GeminiHeat wrote:
An investor purchased a share of non-dividend-paying stock for p dollars on Monday. For a certain number of days, the value of the share increased by r percent per day. After this period of constant increase, the value of the share decreased the next day by q dollars and the investor decided to sell the share at the end of that day for v dollars, which was the value of the share at that time. How many working days after the investor bought the share was the share sold, if \(r=100(\sqrt{\frac{v+q}{p}}-1)\)
A. Two working days later.
B. Three working days later.
C. Four working days later.
D. Five working days later.
E. Six working days later.
GeminiHeatGood Question!
\(r = 100(\sqrt{\frac{v+q}{p}}-1)\)
\(r = 100(\sqrt{\frac{v+q}{p}}) - 100\)
\(r + 100 = 100(\sqrt{\frac{v+q}{p}})\)
\(\frac{(r + 100)}{100} = (\sqrt{\frac{v+q}{p}})\)
\(1 + \frac{r}{100} = (\sqrt{\frac{v+q}{p}})\)\
Squaring both sides;
\((1 + \frac{r}{100})^2 = \frac{v + q}{p}\)
\(v = p(1 + \frac{r}{100})^2 - q\)
Since the expression \((1 + \frac{r}{100})^2\) has a power of \(2\), the value must have increased for \(2\) days and then decreased on another day!
Therefore, a total of \(3\) days
Hence, option B
Hence, option B