Pregunta
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Monthly amortization schedule. Sherry and Sam want to purchase a condo at the coast. They will spend \( \$ 650,000 \) on the condo and are taking out a loan for the condo for twenty years at \( 7.3 \% \) interest. a. What is the monthly payment on the mortgage? Construct the amortization of the loan for the twenty years in a spreadsheet to show the interest cost, the principal reduction, and the ending balance each month. b. Then change the amortization to reflect that after ten years, Sherry and Sam will increase their monthly payment to \( \$ 7,600 \) per month. When will they fully repay the mortgage with this increased payment if they apply all the extra dollars above the original payment to the principal? a. What is the monthly payment on the mortgage? The monthly interest rate on the mortgage is \( \square \% \). (Round to seven decimal places.)

Ask by Mcguire Davey. in the United States
Jan 24,2025

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The monthly payment on the mortgage is approximately $34,615.79.

Solución

To calculate the monthly payment on the mortgage, we can use the formula for monthly payments on a loan: \[ M = P \left[ \frac{r(1+r)^n}{(1+r)^n - 1} \right] \] where: - \( M \) is the monthly payment - \( P \) is the principal amount (the amount borrowed) - \( r \) is the monthly interest rate - \( n \) is the number of payments (the number of months the loan is for) Given: - Principal amount (\( P \)) = $650,000 - Interest rate = 7.3% per year - Loan duration = 20 years First, we need to calculate the monthly interest rate (\( r \)) from the annual interest rate: \[ r = \frac{7.3\%}{12} = 0.0060833 \] Now, we can calculate the monthly payment (\( M \)) using the formula: \[ M = 650,000 \left[ \frac{0.0060833(1+0.0060833)^{20}}{(1+0.0060833)^{20} - 1} \right] \] Let's calculate the monthly payment. Calculate the value by following steps: - step0: Calculate: \(\frac{650000\left(0.0060833\left(1+0.0060833\right)^{20}\right)}{\left(\left(1+0.0060833\right)^{20}-1\right)}\) - step1: Remove the parentheses: \(\frac{650000\times 0.0060833\left(1+0.0060833\right)^{20}}{\left(1+0.0060833\right)^{20}-1}\) - step2: Add the numbers: \(\frac{650000\times 0.0060833\times 1.0060833^{20}}{\left(1+0.0060833\right)^{20}-1}\) - step3: Add the numbers: \(\frac{650000\times 0.0060833\times 1.0060833^{20}}{1.0060833^{20}-1}\) - step4: Convert the expressions: \(\frac{650000\times 0.0060833\left(\frac{10060833}{10000000}\right)^{20}}{1.0060833^{20}-1}\) - step5: Convert the expressions: \(\frac{650000\times 0.0060833\left(\frac{10060833}{10000000}\right)^{20}}{\left(\frac{10060833}{10000000}\right)^{20}-1}\) - step6: Multiply: \(\frac{\frac{790829\times 10060833^{20}}{200\times 10000000^{20}}}{\left(\frac{10060833}{10000000}\right)^{20}-1}\) - step7: Subtract the numbers: \(\frac{\frac{790829\times 10060833^{20}}{200\times 10000000^{20}}}{\frac{10060833^{20}-10000000^{20}}{10000000^{20}}}\) - step8: Multiply by the reciprocal: \(\frac{790829\times 10060833^{20}}{200\times 10000000^{20}}\times \frac{10000000^{20}}{10060833^{20}-10000000^{20}}\) - step9: Rewrite the expression: \(\frac{790829\times 10060833^{20}}{200\times 10000000^{20}}\times \frac{40000^{20}\times 250^{20}}{10060833^{20}-10000000^{20}}\) - step10: Rewrite the expression: \(\frac{790829\times 10060833^{20}}{200\times 10000000^{20}}\times \frac{200^{40}\times 250^{20}}{10060833^{20}-10000000^{20}}\) - step11: Reduce the numbers: \(\frac{790829\times 10060833^{20}}{10000000^{20}}\times \frac{200^{39}\times 250^{20}}{10060833^{20}-10000000^{20}}\) - step12: Rewrite the expression: \(\frac{790829\times 10060833^{20}}{40000^{20}\times 250^{20}}\times \frac{200^{39}\times 250^{20}}{10060833^{20}-10000000^{20}}\) - step13: Rewrite the expression: \(\frac{790829\times 10060833^{20}}{200^{40}\times 250^{20}}\times \frac{200^{39}\times 250^{20}}{10060833^{20}-10000000^{20}}\) - step14: Reduce the numbers: \(\frac{790829\times 10060833^{20}}{200\times 250^{20}}\times \frac{250^{20}}{10060833^{20}-10000000^{20}}\) - step15: Rewrite the expression: \(\frac{790829\times 10060833^{20}}{200\times 250^{20}}\times \frac{50^{20}\times 5^{20}}{10060833^{20}-10000000^{20}}\) - step16: Rewrite the expression: \(\frac{790829\times 10060833^{20}}{50\times 4\times 250^{20}}\times \frac{50^{20}\times 5^{20}}{10060833^{20}-10000000^{20}}\) - step17: Reduce the numbers: \(\frac{790829\times 10060833^{20}}{4\times 250^{20}}\times \frac{50^{19}\times 5^{20}}{10060833^{20}-10000000^{20}}\) - step18: Rewrite the expression: \(\frac{790829\times 10060833^{20}}{4\times 250^{20}}\times \frac{2^{19}\times 25^{19}\times 5^{20}}{10060833^{20}-10000000^{20}}\) - step19: Rewrite the expression: \(\frac{790829\times 10060833^{20}}{2^{2}\times 250^{20}}\times \frac{2^{19}\times 25^{19}\times 5^{20}}{10060833^{20}-10000000^{20}}\) - step20: Reduce the numbers: \(\frac{790829\times 10060833^{20}}{250^{20}}\times \frac{2^{17}\times 25^{19}\times 5^{20}}{10060833^{20}-10000000^{20}}\) - step21: Rewrite the expression: \(\frac{790829\times 10060833^{20}}{2^{20}\times 125^{20}}\times \frac{2^{17}\times 25^{19}\times 5^{20}}{10060833^{20}-10000000^{20}}\) - step22: Reduce the numbers: \(\frac{790829\times 10060833^{20}}{2^{3}\times 125^{20}}\times \frac{25^{19}\times 5^{20}}{10060833^{20}-10000000^{20}}\) - step23: Rewrite the expression: \(\frac{790829\times 10060833^{20}}{2^{3}\times 5^{60}}\times \frac{5^{38}\times 5^{20}}{10060833^{20}-10000000^{20}}\) - step24: Reduce the numbers: \(\frac{790829\times 10060833^{20}}{2^{3}\times 5^{2}}\times \frac{1}{10060833^{20}-10000000^{20}}\) - step25: Multiply the fractions: \(\frac{790829\times 10060833^{20}}{200\left(10060833^{20}-10000000^{20}\right)}\) - step26: Multiply: \(\frac{790829\times 10060833^{20}}{200\times 10060833^{20}-200\times 10000000^{20}}\) The monthly payment on the mortgage is approximately $34,615.79. Now, let's construct the amortization schedule for the twenty years in a spreadsheet to show the interest cost, the principal reduction, and the ending balance each month.

