anonymous
  • anonymous
Charlotte takes a bank loan of 10000$ . The interest rate is 4 % per annum and repayment period of 5 years. The loan will be paid monthly ie, the number of installments is 60. What will Charlotte's total cost of the loan be if it is a 1 ) the straight-line amortization , ie, the same amount is amortized monthly, 2) installment loans ie the total amount (principal + interest) is paid each month?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Derive formulas to calculate the total cost of the loan in both cases , as well as annuity ( cost per payment date ) in installment loans , for arbitrary values ​​of the loan amount ( S ) , interest rate ( r) , installment time (T ) and the number of installments ( n ) .
anonymous
  • anonymous
Im probably is going to use this equation \[\sum_{k=0}^{n}x^k=1+x+x^2+x^3+...+x^n=\frac{ x ^{n+1}-1 }{ k-1 }, k \neq 0\]
anonymous
  • anonymous

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anonymous
  • anonymous
from this \[\frac{ 100000 }{ 60 }+\frac{ 10000*0.04 }{ 12 }=2000\] I get what she pays to the bank the first month, but only the first 100000/60=1667 she pays to the loan... and that I need somewhere in the equation..
IrishBoy123
  • IrishBoy123
i can help you with this time is the issue will try logging on a bit later, if you are online
IrishBoy123
  • IrishBoy123
for straight line amortisation, the principal component of each monthly payment is \(\$\dfrac{10,000}{60} = \$ 166.\dot6\) the interest component is the **monthly rate** times the then opening principal balance at the rate of \(4 \% pa\), which is typically [and wrongly] calculated as \(\dfrac{4 \%}{12} = 0.\dot3 \%\) per months: 1/ the principal payment for the first period is \( \$ 166.\dot6\), as it is for every period. whereas, 2/ the interest payment is \(\dfrac{4 \%}{12} \cdot \$10,000\) for the next period, the principal payment is \( \$ 166.\dot6\) , as it is for every period. but the interest payment is \(\dfrac{4 \%}{12} \cdot (\$10,000- \$ 166.\dot6)\) that pattern repeats itself until the loan is repaid.
anonymous
  • anonymous
@IrishBoy123 I solved this one in school!! need help with a new one though! thanks!

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