stock example

[lucidblue]

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{(v+q)/p}-1] ?

Two working days later.
Three working days later.
Four working days later.
Five working days later.
Six working days later.

[lucidblue]

i had trouble until i used the formula for interest:

(1 + r/100)^d

d = days held

[supersuj]

Tricky question. My answer is B, which is three working days later.

Value of stock on day of purchase= P dollars
daily interest rate = r/100
No. of days = n

Value of stock after n days = P* (1+ (r/100))^n
on n+1 day, stock price = [P* (1+ (r/100))^n] - q

Given r= 100*[sqrt{v+q)/p}-1]

we get r = 100*sqrt{v+q}/p - 100

so 1+(r/100)= sqrt{v+q}/p

squaring both sides, we get p*(1+(r/100))^2 -q= v

So we know that that n=2, which is 2 working days.

But we need to find n+1, so the correct answer is 3 working days later

[megha57]

Three working days.
Suppose shares increased for n days
v = p(1-r/100)^n -q
(v+q)/p = [1 - ((v+q)/p)^0.5 - 1] ^ n
(v+q)/p = ((v+q)/p)^n/2
((v+q)/p)^(n/2 - 1) = 1
n/2 - 1 = 0
n = 2
Including the day it fall; total number of days should be 3.