As a savings plan for college, when their son Bob was born, the Wilburs deposited $10,000 in an account paying 8% compounded annually. How much will the account be worth when Bob is 18 years old?
P = Po(1+r)^n
Po = $10,000
r = 8%/100% = 0.08 = APR expressed as a decimal.
n = 1comp/yr. * 18yrs. = 18 Compounding
periods.
Solve for P.
To calculate the future value of an account, we can use the compound interest formula:
A = P(1 + r/n)^(nt)
Where:
A = future value of the account
P = principal amount (initial deposit)
r = annual interest rate (in decimal form)
n = number of times interest is compounded per year
t = number of years
In this case:
P = $10,000
r = 8% = 0.08
n = 1 (compounded annually)
t = 18
Plugging these values into the formula, we get:
A = $10,000(1 + 0.08/1)^(1*18)
Simplifying further:
A = $10,000(1.08)^18
Using a calculator:
A ≈ $30,984.92
Therefore, the account will be worth approximately $30,984.92 when Bob is 18 years old.
To find out how much the account will be worth when Bob is 18 years old, we need to use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the final amount
P = the initial principal (the initial deposit)
r = the annual interest rate (in decimal form)
n = the number of times that interest is compounded per year
t = the number of years
In this case, the initial deposit (P) is $10,000, the annual interest rate (r) is 8% (or 0.08 in decimal form), and the interest is compounded annually (n = 1). The number of years (t) is 18.
Plugging these values into the formula, we have:
A = 10,000(1 + 0.08/1)^(1*18)
Calculating the exponent first:
A = 10,000(1 + 0.08)^18
Using a calculator:
A ≈ 10,000(1.08)^18
A ≈ 10,000(3.172947)
Calculating the final amount:
A ≈ $31,729.47
Therefore, the account will be worth approximately $31,729.47 when Bob is 18 years old.