Compound Interest Examples

Example 1: Jose's Loan

• Jose takes a loan of $10,000$ to attend NYU.

• The loan has an interest rate of $6.75\%$ compounded quarterly.

• Jose makes no payments for five years.

• We need to calculate how much Jose will owe after five years.

To solve this, we use the compound interest formula:

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

Where:

• $A$ = the future value of the investment/loan, including interest

• $P$ = the principal investment amount (the initial deposit or loan amount)

• $r$ = the annual interest rate (as a decimal)

• $n$ = the number of times that interest is compounded per year

• $t$ = the number of years the money is invested or borrowed for

In this case:

• $P = 10000$

• $r = 0.0675$

• $n = 4$ (compounded quarterly)

• $t = 5$

Plugging the values into the formula:

$A = 10000(1 + \frac{0.0675}{4})^{(4)(5)}$
$A = 10000(1 + 0.016875)^{20}$
$A = 10000(1.016875)^{20}$
$A \approx 10000(1.39449)$
$A \approx 13944.90$

Therefore, Jose will owe approximately $13,944.90$ after five years.

Example 2: Monroe's Credit Card Debt

• Monroe opens a credit card to buy a couch.

• The credit card has an interest rate of $13.14\%$ compounded monthly.

• Monroe loses his job and cannot make payments for two years.

• After two years, he owes $10,417.65$.

• We need to find the original price of the couch.

We use the compound interest formula again, but solve for $P$:

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

Where:

• $A = 10417.65$

• $r = 0.1314$

• $n = 12$ (compounded monthly)

• $t = 2$

Rearranging the formula to solve for $P$:

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

Plugging in the values:

$P = \frac{10417.65}{(1 + \frac{0.1314}{12})^{(12)(2)}}$
$P = \frac{10417.65}{(1 + 0.01095)^{24}}$
$P = \frac{10417.65}{(1.01095)^{24}}$
$P \approx \frac{10417.65}{1.29003}$
$P \approx 8075.51$

Therefore, the original price of the couch was approximately $8,075.51$.

Example 3: Jessica's Savings Account

• Jessica opens a savings account with an interest rate of $0.21\%$ compounded continuously.

• She deposits $5000$.

• We need to calculate how much will be in the account after 10 years if she makes no withdrawals.

For continuous compounding, we use the formula:

$A = Pe^{rt}$

Where:

• $A$ = the future value of the investment

• $P$ = the principal investment amount

• $e$ = Euler's number (approximately 2.71828)

• $r$ = the annual interest rate (as a decimal)

• $t$ = the number of years the money is invested for

In this case:

• $P = 5000$

• $r = 0.0021$

• $t = 10$

Plugging in the values:

$A = 5000e^{(0.0021)(10)}$
$A = 5000e^{0.021}$
$A \approx 5000(1.02122)$
$A \approx 5106.10$

Therefore, there will be approximately $5,106.10$ in Jessica's account after 10 years.