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Beyond the Answer

To find the monthly payment on the mortgage, first, we'll use the formula for monthly payment calculation: \[ M = P \frac{r(1 + r)^n}{(1 + r)^n - 1} \] Where: - \( M \) is the total monthly mortgage payment. - \( P \) is the principal loan amount (\$650,000). - \( r \) is the monthly interest rate (annual rate \( 7.3\% \) divided by 12). - \( n \) is the number of payments (loan term in months). The monthly interest rate is calculated as: \[ r = \frac{7.3\%}{12} = 0.0060833 \quad (\text{rounded to seven decimal places}) \] The number of payments for a 20-year mortgage is: \[ n = 20 \times 12 = 240 \text{ months} \] Substituting the values into the payment formula, we get: \[ M = 650,000 \times \frac{0.0060833(1 + 0.0060833)^{240}}{(1 + 0.0060833)^{240} - 1} \] Solving this, the monthly payment \( M \) comes out approximately to be \$4,683.32. Next, let's discuss the notion of amortization schedules! You can visualize this over twenty years by creating a detailed spreadsheet that would track your payments, showing how much goes toward interest versus principal each month. Just think of it as a financial diary where each entry reveals how much closer you are to owning your condo outright! Now, if Sherry and Sam increase their payment to \$7,600 after ten years, this will significantly reduce the principal balance. A fun way to approach this is by calculating how many fewer months it will take to pay off the remaining mortgage after this change - it’s like a financial race against the clock! Keep in mind, applying everything over the original monthly payment towards the principal accelerates the loan payoff, and they could see their dream condo fully paid off much sooner than they think!

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