Question

Vanna has just financed the purchase of a home for $200 000. She agreed to repay the loan by making equal monthly blended payments of $3000 each at 4%/a, compounded monthly
How long will it take to repay the loan?
How much will be the final payment?
Determine how much interest she will pay for her loan.
How much sooner would the loan be paid if she made a 15% down payment?
How much would Vanna have saved if she had obtained a loan 3%/a, compounded monthly?

Answer 1

The question describes retiring a dept though annuity. The present value of an annuity, PV, where periodic payments are made for n periods at r% interest rate at t periods per year is given by
$PV=P\left[\frac{1-\left(1+\frac{r}{t}\right)^{-nt}}{\frac{r}{t}}\right]$
where: PV = $200,000; P = $3,000; r = 4%; t = 12

Part A.
To find, n the number of years it will take to repay the loan, we have
$200,000=3000\left[\frac{1-\left(1+\frac{0.04}{12}\right)^{-12n}}{\frac{0.04}{12}}\right] \\ \\ =3000\left[\frac{1-\left(\frac{301}{300}\right)^{-12n}}{\frac{1}{300}}\right]=900,000\left[1-\left(\frac{301}{300}\right)^{-12n}\right] \\ \\ \Rightarrow\left[1-\left(\frac{301}{300}\right)^{-12n}\right]=\frac{2}{9} \\ \\ \Rightarrow\left(\frac{301}{300}\right)^{-12n}=1-\frac{2}{9}=\frac{7}{9} \\ \\ -12n\log{\left(\frac{301}{300}\right)}=\log{\left(\frac{7}{9}\right)}$

$\Rightarrow-12n=\frac{\log{\left(\frac{7}{9}\right)}}{\log{\left(\frac{301}{300}\right)}}=-75.52 \\ \\ \Rightarrow n=\frac{-75.52}{-12}=6.29$

Therefore, it will take 6 years and 4 months to repay the loan.


Part B.
Total amount paid for the loan including interest is given by the formular
$A=P\left[\frac{\left(1+\frac{r}{t}\right)^{nt}-1}{\frac{r}{t}}\right]$
where: P = $3,000; r = 4%; t = 12; n = 6.29332634

$A=3000\left[\frac{\left(1+\frac{0.04}{12}\right)^{6.29332634\times12}-1}{\frac{0.04}{12}}\right] \\ \\ =3000\left[\frac{0.285714286}{0.003333333333}\right]=3000(85.7142858)=\ 257,142.86$

Total amount paid as at the month before the last payment (i.e. after 6 years 3 months) is given by
$A=3000\left[\frac{\left(1+\frac{0.04}{12}\right)^{6.25\times12}-1}{\frac{0.04}{12}}\right] \\ \\ =3000\left[\frac{0.283491702}{0.003333333333}\right]=3000(85.04751061)=\ 255,142.53$

Therefore, the amount paid in the final payment is $257,142.86 - $255,142.53 = $2,000.33 or approximately $2,000


Part C.
The amount of interest she will pay for her loan is given by the total amount payed for the duration of the loan less the loan amount.
Therefore, interest = $257,142.86 - $200,000 = $57,142.86


Part D.
If she made a 15% down payment, then amount to be repaid is given by
$200000 - \frac{15}{100}\times200000=200000-30,000=\ 170,000$

Thus, the number of periods for the repayment is given by
$170,000=3000\left[\frac{1-\left(1+\frac{0.04}{12}\right)^{-12n}}{\frac{0.04}{12}}\right] \\ \\ =3000\left[\frac{1-\left(\frac{301}{300}\right)^{-12n}}{\frac{1}{300}}\right]=900,000\left[1-\left(\frac{301}{300}\right)^{-12n}\right] \\ \\ \Rightarrow\left[1-\left(\frac{301}{300}\right)^{-12n}\right]=\frac{17}{90} \\ \\ \Rightarrow\left(\frac{301}{300}\right)^{-12n}=1-\frac{17}{90}=\frac{73}{90} \\ \\ -12n\log{\left(\frac{301}{300}\right)}=\log{\left(\frac{73}{90}\right)}$

$\Rightarrow-12n=\frac{\log{\left(\frac{73}{90}\right)}}{\log{\left(\frac{301}{300}\right)}}=-62.91 \\ \\ \Rightarrow n=\frac{-62.91}{-12}=5.24$

If she made a 15% down payment, it will take her 5 years 3 months to complete the payment.


Part E.
If she have obtained a loan of 3% / annum compounded monthly. Total payment to be made at 3% interest is given by
$A=3000\left[\frac{\left(1+\frac{0.03}{12}\right)^{6.29332634\times12}-1}{\frac{0.04}{12}}\right] \\ \\ =3000\left[\frac{0.207514582}{0.0025}\right]=3000(83.00583299)= \\ \\ \ 249,017.50$

Therefore, she will save $257,142.86 - $249,017.50 = $8,125.36