Compound interest is interest that is calculated on both the money deposited and the interest earned from that deposit.

The formula for compound interest is \(A=P(1+\frac{r}{n})^{nt}\), where \(A\) represents the final balance after the interest has been calculated for the time, \(t\), in years, on a principal amount, \(P\), at an annual interest rate, \(r\). The number of times in the year that the interest is compounded is \(n\).

### Example 1:

Jasmine deposits $520 into a savings account that has a 3.5% interest rate compounded monthly. What will be the balance of Jasmine’s savings account after two years?

To find the balance after two years, \(A\), we need to use the formula, \(A=P(1+\frac{r}{n})^{nt}\). The principal, \(P\), in this situation is the amount Jasmine used to start her account, $520. The rate, \(r\), as stated in the problem, is 3.5% (or 0.35 as a decimal) and compounded monthly, so \(n=12\). Since we are looking for the balance of the account after two years, 2, is the time, \(t\).

\(A=520(1+\frac{0.035}{12})^{12(2)}\)

\(A=557.65\)

The balance of Jasmine’s account after 2 years is $557.65.

### Example 2:

Lex has $1,780.80 in his savings account that he opened 6 years ago. His account has an annual interest rate of 6.8% compounded annually. How much money did Lex use to open his savings account?

To find the principal, \(P\) we can use the same formula, \(A=P(1+\frac{r}{n})^{nt}\). We have the balance of the account, \(A\), after 6 years, which is $1,780.80. The interest rate, \(r\), is 6.8% (or 0.68 as a decimal) and is compounded annually, so \(n=1\). The time, \(t\), is 6, since we know he opened his account 6 years ago. Plug in the known values into the formula and solve for the missing variable, \(P\).

\(1{,}780.80=P(1+\frac{0.068}{1})^{1(6)}\)

\(1{,}780.80=1.484P\)

\(1{,}200=P\)

The principal amount Lex used to open his account 6 years ago is $1,200.

## Compound Interest Sample Questions

**Question #1:**

A teacher wants to invest $30,000 into an account that compounds annually. The interest rate at this bank is 1.8%. How much money will be in the account after 6 years?

$43,389.35

$35,389.35

$33,389.35

$37,389.35

**Answer:**

Use the compound interest formula to solve this problem.

\(A=P(1+\frac{r}{n})^{nt}\)

From here, simply plug in each value and simplify in order to isolate the variable \(A\).

\(A=30{,}000(1+\frac{0.018}{1})^{1×6}\)

\(A=30{,}000(1.018)^6\)

\(A=$33{,}389.35\)

**Question #2:**

An investment earns 3% each year and is compounded monthly. Calculate the total value after 6 years from an initial investment of $5,000.

$5,114.74

$4,984.74

$5,984.74

$2,984.74

**Answer:**

Once again, use the compound interest formula to solve this problem.

\(A=P(1+\frac{r}{n})^{nt}\)

From here, simply plug in each value and simplify in order to isolate the variable \(A\).

\(A=5,000(1+\frac{0.03}{12}^{12×6}\)

\(A=5,000(1.36)^{72}\)

\(A=$5{,}984.74\)

**Question #3:**

Kristen wants to have $2,000,000 for retirement in 45 years. She invests in a mutual fund and pays 8.5% each year, compounded quarterly. How much should she deposit into the mutual fund initially?

$47,421.08

$35,421.08

$43,421.08

$45,421.08

**Answer:**

Once again, use the compound interest formula to solve this problem.

\(A=P(1+\frac{r}{n}^{nt}\)

From here, simply plug in each value and simplify in order to isolate the variable \(P\).

\(2,000,000=P(1+\frac{0.085}{4})^{4×45}\)

\(2,000,000=P(1.02125)^{180}\)

\(P=$45{,}421.08\)

**Question #4:**

Sean invests $50,000 into an index annuity that averages 6.5% per year, compounded semi-annually. After 9 years how much will be in his account?

$88,918.29

$89,918.29

$68,918.29

$81,918.29

**Answer:**

Once again, use the compound interest formula to solve this problem.

\(A=P(1+\frac{r}{n})^{nt}\)

From here, simply plug in each value and simplify in order to isolate the variable \(A\).

\(A=50{,}000(1+\frac{0.065}{2})^{2×9}\)

\(A=50{,}000(1.0325)^{18}\)

\(A=$88{,}918.29\)

**Question #5:**

Calculate the interest rate for an account that started with $5,000 and now has $13,000 and has been compounded annually for the past 12 years.

9.288%

11.288%

3.288%

8.288%

**Answer:**

Once again, use the compound interest formula to solve this problem.

\(A=P(1+\frac{r}{n})^{nt}\)

From here, simply plug in each value and simplify in order to isolate the variable \(r\).

\(13{,}000=5{,}000(1+\frac{r}{1})^{1×12}\)

\(13{,}000=5{,}000(1+\frac{r}{1})^{12}\)

\(2.6=(1+\frac{r}{1})^{12}\)

To get rid of the exponent 12, find the 12th root of both sides. This means raising both sides of the equation to the power of 112.

\(2.6^{\frac{1}{12}}=((1+\frac{r}{1})^{12})^{\frac{1}{12}}\)

\(1.08288=1+r\)

Subtract 1 from both sides.

\(r=0.08288\) (as a decimal)

\(r=8.288\%\) (as a percent)