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. $$ P M T=\frac{P\left(\frac{r}{n}\right)}{\left[1-\left(1+\frac{r}{n}\right)^{-n t}\right]} $$ Select the correct choice below and fill in the answer box within your choice. (Do not round until the final enswer. 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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To determine which mortgage option is more economical in terms of paying less in interest, we need to calculate the regular payment amount for both options using the given formula and then compare the total interest paid.

Given:
- Mortgage amount: $110,000
- 30-year fixed-rate at 8.5%: $$ r = 0.085 $$, $$ n = 12 $$ (monthly payments), $$ t = 30 $$ years
- 15-year fixed-rate at 8%: $$ r = 0.08 $$, $$ n = 12 $$ (monthly payments), $$ t = 15 $$ years

Let's calculate the regular payment amount for both options using the formula:

$$ P M T = \frac{P\left(\frac{r}{n}\right)}{\left[1-\left(1+\frac{r}{n}\right)^{-n t}\right]} $$

1. For the 30-year fixed-rate at 8.5%:
$$ P M T_{30} = \frac{110,000\left(\frac{0.085}{12}\right)}{\left[1-\left(1+\frac{0.085}{12}\right)^{-12 \times 30}\right]} $$

2. For the 15-year fixed-rate at 8%:
$$ P M T_{15} = \frac{110,000\left(\frac{0.08}{12}\right)}{\left[1-\left(1+\frac{0.08}{12}\right)^{-12 \times 15}\right]} $$

Now, let's calculate the regular payment amounts for both options.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{110000\left(\frac{0.085}{12}\right)}{1-\left(1+\frac{0.085}{12}\right)^{-12\times 30}}$$
- step1: Divide the terms:
  $$\frac{110000\left(\frac{0.085}{12}\right)}{1-\left(1+\frac{17}{2400}\right)^{-12\times 30}}$$
- step2: Add the numbers:
  $$\frac{110000\left(\frac{0.085}{12}\right)}{1-\left(\frac{2417}{2400}\right)^{-12\times 30}}$$
- step3: Divide the terms:
  $$\frac{110000\times \frac{17}{2400}}{1-\left(\frac{2417}{2400}\right)^{-12\times 30}}$$
- step4: Multiply the numbers:
  $$\frac{110000\times \frac{17}{2400}}{1-\left(\frac{2417}{2400}\right)^{-360}}$$
- step5: Multiply the numbers:
  $$\frac{\frac{4675}{6}}{1-\left(\frac{2417}{2400}\right)^{-360}}$$
- step6: Subtract the numbers:
  $$\frac{\frac{4675}{6}}{\frac{2417^{360}-2400^{360}}{2417^{360}}}$$
- step7: Multiply by the reciprocal:
  $$\frac{4675}{6}\times \frac{2417^{360}}{2417^{360}-2400^{360}}$$
- step8: Multiply the fractions:
  $$\frac{4675\times 2417^{360}}{6\left(2417^{360}-2400^{360}\right)}$$
- step9: Multiply:
  $$\frac{4675\times 2417^{360}}{6\times 2417^{360}-6\times 2400^{360}}$$
Calculate or simplify the expression $$ \frac{110000(0.08/12)}{1-(1+0.08/12)^(-12*15)} $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{110000\left(\frac{0.08}{12}\right)}{1-\left(1+\frac{0.08}{12}\right)^{-12\times 15}}$$
- step1: Divide the terms:
  $$\frac{110000\left(\frac{0.08}{12}\right)}{1-\left(1+\frac{1}{150}\right)^{-12\times 15}}$$
- step2: Add the numbers:
  $$\frac{110000\left(\frac{0.08}{12}\right)}{1-\left(\frac{151}{150}\right)^{-12\times 15}}$$
- step3: Divide the terms:
  $$\frac{110000\times \frac{1}{150}}{1-\left(\frac{151}{150}\right)^{-12\times 15}}$$
- step4: Multiply the numbers:
  $$\frac{110000\times \frac{1}{150}}{1-\left(\frac{151}{150}\right)^{-180}}$$
- step5: Multiply the numbers:
  $$\frac{\frac{2200}{3}}{1-\left(\frac{151}{150}\right)^{-180}}$$
- step6: Subtract the numbers:
  $$\frac{\frac{2200}{3}}{\frac{151^{180}-150^{180}}{151^{180}}}$$
- step7: Multiply by the reciprocal:
  $$\frac{2200}{3}\times \frac{151^{180}}{151^{180}-150^{180}}$$
- step8: Multiply the fractions:
  $$\frac{2200\times 151^{180}}{3\left(151^{180}-150^{180}\right)}$$
- step9: Multiply:
  $$\frac{2200\times 151^{180}}{3\times 151^{180}-3\times 150^{180}}$$
The regular payment amount for the 30-year fixed-rate at 8.5% is approximately $845.80, and for the 15-year fixed-rate at 8% is approximately $1051.22.

To determine which option is more economical in terms of paying less in interest, we need to calculate the total interest paid for both options.

Let's calculate the total interest paid for both options.
Calculate the value by following steps:
- step0: Calculate:
  $$110000-845.804832\times 30$$
- step1: Multiply the numbers:
  $$110000-25374.14496$$
- step2: Subtract the numbers:
  $$84625.85504$$
Calculate or simplify the expression $$ 110000 - 1051.217293 * 15 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$110000-1051.217293\times 15$$
- step1: Multiply the numbers:
  $$110000-15768.259395$$
- step2: Subtract the numbers:
  $$94231.740605$$
The total interest paid for the 30-year fixed-rate at 8.5% is approximately $84,625.86, and for the 15-year fixed-rate at 8% is approximately $94,231.74.

Since the total interest paid for the 30-year fixed-rate at 8.5% is less than the total interest paid for the 15-year fixed-rate at 8%, the 30-year fixed-rate at 8.5% is more economical in terms of paying less in interest.

Therefore, the correct answer is:
A. The 30-year 8.5% loan is more economical. The buyer will save approximately $94,231.74 in interest.
A. The 30-year 8.5% loan is more economical. The buyer will save approximately \$94,232 in interest.

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