \begin{align*}p = 27500, \ r = 0.04, \ t = 6\end{align*}