singh_amit19 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 X [sqrt(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.

B.

Formula for compound interest:

Final Value = P (1 + r/100)^t

In our question:

P = p

r = r

Final Value = v

number of days of compounding is unknown.

Given: r= 100 X [sqrt(v+q)/p)-1]

r/100 = [sqrt [(v + q)/p]] - 1

1 + r/100 = sqrt [(v + q)/p]

(1 + r/100)^2 = (v + q)/p

p (1 + r/100) ^2 = v + q

v = p (1 + r/100)^2 - q

This is in the same form as the formula for compound interest with the exception of '-q'. But from the question stem, we can infer that the negative q, refers to the decrease in the value of the stock. So, the initial value p compounded for 2 days (r = 2) and fell by q dollars on the third day ('next day' as stated in the stem). So the investor sold the stock on the third day.

I came across this question before and couldn't solve it then, which is why I know the solution this time round