If Sarah paid the same amount ($31,501) but only 2 years after graduation, she would still owe approximately $26,011.12.
a) To calculate the amount Sarah will owe after 11 years, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A is the final amount (amount owed after 11 years),
P is the principal amount (original loan amount),
r is the annual interest rate (as a decimal),
n is the number of times interest is compounded per year, and
t is the number of years.
Given:
P = $38,876
r = 6.6% = 0.066 (as a decimal)
n = 1 (compounded annually)
t = 11 years
Substituting the values into the formula:
A = $38,876(1 + 0.066/1)^(1*11)
A ≈ $38,876(1.066)^11
A ≈ $38,876(1.8388)
A ≈ $71,470.59
Sarah will owe approximately $71,470.59 after 11 years.
b) To calculate how many times higher the amount she owes after 11 years is compared to the original amount, we can divide the final amount by the principal amount:
Times Higher = A / P
Times Higher = $71,470.59 / $38,876
Times Higher ≈ 1.839
Therefore, the amount she owes after 11 years is approximately 1.839 times higher than the original amount.
c) If Sarah makes a large loan payment of $31,501, we can subtract this payment from the amount she owes after 11 years:
Amount Owed = $71,470.59 - $31,501
Amount Owed ≈ $39,969.59
Therefore, after making a payment of $31,501, Sarah will owe approximately $39,969.59.
d) If Sarah paid the same amount ($31,501) but only 2 years after graduation, we need to calculate the interest accumulated during those 2 years and subtract it from the payment amount. Let's calculate the interest first:
Interest = P(1 + r/n)^(nt) - P
Interest = $38,876(1 + 0.066/1)^(1*2) - $38,876
Interest ≈ $38,876(1.1388) - $38,876
Interest ≈ $44,365.88 - $38,876
Interest ≈ $5,489.88
Now, subtract the interest from the payment amount:
Amount Owed = $31,501 - $5,489.88
Amount Owed ≈ $26,011.12
If Sarah paid the same amount ($31,501) but only 2 years after graduation, she would still owe approximately $26,011.12.