# 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

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

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%

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.
$$2.6^{\frac{1}{12}}=((1+\frac{r}{1})^{12})^{\frac{1}{12}} [latex]1.08288=1+r$$

Subtract 1 from both sides.
$$r=0.08288$$ (as a decimal)
$$r=8.288\%$$ (as a percent)