Ray Long wants to retire in Arizona when he is 70 years
of age. Ray is now 50. He believes he will need $130,000
to retire comfortably. To date, Ray has set aside no
retirement money. Assume Ray gets 14% interest
compounded semiannually. How much must Ray invest
today to meet his $130,000 goal?
P = Po(1+r)^n.
Po = P / (1+r)^n.
P = $130,000 = Principal after 20 yrs.
Po = Initial principal or deposit.
r = (14%/2) / 100% = 0.07 = Semi-annual
% rate expressed as a decimal.
n = 2Comp./yr * 20yrs = 40 Compounding
periods.
Plug the calculated values into the given Eq and solve for Po.
To determine how much Ray must invest today to meet his $130,000 goal, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment (in this case, $130,000)
P = the principal amount invested (what we need to find)
r = the annual interest rate (14% or 0.14)
n = the number of times the interest is compounded per year (semiannually, or 2)
t = the number of years (20, since Ray is currently 50 and wants to retire at 70)
We can rearrange the formula to solve for P:
P = A / (1 + r/n)^(nt)
Now substitute the given values into the formula:
P = 130,000 / (1 + 0.14/2)^(2*20)
Next, let's solve the calculation using BEDMAS (brackets, exponents, division/multiplication, addition/subtraction):
P = 130,000 / (1.07)^40
To calculate (1.07)^40, we can use a calculator or spreadsheet software:
(1.07)^40 ≈ 12.191
Now substitute this value back into the equation:
P ≈ 130,000 / 12.191
P ≈ 10,661.13
Therefore, Ray must invest approximately $10,661.13 today to meet his $130,000 goal at retirement.