What annual rate of interest will enable $800 to grow to $1000 in three years?
cant seem to derive the formula for this one, normal compound interest isnt working for me
800 * (1 + i)^3 = 1000
this is with annual compounding
compound interest: 800(1 + r/100)^3 = 1000
simple interest: 800(1 + 3r/100) = 1000
solve for r and express as a %
To solve this problem, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment ($1000 in this case)
P = the principal amount ($800 in this case)
r = the annual interest rate (the value we are trying to find)
n = the number of times the interest is compounded per year (not specified in the question)
t = the number of years (3 in this case)
Since the question does not provide the compounding frequency 'n', we can assume it to be compounded annually. Now, we can rewrite the formula as:
A = P(1 + r)^t
Substituting the given values, we get:
1000 = 800(1 + r)^3
Now, we need to solve for 'r'.
To isolate the (1 + r) term, we can divide both sides by 800:
1000/800 = (1 + r)^3
Simplifying:
5/4 = (1 + r)^3
To solve for (1 + r), we need to take the cube root of both sides:
∛(5/4) = 1 + r
Finally, subtracting 1 from both sides:
∛(5/4) - 1 = r
Using a calculator, we can evaluate ∛(5/4) - 1 ≈ 0.0606.
Therefore, the annual interest rate that will enable $800 to grow to $1000 in three years is approximately 6.06%.