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singh_amit19: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Mon Aug 06, 2007 9:50 pm

by singh_amit19 » Sat Oct 06, 2007 12:05 am

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.

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Source: Beat The GMAT — Problem Solving | Sat Oct 06, 2007 12:05 am

singh_amit19: Junior | Next Rank: 30 Posts; Posts: 14; Joined: Mon Aug 06, 2007 9:50 pm

by singh_amit19 » Sat Oct 06, 2007 12:10 am

My appr:

I zeroed down to the answer

Number of days till the stock prices increased n = [sqrt(v+q)/p) + 1]

So, Total no. of days: n + 1

????

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kajcha: Master | Next Rank: 500 Posts; Posts: 321; Joined: Tue Aug 28, 2007 5:42 am; Thanked: 1 times

by kajcha » Sat Oct 06, 2007 3:58 am

Q asks after how many days the value of the share will be "v" unit after it was increased at the rate of "r%" for first "n" days and decreased by "q" unit on "n+1" th day. So what you need to find is

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

Now solve the rate "r" given in the Q

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

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

=> r/100+1 = 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 ------ (2)

Compare eqs (1) and (2) => n=2

Means, the value of the share rose for 2 days and decreased on the 3rd day...

So, ANS should be B - 3 working days later

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