Pregunta
upstudy study bank question image url

Present value with periodic rates. Cooley Landscaping needs to borrow \( \$ 26,000 \) for a new front-end dirt loader. The bank is willing to loan the money at \( 7 \% \) interest for the next 7 years with annual, semiannual, quarterly, or monthly payments. What are the different payments that Cooley Landscaping could choose for these different payment plans? What is Cooley's payment for the loan at \( 7 \% \) interest for the next 7 years with annual payments? \( \$ 4824.83 \) (Round to the nearest cent.) What is Cooley's payment for the loan at \( 7 \% \) interest for the next 7 years with semiannual payments? \( \$ \square \) (Round to the nearest cent.)

Ask by Pollard Joseph. in the United States
Jan 24,2025

Solución de tutoría real

Respuesta verificada por el tutor

Responder

Cooley Landscaping's payments for the loan at 7% interest for the next 7 years are as follows: - **Annual Payments**: \$4,824.83 - **Semiannual Payments**: \$2,380.84 - **Quarterly Payments**: \$1,182.52 - **Monthly Payments**: \$392.41

Solución

To calculate the different payment plans for Cooley Landscaping's loan, we can use the formula for the present value of an annuity. The formula is given by: \[ PV = P \times \left( \frac{1 - (1 + r)^{-n}}{r} \right) \] Where: - \( PV \) is the present value (the amount borrowed, which is \( 26,000 \)). - \( P \) is the payment amount per period. - \( r \) is the interest rate per period. - \( n \) is the total number of payments. ### Step 1: Annual Payments For annual payments: - Interest rate \( r = \frac{0.07}{1} = 0.07 \) - Number of payments \( n = 7 \) We can rearrange the formula to solve for \( P \): \[ P = PV \times \left( \frac{r}{1 - (1 + r)^{-n}} \right) \] Substituting the values: \[ P = 26000 \times \left( \frac{0.07}{1 - (1 + 0.07)^{-7}} \right) \] ### Step 2: Semiannual Payments For semiannual payments: - Interest rate \( r = \frac{0.07}{2} = 0.035 \) - Number of payments \( n = 7 \times 2 = 14 \) Using the same formula for \( P \): \[ P = 26000 \times \left( \frac{0.035}{1 - (1 + 0.035)^{-14}} \right) \] ### Step 3: Quarterly Payments For quarterly payments: - Interest rate \( r = \frac{0.07}{4} = 0.0175 \) - Number of payments \( n = 7 \times 4 = 28 \) Using the same formula for \( P \): \[ P = 26000 \times \left( \frac{0.0175}{1 - (1 + 0.0175)^{-28}} \right) \] ### Step 4: Monthly Payments For monthly payments: - Interest rate \( r = \frac{0.07}{12} \approx 0.0058333 \) - Number of payments \( n = 7 \times 12 = 84 \) Using the same formula for \( P \): \[ P = 26000 \times \left( \frac{0.0058333}{1 - (1 + 0.0058333)^{-84}} \right) \] Now, let's calculate the payments for semiannual, quarterly, and monthly payments. Solve the equation by following steps: - step0: Solve for \(P\): \(26000=\frac{P\left(1-\left(1+0.0058333\right)^{-84}\right)}{0.0058333}\) - step1: Simplify: \(26000=\frac{\left(10000000\times 10058333^{84}-10000000^{85}\right)P}{58333\times 10058333^{84}}\) - step2: Swap the sides: \(\frac{\left(10000000\times 10058333^{84}-10000000^{85}\right)P}{58333\times 10058333^{84}}=26000\) - step3: Cross multiply: \(\left(10000000\times 10058333^{84}-10000000^{85}\right)P=58333\times 10058333^{84}\times 26000\) - step4: Simplify the equation: \(\left(10000000\times 10058333^{84}-10000000^{85}\right)P=1516658000\times 10058333^{84}\) - step5: Divide both sides: \(\frac{\left(10000000\times 10058333^{84}-10000000^{85}\right)P}{10000000\times 10058333^{84}-10000000^{85}}=\frac{1516658000\times 10058333^{84}}{10000000\times 10058333^{84}-10000000^{85}}\) - step6: Divide the numbers: \(P=\frac{1516658000\times 10058333^{84}}{10000000\times 10058333^{84}-10000000^{85}}\) Solve the equation \( 26000 = P * (1 - (1 + 0.0175)^{-28}) / 0.0175 \). Solve the equation by following steps: - step0: Solve for \(P\): \(26000=\frac{P\left(1-\left(1+0.0175\right)^{-28}\right)}{0.0175}\) - step1: Simplify: \(26000=\frac{\left(400\times 407^{28}-400^{29}\right)P}{7\times 407^{28}}\) - step2: Swap the sides: \(\frac{\left(400\times 407^{28}-400^{29}\right)P}{7\times 407^{28}}=26000\) - step3: Cross multiply: \(\left(400\times 407^{28}-400^{29}\right)P=7\times 407^{28}\times 26000\) - step4: Simplify the equation: \(\left(400\times 407^{28}-400^{29}\right)P=182000\times 407^{28}\) - step5: Divide both sides: \(\frac{\left(400\times 407^{28}-400^{29}\right)P}{400\times 407^{28}-400^{29}}=\frac{182000\times 407^{28}}{400\times 407^{28}-400^{29}}\) - step6: Divide the numbers: \(P=\frac{182000\times 407^{28}}{400\times 407^{28}-400^{29}}\) Solve the equation \( 26000 = P * (1 - (1 + 0.035)^{-14}) / 0.035 \). Solve the equation by following steps: - step0: Solve for \(P\): \(26000=\frac{P\left(1-\left(1+0.035\right)^{-14}\right)}{0.035}\) - step1: Simplify: \(26000=\frac{\left(200\times 207^{14}-200^{15}\right)P}{7\times 207^{14}}\) - step2: Swap the sides: \(\frac{\left(200\times 207^{14}-200^{15}\right)P}{7\times 207^{14}}=26000\) - step3: Cross multiply: \(\left(200\times 207^{14}-200^{15}\right)P=7\times 207^{14}\times 26000\) - step4: Simplify the equation: \(\left(200\times 207^{14}-200^{15}\right)P=182000\times 207^{14}\) - step5: Divide both sides: \(\frac{\left(200\times 207^{14}-200^{15}\right)P}{200\times 207^{14}-200^{15}}=\frac{182000\times 207^{14}}{200\times 207^{14}-200^{15}}\) - step6: Divide the numbers: \(P=\frac{182000\times 207^{14}}{200\times 207^{14}-200^{15}}\) Calculate or simplify the expression \( 26000 * (0.035 / (1 - (1 + 0.035)^{-14})) \). Calculate the value by following steps: - step0: Calculate: \(26000\left(\frac{0.035}{\left(1-\left(1+0.035\right)^{-14}\right)}\right)\) - step1: Remove the parentheses: \(26000\left(\frac{0.035}{1-\left(1+0.035\right)^{-14}}\right)\) - step2: Add the numbers: \(26000\left(\frac{0.035}{1-1.035^{-14}}\right)\) - step3: Convert the expressions: \(26000\left(\frac{0.035}{1-\left(\frac{207}{200}\right)^{-14}}\right)\) - step4: Subtract the numbers: \(26000\left(\frac{0.035}{\frac{207^{14}-200^{14}}{207^{14}}}\right)\) - step5: Divide the terms: \(26000\times \frac{7\times 207^{14}}{200\times 207^{14}-200^{15}}\) - step6: Multiply: \(\frac{26000\times 7\times 207^{14}}{200\times 207^{14}-200^{15}}\) - step7: Multiply: \(\frac{182000\times 207^{14}}{200\times 207^{14}-200^{15}}\) Calculate or simplify the expression \( 26000 * (0.0175 / (1 - (1 + 0.0175)^{-28})) \). Calculate the value by following steps: - step0: Calculate: \(26000\left(\frac{0.0175}{\left(1-\left(1+0.0175\right)^{-28}\right)}\right)\) - step1: Remove the parentheses: \(26000\left(\frac{0.0175}{1-\left(1+0.0175\right)^{-28}}\right)\) - step2: Add the numbers: \(26000\left(\frac{0.0175}{1-1.0175^{-28}}\right)\) - step3: Convert the expressions: \(26000\left(\frac{0.0175}{1-\left(\frac{407}{400}\right)^{-28}}\right)\) - step4: Subtract the numbers: \(26000\left(\frac{0.0175}{\frac{407^{28}-400^{28}}{407^{28}}}\right)\) - step5: Divide the terms: \(26000\times \frac{7\times 407^{28}}{400\times 407^{28}-400^{29}}\) - step6: Multiply: \(\frac{26000\times 7\times 407^{28}}{400\times 407^{28}-400^{29}}\) - step7: Multiply: \(\frac{182000\times 407^{28}}{400\times 407^{28}-400^{29}}\) Calculate or simplify the expression \( 26000 * (0.0058333 / (1 - (1 + 0.0058333)^{-84})) \). Calculate the value by following steps: - step0: Calculate: \(26000\left(\frac{0.0058333}{\left(1-\left(1+0.0058333\right)^{-84}\right)}\right)\) - step1: Remove the parentheses: \(26000\left(\frac{0.0058333}{1-\left(1+0.0058333\right)^{-84}}\right)\) - step2: Add the numbers: \(26000\left(\frac{0.0058333}{1-1.0058333^{-84}}\right)\) - step3: Convert the expressions: \(26000\left(\frac{0.0058333}{1-\left(\frac{10058333}{10000000}\right)^{-84}}\right)\) - step4: Subtract the numbers: \(26000\left(\frac{0.0058333}{\frac{10058333^{84}-10000000^{84}}{10058333^{84}}}\right)\) - step5: Divide the terms: \(26000\times \frac{58333\times 10058333^{84}}{10000000\times 10058333^{84}-10000000^{85}}\) - step6: Multiply: \(\frac{26000\times 58333\times 10058333^{84}}{10000000\times 10058333^{84}-10000000^{85}}\) - step7: Multiply: \(\frac{1516658000\times 10058333^{84}}{10000000\times 10058333^{84}-10000000^{85}}\) Now that we have calculated the payments for the different payment plans, here are the results: 1. **Semiannual Payments**: - Payment: \( P \approx 2380.84 \) 2. **Quarterly Payments**: - Payment: \( P \approx 1182.52 \) 3. **Monthly Payments**: - Payment: \( P \approx 392.41 \) ### Summary of Payments: - **Annual Payments**: \( \$ 4824.83 \) - **Semiannual Payments**: \( \$ 2380.84 \) - **Quarterly Payments**: \( \$ 1182.52 \) - **Monthly Payments**: \( \$ 392.41 \) All amounts are rounded to the nearest cent.

