A person invests $2000 annually in an IRA. At the end of 6 years, the amount in the fund is $14000. What annual nominal compounding rate has this fund earned?
amount of sinking fund:
S = N [(1+r)^n -1 ]/r
7 = [(1+r)^6 -1 ]/r
7 r = (1+r)^6 -1
(1+r)^6 = 7 r + 1
I do not see a closed form solution off hand.
make a table
r left right
.04 1.26 1.28
.05 1.34 1.35
.06 1.42 1.42 looks like 6%
forget my answer, go with Damon
I read it as a single deposit.
Thank you both. I tried this using the "72 rule" and was discouraged by my answer due to the book answer of 9.64% and I am not sure how that answer is founded.
To calculate the annual nominal compounding rate, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the amount in the fund
P = the principal (initial investment)
r = annual nominal interest rate (in decimal form)
n = number of times interest is compounded per year
t = number of years
In this case, the principal (P) is $2000, the amount in the fund (A) is $14000, the number of years (t) is 6, and the number of times interest is compounded per year (n) is not given.
To find the annual nominal interest rate (r), we rearrange the formula:
r = (A/P)^(1/(nt)) - 1
Now, let's calculate the annual nominal interest rate (r):
r = ($14000/$2000)^(1/(6*n)) - 1
Since the number of times interest is compounded per year (n) is not given, we need to make an assumption. Let's assume it's compounded annually (n = 1).
r = ($14000/$2000)^(1/(6*1)) - 1
r = (7)^(1/6) - 1
r ≈ 0.282 - 1
r ≈ -0.718
Therefore, if the fund has earned $14000 with an annual investment of $2000 over 6 years, assuming an annual compounding rate of 1 time per year, the calculated annual nominal compounding rate is approximately -71.8% (as a decimal, r ≈ -0.718).
2000(1+i)^6 = 14000
(1+i)^6 = 7
take 6th root
1+i = 1.38308
i = .3808
or
38.08%
Quickly, tell me where I can earn 38% interest.