Answer:
$ 3934.38 ( approx )
Step-by-step explanation:
Since, the monthly payment formula of a loan,
![P=\frac{PV(\frac{r}{12})}{1-(1+\frac{r}{12})^{-n}}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BPV%28%5Cfrac%7Br%7D%7B12%7D%29%7D%7B1-%281%2B%5Cfrac%7Br%7D%7B12%7D%29%5E%7B-n%7D%7D)
Where,
PV = present value of loan,
r = annual rate of interest,
n = number of months,
If P = $ 330, r = 1.2% = 0.012,
Number of months in 6 years, n = 12 × 6 = 72
By substituting the values,
![330 = \frac{PV(\frac{0.012}{12})}{1-(1+\frac{0.012}{12})^{-72}}](https://tex.z-dn.net/?f=330%20%3D%20%5Cfrac%7BPV%28%5Cfrac%7B0.012%7D%7B12%7D%29%7D%7B1-%281%2B%5Cfrac%7B0.012%7D%7B12%7D%29%5E%7B-72%7D%7D)
![330 =\frac{PV(0.001)}{1-(1.001)^{-12}}](https://tex.z-dn.net/?f=330%20%3D%5Cfrac%7BPV%280.001%29%7D%7B1-%281.001%29%5E%7B-12%7D%7D)
![\implies PV = 330\times \frac{1-(1.001)^{-12}}{(0.001)}](https://tex.z-dn.net/?f=%5Cimplies%20PV%20%3D%20330%5Ctimes%20%5Cfrac%7B1-%281.001%29%5E%7B-12%7D%7D%7B%280.001%29%7D)
Using calculator,
PV ≈ $ 3934.38
Hence, the value of the most expensive car he can afford would be $ 3934.38 ( approx )