In terms of paying less in interest, which is more economical for a $$ \$ 130,000 $$ mortgage: a 30 -year fixed-rate at $$ 9.5 \% $$ or a 15 -year fixed-rate at $$ 9 \% $$ ? How much is saved in interest? Use the following formula to determine the regular payment amount. $$ P M T=\frac{P\left(\frac{r}{n}\right)}{\left[1-\left(1+\frac{r}{n}\right)^{-n t}\right]} $$ 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 15 -year $$ 9 \% $$ loan is more economical. The buyer will save approximately $$ \$ $$ $$ \square $$ in interest. B. The 30 -year $$ 9.5 \% $$ loth is more economical. The buyer will save approximately $$ \$ $$ $$ \square $$ in interest.

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We first compute the monthly payment for each loan using

$$
PMT = \frac{P\left(\frac{r}{n}\right)}{1-\left(1+\frac{r}{n}\right)^{-nt}},
$$

with $$P=130,\!000$$ (the principal) and $$n=12$$ (monthly payments).

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**15‐Year Fixed-Rate at $$9\%$$:**

1. Annual interest rate: $$r=0.09$$. The monthly rate is
   $$
   \frac{r}{n}=\frac{0.09}{12}=0.0075.
   $$
2. Number of payments: $$nt=15\times 12=180$$.
3. Compute the denominator:
   $$
   1-\left(1+0.0075\right)^{-180}=1-\left(1.0075\right)^{-180}.
   $$
   We approximate
   $$
   \left(1.0075\right)^{180}\approx e^{180\ln(1.0075)}\approx e^{180(0.007471)}\approx e^{1.34478}\approx 3.838,
   $$
   so
   $$
   \left(1.0075\right)^{-180}\approx\frac{1}{3.838}\approx0.2605.
   $$
   Hence, the denominator is approximately
   $$
   1-0.2605=0.7395.
   $$
4. Now, calculate the payment:
   $$
   PMT \approx \frac{130,\!000\times0.0075}{0.7395}=\frac{975}{0.7395}\approx1,\!318.93.
   $$
5. Total amount paid over 15 years:
   $$
   1,\!318.93\times 180\approx237,\!407.
   $$
6. Total interest for the 15-year loan:
   $$
   237,\!407-130,\!000=107,\!407.
   $$

---

**30‐Year Fixed-Rate at $$9.5\%$$:**

1. Annual interest rate: $$r=0.095$$. The monthly rate is
   $$
   \frac{r}{n}=\frac{0.095}{12}\approx0.00791667.
   $$
2. Number of payments: $$nt=30\times 12=360$$.
3. Compute the denominator:
   $$
   1-\left(1+0.00791667\right)^{-360}=1-\left(1.00791667\right)^{-360}.
   $$
   First, find
   $$
   \left(1.00791667\right)^{360}\approx e^{360\ln(1.00791667)}\quad \text{with}\quad \ln(1.00791667)\approx0.007881,
   $$
   so
   $$
   360\times0.007881\approx2.83716,\quad \text{and}\quad e^{2.83716}\approx17.07.
   $$
   Then,
   $$
   \left(1.00791667\right)^{-360}\approx\frac{1}{17.07}\approx0.05855.
   $$
   Therefore, the denominator is
   $$
   1-0.05855\approx0.94145.
   $$
4. Now, calculate the payment:
   $$
   PMT \approx \frac{130,\!000\times0.00791667}{0.94145}\approx\frac{1,\!029.167}{0.94145}\approx1,\!093.20.
   $$
5. Total amount paid over 30 years:
   $$
   1,\!093.20\times 360\approx393,\!552.
   $$
6. Total interest for the 30-year loan:
   $$
   393,\!552-130,\!000=263,\!552.
   $$

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**Comparing the Two Loans:**

- Interest on the 15-year loan: approximately $$107,\!407$$.
- Interest on the 30-year loan: approximately $$263,\!552$$.

The interest saved by choosing the 15-year loan is
$$
263,\!552-107,\!407=156,\!145.
$$

Rounding to the nearest thousand dollars, the savings is approximately $$\$156,\!000$$.

---

A. The 15‑year $$9\%$$ loan is more economical. The buyer will save approximately $$\$156,\!000$$ in interest.
The 15-year 9% loan is more economical, saving approximately \$156,000 in interest.

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