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Sagot :
ANSWER:
Loan A requires Bryce to pay less by $1,133.52
EXPLANATION:
For bank A, we're given;
Length of loan in years(t) = 6 years
Annual interest rate(r) = 8% = 8/100 = 0.08
Loan payment(PMT) = $497.94
Number of compounds per year(n) = 12
We can go ahead and determine the loan amount using the below formula as seen below;
[tex]\begin{gathered} Loan\text{ }amount,P_0=\frac{PMT(1-(1+\frac{r}{n})^{-nt})}{\frac{r}{n}} \\ \\ =\frac{497.94(1-(1+\frac{0.08}{12})^{-12*6}}{\frac{0.08}{12}} \\ \\ =\text{ \$}28399.77 \end{gathered}[/tex]So we can see that Bryce will pay back;
[tex]\begin{gathered} Bank\text{ }A\text{ }Loan\text{ }Payback=497.94(12*6) \\ \\ =497.94\left(72\right) \\ \\ =\text{\$}35,851.68 \end{gathered}[/tex]For bank B, we're given;
Length of loan in years(t) = 10 years
Annual interest rate(r) = 5.5% = 5.5/100 = 0.055
Loan payment(PMT) = $308.21
Number of compounds per year(n) = 12
We can go ahead and determine the loan amount using the below formula as seen below;
[tex]\begin{gathered} Loan\text{ }amount,P_0=\frac{PMT(1-(1+\frac{r}{n})^{-nt})}{\frac{r}{n}} \\ \\ =\frac{308.21(1-(1+\frac{0.055}{12})^{-12*10}}{\frac{0.055}{12}} \\ \\ =\text{ \$}28399.57 \end{gathered}[/tex]So we can see that Bryce will pay back;
[tex]\begin{gathered} Bank\text{ B }Loan\text{ }Payback=308.21(12*10) \\ \\ =308.21\left(120\right) \\ \\ =\text{\$}36985.20 \end{gathered}[/tex]We can see that Bryce will pay back $35,851.68 for Loan A and $36,985.20 for Loan B, so Loan A requires Bryce to pay less and by $1,133.52
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