Question

Microsoft's stock price peaked at 6118% of its IPO price more than 13 years after the IPO† Suppose that $40,000 invested in Microsoft at its IPO price had been worth $2,400,000 (6000% of the IPO price) after exactly 13 years. What interest rate, compounded annually, does this represent? (Round your answer to two decimal places.)

Answer 1

Answer:

37.02%

Step-by-step explanation:

The formula of compound interest we will use is:

$FV=PV(1+r)^t$

Where

FV is the future value [2,400,000 in this case]

PV is the present value [40,000 was invested]

r is the rate of interest per year (what we want to find)

t is the time in years [13 years]

Substituting and solving for r:

$FV=PV(1+r)^t\\2,400,000=40,000(1+r)^{13}\\60=(1+r)^{13}\\1+r=\sqrt[13]{60}\\ 1+r=1.3702\\r=0.3702$

Thus, rate of interest was 37.02% (rounded to 2 decimal places)