QuestionQuestion 4 $\$ 4000$ are invested in a bank account at an interest rate of 5 percent per year. Find the amount in the bank after 11 years if interest is compounded annually. $\square$ Find the amount in the bank after 11 years if interest is compounded quarterly. $\square$ Find the amount in the bank after 11 years if interest is compounded monthly. $\square$ Finally, find the amount in the bank after 11 years if interest is compounded continuously. $\square$

Studdy Solution

STEP 1

1. The principal amount invested is $4000.
2. The annual interest rate is 5%.
3. The time period for the investment is 11 years.
4. We need to calculate the future value of the investment for different compounding frequencies: annually, quarterly, monthly, and continuously.

STEP 2

1. Calculate the amount with annual compounding.
2. Calculate the amount with quarterly compounding.
3. Calculate the amount with monthly compounding.
4. Calculate the amount with continuous compounding.

STEP 3

Calculate the amount with annual compounding.
The formula for compound interest is given by:
$A = P \left(1 + \frac{r}{n}\right)^{nt}$
Where: - $A$ is the amount of money accumulated after n years, including interest. - $P$ is the principal amount ($4000). - \( r \) is the annual interest rate (decimal) (0.05). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for in years (11).
For annual compounding, $n = 1$ .
$A = 4000 \left(1 + \frac{0.05}{1}\right)^{1 \times 11}$ $A = 4000 \left(1.05\right)^{11}$
Calculate $A$ :
$A \approx 4000 \times 1.7137$ $A \approx 6854.80$
The amount in the bank after 11 years with annual compounding is approximately \$6854.80.

STEP 4

Calculate the amount with quarterly compounding.
For quarterly compounding, $n = 4$ .
$A = 4000 \left(1 + \frac{0.05}{4}\right)^{4 \times 11}$ $A = 4000 \left(1.0125\right)^{44}$
Calculate $A$ :
$A \approx 4000 \times 1.7137$ $A \approx 6921.59$
The amount in the bank after 11 years with quarterly compounding is approximately \$6921.59.

STEP 5

Calculate the amount with monthly compounding.
For monthly compounding, $n = 12$ .
$A = 4000 \left(1 + \frac{0.05}{12}\right)^{12 \times 11}$ $A = 4000 \left(1.0041667\right)^{132}$
Calculate $A$ :
$A \approx 4000 \times 1.7161$ $A \approx 6940.95$
The amount in the bank after 11 years with monthly compounding is approximately \$6940.95.

STEP 6

Calculate the amount with continuous compounding.
The formula for continuous compounding is given by:
$A = Pe^{rt}$
Where: - $e$ is the base of the natural logarithm. - $r$ is the annual interest rate (decimal) (0.05). - $t$ is the time the money is invested for in years (11).
$A = 4000 \times e^{0.05 \times 11}$
Calculate $A$ :
$A \approx 4000 \times e^{0.55}$ $A \approx 4000 \times 1.7333$ $A \approx 6933.20$
The amount in the bank after 11 years with continuous compounding is approximately \$6933.20.

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