For a stúdent loan of $$ \$ 20,900 $$ at $$ 2.95 \% $$ for 15 years, complete the following. (a) Calculate the monthly payment necessary to amortize the loan amount. (b) Find the amount of money saved over the lifetime of the loan if an additional $$ \$ 100 $$ is added to the monthly payment.

Question

UpStudy AI · Accepted Answer

To solve the problem, we will follow these steps:

### Part (a): Calculate the monthly payment necessary to amortize the loan amount.

The formula for the monthly payment $$ M $$ on an amortized loan is given by:

$$
M = P \frac{r(1 + r)^n}{(1 + r)^n - 1}
$$

Where:
- $$ P $$ is the loan amount (principal),
- $$ r $$ is the monthly interest rate (annual rate divided by 12),
- $$ n $$ is the total number of payments (loan term in months).

**Given:**
- $$ P = 20900 $$
- Annual interest rate = $$ 2.95\% $$ or $$ 0.0295 $$
- Loan term = $$ 15 $$ years

**Calculating the values:**
1. Convert the annual interest rate to a monthly interest rate:
   $$
   r = \frac{0.0295}{12}
   $$
2. Calculate the total number of payments:
   $$
   n = 15 \times 12
   $$

Now, let's calculate $$ r $$ and $$ n $$ and then use them to find $$ M $$.

### Part (b): Find the amount of money saved over the lifetime of the loan if an additional $$ \$ 100 $$ is added to the monthly payment.

To find the new monthly payment with the additional $$ \$ 100 $$, we will calculate the new payment $$ M' $$ as follows:

$$
M' = M + 100
$$

Next, we will calculate the total payment over the life of the loan for both scenarios and find the difference.

Let's perform the calculations for both parts.
Calculate the value by following steps:
- step0: Calculate:
  $$15\times 12$$
- step1: Multiply the numbers:
  $$180$$
Calculate or simplify the expression $$ 20900 * (0.0295/12) * (1 + 0.0295/12)^(15*12) / ((1 + 0.0295/12)^(15*12) - 1) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{20900\left(\frac{0.0295}{12}\right)\left(1+\frac{0.0295}{12}\right)^{15\times 12}}{\left(\left(1+\frac{0.0295}{12}\right)^{15\times 12}-1\right)}$$
- step1: Remove the parentheses:
  $$\frac{20900\left(\frac{0.0295}{12}\right)\left(1+\frac{0.0295}{12}\right)^{15\times 12}}{\left(1+\frac{0.0295}{12}\right)^{15\times 12}-1}$$
- step2: Divide the terms:
  $$\frac{20900\left(\frac{0.0295}{12}\right)\left(1+\frac{59}{24000}\right)^{15\times 12}}{\left(1+\frac{0.0295}{12}\right)^{15\times 12}-1}$$
- step3: Add the numbers:
  $$\frac{20900\left(\frac{0.0295}{12}\right)\left(\frac{24059}{24000}\right)^{15\times 12}}{\left(1+\frac{0.0295}{12}\right)^{15\times 12}-1}$$
- step4: Divide the terms:
  $$\frac{20900\left(\frac{0.0295}{12}\right)\left(\frac{24059}{24000}\right)^{15\times 12}}{\left(1+\frac{59}{24000}\right)^{15\times 12}-1}$$
- step5: Add the numbers:
  $$\frac{20900\left(\frac{0.0295}{12}\right)\left(\frac{24059}{24000}\right)^{15\times 12}}{\left(\frac{24059}{24000}\right)^{15\times 12}-1}$$
- step6: Divide the terms:
  $$\frac{20900\times \frac{59}{24000}\left(\frac{24059}{24000}\right)^{15\times 12}}{\left(\frac{24059}{24000}\right)^{15\times 12}-1}$$
- step7: Multiply the numbers:
  $$\frac{20900\times \frac{59}{24000}\left(\frac{24059}{24000}\right)^{180}}{\left(\frac{24059}{24000}\right)^{15\times 12}-1}$$
- step8: Multiply the numbers:
  $$\frac{20900\times \frac{59}{24000}\left(\frac{24059}{24000}\right)^{180}}{\left(\frac{24059}{24000}\right)^{180}-1}$$
- step9: Multiply:
  $$\frac{\frac{12331\times 24059^{180}}{240\times 24000^{180}}}{\left(\frac{24059}{24000}\right)^{180}-1}$$
- step10: Subtract the numbers:
  $$\frac{\frac{12331\times 24059^{180}}{240\times 24000^{180}}}{\frac{24059^{180}-24000^{180}}{24000^{180}}}$$
- step11: Multiply by the reciprocal:
  $$\frac{12331\times 24059^{180}}{240\times 24000^{180}}\times \frac{24000^{180}}{24059^{180}-24000^{180}}$$
- step12: Rewrite the expression:
  $$\frac{12331\times 24059^{180}}{240\times 24000^{180}}\times \frac{240^{180}\times 100^{180}}{24059^{180}-24000^{180}}$$
- step13: Reduce the numbers:
  $$\frac{12331\times 24059^{180}}{24000^{180}}\times \frac{240^{179}\times 100^{180}}{24059^{180}-24000^{180}}$$
- step14: Rewrite the expression:
  $$\frac{12331\times 24059^{180}}{240^{180}\times 100^{180}}\times \frac{240^{179}\times 100^{180}}{24059^{180}-24000^{180}}$$
- step15: Reduce the numbers:
  $$\frac{12331\times 24059^{180}}{240\times 100^{180}}\times \frac{100^{180}}{24059^{180}-24000^{180}}$$
- step16: Rewrite the expression:
  $$\frac{12331\times 24059^{180}}{240\times 100^{180}}\times \frac{20^{180}\times 5^{180}}{24059^{180}-24000^{180}}$$
- step17: Rewrite the expression:
  $$\frac{12331\times 24059^{180}}{20\times 12\times 100^{180}}\times \frac{20^{180}\times 5^{180}}{24059^{180}-24000^{180}}$$
- step18: Reduce the numbers:
  $$\frac{12331\times 24059^{180}}{12\times 100^{180}}\times \frac{20^{179}\times 5^{180}}{24059^{180}-24000^{180}}$$
- step19: Rewrite the expression:
  $$\frac{12331\times 24059^{180}}{12\times 100^{180}}\times \frac{4^{179}\times 5^{179}\times 5^{180}}{24059^{180}-24000^{180}}$$
- step20: Rewrite the expression:
  $$\frac{12331\times 24059^{180}}{4\times 3\times 100^{180}}\times \frac{4^{179}\times 5^{179}\times 5^{180}}{24059^{180}-24000^{180}}$$
- step21: Reduce the numbers:
  $$\frac{12331\times 24059^{180}}{3\times 100^{180}}\times \frac{4^{178}\times 5^{179}\times 5^{180}}{24059^{180}-24000^{180}}$$
- step22: Rewrite the expression:
  $$\frac{12331\times 24059^{180}}{3\times 4^{180}\times 25^{180}}\times \frac{4^{178}\times 5^{179}\times 5^{180}}{24059^{180}-24000^{180}}$$
- step23: Reduce the numbers:
  $$\frac{12331\times 24059^{180}}{3\times 4^{2}\times 25^{180}}\times \frac{5^{179}\times 5^{180}}{24059^{180}-24000^{180}}$$
- step24: Rewrite the expression:
  $$\frac{12331\times 24059^{180}}{3\times 4^{2}\times 5^{360}}\times \frac{5^{179}\times 5^{180}}{24059^{180}-24000^{180}}$$
- step25: Reduce the numbers:
  $$\frac{12331\times 24059^{180}}{3\times 4^{2}\times 5}\times \frac{1}{24059^{180}-24000^{180}}$$
- step26: Multiply the fractions:
  $$\frac{12331\times 24059^{180}}{240\left(24059^{180}-24000^{180}\right)}$$
- step27: Multiply:
  $$\frac{12331\times 24059^{180}}{240\times 24059^{180}-240\times 24000^{180}}$$
Calculate or simplify the expression $$ 0.0295/12 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{0.0295}{12}$$
- step1: Convert the expressions:
  $$\frac{\frac{59}{2000}}{12}$$
- step2: Multiply by the reciprocal:
  $$\frac{59}{2000}\times \frac{1}{12}$$
- step3: Multiply the fractions:
  $$\frac{59}{2000\times 12}$$
- step4: Multiply:
  $$\frac{59}{24000}$$
Calculate or simplify the expression $$ 180 * 143.829506 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$180\times 143.829506$$
- step1: Multiply the numbers:
  $$25889.31108$$
Calculate or simplify the expression $$ (143.829506 + 100) * 180 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(143.829506+100\right)\times 180$$
- step1: Add the numbers:
  $$243.829506\times 180$$
- step2: Multiply the numbers:
  $$43889.31108$$
### Part (a): Monthly Payment Calculation

