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Question 1107419: Angela wants to have $2400 saved by the end of 3 years. The current rate for a savings account is 4.5%. How much will she need to invest to start?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! this depends on whether it is simple interest or compound interest and, if compound interest, the number of compounding periods per year.
with simple interest, the formula is f = p + (p*r*n).
f is the future value.
p is the present value.
r is the interest rate per time period expressed as a decimal, not a percent.
n is the number of time periods.
your time periods are years.
f = 2400
p = what you want to find.
r = .045 per year
n = 3 years.
the formula becomes 2400 = p + p * .045 * 3
simplify this to get 2400 = p + p * .135
simplify further to get 2400 = 1.135 * p
divide both sides of this equation by 1.135 to get p = 2400/1.135 = 2114.537445.
if your savings account compounds the interest, then the amount you need to invest depends on the number of compounding periods per year.
i'll assume 1 compounding period per year for annual compounding.
i'll then assume 12 compounding periods per year for monthly compounding.
you can also have quarterly compounding and semi-annual compounding and daily compounding and continous compounding and possibly others.
without specifying, i don't know whether you are using simple interest or compounding interest or how many compounding periods per year, if you are using compound interest.
sorry to make it seem more complicated than it should be, but that information should have been supplied in order to avoid the confusion.
the compounding formula is f = p * (1 + r/c) ^ (n*c)
f is the future value.
p is the present value.
r is the annual interest rate.
c is the number of compounding periods per year.
n is the number of years.
with annual compounding, your formula becomes 2400 = p * (1 + .045/1) ^ (3*1).
this simplifies to 2400 = p * (1 + .045) ^ 3.
solve for p to get p = 2400 / (1 + .045) ^ 3).
you will get p = 2103.11185.
with monthly compounding, your formula becomes 2400 = p * (1 + .045/12) ^ (3*12).
solve for p to get p = 2400 / ((1 + .045/12) ^ (12 * 3))
you will get p = 2097.447714.
she would need to invest one of the following, or other amounts, depending on the number of compounding periods per year.
2114.537445 with simple interest.
2103.11185 with compound interest compounded once a year (annual).
2097.447714 with compound interest compounded 12 times a year (monthly).
other compounding periods per year will give you a different amount that needs to be invested to be equal to 2400 in 3 years.
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