Question

If 20600 dollars is invested at an interest rate of 10 percent per year, find the value of the investment at the end of 5 years for the following compounding methods, to the nearest cent.

(a) Annual: $

(b) Semiannual: $

(c) Monthly: $

(d) Daily: $

Answer 1

Answer:

a) The value of the investment is $33176.506

b) The value of the investment is $33555.53

c) The value of the investment is $33893.36

d) The value of the investment is $33961.33

Step-by-step explanation:

This is a compound interest problem

Compound interest formula:

The compound interest formula is given by:

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

A: Amount of money(Balance)

P: Principal(Initial deposit)

r: interest rate(as a decimal value)

n: number of times that interest is compounded per unit t

t: time the money is invested or borrowed for.

In our problem, we have:

A: the value we want to find

P = 20600(the value invested)

r = 0.1

n: Will change for each letter

t = 5.

a) If the interest is compounded anually, n = 1. So.

$A = 20600(1 + \frac{0.1}{1})^{1*5}$

$A = 33176.506$

The value of the investment is $33176.506

b) If the interest is compounded semianually, it happens twice a year, which means n = 2. So:

$A = 20600(1 + \frac{0.1}{2})^{2*5}$

$A = 33555.23$

The value of the investment is $33555.53

c) If the interest is compounded monthly, it happens 12 times a year, which means n = 12. So:

$A = 20600(1 + \frac{0.1}{12})^{12*5}$

$A = 33893.36$

The value of the investment is $33893.36

d) If the interest is compounded monthly, it happens 365 times a year, which means n = 365. So:

$A = 20600(1 + \frac{0.1}{365})^{365*5}$

$A = 33961.33$

The value of the investment is $33961.33