If 32000 dollars is invested at an interest rate of 9 percent per year, find the value of the investment at the end of 5 years for the following compounding met

Question

Annual:

 

Semiannual:

 

Monthly:

 

Daily:

 

Continuously:

Answer 1

Annual: P(5) = 32000(1+0.09)^5 = $49235.97
 
Semiannual: P(5) = 32000(1+0.09/2)^(2*5) = $49695.02
 
Monthly: P(5) = 32000(1+0.09/12)^(12*5) = $50101.79
 
Daily: P(5) = 32000(1+0.09/365)^(365*5) = $50183.21
 
Continuously: P(5) = 32000e^(5*0.09) = $50185.99

Answer 2

I assume you're having trouble with your interest rate formula, which is Interest= principle x rate x time.

 

 

Here's a different (longer) way to work the problem.

 

Start by finding out how much 9% of $32,000 (per every 12 months) works out to be.

 

(multiply $32,000 by 0.09, or 9% = x)

 

Interest rate will be added to the total dollar amount. Be sure that you ONLY add this full amount to the annual figure.

 

You can divide X (the amount of interest in dollars) by 12 months to find the monthly interest payment.

 

You can divide X (the amount of interest in dollars) by 6 months to find the semiannual payment.

 

Take care when you reach 'daily.' You'll need to change months to days, which should give you a pretty small (think decimals) number. ;-)

Answer 3

32000(1.09)⁵                   =49235.96656
32000(1+.09/2)¹⁰            =49695.0215
32000(1+.09/12)⁶⁰          =50101.79286
32000(1+.09/365)^(5×365)=50183.20618  This is the only one I'm not sure of.
32000e^(.09×5)                   =50185.98994

Answer 4

A = P(1 + (r/n)^nt

 

Where P = principal amount (the initial amount you borrow or deposit)
r = annual rate of interest (as a decimal)
t = number of years the amount is deposited or borrowed for.
A = amount of money accumulated after n years, including interest.
n = number of times the interest is compounded per year 

 

Annual: n = 1

 

A = P(1 + (r/n)^nt = [($32,000)(1 + (0.09)/(1)]⁽¹⁾⁽⁵⁾ = [($32,000)(1 + 0.09)]⁵ = $49,235.97

 

Semiannual: n = 2

 

A = P(1 + (r/n)^nt = [($32,000)(1 + (0.09)/(2)]⁽²⁾⁽⁵⁾ = [($32,000)(1 + 0.045)]¹⁰ = $49,695.02

 

Monthly: n = 12

 

A = P(1 + (r/n)^nt = [($32,000)(1 + (0.09)/(12)]^(12)(5) = $50,101.79

 

Daily: n = 365 days (unless it's a leap year)

 

A = P(1 + (r/n)^nt = [($32,000)(1 + (0.09)/(365)]^(365)(5) = [($32,000)(1 + 2.466 x 10^-4)]¹⁸²⁵

A = $50,183.21

 

Continuous:

 

A = Pe^rt = ($32,000)e^(0.09)(5) = ($32,000)e^0.45 = $50,185.99