1. **Loan Amount (P)**: $$ 20900 $$
2. **Monthly Interest Rate (r)**: 
   $$
   r = \frac{0.0295}{12} \approx 0.002458
   $$
3. **Total Number of Payments (n)**: 
   $$
   n = 15 \times 12 = 180
   $$
4. **Monthly Payment (M)**:
   $$
   M = 20900 \frac{0.002458(1 + 0.002458)^{180}}{(1 + 0.002458)^{180} - 1} \approx 143.83
   $$

Thus, the monthly payment necessary to amortize the loan amount is approximately **\$143.83**.

### Part (b): Savings Calculation with Additional Payment

1. **New Monthly Payment (M')**:
   $$
   M' = M + 100 = 143.83 + 100 = 243.83
   $$
2. **Total Payment with Original Monthly Payment**:
   $$
   \text{Total Payment} = M \times n = 143.83 \times 180 \approx 25889.31
   $$
3. **Total Payment with New Monthly Payment**:
   $$
   \text{Total Payment with New Payment} = M' \times n = 243.83 \times 180 \approx 43889.31
   $$
4. **Savings**:
   $$
   \text{Savings} = \text{Total Payment with New Payment} - \text{Total Payment} = 43889.31 - 25889.31 = 18000
   $$

Thus, the amount of money saved over the lifetime of the loan if an additional \$100 is added to the monthly payment is approximately **\$18,000**.

### Summary of Results:
- **Monthly Payment (Part a)**: \$143.83
- **Savings with Additional Payment (Part b)**: \$18,000
**Part (a):** The monthly payment is approximately \$143.83. **Part (b):** Adding \$100 to the monthly payment saves \$18,000 over the loan's lifetime.

For a stúdent loan of \( \$ 20,900 \) at \( 2.95 \% \) for 15 years, complete the following. (a) Calculate the monthly payment necessary to amortize the loan amount. (b) Find the amount of money saved over the lifetime of the loan if an additional \( \$ 100 \) is added to the monthly payment.

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Pre Calculus Questions