SOLUTION: Jim has $36,000 in subsidized student loans at 4.3% when he graduates. He starts paying these loans off six months after he graduates. How much does he owe at this time?

Question 1157159: Jim has $36,000 in subsidized student loans at 4.3% when he graduates. He starts paying these loans off six months after he graduates. How much does he owe at this time?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the loan is 36,000 and will earn interest for 6 months.
how much he will owe 6 months from now is dependent on the conditions of the loan.
you would use the following formulas, depending on the condition of the loan.
if it is a simple interest, then the formula to use is f = p * r * n + p.
f is the future value
p is the present value
r is the interest rate per time period
n is the number of time periods
the formula becomes f = 36,000 * .043 * .5 + 36,000 = 36,774.
if it is compound interest, then it depends on the number of compounding periods per year.
if annual compounding, then r = .043 and n = .5
if semi-annual, then r = .043/2 and n = .5 * 2
if quarterly, then r = .043/4 and n = .5 * 4
if monthly, then r = .043/12 and n = .5 * 12
since monthly compounding is the most likely, then i'll show you how that's done and you can figure out how to do the others if you're so inclined.
the formula is f = p * (1 + r) ^ n
r is the future value
p is the present value
r is the interest rate per time period.
n is the number of time periods.
with annual compounding the annual interest rate and the number of year is left as is and the formula becomes f = 36,000 * (1 + .043) ^ .5 = 36,765.85372 which rounds to 36,765.85.
with monthly compounding, the annual interest rate is divided by 12 and the number of annual time periods is multiplied by 12.
the formula becomes f = p * (1 + r/12) ^ (n * 12)
this becomes f = 36,000 * (1 + .043/12) ^ (.5 * 12)
solve for f to get f = 36780.96697 which rounds to 36,780.97.
the more compounding periods per year, the higher the future value will be until it gets to the ultimate number of compounding periods per year, which is continuous compounding.
there is another formula for that.
it is f = p * e ^ (r * n)
here, it doesn't matter how many compounding periods per year, since the future value will be the same regardless, so it's easiest to use annual rate and number of years.
the formula becomes f = 36,000 * e ^ ( .043 * .5) = 36,782.38045 which rounds to 36,782.38.
the letter e stands for the scientific constant of 2.718281828.....

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