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?

My thoughts for i) I am thinking we have to use the present value formula which is:


So is PV here $130000?$ Is $i = 0.065?$? Can we take $n = 30?$ I have checked with my classmates and most of them seemed to have used Excel, but I was wondering how we can calculate this without Excel.

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?

  • $\begingroup$ Don't you mean $i=.045?$ And aren't the mortgage payments monthly? I would think you take $i={.045\over12}$ and $n=360.$ $\endgroup$
    – saulspatz
    Commented Dec 5, 2018 at 5:12

2 Answers 2


$130,000$ 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: $$FV=130,000\cdot (1+\frac{0.045}{12})^{12\cdot 30}=500,200.75.$$ 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: $$i) \ \ FV=R\cdot \frac{(1+\frac {0.045}{12})^{12\cdot 30}-1}{\frac{0.045}{12}}=500,200.75 \Rightarrow R=658.69.$$
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 $R=658.69$: $$ii) \ FV=658.69\cdot \frac{(1+\frac{0.045}{12})^{120}-1}{\frac{0.045}{12}}=99,592.66;\\ FV=130,000\cdot (1+\frac{0.045}{12})^{120}=203,709.06\\ 203,709.06-99,592.66=104,116.4.$$ 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?

  • $\begingroup$ Ok thank you, so I get all this, with regards to finding the interest, would it be the future value at 10 years minus the amount paid monthly without interest,for instance: $$iii)96,942.16-641.16(120)=20002.92$$ $\endgroup$ Commented Dec 6, 2018 at 4:15
  • $\begingroup$ Also one thing I do not understand is, despite your working being very clear, if I input the data on calculator.net/…, I get the monthly payment of $658.69$, so I was wondering why the answers show different results despite the math making sense$?$ $\endgroup$ Commented Dec 6, 2018 at 4:27
  • $\begingroup$ See my update. The discrepancies in numbers are due to rounding differences. $\endgroup$
    – farruhota
    Commented Dec 6, 2018 at 8:29

Using the PV, it's probably easier.

$L=130,000$, $n=12\times 30=360$ ($30$ years), $i^{(12)}=4.5\%$, $m=12\times 10$ ($10$ years), $j^{(12)}=3.125\%$, $N=12\times 30=360$ ($30$ 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 $R$ $$ L=R\times a_{\overline{n}|i}\quad\Longrightarrow\quad R=\frac{L}{a_{\overline{n}|i}}=\frac{130,000}{197.36}=658.69 $$
  2. The remaining balance after $10$ years $B$ is $$ B=R\times a_{\overline{n-m}|i}=658.69\times 158.07=104,116.27 $$
  3. The interest paid is $$ I=R\times(m-a_{\overline{m}|i})=658.69\times (120-96.49)=15,486.27 $$
  4. The monthly payment on the refinance $R'$ is $$ B=R'\times a_{\overline{N}|j}\quad\Longrightarrow\quad R'=\frac{B}{a_{\overline{N}|j}}=\frac{104,116.27}{233.44}=446.01 $$
  5. The interest paid over the refinance is $$ I'=R'\times(N-a_{\overline{N}|j})=R'\times N-B=446.01\times 360-104,116.27=56,446.80 $$
  6. The total interest over $40$ years is $$ I_{\text{total}}=I+I'=71,933.07 $$

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