|
Question 1187689: if savings in an account are to be at least tripled over 20 years with an interest rate of 5.55% p.a, how often should the interest be compounded? Give your answer in months.
The answer from textbook: at least every 4 months
I calculated my own answer and is 1.0169 something. Can someone please show me how the question should be solved? Thank you.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i'm not sure there's a formula that will let you get the answer directly.
it appears to be an iterative approach.
the formula to use is f = p * (1 + r/c) ^ (n*c)
f is the future value which will be 3.
p is the present value which will be 1.
r is the interest rate per year (percent divided by 100).
c is the number of compounding periods per year.
n is the number of years.
if the number of compounding periods per year is 1, then the formula becomes:
f = 1 * (1 + .0555/1) ^ (20 * 1)
you get f = 2.945538834.
if the number of compounding periods per year is 2, then the formula becomes:
f = 1 * (1 + .0555/2) ^ (20 * 2)
you get f = 2.988817642.
if the number of compounding periods per year is 3, then the formula becomes:
f = 1 * (1 + .0555/3) ^ (20 * 3)
you get f = 3.003737664.
if the number of compounding periods per year is 4, then the formula becomes:
f = 1 * (1 + .0555/4) ^ (20*4)
you get f = 3.011293958.
if the number of compounding periods per year is 5, then the formula becomes:
f = 1 * (1 + .0555/5) ^ (20*5)
you get f = 3.01585904
it looks like 3 compounding periods per year is about where it will be.
that's the closest to f = 3.
3 compounding periods per year is equal to 4 months each because 3 * 4 = 12.
i tried to do it directly by getting 3 = 1 * (1 + .0555/x) ^ (20*x), but couldn't solve that by logarithms because x was in the log function itself and also in the exponent.
i then tried to solve it graphically.
i got x = 2.668.
when x = 2.668, the formula becomes y = 1 * (1 + .0555/2.668) ^ (20 * 2.668).
you get f = 3.000000595.
that's much closer, and would be even closer if the graphing software didn't rounded the answer to 3 decimal digits.
here's what the graphing software output looks like.
the graphing software i used online was at desmos.com.
i also used my ti-84 plus graphing software.
that gave me x = 2.6679529.
when x = 2.6679529, y = (1 + .0555/2.6679529) ^ (20 * 2.6679529) gave me:
f = 3.
there may be a way to get it directly through formula, but i wasn't able to do it.
|
|
|
| |
