# 5.18: Solve Real World Problems Involving Compound Interest

**At Grade**Created by: CK-12

**Practice**Simple and Compound Interest

James is investing $15,000 in the bank. The investment has an interest rate of 6% compounded monthly. After ten years, how much will James have made?

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

### Compound Interest

Interest is important. Understanding interest helps you to make the best decisions. In most cases, in the real world interest is calculated with the **compound interest** formula. The compound interest formula is:

\begin{align*}A = P (1 + \frac{r}{n})^{nt}\end{align*}

Where \begin{align*}A\end{align*} is the amount, \begin{align*}P\end{align*} is the principal amount, \begin{align*}r\end{align*} is the interest rate, \begin{align*}n\end{align*} is the number of times per year the interest is compounded, and \begin{align*}t\end{align*} is the number of years.

Basically, this formula accounts for the fact that as you invest and earn interest, your balance grows. You are not only due interest, then, on your original balance, but on the new balance which includes the first installment(s) of interest. You get paid interest on the interest.

Let’s look at an example.

You invest $100 for 3 years at 10% interest compounded yearly. How much will you have at the end of the three years?

First, write down what you know.

\begin{align*}\begin{array}{rcl} A&=& ? \\ P&=& 100\\ r&=& 10\% = 0.10 \\ n&=& 1 \\ t&=& 3 \\ \end{array}\end{align*}

Next, fill in what you know into the compound interest formula.

\begin{align*}\begin{array}{rcl} A&=& P \left(1+ \frac{r}{n}\right)^{nt}\\ A&=& 100\left(1+ \frac{0.10}{1}\right)^{1\times 3} \\ A&=& 100(1+0.10) ^ 3 \\ \end{array}\end{align*}

Then, solve for your unknown.

\begin{align*}\begin{array}{rcl} A&=& 100(1+0.10)^ 3 \\ A&=& 100(1.10) ^ 3 \\ A &=& 133.10 \end{array}\end{align*}

The answer is 133.10.

In three years, your $100 has grown to $133.10.

Let’s see how different it might be using the different interest formulas for the same amount of time at the same rate.

You want to invest $20,000 for the next 20 years. You have two options for your investment. Bank X offers simple interest at a rate of 8%. Bank Y uses compound interest at a rate of 8% compounded yearly. Which should you choose?

First, start with the simple interest formula \begin{align*}(I = PRT)\end{align*}. Write down what you know.

\begin{align*}\begin{array}{rcl} I&=& ? \\ P&=& 20000\\ r&=& 8\% = 0.08 \\ t&=& 20 \ \text{years} \\ \end{array}\end{align*}

Next, fill in what you know into the simple interest formula and solve for \begin{align*}I\end{align*}.

\begin{align*}\begin{array}{rcl} I&=& PRT \\ I&=& (20000)(0.08)(20) \\ I&=& 32000 \end{array}\end{align*}

Then, add this interest to your original investment.

\begin{align*}\begin{array}{rcl} \text{balance} &=& 20000 + 32000 \\ \text{balance} &=& 52000 \end{array}\end{align*}

The answer is 52000.

Using simple interest, you would have $52,000 in 20 years.

Now let’s use the compound interest formula.

First, write down what you know.

\begin{align*}\begin{array}{rcl} A&=& ? \\ P&=& 20000\\ r&=& 8\% = 0.08 \\ n&=& 1 \\ t&=& 20 \\ \end{array}\end{align*}

Next, fill in what you know into the compound interest formula and solve for \begin{align*}A\end{align*}.

\begin{align*}\begin{array}{rcl} A&=& P \left(1+ \frac{r}{n}\right)^{nt}\\ A&=& 20000\left(1+ \frac{0.08}{1}\right)^{1\times 20} \\ A&=& 20000(1+0.08) ^ {20} \\ A&=& 93219.14 \end{array}\end{align*}

The answer is 93219.14.

Using compound interest, you would have $93,219.14 in 20 years.

You should choose Bank Y to put your money into an investment.

### Examples

#### Example 1

Earlier, you were given a problem about James and his investment.

James is investing $15,000 for ten years at an interest rate of 6% compounded monthly.

First, write down what you know.

\begin{align*}\begin{array}{rcl} A&=& ? \\ P&=& 15000\\ r&=& 6\% = 0.06 \\ n&=& 12 \\ t&=& 10 \\ \end{array}\end{align*}

Next, fill in what you know into the compound interest formula and solve for \begin{align*}A\end{align*}.

\begin{align*}\begin{array}{rcl} A&=& P\left(1+ \frac{r}{n}\right)^{nt}\\ A&=& 15000\left(1+ \frac{0.06}{12}\right)^{12\times 10} \\ A&=& 15000(1.005) ^ {120} \\ A&=& 15000(1.819397) \\ A&=& 27290.95 \\ \end{array}\end{align*}

The answer is 27290.95.

