Sagot :
Increasing the number of installments paid to 13 based on the installments
calculated for 12, reduces the cost of interest paid on the mortgage.
- The amount of extra payment is the amount of a month's payment which is; $1,049.2
- The amount paid by paying 1/12 of the 13 payments per month is; [tex]\underline{\$ 1136.\overline 3}[/tex]
Reasons:
First Part
The monthly payment formula is given as follows;
[tex]M = \dfrac{P \cdot \left(\dfrac{r}{12} \right) \cdot \left(1+\dfrac{r}{12} \right)^n }{\left(1+\dfrac{r}{12} \right)^n - 1}[/tex]
Therefore, we have;
[tex]M = \dfrac{175,000 \cdot \left(\dfrac{0.06}{12} \right) \cdot \left(1+\dfrac{0.06}{12} \right)^{12\times 30} }{\left(1+\dfrac{0.06}{12} \right)^{12 \times 30} - 1} \approx 1049.2[/tex]
The monthly payment per month on the mortgage ≈ $1,049.2
To make an extra monthly payment, a month payment can be added, therefore, the amount of extra monthly payment is; $1,049.2
Second part:
Paying 1/12 of the extra monthly payment per month gives;
[tex]\displaystyle Monthly \ payment = 1049.2 + \frac{1049.2}{12} = 1136.6\overline 3[/tex]
The amount of monthly payment with the extra month included in the 12 payments per year is therefore;
1/12 of Payment per month = [tex]\underline{\$ 1136.\overline 3}[/tex]
Savings in interest cost;
The total value of the payment using 12 payment per year is therefore;
Total payment ≈ 12 × 30 × 1049.2 = 377,712
The total value of the payment using 12 payment per year is $377,712
With extra monthly payment, we have;
[tex]1049.2 \approx \dfrac{175,000 \cdot \left(\dfrac{0.06}{13} \right) \cdot \left(1+\dfrac{0.06}{13} \right)^{n} }{\left(1+\dfrac{0.06}{13} \right)^{n} - 1}[/tex]
Which, by using a graphing calculator, gives, n ≈ 318.99 ≈ 319
The total number of payment, n ≈ 319 months
The total value of the payment using 13 payments = 319 × 1049.2 = 334694.8
The total payment using 13 payments instead of 12 is $334,694.8
The amount saved in interest cost is $377,712 - $334,694.8 = $43,017.2
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