In terms of paying less in interest, which is more economical for a $$ \$ 110,000 $$ mortgage: a 30 -year fixed-rate at $$ 8.5 \% $$ or a 15 -year fixed-rate at $$ 8 \% $$ ? How much is saved in interest? Use the following formula to determine the regular payment amount. PMT $$ =\frac{P\left(\frac{r}{n}\right)}{\left[1-\left(1+\frac{r}{n}\right)^{-n t}\right]} $$ Select the correct choice below and fill in the answer box within your choice. (Do not round until the final answer. Then round to the nearest thousand dollars.) A. The 30-year $$ 8.5 \% $$ loan is more economical. The buyer will save approximately $$ \$ \square $$ in interest. B. The 15-year $$ 8 \% $$ loan is more economical. The buyer will save approximately $$ \$ \square $$ in interest.

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We begin by computing the regular payment (PMT) using  
$$
\text{PMT} = \frac{P\left(\frac{r}{n}\right)}{1-\left(1+\frac{r}{n}\right)^{-nt}},
$$
where  
• $$P=110\,000$$ is the principal,  
• $$r$$ is the annual interest rate,  
• $$n=12$$ is the number of payments per year, and  
• $$t$$ is the term in years.

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**Case 1: 30‐year fixed-rate at $$8.5\%$$**

1. The monthly interest rate is  
   $$
   \frac{r}{n}=\frac{0.085}{12}\approx0.00708333.
   $$
2. The total number of payments is  
   $$
   nt=12\times30=360.
   $$
3. The payment is calculated by  
   $$
   \text{PMT}=\frac{110\,000\times0.00708333}{1-\left(1+0.00708333\right)^{-360}}.
   $$
4. Compute the denominator:
   - Calculate $$1+0.00708333\approx1.00708333$$.
   - Raise to the power $$-360$$:  
     $$
     \left(1.00708333\right)^{-360}=\frac{1}{\left(1.00708333\right)^{360}}.
     $$
     Using logarithms,  
     $$
     360\ln(1.00708333)\approx360\times0.007058\approx2.54088,
     $$
     so  
     $$
     \left(1.00708333\right)^{360}\approx e^{2.54088}\approx12.70.
     $$
     Thus,  
     $$
     \left(1.00708333\right)^{-360}\approx\frac{1}{12.70}\approx0.07874.
     $$
   - The denominator becomes  
     $$
     1-0.07874=0.92126.
     $$
5. Then calculate the PMT:  
   $$
   \text{PMT}\approx\frac{110\,000\times0.00708333}{0.92126}\approx\frac{778.17}{0.92126}\approx844.48.
   $$
6. The total amount paid over 30 years is  
   $$
   844.48\times360\approx304\,113.
   $$
7. Therefore, the total interest paid is  
   $$
   304\,113-110\,000\approx194\,113.
   $$

---

**Case 2: 15‐year fixed-rate at $$8\%$$**

1. The monthly interest rate is  
   $$
   \frac{r}{n}=\frac{0.08}{12}\approx0.00666667.
   $$
2. The total number of payments is  
   $$
   nt=12\times15=180.
   $$
3. The payment is  
   $$
   \text{PMT}=\frac{110\,000\times0.00666667}{1-\left(1+0.00666667\right)^{-180}}.
   $$
4. Compute the denominator:
   - Calculate $$1+0.00666667\approx1.00666667$$.
   - Compute $$\left(1.00666667\right)^{180}$$:  
     $$
     180\ln(1.00666667)\approx180\times0.006622\approx1.19196,
     $$
     so  
     $$
     \left(1.00666667\right)^{180}\approx e^{1.19196}\approx3.293.
     $$
   - Then,  
     $$
     \left(1.00666667\right)^{-180}\approx\frac{1}{3.293}\approx0.3038,
     $$
     and the denominator is  
     $$
     1-0.3038=0.6962.
     $$
5. Now, compute the PMT:  
   $$
   110\,000\times0.00666667\approx733.33,
   $$
   so  
   $$
   \text{PMT}\approx\frac{733.33}{0.6962}\approx1053.8.
   $$
6. The total amount paid over 15 years is  
   $$
   1053.8\times180\approx189\,684.
   $$
7. Thus, the total interest paid is  
   $$
   189\,684-110\,000\approx79\,684.
   $$

---

**Interest Savings Comparison**

The difference in total interest between the 30-year and the 15-year loans is  
$$
194\,113-79\,684\approx114\,429.
$$

When rounding to the nearest thousand dollars, the saving is approximately $$\$114\,000$$.

---

B. The 15-year $$8\%$$ loan is more economical. The buyer will save approximately $$\$114\,000$$ in interest.
The 15-year 8% loan is more economical, saving approximately \$114,000 in interest.

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