SOLUTION: Samantha takes $28000.00 to the bank to invest into a term deposit.the bank gives her three option of interest rate with different compounding. option 1

SOLUTION: Samantha takes $28000.00 to the bank to invest into a term deposit.the bank gives her three option of interest rate with different compounding. option 1 - simple interest type @

Question 1042015: Samantha takes $28000.00 to the bank to invest into a term deposit.the bank gives her three option of interest rate with different compounding.
option 1 - simple interest type @ 10.2 % annually
option 2 - compounded interest type @ 10% annually
option 3 - compounded interest type @ 9.8% daily
compute her total amount after 30 months for all three option.conclude which option she should take to get maximum return on her investment.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
her present amount is 28000
she will be investing for 30 months.

formula for simple interest is i = p * r * n.
i is the interest.
p is the present amount.
r is the interest rate per time period.
n is the number of time periods.

formula for compound interest is i = f - p.
first you find the future value and then you find the interest.
formula for future value is f = p * (1 + r)^n.
f is the future value
p is the present amount
r is the interest rate per time period
n is the number of time periods
i = interest

in the formulas you need to use interest rate as a decimal and not interest rate as a percent.
interest rate as a decimal is equal to interest rate as a percent divided by 100.

option 1 is simple interest type at 10.2% annually.
formula is i = p * r * n.
p = 28000.
r = 10.2% per year divided by 100 = .102 per year divided by 12 = .0085 per month.
n = 30 months.
formula becomes:
i = 28000 * .0085 * 30 = 7140.
interest is equal to 7140.

option 2 is compound interest type at 10% compounded annually.
p = 28000.
r = 10% per year divided by 100 = .10 per year.
n = 30 months divided by 12 = 2.5 years.
formula of f = p * (1 + r)^n becomes:
f = 28000 * (1.10)^2.5
solve for f to get f = 35533.64
i = f - p
this says that the interest is equal to the future value minus the present amount.
formula becomes:
i = 35533.64 - 28000 = 7533.64
interest is 7533.64

option 3 is compound interest type at 9.8% compounded daily.
here it gets sticky because the number of days in a year is not exactly 365 and the number of days in a month is not exactly 30.
you have to make some assumptions unless you are told what to assume is the number of days in a year and, consequently, the number of days in a month.

you can use 365 days in a year or 360 days in a year or 365.25 days in a year, or if you really want to get fancy, some other number that takes into account you have a leap year every 4 years and some other adjustment every 100 years.

all of them will yield an answer that should be very close to each other and should be consistent.

consistent means that the number is either the greater of each of the other two or not the greater of each of the other two.

if we assume 360 days in a year, we would get:

p = 28000
r = .098 per year divided by 360 = .000272222 per day
n = 30 months * 360/12 = 900 days.
formula of f = p * (1 + r)^n becomes:
f = 28000 * (1.000272222)^900
solve for f to get f = 35772.20
i = f - p
i = 7772.20

if we assume 365 days in a year, we would get:

p = 28000
r = .098 per year divided by 365 = .000268493 per day.
n = 30 months * 365/12 = 912.5 days.
formula of f = p * (1 + r)^n becomes:
f = 28000 * (1.000268493)^912.5
solve for f to get f = 35772.2204
i = f - p
i = 7772.22

your interest for the 3 options are:

7140 for option 1.
7533.64 for option 2.
7772.20 for option 3 assuming 360 days in a year.
7772.22 for option 3 assuming 365 days in a year.

option 3 is the winner whether or not you assumed 360 days in a year or 365 days in a year.

option 3 would still be the winner even if you assumed 365.25 days in a year or something even more exotic, so i didn't bother to calculate using those.

they should really have told you how many days in a year to assume.
this takes away the ambiguity.
not telling you makes you work harder that you need to.
in this case, it didn't matter.
either assumption would have led to the same conclusion.

the more times you compound, the greater the effective rate, given that the present amount is the same and the annual interest rate is the same.
the number of times you compound in a year reaches a limit which culminates into the continuous compounding formula of f = p * e^(r * n).
f is the future value
p is the present amount
e is the scientific constant of 2.718281828.....
r is the interest rate per time period.
n is the number of time periods.
in this formula, it doesn't matter whether the time period is in years or months of days or whatever because r*n will always be the same.
for example:
if r is 12% per year and n is 1 year, then r*n becomes .12 * 1 = .12
divide 12% per year by 12 and you get 1% per month.
multiply 1 year by 12 and you get 12 months.
r * n becomes .01 * 12 = .12.
same number.

if you had used the continuous compounding formula for option 3, your answer would have been f = 28000 * e^(.098 * 2.5) = 35773.39 which would have resulted in i = 7773.39.

7773.39 is not much greater than 7772.22
this tells you that assuming daily compounding gives you a close approximation to continuous compounding.

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