SOLUTION: Compute the accumulated values at the end of 10 years if P100 is invested at the rate of 12% per year compounded
(a.) annually
(b.) semi annually
(c.) quarterly
(d.) bi-mont
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(a.) annually
(b.) semi annually
(c.) quarterly
(d.) bi-mont
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Question 1118760: Compute the accumulated values at the end of 10 years if P100 is invested at the rate of 12% per year compounded
(a.) annually
(b.) semi annually
(c.) quarterly
(d.) bi-monthly
(e.) daily
You can put this solution on YOUR website! your annual interest rate is 12% per year.
divide that by 100 to get an annual interest rate of .12 per year.
i'll assume bi-monthly equals once every 2 months.
the formula is f = p * (1 + r) ^ n
f is the future value
p is the present value
r is the interest rate per time period
n is the number of time periods.
in your problem:
p = 100
r = .12 per year
n = 10 years
solution to your problem for each of the scenarios is shown below.
Compute the accumulated values at the end of 10 years if P100 is invested at the rate of 12% per year compounded ...
(a.) annually
number of compounding periods per year = 1
interest rate per time period = .12 / 1 = .12
number of time periods = 10 * 1 = 10
f = p * (1 + r) ^ n becomes f = 100 * (1 + .12) ^ 10 which becomes f = 310.5848208.
(b.) semi annually
number of compounding periods per year = 2
interest rate per time period = .12 / 2 = .06
number of time periods = 10 * 2 = 20
f = p * (1 + r) ^ n becomes f = 100 * (1 + .06) ^ 20 which becomes f = 320.7135472
(c.) quarterly
number of compounding periods per year = 4
interest rate per time period = .12 / 4 = .03
number of time periods = 10 * 4 = 40
f = p * (1 + r) ^ n becomes f = 100 * (1 + .03) ^ 40 which becomes f = 326.2037792
(d.) bi-monthly
bi-monthly is assumed to mean every 2 months.
number of compounding periods per year = 6
interest rate per time period = .12 / 6 = .02
number of time periods = 10 * 6 = 60
f = p * (1 + r) ^ n becomes f = 100 * (1 + .02) ^ 60 which becomes f = 328.1030788
(e.) daily
365 days in a year is assumed.
number of compounding periods per year = 365
interest rate per time period = .12 / 365 = 3.287671233 * 10^-4
number of time periods = 10 * 365 = 3650
f = p * (1 + r) ^ n becomes f = 100 * (1 + 3.287671233 * 10^-4) ^ 3650 which becomes f = 331.9462204
