|
Question 1102759: 4. $1000 is invested at 0.025% per annum interest, compounded quarterly.
a. Calculate the value of the investment at the end of 12 years.
b. Calculate the minimum number of years required for the value of the investment to exceed $10,000.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Compound interest formula is P=Po(1+r/n)^nt
P=1000(1+(0.00025)/4)^48=$1003.
Perhaps it is 2.5% interest or 0.025
Then it is 1000(1+.025/4)^48=$1348.60.
Using the 2.5% interest
10000=1000(1+.025/4)^4t, where t is number of years
10=1.00625^4t
ln both sides
2.303=4t*ln(1.00625)
369.56 compoundings=4t
t=92.39 years round to 93 years
rule of 72 says 72/2.5=doubling time in years which is about 28 years.
doubles a second time at 56 years (to $4000) and again at 84 years (to $8000), so that 93 years is a reasonable answer.
(for the 0.00025 interest, it would be 9210 years.)
|
|
|
| |
