Question

A person who filed bankruptcy in the past is able to get a 30-year mortgage loan at a rate that is 6% higher than what they could have received if they had not filed. The interest rate this person pays on a $150,000 loan is 11%, compounded monthly.

Assume the person could have received the lower interest rate on the loan and saved all of the difference in the payments for the first 10 years of the loan. If this person invested this total amount in an account paying simple interest at the rate of 2.5%, how much money would have accumulated in interest by the time the mortgage is paid off?
a.
$37,395.18
b.
$74,790.37
c.
$623.25
d.
$3,739.52

Answer 1

Answer: a. $37,395.18

Work:
- Monthly rates: r_high = 0.11/12, r_low = 0.05/12, n = 360.
- Monthly payment formula: M = L * r / (1 − (1+r)^−n).

Compute the two payments and their difference:
- M_high ≈ 150000*(0.11/12)/(1−(1+0.11/12)^−360)
- M_low ≈ 150000*(0.05/12)/(1−(1+0.05/12)^−360)
- M_high − M_low ≈ $623.25 per month.

Saved over the first 10 years: 623.25 * 120 = $74,790.37.

Invested at simple interest 2.5% for the remaining 20 years: interest = 74,790.37 * 0.025 * 20 = $37,395.18.