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

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An investor purchased a share of non-dividend-paying stock [#permalink] New post 14 Apr 2007, 16:55
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31% (02:52) correct 68% (03:02) wrong based on 16 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(\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.
[Reveal] Spoiler: OA
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Re: Another 700 level Question - Percents [#permalink] New post 15 Apr 2007, 22:54
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Witchiegrlie 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
(see image)

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


Okay let d=total no. of days for which the price of the stock increases at r% a day.

Now form the question stem we can make the following equation.

v= p [1 + r/100]^d - q
there fore (v+q)/p = [ 1 + r/100 ]
But form the question stem we have r= 100 [ sqrt [ ( v+q)/p] ]
Therefore (v+q)/p= ( 1+ r/100)^d=( r/100 +1)^2

Therefore d=2. So the trader sold the stock after 2+1= 3 days.

Javed.


Cheers!
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 [#permalink] New post 29 Apr 2007, 01:02
B) 3 days

- the price increased by r/100 per day over 'x' days -> p * [1+(r/100)]^x
- the price was reduced by 'q' on day 'x+1' -> p * [1+(r/100)]^x - q
- the selling price was 'v' -> v = p * [1+(r/100)]^x - q

v+q/p = [1+(r/100)]^x
if we compare the above formula with the one given, we will note that x=2, therefore this person sold the share in day x+1 = 3 days later.
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Re: PS-MGMAT CAT 1 [#permalink] New post 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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 [#permalink] New post 08 Dec 2007, 10:23
querio wrote:
B) 3 days

- the price increased by r/100 per day over 'x' days -> p * [1+(r/100)]^x
- the price was reduced by 'q' on day 'x+1' -> p * [1+(r/100)]^x - q
- the selling price was 'v' -> v = p * [1+(r/100)]^x - q

v+q/p = [1+(r/100)]^x
if we compare the above formula with the one given, we will note that x=2, therefore this person sold the share in day x+1 = 3 days later.


So the trick is to know the interest compounding formula and set it equal to V.
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Re: An investor purchased a share of non-dividend-paying stock [#permalink] New post 20 Feb 2012, 05:05
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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] New post 13 Jun 2014, 13:43
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Re: An investor purchased a share of non-dividend-paying stock   [#permalink] 13 Jun 2014, 13:43
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