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# An investor purchased a share of non-dividend-paying stock

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Senior Manager
Joined: 11 Sep 2005
Posts: 332
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An investor purchased a share of non-dividend-paying stock [#permalink]  06 Oct 2007, 00:01
00:00

Difficulty:

5% (low)

Question Stats:

50% (04:26) correct 50% (02:15) wrong based on 6 sessions
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.
Senior Manager
Joined: 11 Sep 2005
Posts: 332
Followers: 1

Kudos [?]: 6 [0], given: 0

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

????
Director
Joined: 09 Aug 2006
Posts: 770
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Re: PS-MGMAT CAT 1 [#permalink]  06 Oct 2007, 05:16
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
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Joined: 07 Aug 2011
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Concentration: Strategy, Marketing
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Re: An investor purchased a share of non-dividend-paying stock [#permalink]  20 Feb 2012, 05:05
1
KUDOS
v = p (1 + \frac{r}{100})^n - q

r = 100\sqrt{\frac{v+q}{p}} -1

v+q = p(\sqrt{\frac{v+q}{p}})^n

p^n^/^2(v+q) = p(v+q)^n^/^2

n = 2

so, the value increased for two days and the investor sold the third day, the same day when the price fell.
ans: b
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Re: An investor purchased a share of non-dividend-paying stock   [#permalink] 20 Feb 2012, 05:05
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