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

First compute the monthly payments at the two interest rates.

Monthly rate i1 = 0.11/12 = 0.0091666667, i2 = 0.05/12 = 0.0041666667, n = 360.
Monthly payment formula: M = P * i / (1 - (1+i)^(-n))

M1 (11%) = 150000*(0.11/12)/(1-(1+0.11/12)^(-360)) ≈ $1,428.28
M2 (5%) = 150000*(0.05/12)/(1-(1+0.05/12)^(-360)) ≈ $805.10

Monthly difference Δ = M1 − M2 ≈ $623.18.
Saved for first 10 years: total saved = 120·Δ.

They then invest that total for the remaining 20 years at simple interest 2.5%:
Interest = (total saved) * 0.025 * 20 = (120·Δ) * 0.5 = 60·Δ ≈ 60·623.18 ≈ $37,395.18.

Answer: a. $37,395.18