Bobby needs to save $10,000 for school in the next two years. He found a bank that offers a 10% interest rate compounded annually. What does he need to deposit at the beginning of the year to have enough money for school?
x(1.10)^2 = 10000
x = 10000/1.1^2 = 8264.46
To find out how much Bobby needs to deposit at the beginning of the year to have enough money for school, we can use the formula for compound interest. The formula is:
A = P(1 + r/n)^(nt)
Where:
A = the amount of money Bobby will have at the end of the two years (in this case, $10,000)
P = the principal amount (the amount Bobby needs to deposit at the beginning of the year)
r = the annual interest rate (10% or 0.10 as a decimal)
n = the number of times the interest is compounded per year (in this case, it's compounded annually, so n = 1)
t = the number of years (in this case, 2)
We can rearrange the formula to solve for P:
P = A / (1 + r/n)^(nt)
Now we can plug in the values:
P = $10,000 / (1 + 0.10/1)^(1*2)
P = $10,000 / (1.10)^2
P = $10,000 / 1.21
P ≈ $8,264.46
Therefore, Bobby needs to deposit approximately $8,264.46 at the beginning of the year to have enough money for school.