James will have $27290.95 in 10 years.

#### Example 2

A fireman invests $40,000 in a retirement account for 2 years. The interest rate is 6%. The interest is compounded monthly. What will his final balance be?

First, write down what you know.

\begin{align*}\begin{array}{rcl} A&=& ? \\ P&=& 40000\\ r&=& 6\% = 0.06 \\ n&=& 12 \\ t&=& 2 \\ \end{array}\end{align*}

Next, fill in what you know into the compound interest formula and solve for \begin{align*}A\end{align*}.

\begin{align*}\begin{array}{rcl} A&=& P\left(1+ \frac{r}{n}\right)^{nt}\\ A&=& 40000\left(1+ \frac{0.06}{12}\right)^{12\times 2} \\ A&=& 40000(1+0.005) ^ {24} \\ A &=& 40000(1.12716) \\ A&=& 45086.39 \end{array}\end{align*}

The answer is 45086.39.

Using compound interest, the fireman would have $45,086.39 in 2 years.

#### Example 3

Calculate the amount of this investment after 5 years with interest compounded yearly.

Principal = $3000

Rate = 4%

First, write down what you know.

\begin{align*}\begin{array}{rcl} A&=& ? \\ P&=& 3000\\ r&=& 4\% = 0.04 \\ n&=& 1 \\ t&=& 5 \\ \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} A&=& P\left(1+ \frac{r}{n}\right)^{nt}\\ A&=& 3000\left(1+ \frac{0.04}{1}\right)^{1\times 5} \\ A&=& 3000(1.04) ^ {5} \\ A&=& 3649.96 \end{array}\end{align*}

The answer is 3649.96.

Therefore using compound interest, you would have $3649.96 in 5 years.

#### Example 4

Calculate the amount of this investment after 5 years with interest compounded every two months.

Principal = $5000

Rate = 3%

First, write down what you know.

\begin{align*}\begin{array}{rcl} A&=& ? \\ P&=& 5000\\ r&=& 3\% = 0.03 \\ n&=& 6 \\ t&=& 5 \\ \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} A&=& P\left(1+ \frac{r}{n}\right)^{nt}\\ A&=& 5000\left(1+ \frac{0.03}{6}\right)^{6\times 5} \\ A&=& 5000(1.005) ^ {30} \\ A&=& 5000(1.1614) \\ A&=& 5807.00 \end{array}\end{align*}

The answer is 5807.00.

Using compound interest, you would have $5807.00 in 5 years.

#### Example 5

Calculate the amount of this investment after 5 years with interest compounded quarterly.

Principal = $12,000

Rate = 9%

First, write down what you know. Note that quarterly is every 3 months (4 times a year).

\begin{align*}\begin{array}{rcl} A&=& ? \\ P&=& 12000\\ r&=& 9\% = 0.09 \\ n&=& 4 \\ t&=& 5 \\ \end{array}\end{align*}

\begin{align*}\begin{array}{rcl} A&=& P\left(1+ \frac{r}{n}\right)^{nt}\\ A&=& 12000\left(1+ \frac{0.09}{4}\right)^{4\times 5} \\ A&=& 12000(1.0225) ^ {20} \\ A&=& 12000(1.5605) \\ A&=& 18726.11 \\ \end{array}\end{align*}

The answer is 18726.11.

Using compound interest, you would have $18726.11 in 5 years.

### Review

Calculate the simple interest by using I =PRT.

1. Principal = $2000, Rate = 5%, Time = 3 years

2. Principal = $12,000, Rate = 4%, Time = 2 years

3. Principal = $10,000, Rate = 5%, Time = 5 years

4. Principal = $30,000, Rate = 2.5%, Time = 10 years

5. Principal = $12,500, Rate = 3%, Time = 8 years

6. Principal = $34,500, Rate = 4%, Time = 10 years

7. Principal = $16,000, Rate = 3%, Time = 5 years

8. Principal = $120,000, Rate = 5%, Time = 4 years

Calculate the following compound interest calculated yearly.

9. Principal = $3000, Rate = 4%

10. Principal = $5000, Rate = 3%

11. Principal = $12,000, Rate = 2%

12. Principal = $34,000, Rate = 5%

13. Principal = $18,000, Rate = 3%

14. Principal = $7800, Rate = 4%

15. Principal = $8500, Rate = 3%

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 5.18.

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

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