Revisado y aprobado por el equipo de tutoría de UpStudy

error msg
Explicar
Simplifique esta solución

Bonus Knowledge

To calculate Cooley Landscaping's payment for the loan with semiannual payments, we can use the formula for the payment of an annuity. The semiannual interest rate will be half of the annual rate, which is \( \frac{7\%}{2} = 3.5\%\) or \( 0.035 \) in decimal form. The number of payments over 7 years with semiannual payments will be \( 7 \times 2 = 14 \) payments. Using the annuity payment formula: \[ PMT = \frac{P \cdot r}{1 - (1 + r)^{-n}} \] Where: - \( PMT \) = payment per period - \( P \) = principal amount = \( 26,000 \) - \( r \) = interest rate per period = \( 0.035 \) - \( n \) = total number of payments = \( 14 \) Plugging in the numbers: \[ PMT = \frac{26000 \cdot 0.035}{1 - (1 + 0.035)^{-14}} \approx \frac{910}{1 - (1.035)^{-14}} \approx \frac{910}{0.49316} \approx 1846.39 \] So, Cooley’s payment for the semiannual loan will be approximately \( \$ 1846.39 \) (rounded to the nearest cent). Now, let's figure out what those different payments look like for quarterly and monthly payments as well! For quarterly payments, the interest rate will be \( \frac{7\%}{4} = 1.75\%\) or \( 0.0175 \). The number of payments will be \( 7 \times 4 = 28 \). Using the same annuity formula: \[ PMT = \frac{26000 \cdot 0.0175}{1 - (1 + 0.0175)^{-28}} \] Calculating this will give us the quarterly payment amount. For monthly payments, the interest rate would be \( \frac{7\%}{12} \) and the number of payments will be \( 7 \times 12 = 84 \), so you can apply the same method to find that out. It’s like creating a fun little payment puzzle for everyone involved!

preguntas relacionadas

Latest Economics Questions

¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde Hazte Premium
Estudiar puede ser una verdadera lucha
¿Por qué no estudiarlo en UpStudy?
Seleccione su plan a continuación
Prima

Puedes disfrutar

Empieza ahora
  • Explicaciones paso a paso
  • Tutores expertos en vivo 24/7
  • Número ilimitado de preguntas
  • Sin interrupciones
  • Acceso completo a Respuesta y Solución
  • Acceso completo al chat de PDF, al chat de UpStudy y al chat de navegación
Básico

Totalmente gratis pero limitado

  • Solución limitada
Bienvenido a ¡Estudia ahora!
Inicie sesión para continuar con el recorrido de Thoth AI Chat
Continuar con correo electrónico
O continuar con
Al hacer clic en "Iniciar sesión", acepta nuestros términos y condiciones. Términos de Uso & Política de privacidad