Answer:
[tex]\$ 433.35[/tex] of the second payment applies to the principal balance
Step-by-step explanation:
Hi
Known data
The first step is to find the monthly interest, so we use [tex]1+i_{Y}=(1+i_{M})^{12}[/tex] to do it. Then we have, [tex]1+0.0435=(1+i_{M})^{12}[/tex],
[tex]i_{M}=\sqrt[12]{1.0435}-1=1.0035-1=0.0035[/tex]
the second step is to find the value of the equal monthly payments with the formula below.
[tex]K=\frac{P}{\frac{1-(1+i_{M})^{-n}}{i_{M}} } =\frac{162000}{\frac{1-(1+0.0035)^{-240}}{0.0035} }=998.84[/tex]
Third, we need to calculate the capital subscription in the two first periods
[tex]I_{1}=P*i_{M}=162000(0.0035)=567\\CS_{1}=K-I_{1}=998.84-567=431.84\\B_{1}=P-CS_{1}=162000-431.84=161568.15[/tex]
[tex]I_{2}=B_{1}*i_{M}=161568.15(0.0035)=565.48\\CS_{2}=K-I_{2}=998.84-565.48=433.35\\\left[CS_{2}=433.35\right][/tex]