Question

Say you buy an house as an investment for 250000$ (assume that you did not need a mortgage). You estimate that the house will increase in value continuously by 31250$ per year. At any time in the future you can sell the house and invest the money in a fund with a yearly interest rate of 8.5% compounded bi-monthly.
If you want to maximize your return, after how many years should you sell the house? Report your answer to 1 decimal place.

Answer 1

Answer:

3.3 years  

Step-by-step explanation:

Let the price of the house  when it is being sold be =  x

The Annual return received from the fund after a year that house is being sold is : A -  P

where ;

P = current market price

A = price after one year

A is given by the formula:

$A = P (1+ \frac{r}{n} )^{{n}{t} }$

where ; P = x

r = 8.5% = 0.085

n = 12

t = 1

$A= x(1+\frac{0.085}{2})^{12*1}$

$A= x(1.00708333)^{12}$

A = 1.08839 x

The annual return is A - P = 1.08839 x - x

= 0.08839x

However, the house should be sold when this return is equivalent to the annual increase in value of the house

∴ 0.08839x = 31250

$x = \frac{31250}{0.08839} \\ \\ x = 353546.78$

Thus , the current price (x) = $353546.78

Profit till that time = Current price - Initial Price

= (353546.78- 250000)$

= $103546.78

The time taken for this much profit to accumulate = $\frac{Total \ profit}{Annual \ profit}$

= $\frac{103546.78}{31250}$

= 3.3135

≅ 3.3 years                 ( to one decimal place)