Problem 7:
We determine the time as follows:
*We can proceed using the following expression:
[tex]t=\frac{\ln (\frac{m}{p})-\ln (\frac{m}{p}-\frac{r}{n})}{n\ln (1+\frac{r}{n})}[/tex]Here, t is the time it will take to pay, m is the maximum she can afford to pay each month, p is the base loan value, r is the interest rate, n is the number of periods. Now we replace:
[tex]t=\frac{\ln(\frac{500}{20000})-\ln(\frac{500}{20000}-\frac{0.071}{12})}{12\ln(1+\frac{0.071}{12})}\Rightarrow t\approx3.8[/tex]So, she will take approximately 3.8 years to pay up the loan.
Problem 8:
We determine the time he has as follows:
We use the expression:
[tex]t=\frac{\ln (\frac{m}{p})-\ln (\frac{m}{p}-\frac{r}{n})}{n\ln (1+\frac{r}{n})}[/tex]Here, t is the time it will take to pay, m is the maximum he can afford to pay each month, p is the base loan value, r is APR, n is the number of periods. Now we replace:
[tex]t=\frac{\ln(\frac{400}{14000})-\ln(\frac{400}{14000}-\frac{0.068}{12})}{12\ln(1+\frac{0.068}{12})}\Rightarrow t\approx3.3[/tex]So, he will take approximately 3.3 years to pay the loan.
Problem 10:
We determine the amount he will have to pay as follows:
*We use the following expression:
[tex]V=P(1+n)^t[/tex]Here V is the value to obtain, P is the original amount, n is the interest rate and t is the number of periods, now we replace:
[tex]V=(22000)(1-0.0485)^{15}\Rightarrow V\approx44766.09[/tex]So, after 15 years he will have to pay approximately $44766.09.