Cfrgtkky

So I am actually confused with how to do loan mortgage problems in my textbook. I was wondering if someone can give me some insights to solving this problem:

Problem: Ten years ago the Peter's bought a house, taking out a 30 year mortgage for $130,000 at 4.5%. This year (exactly 10 years later) they refinanced the house, taking out a new 30 year mortgage for the remaining balance at 3.125%.

i) What was the monthly payment on the original 4.5% mortgage?

ii) What was the remaining balance after 10 years (the amount they then refinanced)?

iii) How much interest did they pay during those first 10 years?

iv) What is the monthly payment on the refinance 3.125% mortgage?

v) How much interest will they pay over the 30 year term of the refinance?

vi) How much total interest will they pay over the full 40 years the Peter's have a loan for the house?

My thoughts, for the first problem: I am thinking we have to use the present value formula which is:

$$PV=Rtimesfrac{1-(1+i)^{-n}}{i}$$

So is PV here $130000?$ and $i = 0.065?$, and can we take $n = 30?$ I have checked with my classmates and most them seemed to have used excel, but I was wondering how we can do this using mathematics. I would appreciate the help.

$130,000$ is the present value of the $30$ year mortgage. The family took this loan under $4.5%$ interest for $30$ years. On the one hand, the future value of the mortgage is:
$$FV=130,000cdot (1+frac{0.045}{12})^{12cdot 30}=500,200.75.$$
It implies the family must repay this much after $30$ years. However, on the other hand, the family is going to regularly pay $R$ amount every month. Then:
$$i) FV=Rcdot frac{(1+frac {0.045}{12})^{12cdot 30}-1}{frac{0.045}{12}}=500,200.75 Rightarrow R=658.69.$$

In MS Excel enter: "$=-PMT(0.045/12;360;130000;1)$". But they stopped paying after $10$ years. So, we must calculate the future value of the mortgage after $10$ years with $R=658.69$:
$$ii) FV=658.69cdot frac{(1+frac{0.045}{12})^{120}-1}{frac{0.045}{12}}=99,592.66;\
FV=130,000cdot (1+frac{0.045}{12})^{120}=203,709.06\
203,709.06-99,592.66=104,116.4.$$
This much money left on the balance to be paid in the remaining $20$ years. Note: The discrepancies are due to rounding.

Can you continue with the rest?

Using the PV, it's probably easier.

$L=130,000$, $n=12times 30=360$ ($30$ years), $i^{(12)}=4.5%$, $m=12times 10$ ($10$ years), $j^{(12)}=3.125%$, $N=12times 30=360$ ($30$ years).

Put $i=frac{i^{(12)}}{12}$ and $j=frac{j^{(12)}}{12}$ and $a_{overline{p}|k}=frac{1-(1+k)^{-p}}{k}$.

So we have:

1. The monthly payment $R$
   $$
   L=Rtimes a_{overline{n}|i}quadLongrightarrowquad R=frac{L}{a_{overline{n}|i}}=frac{130,000}{197.36}=658.69
   $$
   


2. The remaining balance after $10$ years $B$ is
   $$
   B=Rtimes a_{overline{n-m}|i}=658.69times 158.07=104,116.27
   $$
   


3. The interest paid is
   $$
   I=Rtimes(m-a_{overline{m}|i})=658.69times (120-96.49)=15,486.27
   $$
   


4. The monthly payment on the refinance $R'$ is
   $$
   B=R'times a_{overline{N}|j}quadLongrightarrowquad R'=frac{B}{a_{overline{N}|j}}=frac{104,116.27}{233.44}=446.01
   $$
   


5. The interest paid over the refinance is
   $$
   I'=R'times(N-a_{overline{N}|j})=R'times N-B=446.01times 360-104,116.27=56,446.80
   $$
   


6. The total interest over $40$ years is
   $$
   I_{text{total}}=I+I'=71,933.07
   $$