$$ 1 \leftarrow $$ Use PMT $$ =\frac{P\left(\frac{r}{n}\right)}{\left[1-\left(1+\frac{r}{n}\right)^{-n t}\right]} $$ to determine the regular payment amount, rounded to the nearest dollar. In terms of paying less in interest, which is more economical for a $$ \$ 250,000 $$ mortgage: a 30 -year fixed-rate at $$ 7 \% $$ or a 20 -year fixed- rate at $$ 6.5 \% $$ ? How much is saved in interest? 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 $$ 7 \% $$ loan is more economical. The buyer will save approximately $$ \$ \square $$ in interest. B. The 20 -year $$ 6.5 \% $$ loan is more eponomical. The buyer will save approximately $$ \$ \square $$ in interest.

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**Step 1. Calculate the monthly payment for the 30‑year, 7% loan**

For a principal $$ P = 250,\!000 $$, annual interest rate $$ r = 0.07 $$, compounded monthly ($$ n=12 $$), and term $$ t = 30 $$ years, the number of monthly payments is
$$
N = n \cdot t = 12 \cdot 30 = 360.
$$
The monthly interest rate is
$$
\frac{r}{n} = \frac{0.07}{12} \approx 0.00583333.
$$
Using the formula
$$
\text{PMT} = \frac{P\left(\frac{r}{n}\right)}{1-\left(1+\frac{r}{n}\right)^{-n t}},
$$
the monthly payment is
$$
\text{PMT}_{30} = \frac{250,\!000 \times 0.00583333}{1-\left(1.00583333\right)^{-360}}.
$$
First, compute the numerator:
$$
250,\!000 \times 0.00583333 \approx 1,\!458.33.
$$
Then, compute the denominator. Calculate
$$
\left(1.00583333\right)^{360} \quad \text{so that} \quad \left(1.00583333\right)^{-360} \approx e^{-360\ln(1.00583333)}.
$$
Since 
$$
\ln(1.00583333) \approx 0.005816,
$$
we have
$$
360 \times 0.005816 \approx 2.09376 \quad \text{and} \quad e^{-2.09376} \approx 0.1237.
$$
Thus, the denominator is
$$
1 - 0.1237 = 0.8763.
$$
Now, the monthly payment is approximately
$$
\text{PMT}_{30} \approx \frac{1,\!458.33}{0.8763} \approx 1,\!664 \text{ dollars (rounded to the nearest dollar)}.
$$

The total amount paid over 360 months is
$$
1,\!664 \times 360 \approx 599,\!184,
$$
so the total interest paid is
$$
599,\!184 - 250,\!000 \approx 349,\!184.
$$

**Step 2. Calculate the monthly payment for the 20‑year, 6.5% loan**

For a principal $$ P = 250,\!000 $$, annual interest rate $$ r = 0.065 $$, compounded monthly ($$ n=12 $$), and term $$ t = 20 $$ years, the number of monthly payments is
$$
N = 12 \cdot 20 = 240.
$$
The monthly interest rate is
$$
\frac{r}{n} = \frac{0.065}{12} \approx 0.00541667.
$$
The monthly payment is
$$
\text{PMT}_{20} = \frac{250,\!000 \times 0.00541667}{1-\left(1.00541667\right)^{-240}}.
$$
First, compute the numerator:
$$
250,\!000 \times 0.00541667 \approx 1,\!354.17.
$$
For the denominator, calculate
$$
\ln(1.00541667) \approx 0.005399,
$$
so that
$$
240 \times 0.005399 \approx 1.29576 \quad \text{and} \quad \left(1.00541667\right)^{-240} \approx e^{-1.29576} \approx 0.274.
$$
Thus, the denominator is
$$
1-0.274 = 0.726.
$$
Then, the monthly payment is approximately
$$
\text{PMT}_{20} \approx \frac{1,\!354.17}{0.726} \approx 1,\!865 \text{ dollars (rounded to the nearest dollar)}.
$$

The total amount paid over 240 months is
$$
1,\!865 \times 240 \approx 447,\!600,
$$
so the total interest paid is
$$
447,\!600 - 250,\!000 \approx 197,\!600.
$$

**Step 3. Compare the total interest paid and determine the savings**

The interest for the 30‑year loan is approximately
$$
349,\!184,
$$
and for the 20‑year loan is approximately
$$
197,\!600.
$$
The interest saved by choosing the 20‑year at 6.5% over the 30‑year at 7% is
$$
349,\!184 - 197,\!600 \approx 151,\!584.
$$
Rounded to the nearest thousand dollars, the interest saved is about
$$
152,\!000 \text{ dollars}.
$$

**Answer:**  
B. The 20‑year $$6.5\%$$ loan is more economical. The buyer will save approximately $$ \$152,\!000 $$ in interest.
The 20-year 6.5% loan is more economical, saving approximately \$152,000 in interest.

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