Question

The price of a car that you want is $70,000 today. its price is expected to increase by $3300 per year. you now have $35,000 in an investment which is earning 15% per year. the number of years before you have enough to buy the car without borrowing any money is closest to:

Answer 1

Let n=number of years

Price of car
= 70000*(3300n)

Value of investment
= 35000*(1+0.15)^n

To afford to buy the car, 
Value of investment >= Price of car, or
35000(1.15)^n >=70000*3300n
=>
35000(1.15)^n / 70000  >=3300n 
=>
1.15^n >=6600n,
solve using Newton's approximation.
f(n)=1.15^n-6600n = 0
f'(n)=0.13976 (1.15^n)-6600
try n0=100
n1=n0-f(n0)/f'(n0)
=100-f(100)/f'(100)
=96.74
n2=n1-f(n1)/f'(n1)
=95.65
n3=n2-f(n2)/f'(n2)
=95.55

So the answer is it takes 96 years to have enough money to buy the car without borrowing.