**Compound Interest Formula**

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*.

**Compound Interest Formula Sample Questions**

**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*.

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

\(A=557.65\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\) the balance of Jasmine’s account after 2 years

**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, $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*.

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

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

\(1{,}200=P\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\) the principal amount Lex used to open his account 6 years ago

**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}/latex]

[latex]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*.

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

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

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}}

Subtract 1 from both sides.

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

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