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# 5.17: Solve Real World Problems Involving Simple Interest

Difficulty Level: Basic Created by: CK-12
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Practice Simple Interest

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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.

### Interest

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 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 borrow or save. In order to calculate how much interest you will make, you can use the formula:

I=PRT\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. prt===$50003% or 0.032 years\begin{align*}\begin{array}{rcl} p &=& \ 5000 \\ r &=& 3 \% \ \text{or} \ 0.03 \\ t &=& 2 \ \text{years} \end{array}\end{align*}

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

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

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

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.

Ipr===10000200005.6% or 0.056\begin{align*}\begin{array}{rcl} I &=& 10000 \\ p &=& \ 20000 \\ r &=& 5.6 \% \ \text{or} \ 0.056 \end{array}\end{align*} I1000010000===PRT(20000)(0.056)(t)1120 t\begin{align*}\begin{array}{rcl} I &=& PRT \\ 10000 &=& (20000)(0.056)(t) \\ 10000 &=& 1120 \ t \end{array}\end{align*} Next, solve for t\begin{align*}t\end{align*} by dividing both sides by 1120. 10000100001120t=== 1120 t1120 t11208.93\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 (I=PRT)\begin{align*}(I = PRT)\end{align*} to find any of the missing variables if you are given values of the others. You have just used it to solve for I\begin{align*}I\end{align*} and t\begin{align*}t\end{align*}. Let’s try another 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?

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

prt===30003.9% or 0.0395 years\begin{align*}\begin{array}{rcl} p &=& \ 3000 \\ r &=& 3.9 \% \ \text{or} \ 0.039 \\ t &=& 5 \ \text{years} \end{array}\end{align*} II==PRT(3000)(0.039)(5)\begin{align*}\begin{array}{rcl} I &=& PRT \\ I &=& (3000)(0.039)(5) \end{array}\end{align*} Next, solve for I\begin{align*}I\end{align*} II==(3000)(0.039)(5)585\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. balance==3000+5853585\begin{align*}\begin{array}{rcl} \text{balance} &=& 3000 + 585 \\ &=& 3585 \end{array}\end{align*} The answer is 3585. Jessica’s balance will be3585 in the five years.

### Examples

#### Example 1

Earlier, you were given a problem about Nisi putting in $4000 in the bank. 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.

prt===40004.0% or 0.042 years\begin{align*}\begin{array}{rcl} p &=& \ 4000 \\ r &=& 4.0 \% \ \text{or} \ 0.04 \\ t &=& 2 \ \text{years} \end{array}\end{align*} II==PRT(4000)(0.09)(2)\begin{align*}\begin{array}{rcl} I &=& PRT \\ I &=& (4000)(0.09)(2) \end{array}\end{align*} Next, solve for I\begin{align*}I\end{align*}. II==(4000)(0.04)(2)320\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. balance==4000+3204320\begin{align*}\begin{array}{rcl} \text{balance} &=& 4000 + 320 \\ &=& 4320 \end{array}\end{align*} The answer is 4320. After two years, Nisi will have4320.

#### Example 2

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.

Ipt===$21450$2200015 years\begin{align*}\begin{array}{rcl} I &=& \ 21450 \\ p &=& \ 22000 \\ t &=& 15 \ \text{years} \end{array}\end{align*}

I2145021450===PRT(22000)(r)(15)330000 r\begin{align*}\begin{array}{rcl} I &=& PRT \\ 21450 &=& (22000)(r)(15) \\ 21450 &=& 330000 \ r \end{array}\end{align*}

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

2145021450330000r=== 330000 r330000 r3300000.065\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 r\begin{align*}r\end{align*} by 100 since interest rates are reported as percentages.

rr==0.065×10065%\begin{align*}\begin{array}{rcl} r &=& 0.065 \times 100 \\ r &=& 65 \% \end{array}\end{align*}

The nurse invested at 6.5% interest.

#### Example 3

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. prt===$150004.5% or 0.0456 years\begin{align*}\begin{array}{rcl} p &=& \ 15000 \\ r &=& 4.5 \% \ \text{or} \ 0.045 \\ t &=& 6 \ \text{years} \end{array}\end{align*}

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

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

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

#### Example 5

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 owe8113 after one year.

### Review

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*}

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### Vocabulary Language: English

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 $t$.

Interest

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

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.

Rate

The rate is the percentage at which interest accrues.

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.

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